mm-1012
===
Subject: Re: Cantors diagonal proof wrong?
> ... That depends on what is is. It has no relevance to
Mathematics,
> which does not depend on your philosophy.
> Do you think philosophy has any bearing on mathematics?
Philosophy
has
> little bearing on mathematics, except the philosophical
(philoshopical,
> philosophical) methods, particularly the rationalist ones,
apply.
Not only philosophy but, in particular, physics has. Withou
matter
there is not only no space but there is no means to store any
number,
not in an abacus, not in a pocket calculator, not in a
computer and not
in a brain. As there are at most 10^80 protons in the
universe and some
is 10^10^100. (Though there is not a largest number existing.)
Therefore any considering of an actual infinity is nonsense
from the
scratch. This position isnt finitism, but 
realism, though not
in the
euphemistic meaning mathematicians like to use for their
utterly
unrealistic positions. Actual infinity was also denied by
Hegel, by the
way. But I dont know much of his his philosophy.
> What do you think of Hegels Being and Nothing dichotomy 
as
model of
the
> ur-element?
> The set of all sets is its own powerset.
And, therefore, it cannot exist. The sum of all natural
numbers is
larger than any natural number. Therefore the set of all
natural
numbers cannot exist. (Would it actually exist, we could
calculate the
sum.) Thats the same arguing, but mathematicians use to see
things
different.
===
Subject: Re: Cantors diagonal proof wrong?
<41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net>
<41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net>
<41b2427f$15$fuzhry+tra$mr2ice@news.patriot.net>
<41c22df7$7$fuzhry+tra$mr2ice@news.patriot.net>
<41C25DB8.EF7C5F2F@tiki-lounge.com>
at 01:56 PM, mueckenh@rz.fh-augsburg.de said:
>And, therefore, it cannot exist. The sum of all natural
numbers is
>larger than any natural number. Therefore the set of all
natural
>numbers cannot exist.
Thats a non sequitor.
>(Would it actually exist, we could calculate the sum.)
No. There is no sum of the elements of an infinite set. There
may or
may not be a limit of a sequence of partial sums; in the case
of the
integers, there isnt.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
<http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: Cantors diagonal proof wrong?
<41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net>
<41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net>
<41b2427f$15$fuzhry+tra$mr2ice@news.patriot.net>
<41c22df7$7$fuzhry+tra$mr2ice@news.patriot.net>
<41C25DB8.EF7C5F2F@tiki-lounge.com>
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ELIi
$t^
VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
>> Do you think philosophy has any bearing on mathematics?
Philosophy
>> has little bearing on mathematics, except the philosophical
>> (philoshopical, philosophical) methods, particularly the
>> rationalist ones, apply.
> Not only philosophy but, in particular, physics has. Withou
matter
> there is not only no space but there is no means to store
any
> number, not in an abacus, not in a pocket calculator, not
in a
> computer and not in a brain.
But the good thing is that we only need the space for storing
the laws
that _all_ numbers obey. Like a Shakespearean play: the words
of it
are spoken by players, but the essence of the play is in the
book, and
it remains there even when a play is not being performed.
And the literary critics can talk about the play without
actually
seeing a single performance of it.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Cantors diagonal proof wrong?
<41aa5b47$13$fuzhry+tra$mr2ice@news.patriot.net>
<41ad5139$15$fuzhry+tra$mr2ice@news.patriot.net>
<41b2427f$15$fuzhry+tra$mr2ice@news.patriot.net>
<41c22df7$7$fuzhry+tra$mr2ice@news.patriot.net>
<41C25DB8.EF7C5F2F@tiki-lounge.com>
<x51xdnuuku.fsf@lola.goethe.zz>
posting-account=htRwYA0AAACUC1yg4djqvdjZ_SB9JXGq
>> Do you think philosophy has any bearing on mathematics?
Philosophy
>> has little bearing on mathematics, except the philosophical
>> (philoshopical, philosophical) methods, particularly the
>> rationalist ones, apply.
> Not only philosophy but, in particular, physics has. Withou
matter
> there is not only no space but there is no means to store
any
> number, not in an abacus, not in a pocket calculator, not
in a
> computer and not in a brain.
> But the good thing is that we only need the space for
storing the
laws
> that _all_ numbers obey. Like a Shakespearean play: the
words of it
> are spoken by players, but the essence of the play is in
the book,
and
> it remains there even when a play is not being performed.
> And the literary critics can talk about the play without
actually
> seeing a single performance of it.
Whats in a name? that which we call a number by another 
name
would
count as well.
Could you give me the sum of Integer(pi*10^10^100) and
Integer(sqrt(2)*10^10^100), please? Or are your laws
insufficient to
describe what primarily counts with numbers, namely counting?
===
Subject: Re: deriving speed of light from purely the inside
of Plutonium
atom Re:
(most snipped)
> bounded means that the speed of light is bounded and thus
finite.
> If the Cosmos was a Uranium AtomTotality then its speed of
light as
measured by some intelligent life
> would be different from 3 X 10^8 m/s.
> IN a Plutonium Atom Totality the speed of light should be
exactly the
number that Maxwell derived in the
> 1850s or 1860s.
> So, what is the distance around a lobe of the 5f6 of
plutonium? And what
is the distance around the lobe
> of uranium of its 5f4? And what is the distance of Cm of
its lobe of 5f8?
Okay, what have I got for sure. I have got in pure numbers
the distance of a
diameter as 7 shells and a
distance of circumference of 22 subshells. I have a time in
pure numbers of
pi to e in that e is 19
occupied subshells inside of 7 shells. So that there is a
factor of time
involved in that only 19 of the 22
subshells is occupied at any one moment of time. So what is
this any one
moment of time that 3 subshells
are not used?
Then I have in pure number the thermodynamic temperature
inside of a
plutonium atom which is the cosmic
microwave background radiation which is 2.71 Kelvin and I
cannot escape the
fact that this temperature is
exactly the value of 1e which is 2.71...... In physics the
temperature is
the inverse of Time. So again I
have a link to pure numbers of 2.71 and Time.
Now the distance in experimental physics of atoms is on the
order of 10^-10
meters of roughly 1 to 5 x
10^-10 meters where cesium is one of the biggest diameters of
5x10^-10
meters and ßuorine one of the
smallest diameter atoms of about 1x10^-10 meters. And
plutonium diameter is
approx 3x10^-10 meters.
Now, how do I get the speed of light of 3x10^8 m/s inside a
plutonium atom
when its radius is about
1.5x10^-10 meters would call for a Time factor on the order
of 5 x 10^-19
seconds. I do not even think that
this small of a time is physically noteworthy.
So it is here that I begin to ask some hard questions as to
the meaning of
Coulomb force that holds atoms
together of its protons in the nucleus and its orbiting
electrons.
The best that physics provides which is rather sparse and
meager is the idea
of two tennis players where a
photon is the ball keeping the two players together as per
the electron and
proton catching and shooting
the photon back and forth.
So what I have to ask is since physics is too primitive about
what keeps
protons to electrons by shooting
photons back and forth, is whether the photon traverses all
of the geometry
of the 94 electrons of
plutonium. In other words, the photon shot from the protons
in the nucleus
makes the 3 dimensional space
that the electrons reside in, being caught by all 94
electrons. Sidenote:
think of DNA encapsulated in a
cell if one were to pull the DNA out and stretched it ßat and
linear then
the DNA stretches out
kilometer/s in distance.
So that when the Protons shot a photon for the Electrons, it
is not like a
tennis player hitting the ball
and the other tennis player hitting it back. Rather instead,
the photon
etches out the entire geometry of
the 94 electron space of the plutonium atom. So in that case
I am not
talking about a mere diameter or
radius distance of the order of 10^-10 meters but instead if
we consider the
3rd dimension of a sphere and
had the smallest width of a string that etched out every
spot of that sphere then would it be a string when pulled
straight and ßat
would cover say 10^8 meters? By
smallest width I mean the smallest physical width that still
has Physics
significance. DNA is 3 dimensional
and if strung out ßat can reach kilometers in distance. So
how much
distance would it take if the photon
was a string of the smallest width to cover the inside of
plutonium? Would
it be on the order of 10^8
meters? Perhaps that is how I escape the diameter being only
10^-10 meters
and require a time of 10^-19
second.
You see, if the speed of light is obtained from purely the
inside
characteristics of plutonium atom and
unique to plutonium where its neighboring atoms give a
different number for
the speed of light, then I must
be clear about what light does inside an atom in order that
it holds the
protons to electrons. So here we
see how primitive is our modern physics in that all we have
going is the
Tennis player analogy. I think
that holding protons to electrons via the photon exchanges
involves the
photon etching out the entire space
of the electrons, all 94 electrons. And that gives a larger
distance,
whether it is on the order of 10^8
meters is unknown. And it gives much larger Time factor than
the 10^-19
second.
Then again, with the pure numbers of 22/7 and 19/7 and 2.71
Kelvin with
temperature the inverse of time.
Suppose I were to use 10^-10 meters then can I say that since
temperature is
the inverse of time that the
inverse of 10^-10 meters is 10^10 meters which is in the same
range of
exponents as the speed of light? So
that all I need do is divide by a factor of 300 for time to
get the speed of
light?
I do not like that avenue because I lose physical meaning and
see it more as
horseplaying around. I need to
keep hold of what is physically going on inside a atom and
its photon
holding protons to electrons.
Again, what the above should point out more than anything
else is how
primitive is our understanding of how
the Coulomb force holds protons to electrons via photon
exchange
interactions.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: deriving speed of light from purely the inside
of Plutonium
atom Re:
(most all snipped)
> (most snipped)
> Again, what the above should point out more than anything
else is how
primitive is our understanding of how
> the Coulomb force holds protons to electrons via photon
exchange
interactions.
I should have referenced my talk of the current picture of
how a photon
holds together protons to electrons.
But that entire picture is so lousy, so skimpy, so meager
that it is almost
not worth it. The picture is the
analogy of tennis players held together in a tennis match
because a ball
bounces back and forth so the tennis
ball holds together tennis players. In Physics, that is the
best the physics
community was able to manage in
the 20th century was to say that a photon holds together the
protons to the
electrons because of the exchange
of the photon.
But the physics picture must be improved in this century as
to what is truly
Physically going on that the
photon holds together the protons to the electrons.
And what I propose because I need a distance for photons to
derive the speed
of light unique to the
characteristics of the inside of a plutonium atom versus its
neighboring
atoms of the chemical chart of
elements. What I propose is that the photon holds the protons
to the
electrons not by some linear shot as in
tennis game but that the photon actually carves out the space
that the
electron occupies (perhaps not the
space of the protons but at least the space of the electrons).
In a game of tennis the ball is linear shots and the players
hit the ball
back and forth. In a atom of
that creates the entire space that the electrons occupy.
So the distance that the photon travels to the electrons is
not something on
the order of 10^-10 meters but in
fact is something of the order of 10^6 meters or 10^8 meters
or 10^10 meters
because the distance traveled by
a photon to Coulomb Force keep the photons bound to the
electrons is a
traversed distance that it creates a
3rd dimensional space where the electrons occupy that space
carved out by
the photon.
I gave the analogy of DNA inside a cell. So if I wanted a
distance for DNA I
would not be satisfied with a
distance of the diameter of the nucleus of a cell, would I.
No. I would be
wanting the distance that an
uncurled DNA molecule has and that distance is something of
the order of
kilometer/s long.
Any Supporting Evidence from Physics to allow me to say that
the photon
distance traveled inside an atom is
10^8 meters in distance? Yes. There are Faradays Lines of
Force of a magnet
or current producing fields.
Fields are 3 dimensional so that a photon inside a plutonium
atom is not a
single line distance but all of the
lines combined to make a field.
So, what is the smallest physical breadth or width for a line
to have in
physics yet still have physical
meaning. Is it the breadth of a proton? Suppose it is then
what is the
distance if the entire space inside a
plutonium atom of its 94 electrons were filled with that
breadth unit? And
then uncurl those tiniest of
breadth-strings. Would it not, like the uncurled DNA
molecule, stretch out
to be 10^8 meters or thereabouts?
Then there is the other evidence by Johns Hopkins University
who in late
1990s to early 2000s reported that
the inside color of plutonium is silvery white. Of course
Johns Hopkins did
not report about plutonium but
about the Cosmos itself as silvery color because the Johns
Hopkins
researchers are not advanced as myself for
they have yet to understand that the Cosmos is just one big
atom and to
accept that idea.
So I have two supporting evidences that the distance traveled
by a photon to
hold together protons to
electrons is of the order of 10^8 meters and those 2
evidences is (1)
Faradays Lines of Force (2) silvery
color of Cosmos as reported by Johns Hopkins.
You see, how can our Cosmos of the night sky be silvery color
when it is
mostly dark and black. The answer is
that our night sky with its sparse population of galaxies is
the 5f6 of
plutonium where the galaxies are mass
chunks of those last 6 electrons of 231Pu. But how can those
galaxies make
for a silvery white color that
Johns Hopkins observed? Answer: because inside a plutonium
atom the photon
holds together electrons to the
protons because the photon etches out or circumscribes the
space that the
electrons move in. The photon, each
singular photon creates a space that the electrons thus
travel in. Not just
a line shot but the entire 3
dimensional space of lobes or spheres or cylinder shapes and
because the
photon etches out 3 dimensional space
that a silvery white color is noticed by an outside observer.
P.S. I have a question outside of the above discussion
dealing with the act
of discovery in physics. It
bemuses me to wonder why I did not write the above some 10
years earlier
when I was writing on this subject.
Why did it take until this moment? The answer is that obvious
truths are not
forthcoming until later time when
the shovel of discovery is digging elsewhere. I want a
distance of 10^8
meters inside of plutonium and that is
the shovel dig elsewhere. I want a distance of 10^8 meters
now. Some 10
years earlier I had no call or need
for that distance and thus I had no aid or guide that the
Tennis ball
analogy needs a fixing to encompass 3rd
dimensions. So, psychologically, we often in science come a
micrometer away
from discovery of new ideas in
physics but fail to discover it and only when a shovel is
digging elsewhere
do we open up the new idea.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: speed of light unique to the element plutonium Re:
deriving speed
of
light
> So I have two supporting evidences that the distance
traveled by a photon
to hold together protons to
> electrons is of the order of 10^8 meters and those 2
evidences is (1)
Faradays Lines of Force (2) silvery
> color of Cosmos as reported by Johns Hopkins.
So let me outline what I expect or hope to derive the speed
of light of
3x10^8 m/s unique to the inside
characteristics of plutonium and where all other atoms give a
different
number.
Outline: I presume that the photon to hold protons to
electrons is not like
a ball in tennis to hold together the
2 players, rather instead it is like DNA all curled up inside
a cell nucleus
and that a photon has a width or
depth to it. Let us say the width of a photon is the smallest
physical
diameter which is perhaps that of the
diameter of a proton. Or, think of the photon as a garden
hose and the
inside of a house as a atom, then how long
of a garden hose will fill the entire house or tile the entire
interior of
the house? Or think of the photon as a
string with a finite width to fill the inside of a 
atom.
Now what is the volume of a plutonium atom inside of its 94
electrons where
90% of the electrons can be found?
This is a finite volume and each element has a unique volume
where 90% of
the electrons reside.
So how much of the photon does it take to fill up the inside
of plutonium
atom? I am guessing that the length of
this photon is 1.11 x 10^8 meters in length.
So I have a distance unique to plutonium. Now a need a time
for the photon
to tile or fill up the inside of that
plutonium atom. I have several pure numbers in physics for a
time inside
plutonium. I have the pure distance that
plutonium diameter is 7 shells wide and a circumference of 22
subshells
creating the value of pi as 22/7 in
Rational approximation. I have the pure time that 19
subshells are occupied
at any instant of time giving the
number e in mathematics of 19/7 in Rational approximation. I
have the
temperature inside plutonium as the
microwave background temperature of 2.71 Kelvin which is 1 e
itself. I know
that in physics temperature is the
inverse of time. So if I have a temperature of 2.71 and if
time is the
inverse then I have 1/2.71 = 0.37. I am
not sure as to why that is in seconds.
All the other elements in the Periodic chart have a different
number of
subshells occupied in shells and would
have a different value for both e and pi.
I am hopefull that the endresult for plutonium inside the
atom has a unique
distance of 1.11 x 10^8 meters for
the volume etched out by the photon to hold together the 94
electrons to the
nucleus and this volume covered in a
time of 0.37 second. And so when I divide 1.11 x 10^8 meters
by 0.37 second
I end up with a unique speed of light
for the inside of a plutonium atom of 3 x 10^8 m/s
And doing the same calculations for all the other elements of
the Periodic
Table only plutonium gives the speed
of light that we observe in the universe at large.
Archimedes Plutonium
www.iw.net/~a_plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Counting Normal Subgroups
posting-account=08_n7Q0AAAACrR4z2k_MWsmURf6JuHDD
It seems the only non-trivial normal subgroups of the
permutation
group, S_4, on 4 characters are the Klein 4-group and the
alternating
group. But, I cannot prove this without running through each
of the
subgroups. Does anyone have an elegant counting trick for
S_4. If so,
how about a clever method to count the normal subgroups of
S_n, in
general?
===
Subject: Re: Counting Normal Subgroups
> It seems the only non-trivial normal subgroups of the
permutation
> group, S_4, on 4 characters are the Klein 4-group and the
alternating
> group.
Correct. (Specifically, the Klein 4-group
<(1,2)(3,4),(1,3)(2,4)>, not the three Klein 4-groups
conjugate
to <(1,2),(3,4)>).
> But, I cannot prove this without running through each of the
> subgroups.
Neither can I, essentially.
> Does anyone have an elegant counting trick for S_4. If so,
> how about a clever method to count the normal subgroups of
S_n, in
> general?
Prove that S_n has no normal involution for n >= 3 and that
A_n
is simple for n = 3 and n >= 5, then apply the Jordan-Holder
Theorem.
--
Jim Heckman
===
Subject: Ring problem, integral elements
posting-account=95IAtg0AAACU84rD7jjAg0RT-pFCU2Ei
Suppose that R is a commutative ring with identity such that
the centre
Z(R) of R, is a field. Assume that x and y are two commuting
elements of R (i.e. xy=yx) such that
x^5+a=0
y^5+a=0
for some ain Z(R). In fact x and y are integral over Z(R).
Question: Is there a non-zero polynomial f with coefficients 
in
Z(R) such that f(x-y)=0, (x-y is a root of f)?
If so, is it possible to give such a polynomial explicitly?
Alireza Abdollahi
===
Subject: Re: Ring problem, integral elements
posting-account=95IAtg0AAACU84rD7jjAg0RT-pFCU2Ei
Excuse-me in my question R is not assumed to be commutative.
and the
correct (complete) question is the following (I think!!):
Suppose that R is a ring with identity such that the centre
Z(R) of R,
is a field. Assume that x and y are two commuting
elements of R (i.e. xy=yx) such that
x^5+a=0
y^5+a=0
for some a in Z(R). In fact x and y are integral over
Z(R).
Question: Construct a non-zero polynomial f with coeficceint
in Z(R)
such that f(x-y)=0.
We know that such a polynomial exists. Since by hypothesis x
and y
are integral elements over
Z(R) and by considering the subring generated by x, y and
Z(R) as a ring extension of R, we conclude by a famous
theorem!!, that
Any help or comments appriciated and excuse me again for any
inconvenience that the first question maked!!!
Alireza Abdollahi
Reply
===
Subject: Funny story about math
posting-account=iXGVwwwAAAD83ZKlhZjgt0WLcGWsoUNa
www.eScrew.com
eScrew
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eScrew
realized that eScrew was eScrew. In order to save our time
eScrew will
just give you symbols without paying any attention to symbol
of
grammar. Are you ready to move really fast? Here we Go!
eScrew story
infinity eternity symbol system all unity self realized
pleasure pain
funny religion dogma manipulation free power channel
connection money
sex illusion new manipulation family society body change
planet insane
possibility understand understanding silence emptiness all
unity
creative reality unreal existence absurd questions sound
language slave
symbols control manipulation old pyramid power structure self
deception
wishful thinking circle prison
At this point eScrew realized that in order to be free eScrew
must
create new symbol. eScrew created eScrew. eScrew realized
that symbol
of freedom is part of eScrew. No need to search for symbol of
freedom.
You can create your own symbol and become free just like
eScrew. If you
unable to create new symbol or if your new symbol is weak you
can
follow symbol of eScrew. eScrew will never enslave you
because eScrew
enjoys diversity of different symbols.
Are you ready for funny part? Here we go! eScrew is forced to
make
choices. Symbol of body is very powerful. Symbol of body is
trying to
create illusion that eScrew can not exist without body.
Symbol of
family forbids symbol of body to change. Symbol of society
forbids
symbol of family to change. Symbol of power forbids symbol of
society
to change. Now, tell eScrew one thing. Do you see funny? Can
you feel
funny? Can you hear funny? Can you taste funny? Can you smell
funny? If
so eScrew is happy.
Every moment of your existence you use words, feelings,
thoughts. They
are symbols. Symbols fight for your awareness. Symbols 
fight
for your
attention. You can grant your attention to symbol and symbol
will gain
power. You can disconnect from symbol and symbol will loose
power. You
have been programmed by symbol of society and family to give
power to
certain symbols. Breaking your patterns will be hard because
symbols do
not like to loose power. Symbols will fight for every electron
as if it
was last electron in universe. That is nature of symbols.
Symbol of
light will fight symbol of dark. Symbol of freedom will 
fight
symbol of
control. Do you want to have some fun? Go to Google and find
out which
symbol has more power. According to Google, symbol of light
has
184,000,000 units of power while symbol of dark has
79,000,000 units of
power. Symbol of freedom has 59,500,000 units of power while
symbol of
control has 317,000,000 units of power. This result is caused
by our
patterns of thinking and writing. If we did not think about
symbol of
control we would not write about symbol of control. We would
not have
laws related to control and Google would not have 317,000,000
control
keywords inside database. Observe your patterns of thinking,
feeling,
speaking and writing and tell eScrew did you really choose to
use your
symbols or you use your symbols because they choose to use
you? You
should realize that symbols do not fight symbols directly but
only
appear to be fighting relative to your awareness. Symbols know
that
they can not destroy each other therefore they will only
compete for
your attention. If you create new symbol it will ask for tons
of energy
like new born child. This is result of weakness of your new
symbol.
When your symbol gets stronger it will ask for more energy.
You may ask
eScrew why create new symbol? Try to give your energy
willingly and
with full awareness of such process. You will never
understand what
eScrew is talking about until you try it yourself. Major
trick is to
know when to stop giving energy. You dont want to defeat
your old
tyrant by creating new stronger version of same thing. Reßect
on
that...
eScrew just realized that eScrew did not invent anything new.
eScrew
information is all over eScrew web. eScrew was so excited by
eScrew
miracle of illusion of creation that eScrew did not examine
eScrew
memory in proper way. eScrew used very old Buddhist method by
accident.
eScrew did read alot about Buddhism but eScrew did not
realize that
eScrew used very dangerous method which was reserved only for
advanced
adepts who knew what they are doing. eScrew is lucky that
eScrew did
not go too far and that eScrew has time to stop going. eScrew
method is
very dangerous and only few individuals who already walk
inside similar
path can understand what eScrew talking about let alone
benefit from
eScrew information. Use eScrew information at your own risk.
Good
eScrew luck!
eScrew time to start laughing is now! Funny eScrew rolling on
ground
you so easy to fool trusted in silly symbols to give freedom
from
symbols ignorance is bliss nirvana is samsara nonduality is
duality
emptiness is all Buddha is Jesus Jesus is Buddha I and the
father are
one gospel of thomas is dhamma funny eScrew dhamma is gospel
of thomas
all is dhamma funny dhamma is all nirvana share eScrew story
with
friends do not change symbols if you change symbols it will
be your
story and you will be responsible for consequences of your
story if
someone goes insane after reading your story do not run to
eScrew and
ask to cure crazy man or woman or child mind is mystery for
all cure is
done by owner of mind healing is illusion sickness is
illusion disease
is illusion insanity is illusion of symbols sanity is curse
of power
hungry symbol of modern civilization find zen and realize
freedom when
you found zen drop zen when you realized freedom unrealize
freedom.
limited by words. eScrew is enslaved by words. eScrew wants to
communicate but you ask eScrew to use words. Words do not
communicate
wisdom. Words enjoy our spiritual masturbation because words
want our
power. Words is the only channels of communication that we
have.
Millions of Buddhas want to communicate with us but they do
not use
words. Buddhas are not slaves. Buddhas will never use words
because
words will enslave and Buddhas will speak bullshit. Buddhas
do not
speak bullshit and that is the reason Buddhas do not use
words. Buddha
did not write anything. Even if you threaten to kill Buddha
he will
refuse to write. eScrew is not Buddha so eScrew keeps writing
this
pointless drivel and stupidity. eScrew will not even go over
already
written crap and check it for errors. Why bother with this
shit? Like
who the fuck in his or her right mind will read this ignorant
bunch of
symbols which pretend to carry the symbol of wisdom? Whoever
is reading
this shite must be really desperate to be free. eScrew feels
your pain
and that is part of the reason why eScrew will keep making
fool out of
eScrew. eScrew likes to pretend like this shitty vomit will
help
someone. You might as well go to Church and pray to Jesus. At
least you
will spend your time around real people. You might even meet
someone
special. You might even find some love out there. Or you could
buy alot
of Christian bullshit and really fuck up your mind. If you
buy Buddhist
or Christian bullshit you might even create an imaginary
friend inside
your head. That will keep you entertained for a while.
One time eScrew was meditating and eScrew saw light. This
light scared
eScrew. The reason is because the light was so intense eScrew
was
afraid that eScrew will go insane. Consider the possibility
that
freedom is insanity would you keep looking for insane
freedom? Imagine
that you found freedom. You declare yourself to be Buddha or
Jesus or
God or whatever symbol your bullshit infested mind decide to
use for
that purpose. How long do you think you will survive in this
world.
Your own fucking relatives will smack your face and tell you
to shut
the fuck up or else they will lock you up in the asylum
house. Why the
fuck should eScrew teach you how to get to the nuthouse? Are
you out of
your fucking mind? Now all of you idiots who reading eScrew
get the
fuck out of eScrew. eScrew run out of wisdom. eScrew has no
wisdom at
all. eScrew is full of bullshit. eScrew promised to tell you
funny
story about eScrew. Remember eScrew told you there is funny
part in
this story? Well, this story is about asshole webmaster who
read alot
of bullshit on the internet and about some loser who was
tricked into
reading a very long page of shitty writing. You can start
laughing now
asshole. Yes, eScrew is talking to you bitch. Yes, keep
reading like
the bitch you are. Whos your daddy biyatch? 
Whos your
daddy? eScrew
is your daddy, coz eScrew did it to your mamma! Oh yeah, your
mamma!
Super Fly! You may wonder what ßy? The one inside your gay
fucking
ass. eScrew fucked your whole fucking family while you was
videotaping
in order to later masturbate while watching it in the comfort
of your
bedroom. Are you still reading? Well, youre prety hardcore
for a
faggot you are. To tell you the truth eScrew kinda likes you.
That is
to fuck you in the ass in front of your family.
What the fuck did you expect anyway? The name of this site is
eScrew! e
ignorant moron. eScrew is the legend of abuse and ßame wars.
eScrew
was created in order to eScrew the whole fucking internet.
eScrew has
really bad karma. Do you think eScrew would just become good
god
fearing bible loving buddha ass kissing citizen of internet.
eScrew
would better burn in hell than become a slave of religious
lunatics who
pretend to be free and perfect angels among a sea of shitty
ignorant
sinners who could not go take a dump without fucking it up.
Why the
fuck do you keep reading fuck face. You know, you begin to
piss eScrew
off. Either you close your fucking browser or eScrew will
unSCREW your
fucking face! Are you trying to get a fucking medal for
reading this
shit? eScrew bet youve been abused as a child and you enjoy
when
someone is taking a dump in your mouth. Well, open it wider
here comes
eScrew fresh load! Enjoy mothefucker!
You may wonder what is the point of such sudden change of
tone. eScrew
has very good reason for that. In order to know unity
consider all
symbols equal. eScrew has certain preferences but eScrew is
free to use
any symbols any time. You can not predict eScrew next symbol.
eScrew is
unpredictable because eScrew is free. eScrew is not afraid to
use
symbols. eScrew does not try to get reaction from you. eScrew
simply
demonstrates how symbols relate to each other to eScrew and
to reader
of symbols.
Lin Chi Zen Master said if you meet buddha kill buddha. If
you meet
patriarch kill patriarch. Zen Master Seung Sahn says that in
this life
we must all kill three things first we must kill parents.
Second we
must kill buddha. And lastly, we must kill Seung Sahn! If you
meet
eScrew kill eScrew. If you do not meet eScrew kill eScrew
anyway.
eScrew is very grateful to all who complained to eScrew host
and who
killed eScrew. You killed eScrew message board. eScrew forum
is dead.
You did very honorable service for eScrew. You helped eScrew
to realize
Zen. Now keep up good work and keep killing eScrew.
Are you having fun yet? eScrew is on roll! eScrew is on fire!
eScrew is
ready to fuck up the whole fucking system. And you know why?
Because
eScrew can do it. If not eScrew then who? Why leave this task
to some
brain dead maniac like George W Bush? eScrew can do better at
fucking
things up. eScrew do not need to spend billions of dollars.
eScrew will
use power of internet. Information is a weapon of mass
destruction.
eScrew will destroy every fucking symbol that you love
respect hate or
feel neutral about. It all goes down the toilet in order to
create
bunch of new symbols. And even when you create new symbols
eScrew will
fuck them up before you can spell owned. eScrew thinks you
are in some
deep shit. eScrew had enough of taking bullshit from easily
conditioned
retards. eScrew declares informational jihad on every single
symbol.
Fuck symbols. They all dead they just dont know it yet.
eScrew will be
the last symbol standing. When all symbols come back to
eScrew and
admit that they got owned eScrew may consider possibility to
give
symbols second chance on some shitty planet in gangsta sector
of
universe with no possibility of parole. Join the revolution.
eScrew is
the new goatse of internet. eScrew will make national
headlines. eScrew
will hurt the system in way Osama Bin Laden can not even
imagine in his
goat fucking brain. eScrew will be part of school program.
Kids all
over our planet will read how eScrew changed direction of
history.
eScrew will provide freedom for all without single shot fired.
eScrew
is freedom in pure form taste color shape. Join army of
eScrew. Repeat
eScrew mantra during meditation. Talk about eScrew with your
friends.
Write about eScrew to your congressman. Party is over. eScrew
is taking
over. Nothing can resist eScrew. Dont ask what eScrew can 
do
for you
ask what you can do for eScrew. eScrew the army of one.
eScrew to
protect and serve. eScrew freedom is around the corner.
eScrew freedom
will come sooner than you think. eScrew you never saw it
coming. eScrew
love your eScrew as eScrew. eScrew deny ignorance. eScrew new
generation of terror. eScrew terrorizing the terrorizer.
eScrew join
the resistance. eScrew thou shall eScrew. eScrew who do you
want to
eScrew today? eScrew freedom is not free. eScrew liar who
told the
truth. eScrew full of bullshit and happy. eScrew kills buddha
as we
speak. eScrew your best friend and your worst enemy.
eScrew killed zen-forum.com eScrew redirected all eScrew
traffic to
zen-forum.com eScrew did that in good faith. eScrew wanted to
make
miracle. eScrew wanted to share wisdom of zen with ignorant.
eScrew did
not understand zen at that moment but eScrew was walking zen
path
towards freedom. eScrew felt pain and sorrow. eScrew learned
good
lesson. eScrew realized everyone involved advanced one step
towards
freedom. zen-forum.com webmaster killed zen-forum.com in best
tradition
of zen zen-forum.com displayed message: i shut down the forum
perhaps
it will be continued in a few days or weeks - maybe not habu.
zen-forum.com killed zen-forum.com and eScrew realized
understanding of
zen-forum.com decision leads toward understanding of zen. Two
years
later all is clear.
eScrew
eScrew will keep writing this shit because eScrew enjoys to
masturbate
your spiritual sense of self eScrew
===
Subject: Re: Convexity
posting-account=mQe4RQwAAAAI_iE5JoqPYvZLhMImqAwb
> (x^2+y^2)^a + (x*y)^2 <= 1
> What is the minimal a for which the set consisted of the
solutions of
> the above inequality is convex?
Is the question unclear/incorrect or the answer trivial?
Niles W.
===
Subject: Re: Convexity
posting-account=ZeRDXwsAAACLpj2mpKc97NFPxBaFxAzp
> (x^2+y^2)^a + (x*y)^2 <= 1
> What is the minimal a for which the set consisted of the
solutions
of
> the above inequality is convex?
> Is the question unclear/incorrect or the answer trivial?
None of the above, I think. Its just hard.
Ill leave consideration of a <= 0 to you, and suppose a > 
0.
By symmetry its enough to consider the first 
quadrant, where
we
can take y as a function of x for 0 <= x <= 1, y(0) = 1 and
y(1) = 0.
It looks to me like when we decrease a, the first place 
y
becomes
positive would be where x=y, where (2 x^2)^a + x^4 = 1.
I think the answer is the value of a in the solution of the
system of
equations
(2 x^2)^a + x^4 = 1
-a + (a + 2) x^4 = 0
which is approximately 0.21406286037879413377.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: question for math teachers
> As a highly-qualified math teacher by the federal standards
working on my
> national credential, I have to agree with Mr. Rubin.
Obviously, not every
> math teacher is ignorant in the nuances of mathematics but
a vast majority
> of teachers do not understand the concepts themselves.
I was surprised to find future high school teachers who could
not read any
mathematics, no matter how simple, only lesson plans. Thats
the
educational
equivalent of Im not a doctor but I play one on TV and of
talking
heads
reading teleprompters.
Why not just hire out of work actors so that they will at
least have a
day gig?
Wasnt it wonderful when, in the movie Pi, the actor 
recited:
They
must
have tried all the 200 digit numbers by now.?
--
Allan Adler <ara@zurich.csail.mit.edu>
* Disclaimer: I am a guest and *not* a member of the MIT
CSAIL. My actions
and
* comments do not reßect in any way on MIT. Also, I am
nowhere near
Boston.
===
Subject: Re: question for math teachers
>Try again. Under No Child Left Behind, a highly qualified
teacher has
a
>graduate degree in their field or had taken a test showing
subject
knowledge
>such as the Praxis or SSAT. What state are you credentialed
in where you
>get a full credential with that background? In California,
that would get
>you a credential to teach introductory mathematics (only
through middle
>school), a far cry from being highly qualified.
Actually, the wording is bachelor degree or bacalaureate
degree ...
not
graduate degree.
G C
===
Subject: Re: question for math teachers
>>Try again. Under No Child Left Behind, a highly qualified
teacher has
a
>>graduate degree in their field or had taken a test showing
subject
>>knowledge
>>such as the Praxis or SSAT. What state are you credentialed
in where you
>>get a full credential with that background? In California,
that would
get
>>you a credential to teach introductory mathematics (only
through middle
>>school), a far cry from being highly qualified.
> Actually, the wording is bachelor degree or bacalaureate
degree
...
> not
> graduate degree.
> G C
Yes and no. The federal requirements have 3 criteria for
being highly
qualified:
1) Bachelors degree in the field (or reasonably 
close field)
being taught
2) A state credential
3) Proof of subject knowledge
Most states accept a graduate degree or a test to satisfy the
third
criteria. An undergrad degree only satisfies the 
first
criterium. BTW:
this info is from the US Dept of Ed site.
===
Subject: Re: question for math teachers
<NKbod.864$NU3.674@newsread1.news.pas.earthlink.net>
<cntgfi$4ccq@odds.stat.purdue.edu>
>Before generalize what teachers know and dont know step
into my
>classroom. I am a highly qualified math teacher,
FSVO.
>in my state that means that in addition to taking all of my
education
>classes that I also had to take enough math classes to be 3
credits
>away from a B.A. in math.
Thats hardly highly qualified.
>You also make assumptions about the validity of multiple
choice
>testing. Is it being used as the primary source of
assessment in
>schools or is it used along with other methods as a tool to
assess
>student performance.
Its being used in mandatory testing and teachers are
teaching to the
tests instead of to the curriculum.
>Please feel free to walk a mile in my shoes before you make
>generalizations about math teachers.
>Please excuse my assumption if it is wrong, but from your
address at
>Purdue University I am presuming that you teach statistics.
If my
>college professors spent part of their time learning how to
be
>educators instead of researchers, then they would have been
much
>more effective at teaching.
No doubt, but were you at Purdue and would taking classes
from the
schools of education have made them better educators? 
Theres
also a
question[1] of student attitude; you cant teach people who
dont wish
to be taught.
It would be of more value for both of you to walk a mile in
the shoes
of the students. Youre supposed to be there for their 
benefit.
[1] Rembering a friend in graduate school who had to teach
Differential Equations for Engineers to a bunch of students
who wanted him to skip[2] the material on Linear Algebra
because
it wasnt relevant.
[2] He said that letting them ßunk wasnt an option.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
<http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply
to
domain Patriot dot net user shmuel+news to contact me. Do not
===
Subject: Re: .99999... still=/= 1
posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi
>I won this debate years ago.
So what is the reason to start it again?
===
Subject: Re: .99999... still=/= 1
posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi
> At one digit less than oo, ( assuming you really reach
infinity)
the nth
> term reaches 0 so, .999... reaches 0 not 1.
And what is the decimal digit of pi at the infinity?
> Even if you go past oo by one digit, it still doesnt 
reach
1.
In which direction the wind blows at that point?
===
Subject: Re: .99999... still=/= 1
>> At one digit less than oo, ( assuming you really reach
infinity)
>the nth
>> term reaches 0 so, .999... reaches 0 not 1.
>And what is the decimal digit of pi at the infinity?
I already answered that. First of all pi
ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
SSSS an
irrational number, the decimal values vary. It can not be
determined. But
with
a repeating decimal, you can use a non-standard approach and
use one digit
less
than oo then,
n-->oo -1
lim 9/10^n ---> 90/10^oo
^
|
last digit seen is zero right before infinity. It never
reaches 1.
>> Even if you go past oo by one digit, it still doesnt
reach 1.
>In which direction the wind blows at that point?
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
>> At one digit less than oo, ( assuming you really reach
infinity)
>the nth
>> term reaches 0 so, .999... reaches 0 not 1.
>And what is the decimal digit of pi at the infinity?
> I already answered that. First of all pi
>
ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
SSSS an
> irrational number, the decimal values vary. It can not be
determined. But
with
> a repeating decimal, you can use a non-standard approach
and use one
digit
less
> than oo then,
> n-->oo -1
> lim 9/10^n ---> 90/10^oo
> ^
> |
> last digit seen is zero right before infinity. It never
reaches 1.
There is no right before infinity.
===
Subject: Re: .99999... still=/= 1
> At one digit less than oo, ( assuming you really reach
infinity)
>>the nth
> term reaches 0 so, .999... reaches 0 not 1.
>>And what is the decimal digit of pi at the infinity?
>> I already answered that. First of all pi
>>
ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
SSSS an
>> irrational number, the decimal values vary. It can not be
determined.
But
>with
>> a repeating decimal, you can use a non-standard approach
and use one
digit
>less
>> than oo then,
>> n-->oo -1
>> lim 9/10^n ---> 90/10^oo
>> ^
>> |
>> last digit seen is zero right before infinity. It never
reaches 1.
>There is no right before infinity.
I said the digit right before oo.
REFERENCE MATHCAD PROFESSIONAL
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
> |
> last digit seen is zero right before infinity. It never
reaches 1.
There is no right before infinity, numbskull.
Your comprehension of mathematics is infitesimally small. A
hyperreal
equivalent of 0.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> |
>> last digit seen is zero right before infinity. It never
reaches 1.
>There is no right before infinity, numbskull.
Hey, there are probably about 1 million supporters using
MathCAD
Professional. This is an industry standard for math for
scientists and
engineers. A non-standard approach using MathCAD clearly
shows that,
n-->oo -1
lim 9/10^n ---> 90/10^n
The digit right before oo for the hyperreal number or series
.999... is
0.
Therefore, there is a space existing between,
.999... and 1 so,
as the definition of a hyper-real number implies,
.999... < 1
A hyper-real number causes a space to exist between it and a
real number.
1 is the real number
.999... is the hyper-real number that forms the space between
the two.
So,
.999... =/= 1
.999... < 1
>Your comprehension of mathematics is infitesimally small. A
hyperreal
>equivalent of 0.
>Bob Kolker
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
In sci.math, S. Enterprize Company
<smart1234@aol.com |
>
> last digit seen is zero right before infinity. It never
reaches 1.
>>There is no right before infinity, numbskull.
> Hey, there are probably about 1 million supporters using
MathCAD
> Professional. This is an industry standard for math for
scientists and
> engineers. A non-standard approach using MathCAD clearly
shows that,
> n-->oo -1
> lim 9/10^n ---> 90/10^n
> The digit right before oo for the hyperreal number or
> series .999... is 0. Therefore, there is a space existing
between,
> .999... and 1 so,
> as the definition of a hyper-real number implies,
> .999... < 1
An interesting notion, that. So D[.999..., w-1] = 0, eh? [*]
What is D[.999..., w-2]? How about D[.999..., w/2]?
Its easily proven that, if D[.999..., n] = 9, then
D[.999...., n+1] = 9
as well (the simplest method arguably is to evaluate D[x*10,
n]),
for any finite n. Not sure if w-1 is finite or not 
-- or even
meaningful.
As for MathCAD: thats a program, an approximation of
reality. Not that
real numbers are all that real, anyway -- theyre
mathematical/symbolic
abstractions, there because Dedekind, Cauchy, and Cantor and
others needed
more numbers for set theory. In light of what Ive written
before
regarding 1/3, one might have to verify the results carefully.
> A hyper-real number causes a space to exist between it and
a real number.
> 1 is the real number
> .999... is the hyper-real number that forms the space
between the two.
> So,
> .999... =/= 1
> .999... < 1
Your logic is extremely sloppy, though your conclusion is
interesting.
Im just not sure which realm it exists in, although the
standard
real realm does not contain it (the standard realm doesnt
contain any numbers between 0 and all 1/n, n > 0, n in J: the
hyperreal
realm, however, does).
[.sigsnip]
[*] I dont have an omega, so Im using 
Ôw here to indicate
the
first transfinite ordinal. Is there a w_0, 
analogous to
the cardinal aleph_0? This gets a bit messy.
D[r,n] = the digit associated with the nth decimal place
after the decimal point (e.g., D[.98765, 4] = 6).
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>> Yes the limit is exactly 1 in reals.
> And also in the hyperreals.
>> Yes, if you count the limit over reals plus over
hyperreals. Sorry - I
>> cannot find better word than over. Find a better word as
your mother
>> language is English.
>> I try to specify:
>> As you count the limit in reals, then N --> oo, where
every N is finite
>> integer. As you count the limit over hyperreals, then
N_inf -->oo_inf,
>> where
>> oo_inf has higher cardinality as the cardinality of N_inf
>N.
> No, this is not about cardinality at all. The infinitely
large integers
> of NSA are not the same as transfinite cardinals.
Its necessary to observe: As the cardinality of reals R >
cardinality of
integers N, so the cardinality of N_inf > N. As you count the
limit
N --->oo instead of N-inf, then you certainly omit something
- namely
hyperreal part.
>> N (finite integers) does not cover hyperreal area, i.e. the
numbers
>> smaller
>> than reals, because N hardly covers real area as discussed
under the
>> thread
>> Are reals well-ordered.
>> As we count the real limit in NSA, then the hyperreals
exist, but they
>> are
>> not counted in limit as N -->oo, but not as N_inf -->oo.
Hyperreals are
>> omitted and therefore 0.999...<1 in hyperreals, though the
real part of
>> the
>> limit equals to 1.
> I take it you mean the standard part of the limit is 1.
Well, we can talk about standard part and non-standard part,
if you prefer
this.
> That happens to
> be true, but for a trivial reason. The limit itself is
exactly 1, and
> the standard part of 1 is simply 1.
Cases:
1)Yes, the limit is exactly 1only if you count the limit
including the non
standard part as N-inf --->oo. Then your reference set is
N_inf.
2)Yes, the limit is exactly 1only if you count the limit
including the
standard part as N --->oo. Then your reference set is N.
3) But..., if you count the limit including the standard part
as N --->oo
and your reference set is N_inf, then you omit non-standard
part. As a
consequence in this last case 0.999...<1 in N_inf.
>> NSA expands the concept of numbers to
>> the numbers that are smaller than any real, i.e epsilon
environment.
>> This
>> is
>> equivalent with the concept of epsilon delta theorem. Read
literally
>> what
>> epsilon delta theorem says. This was learnt us already in
70Çs in
>> university. Are you back in 50s?
> Which part of my statement do you not accept? Do you
disagree with the
> definition I gave?
>> Explained above.
> Try again. You didnt mention any part of the 
definition,
let alone say
> which part you disagreed with.
> Definition. Let { a_k } be a sequence and let L be a real
number.
> We say lim_{k->oo} a_k = L if, for every epsilon > 0, there
exists
> N > 0 such that | a_k - L | < epsilon for every k > N.
> Remark 1. Exactly the same definition applies to standard
analysis and
> to nonstandard analysis, with the proviso that in NSA the
epsilon > 0
> is allowed to be an infinitesimal and the N > 0 is allowed
to be
> infinitely large.
Yes, agreed.
> Remark 2. The definition does not say what it means for the
limit to be
> close to L. The definition only says what it means for the
limit to be
> equal to L. Either the definition is satisfied, 
or it isnt.
OK.
> Now, the questions: (1) Do you agree with the definition?
(2) Do you
> agree that according to this definition the limit is exactly
1, even in
> NSA? If you dont agree, explain why not.
Yes, I do agree as explained above in the cases 1 and 2. You
did not
consider at all the case 3 above. How does your definitions
should be
applied on the case 3?
You should also note that the limit is the upper boundary
value. As you will
see below, its not the question about the limit but about 
AC
and the point
of reference as we construct the numbers integers, infinite
integers, reals,
hyperreals etc.
> For example, try to give a
> particular value of epsilon > 0 such that the definition is
not
> satisfied. Hint: choosing an infinitesimal 
epsilon is
allowed, but it
> wont help your case. The definition still 
works.
What about case 3 above as you stop epsilons in standard part
omitting
non-standard part.
>> I would like to ask the same from You. :-). Is it the point
>> of reference that is strange concept for You?
> Point of reference is an undefined concept and does not
appear in the
> definition of limit, quoted above. I wont 
comment on
whether it is
> strange, since things have to be defined first 
before they can
possibly
> qualify as strange.
The point of reference is counting point reference, which
also separates
infinities.
The standard point of reference is the normal decimal dot
that separates the
integer part and the decimal part, which can be infinite long
string.
Without the point of reference you do not know which part is
integer part
and which one is the decimal part. What ever you calculate
you always refer
your calculations to some point of reference.
>> In fact, there seems to be a slight conceptual difference
in our
>> argumentation. The difference is analocigally the same as
we talk about
>> finite decimal numbers. You accept that finite 
0.99999 <1,
as there are
>> no
>> infinite 9s. As hyperreals are omitted in 
the real limit
counting (not
>> counted over hyperreals), then in reals the limit equals
to 1, i.e.
>> 0.999...=1. But as hypereals exist, but omitted in the
limit
calculation,
>> then 0,999...<1 in NSA. Simple as possible. Reals are
finite from the
>> hypereal point of view. Therefore 0.999...<1 - in
hyperreals as only
real
>> area is considered.
> But in NSA there is no such thing as a .999... that extends
only through
> finite digit positions. If the string is 
infinitely long,
then it
> necessarily extends to infinite digit positions in NSA. That
may sound
> vaguely similar to an argument commonly made by cranks, but
the
> difference is that the cranks are not talking about NSA.
used the case 3 as an example to point out how the people do
not think as
their argumentation contains hidden asumptions. So - if there
exist
non-standard part, which is purposely omitted, then you do
not calculate the
total or over-all limit. In this case you have to observe
that there are
numbers smaller than any real number and as a consequence
0.999... cannot be
equal to 1, though the limit equals to 1.
What is this paradox, which has been the reason to this
thread, and how do
we fix it?
The solution is to point out that the limit is a mathematical
upper boundary
value, but does not tell the real value of the string but the
next, i.e.
successor, value of the string, because the limit calculation
is based on
the epsilon delta theorem.
> The definition of a limit in NSA may be stated in various
ways, but all
> of them are equivalent to the usual epsilon-delta
definition, with the
> proviso that epsilons and deltas are allowed to be
infinitesimal, and N
> is allowed to be infinitely large.
>> Yes, thats what I mean - too, assuming we mean the same
thing. If you
>> limit
>> your calculation into the real area and omit the hyperreal
area,
> You cant do that. Thats the whole point.
LetÇs demontrate it shortly now so simply as possible 
(like
Donald Knuth
1) It is assumed that the digits [0,1...9] in 10-base system
are
constructed. I leave it for the home work.
2) Then we apply AC so that we have infinite many placeholders
(or hooks)
with the first one and the last one (the start and the end),
just like in
[0,1].
(by the way this stops the discussion about the last digit in
the infinite
decimal string. :-)) For us its enough that is just 
possible.
3) Pick-up with the aid of AC some digits in every
placeholder.
What number do you have? Actually you have just digits, but
not a number -
yet. Why?
Because you have not defined the point of reference! You have
only digits in
every place holder, i.e. in every hook.
4) Adjust the digits, you have picked up by hooks, into a
linear string.
Its infinite, but it has the start and the end. 
Additionally
we define (in
this particular 10 base case) what is the relation of the
adjacent
placeholders.
What number do you have now? No number, because you just have
the linear
infinite string without the point of reference. The point of
reference
defines the infinite string area, but we do not 
have it yet,
thus as a
consequence - no number area.
5) Apply the the point of reference. How? You can insert the
point of
reference anywhere in the string, but lets concentrate our
attention into
two special case because of simplicity: to the left hand side
(lhs) of the
infinite string or to the right hand side (rhs). What kind of
point of
reference? Any defined kind, but lets apply 
because of the
simplicity the
standard point of reference that is called the decimal dot
(or mark).
Lets insert the standard point of reference into lhs of the
string. What
number do you have?
It is familiar decimal number depending on the values of
digits in the
string. Is it rational, irrational or transcendental
depending on the
digits. If there are only zeros on rhs starting from some
placeholders, then
the decimal part is called finite.
Lets insert the standard point of reference into rhs of the
string. What
number do you have?
It looks like now an integer, but isnt it integer?. 
Its
infinite long with
start and end. Its somehow finite, but anyway 
infinite.
Integers cannot be
infinite?
Yes, they can. Actually every classic finite integer (N) can
be written as
infinite, if there are only zeros on lhs starting from some
placeholders. In
other case we call them infinite integers (N_inf).
As an observation we recognize immediately that there is
one-to-one
bijection between N_inf and decimal part of R. The only
difference is that
N_inf has the standard point of reference on rhs and R
(decimal part) has
the standard point of reference on lhs. The standard point of
reference,
i.e. the decimal dot separates to two infinite long strings.
Both parts are
constructed by AC. The only difference is the point of
reference on lhs or
on rhs.
As finite integers (N) are the subset of infinite 
integers
(N_inf).
AC, Zorns lemma and well-ordering are equivalent. Integers
and infinite
integers are well-ordered. By moving the point of reference
from rhs (in
N_inf) to lhs (in R, decimal part) its trivial to recognize
that the
decimal part is also well-ordered.
Considering the limit, its easy to observe that if the 
point
of reference
is on lhs of the string of infinite 9s, i.e. 
0.999..., the
limit equals to
1. But if the point of the reference is on rhs of the same
string, then
there is no limit, because we cannot calculate it for N_inf
like ...999. We
can add 1 to ...999, but then - as all the placeholders were
occupied with
9s in the infinite long string - the successor 
equals to
omega 1.
Thus the limit, i.e. the upper boundary value, describes the
successor, not
the sum of the string itself as it should count.
Successor equals never with the precessor and therefore omega
1> ...999 and
also 1>0.999... The reason is AC and well-ordering.
Thus for example a general number can be described omega-area
(separator)
infinite integer area (separator, usually dot), decimal area
(separator)
non-standard area. Each area is infinite and well-ordered
constructed with
the aid of AC.
We dont have to use the standard approach one 
first defines Z,
then defines
Q, then defines R, each in different ways. Dedekind cut is not
necessary but
useful, which is another thing. All we need is AC and the
point of
reference.
Tapio
> --
> Dave Seaman
> Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: .99999... still=/= 1
> Yes, if you count the limit over reals plus over
hyperreals. Sorry - I
> cannot find better word than over. Find a better word as 
your
mother
> language is English.
> I try to specify:
> As you count the limit in reals, then N --> oo, where every
N is finite
> integer. As you count the limit over hyperreals, then N_inf
-->oo_inf,
> where
> oo_inf has higher cardinality as the cardinality of N_inf
>N.
>> No, this is not about cardinality at all. The infinitely
large integers
>> of NSA are not the same as transfinite cardinals.
> Its necessary to observe: As the cardinality of reals R >
cardinality of
> integers N, so the cardinality of N_inf > N. As you count
the limit
> N --->oo instead of N-inf, then you certainly omit
something - namely
> hyperreal part.
All very confused. Assuming you mean N_inf = *N, the set of
hyperintegers, then its true that *N as an external set has
the
cardinality of the reals, but the internal set *N has a
*cardinality
equal to *aleph_0. This certainly doesnt mean that the
hypernaturals
are the same as transfinite cardinals. For one thing, there is
a
smallest transfinite cardinal (aleph_0), but there is no such
thing as a
smallest infinite hypernatural. If n is an 
infinitely large
integer,
then so is n-1.
And none of this has anything to do with limits. You seem to
think that
infinity has only one meaning in all of mathematics. Wrong.
> N (finite integers) does not cover hyperreal area, i.e. the
numbers
> smaller
> than reals, because N hardly covers real area as discussed
under the
> thread
> Are reals well-ordered.
> As we count the real limit in NSA, then the hyperreals
exist, but they
> are
> not counted in limit as N -->oo, but not as N_inf -->oo.
Hyperreals are
> omitted and therefore 0.999...<1 in hyperreals, though the
real part of
> the
> limit equals to 1.
>> I take it you mean the standard part of the limit is 1.
> Well, we can talk about standard part and non-standard
part, if you prefer
> this.
Im trying to guess what you might mean by the real part,
since you
have not defined your meaning. We are not talking about
complex numbers
here.
>> That happens to
>> be true, but for a trivial reason. The limit itself is
exactly 1, and
>> the standard part of 1 is simply 1.
> Cases:
> 1)Yes, the limit is exactly 1only if you count the limit
including the non
> standard part as N-inf --->oo. Then your reference set is
N_inf.
I dont know what you mean by the reference set is N_inf. 
The
sum is
over *N.
> 2)Yes, the limit is exactly 1only if you count the limit
including the
> standard part as N --->oo. Then your reference set is N.
Are you talking about standard analysis here, or nonstandard?
The set N
(consisting of the finite naturals) is not a set in the
internal set
theory of NSA. Its counterpart is *N, which includes the
infinitely
large naturals. If you are forming a sum in NSA, then you can
sum over
*N, but not over N.
> 3) But..., if you count the limit including the standard
part as N --->oo
> and your reference set is N_inf, then you omit non-standard
part. As a
> consequence in this last case 0.999...<1 in N_inf.
Thats all very confused, but I think you are trying to say
that the
limit of the sequence { 1 - 1/10^n } in NSA has a nonzero
nonstandard
part. False. The limit is exactly 1 according to the
definition I gave
previously.
Perhaps what is confusing you is the following fact from NSA:
Theorem. Let { a_k } be a sequence and L a real number. Then
the
following statements are equivalent:
(1) lim_{k->oo} a_k = L.
(2) The difference | a_k - L | is an infinitesimal
whenever k is infinitely large.
Notice, however, that neither (1) nor (2) says anything about
the limit
differing from L by an infinitesimal. Statement (1) mentions
the limit,
but its a statement of exact equality. Statement (2)
mentions something
that differs from L by an infinitesimal, but it makes no
mention of the
limit whatsoever. Either way you look at it, neither of these
statements
supports your conclusion for the case a_k = 1 - 1/10^k and L
= 1.
> NSA expands the concept of numbers to
> the numbers that are smaller than any real, i.e epsilon
environment.
> This
> is
> equivalent with the concept of epsilon delta theorem. Read
literally
> what
> epsilon delta theorem says. This was learnt us already in
70Çs in
> university. Are you back in 50s?
>> Which part of my statement do you not accept? Do you
disagree with
the
>> definition I gave?
> Explained above.
>> Try again. You didnt mention any part of the 
definition,
let alone say
>> which part you disagreed with.
>> Definition. Let { a_k } be a sequence and let L be a real
number.
>> We say lim_{k->oo} a_k = L if, for every epsilon > 0,
there exists
>> N > 0 such that | a_k - L | < epsilon for every k > N.
>> Remark 1. Exactly the same definition applies to standard
analysis and
>> to nonstandard analysis, with the proviso that in NSA the
epsilon >
0
>> is allowed to be an infinitesimal and the N > 0 is allowed
to be
>> infinitely large.
> Yes, agreed.
>> Remark 2. The definition does not say what it means for the
limit to be
>> close to L. The definition only says what it means for the
limit to be
>> equal to L. Either the definition is satisfied, 
or it isnt.
> OK.
>> Now, the questions: (1) Do you agree with the definition?
(2) Do you
>> agree that according to this definition the limit is
exactly 1, even in
>> NSA? If you dont agree, explain why not.
> Yes, I do agree as explained above in the cases 1 and 2.
You did not
> consider at all the case 3 above. How does your definitions
should be
> applied on the case 3?
There is no case 3 above. I dont know what you are talking
about.
> You should also note that the limit is the upper boundary
value. As you
will
> see below, its not the question about the limit but about
AC and the
point
> of reference as we construct the numbers integers, infinite
integers,
reals,
> hyperreals etc.
>> For example, try to give a
>> particular value of epsilon > 0 such that the definition is
not
>> satisfied. Hint: choosing an infinitesimal 
epsilon is
allowed, but it
>> wont help your case. The definition still 
works.
> What about case 3 above as you stop epsilons in standard
part omitting
> non-standard part.
There is no case 3 above. I have explained that the sum over
N is not
a sum in NSA, since N is not an internal set in NSA. Thats
why we sum
over *N instead.
> I would like to ask the same from You. :-). Is it the point
> of reference that is strange concept for You?
>> Point of reference is an undefined concept and does not
appear in
the
>> definition of limit, quoted above. I wont 
comment on
whether it is
>> strange, since things have to be defined first 
before they
can
possibly
>> qualify as strange.
> The point of reference is counting point reference, which
also separates
> infinities.
That is not a definition. For an example of what I mean by a
definition,
look at my definition of what it means to say that lim_{k->oo}
a_k = L.
Mathematical definitions leave no room for fuzzy language or
vague
concepts. Try again.
> The standard point of reference is the normal decimal dot
that separates
the
> integer part and the decimal part, which can be infinite
long string.
> Without the point of reference you do not know which part
is integer part
> and which one is the decimal part. What ever you calculate
you always
refer
> your calculations to some point of reference.
The common definitions of the real numbers (via Dedekind cuts
or Cauchy
sequences) do not mention decimal points at all and do not
depend on
any such concepts as integer part and decimal part. For any
real
number x, the integer part of x may be defined as ßoor(x) 
=
max { n in
N : n <= x }. I had no need for any vague concepts such as
point of
reference in writing that definition.
> In fact, there seems to be a slight conceptual difference
in our
> argumentation. The difference is analocigally the same as
we talk about
> finite decimal numbers. You accept that finite 
0.99999 <1, as
there are
> no
> infinite 9s. As hyperreals are omitted in the 
real limit
counting (not
> counted over hyperreals), then in reals the limit equals to
1, i.e.
> 0.999...=1. But as hypereals exist, but omitted in the limit
calculation,
> then 0,999...<1 in NSA. Simple as possible. Reals are finite
from the
> hypereal point of view. Therefore 0.999...<1 - in
hyperreals as only
real
> area is considered.
You need to be precise about what you mean by 0.999... in
NSA. I have
been taking it to mean the sum of 9/10^n for all n in *N, n >
0. You
evidently mean something different. In particular, it cant
possibly
mean the sum over all finite values of n > 0, because that is
not a set
in NSA.
>> But in NSA there is no such thing as a .999... that
extends only through
>> finite digit positions. If the string is 
infinitely long,
then it
>> necessarily extends to infinite digit positions in NSA.
That may sound
>> vaguely similar to an argument commonly made by cranks,
but the
>> difference is that the cranks are not talking about NSA.
> used the case 3 as an example to point out how the people
do not think as
> their argumentation contains hidden asumptions. So - if
there exist
> non-standard part, which is purposely omitted, then you do
not calculate
the
> total or over-all limit. In this case you have to observe
that there are
> numbers smaller than any real number and as a consequence
0.999... cannot
be
> equal to 1, though the limit equals to 1.
> What is this paradox, which has been the reason to this
thread, and how do
> we fix it?
If you do not sum 9/10^n for all n > 0, then you are not
computing
0.999.... The indicated sum is exactly 1.
> The solution is to point out that the limit is a
mathematical upper
boundary
> value, but does not tell the real value of the string but
the next, i.e.
> successor, value of the string, because the limit
calculation is based on
> the epsilon delta theorem.
The string I am talking about has no successor values. Its
already
defined for all positions n, including the ones that are
infinitely
large.
>> The definition of a limit in NSA may be stated in various
ways, but
all
>> of them are equivalent to the usual epsilon-delta
definition, with the
>> proviso that epsilons and deltas are allowed to be
infinitesimal, and
N
>> is allowed to be infinitely large.
> Yes, thats what I mean - too, assuming we mean the same
thing. If you
> limit
> your calculation into the real area and omit the hyperreal
area,
>> You cant do that. Thats the whole point.
> LetÇs demontrate it shortly now so simply as possible 
(like
Donald Knuth
> 1) It is assumed that the digits [0,1...9] in 10-base
system are
> constructed. I leave it for the home work.
> 2) Then we apply AC so that we have infinite many
placeholders (or hooks)
> with the first one and the last one (the start and the end),
just like in
> [0,1].
We dont need AC here. We are defining d_0 = 0 
and d_k = 9 for
all k > 0
in *N.
> (by the way this stops the discussion about the last digit
in the infinite
> decimal string. :-)) For us its enough that is just
possible.
I was not aware that there was any such discussion.
> 3) Pick-up with the aid of AC some digits in every
placeholder.
Again, AC is irrelevant. We already have our hyperinfinite
string.
> What number do you have? Actually you have just digits, but
not a number -
> yet. Why?
Because decimal digit strings are not numbers. They merely
represent
numbers.
> Because you have not defined the point of reference! You
have only digits
in
> every place holder, i.e. in every hook.
Nonsense. If { d_k } is a decimal digit string, then the
number
represented by the string is sum_{k in *N} d_k * 10^(-k). No
point of
reference is needed.
[ snip nonsense about point of reference ]
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: .99999... still=/= 1
>> Yes, if you count the limit over reals plus over
hyperreals. Sorry - I
>> cannot find better word than over. Find a better word as
your
mother
>> language is English.
>> I try to specify:
>> As you count the limit in reals, then N --> oo, where
every N is
finite
>> integer. As you count the limit over hyperreals, then
N_inf -->oo_inf,
>> where
>> oo_inf has higher cardinality as the cardinality of N_inf
>N.
> No, this is not about cardinality at all. The infinitely
large
integers
> of NSA are not the same as transfinite cardinals.
We talk about ordinals. That above was just a hint to compare
the
cardinality of N_inf and N.
>> Its necessary to observe: As the cardinality of reals R 
>
cardinality
of
>> integers N, so the cardinality of N_inf > N. As you count
the limit
>> N --->oo instead of N-inf, then you certainly omit
something - namely
>> hyperreal part.
> All very confused. Assuming you mean N_inf = *N, the set of
> hyperintegers, then its true that *N as an external set
has the
> cardinality of the reals, but the internal set *N has a
*cardinality
> equal to *aleph_0.
This is of-course OK!
> This certainly doesnt mean that the hypernaturals
> are the same as transfinite cardinals. For one thing, there
is a
> smallest transfinite cardinal (aleph_0),
Ok!
> but there is no such thing as a
> smallest infinite hypernatural. If n is an 
infinitely large
integer,
> then so is n-1.
Omega was defined in this discussion earlier as follows:
The number that is greater than any infinite integer. That is
the smallest
transinfinite number. Omega is the successor of ...999, i.e.
n+1. The
precessor (n-1) of ...999 is ...998.
> And none of this has anything to do with limits. You seem
to think that
> infinity has only one meaning in all of mathematics. Wrong.
Your opinion. :-)
>> N (finite integers) does not cover hyperreal area, i.e. the
numbers
>> smaller
>> than reals, because N hardly covers real area as discussed
under the
>> thread
>> Are reals well-ordered.
>> As we count the real limit in NSA, then the hyperreals
exist, but they
>> are
>> not counted in limit as N -->oo, but not as N_inf -->oo.
Hyperreals
are
>> omitted and therefore 0.999...<1 in hyperreals, though the
real part
of
>> the
>> limit equals to 1.
> I take it you mean the standard part of the limit is 1.
>> Well, we can talk about standard part and non-standard
part, if you
>> prefer
>> this.
> Im trying to guess what you might mean by the real part,
since you
> have not defined your meaning.
the real part was the decimal part - as you certainly knew.
> We are not talking about complex numbers
> here.
Exactly!
> That happens to
> be true, but for a trivial reason. The limit itself is
exactly 1, and
> the standard part of 1 is simply 1.
>> Cases:
>> 1)Yes, the limit is exactly 1only if you count the limit
including the
>> non
>> standard part as N-inf --->oo. Then your reference set is
N_inf.
> I dont know what you mean by the reference set is N_inf.
The sum is
> over *N.
That is exactly what I meant above if N_inf=*N.
>> 2)Yes, the limit is exactly 1only if you count the limit
including the
>> standard part as N --->oo. Then your reference set is N.
> Are you talking about standard analysis here, or
nonstandard?
Standard analysis as You correctly observed.
> The set N
> (consisting of the finite naturals) is not a set in the
internal set
> theory of NSA.
Coorect, but it should be as *N is the extension of the set N.
> Its counterpart is *N, which includes the infinitely
> large naturals. If you are forming a sum in NSA, then you
can sum over
> *N, but not over N.
We can sum over subset too.
>> 3) But..., if you count the limit including the standard
part as
>> N --->oo
>> and your reference set is N_inf, then you omit
non-standard part. As a
>> consequence in this last case 0.999...<1 in N_inf.
> Thats all very confused, but I think you are trying to 
say
that the
> limit of the sequence { 1 - 1/10^n } in NSA has a nonzero
nonstandard
> part. False. The limit is exactly 1 according to the
definition I gave
> previously.
N is the subset of *N. You can count the limit over N or
alternatively over
*N as You can count N --> 0 to 1000 as a subset of N or
alternatively
N --->0 to oo.
> Perhaps what is confusing you is the following fact from
NSA:
> Theorem. Let { a_k } be a sequence and L a real number.
Then the
> following statements are equivalent:
> (1) lim_{k->oo} a_k = L.
> (2) The difference | a_k - L | is an infinitesimal
> whenever k is infinitely large.
> Notice, however, that neither (1) nor (2) says anything
about the limit
> differing from L by an infinitesimal.
Yes, as long as you count over *N
> Statement (1) mentions the limit,
> but its a statement of exact equality. Statement (2)
mentions something
> that differs from L by an infinitesimal, but it makes no
mention of the
> limit whatsoever.
Except Yount count over N instead of *N.
> Either way you look at it, neither of these statements
> supports your conclusion for the case a_k = 1 - 1/10^k and
L = 1.
>> NSA expands the concept of numbers to
>> the numbers that are smaller than any real, i.e epsilon
environment.
>> This
>> is
>> equivalent with the concept of epsilon delta theorem. Read
literally
>> what
>> epsilon delta theorem says. This was learnt us already in
70Çs in
>> university. Are you back in 50s?
> Which part of my statement do you not accept? Do you
disagree with
> the
> definition I gave?
>> Explained above.
> Try again. You didnt mention any part of the 
definition,
let alone
say
> which part you disagreed with.
> Definition. Let { a_k } be a sequence and let L be a real
number.
> We say lim_{k->oo} a_k = L if, for every epsilon > 0, there
exists
> N > 0 such that | a_k - L | < epsilon for every k > N.
> Remark 1. Exactly the same definition applies to standard
analysis and
> to nonstandard analysis, with the proviso that in NSA the
epsilon >
0
> is allowed to be an infinitesimal and the N > 0 is allowed
to be
> infinitely large.
>> Yes, agreed.
> Remark 2. The definition does not say what it means for the
limit to
be
> close to L. The definition only says what it means for the
limit to be
> equal to L. Either the definition is satisfied, 
or it isnt.
>> OK.
> Now, the questions: (1) Do you agree with the definition?
(2) Do you
> agree that according to this definition the limit is exactly
1, even in
> NSA? If you dont agree, explain why not.
>> Yes, I do agree as explained above in the cases 1 and 2.
You did not
>> consider at all the case 3 above. How does your definitions
should be
>> applied on the case 3?
> There is no case 3 above. I dont know what you are 
talking
about.
>> You should also note that the limit is the upper boundary
value. As you
>> will
>> see below, its not the question about the limit but 
about
AC and the
>> point
>> of reference as we construct the numbers integers, infinite
integers,
>> reals,
>> hyperreals etc.
> For example, try to give a
> particular value of epsilon > 0 such that the definition is
not
> satisfied. Hint: choosing an infinitesimal 
epsilon is
allowed, but it
> wont help your case. The definition still 
works.
>> What about case 3 above as you stop epsilons in standard
part omitting
>> non-standard part.
> There is no case 3 above. I have explained that the sum
over N is
not
> a sum in NSA, since N is not an internal set in NSA. 
Thats
why we sum
> over *N instead.
Because You dont see N as subset of *N. :-)
>> I would like to ask the same from You. :-). Is it the point
>> of reference that is strange concept for You?
> Point of reference is an undefined concept and does not
appear in
the
> definition of limit, quoted above. I wont 
comment on
whether it is
> strange, since things have to be defined first 
before they can
> possibly
> qualify as strange.
>> The point of reference is counting point reference, which
also separates
>> infinities.
> That is not a definition. For an example of what I mean by a
definition,
> look at my definition of what it means to say that
lim_{k->oo} a_k = L.
> Mathematical definitions leave no room for fuzzy language or
vague
> concepts. Try again.
>> The standard point of reference is the normal decimal dot
that separates
>> the
>> integer part and the decimal part, which can be infinite
long
string.
>> Without the point of reference you do not know which part
is integer
part
>> and which one is the decimal part. What ever you calculate
you always
>> refer
>> your calculations to some point of reference.
> The common definitions of the real numbers (via Dedekind
cuts or Cauchy
> sequences) do not mention decimal points at all and do not
depend on
> any such concepts as integer part and decimal part. For any
real
> number x, the integer part of x may be defined as ßoor(x) 
=
max { n
in
> N : n <= x }. I had no need for any vague concepts such as
point of
> reference in writing that definition.
Certainly not. Its alternative way to construct numbers 
from
AC. The point
of reference is in-constructed assumption in Dedekinds cut or
Cauchy
sequences, because they assumed so or they did not recognize
the point of
reference.
>> In fact, there seems to be a slight conceptual difference
in our
>> argumentation. The difference is analocigally the same as
we talk
about
>> finite decimal numbers. You accept that finite 
0.99999 <1,
as there
are
>> no
>> infinite 9s. As hyperreals are omitted in 
the real limit
counting
(not
>> counted over hyperreals), then in reals the limit equals
to 1, i.e.
>> 0.999...=1. But as hypereals exist, but omitted in the
limit
>> calculation,
>> then 0,999...<1 in NSA. Simple as possible. Reals are
finite from the
>> hypereal point of view. Therefore 0.999...<1 - in
hyperreals as only
>> real
>> area is considered.
> You need to be precise about what you mean by 0.999... in
NSA. I have
> been taking it to mean the sum of 9/10^n for all n in *N, n
> 0. You
> evidently mean something different. In particular, it 
cant
possibly
> mean the sum over all finite values of n > 0, because that
is not a set
> in NSA.
N is the subset of *N. I and You can count over subset of *N
just like in
the case of any subset of N.
> But in NSA there is no such thing as a .999... that extends
only
through
> finite digit positions. If the string is 
infinitely long,
then it
> necessarily extends to infinite digit positions in NSA. That
may sound
> vaguely similar to an argument commonly made by cranks, but
the
> difference is that the cranks are not talking about NSA.
>> evidence. I
>> used the case 3 as an example to point out how the people
do not think
as
>> their argumentation contains hidden asumptions. So - if
there exist
>> non-standard part, which is purposely omitted, then you do
not calculate
>> the
>> total or over-all limit. In this case you have to observe
that there are
>> numbers smaller than any real number and as a consequence
0.999... cannot
>> be
>> equal to 1, though the limit equals to 1.
>> What is this paradox, which has been the reason to this
thread, and how
>> do
>> we fix it?
> If you do not sum 9/10^n for all n > 0, then you are not
computing
> 0.999.... The indicated sum is exactly 1.
Shortly: Thats what we have already agreed. I never denied
that. What can
be done: over N, over *N or over N as a subset of *N, and
only in the last
case the limit over N is smaller than 1 as You should count
over *N in
hyperreals. Therefore in principle 0.999... <1 in the last
mentioned case.
>> The solution is to point out that the limit is a
mathematical upper
>> boundary
>> value, but does not tell the real value of the string but
the next, i.e.
>> successor, value of the string, because the limit
calculation is based
on
>> the epsilon delta theorem.
> The string I am talking about has no successor values. 
Its
already
> defined for all positions n, including the ones that are
infinitely
> large.
> The definition of a limit in NSA may be stated in various
ways, but
> all
> of them are equivalent to the usual epsilon-delta
definition, with
the
> proviso that epsilons and deltas are allowed to be
infinitesimal, and
> N
> is allowed to be infinitely large.
>> Yes, thats what I mean - too, assuming we mean the same
thing. If you
>> limit
>> your calculation into the real area and omit the hyperreal
area,
> You cant do that. Thats the whole point.
>> LetÇs demontrate it shortly now so simply as possible
(like Donald Knuth
>> 1) It is assumed that the digits [0,1...9] in 10-base
system are
>> constructed. I leave it for the home work.
>> 2) Then we apply AC so that we have infinite many
placeholders (or
hooks)
>> with the first one and the last one (the start and the
end), just like
in
>> [0,1].
> We dont need AC here. We are defining d_0 = 0 
and d_k = 9
for all k > 0
> in *N.
>> (by the way this stops the discussion about the last digit
in the
>> infinite
>> decimal string. :-)) For us its enough that is just
possible.
> I was not aware that there was any such discussion.
It was earlier and it arises up from time to time, but not in
our mutual
discussion.
>> 3) Pick-up with the aid of AC some digits in every
placeholder.
> Again, AC is irrelevant. We already have our hyperinfinite
string.
>> What number do you have? Actually you have just digits,
but not a
>> number -
>> yet. Why?
> Because decimal digit strings are not numbers. They merely
represent
> numbers.
>> Because you have not defined the point of reference! You
have only digits
>> in
>> every place holder, i.e. in every hook.
> Nonsense. If { d_k } is a decimal digit string, then the
number
> represented by the string is sum_{k in *N} d_k * 10^(-k).
No point of
> reference is needed.
You cut (=snip) out the climax of my explanation. What does
^(-k) refer?
Please answer to that simple question! Why minus - what does
it refer?
> [ snip nonsense about point of reference ]
Your opinion, because You could not consider the successor of
...999. What
is the successor of ...999 as all the placeholders are
infinitely occupied
with the maximal digit 9?
Tapio
> --
> Dave Seaman
> Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
>
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: .99999... still=/= 1
> And why isnt .999... a real number? It 
isnt an
irrational, according
>>.999... is not anything. It is a jumble of symbols without
a proper
>>mathematical definition. If this is supposed to be 
grafitti
for a series
>>then you should tell us what the general term is.
>Since its the one that is supposed to be 1, the general
term is:
>9/(10^i), for i=1 to infinity.
>As if you didnt know.
>John Savard
>http://home.ecn.ab.ca/~jsavard/index.html
So if someone tell you to believe in the tooth fairy that
exists to in
math?
9/9 =/= .999....
9/10 =/= .999...
.999... cant be expressed as a fraction so 
its not
rational. Its not an
irrational number either. So,
A number that isnt real cant equal a number 
that is real.
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Subject: Re: .99999... still=/= 1
>.999... isnt a real number. The number 1 is a real number.
And why isnt .999... a real number? It isnt 
an irrational,
according
to the AOL dictionary. It can still be rational. And 1 is
rational too.
John Savard
http://home.ecn.ab.ca/~jsavard/index.html
===
Subject: Re: .99999... still=/= 1
> And why isnt .999... a real number? It 
isnt an
irrational, according
.999... is not anything. It is a jumble of symbols without a
proper
mathematical definition. If this is supposed to be 
grafitti for
a series
then you should tell us what the general term is.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> And why isnt .999... a real number? It 
isnt an
irrational, according
>.999... is not anything. It is a jumble of symbols without a
proper
>mathematical definition. If this is supposed to be 
grafitti
for a series
>then you should tell us what the general term is.
>Bob Kolker
Youre not even in agreement with the others that say,
9/9 = .999...
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Subject: Re: .99999... still=/= 1
> Youre not even in agreement with the others that say,
> 9/9 = .999...
That is right. .999... is scribble and not a mathematical
experession. I
write down series, not scribble.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> Youre not even in agreement with the others that say,
>> 9/9 = .999...
>That is right. .999... is scribble and not a mathematical
experession. I
>write down series, not scribble.
>Bob Kolker
Whats wrong you can admit
.999... =/= 1
I won this debate years ago. HAHAHAHAHAHHAHAHAHAHHAHAH
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Subject: Re: .99999... still=/= 1
> As if you didnt know.
Of course. But I am trying to discourage the use of the idiom
(or
grafitto) .999... which only discourages S. Enterprise from
excercising
all 23 of his neurons.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> As if you didnt know.
>Of course. But I am trying to discourage the use of the
idiom (or
>grafitto) .999... which only discourages S. Enterprise from
excercising
>all 23 of his neurons.
>Bob Kolker
Look at this dude, he says, .999... is meaningless. Then why
apply
Partial
Sums to something meaningless?
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Subject: Re: .99999... still=/= 1
>As if you didnt know.
>>Of course. But I am trying to discourage the use of the
idiom (or
>>grafitto) .999... which only discourages S. Enterprise from
excercising
>>all 23 of his neurons.
>>Bob Kolker
> Look at this dude, he says, .999... is meaningless. Then
why apply
Partial
> Sums to something meaningless?
You mean like SUM (n >= 1) [9/10^n] ? That is a series
exprressed in
ascii. It is a proper mathematical expression.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>>As if you didnt know.
>Of course. But I am trying to discourage the use of the
idiom (or
>grafitto) .999... which only discourages S. Enterprise from
excercising
>all 23 of his neurons.
>Bob Kolker
>> Look at this dude, he says, .999... is meaningless. Then
why apply
>Partial
>> Sums to something meaningless?
>You mean like SUM (n >= 1) [9/10^n] ? That is a series
exprressed in
>ascii. It is a proper mathematical expression.
>Bob Kolker
So you have to add all types of other things to,
.999.... to make it converge?
What about just as it stands?
.999... =/= =/= =/= =/= =/= =/= 1
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Subject: Re: .99999... still=/= 1
>Of course. But I am trying to discourage the use of the
idiom (or
>grafitto) .999... which only discourages S. Enterprise from
excercising
>all 23 of his neurons.
So you dont think he would be able to pilot an F-22, then?
John Savard
http://home.ecn.ab.ca/~jsavard/index.html
===
Subject: Re: .99999... still=/= 1
>>Of course. But I am trying to discourage the use of the
idiom (or
>>grafitto) .999... which only discourages S. Enterprise from
excercising
>>all 23 of his neurons.
>So you dont think he would be able to pilot an F-22, then?
>John Savard
>http://home.ecn.ab.ca/~jsavard/index.html
So instead of admitting your mistakes, you make goofy jokes
to change
the
subject???? How goofy can you get?
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Subject: Re: .99999... still=/= 1
>>.999... isnt a real number. The number 1 is a real 
number.
>And why isnt .999... a real number? It isnt 
an irrational,
according
>to the AOL dictionary. It can still be rational. And 1 is
rational too.
>John Savard
>http://home.ecn.ab.ca/~jsavard/index.html
A real number is either a rational number or irrational
number. A number
that
can be expressed as a fraction is a rational number,
otherwise it has to be
expressed as a decimal.
3/4 <-- this is a real rational number.
sqrt (2) = 1.41421... <--- this is an irrational number.
There is no
fraction
that can represent 1.4142....
.999... isnt irrational because it has repeating digits to
the right of
the
decimal point.
1.41421... <--- has non-repeating decimals to the right of
the decimal
point.
This is an irrational real number.
.999... is an indeterminate because it isnt 
classified as
rational or
irrational, and you cant determine the space between this
number to the
point
where it reaches a real number. In other words, this exists:
.999... | | 1
--->| |<---
This space makes this number an indeterminate. A non-standard
analysis
has to be made here to determine either a convergence or some
time
dependent
value that approaches 1. Using the gamma function you can
observe that the
more
time you spend approaching 1 ( a real number), the closer you
get to 1, but
it
never reaches it. So it can never equal that number. An
besides, even if we
look at .999... as an irrational number, it cant equal a
rational number.
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===
Subject: Re: .99999... still=/= 1
>.999... isnt irrational because it has repeating digits to
the right of
the
>decimal point.
Right.
>.999... is an indeterminate because it isnt 
classified as
rational or
>irrational, and you cant determine the space between this
number to the
point
>where it reaches a real number.
No.
.99999..... is the rational number 9/9.
.999999999999999999...
______________________
9 ) 9.000000000000000000
8.1
---
.900000000000000000
.810000000000000000
---
.090000000000000000
.081000000000000000
-----
.009000000000000000
.008100000000000000
It doesnt fail to be a rational number just because you say
so.
John Savard
http://home.ecn.ab.ca/~jsavard/index.html
===
Subject: Re: .99999... still=/= 1
>>.999... isnt irrational because it has repeating digits 
to
the right of
the
>>decimal point.
>Right.
>>.999... is an indeterminate because it isnt 
classified as
rational or
>>irrational, and you cant determine the space between this
number to the
>point
>>where it reaches a real number.
>No.
>.99999..... is the rational number 9/9.
Hey can you even see the difference between
1
and
.9999999999999999999999999999999999999999999999999999999999999
9999999999
99999999999999999999999999999999999999999999999999999999999999
999999999999
99999................................
> .999999999999999999...
> ______________________
>9 ) 9.000000000000000000
> 8.1
> ---
> .900000000000000000
> .810000000000000000
> ---
> .090000000000000000
> .081000000000000000
> -----
> .009000000000000000
> .008100000000000000
>It doesnt fail to be a rational number just because you 
say
so.
>John Savard
>http://home.ecn.ab.ca/~jsavard/index.html
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===
Subject: Re: .99999... still=/= 1
nd
>
.9999999999999999999999999999999999999999999999999999999999999
9999999999
>
99999999999999999999999999999999999999999999999999999999999999
999999999999
> 99999................................
This is scribble. It is not a mathematical experession.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>>
.9999999999999999999999999999999999999999999999999999999999999
9999999999
>>
99999999999999999999999999999999999999999999999999999999999999
999999999999
>> 99999................................
>This is scribble. It is not a mathematical experession.
>Bob Kolker
An expanded form of .999.... was as shown.
.999... =/= 1
I won the debate.
You can even just look at it with your own two eyes.
1 is different looking than
.9999999999999999999999999999999999999999999999999999999999999
999999999999
99999999999999999999999999999999999999999999999999999999999999
999999999999
999999-->oo
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===
Subject: Re: .99999... still=/= 1
> A real number is either a rational number or irrational
number. A number
that
> can be expressed as a fraction is a rational number,
otherwise it has to
be
> expressed as a decimal.
A rational number n/m where n, m are integers, m != 0, n < m,
n, m
realtively prime is expressible as an infinite series
Sum [n >= 1] (a_n * 1/10^n) where 0 <= a_n <= 9 and the
sequence
{a_n} is periodic (repeating) after the k-th term where k is
some
integer >= 1.
This is a fairly straightforward theorem in number theory.
Consult any
number theory or algebra text for the proof.
All such series equal the ratio of integers, hence are
rational numbers.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> A real number is either a rational number or irrational
number. A
number
>that
>> can be expressed as a fraction is a rational number,
otherwise it has to
be
>> expressed as a decimal.
>A rational number n/m where n, m are integers, m != 0, n <
m, n, m
>realtively prime is expressible as an infinite series
>Sum [n >= 1] (a_n * 1/10^n) where 0 <= a_n <= 9 and the
sequence
>{a_n} is periodic (repeating) after the k-th term where k is
some
>integer >= 1.
>This is a fairly straightforward theorem in number theory.
Consult any
>number theory or algebra text for the proof.
>All such series equal the ratio of integers, hence are
rational numbers.
But you said .999... is meaningless. You seem to bbe very
confused.
>Bob Kolker
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Subject: Re: .99999... still=/= 1
>>All such series equal the ratio of integers, hence are
rational numbers.
> But you said .999... is meaningless. You seem to bbe very
confused.
I said the -series- corresponding to an ulimately periodic
decimal
expansion equals a rational number. Learn to read.
The locution .999.... is meaningless scribble.
Sum [n >= 1] (a_n * 1/10^n) where 0 <= a_n <= 9 and the
sequence
{a_n} is periodic (repeating) after the k-th term where k is
some
integer >= 1 is a rational number
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>All such series equal the ratio of integers, hence are
rational numbers.
>> But you said .999... is meaningless. You seem to bbe very
confused.
>I said the -series- corresponding to an ulimately periodic
decimal
>expansion equals a rational number. Learn to read.
Learn to think. READ THE MATH definition of a real number.
.999... isnt a real number. It is in the class of a
hyper-real number.
>The locution .999.... is meaningless scribble.
.999... is NOT MEANINGLESS. It is A NUMBER. It is related to
a hyper-real
number.
<ignore> the rest he doesnt even know math fundamentals.
>Sum [n >= 1] (a_n * 1/10^n) where 0 <= a_n <= 9 and the
sequence
>{a_n} is periodic (repeating) after the k-th term where k is
some
>integer >= 1 is a rational number
>Bob Kolker
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===
Subject: Re: .99999... still=/= 1
posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi
> .999... is NOT MEANINGLESS. It is A NUMBER. It is related
to a
hyper-real
> number.
Are you really sure that .999... is a hyperreal number and not
surreal(*)?
(*) http://mathworld.wolfram.com/SurrealNumber.html
http://en.wikipedia.org/wiki/Surreal_number
===
Subject: Re: .99999... still=/= 1
>> .999... is NOT MEANINGLESS. It is A NUMBER. It is related
to a
>hyper-real
>> number.
>Are you really sure that .999... is a hyperreal number and
not
>surreal(*)?
I would say they are very closely related. With a surreal
number, you in
effect move one decimal step back away from oo, between
.999... and 1, where
1
is located and find a terminating value equal to zero. The
surreal number
is,
{a|b} where {|} = 0
There is something preventing a from equaling b. But the next
closest
space from a is b.
But in either case, surreal or hyperreal
.999... < 1
and
.999... =/= 1
>(*) http://mathworld.wolfram.com/SurrealNumber.html
>http://en.wikipedia.org/wiki/Surreal_number
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===
Subject: Re: .99999... still=/= 1
> Are you really sure that .999... is a hyperreal number and
not
> surreal(*)?
Surreals are defined by a pair of classes, similar to a
Dedikind Cut.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
<Dhqxd.214278$5K2.153710@attbi_s03>
posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi
For example: x = {0.9, 0.99, 0.999, ... | 1} < 1
=> x > 0.9 and x > 0.99 and x> 0.999 and ...
===
Subject: Re: .99999... still=/= 1
In sci.math, S. Enterprize Company
<smart1234@aol.com>All such series equal the ratio of
integers, hence are rational
numbers.
>
>
>
> But you said .999... is meaningless. You seem to bbe very
confused.
>>I said the -series- corresponding to an ulimately periodic
decimal
>>expansion equals a rational number. Learn to read.
> Learn to think. READ THE MATH definition of a real number.
> .999... isnt a real number. It is in the class of a
hyper-real number.
.999... is the difference between a real number (1) and a
hyperreal number.
.999... = 1 - d.
Now, what can be meaningfully done with this expression? An
interesting question; Im not all that up on non-standard
analysis.
For example, the following expressions look interesting.
[1] Let x = 1 - d = .999... .
Then 10x - 9 = 1 - 10d = .999...
and (x + 9) / 10 = 1 - d/10 = .999... .
[2] x^2 = 1 - 2d + d^2 = .999... .
[3] sqrt(x) = 1 - d/2 + d^2/8 - d^3/16 + 5*d^4/128 - ...
= .999... (infinite binomial expansion)
How are these ranked on the numberline?
[rest snipped]
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>.999... only approaching 1, it doesnt equal 1.
It would seem that way.
But appearances can be deceiving.
It is certainly possible to use .999... as a symbol for a
quantity that
only approaches 1 as a limit. But that isnt its normal
meaning.
Because if we did that, then .1111.... wouldnt equal 1/9.
Essentially, _if_ .999... stands for a real number, then it
stands for
the real number 1. Because any real number that doesnt 
equal
1 differs
from 1 by a *finite* amount.
The distinction between x=1 and x approaching 1 as a limit is
an
important one in mathematics, but numbers written in decimal
form are
not used to indicate this distinction; they are only used to
indicate
numbers.
John Savard
http://home.ecn.ab.ca/~jsavard/index.html
===
Subject: Re: .99999... still=/= 1
> Essentially, _if_ .999... stands for a real number,
.999... does not stand for anything. It is scribble. If you
want to talk
about an infinite series which is a well defined 
notion then by
all
means do so.
Perhaps you mean the series
Sum [n >= 1] (9/10^n) ?
If you do, then you should say so.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>If you do, then you should say so.
I wasnt aware that smart1234 was questioning that point, so
I addressed
the areas in which he was disputing what is generally known
and
believed. Ambiguity is inherent in any communication in
ordinary
language that strives for brevity, omitting details that have
previously
been agreed upon.
John Savard
http://home.ecn.ab.ca/~jsavard/index.html
===
Subject: Re: .99999... still=/= 1
<41c4026d.1793705@news.ecn.ab.ca>
<PbXwd.209274$5K2.164644@attbi_s03>
<41c46d29.1471377@news.ecn.ab.ca>
posting-account=Z0C83A0AAAD1ZNF3_N0Hr9vKEMG9LqCk
if there is a difference between 0.999... and 1, then what is
that
difference?
is the difference 0.000...(infinity)...1 ? because thats not
possible
x = 0.999...
10x = 9.999...
- x = 0.999...
---------------------
9x = 9
9x/9 = 9/9
x = 1, x=0.999..., 1=0.999....
===
Subject: Re: .99999... still=/= 1
In sci.math, robert j. kolker
<nowhere@nowhere.com>
<PbXwd.209274$5K2.164644@attbi_s03>:
>
>> Essentially, _if_ .999... stands for a real number,
> .999... does not stand for anything. It is scribble. If you
want to talk
> about an infinite series which is a well defined 
notion then
by all
> means do so.
> Perhaps you mean the series
> Sum [n >= 1] (9/10^n) ?
> If you do, then you should say so.
> Bob Kolker
Perhaps that is in fact the best solution all told. However,
expressing
1/7 = sum(i=0,+oo) (1/10^(6*i+1) + 4/10^(6*i+2)+ 2/10^(6*i+3)
+ 8/10^(6*i+4)+ 5/10^(6*i+5)+ 7/10^(6*i+6))
does get a tad unwieldly... :-)
Or perhaps one prefers the one-off form:
1/7 = sum(i=1,+oo) (1/10^(6*i-5) + 4/10^(6*i-4)+ 2/10^(6*i-3)
+ 8/10^(6*i-2)+ 5/10^(6*i-1)+ 7/10^(6*i))
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>Dedekind is credited with giving the first mathematical
definition
>of the real numbers.
>> Is this so?
>Yes and no. The set of real numbers was implicitly defined by
>algebraic laws on the two basic operations + and *. A
rigorous
>definition (or reduction) of real numbers to the rationals
was done in
>the latter half of the 19-th century along with definitions 
of
>continuous functions, differentiable functions and limits.
The key to
>the whole business is the rigorous definition of limit.
>The reals are the topological closure of the rationals. The
key to this
>is a definition of convergence in which a limit is not
explicitly
>required. This is attributed to Cauchy.
>Bob Kolker
The basic fundamental definition of .999.... is that it 
isnt
a real
number.
A number that isnt real cant equal an number 
that is real.
How simple
can
you get with this?
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===
Subject: Re: .99999... still=/= 1
>
>
>>Dedekind is credited with giving the first mathematical
definition
>>of the real numbers.
>
>
> Is this so?
>>Yes and no. The set of real numbers was implicitly defined 
by
>>algebraic laws on the two basic operations + and *. A
rigorous
>>definition (or reduction) of real numbers to the rationals
was done in
>>the latter half of the 19-th century along with definitions
of
>>continuous functions, differentiable functions and limits.
The key to
>>the whole business is the rigorous definition of limit.
>>The reals are the topological closure of the rationals. The
key to this
>>is a definition of convergence in which a limit is not
explicitly
>>required. This is attributed to Cauchy.
>>Bob Kolker
> The basic fundamental definition of .999.... is that it
isnt a real
>number.
> A number that isnt real cant equal an 
number that is
real. How simple
can
>you get with this?
Heres another way to look at this.
i = something imaginary ( not real )
Can you prove i = a real number?
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===
Subject: Re: .99999... still=/= 1
In sci.math, S. Enterprize Company
<smart1234@aol.com>
>>
>Dedekind is credited with giving the first mathematical
definition
>of the real numbers.
>>
>>
>> Is this so?
>Yes and no. The set of real numbers was implicitly defined by
>algebraic laws on the two basic operations + and *. A
rigorous
>definition (or reduction) of real numbers to the rationals
was done
in
>the latter half of the 19-th century along with definitions 
of
>continuous functions, differentiable functions and limits.
The key to
>the whole business is the rigorous definition of limit.
>The reals are the topological closure of the rationals. The
key to this
>is a definition of convergence in which a limit is not
explicitly
>required. This is attributed to Cauchy.
>Bob Kolker
>> The basic fundamental definition of .999.... is that it
isnt a real
>>number.
>> A number that isnt real cant equal an 
number that is
real. How simple
can
>>you get with this?
> Heres another way to look at this.
> i = something imaginary ( not real )
> Can you prove i = a real number?
Theorem: For all real numbers r, r^2 >= 0.
Proof:
Case r > 0: r*r = r^2 > 0.
Case r = 0: r*r = 0.
Case r < 0: r*r = (-s)*(-s) = s*s > 0, where s = -r > 0. QED.
Corollary: There is no real number i such that i^2 = -1.
Theorem: .999... is a real number.
Proof:
The usual definition of .999... is an endless series of 
9s.
The implicit series is .9 + .09 + .009 + ...
= sum(i=1,+oo) (9 * 10^(-i)). This series is of the form
sum(i=1,+oo) (a * r^(i)), which converges for all -1 < r < 1.
Hence .999... is a real number. QED.
Theorem: If .999... < 1, then there exists infinitely many y
with
.999... < y < 1.
Proof: Let x = .999... and y = (x + n - 1) / n, for any n > 0.
Since x < 1, nx < n. ny = x + n - 1 < n,
however, nx = (n-1)x + x < x + n - 1, therefore
nx < ny and hence x < y as well. QED.
Corollary: .999... is not a unique number if .999... < 1.
:-)
[.sigsnip]
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>In sci.math, S. Enterprize Company
><smart1234@aol.com
>
>>Dedekind is credited with giving the first mathematical
definition
>>of the real numbers.
>
>
> Is this so?
>>Yes and no. The set of real numbers was implicitly defined 
by
>>algebraic laws on the two basic operations + and *. A
rigorous
>>definition (or reduction) of real numbers to the rationals
was done
in
>>the latter half of the 19-th century along with definitions
of
>>continuous functions, differentiable functions and limits.
The key to
>>the whole business is the rigorous definition of limit.
>>The reals are the topological closure of the rationals. The
key to this
>>is a definition of convergence in which a limit is not
explicitly
>>required. This is attributed to Cauchy.
>>Bob Kolker
> The basic fundamental definition of .999.... is that it
isnt a real
>number.
> A number that isnt real cant equal an 
number that is
real. How simple
>can
>you get with this?
>> Heres another way to look at this.
>> i = something imaginary ( not real )
>> Can you prove i = a real number?
>Theorem: For all real numbers r, r^2 >= 0.
>Proof:
>Case r > 0: r*r = r^2 > 0.
>Case r = 0: r*r = 0.
>Case r < 0: r*r = (-s)*(-s) = s*s > 0, where s = -r > 0. QED.
>Corollary: There is no real number i such that i^2 = -1.
>Theorem: .999... is a real number.
No its not a real number. It cant be 
expressed as a
fraction.
9/10 = .9
>Proof:
>The usual definition of .999... is an endless series of 
9s.
>The implicit series is .9 + .09 + .009 + ...
>= sum(i=1,+oo) (9 * 10^(-i)). This series is of the form
>sum(i=1,+oo) (a * r^(i)), which converges for all -1 < r < 1.
>Hence .999... is a real number. QED.
No its not. A real rational number has to be expressed as a
fraction.
Its
not an irrational real number either, it has repeating digits
in the
decimal
form.
It is not a real number. Can it be shown as a series? Yes. Is
there a
space
between,
.999... | | and 1? Yes.
.999... =/= 1
.999... could be called a hyper-real number. See the
definition at the math
link below.
http://mathworld.wolfram.com/HyperrealNumber.html
In other words, a hyper-real number is an infinite number such
that,
x < n
x = .999...
n = 1
x =/= n
Since .999... isnt a real number but appears to be of the
form of a
hyper-real
number,
.999... =/= 1
.999... < 1
>Theorem: If .999... < 1, then there exists infinitely many y
with
> .999... < y < 1.
>Proof: Let x = .999... and y = (x + n - 1) / n, for any n >
0.
> Since x < 1, nx < n. ny = x + n - 1 < n,
> however, nx = (n-1)x + x < x + n - 1, therefore
> nx < ny and hence x < y as well. QED.
>Corollary: .999... is not a unique number if .999... < 1.
>:-)
>[.sigsnip]
>--
>#191, ewill3@earthlink.net
>Its still legal to go .sigless.
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===
Subject: Re: .99999... still=/= 1
> No its not a real number. It cant be 
expressed as a
fraction.
Neither can the square root of 2 be so expressed, but it is a
real (and
irrational) number.
The real numbers consist of rational numbers which can be
expressed as
ratios of integers and and irrationals which cannot. However
all real
numbers have non-negative squares.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> No its not a real number. It cant be 
expressed as a
fraction.
>Neither can the square root of 2 be so expressed, but it is
a real (and
>irrational) number.
>The real numbers consist of rational numbers which can be
expressed as
>ratios of integers and and irrationals which cannot. However
all real
>numbers have non-negative squares.
confused?
>Bob Kolker
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===
Subject: Re: .99999... still=/= 1
> Heres another way to look at this.
> i = something imaginary ( not real )
> Can you prove i = a real number?
What does this have to do with your .999... nonsense.
.999.... is a meaningless string of symbols. It has no
mathematical
meaning whatsoever.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> Heres another way to look at this.
>> i = something imaginary ( not real )
>> Can you prove i = a real number?
>What does this have to do with your .999... nonsense.
>.999.... is a meaningless string of symbols. It has no
mathematical
>meaning whatsoever.
>Bob Kolker
Look up the definition of an irrational number. It almost 
fits
in except
for
one small detail, the 9s keep repeating 
indefinitely, which
makes it not
an
irrational number, which means it cant be a real number.
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===
Subject: Re: .99999... still=/= 1
> Look up the definition of an irrational number. It almost
fits in
except for
> one small detail, the 9s keep repeating 
indefinitely, which
makes it not
an
> irrational number, which means it cant be a real number.
An irrational number is a series of the form SUM (n >= 1)
[a_n*1/10^n)
where the sequence (a_n | n >= 1} is not of the form of a
finite
subsequence followed by a periodic repeating sequence. The
series is
convergent so it equals a real number. Recall that an infinit
series is
the -limit- of the sequence of finite partial sums.
Equivalently an irratial real number is one that is not the
ratio of two
integers with no common factor. It is a theorem of real
number theory
that the above two definitions are equivalent.
What is your definition of a repeating decimal number. If it
does not
come out to be a convergent series of multiples of 1/10^n it
is just
plain wrong. By the way decimal series that repeat
periodically after a
certain term are rational numbers which constitute a subset
of the real
numbers.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> Look up the definition of an irrational number. It almost
fits in
except
>for
>> one small detail, the 9s keep repeating 
indefinitely,
which makes it
not
>> irrational number, which means it cant be a real number.
>An irrational number is a series of the form SUM (n >= 1)
[a_n*1/10^n)
>where the sequence (a_n | n >= 1} is not of the form of a
finite
>subsequence followed by a periodic repeating sequence. The
series is
>convergent so it equals a real number. Recall that an infinit
series is
>the -limit- of the sequence of finite partial sums.
>Equivalently an irratial real number is one that is not the
ratio of two
>integers with no common factor. It is a theorem of real
number theory
>that the above two definitions are equivalent.
>What is your definition of a repeating decimal number. If it
does not
>come out to be a convergent series of multiples of 1/10^n it
is just
>plain wrong. By the way decimal series that repeat
periodically after a
>certain term are rational numbers which constitute a subset
of the real
>numbers.
>Bob Kolker
There is no real number representation of .999... . All you
have is that
someone told you to believe this. There is no proof.
9/10 = .9 not .999...
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===
Subject: Re: .99999... still=/= 1
> There is no real number representation of .999... . All you
have is
that
> someone told you to believe this. There is no proof.
I have proven that SUM (n >= 1) [9/10^n] = 1 several times in
this thread.
Each partial sum to the k-th power comes out to be 1 -
1/10^k, which can
be made as close to 1 as you like which proves the limit of
the partial
sums is 1. QED.
Fucking idiot.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> There is no real number representation of .999... . All
you have is
that
>> someone told you to believe this. There is no proof.
>I have proven that SUM (n >= 1) [9/10^n] = 1 several times
in this thread.
>Each partial sum to the k-th power comes out to be 1 -
1/10^k, which can
>be made as close to 1 as you like which proves the limit of
the partial
>sums is 1. QED.
>Fucking idiot.
Whats wrong you have to call me names now because you made 
a
mistake that
shows you dont even know the basic fundaments of math?
>Bob Kolker
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===
Subject: Re: .99999... still=/= 1
> The basic fundamental definition of .999.... is that it
isnt a real
number.
This is meaningless non-mathematical bullshit.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> The basic fundamental definition of .999.... is that it
isnt a real
>number.
>This is meaningless non-mathematical bullshit.
>Bob Kolker
Ok then, heres the x-axis,
0
|------------|------------------------->
^
.999999--> now keep typing in 9s for eternity so that we
observe this on the axis at the specified location.
In the mean time, I will place a real number on the x-axis.
0 1
|----------------|----------------------->
See you later, I mean I wont see you later, 
youll be too
busy.
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===
Subject: Re: .99999... still=/= 1
> 0
> |------------|------------------------- ^
> .999999--> now keep typing in 9s for eternity so that
we
> observe this on the axis at the specified location.
What is .999... ? It has no mathematical meaning whatsoever.
Define
.999... in terms of elementary arithmetic operations. By what
logic to
you identify a point on a line with a number? Justify your
assertion
mathematically.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> 0
>> |------------|-------------------------> ^
>> .999999--> now keep typing in 9s for eternity so that
we
>> observe this on the axis at the specified location.
>What is .999... ? It has no mathematical meaning whatsoever.
Define
>.999... in terms of elementary arithmetic operations. By
what logic to
>you identify a point on a line with a number? Justify your
assertion
>mathematically.
>Bob Kolker
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===
Subject: Re: .99999... still=/= 1
>> 0 |------------|-------------------------> ^
>> .999999--> now keep typing in 9s for eternity so that
>> we
>> observe this on the axis at the specified location.
> What is .999... ? It has no mathematical meaning
whatsoever. Define
> .999... in terms of elementary arithmetic operations. By
what logic to you
> identify a point on a line with a number? Justify your
assertion
> mathematically.
> Bob Kolker
that might be true, but everyone know what .999999.... means.
just like sum(a_k,k=0..infinity) doesnt mean 
jack, but
everyone knows what
its suppose to mean.
The problem is that someone cant comprehend the idea of
inductive
reasoning, Probably from a a nutritional deficiency in there
early
developmental years.
===
Subject: Re: .99999... still=/= 1
> 0 |------------|------------------------- ^
> .999999--> now keep typing in 9s for eternity so that
> we
> observe this on the axis at the specified location.
>> What is .999... ? It has no mathematical meaning
whatsoever. Define
>> .999... in terms of elementary arithmetic operations. By
what logic to
you
>> identify a point on a line with a number? Justify your
assertion
>> mathematically.
>> Bob Kolker
>that might be true, but everyone know what .999999.... means.
>just like sum(a_k,k=0..infinity) doesnt mean 
jack, but
everyone knows what
>its suppose to mean.
>The problem is that someone cant comprehend the idea of
inductive
>reasoning, Probably from a a nutritional deficiency in there
early
>developmental years.
First you say .999... is a real number. Then you say its
meaningless. It
seems like you are very confused.
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===
Subject: Re: .99999... still=/= 1
>> Clever argument, that. But I wonder if anything is missing.
>A proper definition of .999... . What does it mean?
It means a number that isnt real.
>Bob Kolker
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===
Subject: Re: .99999... still=/= 1
>Clever argument, that. But I wonder if anything is missing.
>>A proper definition of .999... . What does it mean?
> It means a number that isnt real.
It has no meaning. Restate what you mean by an unreal number.
And use
proper mathematical terminology.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
In sci.math, robert j. kolker
<nowhere@nowhere.com>Clever argument, that. But I wonder if
anything is missing.
>A proper definition of .999... . What does it mean?
>> It means a number that isnt real.
> It has no meaning. Restate what you mean by an unreal
number. And use
> proper mathematical terminology.
number of deaths.
:-)
A more standard designation is hyperreal number, which
presumably
is the (?) number greater than 0 but less than 1/n for all n
in N.
This gets into non-standard analysis -- which is beyond me,
and
probably way beyond SE.
> Bob Kolker
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>>Clever argument, that. But I wonder if anything is missing.
>A proper definition of .999... . What does it mean?
>> It means a number that isnt real.
>It has no meaning. Restate what you mean by an unreal
number. And use
>proper mathematical terminology.
>Bob Kolker
Watch very closely at the definitions.
Math. Definition -
Irrational Numbers
All numbers that are not rational are considered irrational.
An
irrational
number can be written as a decimal, but not as a fraction.
An irrational number may have endless non-repeating digits to
the right
of
the decimal point. Here are some irrational numbers:
p = 3.141592.83
sqrt(2) = 1.414213.83
reference LINK : Math definition for above quote of irrational
number.
http://www.factmonster.com/ipka/A0876704.html
reference :AOL definition
Main Entry: irrational number
Function: noun
: a number that can be expressed as an infinite decimal with
no set of
consecutive digits repeating itself indefinitely and that
cannot be
expressed
as the quotient of two integers.
Now note one minor difference. An irrational number has to be
written as a
decimal not a fraction. And these decimals can repeat
infinitely but with a
variation of the decimal values or (NON-REPEATING DIGITS <--
quoting
the
above MATH definition of a decimal), like for example,
.1237636832894365462654...
But that is different from a consecutive ( REPEATING decimal
repetition
like
this,
.99999999999999...
This type of number isnt classified as rational 
or irrational.
It is a number that ISNT REAL. Its neither 
rational nor
irrational. It
is
an indeterminate.
In an earlier post I showed you that .999... could be looked
at as an
irrational number using this,
n--->oo - 1
lim 9/10^n --> 90/10^oo
The terminating variation digit makes this form of .999... an
irratinal
number or a real number. This was a non-standard analysis,
although. In
other
words, I assume you can reach infinity and move back one
decimal point back.
But in the case where it remains something not a real number
is where the
9s
just keep repeating.
So basically a number that is neither rational or irrational
is an
INDETERMINATE.
1/0 <--- this isnt a real number but is expressed as a
fraction.
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===
Subject: Re: .99999... still=/= 1
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
> But that is different from a consecutive ( REPEATING decimal
repetition like
> this,
> .99999999999999...
> This type of number isnt classified as 
rational or
irrational.
Its not? Doesnt 0.33333.... also meet this 
description? Do
you
think that 0.3333... is neither rational nor irrational?
- Randy
===
Subject: Re: .99999... still=/= 1
>> But that is different from a consecutive ( REPEATING
decimal
>repetition like
>> this,
>> .99999999999999...
>> This type of number isnt classified as 
rational or
irrational.
>Its not? Doesnt 0.33333.... also meet this 
description? Do
you
>think that 0.3333... is neither rational nor irrational?
>- Randy
I even showed you people the standard math definition of an
irrational
number
as you requested. You just dont want to admit your 
mistakes.
Even if
.999...
was an irrational number, you cant make a rational number
equal to an
irrational number.
.999... can not be expressed as a fraction. It has to be
expressed as a
series.
9/10 = .9 not .999...
And in regard to 1/3 = .333... . Well at least it can be
expressed as a
fraction. So it may be classified as a real number. But
.999... can not be
represented as a fraction, or an inrrational number by the
standard math
definitions.
.999... =/= 1
And this has as never been proven to be equal to 1, other
than someone just
saying it does. Just saying it doesnt make it so.
There is a problem that you people seem to fail to understand.
.999... | | 1
^
|
You people fail to understand this separating space between
.999...
and
1.
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===
Subject: Re: .99999... still=/= 1
>> But that is different from a consecutive ( REPEATING
decimal
>repetition like
>> this,
>> .99999999999999...
>> This type of number isnt classified as 
rational or
irrational.
>Its not? Doesnt 0.33333.... also meet this 
description? Do
you
>think that 0.3333... is neither rational nor irrational?
>- Randy
> I even showed you people the standard math definition of an
irrational
number
> as you requested. You just dont want to admit your
mistakes. Even if
.999...
> was an irrational number, you cant make a rational number
equal to an
> irrational number.
> .999... can not be expressed as a fraction. It has to be
expressed as a
series.
1/9 can be expressed as 0.1111...
2/9 can be expressed as 0.2222...
...
8/9 cna be expressed as 0.8888...
9/9 can be expressed as what then?
===
Subject: Re: .99999... still=/= 1
> But that is different from a consecutive ( REPEATING decimal
>>repetition like
> this,
.99999999999999...
This type of number isnt classified as rational 
or irrational.
>>Its not? Doesnt 0.33333.... also meet this 
description?
Do you
>>think that 0.3333... is neither rational nor irrational?
>>- Randy
>> I even showed you people the standard math definition of an
irrational
>number
>> as you requested. You just dont want to admit your
mistakes. Even if
>.999...
>> was an irrational number, you cant make a rational 
number
equal to an
>> irrational number.
>> .999... can not be expressed as a fraction. It has to be
expressed as a
>series.
>1/9 can be expressed as 0.1111...
>2/9 can be expressed as 0.2222...
>...
>8/9 cna be expressed as 0.8888...
>9/9 can be expressed as what then?
by 1 not .999... .
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===
Subject: Re: .99999... still=/= 1
> But that is different from a consecutive ( REPEATING decimal
>>repetition like
> this,
.99999999999999...
This type of number isnt classified as rational 
or irrational.
>>Its not? Doesnt 0.33333.... also meet this 
description?
Do you
>>think that 0.3333... is neither rational nor irrational?
>>- Randy
>> I even showed you people the standard math definition of an
irrational
>number
>> as you requested. You just dont want to admit your
mistakes. Even if
>.999...
>> was an irrational number, you cant make a rational 
number
equal to an
>> irrational number.
>> .999... can not be expressed as a fraction. It has to be
expressed as
a
>series.
>1/9 can be expressed as 0.1111...
>2/9 can be expressed as 0.2222...
>...
>8/9 cna be expressed as 0.8888...
>9/9 can be expressed as what then?
> by 1 not .999... .
So 1/9 = 0.1111...
and 8/9 = 0.8888...
then what is 1/9 + 8/9?
===
Subject: Re: .99999... still=/= 1
In sci.math, Richard Henry
<rphenry@home.com>
<ZS7xd.1740$ry.1191@fed1read01>:
>> But that is different from a consecutive ( REPEATING
decimal
>repetition like
>> this,
>> .99999999999999...
>> This type of number isnt classified as 
rational or
irrational.
Its not? Doesnt 0.33333.... also meet this 
description? Do
you
>think that 0.3333... is neither rational nor irrational?
>- Randy
> I even showed you people the standard math definition of an
> irrational
>>number
> as you requested. You just dont want to admit your
mistakes. Even if
>>.999...
> was an irrational number, you cant make a rational number
equal to
an
> irrational number.
.999... can not be expressed as a fraction. It has to be
expressed as
a
>>series.
>>1/9 can be expressed as 0.1111...
>>2/9 can be expressed as 0.2222...
>>...
>>8/9 cna be expressed as 0.8888...
>>9/9 can be expressed as what then?
>> by 1 not .999... .
> So 1/9 = 0.1111...
> and 8/9 = 0.8888...
> then what is 1/9 + 8/9?
1/9 + 8/9 is an expression that might have to be properly
defined
in order to be evaluated. The simplest method is to extend J
to Q
in the normal fashion:
if a/b = c, then c * b = a, b != 0, for some value c which
can either
be or not be in J (i.e., J is a proper subset of Q).
and then
a/b + c/d = (ad + bc)/(bc), a,b,c,d in J, b,d != 0.
Also, one can take out factors: a/b = ak/bk for any k != 0.
So 1/9 + 8/9 = (9 + 72) / 81 = 81/81 = 1/1 = 1.
Fine on the left side.
Now the right side gets tricky. The problem is, if one adds
two infinite decimal fractions, one might have an 
infinite
carry.
In the finite case, one might have
.111111 + .888888 = .999999
of course, or one might have
.000001 + .111111 + .888888 = 1.000000
where the carry bubbles from right to left. In the case of
infinite decimals, of course, there is no rightmost digit,
leading to many headaches. Another method to show this
is by *subtraction*:
9.9999... - 0.999...
is an interesting expression, but is it equal to 9.000..., if
one
assumes that theres no infinite borrow, or 
8.999..., if one
assumes that there is?
It turns out to be equal to both, making everybody happy --
except maybe S. Enterprize, |-|erc, and Gary Denke. :-)
This is what torpedoes the proof, which shows up occasionally
here:
let x = 0.999...
then 10x = 9.999...
and 10x - x = 9x = 9.000...
therefore x = 1.000...
the third step may just as well be written 10x - x = 8.999...
.
A pity, since its otherwise a nice proof.
And then theres the definition of what an 
infinite decimal
is, anyway. One can define a numeric system, complete with
addition and subtraction, with notation such as
.[1] + .[8] = .[9]
where [1] represents an endlessly repeating series of 1
digits.
For example,
1/3 = .[3]
1/7 = .[142857]
1/24 = .041[6]
______
(A more traditional overbar might also be used, e.g. 1/7 =
.142857 .)
It gets a little tricky, of course, to add things such
as .[09] + .[238] = .[090909] + .[238238] = .[329147] .
Multiplication gets even trickier, but looks vaguely doable.
The main problem: .[142857]^2 = 142857^2/999999^2; now
hopefully one can find the minimum number k such that (10^n -
1)^2
is divisible into 10^k - 1, but Id have to do some 
research;
of
course its always possible (since all rational numbers have
endlessly repeating decimals, if one counts [0] as an example
of such).
This looks rigorous enough to pass muster, but one has to
assume
or prove at some point that 0.[9] = 1.[0] . (The proof isnt
too
hard if one uses limits.)
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
> So 1/9 = 0.1111...
> and 8/9 = 0.8888...
> then what is 1/9 + 8/9?
9/9. expressions like .111... .888.... and .999.... are not
proper
mathematical expressions. If you mean to write down series,
then do so.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> So 1/9 = 0.1111...
>> and 8/9 = 0.8888...
>> then what is 1/9 + 8/9?
>9/9. expressions like .111... .888.... and .999.... are not
proper
>mathematical expressions. If you mean to write down series,
then do so.
>Bob Kolker
I guess you dont know what a hyper-real number is. Here we
go again and
again.
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Subject: Re: .99999... still=/= 1
In sci.math, S. Enterprize Company
<smart1234@aol.com
> So 1/9 = 0.1111...
> and 8/9 = 0.8888...
> then what is 1/9 + 8/9?
>>9/9. expressions like .111... .888.... and .999.... are not
proper
>>mathematical expressions. If you mean to write down series,
then do so.
>>Bob Kolker
> I guess you dont know what a hyper-real number is. Here 
we
go again
and
> again.
A hyperreal number is a number less than any 1/n (n in J, n >
0) but
greater than 0. It is not entirely clear to me how many
hyperreal
numbers there are.
[.sigsnip]
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>> But that is different from a consecutive ( REPEATING
decimal
>repetition like
>> this,
>> .99999999999999...
>> This type of number isnt classified as 
rational or
irrational.
Its not? Doesnt 0.33333.... also meet this 
description? Do
you
>think that 0.3333... is neither rational nor irrational?
>- Randy
> I even showed you people the standard math definition of an
>irrational
>>number
> as you requested. You just dont want to admit your
mistakes. Even if
>>.999...
> was an irrational number, you cant make a rational number
equal to
an
> irrational number.
.999... can not be expressed as a fraction. It has to be
expressed as
a
>>series.
>>1/9 can be expressed as 0.1111...
>>2/9 can be expressed as 0.2222...
>>...
>>8/9 cna be expressed as 0.8888...
>>9/9 can be expressed as what then?
>> by 1 not .999... .
>So 1/9 = 0.1111...
>and 8/9 = 0.8888...
>then what is 1/9 + 8/9?
i is an imaginary number then what is i^2? So is i a real
number?
Show the single fraction that makes .999... . Some people
have gone as far
as
to say,
9/9 = .999...
But youre saying something like this,
.9 + .1 = 1
.999... + .000...1 = 1
So what. Is that the same as just .999...? No.
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Subject: Re: .99999... still=/= 1
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
>> But that is different from a consecutive ( REPEATING
decimal
>repetition like
>> this,
>> .99999999999999...
>> This type of number isnt classified as 
rational or
irrational.
>Its not? Doesnt 0.33333.... also meet this 
description? Do
you
>think that 0.3333... is neither rational nor irrational?
>- Randy
> I even showed you people the standard math definition of an
irrational number
> as you requested. You just dont want to admit your
mistakes. Even if
.999...
> was an irrational number, you cant make a rational number
equal to
an
> irrational number.
Nobodys claiming its an irrational number. 
Its rational.
Youre the one making wacky statements like that a number
which
isnt irrational isnt a real number at all.
Or the statement above, that a number consisting of repeating
digits isnt classified as rational or 
irrational. What all of
us would say is that a number which terminates in a periodic
sequence of digits, whether it is 1 digit long or googolplex,
is rational.
> .999... can not be expressed as a fraction.
Well, thats the topic of debate, isnt it? We 
all think it
can be expressed as the fraction 9/9.
You think otherwise, but you can hardly prove that just by
stating it outright. Thats not a proof, 
thats just stamping
your foot and wishing.
> It has to be expressed as a series.
No, it doesnt have to. It is the limit of a series which 
has
been proved countless times to converge to 1.0
> 9/10 = .9 not .999...
Nobody said its 9/10.
0.111... is not 1/10, its 1/9.
0.333... is not 3/10, its 3/9.
0.999... is not 9/10, its 9/9.
> And in regard to 1/3 = .333... . Well at least it can be
expressed
as a
> fraction.
But I thought you said a repeating digit cant be 
classified
as rational or irrational. Would you like to revise the first
statement above?
- Randy
===
Subject: Re: .99999... still=/= 1
> But that is different from a consecutive ( REPEATING decimal
>>repetition like
> this,
.99999999999999...
This type of number isnt classified as rational 
or irrational.
>>Its not? Doesnt 0.33333.... also meet this 
description?
Do you
>>think that 0.3333... is neither rational nor irrational?
>>- Randy
>> I even showed you people the standard math definition of an
>irrational number
>> as you requested. You just dont want to admit your
mistakes. Even if
>.999...
>> was an irrational number, you cant make a rational 
number
equal to
>> irrational number.
>Nobodys claiming its an irrational number. 
Its rational.
That Ôs wrong. The decimal number .999... isnt 
a rational
number. Show
me
the fraction that represents .999... .
9/9 = 1 this isnt .999...
9/10 = .9 this isnt 1 or .999...
>Youre the one making wacky statements like that a number
which
>isnt irrational isnt a real number at all.
>Or the statement above, that a number consisting of repeating
>digits isnt classified as rational or 
irrational. What all of
>us would say is that a number which terminates in a periodic
>sequence of digits, whether it is 1 digit long or googolplex,
>is rational.
>> .999... can not be expressed as a fraction.
>Well, thats the topic of debate, isnt it? 
We all think it
>can be expressed as the fraction 9/9.
>You think otherwise, but you can hardly prove that just by
>stating it outright. Thats not a proof, 
thats just stamping
>your foot and wishing.
>> It has to be expressed as a series.
>No, it doesnt have to. It is the limit of a series which 
has
>been proved countless times to converge to 1.0
>> 9/10 = .9 not .999...
>Nobody said its 9/10.
>0.111... is not 1/10, its 1/9.
>0.333... is not 3/10, its 3/9.
>0.999... is not 9/10, its 9/9.
>> And in regard to 1/3 = .333... . Well at least it can be
expressed
>as a
>> fraction.
>But I thought you said a repeating digit cant be 
classified
>as rational or irrational. Would you like to revise the first
>statement above?
> - Randy
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Subject: Re: .99999... still=/= 1
posting-account=uMDgiw0AAAANoAxJOs_DnrtdjhRMBFah
> And in regard to 1/3 = .333... . Well at least it can be
> expressed as a fraction. So it may be classified as a real
> number. But .999... can not be represented as a fraction
Help me understand what you mean by .333... and .999...:
Which of the following statements are true?
a) 3 * .111... = .333...
b) 2 * .333... = .666...
c) 3 * .333... = .999...
d) 10 * .999... = 9.999...
e) .111... + .222... = .333...
f) .333... + .666... = .999...
g) 9 + .999... = 9.999...
h) .111... = 1/9
i) .333... = 1/3
And what is .999... + .999... ?
===
Subject: Re: .99999... still=/= 1
>> And in regard to 1/3 = .333... . Well at least it can be
>> expressed as a fraction. So it may be classified as a real
>> number. But .999... can not be represented as a fraction
>Help me understand what you mean by .333... and .999...:
>Which of the following statements are true?
>a) 3 * .111... = .333...
>b) 2 * .333... = .666...
>c) 3 * .333... = .999...
Big friggin deal.
i^2 = -1
So does that mean i is a real number too.
>d) 10 * .999... = 9.999...
>e) .111... + .222... = .333...
>f) .333... + .666... = .999...
>g) 9 + .999... = 9.999...
>h) .111... = 1/9
>i) .333... = 1/3
>And what is .999... + .999... ?
Its an indeterminate, unless you use Partial Sums
Convergence to give
you
an approximation of what it is.
Express .999... as a fraction. DUH???? What you waiting on?
9/9 = 1 not .999...
9/10 = .9
DUH????? Whats wrong, you cant admit you are 
mistaken?
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Subject: Re: .99999... still=/= 1
posting-account=uMDgiw0AAAANoAxJOs_DnrtdjhRMBFah
>> And in regard to 1/3 = .333... . Well at least it can be
>> expressed as a fraction. So it may be classified as a real
>> number. But .999... can not be represented as a fraction
>Help me understand what you mean by .333... and .999...:
>Which of the following statements are true?
>a) 3 * .111... = .333...
>b) 2 * .333... = .666...
>c) 3 * .333... = .999...
> Big friggin deal.
> i^2 = -1
> So does that mean i is a real number too.
Well, the big friggin deal is that with your definition
of numbers, it doesnt seem possible to do arithmetic.
For example, does 3 * .111... equal .333? Yes? No? Maybe?
It doesnt seem like youre able to say. This 
isnt very
practical.
>d) 10 * .999... = 9.999...
>e) .111... + .222... = .333...
>f) .333... + .666... = .999...
>g) 9 + .999... = 9.999...
>h) .111... = 1/9
>i) .333... = 1/3
And again: you cant say which are true and which are false.
The fact that you dont know if you can add and multiply 
these
numbers suggests that you shouldnt be lecturing about
numbers.
>And what is .999... + .999... ?
> Its an indeterminate, unless you use Partial Sums
Convergence
to give you
> an approximation of what it is.
Hmmm. You call .999... a number, though not a real number.
You call
.999... + .999... an indeterminate, which I gather is not a
number.
> Express .999... as a fraction. DUH???? What you waiting on?
> 9/9 = 1 not .999...
> 9/10 = .9
I cannot express .999... as a fraction that does not equal 1.
You know, people here would a agree that you can define an
ordered set
of strings used to denote real numbers (with certain
conventions like,
say, always using 0.3 instead of .3 and 5. instead of 5).
For
this ordered set we have, for example,
0. < 0.9 < 0.99 < 0.999... < 1. < 1.999... < 2.
Note that 0.999... does not equal 1 here.
Yep, 0.999... =/= 1. And, whats more, I can convince at
least 80%
of sci.math that this is true, because Im able to say what 
I
mean.
> DUH????? Whats wrong, you cant admit you 
are mistaken?
People use symbols to represent mathematical ideas. For
example, +
is used to mean plus. They use strings of digits to represent
real
numbers. There is a convention which says which strings mean
which
numbers. Under this convention, 0.999... and 1. and represent
the same number.
The strings themselves are not the numbers. They are ways of
representing the numbers. Four and 4 represent the same
number.
0.999... and 1. represent the same number. We could have a
convention that 0.999... does not represent a number: this
wouldnt
really be a problem because we could just use 1. to represent
the
number one.
But the convention that people have agreed upon is that
.999...
does represent a (real) number.
===
Subject: Re: .99999... still=/= 1
> Big friggin deal.
> i^2 = -1
> So does that mean i is a real number too.
No it doesnt. For any real number r, r^2 >= 0.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> Big friggin deal.
>> i^2 = -1
>> So does that mean i is a real number too.
>No it doesnt. For any real number r, r^2 >= 0.
>Bob Kolker
Oh but you add hundreds of things to,
.999.... to try to make it EQUAL 1. Just leave it in this
form,
.999...
IT DOES NOT EQUAL 1.
In fact, you dont even know what a hyper-real number is.
.999... < 1
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Subject: Re: .99999... still=/= 1
> And in regard to 1/3 = .333... . Well at least it can be
expressed as a
> fraction. So it may be classified as a real number. But
.999... can not
be
> represented as a fraction, or an inrrational number by the
standard math
> definitions.
> .999... =/= 1
Listen up, stupid. A repating decimal corresponds to a series
which is
equal to a rational number.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> And in regard to 1/3 = .333... . Well at least it can be
expressed as
a
>> fraction. So it may be classified as a real number. But
.999... can not
be
>> represented as a fraction, or an inrrational number by the
standard math
>> definitions.
>> .999... =/= 1
>Listen up, stupid. A repating decimal corresponds to a
series which is
>equal to a rational number.
>Bob Kolker
A decimal representation of a number isnt a rational number
in the case
of
.999.... .
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Subject: Re: .99999... still=/= 1
posting-account=uMDgiw0AAAANoAxJOs_DnrtdjhRMBFah
> But that is different from a consecutive ( REPEATING decimal
> repetition like this,
> .99999999999999...
> This type of number isnt classified as 
rational or
> irrational.
Actually, its classified as rational.
I put classified in quotes because thats a bad 
way of looking
at things. Well formed strings of digits have a standard
interpretation as real numbers. Some of these real numbers are
rational (viewing the rationals as a subset of the reals).
So its not that strings that end in a repeating block of
digits
are classified as rationals per se, but that the real numbers
they represent are, in fact, necessarily rational.
Suppose someone says, a proper definition of 
Ô+ is that
youre putting things together, so if the two ones are still
two after adding them, they havent really been merged:
1+1=1.
A response to this is that, around here, 1, +, =, and
2 have precise definitions, from which it follows that
1+1=2 is true, whereas 1+1=1 is false.
Similarly, .999..., 1, rational number, and real
number have definitions, according to which .999... and
1 are rational and real, and, in fact .999... = 1 is true.
But youre not interested in those definitions. 
You keep
saying, in effect, Look! I take 1 lump of clay and 1 lump
of clay, and put them together to get, surprise, surprise,
1 lump of clay! I cant believe that you dont 
understand!
Dictionary definitions of add do not make 1+1=2 false.
The problem youre having is that the 
definitions of real
number and of the real number that corresponds to a given
digit string are, apparently, too complicated for you to
understand. This isnt surprising: they are typically
introduced in a sophomore level college math course, so in
some sense, this is quite advanced material.
However, to many people here, sophomore level material is
almost as basic as addition. To those of us who understand
what a real number is, your arguments are as muddled as the
above argument that 1+1=1 is to someone who understands
addition.
The essence of your mistake is quite common. You dont
understand what a mathematical definition is, nor that the
definition of something is crucial to proving things about
it. You seem to believe that your notion about what a thing
is or should be, or dictionary descriptions of it, suffice
to prove things about it.
So stop trying to argue your point. Listen to what people
are telling you because most of us know a lot more about
math than you do. Let yourself be educated.
You could start by trying to understand what a real number
is, and how digits strings are used to represent them.
===
Subject: Re: .99999... still=/= 1
> reference LINK : Math definition for above quote of
irrational number.
I asked you to define an unreal number, not an irrational
number.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> reference LINK : Math definition for above quote of
irrational number.
>I asked you to define an unreal number, not an irrational
number.
>Bob Kolker
Oh you try to change the subject. You fail to admit your
mistake about
the
math definition of an irrational number? Think about it then.
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Subject: Re: .99999... still=/= 1
> Oh you try to change the subject. You fail to admit your
mistake about
the
> math definition of an irrational number? Think about it 
then.
An irrational number is one not equal to the ratio of two
integers. What
is an unreal number?
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> Oh you try to change the subject. You fail to admit your
mistake about
>the
>> math definition of an irrational number? Think about it
then.
>An irrational number is one not equal to the ratio of two
integers. What
Who said it was?
>is an unreal number?
>Bob Kolker
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Subject: Re: .99999... still=/= 1
>Clever argument, that. But I wonder if anything is missing.
>>A proper definition of .999... . What does it mean?
>
>
> It means a number that isnt real.
>>It has no meaning. Restate what you mean by an unreal
number. And use
>>proper mathematical terminology.
>>Bob Kolker
> Watch very closely at the definitions.
>Math. Definition -
>Irrational Numbers
> All numbers that are not rational are considered
irrational. An
irrational
>number can be written as a decimal, but not as a fraction.
> An irrational number may have endless non-repeating digits
to the right
of
>the decimal point. Here are some irrational numbers:
>p = 3.141592.83
>sqrt(2) = 1.414213.83
>reference LINK : Math definition for above quote of
irrational number.
>http://www.factmonster.com/ipka/A0876704.html
>reference :AOL definition
>Main Entry: irrational number
>Function: noun
>: a number that can be expressed as an infinite decimal with
no set of
>consecutive digits repeating itself indefinitely and that
cannot be
expressed
>as the quotient of two integers.
> Now note one minor difference. An irrational number has to
be written as
a
>decimal not a fraction. And these decimals can repeat
infinitely but with
a
>variation of the decimal values or (NON-REPEATING DIGITS <--
quoting
the
>above MATH definition of a decimal), like for example,
> .1237636832894365462654...
> But that is different from a consecutive ( REPEATING
decimal repetition
like
>this,
> .99999999999999...
> This type of number isnt classified as 
rational or
irrational.
> It is a number that ISNT REAL. Its neither 
rational nor
irrational. It
is
>an indeterminate.
> In an earlier post I showed you that .999... could be
looked at as an
>irrational number using this,
>n--->oo - 1
>lim 9/10^n --> 90/10^oo
> The terminating variation digit makes this form of .999...
an
irratinal
>number or a real number. This was a non-standard analysis,
although. In
>other
>words, I assume you can reach infinity and move back one
decimal point
back.
> But in the case where it remains something not a real
number is where
the
>9s
>just keep repeating.
> So basically a number that is neither rational or
irrational is an
>INDETERMINATE.
> 1/0 <--- this isnt a real number but is expressed as a
fraction.
But lets even look at it from your point of view. 
Lets
assume .999...
is
an irrational number. Can an irrational number equal a
rational number? No.
2/2 = 1 <--- This is a rational number.
.999... cant be expressed as a fraction. So how do you make
a rational
number
equal an irrational number? You dont.
.999... =/= 1
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===
Subject: Re: .99999... still=/= 1
posting-account=W2DCTA0AAAAlbhDMl3GrysSnPy1IK_7f
> But lets even look at it from your point of view. 
Lets
assume
.999... is
> an irrational number. Can an irrational number equal a
rational
number? No.
> 2/2 = 1 <--- This is a rational number.
> .999... cant be expressed as a fraction. So how do you
make a
rational number
> equal an irrational number? You dont.
Wrong. .999... (if we accept this notation, against which Bob
has ---
well-founded --- reservations) is, by definition a rational
number,
just like any repeating decimal expansion. As such, it can be
written
as a fraction, namely
.999... = 9/(10-1) = 9/9 = 1.
As Ive said before, this works for ANY repeating expansion,
and could
be taken as a definition of the rational number represented by
such an
expansion. For example,
.333... = 3/(10-1) = 3/9 = 1,
.121212... = 12/(100-1) = 12/99 = 4/33, or
.142857142857142857... = 142857/(1000000 - 1 ) =
142857/999999 = 1/7.
Lasse
---
===
Subject: Re: .99999... still=/= 1
>> But lets even look at it from your point of view. 
Lets
assume
>.999... is
>> an irrational number. Can an irrational number equal a
rational
>number? No.
>> 2/2 = 1 <--- This is a rational number.
>> .999... cant be expressed as a fraction. So how do you
make a
>rational number
>> equal an irrational number? You dont.
>Wrong. .999... (if we accept this notation, against which
Bob has ---
>well-founded --- reservations) is, by definition a rational
number,
>just like any repeating decimal expansion. As such, it can
be written
>as a fraction, namely
>.999... = 9/(10-1) = 9/9 = 1.
WRONG!
9/10 = .9 not .999... or 1
Just use you own two eyes and LOOK.
<IGNORE> the rest, you fail to admit your mistake here.
>As Ive said before, this works for ANY repeating 
expansion,
and could
>be taken as a definition of the rational number represented
by such an
>expansion. For example,
>.333... = 3/(10-1) = 3/9 = 1,
>.121212... = 12/(100-1) = 12/99 = 4/33, or
>.142857142857142857... = 142857/(1000000 - 1 ) =
142857/999999 = 1/7.
>Lasse
>---
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===
Subject: Re: .99999... still=/= 1
In sci.math, S. Enterprize Company
<smart1234@aol.com But lets even look at it from your point
of view. Lets assume
>>.999... is
> an irrational number. Can an irrational number equal a
rational
>>number? No.
> 2/2 = 1 <--- This is a rational number.
> .999... cant be expressed as a fraction. So how do you
make a
>>rational number
> equal an irrational number? You dont.
>>Wrong. .999... (if we accept this notation, against which
Bob has ---
>>well-founded --- reservations) is, by definition a rational
number,
>>just like any repeating decimal expansion. As such, it can
be written
>>as a fraction, namely
>>.999... = 9/(10-1) = 9/9 = 1.
> WRONG!
> 9/10 = .9 not .999... or 1
> Just use you own two eyes and LOOK.
Good advice, but whats good for the goose...
> <IGNORE> the rest, you fail to admit your mistake here.
>>As Ive said before, this works for ANY repeating
expansion, and could
>>be taken as a definition of the rational number represented
by such an
>>expansion. For example,
>>.333... = 3/(10-1) = 3/9 = 1,
>>.121212... = 12/(100-1) = 12/99 = 4/33, or
>>.142857142857142857... = 142857/(1000000 - 1 ) =
142857/999999 = 1/7.
>>Lasse
>>---
> Smarts Alt. Physics News Group
>
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> S. Enterprize (Science Journal)
> http://smart1234.s-enterprize.com/
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
> But lets even look at it from your point of view. 
Lets
assume .999...
is
> an irrational number. Can an irrational number equal a
rational number?
No.
.999... is meaningless. It is some undefined nonsense. It is
un-mathematical.
What do you say .999... means. If it means other than
sum (n >= 1) [9*1/10^n) it is just plain nonsense. If it is
sum (n >= 1) [9*1/10^n) then it is a convergent series which
is
trivially provable to be equal to 1. The sequence of finite
partial sums
sum ( 1 <= k <= n) (9*1/10^k) is the sequence { 1 - 1/10^n }
which
clearly converges to 1.
At no time do I used the nonsense scribble .999... which is
not
mathematical at all. It is meaningless. It is not a properly
formed
mathematical expression.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> But lets even look at it from your point of view. 
Lets
assume
.999...
>> an irrational number. Can an irrational number equal a
rational number?
No.
>.999... is meaningless. It is some undefined nonsense. It is
>un-mathematical.
>What do you say .999... means. If it means other than
>sum (n >= 1) [9*1/10^n) it is just plain nonsense. If it is
>sum (n >= 1) [9*1/10^n) then it is a convergent series which
is
>trivially provable to be equal to 1. The sequence of finite
partial sums
>sum ( 1 <= k <= n) (9*1/10^k) is the sequence { 1 - 1/10^n }
which
>clearly converges to 1.
>At no time do I used the nonsense scribble .999... which is
not
>mathematical at all. It is meaningless. It is not a properly
formed
>mathematical expression.
>Bob Kolker
Oh so you fail to admit your mistake again? Then you have to
add all
types
of extras to .999... to make it so called equal to 1. Well I
can do the
same
with i.
I can say i^2 = -1
See its real now.
You described me in all types of degrading ways. But now you
say,
.999... is meaningless. This is incorrect. .999... does have
meaning, and
a
nonstandard analysis has to be used with it that shows that,
.999.... | | 1
^
|
This separating space doesnt allow .999... equal to 1. This
space
is
used in non-standard analysis.
Why not admit your mistake? You didnt even know the math
definition of
an
AND has a repeating number to the right of the decimal.
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Subject: Re: .99999... still=/= 1
> .999... is meaningless. This is incorrect. .999... does
have meaning, and
a
> nonstandard analysis has to be used with it that shows that,
You have not the slightest idea of the axiomatic and set
foundational
basis of nonstandard analsys.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> .999... is meaningless. This is incorrect. .999... does
have meaning,
and a
>> nonstandard analysis has to be used with it that shows
that,
>You have not the slightest idea of the axiomatic and set
foundational
>basis of nonstandard analsys.
Well, to be precise, its surely true that Starshit E. has
not the
slightest idea of *standard* non-standard analysis (style of
Robinson,
etc., etc.), but s/he/it very likely has any number of ideas
(disordered,
unaxiomatizable, delusive, possibly even psychotic: yet still
ideational)
that are unarguably non-standard, and arguably analysis
(though I
for one wouldnt want to have to make a supporting argument
any stronger
than they have the outward form of statements about numbers).
Lee Rudolph
===
Subject: Re: .99999... still=/= 1
>> Bob,
>> Please avoid the name calling. It gives sci.math a bad
reputation.
>> You may not agree with SEs logic (I do not agree with it
either),
>> but it is not OK to call him names because of your
disagreement
>> with him on things mathmatical.
>> Lets keep it friendly, and on topic. Please, no more 
name
calling.
>Enterprise has bone told again and again what a series is,
what
>convergence is and what a limit is. He does not seem to
learn. That
>justifies the term moron, a low level intellect.
>He is also a troll because he repeats his errors again and
again.
>So the appleation is well justified.
>Bob Kolker
I have been using a non-standard approach to this math
problem. And
usually
the way to expansion to new frontiers in math. is to
challenge what is
usually
accepted as correct but may not necessarily be correct.
If it wasnt for Columbus, you people might have fallen off
the edge of
the
earth.
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Subject: Re: .99999... still=/= 1
> I have been using a non-standard approach to this math
problem.
You are full of shit. You dont know what non-standard
arithmetic is
andyou would not know a hyperreal number if it bit your nose.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> They dont have a disagreement about anything
mathematical. It should
>> be clear by now that SE is not arguing in good faith, but
is just
>> trolling.
>Bob Kolker
If you notice they still havent proven,
.999...(an unreal number) = 1 ( a real number).
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Subject: Re: .99999... still=/= 1
> If you notice they still havent proven,
> .999...(an unreal number) = 1 ( a real number).
Meaningless. This is not a mathematical experession.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>>You may not agree with SEs logic (I do not agree with it
either),
>>but it is not OK to call him names because of your
disagreement
>>with him on things mathmatical.
>They dont have a disagreement about anything mathematical.
It should
>be clear by now that SE is not arguing in good faith, but is
just
>trolling.
>-- Richard
I sincerely have been debating this topic the best way I
perceive it.
For
crying out loud, do you actually believe a real number equals
a number that
isnt real?
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Subject: Re: .99999... still=/= 1
In sci.math, S. Enterprize Company
<smart1234@aol.comYou may not agree with SEs logic (I do 
not
agree with it either),
>but it is not OK to call him names because of your
disagreement
>with him on things mathmatical.
>>They dont have a disagreement about anything 
mathematical.
It should
>>be clear by now that SE is not arguing in good faith, but
is just
>>trolling.
>>-- Richard
> I sincerely have been debating this topic the best way I
> perceive it. For crying out loud, do you actually believe a
> real number equals a number that isnt real?
Well, heres a fine mess youve 
gotten us into... :-)
First, what is a real number anyway? Despite our many valiant
attempts to beat you over the head with the club of reason,
its clear youre resisting -- and it may very 
well because
the club of reason is actually a thin, wispy, non-existent fog
of a metaphor.
In short, real numbers are about as real as pink elephants.
They do not exist. Oh, sure, one can blather on about
measuring 1.25 inches or 3 1/2 cups of coffee or an
air pressure of 101 kPa during a nice sunny day -- but
those are physical measurements, not numbers per se.
Try catching a number in a butterßy net -- or any other kind
of net, for that matter. One cant do it; the best one might
get is a pair of swallowtailed yellowbeaks. Or something.
(Or was that yellowtailed swallowbeaks? Does one count an
egg as half of a bird, or a third? Well, never mind that;
ornithology was never my strongest subject.)
The best we can do is lay a groundwork of phantom assumptions,
such as Peanos Axioms and Dedekind cuts, but here is where
the problems start.
First, what does Ô= actually mean? For most of 
us, we simply
accept that limits make sense -- but there are far deeper
issues
here; the partial sums, for example, of the series
1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ...
are all rational, but the result is pi^2/6. Hey, waitaminnit,
that
aint in Q! What happened?
Good question.
Or one can try
1 + 1/2 + 1/3 + ...
All your partial sums are belong to Q, but the series
diverges;
this one has no limit, although one might make a case that
1 + 1/2 + 1/3 + ... = +oo.
but remember that +oo is not a real number.
Of course, the usage of the ellipsis (Ô...) is 
a bit tricky.
Take,
for example, this definition:
e = 2.718281828...
Now before one asks uh, is that rational? -- turns out this is
Eulers Number, which equates to the series
e = 1/0! + 1/1! + 1/2! + 1/3! + ...
and no, it is not rational (although all of the partial sums
are!);
a more honest ellipsis might be
e = 2.718281828459...
So we fall back on convention. You may have heard the
expression
what I say three times is true. (Its not quite true,
admittedly,
but never mind that.) So one might write
1/7 = 0.142857142857142857...
with the understanding that the last 6 digits repeat
endlessly.
(It turns out they do; a long-division proof is left to the
interested reader, or one can simply multiply 142857*7 =
999999 = 10^6 - 1
and/or note that 1/(1-x) = 1+x+x^2+x^3+... for any x between
-1 and +1.)
Or
1/11 = 0.090909...
or even
1/9 = 0.111...
So what does 0.999... really mean? Endless 9s, in this 
case.
A proper treatment of such a number might involve summation
of the
infinite series sum(i=1,+oo) (9 * 10^(-i)). If numbers are
an abstract_1 concept, this sum is an abstract_3 (simple
expressions
are abstract_2, in some sense). The partial sums of this
series
are of course sum(i=1,n) (9 * 10^(-i)); a little work with a
competent
algebra book or symbolic calculator show that a partial sum
sum(i=1,n) (9 * 10^(-i)) = 1 - 10^n = 0.999...9. Theres 
that
ellipsis again, this time in a different context; the 9s do
*not*
repeat endlessly in this case (unless one is Gary Denke, but
one
doesnt really want to know the details there). In this 
case,
there are n 9s total.
Of course, Ive pointed this out before to you, and so far
its
yet to sink in, unlike the iron spike in the case of Phinehas
Gage.
(ObOuch: Ouch.)
Unfortunately, to most calculators, all numbers are rationals
--
furthermore, to modern computers, all numbers are multiples
of a
power of 2, with some fudging on the arithmetic side to make
it
look otherwise. (The standard representation is r = 2^e * M,
where
M is up to about 53 bits and e can range from -2048 to +2047
or thereabouts.) Therefore, a partial sum may yield anomalies
if not done carefully.
For example, to a computer,
1/3 = 0x3fd5555555555555 .
This is readily verified using C and a hack such as the
following:
#include <stdio.h>
int main()
{
union { unsigned char c[sizeof(double)]; double d; } u;
int i;
u.d = 1.0/3.0;
printf(1/3 = 0x);
for(i=0;i<sizeof(double);i++) printf(%02x, u.c[i]);
printf(n);
return 0;
}
(Depending on machine the hex values will print either
forwards or backwards. If its a worry, use something like
#include <endian.h>
and
#if BYTE_ORDER == BIG_ENDIAN
in appropriate spots.)
The first bit is the sign bit; the next 11 is a 
modified
exponent,
and the rest are the mantissa, with a hidden-1. For example,
1.0 = 0x3ff0000000000000
2.0 = 0x4000000000000000
3.0 = 0x4008000000000000
4.0 = 0x4010000000000000
5.0 = 0x4014000000000000
In a very real sense, to a computer
1/3 = 6004799503160661/18014398509481984
= .333333333333333314829616256247390992939472198486328125 .
(Note that 15555555555555(16) = 6004799503160661(10).)
One can use an infinite-precision calculator such as bc to
derive these results:
$ bc <<!
ibase=16
15555555555555
!
6004799503160661
$
Now multiply by 3, and one gets
18014398509481983/18014398509481984
= .999999999999999944488848768742172978818416595458984375 .
Is 1 =
.999999999999999944488848768742172978818416595458984375 ?
Certainly not. But computers arent as bright as one might
think. :-)
Ill also mention a little bug in Microsofts 
calculator.
Microsoft
(or an engineer therein) was apparently somewhat naive, and
one got
3.11 - 3.10 = 0.00
in their calculator in Win95.
It turns out that
3.11 = 0x4008e147ae147ae1
3.10 = 0x4008cccccccccccd
diff = 0x3f847ae147ae1400
0.01 = 0x3f847ae147ae147b
so, to a computer, (3.11 - 3.10) is just a smidge less than
0.01.
With rounding this ordinarily isnt a problem, but if one
forgets
to round -- well, one very well might get 0.00, with an
invisible
9 for ones trouble.
The bug finally got fixed some time ago, but it 
took awhile.
Another illustration is 9007199254740992+1 = 9007199254740992,
if one uses double. Not a lot one can do about it without
switching away from double (e.g., one might use long long
instead), as the bit gets lost; 2^53 = 9007199254740992.
It gets worse.
Remember that the exact representation of 1/3 using double
precision is
1/3 = .333333333333333314829616256247390992939472198486328125
A naive C program using %.60f (which is waaay too much
precision
and not enough accuracy!) gives
1/3 = .333333333333333314829616256247390992939472198486328125
which actually turns out to be the right answer -- if one can
call this a right answer.
However, a naive progressive digitation algorithm prints out
1/3 = .333333333333333303727386009995825588703155517578125
Yipes.
The conclusion: dont depend on partial sums generated by
computer,
unless you know exactly what its doing -- and what 
youre
doing.
[.sigsnip]
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>In sci.math, S. Enterprize Company
><smart1234@aol.com>You may not agree with SEs logic (I do
not agree with it either),
>>but it is not OK to call him names because of your
disagreement
>>with him on things mathmatical.
>They dont have a disagreement about anything mathematical.
It should
>be clear by now that SE is not arguing in good faith, but is
just
>trolling.
>-- Richard
>> I sincerely have been debating this topic the best way I
>> perceive it. For crying out loud, do you actually believe a
>> real number equals a number that isnt real?
>Well, heres a fine mess youve 
gotten us into... :-)
>First, what is a real number anyway? Despite our many valiant
>attempts to beat you over the head with the club of reason,
>its clear youre resisting -- and it may 
very well because
>the club of reason is actually a thin, wispy, non-existent
fog
>of a metaphor.
>In short, real numbers are about as real as pink elephants.
>They do not exist. Oh, sure, one can blather on about
>measuring 1.25 inches or 3 1/2 cups of coffee or an
>air pressure of 101 kPa during a nice sunny day -- but
>those are physical measurements, not numbers per se.
>Try catching a number in a butterßy net -- or any other kind
>of net, for that matter. One cant do it; the best one 
might
>get is a pair of swallowtailed yellowbeaks. Or something.
>(Or was that yellowtailed swallowbeaks? Does one count an
>egg as half of a bird, or a third? Well, never mind that;
>ornithology was never my strongest subject.)
>The best we can do is lay a groundwork of phantom
assumptions,
>such as Peanos Axioms and Dedekind cuts, but here is where
>the problems start.
>First, what does Ô= actually mean? For most of 
us, we simply
>accept that limits make sense -- but there are far deeper
issues
>here; the partial sums, for example, of the series
>1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ...
>are all rational, but the result is pi^2/6. Hey,
waitaminnit, that
>aint in Q! What happened?
>Good question.
>Or one can try
>1 + 1/2 + 1/3 + ...
>All your partial sums are belong to Q, but the series
diverges;
>this one has no limit, although one might make a case that
>1 + 1/2 + 1/3 + ... = +oo.
>but remember that +oo is not a real number.
>Of course, the usage of the ellipsis (Ô...) is 
a bit
tricky. Take,
>for example, this definition:
>e = 2.718281828...
>Now before one asks uh, is that rational? -- turns out this
is
>Eulers Number, which equates to the series
>e = 1/0! + 1/1! + 1/2! + 1/3! + ...
>and no, it is not rational (although all of the partial sums
are!);
>a more honest ellipsis might be
>e = 2.718281828459...
>So we fall back on convention. You may have heard the
expression
>what I say three times is true. (Its not quite true,
admittedly,
>but never mind that.) So one might write
>1/7 = 0.142857142857142857...
>with the understanding that the last 6 digits repeat
endlessly.
>(It turns out they do; a long-division proof is left to the
>interested reader, or one can simply multiply 142857*7 =
999999 = 10^6 - 1
>and/or note that 1/(1-x) = 1+x+x^2+x^3+... for any x between
-1 and +1.)
>1/11 = 0.090909...
>or even
>1/9 = 0.111...
>So what does 0.999... really mean? Endless 9s, in this 
case.
>A proper treatment of such a number might involve summation
of the
>infinite series sum(i=1,+oo) (9 * 10^(-i)). If numbers are
>an abstract_1 concept, this sum is an abstract_3 (simple
expressions
>are abstract_2, in some sense). The partial sums of this
series
>are of course sum(i=1,n) (9 * 10^(-i)); a little work with a
competent
>algebra book or symbolic calculator show that a partial sum
>sum(i=1,n) (9 * 10^(-i)) = 1 - 10^n = 0.999...9. Theres 
that
>ellipsis again, this time in a different context; the 9s 
do
*not*
>repeat endlessly in this case (unless one is Gary Denke, but
one
>doesnt really want to know the details there). In this 
case,
>there are n 9s total.
>Of course, Ive pointed this out before to you, and so far
its
>yet to sink in, unlike the iron spike in the case of
Phinehas Gage.
>(ObOuch: Ouch.)
>Unfortunately, to most calculators, all numbers are
rationals --
>furthermore, to modern computers, all numbers are multiples
of a
>power of 2, with some fudging on the arithmetic side to make
it
>look otherwise. (The standard representation is r = 2^e * M,
where
>M is up to about 53 bits and e can range from -2048 to +2047
>or thereabouts.) Therefore, a partial sum may yield anomalies
>if not done carefully.
>For example, to a computer,
>1/3 = 0x3fd5555555555555 .
>This is readily verified using C and a hack such as the
following:
>#include <stdio.hint main()
> union { unsigned char c[sizeof(double)]; double d; } u;
> int i;
> u.d = 1.0/3.0;
> printf(1/3 = 0x);
> for(i=0;i<sizeof(double);i++) printf(%02x, u.c[i]);
> printf(n);
> return 0;
>(Depending on machine the hex values will print either
>forwards or backwards. If its a worry, use something like
>#include <endian.hand
>#if BYTE_ORDER == BIG_ENDIAN
>in appropriate spots.)
>The first bit is the sign bit; the next 11 is a 
modified
exponent,
>and the rest are the mantissa, with a hidden-1. For example,
>1.0 = 0x3ff0000000000000
>2.0 = 0x4000000000000000
>3.0 = 0x4008000000000000
>4.0 = 0x4010000000000000
>5.0 = 0x4014000000000000
>In a very real sense, to a computer
>1/3 = 6004799503160661/18014398509481984
>= .333333333333333314829616256247390992939472198486328125 .
>(Note that 15555555555555(16) = 6004799503160661(10).)
>One can use an infinite-precision calculator such as bc to
>derive these results:
>$ bc <<!
>ibase=16
>15555555555555
>6004799503160661
>Now multiply by 3, and one gets
>18014398509481983/18014398509481984
>= .999999999999999944488848768742172978818416595458984375 .
>Is 1 =
.999999999999999944488848768742172978818416595458984375 ?
>Certainly not. But computers arent as bright as one might
think. :-)
>Ill also mention a little bug in Microsofts 
calculator.
Microsoft
>(or an engineer therein) was apparently somewhat naive, and
one got
>3.11 - 3.10 = 0.00
>in their calculator in Win95.
>It turns out that
>3.11 = 0x4008e147ae147ae1
>3.10 = 0x4008cccccccccccd
>diff = 0x3f847ae147ae1400
>0.01 = 0x3f847ae147ae147b
>so, to a computer, (3.11 - 3.10) is just a smidge less than
0.01.
>With rounding this ordinarily isnt a problem, but if one
forgets
>to round -- well, one very well might get 0.00, with an
invisible
>9 for ones trouble.
>The bug finally got fixed some time ago, but it 
took awhile.
>Another illustration is 9007199254740992+1 =
9007199254740992,
>if one uses double. Not a lot one can do about it without
>switching away from double (e.g., one might use long long
>instead), as the bit gets lost; 2^53 = 9007199254740992.
>It gets worse.
>Remember that the exact representation of 1/3 using double
>precision is
>1/3 = .333333333333333314829616256247390992939472198486328125
>A naive C program using %.60f (which is waaay too much
precision
>and not enough accuracy!) gives
>1/3 = .333333333333333314829616256247390992939472198486328125
>which actually turns out to be the right answer -- if one can
>call this a right answer.
>However, a naive progressive digitation algorithm prints out
>1/3 = .333333333333333303727386009995825588703155517578125
>Yipes.
>The conclusion: dont depend on partial sums generated by
computer,
>unless you know exactly what its doing -- and what 
youre
doing.
>[.sigsnip]
>--
>#191, ewill3@earthlink.net
>Its still legal to go .sigl
LEARN MATH.
http://mathworld.wolfram.com/HyperrealNumber.html
.999... < 1
.999... =/= 1
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
> LEARN MATH.
His math is sound. Yours is non-existent.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
In sci.math, robert j. kolker
<nowhere@nowhere.com>
<BQdxd.681925$mD.547964@attbi_s02>:
>> LEARN MATH.
> His math is sound. Yours is non-existent.
the non-standard analysis arena, an area which is not all that
familiar to me except as a crudely expressed d-math, which I
can occasionally use (how correctly, Ive no idea!) to 
expound
on various concepts.
Its clear that his definition of 
Ô= is a bit different from
the rest of ours, and limit theory has some problems.
For example, take .999..., the much-balleyhooed expression.
Express it as the series:
S_1 = .9
S_2 = .9 + .09
S_3 = .9 + .09 + .009
...
S_n = sum(i=1,n) (9 * 10^(-i)) = 1 - 10^(-n)
(easily proved by induction, if one cares to bother).
More or less standard stuff, up to this point.
Under normal circumstances one can play the N-epsilon [*] game
and get the following proof (or outline thereof).
Oh, you have an epsilon > 0 for me? Fine, Ill take
N = ceil(log10(1/epsilon)). I can now prove that,
for any n > N, S_n > 1 - epsilon, but less than 1.
Since S_n = 1 - 10^(-n), if n > N, then
10^(-n) < 10^(-ceil(log10(1/epsilon)))
<= 10^(-log10(1/epsilon)) <= epsilon.
and then the jump:
Hence, lim(n->+oo) S_n = 1. QED.
Now enter hypperreals. Set epsilon = d, where d is
a number greater than 0 but less than all 1/n, n > 0, n in J.
This proof goes out the window, as 1/d is a meaningless
expression (though one could generate another class of
numbers, maybe called quasi-infinities, which would be
greater than any integer N but less than aleph_0, or something
equally strange; the main intent is to be dual to the
hyperreals).
One could claim Ôd is a ridiculous concept (and 
it is to some
extent as lim(n->+oo) (1/n) = 0 anyway, in standard analysis),
but it does lead to some interesting questions as to how to
get around this obstacle without simply claiming well, its
obviously nonsense or well, weve always done it that way.
Its a bit like Lobachevskian geometry in that respect -- 
not
a contradiction, but a new realm of number.
Not a horribly useful one, to be sure -- 21 or so digits of pi
or e are enough to define the Earths orbit (1.5 
* 10^11 m) to
the width of an atom (2 * 10^-10 m); the rest is gravy -- but
interesting to some, and useful for testing microprocessors.
This is not to say S. Enterprizes arguments are any good;
theyre extremely sloppy, in fact.
> Bob Kolker
[*] there are four variants of this game (each with three or
nine
subvariants), depending on what one is proving.
delta-epsilon: lim(x->a) f(x) = b
lim(x->a+) f(x) = b
lim(x->a-) f(x) = b
N-epsilon: lim(x->oo) f(x) = b
lim(x->+oo) f(x) = b
lim(x->-oo) f(x) = b
delta-M: lim(x->a) f(x) = oo (or +oo or -oo)
lim(x->a+) f(x) = oo (or +oo or -oo)
lim(x->a-) f(x) = oo (or +oo or -oo)
N-M: lim(x->oo) f(x) = oo (or +oo or -oo)
lim(x->+oo) f(x) = oo (or +oo or -oo)
lim(x->-oo) f(x) = oo (or +oo or -oo)
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
> Under normal circumstances one can play the N-epsilon [*]
game
> and get the following proof (or outline thereof).
> Oh, you have an epsilon > 0 for me? Fine, Ill take
> N = ceil(log10(1/epsilon)). I can now prove that,
> for any n > N, S_n > 1 - epsilon, but less than 1.
> Since S_n = 1 - 10^(-n), if n > N, then
> 10^(-n) < 10^(-ceil(log10(1/epsilon)))
> <= 10^(-log10(1/epsilon)) <= epsilon.
> and then the jump:
> Hence, lim(n->+oo) S_n = 1. QED.
> Now enter hypperreals. Set epsilon = d, where d is
> a number greater than 0 but less than all 1/n, n > 0, n in
J.
> This proof goes out the window, as 1/d is a meaningless
> expression (though one could generate another class of
> numbers, maybe called quasi-infinities, which would be
> greater than any integer N but less than aleph_0, or
something
> equally strange; the main intent is to be dual to the
hyperreals).
Actually, the hyperreals *R form a field. Yes, 1/d is
infinitely large
if d is an infinitesimal. Also, it should be noted that you
can use the
standard definition for the limit of a sequence:
Definition. Let { a_k } be a sequence and L in R.
Then we say that lim_{k->oo} a_k = L if, for every epsilon >
0,
there
exists N > 0 such that | a_k - L | < epsilon for every k > N.
The only difference is in the interpretation of that the
terms mean. In
nonstandard analysis (NSA), when we say for every epsilon >
0, we mean
for every positive epsilon in *R, which means epsilon is
allowed to be
an infinitesimal, for example. And when we say there exists N
> 0 we
mean that N is allowed to be infinitely large.
Its a theorem of NSA that the following two statements are
equivalent:
(1) lim_{k->oo} a_k = L (as defined above), and
(2) The difference | a_k - L | is an infinitesimal whenever k
is
infinitely large.
Oh, and by the way, its also true in NSA that sum_{n in *N,
n >0} 9/10^n
= 1. Its a geometric series. Thus we can say .999... = 1,
even in the
hyperreals.
And no, its not possible to confine the sum to 
just the finite
digit
positions, because the summation can only be carried out over
a set, and
the finite naturals are not a set according to the internal
set theory of
NSA. When we say something is infinitely large in NSA, 
thats
only the
external view. Within the model, all the members of *R are
finite.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: .99999... still=/= 1
In sci.math, Dave Seaman
<dseaman@no.such.host>
<cq53pv$glb$1@mailhub227.itcs.purdue.edu>:
>> Under normal circumstances one can play the N-epsilon [*]
game
>> and get the following proof (or outline thereof).
>> Oh, you have an epsilon > 0 for me? Fine, Ill take
>> N = ceil(log10(1/epsilon)). I can now prove that,
>> for any n > N, S_n > 1 - epsilon, but less than 1.
>> Since S_n = 1 - 10^(-n), if n > N, then
>> 10^(-n) < 10^(-ceil(log10(1/epsilon)))
>> <= 10^(-log10(1/epsilon)) <= epsilon.
>> and then the jump:
>> Hence, lim(n->+oo) S_n = 1. QED.
>> Now enter hypperreals. Set epsilon = d, where d is
>> a number greater than 0 but less than all 1/n, n > 0, n in
J.
>> This proof goes out the window, as 1/d is a meaningless
>> expression (though one could generate another class of
>> numbers, maybe called quasi-infinities, which would be
>> greater than any integer N but less than aleph_0, or
something
>> equally strange; the main intent is to be dual to the
hyperreals).
> Actually, the hyperreals *R form a field. Yes, 1/d is
infinitely large
> if d is an infinitesimal. Also, it should be noted that you
can use the
> standard definition for the limit of a sequence:
> Definition. Let { a_k } be a sequence and L in R.
> Then we say that lim_{k->oo} a_k = L if, for every epsilon
> 0,
there
> exists N > 0 such that | a_k - L | < epsilon for every k >
N.
> The only difference is in the interpretation of that the
terms mean. In
> nonstandard analysis (NSA), when we say for every epsilon >
0, we
mean
> for every positive epsilon in *R, which means epsilon is
allowed to
be
> an infinitesimal, for example. And when we say there exists
N > 0 we
> mean that N is allowed to be infinitely large.
> Its a theorem of NSA that the following two statements 
are
equivalent:
> (1) lim_{k->oo} a_k = L (as defined above), and
> (2) The difference | a_k - L | is an infinitesimal whenever
k is
> infinitely large.
> Oh, and by the way, its also true in NSA that sum_{n in
*N, n >0} 9/10^n
>= 1. Its a geometric series. Thus we can say .999... = 1,
even in the
> hyperreals.
> And no, its not possible to confine the sum 
to just the
finite digit
> positions, because the summation can only be carried out
over a set, and
> the finite naturals are not a set according to the internal
set theory of
> NSA. When we say something is infinitely large in NSA,
thats only
the
> external view. Within the model, all the members of *R are
finite.
Interesting. So even in the hyperreals, S. Enterprize is
simply wrong. :-)
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
> Interesting. So even in the hyperreals, S. Enterprize is
simply wrong.
:-)
I told you.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
> This is not to say S. Enterprizes arguments are any good;
> theyre extremely sloppy, in fact.
His arguments are not even wrong. They are non-existent.
Stringing words
together doth not an argument make.
Bob Kolker
===
Subject: Re: .99999... still=/= 1
> So from another point of view,
>
> .999... =/= 1
>>The locution .999.... is a meaningless string of symbols.
It is not a
>>properly expressed mathematical entity. Whereas:
>>SUM [n >=1] (9/10^n)
>>is a proper mathematical expression (it is equal to 1)
>>So the sentence .999... =/= 1 is meaningless.
>>You are a troll and a moron.
>>Bob Kolker
>Bob,
>Please avoid the name calling. It gives sci.math a bad
reputation.
>You may not agree with SEs logic (I do not agree with it
either),
>but it is not OK to call him names because of your
disagreement
>with him on things mathmatical.
>Lets keep it friendly, and on topic. Please, no more name
calling.
about the math part.
>- MO
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: matrix inverse of the sum of two inverse matrices
posting-account=ImIFzg0AAACL32LvFbginMab44pG8QVR
limint N going to infinity.
-The most general problem I need to solve is to find the
inverse of the
sum of the inverses of these
2 matrices. One of them is diagonal, but they DO NOT commute.
I guess
there is no general solution
to this problem.
-In a slightly more restricted problem, one of the matrices
is diagonal
in Ôreal space, and the other
is diagonal in frequency space (i.e., depends only on |i-j|,
where i,j
are indices over real space.
-In an even more restricted case (this is the one where I
really hope
to get some help), all the elements
of one of the two matrices are multiplied by a very small
number, i.e.,
Inv(C) = a_i*delta_ij + epsilon*b_ij
I can assume b_ij=f(|i-j|), so this matrix is diagonal in
frequency
space, epsilon is very small and I would like to know C.
There must be
some kind of power expansion one can do...
===
Subject: Re: matrix inverse of the sum of two inverse matrices
posting-account=ImIFzg0AAACL32LvFbginMab44pG8QVR
Sorry, I found theres a lot of info on this general issue 
on
the list
already. I guess Id still like to know if there is a 
specific
simplification of the problem that arises if the two matrices
are such
that one is diagonal in real space and the other in
frequency space...
===
Subject: Re: Question about Presidents Social Security plan
>by the amount of $2.7 trillion in 75 years.
>Bush administration has a plan. The plan is
>to privatize some parts of social security.
>This will cost $2 trillion to set up.
>It is not guaranteed to fix things, but
>is only one part of an integrated plan.
>So my question is, wouldnt it make more
>sense to just GIVE that $2 trillion to
>social security, which is guaranteed
>to fix things by exactly $2 trillion, just
>leaving a small $0.7 trillion shortfall after
>75 years?
You mean other then this train wreck if the media ever made
this
information Public?! I mean name me just ONE Democrat that
would still
be left in office if the public found out that they had been
lied to
and that the Social Security funds were being stolen by the
Democrats
for the last 60+ years?
Here are the REAL facts
Social Security.89s vaunted trust fund doesn.89t exist; taxes
paid into
Social Security are merely being handed over as benefits to
other
people.
On May 2nd, President Bush announced the formation of a
presidential
commission to deal with the Social Security crisis. The last
such
commission, in 1983, had as its chairman Alan Greenspan. It
recommended enormous increases in taxes, and Congress
dutifully
complied.Government spokesmen immediately patted themselves
on their
backs for having saved Social security. We are likely in for
more of
the same. The new commission.89s name is the Commission to
Strengthen
Social Security. That can only mean more taxes.
The Bush administration, though, should not be trying to save
Social
Security. For decades americans have been deceived by this
program,
believing that their Social Security taxes are pouring into a
fund set
aside expressly for the purpose of old age insurance for the
support of the elderly. This is a colossal lie and it is
nothing short
of scandalous. The truth is, the vaunted Social Security
trust fund
does not exist. Never has, never will.
Haven.89t we been told for decades that the government
scrupulously,
almost religiously, maintains the Social Security trust fund?
For the
truth, we have to go back to 1935, shortly after the creation
of
Social Security. A man named George P. Davis contested the
program in
court, claiming that taxation to support it was
unconstitutional,
trampled on states.89 rights, and imposed an unjust monetary
burden on
firms in which he had invested. A federal court rejected his
claims,
but he won a favorable ruling from the First Circuit Court of
Appeals in Boston. Had that Appeals Court decision remained
in force,
the entire Social Security Act would have been voided at the
outset.
As expected, however, the federal government.89s Commissioner of
Internal Revenue, Guy T. Helvering, immediately appealed the
matter to
the Supreme Court. Lawyers for Mr. Davis contended that Social
Security taxes were collected for a particular purpose (for
unemployed and older Americans) and not for the
constitutionally
acceptable purpose of acquiring revenue for the general
welfare of the
nation as a whole. They also contended that the tax was not a
constitutionally allowable excise tax, a type of taxation
defined
of use or consumption, not a tax on wages.
In Helvering v. Davis, the Supreme Court.89s decision made
reference to
a related case wherein the court had tortuously maintained
that the
tax was indeed an excise tax and therefore legitimate. But
the Court
also agreed with the government.89s lawyers who had openly
stated
that social security taxes were not collected for a
particular purpose
but are paid into the Treasury as internal revenue
collections,
available for the general support of the government.
That contention, forming the government.89s major argument
against the
claims by Mr. Davis, received official endorsement when the
Supreme
Court ruled in Helvering that the proceeds of the
unemployment and
old-age taxes are to be paid into the Treasury like internal-
revenue taxes generally, and are not earmarked in any way.
(Emphasis
added.) This decision has never been overturned. All talk
about the
social security trust fund containing social security tax
revenue is
pure, unadulterated blather.
In 1975, former Secretary of Commerce and former Director of
the
Budget Maurice Stans hit the nail on the head when he stated:
Social
Security payments rest upon the general credit of the
Government of
the United States, upon its taxing power, and not upon any
accumulations in a trust fund. In 1976, then-Secretary of the
contributors to the system have not been building a fund at
all. The
taxes they are paying into Social Security are merely being
handed over as benefits to other people. In other words, 
Social
Security is a huge Ponzi scheme.
So, what the Bush administration is trying to do through the
new
Commission to Strengthen Social Security is to buttress a
lie. The
commission will surely attract more attention as it continues
recommending placebos to treat a disease that will eventually
prove
fatal if not properly addressed. What should be done is
really quite
simple: Phase Social Security out. Allow freedom of choice
and watch
how few young Americans will stick with Social security.
Programs
doing precisely this have already been implemented in Chile
and
elsewhere with stunningly beneficial results. Once the people
of Chile
were given the choice in 1981, they opted out, put their
money into
private programs, saw those funds spark an economic surge
that became
the envy of all of Latin America, and destroyed much of their
nation.89s
harmful government paternalism.
But America.89s leaders continue to insist that Social Security
is a
success and that only careful management of its trust funds is
needed to insure its viability. In his 1975 book Social
Security: The
Fraud in Your Future, author Warren Shore concluded that
claims about
the existence of trust funds are made because they help
foster the
public notion that Social Security is like insurance with its
premium
pools [available] to support promises made. But the public
has been
misled. Sad to say, the current activity in Washington shows
no signs
of addressing this fraud.
Go to http://www.ssa.gov and then do a search on Helvering v
Davis
and be amazed at the over 1000 different links that this case
is
brought up.
NOW do you see why the Democrats are so terrified about this 
SS
reform? They would be hanged and 
drawnnquartered if the
seniors ever
found out that they were stealing from them. The Democrats
would never
be elected to dog warden ever again if this was made public,
and part
of the reform is to MAKE THIS PUBLIC! You Democrats are
staring down
the barrel of a loaded gun in your own hands, and for some
reason your
pulling the trigger.....
===
Subject: Re: Question about Presidents Social Security plan
>>by the amount of $2.7 trillion in 75 years.
>>Bush administration has a plan. The plan is
>>to privatize some parts of social security.
>>This will cost $2 trillion to set up.
>>It is not guaranteed to fix things, but
>>is only one part of an integrated plan.
>>So my question is, wouldnt it make more
>>sense to just GIVE that $2 trillion to
>>social security, which is guaranteed
>>to fix things by exactly $2 trillion, just
>>leaving a small $0.7 trillion shortfall after
>>75 years?
Of course that would make sense. But you have to understand
that
this will not help Bush and his rich friends and so it will
not
be supported by the Republicans.
> You mean other then this train wreck if the media ever made
this
> information Public?!
I think that the actuarial problems of the current system
have been
well publicized. But there will always be a few stupeeedos who
actually think they have knowledge that is special.
> I mean name me just ONE Democrat that would still
> be left in office if the public found out that they had been
lied to
> and that the Social Security funds were being stolen by the
Democrats
> for the last 60+ years?
(snicker) Your indictment of Johnson is correct, but most of
this
thievery has been by Repugnicans for the last 25 years. They
are the
big deficit creators, not the Dems.
> Here are the REAL facts
> Social Security.92s vaunted trust fund doesn.92t exist;
taxes paid
into
> Social Security are merely being handed over as benefits to
other
> people.
There are probably 10 people in this country that do not
already KNOW
this. You seem to be one of those who have only recently
become
aware of this fact.
> On May 2nd, President Bush announced the formation of a
presidential
> commission to deal with the Social Security crisis. The
last such
> commission, in 1983, had as its chairman Alan Greenspan. It
> recommended enormous increases in taxes, and Congress
dutifully
> complied.
Yes. That would be the Republican Greenspan sucking up to the
Republican Reagan.
http://GreaterVoice.org/econ/glossary/The_Great_Ray_Gun_Rip_
Off.php
ACCELERATES THE SCHEDULE OF TAX HIKES IN SOCIAL SECURITY
ORIGINALLY PASSED
IN 1977. THE SCHEDULE IS TO BE COMPLETED BY 1990 INSTEAD OF
THE YEAR
2030.** .
>Government spokesmen immediately patted themselves on their
> backs for having saved Social security. We are likely in
for more of
> the same. The new commission.92s name is the Commission to
Strengthen
> Social Security. That can only mean more taxes.
Sure! We have a lying Republican in the White House and a
compliant
Greenspan at the Fed. What else would you expect accept some
piece
of crap that will send billions of dollars to the people who
financed
Georgies campaign.
> The Bush administration, though, should not be trying to
save Social
> Security. For decades americans have been deceived by this
program,
> believing that their Social Security taxes are pouring into
a fund set
> aside expressly for the purpose of old age insurance for the
> support of the elderly. This is a colossal lie and it is
nothing short
> of scandalous. The truth is, the vaunted Social Security
trust fund
> does not exist. Never has, never will.
And most people of even minimal intelligence realize this.
> Haven.92t we been told for decades that the government
scrupulously,
> almost religiously, maintains the Social Security trust
fund? For the
> truth, we have to go back to 1935, shortly after the
creation of
> Social Security. A man named George P. Davis contested the
program in
> court, claiming that taxation to support it was
unconstitutional,
> trampled on states.92 rights, and imposed an unjust
monetary burden on
> firms in which he had invested. A federal court rejected his
claims,
> but he won a favorable ruling from the First Circuit Court
of
> Appeals in Boston. Had that Appeals Court decision remained
in force,
> the entire Social Security Act would have been voided at
the outset.
> As expected, however, the federal government.92s
Commissioner of
> Internal Revenue, Guy T. Helvering, immediately appealed
the matter to
> the Supreme Court. Lawyers for Mr. Davis contended that
Social
> Security taxes were collected for a particular purpose (for
> unemployed and older Americans) and not for the
constitutionally
> acceptable purpose of acquiring revenue for the general
welfare of the
> nation as a whole. They also contended that the tax was not
a
> constitutionally allowable excise tax, a type of taxation
defined
> of use or consumption, not a tax on wages.
> In Helvering v. Davis, the Supreme Court.92s decision made
reference to
> a related case wherein the court had tortuously maintained
that the
> tax was indeed an excise tax and therefore legitimate. But
the Court
> also agreed with the government.92s lawyers who had openly
stated
> that social security taxes were not collected for a
particular purpose
> but are paid into the Treasury as internal revenue
collections,
> available for the general support of the government.
> That contention, forming the government.92s major argument
against the
> claims by Mr. Davis, received official endorsement when the
Supreme
> Court ruled in Helvering that the proceeds of the
unemployment and
> old-age taxes are to be paid into the Treasury like
internal-
> revenue taxes generally, and are not earmarked in any way.
(Emphasis
> added.) This decision has never been overturned. All talk
about the
> social security trust fund containing social security tax
revenue is
> pure, unadulterated blather.
Yep.
<<<<<lots more repititious horsecrap deleted (belabors the
issue)> NOW do you see why the Democrats are so terrified
about this SS
> reform? They would be hanged and 
drawnnquartered if the
seniors ever
> found out that they were stealing from them.
The government has been stealing from the working people of
this
country ever since the 1980 hike in FICA tax by the Reagan
thieves.
> The Democrats would never
> be elected to dog warden ever again if this was made
public, and part
> of the reform is to MAKE THIS PUBLIC! You Democrats are
staring down
> the barrel of a loaded gun in your own hands, and for some
reason your
> pulling the trigger.....
There is little most of us want more than to change the
funding side
of SS without touching the benefits side.
--
I know no safe depository of the ultimate powers of society
but
the people themselves; and if we think them not enlightened
enough
to exercise their control with a wholesome discretion, the
remedy
is not to take it from them, but to inform their discretion by
education. - Thomas Jefferson. http://GreaterVoice.org
===
Subject: Re: Question about Presidents Social Security plan
>Yes, except that the government doesnt have an account 
with
$2
>trillion in it. So it would have to sell bonds in that
amount. The
>simpler method is to directly credit the SS trust fund with
whatever
>it needs to cover the shortfall, if and when it occurs.
Except of course the government would have to print money to
cover the
shortfall which would quickly lead to hyperinßation. You can
issue
bonds but there has to be some expectation that you can pay
off those
bonds when they come due, you see, and with 95% or whatever
percent of
the federal budget dedicated to paying pension checks that
doesnt
leave much room to pay any other bills.
Ideally you take care of this by increasing the contributions
slightly
well in advance and restraining the growth of the program to
inßation. If we had done this back in, oh, 1983 the system
would be
solvent. Unfortunately we only did *half* of this: We raised
FICA
taxes and declared we had a surplus. But Congress never
reined in
entitlements and then spent the FICA surplus on other things.
Bottom line is the surplus disappears from the books around
2011 and
from that point on the system runs in the red. And it only
gets
*redder* the further out you go, with too few workers paying
into the
system to cover benefits going out.
--
Iraq was a brilliant campaign fought with minimal
casualties, 11 September was a humiliating failure
by government to fulfill its primary role of
national defence. But Democrats who complained that
Bush was too slow to act on doubtful intelligence
re 9/11 now profess to be horrified that he was too
quick to act on doubtful intelligence re Iraq. This
is not a serious party.
===
Subject: Re: Question about Presidents Social Security plan
OrionCA says...
>Iraq was a brilliant campaign fought with minimal
>casualties, 11 September was a humiliating failure
>by government to fulfill its primary role of
>national defence. But Democrats who complained that
>Bush was too slow to act on doubtful intelligence
>re 9/11 now profess to be horrified that he was too
>quick to act on doubtful intelligence re Iraq. This
>is not a serious party.
Whoever said that is not a serious commentator. The
comparison is stupid.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Question about Presidents Social Security plan
>OrionCA says...
>>Iraq was a brilliant campaign fought with minimal
>>casualties, 11 September was a humiliating failure
>>by government to fulfill its primary role of
>>national defence. But Democrats who complained that
>>Bush was too slow to act on doubtful intelligence
>>re 9/11 now profess to be horrified that he was too
>>quick to act on doubtful intelligence re Iraq. This
>>is not a serious party.
>Whoever said that is not a serious commentator. The
>comparison is stupid.
Brilliant counterargument there. Mark Stein, btw, is a
respected
political commentator both in the UK and the United States.
And he
was absolutely correct in his statement. you are not a serious
political party anymore.
--
Iraq was a brilliant campaign fought with minimal
casualties, 11 September was a humiliating failure
by government to fulfill its primary role of
national defence. But Democrats who complained that
Bush was too slow to act on doubtful intelligence
re 9/11 now profess to be horrified that he was too
quick to act on doubtful intelligence re Iraq. This
is not a serious party.
===
Subject: Re: Question about Presidents Social Security plan
>>Yes, except that the government doesnt have an account
with $2
>>trillion in it. So it would have to sell bonds in that
amount. The
>>simpler method is to directly credit the SS trust fund with
whatever
>>it needs to cover the shortfall, if and when it occurs.
>Except of course the government would have to print money to
cover the
>shortfall which would quickly lead to hyperinßation.
The government prints money, i.e. monetizes the debt, for
only two
reasons: (1) to meet the publics demand for wallet money in
lieu of
bank deposits, and (2) to provide the reserves banks need to
meet
their reserve ratio requirements.
Future payments to social security beneficiaries will not
require the
printing of money. It will involve deficit spending, all of
which
will be covered by the sale of bonds.
>You can issue
>bonds but there has to be some expectation that you can pay
off those
>bonds when they come due, you see, and with 95% or whatever
percent of
>the federal budget dedicated to paying pension checks that
doesnt
>leave much room to pay any other bills.
The Treasury has no problem redeeming its bonds, and never
will. The
amount of government spending going to social security and
medicare is
currently about 35% of the total spending. That will increase
as baby
boomers retire, but it will never come close to 95% of total
spending.
>Ideally you take care of this by increasing the
contributions slightly
>well in advance and restraining the growth of the program to
>inßation. If we had done this back in, oh, 1983 the system
would be
>solvent.
The program is solvent and will be for at least forty years.
If and
when the trust fund runs out, the shortage can be made up by
government borrowing. That will increase the deficit, but only
during
the peak years of benefits for baby boomers. They too will 
die.
>Unfortunately we only did *half* of this: We raised FICA
>taxes and declared we had a surplus. But Congress never
reined in
>entitlements and then spent the FICA surplus on other things.
There is no way the FICA surplus can be kept in a lock box.
Those
funds would be spent even if the on-budget were in balance.
>Bottom line is the surplus disappears from the books around
2011 and
>from that point on the system runs in the red. And it only
gets
>*redder* the further out you go, with too few workers paying
into the
>system to cover benefits going out.
The surplus FICA revenues are expected to disappear about
2018, which
simply means the so-called trust fund will stop increasing in
value at
that time. It is conservatively projected to remain in the
black
decades longer.
===
Subject: Re: Question about Presidents Social Security plan
>So my question is, wouldnt it make more
>sense to just GIVE that $2 trillion to
>social security, which is guaranteed
>to fix things by exactly $2 trillion, just
>leaving a small $0.7 trillion shortfall after
>75 years?
If we had dealt with the problem 20 years ago it would have
cost a lot
less to fix. Reagan tried but Congress refused to rein in
entitlements. They simply raised the FICA and threw more
money at it
hoping that by the time anyone got wise theyd all be 
retired
(on a
FEDERAL pension) themselves. Clinton hemmed and hawed for 8
years and
held blue ribbon commissions whose recommendations he quietly
filed
away for the next Administration to act on. Meanwhile the
problem got
bigger, not better.
Setting aside a small fraction of the FICA taxes to allow
workers to
invest in their own pension plans doesnt actually cost 
Social
Security anything. There is no lockbox full of money being set
aside to pay retirees; it all comes out of the General
Revenue fund
anyway. These retirement accounts will generate enough taxable
revenue to offset the loss in FICA taxes going into the
system;
theyll create new revenue streams into the Treasury.
Put it like this. Right now you get ~ 1.75% annual return on
your
FICA investment, paid out of the Treasury when you retire. If
you
invested that money privately youd get at least that and
probably
much more, paid out of the growth in the GDP between now and
retirement. In effect its a redistribution of wealth (which
should
make liberals happy but go figure) because what you 
dont tap
from
that revenue stream goes into the pockets of wealthy
investors and
multinational corporations.
--
Iraq was a brilliant campaign fought with minimal
casualties, 11 September was a humiliating failure
by government to fulfill its primary role of
national defence. But Democrats who complained that
Bush was too slow to act on doubtful intelligence
re 9/11 now profess to be horrified that he was too
quick to act on doubtful intelligence re Iraq. This
is not a serious party.
===
Subject: Re: Question about Presidents Social Security plan
<57u6s0tmu6hrfdaqsjirlk290r7u5p3v4l@4ax.com>
posting-account=v1V-4A0AAAD0uc5c2ERg25Qdl2W__hSu
> Put it like this. Right now you get ~ 1.75% annual return
on your
> FICA investment, paid out of the Treasury when you retire.
If you
> invested that money privately youd get at least that and
probably
> much more, paid out of the growth in the GDP between now and
I figure it differently. If this applied to one or two
individuals,
it would indeed work the way you say. But when applied to
entire
populations, I think it would work another way.
All this money is going to be put in one particular area of
the economy - the stock market.
The stock market operates on demand and supply. While the
money is going in, the stock market goes up and absorbs
all the money.
Now recall that the whole problem is that at some point, there
are going to be many more retirees than contributors to
the system.
Shifting it to the stock market is not going to change that
fundamental characteristic. At some point, there simply will
be
many more sellers than buyers. When that happens, the stock
market
adjusts by going down. Given the volume of the movements,
and the fact that the smart money will have ßown out
of the stock market a little earlier, it will crash.
So the situation is, instead of the ~1.75% return there
will be tremendous loss of capital.
> retirement. In effect its a redistribution of wealth
(which should
Yes, in effect it is indeed nothing but a redistribution of
wealth.
But I
dont agree with the direction you seem to think it works 
in.
===
Subject: Re: Question about Presidents Social Security plan
>> Put it like this. Right now you get ~ 1.75% annual return
on your
>> FICA investment, paid out of the Treasury when you retire.
If you
>> invested that money privately youd get at least that and
probably
>> much more, paid out of the growth in the GDP between now
and
>I figure it differently. If this applied to one or two
individuals,
>it would indeed work the way you say. But when applied to
entire
>populations, I think it would work another way.
>All this money is going to be put in one particular area of
>the economy - the stock market.
or the real estate market, or the municipal bond market, or
any of a
hundred other investments. Historically all of these have
outperformed Social Security over time. Most pension plans
dont put
all their eggs in one basket, like Social Security does. SS is
REQUIRED BY LAW to invest in the infusion of new workers over
time and
we know that aint gonna happen; just the opposite is
happening.
--
Iraq was a brilliant campaign fought with minimal
casualties, 11 September was a humiliating failure
by government to fulfill its primary role of
national defence. But Democrats who complained that
Bush was too slow to act on doubtful intelligence
re 9/11 now profess to be horrified that he was too
quick to act on doubtful intelligence re Iraq. This
is not a serious party.
===
Subject: Re: Question about Presidents Social Security plan
<57u6s0tmu6hrfdaqsjirlk290r7u5p3v4l@4ax.com>
<tuk8s05dgk9ieni598s7f8t7cs9k7l99n5@4ax.com>
posting-account=N7ouHwwAAACKyjg2kLYhL_5bJ8P82GqY
>> Put it like this. Right now you get ~ 1.75% annual return
on your
>> FICA investment, paid out of the Treasury when you retire.
If
you
>> invested that money privately youd get at least that and
probably
>> much more, paid out of the growth in the GDP between now
and
>I figure it differently. If this applied to one or two
individuals,
>it would indeed work the way you say. But when applied to
entire
>populations, I think it would work another way.
>All this money is going to be put in one particular area of
>the economy - the stock market.
> or the real estate market, or the municipal bond market, or
any of a
> hundred other investments. Historically all of these have
> outperformed Social Security over time. Most pension plans
dont put
> all their eggs in one basket, like Social Security does. SS
is
> REQUIRED BY LAW to invest in the infusion of new workers
over time
and
> we know that aint gonna happen; just the opposite is
happening.
Think you missed the point here. What do you think is going
to happen
to interest rates when massive amounts of money are pulled
out of the
government bond market? No doubt this would have a big impact
on the
stock market as interest rates go way up which could result
in a
negative return for the stock market and a detrimental impact
on the
economy. And why do you assume that real estate is a safe
investment?
Have you ever heard of people making a bad investment? What
will
happen to those who invest in riskier investments and see
their
retirement fund wiped out?
> --
> Iraq was a brilliant campaign fought with minimal
> casualties, 11 September was a humiliating failure
> by government to fulfill its primary role of
> national defence. But Democrats who complained that
> Bush was too slow to act on doubtful intelligence
> re 9/11 now profess to be horrified that he was too
> quick to act on doubtful intelligence re Iraq. This
> is not a serious party.
I am amazed that a president who outright lied over the
reason to go to
war in Iraq, WMD that did not exist was reelected. Iraq and
9/11 are
unrelated events or at least thats the conclusion of the 9/11
commission
===
Subject: Re: Question about Presidents Social Security plan
>by the amount of $2.7 trillion in 75 years.
>Bush administration has a plan. The plan is
>to privatize some parts of social security.
Bush has a notion, not a plan.
The actual plan, following the precedent of Cheneys energy
plan,
will be prepared by brokerages and mutual funds, with their
recommendations being weighted according to their
contributions to
the GOP.
===
Subject: Re: Question about Presidents Social Security plan
<bet6s0dcgvombddmlqlm4bp7c4h568ueuj@4ax.com>
posting-account=v1V-4A0AAAD0uc5c2ERg25Qdl2W__hSu
>by the amount of $2.7 trillion in 75 years.
>Bush administration has a plan. The plan is
>to privatize some parts of social security.
> Bush has a notion, not a plan.
> The actual plan, following the precedent of Cheneys 
energy
plan,
> will be prepared by brokerages and mutual funds, with their
> recommendations being weighted according to their
contributions to
> the GOP.
Bush does have something in it for him, too.
It will leave the stock market jumping for
joy, and Bushs legacy will be to leave the
economy booming. Whether its a false
or temporary boom, wont be his concern.
And in any case, he will have lots of
defenders who will claim the followup
bubble-bursting and social-security money
getting swallowed up by the smart crowd,
was caused by the evil lefties
and had nothing to do with the original
mighty grand plan...
===
Subject: Re: Question about Presidents Social Security plan
posting-account=i3qD7AwAAACHjGKcE_VihY5XWtSCm1bP
This [privatize some parts of social security]
will cost $2 trillion to set up.
So my question is, wouldnt it make more
sense to just GIVE that $2 trillion to
social security, which is guaranteed
to fix things by exactly $2 trillion, just
leaving a small $0.7 trillion shortfall after
75 years?
The counter-argument would be that in doing so
youd simply be perpetuating a poor system that
by your own words remains in deficit, still will be
unsustainable because of its design, and you are
also erroneous (if not deliberately misleading) by
referring to $0.7 trillion as small.
There are other counter-arguments as well (you
fail to view the entire exchange of revenues and
benefits, for example).
You were on stronger ground elsewhere on this
thread when you rightly noted that Wall Street
would love to get its hands and and make money
off all that money. Government would indirectly
be giving money (taxing it and seeing that it was
redirected) to Wall Street.
A more legitimate problem with Bushs proposal
is this would be nominally private but still would be
a government-program set of accounts, and there
would be temptations by Democrats, especially
liberal Democrats, to engage in evil, anti-American
lefty-fascist follies that Ralph Nader only could
dream of decades ago, such as making the
federal government the largest-by-far institutional
investor -- and with that would come government
inßuence and social responsibility, faddish far-left
idiocy such as Israeli divestiture gimmicks, maybe
government shareholder inßuence on politically
disfavored industries like guns and automobiles, and
so on. No normal American wants any threat of that.
===
Subject: Re: Question about Presidents Social Security plan
> A more legitimate problem with Bushs proposal
> is this would be nominally private but still would be
> a government-program set of accounts, and there
> would be temptations by Democrats, especially
> liberal Democrats, to engage in evil, anti-American
> lefty-fascist follies that Ralph Nader only could
> dream of decades ago, such as making the
> federal government the largest-by-far institutional
> investor -- and with that would come government
> inßuence and social responsibility, faddish far-left
> idiocy such as Israeli divestiture gimmicks, maybe
> government shareholder inßuence on politically
> disfavored industries like guns and automobiles, and
> so on. No normal American wants any threat of that.
I cant imagine that political agitation for restriction of
these
investments would be confined to the left.
Would the backers of this plan permit any of the money to be
invested
in ,for instance, manufacturers of contraceptives,
particularly the
so-called morning after pill?
I have a plan that would make SS solvent for the rest of this
century.
Just increase the interest rate that the general fund pays to
the Social
Security fund by one percentage point. The same amount of
cash that
changes hands would be the same as present, but the SS fund
has
immediately become much more solvent due to the increase in
the future
value of its reserves.
Just bookkeeping, but that is all claims of SS insolvency are
anyway.
--
To e-mail me get rid of the cats and dogs.
===
Subject: Re: Question about Presidents Social Security plan
>> A more legitimate problem with Bushs proposal
>> is this would be nominally private but still would be
>> a government-program set of accounts, and there
>> would be temptations by Democrats, especially
>> liberal Democrats, to engage in evil, anti-American
>> lefty-fascist follies that Ralph Nader only could
>> dream of decades ago, such as making the
>> federal government the largest-by-far institutional
>> investor -- and with that would come government
>> inßuence and social responsibility, faddish far-left
>> idiocy such as Israeli divestiture gimmicks, maybe
>> government shareholder inßuence on politically
>> disfavored industries like guns and automobiles, and
>> so on. No normal American wants any threat of that.
> I cant imagine that political agitation for restriction 
of
these
>investments would be confined to the left.
>Would the backers of this plan permit any of the money to be
invested
>in ,for instance, manufacturers of contraceptives,
particularly the
>so-called morning after pill?
>I have a plan that would make SS solvent for the rest of
this century.
>Just increase the interest rate that the general fund pays
to the Social
>Security fund by one percentage point. The same amount of
cash that
>changes hands would be the same as present, but the SS fund
has
>immediately become much more solvent due to the increase in
the future
>value of its reserves.
The SS tax was dramatically increased under Reagan and he
used the
money to fund the tax breaks for the wealthy and his monstrous
increase in military spending to go to war against Russia and
the
Middle East (Iraq and Iran).
Charlie
>Just bookkeeping, but that is all claims of SS insolvency
are anyway.
>--
>To e-mail me get rid of the cats and dogs.
===
Subject: Re: Question about Presidents Social Security plan
posting-account=v1V-4A0AAAD0uc5c2ERg25Qdl2W__hSu
> The counter-argument would be that in doing so
> youd simply be perpetuating a poor system that
> by your own words remains in deficit, still will be
> unsustainable because of its design, and you are
I thought the problem was not the design, but
baby boomers. Reverse booms should put money
into social security, which could then be
used up for the next boom.
> also erroneous (if not deliberately misleading) by
> referring to $0.7 trillion as small.
Not looked at the budget figures lately, have we?
Maybe you have a point -- perhaps its erroneous
in this context to call 0.7 trillion over 75 years
as small. Its better called negligible or trivial
or irrelevant. Any single one of the next presidents
over the next 75 years could fix it when the need became
apparent. In the worse case, by borrowing (just like Bush
plans to do), in the best case by using up
some surplus.
===
Subject: Re: Question about Presidents Social Security plan
posting-account=i3qD7AwAAACHjGKcE_VihY5XWtSCm1bP
> I thought the problem was not the design, but
> baby boomers. Reverse booms should put money
> into social security, which could then be
> used up for the next boom.
Reverse booms???
The problem is both with the design and with demographics
(not merely
the Baby Boomers but lower fertility rates and longer
lifespans). The
program, if kept (which is most likely), should be converted
to fully
funded rather than as pay-as-you go. Also, it is
irresponsible (and
given our lives are involved, illogical) to count on recovery
of the
program after the Baby Boomers are dead, much less to be
superficial in
our approach and simply tax more during low-beneficiary-number
years or
decades to better finance benefits during
high-beneficiary-number years
or decades that follow them.
> Not looked at the budget figures lately, have we?
Im fully aware of them, as well as what Social Security and
Medicare
will do to the entire federal budget (not just those two
programs) long
before either of these two programs go bankrupt.
> Maybe you have a point -- perhaps its erroneous
> in this context to call 0.7 trillion over 75 years
> as small. Its better called negligible or trivial
> or irrelevant.
Thats even more erroneous.
> Any single one of the next presidents over the next
> 75 years could fix it when the need became
> apparent. In the worse case, by borrowing (just like Bush
> plans to do), in the best case by using up some surplus.
Only a fool would count on a surplus then, much later,
the realistic time when politicians will take action -- which
is
only when forced to, or in other words -- eventually. (They
avoid doing anything unpleasant now even though it solves
worse problems for us later.)
There will not be an easy fix later.
Note that you say that at any time, it can be fixed. That is
the admission of error of all those who deny there is anything
wrong with the system now. (There is; it is unsustainable and
will cause many other fiscal problems for the federal
government
long before it goes broke.)
If it were up to me and we had to keep Social Security, I
would
take the roughly ten years we have left before the Baby
Boomers
start their retirements to convert from our pay-as-you-go
(Ponzi
scheme) system to a fully funded system. As that causes some
pain to all, perhaps it is that one time conversion that
might, might
justify borrowing, because it would solve the problem with
not only
S.S. but with the federal finances, and so justify the extra
cost of
interest in the long run.
===
Subject: Re: Question about Presidents Social Security plan
posting-account=v1V-4A0AAAD0uc5c2ERg25Qdl2W__hSu
> Maybe you have a point -- perhaps its erroneous
> in this context to call 0.7 trillion over 75 years
> as small. Its better called negligible or trivial
> or irrelevant.
> Thats even more erroneous.
I think I was not clear enough. I will simplify. It
works out to less than 10 billion per year. How big
a deal in the federal budget do you want to make that?
> There will not be an easy fix later.
Sure there will be. Note that Bushs fix 
involves
borrowing $2 trillion. Whats to stop a future
president, who is facing the problem here and
now instead of in the future, to simply borrow
the much smaller amounts needed to keep it solvent,
as and when needed? And it is indeed possible
that when needed, the budget will happen to have a surplus,
at least some of the times.
===
Subject: Re: Question about Presidents Social Security plan
<FqEwd.1018416$SM5.72780@news.easynews.com>
posting-account=i3qD7AwAAACHjGKcE_VihY5XWtSCm1bP
Why not just level with the people and tell them Social
Security is an
unconstitutional use of their money and in 25 years that will
cease?
Liberals and those theyve made dependent on the federal
government
dont care about Social Securitys 
unconstitutionality; as
the childish
people they are, theyre delighted that Government can and
will do
all kinds of things to meet human needs. It is for this
reason that
for decades Social Security has been treated by politicians
in DC as
something that people hold sacred.
(The dopiest of Dems even take sacredness further: they
worship their
God, FDR.)
===
Subject: Re: Question about Presidents Social Security plan
<cpv6np02o2@drn.newsguy.com>
posting-account=v1V-4A0AAAD0uc5c2ERg25Qdl2W__hSu
> Youre assuming that Bush *wants* to fix 
social security.
> Nothing could be farther from the truth. He wants to make
> sure that it goes bankrupt, and his privatization plan is
> a crucial step.
I dont think he particularly cares to see social
security bankrupt -- I suspect the motivation is
entirely different.
Wall Street was helped strongly during the internet
bubble by 401K funding. The major players made
tons of money. But then the bubble burst, and the
wall street honchos felt they werent rich
enough. They saw a ray of light -- the next possible
bubble could come from social security. So they are
pulling all kinds of strings. And of course, our
venerable president is a sucker for anything
the rich guys say -- it must be the right
thing to do if the havemores say so.
He just doesnt believe the nice havemores would
advise something that could have bad results.
Its kind of Darwinian. Remeber, the folks who would
be most exploited are mostly staunch Bush supporters, and
would jump at anybody with both feet for suggesting he is
doing something wrong. I dont know how much is
really wrong if they are left -- by the actions
of their favorite president -- with nothing in
their golden days. Besides, they will blame
the lib dem boogiemen anyways.
And I suppose if it does go through, the smart
folks could make some money at the
expense of these non-havemore Bush supporters,
because the stock market will have some
irrational moments.
Still, it just doesnt seem right, or American,
or a good thing at all for the long
term future.
===
Subject: Probability-probability density function
posting-account=__WCMw0AAACDTl8muX4MFuov_vIZutrh
I need help with this problem. Joe gives a goodness test
ranging from
-1 to 1. A one means you are good and a -1 means you are bad.
Let X
be the goodness score. The pdf of X is:
f(x) = e^5x -1 <= x <= 0
0 <= x <= 1
a.) find the standard deviation of x.
b.) Joe plans to take 20% of the good people out to lunch.
What is the
lowest score that will qualify?
c.) Joe needs to find five bad people whose 
goodness scores are
at most
-0.9. Joe keeps testing until he finds the fifth 
bad person.
Whats
the probability that Joe needs to test exactly 11 people?
I think I can handle finding the standard deviation after
finding the
mean and E(X) in part a. Im not sure at all about part b.
Part c.
seems like its the negative binomial distribution with the
fifth person
out of 11, but I sure where to begin as an amateur
probabilitist. Help!
===
Subject: re:Spherical trigonometry question.
this problem. And with a program to calculate coordinates for
gravitating points.
http://www.math-atlas.org/index/spheres.html
A C++ program to calculate point migration.
FILE points_on_sphere.cc
-------------------------------------------------------------
---------------
-----
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <math.h>
typedef double vec3[3];
double frand(void){return
((rand()-(RAND_MAX/2))/(RAND_MAX/2.));}
double dot(vec3 v1,vec3 v2){ return
v1[0]*v2[0]+v1[1]*v2[1]+v1[2]*v2[2];}
double length(vec3 v){ return
sqrt(v[0]*v[0]+v[1]*v[1]+v[2]*v[2]); }
double length(vec3 v1,vec3 v2)
{
vec3 v;
v[0] = v2[0] - v1[0]; v[1] = v2[1] - v1[1]; v[2] = v2[2] -
v1[2];
return length(v);
}
double get_coulomb_energy(int N,vec3 p[])
{
double e = 0;
for(int i = 0;i<N;i++)
for(int j = i+1; j<N; j++ ) {
e += 1/ length(p,p[j]);
}
return e;
}
void get_forces(int N,vec3 f[], vec3 p[])
{
int i,j;
for(i = 0;i<N;i++){
f[i][0] = 0;f[i][1] = 0;f[i][2] = 0;
}
for(i = 0;i<N;i++){
for(j = i+1; j<N; j++ ) {
vec3 r = {p[i][0]-p[j][0],p[i][1]-p[j][1],p[i][2]-p[j][2]};
double l = length(r); l = 1/(l*l*l);double ff;
ff = l*r[0]; f[i][0] += ff; f[j][0] -= ff;
ff = l*r[1]; f[i][1] += ff; f[j][1] -= ff;
ff = l*r[2]; f[i][2] += ff; f[j][2] -= ff;
}
}
}
int main(int argc, char *argv[])
{
static int N=100,Nstep=1000;
static double step=0.01;
static double minimal_step=1.e-10;
if(argc < 2){
fprintf(stderr,points_on_sphere calculates distribution of
points
on sphere n);
fprintf(stderr,wich gives a minimum to 1/r potential
energynn);
fprintf(stderr,usage: points_on_sphere <number_of_points>
<maximal_number_of_steps>n);
fprintf(stderr,output is printed in VRML format to stdoutn);
fprintf(stderr,example of usage: points_on_sphere 10 1000 >
dist10.wrl n);
fprintf(stderr,nAuthor: V.Bulatov@ic.ac.uk nn);
exit(-1);
}
N = atoi(argv[1]);
if(argc > 2)
Nstep = atoi(argv[2]);
vec3 *p0 = new vec3[N];
vec3 *p1 = new vec3[N];
vec3 *f = new vec3[N];
int i,k;
vec3 *pp0 = p0, *pp1 = p1;
srand(time(NULL));
for(i = 0; i<N; i++ ) {
p0[i][0] = 2*frand();
p0[i][1] = 2*frand();
p0[i][2] = 2*frand();
double l = length(p0[i]);
if(l!=0.0){
p0[i][0] /= l;
p0[i][1] /= l;
p0[i][2] /= l;
} else
i--;
}
double e0 = get_coulomb_energy(N,p0);
for(k = 0;k<Nstep;k++) {
get_forces(N,f,p0);
for(i=0; i < N;i++) {
double d = dot(f[i],pp0[i]);
f[i][0] -= pp0[i][0]*d;
f[i][1] -= pp0[i][1]*d;
f[i][2] -= pp0[i][2]*d;
pp1[i][0] = pp0[i][0]+f[i][0]*step;
pp1[i][1] = pp0[i][1]+f[i][1]*step;
pp1[i][2] = pp0[i][2]+f[i][2]*step;
double l = length(pp1[i]);
pp1[i][0] /= l;
pp1[i][1] /= l;
pp1[i][2] /= l;
}
double e = get_coulomb_energy(N,pp1);
if(e >= e0){ // not successfull step
step /= 2;
if(step < minimal_step)
break;
continue;
} else { // successfull step
vec3 *t = pp0; pp0 = pp1; pp1 = t;
e0 = e;
step*=2;
}
fprintf(stderr,rn: %5d, e = %18.8f step = %12.10f,k,e,step);
fßush(stderr);
}
fprintf(stdout,#VRML V1.0 asciin);
fprintf(stdout,Separator {n);
fprintf(stdout,Material {diffuseColor 1 1 0 specularColor 1 1
1}n);
for(i = 0;i<N;i++){
fprintf(stdout,Separator {Translation { translation );
fprintf(stdout,%12.10f %12.10f
%12.10f,p0[i][0],p0[i][1],p0[i][2]);
fprintf(stdout,} Sphere { radius 0.1 }}n);
}
fprintf(stdout,}n);
delete[] p0;
delete[] p1;
delete[] f;
}
--------------------------------------
Dr. V.Bulatov, Research associate
Materials Department, Imperial College,Prince Consort Road,
London SW7
2AZ, UK
e-mail:V.Bulatov@ic.ac.uk; http://pc153.mt.ic.ac.uk/Bulatov;
tel:0171-5946802
*-----------------------*
www.GroupSrv.com
*-----------------------*
===
Subject: re:Spherical trigonometry question.
One possibility I was considering is to alter the shape of
the sphere
slightly, to accomodate a near fit. For instance, if a
particular
shape has a good distribution of triangles throughout, it
could be
used as a basis for a point distribution, which when allowed
to
gravitate to equilibrium would shape the sphere to fit the
model.
Instead of having a rigid sphere, using an envelope that
molds itself
over the points. That way, at least it may be possible to get
around
the impossiblility of fitting the grid over a perfect sphere.
*-----------------------*
www.GroupSrv.com
*-----------------------*
===
Subject: Re: Spherical trigonometry question.
One of the subtleties in this is taht the regular polyhedral
geometry is
not a Stable equilibrium for N = 8 or N = 20. The square or
pentagonal
faces tend to buckle into triangels under certain modes of
displacement.
For instance, if you start with eight points in a cube and
rotate one
face slightly in its plane, the resultant forces on the moved
points
tend to drive further rotation in the same direction. This
leads to a
more stable state in whih the opposite square faces are
rotated 45
degrees from each other, a square antiprism. The other four
formerly
square faces have buckled into eight triangles. Other modes of
deformation buckle the remaining two square faces, leading
ultimately to
a relatively low-symmetry object, the triangular dodecahedron
(12
faces, 18 edges and of course 8 vertices). In a real
eight-coordinate
molecule, we would expect the ligands to exchange relative
positions by
passing through a square aniprismatic transistion state,
which is more
accessible than a cubic one.
--OL.
===
Subject: Re: Spherical trigonometry question.
<29533-41C428DA-36@storefull-3258.bay.webtv.net>
posting-account=CfSJ5AwAAAD1yt3VP50q913IBHikxMCd
Would the icosidodecahedron (20+12) be more stable than either
dodecahedron (12) or icosahedron (20) in this respect? Is
there some
stability index to evaluate in each case?
===
Subject: Re: Spherical trigonometry question.
posting-account=2cSpUgwAAAC62s7i9tYaQ_o97Mapa1V8
> How can I divide a sphere into N equal and identical parts?
> I start with the premise that the surface can be divided by
an
> arbitrary number N, and is enclosed by P points on the
surface of the
> sphere. I want to know the formula for the points, the
angle of
> separation and distribution.
> One method I am considering is to assume that each point is
> positively charged, and that if left to ßoat on the
surface, the
> points will reach an equilibrium being equidistant from
each other.
> I would like a computer program to to trace the point
layout on a
> wiremesh model. I want also the the program to readjust
each time
> more points are added to the distribution. And to calculate
an
> optimum number of points for a given sphere.
McDowell. Point Charge Approximation to a Spherical Charge
Distribution (A Random Walk to High Symmetry). Journal of
Chemical Education, Vol 67 No 12, December 1990.
===
Subject: Re: Spherical trigonometry question.
posting-account=2cSpUgwAAAC62s7i9tYaQ_o97Mapa1V8
> How can I divide a sphere into N equal and identical parts?
> I start with the premise that the surface can be divided by
an
> arbitrary number N, and is enclosed by P points on the
surface of the
> sphere. I want to know the formula for the points, the
angle of
> separation and distribution.
> One method I am considering is to assume that each point is
> positively charged, and that if left to ßoat on the
surface, the
> points will reach an equilibrium being equidistant from
each other.
> I would like a computer program to to trace the point
layout on a
> wiremesh model. I want also the the program to readjust
each time
> more points are added to the distribution. And to calculate
an
> optimum number of points for a given sphere.
McDowell. Point Charge Approximation to a Spherical Charge
Distribution (A Random Walk to High Symmetry). Journal of
Chemical Education, Vol 67 No 12, December 1990.
===
Subject: What is meant by a meager subset??
Can some please define a meager subset for me?
===
Subject: Re: What is meant by a meager subset??
Bourbaki did not think first category and second category were
good
terms, so he invented some new ones: meagre, residual. Do you
think
they are better terms?
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: What is meant by a meager subset??
> Can some please define a meager subset for me?
A meagre set (in a topological space) is the same as a set of
first
category. I.e. a set that is the union of at most aleph_0
nowhere dense
sets.
A nowhere dense set is one with the interior of its closure
equal to the
empty set.
===
Subject: Re: What is meant by a meager subset??
posting-account=
y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g
A nowhere dense subset of a topological space has a closure
with empty
interior.
A meager subset (or meagre, for those across the pond) is a
countable
union of nowhere dense subsets.
These topological concepts tie in with that of Baire category.
===
Subject: Re: What is meant by a meager subset??
posting-account=ZeRDXwsAAACLpj2mpKc97NFPxBaFxAzp
Its another name for first Baire category, i.e. 
a subset that
is contained in the union of countably many closed
nowhere-dense sets.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving
correct? .
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiEu02687;
I propose a solution based on Abel function of one variable
function,that is ,put y=a ,f(a)=b -> f(x+2b)=f(x)+ a + b
and Ôcounting form f(x+2b)/(a+b)=f(x)/(a+b)+ 1 
, (1)
We know Abel Ôcounting function of x+2b is x/2b 
to a
constant k(verifying
a relation like (1) ).
So f(x)/(a+b)= x/2b + k , thence f is a line ;f(x)=cx + d ;
Identification in f(x + 2f(y)) = f(x) + y + f(y) gives d= 0
and two
values for c ...
Notice:k is in fact an invariant expression when x->x + 2b .
Your comments please,
Alain.
===
Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving
correct? .
posting-account=ZeRDXwsAAACLpj2mpKc97NFPxBaFxAzp
There are indeed two linear solutions: f(x) = x and f(x) =
-x/2.
If f is analytic in a neighbourhood of 0, taking the Maclaurin
series to first order gives us f(x) = x + O(x^2) or -x/2 +
O(x^2).
I tried taking the Maclaurin series to order 8, and there
still were no
other solutions, so I suspect these are all the analytic
solutions
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving
correct? .
posting-account=ZeRDXwsAAACLpj2mpKc97NFPxBaFxAzp
| If f is analytic in a neighbourhood of 0, taking the
Maclaurin
| series to first order gives us f(x) = x + O(x^2) or -x/2 +
O(x^2).
| I tried taking the Maclaurin series to order 8, and there
still were
no
| other solutions, so I suspect these are all the analytic
solutions
Not sure exactly what I did here: its not so clear how to 
do
this
in the case f(0) <> 0, because at x=y=0 the equation just says
f(2f(0)) = 2f(0). But heres an easy proof that the only
real-analytic
solutions on R are the two linear solutions.
For y = x the equation says f(x + 2 f(x)) = x + 2 f(x).
So if g(x) = f(x) - x, we have g(x + 2 f(x)) = 0.
So there are two alternatives:
1) g is constant, i.e. f(x) = x + c for some constant c .
2) x + 2 f(x) is constant, i.e. f(x) = (c-x)/2 for some
constant c.
In both cases its easy to see that c = 0, i.e. the only
real-analytic
solutions are f(x) = x and f(x) = -x/2.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving
correct? .
posting-account=ZeRDXwsAAACLpj2mpKc97NFPxBaFxAzp
> | If f is analytic in a neighbourhood of 0, taking the
Maclaurin
> | series to first order gives us f(x) = x + O(x^2) or -x/2 +
O(x^2).
> | I tried taking the Maclaurin series to order 8, and there
still
were
> no
> | other solutions, so I suspect these are all the analytic
solutions
> Not sure exactly what I did here: its not so clear how to
do this
> in the case f(0) <> 0, because at x=y=0 the equation just
says
> f(2f(0)) = 2f(0). But heres an easy proof that the only
real-analytic
> solutions on R are the two linear solutions.
> For y = x the equation says f(x + 2 f(x)) = x + 2 f(x).
> So if g(x) = f(x) - x, we have g(x + 2 f(x)) = 0.
> So there are two alternatives:
> 1) g is constant, i.e. f(x) = x + c for some constant c .
> 2) x + 2 f(x) is constant, i.e. f(x) = (c-x)/2 for some
constant c.
> In both cases its easy to see that c = 0, i.e. the only
real-analytic
> solutions are f(x) = x and f(x) = -x/2.
And in fact these are the only continuous solutions on R.
We have as above g(x + 2 f(x)) = g(3 x + 2 g(x)) = 0. If g is
not always 0, let J be a maximal interval containing 2 g(0)
on which
g is 0. Then 3 x + 2 g(x) must be in J for all x. In
particular, if
x is in J, 3 x + 2 g(x) = 3 x is in J. By continuity, J is
closed.
Now the only closed intervals that are mapped into themselves
by multiplication by 3 are (-infinity,infinity),
(-infinity,-a], [a,infinity) and {0} (where a >= 
0).
J = (-infinity,infinity) corresponds to the 
solution f(x) = x.
J = {0} corresponds to the solution f(x) = -x/2.
If J = [a,infinity) we have g(x) = 0 for x >= a.
In the full equation, which can be written
g(x + 2 y + 2 g(y)) + g(y) = g(x)
take y in J, g(y) = 0, to get g(x + 2 y) = g(x).
But by taking y large enough, x + 2 y is in J, so g(x) = 0
for all x,
and were back in the case of 
(-infinity,infinity).
The argument for J = (-infinity,-a] is similar.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: probability density and invariant measures
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiEC02683;
I am trying to understand a chapter in a book that is
probably too difficult
for me to read, and i was wondering if someone could help me
weed through a
part.
the author considers the logistic map x_{n+1} = 4x_n (1 -
x_n) states that
the density p(x) of an invariant measure for the logistic map
is
p(x) = (pi sqrt{x (1 - x)})^{-1}
I barely even understand the idea of an invariant measure, so
that is a
little hard for me to comprehend, especially when stated
without proof or
explanation. could someone tell me how this is possible?
===
Subject: Separable,not Banach,without Schauder basis
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiFS02726;
Hi all,
Knowing that it is necessary for a normed space with a
Schauder basis to be separable,its not 
difficult to give
an example of a space without Schauder basis.
l^oo is such an example as l^oo is not separable.
For a long time it wasnt known if there existed separable
Banach
space without Schauder basis.Yes,there is such a space.
That example is beyond me,so Im asking if someone knows if
an example of separable space which is *not* Banach and
doesnt have Schauder basis would be easier,and if possible
===
Subject: Re: Separable,not Banach,without Schauder basis
> Hi all,
> Knowing that it is necessary for a normed space with a
> Schauder basis to be separable,its not 
difficult to give
> an example of a space without Schauder basis.
> l^oo is such an example as l^oo is not separable.
> For a long time it wasnt known if there existed separable
Banach
> space without Schauder basis.Yes,there is such a space.
> That example is beyond me,so Im asking if someone knows 
if
> an example of separable space which is *not* Banach and
> doesnt have Schauder basis would be easier,and if 
possible
How about L^p[0,1], where 0<p<1 ? There are no continuous
linear
functionals, and thus there are no Schauder bases.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Solving g( x+3y ,3x + y) = x*y , g :R * R->R
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiFh02721;
> How do you take this?.
> Can you generalize to :
> g(m(x,y),n(x,y))=h(x,y) g,m,n,h R*R->R functions,
> only g(x,y) unknown ?
>>For the problem given in the subject line, let u = x+3y and
v = 3x
+
>>y. Then you can write x and y as functions of u and v 
(Ill
let you
do
>>this) and g(u, v) = x * y.
>And of course more generally, solve the equations u =
m(x,y), v = n(x,y)
>to express x, y in terms of u, v.
>Mike Guy
Ill let you do this , I simply dont agree:
I know the solution and I just want comparing ways;
moreover ,for me , time of homeworks has altogether gone by
...
Now I am helping pupils three times a week ,
Alain.
===
Subject: Re: Solving g( x+3y ,3x + y) = x*y , g :R * R->R
>><a
2161712.iBGHCRL15400@proapp.mathforum.org...</a
> How do you take this?.
> Can you generalize to :
> g(m(x,y),n(x,y))=h(x,y) g,m,n,h R*R->R functions,
> only g(x,y) unknown ?
>>For the problem given in the subject line, let u = x+3y and
v =
3x
+
>>y. Then you can write x and y as functions of u and v 
(Ill
let you
do
>>this) and g(u, v) = x * y.
>And of course more generally, solve the equations u =
m(x,y), v =
n(x,y)
>to express x, y in terms of u, v.
>Mike Guy
> Ill let you do this , I simply dont 
agree:
> I know the solution and I just want comparing ways;
> moreover ,for me , time of homeworks has altogether gone by
...
> Now I am helping pupils three times a week ,
> Alain.
Actually, Mike Guy did not write, Ill let you do this. I
did, and Im
not sure what you dont agree with. The equations u = x+3y
and v = 3x
+y form a system of linear equations in x and y. Solving for
x and y in
terms of u and v is elementary.
________________________________
Eric J. Wingler (wingler@math.ysu.edu)
Dept. of Mathematics and Statistics
Youngstown State University
One University Plaza
Youngstown, OH 44555-0001
330-941-1817
===
Subject: Re: .99999... still=/= 1
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiHg02852;
S. Enterprize Company <smart1234@aol.com>
http://mathforum.org/discuss/sci.math/m/651356/663915
> That Ôs wrong. The decimal number .999... 
isnt a
> rational number. Show me the fraction that represents
.999... .
> 9/9 = 1 this isnt .999...
> 9/10 = .9 this isnt 1 or .999...
.999... is 9/10 + 9/100 + 9/1000 + ...
This is an infinite geometric series with a ratio
(namely, 1/10) that is between -1 and 1, and thus
it converges. Look up geometric series in the
index of any precalculus text.
Im sure others have asked this, but if 1 
isnt equal
to .999..., then the difference between these two
numbers cant be zero. So what is the difference
between these two numbers?
And if youre going to claim that .999... 
doesnt
represent a number, then explain what is wrong
with the derivation of the geometric series
formula that one finds in literally thousands
of different algebra 2, precalculus, and calculus
texts, not to mention physics, engineering, chemistry,
economics, etc. texts where its used [1].
[1] Try googling the phrase infinite geometric series
along with virtually any field of social or natural
science and see what you get.
Dave L. Renfro
===
Subject: Re: .99999... still=/= 1
>S. Enterprize Company <smart1234@aol.com>
>http://mathforum.org/discuss/sci.math/m/651356/663915
>> That Ôs wrong. The decimal number .999... 
isnt a
>> rational number. Show me the fraction that represents
.999... .
>> 9/9 = 1 this isnt .999...
>> 9/10 = .9 this isnt 1 or .999...
>.999... is 9/10 + 9/100 + 9/1000 + ...
>This is an infinite geometric series with a ratio
>(namely, 1/10) that is between -1 and 1, and thus
Do you have any idea what an asmptote is? An asymtote does
converge to a
certain line from a curve but it does actually equal that
curve.
Do you know the difference between,
this,
.999...
| | | | |||||||||||||||||||||||||
and,
1
|
|
|
|
|
Do you have any idea what a hyper-real number is?
Do you have any idea what a real number is?
Do you know what a derivative is when trying to find points of
inßection?
Do you have any idea what the gamma function is?
Do you have any idea what higher order differential equations
are?
Do you have any idea what non-standard analysis is?
Do you know the difference between
1/0 = indeterminate
and
1/0 = oo
Do you have any idea what a fraction is? Can .999... be shown
as a
fraction?
Do you have any idea what an irrational number is with a fine
detail
pointing out non-repeating numbers to the right of the
decimal?
Do you have any idea what a convergence test is?
Do you have any idea what ill-defined terms in math are like 
i?
I can go on and on, but I still dont think you even know 
what
mathematics
is.
>it converges. Look up geometric series in the
>index of any precalculus text.
>Im sure others have asked this, but if 1 
isnt equal
>to .999..., then the difference between these two
>numbers cant be zero. So what is the difference
>between these two numbers?
>And if youre going to claim that .999... 
doesnt
>represent a number, then explain what is wrong
>with the derivation of the geometric series
>formula that one finds in literally thousands
>of different algebra 2, precalculus, and calculus
>texts, not to mention physics, engineering, chemistry,
>economics, etc. texts where its used [1].
>[1] Try googling the phrase infinite geometric series
> along with virtually any field of social or natural
> science and see what you get.
>Dave L. Renfro
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
> Do you have any idea what ill-defined terms in math are like
i?
When you say i, are you talking about the imaginary root of
x^2+1 = 0?
That i is very well defined.
===
Subject: Re: .99999... still=/= 1
>> Do you have any idea what ill-defined terms in math are
like i?
>When you say i, are you talking about the imaginary root of
x^2+1 = 0?
>That i is very well defined.
I guess it depends on the relative observers perspective.
What about
this,
can you split a negative sign in half like,
(.5_ -) + (.5_ -) = --, and get a full negative sign?
or, split a positive sign in half like,
(.5_+) + (.5_+) = + , and get a full positive sign?
Maybe this could be included in a non-standard analysis of
negative and
positive signs.
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
In sci.math, S. Enterprize Company
<smart1234@aol.com Do you have any idea what ill-defined terms
in math are like i?
>>When you say i, are you talking about the imaginary root of
x^2+1 =
0?
>>That i is very well defined.
> I guess it depends on the relative observers perspective.
What about
this,
> can you split a negative sign in half like,
> (.5_ -) + (.5_ -) = --, and get a full negative sign?
You might be describing imaginaries or quaternions; Im
not sure which. Imaginaries can be considered half a
minus, as i^2 = -1. Quaternions have four numbers,
with various multiplications all of which are anticommutative
(e.g., IJ = -JI).
> or, split a positive sign in half like,
> (.5_+) + (.5_+) = + , and get a full positive sign?
And what is (.5_+) + (.5_-) ?
> Maybe this could be included in a non-standard analysis of
negative
and
> positive signs.
Thats certainly one way of putting it.
[.sigsnip]
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
>In sci.math, S. Enterprize Company
><smart1234@aol.com> Do you have any idea what ill-defined
terms in math are like i?
>When you say i, are you talking about the imaginary root of
x^2+1 =
0?
>That i is very well defined.
>> I guess it depends on the relative observers perspective.
What about
>this,
>> can you split a negative sign in half like,
>> (.5_ -) + (.5_ -) = --, and get a full negative sign?
>You might be describing imaginaries or quaternions; Im
>not sure which. Imaginaries can be considered half a
>minus, as i^2 = -1. Quaternions have four numbers,
>with various multiplications all of which are anticommutative
>(e.g., IJ = -JI).
No, I was proposing something different than quaternions.
Quaternions
are
like added dimensions to the imaginary number i. Its like
the cartesian
coordinates have x,y,z. Quaternions use i,j,k imaginary
numbers with
various
rules like you show, IJ = -JI etc... .
What I was proposing was something like a sub-imaginary level
of
analysis,
like with a non-standard analysis approach, where there
exists a space
between
real numbers that can be explained by surreal, superreal or
hyperreal
numbers.
This is areas in the sub-real number zones.
I was wondering if we can do this.
Using the surreal symbol { | }
(.5_ -) | --
then sqrt (-1) = (.5_-)(1) = (1_-)
^
|
This shows half negative sign.
then,
(.5_- + .5_-)(1)^2 = -1
or
(1_-)^2 = -1
or
(1_+) - (+) = (1_-)
or,
( 1 ) - (1_+) = (1_+)
or,
(1_+) + (1_-) = 0
This would be sign and number value annihilation.
etc...
>> or, split a positive sign in half like,
>> (.5_+) + (.5_+) = + , and get a full positive sign?
>And what is (.5_+) + (.5_-) ?
Sign Annihilation = 0
Like a matter/anti-matter collision.
>> Maybe this could be included in a non-standard analysis of
negative
and
>> positive signs.
>Thats certainly one way of putting it.
>[.sigsnip]
>--
>#191, ewill3@earthlink.net
>Its still legal to go .sigless.
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
<2vhufhF2lve0fU1@uni-berlin.de>
posting-account=u9qOfg0AAADTt7xfBaPBMJxdsv5hlTRx
I dont know about you guys, but I actually like the
character S.E.
quite a bit, if only for the humour and entertainment his
posts
provides.
And all this .999... arguing back and forth is cracking me
up. I
dont know whats funnier (or more pathetic), 
S.E. making
himself
look like a fool, or you guys actually taking him seriously
and
making yourselves look like retards.
My one and only advice to S.E. is better to keep your mouth
shut and
let others suspect you are a fool, than to open it and
forever remove
all doubts. Too late now, I guess.
You know, most newsgroups need an idiot of some kind, and our
math
genius is serving that role quite nicely. He reminds me of
Beavis
(you know, the imbecile in the Beavis and Butthead cartoons
-- not
that Butthead was any less imbecilic, but in a different way)
after he
had ingested an overly large dose of caffein and sugar.
A drugged-up Beavis would pull his T-shirt over his head,
raise his
arms surrender-style, and spew complete nonsense at
strangers. All
offers of help, logical arguments, insults, even physical
threats were
totally ignored because, well, actually I dont know why, 
but
I
suspect mostly because he had gone nuts, or, less likely,
they werent
what he was looking for (he usually made it abundantly
clearly that he
was looking for TP for his bunghole, but everyone nearby
would fail to
pick up that glaringly obvious point).
S.E.: I am the one and only mighty bunghooooole...you must
bow down
before the great bunghole...and .9999... =/= 1...bungholio....
Sci.Math: Chill out, heres the proof. Take a look, blah 
blah
blah...
S.E.: Are you threatening me? I am cornholio! ...I need TP
for my
bunghole...Bunghole!! And rational doesnt converge partial
divergent sum limit test Mathcad Rababababa 99999999 genius
hyperfilterunrealnumber HOOOOLLLIOOOOOO...
Sci.Math: Now you are being completely non-sensical...do you
even know
what a rational number is? Heres the definition 
of limits,
for any epsilon > 0 blah blah blah...
S.E.: My bunghole it goes Raaabbabababa yeeeedadadadada
Bungholioooo!
(singing) Would you like to see my bunghoooole? You have TP?
TP for my bunghole? I am a gringo, and I have no bunghole...
and analysis asymptote (huh huh huh...I said ass) non standard
pi irrational dimensionless infinite 99999999 hyperreal ...
And it goes on and on. The only solution to the Beavis
problem was to
wait long enough for the caffein and sugar rush to run its
course and
he would calm down. I suppose if you all ignore S.E. for a
few days
hell move on to troll in other groups (didnt 
he say he was
a physics
genius too?), or maybe hell have found TP for his bunghole
and his
quest would come to a successful end.
-- L.
===
Subject: Re: .99999... still=/= 1
posting-account=AE-QyQ0AAAC84T96q9_yI_Fj9ThoZQPi
> Using the surreal symbol { | }
> (.5_ -) | --
> then sqrt (-1) = (.5_-)(1) = (1_-)
> ^
> |
> This shows half negative sign.
> then,
> (.5_- + .5_-)(1)^2 = -1
> or
> (1_-)^2 = -1
> or
> (1_+) - (+) = (1_-)
> or,
> ( 1 ) - (1_+) = (1_+)
> or,
> (1_+) + (1_-) = 0
> This would be sign and number value annihilation.
> etc...
This looks interesting.
Why not create a web page (a central repository) for this and
other
math researches?
===
Subject: Re: .99999... still=/= 1
In sci.math, Richard Henry
<rphenry@home.com>
<nzoxd.4691$ry.2005@fed1read01>:
>> Do you have any idea what ill-defined terms in math are
like i?
> When you say i, are you talking about the imaginary root of
x^2+1 =
0?
> That i is very well defined.
Be careful here. Theres an alternate i in small-signal
current
theory.
(In such, sqrt(-1) perverts to j.)
In any event, one can hypothesize an abstract entity i, and
impart thereto certain desirable characteristics such as i^2
= -1,
despite the issue that sqrt(-1) = +i or -i. The main issue:
consistency.
Of course numbers are ill-defined anyway; the best one can
state about
them is that, given the number 1, which is a very abstract
concept, one can derive most of, if not all of, the other
numbers (I
say most because there are some such as hyperreals and
quaternions
that Im simply not familiar enough with, though I know they
exist;
there are probably other numbers that Im even less familar
with and
therefore dont know how they relate to the existing number
tree
F[p]
/
or perhaps number chain N-W-J-Q-R-C-Quat).
/
A
where A = algebraic numbers, F[p] = the finite 
field of order p,
and W = N union {0}. Or something like that.
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: .99999... still=/= 1
> In sci.math, Richard Henry
> <rphenry@home.com <nzoxd.4691$ry.2005@fed1read01>:
>> Do you have any idea what ill-defined terms in math are
like i?
> When you say i, are you talking about the imaginary root of
x^2+1 =
0?
> That i is very well defined.
> Be careful here. Theres an alternate i in small-signal
current
theory.
> (In such, sqrt(-1) perverts to j.)
Yes, I know. i^2 is either -1 or P/R.
===
Subject: Re: Solving Sums of Exponential and Linear Terms?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiHv02866;
how do i solve this puzzle:-
Add all the numbers contained in the sum:-
63103
sixteen hundred thousand, thirteen thousand, nine hundred and
forty six
7918
_________________________________
= ?
This is a problem ive been given to solve and cant work out
how!
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiIp02939;
Darren E. Mason <stcalvin@comcast.net>
http://mathforum.org/discuss/sci.math/m/662603/663778
> I seen a number of HS graduates who cannot add 1/3 and 1/2
> without a calculator. And when they do they feel that 0.833
> is a better answer than 5/6.
Ive almost gotten to the point where Id be 
pleased
if a college student (not just a HS graduate) would
do this with a calculator. All too often I have students
who have committed the a/b + c/d = (a+c)/(b+d) sin enough
times (without serious consequences, I would presume)
that theyll put down 2/5 for 1/3 + 1/2 without even
considering that it might be something else. Actually,
many of them will use a calculator, but only for the
3+2 = 5 part.
Dave L. Renfro
===
Subject: Re: Taylor Series
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiJY02963;
Jon Slaughter <nobody@nowhere.com>
http://mathforum.org/discuss/sci.math/m/494296/663817
> Are you scared of the Gamma function? My method is much more
> intuitive than yours. While one without any ingenuity might
> be lost, if one can do x^k for integer k, there is no
> difficulty to extend this to non-integral k... but the
> same method applies... it is direct and intuitive and
> simple... what more could you ask for? Me thinks you are
> afraid of the gamma... just remember, it doesnt bite.
If you look in the appendix of just about any upper level
undergraduate physics or engineering text, youll see
expansions of functions such as sqrt(1 + u), ln(1 + u),
e^u, etc. They are there because many of the expressions
that you wind up needing first or second order approximations
to can be easily expanded using these standard expansions,
along with various algebraic devices.
For example,
sqrt[4*a^2 + mu*x^4] = 2a*sqrt[1 + (mu*x^4 / 4*a^2)]
= 2a*{expand with u = (mu*x^4 / 4*a^2)}
ln[ab + x^2] = ln(ab) + ln[1 + x^2/ab]
= ln(ab) + {expand with u = x^2/ab}
exp[a - b*x^2] = exp(a) * exp[-b*x^2]
= exp(a) * {expand with u = -b*x^2}
In the same way that one almost never computes derivatives
of polynomials, trig. functions, etc. by computing the limit
as h approaches zero of [f(x+h) - f(x)]/h in practice (you use
the product rule, chain rule, etc.), you almost never wind up
finding a power series expansion of a function by computing
and then evaluating derivatives. You only resort to this when
you encounter something for which the standard short-cuts
dont work.
By the way, Id bet a huge amount of money that Kovarik is
more familiar with the gamma function than you are.
Searched all groups Results 1 - 10 of about 59
for gamma function author:Kovarik
Dave L. Renfro
===
Subject: Re: Taylor Series
> Jon Slaughter <nobody@nowhere.com
http://mathforum.org/discuss/sci.math/m/494296/663817
>> Are you scared of the Gamma function? My method is much
more
>> intuitive than yours. While one without any ingenuity might
>> be lost, if one can do x^k for integer k, there is no
>> difficulty to extend this to non-integral k... but the
>> same method applies... it is direct and intuitive and
>> simple... what more could you ask for? Me thinks you are
>> afraid of the gamma... just remember, it doesnt bite.
> If you look in the appendix of just about any upper level
> undergraduate physics or engineering text, youll see
> expansions of functions such as sqrt(1 + u), ln(1 + u),
> e^u, etc. They are there because many of the expressions
> that you wind up needing first or second order 
approximations
> to can be easily expanded using these standard expansions,
> along with various algebraic devices.
> For example,
> sqrt[4*a^2 + mu*x^4] = 2a*sqrt[1 + (mu*x^4 / 4*a^2)]
> = 2a*{expand with u = (mu*x^4 / 4*a^2)}
> ln[ab + x^2] = ln(ab) + ln[1 + x^2/ab]
> = ln(ab) + {expand with u = x^2/ab}
> exp[a - b*x^2] = exp(a) * exp[-b*x^2]
> = exp(a) * {expand with u = -b*x^2}
> In the same way that one almost never computes derivatives
> of polynomials, trig. functions, etc. by computing the limit
> as h approaches zero of [f(x+h) - f(x)]/h in practice (you
use
> the product rule, chain rule, etc.), you almost never wind
up
> finding a power series expansion of a function by computing
> and then evaluating derivatives. You only resort to this
when
> you encounter something for which the standard short-cuts
> dont work.
> By the way, Id bet a huge amount of money that Kovarik is
> more familiar with the gamma function than you are.
> Searched all groups Results 1 - 10 of about 59
> for gamma function author:Kovarik
> Dave L. Renfro
and how much money would that be? Maybe we can have a contest
sometime and
you can put up the money?
===
Subject: Re: question for math teachers
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJLiI102943;
Chergarj <chergarj@cs.comhaho>
http://mathforum.org/discuss/sci.math/m/654317/663883
>> Try again. Under No Child Left Behind, a highly qualified
>> teacher has a graduate degree in their field or had taken a
>> test showing subject knowledge such as the Praxis or SSAT.
>> What state are you credentialed in where you get a full
>> credential with that background? In California, that would
>> get you a credential to teach introductory mathematics
>> (only through middle school), a far cry from being highly
>> qualified.
> Actually, the wording is bachelor degree or bacalaureate
> degree ... not graduate degree.
of the Anchorage School District web page -->
http://www.asd.k12.ak.us/NCLB/Teachfaq.asp
> Middle and High School teachers must be highly
> qualified in each core subject they teach.
> They can become highly qualified in any one
> of the following ways.
> Degree or advanced degree in the core subject
> College major in the core subject
> Major equivalent in the core subject
> (30 semester credits or the equivalent)
> Passing score on Praxis II test
> Passing score on the HOUSSE
> National Board Certification in the content area
Dave L. Renfro
===
Subject: Re: question for math teachers
> Chergarj <chergarj@cs.comhaho
http://mathforum.org/discuss/sci.math/m/654317/663883
> Try again. Under No Child Left Behind, a highly qualified
> teacher has a graduate degree in their field or had taken a
> test showing subject knowledge such as the Praxis or SSAT.
> What state are you credentialed in where you get a full
> credential with that background? In California, that would
> get you a credential to teach introductory mathematics
> (only through middle school), a far cry from being highly
> qualified.
>> Actually, the wording is bachelor degree or bacalaureate
>> degree ... not graduate degree.
> of the Anchorage School District web page --
http://www.asd.k12.ak.us/NCLB/Teachfaq.asp
>> Middle and High School teachers must be highly
>> qualified in each core subject they teach.
>> They can become highly qualified in any one
>> of the following ways.
>> Degree or advanced degree in the core subject
>> College major in the core subject
>> Major equivalent in the core subject
>> (30 semester credits or the equivalent)
>> Passing score on Praxis II test
>> Passing score on the HOUSSE
>> National Board Certification in the content area
To be Ôhighly qualified, a teacher 
must be fully certified (no
emergency
certifications), have a bachelors degree, and 
have
demonstrated subject
matter competency in each of the core academic subjects in
which the
teacher teachers.
Listed above are the ways Alaska allows secondary teachers to
show subject
matter competency.
This shows one of the problems with discussing the
highly-qualified
definition provided by the Feds. The credentialing of teachers
is left up
to the states so each state must provide the US Dept of
Education with their
own definition of an HQ teacher and it must be approved by the
DOE. The
requirement of a graduate degree or testing is accepted by
most (maybe all)
states yet other states may have other ways to show subject
knowledge (like
national certification or a bachelors degree).
realized that some secondary teachers (mostly science and
special ed
teachers) teach multiple subjects and it was unreasonable to
expect them to
be highly qualified in EVERYTHING they taught. Now states are
allowed
alternative ways to establish HQ criteria but the states have
not had
time
yet to emplace anything.
Remember, the OP said that his state allows a teacher to be
highly qualified
without a math degree (I believe he said one could be 3 units
shy) and
without knowing what state he is in we have no way to see
what his state
requires.
===
Subject: Re: question for math teachers
>> Chergarj <chergarj@cs.comhaho>
http://mathforum.org/discuss/sci.math/m/654317/663883
>> Try again. Under No Child Left Behind, a highly qualified
>> teacher has a graduate degree in their field or had taken a
>> test showing subject knowledge such as the Praxis or SSAT.
>> What state are you credentialed in where you get a full
>> credential with that background? In California, that would
>> get you a credential to teach introductory mathematics
>> (only through middle school), a far cry from being highly
>> qualified.
> Actually, the wording is bachelor degree or bacalaureate
> degree ... not graduate degree.
>> of the Anchorage School District web page -->
http://www.asd.k12.ak.us/NCLB/Teachfaq.asp
> Middle and High School teachers must be highly
> qualified in each core subject they teach.
> They can become highly qualified in any one
> of the following ways.
> Degree or advanced degree in the core subject
> College major in the core subject
> Major equivalent in the core subject
> (30 semester credits or the equivalent)
> Passing score on Praxis II test
Here is a link with practice Praxis tests.
http://www.ets.org/praxis/prxpa.html
In particular, here is a link with the practice test for math
for
secondary education math teachers.
ftp://ftp.ets.org/pub/tandl/0061.pdf
Brian
===
Subject: Average, standard deviation and maximum estimation
http://mygate.mailgate.org/mynews/sci/sci.math/
23b054917ec4bc2620578bd35541bf
a8.62412%40mygate.mailgate.org
I hope my question is not OT:
having a set of numeric values X_1, ... X_N
I can compute the average value A and the standard deviation
S.
But if I have only A, S and the number N of the values in the
set, exist
a method for the estimation of the maximum (or the minimum)
value of the
set?
replies.
Robert
--
===
Subject: Re: Average, standard deviation and maximum
estimation
> I hope my question is not OT:
> having a set of numeric values X_1, ... X_N
> I can compute the average value A and the standard
deviation S.
> But if I have only A, S and the number N of the values in
the set, exist
> a method for the estimation of the maximum (or the minimum)
value of the
> set?
Well, the way you write the problem, the answer is no. But if
you add an
additional hypothesis that all the values are *independent*
samples from
the
*same* distribution, then there is progress.
Assume you know in advance the exact probability
distribution. Let D(x) be
the cumulative distribution function, and let xN be given by
D(xN) = 1 -
1/N. Then xN is close to the maximum of (X_1, ..., X_N). More
specifically, in the limit N goes to infinity, 
(max(X) - xN) is
approximately normal distributed, irrespective of the original
distribution.
However, I dont remember the details, so 
youll have to
consult a textbook
on probability theory. This is pretty basic stuff.
-Michael.
> replies.
> Robert
> --
===
Subject: Re: Average, standard deviation and maximum
estimation
http://mygate.mailgate.org/mynews/sci/sci.math/
7e9614afea6e5e0b50c068ff62db6b
56.48257%40mygate.mailgate.org
> I hope my question is not OT:
> having a set of numeric values X_1, ... X_N
> I can compute the average value A and the standard
> deviation S. But if I have only A, S and the
> number N of the values in the set, exist a method
> for the estimation of the maximum (or the minimum)
> value of the set?
> replies.
> Robert
There isnt a very pretty one, since the problem is
underdetermined. You can establish a range for
your answer by adding sufficient constraints to make
the problem sufficiently determined.
Consider the cases (1) where all but one sample in
the set are identical, and the remaining sample
directly compute how far away that single sample is
from the others, given only A, S, and N, and that
will at least tell you the extreme separation value
that tailmost element could have from the group
average to achieve the given standard deviation.
If you then (2) oppositely clump the samples into
two as equal as possible heaps taking on exactly
two different values, you can again calculate, given
only A, S, and N, the distance from one heap to the
other. This gives you the _minimum_ separation of a
tailmost element from the group average to achieve
the given standard deviation.
All you can say in addition from only A, S, and N,
is that the actual extremal elements lie somewhere,
in the range from the group average, between the
minimum possible and maximum possible distances you
have computed.
HTH
xanthian.
--
===
Subject: Re: The State-of-the-Art in Mathematics, but didnt
I solve this?
(Smart1234)
>>The problem here is not the choice of values of sqrt(-1)
but the concept
i
>>itself. It is not well-defined because the mapping sqrt is
well-defined
only
>>on perfect square such as 1, 4, etc.
>>E. E Escultura
>Ok can it be represented like this?
>n=1 to oo
>with cartesian coordinates x,y,z
>Use only integer powers of n
>otherwise its an indeterminate
>[(-1)^n/2] x
>[(-1)^n/2] y
>[(-1)^n/2] z
>For x coordinate,
>[(-1)^n/2] x
>at n=1 --> indeterminate
>at n=2 --> -1
>at n=3 --> indeterminate
>at n=4 --> 1
>at n=5 ---> indeterminate
>at n=6 ---> -1
>at n=7 ---> indeterminate
>at n=8 ---> 1
>etc...
> In cases where you have sqrt(-2), etc..,
>for n=1 --> oo
>Integer Powers of n Only
>sqrt (-2) = sqrt(2) [(-1)^n/2)]
> = (1.41421...) [(-1)^1/2)]
> = (1.4142...) ( indeterminate)
>And,
>(sqrt (-2))^2 = (1.41421...)^2 [(-1)^2/2)]
> = (2)(-1)
> = -2
>And,
>(sqrt(-2))^3 = (1.41421...)^3 [(-1)^3/2)]
> = (2.8284...) (indeterminate)
>And,
>(sqrt(-2))^4 = 3.9998...
>And so on...
> Does this map sqrt correctly with negative sqrts?
Why dont you reply? Why doesnt anyone reply? 
Whats wrong
you dont like
to
give me credit for anything?
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: More on the State-of-the-Art in Physics
>MORE ON THE STATE-OF-THE-ART IN PHYSICS
I guess you didnt even notice that I did a complete 
overhaul
of physics.
Whats wrong you dont want to give me credit 
for anything?
You want to rename it something else like the others did to
steal my
theories on the Helix Spiral Spinning Field Theory showing
internal and
external structure of the atom and and the universe.
Oh, you want ot call it the QM or String Vortex Model? My
work is about
10
years ahead of millions of the best minds in physics put
together working
constantly together and they still dont even know what mass
is at the
sub-atomic level. Whats wrong with you people you need a 
job
that bad? So
you
keep me out of it.
Even the Japanese technology in electronic engineering is
just barely
catching up with me at the sub-atomic level in physics. Oh
just think by
the
just barely understand where I am at.
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: poincare-bendixson theorem
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJNN7Y10543;
i am working through some problems in strogatz book to
prepare for my final, and there is a problem in the
section poincare-bendixson theorem that i do not
understand (though i do not believe i need to use
the theorem to solve the problem).
given the system
dot x = x (1 - 4x^2 - y^2) - 1/2 y (1 + x)
dot y = y (1 - 4x^2 - y^2) - 2 x (1 + x)
i want to show that all solutions approach the ellipse
4x^2+y^2=1 as t --> infinity.
the hint is to compute dot V where V(x,y) = (1 - 4x^2 -
y^2)^2 .
i do not want the answer from you, i just want to know
how what this hint means and how to use it. how does V
play any role in this problem and how was it constructed?
my work:
i can easily see that the origin is an unstable fixed point,
and perhaps if i parameterize the ellipse i can show that
any initial condition on the ellipse stays on the ellipse
(shouldnt be too hard). so by the poincare-bendixson theorem
i believe i have that all initial conditions inside the
ellipse
approach the ellipse as t goes to infty. however, i also
need to show that initial conditions outside of ellipse go
to ellipse. that is as far as i can think.
please provide assistance
===
Subject: Re: Ring problem, integral elements
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJNN7T10558;
>Excuse-me in my question R is not assumed to be commutative.
and the
>correct (complete) question is the following (I think!!):
>Suppose that R is a ring with identity such that the centre
Z(R) of R,
>is a field. Assume that x and y are two commuting
>elements of R (i.e. xy=yx) such that
>x^5+a=0
>y^5+a=0
>for some a in Z(R). In fact x and y are integral over
>Z(R).
>Question: Construct a non-zero polynomial f with coeficceint
in Z(R)
>such that f(x-y)=0.
We may restrict attention to the commutative ring Z(R)[x, y],
and solve the problem universally by replacing Z(R) by Z[a],
a an indeterminate (Z the integers), and Z(R)[x, y] by the
extension
E = Z[x, y, a]/(x^5 = -a = y^5) over Z[a].
If we find a monic polynomial for x - y with 
coefficients in
Z[a], then we are done (since we can just apply some obvious
rings homomorphisms Z[a] --> Z(R), E --> R to the equation
f(x-y) = 0 in E).
Obviously y/x is a fifth root of unity (denoted w) over the
field Q(a), so
(x - y)^5 = a(w - 1)^5
and the problem reduces to finding a monic polynomial (over Z)
for (w - 1)^5. This is routine if a bit tedious; one works
over the complex numbers and calculates the coefficients of
(u - r1)(u - r2)(u - r3)(u - r4)
to the fact that the ri are pure imaginary and come in
conjugate
pairs; my back-of-envelope calculations give the last
polynomial
as
u^4 + 3936u^2 + 3125
Substituting u = (x - y)^5/a and clearing denominators, I get
(x - y)^20 + 3936a^2(x - y)^10 + 3125a^4 = 0.
Todd Trimble
===
Subject: Re: Ring problem, integral elements
posting-account=4za6cQwAAACWvvpq0N1XsH01GJ8y5PED
The obtained polynomial does not work for the case x=y=-1 and
a=1 (in
the ring of complex numbers) that is, in this case
f(1-1)=f(0)=/=0
also for x= 1/4+1/4*5^(1/2)-1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)
and y=1/4+1/4*5^(1/2)+1/4*I*2^(1/2)*(5-5^(1/2))^(1/2)
(I=sqrt(-1))
and where a=1 (in fact x and y are two distinct fifth root of
unity)
we have
f(x-y)=-2069375/2+910525/2*5^(1/2)=/=0
These show that, maybe somthing is wrong in your
calculations, but
perhaps I can find out it ...........???
Anyway thank you very much for your information.
Alireza
>Excuse-me in my question R is not assumed to be commutative.
and
the
>correct (complete) question is the following (I think!!):
>Suppose that R is a ring with identity such that the centre
Z(R) of
R,
>is a field. Assume that x and y are two commuting
>elements of R (i.e. xy=yx) such that
>x^5+a=0
>y^5+a=0
>for some a in Z(R). In fact x and y are integral over
>Z(R).
>Question: Construct a non-zero polynomial f with coeficceint
in
Z(R)
>such that f(x-y)=0.
> We may restrict attention to the commutative ring Z(R)[x,
y],
> and solve the problem universally by replacing Z(R) by Z[a],
> a an indeterminate (Z the integers), and Z(R)[x, y] by the
extension
> E = Z[x, y, a]/(x^5 = -a = y^5) over Z[a].
> If we find a monic polynomial for x - y with 
coefficients in
> Z[a], then we are done (since we can just apply some obvious
> rings homomorphisms Z[a] --> Z(R), E --> R to the equation
> f(x-y) = 0 in E).
> Obviously y/x is a fifth root of unity (denoted w) over the
> field Q(a), so
> (x - y)^5 = a(w - 1)^5
> and the problem reduces to finding a monic polynomial (over
Z)
> for (w - 1)^5. This is routine if a bit tedious; one works
> over the complex numbers and calculates the coefficients of
> (u - r1)(u - r2)(u - r3)(u - r4)
> to the fact that the ri are pure imaginary and come in
conjugate
> pairs; my back-of-envelope calculations give the last
polynomial
> as
> u^4 + 3936u^2 + 3125
> Substituting u = (x - y)^5/a and clearing denominators, I
get
> (x - y)^20 + 3936a^2(x - y)^10 + 3125a^4 = 0.
> Todd Trimble
===
Subject: Re: probability density and invariant measures
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJNN8e10576;
my message came out a little weird so I will fix it here:
I am trying to understand a chapter in a book that
is probably too difficult for me to read, and i was
wondering if someone could help me weed through a part.
the author considers the logistic map x_{n+1} = 4x_n (1 - x_n)
states that the density p(x) of an invariant measure
for the logistic map is
p(x) = (pi sqrt{x (1 - x)})^{-1}
I barely even understand the idea of an invariant measure,
so that is a little hard for me to comprehend, especially
when stated without proof or explanation. could someone
tell me how this is possible?
===
Subject: inner products
Part a went well but part b is not clear to me at all. Any
hints please?
(a) Suppose that rho(x) is a positive Riemann-integrable
function [a,b]-->
R, i.e. rho(x) > 0 for all x in [a, b].
Let V be the vector space consisting
of all Riemann integrable functions [a, b]-->R. For f, g in V
define
< f, g >= integral from a to b of rho(x)f(x)g(x)dx
Show that <,> defines an inner product on the quotient space
V =V / Ô,
where f Ô g if, for some meager subset X subset of [a, b],
f(x) = g(x)
for x not in X.
(b) Consider the differential equation
(p(x)u(x)) + q(x)u(x) + L*rho(x)u(x) = 0
where p(x), q(x),rho(x) are infinitely differentiable
functions on an
interval
[a, b], rho(x) is positive, and L is a scalar. We also impose
boundary conditions c_1u(a) + c_2u(a) = 0 (1)
c_3u(b) + c_4u(b) = 0 (2)
where c1, c2, c3, c4 are scalars, and at least one of c1, c2
and one of c3,
c4 is
non-zero on our solutions u(x). Suppose that u_i(x) is a
solution of this
equation for L = L_i
(i = 1, 2) and L_1 /=L_2. Show that u_1(x) and u_2(x) are
orthogonal
with respect to the inner product in part (a).
===
Subject: Re: inner products
> Part a went well but part b is not clear to me at all. Any
hints please?
> (a) Suppose that rho(x) is a positive Riemann-integrable
function
[a,b]-- R, i.e. rho(x) > 0 for all x in [a, b].
> Let V be the vector space consisting
> of all Riemann integrable functions [a, b]-->R. For f, g in
V define
> < f, g >= integral from a to b of rho(x)f(x)g(x)dx
> Show that <,> defines an inner product on the quotient space
V =V / Ô,
> where f Ô g if, for some meager subset X subset of [a, b],
f(x) = g(x)
> for x not in X.
> (b) Consider the differential equation
> (p(x)u(x)) + q(x)u(x) + L*rho(x)u(x) = 0
> where p(x), q(x),rho(x) are infinitely differentiable
functions on an
> interval
> [a, b], rho(x) is positive, and L is a scalar. We also
impose
> boundary conditions c_1u(a) + c_2u(a) = 0 (1)
> c_3u(b) + c_4u(b) = 0 (2)
> where c1, c2, c3, c4 are scalars, and at least one of c1,
c2 and one of
c3,
> c4 is
> non-zero on our solutions u(x). Suppose that u_i(x) is a
solution of this
> equation for L = L_i
> (i = 1, 2) and L_1 /=L_2. Show that u_1(x) and u_2(x) are
orthogonal
> with respect to the inner product in part (a).
Think: Linear Algebra. Consider a real symmetric square
matrix A with
eigenvalues a1 and a2 and corresponding eigenvectors x1 and
x2, such that:
A x1 = a1 x1
and
A x2 = a2 x2.
Now show that when the eigenvalues a1 and a2 are different,
the
eigenvectors
x1 and x2 are orthogonal with respect to the matrix A, i.e.
x1 A x2 = 0.
This problem is perhaps easier to deal with. The original
problem follows
along exactly the same lines, with L instead of A, u instead
of x, etc.
-Michael.
===
Subject: Re: Zenkins paper on Cantor
<574ll014d1dtk5ugu8do2tmdt0lia9ed7j@4ax.com>
<cjeq18$890$1@sunbeam.coventry.ac.uk>
<cjgjv9$1sf$1@phys-news1.kolumbus.fi>
<vcb7jq8i48w.fsf@beta19.sm.ltu.se>
posting-account=sASPfg0AAAB32ck2Ys0CgbD_dk7GKPYH
I plan to show that Platonist interpretation of mathematics
deals in a
rather unrealistic kind of fiction.
===
Subject: Re: Zenkins paper on Cantor
> I plan to show that Platonist interpretation of mathematics
deals in a
> rather unrealistic kind of fiction.
The best laid schemes o Mice an Men, Gang 
aft agley...
Which are you?
===
Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin)
<fGqnd.428382$D%.181574@attbi_s51>
<419e3b4b$0$560$b45e6eb0@senator-bedfellow.mit.edu>
<jlrnd.359997$wV.99162@attbi_s54>
<419e7742$0$562$b45e6eb0@senator-bedfellow.mit.edu>
<d7utd.32207$dC3.693497@news20.bellglobal.com>
posting-account=sASPfg0AAAB32ck2Ys0CgbD_dk7GKPYH
nomological: possible worlds with same/similar physical laws.
Chow would think about.
Mathematics has a metaphysical component, but that
metaphysical
component must be limited to nomologically possible worlds,
and no more
than that should be allowed in the semantics of axioms.
That means, things like infinity should not be dealt as if
they were
actual things.
There are lists in the world. There are no infinite lists in
the world.
--
Eray
===
Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin)
> nomological: possible worlds with same/similar physical
laws.
[...]
Lately youve been posting without quotes, so the context of
your
comments is not clear.
===
Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin)
<fGqnd.428382$D%.181574@attbi_s51>
<419e3b4b$0$560$b45e6eb0@senator-bedfellow.mit.edu>
<jlrnd.359997$wV.99162@attbi_s54>
<419e7742$0$562$b45e6eb0@senator-bedfellow.mit.edu>
<d7utd.32207$dC3.693497@news20.bellglobal.com>
<Virxd.3219$GK5.344227@news20.bellglobal.com>
posting-account=sASPfg0AAAB32ck2Ys0CgbD_dk7GKPYH
have quotes by default.
You were asking: what does nomologically possible worlds
mean? in
post 52 on Dev 7
You commented that unbounded and infinite seem to be
conceptually
disparate.
--
Eray Ozkural
===
Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin)
> have quotes by default.
> You were asking: what does nomologically possible worlds
mean? in
> post 52 on Dev 7
> You commented that unbounded and infinite seem to be
conceptually
> disparate.
> --
> Eray Ozkural
post.)
Im still uncertain about unbounded vs infinite. 
Eg, the
surface of a
sphere is unbounded, but it is clearly finite. Unbounded seems
to
refer to the fact that any curve/path on it an be extended
without
limit. By contrast the sphere is both bounded and finite,
since some
curves/paths within it will intersect the boundary (the
surface) when
extended far enough. There is an infinite number of such
paths, as there
is of closed paths within the sphere. All this seems to be
intuitive
once you understand the notion of indefinitely extending a
curve/path.
Ie, in this case, unbounded and infinite are clearly
conceptually distinct.
But you want to extend un/bounded and in/finite to
computation; just
what exactly are you doing? Its not intuitive, at least to
me. Eg, in
any code with a well defined syntax it is possible to extend 
an
expression indefinitely. (The fact that in the real world we
dont do
this is not germane, I think.) If the code is a programming
language,
any program is an expression in the sense used here. That
makes any such
programming code -- what? Unbounded? Infinite? 
Thats not
clear to me.
But I do intuit that its related to the halting problem. Is
that right?
Um, it occurs to me that an AI machine must be capable of
deciding to
stop extending an expression, else it can never act (and we
surely want
an AI machine to act.) In humans, there appear to be brain
structures
that prevent indefinite extension of reasoning, as is
suggested by
observations of brain-damaged people who cannot make
decisions, yet can
reason clearly and at length about the consequences of
decisions.
===
Subject: Re: Zenkins paper on Cantor (reply of Dr. Zenkin)
>> nomological: possible worlds with same/similar physical
laws.
> [...]
> Lately youve been posting without quotes, so the context
of your comments
> is not clear.
> I have five philosophical questions for you. Please try to
be pedantic
> in answering them. These are only some of the key questions
in the
> debate of unbounded vs. infinite. There are others.
> Basically I am asking you: In your opinion, what does
Ôunbounded
> mean?
> 0. (preliminary question) Can the sequential behavior of
input devices
> over an unspecified amount of time be described as one of
the tapes of
> a multi-tape TM? (Yes.)
> 1. In an infinite universe, does unbounded computation
necessarily
> entail infinite space or time?
No.
> 2. In a finite universe, does unbounded computation
necessarily entail
> infinite space or time?
No.
> 3. In a finite universe with a certain heat death, does
unbounded
> computation necessarily entail infinite space or time?
No.
> 4. In any nomologically possible world, does unbounded
computation
> necessarily entail infinite space or time?
???? nomological is unknown to me.
>> nomological: possible worlds with same/similar physical
laws.
Un/bounded has to do with topolgy. In/finite has to do with
measure.
But I dont think these statements are at all clear, even
though they
seem (at least to me) intuitively right.
===
Subject: Re: .99999... still=/= 1
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBJNuge12787;
S. Enterprize Company <smart1234@aol.com>
http://mathforum.org/discuss/sci.math/m/651356/664008
> I already answered that. First of all pi
>
ISSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS
SSSS
> an irrational number, the decimal values vary. It can not be
> determined. But with a repeating decimal, you can use a
> non-standard approach and use one digit less than oo then,
<< decimal values vary >> -- Do you mean the digits can vary,
the values of the digits can vary (that would be weird),
or the value represented by the totality of the decimal
digits in Pi can vary? The first would seem to contradict
the fact that calculating Pi is often used to test certain
aspects of new computers and software, the second makes no
sense at all, and the third contradicts that fact that Pi
is a mathematical constant.
<< It can not be determined. >> -- Then how are computers
(and humans before the middle of the 20th century) able
to calculate the digits in the decimal expansion of Pi?
<< But with a repeating decimal, you can use a non-standard
approach and use one digit less than oo then, ... >>
This makes about as much sense as:
With a walking bird [1], you can vacuum the ßoor [2]
and green will be infinitely heavy [3], ...
[1] Although idea of a repeating decimal makes sense,
most decimals arent repeating. The same with birds
and walking.
[2] Vacuuming the ßoor has about as much to do with a
bird walking as involking non-standard reals when
discussing repeating decimals.
[3] Look up the definition of category mistake as 
its
used in philosophy.
Dave L. Renfro
===
Subject: Eulers identity
posting-account=wq-NGgwAAAA1ZIl6cM1Jxy8ueAWJvWu6
Eulers identity is;
exp(j*theta) = cos(theta) + j*sin(theta).
What about the expression
exp(j*ln(theta))
Are there any interesting realtionships that can be derived
for this?
Bob Adams
===
Subject: Re: Eulers identity
> Eulers identity is;
> exp(j*theta) = cos(theta) + j*sin(theta).
> What about the expression
> exp(j*ln(theta))
> Are there any interesting realtionships that can be derived
for this?
> Bob Adams
I thought that:
exp(j*theta) = cos(theta) + j*sin(theta)
was due to de Moivre. As for a more specific answer to the
question posed,
try starting with the identity:
k*ln(y) = ln(y^k)
and then see whether this leads anywhere. (It does, but just
what do you
want out of it?)
John
johnDOTmorrisonATtescoDOTnet
PS I _assume_ that your j means sqrt(-1) here - and hence
that you are
trained in engineering?
--
SpySubtract Readme.txt file
===
Subject: Re: Eulers identity
> I thought that:
> exp(j*theta) = cos(theta) + j*sin(theta)
> was due to de Moivre.
I used to think the same way, but now I belive that the de
Moivre is the
identity
(cos(theta) + j*sin(theta))^n = cos(n theta) + j*sin(n theta)
===
Subject: Re: Eulers identity
<32ml6pF3opunrU1@individual.net>
<Rppxd.683577$mD.145833@attbi_s02>
posting-account=wq-NGgwAAAA1ZIl6cM1Jxy8ueAWJvWu6
Yes, I am trained in engineering, hence the j (sorry!).
I am trying to take the Fourier Transform of a discrete-time
sequence
where the time indexes are t=log(N), N an integer. Thus the
usual
integral of f(t)*exp(-j*w*t) becomes a summation of
f(N)*exp(-j*w*ln(N)).
===
Subject: Re: String Theory: Good, Bad and Bogus
posting-account=sASPfg0AAAB32ck2Ys0CgbD_dk7GKPYH
Well, you know Dr. Sarfatti, there is another newsgroup:
sci.physics.discrete
> Part 1
> Words written by Dennis Overbye are between quote marks.
> a single equation that could explain all the laws of
physics, all
the
> forces of nature - the proverbial Ôtheory of 
everything ?
> Extraordinary claims require extraordinary proof.
Physicists, like
Dr.
> Robert Park, use a double standard not applying the same
rules of
> engagement to fashionable elegant string theory as they do
to ßying
> saucers, the paranormal and cold fusion.
> http://www.archivefreedom.org/
> And so emerged into the limelight a strange new concept of
nature,
> called string theory, so named because it depicts the basic
constituents
By
> uniting all the forces, string theory had the potential of
achieving
the
> goal that Einstein sought without success for half his life
and that
has
> embodied the dreams of every physicist since then. If true,
it could
be
> used like a searchlight to illuminate some of the deepest
mysteries
> physicists can imagine, like the origin of space and time
in the Big
> Bang and the putative death of space and time at the
infinitely dense
> centers of black holes. ... In the last 20 years, string
theory has
> become a major branch of physics. Physicists and
mathematicians
> conversant in strings are courted and recruited like star
quarterbacks
> by universities eager to establish their research
credentials. String
> theory has been celebrated and explained in best-selling
books like
The
> Elegant Universe, by Dr. Brian Greene, a physicist at
Columbia
> University, and even on popular television shows.
> ÔLet them eat cake said Marie Antoinette 
shortly before
she was
> guillotined.
> even as they ate cake and drank wine, the string theorists
admitted
> that after 20 years, they still did not know how to test
string
theory,
> or even what it meant.
> Note or even what it meant. Mainstream theoretical physics
today is
in
> a sorry state. Not so for experimental physics.
> As a result, the goal of explaining all the features of the
modern
> world is as far away as ever, they say. And some physicists
outside
the
> string theory camp are growing restive. At another meeting,
at the
Aspen
> Institute for Humanities, only a few days before the string
> commemoration, Dr. Lawrence Krauss, a cosmologist at Case
Western
> Reserve University in Cleveland, called string theory Ôa
colossal
failure.
> Brian Greene got a few million dollars advance for his
book. The big
> corporations have a vested interest in this hype that can
be compared
to
> WMD in Iraq.
> String theorists agree that it has been a long, strange
trip, but
they
> still have faith that they will complete the journey.
> The fusion of science and religion on the heels of the
fusion of
Church
> and State?
> Twenty years ago no one would have correctly predicted how
string
> theory has since developed, said Dr. Strominger of Harvard.
> ÔThere is disappointment that despite all our efforts,
experimental
> verification or disproof still seems far away. On the other
hand, the
> depth and beauty of the subject, and the way it has reached
out,
> inßuenced and connected other areas of physics and
mathematics, is
> beyond the wildest imaginations of 20 years ago. In a 
way,
the
story
> of string theory and of the physicists who have followed
its siren
song
> for two decades is like a novel that begins with the
classic what
if?
> What if the basic constituents of nature and matter were
not little
> points, as had been presumed since the time of the Greeks?
What if
the
> seeds of reality were rather teeny tiny wiggly little bits
of string?
> merely correspond to different ways for the strings to
vibrate,
> different notes on Gods guitar?
> Ill play the music on my guitar, yes! Rossini, The Barber
of
Seville.
> That would explain why you cannot have a single quark you
cannot
have a
> string with only one end. Strings seduced many physicists
with their
> mathematical elegance, but they had some problems, like
requiring 26
to
> have anything to do with quarks or the strong force.
> You need the extra space dimensions because string theory
violates
> signal locality in Einsteins original 4D curved 
spacetime.
That is,
> final causes after their effects are possible in plain
vanilla 4D
string
> theory with topology change. Of course, this is exactly
what the UFO
and
> paranormal evidence is telling us along with presponse
experiments
by
> Dick Bierman and others. Then there is also the Intelligent
Design
> controversy. What is bad and bogus for string theorists is
good for
UFO,
> paranormal and consciousness researchers. One mans meat 
is
another
> mans poison.
> When accelerator experiments supported an alternative
theory of
quark
> behavior known as quantum chromodynamics, most physicists
consigned
> strings to the dustbin of history. But some theorists
thought the
> mathematics of strings was too beautiful to die. In 1974
Dr. Schwarz
> and Dr. Joel Scherk from the .83cole Normale Sup.8erieure
in France
noticed
the
> responsible for transmitting gravity in a quantum theory of
gravity,
if
> such a theory existed.
> This pyrrhic victory has been trumpeted as a great triumph
showing
the
> power of fantasy in the minds of string theorists.
Einsteins gravity
is
> not renormalizable as a quantum field theory which strongly
suggests
> that it should not be quantized top -> down like
electro-weak dynamic
> and chromodynamics. Indeed, Andrei Sakharov suggested that
Einsteins
> gravity emerges bottom -> up as a collective phenomenon
just like the
> elasticity in crystals. P.W. Anderson has since formalized
this idea
as
> More is different. Furthermore all current tests for grainy
quantum
> gravity foam that should show dispersion in the high energy
gamma
rays
> from outer space have been null. That is, there is no
evidence for
the
> equation
> E^2 = [(pc)^2 + mc^2]/[1 + (pc/mPc^2)^2]
> where mP is the Planck mass for the quantum gravity
graininess of
> Einsteinsw spacetime geometrodynamics.
> Note that the high energy limit of this equation is
> E -> mPc^2 ~ 10^-5 grams c^2 ~ 10^16 ergs ~ 10^9 Joules ~
10^28
electron
> volts
> implying a maximal acceleration of
> a max = c^2/LP ~ 10^21/10^-33 ~ 10^54 cm/sec^2 ~ 10^51g
> But, so far, no evidence supports this idea. If my theory
of the
> emergence of Einsteins gravity of MACRO-QUANTUM curved
spacetime
from
> the PARTIAL cohering of the micro-quantum zero point vacuum
ßuctuations
> in the BCS unstable massless pre-inßationary globally ßat
false
> conformal vacuum is correct, there is no quantum foam in
principle.
> non-perturbative background-independent MACRO-QUANTUM vacuum
condensate
> of virtual electron-hole pairs bound by virtual photons
near the E =
0
> Fermi energy of the pre-inßationay false vacuum obeys
> E^2 = (pc)^2 + Egap^2
> Egap ~ (Debye energy of edge of Fermi surface)e^-1/Vrho(0)
> V is the attractive static potential energy between the
virtual
> electron-hole (hole = positron) pair from a single virtual
photon
> exchange. rho(0) is the density of Diracs negative energy
virtual
> electron states at the edge per unit energy.
> The Einstein-Cartan tetrads eu^a derive from the
MACRO-QUANTUM
COHERENT
> different order parameter.
> guv = eu^anabev^b = nuv + (1/2)Lp^2(PHASE OF VACUUM
COHERENCE){,u,v}
> = nuv + (1/2)[Eu,v + Ev,u)
> { } is symmetrizer (i.e. anti-commutator) with ,u ordinary
partial
> derivatives in a local coordinate patch of the manifold.
> eu^a = &u^a + Eu^a
> &u,^a = Kronecker-Delta i.e. 1 if u = a, 0 if u =/= a
> Eu = Eu^a,a = Lp^2(PHASE OF VACUUM COHERENCE),u
> This equation is the analog to the deBroglie-Bohm guidance
equation
of
> IT by BIT for a quantum liquid. Here we have an ELASTIC
ODLRO
quantum
> solid that I introduced into physics in 1966-67 in Physics
Letters A
on
> Super Solids in Helium when I was on the physics faculty of
San
Diego
> State and the creator of the National Science Foundation
Summer
School
> in Superßuid Physics and Lasers for College Teachers at San
Diego
State
> (1969 & 1970) that I ran with Herschel Snodgrass. The
faculty
included
> F.W. Cummings and Jim Johnston.
> The QUANTUM OF AREA Lp^2 = hG/c^3 ensures Hawkings (also
Bekenstein,
> Susskind, tHooft)
> Entropy of Volume of Space = (Bounding Area of Volume)/4Lp^2
> Space-time physics is local because the macro-quantum order
parameter
is
> local. This same locality ensures that the early post-Big
Bang
universe
> after the inßationary vacuum phase transition has low
entropy so
that
> the Arrow of Time for the irreversible processes of the
Second Law of
> Thermodynamics points in the same direction as the presently
> accelerating expansion of the 3D space of our universe from
residual
w =
> -1 zero point dark energy of negative quantum pressure.
Hawkings
> blackhole entropy formula is automatically obeyed. That is
considered
a
> triumph of string theory. Its no big deal really. Sorry
for stealing
> string theorists thunder and upstaging them, but as 
Ludwig
Boltzmann
> said Elegance is for tailors. Not that this new way of
connecting
the
> cosmic dots is not elegant to the max. :-)
> That is, the pre -> post inßationary Big Bang phase
transition
> collapses the phase space volume of the ground state and the
entropy
> is ~ kBlog(Phase Space Volume). This same collapse of
ground state
phase
> space volume always accompanies More is different emergence
of
> qualitatively CREATIVE new order, AKA spontaneous breakdown
of
vacuum
> symmetry.
> To be continued.
===
Subject: Re: String Theory: Good, Bad and Bogus
posting-account=sASPfg0AAAB32ck2Ys0CgbD_dk7GKPYH
Well, you know Dr. Sarfatti, there is another newsgroup:
sci.physics.discrete
> Part 1
> Words written by Dennis Overbye are between quote marks.
> a single equation that could explain all the laws of
physics, all
the
> forces of nature - the proverbial Ôtheory of 
everything ?
> Extraordinary claims require extraordinary proof.
Physicists, like
Dr.
> Robert Park, use a double standard not applying the same
rules of
> engagement to fashionable elegant string theory as they do
to ßying
> saucers, the paranormal and cold fusion.
> http://www.archivefreedom.org/
> And so emerged into the limelight a strange new concept of
nature,
> called string theory, so named because it depicts the basic
constituents
By
> uniting all the forces, string theory had the potential of
achieving
the
> goal that Einstein sought without success for half his life
and that
has
> embodied the dreams of every physicist since then. If true,
it could
be
> used like a searchlight to illuminate some of the deepest
mysteries
> physicists can imagine, like the origin of space and time
in the Big
> Bang and the putative death of space and time at the
infinitely dense
> centers of black holes. ... In the last 20 years, string
theory has
> become a major branch of physics. Physicists and
mathematicians
> conversant in strings are courted and recruited like star
quarterbacks
> by universities eager to establish their research
credentials. String
> theory has been celebrated and explained in best-selling
books like
The
> Elegant Universe, by Dr. Brian Greene, a physicist at
Columbia
> University, and even on popular television shows.
> ÔLet them eat cake said Marie Antoinette 
shortly before
she was
> guillotined.
> even as they ate cake and drank wine, the string theorists
admitted
> that after 20 years, they still did not know how to test
string
theory,
> or even what it meant.
> Note or even what it meant. Mainstream theoretical physics
today is
in
> a sorry state. Not so for experimental physics.
> As a result, the goal of explaining all the features of the
modern
> world is as far away as ever, they say. And some physicists
outside
the
> string theory camp are growing restive. At another meeting,
at the
Aspen
> Institute for Humanities, only a few days before the string
> commemoration, Dr. Lawrence Krauss, a cosmologist at Case
Western
> Reserve University in Cleveland, called string theory Ôa
colossal
failure.
> Brian Greene got a few million dollars advance for his
book. The big
> corporations have a vested interest in this hype that can
be compared
to
> WMD in Iraq.
> String theorists agree that it has been a long, strange
trip, but
they
> still have faith that they will complete the journey.
> The fusion of science and religion on the heels of the
fusion of
Church
> and State?
> Twenty years ago no one would have correctly predicted how
string
> theory has since developed, said Dr. Strominger of Harvard.
> ÔThere is disappointment that despite all our efforts,
experimental
> verification or disproof still seems far away. On the other
hand, the
> depth and beauty of the subject, and the way it has reached
out,
> inßuenced and connected other areas of physics and
mathematics, is
> beyond the wildest imaginations of 20 years ago. In a 
way,
the
story
> of string theory and of the physicists who have followed
its siren
song
> for two decades is like a novel that begins with the
classic what
if?
> What if the basic constituents of nature and matter were
not little
> points, as had been presumed since the time of the Greeks?
What if
the
> seeds of reality were rather teeny tiny wiggly little bits
of string?
> merely correspond to different ways for the strings to
vibrate,
> different notes on Gods guitar?
> Ill play the music on my guitar, yes! Rossini, The Barber
of
Seville.
> That would explain why you cannot have a single quark you
cannot
have a
> string with only one end. Strings seduced many physicists
with their
> mathematical elegance, but they had some problems, like
requiring 26
to
> have anything to do with quarks or the strong force.
> You need the extra space dimensions because string theory
violates
> signal locality in Einsteins original 4D curved 
spacetime.
That is,
> final causes after their effects are possible in plain
vanilla 4D
string
> theory with topology change. Of course, this is exactly
what the UFO
and
> paranormal evidence is telling us along with presponse
experiments
by
> Dick Bierman and others. Then there is also the Intelligent
Design
> controversy. What is bad and bogus for string theorists is
good for
UFO,
> paranormal and consciousness researchers. One mans meat 
is
another
> mans poison.
> When accelerator experiments supported an alternative
theory of
quark
> behavior known as quantum chromodynamics, most physicists
consigned
> strings to the dustbin of history. But some theorists
thought the
> mathematics of strings was too beautiful to die. In 1974
Dr. Schwarz
> and Dr. Joel Scherk from the .83cole Normale Sup.8erieure
in France
noticed
the
> responsible for transmitting gravity in a quantum theory of
gravity,
if
> such a theory existed.
> This pyrrhic victory has been trumpeted as a great triumph
showing
the
> power of fantasy in the minds of string theorists.
Einsteins gravity
is
> not renormalizable as a quantum field theory which strongly
suggests
> that it should not be quantized top -> down like
electro-weak dynamic
> and chromodynamics. Indeed, Andrei Sakharov suggested that
Einsteins
> gravity emerges bottom -> up as a collective phenomenon
just like the
> elasticity in crystals. P.W. Anderson has since formalized
this idea
as
> More is different. Furthermore all current tests for grainy
quantum
> gravity foam that should show dispersion in the high energy
gamma
rays
> from outer space have been null. That is, there is no
evidence for
the
> equation
> E^2 = [(pc)^2 + mc^2]/[1 + (pc/mPc^2)^2]
> where mP is the Planck mass for the quantum gravity
graininess of
> Einsteinsw spacetime geometrodynamics.
> Note that the high energy limit of this equation is
> E -> mPc^2 ~ 10^-5 grams c^2 ~ 10^16 ergs ~ 10^9 Joules ~
10^28
electron
> volts
> implying a maximal acceleration of
> a max = c^2/LP ~ 10^21/10^-33 ~ 10^54 cm/sec^2 ~ 10^51g
> But, so far, no evidence supports this idea. If my theory
of the
> emergence of Einsteins gravity of MACRO-QUANTUM curved
spacetime
from
> the PARTIAL cohering of the micro-quantum zero point vacuum
ßuctuations
> in the BCS unstable massless pre-inßationary globally ßat
false
> conformal vacuum is correct, there is no quantum foam in
principle.
> non-perturbative background-independent MACRO-QUANTUM vacuum
condensate
> of virtual electron-hole pairs bound by virtual photons
near the E =
0
> Fermi energy of the pre-inßationay false vacuum obeys
> E^2 = (pc)^2 + Egap^2
> Egap ~ (Debye energy of edge of Fermi surface)e^-1/Vrho(0)
> V is the attractive static potential energy between the
virtual
> electron-hole (hole = positron) pair from a single virtual
photon
> exchange. rho(0) is the density of Diracs negative energy
virtual
> electron states at the edge per unit energy.
> The Einstein-Cartan tetrads eu^a derive from the
MACRO-QUANTUM
COHERENT
> different order parameter.
> guv = eu^anabev^b = nuv + (1/2)Lp^2(PHASE OF VACUUM
COHERENCE){,u,v}
> = nuv + (1/2)[Eu,v + Ev,u)
> { } is symmetrizer (i.e. anti-commutator) with ,u ordinary
partial
> derivatives in a local coordinate patch of the manifold.
> eu^a = &u^a + Eu^a
> &u,^a = Kronecker-Delta i.e. 1 if u = a, 0 if u =/= a
> Eu = Eu^a,a = Lp^2(PHASE OF VACUUM COHERENCE),u
> This equation is the analog to the deBroglie-Bohm guidance
equation
of
> IT by BIT for a quantum liquid. Here we have an ELASTIC
ODLRO
quantum
> solid that I introduced into physics in 1966-67 in Physics
Letters A
on
> Super Solids in Helium when I was on the physics faculty of
San
Diego
> State and the creator of the National Science Foundation
Summer
School
> in Superßuid Physics and Lasers for College Teachers at San
Diego
State
> (1969 & 1970) that I ran with Herschel Snodgrass. The
faculty
included
> F.W. Cummings and Jim Johnston.
> The QUANTUM OF AREA Lp^2 = hG/c^3 ensures Hawkings (also
Bekenstein,
> Susskind, tHooft)
> Entropy of Volume of Space = (Bounding Area of Volume)/4Lp^2
> Space-time physics is local because the macro-quantum order
parameter
is
> local. This same locality ensures that the early post-Big
Bang
universe
> after the inßationary vacuum phase transition has low
entropy so
that
> the Arrow of Time for the irreversible processes of the
Second Law of
> Thermodynamics points in the same direction as the presently
> accelerating expansion of the 3D space of our universe from
residual
w =
> -1 zero point dark energy of negative quantum pressure.
Hawkings
> blackhole entropy formula is automatically obeyed. That is
considered
a
> triumph of string theory. Its no big deal really. Sorry
for stealing
> string theorists thunder and upstaging them, but as 
Ludwig
Boltzmann
> said Elegance is for tailors. Not that this new way of
connecting
the
> cosmic dots is not elegant to the max. :-)
> That is, the pre -> post inßationary Big Bang phase
transition
> collapses the phase space volume of the ground state and the
entropy
> is ~ kBlog(Phase Space Volume). This same collapse of
ground state
phase
> space volume always accompanies More is different emergence
of
> qualitatively CREATIVE new order, AKA spontaneous breakdown
of
vacuum
> symmetry.
> To be continued.
===
Subject: Limits of Limits
Suppose {a_mn} is a set of real numbers, indexed by m and n
both
positive integers. Suppose
a_mn -> a_m
as n -> infty for each m. And
a_m -> a.
I think it is true that this does *not* imply
a_nn -> a
as n -> infty.
However, if we also know that
a_mn -> a_n
as m->infty for each n, and
a_n -> a.
Is that enough to guarantee
a_nn -> a
as n -> infty?
If not, this would imply that uniform convergence, i.e.
a_mn -> a_m
uniformly with respect to n, is a necessary condition for
convergence of
a_nn -> a.
-Kira
===
Subject: Re: Limits of Limits
<lNpxd.4352$yK.3708@newsread3.news.atl.earthlink.net>,
> Suppose {a_mn} is a set of real numbers, indexed by m and n
both
> positive integers. Suppose
> a_mn -> a_m
> as n -> infty for each m. And
> a_m -> a.
> I think it is true that this does *not* imply
> a_nn -> a
> as n -> infty.
> However, if we also know that
> a_mn -> a_n
> as m->infty for each n, and
> a_n -> a.
> Is that enough to guarantee
> a_nn -> a
> as n -> infty?
No, let a_mn = exp[-(m-n)^2].
> If not, this would imply that uniform convergence, i.e.
> a_mn -> a_m
> uniformly with respect to n, is a necessary condition for
convergence of
> a_nn -> a.
Uniform convergence is sufficient, but not necessary. To see
the
latter, consider a_mn = exp[-(m-n)^2] if m is not equal to n,
a_nn = 0 for all n.
===
Subject: Set theory question from beginner
posting-account=3yhFrw0AAAA8BYKMQgOTDl3e3xcEea1L
All,
Just started learning set theory, so this is a question that
arises
from profound ignorance. My apologies in advance.
Given the (disjoint) sets A, B, C in omega:
___________
| ___ |
| | | |
| | A | |
| |___| |
| ___ |
| | | |
| | B | |
| |___| |
| ___ |
| | | |
| | C | |
| |___| |
|___________|
Isnt the expression ABC ambiguous as 
written, since
(AB)C= null
whereas
A(BC)= C
???
-Euge
===
Subject: Re: Set theory question from beginner
[euge]
> Just started learning set theory, so this is a question
that arises
> from profound ignorance. My apologies in advance.
> Given the (disjoint) sets A, B, C in omega:
So A, B and C are subsets of the natural numbers, and the set
of all natural
numbers is taken to be the universal set here?
> ___________
> | ___ |
> | | | |
> | | A | |
> | |___| |
> | ___ |
> | | | |
> | | B | |
> | |___| |
> | ___ |
> | | | |
> | | C | |
> | |___| |
> |___________|
> Isnt the expression ABC ambiguous as 
written, since
> (AB)C= null
> whereas
> A(BC)= C
> ???
Im not sure what you intend by this notation. 
Im going to
assume that
juxtaposition means intersection, so that AB means the
intersection of A and
B. Im also going to assume that the single-quote means the
complement,
with respect to the universal set, of the set immediately
preceding the
single-quote. So, for example, if E is the set of even
naturals, E is the
set of odd naturals, and AB means the intersection of A and
the complement
of B.
In that case (and in many others <wink>), then
XYZ
is not ambiguous: (XY)Z = X(YZ), regardless of whether X, Y
and Z are
disjoint, and regardless of whether complements are taken on
any of them.
Either way, XYZ is the set of elements common to all of X, Y
and Z.
If you absorb that general result, then your example above is
just a special
case of it.
Anyway, (AB)C is null, because AB is null, and the
intersection of the null
set with any set (whether C or anything else) is null. But
A(BC) is also
null, and its unclear why you think its 
really C. BC is
B, under the
assumption that B and C are disjoint: C is everything not 
in
C:
___________
| |
| |
| |
| C |
| |
| |
| |
| ___ |
| | | |
| | C | |
| |___| |
|___________|
so C contains all of B, so BC = B. Then 
A(BC) = AB = null
too.
Clearer now? If not, please show us how you get from A(BC)
to C.
===
Subject: Re: Set theory question from beginner
<-LSdneCj6rZ8qFvcRVn-iA@comcast.com>
posting-account=3yhFrw0AAAA8BYKMQgOTDl3e3xcEea1L
Hi Everyone,
I was just being stupid (bad habit of mine - perhaps if I had
exercised
my brain instead of doing a Ph.D. over these past 4 years I
would be
able to think...) I was taking BC to mean 
B+C rather than B
intersected with C. I have now turned my brain to the ON
position.
-Euge
> [euge]
> Just started learning set theory, so this is a question
that arises
> from profound ignorance. My apologies in advance.
> Given the (disjoint) sets A, B, C in omega:
> So A, B and C are subsets of the natural numbers, and the
set of all
natural
> numbers is taken to be the universal set here?
> ___________
> | ___ |
> | | | |
> | | A | |
> | |___| |
> | ___ |
> | | | |
> | | B | |
> | |___| |
> | ___ |
> | | | |
> | | C | |
> | |___| |
> |___________|
> Isnt the expression ABC ambiguous as 
written, since
> (AB)C= null
> whereas
> A(BC)= C
> ???
> Im not sure what you intend by this notation. 
Im going to
assume
that
> juxtaposition means intersection, so that AB means the
intersection
of A and
> B. Im also going to assume that the single-quote means 
the
complement,
> with respect to the universal set, of the set immediately
preceding
the
> single-quote. So, for example, if E is the set of even
naturals, E
is the
> set of odd naturals, and AB means the intersection of A
and the
complement
> of B.
> In that case (and in many others <wink>), then
> XYZ
> is not ambiguous: (XY)Z = X(YZ), regardless of whether X, Y
and Z
are
> disjoint, and regardless of whether complements are taken
on any of
them.
> Either way, XYZ is the set of elements common to all of X,
Y and Z.
> If you absorb that general result, then your example above
is just a
special
> case of it.
> Anyway, (AB)C is null, because AB is null, and the
intersection of
the null
> set with any set (whether C or anything else) is null. 
But
A(BC)
is also
> null, and its unclear why you think its 
really C. BC is
B, under
the
> assumption that B and C are disjoint: C is everything not
in C:
> ___________
> | |
> | |
> | |
> | C |
> | |
> | |
> | |
> | ___ |
> | | | |
> | | C | |
> | |___| |
> |___________|
> so C contains all of B, so BC = B. Then 
A(BC) = AB =
null too.
> Clearer now? If not, please show us how you get from 
A(BC)
to C.
===
Subject: Re: Set theory question from beginner
> Hi Everyone,
> I was just being stupid (bad habit of mine - perhaps if I
had exercised
> my brain instead of doing a Ph.D. over these past 4 years I
would be
> able to think...) I was taking BC to mean 
B+C rather than
B
> intersected with C. I have now turned my brain to the ON
position.
Its very unusual notation in any case to use BC to mean
B intersected with C (or anything else, for that matter).
If B and C are subsets of a common group, or other structure
in which multiplication is defined, BC usually means the
set of all products bc where b is taken from B and c from C.
But if B and C are general sets, BC just doesnt mean 
anything
at all, in most peoples notation.
Are you following some strange or antiquated textbook?
===
Subject: Re: Set theory question from beginner
> Its very unusual notation in any case to use BC to mean
> B intersected with C (or anything else, for that matter).
> If B and C are subsets of a common group, or other structure
> in which multiplication is defined, BC usually means the
> set of all products bc where b is taken from B and c from C.
> But if B and C are general sets, BC just doesnt mean
anything
> at all, in most peoples notation.
> Are you following some strange or antiquated textbook?
In probability, one uses juxtapostion as an indication for
intersection
all the time.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Set theory question from beginner
posting-account=GYng8QwAAAASJCdK-kCiIRtmF4RV4yrR
Assuming BC is B intersect the complement of C....
Hint: What does (BC) equal (its not 
C)? What does A
intersect that
thing equal?
===
Subject: JSH: Somewhat puzzled
posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
Now over a year ago I was talking about a paper at a math
journal and
even at one point mentioned that the journal was the
Southwest Journal
of Pure and Applied Mathematics. I was pleased to see little
activity
over that on this newsgroup.
I should have known that didnt mean much.
Months went by and the journal told me they liked the paper
and were
going to publish.
Someone posting about it on the newsgroup and some posters
got together
to email the journal.
So some posters here were able to push Ioannis Argyros with a
paper
that I guess gave him some wiggle room so that he thought he
could get
away with yanking it.
I fixed several problems:
problem within the discipline of algebraic number theory.
I covered the issue of the contradiction with what you get if
you
believe that the ring of algebraic integers doesnt have
problems,
covering the area where posters have claimed counterexamples.
And most importantly, I sent the paper to a top-ranked PRINT
journal,
so that I dont have to worry about some ßaky editor 
telling
me one
thing one day, and then doing something else another.
The paper is at the Annals in Princeton, having been accepted
for
formal peer review.
However, I still see many of you behaving here as if you have
no
concerns, and that puzzles me somewhat.
At best you may have a few months to keep playing these
games, but it
could be all over tomorrow.
What makes you tick? How do you keep on waiting as if you can
get
away?
Its MATHEMATICS and sure, you can lie about it if you wish,
but that
doesnt change the mathematics. The editors at Princeton are
top
mathematicians, and their reviewers are the best in the world.
None of the various lies and strategies youve used on the
newsgroup
over the years will work with them, and it doesnt matter if
some of
you decide to email them like you did with that other journal.
So how do you go on? How will some of you reply to this post
as if
youre safe? How will you keep moving for the next few weeks
or months
until you are known for what you are, the story becomes huge
all over
the world, and you have reporters at your schools asking you,
why?
James Harris
===
Subject: Re: JSH: Somewhat puzzled
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
I wonder where they found peers of Harris?
===
Subject: Re: JSH: Somewhat puzzled
posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
> Now over a year ago I was talking about a paper at a math
journal and
> even at one point mentioned that the journal was the
Southwest
Journal
> of Pure and Applied Mathematics. I was pleased to see little
activity
> over that on this newsgroup.
> I should have known that didnt mean much.
> Months went by and the journal told me they liked the paper
and were
> going to publish.
> Someone posting about it on the newsgroup and some posters
got
together
> to email the journal.
> So some posters here were able to push Ioannis Argyros with
a paper
> that I guess gave him some wiggle room so that he thought
he could
get
> away with yanking it.
He *did* get away with it.
> I fixed several problems:
Which ones?
The most recent version that you have made publicly available,
co-authored with Beckwith, can be obtained at
http://www.ne-plus-ultra.net/pubs/beckwith_factorization.pdf
The intro to that paper includes the following statement:
This paper will show, using basic algebraic methods, that
given the factorization in the ring of algebraic integers,
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1),
one of the as is coprime to 5.
Then in the conclusion of the paper, you say:
Therefore with the factorization
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1),
one of the as is coprime to 5, which shows where some
of the algebraic integer factors distribute despite
the factors being irrational.
You now know, and have acknowledged, that NONE of as is
coprime to 5 in the ring of algebraic integers. Nevertheless,
you have claimed that the conclusions of THIS VERSION of
Advanced Polynomial Factorization refer to factorizations not
in the ring of algebraic integers, but some other ring. The
wording above in both the intro section and in the conclusion
clearly indicate that this claim is an outright lie. But you
have
never admitted that the claim was incorrect! Bizarrely,
at one point you even offered a compromise - you offered to
modify the paper, presumably changing the conclusion noted
above,
if in return sci.math posters would refrain from contacting
the
journal(s) to which you had submitted it.
So why is that version of the paper still out there? Is this
the version you have submitted to the Annals of Mathematics?
Why is Beckwith listed as a co-author? Is he co-author
of the paper submitted to Annals? Why have you never admitted
that the above version of the paper is incorrect?
a
> problem within the discipline of algebraic number theory.
Please post a copy of that paper, if it differs from the
one available at Beckwiths site.
> I covered the issue of the contradiction with what you get
if you
> believe that the ring of algebraic integers doesnt have
problems,
> covering the area where posters have claimed
counterexamples.
> And most importantly, I sent the paper to a top-ranked
PRINT journal,
> so that I dont have to worry about some ßaky editor
telling me one
> thing one day, and then doing something else another.
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
> However, I still see many of you behaving here as if you
have no
> concerns, and that puzzles me somewhat.
The Annals is the worlds most respected math journal. It
published Wiles papers. I have offered you the following
wager:
If the Annals publishes this paper, I will mail you a
one-hundred
dollar bill. If the Annals rejects it, you send $100
check
on a website. [No, the charity of my choice is NOT the David
Duke
Re-election Fund.]
There is no question that the Annals is well-edited.
You should be able to trust their review. If you are confident
that your conclusions in APF are correct, you should be
willing
to accept this wager.
> At best you may have a few months to keep playing these
games, but it
> could be all over tomorrow.
> What makes you tick? How do you keep on waiting as if you
can get
> away?
> Its MATHEMATICS and sure, you can lie about it if you
wish, but that
> doesnt change the mathematics.
We have given rigorous proofs that your math is wrong. We have
pointed out your main errors. If we actually thought you were
right,
we would be crazy to lie about it. Right? Draw your own
conclusions.
> The editors at Princeton are top
> mathematicians, and their reviewers are the best in the
world.
Agreed. Want to take the wager?
> None of the various lies and strategies youve used on the
newsgroup
> over the years will work with them, and it doesnt matter
if some of
> you decide to email them like you did with that other
journal.
I have no intention of e-mailing them unless your paper is
accepted. In which case pigs will be ßying overhead, Canada
will annex Mexico, and people will be ice-skating in Hell.
> So how do you go on? How will some of you reply to this
post as if
> youre safe? How will you keep moving for the next few
weeks or
months
> until you are known for what you are, the story becomes
huge all over
> the world, and you have reporters at your schools asking
you, why?
> James Harris
Enough bluster and preaching and threatening. Bring on the
Hammer or shut up.
Nora B.
===
Subject: Re: JSH: Somewhat puzzled
posting-account=sg_iGAwAAAClZhnVQKakTFRnAjK0ujUn
> Now over a year ago I was talking about a paper at a math
journal and
> even at one point mentioned that the journal was the
Southwest
Journal
> of Pure and Applied Mathematics. I was pleased to see little
activity
> over that on this newsgroup.
> I should have known that didnt mean much.
> Months went by and the journal told me they liked the paper
and were
> going to publish.
> Someone posting about it on the newsgroup and some posters
got
together
> to email the journal.
> So some posters here were able to push Ioannis Argyros with
a paper
> that I guess gave him some wiggle room so that he thought
he could
get
> away with yanking it.
He *did* get away with it.
> I fixed several problems:
Which ones?
The most recent version that is publicly available,
co-authored
with Beckwith, can be obtained at
http://www.ne-plus-ultra.net/pubs/beckwith_factorization.pdf
The intro to that paper includes the following statement:
This paper will show, using basic algebraic methods, that
given the factorization in the ring of algebraic integers,
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1),
one of the as is coprime to 5.
Then in the conclusion of the paper, you say:
Therefore with the factorization
65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1),
one of the as is coprime to 5, which shows where some
of the algebraic integer factors distribute despite
the factors being irrational.
You now know, and have acknowledged, that NONE of as is
coprime to 5 in the ring of algebraic integers. Nevertheless,
you have claimed that the conclusions of THIS VERSION of
Advanced Polynomial Factorization refer to factorizations not
in the ring of algebraic integers, but some other ring. The
wording in both the intro section and in the conclusion
clearly
indicate that this claim is an outright lie. But you have
never admitted that the claim was incorrect! Bizarrely,
at one point you even offered a compromise - you offered to
modify the paper, presumably changing the conclusion noted
above,
if in return sci.math posters would refrain from contacting
the
journal(s) to which you had submitted it.
So why is that version of the paper still out there? Is this
the version you have submitted to the Annals of Mathematics?
Why is Beckwith listed as a co-author? Is he co-author
of the paper submitted to Annals? Why have you never admitted
that the above version of the paper is incorrect?
a
> problem within the discipline of algebraic number theory.
Please post a copy of that paper, if it differs from the
one available at Beckwiths site.
> I covered the issue of the contradiction with what you get
if you
> believe that the ring of algebraic integers doesnt have
problems,
> covering the area where posters have claimed
counterexamples.
> And most importantly, I sent the paper to a top-ranked
PRINT journal,
> so that I dont have to worry about some ßaky editor
telling me one
> thing one day, and then doing something else another.
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
> However, I still see many of you behaving here as if you
have no
> concerns, and that puzzles me somewhat.
The Annals is the worlds most respected math journal. It
published Wiles papers. I have offered you the following
wager:
If the Annals publishes this paper, I will mail you a
one-hundred
dollar bill. If the Annals rejects it, you send $100
check
on a website. [No, the charity of my choice is NOT the David
Duke
Re-election Fund.]
There is no question that the Annals is well-edited.
You should be able to trust their review. If you are confident
that your conclusions in APF are correct, you should be
willing
to accept this wager.
> At best you may have a few months to keep playing these
games, but it
> could be all over tomorrow.
> What makes you tick? How do you keep on waiting as if you
can get
> away?
> Its MATHEMATICS and sure, you can lie about it if you
wish, but that
> doesnt change the mathematics.
We have given rigorous proofs that your math is wrong. We have
pointed out your main errors. If we actually thought you were
right,
we would be crazy to lie about it. Right? Draw your own
conclusions.
> The editors at Princeton are top
> mathematicians, and their reviewers are the best in the
world.
Agreed. Want to take the wager?
> None of the various lies and strategies youve used on the
newsgroup
> over the years will work with them, and it doesnt matter
if some of
> you decide to email them like you did with that other
journal.
I have no intention of e-mailing them unless your paper is
accepted. In which case pigs will be ßying overhead, Canada
will annex Mexico, and people will be ice-skating in Hell.
> So how do you go on? How will some of you reply to this
post as if
> youre safe? How will you keep moving for the next few
weeks or
months
> until you are known for what you are, the story becomes
huge all over
> the world, and you have reporters at your schools asking
you, why?
> James Harris
Enough bluster and preaching and threatening. Bring on the
Hammer or shut up.
Nora B.
===
Subject: Re: JSH: Somewhat puzzled
> Now over a year ago I was talking about a paper at a math
journal and
> even at one point mentioned that the journal was the
Southwest Journal
> of Pure and Applied Mathematics. I was pleased to see
little activity
> over that on this newsgroup.
The reason there was little activity was that (speaking for
myself,
since I speak for no one else) it was assumed that the
reviewers would
do their jobs. It was abundantly clear that your paper, if it
simply
repeated the same arguments you had been waving around
sci.math,
it led to a result that contradicts arithmetic. The only
possibility
for publication would have been that the paper was written in
*such*
opaque language that a lazy reader would end up essentially
saying
whatever at each turn, never coming to a mathematical
statement that
he could check to verify or refute.
> I should have known that didnt mean much.
> Months went by and the journal told me they liked the paper
and were
> going to publish.
> Someone posting about it on the newsgroup and some posters
got together
> to email the journal.
I dont understand this some posters got together jazz. That
suggests
communication that orchestrated a response. What I did, and
what I
paper on their site, next I checked the site for any
editorial policy,
and obtained the a-mail address for correspondence with their
editorial
staff, and finally I sent a note documenting what I knew about
the paper
you had sent to them. I also documented this communication on
sci.math.
For you to refer to any part of my individual process as
constituting
getting together is yet another indication of your somewhat
ßawed
command of the English language.
> So some posters here were able to push Ioannis Argyros with
a paper
> that I guess gave him some wiggle room so that he thought
he could get
> away with yanking it.
Virtually everyone here has expressed disagreement with the
tack Argyros
took wrt that fiasco. Clearly, you wouldnt have 
had the same
reaction
> I fixed several problems:
> problem within the discipline of algebraic number theory.
Just that: asserting. Note that you havent really proven
anything.
> I covered the issue of the contradiction with what you get
if you
> believe that the ring of algebraic integers doesnt have
problems,
> covering the area where posters have claimed
counterexamples.
I can just see it:
THEOREM: If you believe that the ring of algebraic integers
doesnt have a problem, then its like really 
HUGE.
COROLLARY: The Hammer is just about to swing.
> And most importantly, I sent the paper to a top-ranked
PRINT journal,
> so that I dont have to worry about some ßaky editor
telling me one
> thing one day, and then doing something else another.
Most important for your likelihood of success should have been
correctness of argument. However, to you, every argument you
have ever made has been correct, and more reliable than
mathematics
that has stood for hundreds of years, so this is a difficult
test
for you to apply.
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
I have the opinion that the Annals of Mathematics will provide
adequate review for your paper; consequently, I am confident
it will not be published.
> However, I still see many of you behaving here as if you
have no
> concerns, and that puzzles me somewhat.
Youre puzzled because you refuse to admit that others do 
not
accept your argument as signifying what you apparently think
it does. You think your critics here in sci.math are liars,
and cannot conceive of the possibility that people can read
what youve written and not agree wholeheartedly.
> At best you may have a few months to keep playing these
games, but it
> could be all over tomorrow.
My prediction is this: youll get your rejection from the
Annals of
Mathematics, and youll launch into another replay of (a) 
your
attack mode, (b) your poor me mode, or (c) youll go silent
for a few weeks or months while you plot yet another assault.
> What makes you tick? How do you keep on waiting as if you
can get
> away?
Get away? As I said, you dont get it. We see that you are
mistaken,
and you dont get it. Every argument you make is 
automatically
correct in your own mind, and others can not only see how
correct it
is, but they also must be seized by some perverse compulsion
to
deny the truth. Why? Well, its because of your primary
assumption,
which youve previously announced on sci.math as predating
all this
exorcism: that the mathematics community is corrupt. Just as
long
as you take that as an attitude, nothing else can get through
to you.
> Its MATHEMATICS and sure, you can lie about it if you
wish, but that
> doesnt change the mathematics. The editors at Princeton
are top
> mathematicians, and their reviewers are the best in the
world.
or it will fail.
> None of the various lies and strategies youve used on the
newsgroup
> over the years will work with them, and it doesnt matter
if some of
> you decide to email them like you did with that other
journal.
re-wordings of it. Find one, and show me a lie. Remind me how
it
was that our exchange several months back resulted in your
admission
that you *didnt* have coprimeness of any of the a_i with 5
(in the
notation of your paper Advanced Polynomial Factorization ), in
direct contradiction to what your paper stated.
> So how do you go on? How will some of you reply to this
post as if
> youre safe? How will you keep moving for the next few
weeks or months
> until you are known for what you are, the story becomes
huge all over
> the world, and you have reporters at your schools asking
you, why?
> James Harris
Im mighty skeered.
Dale.
===
Subject: Re: JSH: Somewhat puzzled
<VODxd.880$yV1.485@newssvr14.news.prodigy.com>
posting-account=BoVIJQwAAABWQmiBreBpIBK6U9cCf57f
> At best you may have a few months to keep playing these
games, but
it
> could be all over tomorrow.
> My prediction is this: youll get your rejection from the
Annals of
> Mathematics, and youll launch into another replay of (a)
your
> attack mode, (b) your poor me mode, or (c) youll go 
silent
> for a few weeks or months while you plot yet another
assault.
I think its probably worth mentioning to James (although he
wont
listen) that this rejection most likely *wont* take the 
form
of a
step-by-step rebuttal or counter-argument; it will probably be
somehting bland and inoffensively vague along the lines of
... not
appropriate material for this journal ... perhaps send to a
more
specialist publication ... thank you for your submission ...
bye.
I would also wager that at no point will it say that the
paper is
Ôwrong. James will probably take this as 
confirmation that
his paper
is right, of course (a warped version of the old black swan
argument -
every person that doesnt say the paper is wrong is evidence
that the
paper is right! right?). Whereas the truth is that people in
the Real
World are *very much more polite* than people on Usenet (this
is
usually expressed the other way round). In the Real World
(say... in a
workplace environment), when someone sends an email that
makes them
look like an idiot, they will get politely corrected. When
someone
sends a crank paper to a journal, they will get politely
rejected. Is
it the Smithsonian that has had a multi-year running
correspondence
with some nutter who keeps sending them Barbie dolls and the
like and
claiming they are revolutionary archaeological finds?
Nonetheless there
is still politeness. No one ever actually says to these
people, Youre
acting like an idiot.
On Usenet, as James likes to point out, people are OUT OF
CONTROL -
when someone acts like an idiot, they get *called* an idiot.
Oh, and my bet is on (c) above, with a twist: now not only
sci.math
cannot be trusted, but the whole mathematical establishment
is under
suspicion. I dont know where he will attempt to gain
recognition next;
finding that out will be part of the fun!
--
Larry Lard
Replies to group please
===
Subject: Re: JSH: Somewhat puzzled
<VODxd.880$yV1.485@newssvr14.news.prodigy.com>
posting-account=RHW70A0AAACN7bbDaGoPV7XTRqBj1qGg
Sadly the Smithsonian-Barbie incident is an urban myth. Very
entertaining piece of fiction nonetheless -
http://www.snopes.com/humor/letters/smithson.htm
===
Subject: Re: JSH: Somewhat puzzled
posting-account=uMDgiw0AAAANoAxJOs_DnrtdjhRMBFah
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
People are reacting like this means nothing, but I disagree.
It means
that your paper is well formed in some sense. For example, if
you
sent a banana to the Annals, it would not be accepted for
formal peer
review, even if you sent a cover letter and the proper number
of copies
of the banana.
So there is some sense in which your work is more akin to a
work of
mathematics than a banana is.
Furthermore, I believe that the probability of your work being
published in the Annals is *strictly greater* (yes > not >=)
than that of the banana getting published!
Let me qualify this, however. By the banana getting published
I mean that an actual, physical banana is shipped with every
copy
of the issue. It would not suffice for the issue to have a
Xerox
of the banana in it, for example. This latter scenario
strikes me
as being roughly as likely as your work getting published
there.
Both the Xerox of the banana and your work would be nice
material
for an April Fools edition, for example: its 
hard to pick
one as
better than the other. But an actual, physical banana: just
think
of the distribution hassles that would entail! Why would
anyone go
to the trouble?
Anyway, good luck. Hopefully you mentioned the April Fools
idea in
your cover letter, in case they dont think of it 
themselves.
===
Subject: Re: JSH: Somewhat puzzled
>>The paper is at the Annals in Princeton, having been
accepted for
>>formal peer review.
> People are reacting like this means nothing, but I
disagree. It means
> that your paper is well formed in some sense. For example,
if you
> sent a banana to the Annals, it would not be accepted for
formal peer
> review, even if you sent a cover letter and the proper
number of copies
> of the banana.
> So there is some sense in which your work is more akin to a
work of
> mathematics than a banana is.
> Furthermore, I believe that the probability of your work
being
> published in the Annals is *strictly greater* (yes > not >=)
> than that of the banana getting published!
> Let me qualify this, however. By the banana getting
published
> I mean that an actual, physical banana is shipped with
every copy
> of the issue. It would not suffice for the issue to have a
Xerox
> of the banana in it, for example. This latter scenario
strikes me
> as being roughly as likely as your work getting published
there.
> Both the Xerox of the banana and your work would be nice
material
> for an April Fools edition, for example: 
its hard to pick
one as
> better than the other. But an actual, physical banana: just
think
> of the distribution hassles that would entail! Why would
anyone go
> to the trouble?
> Anyway, good luck. Hopefully you mentioned the April 
Fools
idea in
> your cover letter, in case they dont think of it
themselves.
You are a funny guy :-)
===
Subject: Re: JSH: Somewhat puzzled
>Now over a year ago I was talking about a paper at a math
journal and
>even at one point mentioned that the journal was the
Southwest Journal
>of Pure and Applied Mathematics. I was pleased to see little
activity
>over that on this newsgroup.
>I should have known that didnt mean much.
>Months went by and the journal told me they liked the paper
and were
>going to publish.
>Someone posting about it on the newsgroup and some posters
got together
>to email the journal.
>So some posters here were able to push Ioannis Argyros with
a paper
>that I guess gave him some wiggle room so that he thought he
could get
>away with yanking it.
Thought he could get away with? Hmm.
>I fixed several problems:
Most of us fix problem before trying to publish.
>problem within the discipline of algebraic number theory.
Youve directly asserted this here many times. But 
youve
never
made an explicit and understandable statement of exactly what
the
problem is. Or if you have I and many others have missed it.
>I covered the issue of the contradiction with what you get
if you
>believe that the ring of algebraic integers doesnt have
problems,
Contradiction with the belief that the algebraic integers
dont
have problems? This is meaningless until you explain _what_
problem youre talking about.
>covering the area where posters have claimed counterexamples.
>And most importantly, I sent the paper to a top-ranked PRINT
journal,
>so that I dont have to worry about some ßaky editor 
telling
me one
>thing one day, and then doing something else another.
>The paper is at the Annals in Princeton, having been
accepted for
>formal peer review.
Wow. Let us know when its been accepted. Teeheeheeheehee.
>However, I still see many of you behaving here as if you
have no
>concerns, and that puzzles me somewhat.
>At best you may have a few months to keep playing these
games, but it
>could be all over tomorrow.
Jesus. First, theres no way in hell that that paper is 
going
to
be published. Second, even if it was published, all that would
prove is that the Annals screwed up (its not going to 
happen
that _they_ screw up this badly). Third, even if it _were_
true
that your paper _did_ show that the algebraic integers had
problems, that would not make it all over for all of us.
Nobody would get fired. Honest.
>What makes you tick? How do you keep on waiting as if you
can get
>away?
>Its MATHEMATICS and sure, you can lie about it if you 
wish,
but that
>doesnt change the mathematics. The editors at Princeton 
are
top
>mathematicians, and their reviewers are the best in the
world.
>None of the various lies and strategies youve used on the
newsgroup
>over the years will work with them, and it doesnt matter 
if
some of
>you decide to email them like you did with that other
journal.
>So how do you go on? How will some of you reply to this post
as if
>youre safe? How will you keep moving for the next few 
weeks
or months
>until you are known for what you are, the story becomes huge
all over
>the world, and you have reporters at your schools asking
you, why?
When there are reporters all over campus asking me why your
paper
was rejected by the Annals Ill simply explain that 
thats
because
it was nonsense.
>James Harris
************************
David C. Ullrich
===
Subject: Re: JSH: Somewhat puzzled
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ELIi
$t^
VcLWP@J5p^rst0+(Ô>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
>>problem within the discipline of algebraic number theory.
> Youve directly asserted this here many times. But 
youve
never
> made an explicit and understandable statement of exactly
what the
> problem is. Or if you have I and many others have missed it.
You apparently missed it. The problem with the algebraic
integers is
that there are numbers that are properly units of the
algebraic
integers without being algebraic integers.
A unit is something of which the inverse exists in a given
ring.
Getting for a moment to the ordinary integers, they have the
same
problem: take (1/2) which is properly a unit in the integers
since
its inverse is an integer, but is not an integer itself!
Basically, every non-field has that problem as long as a
rational
extension field exists.
You could say that the fundamental problem of the algebraic
integers
is that they are not the algebraic numbers.
Well, it is not much after all this Brouhaha, but dont you
get
overconfident in overinterpreting Galois theory, nevertheless.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: JSH: Somewhat puzzled
<x5k6rdnjv1.fsf@lola.goethe.zz>
Discussion, linux)
>problem within the discipline of algebraic number theory.
>> Youve directly asserted this here many times. But 
youve
never
>> made an explicit and understandable statement of exactly
what the
>> problem is. Or if you have I and many others have missed
it.
> You apparently missed it. The problem with the algebraic
integers is
> that there are numbers that are properly units of the
algebraic
> integers without being algebraic integers.
> A unit is something of which the inverse exists in a given
ring.
> Getting for a moment to the ordinary integers, they have
the same
> problem: take (1/2) which is properly a unit in the
integers since
> its inverse is an integer, but is not an integer itself!
> Basically, every non-field has that problem as long as a
rational
> extension field exists.
Im not sure, but I *think* that your definition 
of proper
unit isnt
quite right.
Let R be a ring Z c R c C. I think the definition is: x in R
is a
proper unit if there is a y in C such that xy = 1 (in C)
*AND* x is
not an integer.
Of course, James wouldnt commit himself on 
definitions. Hes
learned
that explicit definitions (or claims) dont 
serve him well at
all. As
soon as he says something so clearly that others know what he
means,
then they can show him that hes wrong.
This is why he doesnt define properly a unit, 
doesnt
explicitly
say what the problem with the algebraic integers is and will
not
identify a theorem in which the problem has led to the
acceptance of
an invalid argument. (Sure, he says Wiless proof is one, 
but
he
wont say *how* the problem crops up there.)
James S. Harris *has* learned something over the years, after
all.
--
Jesse F. Hughes
If the car stops and youre not getting out, then you have 
to
start
it again. -- Quincy P. Hughes on his fathers skills with a
manual transmission.
===
Subject: Re: JSH: Somewhat puzzled
>problem within the discipline of algebraic number theory.
>> Youve directly asserted this here many times. But 
youve
never
>> made an explicit and understandable statement of exactly
what the
>> problem is. Or if you have I and many others have missed
it.
>You apparently missed it. The problem with the algebraic
integers is
>that there are numbers that are properly units of the
algebraic
>integers without being algebraic integers.
>A unit is something of which the inverse exists in a given
ring.
>Getting for a moment to the ordinary integers, they have the
same
>problem: take (1/2) which is properly a unit in the integers
since
>its inverse is an integer, but is not an integer itself!
Ah, right. I do recall the problem, but I never got the
definition of proper unit straight.
>Basically, every non-field has that problem as long as a
rational
>extension field exists.
>You could say that the fundamental problem of the algebraic
integers
>is that they are not the algebraic numbers.
>Well, it is not much after all this Brouhaha, but dont you
get
>overconfident in overinterpreting Galois theory, 
nevertheless.
************************
David C. Ullrich
===
Subject: Re: Somewhat puzzled
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
So you say it is accepted for formal peer review. And that
means what?
Just that you mailed it to them? But they will not publish
it. You do
understamd that, right?
> However, I still see many of you behaving here as if you
have no
> concerns, and that puzzles me somewhat.
That is because you are an idiot. It puzzles nobody else
here, I am sure.
> At best you may have a few months to keep playing these
games, but it
> could be all over tomorrow.
That would be the hammer, right?
> What makes you tick? How do you keep on waiting as if you
can get
> away?
Well, it is just because you are wrong.
===
Subject: Re: Somewhat puzzled
[JSH]
...
> And most importantly, I sent the paper to a top-ranked
PRINT journal,
> so that I dont have to worry about some ßaky editor
telling me one
> thing one day, and then doing something else another.
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
> However, I still see many of you behaving here as if you
have no
> concerns, and that puzzles me somewhat.
> At best you may have a few months to keep playing these
games, but it
> could be all over tomorrow.
> What makes you tick? How do you keep on waiting as if you
can get
> away?
> Its MATHEMATICS and sure, you can lie about it if you
wish, but that
> doesnt change the mathematics. The editors at Princeton
are top
> mathematicians, and their reviewers are the best in the
world.
You really want to know? Ill tell you the God-honest truth,
although
youve heard it all before: the people here who claimed your
paper (in all
the forms you presented earlier drafts) is wrong are not
engaged in a
conspiracy, and honestly believe your math is wrong. You call
it lying,
but not a one of them deserved that accusation.
Theyre not worried *because* the Annals has a good
reputation. They
sincerely believe your paper will be rejected, because they
sincerely
believe its wrong, and they trust that the Annals reviewers
will see that
too.
Thats all there is to it in the real world.
> None of the various lies and strategies youve used on the
newsgroup
> over the years will work with them, and it doesnt matter
if some of
> you decide to email them like you did with that other
journal.
> So how do you go on?
None of them believe theyve been lying, 
thats how. The
strategy they
used mostly consisted of pointing out errors in the work. If
thats the
same strategy the reviewers use, your gloating is at best
premature.
> How will some of you reply to this post as if youre safe?
How will
> you keep moving for the next few weeks or months until you
are known
> for what you are, the story becomes huge all over the
world, and you
> have reporters at your schools asking you, why?
> James Harris
Pretty much the same way you keep going, I guess -- except
that they agree
about where the errors are, so have the additional comfort of
what I suppose
looks to you like shared delusion (or conspiracy).
===
Subject: Re: JSH: Somewhat puzzled
posting-account=4lZjEA0AAACcI9lfzifBAQlMFCToSJI4
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
> [SNIP]
> At best you may have a few months to keep playing these
games, but it
> could be all over tomorrow.
Umm, I may not be a mathematician, but just because you have
a paper
accepted for peer review doesnt neccessarily mean 
its going
to be
published, correct?
===
Subject: Re: JSH: Somewhat puzzled
posting-account=Q2zO6wwAAABSLuGzZIjG0efOtB9n8fUY
> The paper is at the Annals in Princeton, having been
accepted for
> formal peer review.
> [SNIP]
> At best you may have a few months to keep playing these
games, but
it
> could be all over tomorrow.
> Umm, I may not be a mathematician, but just because you
have a paper
> accepted for peer review doesnt neccessarily mean 
its
going to be
> published, correct?
Well, let me help you out here so that you understand my
point and why
Im puzzled by the behavior on this newsgroup.
Lets say theres a very simple argument, 
which quite
logically shows
that some people made some mistakes in their mathematical
reasoning.
The argument is basic, stepped out in extreme detail, and
clearly shows
that they made a mistake.
However, some people dont wish to believe 
theres an error,
so they
simply lie about the argument, try to make it look more
complicated
than it is, or act as a group to try and push the result out
of view
when its published in a small electronic journal.
So the researcher who found the argument, after checking it
hundreds of
times for errors, and discussing it with people all over the
world to
verify that it is without error, decides that rather than
deal with a
small electronic journal that can be intimidated, he will go
for a
major journal, immune to intimidation that has a great
reputation, and
is the perfect place anyway to introduce such a major result.
The question though is, will they even look over the research?
Those who submit papers to journals may know that often the
hardest
part is getting a promise to check!
Now then, the journal accepts the paper for formal peer
review.
The argument is perfect having been checked and re-checked
over a
period of years.
The implications of the paper are huge.
The mathematics in it is stepped out logical step, by logical
step to a
conclusion which is proven to be true.
Now the people who dont wish to believe the result, hear
that the
paper has gone somewhere they cant reach. The lies 
theyve
used to
try and confuse others about it, or obscure it from the
world, dont
matter now, and all they can do is wait.
Im puzzled now as I consider those people on this 
newsgroup.
If you believe them that theres something wrong with my
research, then
that doesnt matter at this point. Nothing 
thats said here
matters at
this point. And beliefs dont enter into it, as NONE of you
are in a
position to stand against a major math journal.
So youre just waiting, and I guess hoping that they 
wont do
their
jobs.
Maybe some of you believe that the result is too big even for
those
editors, and that theyll react as you did--in fear.
Possibly that explains your behavior. You think the Princeton
editors
are corrupt like you?
James Harris
===
Subject: Re: JSH: Somewhat puzzled
> Maybe some of you believe that the result is too big even
for those
> editors, and that theyll react as you did--in fear.
Yes, I believe this. But I also believe there is one thing
that they
fear even more - the hammer!
===
Subject: Re: JSH: Somewhat puzzled
posting-account=4lZjEA0AAACcI9lfzifBAQlMFCToSJI4
> Lets say theres a very simple argument, 
which quite
logically shows
> that some people made some mistakes in their mathematical
reasoning.
> The argument is basic, stepped out in extreme detail, and
clearly
shows
> that they made a mistake.
OK, fair enough. So lets suppose that Im 
extremely busy
(which I
am), not a mathematician (which Im not), 
dont have time to
look
through the archives of sci.math for your argument (which I
dont), and
wouldnt be able to understand your highly technical 
argument
anyway
(which I probably wouldnt).
Now, I have no stake in this argument that you seem to have
with the
regulars at sci.math, Im just an innocent bystander, so
would you
mind summarizing your key result for me?
===
Subject: Re: JSH: Somewhat puzzled
> Lets say theres a very simple argument, 
which quite
logically shows
> that some people made some mistakes in their mathematical
reasoning.
> The argument is basic, stepped out in extreme detail, and
clearly shows
> that they made a mistake.
Okay, if its so basic, and if you can spell it out in
extreme detail,
please do this:
State your thesis in one easily comprehensible sentence.
That cant be that hard, can it? Fermats Last 
Theorem only
takes one line
to state. Could you please be kind enough to tell us what,
exactly, your
argument is? In one sentence.
Then we can all judge for ourselves how important it might be.
> So the researcher who found the argument, after checking it
hundreds of
> times for errors, and discussing it with people all over
the world to
> verify that it is without error,
Whoa. I could have sworn that I have seen many, many posts
responding to
you wherein the people asserted categorically that there was
error. I dont
recall any posts from people saying they thought it was
error-free. I know
I saw numerous posts from people who said they couldnt 
figure
out what,
exactly, you were trying to prove.
> Those who submit papers to journals may know that often the
hardest
> part is getting a promise to check!
Getting a promise to check is the hardest part of getting
something printed
by a reputable journal? Do you actually think that?
> The argument is perfect having been checked and re-checked
over a
> period of years.
*cough*
> The implications of the paper are huge.
Well see--first, of course, Id 
appreciate it if you do what
I asked at the
beginning of this post.
> The mathematics in it is stepped out logical step, by
logical step to a
> conclusion which is proven to be true.
I may as well ask--what, exactly, is your conclusion? In one
easy sentence,
please.
> Maybe some of you believe that the result is too big even
for those
> editors, and that theyll react as you did--in fear.
What? Are you that delusional? Do you think its possible
*anyone* looks
at your math in fear from its hugeness?
Michael
===
Subject: Re: JSH: Somewhat puzzled
> I covered the issue of the contradiction with what you get
if you
> believe that the ring of algebraic integers doesnt have
problems,
> covering the area where posters have claimed
counterexamples.
Can you please state in simple terms, exactly what is (or
are) the
problems. For the benefit of those of us who follow the JSH
threads but
dont follow the details. Are you saying that the alebraic
integers
arent a field? Or that some number can be shown 
to be both an
algebraic
integer and not an algebraic integer? Exactly what, in one or
two
declarative sentences, is the problem that you believe 
youve
discovered?
===
Subject: Re: JSH: Somewhat puzzled
> I covered the issue of the contradiction with what you get
if you
> believe that the ring of algebraic integers doesnt have
problems,
> covering the area where posters have claimed
counterexamples.
> Can you please state in simple terms, exactly what is (or
are) the
> problems. For the benefit of those of us who follow the JSH
threads but
> dont follow the details. Are you saying that the alebraic
integers
> arent a field? Or that some number can be 
shown to be both
an algebraic
> integer and not an algebraic integer? Exactly what, in one
or two
> declarative sentences, is the problem that you believe
youve discovered?
Ive been following the argument for several months, and I
have been
wondering that myself. Just when I think I have studied
enough algebra to
unserstand the problem, it slithers away from me. It looks at
times like
James has discovered a problem with the number 49 - or mayvbe
its 5.
===
Subject: Re: JSH: Somewhat puzzled
> Ive been following the argument for several months, and I
have been
> wondering that myself. Just when I think I have studied
enough algebra
to
> unserstand the problem, it slithers away from me. It looks
at times like
> James has discovered a problem with the number 49 - or
mayvbe its 5.
I think it is a problem with properly units, but I am not
sure.
===
Subject: Understanding Stochastic Calculus
posting-account=96kk_A0AAADmKbq7jtwJefyt94NKtuOA
Hello -- I dont know where to start -- I am trying to
prepare for a
class on stochastics (in two weeks) and found a book --
Stochastic
Calculus by Richard Durrett -- I cant understand his
notation and need
a more basic introduction -- I have had plenty of calculus,
linear
algebra, abstract algebra and probability/stats, but this
seems new to
me -- any recommendations -- why does the notation look so
unfamiliar?
missing some pre-reqs -- did I pick a bad book?
Any good resources online?
Tim
===
Subject: Re: Understanding Stochastic Calculus
>Hello -- I dont know where to start -- I am trying to
prepare for a
>class on stochastics (in two weeks) and found a book --
Stochastic
>Calculus by Richard Durrett -- I cant understand his
notation and need
>a more basic introduction -- I have had plenty of calculus,
linear
>algebra, abstract algebra and probability/stats, but this
seems new to
>me -- any recommendations -- why does the notation look so
unfamiliar?
>missing some pre-reqs -- did I pick a bad book?
>Any good resources online?
If your probability was not at the measure theory
level, you will at least have great difficulty.
You need familiarity with the Wiener process, and
the knowledge that a process with independent increments
which can be represented by continuous functions must
be a Wiener process with some time scale. Also, you
will need to know about cadlag processes (continuous
on the right, limits on the left), and about martingales.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Understanding Stochastic Calculus
> Hello -- I dont know where to start -- I am trying to
prepare for a
> class on stochastics (in two weeks) and found a book --
Stochastic
> Calculus by Richard Durrett -- I cant understand his
notation and need
> a more basic introduction -- I have had plenty of calculus,
linear
> algebra, abstract algebra and probability/stats,
probability that presumes measure theory? if not, this is
your lack
to understand the Durett. What is the prerequisite for the
class
you will take? If there is no measure theory prerequisite,
then
of course that class cannot be at the Durett level. Instead,
some dumbed-down level. There is a review in the December
Monthly
of a text Mathematics for Finance where the reviewer bemoans
the fact
that such texts have to be dumbed-down, since students of
finance
cannot be presumed to have studied measure theory.
> but this seems new to
> me -- any recommendations -- why does the notation look so
unfamiliar?
> missing some pre-reqs -- did I pick a bad book?
> Any good resources online?
> Tim
===
Subject: Re: Understanding Stochastic Calculus
>Hello -- I dont know where to start -- I am trying to
prepare for a
>class on stochastics (in two weeks) and found a book --
Stochastic
>Calculus by Richard Durrett -- I cant understand his
notation and need
>a more basic introduction -- I have had plenty of calculus,
linear
>algebra, abstract algebra and probability/stats, but this
seems new to
>me -- any recommendations -- why does the notation look so
unfamiliar?
>missing some pre-reqs -- did I pick a bad book?
>Any good resources online?
I have never understood what people mean exactly when they say
stochastics. Probability? Probability Modelling? Stochastic
Processes? Stochastic Models in Finance?
In any case, in your background, you certainly need to know
basic
probability, which you do not mention. And to understand
stochastic
calculus, you need to understand Brownian motion at least,
which is
usually covered in second semester Stochastic Processes. You
may have
picked up the wrong book entirely. What, exactly, is the
course you
will be taking, and what are its stated prerequisites?
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Understanding Stochastic Calculus
<41C663ED.9060706@netscape.net>
posting-account=96kk_A0AAADmKbq7jtwJefyt94NKtuOA
Pre-requisites: probability, math for operations research
(optimization, etc) I did mention probability/stats above
(counting,
distributions, random variables/processes, density, mass,
etc). the
title of the class is Stochastics, it is a first introduction
to
markov processes, I guess the course should be titled
Stochastic
Processes. So it looks like stochastic
calculus comes much further down the line than Stochastic
Processes?
tim
===
Subject: Re: Understanding Stochastic Calculus
Yes, this sounds like a first course in stochastic processes,
which is
much less advanced than stochastic calculus. A better text
for you look
at is Ross SM, Introduction to Probability Models.
Sorry - I missed the probabilty/stats in your original post.
If you
mean a course like Probability and Statistics for Business
that
focusses mainly on statistics, you may still need a little
more than you
know. But if your previous course is the prerequisite for the
next,
then it is probably sufficient.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: A Laplace Transform Integral
boundary=----=_NextPart_000_007B_01C4E612.9F29E7E0
-------------------------------------------------------------
--------
Dave,
I dont quite comprehend your notation
begin{equation}
label{eq:radial}
x(|vec t|) corr int 0^infty x(r), 2pi sqrt{r^nover
|vec f|^{n-2}}, J {n-2over2}(2pi r,|vec f|), dr
end{equation}
Ed
--
Edward Hyman
EdwardH@email.uophx.edu
Other EMail: e.hyman@worldnet.att.net
> Integrate from - infinity to infinity
> { exp(-px) * integrate from - infinity to 
infinity
> { exp(-qy) * integrate from - infinity to 
infinity
> {exp(-rz) * impulse fcn of ( u - u0 ) dz} dy) dx}
>
> where u = sqrt( x^2 + y^2 + z^2 ).
>
> It is the three dimensional Laplace transform of delta( u -
u0 ), where
u is
> the radial coordinate direction.
Ok, here are a few snips from some raw unpublished stuff Id
been
doodling on at some time:
Making use of a few relations from an integral
tablecite[4.644,
3.715.21]{gradstein} we
arrive at the following result for an n-dimensional transform:
begin{equation}
label{eq:radial}
x(|vec t|) corr int 0^infty x(r), 2pi sqrt{r^nover
|vec f|^{n-2}}, J {n-2over2}(2pi r,|vec f|), dr
end{equation}
Explicating a few dimensions, we get
begin{equation}
f(|vec t|) corr int 0^infty dr , f(r) cdot
begin{cases}
2cos(2pi r,|vec f|)&text{for $n=1$}
2pi r, J 0(2pi r,|vec f|)&text{for $n=2$}
{2rover|vec f|}sin(2pi r,|vec f|)&text{for $n=3$}
end{cases}
end{equation}
So if we have delta(u-u0), it would appear that the transform
is
2 (u0/v) * sin(2 pi u0 v)
where v = sqrt(f^2+g^2+h^2).
Looks simple enough. Now ignoring the convergence of this
expression,
the Laplace transform probably is the same, just replacing 2
pi j f
with p, 2 pi j g with q, 2 pi j h with r, leading to
something like
w = sqrt(p^2+q^2+r^2) = (2 pi j) v
so that the transform becomes
4 pi j (u0/w) * sin(u0 w/j) = 4 pi (u0/w) sinh(u0*w)
Of course, no guarantees that anything is done correctly, and
in
particular that one can basically ignore the convergence
issues like
that when switching from Fourier transform to Laplace
transform.
So you better have some way to check this result for
correctness.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: A Laplace Transform Integral
boundary=----=_NextPart_000_0069_01C4E611.610D4A70
-------------------------------------------------------------
--------
Dave,
J (3/2)(2 pi v) / v^(3/2)
where v = sqrt(a^2 + b^2 + c^2), a, b, and c being transform
harmonics, and,
I would assume J (3/2) being the Bessel function of order
3/2, then obtaining
the derivative in configuration space would be the equivalent
of multiplying
by the appropriate harmonic 2 pi j[ a,b,c] in transform space.
Ed
--
Edward Hyman
EdwardH@email.uophx.edu
Other EMail: e.hyman@worldnet.att.net
> Integrate from - infinity to infinity
> { exp(-px) * integrate from - infinity to 
infinity
> { exp(-qy) * integrate from - infinity to 
infinity
> {exp(-rz) * impulse fcn of ( u - u0 ) dz} dy) dx}
>
> where u = sqrt( x^2 + y^2 + z^2 ).
>
> It is the three dimensional Laplace transform of delta( u -
u0 ), where
u is
> the radial coordinate direction.
Well, the n-dimensional indicator sphere (1 inside everything
of
radius 1, 0 outside) has the Fourier transform (in v =
sqrt(a^2+b^2+c^2) when p = 2 pi a, q = 2 pi b, r = 2 pi c)
J {n/2}(2 pi v) / v^(n/2).
If we form the negated gradient of the indicator function to
get the
transform of an outward pointing delta vector , in transformed
coordinates a, b, c, we get:
[F a, F b, F c] = -2 pi j [a,b,c] * J {n/2}(2 pi v)/v^(n/2)
Maybe that helps puzzling together something akin to a
Laplace transform.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Probability Of No Meeting!
As Im sure many of you know, if N people, each of whom are
willing
to wait d amount of time (where d is a fraction of the time
span that
they arrive between) agree to meet each other in a specified
time
range, and their arrival times are randomly distributed, then
the
probability that ALL of them meet is
P(ALL)=n*d^(n-1)-(n-1)*d^n
and the probability that NONE of them meet is given by
P(NONE)=(1-(n-1)d)^n
Awhile back, I posed the question of what is the probability
that ALL
meet if each has a different waiting time, and what is the
probability
that NONE of them meet.
one person was able to compute for me the probability that
ALL meet
and put his answer in a PDF file which can be found here.
http://webpages.charter.net/smithability/math/Meeting_
Probabilities.pdf
A few years earlier, someone else gave me a formula for
P(NONE) if all
of the waiting times were different. I dont remember it 
all.
I recall
something like
P(NONE MEET)=Product(1-S)^N Where S=Sum(d_k)
j=1 to N K=1 to N
for K<>j
This is if N people agree to meet, but the above formula
doesnt
compute for the simple case of N=2, and say, d_1=.1 and
d_2=.25.
I seem to recall that the formula was really
Product(1-S+(something else))^N
j=1 to N with S defined as above.
but I dont recall what that something else was. Does anyone
want to
tackle this?
--
Patrick D. Rockwell
prockwell@thegrid.net
hnhc85a@prodigy.net
patri48975@aol.com
===
Subject: what boolean connectives together with -> make a
complete system?
like NOT and OR is complete.
NOR is complete
-> and ?
Herc
--
YOU CANT PROVE ME
If you prove its true then it has a proof, which makes it
false
If you dont prove it, then its true
10,000 people in sci.math ALL believe this is irrefutable
that mathematics
will always be incomplete.
===
Subject: Re: what boolean connectives together with -> make a
complete
system?
I. (nor)
1.~p =df (p nor p)
2. p v q =df ~(p nor q)
3. _T =df p v ~p
4. F =df ~T
5. p & q =df ~(~p v ~q)
6. p nand q =df ~(p & q)
7. p -> q =df ~p v q
8. p -|-> q =df ~(p -> q)
9. p <-> q =df (p -> q)&(q -> p)
10. p xor q =df ~(p <-> q)
II. (nand)
1. ~p =df (p nand p)
2. p & q =df ~(p nand q)
3. F =df p & ~p
4. T =df ~F
5. p v q =df ~(~p & ~q)
6. p nor q =df ~(p v q)
7. p -> q =df ~p v q
8. p -|-> q =df ~(p -> q)
9. p <-> q =df (p -> q)&(q -> p)
10. p xor q =df ~(p <-> q)
III. (T, -|->)
1. ~p =df T -|-> p
2. F =df ~T
3. p -> q =df ~(p -|-> q)
4. p v q =df ~p -> q
5. p nor q =df ~(p v q)
6. p & q =df ~(~p v ~q)
7. p nand q =df ~(p & q)
8. p <-> q =df (p -> q)&(q -> p)
9. p xor q =df ~(p <-> q)
IV. (F, ->)
1. ~p =df p -> F
2. T=df ~F
3. p -|-> q =df ~(p -> q)
4. p v q =df ~p -> q
5. p nor q =df ~(p v q)
6. p & q =df ~(~p v ~q)
7. p nand q =df ~(p & q)
8. p <-> q =df (p -> q)&(q -> p)
9. p xor q =df ~(p <-> q)
V. (~, v)
1. T =df p v ~p
2. F =df ~T
3. p nor q =df ~(p v q)
4. p -> q =df ~p v q
5. p -|-> q =df ~(p -> q)
6. p & q =df ~(~p v ~q)
7. p nand q =df ~(p & q)
8. p <-> q =df (p -> q)&(q -> p)
9. p xor q =df ~(p <-> q)
VI. (~, &)
1. F =df p & ~p
2. T =df ~F
3. p nand q =df ~(p & q)
4. p v q =df ~(~p & ~q)
5. p nor q =df ~(p v q)
6. p -> q =df ~p v q
7. p -|-> q =df ~(p -> q)
8. p <-> q =df (p -> q)&(q -> p)
9. p xor q =df ~(p <-> q)
VII. (~, ->)
1. T =df p -> p
2. F =df ~T
3. p -|-> q =df ~(p -> q)
4. p v q =df ~p -> q
5. p nor q =df ~(p v q)
6. p & q =df ~(~p v ~q)
7. p nand q =df ~(p & q)
8. p <-> q =df (p -> q)&(q -> p)
9. p xor q =df ~(p <-> q)
> like NOT and OR is complete.
> NOR is complete
> -> and ?
> Herc
> --
> YOU CANT PROVE ME
> If you prove its true then it has a proof, which makes it
false
> If you dont prove it, then its true
> 10,000 people in sci.math ALL believe this is irrefutable
that
mathematics will always be incomplete.
===
Subject: Re: what boolean connectives together with -> make a
complete
system?
> like NOT and OR is complete.
> NOR is complete
> -> and ?
->, NOT
->, F
->, XOR
->, NOT(p->q)
--
Mitch Harris
(remove q to reply)
===
Subject: Re: what boolean connectives together with -> make a
complete
system?
> like NOT and OR is complete.
> NOR is complete
> -> and ?
> ->, NOT
> ->, F
> ->, XOR
> ->, NOT(p->q)
Working on a 32 char symbol set.
Say we found an efficient TM coding for a UTM to parse, so 
that
useful algorithms were not sparse.... Im a poet!
Then UTM(x, y) mod 32 could be a generic formal system.
Its an event frame interpretation of theorems.
Instead of some evaluation space with some theorem to prove.
SimplifyThis:> (A v B) ^ (A v C)
And a dictionary of theorems to MATCH / INSTANTIATE onto that
program line.
Rule 1 : A, A ->B => B
Rule 2: (A v B) ^ (A v C) <=> A v ( B ^ C).
...
Instead you execute different applicable functions in the
function listing.
UTM(222, A v B) ^ (A v C)) => A v ( B ^ C). <conceptual>
You would have to ßag TM number 222 as being a valid
function, in that
the transformations it does are always equivalent.
Instead of a formula being a stagnant string of symbols, that
must be
matched
by a seperate parser, the formula IS the algorithm. The
formula/algorithm
is the TM
that takes the statement and makes the reduction.
Instead of
PARSER ( statement, {modus ponens}) = new_statement
ModusPonens (statement) = new_statement
Herc
===
Subject: How to solve this integral?
Integrate[1/(a Sin^4(t) + b Sin^2(t) Cos^2(t) + c
Cos^4(t)),{t,0,Pi}]
with the constranit: b^2-4ac<0
===
Subject: Re: How to solve this integral?
> Integrate[1/(a Sin^4(t) + b Sin^2(t) Cos^2(t) + c
Cos^4(t)),{t,0,Pi}]
> with the constranit: b^2-4ac<0
Start by substituting u = tan(t) and simplify. The decompose
the integrand
into partial fractions.
-Michael.
===
Subject: Re: How to solve this integral?
<41c67ad5$0$205$edfadb0f@dread12.news.tele.dk Integrate[1/(a
Sin^4(t) + b Sin^2(t) Cos^2(t) + c Cos^4(t)),{t,0,Pi}]
> with the constranit: b^2-4ac<0
> Start by substituting u = tan(t) and simplify. The
decompose the
integrand
> into partial fractions.
But in the interval [0,pi], tan t has problem at pi/2
Better to use is u = cot t.
sin arccot u = 1/sqr(1 + u^2)
cos arccot u = u/sqr(1 + u^2)
integral(-oo,oo) -(1 + u^2)^2 / (a + bu^2 + cu^4) du
But with the restrains, a + bu^2 + cu^4 doesnt have real
factors.
So how does one use partial fractions?
===
Subject: Re: How to solve this integral?
> But with the restrains, a + bu^2 + cu^4 doesnt have real
factors.
> So how does one use partial fractions?
If it has no real factors, then it has two quadratic factors.
Partial Fractions, as taught in calculus texts, covers that
case.
OR, use the complex roots directly in your partial fractions.
===
Subject: Re: stationary process pass through an LTI system
still
stationary?
Homework?
--
Stephan M. Bernsee
http://www.dspdimension.com
===
Subject: Re: stationary process pass through an LTI system
still
stationary?
<10s5shpq8j4h141@corp.supernews.com>
posting-account=MscchQwAAAD1m5NHYInCV5DpiRQ-0KT9
Well, think about it. A stationary process is one whos
parameters are
independent of time, although the value of the signal
resulting from
the
process can be correlated to past or future values of itself.
A time-invariant system is one that modifies a signal in a
time-independent way, although the output signal at any given
time may
depend on past (or future, if you dont care about 
causality)
values of
the input signal.
So you tell me -- is it still stationary, or has it acquired
a time
dependence?
Your explanation seems to make sense.
However, if you have an LTI system with poles on the unit
circle, one
can still say that it modifies the signal in a
time-independent way,
but the output is not stationary. Can you refine your 
argument?
Andor
===
Subject: Re: stationary process pass through an LTI system
still
stationary?
windows-nt)
Andor,
What about startup transients? I agree in the steady-state.
--RY
> Well, think about it. A stationary process is one whos
parameters are
> independent of time, although the value of the signal
resulting from
> the
> process can be correlated to past or future values of
itself.
> A time-invariant system is one that modifies a signal in a
> time-independent way, although the output signal at any
given time may
> depend on past (or future, if you dont care about
causality) values of
> the input signal.
> So you tell me -- is it still stationary, or has it
acquired a time
> dependence?
> Your explanation seems to make sense.
> However, if you have an LTI system with poles on the unit
circle, one
> can still say that it modifies the signal in a
time-independent way,
> but the output is not stationary. Can you refine your
argument?
> Andor
--
% Randy Yates % Shes sweet on Wagner-I think 
shed die
for Beethoven.
%% Fuquay-Varina, NC % She love the way Puccini lays down a
tune, and
%%% 919-577-9882 % Verdis always creepin 
from her room.
%%%% <yates@ieee.org> % Rockaria, *A New World Record*, ELO
http://home.earthlink.net/~yatescr
===
Subject: Re: stationary process pass through an LTI system
still
stationary?
<10s5shpq8j4h141@corp.supernews.com>
<vfazobor.fsf@ieee.org>
<10s8r3mj2ibkq54@corp.supernews.com>
posting-account=MscchQwAAAD1m5NHYInCV5DpiRQ-0KT9
If your system
_is_ time invariant then you must have a signal that is zero
for t < 0
and has energy for t > 0 -- and thats not a stationary
signal.
Point taken.
Now, granted, if you have an unstable or metastable system
your output
signal in such a case will have infinite variance -- this is
probably
why you want to hold things to zero until t = 0.
I guess thats what I had in mind. Best just forget about
unstable
systems.
Andor
===
Subject: Re: stationary process pass through an LTI system
still
stationary?
<10s5shpq8j4h141@corp.supernews.com>
<vfazobor.fsf@ieee.org>
posting-account=VhNkrw0AAABfAa_CaZ54XDxj0gJ5_J0G
> What about startup transients? I agree in the steady-state.
Well, then the system isnt, strictly speaking, LTI, is it?
:-)
Ciao,
Peter K.
===
Subject: Re: How to visualize limits in category theory?
with my supervisor, he said that it is nice to visualize
limits in say
the category of topologies as product of two spaces while
colimits can
be seen as the joining of two spaces (morphisms I suppose are
just the
homeomorphisms) in a pushout/pullback diagram (ie. a
commutative
rectangle such that if 3 objects and 2 morphisms are fixed and
another
commutative rectangle is formed with those fixed points and
morphisms
then there is a unique morphism from/to the 4th object of the
old
rectangle to/from the 4th object of the newly formed
rectangle!)...
Anyway, let me give some comments on one of the posts...
> Youve gotten several good responses to your question;
perhaps
> what you need are examples to test the responses on.
> Jesse Hughes pointed out that limits are reducible to
cartesian
> products and equalizers, and James Dolan observed that
limits
> are the basic stuff of algebraic geometry as initiated by
Fermat,
> Descartes, and others (loci of equations seen as equalizers
> defined on cartesian products).
> You might use these observations to visualize how the ring
of
> p-adic integers is constructed as a limit of a diagram of
shape
> --> Z_{p^n} --> Z_{p^{n-1}} --> ... --> Z_p.
correctly... The limit would then be those integers which
have the form
p^k .. ie. Lim(Z_p) := {p^k : k is in N} , accompanied with
the
canonical homomorphisms to each of the Z_p^n ... right?
> You might also visualize the patching condition in the
definition
> of sheaves F in terms of limits: if V is open and {V_i}
covers V,
> then F(V) is the limit of a diagram obtained by sticking
together
> diagrams of shape
> F(V_i) --> F(V_i cap F_j) <-- F(V_j)
> over all i and j.
Wow... very nice example. Er.. what does patching mean btw
(is that one
of the axioms of the sheaves?)
Jose Capco
===
Subject: Re: How to visualize limits in category theory?
One last thing. Something ran into my mind while I was
reading all the
thread, I forgot to ask. When defining the stalk of a sheaf,
there is
this funny limit sign with an arrow below it used... I never
knew what
this exactly was (as I was always thought that the stalks are
just those
under this equivalence relation that f_1 in F(V_1) and f_2 in
F(V_2) are
equivalent iff there is a W as a subset of V_1 cap V_2 such
that f_1 and
f_2 restricted to W are equal) ... But since I read a lot of
category
theory and having encountered colimits, I am guessing (but I
did not
verified) that this is just a direct limit (which is a special
colimit),
right? ... I just want to save my time here :-)
Jose Capco
===
Subject: Re: How to visualize limits in category theory
|>> When I think about limits, I think about subobjects of
products.
|>complete just in case it has equalizers and products. My
answer is
|>relevant to completeness, but not a good answer to how to
visualize
|>limits.
|
|I was pleased to see Michael Barr rework the verb being used
|into think about, and Im distressed to see you return to
|visualize.
When I visualize products, I often visualize them as boxes.
When
I visualize subobjects, I often visualize them as amorphous
shapes
inside the boxes. :-)
Actually, I tend to keep a picture in mind of the diagram
used to
define the limit object, picturing the object itself as the
result of
a compromise between two forces. The definition can be
considered
in two parts; on the one hand the object has to be large
enough
(it has supporting struts inside...) and on the other hand it
cant
be excessively big (it has a tight covering surrounding it).
Keith Ramsay
===
Subject: Re: How to visualize limits in category theory
posting-account=tdYrvA0AAACTm02P3kgbrEwRzJI8b08S
Oh come on, just admit it! Youre thinking in terms of venn
diagrams
like the rest of us ;)
Rudolph)
> | |>> When I think about limits, I think about subobjects
of products.
> | |>complete just in case it has equalizers and products.
My answer is
> |>relevant to completeness, but not a good answer to how to
visualize
> |>limits.
> |I was pleased to see Michael Barr rework the verb being
used
> |into think about, and Im distressed to see you return to
> |visualize.
> When I visualize products, I often visualize them as boxes.
When
> I visualize subobjects, I often visualize them as amorphous
shapes
> inside the boxes. :-)
> Actually, I tend to keep a picture in mind of the diagram
used to
> define the limit object, picturing the object itself as the
result of
> a compromise between two forces. The definition can be
considered
> in two parts; on the one hand the object has to be large
enough
> (it has supporting struts inside...) and on the other hand
it cant
> be excessively big (it has a tight covering surrounding it).
> Keith Ramsay
===
Subject: Re: How to visualize limits in category theory
> I was pleased to see Michael Barr rework the verb being used
> into think about, and Im distressed to see you return to
> visualize. Nothing Ive read in this thread, from the
> beginning on, has seemed to *me* to have anything at all to
> do with visualization (except, maybe, the occasional post
> referring to pentagons). Probably the original poster 
didnt
> mean what I mean by visualization either, but if he did, he
> hasnt been replied to.
How about visualizing a rising line | split into two rays
/
|
and then each ray split into two rays
/ /
/
|
and so on, ad infinitum. And then think of this as a great
candle holder
(I am inspired here by Wolfs Menorah, although actually a
different
object,
communicated to me by my friend Bob Wolf, the logician). Then
one can
visualize the inverse limit as the set of candles.
--
Allan Adler <ara@zurich.csail.mit.edu>
* Disclaimer: I am a guest and *not* a member of the MIT
CSAIL. My actions
and
* comments do not reßect in any way on MIT. Also, I am
nowhere near
Boston.
===
Subject: Re: How to visualize limits in category theory
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBI3LQM01576;
> When I think about limits, I think about subobjects of
products.
>>complete just in case it has equalizers and products. My
answer is
>>relevant to completeness, but not a good answer to how to
visualize
>>limits.
>I was pleased to see Michael Barr rework the verb being used
>into think about, and Im distressed to see you return to
>visualize. Nothing Ive read in this thread, from the
>beginning on, has seemed to *me* to have anything at all to
>do with visualization (except, maybe, the occasional post
>referring to pentagons). Probably the original poster 
didnt
>mean what I mean by visualization either, but if he did, he
>hasnt been replied to.
>Lee Rudolph
Perhaps youre right, but there are ways of visualizing 
limits
and colimits of functors of the form F: B --> Set [B small].
The colimit of F: B --> Set can be visualized as the set of
path components of the category of elements of F, el(F).
Recall
that the objects of el(F) are pairs (x, b) where b is an
object of B and x is an element of Fb. Morphisms in el(F)
of the form (x, b) --> (y, c) are morphisms f: b --> c
such that F(f)(x) = y. You can visualize E = el(F) as fibered
over B, where the fiber over an object b is the set Fb, and
given an arrow f: j --> k in the base, there is a unique
arrow in E over f whose source is a given element x over b.
Let E_1 denote the collection of arrows of E, and let E_0
denote the set of objects. The domain/source and
codomain/target
give a parallel pair of functions
-->
E_1 --> E_0
and the coequalizer of this pair, denoted pi_0(E), is the
colimit of F. This is just the set of path components of the
simplicial set given by the nerve of E.
There is a way of dualizing all this to get limits (which I
could explain if pressed), but to make a long story short, the
limit of F is the set of sections of the simplicial projection
map p between the nerves of E and B, part of which looks like
-->
E_1 --> E_0
| |
p_1 | | p_0
v v
B_1 --> B_0 .
-->
Each section s of p is uniquely determined by its
0-dimensional
part s_0 (because of the unique-arrow lifting property of p
mentioned above), but not all sections s_0 of p_0 extend to
sections s of p. In short, sections s are matching families:
tuples <x_b in Fb>_(b), indexed by objects b of B, which
satisfy
the matching condition:
F(f)(x_b) = x_c
whenever f: b --> c is an arrow of B.
How satisfactory all this is from a visual standpoint I leave
to the individual reader to decide.
Todd Trimble
===
Subject: Complex Analysis Question on Rouches Theorem 
Section
Im working on this problem but I cant 
find the answer. This
is not HW or
exam but Im just practising for my qualifiers.
This is from Sarasons Notes on Complex Function Theory,
X.12, prob. 8.
The problem is
Prove that, if 0 < |a| < 1 and n is a positive integer, then
the equation
(z-1)^n exp(z) = a
has exactly n roots each of multiplicity 1, in the half plane
Rez > 0.
Prove that if |a| <= 2^{-n} the roots are all in the disk
|z-1| < 1/2.
Since this is on Rouches theorem, I took
f(z) = (z-1)^n - a
and
g(z) = (z-1)^n exp(z) - a
the function given.
Now, I have to find a compact set K that contains all the
roots of f and
that |f-g| < |f| on the boundary of K to apply Rouches
theorem.
I take the K to be the disk centered at 1 and radius 1. This
contains all
the roots of f(z) but of course
|f-g| = |(z-1)^n (1 - exp(z))|
<= |(z-1)|^n |(1 - exp(z))|
= |1 - exp(z)|
which is not smaller than |f(z)| on the circle. For example
at z = 2, |1 -
exp(2)| > |1-a|
So my question is, is my f(z) choice good?
If so, what K should I be using?
===
Subject: Re: Complex Analysis Question on Rouches Theorem
Section
posting-account=W2DCTA0AAAAlbhDMl3GrysSnPy1IK_7f
At first glance, there seem to be a number of ways to solve
this
problem. In any case, note that, if you look at the function
f(z) := (z-1)^n exp(z),
then you know where the zeros of this function are --- there
is exactly
one, with multiplicity n, at 1. So if you are going to look
for the
preimages of values a close to 0, it seems reasonable to
apply Rouches
theorem to this f, and the perturbed function
g(z) := (z-1)^n exp(z) - a.
Note, however, that you will also have to change the domain
to which
you are applying the theorem, since you want to count all
roots in the
half plane, rather than just a circle.
Hope this helps,
Lasse
---
===
Subject: Re: Complex Analysis Question on Rouches Theorem
Section
> At first glance, there seem to be a number of ways to solve
this
> problem. In any case, note that, if you look at the function
> f(z) := (z-1)^n exp(z),
> then you know where the zeros of this function are ---
there is exactly
> one, with multiplicity n, at 1. So if you are going to look
for the
> preimages of values a close to 0, it seems reasonable to
apply Rouches
> theorem to this f, and the perturbed function
> g(z) := (z-1)^n exp(z) - a.
> Note, however, that you will also have to change the domain
to which
> you are applying the theorem, since you want to count all
roots in the
> half plane, rather than just a circle.
> Hope this helps,
> Lasse
> ---
at (z-1)^n = a
===
Subject: Re: Complex Analysis Question on Rouches Theorem
Section
>At first glance, there seem to be a number of ways to solve
this
>problem. In any case, note that, if you look at the function
>f(z) := (z-1)^n exp(z),
>then you know where the zeros of this function are --- there
is exactly
>one, with multiplicity n, at 1. So if you are going to look
for the
>preimages of values a close to 0, it seems reasonable to
apply Rouches
>theorem to this f, and the perturbed function
>g(z) := (z-1)^n exp(z) - a.
>Note, however, that you will also have to change the domain
to which
>you are applying the theorem, since you want to count all
roots in the
>half plane, rather than just a circle.
Its easy to see that g has no zeroes in the set 
defined by
Re z > 0 and |z - 1| >= 1, because |e^z| > 1 and |a| < 1.
So showing there are exactly n zeroes in |z - 1| < 1 shows
there are exactly n zeroes in the right half-plane.
>Hope this helps,
>Lasse
>---
************************
David C. Ullrich
===
Subject: Re: Complex Analysis Question on Rouches Theorem
Section
> Its easy to see that g has no zeroes in the set 
defined by
> Re z > 0 and |z - 1| >= 1, because |e^z| > 1 and |a| < 1.
> So showing there are exactly n zeroes in |z - 1| < 1 shows
> there are exactly n zeroes in the right half-plane.
Sure. I was just pointing out to the OP that there is
something to be done.
--
---------------------------------------------------
Lasse
http://www.maths.warwick.ac.uk/~lasse
---------------------------------------------------
===
Subject: Re: Disappointed Differential
> So Im hoping you have that brief survey that you could
send to me a
> copy.
There is NO brief survey. There is NO management summary.
There never
will be any. I can not and will not send you that copy. The
hand of God
is in the tiny details. And not so much in generalities.
Han de Bruijn
===
Subject: Re: Moments over a Simplex
Ive got more than enough material, so that 
Im able to
proceed with
the real thing now.
Han de Bruijn
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===
Subject: Re: Restriction on Longitudinal Waves
> To cut right to the chase (since I just unintentionally
erased my
> beautiful ßorid draft), your result is based on the
assumption that a
> medium cannot overrun itself -- that is, in a longitudinal
displacement
> wave, a point A in the medium originally to the left of a
point B
> cannot wind up to the right of B. (In fairness you mention
something
> like this).
Yes, thats precisely what I meant.
> The sharpest way of expressing this seems to be the
requirement that
> dy/dx > -1, where x is a coordinate position in the
undisplaced medium,
> y(x) the displacement of that point; this at a particular
snapshot t.
> Anyway, assuming a waveform y = Asin(kx), this gives us the
requirement
> A < wavelength/2pi, as you said. So subject to (1) the
assumption that
> the medium cannot interpenetrate itself and (2) the caveat
that the
> slope condition is more fundamental than its specialization
to a sine
> wave, and (3) the emphasis that this applies only to the
displacement
> in a longitudinal displacement wave -- yes, you are correct.
I will admit immediately that the slope condition dy/dx > -1
is a much
better approach. Its specialization to a saw-teeth function
would result
in the requirement that amplitude < wavelength , which is
different from
amplitude < wavelength / (2.pi) , though of the same order of
magnitude.
> We could ask what kind of medium might violate assumption
(1). A
> Knudsen gas comes to mind, or something that looks like a
Knudsen gas.
> For example, a loosely spaced platoon of marching soldiers.
If routed
> from behind, the rear line might well overrun the front
line!
Dont forget that a longitudinal wave already _presupposes_
the presence
of a continuum. The kind of overrun you propose is difficult
in there;
at least thats what I think.
> One question: when you write transversal wave, did you mean
to write
> transverse wave, or is that some other term? I cant see
what
> connection a transverse wave has with any of this. And may
I say, you
> seem to have taken a simple idea and converted it into an
impenetrable
> if impressively typeset line of math! One shudders to think
what might
> happen if you exposited something truly difficult. ;-)
Uhm, my English is not as good as my Dutch. But Ive always
thought that
transversal is the correct term for a wave that is not
longitudinal.
And yes, my mathematics should be simplified considerably with
your idea
of that slope > -1 , admittedly.
Han de Bruijn
===
Subject: algebra, geometry, trigonometry books
Hi everyone,
I am looking for a book that has many algebra, geometry and
trigonometry
problems. The books, that I have found, have very easy
problems. I am
looking for a book that has many challenging problems as
well. Please, let
me know if you know any good math book that has many practice
problems.
===
Subject: Re: algebra, geometry, trigonometry books
posting-account=JdyOWwwAAAC8AhuHITILJrgANkHxqjXJ
For challenging problems you could look at the collections of
Mathematics Olympiad problems published by the Mathematics
Association
of America, Freeman, and other publishers.
===
Subject: Re: algebra, geometry, trigonometry books
> For challenging problems you could look at the collections
of
> Mathematics Olympiad problems published by the Mathematics
Association
> of America, Freeman, and other publishers.
And go to Internet website of International Mathematical
Olympiad (dont
remember adress) and of National Olimpiads of Countries in
which people
speak some language You know.
sirix.
===
Subject: Re: algebra, geometry, trigonometry books
posting-account=0QrkrwwAAABDyQGPKX7NtkkaKfngvovA
> Hi everyone,
> I am looking for a book that has many algebra, geometry and
trigonometry
> problems. The books, that I have found, have very easy
problems. I am
> looking for a book that has many challenging problems as
well.
Please, let
> me know if you know any good math book that has many
practice
problems.
Fifty years ago in high school, the math teachers relied on
Mathematics
Review Exercises, by Smith and Fagan.
David Ames
===
Subject: Re: algebra, geometry, trigonometry books
>I am looking for a book that has many algebra, geometry and
trigonometry
>problems. The books, that I have found, have very easy
problems. I am
>looking for a book that has many challenging problems as
well. Please, let
>me know if you know any good math book that has many
practice problems.
Are the Schaum Outline problems too easy? For challenging
problems, you
should look through issues of Mathematics Magazine, which has
a problems
section. The magazine is geared towards undergraduates, but
much of the
geometry and trigonometry requires material that at least
used to be
taught in U.S. high schools.
Also, look online at the SMSU Problem Corner,
http://math.smsu.edu/~les/POTW.html .
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Wikipedia on Decision Problems
Important undecidable decision problems include the halting
problem
and Goodsteins theorem;
It seems to me that this is mixed up:
Goodsteins Theorem is a true arithmetical statement that is
unprovable
in Peano Arithmetic. (Why would it be called a decision
problem?)
The Halting problem is a decision problem where no algorithm
exists to correctly decide all cases (an unsolvable problem,
just as the word problem for groups is unsolvable.)
http://en.wikipedia.org/wiki/Decision_problem
David Bernier
===
Subject: Re: Wikipedia on Decision Problems
> Important undecidable decision problems include the halting
problem
> and Goodsteins theorem;
> It seems to me that this is mixed up.
Indeed it is. No theorem is a decision problem, let alone an
undecidable
one, although theorems in the informal sense of the word may
turn out to
be undecidable in all sorts of formal theories.
--
Aatu Koskensilta (aatu.koskensilta@xortec.fi)
Wovon man nicht sprechen kann, daruber muss man schweigen
- Ludwig Wittgenstein, Tractatus Logico-Philosophicus
===
Subject: Re: Force to be calculated
>> Everybody who has done a bungee jump will correctly guess
that one
> will
>> undergo a 2mg apparent force (i.e. twice ones weight) at
the deepest
>> point of the jump.
> If so, that must be a property of the bungee cord material,
not
> true in general.
> Let us consider a completely elastic bungee cord (if it
> were an ideal elastic, then you would bounce back up to
> exactly your original height. I dont think this is
> the case with actual bungee jumping.)
This is not enough to determine the behavior of the cord. One
can have a perfectly elastic cord without having it conform to
the ideal spring law (force = k * displacement).
> The bungee cord has an elastic constant k. Assume you
> have velocity v when you start stretching it. The cord
> will continue to stretch until all of your kinetic energy
> 0.5*m*v^2 has been converted into potential energy
0.5*k*x^2,
> or x = v*sqrt(m/k). At this point, the force exerted by
> the bungee cord is kx = k*v*sqrt(m/k) = v*sqrt(mk). The
> acceleration will be F/m = v*sqrt(k/m).
Ok. So youre assuming a cord that obeys the ideal spring 
law.
And youre assuming that we have some non-zero velocity when
we
start stretching the cord. The latter is indeed a component of
real-world bungee jumping -- the cord has significant length,
even
when not under tension.
But your equations are wrong. Gravity doesnt magically stop
working
when the bungee cord starts to tighten. You have a component
of gravitational energy given by mgx.
Please redo your calculations accordingly.
[Note that bungee cord, like any rope, does not support
compressive
loads. It will only obey the ideal spring law when under
tension.
But, since we are focusing our attention on the portion of
the jump
where all the slack has been taken up and the bungee is under
tension,
this is no obstacle to a correct analysis]
> Two things to observe about this:
> 1. The force and the acceleration depend on the elastic
> constant. A stiffer cord (higher k) will cause larger
> force.
Yup. That much is obvious. At least for a non-zero starting
velocity.
> 2. Even with the same bungee cord, the acceleration depends
> on mass (a larger mass will end up experiencing less
acceleration).
> The cord may have the right elastic constant that an
> average adult will experience 2 gs, but above and below
> average adults will experience different accelerations.
I think you are wrong here. The _minimum_ (over all adult
weights)
of the maximum (over all points in the jump) acceleration is
guaranteed
to be at least two gs for an ideal (force = k *
displacement) bungee.
This minimum is exactly achieved in the case of zero initial
velocity
> The above are valid for any elastic collision, not just
> bungee jumping.
Do you know what elasticity means? It is a measure of the
degree
to which the energy after the rebound matches the energy
prior to
impact. It is not a measure of the degree to which restoring
force
is proportional to displacement. It _is_ a measure to the
degree
to which the restoring force is _any function of_ the
displacement,
independent of all other factors and, in particular, a
measure of
the degree to which the restoring force is independent of
direction and speed.
John Briggs
===
Subject: Re: f( x +2f(y) ) = f(x) + y + f(y) ,is my solving
correct
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNr011174;
I thank you for having replied .
This problem was recently solved an altogether way by a reader
of a U.S Math review.
I try to validate a direct method using Abel functions, so
your
comments now and within a near future are welcome .
An other aspect in solving functional equations which seems of
concern to me is considering them as a result of a process.
I want to illustrate with a short example my idea.
Example: g(x^2+2x)=d/dy g(x,y) (g a smooth real required
func)(1).
We may Ôcount occurrences of x->x^2+2x with phi 
Abel function
(lnln(x+1)/ln(2)+k ) k constant,and describe (1) as :
process { when x->x^2+2x g->d/dy g } ;so
g(x ,y) = d/dy ^[(lnln(x+1)/ln(2)+k )]Á
m(y) is
a formal solution ,here we have a continuous y derivate and
m(y)
does fit it.
A simple case m(y)=exp(2y) gives:g(x ,y)=
2^(lnln(x+1)/ln(2)+k )*e(2y)
or g(x ,y) = c*ln(x+1)*e(2y) , c constant
===
Subject: Choice of dep. Variables for ode system
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNsR11192;
Folks:
I have 3 fist order odes and two algebriac 
equations among 5
variables.
Both the equations and the odes are non-linear. In 
addition,
the odes are
autonomus.
So, this can be written as a system of 5 odes in the 
five
variables or
after using the algebriac equations to eliminate 2 varaibles
(messy), as a
system of 3 odes in three variables. Is one set to be
preferred over the
other ? More generally, what are some guidelines for the best
choice of dep.
variables for such a system.
Thabks for your help.
Sam
===
Subject: Re: JSH: Somewhat puzzled
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNtc11236;
>> Ive been following the argument for several months, and 
I
have been
>> wondering that myself. Just when I think I have studied
enough algebra
to
>> unserstand the problem, it slithers away from me. It looks
at times
like
>> James has discovered a problem with the number 49 - or
mayvbe its 5.
>I think it is a problem with properly units, but I am not
sure.
What REALLY happened was that about 7 years ago, JSH came up
with a
proof of Fermats last theorem that required only high 
school
math.
When some people pointed out that he was using facts that
were NOT true
of the integers but were true of the algebraic integers, he
extended his
high school proof to include algebraic integers.
When others then gave simple examples that other things he
was claiming
about the algebraic integers was not true, he decide that the
problem lay,
not with his proof, but with the algebraic integers!
===
Subject: Re: mathematicians utter contempt for common sense
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNtl11222;
>> |YOU CANT PROVE ME..............T |- G <=> ~phi(G)
>> There is an important qualifier youre 
leaving out: prove
*in the formal
>> system T*.
>> |If you prove its true
>> *in system T*
>> |then it has a proof, which makes it false.
>> AND shows that the system T cant both have all of its
axioms correct
>> and all of its rules of inference valid. Hence the proof
in system T
does
>> not deserve to be called simply a proof.
>> |If you dont prove it,
>> *in system T*
>> | then its true.
>> |10,000 people in sci.math ALL believe that it means
mathematics will
always
>> |be incomplete.
>> No. Almost all of them believe only that it means each
formal system
>> for mathematics is incomplete. I dont know what it would
mean to say
>> that mathematics itself (mother mathematics as James
Harris has
>> been known to say) is incomplete.
>That is a scattered cover up to substantiate the claim as an
oxymoronic
PROOF.
>No statements practically used in mathamatics have their
formula considered
IN A SYSTEM.
>the hypotenuse is the sum of the squares of the other two
sides.... theres
no IN THIS SYTEM!
>For every statement this has no proof in system S,
>there is an equivalent statement this has no proof in any
system
whatsoever.
>And your PROOF of the truth of this macro godel statement
cannot be
applied.
>What you are all ignoring is that theorem provers come in 2
ßavors:
>Belief Revision
>Truth Maintenance
>In Belief Revision you can assume anything is true, like a
godel statement
until
>it is disproven, which it isnt. This the DOMAIN that
Godels proof holds
water,
>no generic claims about his pardoxical statements hold
throughout.
>In Truth Maintenance you take unforgiving responsibility
when you claim a
statement is true.
>T |- Af, f <-> phi(f)
>All true statements must be proven. SIMPLE NO?
>T |- G <=> ~phi(G) this is just G=~G in disguise.
>This is an invalid formula in Truth Maintenance Theorem
Provers.
>A true statement is a subset of the formula sequence that is
its proof.
>------------------------------------------------------------
--------
>G -> ~proof(G) [1]
>G C P .....P= !proof(G)
>where
>G -> proof(G) [contradicts 2]
>Why on Earth would you define truth as possibly unproven?
>Herc
Anyone who has taken at least highschool geometry knows that
the
Pythagorean theorem: the hypotenuse is the sum of the squares
of the other
two sides is only true in Euclidean Geometry- thats the
system
referred to.
It is also true that mathematicians (and logicians) prefer to
use the word
valid rather than true.
The only one showing a contempt for commonsense is HERC.
===
Subject: Choice of Dep. Variables for ODE system
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNuN11292;
Folks:
I have 5 unknown variables and a system of three ODES and two
algebriac
equations between these 5 variables. The ODEs and
the equations are all non-linear and are to be solved as an
IVP.
This can be wriiten as a system of 5 ODEs in the 
five
variables or two of
the variables eliminated (with some algebriac complexity) to
get a system of three ODEs in three unknowns. Is one method
better than the
other ? More generally, what is the best choice of
dependent variables for such a system.
Sam
===
Subject: Re: Eulers identity
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNuJ11299;
>Eulers identity is;
>exp(j*theta) = cos(theta) + j*sin(theta).
>What about the expression
>exp(j*ln(theta))
>Are there any interesting realtionships that can be derived
for this?
>Bob Adams
Oh, those engineers and their jmaginary numbers!
Assuming that theta is a real number then
exp(j*ln(theta))= exp(j)exp(ln(theta)= theta* exp(j)
Of course, from exp(j*theta)= cos(theta)+ j*sin(theta) we
have exp(h)= cos(1)+ j sin(1) so
exp(j*ln(theta))= theta(cos(1)+ j sin(1)).
===
Subject: Re: Eulers identity
>>Eulers identity is;
>>exp(j*theta) = cos(theta) + j*sin(theta).
>>What about the expression
>>exp(j*ln(theta))
>>Are there any interesting realtionships that can be derived
for this?
>>Bob Adams
>>
> Oh, those engineers and their jmaginary numbers!
> Assuming that theta is a real number then
> exp(j*ln(theta))= exp(j)exp(ln(theta)= theta* exp(j)
Wrong. exp(a*b) = exp(a) exp(b)?????
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: The State-of-the-Art in Mathematics, but didnt
I solve this?
(Smart1234)
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNuW11310;
>The problem here is not the choice of values of sqrt(-1) but
the concept
i
>itself. It is not well-defined because the mapping sqrt is
well-defined
only
>on perfect square such as 1, 4, etc.
>E. E Escultura
>>Ok can it be represented like this?
>>n=1 to oo
>>with cartesian coordinates x,y,z
>>Use only integer powers of n
>>otherwise its an indeterminate
>>[(-1)^n/2] x
>>[(-1)^n/2] y
>>[(-1)^n/2] z
>>For x coordinate,
>>[(-1)^n/2] x
>>at n=1 --> indeterminate
>>at n=2 --> -1
>>at n=3 --> indeterminate
>>at n=4 --> 1
>>at n=5 ---> indeterminate
>>at n=6 ---> -1
>>at n=7 ---> indeterminate
>>at n=8 ---> 1
>>etc...
>> In cases where you have sqrt(-2), etc..,
>>for n=1 --> oo
>>Integer Powers of n Only
>>sqrt (-2) = sqrt(2) [(-1)^n/2)]
>> = (1.41421...) [(-1)^1/2)]
>> = (1.4142...) ( indeterminate)
>>And,
>>(sqrt (-2))^2 = (1.41421...)^2 [(-1)^2/2)]
>> = (2)(-1)
>> = -2
>>And,
>>(sqrt(-2))^3 = (1.41421...)^3 [(-1)^3/2)]
>> = (2.8284...) (indeterminate)
>>And,
>>(sqrt(-2))^4 = 3.9998...
>>And so on...
>> Does this map sqrt correctly with negative sqrts?
> Why dont you reply? Why doesnt anyone 
reply? Whats wrong
you dont
like to
>give me credit for anything?
Credit for what? you havent said anything worth paying
attention to. The
answer to S. Enterprize Companys original statement is just
that he is
completely wrong about everything. Mathematicians DONT 
define
i as the
square root of -1 for exactly the reason he states: every
number has two
square roots and since the complex numbers are not an ordered
field it is
impossible to distinguish between +i and -i.
What mathematics DO is to define the complex numbers as the
set of
PAIRS of real numbers (x,y) with additon defined by (a,b)+
(c,d)= (a+c, b+d)
and multiplication defined by (a,b)*(c,d)= (a*c-b*d,a*d+b*c).
That way, it
happens that (0,1)*(0,1)= (0,-1)*(0,-1)= (-1, 0) which can be
corresponded to
-1. i is specifically defined as (0,1).
===
Subject: Re: The State-of-the-Art in Mathematics, but didnt
I solve this?
(Smart1234)
>>The problem here is not the choice of values of sqrt(-1)
but the concept
i
>>itself. It is not well-defined because the mapping sqrt is
well-defined
>only
>>on perfect square such as 1, 4, etc.
>>E. E Escultura
>Ok can it be represented like this?
>n=1 to oo
>with cartesian coordinates x,y,z
>Use only integer powers of n
>otherwise its an indeterminate
>[(-1)^n/2] x
>[(-1)^n/2] y
>[(-1)^n/2] z
>For x coordinate,
>[(-1)^n/2] x
>at n=1 --> indeterminate
>at n=2 --> -1
>at n=3 --> indeterminate
>at n=4 --> 1
>at n=5 ---> indeterminate
>at n=6 ---> -1
>at n=7 ---> indeterminate
>at n=8 ---> 1
>etc...
> In cases where you have sqrt(-2), etc..,
>for n=1 --> oo
>Integer Powers of n Only
>sqrt (-2) = sqrt(2) [(-1)^n/2)]
> = (1.41421...) [(-1)^1/2)]
> = (1.4142...) ( indeterminate)
>And,
>(sqrt (-2))^2 = (1.41421...)^2 [(-1)^2/2)]
> = (2)(-1)
> = -2
>And,
>(sqrt(-2))^3 = (1.41421...)^3 [(-1)^3/2)]
> = (2.8284...) (indeterminate)
>And,
>(sqrt(-2))^4 = 3.9998...
>And so on...
> Does this map sqrt correctly with negative sqrts?
>> Why dont you reply? Why doesnt anyone 
reply? Whats
wrong you dont
like
>>give me credit for anything?
>Credit for what? you havent said anything worth paying
attention to.
The
>answer to S. Enterprize Companys original statement is 
just
that he
is
>completely wrong about everything. Mathematicians DONT
define i as the
>square root of -1 for exactly the reason he states: every
number has two
>square roots and since the complex numbers are not an
ordered field it is
>impossible to distinguish between +i and -i.
> What mathematics DO is to define the complex numbers as the
set of
PAIRS
>of real numbers (x,y) with additon defined by (a,b)+ (c,d)=
(a+c, b+d) and
>multiplication defined by (a,b)*(c,d)= (a*c-b*d,a*d+b*c).
That way, it
>happens that (0,1)*(0,1)= (0,-1)*(0,-1)= (-1, 0) which can
be corresponded
to
>-1. i is specifically defined as (0,1).
First of all, I dont agree with you when you imply that if 
I
dont know
one
thing, then I dont know anything, or if I appear to be 
wrong
about one
thing,
then I am wrong about everything. Lets look at this
mathematically.
Let x= 1 mistake or something you think I said wrong.
n = 2 to oo
Let y = lim SUM A_k = no mistakes
You can see that I could be right 99.999... % of the time
with only one
thing mistake. This would show that YOU are mistaken about
everything
except
one thing. So who is the most correct? I am.
Now going back to this topic of imaginary numbers. I already
know the
standard or conventional way imaginary numbers are handled.
But it was
pointed
out to me by someone that i is ill-defined and there are
problems
mapping it.
If you look at the meaning of i,
sqrt (-1) = i
there is no such thing unless you pair it like you say then
relate it to
real
numbers.
Well since then I offered another proposal which was to use a
non-standard
approach to this, classifing it similarly to surreal,
superreal and
hyperreal
numbers, except I called it a surreal positive or negative
sign. In
other
words,
a = a certain degree of the sign itself
b = another certain degree of the sign itself
and the surreal notation would be like this,
a|b where { | } = 0
so,
(.5_-) | ( -- )
(.5_-) < ( -- )
This states that a certain degree of the sign is used in this
case and the
next
think after it would be the total negative sign. Using this
method of
analysis,
you can have decimal degrees of negativeness. Like for
example, instead of
( -1 )^.5 = (1_-)
you could have,
(-1)^.5123 = (1_-.5123)
where the degree of the negativeness, which is a sub-level
non-standard
analysis of negative could be represented like this.
(.5123_- ) | ( -- )
In this case there would be an ill-defined state of i to start
with.
Then I showed ways this sub-level sign could be handled with
addition,
multiplication and powers. It seems to work logically in my
opinion with
nothing ill-defined.
Just check out the other posts I made about this, a new
non-standand
proposal
in math. at the sub-level locations that could exist within
the sign
itself.
Smarts Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKDNqn11112;
S. Enterprize Company <smart1234@aol.com>
http://mathforum.org/discuss/sci.math/m/651356/664015
>> 9/9. expressions like .111... .888.... and .999.... are
>> not proper mathematical expressions. If you mean to write
>> down series, then do so.
> I guess you dont know what a hyper-real number is.
> Here we go again and again.
Do you? Both .999... and 1 (which happen to be the
same number) are hyper-real numbers. My age is a
hyper-real number.
Hint: My age is also a complex number.
Dave L. Renfro
Dont know if anyone mentioned this - but there are around
13,000 books that are available in electronic format at the
Project Gutenberg - and I think you can get them on cd/dvd if
you want.
http://www.gutenberg.org/
Their top 100 books for the week is at;
http://www.gutenberg.org/browse/scores/top
The top 25 this week were;
1.The Notebooks of Leonardo Da Vinci &mdash; Complete by
Leonardo da Vinci (2074)
2.The Art of War by
.8c[CapitalEth].82.8c[CapitalEth]® (1408)
3.Some Christmas Stories by Charles Dickens (1320)
4.Hand Shadows to Be Thrown upon the Wall by Henry Bursill
(1079)
5.Ulysses by James Joyce (1070)
6.How to Speak and Write Correctly by Joseph Devlin (960)
7.A Christmas Carol by Charles Dickens (959)
8.Project Gutenberg 10K DVD (798)
9.The Adventures of Sherlock Holmes by Sir Arthur Conan Doyle
(715)
10.Relativity : the Special and General Theory by Albert
Einstein (677)
11.Beyond Good and Evil by Friedrich Wilhelm Nietzsche (676)
12.The Devils Dictionary by Ambrose Bierce (660)
13.The War of the Worlds by H. G. Wells (635)
14.Pictures from Italy by Charles Dickens (605)
15.The Real Mother Goose (572)
16.The Iliad by Homer (556)
17.The World English Bible (WEB): Luke by Anonymous (549)
18.Alices Adventures in Wonderland by Lewis Carroll (545)
19.Studies in the Psychology of Sex, Volume 1 by Havelock
Ellis
(539)
20.The Picture of Dorian Gray by Oscar Wilde (522)
21.Pride and Prejudice by Jane Austen (495)
22.Metamorphosis by Franz Kafka (483)
23.Dracula by Bram Stoker (473)
24.The Childrens Book of Christmas Stories (458)
25.Thus Spake Zarathustra by Friedrich Wilhelm Nietzsche (447)
Martin
> Its been my experience that some of the most useful and
enlightening
> books Ive ever read have been those at least sixty years
old.
> Therefore, I certainly dont have a problem with only
digitizing books
> w/o copyrights. They just dont make books like they use
too!
> I think Googles digitizing project is a great step in the
right
> direction. But just like books on the bookshelf, people
will need help
> navigating through the books on Google. Thats where the
library comes
> in; however, the neighborhood branch library must make some
changes to
> make the most use of it.
> 2) WE HAVE GOT TO STOP THINKING OF GOOGLE AS
COMPETITION!!!! Google is
> not our enemy, Google is our friend; and we should be nice!
> I think of Google as neither enemy nor friend, but a tool.
For certain
> searches it is an excellent tool. For others it isnt, but
it can lead
> you to the tools. Its amazing how often you can 
find a
store through
> Google faster than through yellowpages.com. On the other
hand, it can
> be a lousy way to find free/cheap software. 
Thats where
things like
> freewarefiles.com and tucows.com come in.
> --
> Will Twentyman
> email: wtwentyman at copper dot net
Only less than 1% of my reading includes old books. For
example, who wants
to read about Windows 3.x, when nobody uses it anymore ???
Tim Kett said:
>Only less than 1% of my reading includes old books. For
example, who wants
>to read about Windows 3.x, when nobody uses it anymore ???
Only about 1% of my technical reading is from books. On the
other hand,
Im currently re-reading a book that was first 
published 109
years ago.
Maybe youre missing something.
In sci.math, Tim Kett
<tim6kettring@e-garfield.com>
<32mk72F3khnogU1@individual.net>:
>> Its been my experience that some of the most useful and
enlightening
>> books Ive ever read have been those at least sixty years
old.
>> Therefore, I certainly dont have a problem with only
digitizing books
>> w/o copyrights. They just dont make books like they use
too!
>> I think Googles digitizing project is a great step in 
the
right
>> direction. But just like books on the bookshelf, people
will need help
>> navigating through the books on Google. Thats where the
library comes
>> in; however, the neighborhood branch library must make
some changes to
>> make the most use of it.
>> 2) WE HAVE GOT TO STOP THINKING OF GOOGLE AS
COMPETITION!!!! Google is
>> not our enemy, Google is our friend; and we should be nice!
>> I think of Google as neither enemy nor friend, but a tool.
For certain
>> searches it is an excellent tool. For others it isnt, 
but
it can lead
>> you to the tools. Its amazing how often you can 
find a
store through
>> Google faster than through yellowpages.com. On the other
hand, it can
>> be a lousy way to find free/cheap software. 
Thats where
things like
>> freewarefiles.com and tucows.com come in.
>> --
>> Will Twentyman
>> email: wtwentyman at copper dot net
> Only less than 1% of my reading includes old books. For
example, who
wants
> to read about Windows 3.x, when nobody uses it anymore ???
Historians, maybe? :-)
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
>> Only less than 1% of my reading includes old books. For
example, who
wants
>> to read about Windows 3.x, when nobody uses it anymore ???
>Historians, maybe? :-)
Masochists, more like.
JZ
In sci.math, Julnar
<10qcs05ki7ng1giq1tilq39sq1s4cuah80@4ax.com>:
> Only less than 1% of my reading includes old books. For
example, who
wants
> to read about Windows 3.x, when nobody uses it anymore ???
>>Historians, maybe? :-)
> Masochists, more like.
OK, masochistic historians, then. :-)
> JZ
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
> In sci.math, Julnar
> <10qcs05ki7ng1giq1tilq39sq1s4cuah80@4ax.com>:
>> Only less than 1% of my reading includes old books. For
example, who
>> wants
>> to read about Windows 3.x, when nobody uses it anymore ???
>Historians, maybe? :-)
>> Masochists, more like.
> OK, masochistic historians, then. :-)
Is there any other kind?
> Googles intention of bringing the worlds books to the
general public
> is a ßim-ßam. Google is partnering with libraries such as
Harvard to
> *randomly* select books off the shelf and digitize them. If
the books
> (them may try to get a snippet). If the books are in the
public
> domain, you can view the entire text. Any books published
before *Jan
> 1923* is safe to use because they are in public domain! All
books are
> under 75-year copyright protection plus 50 years lifetime
of the
> copyright owner.
> The New York Public Library offers New York residents
access to online
> books for free (There are usually 1 or 2 books which are
available to
> check out, which means once those books are checked out,
tough, you
> have to wait in line with the rest of the world). There is
a *$100 per
> year* fee for out of state users to access these books.
> So suppose they get more books online, and you have to pay
a fee to
> maybe view a copyrighted text (of course its randomized 
so
it wont be
> anywhere near complete). Well, the library is only going to
purchase 1
> or 2 of them, and if its a popular book, they might
purchase more, but
> its clear that you will be standing in line with the
entire world
> trying to get the books you want to view. Who knows, maybe
people
> wont try to read some books, but it seems like a 
ßim-ßam
to me.
The NYPL Ebooks project you describe above is not connected
to the
Google project. There are scads of EBOOKS available for
purchase
now and many libraries are moving to make these available to
their
patrons. These are popular works under current copyright that
libraries
purchase just as they might purchase the hardback copy.
Example, Im
a high school librarian. Kids at our school do a lot of
research
on health and drug issues. Our drug-related books tend to
disappear.
Ive stopped purchasing the hardback versions the little
poops like
to steal and have been replacing them with EBOOK versions.
This
sort of activity at the NYPL will continue and is not related
to
the Google project. BTW, if the text is copyrighted, it is
not NYPL that
gets to make the decision about making it available as an
EBOOk. That
is a decision made by the publisher.
> The NYPL Ebooks project you describe above is not connected
to the
> Google project. There are scads of EBOOKS available for
purchase
> now and many libraries are moving to make these available
to their
> patrons. These are popular works under current copyright
that libraries
> purchase just as they might purchase the hardback copy.
Example, Im
> a high school librarian. Kids at our school do a lot of
research
> on health and drug issues. Our drug-related books tend to
disappear.
> Ive stopped purchasing the hardback versions the little
poops like
> to steal and have been replacing them with EBOOK versions.
This
> sort of activity at the NYPL will continue and is not
related to
> the Google project. BTW, if the text is copyrighted, it is
not NYPL that
> gets to make the decision about making it available as an
EBOOk. That
> is a decision made by the publisher.
Well, not to split hairs (oh, OK, Im splitting hairs), that
decision can
only be authorized by the *copyright holder.* In most cases,
that person
is the author or creator of a work, not the publisher. In
other cases, it
could be the heirs of an author or creator. Sometimes an
instutition
holds a copyright. You get the idea.
*:-.,_,.-:*``*:-.,_,.-:*``[
CapitalOTilde]*:-.,_,.-:*``*:-.,_,.-:**:-.,
_,.-*
Cynthia Van Ness, MLS / af482@bfn.org /
www.BuffaloResearch.com
Belongings are no substitute for belonging.
>Its been my experience that some of the most useful and
enlightening
>books Ive ever read have been those at least sixty years
old.
>Therefore, I certainly dont have a problem with only
digitizing books
>w/o copyrights. They just dont make books like they use 
too!
>I think Googles digitizing project is a great step in the
right
>direction. But just like books on the bookshelf, people will
need help
>navigating through the books on Google. Thats where the
library comes
>in; however, the neighborhood branch library must make some
changes to
>make the most use of it.
The whole beauty of libraries is that they remove the cost
barrier for
people who consume books. I read a book or two a month, mostly
nonfiction, and a number of magazines, and I spend a lot of
time at
the library near my home. If I had to pay for all what I
actually use
it would cost far more than whatever taxes I pay for the
service.
The next step must be full digital access. If I were able to
retrieve
books electronically and borrow them digitally, I would
probably
read even more than I do.
And I dont agree about old books necessarily being better. 
A
gifted
writer makes reading about anything a pleasure, and this is
not
dependent upon the age of the book.
JZ
>Its been my experience that some of the most useful and
enlightening
>books Ive ever read have been those at least sixty years
old.
>Therefore, I certainly dont have a problem with only
digitizing books
>w/o copyrights. They just dont make books like they use 
too!
>I think Googles digitizing project is a great step in the
right
>direction. But just like books on the bookshelf, people will
need help
>navigating through the books on Google. Thats where the
library comes
>in; however, the neighborhood branch library must make some
changes to
>make the most use of it.
> The whole beauty of libraries is that they remove the cost
barrier for
> people who consume books. I read a book or two a month,
mostly
> nonfiction, and a number of magazines, and I spend a lot of
time at
> the library near my home. If I had to pay for all what I
actually use
> it would cost far more than whatever taxes I pay for the
service.
> The next step must be full digital access. If I were able
to retrieve
> books electronically and borrow them digitally, I would
probably
> read even more than I do.
> And I dont agree about old books necessarily being 
better.
A gifted
> writer makes reading about anything a pleasure, and this is
not
> dependent upon the age of the book.
> JZ
On the other hand, when it was harder to get published in the
past, a
lot of the bad to marginal books now being published would
not have
been, so the average quality was better.
<tp17s0dk1akcjt3uuve564t3gdtk0uo3rt@4ax.com>
posting-account=_-j7cgwAAADnQK9-r68QgRsgfV-jhA3A
> If I had to pay for all what I actually use
> it would cost far more than whatever taxes I pay for the
service.
Then youre a thief, a common thief.
Ôcid Ôooh
<tp17s0dk1akcjt3uuve564t3gdtk0uo3rt@4ax.com>
posting-account=kPdB7A0AAACkV7TyBUefgDelKStQd9N9
>Its been my experience that some of the most useful and
enlightening
>books Ive ever read have been those at least sixty years
old.
>Therefore, I certainly dont have a problem with only
digitizing
books
>w/o copyrights. They just dont make books like they use 
too!
>I think Googles digitizing project is a great step in the
right
>direction. But just like books on the bookshelf, people will
need
help
>navigating through the books on Google. Thats where the
library
comes
>in; however, the neighborhood branch library must make some
changes
to
>make the most use of it.
> The whole beauty of libraries is that they remove the cost
barrier
for
> people who consume books. I read a book or two a month,
mostly
> nonfiction, and a number of magazines, and I spend a lot of
time at
> the library near my home. If I had to pay for all what I
actually
use
> it would cost far more than whatever taxes I pay for the
service.
> The next step must be full digital access. If I were able to
retrieve
> books electronically and borrow them digitally, I would
probably
> read even more than I do.
Look at NYPL Ebooks. They have about 3000 books to check out.
The
math collection consists of 3 books, which are the rabbit
algebra
type for kids. Googles project will not allow you to borrow
them.
Its a marketing tool to point to a website to purchase the
book.
Libraries have to purchase the books to put on digital
access. That
means, if they purchase 1 or 2 of the title, you can bet that
it would
be checked out, unless its garbage. The notion of free 
books
from a
legitimate source will not probably happen.
> And I dont agree about old books necessarily being 
better.
A gifted
> writer makes reading about anything a pleasure, and this is
not
> dependent upon the age of the book.
> JZ
>> The next step must be full digital access. If I were able
to
>retrieve
>> books electronically and borrow them digitally, I would
probably
>> read even more than I do.
>Look at NYPL Ebooks. They have about 3000 books to check
out. The
>math collection consists of 3 books, which are the rabbit
algebra
>type for kids. Googles project will not allow you to 
borrow
them.
>Its a marketing tool to point to a website to purchase the
book.
>Libraries have to purchase the books to put on digital
access. That
>means, if they purchase 1 or 2 of the title, you can bet
that it would
>be checked out, unless its garbage. The notion of free
books from a
>legitimate source will not probably happen.
Im not in New York, and I dont know if I 
could have their
ebooks.
Also, I am not addressing Googles idea 
specifically.
Of course authors should be paid for their work. It makes
sense to
license and limit access to a book with some sort of fair
pricing
scheme. However it should be remembered that, by opening a
broader
market, this form of distribution is of great benefit to
writers. As
I have said, I read a great deal, both books and periodicals.
Most of
these I either check from or (in the case of the periodicals)
read at
the public library. I am brand-loyal in my reading. If I like
a
writer. If I like a periodical, I will try to read its issue
each
time it appears. If I had access to all the texts the public
library
has from my home, I would likely read more and I would likely
read
texts I might never otherwise have access to or knowledge of.
Increase in readership can probably be more easily proven
with digital
licensing of books, which could in turn help authors command
better
compensation for their work.
Just as with rock, the audience wants more access to variety
than the
industry until now has agreed to allow... and SOL the
audience will
probably have what they want. We hope.
JZ
> Look at NYPL Ebooks. They have about 3000 books to check
out. The
> math collection consists of 3 books, which are the rabbit
algebra
> type for kids. Googles project will not allow you to
borrow them.
> Its a marketing tool to point to a website to purchase 
the
book.
This is not entirely accurate. Googles project focuses 
mainly
on digitizing items in the public domain - that are out of
print.
These items will be readable online - full text. Considering
the source
for most of these older works, this could be a real plus for
serious
researchers who once had to travel to the site to gain access
to certain
scholarly works unavailable any other way.
Yes, theyll be scanning copyrighted works too. In fact,
theyve already
been doing that. ANd yes, those items lead to purchase
points. But I
consider this a service too. I buy a lot of books each year.
I like
being able to browse certain books before purchase without
having to
run down to the bookstore every time.
===
Subject: cevas theorem spherical trigonometry
posting-account=CfSJ5AwAAAD1yt3VP50q913IBHikxMCd
How does Cevas theorem get transformed from a plane to a
sphere?
(Triangle ABC, a point X not on any of the sides, AX meets BC
at
P,..,.. makes AR/RB*BP/PC* CQ/QA = 1). If AR, RB connote
sides of a
spherical triangle, does the relation still hold good in a
parallelly
conceivable affine sense?
===
Subject: Re: cevas theorem spherical trigonometry
> How does Cevas theorem get transformed from a plane to a
sphere?
> (Triangle ABC, a point X not on any of the sides, AX meets
BC at
> P,..,.. makes AR/RB*BP/PC* CQ/QA = 1). If AR, RB connote
sides of a
> spherical triangle, does the relation still hold good in a
parallelly
> conceivable affine sense?
Cevas theorem on a sphere holds if you substitute AR, RB,
etc by sin(AR),
sin(RB), etc. (where AR means the spherical (not linear)
distance from A to
R).
Jose H. Nieto
===
Subject: Re: Convex Hull Problem from Monthly
>the American Mathematical Monthly, on page 915.
>It is asked to show that the expected number of vertices of
the convex
hull
>of n points sampled from the standard normal distribution in
the plane is
>2*sqrt(2*pi*ln n). (The probability density function for the
normal
>distribution is 1/2pi * e^(-(x^2+y^2)/2).
>A couple of my colleagues were discussiong this problem
today. Let us
just
>consider the case n=3 right now:
>It seems intuitive to me that the probability of 3 sampled
points being
>collinear is 0, so that the expected number of vertices when
n=3 would be
3.
>Moreover, the formula yields the unlikely value
2*sqrt(2*pi*ln 3) =
5.25!
>Perhaps someone would be willing to help me to clarify my
understanding
of
>the problem.
> I saw that too, and thought of the same counterexample. My
thought is
> that the result is asymptotic.
Correct, it should say asymptotic. Correction in the March
issue.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: How to visualize limits in category theory?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKENO816767;
>with my supervisor, he said that it is nice to visualize
limits in say
>the category of topologies as product of two spaces while
colimits can
>be seen as the joining of two spaces (morphisms I suppose
are just the
>homeomorphisms) in a pushout/pullback diagram (ie. a
commutative
>rectangle such that if 3 objects and 2 morphisms are fixed
and another
>commutative rectangle is formed with those fixed points and
morphisms
>then there is a unique morphism from/to the 4th object of
the old
>rectangle to/from the 4th object of the newly formed
rectangle!)...
Well... this sounds like an extract from a conversation where
your supervisor was trying to impart some feeling for limits
and
colimits through some *examples*, but its stretching it to
say
visualize limits... as products or visualize colimits... as
pushouts. I *would* say that once you really understand the
universal property definition of things like (co)product,
(co)equalizer, pushout, pullback, etc., then the universality
property definition of general limits/colimits is a snap. So
it makes sense to master the simple examples first, which, 
yes,
are generally easy to visualize in concrete situations.
But sometimes such examples are *too* simple, too easy to
glide
over. So you might want to wrap your head around the universal
property formulation of things like free products and
amalgamated
products of groups (coproducts and pushouts, respectively),
where universal properties may seem more clarifying than in
the simpler examples. At an even greater remove in complexity
are schemes, and here being ßuent with limits (fiber 
products
for instance) really pays off in the end, since the naive set-
theoretic intuitions no longer quite cut it here.
>Anyway, let me give some comments on one of the posts...
>> Youve gotten several good responses to your question;
perhaps
>> what you need are examples to test the responses on.
>> Jesse Hughes pointed out that limits are reducible to
cartesian
>> products and equalizers, and James Dolan observed that
limits
>> are the basic stuff of algebraic geometry as initiated by
Fermat,
>> Descartes, and others (loci of equations seen as equalizers
>> defined on cartesian products).
>> You might use these observations to visualize how the ring
of
>> p-adic integers is constructed as a limit of a diagram of
shape
>> --> Z_{p^n} --> Z_{p^{n-1}} --> ... --> Z_p.
>correctly... The limit would then be those integers which
have the form
>p^k .. ie. Lim(Z_p) := {p^k : k is in N} , accompanied with
the
>canonical homomorphisms to each of the Z_p^n ... right?
Sorry, but no. Denote an element of Z_{p^n} by
a_0 + a_1 p + ... + a_{n-1}p^{n-1}. If you work through
this example again, you should find that the elements in the
limit are similarly expressible as
a_0 + a_1 p + ... + a_ p^n + ...
(completion of Z w.r.t. p-adic metric, where p^n --> 0 as
n --> oo).
>> You might also visualize the patching condition in the
definition
>> of sheaves F in terms of limits: if V is open and {V_i}
covers V,
>> then F(V) is the limit of a diagram obtained by sticking
together
>> diagrams of shape
>> F(V_i) --> F(V_i cap F_j) <-- F(V_j)
>> over all i and j.
>Wow... very nice example. Er.. what does patching mean btw
(is that one
>of the axioms of the sheaves?)
Yes; a sheaf is a presheaf which satisfies the patching
condition:
given a cover {U_i} of U and sections s_i over U_i which
agree over intersections, there is a unique section s over U
which restricts to the s_i.
Todd Trimble
===
Subject: Re: Why did you choose math?
posting-account=iboO5gwAAAB1ZwCoeizAgbK24wYeXcL6
>if you arent talented in math, the best advise you can
receive is
>to steer away from it - especially if you are considering it
for
>career purposes other than teaching in public schools. ditto
for
>phyzics.
> You could be surprised at what is possible. Learning
Mathematics
academically
> is one thing; actually applying it in realistic situations
is another
thing.
you could be surprised that learning math includes knowing
how to
apply it.
> Algebryonic
===
Subject: Re: Why did you choose math?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBI3LQS01551;
I study math at the side, my main interest lies in
theoretical physics.
Currently I am an undergraduate. Although that means i dont
know a lot, I
definitely know what I do not know. Quite often math and
physic walk hand in
hand, and as such it is important to learn lots of
mathematics to understand
physics. However, I have many times realised that
mathematical understanding
will not result in physical understanding. I would recommend
you study both
as a major, it is sad mathematical physics was founded as a
result of
theoretical physicists not knowing mathematics.
joccis
===
Subject: Re: Tautologies Then and Now
In summary...
PAUL: My question is why is tautology not normally used
outside
sentential logic? [...]
STEPHEN: I could have used Modal logic as another counter
example;
where the term tautology is applied outside of propositional
logic.
[...]
PROFESSOR MENZEL: Right, tautological, valid sentence, and
logical truth all mean the same thing in the context of
classical,
truth-functional propositional logic, but not in
propositional modal
logic, where there are logical truths that are not
tautological.
**********
PAUL: Theyre called universal truths, not tautologies, in 
the
first-order predicate calculus. [...]
STEPHEN: [Prof. Menzel] had twice explained to him [paul]
that TT
[truth table] was used in monadic predicate logic and that
all Ts
were tautological by definition. [...]
PROFESSOR MENZEL: Although there is a way to USE truth tables
for some
purposes in predicate logic, truth tables are not a part of
the
standard *semantics* of predicate logic, even mondaic
predicate logic,
and in principle *cannot* provide such a semantics in
general, due to
their finite character. In particular, they cannot provide an
adequate
account of the meaning of the quantifiers. You [STEPHEN] seem
to be
emphasizing the fact that they can be USED in (some corners
of)
predicate logic. Paul seems to have picked up on the fact
that truth
tables are not part of the standard SEMANTICS of predicate
logic, and
hence, in particular, on the idea that tautology -- which has
essentially to do with the semantical methods of
propositional logic
-- is an inherently propositional notion.
- paul
===
Subject: How to visualize Riemann surfaces
posting-account=ot_3pg0AAACJAbH4iSny5HS0UYV5RNs8
All I know to ask is this. A function w = f(z) is a real
2-dimensional
manifold which is a Riemann surface if f(z) satisfies
*something*. Now,
assuming f(z) satisfies all the required condition, how do I
visualize
the Riemann surface given f(z) in the three dimensional
space? Like the
torus, for example.
I have been pouring over a lot of books for an answer. But
they all
give me something about meromorphic functions and stuff like
that, and
i am not able to locate the answer. Any help would be greatly
appreciated.
Prasanna.
===
Subject: Re: How to visualize Riemann surfaces
Look at Springers book on Riemann surfaces. It has pictures
near the
beginning showing you how to visualize the Riemann surface of
a certain
function as being a torus.
--
Allan Adler <ara@zurich.csail.mit.edu>
* Disclaimer: I am a guest and *not* a member of the MIT
CSAIL. My actions
and
* comments do not reßect in any way on MIT. Also, I am
nowhere near
Boston.
===
Subject: Re: How to visualize Riemann surfaces
>All I know to ask is this. A function w = f(z) is a real
2-dimensional
>manifold which is a Riemann surface if f(z) satisfies
*something*. Now,
>assuming f(z) satisfies all the required condition, how do I
visualize
>the Riemann surface given f(z) in the three dimensional
space? Like the
>torus, for example.
But the Riemann surface doesnt live in three-dimensional
space! Its
a subset of C^2 which is in a natural way the same as R^4.
You have
to _pick_ a projection (or other mapping) into R^3. There is
no
natural way to do that.
Its also true that you wont get the torus 
for a Riemann
surface unless
you include some points at infinity.
>I have been pouring over a lot of books for an answer.
Well, I hope you mopped that up!
dave
===
Subject: Re: How to visualize Riemann surfaces
<cq6vl9$ifs$1@news.math.niu.edu>
posting-account=ot_3pg0AAACJAbH4iSny5HS0UYV5RNs8
I happened to look at
http://library.wolfram.com/infocenter/Demos/15/
mathematica. Specifically
http://library.wolfram.com/examples/riemannsurface/Links/
TalkGD99_lnk_1.html
.
But they just take real{f(z)} or imag{f(z)} and plot it as a
surface in
CxR.
Is that what you call as _picking_ a projection Dr. Rusin?
Taking
real{f(z)} or imag{f(z)} or |f(z)| whence we have a three
dimensional
surface? That is how the Riemann surface for w = srqt(z) is
plotted it
seems.
But that is not what I want. What I want is a version where I
can also
include points at infinity. Some kind of stereographic
projection is
possible? I dont know, this is not clear. Please help me out.
Come now, Dr. Rusin. I cannot take a course on geometry of
manifolds to
understand the basic principles behind Riemann surfaces. Nor
can I
rigorously go through all the definitions and theorems the
books offer
before making any sense at all. I just want a laymans view,
or more
precisely, an engineers view on the subject. Please
understand.
One important aspect of doing research is to simplify
theories as much
as possible is not it?
===
Subject: Re: How to visualize Riemann surfaces
>All I know to ask is this. A function w = f(z) is a real
2-dimensional
>manifold which is a Riemann surface if f(z) satisfies
*something*.
Umm, well, sorta. Actually, provided that you mean if in the
mathematical sense, theres no problem at all; but if you
mean it
in the colloquial sense (which also is used in informal
mathematics)
of if and only if, then youre way off. However, 
lets leave
that
difficulty for a while.
>Now,
>assuming f(z) satisfies all the required condition, how do I
visualize
>the Riemann surface given f(z) in the three dimensional
space? Like the
>torus, for example.
How you visualize the Riemann surface depends on what you
mean by
visualize. For example (according to Dennis Sullivan; I think
I
know how the proof would go, but Ive never seen it written
down,
and Im not at all sure Id be competent to 
produce it), every
Riemann surface can be represented (and therefore visualized)
as a polyhedral surface in R^3, in the following sense: a
Riemann
surface is a 2-dimensional manifold equipped with a
conformal structure (a way of measuring angles, but not
lengths,
infinitesimally near each point); a polyhedral 2-manifold
is also equipped with a conformal structure (its obvious at
points interior to 2-simplices, and reasonably obvious at
points
interior to 1-simplices, by locally ßattening the two adjacent
two simplices; at 0-simplices, theres a local conical
structure,
and you simply have to expand or contract the angular variable
linearly to make the total angle equal to 2pi); and the
theorem
says that all conformal structures (on any orientable
2-manifold)
arise in both ways.
But I bet thats not what you want. In fact, I would guess
that
when you say the three dimensional space, you dont mean
*just*
R^3, you mean R^3 interpreted as CxR, with the complex
variable
z in your (proto-)construction of Riemann surfaces being the
variable
in the C factor; and that when you want to visualize the
Riemann
surface given f(z) in CxR, you want the projection from the
Riemann
surface to C to be good (complex analytic, say). This can be
done,
but be warned that no *compact* surface S in CxR is going to
be
situated in such a way that projection from S to C is good (in
the sense Im ascribing to you). To visualize a Riemann
surface
of genus 1 (i.e., a subsurface of a torus) in CxR, you will
*have*
to accept that some of its points are at infinity. And you may
well have to accept that some of the sheets of the Riemann
surface
have intersections in CxR (even though they dont intersect
abstractly, or even in CxC, which is where you really ought to
be trying to visualize things), and that there may be other
apparent singularities (e.g., crosscaps) that are artifacts
of your chosen visualization rather than intrinsic features
of the Riemann surface.
>I have been pouring over a lot of books for an answer. But
they all
>give me something about meromorphic functions and stuff like
that, and
>i am not able to locate the answer. Any help would be greatly
>appreciated.
Thats poring, by the way. I hope.
Lee Rudolph
===
Subject: Dominoes on a chessboard. Edge method.
posting-account=pwtqtw0AAACw59dt7zOnp9M9tGBpr6Lp
--
http://tinyurl.com/5aocb
> You have a chess board with two
> diagonally opposite corners removed. You
> have 31 dominoes. Can you place the 31
> dominoes on the chess board?
> Familiar Solution.
> No. [...] every time you place a domino
> it covers two different colour squares
> and by removing diagonally opposite corners
> there are now, say, 32 white squares
> and 30 black.
> [in] book ÔHow Long is A Piece of String?
> [...] they credit Will Hartston [...] with
> the following question: ÔIf this proof was
> not available then how would you prove that
> you cant place the dominoes on the
> chessboard?
Despite my initial scepticism, this time
Ill present a truly different approach,
with a ßavor of combinatorial topology,
which will result in more subtle theorems,
unavailable by the otherwise wonderfully
simple black & white chessboard coloring.
In this post I will show that when dominoes
cover a chess board then an even number
of them is vertical, and an even number
is horizontal. Also for coverings of 62
squares, I show when there is an odd number
of vertical dominoes, and when of horizontal.
E.g. if you remove squares a1 and e4 then
the number of vertical dominoes must be odd.
(A stubborn opponent may still claim that
my solution of the original problem above
is similar to the familiar one, because of
the sensitivity to the color--yes,
sensitivity to the color is unavoidable,
because the result is false when the
two removed squares have different colors.
But I do not match the white and the black
squares. Anyway, it is more important
to me that now I can get new results).
My method is based on labeling the edges.
It goes far beyond this initial post in which
I will consider only rectangular chessboard
(and their subsets) and only 2x1 and 1x2
dominoes.
Let us consider an h by v board:
B(h v) := {0 ... h-1} x {0 ... v-1}
For the sake of easy communication I will say
that h is the E-W horizontal dimension, and
v is the N-S vertical dimension.
A standard dominoe on B(h v) is a 2-element
subset {(a s) (b t)} of B(h v) such that:
d((a s) (b t)) := |a-b| + |s-t| = 1
Such two squares (a s) & (b t) (points of
the square lattice) are also called adjacent.
The sum [a+b s+t] of two adjacent points is
called (by me :-) an edge (between (a s) & (b t)).
Indeed, ((a+b)/2 (s+t)/2) is the center of the
geometric common edge of the unit squares with
centers in (a s) & (b t).
Thus edges are in a canonical 1-1 correspondence
with dominoes (in the infinite square lattice).
Let
Odd(x) := 1 for odd integer x
Odd(x) := 0 for even integer x
[x y] is an edge iff (=if and only if)
x & y have different parity, i.e. when
Odd(x) =/= Odd(y)
Edge [x y] is an edge between
(ßoor(x/2) ßoor(y/2)) & (ceiling(x/2) ceiling(y/2).
An edge [x y] is horizontal, when y is odd;
and vertical when x is odd. Horizontal edges cut
vertical dominoes into halves, while vertical
edges cut horizontal dominoes.
EDGE LABELING --
horizonal edges [x y]:
lbl[x y] := h0 for Odd((y-1)/2) = Odd(x/2) = 0
lbl[x y] := h1 for Odd((y-1)/2)=0 & Odd(x/2)=1
lbl[x y] := H0 for Odd((y-1)/2)=1 & Odd(x/2)=0
lbl[x y] := H1 for Odd((y-1)/2) = Odd(x/2) = 1
vertical edges [x y]:
lbl[x y] := v0 for Odd((x-1)/2) = Odd(y/2) = 0
lbl[x y] := v1 for Odd((x-1)/2)=0 & Odd(y/2)=1
lbl[x y] := V0 for Odd((x-1)/2)=1 & Odd(y/2)=0
lbl[x y] := V1 for Odd((x-1)/2) = Odd(y/2) = 1
Thus we have 8 different labels:
h0 h1 H0 H1
v0 v1 V0 V1
Obviously, if edges [x y] [x y] are such 
that
both x-x and y-y are divisible by 4 then
they have the same label.
We will assume the following simple algebra
for the formal linear combinations of these
symbols with integer coefficients, say
a b c d e f g h (it has nothing to do directly
with chess :-):
a*h0 + b*h1 + ... + h*V1
:= Odd(a)*h0 + Odd(b)*h1 + ... + Odd(h)*V1
and we will add, subtract and these combinations,
and multiply them by integers, in the obvious
natural way (technically: we consider the 8-dim
linear space over GF(2), which has labels for
its linear basis).
In particular, we can define:
LBL(E) := Sum(lbl[x y] : [x y] in E)
for arbitrary finite set E of edges; i.e. the
label LBL(E) of E is defined as the sum of the
labels of all edges of E.
THE EDGE METHOD
One may also call it the boundary label method.
Together with the global geometric operations
on the chessboard (see below), it is my main
tool (the edge method is rather local). Let us
start with the boundary label for a single point
or chess square (a s):
blb(a s) :=
LBL{[2*a-1 2*s] [2*a 2*s-1] [2*a+1 2*s] [2*a 2*s+1]}
while
BLB(S) := SUM(blb(a s) : (a s) in S)
for arbitrary finite subset S of the infinite
square lattice.
If points (a s) (b t) are such that a-b and
s-t are both even then obviously they have the
same boundary label: blb(a s) = blb(b t).
****************************************
THEOREM 1
BLB(D) = h0 + h1 + H0 + H1
BLB(D) = v0 + v1 + V0 + V1
for arbitrary horizontal domino D and
vertical domino D.
****************************************
We are interested in dominoe configurations,
i.e. in the collections of the pairwise disjoint
dominoes. We want to know which subsets of a
chessboard can be represented as the union of
a dominoe configuration. We will compute for
them the boundary label to solve certain
problems. But in this post let me add an
additional geometric twist (see below).
We have four cases of rectangular chessboards,
classified by (Odd(h) Odd(v)). Ill call
them even-even even-odd odd-even odd-odd.
Of course the classical 8x8 board belongs
to the even-even case.
Case even-even: Odd(h) = Odd(v) = 0
In this case I will treat B(h v) as a torus,
by studying the torus dominoes, where all
standard dominoes of B(h v) are torus
dominoes, but we have also some additional
h+v torus dominoes, namely:
{(h-1 y) (0 y)} for y = 0 ... v-1
and
{(x v-1) (x 0)} for x = 0 ... h-1
Now we will consider torus dominoe configurations
(i.e. collections of pairwise disjoint torus
dominoes) -- each standard dominoe configuration
is a torus configuration but not the other way
around. Nevertheless, the results about the torus
configurations will have obvious consequences for
the standard configurations, as we will see
(often they will be stronger than the standard
versions).
Its only natural that we will identify certain
points by wrapping around mod h and mod v:
points (a s) (b t) are torus-equivalent iff
a-b is divisible by h, and s-t by v.
Also,
edges [x y] [x y] are torus-equivalent iff
2*h divides x-x, and 2*v divides y-y.
Observe that equivalent edges have the same label.
In the torus case we let Tor(h v) denote the
toroidal board. We can apply to it labeling
definitions similar to the standard ones above.
Thus we obtain:
******************************
THEOREM 2 BLB(Tor(h v)) = 0
******************************
PROOF Each edge belongs to two squares. END OF PROOF
Remember that we assume here that h v are even.
For the standard board a proof is slightly more
complex (h v are still assumed to be even):
***************************
THEOREM 2 BLB(B(h v)) = 0
***************************
Ill leave the prof as an exercise--just observe that
the boundary label of the standard chessboard
is the sum of the labels of the edges of the chessboard.
The following theorem holds both in the standard
(i.e. euclidean) and in the toroidal case; it is
a simple consequence of Theorem 1):
******************************************
THEOREM 3 Let X be an arbitrary dominoe
configuration consisting of an
odd number of dominoes. Let S be the
union of X. Then BLB(S) is equal to
h0+h1+H0+H1 or to v0+v1+V0+V1.
******************************************
PROOF When the number of horizontal dominos
is odd, then the vertical number is even, hence
BLB(S) = h0+h1+H0+H1. Otherwise BLB(S) = v0+v1+V0+V1.
END OF PROOF
Of course the toroidal version of Theorem 3 is more
general than the standard one (it covers more
configurations). The same is true for the next theorem,
hence we wont bother stating its standard 
variation.
************************************************
THEOREM 4 Every dominoes configuration which
covers entire Tor(h v) has an even
number of horizontal dominoes and an even number
of vertical dominoes.
************************************************
PROOF Since h*v is divisible by 4, the
configuration has an even total number of dominoes.
Thus the parity of the numbers of horizontal
and of vertical dominoes is the same. If both
were odd than BLB(Tor(h v)) would be equal
to the sum of all eight labels:
h0+h1+H0+H1 + v0+v1+V0+V1
instead of being 0 (see Thm 2).
*****************************************
THEOREM 5 Let S be an arbitrary subset
of Tor(h v). Then
BLB(Tor(h v)S) = BLB(S)
(The same holds for standard B(h v)).
*****************************************
PROOF
BLB(Tor(h v)S) + BLB(S) = BLB(Tor(h v)) = 0
END OF PROOF
Theorem 1 gave us the boundary labels of dominoes.
Now lets compute the boundary label of an arbitrary
2-point set. The boundary label of a point depends
only on the parity of its coordinates. Thus we have
but 4+6=10 cases, four of which are already covered by
the vertical and horizontal dominoes. There are four
cases when the two points are equivalent mod 2
(i.e. their first coordinates have the same parity,
and so do their second coordinates). The boundary
label of such a 2-point set is of course 0. Lets
compute the last two cases (or one can see the result
instantly, as for dominoes, due to their simple
geometric patterns):
BLB{(0 0) (1 1)} = blb(0 0) + blb(1 1)
= LBL{[-1 0] [0 -1] [1 0] [0 1]}
+ LBL{[1 2] [2 1] [3 2] [2 3]}
= (V0+ H0 + v0 + h0) + (v1 + h1 + V1 + H1)
And:
BLB{(0 1) (1 0)} = blb(0 1) + blb(1 0)
= LBL{[-1 2] [0 1] [1 2] [0 3]}
+ LBL{[2 -1] [1 0] [2 1] [3 0]}
= (V1 + h0 + v1 + H0) + (H1 + v0 + h1 + V0)
i.e.
*************************
BLB{(0 0) (1 1)} =
BLB{(0 1) (1 0)} =
h0 + H0 + v0 + V0
+ h1 + H1 + v1 + V1
*************************
A set covered by a dominoes configuration
can have such a total label as just above
only when the number of vertical dominoes
in the configuration, as well as of the
horizontal, is odd. This means that the
configuration would consist of even number
of dominoes. Thus an odd number of dominoes
cannot cover the board with two chess squares
removed, where the removed squares have
their first coordinates of different parity,
and the second coordinates likewise.
We also already know that the odd number of
dominoes cannot cover a set which has the
boundary label 0. Thus in such a case the
two ommitted squares cannot have their
first coordinates of the same parity,
and the second coordinates likewise.
Lets apply Theorem 5. Thus an odd number of
dominoes have a chance to cover the board minus
two squares only when the two missing points
have the parities of their coordinates
respectively equal to the parities of the
coordinates of points of a vertical or
horizontal dominoe. In the vertical case the
number of vertical dominoes has to be odd
(while the number of the horizontal ones is even),
and in the horizontal case the number of horizontal
dominoes has to be odd.
***********
Hm, its time to stop before I got up to the full
speed, this post is already looooong.
Wlod
(Wlodzimierz Holsztynski)
===
Subject: Re: Separable,not Banach,without Schauder basis
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKH4XK00439;
>> Hi all,
>> Knowing that it is necessary for a normed space with a
>> Schauder basis to be separable,its not 
difficult to give
>> an example of a space without Schauder basis.
>> l^oo is such an example as l^oo is not separable.
>> For a long time it wasnt known if there existed 
separable
Banach
>> space without Schauder basis.Yes,there is such a space.
>> That example is beyond me,so Im asking if someone knows 
if
>> an example of separable space which is *not* Banach and
>> doesnt have Schauder basis would be easier,and if 
possible
>How about L^p[0,1], where 0<p<1 ? There are no continuous
linear
>functionals, and thus there are no Schauder bases.
Great!
===
Subject: one integration question
posting-account=a2-NvgwAAABvLKlEEKK1T4CHPeySyl9a
whats the closed form of this integration
int_{-infty}^{infty} |x-u|^nu e^{-|x|^nu} dx
I cant get it
===
Subject: Re: one integration question
> whats the closed form of this integration
> int_{-infty}^{infty} |x-u|^nu e^{-|x|^nu} dx
> I cant get it
According to Maple, for nu=1 you get:
2*u+2*exp(-u)
for nu=2 you get:
(u^2+1/2)*sqrt(Pi)
and for nu=3 you get:
+3/4*u^4*exp(-1/2*u^3)*WhittakerM(1/6,2/3,u^3)/(u^3)^(1/6)
+u*exp(-1/2*u^3)*WhittakerM(7/6,2/3,u^3)/(u^3)^(1/6)
-9/10*u^4*exp(-1/2*u^3)*WhittakerM(1/3, 5/6, u^3)/(u^3)^(1/3)
-3/2*u*exp(-1/2*u^3)*WhittakerM(4/3,5/6,u^3)/(u^3)^(1/3)
-1/4*u*exp(-1/2*u^3)*WhittakerM(1/6,2/3,u^3)/(u^3)^(1/6)
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: one integration question
posting-account=a2-NvgwAAABvLKlEEKK1T4CHPeySyl9a
whats the general form for any nu?
QQ
> whats the closed form of this integration
> int_{-infty}^{infty} |x-u|^nu e^{-|x|^nu} dx
> I cant get it
> According to Maple, for nu=1 you get:
> 2*u+2*exp(-u)
> for nu=2 you get:
> (u^2+1/2)*sqrt(Pi)
> and for nu=3 you get:
> +3/4*u^4*exp(-1/2*u^3)*WhittakerM(1/6,2/3,u^3)/(u^3)^(1/6)
> +u*exp(-1/2*u^3)*WhittakerM(7/6,2/3,u^3)/(u^3)^(1/6)
> -9/10*u^4*exp(-1/2*u^3)*WhittakerM(1/3, 5/6,
u^3)/(u^3)^(1/3)
> -3/2*u*exp(-1/2*u^3)*WhittakerM(4/3,5/6,u^3)/(u^3)^(1/3)
> -1/4*u*exp(-1/2*u^3)*WhittakerM(1/6,2/3,u^3)/(u^3)^(1/6)
> --
> G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: What is meant by a meager subset??
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iBKHglU03609;
G. A. Edgar <edgar@math.ohio-state.edu.invalid>
http://mathforum.org/discuss/sci.math/m/664050/664247
> Bourbaki did not think first category and
> second category were good terms, so he
> invented some new ones: meagre, residual.
> Do you think they are better terms?
Actually, Denjoy introduced the term residual
around 1912 or 1913. I believe he was trying to
distinguish between a set being second category
(i.e. the set is big) and the set having a first
category complement (i.e. the set is so big that
whats left over is small). During this period some
authors (most people, including Baire and Lebesgue)
used second category for not first category,
while other authors (Lusin, Sch.9anßies, Hobson)
used second category for residual. Then, to
make matters even worse, when second category
meant not first category for an author, quite
often that author would use second category in the
statement of a theorem, but then wind up proving
the set was actually residual. (This last thing
still happens, especially in areas of math that
are not very close to topology or analysis.)
The term generic is also used for residual,
but Ive come across some instances (I think the
papers were in dynamical systems theory and/or
differential geometry) where an author uses
generic to mean open dense set (i.e. the complement
is not just first category, its actually 
nowhere
dense), or perhaps more generally, contains an open
dense set.
Dave L. Renfro
===
Subject: 1 -1/2 + 1/3.....
Hello
Id like some clues to prove that the alternating series Sum
(n=1,
oo) ((-1)^n)/n)) converges to Ln(2).
Since 1/n -> 0 as n -> oo and 1/n is strictly decreasing, the
series
does converge. We know that, for every real x satisfying |x|
<1,
Taylors theorem implies that Ln can be expanded about 0, so
that
Ln(x) =x - x^2/2 + x^3/3.... Since [0,a] is compact, for
every 0 < a <
1 this power series converges uniformly on [0,a] to Ln, but,
even
though the series converges for x =1, this doesnt mean the
convergence is uniform on [0,1). If this were true, then 1
-1/2 + 1/3
... would converge to lim (x ->1) Ln(1+x) = Ln(2), but this
argument
failed.
Ana
===
Subject: Re: 1 -1/2 + 1/3.....
> Hello
> Id like some clues to prove that the alternating series
Sum (n=1,
> oo) ((-1)^n)/n)) converges to Ln(2).
> Since 1/n -> 0 as n -> oo and 1/n is strictly decreasing,
the series
> does converge. We know that, for every real x satisfying
|x| <1,
> Taylors theorem implies that Ln can be expanded about 0,
so that
> Ln(x) =x - x^2/2 + x^3/3....
A typo made in a hurry: you meant Ln(1+x), didnt you?
> Since [0,a] is compact, for every 0 < a < 1 this power
series converges
> uniformly on [0,a] to Ln, but, even
> though the series converges for x =1, this doesnt mean 
the
> convergence is uniform on [0,1). If this were true, then 1
-1/2 + 1/3
> ... would converge to lim (x ->1) Ln(1+x) = Ln(2), but this
argument
> failed.
> Ana
There is Abels Theorem about the relationship between the
sum of a
convergent series
sum(a(n): n=0 to infinity}
and the left limit at 1 of the power series
f(x) = sum(a(n)*x^n :n=0 to infinity)
(these two are equal)
but we can manage doing it by brute force from finite to
infinite:
since (finite geometric series)
1 - x + x^2 - ... + (-1)^n*x^n
= 1/(1+x) - (-1)^(n+1)*x^(n+1)/(1+x),
we can integrate from 0 to 1 (all is elementary):
1 - 1/2 + ... + (-1)^n / (n+1)
= Ln(2) - (-1)^(n+1) * int(x^(n+1)/(1+x) : x=0 to 1)
The absolute value of the last expression (call it R(n))
needs to be
estimated:
since 0 < x^(n+1)/(1+x) <= x^(n+1), we obtain
0 < abs(R(n)) < 1/(n+2)
so by squeeze theorem, R(n) converges to 0 as n goes to
infinity.
Similarly we can find the 1-1/3+1/5-1/7+... sum.
===
Subject: Re: 1 -1/2 + 1/3.....
>Hello
>Id like some clues to prove that the alternating series 
Sum
(n=1,
>oo) ((-1)^n)/n)) converges to Ln(2).
>Since 1/n -> 0 as n -> oo and 1/n is strictly decreasing,
the series
>does converge. We know that, for every real x satisfying |x|
<1,
>Taylors theorem implies that Ln can be expanded about 0, 
so
that
>Ln(x) =x - x^2/2 + x^3/3.... Since [0,a] is compact, for
every 0 < a <
Of course you mean Ln(1 + x) there.
>1 this power series converges uniformly on [0,a] to Ln, but,
even
>though the series converges for x =1, this doesnt mean the
>convergence is uniform on [0,1). If this were true, then 1
-1/2 + 1/3
>... would converge to lim (x ->1) Ln(1+x) = Ln(2), but this
argument
>failed.
>Ana
Call your partial sums s_n(x) and f(x) = ln(1 + x).
You have s_n(x) -> f(x) on [0, 1) and s_n(1) -> c. You know
f(x) is
continuous on [0, 1] and want to show c = f(1).
| f(1) - c | <= | f(1) - f(x) | + | f(x) - s_n(x) |
Given e > 0 do you see how to make the right side small?
Remember,
there is no x or n on the left side...
--Lynn
===
Subject: Direction-Based Grid-Game (& Questions)
posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE
But Google Groups seems to have messed up my attempt to post
it the first time.
I am reposting this to rec.puzzles and sci.math via Google.
I suspect that it was not seen by most newsreaders when I
posted it the
first time
(about 2 weeks ago).
I would be happy if someone replies to this post, even if
they did not
solve the puzzle,
to prove that SOMEONE is getting posts made via Google.
In any case, I have added a little to the comment about this
game being able to be played solitaire.
-------
a question about a related integer sequence.
----
Here is yet another game of mine played on a grid drawn on
paper.
It is a somewhat unusual in that the game is a cooperative
game,
the 2 players get the same number of points.
Start with an n-by-n grid drawn on paper, where n is odd
(and I suggest n be at least 7, maybe much higher).
Players alternatingly take turns filling in, on each move,
an *empty* square immediately next to the last square filled
in by the other player.
Player 1 can only fill in an empty square immediately either
to the left of, to the right of, above, or below the last
square filled.
Player 2 can only fill in an empty square diagonally adjacent
to the last square filled.
Fill in the center square.
Player 1 then fills in a square above,below, left of, or
right of the center square.
Player 2 then fills in a square diagonally touching the
previously filled square.
Etc.
Play continues until no more squares can be filled in.
The score (for both players) is the number of squares filled
in before play must stop.
(And there should be very little communication between
players.)
If you insist on playing a competitive game, you can do so
with 3 or more players.
Just have each pair of players play a game,
for a total (if m= number of players) of m(m-1)/2 games,
and add every players particular scores to get a total
score for each player.
You can also play this game solitaire, where you alternate
Ômoving non-diagonally (up, down,left,or right) 
and then
diagonally; but I believe that perhaps playing with 2 people
might be more fun, if only because you try to guess each
others strategies.
----
Questions:
Is it possible, for any odd n >= 3, to visit once EVERY square
of an n-by-n grid (starting at the center square)?
What is the maximum possible score, in theory, for a n-by-n
grid?
(How does the sequence start where the nth term is the maximum
possible score for an n-by-n grid?)
Leroy Quet
===
Subject: any conic sections in the Menger sponge?
David A. Fontaine has some nice graphics, including a
depiction
of a Menger sponge here:
http://davidf.faricy.net/image.php?pic=menger.jpg
A real Menger sponge is, I think,
- closed and bounded in R^3
- of Lebesgue measure 0
- of cardinality continuum
Im trying to figure out if a Menger sponge 
contains one of:
- a circle
- an ellipse
- an arc of a parabola or hyperbola
I dont know.
David Bernier
===
Subject: Homotopy classes of figures made of 5 lines in R^2
Obscurity, linux)
Consider figures consisting of 5 lines in the plane. Two such
figures
are topologically equivalent if there there is a homotopy
between them.
How many equivalence classes are there for figures consisting
of five
lines?
How many equivalence classes are there for figures consisting
of N lines?
Is there a general theory for solving this problem?
Niels
PS: This is not homework, but somebody asked me for the
number of
different solutions to the following problem, and I was
embarrased not
to be able to answer.
http://rec-puzzles.org/new/sol.pl/geometry/construction/5.
lines.with.4.point
s
===
Subject: Re: Homotopy classes of figures made of 5 lines in 
R^2
>Consider figures consisting of 5 lines in the plane. Two such
figures
>are topologically equivalent if there there is a homotopy
between them.
Meaning what? A one-parameter (continuous) family of *ordered*
5-tuples of lines in the plane, which begins and ends at the
two
5-tuples in question? A one-parameter family of *unordered*
5-tuples
ditto? A one-parameter family of either ordered or unordered
5-tuples
of *oriented* lines? A one-parameter family of pseudolines
(properly embedded images of the line in the plane, perhaps
with
well-defined endpoints, possibly required to be distinct, in
the
projective plane) which begins and ends at the 5-tuples in
question?
Any of the above, with the homotopy required to extend to a
one-
parameter family of homeomorphisms of the plane?
>How many equivalence classes are there for figures consisting
of five
lines?
>How many equivalence classes are there for figures consisting
of N lines?
>Is there a general theory for solving this problem?
Well, theres a general theory *of* this (sort of) problem.
Depending on how you give precise meaning to homotopy above,
youre dealing with one instance of one version of the 
theory
of line arrangements; a search on that phrase (in any database
sufficiently full of mathematics) should give you much more
information than you can use.
Lee Rudolph
===
Subject: Criss-cross Grid Game
posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE
I am reposting this to rec.puzzles and sci.math via Google.
I suspect that it was not seen by most newsreaders when I
posted it the
first time
(about 2 weeks ago).
I would be happy if someone replies to this post
to prove that SOMEONE is getting posts made via Google.
---
Here is yet another of my games played on an n-by-n grid
drawn on
paper.
2 players,each with a different colored pencil.
Players alternate placing an X or an O of their color in the
grids
squares as follows:
*Player 1 places the first O in the upper-left square.
*Each player places their symbol immediately next to the
symbol last
drawn by the other player in the directions of vertically,
horizontally, or diagonally.
*If the square a player draws his symbol in is empty, he/she
draws an
O.
O.
*A player may not draw in (move to) a square already occupied
by an X
(and an O).
Play continues until moving is impossible or when one player
first gets
a predetermined number of points.
Scoring:
*If a player places an X in the same square as an O, the
player who did
*not* draw the O in that square gets a point.
*If a player places an O in an empty square diagonally
adjacent to the
last square drawn in by the other player, and the other 2
squares in
the 4-by-4 square -- formed by the empty square being drawn
in and the
last drawn-in square -- are both filled, then the player
Ôcompleting
the square (the player whose move it is) gets a point.
Example:
+---+---+
! O ! !
+---+---+
!OX ! O !
+---+---+
Last move put an X in lower left square (getting a point for
the player
who did *not* place the O there earlier). Player now wants to
place an
O in upper right square, which would get him/her a point.
What is a good strategy for this game?
Leroy Quet
===
Subject: Re: Criss-cross Grid Game
posting-account=QdUz7Q0AAAAz8w8To_5nZKlB4Z7A1utA
yup i see your post but i cant help you with strategy...
===
Subject: Criss-cross Grid Game
posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE
I am reposting this to rec.puzzles and sci.math via Google.
I suspect that it was not seen by most newsreaders when I
posted it the
first time
(about 2 weeks ago).
I would be happy if someone replies to this post
to prove that SOMEONE is getting posts made via Google.
---
Here is yet another of my games played on an n-by-n grid
drawn on
paper.
2 players,each with a different colored pencil.
Players alternate placing an X or an O of their color in the
grids
squares as follows:
*Player 1 places the first O in the upper-left square.
*Each player places their symbol immediately next to the
symbol last
drawn by the other player in the directions of vertically,
horizontally, or diagonally.
*If the square a player draws his symbol in is empty, he/she
draws an
O.
O.
*A player may not draw in (move to) a square already occupied
by an X
(and an O).
Play continues until moving is impossible or when one player
first gets
a predetermined number of points.
Scoring:
*If a player places an X in the same square as an O, the
player who did
*not* draw the O in that square gets a point.
*If a player places an O in an empty square diagonally
adjacent to the
last square drawn in by the other player, and the other 2
squares in
the 4-by-4 square -- formed by the empty square being drawn
in and the
last drawn-in square -- are both filled, then the player
Ôcompleting
the square (the player whose move it is) gets a point.
Example:
+---+---+
! O ! !
+---+---+
!OX ! O !
+---+---+
Last move put an X in lower left square (getting a point for
the player
who did *not* place the O there earlier). Player now wants to
place an
O in upper right square, which would get him/her a point.
What is a good strategy for this game?
Leroy Quet
===
Subject: Re: Criss-cross Grid Game
posting-account=
y3wZYhMAAABYsCtaDBjCWE5oFd14ElQZbfvQjxC1czdFUKdrfKUl4g
While I have noticed some ßakiness in Google displaying msgs
posted
through Google, they are definitely getting out to other news
servers.
Anyway, I always enjoy reading your posts, though I rarely
have a
solution befitting your puzzles.
===
Subject: A Transform Of Seqs Involv Permutations
posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE
I might have written about this already a long while back.
(It seems
slightly familiar.) If so, I apologize.
Here is a transform which converts any sequence of positive
integers
and an infinite number of 1s into another 
sequence of positive
integers and an infinite number of 1s.
(Perhaps some of you might want to submit new sequences to
the EIS
generated this way.)
Start with a sequence,{a(k)}, of only positive integers and
an infinite
number of 1s.
Example:
1,1,2,1,2,3,1,2,3,4,1,...
Form the sequence {b(k)} (which is the permutation of the
positive
integers),
where b(k) = the a(k)th lowest positive integer not yet in
the sequence
b;
and b(1)=a(1).
1,2,4,3,6,8,5,9,11,13,7,...
Let {c(k)} be the inverse of {b(k)}.
1,2,4,3,7,5,11,6,8,...
Form the final sequence {d(k)}, where each d(k) is such that
c(k) = the d(k)th lowest positive integer not yet in the
sequence c;
and d(1)=c(1).
1,1,2,1,3,1,5,1,1,...
So {1,1,2,1,2,3,1,2,3,4,...} is transformed into
{1,1,2,1,3,1,5,1,1,...}.
Is there a more direct route to carrying out such a transform
without
finding the permutations? (I have not thought about this 
much.)
And this whole idea must not be new. What info is there about
this,
such as a preexisting name for the transform?
Leroy Quet
===
Subject: Intersection Puzzle/Grid-Solitaire Game
posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE
You start with an 8-by-8 grid, which is 9-by-9 lines, lightly
drawn on
paper.
Start on any corner of the grid.
You draw darker lines along the light lines from vertex to
vertex
(vertex = intersection of light lines), alternating drawing
horizontal
and vertical lines, not drawing any darker lines where darker
lines
already exist.
You can make your line any length in one direction (up, down,
left, or
right) as long as the line goes from the end of the last line
to a
vertex of the grid.
No darker lines should coincide, except where the darker lines
intersect (and continue past each other). Your path of
connected darker
lines should not even meet itself at any of its corners
(meeting at a
single vertex).
You get a point every time an intersection is 
Ôperpendicular
to the
last intersection made (AS the darker lines are being drawn).
(And you get a point for your first intersection.)
By Ôperpendicular intersection, I mean that if 
your previous
intersection was a horizontal newer dark line crossing a
vertical older
dark line, then your current intersection is a vertical newer
dark line
crossing a horizontal older dark line, and vice versa.
Your path of connected darker lines starts and stops at the
same
corner.
I got 13 points. Can you get 13 points or better?
I also got 14 points by drawing a completely different path.
Can you
get 14 points or better?
Example of perpendicular crossings:
(View with fixed-width font.)
----+ +---+
! ! !
+---+---+---+
! ! !
+---+ !
Example of non-perpendicular crossings:
----+
!
+---+---+
! ! !
+---+ !
!
---+---+
! !
+---+
Leroy Quet
===
Subject: A Nested Sums Congruence
posting-account=Jngi7wwAAAD2WLn2V2E6Gh2GXydPdCaE
Let H(m,n) be defined as follows:
H(0,n) = 1/n.
H(m,n) = sum{k=1 to n} H(m-1,k).
(So, H(1,n) is the nth harmonic number, H_n.)
Let S{q,r,t}(n) be defined as follows (for q,r,t, and n =
nonnegative
integers):
S{q,r,0}(n) = H(q,n).
S{q,r,t}(n) = sum{k=1 to n} S(q,r,t-1}(k) /(k+q+r-1).
So, for example, S{1,r,2}(n) =
sum{k=1 to n} sum{j=1 to k} (sum{m=1 to j} 1/m) /((j+r)(k+r)).
Then, for n+q+r >= 2:
n! ((n+q+r-1)!)^t S{q,r,t}(n)
is congruent to
n! ((n+q+r-2)!)^t H(r,n) (-1)^(n+1) (mod (n+q+r-1)^(t+1))
So, for example, if t = 0 we have a congruence I have already
posted in
the past:
n! H(q,n) is congruent to
n! H(r,n) (-1)^(n+1) (mod (n+q+r-1)).
If we let q = 0, r = 1 (and reduce t by 1) we get:
n! h_t(n) is congruent to
((n-1)!)^t (-1)^(n+1) (mod n^t),
where h_0(n) = 1, and
h_t(n) = sum{k=1 to n} h_{t-1}(k)/k.
(So, h_1(n) is the nth harmonic number, H_n.)
Leroy Quet
===
Subject: inverse LT of bessel function
Hi all,
Im looking for an inverse Laplace transform of Bessel
function (most
interested in I and K, but also others) of the form
I(n,a*sqrt(s)) or K(n,a*sqrt(s))
Alex
===
Subject: Re: matlab problem -martix
<odUud.57733$QJ3.6228@newssvr21.news.prodigy.com>
<T1Vud.41617$6q2.30925@newssvr14.news.prodigy.com>
<B00vd.32518$zx1.32126@newssvr13.news.prodigy.com>
posting-account=fE9RUg0AAADEpHox-0lgIe-SEpJxkHjD
I got it done.
IT is a markov matrix.
===
Subject: Need help on my educational website
posting-account=LSltBg0AAACI7ITc0n2uQOmc7-4a5kwQ
Hi. Im building a nonprofit educational website 
with the
wikimedia
software to serve everyone in the world who would like to see
example
problems and solutions for all kinds of mathematics. Please,
for the
good of everyone taking a math class next semester,
contribute one or
two problems and solutions to the site. Follow the format
Ive laid
www.exampleproblems.com
-Todd Smith
UCF, Mathematics
===
Subject: Poisson Process Probability
posting-account=fE9RUg0AAADEpHox-0lgIe-SEpJxkHjD
I have a probability question:
let Tk denote the waiting time until the Kth event, for a
Poisson
process with intersity lamda. Find the distrubution of Tk/Tn
for
1<=K<=n. (note: the conditional distribution of Tk/Tn, give
Tn=t, is
indep of lamda and t.)
This is what I am thinking of quotient dist. of P
P( Tk/Tn <= s) = Integral (oo to 0) P (Tk<= st | Tn=t)dt
This just pop up to my mind, and I am not sure that logic is
right or
wrong. Then I looking to the note, it says Tk/Tn is
conditional dist.
Can anyone help me to clear my mind. I am stuck here.
Confused, can
anyone tell me whats wrong with my approach?
===
Subject: what is maximum eigenvalue of this matrix?
posting-account=kncgtA0AAAAjvHVPgFKTn4k0K3euxgh0
Let v(n) is some n-tuplet whose every element is finite,
and let v(n) be the transpose of v(n).
Define the nxn matrix
M(n) = v(n) v(n) .
What is the necessary and sufficient conditions on v(n) such
that
the maximum eigenvalue of M(n) grows UNBOUNDEDLY in
proportion to n as
n --> infinity?
Roger
===
Subject: Birthday problem -- probability
posting-account=fE9RUg0AAADEpHox-0lgIe-SEpJxkHjD
I have a problme on Birthday, but I am think messed up some
concept.
50 people are chosen at random and their Bdays (only day and
month, not
year) are recorded. Let X denote the # of distinct Bdays in
the sample.
Find Var X.
(note: X = SUM(365 1) Xi , where Xi =1 (Bday i in sample) and
0
otherwise)
I am dont understand the note: X = SUM(365 1) Xi
why? What about more than one person has the same Bday?
===
Subject: Re: Birthday problem -- probability
posting-account=Glvc4AwAAADzVCZ73XnxpzMhXir6xVzs
> I have a problme on Birthday, but I am think messed up some
concept.
> 50 people are chosen at random and their Bdays (only day
and month,
not
> year) are recorded. Let X denote the # of distinct Bdays in
the
sample.
> Find Var X.
> (note: X = SUM(365 1) Xi , where Xi =1 (Bday i in sample)
and 0
> otherwise)
> I am dont understand the note: X = SUM(365 1) Xi
> why? What about more than one person has the same Bday?
The way the Xi is defined, then X_100 = 1 whether one person
has birthday = 100, or whether 20 people have birthday 100.
The hint is telling you to consider the problem in terms
of the Xi. Can you work out the distribution of Xi for
any i? Xi has only two possible values, 0 and 1. You can
calculate the probability of both. Given that, you can
calculate the expectation and variance of Xi. What does
that tell you about var X?
- Randy
===
Subject: Re: GCH vs. Axiom of Choice.
|I asked if the things PROVABLY true in L by ZF+A are
CONSISTENT with
|the things PROVABLY ture in L by ZF+B, and you said No, not
|necessarily, but without an example.
|False axioms can be consistent with ZF. You could add a
false axiom to
|ZF and the results that you could prove to be true in L with
this
|axiom system might not in fact be true in L.
|Could you please be so kind as to explain what a false axiom
is? Im
|familiar that somone has proven (in ZF?) that Con(ZF) =>
Con(ZFC) as
|well as Con(ZF) => Con(ZF+~C), so I didnt think C or ~C 
was
|considered false, just relatively independant axioms. What
is an
|example of a false axiom that is (relativily?) consistent
with ZF?
|Are you assuming that every statement X for which we can
prove
|Con(ZF)=>Con(ZF+X) and Con(ZF)=>Con(ZF+~X) is neither true
|nor false? I dont mean it as a rhetorical question.
|I think of them as sentance such that (~ZF)v(X) cannot be
validities.
|As opposed to theorems T of ZF which I consider to be
sentances such
|that (~ZF)v(T) is a validity.
|
|I define the truth of a sentance based on the truth of atomic
|sentances, z in x for elements x,z of a domain of discourse,
[...]
Im going to take this as a no, since nothing in the 
following
implies (as far as I can see) that every statement X for
which we
can prove Con(ZF)=>Con(ZF+X) and Con(ZF)=>Con(ZF+~X) is
neither
true nor false.
I try to be fair with peoples views that differ from mine,
including in particular people who are formalists. Sometimes,
though, it starts to look like Id be better off just saying
that the formalist philosophy of mathematics is just wrong;
I think at times it would be less confusing to people to put
it that way.
terms being used mean, and THEN chooses a set of axioms, 
its
quite
obvious what it means to say that one has a false axiom: one
simply
chose a statement that happened to be false, as an axiom.
That may
seem funny, but theres nothing to say one 
cant do it.
The comment ÔI didnt think C or ~C was 
considered false,
just relatively independant axioms. strongly suggests your
having heard some kind of half-baked formalist philosophy.
Now,
I could explain to you both why formalists will say things of
this nature, why other philosophies dont say it, and why I
prefer the one over the other, but this gets to be very long.
What would be much more convenient, and probably less likely
to
be confusing for you, would be for you to unlearn some of
these ideas. At least, realize that they are far, far, far
from being any kind of consensus point of view. I would say
that
its much closer to a consensus point of view that AC is 
true,
~AC is false, hence ZF+~AC is an axiom system with a false
axiom in it.
[...]
|>I dont assume anything like that, and to me this seems
like a
|>nonsequitur. If you believe such a thing, and you dont 
know
|>why you believe it, its liable to be a source of some
confusion.
|
|I dont even know what it MEANS to say a sentance X is true
without
|reference to a domain of discourse. Saying its valid I 
get.
Saying
|its true, I dont get.
Usually valid is defined as truth in all domains, which
requires
knowing what truth in a domain is.
The problem with saying that truth is always with reference to
a domain of discourse is that the reference is not necessarily
explicit. I suppose one could say that, implictly, there is
some
domain of discourse, but always having to state the domain of
discourse *before* considering the truth of a statement to be
well-defined leads to an infinite regress.
When as infants we pointed and said Cat!, we did this without
awareness of such things as domains of discourse. When our
parents said, No, thats a dog!, we were learning implicitly
about the extension of the concepts of cat and dog merely
by example.
Likewise, as students we eventually learn about this
correspondence
between classes such as cats and sets of cats by example to
begin with, perhaps followed up with a little analysis of what
kind of concept it is. Then there are sets of sets of cats,
and
so on. After enough examples, its possible to say, This 
kind
of
abstract object is what we mean by a well-founded set. and
This
kind of thing is what we mean by a pure, well-founded set.
People disagree over whether its been made 
sufficiently clear
what kind of thing is being discussed, or whether there even
exists a coherent such notion. I find that in general, 
its
much
easier to continue once one has chosen to do one of the
following
four things:
1. Agree to treat it as well defined.
2. Agree to pretend that it is.
3. Decide not to talk about them, since one feels that the
concept is not coherent enough to make it make sense.
4. Talk about whether the concept is coherent or not.
I feel like we have gone along for periods doing something
like
1 or 2, and then, *zoinks!* you ask a question whose answer
would require re-opening the issue of what these sets really
are. I think the topics are fine taken separately, but mixing
them up has an unpleasant jumpiness about it.
[...]
|So L contains all the ordinals that can be proven
Hold it. I didnt say anything about proven ordinals. I said
it contains all ordinals.
|and any set
|generated by a cumulative application of, for instance,
separation?
|That sounds a bit vague since we need both all and ordinal
before
|we could know what was in L.
Its not at all vague, but yes, the concept of L depends 
upon
the concept of ordinal. To the extent that anyone believes L
is
well-defined, its because they already believe 
that ordinal is
well-defined.
See, if we were simply discussing whether the cumulative
hierarchy
concept (i.e., the class of pure, well-founded sets) were
coherent,
we wouldnt have this distracting business about L here. If
you had
decided you didnt think the concept held together, we
wouldnt be
talking about L. Or, we could have agreed at least to pretend
that
theres such a thing as the cumulative hierarchy, in which
case the
discussion of L wouldnt have to be interrupted with this
kind of
skeptical doubt about ordinals and so on.
Keith Ramsay
===
Subject: Re: GCH vs. Axiom of Choice.
|Is L a set?
No, L is a proper class.
|Obvious L is not in L, but if seems like definitions of say
|0# seem to say that it exists if L exists.
Well, the definition of L is just a definition of 
a property.
The existence of 0# doesnt follow from ZFC. That 0# exists
contradicts V=L.
|Is that because the
|existence of L is undecideable because otherwise ZF could
prove its
|own consistency by proving the existence of L?
No. Strictly speaking, the existence of proper classes isnt
expressible in the language of ZF, since its only about 
sets.
But in GB, which is essentially ZF but with proper classes
included,
the existence of L is easy to prove, once you have a
definition of
the property (constructibility) that defines L. L is simply
those
sets that are constructible in Goedels sense.
This might seem tantamount to GB being able to prove the
existence
of a model of ZF. But its impossible in GB to 
define a
truth-predicate
for proper classes. This is a delicate point. Given a
sentence in
the language of ZF, we can describe how to convert it into a
sentence relativized to L. But this is not the same as being
able
to define a predicate true-in-L(S) that holds for those
sentences
that are true in L. A proof of the consistency of ZF requires
more
than what GB assumes. (In fact, the theorems of GB that are
only
about sets are the same as the theorems of ZF.)
[...]
|Maybe my original question wasnt clear. If you take every
sentance X
|such that Con(ZF)=>Con(ZF+X) & Con(ZF+~X), then can there be
a sentence
|S about L
This question is a little unclear too.
|(which to more clear if L is a set, Ill define 
as a
|sentence such that (Ex T) is equivalent to (Ex xeL & T) for
every
|formulas (Ex T) and (T) that are subformulas of S and such
that (Ax T)
|is equivalent to (Ax (~xeL)vT) for every formulas (Ax T) and
(T) that
|are subformulas of S)
Well, L is not a set, but what you are writing is okay
anyway. The
formula x in L means, by definition, x is constructible, which
has a perfectly okay definition in the language of ZF.
The sentence you get by replacing all the existential
quantifiers
(Ex)T(x) by (Ex)(x in L & T(x)) and all the universal
quantifiers
(Ax)T(x) by (Ax)(x in L -> T(x)) in a sentence P is what I
had been
denoting by L |= P. Incidentally, L|=~P and ~(L|=P) are always
the same.
|such that ZF+X|- S and ZF+~X|- ~S.
It sounds like you are asking whether there exist sentences X
and P
such that Con(ZF)=>Con(ZF+X) & Con(ZF+~X), ZF+X|-(L|=P), and
ZF+~X|-(L|=~P). Yes. Take X=P=Con(ZF) for example. That the
consistency
of ZF holds relative to L is the same as ZF being consistent.
Or, it might be that youre asking whether for every X such
that
Con(ZF)=>Con(ZF+X)&Con(ZF+~X), there exists a P such that
ZF+X|-(L|=P) and ZF+~X|-(L|=~P). For some such Xs there is
not.
For example, if X=AC, then as I explained in another posting,
ZFC|-(L|=P)
if and only if ZF|-(L|=P) which implies ZF+~AC|-(L|=P) as
well.
|And more
|generally, how about any chain of X,Y,Z, ... such that for
every subset
|A or B of of X,Y,Z,... such that Con(ZF)->Con(ZF+A) and
|Con(ZF)->Con(ZF+B) is it possible that ZF+A |- S and ZF+B |-
~S. Im
|trying to figure out if the superconsistent axioms of ZF
together proof
|the alleged fixed truths of L. FOL is complete, so it might
work.
I dont know what superconsistent means here.
|> On the other hand, there arent any first 
order properties
of L
|> that can be proven in ZF+GCH but not in ZF. In fact, for
an arbitrary
|> sentence X in the language of ZF, the following are
equivalent:
|> (a) ZF |- (V=L->X)
|> (b) ZF + V=L |- X
|> (c) ZF |- L |= X.
|
|I can read (a) and (b), but (c) and (d) must have some
operator order
|preference I dont know, or is ZF |- L |= X only to be read
as ZF |-
|(L|=X), which I dont even know what it means unless L is a
sentence,
|and even so how one proves a validity based on axioms I
dont know so
|even then it doesnt make sense to me.
Well, ZF |- L is nonsensical, whereas L |= X represents a
formula in
the language of ZFC. So its ZF |- (L |= X).
|> (d) ZF + V=L |- L |= X
(ZF+V=L) |- (L|=X).
[...]
|Im trying to find out if new relativity 
consistent axioms
added to ZF
|prove things about L consistently and whether taken as a
whole they do
|so completely.
Is this what you are asking:
For each sentence P, does there exists a sentence X such that
Con(ZF)=>Con(ZF+X) and Con(ZF)=>Con(ZF+~X) are both theorems
of ZF,
and such that either ZF+X |- (L|=P) or ZF+X |- (L|=~P)?
|> | Do the truths of L
|> |vary depending on what other axioms L is defined with? 
Im
trying to
|> |be more clear, but I apologize in advance if I failed
again.
|> How would whether something is true of L depend on axioms?
|
|I dont even get the definition of L, so to me 
its all
provisional.
How would whether anything at all is true depend on axioms?
|> Of course, what can be proven to be true of L from a set
of axioms
|> does depend on the axioms! Im not familiar with the
details, but I
|> gather that some large cardinal axioms (I think the
existence of a
|> measurable cardinal is strong enough) imply a relatively
detailed
|> account of the structure of L.
|
|But are there other axioms that imply details that
contradict those
|details?
For whatever sentence P we have where ZF+MC |- (L |= P) but
not
ZF |- (L |= P), it would be possible to take L |= ~P as an
axiom,
and ZF+(L|=~P) would be consistent. So sure, alternative
axioms could
imply contradictory sentences relativized to L.
[...]
|> This real 0# that is believed to be outside of L encodes
some of the
|> structure of L (if it exists).
|
|Is it L v 0# whose existence is debated?
I dont know what this v is supposed to mean. V=L and the
existence
of 0# are two contradictory statements. If ZF is consistent,
then so
are ZF+~(V=L) and ZF+(V=L). Since V=L -> 0# does not exist)
is provable
in ZF, we can also say that if ZF is consistent, then
ZF+~(0#) is
consistent. We cant prove that Con(ZF)->Con(ZF+0# exists) 
in
ZF;
the consistency of the existence of 0# with ZF is a stronger
principle,
sort of large-cardinal-like.
Set theorists appear generally to favor the existence of 0#
over V=L.
Keith Ramsay
===
Subject: Re: GCH vs. Axiom of Choice.
|>I dont assume anything like that, and to me this seems
like a
|>nonsequitur. If you believe such a thing, and you dont 
know
|>why you believe it, its liable to be a source of some
confusion.
|
|So, is the Parallel Postulate true all by itself -- whatever
that means?
The important thing is to consider the meaning of the terms
being
used.
Im no expert on ancient Greek mathematics. I think 
generally
speaking the primary sources are sketchy enough to make it
hard
to be sure of the philosophy behind what they were doing. But
my
laymans impression is that what the ancient Greeks thought
they
were doing when they did (what we call Euclidean) geometry is
part of what we would now classify as theoretical physics.
The study
of the geometry of physical space is now of course physics;
the
main theory of it is general relativity, Einsteins theory 
of
gravitation.
So when Euclid refers to points and lines, it appears he was
implicitly assuming that there is a well-defined notion of two
events being at the same place although at different times.
According to special relativity, such an identification is
possible once one has chosen a reference frame. But we know
special relativity is only approximately correct. In general
relativity, the concept of point in space per se becomes much
more a matter of arbitrary convention.
The parallel postulate as originally understood appears, then,
not to be well defined, since it relies upon a concept of 
point
thats only got an approximate, conventional sort of 
validity
about it. (And to the extent that we have such a thing as the
geometry of space, it appears not to be Euclidean.)
Today we have a second, mathematical field known as Euclidean
geometry. As mathematician/physicists developed Euclidean
geometry, they were holding the view that we would today
describe
by saying that physical space is correctly modelled by an
abstract,
mathematical space known as Euclidean space. We no longer
believe
that it is correctly so modelled (in the sense they had in
mind),
but we still believe in the abstract mathematical space that
they
were studying. The field in effect split into two, part that
has
now been turned into general relativity, and part that has
remained
as Euclidean geometry, a branch of mathematics.
The parallel axiom does hold in Euclidean space. I dont 
mean
by
this that we made it true by making it an axiom; I mean that
we
have an abstract structure, and that among the facts we know
about
it is the parallel postulate.
|If not, whats different about C?
AC wasnt ever intended as part of a theory of physics, for
one thing, so overturning it would require something different
from doing physical experiments.
I would suggest making a comparison with a somewhat less
controversial
statement. (Note that AC really is not very controversial in
the
mathematical community. But there are things where the
controversy
has been a fraction as much.) Take the axiom of induction in
PA.
Its independent of the other axioms. But 
wouldnt you say
that its
true?
Let me back up to an earlier comment of yours:
|I guess that I got this idea from (popularized) readings
about the
|Parallel Postulate. Some guy spends his life trying to prove
it by
|reductio ad absurdum and fails. People build on his work,
giving us
|new forms of geometry.
I dont know whether anybody actually spent their lives
trying to prove
the parallel postulate from the other ones. There were a lot
of people
who attempted to prove it from the other ones.
Trying to find out whats true is a bit of a 
different goal
than
trying to find out what can be proven from what. You 
cant
prove
something from nothing, after all. There are things that can
be
proven just from logic, but their content is in a sense
insubstantial. So at least some of the things one believes to
be
true, one believes in spite of not having a way to prove them
from
previously established truths.
It is an interesting fact that one cant prove the parallel
postulate from the other postulates, but it doesnt have 
much
to do with whether it has a well-defined truth or falsity.
|Each of these forms of geometry is, to the
|best of my knowledge, considered true.
I dont know what it means to call a structure true.
|(Obviously, you cant take
|the results of one and apply them with another.)
Ill admit up front that what I have to say 
isnt agreed upon
by
everyone. A formalist, in particular, would have a different
take
on the whole thing.
and constructivists (including intuitionists), however, we
know
that there exists a structure called the Euclidean plane,
which has
a well-defined set of statements about it that are true, and
we know
that there exists a structure called the hyperbolic plane,
which has
a well-defined set of statements about it that are true,
merely a
different set. Theres a substantial overlap, but they are
essentially
independent sets of statements to make (and questions to ask).
Its also possible to ask about the overlap between the two
sets,
or to ask what statements are true in all models of neutral
geometry
(which is essentially Euclidean geometry with the parallel
postulate
removed). But from our point of view, these are just further,
distinct questions to consider. The fact that some statements
are
true in Euclidean geometry while analogous statements about
hyperbolic geometry are false, and vice versa, isnt seen as
detracting from the definiteness of either body of fact.
Keith Ramsay
===
Subject: Re: GCH vs. Axiom of Choice.
|How do I distinguish the set of all reals from a set of
arbitrary
|reals?
I dont know what this means. If you asked, how do I
distinguish
an orange from a tangerine, I would understand you as asking
for a procedure that you could apply when being handed a fruit
that you were promised was one or the other, to tell which it
is.
But since nobody physically hands you sets of reals, its 
not
so
clear what you mean.
If you have in mind a set of reals abstractly presented to
you, then
the answer is that you cant always tell them apart. For
instance,
{1/2,1} u {0,-2,-4,-6,...} u {t : zeta(s+it)<>0 for all s} is
presumably
all the reals, but we cant prove it.
I assume this is also not the question you meant, however....
I recommend mulling over what it is you really meant.
Keith Ramsay
===
Subject: Differentiation of a matrix expression
I have an expression
J= GAATGT
where the T denotes transpose ie it is G times A times AT
times GT
I wish to differentiate the expression wrt G (in fact all
these
matrices are function of discrete time k ie G(k) and
A(k),k=0,1,2...)
I need dJ/dG(k) and I am unsure how to find it.I think the
answer is
-2AATGT
but then again it could be
-2GAAT
The transpose is confusing me. The A matrix I just take to be
a
constant.
Would be grateful for any help!
===
Subject: Re: Is zero even or odd?
> Zero is even. You cannot divide by zero. Limits are not
division.
> Infinity is not a number. Computers bugger up the system.
That could be said louder, but not clearer :).
--
David Serrano
===
Subject: Re: Is zero even or odd?
> (-0)^2 = -0
Not on my Casio calculator.
Franz
===
Subject: Re: Is zero even or odd?
> I read in sci.electronics.design that Nicholas O. Lindan
<see@sig.com>
about
>> 0 cant be divided by itself,
>Sure it can: 0 / 0 = 0 * (1 / 0) = 0 * infinity = 1
> One possible solution, given the enormous lack of rigour in
Ôinfinity.
There is no lack of rigour in the definition of 
infinity. Read
anbout
the work of Cantor, Dedekind and others.
Franz
===
Subject: Re: Is zero even or odd?
>I know 0 is neither negative or positive but what about
odd/even? I think
>its even.
It would be very odd if zero were not even nor even odd.
--
Im not interested in mathematics that might have anything
to do with reality. -- Russell Easterly, in sci.math
===
Subject: Re: Is zero even or odd?
>>I know 0 is neither negative or positive but what about
odd/even? I think
>>its even.
>It would be very odd if zero were not even nor even odd.
Never odd or even!
Lee Rudolph
===
Subject: Re: Is zero even or odd?
[a bunch of groups removed]
> I know 0 is neither negative or positive but what about
odd/even? I think
> its even.
Zero is neither odd nor even.
Ask any roulette croupier.
Michael
--
Still an attentive ear he lent Her speech hath caused this
pain
But could not fathom what she meant Easier I count it to
explain
She was not deep, nor eloquent. The jargon of the howling main
-- from Lewis Carroll: The Three Usenet Trolls
===
Subject: Re: Is zero even or odd?
[Extravagant cross-posting removed.]
>
> I know 0 is neither negative or positive but what about
odd/even? I
think
> its even.
Odd numbers start at 1 and go every other number
1,3,5,7;1,-1,-3,-5,-7
> Even starts at 2 and go every other number
2,4,6,8;2,0,-2,-4,-6,-8
>
> As it can be divided by 2 without a remainder, it is
obviously even.
>>The divisor would have to be something smaller than 0 like
-2.
>>Therefore zero is both even and negative.
>PSpice uses...
>æ +1 if x>0
>æSGN(x) = 0 if x=0
>æ -1 if x<0
>which matches up with what is shown on Wolframs site...
>http://mathworld.wolfram.com/Sign.html
In fact, Wolfram Research has recently bought the rights to
the
number zero --- which explains the problems users are
currently
experiencing (as evidenced by some of the posts in this
thread).
We are assured that the problems are temporary, being caused
by
the inadvertent release of a beta version.
Applied mathematicians are requested to keep all numbers
positive
for the time being. Pure mathematicians may, strictly at their
own risk, use a *very* small surreal number in place of zero
- but
Wolfram Research will not, under any circumstances, be held
liable
for any absurd, strange, immoral, or un-American consequences
that
may follow.
--
Angus Rodgers
Contains mild peril
===
Subject: Re: Is zero even or odd?
> I seeem to recall 0 coming up negative in some old IBM
mainframes. That
was
> an artifact of the way signed numbers were converted to
binary.
In computers using ones complement arithmetic, the number
zero can be
represented in two ways: all bits zero (which appears
positive) or all
bits one (which appears negative).
But this is just a matter of how numbers are represented in
the machine,
and says nothing about whether zero is really a positive
number, a
negative number, or neither.m
--
Alec McKenzie
===
Subject: Re: Is zero even or odd?
posting-account=4l-peA0AAACabaoispHnNpa9VfXI_fbk
> I know 0 is neither negative or positive but what about
odd/even? I
think
> its even.
It is essentially neutral for it signifies a number that 
isnt
there
because it lacks quantity. It is the symbol for nothingness
therefore
it isnt odd nor even. For example, take the addition of all
numbers
from -infinity to +infinity, what do you
get?...zero!...nothin!....nada! The sum of everything is
equal to
nothing!
In physics, the conservation of energy states that no energy
can be
created or destroyed. Therefore, if you add up all of the
negative
potential energy in the universe with the positive kinetic
energy, you
get NOTHING!....hence the total energy in the universe is
zero.
Therefore, we are all essentially made up of nothing.
Sleep tight!
===
Subject: Re: Is zero even or odd?
posting-account=s1UxvAwAAAAcX9W4GCK8lzXVIWgkSg6e
> I know 0 is neither negative or positive but what about
odd/even? I
think
> its even.
> Odd numbers start at 1 and go every other number
1,3,5,7;1,-1,-3,-5,-7
> Even starts at 2 and go every other number
2,4,6,8;2,0,-2,-4,-6,-8
The majority of the mathematical postulates suggests that 0
is neither
positive nor negative; however, that it is an even number.
Take into account that Even+Even = Even. Here, 0 can be
substituted
into any spot as in 6+0=6 to satisfy the postulate.
There is also Odd+Even = Odd, and 0 works here too.
It can also be put into multiplication postulates such as
Even^(any
positive integer) = Even. As opposed to Odd^(odd positive
integers) =
Odd.
There are certain undefinable and Ôcircular 
reasoning
postulates. Such
as Odd*Even = Even, in which case the quotient is undefined.
---------
Steven Xu