mm-369
> It seems to me that
> Becquerel, who discovered fission,
> and Fermi who discovered how to make the slow neutrons
> needed to achieve enough fission,
> and Frisch who first understood fission,
> and the unnamed explosives technician
> who finally convinced the scientists to
> use a shaped charge to achieve critical mass,
> played the most important roles in the development
> of the fission bomb, and not to mention,
> General Groves who direc the project,
> Lawrence who worked out the isotope separation schemes,
> and the scientists, engineers and technicians at
> Tennessee and Washington plants
> who found ways to make enough fissionable material
> to make a few bombs.
> Note that Fermi, the Italian Navigator was Italian,
> not Jewish, as Glazier asserts,
> and that Klaus Fuchs was the spy who passed
> American and British atomic bomb secrets
> to the Russians.
> --
> Tom Potter http://tompotter.us
You once again demonstrate your stupidity. Fermi was not
Jewish but
Laura Fermi (his wife) was. Klaus Fuchs was not Jewish, he was
the son
of a Lutheran minister. So having spilled your stupidity out
for all to
see we can infer that all the other stuff is just as stupid
and wrong as
this attempt.
FK
===
Subject: Re: Fundamental Reason for High Achievements of Jews
>Wonderfully amusing that your claim that the Jews who migra
to
Israel
>>from the USSR are Bolsheviks.
A war-for-profit instigator by any other name,
> would be a war-for-profit instigator.
I trust that fkasner will explain to us
> why over one million Russian Jews migra to Israel,
> and thousands to New York,
> after the native Russians began to regain control of their
nation?
Of course, it was because the Russian were/are anti-Semitic,
> and NOT because the Bolsheviks had raped Russia,
> and destroyed its' viability,
> and that their standard of living had slipped below
> that of their free world brothers,
> and that as the Native Russians were regaining control of
their nation,
> it was becoming even harder for them to enjoy a high life
style.
Yeah!
> That's the ticket!
> It was because the Russians are anti-Semitic!
Like the Germans, the Muslims, the French, the Italians,
> most Americans, Asians, etc., the Russians are anti-Semitic!
> Yeah!
> That's the ticket!
--
> Tom Potter http://tompotter.us

> I have already supplied you with historical facts that you
do not
> address whereas you keep insisting that I have to do the
investigations
> that demonstrate the things you claim to be true. Please
note that the
> Jews who left the Soviet Union when it collapsed dis so
because they
> finally were able to leave the Soviet Union. Prior to that
they were
> generally not allowed to leave. Very few got out before the
USSR
> listening? Or are you just dreaming about other specious
claims that you
> say are obvious truths but you refuse to deal with in detail
or produce
> documentary proof.
> Note also that few Russians who were Jews were permit to
join the
> Communist Party. How then can you insist on calling them
Bolsheviks. In
> fact that term is applicable to those members of the
Communist Party who
> continued to favor revolution rather than deal with the
Kerensky led new
> Duma and its socialist flavored attempt at a constitutional
monarchy.
> But then again you probably never heard of Kerensky. Whereas
I and
> several hundred others heard him speak at the University of
Chicago and
> the Communists were dumbfounded and were silent during the
question
> period. It seems he was an element of history they couldn't
fabricate
> quickly enough. Similar to you. Your similarity to the
Communists is
> quite striking. If we merely substitute the word Capitalist
for your
> use of the word Jew we see the way you have absorbed their
political
> and thought processes in toto. You have revealed yourself
thoroughly,
> Comrade.
> FK
As can be seen by studying FACTUAL history,
rather than self-serving spin,
most of the central players in the overthrow of the
Russian government, and the massacre of the Russian royal
family,
were Bolsheviks, and most of the Bolsheviks were Jews,
and most of the power positions in Russia
were held by these same people for many years.
It was only when the native Russians,
loyal to Russia, and life, liberty and happiness
for ALL people begin to regain control of the
Russian government that the people who raped Russia
began to leave.
And note in the case currently in the news,
that of the battle between Putin and Khodorkovsky,
that the Russians are still trying to keep the
remnants of the old order from recapturing control of their
nation,
using greedy, immoral, criminal tactics.
As Putin said yesterday
Everyone should be equal before the law,
irrespective of how many billions of dollars a person has on
his personal
or
corporate account. -
Otherwise, we will never teach and force anyone to pay taxes -
and defeat organized crime and corruption.
And as Satarov said regarding the same matter,
The [oligarchs] are paying the price now for placing the wrong
bet [by
backing Putin] in 2000,
The [oligarchs] are just Bolsheviks by another name,
and using another greedy, immoral tactic
to steal the life blood of the Russian people.
As can be seen,
because of their small numbers,
it is necessary for the people who
co-opt governments so they can steal billions from the masses,
to put a front man like FDR, Bush, Blair, Stalin, Putin, etc,
in charge, and even though they think they can control
their front man, sometimes one arises,
that turns on them,
usually to protect their own ass from these evil criminals.
I dare say that the will really hit the fan in Russia,
and quickly spread around the world,
and that the peoples of the world will begin
to come down on the evil criminals who
buy and intimidate their leaders,
and instigate war for power and riches.
Can you imagine what will happen in America
when tens of thousands of disillusioned,
servicemen come home and think about
what has happened to them, their families,
and their country???
--
Tom Potter http://tompotter.us
===
Subject: Re: Fundamental Reason for High Achievements of Jews
>>Wonderfully amusing that your claim that the Jews who migra
to
Israel
>>from the USSR are Bolsheviks.
>A war-for-profit instigator by any other name,
>would be a war-for-profit instigator.
>I trust that fkasner will explain to us
>why over one million Russian Jews migra to Israel,
>and thousands to New York,
>after the native Russians began to regain control of their
nation?
>Of course, it was because the Russian were/are anti-Semitic,
>and NOT because the Bolsheviks had raped Russia,
>and destroyed its' viability,
>and that their standard of living had slipped below
>that of their free world brothers,
>and that as the Native Russians were regaining control of
their nation,
>it was becoming even harder for them to enjoy a high life
style.
>Yeah!
>That's the ticket!
>It was because the Russians are anti-Semitic!
>Like the Germans, the Muslims, the French, the Italians,
>most Americans, Asians, etc., the Russians are anti-Semitic!
>Yeah!
>That's the ticket!
>--
>Tom Potter http://tompotter.us
>>I have already supplied you with historical facts that you
do not
>>address whereas you keep insisting that I have to do the
investigations
>>that demonstrate the things you claim to be true. Please
note that the
>>Jews who left the Soviet Union when it collapsed dis so
because they
>>finally were able to leave the Soviet Union. Prior to that
they were
>>generally not allowed to leave. Very few got out before the
USSR
>>listening? Or are you just dreaming about other specious
claims that you
>>say are obvious truths but you refuse to deal with in detail
or produce
>>documentary proof.
>>Note also that few Russians who were Jews were permit to
join the
>>Communist Party. How then can you insist on calling them
Bolsheviks. In
>>fact that term is applicable to those members of the
Communist Party who
>>continued to favor revolution rather than deal with the
Kerensky led new
>>Duma and its socialist flavored attempt at a constitutional
monarchy.
>>But then again you probably never heard of Kerensky. Whereas
I and
>>several hundred others heard him speak at the University of
Chicago and
>>the Communists were dumbfounded and were silent during the
question
>>period. It seems he was an element of history they couldn't
fabricate
>>quickly enough. Similar to you. Your similarity to the
Communists is
>>quite striking. If we merely substitute the word Capitalist
for your
>>use of the word Jew we see the way you have absorbed their
political
>>and thought processes in toto. You have revealed yourself
thoroughly,
>>Comrade.
>>FK
> As can be seen by studying FACTUAL history,
> rather than self-serving spin,
> most of the central players in the overthrow of the
> Russian government, and the massacre of the Russian royal
family,
> were Bolsheviks, and most of the Bolsheviks were Jews,
> and most of the power positions in Russia
> were held by these same people for many years.
> It was only when the native Russians,
> loyal to Russia, and life, liberty and happiness
> for ALL people begin to regain control of the
> Russian government that the people who raped Russia
> began to leave.
> And note in the case currently in the news,
> that of the battle between Putin and Khodorkovsky,
> that the Russians are still trying to keep the
> remnants of the old order from recapturing control of their
nation,
> using greedy, immoral, criminal tactics.
> As Putin said yesterday
Everyone should be equal before the law,
> irrespective of how many billions of dollars a person has on
his personal
or
> corporate account. -
> Otherwise, we will never teach and force anyone to pay taxes
-
> and defeat organized crime and corruption.
> And as Satarov said regarding the same matter,
The [oligarchs] are paying the price now for placing the wrong
bet [by
> backing Putin] in 2000,
> The [oligarchs] are just Bolsheviks by another name,
> and using another greedy, immoral tactic
> to steal the life blood of the Russian people.
> As can be seen,
> because of their small numbers,
> it is necessary for the people who
> co-opt governments so they can steal billions from the
masses,
> to put a front man like FDR, Bush, Blair, Stalin, Putin, etc,
> in charge, and even though they think they can control
> their front man, sometimes one arises,
> that turns on them,
> usually to protect their own ass from these evil criminals.
> I dare say that the will really hit the fan in Russia,
> and quickly spread around the world,
> and that the peoples of the world will begin
> to come down on the evil criminals who
> buy and intimidate their leaders,
> and instigate war for power and riches.
> Can you imagine what will happen in America
> when tens of thousands of disillusioned,
> servicemen come home and think about
> what has happened to them, their families,
> and their country???
> --
> Tom Potter http://tompotter.us
Well you have finally evidenced your lying self. Your claims
about the
Russian revolution of 1917 and the Russian government
subsequent to then
are totally specious. You are clearly in the best traditions of
Goebels: tell a lie often enogh and maybe many will begin to
believe it.
In your case the lying is consistent. The claim that most of
the
Bolsheviks were Jewish is totally specious. The claim that
Jews were
abundant in the government of the USSR is equally specious.
You are one
impossible liar.
FK
===
Subject: Re: Fundamental Reason for High Achievements of Jews

>

Hey, pot, (as in illegal substance or pot [quite applicable
for
>this troll], your are becoming quite old. The Protocols of
the
Elders
>>Whatever ... take this irrelevant rubbish out of our
newsgroup. Thanks
>
It usually helps (though it probably won't in this case) if you
> identify the group you are reading. As far as I can tell,
there aren't
> any alt.psychic regulars contributing to this thread.
> Actually, Bob, I'm with you on this. It has absolutely
nothing to do
> with chemistry at all. It should be expunged. I promise from
now on not
> to respond to that ass.
> FK
Actually fkasner,
I see no reason for you to call Lloyd Parker an ass,
just because he responds to posts about
the worlds' most serious problem,
the instigation of conflict and war for power and wealth.
As can be seen, the same people who instiga
the class wars of the 1900's for power and wealth
are instigating the religious wars of the 2000's,
as the loot from their class wars is almost gone,
and as can be seen by their escalating efforts,
they are desperate to get the religious wars in high gear.
Who instigates conflict and war for power and wealth?
Just open your eyes, and your mind,
and when you see two large groups of folks
fighting, note who is in the eye of the hurricane,
and working both side of the fence.
And most importantly,
note who ends up profiting from the misery of others.
Follow the money!
--
Tom Potter http://tompotter.us
===
Subject: Re: Fundamental Reason for High Achievements of Jews
--------------------------------------------------------------
---------
> It was so much easier to blame it on Them. It was bleakly
depressing to
> think that They were Us. If it was Them, then nothing was
anyones fault.
> If it was Us, what did that make Me ? After all, Im one of
Us. I must be.
> Ive certainly never thought of myself as one of Them.
No-one ever thinks
> of themselves as one of Them. Were always one of Us.
Its Them that do
> the bad things. <=> Terry Pratchett. Jingo.
Bruce Sinclair makes a good point with his sig!
As can be seen from studying history,
the people who instigate conflict and war for power and wealth,
used the Them vs. Us tactic to instigate their class wars of
the
1900's,
and they are using the same tactic to instigate the religious
wars of the
2000's.
The Them in their class wars were the folks who had more money
than
Us,
and the Them is their class wars are the Muslims
The people that instigate war for power and wealth,
don't care what defines Them or Us,
Rich/poor, Muslim/non-Muslim, North/South,
Protestant/Catholic, etc.
They have always exploi differences between folks
in order to get them fighting, and
as long as differences exist between people,
they will use these differences to instigate conflict and war.
Look around with open eyes and an open mind,
and when you see conflict between two large groups of folks,
note who is in the eye of the hurricane,
and working both sides of the conflict.
The world is too small,
and the stakes are too high
to allow people to instigate conflict and war
for power and wealth.
--
Tom Potter http://tompotter.us
===
Subject: Re: Fundamental Reason for High Achievements of Jews
>>
--------------------------------------------------------------
---------
>> It was so much easier to blame it on Them. It was bleakly
depressing to
>> think that They were Us. If it was Them, then nothing was
anyones fault.
>> If it was Us, what did that make Me ? After all, Im one
of Us. I must be.
>> Ive certainly never thought of myself as one of Them.
No-one ever thinks
>> of themselves as one of Them. Were always one of Us.
Its Them that do
>> the bad things. <=> Terry Pratchett. Jingo.
Bruce Sinclair makes a good point with his sig!
Not so ... Terry Pratchet makes it :)
I still say ... make points elsewhere :) :)
Thanks
--------------------------------------------------------------
---------
It was so much easier to blame it on Them. It was bleakly
depressing to
think that They were Us. If it was Them, then nothing was
anyones fault.
If it was Us, what did that make Me ? After all, Im one of
Us. I must be.
Ive certainly never thought of myself as one of Them. No-one
ever thinks
of themselves as one of Them. Were always one of Us.
Its Them that do
the bad things. <=> Terry Pratchett. Jingo.
===
Subject: re: Brian Greene's NOVA Elegant Universe
ref. http://qedcorp.com/APS/StarGate1.mov
Jack,
I also watched the NOVA program on string theory last night.
The answer
to Steven Weinberg's question about the cosmological constant
(as the
energy of empty space) may have been implicit in a key point
the program.
Weinberg says: If you try to calculate the energy in empty
space,
taking into account only fluctuations in fields of wavelengths
where we
understand the physics, you get an incredibly large energy,
much too
large to possibly fit what we know about the expansion of the
universe.
For me the most telling moment in Brian Greene's discussion of
string
theory was when he illustra the quieting down of the
fluctuations of
space caused by the fluctuations of the strings.
Yes, I noticed that. That is what I also get with my much
simpler model
Cosmological Constant = (Planck Area)^-1[1 - (Planck
Volume)(Vacuum
Coherence)^2]
Also Einstein's c-number emergent MACRO-QUANTUM coherent
Andrei Sakharov
metric elasticity tensor:
guv = Minkowski tensor + (Planck Area)(Goldstone Phase)(,u,v)
( ) is symmetrizer of partial derivatives ,u & ,v
Basic torsion field is
Suv = Planck Area)(Goldstone Phase)[,u,v]
[ ] is anti-symmetrizer (same notation as Penrose & Rindler)
What is the math idea of what Brian was alluding to? Oh you
mean finite
length of string?
But that is still too big a number when
Cosmological Constant ~ 1/(String Length)^2
The idea that the zero-point energy of spacetime (based on the
Heisenberg uncertainty principle) becomes enormous enough to
break
spacetime into a quantum foam is based on a quantum field
theory view of
entities).
Yes, but with my Vacuum Coherence Field = 0 implicitly assumed.
[Here's a question for the pessimistic Sheldon Glashow: If it
takes an
accelerator the size of the Milky Way galaxy to see something
as small
as a Planck scale (10^-33 cm) string, how big an accelerator
do we need
theory)? After all a point is infinitely smaller than 10^-33
cm. We
see their effects at a scale of around 10^-18 cm -- much
larger than the
point-like scale which is infinitesimal.]
Not really a fair question.
The idea is that the electron is a micro-geon (e.g. Ya
Burinski).
Imagine a ring singularity at radius
e^2/mc^2 ~ 1 fermi
perhaps with thickness 1 Newtonian Planck area.
Actually the core of this ring is exotic vacuum dark matter
with
positive pressure holding the charge and compensating the
rotations. The
plasma cloud of virtual electron-positron pairs and virtual
photons
extends out to h/mc ~ 10^-11 cm. This is the low energy
picture. Hit the
electron with huge momentum transfers p and I think the
enormous
space-warp at the micro-scale will make the electron appear
more and
transfers in high-energy scattering. Similarly for quarks
inside the
hadrons. Basically the radius stays at e^2/mc^ and the
circumference
of the ring shrinks relative to fixed radius. For simplest toy
Schwartzschild
effective size of electron (neglecting charge and spin just
show the
idea) = (e^2/2pimc^2)(1 - 2G*mp/hc^2)^1/2
where G* ~ 10^40G(Newton)
Then use blackhole-string duality to get to your string
picture?
approximation to the tiny (10^-33 cm sized) strings. This
entails the
further idea that when one looks at regions of space
approaching 10^-33
cm (the Planck length), rather than space fluctuating so
violently that
it breaks up in a quantum foam, the fluctuations are tame --in
fact
theory--contra to Sheldon Glashow very negative comments
concerning the
verifiability of string theory).
I need to see how the math of that actually works. I have some
string
theory books like Polchinski. Is it in there?
Moreover--and this is a very big idea--when one looks at
regions of
space approaching the Planck scale, the spacetime goes
hyperdimensional.
So it is in effect the tiny unseen dimensions that tame
quantum foam (as
calcula in quantum field theory).
What is the actual string theory formula for Einstein's
cosmological
constant? I have one above in my simple picture which also
explains both
dark energy and dark matter as w = -1 exotic vacuum phases on
the large
scale.
This is actually ancient history in string theory, so ancient
that it
is usually forgotten (if ever noticed). Thirty years ago Lars
Brink and
H.B. Nielsen did a zero-point energy calculation on the
strings (both
the bosonic 26 dimensional variety and the 10 dimensional
superstring
variety). These dimensions were already known by 1973, but by
way of
very abstract calculations. Brink and Nielsen wan a more
physically
intuitive picture of the hidden dimensions. They assumed only
that the
zero-point energy fluctuations would be absorbed by the
harmonic
fluctuations of the strings.
Sounds like the harmonics play a role similar to my
MACRO-QUANTUM
Vacuum Coherence Field.
Remember however that I derive Einstein's classical GR
equations from
modulating the Goldstone Phase of the More is different Vacuum
Coherence Field which is a giant vacuum local wave not tiny
little
harmonics.
I also derive dark energy and dark matter from modulating the
Higgs
amplitude part of the Vacuum Coherence Field.
In polar representation
Vacuum Coherence Field = (Higgs Amplitude Field)e^i(Goldstone
Phase Field)
The topological defects, i.e. Goldstone phase singularities,
are well
known to be 1D strings in this case of O(2) internal symmetry
in 3D space/
Einstein's guv is from local gauge invariance compensating
fields on the
4-parameter translation subgroup of the 15 parameter Conformal
Group.
Dark energy/matter exotic vacuum field is from local gauge
invariance
compensating field on the 4-parameter mirror translation
subgroup of the
15 parameter Conformal Group. (like Tony Smith's conformal
gravitons)
Shipov's torsion field is from local gauge invariance
compensating field
on 6 parameter Lorentz O(1,3) subgroup of the Conformal Group.
Also the dilation field from the 1 parameter sub-group of the
conformal
group hence R(t) = e^Ht in the inflation phase.
*Note according to Rindler and Penrose all sorts of stuff
breaks down
when torsion =/= 0
e.g. no longer d^2 = 0 for exterior differential Cartan forms.
No longer general Stoke's theorem
Integral over domain of d(form) = Integral over boundary of
the form
breaks down as does
Boundary of a boundary vanishes for the topology of fields as
Wheeler
tells IT FROM BIT.
Note also that 4-World Vectors (first rank tensors of
Conformal Group)
are 2qubit strings of the Bell Basis of quantum cryptography.
That's all
in Penrose and Rindler though they did not mention qubits back
in the
80's when the book was written - hence the IT FROM BIT as
Penrose spinor
(quBIT) substratum of IT space-time!
What they left to be calcula from this premise was the
dimensionality of spacetime in order to remain consistent with
this
premise. Lo and behold -- the 26-d bosonic dimensions and the
10-d
superstring dimensions (which include fermions) drop out of
their string
zero-point energy calculation.
Is the idea that they impose a vanishing zero point energy
density from
all strings and from that constraint get the extra-dimensions
of space?
The extra dimensions of space, if large, may amp up G(Newton)
to G* by
40 powers of ten at the fermi scale in the 3 large ordinary
dimensions?
Ref: L. Brink and H.B. Nielsen, A Simple Physical
Interpretation of
the Critical Dimension of Space-Time in Dual Models, *Physics
Letters*,
45B:4, 332-336 (6 Aug 1973). This paper is also included as
paper #9 in
the anthology edi by John H. Schwarz, *Superstrings: the first
15
years of superstring theory* Vol. I (World Scientific, 1985).
BTW: There may in fact already be observational evidence for
the string
theory taming of the violent (quantum field theory) quantum
foam. Of
course, since the observations were made by astronomers (who
probably
don't study the fine points of string theory ;-), this
evidence is not
tou as relevant to string theory.
Again I need to see a string theory formula that does same
thing as my
formulas
Cosmological Constant = (Planck Area)^-1[1 - (Planck
Volume)(Vacuum
Coherence)^2]
Also Einstein's c-number emergent MACRO-QUANTUM coherent
Sakharov
metric elasticity tensor
guv = Minkowski tensor + (Planck Area)(Goldstone Phase)(,u,v)
( ) is symmetrizer of partial derivatives ,u & ,v
Basic torsion field is
Suv = Planck Area)(Goldstone Phase)[,u,v]
[ ] is anti-symmetrizer (same notation as Penrose & Rindler)
The title of this news item from the University of Alabama in
Huntsville is ironically misleading. The title is Does sharp
image of
distant galaxy shred the fabric of space and time? Actually,
Richard
Lieu and Lloyd Hillman (at Huntsville) expec the Hubble
telescope
pictures of the Airy disks (or rings) genera by distant
galaxies to
show a quantization of time -- which idea is based on the
shredding of
spacetime at the Planck scale). The sharpness of the Airy
disks was
interpre to mean that time is not quantized at the Planck
scale.
This should count as indirect evidence for the string theory
taming
of the fluctuations of spacetime--via the hidden dimensions of
string
theory.
Or my formula, which is simpler.
Thus this could be the first indirect evidence for the hidden
dimensions.
The sharp airy disks from distant galaxies was also confirmed
by the
independent observations (of different galaxies) by Roberto
Ragozzoni of
the Max Planck Institute (also using the Hubble telescope).
And this
confirmation led to the short news item in the September issue
of
*Astronomy*.
calculations mentioned above. I don't know if they will
publish my
letter. Anyway I sent you a copy of my email letter to
*Astronomy* on
August 11. And you sent it on to several other people -- so
the idea is
out there. Don't know if its significance understood.
Nuff said ;-)
Saul-Paul
----------
Subject: Re: Nova's presentation of Brian Green's The Elegant
Universe
Happened to catch part of this string theory show last night
and I'm
not sure that simplistic shows for laymen, like me, are that
helpful
if they don't go as far as the meaningful explanations that
your web
site presents. Anyway, all the big stars were on the show and I
thought this quote from Steven Weinberg was interesting, from
Nova's
website, especially in light of your theory:
NOVA: What do you see as string theory's greatest failure?
Weinberg: A disappointing aspect of string theory is that it
has so
far failed to shed any light at all on what is probably the
biggest
outstanding problem in the physics of what we can actually see
in
nature -- the failure to understand the energy of empty space,
the
so-called cosmological constant. If you try to calculate the
energy in
empty space, taking into account only fluctuations in fields of
wavelengths where we understand the physics, you get an
incredibly
large energy, much too large to possibly fit what we know
about the
expansion of the universe. There must be some complica
cancellations that make the energy in empty space very small.
Yes, they do not have the idea of vacuum coherence damping
down the
random zero point fluctuations -- oil on the waters.
/zpf = Lp*^-2[1 - Lp*^3Higgs^2]
The nice thing about the additional hologram idea is that
Lp* = Lp^2/3(c/H)^1/3
gives Lp* ~ 1 Gev scale i.e. 10^-13 cm
the bad thing is that Lp* would seem to be time dependent,
which would
I think be falsified because it would say that the mass scale
of
lepto-quarks is weakening.
This would, it appears, contradict the observations of old
stars at
least. In any case this is a problem I do not yet understand.
My theory
does not depend on the hologram idea fortunately but it is a
nice
addition to it, if I can resolve this issue.
===
Subject: two number-theoretical limits (& bonus sum)
Let d(k) = the number of positive divisors of k.
then, does:
limit{m -> oo}
(1/m) (sum{k=1 to m} d(k)) - ln(m) =
2*c -1,
where c = Euler's constant (.5772...)?
And rela, does:
limit{m -> oo}
(1/m) sum{k=2 to m} (m(mod k))/k =
1 -c,
where 0 <= m(mod k) <= k-1 ?
I am doubting that these, at least the first, is true, but only
because the limit does not look familiar. (And such a limit
*must* be
well-known.)
I derived (derived) these limits from, in part,
sum{k=1 to m} sum{j|k, j <= sqrt(k)} a(j) =
sum{1 <= k <= sqrt(m)} a(k) (floor(m/k) +1 -k)
So, hopefully this sum.identity is correct anyhow.
(I set a(k) = 1, of course, to help get limits. I then found a
limit
involving the number of positive divisors of k, where each
divisor is
<= sqrt(k).)
(limit (1/m)(sum{k=1 to m} d'(k)) - ln(m)/2 = c -1/2,
where d'(k) = ceiling(d(k)/2) = # divisors <= sqrt(k).)
Leroy Quet
===
Subject: Re: two number-theoretical limits (& bonus sum)
> Let d(k) = the number of positive divisors of k.
> then, does:
> limit{m -> oo}
> (1/m) (sum{k=1 to m} d(k)) - ln(m) =
> 2*c -1,
> where c = Euler's constant (.5772...)?
> And rela, does:
> limit{m -> oo}
> (1/m) sum{k=2 to m} (m(mod k))/k =
> 1 -c,
> where 0 <= m(mod k) <= k-1 ?
> I am doubting that these, at least the first, is true, but
only
> because the limit does not look familiar. (And such a limit
*must* be
> well-known.)
> I derived (derived) these limits from, in part,
> sum{k=1 to m} sum{j|k, j <= sqrt(k)} a(j) =
> sum{1 <= k <= sqrt(m)} a(k) (floor(m/k) +1 -k)
> So, hopefully this sum.identity is correct anyhow.
> (I set a(k) = 1, of course, to help get limits. I then found
a limit
> involving the number of positive divisors of k, where each
divisor is
> <= sqrt(k).)
> (limit (1/m)(sum{k=1 to m} d'(k)) - ln(m)/2 = c -1/2,
> where d'(k) = ceiling(d(k)/2) = # divisors <= sqrt(k).)
>
The first part is correct. It is proved in good old Hardy and
Wright,
theorem 320 in the form
d(1) + d(2) + ... + d(n) = n log n + 2 gamma - 1)n +
O(sqrt(n)).
Don't know about part 2.
Martin Cohen
===
Subject: Re: Integral
> You can, you just don't know a name for it.
>> It's called a Sine Integral :
>> Si(z)= int_0^z sin(x)/x dx
>Give it a name and it's solved, yup. It doesn't really
address the
>OP's question though, does it. He obviously wants to know how
to work
>out the integral. He has probably found that the methods he
has
>learned don't seem to work and wants to know if there is a
clever
>method he hasn't figured out yet. And the answer for him is
no, there
>isn't any way to work it.
>Of course, giving the antiderivative a name and using that
for the
>answer has its uses, somewhat like in the medical profession
when some
>mysterious disease shows up. They may not know what causes
it, how it
>spreads, how to treat it, or anything else about it, but they
can give
>it a name. Our tests indicate that you have di-flukus. Feel
better
>now?
> But once you have the name, you can look it up and see what
is known
> about it. And for functions such as Si, there is in fact a
lot of
> information that can be found in standard reference works
> or in computer algebra systems such as Maple or Mathematica.
It's also a standard (or at least common) exercise when
introducing
asymptotic expansions.
Jon Miller
===
Subject: Re: Integral
You can't.
You can, you just don't know a name for it.
> I certainly can't, and I don't think you can either.
> (This is semantics)
Does that mean you can't take a square root either? If you're
willing to
agree to that, then I'm willing to agree that you can't
compute lots of
things, but can only approximate them.
There are lots of ways to deal with the Si(x). As with roots
of integers.
Jon Miller
===
Subject: Re: Integral
| I certainly can't, and I don't think you can either.
| (This is semantics)
| I hate it when people say you can't integrate something,
just because
the
| result is beyond the reach of elementary functions.
One reason to be careful of saying you can't integrate that is
that
there exist nonintegrable functions, in the sense that there
doesn't
exist an integral, whether elementary or not, and one doesn't
want to
confuse the two. To me that's a good enough reason not ever to
say you
can't integrate a function unless either it's nonintegrable or
they
know I mean not in some specific form.
Another reason is that you often get a chance to remind the
questioner
of the relative nature of the concept of being able to do an
integral,
and the arbitrariness of such standards as being able to do it
in
elementary terms. Sometimes people really do want to get the
integral
in terms of Bessel functions and so on.
===
Subject: Geometry of 2nd derivative of a Vector function
charset=iso-8859-1
If we have components of a vector each parameterized by two
variables so
that this parameterization defines a surface, what is the
geometric
interpretation of the 2nd derivative of that vector function
with respect
to
each parameter? I know that the second derivative of a space
curve with
respect to its one parameter is a vector normal to the curve.
Is the 2nd
derivative of a surface normal to the surface? Instead of the
2nd
derivative
with respect to one parameter, I'm wondering about the 2nd
derivative with
respect to once each of two parameters, d^2x/dsdt. (NOT
d^2x/dt^2)
Thanks.
===
Subject: Re: Geometry of 2nd derivative of a Vector function
> If we have components of a vector each parameterized by two
variables so
> that this parameterization defines a surface, what is the
geometric
> interpretation of the 2nd derivative of that vector function
with respect
to
> each parameter? I know that the second derivative of a space
curve with
> respect to its one parameter is a vector normal to the curve.
It is? The thing I would write down with that description
is d^f/ds^2, which is a scalar.
> Is the 2nd
> derivative of a surface normal to the surface? Instead of
the 2nd
derivative
> with respect to one parameter, I'm wondering about the 2nd
derivative
with
> respect to once each of two parameters, d^2x/dsdt. (NOT
d^2x/dt^2)
A function f:R^2 -> R, i.e. a scalar function f(x,y) of
x and y, describes a surface.
The second derivative of this function is called the
Hessian. It is a 2 x 2 matrix whose (i,j)-th component
is d^f/(dx_i dx_j). That is, the diagonal elements are
d^2f/dx^2 and d^f/dy^2, and the off diagonal elements are
both d^f/(dx dy).
The eigenvalues and eigenvectors of this matrix have some
physical meaning in terms of the convexity of the
function. For instance, if the eigenvalues are all
positive (the Hessian is positive definite) then the
function is concave upward in all directions.
You're asking about a vector valued function, which means
you have one Hessian for each component. And you're also
asking about only the off-diagonal elements of the
Hessians. I'm not sure what interpretation to put
on that. You can take the two scalars d^2f_x/(dx dy)
and d^2f_y/(dx dy) and interpret them as a two-vector.
But that vector doesn't have any fixed relationship
with the surface(s) f_x, f_y.
- Randy
===
Subject: Re: Geometry of 2nd derivative of a Vector function
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>It is? The thing I would write down with that description is
>d^f/ds^2, which is a scalar.
What is f? Presumably you mean a mapping of f: R -> R, but he
specified a 3-vector function.
>A function f:R^2 -> R, i.e. a scalar function f(x,y) of x and
y,
>describes a surface.
But it's not the only way to describe one, and it doesn't work
for all
surfaces.
>You can take the two scalars d^2f_x/(dx dy)
>and d^2f_y/(dx dy) and interpret them as a two-vector.
>But that vector doesn't have any fixed relationship
>with the surface(s) f_x, f_y.
ITYM the surface f_x, f_y, f_z.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolici bulk E-mail will be 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 reply to spamtrap@library.lspace.org
===
Subject: Re: Geometry of 2nd derivative of a Vector function
> If we have components of a vector each parameterized by two
variables
so
> that this parameterization defines a surface, what is the
geometric
> interpretation of the 2nd derivative of that vector function
with
respect to
> each parameter? I know that the second derivative of a space
curve with
> respect to its one parameter is a vector normal to the curve.
> It is? The thing I would write down with that description
> is d^f/ds^2, which is a scalar.
But a space curve would look like s -> (x(s), y(s), z(s)), s
real. Its
second derivative is (x''(s), y''(s), z''(s)). I do not know
why the OP
thinks this is always normal to the curve; obviously it need
not be.
===
Subject: Reversing Game's Direction (was:Sequence Game
(Multiplying
Divisors))
I have pas my entire original post below.
Following this (sorry, long) original post is my current topic.
> Here is a game based on dividing/multiplying (and may have
some, if
> only a little, teaching value).
> Possible name for game: Nonconformity
> 2 players.
> (Rules explained from the viewpoint of one player, you.)
> Each player gets either (in one variation) their own
> randomly-genera starting positive integer (m, for you, and
n, for
> your opponent), or (in the other variation) both players get
the same
> randomly-genera positive integer m.
> m and n are >= 2.
> Each player has a sequence of positive integers ({a(j)} for
you,
> {b(j)} for your opponent).
> a(1) = m.
> b(1) = n (or m).
> At each move, each player adds one new positive integer to
their
> sequence. This integer is picked secretly (ie. the k_th
element of
> each player's sequence is not influenced by what the other
player
> picks for his/her k_th element).
> But after the integers are picked, these integers are
revealed. (ie.
> At each move, after the integers are picked, each sequence
{as much of
> each sequence which has been comple} is then known to both
> players.)
> So, the restrictions on the picked integers:
> 1) Each integer is >= 2, and
> 2) a(k+1) is a positive multiple of any divisor >= 2 of a(k).
> (ie. a(k+1) is a positive multiple of any prime dividing
a(k).)
> (And likewise, b(k+1) is a positive multiple of any divisor
>= 2 of
> b(k).)
> 3) Each a(k) is not among {a(1), a(2), a(3),...,a(k-1)}.
> (And likewise, b(k) is not among the previous terms of the
> b-sequence.)
> But the reason for the possible name (Nonconformity) is
apparent
> (somewhat) in this game's scoring:
> You, for example, get a point as follows:
> One point for each a(k) where:
> min(b(j)) < a(k) < max(b(j)), for 1 <= j <= k-1.
> and a(k) is not among {b(1), b(2), b(3),...,b(k-1)}.
> (Likewise, your opponent gets a point for the
counter-situation {where
> all a's and b's are exchanged}.)
> So, the scoring is based upon each a(k) and b(k),
> and upon how these terms relate to the sequences up to
a(k-1) and
> b(k-1).
> Players play a predetermined number of moves.
> Sample game (with only 8 moves/player):
> m=6, n=4:
> player 1: 6, 4, 8, 10, 12, 14, 7, 21
> player 2: 4, 2, 6, 3, 9, 12, 14, 7
> In this (lame) example, player 1 gets one point for his/her
7, and
> player 2 gets a point for his/her 9.
> (A tie.)
> (Of course, a game with more moves would be more interesting
and less
> likely to tie.)
 I have played some a sample 'game' or two where I was my own
opponent
> (see above example). It seems as if there is a little
stategy, at
> least, to playing this.
> What would be some good stategies for playing this game?
> I have purposefully designed it so that the game gives no
inherent
> advantage to either player (other than the
advantages/disadvantages
> which occur by chance if each player has a different starting
> integer).

I have been wondering today about games played backwards in
time, but
maintaining the same winner and the same difference between the
player's scores, if not the same scores themselves, after being
played backwards as opposed to forwards.
For example, using the above game:
I have noticed that this game can be played (almost...)
backwards in
time.
First, as a(k+1) is a multiple of a prime dividing a(k),
a(k-1) is a
multiple of a prime dividing a(k) also.
One aspect of scoring in the reverse game would be that
players would
previously by his/her opponent.
(previously refers to the backwards-people's perspective.)
(the above rule helps keeps the same DIFFERENCE between
players'
scores.)
A problem occurs when finding the converse of the min/max-rule.
For here, I GUESS, we would want each player to try to keep
his/her
min/max to AVOID his/her opponent's integers.
So, players might simply play arbitrarly large integers,
perhaps.
Now, when I refer to the reversal of a game, I, in this case,
assume
that the score-differences would remain the same if the
sequences are
genera in isolation, without any knowledge of the other
sequence.
But even without knowlege of the other sequence, players
playing a
strict reverse-analog of my original game might still play
arbitrarily
large integers.
So, perhaps the min/max-rule could be altered in the
reverse-game so
as to, as in the original game, encourage the integers to
remain in a
reasonable range.
I have not been able to think about this much tonight, and I
am in a
hurry to get off-line. So I have been relying on notes.
Hopefully this
is clear. It is meant to be subject to one's own interpration,
in a
way, anyhow...because it is meant mostly to only be a
starting-point
for a discussion..
Leroy Quet
===
Subject: Re: Transcendental numbers
> I know that rational and irrational numbers are everywhere
dense in
> the real number continuum, but what about transcendental
numbers?
> Do
> intervals exist in the real number continuum that contain no
> transcendental numbers? Has this problem even been solved and
is
> there
> a proof? (Maybe this belongs in another math group?)
> There are more transcententals in any real interval than
rational
> or even algebraic numbers. Transcendentals are , if anything,
more
> dense, since there are uncontably many in any real interval
whereas
> there are only countably many algebraics (including the
rationals).
> By what mechanism can you count the algebraics?
How many polynomials (with rational coefficients) of degree
one are
> there? They are of the form a_0 + a_1*x. There are countably
many
> possibilities for the constant term a_0, and countably many
> possibilities for a_1, the coefficient of x. Countable times
countable
> is countable, so there are countably many degree 1 polys.
Each has
1
> root, so there are countably many numbers that are roots of
degree
1
> polys.
How many of degree two? Same logic applied to a_0 + a_1*x +
a_2*x^2.
> Countable times countable times countable is countable, so
there
are
> countably many polys. Each poly has two roots, so that's
countably
many
> possible roots.
Dot dot dot . . . there are countably many reals that are
roots of
polys
> of degree 1, of degree 2, . . ., of degree n, . . .
Adding them all up, we get a countable sum of countable
numbers,
which
> is countable.
> One tiny thing you left that makes a world of difference: a
polynomial
must
> be of finite degree. Otherwise you would be claiming the R is
countable.
Well, there are infinitely many orders of finite-ordered
polynomials.
 Makes no sense.
> Then there are infinitely many coefficients.
Then, what you are looking at there is N x N x N x ....
 NxNxN . . . is not the polynomials, it's the ring of formal
power
> series. The polynomials are N+N+N+... where '+' means direct
sum. In
> an infinite direct sum of rings, all but finitely many terms
of any
> element are zero. That's what makes polynomials polynomials.
Otherwise
> they'd be power series.
> There are those and then some.
It's been discussed here that that's uncountable. That means
here
> that there may be a bijection between N x N x N x ... x N x
... and R.
 Well yes there is, but that has nothing to do with algebraic
numbers or
> polynomials.
> Do you say the algebraics are countable?
> Yes.
> There exists a one-to-one
> function from them to the Cartesian product N and itself
infinitely
> many times.
> No, that's not true. The direct sum of countably many copies
of N is
> countable. The direct product is not.
> The Cartesian product of N and itself infinitely many
> times is uncountable, just ask Ullrich or Virgil.
 I agree. But the Cartesian product of countably many copies
of N gives
> you the formal power series, not polynomials.
> What the hell, you might say.
 Yeah, you got me there.
> Consider the polynomials, in particular the polynomial with
integer
> coefficients.
a_0
That's a constant, or a_0 x^0. There is one for each integer,
> Z(-oo,oo).
Then let's consider another coefficient. Let's only worry
about the
> positive integers, Z+.
a_1 x^1 + a_0 x^0.
Then we have the Cartesian product Z+ x Z+ for each possible
value of
> a_1 and a_0, representing a subset of the polynomials of
order 1 with
> integer coefficients.
1 x^1 + 1 x^0
> 1 x^1 + 2 x^0
> 1 x^1 + 3 x^0
> 1 x^1 + ...
> 2 x^1 + 1 x^0
> 2 x^1 + ...
> ...
Keep in mind that any polynomial a_1 x_1 + a_0 x^0 has roots
that are
> roots of (a_1 x_1 + a_0 x^0) *r for real r, eg for
coefficients
> (a_1, a_2)=(1, 3) that polynomial has the same roots as for
(2, 6),
> (3, 9), (4, 12), etc.
Anyways, to represent a polynomial with two positive integer
> coefficients we have the Cartesian product of Z+ x Z+.
Then, go on to the third coefficient, etcetera. You might
immediately
> notice that that set of polynomials can be represen by Z+ x
Z+ x Z+
> ..., one coordinate for each coefficient.
The polynomial has finite rank, or order, degree, and its rank
can
> range from zero to infinity. That's very similar to an
integer, an
> integer is finite, but there are infinitely many of them. The
> polynomials rank is an integer.
So what we get then to represent only all polynomials with
positive
> integer coefficients is the Cartesian product Z+ x Z+ x Z+ x
...,
> which has been claimed to be uncountable.
What's up with that?
> You are confused about the distinction between a direct
product and a
> direct sum. In a direct sum, all but finitely terms are
zero. So for
> example in the countable direct sum of N,
(1,1,1,1,0,0,0,0,0,0,0...) is
> an element, but (1,1,1,1,1...) is not.
Thanks for your reply although I disagree with you.
Look at the last example, about (1, 1, 1, 1, ...) not
representing a
polynomial, I agree for the same reason 1111... is not an
integer,
because polynomials have finite degree or order.
The coefficient coordinates: (1), (1, 1), (1, 1, 1), ... each
represent polynomials.
The formal power series here seems to be something like:
oo
--
a_i x^i
--
i=1
where a polynomial is of the form
n
--
a_i x^i
--
i=1
where n is the finite order of the polynomial, or even
oo
--
a_i x^i
--
i=1
where finitely many of the a_i's are non-zero.
There is no distinction between the polynomial and the power
series
except that in the polynomial only finitely many a_i's are
non-zero.
For a formal power series to converge a_i must be less than
1/x_i
except for finitely many values, there are more conditions
than that.
A polynomial is finite where each a_i is finite.
So we get to that a polynomial has a finite number of terms,
the
number of terms is a natural integer. An integer is a finite
integer
but there are infinitely many of them.
In talking about a polynomial then of the infinite coordinates
of N x
N x N x ..., eg (a_0, a_1, a_2, ...) only finitely many
coordinates
are non-zero _for a given polynomial_. Yet, there exist
polynomials
for each a_i such that a_i is the only non-zero term. That is
to say,
for each polynomial with only finitely many non-zero terms
a_i, there
are infinitely many other polynomials such that among them all
infinitely many have non-zero a_i's where a polynomial has
zero a_i's.
In the case of considering the real roots of any polynomial
with
integer coefficients, the algebraic numbers rational and
irrational,
it is necessary to consider each possible infinite list of
coordinates.
One way to approach this is to just have a list of coordinates
for
each non-zero term so that the list is finite. For example a
listing
of the coordinates with exactly one non-zero term would be Z+,
for the
term, cross Z+, for what index it is. For two coordinates it
could be
represen as {a_i, a_j} E Z+ for the two terms, and (i, j) E Z+
x
Z+.
Each polynomial with integer coefficients is represen in Z x Z
x Z
x ..., as is each power series with integer coefficients. Any
element
of Z x Z x Z x ... that represents a polynomial has only
finitely many
non-zero coordinates, otherwise it represents a power series.
Fish fry: baby fish. Fish fry: event of seared fish flesh, most
often cooked in oil, or pan-fried, cooked against the pan. I
fry
catfish fillets at 375 in half an inch of oil breaded with
prepackaged
cornmeal mix, fry until golden, serve with lemon and diced dill
pickles in mayonnaise, minimalist fish fry. I maintain a
cichlid,
it's decorative.
So anyways we look at the product of infinitely many copies of
Z as
integer coefficients of power series. We're looking at
polynomials,
for any given polynomial its coordinate contains almost all
zeros,
with only finitely many non-zero elements. I understand that
the
polynomials are a proper subset of the power series.
Yet, there is still the inductive argument that for non-zero
coefficients a_i that a_0 x^0 is a polynomial, as is a_0 x^0 +
a_1
x^1, a_0 x^0 + a_1 x^1 + ... + a_i x^i, etcetera, ad
infinitum, there
always exists another finite i.
Please define direct sum and direct product. Types of product:
inner,
outer, dot, wedge, cross, vector, direct, Cartesian. The dot
and
wedge products are inner and outer products, the cross product
is the
vector product, the Cartesian product is a direct product. The
direct
sum sign looks like the encircled addition sign, the direct
product
sign the encircled multiplication sign.
Back to the inductive argument, Z is not enough to represent
any
polynomial with integer coefficients, neither is Z x Z, Z x Z
x Z,
Z^4, Z^5, etcetera. For any product of finitely many copies of
Z,
there are polynomials not represen, thus requiring Z^N to
represent
all possible polynomials, where that also can represent the
power
series.
So we have that a proper subset of Z^N, that here being the
Cartesian
product of infinitely many copies of Z, represents the
polynomials,
where each coordinate that represents a polynomial has only
finitely
many non-zero elements, those being the coefficients. There's
that,
and that for no finite value n is Z^n sufficient to encode the
integer
coefficients of any and all possible polynomials.
I try to understand direct sum. I look to MathWorld,
http://mathworld.wolfram.com/DirectSum.html, and it says the
direct
sum of two sets of integers A and B are {a+b : a E A and b E
B}. I
deduce from that that the direct sum of two copies of the
naturals is
the naturals, the direct sum of the naturals and Z+ is the
naturals,
the direct sum of two copies of Z+ is Z+{1}, and the direct
sum of
infinitely many copies of Z+ is the empty set. Wait a second,
it
looks like the result is again using the unclear MathWorld set
description, using curly braces there to signify a multiset or
sequence instead of a set. That definition does not seem
clear. I
turn to Wikipedia, http://en.wikipedia.org/wiki/Direct_sum, it
introduces the tensor product. I search for direct sum of sets
and
find http://www.mathreference.com/set,dirsum.html, The Direct
Sum of
Sets. It describes the direct sum much as you do. It says
specifically The direct sum is always nonempty. It doesn't say
very
much beyond that. The direct sum of two copies of the empty
set is
empty, I would think. Please explain direct sum. I can see how
a
subset of the direct product with only finitely many non-zero
elements
in each sequence is useful to describe the polynomials, I just
don't
yet see how you got there from {a+b : a E A and b E B}.
I was thinking that the polynomials' coefficient sequences
would map
to Z^N. They inject into Z^N, a set of infinitely many
sequences of
values of Z, via identity. I thought they might biject yet I
don't
see a method for that.
This leads me to think about roots of power series. If a_0 is
zero a
root is zero, meaning a value of x for which the power series
evaluates to zero, not the expanded power series exponentia to
1/n
for positive integer n>2. If the power series is x-1
multiplied with
itself infinitely many times then a root is one.
This thread is about transcendental numbers. The transcendental
number is not an algebraic number. The algebraic number is the
root
of a polynomial with integer coefficients.
Is not the root of a power series with integer coefficients,
that is
to say, a value for which a power series evaluates to zero,
algebraic?
Probably not always, no. That's saying that the roots of some
power series with integer coefficients that are not
polynomials are
transcendental. Besides that some polynomials have no real
roots.
We want to represent the integer coefficients of all
polynomials. For
no finite n is Z^n sufficient. Instead a proper subset of Z^N
which
is called the direct sum of infinitely many copies of Z with
each
coordinate sequence having only finitely many non-zero values
is used.
For no finite n is Z^n sufficient, the coefficients of any
polynomial
as a sequence is not necessarily an element of Z^n and the set
of all
sequences of coefficients of any polynomial does not inject
into Z^n.
Ross F.
===
Subject: Re: Transcendental numbers
<lengthy prior discussion snippedYou are confused about the
distinction between a direct product and a
> direct sum. In a direct sum, all but finitely terms are
zero. So for
> example in the countable direct sum of N,
(1,1,1,1,0,0,0,0,0,0,0...) is
> an element, but (1,1,1,1,1...) is not.
> Thanks for your reply although I disagree with you.
My statement is an immediate consequence of the definition I
made, so
it's not really subject to being disagreed with. I defined the
direct
sum of countably many copies of N to be a particular *SUBSET*
of the
countable-fold Cartesian product of N. The direct sum is the
subset of
infinite-tuples having only finitely many nonzero terms. In
other
words BY DEFINITION, the direct sum of countably many copies
of N
contains (1,1,1,0,0,0,0,...) but DOES NOT contain
(1,1,1,1,...).
That is not a statement that can be argued with, it's a
definition.
> Look at the last example, about (1, 1, 1, 1, ...) not
representing a
> polynomial, I agree for the same reason 1111... is not an
integer,
> because polynomials have finite degree or order.
Yes, that is correct.
> The coefficient coordinates: (1), (1, 1), (1, 1, 1), ... each
> represent polynomials.
Yes, true.
> The formal power series here seems to be something like:
> oo
> --
> a_i x^i
> --
> i=1
> where a polynomial is of the form
> n
> --
> a_i x^i
> --
> i=1
> where n is the finite order of the polynomial, or even
> oo
> --
> a_i x^i
> --
> i=1
> where finitely many of the a_i's are non-zero.
That's exactly right.
> There is no distinction between the polynomial and the power
series
> except that in the polynomial only finitely many a_i's are
non-zero.
That's like saying the only difference between night and day
is that
it's dark out at night. That is EXACTLY the difference. And the
distinction between polynomials and power series is exactly
that a
polynomial is restric to having finitely many nonzero
coefficients.
This single difference -- which you dismiss as being no
distinction . .
. except, is responsible for the key feature you were
concerned about
in your previous post. The set of polynomials having rational
coefficients is a countable set, and the set of power series
having
rational coefficients is NOT a countable set. It is precisely
the at
most finitely many nonzero coefficients condition that makes
the
countability proof go through for polynomials.
> For a formal power series to converge a_i must be less than
1/x_i
> except for finitely many values, there are more conditions
than that.
> A polynomial is finite where each a_i is finite.
Convergence is totally irrelevant here. We are discussing
merely formal
polynomials and power series, that is expressions of the form
such and
so. That's why we need to remember in the back of our minds
that to
make expressions of the form rigorous, we must actually deal
with
infinite-tuples. We are in the realm of algebra, not analysis.
We
regard polynomials and power series merely as n-vectors or
countable-vectors (to abuse the terminology a bit)
respectively.
> So we get to that a polynomial has a finite number of terms,
the
> number of terms is a natural integer. An integer is a finite
integer
> but there are infinitely many of them.
Yes that's true. There are polynomials of degree 1;
polynomials of
degree 2; polynomials of degree 3; and so forth. However, every
polynomial has some specific degree. Every polynomial has
associa
with it some number n such that if m > n, then the m-th
coefficient is
0. Power series do not have that restriction.
> In talking about a polynomial then of the infinite
coordinates of N x
> N x N x ..., eg (a_0, a_1, a_2, ...) only finitely many
coordinates
> are non-zero _for a given polynomial_. Yet, there exist
polynomials
> for each a_i such that a_i is the only non-zero term.
Yes that's true too. If I understand you, you are thinking of
1, x,
x^2, x^3, x^4, ... Or in vector notation, (1,0,0,0,....),
(0,1,0,0,0)... etc.
That is to say,
> for each polynomial with only finitely many non-zero terms
a_i, there
> are infinitely many other polynomials such that among them
all
> infinitely many have non-zero a_i's where a polynomial has
zero a_i's.
> In the case of considering the real roots of any polynomial
with
> integer coefficients, the algebraic numbers rational and
irrational,
> it is necessary to consider each possible infinite list of
> coordinates.
I think you lost me there. Can you try to state what you mean
more
clearly? You do agree that there are countably many polys (with
rational or integer coefficients) of degree 1; countably many
of degree
2; countably many of degree 3; and in general, countably many
of degree
n. Do you agree with that?
> One way to approach this is to just have a list of
coordinates for
> each non-zero term so that the list is finite. For example a
listing
> of the coordinates with exactly one non-zero term would be
Z+, for the
> term, cross Z+, for what index it is. For two coordinates it
could be
> represen as {a_i, a_j} E Z+ for the two terms, and (i, j) E
Z+ x
> Z+.
I don't know what you mean by a listing of the coordinates
with one
non-zero term.
> Each polynomial with integer coefficients is represen in Z x
Z x Z
> x ..., as is each power series with integer coefficients.
Yes this is very true. The polys are the ones with finitely
many
nonzero coefficients.
Any element
> of Z x Z x Z x ... that represents a polynomial has only
finitely many
> non-zero coordinates, otherwise it represents a power series.
That is exactly right.
> Fish fry: baby fish. Fish fry: event of seared fish flesh,
most
> often cooked in oil, or pan-fried, cooked against the pan. I
fry
> catfish fillets at 375 in half an inch of oil breaded with
prepackaged
> cornmeal mix, fry until golden, serve with lemon and diced
dill
> pickles in mayonnaise, minimalist fish fry. I maintain a
cichlid,
> it's decorative.
I have other fish to fry.
> So anyways we look at the product of infinitely many copies
of Z as
> integer coefficients of power series. We're looking at
polynomials,
> for any given polynomial its coordinate contains almost all
zeros,
> with only finitely many non-zero elements. I understand that
the
> polynomials are a proper subset of the power series.
You seem to understand exactly what I'm saying. I don't
understand your
objection or confusion.
Once you agree that the polys are a proper subset of the power
series,
you can see that the way we defined the polys lets you prove
they are
countable; whereas you cannot apply the same logic to the
power series.
If it helps any, another way to think about this is to
consider power
series with coefficients in {0,1}. That's a FINITE set of
coefficients.
You can actually enumerate specifically every single
polynomial. The
ONLY 1-degree polys are x and x+1. The only 2-degree polys are
x^2,
x^2+x, and x^2+1.
> Yet, there is still the inductive argument that for non-zero
> coefficients a_i that a_0 x^0 is a polynomial,
Well a_0x^0 is in fact just a_0, a constant, or a zero-degree
polynomial. None of these have roots so they're not important
when
counting algebraic numbers.
as is a_0 x^0 + a_1
> x^1, a_0 x^0 + a_1 x^1 + ... + a_i x^i, etcetera, ad
infinitum, there
> always exists another finite i.
Yes there does. Just like when I count 1,2,3,4,5,...n,...
there always
exists another n. But the natural numbers are countable
nevertheless.
> Please define direct sum and direct product.
I've done this several times.
Types of product: inner,
> outer, dot, wedge, cross, vector, direct, Cartesian. The dot
and
> wedge products are inner and outer products, the cross
product is the
> vector product, the Cartesian product is a direct product.
The direct
> sum sign looks like the encircled addition sign, the direct
product
> sign the encircled multiplication sign.
Well we have been a little loose about terminology.
Technically the
Cartesian product of rings (like Z) is defined as the
Cartesian product,
with ring operations inheri coordinate-wise from Z. With this
definition, you will see that the polynomials are a subring of
the power
series. The idea is to define the algebraic structure of the
direct
product (or sum) in terms of the ring containing the
coefficients.
> Back to the inductive argument, Z is not enough to represent
any
> polynomial with integer coefficients, neither is Z x Z, Z x
Z x Z,
> Z^4, Z^5, etcetera. For any product of finitely many copies
of Z,
> there are polynomials not represen, thus requiring Z^N to
represent
> all possible polynomials, where that also can represent the
power
> series.
Yes that's true. We start with Z^N. That is the Cartesian
product, and
we can make it into a ring by defining pointwise operations,
just like
adding vectors in analytic geometry. Then we make a DEFINITION
that we
will call a member of the Cartesian product a polynomial just
in case
it happens to have finitely many nonzero terms.
Perhaps your confusion is that you think I'm stating a
conclusion. I'm
not. I'm making a definition. A polynomial is such and so.
Given
that, here's a thingie. Does it fit the description? Then it's
a
polynomial. If not, it's not. So **BY DEFINITION*
(1,1,1,1,1...) is
not a polynomial.
> So we have that a proper subset of Z^N, that here being the
Cartesian
> product of infinitely many copies of Z, represents the
polynomials,
> where each coordinate that represents a polynomial has only
finitely
> many non-zero elements, those being the coefficients.
No, you didn't say that right. A polynomial has only finitely
many
nonzero coordinates. What you said was that each coordinate
that
represents a polynomial ... which makes no sense. A polynomial
is a
sequence of coordinates.
There's that,
> and that for no finite value n is Z^n sufficient to encode
the integer
> coefficients of any and all possible polynomials.
Well that's absolutely true, but so what?
> I try to understand direct sum. I look to MathWorld,
> http://mathworld.wolfram.com/DirectSum.html, and it says the
direct
> sum of two sets of integers A and B are {a+b : a E A and b E
B}.
That's a totally different meaning of the word. My usage is
standard.
Actually as far as I know, Wolfram is here using a nonstandard
meaning
of direct sum. I looked at the page you referenced. It is true
(as the
page says) that the direct sum is the coproduct in the
category of
modules. That's a bit of jargon that tells me that the author
of that
page knows the same definition of direct sum that I do. The
business
about adding a and b together like that is a total error on
the part of
the author of that page.
I
> deduce from that that the direct sum of two copies of the
naturals is
> the naturals, the direct sum of the naturals and Z+ is the
naturals,
> the direct sum of two copies of Z+ is Z+{1}, and the direct
sum of
> infinitely many copies of Z+ is the empty set.
No the Wolfram page is completely in error, unfortunately. It's
surprising.
Wait a second, it
> looks like the result is again using the unclear MathWorld
set
> description, using curly braces there to signify a multiset
or
> sequence instead of a set. That definition does not seem
clear. I
> turn to Wikipedia, http://en.wikipedia.org/wiki/Direct_sum,
it
> introduces the tensor product. I search for direct sum of
sets and
> find http://www.mathreference.com/set,dirsum.html, The
Direct Sum of
> Sets. It describes the direct sum much as you do. It says
> specifically The direct sum is always nonempty. It doesn't
say very
> much beyond that. The direct sum of two copies of the empty
set is
> empty, I would think. Please explain direct sum. I can see
how a
> subset of the direct product with only finitely many
non-zero elements
> in each sequence is useful to describe the polynomials, I
just don't
> yet see how you got there from {a+b : a E A and b E B}.
As I poin out, it does not make sense to speak of the direct
sum of
sets. We defined direct sum by referring to a particular
privileged
element of Z, namely zero. In other words, you can have a
Cartesian
product of SETS, but to define a direct sum you need to use
sets having
a privileged element, such as zero.
I'm sorry Wolfram happens to have prin a brain fart on this
particular topic, but they did.
> I was thinking that the polynomials' coefficient sequences
would map
> to Z^N. They inject into Z^N, a set of infinitely many
sequences of
> values of Z, via identity. I thought they might biject yet I
don't
> see a method for that.
No Ross they don't biject. The polynomials are countable and
the power
series are not. We've been over this about 20 times already.
> This leads me to think about roots of power series.
Well in this discussion it's irrelevant. Power series represent
functions much wilder than polynomials. For example I'm sure
you know
there's a power series for e^x. That doesn't have any roots.
If a_0 is zero a
> root is zero, meaning a value of x for which the power series
> evaluates to zero, not the expanded power series exponentia
to 1/n
> for positive integer n>2. If the power series is x-1
multiplied with
> itself infinitely many times then a root is one.
It's not fruitful to wander off from algebra into analysis
right now.
Considering the roots of power series is a whole 'nother
subject. You
can use power series to represent sin, cosine, arctangent, the
whole zoo
of math functions.
> This thread is about transcendental numbers. The
transcendental
> number is not an algebraic number. The algebraic number is
the root
> of a polynomial with integer coefficients.
Yes. I think this is where I came in to this movie. It would
be more
precise to say that *an* algebraic number is the root of a
poly with
integer coefficients.
> Is not the root of a power series with integer coefficients,
that is
> to say, a value for which a power series evaluates to zero,
algebraic?
No of course not. Consider the power series for sin x. That is
zero at
0, pi, 2pi, 3pi, 4pi, 5pi, ... It's known that pi is a
transcendental
number. There is no polynomial having pi as a root.
> Probably not always, no. That's saying that the roots of some
> power series with integer coefficients that are not
polynomials are
> transcendental. Besides that some polynomials have no real
roots.
Power series are very powerful! (pun intended). Power series
can
represent all kinds of functions such as exp, sin, cosine and
functions
much wilder than that.
> We want to represent the integer coefficients of all
polynomials.
Makes no sense. The integer coefficients are integers.
For
> no finite n is Z^n sufficient.
You have said this previously, to which I replied, yes but so
what?
There certainly are infinitely many polynomials. Countably
infinite.
Instead a proper subset of Z^N which
> is called the direct sum of infinitely many copies of Z with
each
> coordinate sequence having only finitely many non-zero
values is used.
Yes. That's how we DEFINE polynomials.
> For no finite n is Z^n sufficient, the coefficients of any
polynomial
> as a sequence is not necessarily an element of Z^n
If you fix n ahead of time, then Z^n can only express the
polynomials of
degree n or less. That is correct. You've said that several
times
already. But it has no bearing on the fact that the
polynomials are
countable. Just like if I fix N, then the numbers less than N
are not
sufficient to enumerate all the natural numbers.
and the set of all
> sequences of coefficients of any polynomial does not inject
into Z^n.
The set of polynomials does inject into Z^N. It does not
biject. And
you are right, if you choose an n, like say n = 23246699, it
is not
possible to express all polynomials in Z^n. Just the ones
whose degree
is at most 23246699.
You clearly understand what we're talking about, so I'm not
sure what
your point of confusion or disagreement is.
===
Subject: Re: Chained arrow notation experts- 9->9->9->9 sketch?
Regarding Graham's number,here is a diagram that I've seen on
the web-
/ 1) 3^^^^3
|
| 2) 3^^...1)...^^3 [where there are 1) = 3^^^^3 up-arrows]
|
| 3) 3^^...2)...^^3 [where there are 2) up-arrows]
| .
64 levels < .
| .
| .
| 63) 3^^...62)...^^3
|
64) 3^^...63)...^^3 <--- Graham's #
(-the process of so and so many arrows is far from being
mathematically precise and needs to be reworked;however,it
lends
itself to being spacially compact. )
Labelling sections and bays
one section
________________^_______ |
/ |
2 bays |
______________^_________ |
/ |
bay 1 bay 2 |
___^___ ______^________ |
/ / > one section
/ 1) 3^^^^3 |
| 2) 3^..1)..^3 |
| 3) 3^..2)..^3 |
64 levels< . |
| . |
| . |
64) 3^..63)..^3 /
where a bay (OL) includes all levels and a section is all bays
and
their levels.
Using this mechanism, the Conway-Guy expression
a -> b -> ..... x -> y -> z
can be defined in terms of Knuth up-arrows, e.g.
2->3->3->4
8 bays
____________________________^________________________________
/
bay 1 bay 2 bay 3 4-7 bay 8
__^__ _________^__________ _____^_______ __^_ _____^______
/ / / / /
/ 1) 2^3 = 8 / 1) 2^3 / / 1) 2^3
| 2) 2^..1)..^3 | 2) 2^..1)..^3| | 2) 2^..1)..^3
8 | 3) 2^..2)..^3 | . | | .
levels< . | . | 4 | .
| . | . | bays| .
8) 2^..7)..^3 levels< . |here | .
= 2->3->8->2 | . | | .
= 2->3->2->3 ..2^........^3<.....< .
levels| | .
..2^........^3
= 2->3->8->3 bays
= 2->3->2->4
________________________________________________________^_____
/ / 1) 2^3 / 1) 2^3 / / 1) 2^3
| . | . | | .
levels< . | . | | .
| . | . | | .
8) 2^..7)..^3 levels< . | | .
| . | | .
..2^........^3<.....< .
levels| | .
..2^.......^3
= 2->3->3->4
the above can be called 2 section levels
crunch the diagram down to this;dropping bays # 1 and 3 ---
7 bays
___________^_____________
/
1) 2^3 / / 1) 2^3
. | | .
. | | .
8)2^..7)..^3<..< .
levels | | .
..2^.....^3 = 2->3->2->4
bays
___________________^_____
/1) 2^3 / / 1) 2^3
. | | .
. | | .
8)2^..7)..^3<..< .
levels| | .
..2^.....^3 = 2->3->3->4
Following and extending the concept---
> another try at 9->9->9->9---
> 387420488 section bays
>
_______________________________________________^______________
_______________
________________________________
> /
> / 387420488 bays
> | _______________________^_____________________
> |/ 1) 9^9=387420489 / / 1) 9^9
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | . | | .
> | . | | .
> |387420489) 9^..387420488)..^9<..< .
> | levels| | .
> | ..9^........^9
> | bays
> /387420488< ______________________________________^______
> | section|/ .
> | levels | . /
/ 387420488 bays
> | | . |
| _______________________^_____________________
> | | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9
> | |/ 1) 9^9 / / 1) 9^9 |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |
| . | | .
> | | . | | . |
| . | | .
> | | . | | . |
|387420489) 9^..387420488)..^9<..< .
> | |387420489) 9^..387420488)..^9<..< . |
| levels| | .
> | | levels | | . |
| ..9^........^9
> |
..9^.........^9<...< bays
> | section levels |
| ___________________________________^_________
> | |
|/ .
> | |
| .
> | |
| ___________________________________^_________
> | |
|/ 1) 9^9 / / 1) 9^9
> | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
> 387420488< |
| . | | .
> levels of| |
| . | | .
> section | |
|387420489) 9^..387420488)..^9<..< .
> levels | |
| levels | | .
> |
..9^........^9
> |
section bays
> |
______________________________________________________________
_______________
_____________________^_________
> |/
.
> |
.
> |
______________________________________________________________
_______________
_____________________^_________
> |/ / 387420488 bays
> | | _______________________^_____________________
> | |/ 1) 9^9=387420489 / / 1) 9^9
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | | . | | .
> | | . | | .
> | |387420489) 9^..387420488)..^9<..< .
> | | levels| | .
> | | ..9^........^9
> | | bays
> 387420488< ______________________________________^______
> section|/ .
> levels | . /
/ 387420488 bays
> | . |
| _______________________^_____________________
> | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9
> |/ 1) 9^9 / / 1) 9^9 |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |
| . | | .
> | . | | . |
| . | | .
> | . | | . |
|387420489) 9^..387420488)..^9<..< .
> |387420489) 9^..387420488)..^9<..< . |
| levels| | .
> | levels | | . |
| ..9^........^9
>
..9^.........^9<...< bays
> section levels |
| ___________________________________^_________
> |
|/ .
> |
| .
> |
| ___________________________________^_________
> |
|/ 1) 9^9 / / 1) 9^9
> |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
> |
| . | | .
> |
| . | | .
> |
|387420489) 9^..387420488)..^9<..< .
> |
| levels | | .
>
..9^........^9
>
levels of
> 387420488 bays
of section bays section levels
>
______________________________________________________________
_^_____________
______________________________________ |
> /
v
>
v
> 387420488 section
bays v
>
______________________________________________________________
_^_____________
_______________________________ | | v
> /
| | v
> / 387420488 bays
| | v
> | _______________________^_____________________
| | v
> |/ 1) 9^9=387420489 / / 1) 9^9
| | v
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9
| | v
> | . | | .
| | v
> | . | | .
| | v
> |387420489) 9^..387420488)..^9<..< .
| | v
> | levels| | .
| | v
> | ..9^........^9
| | v
> | bays
| | v
> 387420488< ______________________________________^______
| | v
> section|/ .
| | v
> levels | . / /
387420488 bays | | v
> | . | |
_______________________^_____________________ | | v
> | ______________________________________^______ | |/
1) 9^9 / / 1) 9^9 | | v
> |/ 1) 9^9 / / 1) 9^9 | |
2) 9^..1)..^9 | | 2) 9^..1)..^9 | | v
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
. | | . | | v
> | . | | . | |
. | | . | | v
> | . | | . |
|387420489) 9^..387420488)..^9<..< . | | v
> |387420489) 9^..387420488)..^9<..< . | |
levels| | . | | v
> | levels | | . | |
..9^........^9 | | v
> ..9^.........^9<...<
bays | | v
> section levels | |
___________________________________^_________ | | v
> | |/
. | | v
> | |
. | | v
> | |
. | | v
> | |
___________________________________^_________ | | v
> | |/
1) 9^9 / / 1) 9^9 | | v
> | |
2) 9^..1)..^9 | | 2) 9^..1)..^9 | | v
> | |
. | | . | | v
> | |
. | | . | | v
> |
|387420489) 9^..387420488)..^9<..< . | | v
> | |
levels | | . | | v
>
..9^........^9 | | v
>
section bays | | v
>
______________________________________________________________
_______________
_____________________^_________ | | v
> /
. | | v
>
. .
. | |
>
______________________________________________________________
_______________
_____________________^_________ | |
> / / 387420488 bays
| |
> | _______________________^_____________________
| |
> |/ 1) 9^9=387420489 / / 1) 9^9
| |
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9
| |
> | . | | .
| |
> | . | | .
| |
> |387420489) 9^..387420488)..^9<..< .
| |
> | levels| | .
| |
> | ..9^........^9
| |
> | bays
| |
> 387420488< ______________________________________^______
| |
> section|/ .
| |
> levels | . / /
387420488 bays | |
> | . | |
_______________________^_____________________ | |
> | ______________________________________^______ | |/
1) 9^9 / / 1) 9^9 | |
> |/ 1) 9^9 / / 1) 9^9 | |
2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
. | | . | |
> | . | | . | |
. | | . | |
> | . | | . |
|387420489) 9^..387420488)..^9<..< . | |
> |387420489) 9^..387420488)..^9<..< . | |
levels| | . | |
> | levels | | . | |
..9^........^9 | |
> ..9^.........^9<...<
bays | |
> section levels | |
___________________________________^_________ | |
> | |/
. | |
> | |
. | |
> | |
. | |
> | |
___________________________________^_________ | |
> | |/
1) 9^9 / / 1) 9^9 | |
> | |
2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
> | |
. | | . | |
> | |
. | | . | |
> |
|387420489) 9^..387420488)..^9<..< . | |
> | |
levels | | . | |
>
..9^........^9 / /
>
bays of section bays
>
______________________________________________________________
_______________
__________________________^___________________________________
_______________
______________________________________________________________
_______________
____ |
> /
387420488 section bays
|
>
______________________________________________________________
_^_____________
_______________________________
|
> /
|
> / 387420488 bays
|
> | _______________________^_____________________
|
> |/ 1) 9^9=387420489 / / 1) 9^9
/ /
387420488 section bays
|
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9
| |
______________________________________________________________
_^_____________
_______________________________ |
> | . | | .
| |/
|
> | . | | .
| | /
387420488 bays
|
> |387420489) 9^..387420488)..^9<..< .
| | |
_______________________^_____________________
|
> | levels| | .
| | |/ 1)
9^9=387420489 / / 1) 9^9
|
> | ..9^........^9
| | | 2)
9^..1)..^9 | | 2) 9^..1)..^9
|
> | bays
| | |
. | | .
|
> /387420488< ______________________________________^______
| | |
. | | .
|
> | section|/ .
| | |387420489)
9^..387420488)..^9<..< .
|
> | levels | . /
/ 387420488 bays | | |
levels| | .
|
> | | . |
| _______________________^_____________________ | | |
..9^........^9
|
> | | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9 | | |
bays
|
> | |/ 1) 9^9 / / 1) 9^9 |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |387420488<
______________________________________^______
|
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |
| . | | . | | section|/
.
|
> | | . | | . |
| . | | . | | levels |
. / / 387420488 bays
|
> | | . | | . |
|387420489) 9^..387420488)..^9<..< . | | |
. | |
_______________________^_____________________ |
> | |387420489) 9^..387420488)..^9<..< . |
| levels| | . | | |
______________________________________^______ | |/ 1) 9^9
/ / 1) 9^9 |
> | | levels | | . |
| ..9^........^9 | | |/ 1)
9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
> |
..9^.........^9<...< bays | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | .
| | . |
> | section levels |
| ___________________________________^_________ | | |
. | | . | | . | |
. |
> | |
|/ . | | |
. | | . | |387420489) 9^..387420488)..^9<..<
. |
> | |
| . | | |387420489)
9^..387420488)..^9<..< . | | levels| |
. |
> | |
| . | | |
levels | | . | |
..9^........^9 |
> | |
| ___________________________________^_________ | |
..9^.........^9<...<
bays |
> | |
|/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |
> | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
| |/ .
|
> 387420488< |
| . | | . | |
| | .
|
> levels of| |
| . | | . | |
| | .
|
> section | |
|387420489) 9^..387420488)..^9<..< . | |
| |
___________________________________^_________ |
> levels | |
| levels | | . | |
| |/ 1) 9^9 / / 1)
9^9 |
> |
..9^........^9 | |
| | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
> |
section bays | |
| | . | |
. |
> |
______________________________________________________________
_______________
_____________________^_________ | |
| | . | | . |
> |/
. | |
| |387420489) 9^..387420488)..^9<..<
. |
> |
. | |
| | levels | |
. |
> |
. | |
..9^........^9 |
> |
______________________________________________________________
_______________
_____________________^_________ | |
section bays |
> |/ / 387420488 bays
| |
______________________________________________________________
_______________
_____________________^_________ |
> | | _______________________^_____________________
| |/
.
|
> | |/ 1) 9^9=387420489 / / 1) 9^9
| |
.
|
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9
| |
.
|
> | | . | | .
| |
______________________________________________________________
_______________
_____________________^_________ |
> | | . | | .
| |/ /
387420488 bays
|
> | |387420489) 9^..387420488)..^9<..< .
| | |
_______________________^_____________________
|
> | | levels| | .
| | |/ 1)
9^9=387420489 / / 1) 9^9
|
> | | ..9^........^9
| | | 2)
9^..1)..^9 | | 2) 9^..1)..^9
|
> | | bays
| | |
. | | .
|
> |387420488< ______________________________________^______
| | |
. | | .
|
> | section|/ .
| | |387420489)
9^..387420488)..^9<..< .
|
> levels | . /
/ 387420488 bays | | |
levels| | .
|
> | . |
| _______________________^_____________________ | | |
..9^........^9
|
> | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9 | | |
bays
|
> |/ 1) 9^9 / / 1) 9^9 |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |387420488<
______________________________________^______
|
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |
| . | | . | | section|/
.
|
> | . | | . |
| . | | . | | levels |
. / / 387420488 bays
|
> | . | | . |
|387420489) 9^..387420488)..^9<..< . | | |
. | |
_______________________^_____________________ |
> |387420489) 9^..387420488)..^9<..< . |
| levels| | . | | |
______________________________________^______ | |/ 1) 9^9
/ / 1) 9^9 |
> | levels | | . |
| ..9^........^9 | | |/ 1)
9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
>
..9^.........^9<...< bays | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | .
| | . |
> section levels |
| ___________________________________^_________ | | |
. | | . | | . | |
. |
> |
|/ . | | |
. | | . | |387420489) 9^..387420488)..^9<..<
. |
> |
| . | | |387420489)
9^..387420488)..^9<..< . | | levels| |
. |
> |
| . | | |
levels | | . | |
..9^........^9 |
> |
| ___________________________________^_________ | |
..9^.........^9<...<
bays |
> |
|/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |
> |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
| |/ .
|
> |
| . | | . | |
| | .
|
> |
| . | | . | |
| | .
|
> |
|387420489) 9^..387420488)..^9<..< . | |
| |
___________________________________^_________ |
> |
| levels | | . | |
| |/ 1) 9^9 / / 1)
9^9 |
>
..9^........^9 | |
| | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
>
levels of section bays<....<
| | . | |
. |
>
| |
| | . | |
. |
>
| |
| |387420489) 9^..387420488)..^9<..<
. |
>
| |
| | levels | |
. |
>
..9^........^9 |
>
bays
of bays
>
section bays |
>
______________________________________________________________
_______________
______________________________________________________________
_______________
______________________________________________________________
_________^_____
____ |
> /
.
|
>
.
|
>
.
|
>
______________________________________________________________
_______________
______________________________________________________________
_______________
______________________________________________________________
_________^_____
____ |
> /
| 8
>
387420488 section bays
> levels
>
______________________________________________________________
_^_____________
_______________________________
|of
bays
> /
|of sect.
> / 387420488 bays
|bays
> | _______________________^_____________________
|
> |/ 1) 9^9=387420489 / / 1) 9^9
/ /
387420488 section bays
|
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9
| |
______________________________________________________________
_^_____________
_______________________________ |
> | . | | .
| |/
|
> | . | | .
| | /
387420488 bays
|
> |387420489) 9^..387420488)..^9<..< .
| | |
_______________________^_____________________
|
> | levels| | .
| | |/ 1)
9^9=387420489 / / 1) 9^9
|
> | ..9^........^9
| | | 2)
9^..1)..^9 | | 2) 9^..1)..^9
|
> | bays
| | |
. | | .
|
> /387420488< ______________________________________^______
| | |
. | | .
|
> | section|/ .
| | |387420489)
9^..387420488)..^9<..< .
|
> | levels | . /
/ 387420488 bays | | |
levels| | .
|
> | | . |
| _______________________^_____________________ | | |
..9^........^9
|
> | | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9 | | |
bays
|
> | |/ 1) 9^9 / / 1) 9^9 |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |387420488<
______________________________________^______
|
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |
| . | | . | | section|/
.
|
> | | . | | . |
| . | | . | | levels |
. / / 387420488 bays
|
> | | . | | . |
|387420489) 9^..387420488)..^9<..< . | | |
. | |
_______________________^_____________________ |
> | |387420489) 9^..387420488)..^9<..< . |
| levels| | . | | |
______________________________________^______ | |/ 1) 9^9
/ / 1) 9^9 |
> | | levels | | . |
| ..9^........^9 | | |/ 1)
9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
> |
..9^.........^9<...< bays | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | .
| | . |
> | section levels |
| ___________________________________^_________ | | |
. | | . | | . | |
. |
> | |
|/ . | | |
. | | . | |387420489) 9^..387420488)..^9<..<
. |
> | |
| . | | |387420489)
9^..387420488)..^9<..< . | | levels| |
. |
> | |
| . | | |
levels | | . | |
..9^........^9 |
> | |
| ___________________________________^_________ | |
..9^.........^9<...<
bays |
> | |
|/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |
> | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
| |/ .
|
> 387420488< |
| . | | . | |
| | .
|
> levels of| |
| . | | . | |
| | .
|
> section | |
|387420489) 9^..387420488)..^9<..< . | |
| |
___________________________________^_________ |
> levels | |
| levels | | . | |
| |/ 1) 9^9 / / 1)
9^9 |
> |
..9^........^9 | |
| | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
> |
section bays | |
| | . | |
. |
> |
______________________________________________________________
_______________
_____________________^_________ | |
| | . | | . |
> |/
. | |
| |387420489) 9^..387420488)..^9<..<
. |
> |
. | |
| | levels | |
. |
> |
. | |
..9^........^9 |
> |
______________________________________________________________
_______________
_____________________^_________ | |
section bays |
> |/ / 387420488 bays
| |
______________________________________________________________
_______________
_____________________^_________ |
> | | _______________________^_____________________
| |/
.
|
> | |/ 1) 9^9=387420489 / / 1) 9^9
| |
.
|
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9
| |
.
|
> | | . | | .
| |
______________________________________________________________
_______________
_____________________^_________ |
> | | . | | .
| |/ /
387420488 bays
|
> | |387420489) 9^..387420488)..^9<..< .
| | |
_______________________^_____________________
|
> | | levels| | .
| | |/ 1)
9^9=387420489 / / 1) 9^9
|
> | | ..9^........^9
| | | 2)
9^..1)..^9 | | 2) 9^..1)..^9
|
> | | bays
| | |
. | | .
|
> |387420488< ______________________________________^______
| | |
. | | .
|
> | section|/ .
| | |387420489)
9^..387420488)..^9<..< .
|
> levels | . /
/ 387420488 bays | | |
levels| | .
|
> | . |
| _______________________^_____________________ | | |
..9^........^9
|
> | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9 | | |
bays
|
> |/ 1) 9^9 / / 1) 9^9 |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |387420488<
______________________________________^______
|
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |
| . | | . | | section|/
.
|
> | . | | . |
| . | | . | | levels |
. / / 387420488 bays
|
> | . | | . |
|387420489) 9^..387420488)..^9<..< . | | |
. | |
_______________________^_____________________ |
> |387420489) 9^..387420488)..^9<..< . |
| levels| | . | | |
______________________________________^______ | |/ 1) 9^9
/ / 1) 9^9 |
> | levels | | . |
| ..9^........^9 | | |/ 1)
9^9 / / 1) 9^9 | | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
>
..9^.........^9<...< bays | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | .
| | . |
> section levels |
| ___________________________________^_________ | | |
. | | . | | . | |
. |
> |
|/ . | | |
. | | . | |387420489) 9^..387420488)..^9<..<
. |
> |
| . | | |387420489)
9^..387420488)..^9<..< . | | levels| |
. |
> |
| . | | |
levels | | . | |
..9^........^9 |
> |
| ___________________________________^_________ | |
..9^.........^9<...<
bays |
> |
|/ 1) 9^9 / / 1) 9^9 | |
section levels | |
___________________________________^_________ |
> |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
| |/ .
|
> |
| . | | . | |
| | .
|
> |
| . | | . | |
| | .
|
> |
|387420489) 9^..387420488)..^9<..< . | |
| |
___________________________________^_________ |
> |
| levels | | . | |
| |/ 1) 9^9 / / 1)
9^9 |
>
..9^........^9 | |
| | 2) 9^..1)..^9 | |
2) 9^..1)..^9 |
>
levels of section bays<....<
| | . | |
. |
>
| |
| | . | |
. |
>
| |
| |387420489) 9^..387420488)..^9<..<
. |
>
| |
| | levels | |
. |
>
..9^........^9 /
>
|
>
v
>
9->9->9->9
===
Subject: Re: Chained arrow notation experts- 9->9->9->9 sketch?
Regarding Graham's number,here is a diagram that I've seen on
the web-
/ 1) 3^^^^3
|
| 2) 3^^...1)...^^3 [where there are 1) = 3^^^^3 up-arrows]
|
| 3) 3^^...2)...^^3 [where there are 2) up-arrows]
| .
64 levels < .
| .
| .
| 63) 3^^...62)...^^3
|
64) 3^^...63)...^^3 <--- Graham's #
(-the process of so and so many arrows is far from being
mathematically precise and needs to be reworked;however,it
lends
itself to being spacially compact. )
Labelling sections and bays
one section
________________^_______ |
/ |
2 bays |
______________^_________ |
/ |
bay 1 bay 2 |
___^___ ______^________ |
/ / > one section
/ 1) 3^^^^3 |
| 2) 3^..1)..^3 |
| 3) 3^..2)..^3 |
64 levels< . |
| . |
| . |
64) 3^..63)..^3 /
where a bay (OL) includes all levels and a section is all bays
and
their levels.
Using this mechanism, the Conway-Guy expression
a -> b -> ..... x -> y -> z
can be defined in terms of Knuth up-arrows, e.g.
2->3->3->4
8 bays
____________________________^________________________________
/
bay 1 bay 2 bay 3 4-7 bay 8
__^__ _________^__________ _____^_______ __^_ _____^______
/ / / / /
/ 1) 2^3 = 8 / 1) 2^3 / / 1) 2^3
| 2) 2^..1)..^3 | 2) 2^..1)..^3| | 2) 2^..1)..^3
8 | 3) 2^..2)..^3 | . | | .
levels< . | . | 4 | .
| . | . | bays| .
8) 2^..7)..^3 levels< . |here | .
= 2->3->8->2 | . | | .
= 2->3->2->3 ..2^........^3<.....< .
levels| | .
..2^........^3
= 2->3->8->3 bays
= 2->3->2->4
________________________________________________________^_____
/ / 1) 2^3 / 1) 2^3 / / 1) 2^3
| . | . | | .
levels< . | . | | .
| . | . | | .
8) 2^..7)..^3 levels< . | | .
| . | | .
..2^........^3<.....< .
levels| | .
..2^.......^3
= 2->3->3->4
the above can be called 2 section levels
crunch the diagram down to this;dropping bays # 1 and 3 ---
7 bays
___________^_____________
/
1) 2^3 / / 1) 2^3
. | | .
. | | .
8)2^..7)..^3<..< .
levels | | .
..2^.....^3 = 2->3->2->4
bays
___________________^_____
/1) 2^3 / / 1) 2^3
. | | .
. | | .
8)2^..7)..^3<..< .
levels| | .
..2^.....^3 = 2->3->3->4
Following and extending the concept---
> another try at 9->9->9->9---
> 387420488 section bays
>
_______________________________________________^______________
_______________
________________________________
> /
> / 387420488 bays
> | _______________________^_____________________
> |/ 1) 9^9=387420489 / / 1) 9^9
> | 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | . | | .
> | . | | .
> |387420489) 9^..387420488)..^9<..< .
> | levels| | .
> | ..9^........^9
> | bays
> /387420488< ______________________________________^______
> | section|/ .
> | levels | . /
/ 387420488 bays
> | | . |
| _______________________^_____________________
> | | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9
> | |/ 1) 9^9 / / 1) 9^9 |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9 |
| . | | .
> | | . | | . |
| . | | .
> | | . | | . |
|387420489) 9^..387420488)..^9<..< .
> | |387420489) 9^..387420488)..^9<..< . |
| levels| | .
> | | levels | | . |
| ..9^........^9
> |
..9^.........^9<...< bays
> | section levels |
| ___________________________________^_________
> | |
|/ .
> | |
| .
> | |
| ___________________________________^_________
> | |
|/ 1) 9^9 / / 1) 9^9
> | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9
> 387420488< |
| . | | .
> levels of| |
| . | | .
> section | |
|387420489) 9^..387420488)..^9<..< .
> levels | |
| levels | | .
> |
..9^........^9
> |
section bays
> |
______________________________________________________________
_______________
_____________________^_________
> |/
.
> |
.
> |
______________________________________________________________
_______________
_____________________^_________
> |/ / 387420488 bays
> | | _______________________^_____________________
> | |/ 1) 9^9=387420489 / / 1) 9^9
> | | 2) 9^..1)..^9 | | 2) 9^..1)..^9
> | | . | | .
> | | . | | .
> | |387420489) 9^..387420488)..^9<..< .
> | | levels| | .
> | | ..9^........^9
> | | bays
> 387420488< ______________________________________^______
> section|/ .
> levels | . /
/ 387420488 bays
> | . |
| _______________________^_____________________
> | ______________________________________^______ |
|/ 1) 9^9 / / 1) 9^9
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| |/ .
|
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9^9 |
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|
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387420488 section bays
|
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| |
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| |/
|
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| | /
387420488 bays
|
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| | |
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| | |/ 1)
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| | |
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|
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| | |
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|
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|
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|
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______________________________________^______
|
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|
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|
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. | |
_______________________^_____________________ |
> | |387420489) 9^..387420488)..^9<..< . |
| levels| | . | | |
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/ / 1) 9^9 |
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bays |
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section levels | |
___________________________________^_________ |
> | |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
| |/ .
|
> 387420488< |
| . | | . | |
| | .
|
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|
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| |
___________________________________^_________ |
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9^9 |
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section bays | |
| | . | |
. |
> |
______________________________________________________________
_______________
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_______________
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section bays |
> |/ / 387420488 bays
| |
______________________________________________________________
_______________
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| |/
.
|
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| |
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|
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| |/ /
387420488 bays
|
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| | |
_______________________^_____________________
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| | |/ 1)
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| | |
. | | .
|
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| | |
. | | .
|
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9^..387420488)..^9<..< .
|
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|
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| _______________________^_____________________ | | |
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|
> |/ 1) 9^9 / / 1) 9^9 |
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______________________________________^______
|
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|
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|
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. | |
_______________________^_____________________ |
> |387420489) 9^..387420488)..^9<..< . |
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/ / 1) 9^9 |
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>
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| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | | .
| | . |
> section levels |
| ___________________________________^_________ | | |
. | | . | | . | |
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> |
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. |
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| ___________________________________^_________ | |
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bays |
> |
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section levels | |
___________________________________^_________ |
> |
| 2) 9^..1)..^9 | | 2) 9^..1)..^9 | |
| |/ .
|
> |
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|
> |
| . | | . | |
| | .
|
> |
|387420489) 9^..387420488)..^9<..< . | |
| |
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> |
| levels | | . | |
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9^9 |
>
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| | 2) 9^..1)..^9 | |
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>
levels of section bays<....<
| | . | |
. |
>
| |
| | . | |
. |
>
| |
| |387420489) 9^..387420488)..^9<..<
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>
| |
| | levels | |
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>
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>
|
>
v
>
9->9->9->9
===
Subject: Re: Quick Math Guide to core error issues
thank *o*, for small things, like using a vriable consistently
-- another milestone for the Ten Year Programme!
> Oh yeah, I've taken your advice though as I'm using only m
as a
> variable in my recent postings as I'm *really* ready to
finish things
> up.
> it's not a big deal. In any event, unlike with Nora Baron, I
won't
> put quotes around your name.
--il duce d'Enron!
===
Subject: Re: Quick Math Guide to core error issues
Is it me or can anyone follow what JSH is ranting about?
> See:
http://www.google.com/search?q=%22James+Harris%22+site%
3Awww.crank.net
> http://www.crank.net/harris.html
Now that's just mean. I think that Sam Wormley is a worm.
He's a weak, ineffectual and intellectually deficient person,
who
unfortunately has delusions of grandeur, so he posts a lot.
And yes, you may THINK I have delusions of grandeur, but I'm
putting
out my research while Sam Wormley is just being an annoying
twerp.
He's too small to do his own thing, so he attacks others and
post
links as if knowing a link shows you actually know something.
===
Subject: Re: Quick Math Guide to core error issues
<3F918BA4.32121731@mchsi.com>
<3c65f87.0310290340.45918a51@posting.google.com>
Discussion, linux)
>> See:
http://www.google.com/search?q=%22James+Harris%22+site%
3Awww.crank.net
>> http://www.crank.net/harris.html
> Now that's just mean. I think that Sam Wormley is a worm.
Please, try to stay within your role.
Sam Wormley is a worm comments are reserved for Herc.
When you start crossing into Herc's turf, it's difficult for
those of
us keeping score at home. How would you like it if he star
threatening congressional hearings?
--
My proofs are out there.
-- James S. Harris
===
Subject: Re: Quick Math Guide to core error issues
Is it me or can anyone follow what JSH is ranting about?
See:
http://www.google.com/search?q=%22James+Harris%22+site%
3Awww.crank.net
> http://www.crank.net/harris.html
> Now that's just mean. I think that Sam Wormley is a worm.
> He's a weak, ineffectual and intellectually deficient
person, who
> unfortunately has delusions of grandeur, so he posts a lot.
> And yes, you may THINK I have delusions of grandeur, but I'm
putting
> out my research while Sam Wormley is just being an annoying
twerp.
> He's too small to do his own thing, so he attacks others and
post
> links as if knowing a link shows you actually know something.
James, when you attack someone personally, why are you
surprised when
people turn around and attack you personally?
===
Subject: Re: Quick Math Guide to core error issues
Is it me or can anyone follow what JSH is ranting about?
See:
http://www.google.com/search?q=%22James+Harris%22+site%
3Awww.crank.net
> http://www.crank.net/harris.html
Now that's just mean. I think that Sam Wormley is a worm.
He's a weak, ineffectual and intellectually deficient person,
who
> unfortunately has delusions of grandeur, so he posts a lot.
And yes, you may THINK I have delusions of grandeur, but I'm
putting
> out my research while Sam Wormley is just being an annoying
twerp.
He's too small to do his own thing, so he attacks others and
post
> links as if knowing a link shows you actually know something.
 James, when you attack someone personally, why are you
surprised when
> people turn around and attack you personally?
Sam Wormley STAR IT by posting a flame link in MY THREAD!!!
It turns out that Sam Wormley is wormy, and intellectually
deficient,
convinced that he can post links rather than actually know
anything.
If he wants to behave like a twerp, and an intellectual
weakling, then
I have the right to answer.
And my answer is that he's deficient morally and
intellectually, posts
too damn much for what he's saying, and apparently has some
delusions
of grandeur dependent on posting links versus actually knowing
of what
he speaks.
He's a worm. He's inferior. He's mentally deficient.
===
Subject: Re: Explaining math definition problem
correction to the correction, and this is the 2nd ed.
of B&M: he said that Cusa had a rectification
of the circle, but he actually gave a simple argument,
showing that it was not even rational, or
transendental, as we now say. he said,
because the approximations by inscribing/circumscribing polgona
with more & more sides, just get further from the ultimate
shape
of circularity, although the error from pi gets smaller.
It's a different species of shape.
> another error that it made was:
> said that Cusa made an erroneous proof
> that the circle was incomeasurable with teh tetragon
> the question is, do all cases fall to your peculiar method,
or
> did you find a counterexample in Object Ring Theory?
--les ducs d'Enron!
===
Subject: Re: Explaining math definition problem
> Readers can look at the argument, and see what actually is
in it.
> Notice how I'll be strongly emphasizing constant terms all
the way
> down.
> P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078
> which has a constant term that is 1078.
> Well P(x) can also be written out as
> P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x
)(5)(7^2) + 7^3
You can keep on reposting this error, but it only reflects on
you. The
correct factorization
is:
P(x)= 7^2(2401 x^3 - 147 x^2 - 3x) (5^3) + 3(1 + 49 x
)(5)(7^2) + 7^3
Your factorization is wrong. It does not expand to agree with
the original
equation defining
P(x). Do you ever check your work? Or isn't that necessary
with an intellect
as advanced as
yours?
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
--
http://www.crbond.com
===
Subject: Re: Explaining math definition problem
> Readers can look at the argument, and see what actually is
in it.
Notice how I'll be strongly emphasizing constant terms all the
way
> down.
P(x) = 14706125 x^3 - 900375 x^2 + 17640 x + 1078
Hmmm...it looks like C. Bond is actually right, and I have a
sign
problem as it should be -17640 x.
<dele Your factorization is wrong. It does not expand to agree
with the original
equation defining
> P(x). Do you ever check your work? Or isn't that necessary
with an
intellect as advanced as
> yours?
You are correct that I gave the wrong polynomial as it should
be
P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078.
Thanks for pointing that out!!!
===
Subject: Re: Explaining math definition problem
[snip]
> You are correct that I gave the wrong polynomial as it
should be
> P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078.
> Thanks for pointing that out!!!
>
Happy to oblige. No doubt you'll be equally grateful for
future posts which
identify further
errors.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
--
http://www.crbond.com
Subject: Re: Explaining math definition problem
===
>>I parafrase you and ahve a question at the end:
> You don't have my permission to parafrase and I don't find
mocking
> posts of interest.
He wasn't mocking you. English isn't his first language.
> Now I've looked over some of your posts and can rather
easily explain
> what you seem to find significant.
> Actually, part of the reason I haven't been terribly worried
about
> answering you in detail is I've wondered how many people may
begin to
> doubt algebra and think that maybe math is inconsistent if
they can't
> figure out what's happening in your examples.
> I will say that it's quite simple.
> After all, constants *are* constants, and math is
consistent, so there
> has to be a rational explanation.
We all know that. You are treating non-constants as if they
behaved
like constants, however.
Please try to understand what the objections actually are
before
defending yourself.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Explaining math definition problem
>where the a's are roots of
>a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x).
>Notice it *appears* that the constant terms for the three
factors are
>all 7, which can't be right, as the constant term of P(x) is
1078, so
>setting x=0, reveals
Why do you keep talking about constant terms? The a_1, a_2, a_3
do not have constant terms; that concept is meaningless in this
context. If you mean the value at 0, why don't you say so? If
you mean something else, what is the constant term of
sqrt(x+1) + 1?
Peter van Rossum
--
Peter van Rossum, <petervr@nmsu.edu> | Universal law of
linearity: for
all
Dept. of Mathematics, New Mexico | f : R -> R and for all x, y
in
R:
State University, Las Cruces, NM, USA. | f(x + y) = f(x) + f(y)
===
Subject: Re: Explaining math definition problem
> I'm an independent researcher, which means that I use my
*own*
> funding, and my *own* direction to go out and see what
knowledge I can
> obtain. Some of my research has been in the area of
mathematics.
Short, straightforward question:
Let
Q(m) = 7*(m^2*x^2 + 3*x + 7).
Then Q(0) = 7*3*x + 7^2 = 7*(3*x + 7).
Note that Q(0)/7 = 3*x + 7. This is similar to
your polynomial P(m), where you have
P(0) = 7^2*(3*x + 7), or
P(0)/7^2 = (3*x + 7).
Now assume that my polynomial Q(m) is factored in the form
Q(m) = (b1*x + 7)*(b2*x + 7).
As in your P(m), the coefficients b1 and b2 are going to
be functions of m: b1(m) and b2(m).
The fact that Q(0) = 7*(3*x + 7) is consistent with
b1(0) = 0 and b2(0) = 3. Agreed?
So b1(0) is divisible by 7 and b2(0) is relatively
prime to 7.
This is similar to your polynomial P(m), where
a1(0) = a2(0) = 0 and a3(0) = 3, with a1 and a2 being
divisible by 7 and a3 being relatively prime to 7.
It appears to me that my (quadratic) polynomial Q(m)
has all the essential properties that your cubic
polynomial P(m) has. Your arguments regarding
independence of the constant term, etc., should
apply.
So, for my polynomial Q, I would like to see
your answer to the question: for my polynomial Q(m), for m
in general, is b1(m) or b2(m) divisible by 7? Both,
only one, or neither ? If the answer differs from what
you get with P(m), I wonder if you might explain why?
Andrzej
===
Subject: 7th degree polynomial
possible way to solve 7th degree polynomial
http://www.geocities.com/jongiff2000/a9_polyroot_index.html
===
Subject: Re: Wiles' Proof
permission for an emailed response.
As it happens, I just picked up Fermat's Last Theorem For
Amateurs
by Paulo Ribenboim, published by Springer. ISBN 0-387-98508-5.
The book mostly concentrates on various partial solutions. The
epilogue gives a sketch of the proof in a few pages, but
doesn't claim
to do more than make one exci by the idea.
Thomas
===
Subject: converging partial arithmetic means
why is it that lim (x1+...+xn)/n = a, if lim xn = a (lims are
to inf)?
thanks
===
Subject: Re: converging partial arithmetic means
> why is it that lim (x1+...+xn)/n = a, if lim xn = a (lims
are to inf)?
> thanks
The proofs essentially split the xi into two parts, those with
abs(xi-a) > eps and those with abs(xi-a) <= eps.
For any eps > 0, the first part is negligable (sp?), and the
second part
is, for the i <= n, as n gets large, almost all the sequence.
So, the average value is within 2*eps of a (the 2 takes care
of the
first part of the sequence).
For the product, take logs.
Martin Cohen
===
Subject: Re: converging partial arithmetic means
> why is it that lim (x1+...+xn)/n = a, if lim xn = a (lims
are to inf)?
> even more interesting, why lim (x1*...*xn)^(1/n) = a too?
especially
since
> thanks
Cesaro theorem.
===
Subject: Re: converging partial arithmetic means
> why is it that lim (x1+...+xn)/n = a, if lim xn = a (lims
are to inf)?
> even more interesting, why lim (x1*...*xn)^(1/n) = a too?
especially
since
thanks
> Cesaro theorem.
that seems like a much more general theorem, is there anyway
to prove
this case directly?
===
Subject: Re: Multiplying by Five
Interesting but not quite. I recall she taught it in the 7th
grade.
>A long time ago, I vaguely recall an elementary school
teacher telling
>us some trick for quickly multiplying two numbers (2 digits,
I believe)
>with one or both having 5 as the second digit. Anyone recall
the trick?
>Numbers like 35*45 or maybe even 20*35.
> Something like this maybe?
> Let x=35-5=30 and y=45+5=50 then:
> 35*45=(x+5)(y-5)=xy - 5x + 5y -25 = xy +5(y-x) -
25=1500+100-25=1575
===
Subject: Re: Multiplying by Five
That's an interesting one, but I don't believe what I was
think about. WW
> A long time ago, I vaguely recall an elementary school
teacher telling
> us some trick for quickly multiplying two numbers (2 digits,
I believe)
> with one or both having 5 as the second digit. Anyone recall
the
trick?
> Numbers like 35*45 or maybe even 20*35.
> Maybe this isn't what you're thinking of, but there's an
> easy-to-use trick for _squaring_ a number ending in 5:
multiply
> the bit before the 5 by (itself+1), then append 25.
> e.g.
> 15^2 = 225 (1*2=2)
> 25^2 = 625 (2*3*6)
> ...
> 125^2 = 15625 (12*13=156)
> Andrew Taylor
> Cambridge UK
===
Subject: Re: Wiles' Proof
Seems to me there is a niche here. Someone needs to write the
book,
Fermat's Last Theorem for Dummies.
--
(who someday will write Stochastic Optimal Control for Dummies)
===
Subject: Re: Wiles' Proof
> Seems to me there is a niche here. Someone needs to write
the book,
Fermat's Last Theorem for Dummies.
Dr. Herschkorn, for someone with your talent for clarity and
for
explaining things to dummies like me........ It would only
take a few
months' sabbatical, probably. Think of the money and compare to
academic salary!
All the best am
===
Subject: Re: Wiles' Proof
> Seems to me there is a niche here. Someone needs to write
the book,
Fermat's Last Theorem for Dummies.
Hey! I'm not a dummy! :)
But seriously, someone _should_ write a book that outlines the
proof or
the idea of the proof, teaching some of the necesary math from
undergraduate level and up.
/David
===
Subject: Re: Wiles' Proof
I suggest that Dummies (tm) hire the team
of Magadin and Harris.
> But seriously, someone _should_ write a book that outlines
the proof or
> the idea of the proof, teaching some of the necesary math
from
> undergraduate level and up.
--les ducs d'Enron!
http://www.cecaust.com.au/
http://members.tripod.com/~american_almanac/
===
Subject: Re: Wiles' Proof
> But seriously, someone _should_ write a book that outlines
the proof or
> the idea of the proof, teaching some of the necesary math
from
> undergraduate level and up.
> /David
Some of the books mentioned by earlier posters attempt to do
this. But
as was also no, if an author needs to start with undergrad
math and
keep the book to manageable size (say 300 pages or so), he/she
is
going to end up being very sketchy. Possibly a more worthwhile
project
would be a book that explains the proof, to the extent
possible, for
someone who has at least the basic undergraduate math courses.
Those
courses would include:
(1) linear algebra
(2) abstract algebra (groups, rings, fields, including some
Galois
theory)
(3) complex analysis
Note that these are the minimal prerequisites for graduate
courses
such as graduate level algebra (i.e., the material covered in,
say,
Lang's Algebra, though I don't know that I'd recommend Lang's
book
as a textbook for self-study!), algebraic geometry, algebraic
and
analytic number theory, and representation theory. And the
material
covered in the graduate courses provides the background to
_begin_ to
get into the details of Wiles's proof.
I guess what I'm trying to say is that if someone wants
anything
beyond a superficial hand-waving approach to Wiles's proof in
a book
that's less than a couple of thousand pages in length, then
they'll
probably need to start at least at the point of someone with an
undergrad math major degree. So a well-written and worthwhile
Wiles's
Proof for Dummies book would have as its target audience a
rather
select group of dummies! (I've heard math majors called
various names,
but dummy isn't usually one of them.)
None of this is to say that it isn't possible to write an
entertaining
book that gives some vague idea of the proof for someone with
very
little (i.e., less than an undergrad math major) background.
Singh's
book attempts to do that, with some success. But reading these
posts,
I think people want more of the math detail than that.
JHS
===
Subject: Re: Wiles' Proof
... stuff dele ...
> ... So a well-written and worthwhile Wiles's
> Proof for Dummies book would have as its target audience a
rather
> select group of dummies! (I've heard math majors called
various names,
> but dummy isn't usually one of them.)
I'm thinking you haven't paid attention to many of the JSH
threads.
... the rest dele ...
> JHS
Dale
===
Subject: Re: Wiles' Proof
> Seems to me there is a niche here. Someone needs to write
the book,
> Fermat's Last Theorem for Dummies.
 Hey! I'm not a dummy! :)
> But seriously, someone _should_ write a book that outlines
the proof or
> the idea of the proof, teaching some of the necesary math
from
> undergraduate level and up.
> /David
Gouv.90a, Fernando Q.
A marvelous proof.
Amer. Math. Monthly 101 (1994), no. 3, 203--222.
Here's the review:
Recently, Andrew Wiles announced a stunning breakthrough toward
proving the conjecture of Shimura and Taniyama that every
elliptic
curve over the rationals is modular. This announcement has also
genera excitement and interest among the general public,
because of
the connection between the Shimura-Taniyama conjecture and
Fermat's
last theorem.
Intended for an audience of nonspecialists, this paper
communicates
some of the basic ingredients that go into Wiles' work, and
how this
work relates to Fermat's last theorem. The author begins by
discussing
the main actors in the proof: elliptic curves, modular forms,
and the
Shimura-Taniyama conjecture which postulates a fascinating,
mysterious
connection between these two notions. He explains the
fundamental
result of Ribet, based on a construction of Hellegouarch and
Frey and
work of Serre, which gives a link between the Shimura-Taniyama
conjecture and Fermat's last theorem. He then briefly mentions
some of
the ideas behind Wiles' attack on the Shimura-Taniyama
conjecture.
Written in a clear, lively and engaging style, this paper is
ideal for
the cultiva layperson who wishes to be introduced to the
fascinating ideas that may lead to a solution of number
theory's most
famous unsolved problem.
===
Subject: Re: Wiles' Proof
> Gouv.90a, Fernando Q.
A marvelous proof.
> Amer. Math. Monthly 101 (1994), no. 3, 203--222.
/David
===
Subject: Re: Wiles' Proof
Gouv.90a, Fernando Q.
> A marvelous proof.
> Amer. Math. Monthly 101 (1994), no. 3, 203--222.
>
Or, Mathematical Association of America sells back issues.
Maybe Gouvea has it on his web site.
===
Subject: Re: Wiles' Proof
Sure.
> Or, Mathematical Association of America sells back issues.
Okay.
> Maybe Gouvea has it on his web site.
Who?
/David
===
Subject: Re: Wiles' Proof
> Maybe Gouvea has it on his web site.
> Who?
> /David
http://www.colby.edu/personal/f/fqgouvea/
===
Subject: Re: Wiles' Proof
>Sure.
>> Or, Mathematical Association of America sells back issues.
>Okay.
>> Maybe Gouvea has it on his web site.
>Who?
Lee Rudolph
===
Subject: How to prove this?!?!
Any ideas on this?
Prove that:
int{[int[(int(f(t) dt, 0,u),0,v]du],0,x} dv=
(1/2)*int((x-t)^2*f(t) dt,
0,x)
===
Subject: Re: How to prove this?!?!
> Any ideas on this?
> Prove that:
> int{[int[(int(f(t) dt, 0,u),0,v]du],0,x} dv=
(1/2)*int((x-t)^2*f(t) dt,
> 0,x)
Try using integration by parts on the right side to get rid of
those
(x-t)'s.
Have a tolerable existence. Eli
===
Subject: Re: Damped harmonic oscillation
Thanks to both of you,
I don't think the proposals solves the problem or maybe I
don't understand
the correctly.
The oscillation is caused by a non specific up/down motion of
the of the
spring, performed by a person.
I can measure the up/down distance, speed and acceleration and
know all
parametres in the formula but C and phi.
I gess that both C and phi in some way must be derived from the
accelaration.
My idea was to in steps of delta t = t[n]-t[n-1] measure the
distance,
speed
and acceleration and then calculate C and phi, when finally
should be
inser in d(t) = C*[e^(-t/tau)]*cos(o*t - phi).
Best regards
Torben W. Hansen
Denmark
===
Subject: Structure of Frobenius Complement
I have read that Burnside was able to prove that the Frobenius
complement of a Frobenius group had the property that any Sylow
p-subgroup of the group had a unique subgroup of order p.
Therefore,
any Sylow p-subgroup of such a group is cyclic unless p=2 and
the
Sylow 2-subgroup is a generalized quaternion group. Can anyone
give
me a reference (preferably in English) on how Burnside did
this?
---- David
===
Subject: Re: Structure of Frobenius Complement
>I have read that Burnside was able to prove that the Frobenius
>complement of a Frobenius group had the property that any
Sylow
>p-subgroup of the group had a unique subgroup of order p.
Therefore,
>any Sylow p-subgroup of such a group is cyclic unless p=2 and
the
>Sylow 2-subgroup is a generalized quaternion group. Can
anyone give
>me a reference (preferably in English) on how Burnside did
this?
I don't know how it compares with Burnside's proof, but there
is a proof
on page 86 of the book John Dixon & Brian Mortimer,
Permutation Groups
(Springer, 1996)
Derek Holt.
===
Subject: Graduate schools that teach how to teach
I am considering applying to grad school again soon. What I am
looking for is a school that not only teaches mathematics, but
teaches
its students how to become teachers. It also would help if the
department fully funds all or most of its students, as I have
no
desire to increase my student loan debt.
I am interes in becoming a professor of mathematics at a
teaching-orien school. I'm not saying research is unimportant,
but
I would rather be known as a great teacher than a great
researcher.
If anyone has some recommendations as to schools that might
meet my
criteria, please post or privately email me. If I receive any
responses in private email that don't duplicate what's pos,
I'll
summarize those as well.
Thanks!
Subject: Re: Graduate schools that teach how to teach
===
> I am considering applying to grad school again soon. What I
am
> looking for is a school that not only teaches mathematics,
but teaches
> its students how to become teachers. It also would help if
the
> department fully funds all or most of its students, as I
have no
> desire to increase my student loan debt.
> I am interes in becoming a professor of mathematics at a
> teaching-orien school. I'm not saying research is
unimportant, but
> I would rather be known as a great teacher than a great
researcher.
> If anyone has some recommendations as to schools that might
meet my
> criteria, please post or privately email me. If I receive any
> responses in private email that don't duplicate what's pos,
I'll
> summarize those as well.
> Thanks!
Look for schools with graduate degrees in the teaching of
mathematics.
You may be wanting to combine degrees in mathematics with
education. I
suspect that most schools that offer graduate math programs
will have
something in line with what you want.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Graduate schools that teach how to teach
> I am considering applying to grad school again soon. What I
am
> looking for is a school that not only teaches mathematics,
but teaches
> its students how to become teachers. It also would help if
the
> department fully funds all or most of its students, as I
have no
> desire to increase my student loan debt.
> I am interes in becoming a professor of mathematics at a
> teaching-orien school. I'm not saying research is
unimportant, but
> I would rather be known as a great teacher than a great
researcher.
> If anyone has some recommendations as to schools that might
meet my
> criteria, please post or privately email me. If I receive any
> responses in private email that don't duplicate what's pos,
I'll
> summarize those as well.
> Thanks!
A collateral question and approach might be to locate
teaching-orien schools and to ask those math departments if
they
know schools that could meet your needs.
David Ames
===
Subject: prove it
n^4+n^2=6 (mod 7) for all n
Subject: Re: prove it
===
> n^4+n^2=6 (mod 7) for all n
try it for n=0
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: prove it
> n^4+n^2 == 6 (mod 7) for all n
False for integers in 3 of the 7 conguence classes mod 7,
namely
false for n == 0 mod 7, n == 1 mod 7 and n == 6 mod 7.
Could one say that it is just barely more true than false (4
true
equivalence classes to 3 false ones)?
===
Subject: Re: prove it
In sci.math, Euler
<pro272@hotmail.com>
<e64d6097.0310300131.3097b30d@posting.google.com>:
> n^4+n^2=6 (mod 7) for all n
Counterexample: n = 0
n^4+n^2=0
Counterexample: n = 1
n^4+n^2=2
Works: n = 2
n^4+n^2=16+4=20=6 (mod7)
Works: n = 3
n^4+n^2=81+9=90=6 (mod7)
Works: n = 4
n^4+n^2=272=6 (mod7)
Works: n = 5
n^4+n^2=650=6 (mod7)
Counterexample: n = 6 or -1
n^4+n^2=1332=2 (mod7)
Well, 4 out of 7 isn't too bad.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: prove it
> n^4+n^2=6 (mod 7) for all n
Interesting: then x^4 + x^2 - 6 = 0 would be a degree 4
equation
with 7 roots in the field F_7. :-(
--
===
Subject: Re: prove it
> n^4+n^2=6 (mod 7) for all n
Not true. If n=0 (mod 7) then n^4+n^2=0 (mod 7) If n=1 or n=5
(mod 7) then
n^4+n^2=2 (mod 7)
Have a tolerable existence. Eli
===
Subject: Re: prove it
Eli <ecooper@mathstat.umass.edu> escribi.97 en el mensaje
>> n^4+n^2=6 (mod 7) for all n
> Not true. If n=0 (mod 7) then n^4+n^2=0 (mod 7) If n=1 or
n=5 (mod
> 7) then n^4+n^2=2 (mod 7)
> Have a tolerable existence. Eli
You meant if n = 1 or n = 6 = -1 (mod 7), n^4 + n^2 = 2 (mod
7).
And obviously, if n = 0 (mod 7), also n^4 + n^2.
Only for n = 2, 3, 4 or 5, yes, n^4 + n^2 = 6 (mod 7)
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: prove it
> You meant if n = 1 or n = 6 = -1 (mod 7), n^4 + n^2 = 2 (mod
7).
Yes, that is what I meant.
Have a tolerable existence. Eli
===
Subject: Re: prove it
Euler
> n^4+n^2=6 (mod 7) for all n
Something's wrong, Leonhard old chap. A fourth-order
polynomial can have at
most four
zeros (not seven) modulo a prime.
LH
===
Subject: Re: Was: Convergence on a space with no topology
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9TDN4X13238;
>Subject: Was: Convergence on a space with no topology
>Convergence on a space with no topology got cluttered
relatively
>fast. In my opinion, it was getting hard for anyone reading
the
>message for the first time to follow. Therefore, I try to
give the
>main points of what was said some sort of chronological order.
>With all that feedback, you need present revision.
>Skip the blah blah and just get to the punch line.
Ok, but first, is the weather so damn cold where you are, too?
I mean, I almost froze my fingers off a few days ago on
the way to the university...
>Are you reinventing the Hausdorff metric for sets?
>Do you know what the Hausdorff metric for sets is?
For time reasons, I was only able to address the second part
of your message in my last reply.
The Hausdorff metric begins by defining a neighborhood:
if A subset X and r > 0, N_r(A) = {y | d(x,y) < r for some x
in A)
and then procedes to define a distance
function h as:
h(A,B) = inf { r > 0: A subset N_r(B) and B subset N_r(A) }
which has the propeties of a metric if the sets involved are
closed /compact.
Since I have, in no way, defined an open set or neighborhood
- much less a distance function- how can I have reinven
the Hausdorff metric?
The only way I currently see to interpret this question
properly
is:
If A_n -> A, does it not follow that
lim n -> infty h(A_n, A) where h is the Hausdorff
metric and the sets involved are closed / compact, etc.?
(I presume this to be valid on a hunch and would appreciate
any proofs, as I have not been able to prove it yet.)
This because the converse (from lim n -> infty h(A_n, A)
follows
A_n -> A) certainly does not appear to
be true: If A_n -> A then, as was previously poin
out, A_n would converge to all subsets of A, too, were it not
for the additional requirement A_n subset A for all n
(or something similar like: there exists an m such that
A_n subset A forall n > m).
The following does appear to be equivilant...
1. A_n -> A
2. A_n subset A forall n and for all p > 0, p in R for all a
in A
exists q in N for all n > q exists b in A_n: d(a,b) < p
A stricter form of convergence:
1. A_n ->* A
2. A_n subset A forall n and for all p > 0, p in R exists q in
N for
all a in A for all n > q exists b in A_n: d(a,b) < p
To me, the last definition is interesting because it
follows (trivially) from it that if A_n ->* A, then A_n is a
*Cauchy-sequence (if A_n subset A for all n, then we can
obviously
substitute A_m for A in the above condition for all m > q).
Lastly, your question about B*...
Why drop it?
B* was put in to emphasize that we may not always want to
consider
P(X), the set of all subsets of X -which may be insanely
large-,
but some subset P(X) (such as the set of all Cauchy-sequences
in X
or the set of all single points of X, etc.).
===
Subject: Re: Was: Convergence on a space with no topology
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9TDNO913283;
[snip]
>Skip the blah blah and just get to the punch line.
Ok, but first, is the weather so damn cold where you are, too?
I mean, I almost froze my fingers off a few days ago on
the way to the university...
>Are you reinventing the Hausdorff metric for sets?
>Do you know what the Hausdorff metric for sets is?
For time reasons, I was only able to address the second part
of your message in my last reply.
The Hausdorff metric begins by defining a neighborhood:
if A subset X and r > 0, N_r(A) = {y | d(x,y) < r for some x
in A)
and then procedes to define a distance
function h as:
h(A,B) = inf { r > 0: A subset N_r(B) and B subset N_r(A) }
which has the propeties of a metric if the sets involved are
closed /compact.
Since I have, in no way, defined an open set or neighborhood
- much less a distance function- how can I have reinven
the Hausdorff metric?
The only way I currently see to interpret this question
properly
is:
If A_n -> A, does it not follow that
lim n -> infty h(A_n, A)=0 where h is the Hausdorff
metric and the sets involved are closed / compact, etc.?
(I presume this to be valid on a hunch and would appreciate
any proofs, as I have not been able to prove it yet.)
This because the converse (from lim n -> infty h(A_n, A)=0
follows
A_n -> A) certainly does not appear to
be true: If A_n -> A then, as was previously poin
out, A_n would converge to all subsets of A, too, were it not
for the additional requirement A_n subset A for all n
(or something similar like: there exists an m such that
A_n subset A forall n > m).
The following does appear to be equivilant...
1. A_n -> A
2. A_n subset A forall n and for all p > 0, p in R for all a
in A
exists q in N for all n > q exists b in A_n: d(a,b) < p
A stricter form of convergence:
1. A_n ->* A
2. A_n subset A forall n and for all p > 0, p in R exists q in
N for
all a in A for all n > q exists b in A_n: d(a,b) < p
To me, the last definition is interesting because it
follows (trivially) from it that if A_n ->* A, then A_n is a
*Cauchy-sequence (if A_n subset A for all n, then we can
obviously
substitute A_m for A in the above condition for all m > q).
Lastly, your question about B*...
Why drop it?
B* was put in to emphasize that we may not always want to
consider
P(X), the set of all subsets of X -which may be insanely
large-,
but some subset P(X) (such as the set of all Cauchy-sequences
in X
or the set of all single points of X, etc.).
Your message-attacks are getting more and more personal...
>How about studying up on the Hausdorff metric? It's a metric
for P(S),
>based upon the metric for S. Much like what your attempting.
So get
>smart, instead of just stumbling around, find out what others
have already
>complished decades ago that have bearing on your attempts.
Hhhm... if I get smart, to that mean I become more and more
like you?
[snip]
>> If (X,d) is a space whose metric is induced by a norm,
>> then X must be a vector space- since a norm is only defined
>> on a vector space (nevertheless, sorry for not explicitly
stating this).
>Baloney, you said metric space and a norm can be defined for
a group
>making it a topological group or conversely certain metrics
for a
>topological group can induce or be given norms. So if you
want vector
>space, say you want a vector space, a normed vector space or
even an inner
>product space.
>> But your counterexample space {0,1} is not a vector space.
>> Indeed, if you look at my proof of the proposition, you
>> will see that it relies on (t_n a) also being in the space
where
>> t_n > 0 is a scalar...
>It's not? It's a one dimensional vector space over Z_2, that
is with
>scalars integers modulus 2. Yes, Z_2 is a field, a finite
field. As it
>has the discrete topology, it can be given a discrete metric
such as
> d(0,1) = d(1,0) = 1, d(0,0) = d(1,1) = 0.
>For a norm, |0| = 0, |1| = 1 will do.
Ok you got me, in my short career (as a physics student who has
taken the last two years off on fatherly leave)
I have never heard of a normed space defined over a finite
scalar field.
Indeed, sob sob, I have only seen K=R or K=C. Although I
feel very aware of the possibility of definining abstract
vector spaces over
other more general entities. Especially since I have
looked at things like this myself- mainly playing around with
the
quaternions.
C. Dement
Subject: Re: Was: Convergence on a space with no topology
===
Subject: Re: Was: Convergence on a space with no topology
>>Skip the blah blah and just get to the punch line.
>Ok, but first, is the weather so damn cold where you are, too?
>I mean, I almost froze my fingers off a few days ago on
>the way to the university...
How affable. Portland having wonderful greenhouse warming
weather, warm,
sunny, some rain and just today tempeture drop 50-40's. Predic
40-20's. Where's you? BTW, if you'se going @mathform, why such
a
double bland name?
>The Hausdorff metric begins by defining a neighborhood:
>if A subset X and r > 0, N_r(A) = {y | d(x,y) < r for some x
in A)
>and then procedes to define a distance function h as:
>h(A,B) = inf { r > 0: A subset N_r(B) and B subset N_r(A) }
>which has the propeties of a metric if the sets involved are
>closed /compact.
I've got, which as I recall is laborously equivalent to yours:
dh(A,B) = max(sup{ d(a,B) | a in A }, sup{ d(b,A) | b in B })
metric for nonnul closed bounded subsets of metric space (S,d)
>Since I have, in no way, defined an open set or neighborhood
>- much less a distance function- how can I have reinven
>the Hausdorff metric?
You're using a metric here.
>Let (X, d) be a space whose metric is induced by a norm.
>Call (A_n) subset B* subset P(x), the set of all subsets of X,
>a *Cauchy sequence if (R = reals, N = natural numbers)
>For all p > 0, p in R, exists q in N for all n,m > q for all
>a in A_n exists b in A_m: d(a,b) < p.
>The only way I currently see to interpret this question
properly is:
>If A_n -> A, does it not follow that
>lim n -> infty h(A_n, A)=0 where h is the Hausdorff
>metric and the sets involved are closed / compact, etc.?
That's how a metric is supposed to work.
For compact sets I'd expect dh(A,B) = 0 ==> A = B
I recall a set theory definition of A_n -> A like
A = /{ { / Aj | j > k } | k in N }
= /{ { / Aj | j > k } | k in N }
provided both are equal. The limsup, liminf of sets?
>(I presume this to be valid on a hunch and would appreciate
>any proofs, as I have not been able to prove it yet.)
So far you've just an intuitive notion of A_n -> A, nothing
defined.
>This because the converse (from lim n -> infty h(A_n, A)=0
follows
>A_n -> A) certainly does not appear to be true: If A_n -> A
then,
>as was previously poin out, A_n would converge to all subsets
of
>A, too, were it not for the additional requirement A_n subset
A for
>all n (or something similar like: there exists an m such that
>A_n subset A forall n > m).
>The following does appear to be equivilant...
>1. A_n -> A
>2. A_n subset A forall n and for all p > 0, p in R for all a
in A
>exists q in N for all n > q exists b in A_n: d(a,b) < p
Rather limi, that A_n has to approach from below.
A dual, approach from above definition, wants to ensue.
>A stricter form of convergence:
'uniform convergence'?
>1. A_n ->* A
>2. A_n subset A forall n and for all p > 0, p in R exists
>q in N for all a in A for all n > q exists b in A_n: d(a,b) <
p
>To me, the last definition is interesting because it
>follows (trivially) from it that if A_n ->* A, then A_n is a
>*Cauchy-sequence (if A_n subset A for all n, then we can
obviously
>substitute A_m for A in the above condition for all m > q).
>Lastly, your question about B*...
>Why drop it?
It wasn't doing anything, wasn't given any meaning and did
nothing
except briefly get in the way of understanding the definition.
>B* was put in to emphasize that we may not always want to
consider
>P(X), the set of all subsets of X -which may be insanely
large-,
>but some subset P(X) (such as the set of all Cauchy-sequences
in X
>or the set of all single points of X, etc.).
If B* is the be the set of all Cauchy sequences, then it
needs not be in the definition of Cauchy sequence.
----
===
Subject: Re: Pi Calculation ?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9U0ogP16809;
>One way to calculate pi is using the formula for ATAN:
> atan(x) = x^1/1 - x^3/3 + x^5/5 - x^7/7 + x^9/9 ...
I'm working on a targetting project on a robot controlled by a
microcontroller. It has approximately 104 steps per
revolution. Using
integer math only, how might I find atan(), returned as
104-ians instead
of radians or degrees?
- Kipp
===
Subject: Re: Pi Calculation ?
>One way to calculate pi is using the formula for ATAN:
atan(x) = x^1/1 - x^3/3 + x^5/5 - x^7/7 + x^9/9 ...
> I'm working on a targetting project on a robot controlled by
a
> microcontroller. It has approximately 104 steps per
revolution.
> Using integer math only, how might I find atan(), returned as
104-ians instead of radians or degrees?
Lookup table with 104 entries. Precalculate the tables
at bootup time, then just do a search through the
tables when you want to find the angle value.
You'll need a little bit of logic to sort out
which quadrant you're in.
- Randy
===
Subject: Re: Very interesting problem on Real Analysis
> You mean this one, that someone repos just to get it into the
> relevant folder in his newsreader?
>Subject: Re: IT IS A POLYNOMIAL
>Distribution: world
<1998050216380100.MAA06675@ladder01.news.aol.com>
Those links gave me trouble, but here it is again:
http://www.math.niu.edu/~rusin/known-math/98/f.is.poly
I'm still trying to do it with harmonic analysis. Vaguely:
By a change of variable we can switch to a function g on the
interval
[-1,1]. Write
g(x)=sum_n k_n H_n(x) all xin [-1,1]
where the k_n are real constants and the H_n are the Hermite
polynomials,
which form a complete orthogonal system for all the continuous
functions on
[-1,1]. Termwise differentiation is legitimate, and moreover
the derivative
of H_(n+1) is H_n times a constant, which looks convenient.
But I'm not
sure
this will really go anywhere.
Very interesting problem, that's for sure.
LH
===
Subject: Re: Very interesting problem on Real Analysis
Originator: grubb@lola
>Prove that f is polynomial in every component of
V=Union(int(Cn)) and that
>UV has no isola points.
>Then show that UV is empty (or else you'll get a
contradiction).
How do you prove that this is empty? I can see that f has to
be a
polynomial on some interval. It is also easy to see that it is
a polynomial on each of a collection of intervals whose union
has nowhere dense complement. Now what?
--Dan Grubb
===
Subject: Re: Very interesting problem on Real Analysis
>>Prove that f is polynomial in every component of
V=Union(int(Cn)) and
that
>>UV has no isola points.
>>Then show that UV is empty (or else you'll get a
contradiction).
>How do you prove that this is empty? I can see that f has to
be a
>polynomial on some interval. It is also easy to see that it is
>a polynomial on each of a collection of intervals whose union
>has nowhere dense complement. Now what?
He made a later post which was much more complete.
See also the posts by Dor and Israel.
>--Dan Grubb
************************
===
Subject: Re: Very interesting problem on Real Analysis
Originator: grubb@lola
>Prove that f is polynomial in every component of
V=Union(int(Cn)) and
that
>UV has no isola points.
>Then show that UV is empty (or else you'll get a
contradiction).
>>How do you prove that this is empty? I can see that f has to
be a
>>polynomial on some interval. It is also easy to see that it
is
>>a polynomial on each of a collection of intervals whose union
>>has nowhere dense complement. Now what?
>He made a later post which was much more complete.
>See also the posts by Dor and Israel.
Thanks. I saw those *after* I pos this. That'll teach me.
---Dan Grubb
===
Subject: Harmonic oscillation
Dear Ladies and Gentlemen,
Im still and desperately searching for a function of either
velocity or
accelaration ( d(t) = f(v)
or/and d(t) = f(a)), which expresses the damped oscillation
distance d(t)
caused by
varying velocity or accelaration.
I found this formula, which express a mass m in [kg] hanging
in a spring k
in [N/m],
d(t) = C*[e^(-t/tau)]*cos(o*t - phi)
where:
d(t) = distance in [m] to the time t in [s],
C = distance different from neutral position in [m],
o = SQRT(kg/m), resonance frequency in [rad/s]
tau = time constant for damping oscillation
phi = initial phase
The oscillation is caused by a non specific up/down motion of
the of the
spring, performed by a person.
I can measure the up/down distance, speed and acceleration and
I know all
parametres in the formula but C and phi.
I gess that both C and phi in some way must be derived from the
acceleration.
My idea was to measure the distance, speed and acceleration in
steps of
delta t = t[n]-t[n-1] and then step by step calculate C and
phi to be
inser in :
d(t) = C*[e^(-t/tau)]*cos(o*t - phi) where
where: d(t)=f(t)/k, tau=2*m/b, o=SQRT(k/m)
and as far as I know it should be the solution to the
differential
equation
m*d'' + b*d' +k*d = f(t) also mentioned by Anselm
Proschniewski in an
earlier thread.
where: m=mass, b=damping, k=stiffness, f(t)=force,
d=displacement,
d'=speed,
d=acceleration.
I'm not sharp in differential-, integral equations, but have
some basic
knowledge about this and Laplace Transformation. I also know a
part of the
laws in physics as
f=m*a, E=1/2*m*v^2, Hooks law etc, etc... but I can't solve my
problem
Further more I use a math tool from Texas Instrument called
DERIVE 5.06.
What do I miss ?
Best regards
Torben W. Hansen
Denmark
===
Subject: Re: Harmonic oscillation
X-ID: Z6E3SmZJoenaviHY2PBmV1g-4eyaxzB7o1ynPkKmnACpgKiMmygo0w
Torben W. Hansen schrieb:
> Dear Ladies and Gentlemen,
> Im still and desperately searching for a function of either
velocity or
> accelaration ( d(t) = f(v)
> or/and d(t) = f(a)), which expresses the damped oscillation
distance
d(t)
> caused by
> varying velocity or accelaration.
If the frequency is fixed you seem to need a least squares
fit. Take the
functions
d(t)=c0+e^(-t/tau)*(c1*cos(o*t)+c2*cos(o*t))
see, e.g. numerical recipes in C (or your favourite prog.
lang.). phi
is
something like atan2(c1,c2). If freq / damping is to be
determined, the
parameter optimization gets nonlinear. I assume that you have
measured d at
several time offsets?
hth
Klaus
> I found this formula, which express a mass m in [kg] hanging
in a spring
k
> in [N/m],
> d(t) = C*[e^(-t/tau)]*cos(o*t - phi)
> where:
> d(t) = distance in [m] to the time t in [s],
> C = distance different from neutral position in [m],
> o = SQRT(kg/m), resonance frequency in [rad/s]
> tau = time constant for damping oscillation
> phi = initial phase
> The oscillation is caused by a non specific up/down motion
of the of the
> spring, performed by a person.
> I can measure the up/down distance, speed and acceleration
and I know all
> parametres in the formula but C and phi.
> I gess that both C and phi in some way must be derived from
the
> acceleration.
> My idea was to measure the distance, speed and acceleration
in steps of
> delta t = t[n]-t[n-1] and then step by step calculate C and
phi to be
> inser in :
> d(t) = C*[e^(-t/tau)]*cos(o*t - phi) where
> where: d(t)=f(t)/k, tau=2*m/b, o=SQRT(k/m)
> and as far as I know it should be the solution to the
differential
equation
> m*d'' + b*d' +k*d = f(t) also mentioned by Anselm
Proschniewski in an
> earlier thread.
> where: m=mass, b=damping, k=stiffness, f(t)=force,
d=displacement,
d'=speed,
> d=acceleration.
> I'm not sharp in differential-, integral equations, but have
some basic
> knowledge about this and Laplace Transformation. I also know
a part of
the
> laws in physics as
> f=m*a, E=1/2*m*v^2, Hooks law etc, etc... but I can't solve
my problem
> Further more I use a math tool from Texas Instrument called
DERIVE 5.06.
> What do I miss ?
> Best regards
> Torben W. Hansen
> Denmark
===
Subject: Random Number Generator
Is it possible to make an 'ideal' random number generator ? By
'ideal'
I mean that its output will be always random irrespective of
the way
in which its output is sampled..
Jean
===
Subject: Re: Random Number Generator
> Is it possible to make an 'ideal' random number generator ?
By 'ideal'
> I mean that its output will be always random irrespective of
the way
> in which its output is sampled..
> Jean
Early on (the 40s and 50s) there was some talk of embedding a
physical
process
inside the computer to generate random numbers. They were
talking about
vacuum
tubes, but if we think of a radioactive rock the principle
would be the
same.
The rate of radioactive emission would control the generation
of random
numbers,
Bill
===
Subject: Re: Random Number Generator
> Is it possible to make an 'ideal' random number generator ?
By 'ideal'
> I mean that its output will be always random irrespective of
the way
> in which its output is sampled..
> Jean
A *finite* sequence of numbers will never be truly random.
-Michael.
===
Subject: scaling data
Hi all,
Could anyone tell me how to (linearly) rescale a data set on
the interval
[0, 1] to
[0, 0.1]? I've forgotten how to do this.
More generally, given a data set on the interval [a, b], is
there a formula
to rescale it to the interval [x, y]?
Thanks in advance,
--
_|//_
( O-O )
---------------------------o00--(_)--00o----------------------
--------
Colm G. Connolly |
Department of Computer Science |
University College Dublin (UCD) |
Belfield, Dublin 4 |
.83ire / Republic of Ireland |
===
Subject: Re: scaling data
t is in [a,b]
t-a is in [0,b-a]
(t-a)/(b-a) is in [0,1]
(t-a)*(x-y)/(b-a) is in [0,y-x]
x+(t-a)*(x-y)/(b-a) is in [x,y]
f(t)=x+(t-a)*(x-y)/(b-a) is an affine function with f(a)=x and
f(b)=y. Cool
-gs-
> Hi all,
> Could anyone tell me how to (linearly) rescale a data set on
the interval
> [0, 1] to
> [0, 0.1]? I've forgotten how to do this.
> More generally, given a data set on the interval [a, b], is
there a
formula
> to rescale it to the interval [x, y]?
> Thanks in advance,
> --
> _|//_
> ( O-O )
>
---------------------------o00--(_)--00o----------------------
--------
> Colm G. Connolly |
> Department of Computer Science |
> University College Dublin (UCD) |
> Belfield, Dublin 4 |
> .83ire / Republic of Ireland |
===
Subject: Re: scaling data
Thanks a million I see now how to do it, but don't you mean:
t in [a, b]
t-a in [a, b-a]
(t-a)/(b-a) in [0, 1]
((t-a)/(b-a))/(b-a) in [0, y-x]
x + ((t-a)/(b-a))/(b-a) [x, y]
f(t)=x+(t-a)*(x-y)/(b-a) with f(a)=x and f(b)=y.
> t is in [a,b]
> t-a is in [0,b-a]
> (t-a)/(b-a) is in [0,1]
> (t-a)*(x-y)/(b-a) is in [0,y-x]
> x+(t-a)*(x-y)/(b-a) is in [x,y]
> f(t)=x+(t-a)*(x-y)/(b-a) is an affine function with f(a)=x
and f(b)=y.
> Cool
> -gs-
>> Hi all,
>> Could anyone tell me how to (linearly) rescale a data set
on the
interval
>> [0, 1] to
>> [0, 0.1]? I've forgotten how to do this.
>> More generally, given a data set on the interval [a, b], is
there a
> formula
>> to rescale it to the interval [x, y]?
>> Thanks in advance,
>> --
>> _|//_
>> ( O-O )
>>
---------------------------o00--(_)--00o----------------------
--------
>> Colm G. Connolly |
>> Department of Computer Science |
>> University College Dublin (UCD) |
>> Belfield, Dublin 4 |
>> .83ire / Republic of Ireland |
--
_|//_
( O-O )
---------------------------o00--(_)--00o----------------------
--------
Colm G. Connolly |
Department of Computer Science |
University College Dublin (UCD) |
Belfield, Dublin 4 |
.83ire / Republic of Ireland |
===
Subject: Fundamental Theorem of Calculus problem
Why is the derivative of
f(x) = integral (from x to 0) of [cos(xt)/t]dt
not equal to -cos(x^2)/x ?
Switch the order of integration to make integral negative, and
use fund.
theorem of calculus :
d/dx (integral from a to x of [f(t)dt]) = f(x)
John
===
Subject: Re: Fundamental Theorem of Calculus problem
> Why is the derivative of
> f(x) = integral (from x to 0) of [cos(xt)/t]dt
> not equal to -cos(x^2)/x ?
Because:
(1) the integrand cos(xt)/t depends on x;
(2) the integrand becomes unbounded as t -> 0 and
the improper integral diverges.
In neither of these situtations does the fundamental theorem
of calculus
as you have sta it apply.
> Switch the order of integration to make integral negative,
and use fund.
> theorem of calculus :
> d/dx (integral from a to x of [f(t)dt]) = f(x)
> John
--
P.A.C. Smith
'If the Apocalypse comes, beep me.' <*>
http://www.srcf.ucam.org/~pas51
===
Subject: Difference equation resource -- help
Can anyone tell me how to solve a second order difference
equation(DE)
such as
A(n,m)+A(n-1,m)+2*A(n,m-1)-4*A(n-1,m-1)=1
where f(m,n) is known and b,c,d are constant.
Or, show me some online resources discussing multi-variable
DE's?
I suspect it's closely rela to partial differential
equation(PDE)
theories, but *how*?
===
Subject: Re: Difference equation resource -- help
what's f(m,n)?
can also often do separation of variables like in PDE.
A(m,n)=M(m)*N(n) if bc allow
> Can anyone tell me how to solve a second order difference
equation(DE)
> such as
> A(n,m)+A(n-1,m)+2*A(n,m-1)-4*A(n-1,m-1)=1
> where f(m,n) is known and b,c,d are constant.
> Or, show me some online resources discussing multi-variable
DE's?
> I suspect it's closely rela to partial differential
equation(PDE)
> theories, but *how*?
===
Subject: Re: Difference equation resource -- help
> Can anyone tell me how to solve a second order difference
equation(DE)
> such as
> A(n,m)+A(n-1,m)+2*A(n,m-1)-4*A(n-1,m-1)=1
> where f(m,n) is known and b,c,d are constant.
> Or, show me some online resources discussing multi-variable
DE's?
> I suspect it's closely rela to partial differential
equation(PDE)
> theories, but *how*?
Since your equation is linear, you should have good chances.
Start by
considering the homogeneous equation (0 on right hand side).
Insert A(n,m)
=
alpha^n * beta^m, and solve for alpha and beta. Next generate
linear
combinations of these.
You still need to consider boundary conditions, though.
-Michael.
===
Subject: is there a function satisfying this
is there a function
u: N -> N
so that
u(i) + u(j) = even if i <> j
u(i) + u(j) = odd if i = j
?
===
Subject: Re: is there a function satisfying this
> is there a function
> u: N -> N
> so that
> u(i) + u(j) = even if i <> j
> u(i) + u(j) = odd if i = j
Let's see... 1=1, so u(1) + u(1) is odd, but this is 2 times
u(1), even.
No.
===
Subject: Re: is there a function satisfying this
and what if the function is changed into
u: N -> R
?
===
Subject: Re: is there a function satisfying this
>and what if the function is changed into
>u: N -> R
>?
i.e. 2 u(i) is an odd integer, and
u(i) + u(j) is an even integer if i <> j.
Well, if u(i) = x/2 and u(j) = y/2, then (x+y)/2 is an even
integer iff
x+y == 0 mod 4.
Of the odd integers 2 u(1), 2 u(2) and 2 u(3), at least two
are congruent
mod 4; if these are 2 u(i) and 2 u(j), then 2 (u(i)+u(j)) is
congruent to
2 mod 4, and u(i)+u(j) is odd.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: is there a function satisfying this
> and what if the function is changed into
> u: N -> R
> ?
and what is meant by an even real number?
===
Subject: Re: Applying to Grad school (GRE verbal?)
......................
>In reality, I suspect that what is given the MOST weight in
the final
>decision is not the GRE test, but rather the coursework taken
and the
>letters of recommendation. Particularly the latter. As I
recall from
>my own application experience, the GRE test was more a filter
than a
>decision maker: in order for your application to be
considered, you
>should normally score at least this much on each part; and
then it was
>ignored in favor of letters of rec. and so on.
I have seen thousands of applications for graduate students
in statistics, and a fair number in mathematics. The problem
at this time is that there is NO good information, and that
includes the GRE.
The grades in the important undergraduate courses, abstract
algebra and foundations of analysis, are often not given in
a meaningful manner, and it is even the case that the important
parts were not taught; teaching to the level of the bodies in
the classroom, whether they should have even gotten near that
level, has reached this far, and even into graduate courses.
--
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: Applying to Grad school (GRE verbal?)
>Dear all,
>I don't know how many can comment, but I am applying to Math
grad school
and
>was wondering how important the general GRE test is when
applying to the
top
>math schools. In particular, the Verbal section. I am an
American
student,
>and this section still troubles me...The rest of my
application is very
>good. I expect to get in the low to mid 500's on the
verbal...(800 quant
>and 4-5 on the writing section out of 6).
John
It depends on each department. Each has its own admission
criteria.
My experience, after many years in the business (30+),
is that the verbal part s a better predector of grad. math
success
than the other section.
Another excellent predictor of math success is a foreign
language
aptitude test !!!
===
Subject: Re: Applying to Grad school (GRE verbal?)
Thanks for the brilliant insight.
>Dear all,
I don't know how many can comment, but I am applying to Math
grad school
and
>was wondering how important the general GRE test is when
applying to the
top
>math schools. In particular, the Verbal section. I am an
American
student,
>and this section still troubles me...The rest of my
application is very
>good. I expect to get in the low to mid 500's on the
verbal...(800
quant
>and 4-5 on the writing section out of 6).

John
>
It depends on each department. Each has its own admission
criteria.
> My experience, after many years in the business (30+),
> is that the verbal part s a better predector of grad. math
success
> than the other section.
> Another excellent predictor of math success is a foreign
language
> aptitude test !!!
===
Subject: Re: Applying to Grad school (GRE verbal?)
> Dear all,
> I don't know how many can comment, but I am applying to Math
grad school
and
> was wondering how important the general GRE test is when
applying to the
top
> math schools. In particular, the Verbal section. I am an
American
student,
> and this section still troubles me...The rest of my
application is very
> good. I expect to get in the low to mid 500's on the
verbal...(800 quant
> and 4-5 on the writing section out of 6).
the subject test is key.
 John
===
Subject: Condition on commuting matrices, and Lyapunov equation
Dear all,
Do you know what is the characterization of matrices X, Y such
that XY=YX?
Suppose X is given? What can you say about Y in general?
Regarding a version of Lyapunov matrix equation AX=XB, A, B
are known to
have block partitioned structure A=[a b1; 0 c], B=[a b2; 0 c],
can you
name
an easy way to solve for X, with the constraint X=[I 0; x1 x2]
? Do you
just
dispose it entry by entry and look at it a system of equations?
In general, is there any known result of solving Lyapunov
equation with
constraint?
Thank you all,
B. C.
===
Subject: Re: Condition on commuting matrices, and Lyapunov
equation
>Do you know what is the characterization of matrices X, Y
such that XY=YX?
>Suppose X is given? What can you say about Y in general?
These are both square matrices of the same size, say n x n.
Note that for any invertible n x n matrix S, SXS^(-1) and
SYS^(-1)
commute iff X and Y do. So wlog we may assume X is in Jordan
Canonical
Form.
Now since (X-rI)^k and Y commute, Y must leave invariant the
generalized
eigenspaces {x: (X-rI)^k x = 0} for eigenvalues r of X and
positive
integers k. Consider the sets of indices B_1 = {i_1 ... i_1 +
k_1} and
B_2 = {i_2 ... i_2 + k_2} corresponding to two (not necessarily
distinct) Jordan blocks. If the blocks are for different
eigenvalues,
then Y_{ij} = 0 for i in B_1 and j in B_2. If they are for the
same
eigenvalue r, then the (i_1 + j,i_2 + k) entry of Y is 0 if k
< j,
is equal to the (i_1+j-1,i_2+k-1) entry if 1 <= j <= k <=
min(k_1,k_2),
and otherwise is arbitrary.
>Regarding a version of Lyapunov matrix equation AX=XB, A, B
are known to
>have block partitioned structure A=[a b1; 0 c], B=[a b2; 0
c], can you
name
>an easy way to solve for X, with the constraint X=[I 0; x1
x2] ? Do you
just
>dispose it entry by entry and look at it a system of
equations?
I get
[ b1 x1, b1 x2 - b2 ]
AX - XB = [ c x1- x1 a, c x2 - x1 b2 - x2 c ]
If b1 is invertible, then x1 = 0, x2 = b1^(-1) b2, and this x2
must
commute with c. If b1 is not invertible I guess it will be more
complica.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Condition on commuting matrices, and Lyapunov
equation
>Do you know what is the characterization of matrices X, Y
such that
XY=YX?
>Suppose X is given? What can you say about Y in general?
Well, if X is an nxn matrix, then any polynomial Y=p(X)
commutes with X.
Under certain conditions, these are the only Y. Maybe the
condition is
that X has n distinct eigenvalues, or something like that.
===
Subject: Re: Condition on commuting matrices, and Lyapunov
equation
>Well, if X is an nxn matrix, then any polynomial Y=p(X)
commutes with X.
>Under certain conditions, these are the only Y. Maybe the
condition is
>that X has n distinct eigenvalues, or something like that.
I'm pretty sure it's that every Jordan block of X is for a
different
eigenvalue, or equivalently that each eigenvalue of X has
geometric
multiplicity 1 (i.e. dim {x: Xx = rx} = 1).
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: numerical differentiation of oscillating singal
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE4RL05082;
>> It might be helpful to explain what went wrong with the
methods you
>> tried. Is your signal described by an analytic formula, or
by a
>> series of samples?
>Actually, I am processing a signal f(t) and in each step of
time,
> f(t+dt) = G[D[f(t)]]
>here D means the differentiation of f(t) and G is a function
to form
>D[f(t)] to a new signal. However, f(t) is a signal with a
rapidly
>oscillating tail. A common algorithm (such as forward
difference) to
>perform numerical differentiation will lead to wrong result.
For this
>purpose, I adopt a high-order difference formula. It's weird
that the
>high-order formula is even worse than the central difference
formula.
>I also take the fourier method but no help
>D[f(t)] = iff(i*w*F(w));
>w denotes for the angular frequency and ifft denotes the
inverse
>fourier transformation. F(w) is the fourier spectrum of f(t).
It looks like what your problem needs is some control theory.
If your function G is a linear function of the set of samples,
then
it can be modeled with a Laplace transform. Differentiation
becomes
multiplication by s in the Laplace frequency domain. In order
to
stabilize your system, a compensating filter needs to be added.
The goal of the compensation is to ensure that the loop gain
drops
below unity gain before the phase shift reaches 180 degrees,
otherwise the system will oscilate.
In a purely sampled system, the Laplace transform can be
rewritten
using the substitution z = exp(s T), where T is the sample
period.
This is called the Z transform.
If the system is nonlinear, the problem becomes more difficult.
If it is locally close to linear, the compensation can be adjus
along the way. If it is highly nonlinear, some kind of phase
space
method may be needed.
It is hard to go into more detail without knowing what your G
is.
===
Subject: Re: Integral of e^x^2 dx ?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE5AD05173;
>could be expressed as some form of error function
Something like: -i sqrt(Pi)/2 erf(i z)
>> I am an undergrad college student and I ran across a
problem that
>intrigued me. What is the integral of e^x^2 dx ? I asked a
few teachers
>and searched the internet for a little while, but to no
avail. Does
anyone
>out there have any idea how to solve this one? Possibly prove
it too?
I'm
>killing myself not being able to figure it out even after a
week of
>thought!!!
>> Kris
===
Subject: Re: Fibonacci
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE61205237;
Fibonacci numbers can also be produced using hyperbolic
functions:
F(even n) = 2/sqrt(5) sinh(n ln(t))
F(odd n) = 2/sqrt(5) cosh(n ln(t))
Swapping the sinh and cosh produces Lucas numbers:
L(even n) = 2 cosh(n ln(t))
L(odd n) = 2 sinh(n ln(t))
where t = (1 + sqrt(5))/2
===
Subject: Re: minimum foam
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE4ox05134;
>Does anyone disagree that space fills with irregular
tetrahedra
>such that each vertex is shared by 20 tetradrahedra? NO? Okay.
>Then....
>Place a vertex at the center of each tetrahedron in that
maximum
>foam and connect the dots. Each vertex has 4 edges connec.
Voila!
>Space fills with irregular pentagonal dodecahedra. Minimum
foam.
>Yes? No?
>Dick Fischbeck
>East Belfast, Maine
>Randome, LLC
>Subject: irregular p-dodecahedra
>Does anyone know if space fills with irregular pentagonal
dodecahedra?
>Why or why not?
If you go to 4 dimensions, then the tetrahedra can be regular.
What you will find is that exactly 600 of them will close to
form
a polytope known as the 600-cell. The dual is the 120-cell
which
is made out of 120 regular dodecahedra.
If you try to do this with distortion in Euclidean 3-space,
you will
get a distor version of the same result. It is analogous to
trying to tile the plane with triangles, while allowing only
five
around each vertex, or tiling with three pentagons around each
vertex.
===
Subject: Help
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE5eK05202;
Any people can help me on this problem?
M is a m-dimensional submanifold of R^n. Show that the Gauss
map f: M -->
Gr(m,n) is smooth, where Gr(m,n) denotes the set of
m-dimensional subspaces
of R^n (Grassmannian).
Thanks!
libaiyang2000@yahoo.com
===
Subject: request help for wigner ville distribution
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE5Iq05186;
hi..
In wvd formula
there's formula exp(-2*pi*f*tau).
I can't use these formula below
exp(i*x)=cos(x)+i*sin(x) and
exp(-i*x)=cos(x)-i*sin(x)
so I still calculate using exp.
In wvd codes that I write.
If you dont mind
Could you explain to me
how exp(-2*pi*f*tau) to calculate
in wvd formula
This is one thing that
confuse me when I try to write
WVD codes.
could someone give me the codes
to callcuate wvd .
Thank you very much
===
Subject: Re: Glenn Lamb's challenge: Prove 0! = 1
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE52o05162;
i do not under stand what u are saying.give me a mail how you
got 0!=1
===
Subject: Re: prove it
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE6nF05325;
>> n^4+n^2=6 (mod 7) for all n
>Not true. If n=0 (mod 7) then n^4+n^2=0 (mod 7) If n=1 or n=5
(mod 7)
then
>n^4+n^2=2 (mod 7)
>Have a tolerable existence. Eli
Exactly the problem was:
(n^5-n)(n^4+n^2-6)=0 (mod 210) for all n
Thanks
===
Subject: Re: prove it
euler <anonymous@mathforum.org> escribi.97 en el mensaje
> n^4+n^2=6 (mod 7) for all n
Not true. If n=0 (mod 7) then n^4+n^2=0 (mod 7) If n=1 or n=5
(mod 7)
then
>n^4+n^2=2 (mod 7)
Have a tolerable existence. Eli
> Exactly the problem was:
> (n^5-n)(n^4+n^2-6)=0 (mod 210) for all n
> Thanks
210 = 2*3*5*7
You must prove that M(n) = (n^5-n)(n^4+n^2-6) is multiple of
2, 3, 5 and 7
for all n.
For 2 is obvious, each factor is even for all n, eben or odd.
Then M(n) is
always multiple of 4.
For 3: 0^5 - 0 = 1^5 - 1 = 2^5 - 2 = 0 (mod 3)
For 5: n^5 = n (mod 5) by Fermat little theorem.
For 7: 0^5 - 0 = 1^5 - 1 = 0 (mod 7). Also 6^5 - 6 = (-1)^5 -
(-1) = -1 + 1
= 0 (mod 7)
For n = 2, 3, 4 (= - 3) or 5 (= - 2) (mod 7), n^4 + n^2 - 6 =
0 (mod 7), as
it has seen in previous post.
Actually, you can say that (n^5-n)(n^4+n^2-6) is always
multiple of 420.
As
M(2) = 420, gcd(M(n)) = 420, for all n.
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Was: Convergence on a space with no topology
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE7Sa05420;
It was mentioned that:
The following does appear to be equivilant...
1. A_n -> A
2. A_n subset A forall n and for all p > 0, p in R for all a
in A
exists q in N for all n > q exists b in A_n: d(a,b) < p
A stricter form of convergence:
1. A_n ->* A
2. A_n subset A forall n and for all p > 0, p in R exists q in
N for
all a in A for all n > q exists b in A_n: d(a,b) < p
A simple example regarding both types of convergence:
Let N be the natural numbers (no zero) and N_n = {1,2, ...,n}.
It is easy to see that N_n -> N. However, !(N_n ->* N).
Proof:
We have to check:
for all p > 0, p in R exists q in N for
all a in N for all n > q exists b in N_n: |a-b| < p
This is equivilant to (choose p < 1)
exists q in N for
all a in N for all n > q exists b in N_n: a = b
which, in turn, is equivilant to
exists q in N for all a in N for all n > q: a < n + 1
This is not valid for a = q + 2 and n = q + 1, q.e.d.
===
Subject: if n is prime...
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE76k05349;
show that:
a) 3^n+(-2)^n+(-1)^n is multiple of n. for all n prime.
b) 5^n-2(3^n)+1 is also multiple of n for all n prime.
===
Subject: Re: if n is prime...
> show that:
> a) 3^n+(-2)^n+(-1)^n is multiple of n. for all n prime.
> b) 5^n-2(3^n)+1 is also multiple of n for all n prime.
===
Subject: Re: if n is prime...
adrian <fullresponse21@yahoo.com> escribi.97 en el mensaje
> show that:
> a) 3^n+(-2)^n+(-1)^n is multiple of n. for all n prime.
> b) 5^n-2(3^n)+1 is also multiple of n for all n prime.
a) If n = 2, it is true. If n is prime > 2, then n is odd and
(-1)^n = -1
and (-2)^n = - 2^n.
Then, you must to show that
3^n - 2^n = 1 (mod n) (#1)
for all odd prime n
But the Fermat Little Theorem says that a^n = a (mod n) for
all a, if n is
prime. Then #1 reduces to 3 - 2 = 1 (mod n).
b) As before, 5^n-2(3^n)+1 = 5 - 2*3 + 1 = 0 (mod n)
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: if n is prime...
>> Show 3^p + (-2)^p + (-1)^p is a multiple of p, for all p
prime.
= 3 + -2 + -1 = 0 (mod p) by Fermat's little theorem
Below you treat -2 and -1 different from 3.
Do you think a^p = a (mod p) fails if a < 0 ?
Hint: mod p: a = b => a^n = b^n
e.g. p-2 = -2 => (p-2)^n = (-2)^n
> If p = 2, it is true. If p is prime > 2, then p is odd
> and (-1)^p = -1 and (-2)^p = - 2^p.
> Then, you must show (#1) 3^p - 2^p = 1 (mod p) for all odd
prime p
> But the Fermat Little Theorem says that a^p = a (mod p) for
all a,
> if p is prime. Then #1 reduces to 3 - 2 = 1 (mod p).
-Bill Dubuque
===
Subject: Re: if n is prime...
> show that:
> a) 3^n+(-2)^n+(-1)^n is multiple of n. for all n prime.
> b) 5^n-2(3^n)+1 is also multiple of n for all n prime.
HINT
if p is prime a^p mod p = a
Cheers Hanford
===
Subject: Re: Algebraic Closure
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id h9UE7J805391;
>What is the algebraic closure of a finite field?
>I'd be happy just to know the algebraic closure of Z/<2>.
Let K be a finite field and L the algebraic closure of K. Then
we know that
for every integer n there exists exactly one field M within L
that has degree
n over K.
The field M looks like K[X]/pK[X], where p is an arbitrary
irreducibel
polynomial of degree n.
There are algorithms to determine the irreducibel polynomials
over the
field K (for example Berlekamps algorithm).
The field L is the union over all the field M, and these
fields form
an ascending chain.
Every algebraic computation within L typically involves only
finitely
many elements. So you can perform the calculation in one of
the M's,
that are under your control.
So in a sense one can say that the algebraic closure of K is
>known<.
===
Subject: Re: Delay differential equation
>I have a 2nd order DDE
y''(x) + ay'(x) + b(y(x) - y(x-a) = 0
... which he later amended to y''(x) + ay'(x) + b(y(x) -
y(x-c)) = 0
>with 2 bc's
y(0) = 1 and y(x) = 0 when x->Inf
I know y(x) for 0<x<a
>My problem is that to enforce the second bc I need the
solution as
>x->Inf , and I am looking for a analytic solution so I was
wondering if
>there is any shortcuts for getting the solution at x=Large
without going
>through all x<Large.
> If you try y = exp(r t), you'll see that this is a solution
iff
> (r^2 + a r + b) exp(r a) - b = 0.
... which of course becomes (r^2 + a r + b) exp(r c) - b = 0.
> One root of this is r = 0, and
> there are probably infinitely many other roots (mostly
complex).
> To have y -> 0 as x -> infinity, you'll want to use a linear
> combination of the solutions for roots with negative real
part.
Some further thoughts:
I'm assuming a, b, c > 0.
1) If a^2 - 2 b >= 0, I'm pretty sure |r^2 + a r + b| >= b for
Re(r) >= 0, with equality only for r=0. Since |exp(r c)| >= 1
as well, the only root with Re(r) >= 0 is r = 0. If a^2 - 2 b
< 0,
you may have to worry about roots in the right half plane. Of
course, these correspond to solutions y(t) = exp(r t) that
behave
badly as t -> infinity.
2) If y is a solution, let
U(x) = y'(x) + a y(x) + b int_{x-c}^x y(t) dt.
Then U'(x) = y''(x) + a y'(x) + b((y(x) - y(x-c)) = 0, so U is
constant. Now for any solution where y(t) and y'(t) -> 0 as
t -> infinity, we have U(x) -> 0 as x -> infinity and therefore
U(x) = 0. On the other hand, for the constant solution y = 1 we
have U = a + b c. So (at least if a^2 - 2 b >= 0) I think that
lim_{t -> infinity} y(t) = U/(a+bc)
which you can calculate using your known values for 0 <= x <=
c.
3) This must all be well-known in the literature on stability
of delay-differential equations.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Delay differential equation
>>I have a 2nd order DDE
>>y''(x) + ay'(x) + b(y(x) - y(x-a) = 0
>If you try y = exp(r t), you'll see that this is a solution
iff
>(r^2 + a r + b) exp(r a) - b = 0. One root of this is r = 0,
and
>there are probably infinitely many other roots (mostly
complex).
>To have y -> 0 as x -> infinity, you'll want to use a linear
>combination of the solutions for roots with negative real
part.
Your answer doesn't mention the LambertW function. In past
similar
responses, you have evoked it. Does this mean that in this
case with
the second order derivative, it won't work? I think I will
wait for
your answer before I explore any further.
Reference:
http://www.apmaths.uwo.ca/~rcorless/frames/PAPERS/LambertW/
LambertW.ps
see especially page 7
John Bailey
http://home.rochester.rr.com/jbxroads/mailto.html
===
Subject: Re: Delay differential equation
>I have a 2nd order DDE
>y''(x) + ay'(x) + b(y(x) - y(x-a) = 0
... and Stefan later admit that he meant
y''(x) + a y'(x) + b (y(x) - y(x-c)) = 0
>If you try y = exp(r t), you'll see that this is a solution
iff
>(r^2 + a r + b) exp(r a) - b = 0. One root of this is r = 0,
and
>there are probably infinitely many other roots (mostly
complex).
>To have y -> 0 as x -> infinity, you'll want to use a linear
>combination of the solutions for roots with negative real
part.
> Your answer doesn't mention the LambertW function. In past
similar
> responses, you have evoked it. Does this mean that in this
case with
> the second order derivative, it won't work? I think I will
wait for
> your answer before I explore any further.
I could be wrong, but I don't think that the equation
(r^2 + a r + b) exp(r c) - b = 0 can be solved (in general)
using
LambertW. It can be done in the special case a = 2 sqrt(b), in
which
case the roots other than r=0 are 2/c W(-ca/4 exp(ca/4)) - a/2
where
W is any of the branches of LambertW.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Delay differential equation
Thanks for the reply !
> I assume this should be
> y''(x) + ay'(x) + b(y(x) - y(x-a)) = 0
> but do you really want the two a's to be the same?
No they are not the same, my mistake. It should be
y''(x) + ay'(x) + b(y(x) - y(x-c)) = 0
>>with 2 bc's
>>y(0) = 1 and y(x) = 0 when x->Inf
> So actually the boundary condition on the left is y(x) =
(some known
> function) for 0 <= x <a? This will then determine y(x) for
all x,
> as you know because you can solve for each interval in turn.
And then
> it's not a matter of enforcing the second bc, it's either
true or
false.
> But maybe you only know y(x) for 0<x<a up to some unknown
parameter.
Right! The solution for 0<x<c has one unknown constant C_0. To
determine
C_0 I need the second bc.
> If you try y = exp(r t), you'll see that this is a solution
iff
> (r^2 + a r + b) exp(r a) - b = 0. One root of this is r = 0,
and
> there are probably infinitely many other roots (mostly
complex).
> To have y -> 0 as x -> infinity, you'll want to use a linear
> combination of the solutions for roots with negative real
part.
I'll try this approach and see what happens for my case.
Thanks again.
/Stefan
===
Subject: Re: Boolean Algebra - Arithmetic Relationship
> Many years ago when I was a student of osophy I read Popper
writing
> against this Pavlovian idea. The observation (made by the
dog) of A
is
> followed by B does _not_ give rise to the hypothesis A
causes B.
> Can't remember why.
I would warn you against confusing A causes B with the
logician's
if
> A then B. For the logician, if A then B is defined by the
table
A B if A then B
> t t t
> t f f
> f t t
> f f t
where t means true and f means false. The ancient Greeks
discussed
> this and it was taken up again in the late nineteenth
century.
> I agree that the Pavlovian response is an inference, whereas
the
> Logician's assertion that A implies B, is necessarily true by
> definition.
> But isn't this how we learn and develop logical models of our
> environment? I don't think we could live without making
inferences
> such as these from experience.
Popper would say that we have (for whatever reason) the theory
first and
then we test it (or should) against experience. According to
Popper the
theory always comes before the experience.
> Furthermore, isn't all reasoning basically based upon this
idea of
> inference?
> SomeEquation => SomeOtherEquation
> SomeTruth => SomeOtherTruth
> Other than that, it appears variable subtitution is all that
is
> necessary.
> Of course I may be terribly confused and misinformed, but
this seems
> to me to be the way things work.
> I have a deep respect for the basic ideas of Karl Popper,
Thomas Kuhn
> and the Logical Positivists camp in general. I like to learn
more
> about his argument against Pavlovian inferences.
You know, don't you, that Popper was opposed to Kuhn and to
the logical
positivists. For Popper v Pavlov, see the footnote on page 45
of
Realism and the Aim of Science and the works referred to there.
> -Steve
--
G.C.
===
Subject: Re: Understanding a proof (Was: Re: Wiles' Proof)
> I thought that it might be interesting to quote G. H. Hardy
here:
> There is strictly no such thing as mathematical proof;
proofs are what
> Littlewood and I call gas, rhetorical flourishes, devices to
stimulate
> the imaginations of pupils.

If that quotation is exact, then Hardy contradicts himself,
because
in his
A Mathematician's Apology (Speaking of Euclid's proof of the
infinitude of prime numbers), he says:'Two thousand years have
no
written a wrinkle on it.'And afterwards:'...the proof can be
mastered
in an hour by any intelligent reader...'
It is strange...Why Hardy mixed Littelwood in that nonsense?
L.Rodriguez
===
Subject: Re: Understanding a proof
permission for an emailed response.
X-Tom-Swiftie: We're going to sue you for that window system,
Tom said
inexorably
> That makes it a bit difficult for logic students. After all,
suppose
> you're studying a purely formal proof of, say, one of the De
Morgan
> laws. If it's already purely formal, you'll never understand
it,
> since you *can't* right down any more detailed proof.
Huh? That doesn't make any sense at all. De Morgan laws are
hardly
taken as primitive in most systems these days. (I know that
they were
taken as such in Hofstadter's GEB, but that's a special case
indeed.)
Typical fashion these days is to use a natural deduction
system,
which is essentially a formal system for logic which uses no
axioms.
The most elegant have a pair of rules for each connective, one
to
introduce it, one to eliminate it.
So if I want to prove that ~(P & Q) entails ~P v ~Q following
Fitch-style proof in a natural deduction system:
1 | ~(P & Q)
+---
2 | | ~(~P v ~Q)
| +---
3 | | | ~P
| | +---
4 | | | ~P v ~Q [v Intro; 3]
5 | | | <false> [<false> Intro; 2, 4]
| |
6 | | ~~P [~ Intro; 3-5]
7 | | P [~ Elim; 6]
| |
8 | | | ~Q
| | +---
9 | | | ~P v ~Q [v Intro; 8]
10 | | | <false> [<false> Intro; 2, 9]
| |
11 | | ~~Q [~ Intro; 8-10]
12 | | Q [~ Elim; 11]
13 | | P & Q [& Intro; 7, 13]
14 | | <false> [<false> Intro; 1, 13]
|
15 | ~~(~P v ~Q) [~ Intro; 2-14]
16 | ~P v ~Q [~ Elim; 15]
That gives you some of the flavor. Fitch-style involves the
use of
subproofs. You can see above the use of v Intro and & Intro,
and how
they are justified, as well as the rules for introducing ~ and
removing ~. (~ intro is the formalization of indirect proof,
for
example). The & elim rule lets you extract either conjunct.
The rule
for v elim is proof by cases, you need a subproof for each
disjunct,
leading from that disjunct to a common conclusion.
The usual fashion is not to define connectives in terms of
each other,
but rather to give further pairs of rules. So implication has
two
rules, where the intro rule is justified by a subproof, and
the elim
rule is just modus ponens.
Experience in teaching elementary logic to students shows that
natural
deduction systems are much more easily grasped than axiomatic
systems;
the latter lead to the question why that axiom and no other,
and you
are right that if you don't get it, you end up stuck. Whereas
the
natural deduction systems seem much closer to people's unformed
intuitions.
Thomas
===
Subject: Re: Understanding a proof
<3f9d06e2$0$69956$edfadb0f@dread12.news.tele.dk>
<3f9d8023_5@127.0.0.1>
<87k76ql6wu.fsf@phiwumbda.org>
<87k76nqmqo.fsf@becket.becket.net>
Discussion, linux)
>> That makes it a bit difficult for logic students. After
all, suppose
>> you're studying a purely formal proof of, say, one of the
De Morgan
>> laws. If it's already purely formal, you'll never
understand it,
>> since you *can't* right down any more detailed proof.
> Huh? That doesn't make any sense at all. De Morgan laws are
hardly
> taken as primitive in most systems these days. (I know that
they were
> taken as such in Hofstadter's GEB, but that's a special case
> indeed.)
I'm aware of that.
Before we go further, let me quote the passage to which I was
replying. Then we'll use your proof below to make my (not very
deep)
point -- not very deep, hell, it was just a joke. But I'll run
it
into the ground by defending it.
,----
| >>What exactly does it mean to understand a proof?
| > ...
| >
| > To be able to sit down and rewrite the entire proof
| > BUT including one more level of detail for every step.
`----
[snip introduction to natural deduction]
> So if I want to prove that ~(P & Q) entails ~P v ~Q following
> Fitch-style proof in a natural deduction system:
> 1 | ~(P & Q)
> +---
> 2 | | ~(~P v ~Q)
> | +---
> 3 | | | ~P
> | | +---
> 4 | | | ~P v ~Q [v Intro; 3]
> 5 | | | <false> [<false> Intro; 2, 4]
> | |
> 6 | | ~~P [~ Intro; 3-5]
> 7 | | P [~ Elim; 6]
> | |
> 8 | | | ~Q
> | | +---
> 9 | | | ~P v ~Q [v Intro; 8]
> 10 | | | <false> [<false> Intro; 2, 9]
> | |
> 11 | | ~~Q [~ Intro; 8-10]
> 12 | | Q [~ Elim; 11]
> 13 | | P & Q [& Intro; 7, 13]
> 14 | | <false> [<false> Intro; 1, 13]
> |
> 15 | ~~(~P v ~Q) [~ Intro; 2-14]
> 16 | ~P v ~Q [~ Elim; 15]
Now, according to Don Taylor's (now revised) hypothesis, in
order to
claim that I understand your proof, I must be able to rewrite
it with
more detail. I confess that I cannot add any more detail to
your
proof, since each statement is either an assumption or the
immediate
conclusion from a natural deduction rule of inference such
that the
premises are [blah blah blah]. Therefore, it is apparent that
I do
not understand your proof (according to Don's suggestion),
despite the
fact that I can confirm each step and even describe informally
the
general strategy that leads to the formal proof.
Thanks for the time you took to write down part of the proof
of De
Morgan's law, since I think it makes my point that much more
obvious.
I sure as heck wasn't gonna to write it down and evidently I
didn't
get my point across without an explicit example.
[snip the remaining evaluation of natural deduction, since
it's not
that relevant to my comments]
--
We are happy that you agree that customers need to know that
Open
Source is legal and stable, and we heartily agree with that
sentence
of your letter. The others don't seem to make as much sense,
but we
find the dialogue refreshing. -- Linus Torvalds to Darl McBride
===
Subject: Re: Understanding a proof
permission for an emailed response.
Send your questions to ``ASK ZIPPY'', Box 40474, San
Francisco, CA
94140, USA
> Now, according to Don Taylor's (now revised) hypothesis, in
order to
> claim that I understand your proof, I must be able to
rewrite it with
> more detail. I confess that I cannot add any more detail to
your
> proof, since each statement is either an assumption or the
immediate
> conclusion from a natural deduction rule of inference such
that the
> premises are [blah blah blah].
Fair enough! Thanks for toting up the context and repairing my
mistake. Glad my proof could help make your point. ;)
I might add that perhaps Don Taylor is on to something, even
if not
quite right. I hesitate to say exactly how to put it.
I think there is something about the ability to proffer
explanations
of steps as a part of understanding. I have become accustomed
to
thinking there are two different contexts of proving, and
something
gets missed when they are confla:
Proof as convincement is what mathematicians normally do. The
job of
a proof here is to convince another (or in some sense,
oneself) that a
theorem is true. Taking as a harmless simplification that all
theorems are conditional sentences, that is a proof is
designed to
show that the implication in questino is a valid one. A proof
works
by reducing the implication to a series of smaller arguments,
each of
which must be valid. The prover's job is to reduce to a series
of
steps which both prover and provee agree are valid.
We sometimes say that the provee must understand the proof in
order to
judge it. That means that the provee must see each step and
know why
it is a valid inference. Roughly speaking, that means that the
provee
must be able, in turn, to proffer a proof for each step to some
unspecified third party. In that sense, we say that a provee
has
understood a proof when she can rewrite it in more detail.
A second context for proving is what students do in proving
theorems
to instructors. The purpose here is not to convince the
provee, but
rather for the prover to demonstrate their understanding of the
proof. Ideally, this works by some kind of back-and-forth
method, in
which the provee (the instructor) can ask for clarification of
any
step, demanding a reduction of it to sub-steps, and in that
way to
verify to his satisfaction the student's understanding. Here
we are
not concerned with whether a provee understands a proof, but
rather
whether a prover does. Still, the measure is much the same: the
ability to proffer subproofs of the individual steps on demand.
As you rightly note, this bottoms out. Or does it? Unlike the
average student in a logic class, I *can* give an explanation
for why
the natural deduction rules for & are what they are. This is
in some
sense a different order of proof. It's not a more detailed
step in a
proof--in that sense, Don Taylor is mistaken. But it *is* some
kind
of justification. My explanation will be a proof-theoretic
demonstration that the rule is truth-preserving when
considered in the
framework of a Fitch-style proof, given the truth-table that I
take to
be the definition of the meaning of &.
I would say that my students do not actually understand
*really* why
the rules for & are the ones that they are taught until they
begin to
grok at least the soundness demonstrations for propositional
logic.
But that isn't a problem: it's perfectly legitimate for them to
understand some things and not others. Moreover, they are
entitled to
use such steps in their own proofs even if they do not
understand why
we say they are valid steps.
Which brings me to a third kind of understanding of proofs.
This is
what we mean when someone looks at my natural-deduction proof
of a De
Morgan law and can explain the structure of it. Not merely
check the
steps (which is of course mechanizable), and not justify why
the rules
of the system are the right ones, but rather, to explain the
overarching structure of the proof.
We overlook this kind of understanding sometimes, because in
our usual
informal proofs most of the energy is spent in communicating
the
structure. Some teachers are better at it than others, though,
and
many beginning students come away from a proof understanding
every
single step (and are able to offer a subproof of each one) and
having
no clue whatsoever how the whole fits together and how to
characterize
what is really going on inside the proof.
This I think might well be the most important kind of
understanding,
and I think Don Taylor's notion leaves it out entirely.
Anyway, I offer these more sophistica (I hope) reflections to
atone
for my presumption in misreading the previous context and
giving a
pedantic example to folks who hardly needed one.
Thomas
===
Subject: Re: Understanding a proof (Was: Re: Wiles' Proof)
> Surely a novice might make the same objections regarding a
proof
> by an expert, only because the novice regarded some missing
step
> as essential. But the missing step might be obvious to all
experts,
> and hence should be omit on grounds of obscuring the main
idea
> (assuming that the proof was presen in a forum for experts).
So you are saying that determining has he/she understood the
proof
depends
on the *observer*? In other words, a proof can not be judged
objectively,
but depends on who the judge is, or at least the intended
audience, as well
as who the author is...
I would have expec a proof to be evalua completely objectively
and
disjunct from all surroundings, including the author and the
audience. Is
this just being too idealistic?
I mean, I don't like the idea that I present a proof and get
critized for
not having understood it, only to find the same proof presen
by a
well-renowned professional mathematician and subsequently
accep without
comment.
Can we (the readers of sci.math) accept such relativity?
On a second note, maybe I am mixing up two different concepts:
(1)
Validating that a proof is correct, and (2) assessing whether
the author
has
understood the proof. IMVHO, case (1) should be independent of
both author
and reader. Perhaps case (2) should be independent too,
although a
well-renowned professional mathematician might feel disrespec
for having
to explain trivial concepts in excruciating detail
-Michael.
===
Subject: Re: Understanding a proof (Was: Re: Wiles' Proof)
I'd have to issue a proviso:
if you really believe in fuzzy logic (or, what is the same,
the Copenhagen interpretation of QM ... many universes),
then maybe that's possible. but we don't know, Why!
> the simplest proof taht i know of PT is the lunes one,
> Why is the question absurd?
--les ducs d'Enron!
===
Subject: Re: Understanding a proof (Was: Re: Wiles' Proof)
> I'd have to issue a proviso:
> if you really believe in fuzzy logic (or, what is the same,
> the Copenhagen interpretation of QM ... many universes),
> then maybe that's possible. but we don't know, Why!
I don't understand what you're saying.
My point is very simple and real: you have no way of proving
things for
certain without it ultimately being dependant on the goodwill
of humans.
/David
===
Subject: Re: Understanding a proof (Was: Re: Wiles' Proof)
Shmuel (Seymour J.) Metz <spamtrap@library.lspace.org.invalid
at 12:52 PM, Michael Jrgensen <ingen@ukendt.dk>
said:
>To me, understanding means to be able to explain to someone
else.
> There have been cases of fallacious proofs that were accep
for a
> long time. People were able to explain the proof to each
other, but
> the explanations were actually wrong. IMHO they did not
understand
> the purpor proof.
> To me, you understand the proof when you understand each
step along
> the way. That's not as simple as it sounds, since sometimes
a step can
> depend on very subtle distinctions or unsta assumptions.
Perhaps we must reach the conclusion that the only proofs we
can be
certain we understand are the fallacious ones....
--
===
Subject: Re: Definition of limit(???)
> Edgar has poin out that I didn't read the section carefully
> enough - the right definition comes later.
> The two definitions I was talking about are _not_ equivalent.
> The problem is not sequences versus epsilons, the problem
> is whether f(a) should be relevant:
> Say f(x) = 0 for x <> 0, f(0) = 1. Then lim_{x->0} f(x)
should
> be 0, but by one of the definitions above the limit does
> not exist.
> (At least that's what I thought the problem was - in fact
> I missed a tiny bit of notation, the wrong definition
> above is a definition of something other than
> lim_{x->a} f(x).)
In his The Elements of Real Analysis, Robert Bartle defines
two kind
of limits like y I've never seen anyone else do. If f is
defined on a
subset D of a real vector space and has values in another real
vector
space, and a is a limit point of D (actually, Bartle uses the
term
accumulation point), then he defines:
L is the NON-DELE limit of f at a if, for every neighborhood V
of
L, there's a neighborhood U of a, depending on V (Bartle is
very
careful with definitions and likes to emphasize details like
this),
such that f(x) is in V for every x in U cap D (another point
that
Bartle emphasizes).
And then, Bartle defines the DELE limit (the usual concept) l
of f
at a. The only (and important!) difference is at the very end
of the
definition: ...., such that f(x) is in V for every x in U cap
D such
that x<>a. And then Bartle, who likes to define limits and
continuity
in terms of neighborhoods, proves that such definitions are
equivalent
to the eps-delta ones, the first (non dele limit) with |x -a| <
delta and the second (dele limit) with, as usual, 0<|x-a|
delta.
Then Bartle proves that f is continuous at a iff L = l.
Though Bartle is a great author, I think these non-dele and
dele
limit things only cause confusion. I see no point in defining
such
concepts. In my opinion, they add nothing but lenght to the
book.
(Well, Bartle suggests such concepts make it easier to prove
those
theorems about continuity and limits of composite functions)
In my opinion, defining limits in terms of sequences is a bad
idea, at
least in the case of metric spaces. I don't think it's natural.
Artur
===
Subject: Re: Definition of limit(???)
> Ross seems to think that things will be easiest to follow if
he first
> talks about convergent sequences and then bases everything
> else on them. I'm not sure that that's right but I don't
have any
> big complaint with it.
>>That's the trouble with really understanding the subject.
Equivalent
>>definitions are, well, equivalent, so why not just go with
the flow?
This is not really understanding the subject, but
merely being able to prove the theorems. One should
attempt to teach the concepts, and while I believe
convergence is very important, I would make limits of
sequences a special case of the general notion of
convergence, NOT otherwise as is done by Ross. There
are many cases in which I will absolutely not teach the
standard definitions as definitions, as they are often
highly misleading, and this is even in elementary
courses. Many of the usual definitions in probability
hide the concepts quite well; it may be that they are
easy to verify, but that should be as theorems or as
computational procedures, not as definitions.
>>We need people versed in pedagogy to teach courses, so
they'll look at
>>things from an educational point of view.
>Fascinating. The fact that I'm not certain his opinion is
correct but
>I'm also not certain it's wrong means I'm not looking at
things from
>an educational point of view. Huh.
At this time, it seems that those versed in pedagogy
have been taught to avoid starting with concepts. Try
teaching elementary school teachers, or even high school
teachers of mathematics, mathematical concepts. The
pedagogy experts put so much memorization and computation
in the way that concepts are very hard to get.
>Yes, it certainly is true that having people who really
understand
>a subject teach it is a bad idea. (On what planet, exactly?)
If we ever are going to get good education, it will have
to be done by those who have not had courses in pedagogy,
or who at least reject them utterly.
--
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: Definition of limit(???)
===
>>Ross seems to think that things will be easiest to follow if
he first
>>talks about convergent sequences and then bases everything
>>else on them. I'm not sure that that's right but I don't
have any
>>big complaint with it.
>That's the trouble with really understanding the subject.
Equivalent
>definitions are, well, equivalent, so why not just go with
the flow?
> This is not really understanding the subject, but
> merely being able to prove the theorems. One should
> attempt to teach the concepts, and while I believe
> convergence is very important, I would make limits of
> sequences a special case of the general notion of
> convergence, NOT otherwise as is done by Ross. There
> are many cases in which I will absolutely not teach the
standard definitions as definitions, as they are often
> highly misleading, and this is even in elementary
> courses. Many of the usual definitions in probability
> hide the concepts quite well; it may be that they are
> easy to verify, but that should be as theorems or as
> computational procedures, not as definitions.
>We need people versed in pedagogy to teach courses, so
they'll look at
>things from an educational point of view.
>>Fascinating. The fact that I'm not certain his opinion is
correct but
>>I'm also not certain it's wrong means I'm not looking at
things from
>>an educational point of view. Huh.
> At this time, it seems that those versed in pedagogy
> have been taught to avoid starting with concepts. Try
> teaching elementary school teachers, or even high school
> teachers of mathematics, mathematical concepts. The
> pedagogy experts put so much memorization and computation
> in the way that concepts are very hard to get.
The problem, as I see it, is in goals. The goal of people at
the
elementary/high school level is to produce students capable of
crunching
some numbers on a test.
The goal of mathematicians at the college level (I think) is
to promote
understanding of underlying relationships and principles.
These are
contradictory goals. The question becomes: which is the right
goal, and
if the first goal is incorrect, how do we get it fixed?
>>Yes, it certainly is true that having people who really
understand
>>a subject teach it is a bad idea. (On what planet, exactly?)
> If we ever are going to get good education, it will have
> to be done by those who have not had courses in pedagogy,
> or who at least reject them utterly.
Pedagogy needs to dictate the means to the goals, instead of
dictating
the goals. There are too many people out there who have an
idea of how
to teach something, without knowing what the goal is. They
consistently
fail.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Definition of limit(???)
>I'm taeching the beginning epsilon-delta course. Looking
>ahead in the book (Ross, Elementary Analysis: the Theory
>of Calculus) I see the following definition:
>Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for
>every sequence (x_n) in A such that x -> a we have
>f(x_n) -> L.
>At first I thought this must be a typo. But it turns out
>he means it - later when he shows that this definition
>is equivalent to the one in terms of epsilon and delta
>the condition is |x - a| < delta, not 0 < |x-a| < delta.
>I'm shocked. Is this version of the definition actually
>standard in some circles? My impression is that the
>definition is inconsistent with what the students are
>almost certainly going to see in later courses - is this
>correct? (I hate to cause confusion by using a
>definition different from what's in the book unless
>I have a very good reason, but if this definition is
>as rare as it seems to me it is that could be a good
>enough reason - hence the question whether it
>really is extremely uncommon.)
This is valid for every function from a metric space to
another metric space.
It can also be done more generally, and should, but with
convergence of nets instead of sequences. It is very often
the only approach to the topology which is easy to work
with. There is no problem in showing that the definition
is equivalent to the usual one if open sets are used, as
they are in general to define limit.
--
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: Definition of limit(???)
>>I'm taeching the beginning epsilon-delta course. Looking
>>ahead in the book (Ross, Elementary Analysis: the Theory
>>of Calculus) I see the following definition:
>>Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for
>>every sequence (x_n) in A such that x -> a we have
>>f(x_n) -> L.
>>At first I thought this must be a typo. But it turns out
>>he means it - later when he shows that this definition
>>is equivalent to the one in terms of epsilon and delta
>>the condition is |x - a| < delta, not 0 < |x-a| < delta.
>>I'm shocked. Is this version of the definition actually
>>standard in some circles? My impression is that the
>>definition is inconsistent with what the students are
>>almost certainly going to see in later courses - is this
>>correct? (I hate to cause confusion by using a
>>definition different from what's in the book unless
>>I have a very good reason, but if this definition is
>>as rare as it seems to me it is that could be a good
>>enough reason - hence the question whether it
>>really is extremely uncommon.)
>This is valid for every function from a metric space to
>another metric space.
>It can also be done more generally, and should, but with
>convergence of nets instead of sequences. It is very often
>the only approach to the topology which is easy to work
>with. There is no problem in showing that the definition
>is equivalent to the usual one if open sets are used, as
>they are in general to define limit.
If you looked more closely, or read some of the thread,
you'd see that it is not equivalent to the usual definition
of limit. Let f(x) = 0 for x <> 0, f(0) = 1. By the definition
above the limit of f(x) as x->0 does not exist.
************************
===
Subject: Re: Definition of limit(???)
<bnlud2$1370uf$1@ID-154916.news.uni-berlin.de>
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>wouldnt |x-a| < delta, delta > 0 (unless Ross didnt say that
delta
>was > 0), be meaning the same as 0 < |x-a| < delta.
No, because x might equal a. He's concerned with how to define
the
limit in the case where the function is discontinuous at a
point. By
one definition, f(x) does not have a limit at x=a, by the
other it may
but the limit is not f(a).
It's not that critical as long as the definition is used
consistently;
he's worried about the student having to deal with an
inequivalent
definition in a subsequent course. He may also be concerned
with
confusing the student when left handed and right handed limits
come
up.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
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===
Subject: Re: Definition of limit(???)
<bnlud2$1370uf$1@ID-154916.news.uni-berlin.de>
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at 12:41 PM, Jonathan Miller <jonathan.e.miller@comcast.net>
said:
>That's the trouble with really understanding the subject.
Equivalent
>definitions are, well, equivalent, so why not just go with
the flow?
Because they are not equivalent once things get more general.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
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===
Subject: Re: Definition of limit(???)
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>I'm taeching the beginning epsilon-delta course. Looking
ahead in
>the book (Ross, Elementary Analysis: the Theory of Calculus)
I see
>the following definition:
>Suppose that f : A -> R. Then lim_{x->a} f(x) = L if for every
>sequence (x_n) in A such that x -> a we have f(x_n) -> L.
I can certainly see a pedagogical problem with that, for the
students
who eventually take Topology.
>At first I thought this must be a typo. But it turns out he
means it
>- later when he shows that this definition
>is equivalent to the one in terms of epsilon and delta
>the condition is |x - a| < delta, not 0 < |x-a| < delta.
Why does that last matter? 0 <= |x-a| and f(x) = f(x), so the
proofs
go through with either condition.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
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===
Subject: education (was: Definition of limit(???)
> Yes, it certainly is true that having people who really
understand
> a subject teach it is a bad idea. (On what planet, exactly?)
Multiple Choice:
Future high-school mathematics instructors should major in:
(A) education
(B) mathematics
(C) English literature
(D) fundraising
===
Subject: Re: education (was: Definition of limit(???)
<bnlud2$1370uf$1@ID-154916.news.uni-berlin.de>
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>Multiple Choice:
>Future high-school mathematics instructors should major in:
What do you mean by should? Are you asking about what they
need to
do in order to get their certification, or about what they
need to do
in order to 4be able to communicate the subject matter?
> (A) education
Needed to get certification. May well destroy the ability to
teach.
> (B) mathematics
Only if you actually care whether the students learn anything.
> (C) English literature
Sounds good to me. Grammar wouldn't hurt either.
> (D) fundraising
I hope not.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
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Subject: Re: education
===
>>Yes, it certainly is true that having people who really
understand
>>a subject teach it is a bad idea. (On what planet, exactly?)
> Multiple Choice:
> Future high-school mathematics instructors should major in:
> (A) education
> (B) mathematics
> (C) English literature
> (D) fundraising
B. A is useful, but knowing *how* to teaching without knowing
*what* to
teach will fail. This seems to be a problem with some of our
public
schools.
Note: you can know how to teach without having a degree that
proclaims
that.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: education
> Yes, it certainly is true that having people who really
understand
> a subject teach it is a bad idea. (On what planet, exactly?)
>> Multiple Choice:
>> Future high-school mathematics instructors should major in:
>> (A) education
>> (B) mathematics
>> (C) English literature
>> (D) fundraising
> B. A is useful, but knowing *how* to teaching without
knowing *what*
> to teach will fail. This seems to be a problem with some of
our
> public schools.
> Note: you can know how to teach without having a degree that
proclaims
> that.
Unfortunately, many do not realize that talent in and
knowledge of
mathematics (or any other subject) without the ability to
teach is just
as useless in the classroom. This lack of recognition is
particularly
prominent at the university level.
--
===
Subject: Re: education
<bnlud2$1370uf$1@ID-154916.news.uni-berlin.de>
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at 11:03 PM, Stephen J. Herschkorn
<herschko@rutcor.rutgers.edu>
said:
>Unfortunately, many do not realize that talent in and
knowledge of
>mathematics (or any other subject) without the ability to
teach is
>just as useless in the classroom.
Unfortunately, many do not realize that a degree in education
confers
neither an education nor the ability to teach. --
Shmuel (Seymour J.) Metz, SysProg and JOAT
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===
Subject: Re: education
>> Yes, it certainly is true that having people who really
understand
>> a subject teach it is a bad idea. (On what planet, exactly?)
> Multiple Choice:
> Future high-school mathematics instructors should major in:
> (A) education
> (B) mathematics
> (C) English literature
> (D) fundraising
>> B. A is useful, but knowing *how* to teaching without
knowing *what*
>> to teach will fail. This seems to be a problem with some of
our
>> public schools.
>> Note: you can know how to teach without having a degree
that proclaims
>> that.
>Unfortunately, many do not realize that talent in and
knowledge of
>mathematics (or any other subject) without the ability to
teach is just
>as useless in the classroom. This lack of recognition is
particularly
>prominent at the university level.
I am inclined to disagree; there are few who do not have
the ability to teach those who are capable of understanding
and who are willing to understand the subject. Those who
teach in the manner the educationists have messed things up
in elementary and high school, by teaching facts, formulas,
and methods of calculation, continue the damage done to the
students. Give a student a straight cookbook calculus
course, and that student will have difficulty in understanding
what a continuous function or a derivative is.
We have had a posting by a college student who complained
that the instruction was not teaching concepts like the
concept of inner product of two sequences as the sum of
products. Those are not concepts, but formulas, which
add nothing to understanding, but detract much. We would
do better by giving few formulas and lots of concepts,
starting with variable as standing for anything, which
belongs in first grade.
In the moderately distant past, college students all had
the classical Euclidean geometry, which emphasized the
understanding and construction of proofs. Now, few have
had this, but their geometry course was facts and formulas,
and possibly memorization of some proofs. Concepts are
not learned by learning how to calculate answers, and this
even includes arithmetic.
--
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: education
===
>> Yes, it certainly is true that having people who really
understand
>> a subject teach it is a bad idea. (On what planet, exactly?)
> Multiple Choice:
> Future high-school mathematics instructors should major in:
> (A) education
> (B) mathematics
> (C) English literature
> (D) fundraising
>> B. A is useful, but knowing *how* to teaching without
knowing *what*
>> to teach will fail. This seems to be a problem with some of
our
>> public schools.
>> Note: you can know how to teach without having a degree
that proclaims
>> that.
> Unfortunately, many do not realize that talent in and
knowledge of
> mathematics (or any other subject) without the ability to
teach is just
> as useless in the classroom. This lack of recognition is
particularly
> prominent at the university level.
Unfortunately, it is also present in the high school level
with those
bearing a degree that would suggest otherwise. Then I (and
others) get
to pick up the pieces when they hit college. If they get
nailed twice
with poor teachers, they may never learn math.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: education
> Yes, it certainly is true that having people who really
understand
> a subject teach it is a bad idea. (On what planet,
exactly?)
>> Multiple Choice:
>> Future high-school mathematics instructors should major in:
>> (A) education
>> (B) mathematics
>> (C) English literature
>> (D) fundraising
> B. A is useful, but knowing *how* to teaching without
knowing *what*
> to teach will fail. This seems to be a problem with some of
our
> public schools.
> Note: you can know how to teach without having a degree that
proclaims
> that.
>> Unfortunately, many do not realize that talent in and
knowledge of
>> mathematics (or any other subject) without the ability to
teach is just
>> as useless in the classroom. This lack of recognition is
particularly
>> prominent at the university level.
>Unfortunately, it is also present in the high school level
with those
>bearing a degree that would suggest otherwise. Then I (and
others) get
>to pick up the pieces when they hit college. If they get
nailed twice
>with poor teachers, they may never learn math.
We're all making excellent points. What I'm curious about is
whether
anyone agrees with the point that star this: The idea that a
deep
understanding of mathematics is a _bad_ thing for a math
teacher.
(That being the point that I was replying to in the quote at
the
top of this subthread.)
************************
Subject: Re: education
===
> We're all making excellent points. What I'm curious about is
whether
> anyone agrees with the point that star this: The idea that a
deep
> understanding of mathematics is a _bad_ thing for a math
teacher.
> (That being the point that I was replying to in the quote at
the
> top of this subthread.)
I vehemently *disagree* with it. A deep understanding of
mathematics is
necessary for being a math teacher. If you don't have
sufficient depth
of understanding, you cannot guide the students towards what is
important and what is not. You also cannot see the deeper
relationships
that may help them understand what they are learning.
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: education (was: Definition of limit(???)
>> Yes, it certainly is true that having people who really
understand
>> a subject teach it is a bad idea. (On what planet, exactly?)
>Multiple Choice:
>Future high-school mathematics instructors should major in:
> (A) education
> (B) mathematics
> (C) English literature
> (D) fundraising
That strikes me as a fairly tricky question, to tell you the
truth.
But I don't see the relevance of the question to what you
quo. Ask me an easier one:
True or False:
Future high-school teachers should have a sound understanding
of mathematics, in fact not just the math they're going to be
teaching but a little more than that.
I know the answer to _that_ one. (And note that I was replying
to this: That's the trouble with really understanding the
subject,
implying that a really understanding the subject is a bad
thing.
Saying that it's ridiculous to say understanding the subject
you're teaching is a bad thing does not imply that
understanding
the subject is all that's required to teach young children...)
************************
===
Subject: B-Splines Explaination
Hi All,
I am interes in coding in java a b-spline fit of a series of
3d points I
have genera through an image tracking application I have
written for my
final year undergrad project. I am not a maths student, just a
humble
computer science undergrad, therefore I know little about
b-splines.
I have had b-splines mentioned to me as the solution to my 3d
curve fitting
problem.
However, as mentioned above I'm not maths genius, therefore I
am asking
people firstly for a link or for somebody to give me a nice
simple
explaination of what b-splines are all about. Also how I would
go about
setting up the problem with the use of my 3d points in a
computer program.
Matrix, recursion ???etc
And secondly if anybody knows a java implementation of this,
taking into
amount i want to specify the number of inputs i use, that
would be even
better.
Adam
===
Subject: Hard tensor question (kst).
1) Suppose K= |g^uv|, where g^uv is the
contravariant metric tensor, and K is
it's determinant and is a relative tensor.
And then form a covariant tensor K_uv,
by K_uv = K*g_uv where g_uv is a the
covariant metric, would the determinant,
|K_uv| = 1 ?
2) If so, a metric like
g_uv = K_uv +A_u*B_v Eq.(kst1)
has the determinant (set g = |g_uv|)
g = |g_uv| = |K_uv| + |A_u*B_v| therefore,
g = 1 + g*A*B where g*A*B = |A_u*B_v|,
and A*B=g^uv*A_u*B_v
I've assumed g=1/K so far.
3) If so K = 1 - A*B and 1 = K+A*B
Certainly the number 1 is invariant, therefore
the sum K+A*B is invariant.
Thinking along the lines that the covariant
derivative of the metric tensor is zero it
would be interesting to investigate DK,
(D is the absolute derivative with parameter
ds, ds^2 = g_uv dx^u dx^v). ie,
DK = 0 = DA*B + DB*A and find the
asymmetry,
DA*B = - DB*A. Eq.(kst2)
Dividing this by A and B produces
DA/A = - DB/B and then
Dlog A = -Dlog B.
This is easily integra,
log A = - log B + k
(k= the integration constant).
expoteniating produces,
A =exp(log A) = exp(- log B + k)
A = -B*e where e=(+/-)*log k.
Coefficient e is from the integration constant k
and is necessarily invariant, and a constant
that results from k, there seems to be no data
available to assign sign.
Now lets have some fun by considering two
values of e , e =0 and then e=1, (or e=-1).
When e=0 Eq.(kst1) becomes
g_uv = K_uv
and IMO represents an orthogonal space
(orthogonal meaning one where Cartesian
perpendicular base vectors i,j,k are valid).
OTOH, when e=1 then
g_uv = K_uv +A_u*B_v Eq.(kst1)
is valid, because A_u*B_v is non-zero,
and represents a space where the Cartesian
base vectors i,j,k are invalid.
To recap for clarity, If a relative tensor |K_uv| =1
then e is an invariant constant, that defines
departure from Euclidian space.
Sofar, this post is mathematical, but it has very
good physical equivalence.
I've argued in other threads the gravitational
constant is in fact a relative invariant of weight 2,
and have included this as the quantity K in
Eq.(kst1).
Secondly, the geometric quantity e that appears
within the above is physically the fundamental
charge +e or -e.
Finally, I wan to share my current thinking
on the relation of the Gravitational constant K,
and the fundamental charge e in a general metric.
Opinions and corrections welcomed,
flames are ok, if I'm an idiot, (I'm an old man
senility is a fact of life) take your best shot.
Thanks if anyone read this far...
Ken S. Tucker
===
Subject: How did Euler determine Euler's Constant?
I know that :
Euler's constant = Lim (n-> infinity) [ Sum(i/j) from j=1 to n
- ln(n)]
but, I can't see how one actually evaluates this relation to
get the value
of Euler's constant.
Thankx,
MB
===
Subject: Re: How did Euler determine Euler's Constant?
> I know that :
> Euler's constant = Lim (n-> infinity) [ Sum(i/j) from j=1 to
n - ln(n)]
> but, I can't see how one actually evaluates this relation to
get the
value
> of Euler's constant.
I gather that, to you, to determine a constant is to get its
value.
Not everyone shares this way of seeing things.
In the first place, to this day nobody knows the value of
Euler's
constant - we may know it to a thousand or a million or a
billion
decimal places, but that still leaves quite a few decimal
places
that we don't know. Nor do we know how to express it in terms
of
pi and e and square root of 2 and suchlike.
So I'll assume you're asking, how did Euler evaluate the
constant
to however many decimals he got?
Well, how many decimals did he get? Do you know whether he
evalua
it at all? Seems to me you have to answer that one before you
even
ask the question (that I think) you're asking.
But there are a few things Euler might have used, if he did
want to
evaluate the constant to a few decimals. One of them is called
the
Euler-Maclaurin formula, so one presumes Euler had some
familiarity
with it. I recommend looking it up.
And if you really want to know about Euler's constant, there's
a
nice recent book called Gamma, by Havil, which will tell you
heaps
about it.
--
===
Subject: Feigenbaum Number
What is a good introduction?
/David
===
Subject: Re: Feigenbaum Number
> What is a good introduction?
> /David
Here are few web sites where you can get your feet wet:
http://www.fortunecity.com/emachines/e11/86/expmaths.html#
Feigenbaum
http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Chaos/
Chaos.html
http://www.stud.ntnu.no/~berland/math/feigenbaum/
His actual papers on this topic are:
Quantative Universality for a Class of Non-linear
Transformations, J.
Stat. Physics, vol. 19, 25-52, 1978.
The Universal Metric Properties of Nonlinear Transformations,
J. Stat.
Physics, vol. 21, 669-706, 1979
FFeigenbaum, M. J., Quantitative Universality for a Class of
Nonlinear
Transformations, Journal of Statistical Physics 19, 25-52
Feigenbaum, M. J., Quantitative Universality for a Class of
Nonlinear
Transformations, Journal of Statistical Physics 19, 25-52
(1978)
(1978)eigenbaum, M. J., Quantitative Universality for a Class
of
Nonlinear
Transformations, Journal of Statistical Physics 19, 25-52
(1978)
===
Subject: Re: Feigenbaum Number
> His actual papers on this topic are:
Quantative Universality for a Class of Non-linear
Transformations, J.
> Stat. Physics, vol. 19, 25-52, 1978.
The Universal Metric Properties of Nonlinear Transformations,
J.
Stat.
> Physics, vol. 21, 669-706, 1979
> FFeigenbaum, M. J., Quantitative Universality for a Class of
Nonlinear
> Transformations, Journal of Statistical Physics 19, 25-52
> Feigenbaum, M. J., Quantitative Universality for a Class of
Nonlinear
> Transformations, Journal of Statistical Physics 19, 25-52
(1978)
> (1978)eigenbaum, M. J., Quantitative Universality for a
Class of
Nonlinear
> Transformations, Journal of Statistical Physics 19, 25-52
(1978)
Do you think his supervisor counts these as 5 papers?
===
Subject: Limsup, Liminf
If a sequence of reals {s_n} converges, then the corresponding
Limsup
equals its Liminf. These 2, however, appear to be independent
of the
particular ordering of {s_n}. Is it accurate to state, then,
that
Limsup=Liminf does not necessarily imply the convergence of
{s_n}?
Any help/explanation apprecia.
===
Subject: Re: Limsup, Liminf
> If a sequence of reals {s_n} converges, then the
corresponding Limsup
> equals its Liminf. These 2, however, appear to be
independent of the
> particular ordering of {s_n}. Is it accurate to state, then,
that
> Limsup=Liminf does not necessarily imply the convergence of
{s_n}?
> Any help/explanation apprecia.
You can rearrange a SERIES and get a different value, not a
sequence.
--
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Limsup, Liminf
> If a sequence of reals {s_n} converges, then the
corresponding Limsup
> equals its Liminf. These 2, however, appear to be
independent of the
> particular ordering of {s_n}.
So is convergence. If a sequence of reals converges, so does
any
rearrangement of the sequence.
--
===
Subject: Re: Limsup, Liminf
>If a sequence of reals {s_n} converges, then the
corresponding Limsup
>equals its Liminf. These 2, however, appear to be independent
of the
>particular ordering of {s_n}. Is it accurate to state, then,
that
>Limsup=Liminf does not necessarily imply the convergence of
{s_n}?
No, a sequence has a limit if and only if the limsup equals the
liminf.
Independent of the ordering is a little informal, but the
_limit_
of a sequence is _also_ independent of the ordering, in the
same sense in which the liminf and limsup are. (Ie, if (n_j) is
a permutation of the integers then the limit of s_{n_j} is the
same as the limit of s_n.)
>Any help/explanation apprecia.
************************
===
Subject: Re: Limsup, Liminf
> If a sequence of reals {s_n} converges, then the
corresponding Limsup
> equals its Liminf. These 2, however, appear to be
independent of the
> particular ordering of {s_n}. Is it accurate to state, then,
that
> Limsup=Liminf does not necessarily imply the convergence of
{s_n}?
> Any help/explanation apprecia.
Since you're working on the real number line, this simplifies
things a bit.
You can think of the LimSup as the largest limit point in a
sequence of
real
numbers, and similarly, you can think of LimInf as the
smallest limit point
in a sequence of real numbers. If they are equal, then the
sequence has a
unique limit point, and hence converges.
This holds even when LimSup and LimInf are +/- infinity also
(I'll let you
think about that).
MB
===
Subject: Re: Homological algebra
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at 04:02 PM, AAA <math1060@cox.net> said:
>Can someone suggest a good book on Homological algebra??
What are you looking for? Does it have to be selfcontained?
Does it
have to be current?
Take a look at Homology by MacLane. I also vaguely recall a
Homological Algebra by MacLane and Eilenberg, but perhaps I'm
thinking of a different book.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolici bulk E-mail will be 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 reply to spamtrap@library.lspace.org
===
Subject: Re: Homological algebra
Content-transfer-encoding: 8bit
> Take a look at Homology by MacLane. I also vaguely recall a
Homological Algebra by MacLane and Eilenberg, but perhaps I'm
> thinking of a different book.
i have cartan & eilenberg homological algebra, released as a
paperback in the princeton landmarks in math series.
rip
===
Subject: Re: Homological algebra
> Can someone suggest a good book on Homological algebra??
I'm not sure where you are starting from, but I have been
reading Hilton and Wu's _A_Course_in_Modern_Algebra_. The
final chapter introduces homological techniques (Ext and Tor).
This book has much to recommend it, but for now let me just
say that it is a reasonably thin book that starts out nice
and slow--groups--and concentrates basically on
concepts which will lead to homological techniques, without
much else. For example, there is a chapter on category theory
including a discussion of abelian categories and adjoint
functors.
Projective and injective modules are discussed at length.
The major problem with this book is that it is currently being
offered by in the Wiley Classic's Library at approximately
$100,
which I suspect gives them a healthy profit margin. However,
copies
are available on line.
Best wishes,
Mike
===
Subject: Re: Homological algebra
>>Can someone suggest a good book on Homological algebra??
> I'm not sure where you are starting from, but I have been
> reading Hilton and Wu's _A_Course_in_Modern_Algebra_. The
> final chapter introduces homological techniques (Ext and
Tor).
> This book has much to recommend it, but for now let me just
> say that it is a reasonably thin book that starts out nice
> and slow--groups--and concentrates basically on
> concepts which will lead to homological techniques, without
> much else. For example, there is a chapter on category theory
> including a discussion of abelian categories and adjoint
functors.
> Projective and injective modules are discussed at length.
> The major problem with this book is that it is currently
being
> offered by in the Wiley Classic's Library at approximately
$100,
> which I suspect gives them a healthy profit margin. However,
copies
> are available on line.
> Best wishes,
> Mike
That REALLY bugs me when publishers take a classic, photo-copy
it,
and sell it for an outrageous price.
My own (and paid for!) example is Hardy and Wright's classic
An Introcuction to the Theory of Numbers. I got the paperback
edition
in the early 1990's for $32.95. The list price is now $61.95.
This is ABSURD!
Martin Cohen
===
Subject: Re: Grad school: late entry
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>I'd like some advice from people who have been in my
position; esp.
>those who have made it through grad school!
Well, everybody is different, so what worked for me may not
work for
you. Why did you go back to graduate school? Why in
Mathematics? If
you don't love Mathematics, then possibly you might be better
off
rethinking your academic goals.
I went back to grad school just short of 5 years after leaving
college, and was working 7 days a week on top of my studies. I
selec classes based on what I was interes in rather than what
might be useful after graduation. Those were the best years of
my
life.
>I know graduate school is hard for everyone, but is hard work
the
>only solution?
It's an essential ingredient.
>Because at the end of the day, for first year students at
least, it
>is the grades on the exams and the courses that really matter,
>irrespective of mathematical potential.
That attitude is part of your problem. It is the material and
the
insights that matter; the grades are only a way to measure
those. If
you're not there for the sake of learning then you are going
to have
problems. Find topics that you want to learn for their own
sakes, and
work hard on those. If you're not having fun in the process
then
something is wrong.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolici bulk E-mail will be 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 reply to spamtrap@library.lspace.org
===
Subject: Re: Grad school: late entry
> Hallo everyone,
> I worked for about 3 years after college before getting into
grad
> school. (I am currently in the first year). And I find math
so much
> more difficult now, than I ever did; keeping up with the
pace of the
> courses is turning out to be a nightmare. I have forgotten
even some
> elementary results; and find myself stuck in a painstakingly
slow
> iterative process (going back to my old Calc III notes...).
My profs
> want me to speed up things, but I cannot seem to find an
optimal path.
> I'd like some advice from people who have been in my
position; esp.
> those who have made it through grad school!
here is some useful advice: ABANDON SHIP.
exception: you love math for its own sake.
> I know graduate school is hard for everyone, but is hard
work the
> only solution? Is there a way of working smart instead of
hard?
> Because at the end of the day, for first year students at
least, it is
> the grades on the exams and the courses that really matter,
> irrespective of mathematical potential.
> Thanks a lot in advance.
 Frank
> P.S.--I am not sure if there is any wisdom in sampling life
for a
> few years before beginning grad school.
===
Subject: Submartingales and Gambling problem help
To all math/probability scholars, please help with the
following game :
Consider the following game :
Deck of 52 cards. You start out with k dollars. You place a
bet on the
first hand. The game is that you guess whether the next card
will be black
or red. The dealer flips over a card and if you're right, you
receive
whatever you bet, and if you lose, your bet is gone. You keep
playing this
game until all 52 cards run out. The caveat is that after the
first card
is
gone, you can use that information to your advantage for the
next card. So
say the first card was red, then obviously you will bet that
the next card
will be black since you know there are 26 blacks in the deck
and 25 reds.
So this is a favorable game since if you bet p dollars, you
receive p if
you
win, and lose p if you lose (and on some hands your odds are
greater than
50% chance of winning, and the minimum odds of winning any
hand are 50%
since you are allowed to memorize all of the cards that were
already
played)
My question is, how much would you pay to play this game? In
other words,
what is the optimal strategy for this game and what is your
expec
winnings using this strategy?
I can put a minimum bound on the game as follows : If you just
bet 0 on
every hand until the last card, you know what the last card is
since you
memorized all the previous cards. Then you bet k dollars and
you win k
dollars (since you know the last card). So at minimum you can
expect to
win
k dollars, so you should pay k dollars to play the game since
in the end
you gain 2k and end up with k (which is what you star with).
Can I use martingale theory to solve this problem? Is this a
famous
problem? What is the solution?
Thanks in advance,
Michael
===
Subject: Help...Never seen this before
I have been given this math problem and i have never seen this
symbol
before...it would be greatly apprecia if someone would lend
their
advice:
(superscript) 4
Find the value of: (uppercase pi symbol) (2k + 1)
k=0
sorry for the complica layout, i know of no way to insert the
symbols
Thanks
--
tml
===
Subject: Re: Help...Never seen this before
<vaa12059iej4bdbdl1u724hs90ac0ev5q4@4ax.com>:
> I have been given this math problem and i have never seen
this symbol
> before...it would be greatly apprecia if someone would lend
their
> advice:
> (superscript) 4
> Find the value of: (uppercase pi symbol) (2k + 1)
> k=0
> sorry for the complica layout, i know of no way to insert the
> symbols
> Thanks
I believe you are referring to a product symbol (not sure of
the exact
terminology for it).
So, what you've described would mean to multiply all values of
the
expression (2k+1) where k takes on each of the values from 0
up to 4.
so:
[2(0)+1][2(1)+1][2(2)+1][2(3)+1][2(4)+1]
like that.
--
Sheila King
http://www.thinkspot.net/sheila/
http://www.k12groups.org/
--
tml
===
Subject: Re: Help...Never seen this before
X-No-Archive: yes
><vaa12059iej4bdbdl1u724hs90ac0ev5q4@4ax.com>:
>> I have been given this math problem and i have never seen
this symbol
>> before...it would be greatly apprecia if someone would lend
their
>> advice:
>> (superscript) 4
>> Find the value of: (uppercase pi symbol) (2k + 1)
>> k=0
>> sorry for the complica layout, i know of no way to insert
the
>> symbols
>> Thanks
>I believe you are referring to a product symbol (not sure of
the exact
>terminology for it).
>So, what you've described would mean to multiply all values
of the
>expression (2k+1) where k takes on each of the values from 0
up to 4.
>so:
>[2(0)+1][2(1)+1][2(2)+1][2(3)+1][2(4)+1]
>like that.
Sheila, I'm wondering if Potterwasp isn't referring to a
summation,
uppercase sigma. Either that, or educate me as to where this
sort of
operation, a progressive product might be used.
Please.
--
--
tml
===
Subject: Re: Help...Never seen this before
X-No-Archive: yes
><vaa12059iej4bdbdl1u724hs90ac0ev5q4@4ax.com>:
> I have been given this math problem and i have never seen
this symbol
>> before...it would be greatly apprecia if someone would lend
their
>> advice:
(superscript) 4
>> Find the value of: (uppercase pi symbol) (2k + 1)
>> k=0
> Sheila, I'm wondering if Potterwasp isn't referring to a
summation,
> uppercase sigma. Either that, or educate me as to where this
sort of
> operation, a progressive product might be used.
> Please.
Hello Charlie,
Since she said pi and not sigma I doubt she is referring to
summation.
It really is hard to find an example of this on Google, since
I have no
idea what to call it. To my recollection, the best I can
remember is
product. lol. But of course that returns many other results as
well.
In any case I have found an example for you.
http://www.math.rutgers.edu/courses/573A/573-f00/S1/plnt0573.
pdf
See page five of the above document.
For one example, this is used in numerical methods for
interpolation of
polynomials. There are many uses for it in higher mathematics.
--
Sheila King
http://www.thinkspot.net/sheila/
http://www.k12groups.org/
--
tml
===
Subject: Re: Help...Never seen this before
>><vaa12059iej4bdbdl1u724hs90ac0ev5q4@4ax.com>:
>I have been given this math problem and i have never seen
this symbol
> before...it would be greatly apprecia if someone would lend
their
> advice:
(superscript) 4
> Find the value of: (uppercase pi symbol) (2k + 1)
> k=0
>> Sheila, I'm wondering if Potterwasp isn't referring to a
summation,
>> uppercase sigma. Either that, or educate me as to where
this sort of
>> operation, a progressive product might be used.
>> Please.
> Hello Charlie,
> Since she said pi and not sigma I doubt she is referring to
summation.
> It really is hard to find an example of this on Google,
since I have no
> idea what to call it. To my recollection, the best I can
remember is
product. lol. But of course that returns many other results as
well.
> In any case I have found an example for you.
>
http://www.math.rutgers.edu/courses/573A/573-f00/S1/plnt0573.
pdf
> See page five of the above document.
> For one example, this is used in numerical methods for
interpolation of
> polynomials. There are many uses for it in higher
mathematics.
In bioinformatics, we use the product symbol, an uppercase pi,
a lot.
We often have to take the joint probability of many events,
which (if
you assume independence) is the product of the probability of
the
individual events. I also get a lot of products when dealing
with
Dirichlet distributions, which are the most popular prior
distribution
for discrete alphabets in Bayesian statistics, since they form
a
conjugate prior (the posterior probability distribution is
from the
same family of functions as the prior probability
distribution, with a
particularly simple change of parameters).
See
http://www.soe.ucsc.edu/research/compbio/dirichlets/
dirichlet-papers.html
for a pointer to a paper on Dirichlet mixtures that has a lot
of
product symbols.
One also gets products a lot when doing combinatorics, though
many of
them can be re-written in terms of the factorial function and
the
binomial coefficients.
The upper-case pi is a completely standard and extremely useful
notational tool---it should be taught in pre-calculus classes
at
about the same time that ln(x) and exp(x) are taught.
--
Kevin Karplus karplus@soe.ucsc.edu
http://www.soe.ucsc.edu/~karplus
life member (LAB, Adventure Cycling, American Youth Hostels)
Effective Cycling Instructor #218-ck (lapsed)
Professor of Computer Engineering, University of California,
Santa Cruz
Undergraduate and Graduate Director, Bioinformatics
Affiliations for identification only.
--
tml
===
Subject: factoring quadratic equations
how do you factor an equation such as
15s^2 minus 16st plus 4t^2
I hope that that makes sense. Please if someone could just
list the
steps or somehow explain it, that would be very helpful
--
tml
===
Subject: Re: factoring quadratic equations
> how do you factor an equation such as
> 15s^2 minus 16st plus 4t^2
> I hope that that makes sense. Please if someone could just
list the
> steps or somehow explain it, that would be very helpful
Since 15*4 = 60
and
(-10)(-6) = 60
and
(-10) + (-6) = -16
Then re-write your expression as:
15 s^2 + (-10)st + (-6)st + 4 t^2
Now factor by grouping:
5s(3s - 2t) + -2t(3s - 2t)
(3s - 2t)(5s - 2t)
Check the above answer by multiplying it back out together to
see if it
gives the original expression.
Or, some people are just good at guessing/seeing the numbers
needed and can
go pretty much straight from the original expression you
provided to the
answer. I think to some extent, that comes with practice.
--
Sheila King
http://www.thinkspot.net/sheila/
http://www.k12groups.org/
--
tml
===
Subject: Re: factoring quadratic equations
> how do you factor an equation such as
> 15s^2 minus 16st plus 4t^2
> I hope that that makes sense. Please if someone could just
list the
> steps or somehow explain it, that would be very helpful
that is just the product of two binomials in the form of:
(as + bt) x (cs - dt)
to factor this, you will have to create all the possible
combinations of
factors of 15 (for the s term) and 4 (for the t term) or in
another way,
the a and c will be factor pairs of 15 while b and d will be
factor pairs
of 4. It is basically trial and error until you get the right
combo to
multiply out to the one you seek. the work is left for you.
Steve
--
tml
===
Subject: Re: factoring quadratic equations
X-No-Archive: yes
>how do you factor an equation such as
>15s^2 minus 16st plus 4t^2
15s^2 = 15s * s or 5s * 3s
4t^2 = 4t * t or 2t * 2t
It follows that
(15s - 4t)(s - t) = 15s^2 - 19st + 4t^2 ... Nup
(15s - t)(s - 4t) = 15s^2 - 61st + 4t^2 ... Strike Two
(5s - 2t)(3s - 2t) = 15s^2 -16st + 4t^2 ... Eureka!
hth
--
--
tml
===
Subject: Re: MAPLE:Lists of Random numbers
> I need to generate about 8 to lists of 15 to 20 random
> numbers each betwee 0.1 and 0.25 with a defined average
value and a
> defined standard deviation. It is possible to do this with
MAPLE
> vesrsion 9.0 ?
> If not, are ther any programs out there that would do this?
> I would also like to be able to import the lists to a Word
or PDF
> document.
The 'main' problem (as Julian Noble already no) is to specify
your desired distribution since you can not simpliy cut off the
values: you have to say which pdf you want.
For an 'unlimi' usual normal distribution it is simple to get
10 values by mu:=0; sigma:=1; [stats[random,
normald[mu,sigma]](10)];
===
Subject: Re: MAPLE:Lists of Random numbers
>> I need to generate about 8 to lists of 15 to 20 random
>> numbers each betwee 0.1 and 0.25 with a defined average
value and a
>> defined standard deviation. It is possible to do this with
MAPLE
>> vesrsion 9.0 ?
>> If not, are ther any programs out there that would do this?
>> I would also like to be able to import the lists to a Word
or PDF
>> document.
>The 'main' problem (as Julian Noble already no) is to specify
>your desired distribution since you can not simpliy cut off
the
>values: you have to say which pdf you want.
Maple can help you choose the distribution, of course.
>For an 'unlimi' usual normal distribution it is simple to get
>10 values by mu:=0; sigma:=1; [stats[random,
normald[mu,sigma]](10)];
One possibility is to use a transformation of a uniform random
variable.
A two-parameter family will be needed, adjusting the
parameters to get
the desired mean and standard deviation. For example, suppose
we use
a piecewise linear function g whose graph is the polygon from
(0, 0.1) to (c,d) to (1, 0.25).
> a:= 1/10: b:= 1/4:
g:= x -> piecewise(x<c, a+x*(d-a)/c, d+(x-c)/(1-c)*(b-d)):
Calculate E[g(X)] and E[g(X)^2] if X is uniform on the
interval [0,1].
> int(g(x),x=0..1): m1:= normal(%) assuming c>0,c<1;
int(g(x)^2, x=0..1): m2:= normal(%) assuming c>0,c<1;
3 1 1
m1 := - -- c + - d + -
40 2 8
1 7 1 1 1 2
m2 := - -- d c - --- c + -- d + -- + - d
20 400 12 48 3
For example, to get mean 0.2 and standard deviation 0.04:
solve({m1 = 0.2, m2 = 0.2^2 + 0.04^2});
{d = .1960000000, c = .3066666667}
Note that we need 0 <= c <= 1 and a <= d <= b for this to be
valid.
Now to get 15 random values:
> myg:= subs(%,eval(g));
map(myg, [stats[random,uniform](15)]);
[.2099308423, .1798419315, .2463333850, .2221223959,
.1484944004,
.2387900236, .2251865567, .2471118664, .2401995410,
.2426749747,
.2149171000, .1263350463, .1209928651, .2205407466,
.1699889792]
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Decker Quadratic, algebraic integers, done
Turns out there's a rather direct approach to showing a
problem with
the old concepts about the ring of algebraic integers.
Decker, a sci.math poster, pos (reference information at
bottom)
the quadratic
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his a's are roots of
a^2 - (x - 1)a + 7(x^2 + x).
Letting x=2, you have
a_1(2)^2 - a_1(2) + 42 = 0, which gives
a_1(2) = (1 + sqrt(-167))/2 as one of two solutions.
Now consider the quadratic
y^2 - by - 7 = 0, which has as one of its roots
(b + sqrt(b^2 + 28))/2.
Note that root is an algebraic integer factor of 7 for all
algebraic
integers y, and b.
Now consider
(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2
which is
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
and working to eliminate square root terms gives
28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0
and working still further I get
196 z^4 + 28b z^3 + (186b^2 + 2324) z^2 - 168bz + 7056 = 0
and I can divide both sides by 28 to finally get
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0.
Importantly, for any integer b, such that it is irreducible
over Q,
*none* of the solutions for z can be an algebraic integer!
First question may be, after starting out with
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))z
why do I have a quartic and what are the other roots?
Well if I used
(1 - sqrt(-167)) = (b + sqrt(b^2 + 28))z
I'd get the same result, as eliminating the square roots at
each
points creates *two* possible solutions that will work. I
handled two
square roots, so I have four solutions, which are
z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28))
z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28))
z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28))
z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28))
Now then, imagine that there exists some algebraic integer b
for which
it is reducible over Q, then the root will be a fraction with
a 7 in
the denominator.
Let c/7 be such a root, where c is then an integer, then I'd
have
(1 + sqrt(-167)) = (b + sqrt(b^2 + 28))c/7, or
(1 - sqrt(-167)) = (b + sqrt(b^2 + 28))c/7, or
(1 + sqrt(-167)) = (b - sqrt(b^2 + 28))c/7, or
(1 - sqrt(-167)) = (b - sqrt(b^2 + 28))c/7.
but for any of those possibilities, the result on the right
side has
to be coprime to 7,
since (b + sqrt(b^2 + 28))/2 [(b - sqrt(b^2 + 28))/2] = -7.
If you go to b's that are not integers, but are some arbitrary
algebraic integer, you would get a defining polynomial for z
of higher
and higher degree, but it'd still be non-monic, and that
result would
still follow.
The problem is that no matter how high you go, just like with
7z^4 + 5z^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0
your result would be a non-monic polynomial, so for it to ever
have a
rational root, at least one rational root would have to be a
fraction
with a 7 in the denominator, as remember 7 is prime.
Therefore, (1 + sqrt(-167)) has no non-unit factors in common
with 7
in the ring of algebraic integers!!!
QED
===
Subject: Numerical Integration in Maple9
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1CNrqL26895;
===
I am solving a set of 3 PDEs in Maple 9. The variables are:
1. C[O2,agg](r,t)
2. eta[agg](r,t)
3. J[agg](r,t)
I have the appropriate governing equations and boundary
consitions.
One of the boundary conditions for the third PDE is the
following
> BC6 := r=r[agg], J[agg](r,t) =
evalf(Int(4.0*pi*r^2*(diff(C[agg]*epsilon[agg]*epsilon[mem]*
eta[agg](r,t),t)
- n*F*R[orr]),r=0..r[agg])): BC6;
which is involving differential of eta[agg](r,t) with time and
C[O2,agg](r,t). Can this kind of boundary condition be handled
by
Maple 9.
madhu
===
Subject: geometry
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1D3G8N10976;
===
the different type of triangle
===
Subject: nsa
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i1DIVUm15150;
===
in any case ...
en tout cas , cela nous change de l'analyse coventionnelle et
de de tous les Abramowitz Stegun et compagnie ...
!
JP Fortin
===
Subject: Re: Journals and my papers
Dear Newsgroup:
I have been sending my papers for possible publications, well,
they
are really unconventional.
The two papers on Riccati Methods was not suitable for the IMA
journal
on Numerical Analysis, neither for IMA journal of Applied
Math. I
tried EuroMath Bulletin but they said they are not in business
any
more.
So I send it to the electronic Journal of differential
equations.
The paper on Polynomials and Riccati was quickly rejec within
two
days of its submission and I send it to the IMA journal of
applied
mathematics, it was not suitable too.
So I send it to the Central European Journal of Mathematics.
looking for suitable referees. I had reques information about
the
suitability of the papers, still looks like they are searching.
These three papers are very difficult to judge.
I had demonstra a Maple solution of a Riccati in my paper on
Associa Legendre. The Maple solution had a very complex
integral,
once evalua, and it was involving a lot of sum of logarithmic
functions.
My two solutions do not have any integrals, logarithms and even
trigonometric and also no exponential functions.
This is the difference in methods and the use of algorithms.
Certainly
the future applications of methods should open a new package
in the
future symbolic software, treating such solutions.
The other ones in new trigonometric solutions of Bessel,
possibly it
is difficult for them to admit that no past masters saw this
new set.
I hope they could find a suitable referee.
The Journal of Algebra apparently in thinking over the papers.
Well, these are records of the future history of matematics
textbooks.
Sincerely
Dr.Mehran Basti
===
Subject: Four rela papers published by GTP
Originator: israel@math.ubc.ca (Robert Israel)
***** ****** *****
Geometry and Topology Publications are deligh to announce the
publication of four rela papers which jointly make major
advances
in 3-manifold topology, including a proof of property P and
strong
genus bounds for knots.
The four papers are:
(1) A few remarks about symplectic filling
by Yakov Eliashberg
(2) On symplectic fillings
by John B Etnyre
(3) Witten's conjecture and Property P
by P B Kronheimer and T S Mrowka
(4) Holomorphic disks and genus bounds
by Peter Ozsvath and Zoltan Szabo
Full details follow:
(1)
URL:
http://www.maths.warwick.ac.uk/gt/GTVol8/paper6.abs.html
Title:
A few remarks about symplectic filling
Author(s):
Yakov Eliashberg
Abstract:
We show that any compact symplectic manifold (W,omega) with
boundary
embeds as a domain into a closed symplectic manifold, provided
that
there exists a contact plane xi on dW which is weakly
compatible with
omega, i.e. the restriction omega|xi does not vanish and the
contact orientation of dW and its orientation as the boundary
of the
symplectic manifold W coincide. This result provides a useful
tool for
new applications by Ozsvath-Szabo of Seiberg-Witten Floer
homology
theories in three-dimensional topology and has helped complete
the
Kronheimer-Mrowka proof of Property P for knots.
Secondary: 57M50
Keywords:
Contact manifold, symplectic filling, symplectic Lefschetz
fibration,
open book decomposition
Proposed: Robion Kirby
Seconded: Peter Ozsvath, Dieter Kotschick
Author(s) address(es):
Department of Mathematics, Stanford University
Stanford CA 94305-2125, USA
Email: eliash@gauss.stanford.edu
(2)
Algebraic and Geometric Topology
URL:
http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-5.abs.html
Title:
On symplectic fillings
Author(s):
John B Etnyre
Abstract:
In this note we make several observations concerning symplectic
fillings. In particular we show that a (strongly or weakly)
semi-fillable contact structure is fillable and any filling
embeds as
a symplectic domain in a closed symplectic manifold. We also
relate
properties of the open book decomposition of a contact
manifold to its
possible fillings. These results are also useful in proving
property P
295-310] and in showing the contact Heegaard Floer invariant
of a
fillable contact structure does not vanish [P Ozsvath and Z
Szabo,
Secondary: 57M50
Keywords:
Tight, symplectic filling, convexity
Author(s) address(es):
Department of Mathematics, University of Pennsylvania
209 South 33rd St, adelphia, PA 19104-6395, USA
Email: etnyre@math.upenn.edu
URL: http://math.upenn.edu/~etnyre
(3)
URL:
http://www.maths.warwick.ac.uk/gt/GTVol8/paper7.abs.html
Title:
Witten's conjecture and Property P
Author(s):
P B Kronheimer and T S Mrowka
Abstract:
Let K be a non-trivial knot in the 3-sphere and let Y be the
3-manifold obtained by surgery on K with surgery-coefficient
1. Using
tools from gauge theory and symplectic topology, it is shown
that the
fundamental group of Y admits a non-trivial homomorphism to
the group
SO(3). In particular, Y cannot be a homotopy-sphere.
Secondary: 57R17
Keywords:
3-manifold, knot, surgery, homotopy sphere, gauge theory
Proposed: Robion Kirby
Seconded: John Morgan, Ronald Stern
Author(s) address(es):
Department of Mathematics, Harvard University
Cambridge MA 02138, USA
and
Department of Mathematics, Massachusetts Institute of
Technology
Cambridge MA 02139, USA
Email: kronheim@math.harvard.edu, mrowka@math.mit.edu
(4)
URL:
http://www.maths.warwick.ac.uk/gt/GTVol8/paper8.abs.html
Title:
Holomorphic disks and genus bounds
Author(s):
Peter Ozsvath and Zoltan Szabo
Abstract:
We prove that, like the Seiberg-Witten monopole homology, the
Heegaard
Floer homology for a three-manifold determines its Thurston
norm. As a
consequence, we show that knot Floer homology detects the
genus of a
knot. This leads to new proofs of certain results previously
obtained
using Seiberg-Witten monopole Floer homology (in collaboration
with
Kronheimer and Mrowka). It also leads to a purely
Morse-theoretic
interpretation of the genus of a knot. The method of proof
shows that
the canonical element of Heegaard Floer homology associa to a
weakly symplectically fillable contact structure is
non-trivial. In
particular, for certain three-manifolds, Heegaard Floer
homology gives
obstructions to the existence of taut foliations.
Secondary: 57M27, 57N10
Keywords:
Thurston norm, Dehn surgery, Seifert genus, Floer homology,
contact
structures
Proposed: Robion Kirby
Seconded: John Morgan, Ronald Stern
Author(s) address(es):
PO: Department of Mathematics, Columbia University New York,
NY 10025, USA
and
Institute for Advanced Study, Princeton, New Jersey 08540, USA
SZ: Department of Mathematics, Princeton University
Princeton, New Jersey 08544, USA
Email: petero@math.columbia.edu, szabo@math.princeton.edu
===
Subject: Paper published by Geometry and Topology
Originator: israel@math.ubc.ca (Robert Israel)
The following paper has been published:
URL:
http://www.maths.warwick.ac.uk/gt/GTVol8/paper9.abs.html
Title:
Formal groups and stable homotopy of commutative rings
Author(s):
Stefan Schwede
Abstract:
We explain a new relationship between formal group laws and
ring
spectra in stable homotopy theory. We study a ring spectrum
deno DB
which depends on a commutative ring B and is closely rela to
the
topological Andre-Quillen homology of B. We present an explicit
construction which to every 1-dimensional and commutative
formal group
law F over B associates a morphism of ring spectra F_*: HZ -->
DB from
the Eilenberg-MacLane ring spectrum of the integers. We show
that
formal group laws account for all such ring spectrum maps, and
we
identify the space of ring spectrum maps between HZ and DB.
That
description involves formal group law data and the homotopy
units of
the ring spectrum DB.
Secondary: 14L05
Keywords:
Ring spectrum, formal group law, Andre-Quillen homology
Proposed: Bill Dwyer
Seconded: Thomas Goodwillie, Haynes Miller
Author(s) address(es):
Mathematisches Institut, Universitaet Bonn
53115 Bonn, Germany
Email: schwede@math.uni-bonn.de
===
Subject: Paper published by Geometry and Topology
Originator: israel@math.ubc.ca (Robert Israel)
An erratum to an announcement made earlier today is given at
the end.
---
The following paper has been published:
URL:
http://www.maths.warwick.ac.uk/gt/GTVol8/paper10.abs.html
Title:
Extended Bloch group and the Cheeger-Chern-Simons class
Author(s):
Walter D Neumann
Abstract:
We define an extended Bloch group and show it is naturally
isomorphic
to H_3(PSL(2,C)^delta;Z). Using the Rogers dilogarithm
function this
leads to an exact simplicial formula for the universal
Cheeger-Chern-Simons class on this homology group. It also
leads to an
independent proof of the analytic relationship between volume
and
Chern-Simons invariant of hyperbolic 3-manifolds conjectured by
Neumann and Zagier [Topology 1985] and proved by Yoshida
[Invent. Math. 1985] as well as effective formulae for the
Chern-Simons invariant of a hyperbolic 3-manifold.
Secondary: 19E99, 57T99
Keywords:
Extended Bloch group, Cheeger-Chern-Simons class, hyperbolic,
3-manifold
Proposed: Robion Kirby
Seconded: Shigeyuki Morita, Benson Farb
Author(s) address(es):
Department of Mathematics, Barnard College
Columbia University, New York, NY 10027, USA
Email: neumann@math.columbia.edu
Erratum:
=======
Paper 6 (A few remarks about symplectic filling by Yakov
Eliashberg)
announced as part of the set of four papers earlier today, was
proposed by Leonid Polterovich NOT Robion Kirby as announced.
===
Subject: Re: derived tensor product
Originator: israel@math.ubc.ca (Robert Israel)
dir:
>> Let R be a commutative ring. Does there exist a nonzero
object X in
>> the (unbounded) derived category for R so that the derived
tensor
>> product of X with itself is zero? (I think that this is
impossible if
>> R is noetherian, but what if it isn't?) For any n>=2, does
there
>> exist an object X so that the n-fold derived tensor product
of X with
>> itself is zero, but the (n-1)-fold derived tensor product
is not?
>> Here's a rela non-example: notice that as Z-modules, Q/Z
tensor Q/Z
>> is zero. In the derived category of Z, though, I think that
the
>> derived tensor product of Q/Z with itself is isomorphic to
Q/Z in
>> degree 1 (or degree -1, depending on how you grade things).
This is
>> correct, right? (I'm not very good with the derived tensor
product,
>> but it's legitimate to compute it using flat resolutions,
isn't it?)
>> --
>> J. H. Palmieri
>> Dept of Mathematics, Box 354350
mailto:palmieri@math.washington.edu
>> University of Washington
http://www.math.washington.edu/~palmieri/
>> Seattle, WA 98195-4350
>If nobody else does i try a sketch based on my dusty
Hartshorne [H]
Residues and Duality, Chap II  4 and 5.
>For derived tensors (or local hyperTor) i think you need some
left
>boundedness (as you need flat resolutions).
No: you don't need boundedness.
This hypothesis is not necessary in order to derive functors
in the
category of complexes. You can take a look at:
V.K.A.M Guggenheim - J.P. May, On the theory and applications
of
N. Spalstentein, Resolutions of unbounded complexes, Comp.
Math. 65
(1988)
A. Roig, Mod.8fles minimaux et foncteurs d.8eriv.8es, JPAA 91
(1994).
V. Hinich, Homological algebra of homotopy algebras, 25 (1997).
the category of complexes. You can find a note from the author
in
Arxiv.)
Agust.92 Roig
===
Subject: Balls and points with integer coordinates.
Consider R^n, n>1, with the usual Euclidean metric. Call a
point a lattice
point if it has integer coordinates, so the lattice points in
R^n are
simply Z^n.
It is not very hard to show that given any non-negative
integer m
there exist a closed ball B in R^n (usual metric) such that B
contains
exactly m lattice points in its interior. I remember having
seen or
heard somewhere, that the same result is true when the
interior is
replaced with the boundary. My question is the following :
What is known about the coupled problem where one looks for
pairs of
non-negative integers (k,m) such that there exist a closed
ball in R^n
having exactly k lattice points on its boundary and exactly m
in its
interior ??
-Anders
===
Subject: Re: subgroups of linear algebraic groups
>> Let F be a field. Does there exist a subgroup G of GL(n,F)
with the
>> following properties?
>> 1) G acts transitively on the set of lines in F^n (i.e. on
P^{n-1}(F)
>> )
>> 2) Denote by D the subgroup of diagonal matrices, and Z the
subgroup
>> of scalar matrices (i.e. the centre of GL(n,F)). Then G
intersects D
>> only in Z.
>> In other words, I want a group which acts transitively on
lines, but
>> whose diagonal elements act trivially on lines. Does such a
group
>> exist?
A couple of comments on your question. If G acts transitively
on lines,
GZ acts transitively on all the nonzero elements of F^n. So
you may as
well look for G that acts transitively on all the nonzero
elements.
Being diagonal is a property of matrices not of maps, which
means it is
not a very 'natural' question as it stands.
Some of the answers given so far seem to be wrong.
>SL(n,F) will always do it if F is algebraically closed.
Probably will do
>anyway, but wouldn't like to guarantee it if F is finite, say.
If char F is not 2 and n > 2, SL(n,F) contains diagonal
matrices which
are not scalar multiples of the identity. Eg
diag(-1,-1,1,...1).
>There are others too, to see why, just imagine n=3 F=R for
ease of
>visualization - pick two lines, then either rotate about an
axis
>orthogonal to both, or reflect in some plane and you see
O(3,R) and
>SO(3,R) will do.
SO(3,R) contains diag(-1,-1,1).
I think this does work: If there is a field or division ring D
which
contains F and has degree n over F, then the multiplicative
group D' of
D acts linearly on D regarded as a vector space over F, so you
can
identify D' with a subgroup of GL(n,F). It is transitive
because if a,b
in D', ax=b has a solution in D'. If you choose a basis for D
which
contains 1 (the identity of F regarded as a vector in D) then
D' has the
'diagonal implies scalar' property with respect to this basis.
This works for all n and quite a lot of choices of F, all
finite F for
example. For F=R, n=4 the quaternions will do for D. The case
F=R, n=3
remains open. Someone must know the answer to that least.
--
Graham Jones
http://www.visiv.co.uk
Emails to graham@visiv.co.uk may be dele as spam
Please add a j just before the @ to ensure delivery
===
Subject: Re: subgroups of linear algebraic groups
>Let F be a field. Does there exist a subgroup G of GL(n,F)
with the
>following properties?
>1) G acts transitively on the set of lines in F^n (i.e. on
P^{n-1}(F)
>)
>2) Denote by D the subgroup of diagonal matrices, and Z the
subgroup
>of scalar matrices (i.e. the centre of GL(n,F)). Then G
intersects D
>only in Z.
>In other words, I want a group which acts transitively on
lines, but
>whose diagonal elements act trivially on lines. Does such a
group
>exist?
I think not if F is a reasonable topological field (say
non-archimedian).
Sounds like you want a gp H which is anisotropic gp mod center
and acts
transtitively on a
non-compact set S. It seems to me H/stab cong S is a
contradiction.