mm-89
===

For the oneness of a number ?n' :if(n is even) n = n/2;>if(n
is odd) n = 3*n + 1;Keep doing this until n = 1. The amount
of necessary steps for this is>called the oneness of
n.m
===
I am trying to
build fractals generated by the principal of the oneness
of>>a> number.>>For the de?ition of the oneness of a number,
view the internet, or>>below.*It's more commonly called the
Collatz Conjecture or the 3x+1 problem.>>I have been
unseccesful in creating a pattern around the oneness
although>>I have struck on some intriguing details. These
mainly resolve around the>>issue of, if we take the oneness
of the oneness of a number, and keep>>doing that, how long
does it take ?til we reach 1?Depends on how many binary
digits the number has (which is related to the> magnitude of
the number) AND ALSO on the pattern of 1s and 0s in the>
binary representaion (which is NOT related to the magniyude
of the> number).>>(note that for 5 this never happens)It
doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1No, you didn't
listen
well, I take the oneness of the oneness for therecursive
oneness looplength.The oneness of 5 is 5.the oneness of 5 is
5, etc..>>I think I might be able to ?d a pattern in that
since, if I call>>this the recursive oneness loop length (if
anybody can think a better>>name, tell me),It's sometimes
refered to as the stopping time. Others make up their own>
terminology.>> then, the recursive oneness loop length of a
number like>>20000000 is still only 15. I am trying to make a
large-scale 2d map of it,How are you plotting this in 2D?1.
Using complex number representation and messing around with
that.2. Using y as a period of x (y = x/WIDTH, so to speak)--
Quaternion
===
>Message-id:
<cwerb.12363$2y.111726@phobos.telenet-ops.beMessage-id:
<NSdrb.12242$iN2.413788@phobos.telenet-ops.be>I am trying to
build fractals generated by the principal of the oneness
of>a>> number.>For the de?ition of the oneness of a number,
view the internet, or>below.*> It's more commonly called
the Collatz Conjecture or the 3x+1 problem.>I have been
unseccesful in creating a pattern around the oneness
although>I have struck on some intriguing details. These
mainly resolve around the>issue of, if we take the oneness of
the oneness of a number, and keep>doing that, how long does it
take ?til we reach 1?> Depends on how many binary digits
the number has (which is related to the>> magnitude of the
number) AND ALSO on the pattern of 1s and 0s in the>> binary
representaion (which is NOT related to the magniyude of the>>
number).>>(note that for 5 this never happens)> It
doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1No, you didn't
listen
well, I take the oneness of the oneness for the>recursive
oneness looplength.That is a bit hard to follow.>The oneness
of 5 is 5.>the oneness of 5 is 5, etc..Let me see if I got
this straight. If S(n) is the stopping time, you are
then?ding the stopping time of the stopping
time?S(n)S(S(n))S(S(S(n)))etc.until that reaches 1?So if the
stopping time of n is n (as with 5) you get a loop?I haven't
considered that before, any other loops other than 5?
Anything elseinteresting about doing that?I think I might be
able to ?d a pattern in that since, if I call>this the
recursive oneness loop length (if anybody can think a
better>name, tell me),> It's sometimes refered to as the
stopping time. Others make up their own>> terminology.>>
then, the recursive oneness loop length of a number
like>20000000 is still only 15. I am trying to make a
large-scale 2d map of it,> How are you plotting this in
2D?1. Using complex number representation and messing around
with that.>2. Using y as a period of x (y = x/WIDTH, so to
speak)That doesn't help much. Can you be more speci??--
>Quaternion--MensanatorAce of Clubs
===
>>Message-id:
<cwerb.12363$2y.111726@phobos.telenet-ops.be>Message-id:
<NSdrb.12242$iN2.413788@phobos.telenet-ops.be>I am trying to
build fractals generated by the principal of the oneness>>of
a> number.>>For the de?ition of the oneness of a number,
view the internet, or>>below.*> It's more commonly called
the Collatz Conjecture or the 3x+1 problem.>>I have been
unseccesful in creating a pattern around the
oneness>>although I have struck on some intriguing details.
These mainly resolve>>around the issue of, if we take the
oneness of the oneness of a number,>>and keep doing that, how
long does it take ?til we reach 1?> Depends on how many
binary digits the number has (which is related to> the
magnitude of the number) AND ALSO on the pattern of 1s and 0s
in the> binary representaion (which is NOT related to the
magniyude of the> number).>>(note that for 5 this never
happens)> It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1>>No, you
didn't listen well, I take the oneness of the oneness for
the>>recursive oneness looplength.That is a bit hard to
follow.>>The oneness of 5 is 5.>>the oneness of 5 is 5,
etc..Let me see if I got this straight. If S(n) is the
stopping time, you are> then ?ding the stopping time of the
stopping time?S(n)> S(S(n))> S(S(S(n)))> etc.until that
reaches 1?Yes, and the amount of recursive calls needed for
that I call the recursiveoneness loop length. Let's call this
ROLL(n) > So if the stopping time of n is n (as with 5) you
get a loop?Yeah,> I haven't considered that before, any other
loops other than 5? Anything> else interesting about doing
that?5 is not the only one that loops, any number of which
the oneness is 5 willcause a loop. The oneness of 32 is
5.Very few numbers have a oneness of 5. I'm not sure but
something tells methe only next one of which ROLL(n) becomes
in?ite is 2^32. (which causesthe trap 2^32->32->5->5->5->...
of course)And very few numbers n cause ROLL(n) to become
in?ite.Again like I said, it takes a long while for ROLL(n)
to become larger than16. This only happens regularily when n
exceeds 2000000000 (2 billion)You just gave me a cool idea
btw. Suppose you have a number ?s' and you callN(s) the
amount of numbers for which the oneness is ?s'. I'll
try
thatlater. There must be, for each ?s', at least one even and
at least one oddnumber that reaches them? Could those two
systems form two two-dimensionalaxises?Also, I made another
recursive loop that doesS(n)S(S(n))S(S(S(n)))Until this
reaches ?n', not one. Let's call this looplength the
recursiveoneness identity loop length, ROILL.Note that this
only happens for 1,2,3,4,5,7, and 16Also note that in my
de?ition ROLL(1) = ROILL(1) = 6.Since S(1) = 3S(3) = 7S(7) =
16S(16) = 4S(4) = 2S(2) = 1I think this set of numbers might
be interesting, in intuitive terms 3 and 7might be seen as
the ?prime' counterparts of the 2-powers 16 and 4, and 2as
the main axis, being as well prime as the smallest power of
2.Well perhaps you are into intuitive stuff and care to think
about it.Again note that this is my de?ition, some de?itions
would say S(1) = 0.I'm not a believer in that :)>>I think I
might be able to ?d a pattern in that since, if I call>>this
the recursive oneness loop length (if anybody can think a
better>>name, tell me),> It's sometimes refered to as the
stopping time. Others make up their own> terminology.>>
then, the recursive oneness loop length of a number
like>>20000000 is still only 15. I am trying to make a
large-scale 2d map of>>it,> How are you plotting this in
2D?>>1. Using complex number representation and messing
around with that.>>2. Using y as a period of x (y = x/WIDTH,
so to speak)That doesn't help much. Can you be more
speci??That's because called brainstorm coding and
it's a
bit hard to express withwritten language. I did say what is
necessary : I've tried to visualize anypossible patterns in
the one-dimensional series, let's leave it at thatbecause
I've tried many things.As for the complex numbers, I tried
dividing by 2+2i when the complex andreal parts are even, and
multiplying with 3+3i and adding 1+1i when theyare odd. I'm
still experimenting with that..-- Quaternion
===
Does two have
more oneness than one?
===
> Does two have more oneness than
one?In my de?tion, the oneness of one is 3.1->4->2->1takes
three iterationsIt could be that some de?itions say that the
oneness of 1 is 0, since itstarts at the stopping time. It
doesn't matter that much, and I prefer itthis way, where
S(1)=3. And thus larger than S(2)-- Quaternion
===
> Does two
have more oneness than one?In my de?tion, the oneness of one
is 3.1->4->2->1> takes three iterationsIt could be that some
de?itions say that the oneness of 1 is 0, since it> starts
at the stopping time. It doesn't matter that much, and I
prefer it> this way, where S(1)=3. And thus larger than
S(2)--> QuaternionIf you use an index to represent the
collatz-transformations,like C_0 = 1 C_1 = 1*2^1 C_2 = 1*2^2
C_a = 1*2^a C_2,0 = (C_2 -1)/3 C_a,0 = (C_a -1)/3 C_a,b =
C_a,0 * 2^b with the periodic property that C_0 = C_2,0 =
C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1 (which re?the
loop 1-4-2-1....) and so on, then this indicates in a very
natural way the oneness and stopping time of a number: it's
just the sum of all indexes plus the number of them: C_4,0 =
5 Stopping-time = (4+0) + (1) = 5 C_4,1,0 = 3 Stopping-time =
(4+1+0) + (2) = 7 Because of the loop at 1 it is also C_4,0 =
C_2,4,0 = C_2,2,....,2,4,0 = 5 so the stopping time is not
exactly unique. With the additional restriction of allowing
only the smallest number of indexes the de?ition is unique
then. For 1 the stopping time is then also unique, namely 0,
since the representation with the shortest index is C_0 = 1
Gottfried Helms
===
>> Does two have more oneness than
one?> In my de?tion, the oneness of one is 3.>
1->4->2->1>> takes three iterations> It could be that
some de?itions say that the oneness of 1 is 0, since>> it
starts at the stopping time. It doesn't matter that much, and
I prefer>> it this way, where S(1)=3. And thus larger than
S(2)> -->> QuaternionIf you use an index to represent the
collatz-transformations,> like C_0 = 1> C_1 = 1*2^1> C_2 =
1*2^2> C_a = 1*2^a C_2,0 = (C_2 -1)/3> C_a,0 = (C_a -1)/3>
C_a,b = C_a,0 * 2^b with the periodic property that> C_0 =
C_2,0 = C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1 (which re?he l
oop 1-4-2-1....) and so on, then this indicates in a very
natural way the oneness and> stopping time of a number: it's
just the sum of all indexes plus the> number of them:> C_4,0
= 5 Stopping-time = (4+0) + (1) = 5> C_4,1,0 = 3
Stopping-time = (4+1+0) + (2) = 7 Because of the loop at 1 it
is also C_4,0 = C_2,4,0 = C_2,2,....,2,4,0 = 5 so the stopping
time is not exactly unique. With the additional> restriction
of allowing only the smallest number of indexes the>
de?ition is unique then. For 1 the stopping time is then
also unique,> namely 0, since the representation with the
shortest index is C_0 = 1Damn let me take that in tonight,
it's quite complex for me to understandwhat you mean to
represent with ?C_a1,a2,a3..,an' right now and how it can?ht
the logic that if you follow the iteration scheme you get a
bijectionbetween n and the amount of iterations that are
needed to transform it to1..-- Quaternion
===
> Damn let me
take that in tonight, :-)) So I wish, that the nightmares
won't catch you... Gottfried(P.S. I had them already, just
some weeks ago :-)
===
Does two have more oneness than
one?>That seems vague, and depends on context. Is this
ONENESS part of some higheraspect of Mathematics beyond the
typical undergraduate study of Calculus? At a simpler level,
it seems that you could treat any single group of a
givenquantity as ONE WHOLE; so a grouped unit is one whole
set. This allows us to?d a fraction of any number as well as
?d the fraction of 1. G C
===
> It turns out that I can
isolate the current dispute easily enough by> focusing on the
factorizations:Consider(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5
b_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22where
even by inspection you can see that the constant terms are>
separated out, so that you have 1(1)(22) = 22, the constant
term of> the polynomial.I'll add that at x=0, a_1(0) = a_2(0)
= b_3(0) = 0.Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) +
22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)where again
you see that the constant terms match as now you have>
7(7)(22) = 1078, which is again the constant term of the
polynomial.If 22 does not have 7 as a factor, the former
factorization is the> *only* allowed way for 49 to divide
through.(For more detail, like what the a's are,
seehttp://mathforpro?.blogspot.com/where more is
explained.)I have a result that shows a problem with a
de?ition that> mathematicians have used for over a hundred
years, And, what, exactly, is the problem?> and rather than>
face the result which follows from rather basic algebra
mathematicians> are being pussies and running like scared
cowards from the result.Some posters, not professional
mathematicians from what I've gathered,> are at least trying
to stand and ?ht, but because what I have is a> math proof,
their claims are necessarily irrational.I need you to stand
up for the truth, here and now.Let's chase these
mathematician cowards down, and make them face the>
music.After all physics had its challenges and physicists
faced them, while> mathematicians seem to believe that they
can just ignore problems.Let's take ?em out.> James Harris>
http://mathforpro?.blogspot.com/
===
Let's take ?em out.
James Harris Take ?em out? What do you mean by that? Are you
back to your previous> threat of calling out the military or
the CIA and the FBI, or are you> advocating the formation of
a private militia?>I think James wants his vast silent
audience to treat mathematicians to> dinner and a movie. I
suspect that there's an ulterior motive.Physics students are
a cool crowd and I know because I was once aphysics
undergrad, and I'd get into all kinds of
fascinatingdiscussions, as physics is that kind of ?ld.When
you deal with crackpots and cranks, which mathematicians are
now,who have been backed into a logical corner, you go ahead
and deliverthe coup de grace of the irrefutable facts.That
is, you take ?em out.Now then, the crackpots may still
believe their nonsense, butcivilized society is once again
protected by the vigillance of activeand highly intelligent
minds, as the bulk of society can move on withthe truth.I've
isolated mathematicians by a logical argument, having made
amajor discovery--that's what physics people do--in what they
probablythink of as their area alone, and caught them trying
to dodge theresult and its implications.Remember
mathematicians have claimed that pure math was for thebene?
of humanity, but I've shown them trying to hide one of
themost astounding results in math history (one of three,
mind you).So it's time for the physics people to look at the
facts, stand forlogic over social needs, and take ?em
out.James Harris <8Barb.9828$y95.4857@nwrdny01.gnilink.net>
<3c65f87.0311081454.195d813a@posting.google.com>
===
>
Remember mathematicians have claimed that pure math was for
the> bene? of humanity, but I've shown them trying to hide
one of the> most astounding results in math history (one of
three, mind you).Now a lot of people think that James is an
egotist with no sense ofproportion. I hope that this quote
lays that accusation to rest. Aswe can see here, James
exhibits an appropriate sense of humility byallowing
(charitably, I'm sure) that his result is only one of
*three*of the greatest results in math history.Don't leave us
hanging here, James. What are the other two?-- Jesse Hughes If
you really think there's a bug you should report a bug.
Maybeyou're not using it properly... It turns out Luddites
don't know howto use software properly, so you should look
into that. -- Bill Gates
===
>Message-id:
<87smkxylda.fsf@phiwumbda.org> Remember mathematicians have
claimed that pure math was for the>> bene? of humanity, but
I've shown them trying to hide one of the>> most astounding
results in math history (one of three, mind you).Now a lot of
people think that James is an egotist with no sense
of>proportion. I hope that this quote lays that accusation to
rest. As>we can see here, James exhibits an appropriate sense
of humility by>allowing (charitably, I'm sure) that his
result is only one of *three*>of the greatest results in math
history.Don't leave us hanging here, James. What are the other
two?Didn't he solve The Riemann Hypothesis with his prime
counting function?-- >Jesse Hughes >If you really think
there's a bug you should report a bug. Maybe>you're
not using
it properly... It turns out Luddites don't know how>to use
software properly, so you should look into that. -- Bill
Gates--MensanatorAce of Clubs
===
It turns out that I can
isolate the current dispute easily enough by> focusing on the
factorizations: Consider [snipped] Crank Information>
http://www.crank.net/harris.html>
http://www.crank.net/usenet.html>
http://www.google.com/search?q=harris+site%3Awww.crank.net>
http://www.google.com/search?q=%22james+harris%22+site%
3Ausers.pandora.be Readers should please check out *all* the
links Sam the Worm listed. As for the math, notice that with
(5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3
- 18375 x^2 - 360 x + 22 no other factorization works as long
as 7 is not a factor of 22.You have only demonstrated anything
for the case x=0, not ?x' in general.--There are two things
you must never attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
===
I hope you're not
speaking poorly about my Venus lizard folk, Catharsand
all.This may be getting some folks a wee bit off topic from
group focus,or that of my intended agenda upon Venus life
(lizard folk and all),but according to some fairly recent
feedback, I've learned a thing ortwo about our nasty moon, as
a place that I believe we've long neededto establish a lunar
space elevator (LSE) in order to be ef?ientlygetting
ourselves off to visiting the wizard of Oz at Venus L2,
aswell as for reaching out to those thoroughly irradiated to
deathsnowman/snowwoman situated on
Mars:http://guthvenus.tripod.com/gv-cm-ccm-01.htmHere's a
little typical feedback of supposed facts from: Jay
Windley(webmaster@clavius.org)outside the solar system, and
our sun is the primary defense againstlower energy and thus
their secondary effects in the ambient areminimal.My
thoughts:
http://guthvenus.tripod.com/
gv-moon-radiation.htmUnfortunately, I may have incorrectly
utilized the terminology ofcosmic rays, though fortunately, I
never speci?d upon any speci?spectrum of high-energy cosmic
rays, just pointing out that our sunis certainly capable of
tossing out its fair share of far worse thingsthan visible
photons plus IR worth of BTUs and of those nasty
UVs.Obviously a supernovae is worth a thousand fold in terms
of beingnasty, thereby from the far off generated galactic
in?st offer ameasurable degree of such events, and of
the secondary radiation givenoff by all that infamous
clumping lunar dirt should become a fairlydarn good as well
as unobstructed indicator of whatevercosmic/galactic in?ems
fairly odd that shuch measurementsaren't common place,
as where's the justi?ation for not otherwiseproviding this
level of information?The assertions or premise offered by the
likes of Jay Windley, that ofour moon not only lacking an
atmosphere but also without a Van Allenbuffer zone is not
such a bad thing if you're out and about on thelunar surface,
seems somewhat risky if not downright lethal. I mighthave come
into that understanding if we're referring to an
earthshineilluminated lunar surface, but I'm not going so far
if that's of anyfully solar illuminated environment while
wearing a mostly syntheticmoon suit because, we're not
talking about avoiding a 270 nm UV sunburn.Sorry about all my
make-due reverse engineering logic, or lackthereof. I was
simply trying to establish upon the amount of solarradiation
that becomes hard X-Ray class. So exactly how much is it ona
typical lunar day, or how about on a good day as well as on a
trulyDo we suppose that happens to include the likes of the
last couple ofweeks of horri? solar ?ems there should
be some speci? knowledge (excluding Apollo) ofwhat's what
pertaining to the solar illuminated surface, as opposed tothe
absolute lunar nighttime environment and, of something
speci?pertaining to whatever earthshine contributes.This
task of obtaining knowledge is somewhat like my getting a
graspupon the applied energy (thrust or torque) involved in
acceleratingsomething the size and mass of the moon, so that
it accelerates andthereby recedes form Earth at 38 mm/year.As
worthy feedback provided from: Ami Silberman
(silber@mitre.org)The mechanisms for the lunar recession have
been well understood fordecades. In a nutshell, tides cause
friction between the oceans andthe ocean ? which
transfers energy from the solid part of theearth to the
oceans. One of the effects of this friction is that thetidal
bulge is off-center, and is located eastward of the moon.
(Sothe high tide actually occurs when the moon is west of
overhead.) Theresult of the tidal bulge being off center is
that there is a torgueeffect placed on the moon, and this in
turn transfers energy from theearth to the moon. The earth's
spin rate slows, the moon is speeded inits orbit and therefor
moves further away from the earth. (Thistransfer of energy is
essentially a transfer of angular momentum,which is a
conserved quantity.) The historical (over geological
eras)rate of recession has varied due to varying amounts of
tidal frictiondue to shallower or deeper oceans, and the
positions of thecontinents.For the bene? of all my loyal
critics, I've conceded that there's adarn good chance
that
the likes of Tim Thompson has more than a fewvalid points as
to his version of what's what. This following page isjust
another example of my learning from the pros, of accepting
otherinput, which may even including the likes of what Ami
Silberman justpresented, that I'd not be calling ?s
there actually seems to besome considerable worth to at least
Tim's version of the lunarrecession, if I don't say
so
myself.http://guthvenus.tripod.com/earth-moon-energy.htm
===

Unfortunately Scott, this sting/ruse is way so much bigger
than evenyou can imagine. I've posted this folowing in at
least a half dozentopics that I believe are related to
obtaining the truth.This may be getting some folks a wee bit
off topic from group focus,or that of my intended agenda upon
Venus life (lizard folk and all),but according to some fairly
recent feedback, I've learned a thing ortwo about our nasty
moon, as a place that I believe we've long neededto establish
a lunar space elevator (LSE) in order to be ef?ientlygetting
ourselves off to visiting the wizard of Oz at Venus L2,
aswell as for reaching out to those thoroughly irradiated to
deathsnowman/snowwoman situated on
Mars:http://guthvenus.tripod.com/gv-cm-ccm-01.htmHere's a
little typical feedback of supposed facts from: Jay
Windley(webmaster@clavius.org)outside the solar system, and
our sun is the primary defense againstlower energy and thus
their secondary effects in the ambient areminimal.My
thoughts:
http://guthvenus.tripod.com/
gv-moon-radiation.htmUnfortunately, I may have incorrectly
utilized the terminology ofcosmic rays, though fortunately, I
never speci?d upon any speci?spectrum of high-energy cosmic
rays, just pointing out that our sunis certainly capable of
tossing out its fair share of far worse thingsthan visible
photons plus IR worth of BTUs and of those nasty
UVs.Obviously a supernovae is worth a thousand fold in terms
of beingnasty, thereby from the far off generated galactic
in?st offer ameasurable degree of such events, and of
the secondary radiation givenoff by all that infamous
clumping lunar dirt should become a fairlydarn good as well
as unobstructed indicator of whatevercosmic/galactic in?ems
fairly odd that shuch measurementsaren't common place,
as where's the justi?ation for not otherwiseproviding this
level of information?The assertions or premise offered by the
likes of Jay Windley, that ofour moon not only lacking an
atmosphere but also without a Van Allenbuffer zone is not
such a bad thing if you're out and about on thelunar surface,
seems somewhat risky if not downright lethal. I mighthave come
into that understanding if we're referring to an
earthshineilluminated lunar surface, but I'm not going so far
if that's of anyfully solar illuminated environment while
wearing a mostly syntheticmoon suit because, we're not
talking about avoiding a 270 nm UV sunburn.Sorry about all my
make-due reverse engineering logic, or lackthereof. I was
simply trying to establish upon the amount of solarradiation
that becomes hard X-Ray class. So exactly how much is it ona
typical lunar day, or how about on a good day as well as on a
trulyDo we suppose that happens to include the likes of the
last couple ofweeks of horri? solar ?ems there should
be some speci? knowledge (excluding Apollo) ofwhat's what
pertaining to the solar illuminated surface, as opposed tothe
absolute lunar nighttime environment and, of something
speci?pertaining to whatever earthshine contributes.This
task of obtaining knowledge is somewhat like my getting a
graspupon the applied energy (thrust or torque) involved in
acceleratingsomething the size and mass of the moon, so that
it accelerates andthereby recedes form Earth at 38 mm/year.As
worthy feedback provided from: Ami Silberman
(silber@mitre.org)The mechanisms for the lunar recession have
been well understood fordecades. In a nutshell, tides cause
friction between the oceans andthe ocean ? which
transfers energy from the solid part of theearth to the
oceans. One of the effects of this friction is that thetidal
bulge is off-center, and is located eastward of the moon.
(Sothe high tide actually occurs when the moon is west of
overhead.) Theresult of the tidal bulge being off center is
that there is a torgueeffect placed on the moon, and this in
turn transfers energy from theearth to the moon. The earth's
spin rate slows, the moon is speeded inits orbit and therefor
moves further away from the earth. (Thistransfer of energy is
essentially a transfer of angular momentum,which is a
conserved quantity.) The historical (over geological
eras)rate of recession has varied due to varying amounts of
tidal frictiondue to shallower or deeper oceans, and the
positions of thecontinents.For the bene? of all my loyal
critics, I've conceded that there's adarn good chance
that
the likes of Tim Thompson has more than a fewvalid points as
to his version of what's what. This following page isjust
another example of my learning from the pros, of accepting
otherinput, which may even including the likes of what Ami
Silberman justpresented, that I'd not be calling ?s
there actually seems to besome considerable worth to at least
Tim's version of the lunarrecession, if I don't say
so
myself.http://guthvenus.tripod.com/earth-moon-energy.htm
===
I
posted an example earlier today to try and help those of you
who'vebeen so thoroughly confused by posters like Arturo
Magidin, NoraBaron and Dik Winter to see how it all works,
and it turns out thatif you were smart enough you should have
noticed something.Consider(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) =
49(x^3 + 5x^2 + 3x + 2)where the c's *should* be algebraic
integers. I think few of youwould have a problem realizing
that only two of the c's can have 7 asa factor with that
example!!!Now then, you have as a zero of the factorization x
= -7/c_1, so letx= -7/c_1, so you have49(-7^3/c_1^3 +
5(49)c_1^2 - 21/c_1 + 2) = 0which is2c_1^3 - 21 c_1^2 + 245
c_1 - 343 = 0.Please tell me that example is simple enough
that now you'll quit?hting me on my work.My guess is that
polynomial irreducible over Q, but I haven't checkedit since
I recently deleted off my math software because I had
toreinstall Windows XP.Now then, that example should cause
certain posters to postretractions, or they can try and
run.Some of you might want to get legal advice.James
Harrishttp://mathforpro?.blogspot.com/
===
> I posted an
example earlier today to try and help those of you who've>
been so thoroughly confused by posters like Arturo Magidin,
Nora> Baron and Dik Winter to see how it all works, and it
turns out that> if you were smart enough you should have
noticed something.Consider(c_1 x + 7)(c_2 x + 7)( c_3 x + 2)
= > 49(x^3 + 5x^2 + 3x + 2)where the c's *should* be
algebraic integers. I think few of you> would have a problem
realizing that only two of the c's can have 7 as> a factor
with that example!!!Now then, you have as a zero of the
factorization x = -7/c_1, so let> x= -7/c_1, so you
have49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0which
is2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.Please tell me that
example is simple enough that now you'll quit> ?hting me on
my work.My guess is that polynomial irreducible over Q, but I
haven't checked> it since I recently deleted off my math
software because I had to> reinstall Windows XP.Now then,
that example should cause certain posters to post>
retractions, or they can try and run.Some of you might want
to get legal advice. This example is interesting for several
reasons. One, it wasshown almost immediately by more than one
poster that thereis no factorization with integer coef?ients
of the formthat Harris claimed. In spite of that for a while
he did notretract his claim that c_1 etc. *should* be
algebraic integers. Two, he never acknowledged his error.
Three, hestarted a new thread elsewhere on the same topic.
Four, heclaimed in that thread that c_1, etc. - which he
himself described as algebraic *numbers* - could not be
written asthe quotient of two coprime algebraic integers. The
lattercontradicts an old and very deep theorem of Dedekind.
Altogether, a ?sco. Confusion, and one error piled on top of
another. But what is really ironic in this is that Harris
*himself*proved that there is no factorization of the
polynomial above with integer coef?ients - and he used the
identicalmethod that we have used many times to show that his
claimsregarding the factorization of the famous P(x) [i.e.,
thattwo of the a's must be divisible by 7 or 5 or f or
whatever] arefalse. So he evidently now understands and
accepts that. But he does not react to it logically. He
continues to say, I believe, that the a's *should* be
divisible by f,with never a real explanation of *should*. He
clings to aclearly-incorrect argument based on constant
terms. Heignores the fact that if his central argument were
correct, mathematicswould be inconsistent. He concludes I
think that the a's divided by f are objects, however with no
clear method of telling what is an object and what is not,
and no body of theorems regarding objects such as one has for
algebraic integers. Why should anyone care if you prove
something about objects, if no one can tell what an object is
? Actually I think the meaning of *should* is pretty clear.
The a's *should* be divisible by f because, if theyare not,
his entire proof collapses. Thus the true translation of
*should* is: *I desperately want it tobe true!*, because if
it's not, 8 years of work, torment,hope, and dreams of fame
and riches go down the drain! Best summed up as: backwards
logic, and pathos. Plus he is now running away from all
substantive objections.He responds to super?ial issues on
notation and deletessubstantive parts of posts where the real
problems with hislogic are confronted. He is starting new
threads at a frantically high rate, interspersed with rants,
whines, paranoid delusional claims, even legal threats,
posted to several different news groups simultaneously: the
actions of a frightened and desperate person, it would
appear. Nora B. > James Harris>
http://mathforpro?.blogspot.com/
===
... > Four, he > claimed
in that thread that c_1, etc. - which he himself > described
as algebraic *numbers* - could not be written as > the
quotient of two coprime algebraic integers. The latter >
contradicts an old and very deep theorem of Dedekind. The
deepness is not that an algebraic number can be written as
thequotient of two algebraic integer, but that they also can
be chosento be made coprime. That it is possible is because
the algebraicintegers form a gcd ring. That is you can de?e
some (sensible)form of gcd. I do not know whether this was
already known byDedekind. Arturo will probably know (he has
already posted about it).-- dik t. winter, cwi, kruislaan
413, 1098 sj amsterdam, nederland, +31205924131home:
bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
>...> Four, he> claimed in that
thread that c_1, etc. - which he himself > described as
algebraic *numbers* - could not be written as> the quotient
of two coprime algebraic integers. The latter> contradicts an
old and very deep theorem of Dedekind. The deepness is not
that an algebraic number can be written as the>quotient of
two algebraic integer, but that they also can be chosen>to be
made coprime. That it is possible is because the
algebraic>integers form a gcd ring. That is you can de?e
some (sensible)>form of gcd. I do not know whether this was
already known by>Dedekind. Arturo will probably know (he has
already posted about it).In fact, stronger than just a gcd
domain: you can write the gcd as alinear combination of the
two numbers (i.e., a Bezout domain). Yes,this was known by
Dedekind and by Kummer.In Dedekind's _Theory of Algebraic
Numbers_, for example (availablefrom Cambridge Mathematical
Library, translated and introduced by JohnA deeper
investigation will enable us to see that two
nonzero[algebraic] integers a,b have a ->greatest common
divisor<- [emphasisin the original], which ca be put in the
form aa' + bb' where a' andb'
are [algebraic] integers. This
important theorem is not at all easyto prove with the help of
the principles developed so far, but weshall later (section
30) be able to derive it simply from the theoryof ideals.In
section 30, after proving the ?iteness of the class number,
andthat any ideal A raised to the class number is principal
(abouttwo pages), he proves the result [I am changing, as I
did above, somegreek letters into latin letter]: Any two
algebraic integers, a,b have a common divisor d which can be
expressed in the form d= aa'+ bb', where a'
and b' are
likewise algebraic integers. Proof. We assume that the two
numbers a,b are nonzero, otherwise the theorem is evident.
Then it is easy to see that there is a ?ld F of ?ite degree
including both a and b. Let R be the domain of integers of
this ?ld, and let h be the number of classes of ideals. Now
let Ra = AD, Rb = BD, D^h = Rd_1 where D is the greatest
common divisor of the ideals Ra and Rb, and d_1 is in R.
Since a^h and b^h are divisible by D^h, we can write a^h =
a_1*d_1, b^h = b_1*d_1, Ra_1 = A^h, Rb_1 = B^h where a_1,b_1
are likewise in R. Also, since A and B are relatively prime
ideals, D will be the greatest common divisor of Ra_1 and
Rb_1 and, since the number 1 is in R, there will be two
numbers a_2 and b_2 in R satisfying the condition a_1*a_2 +
b_1*b_2 = 1, or a^h*a_2 + b^h*b_2 = d_1. If we now put d_1 =
d^h, then the integer d will be a common divisor of a and b,
since a^h and b^h are divisible by d_1, and hence, since
h>=1, we can put a_2*a^{h-1} = a'*d^{h-1}, b_2b^{h-1} =
b'*d^{h-1}, where a', b' are integers
satisfying the
condition aa'+bb' = d. QED If at least one of the two
numbers
a,b is nonzero, then the number d, and any of its associates,
deserves the name ->greatest<- common divisor of a,b
[emphasis in the original]. If d is a unit, then a,b may be
called ->relatively prime<- [emphasis in the original] and
two such numbers enjoy the characteristic property that any
number m divisible by a and b is also divisible by ab. This
because the equations
m=aa''=bb'', and
1=aa'+bb' imply
m=ab(a'b''+b'a'[Para
graph]), and the converse is equally valid, since
a,b are both nonzero.That's the end of Dedekind's
book.By the
way, I highly recommend Dedekind's book (and
Stillwell'sintroduction). It is not only a very good
introduction to Alg. NumberTheory, but it is amazing how
modern it is for something written over100 years ago. Modulo
a few precisions (Dedekind calls somethingmodules when we
would call them Z-modules, and so on), it could workas a
textbook in a class
today.
===
=======================================

===
=========It's not denial. I'm just very selective
about
what I accept as reality. --- Calvin (Calvin and
Hobbes)
===
======================================

===
=========Arturo
Magidinmagidin@math.berkeley.edu
===
welcome to Platform 3 and
1/7,the short proof of the last theorem of Fermat;Usenet is
just to small to contain it! > Some of you might want to get
legal advice. > But what is really ironic in this is that
Harris *himself*> proved that there is no factorization of
the polynomial > above with integer coef?ients - and he used
the identical> method that we have used many times to show
that his claims> regarding the factorization of the famous
P(x) [i.e., that> two of the a's must be divisible by 7 or 5
or f or whatever] are> false. So he evidently now understands
and accepts that. > Actually I think the meaning of *should*
is pretty > clear. The a's *should* be divisible by f
because, if they> are not, his entire proof collapses. Thus
the true > translation of *should* is: *I desperately want it
to> be true!*, because if it's not, 8 years of work, torment,>
hope, and dreams of fame and riches go down the drain! > Best
summed up as: backwards logic, and pathos. >
http://mathforpro?.blogspot.com/--ils duces
d'Enron!http://larouchepub.com/
===
> Consider > (c_1 x +
7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2) >
where the c's *should* be algebraic integers.Why should they
be algebraic integers?-- dik t. winter, cwi, kruislaan 413,
1098 sj amsterdam, nederland, +31205924131home: bovenover
215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
> I posted an example earlier
today to try and help those of you who've> been so thoroughly
confused by posters like Arturo Magidin, Nora> Baron and Dik
Winter to see how it all works, and it turns out that> if you
were smart enough you should have noticed something. Consider
(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)
where the c's *should* be algebraic integers.Why should they
be? Please provide a proof.--There are two things you must
never attempt to prove: the unprovable --and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
===
In sci.math, C.
Bond<cbond@ix.netcom.com><3FAE78C0.71CE871C@ix.netcom.com>:>
>> I posted an example earlier today to try and help those of
you who've>> been so thoroughly confused by posters like
Arturo Magidin, Nora>> Baron and Dik Winter to see how it all
works, and it turns out that>> if you were smart enough you
should have noticed something.>> Consider>> (c_1 x + 7)(c_2 x
+ 7)( c_3 x + 2) =>> 49(x^3 + 5x^2 + 3x + 2)>> where the c's
*should* be algebraic integers.Why should they be? Please
provide a proof.If he does provide a valid proof I'll be very
surprised; I'vealready proven that they're not.
:-)(How I
prove that my proof is valid is not entirely clear butI think
I can at least explain most of my steps. :-) )[.sigsnip]--
#191, ewill3@earthlink.netIt's still legal to go
.sigless.
===
> I posted an example earlier today to try and
help those of you who've> been so thoroughly confused by
posters like Arturo Magidin, Nora> Baron and Dik Winter to
see how it all works, and it turns out that> if you were
smart enough you should have noticed something.Consider(c_1 x
+ 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)where
the c's *should* be algebraic integers. No, you're
completely
off base on this. There is nofactorization of the form you
propose where c1, c2, andc3 are algebraic integers. This is
easy to see. First,note that the roots of [1] R(x) = x^3 +
5*x^2 + 3*x + 2are all algebraic integers - say, a, b, and c
- andtheir product is -2: a*b*c = -2.Since R(x) is
irreducible, a, b, and c are non-units.Now you can write your
original polynomial as 49*(x - a)*(x - b)*(x - c),and then
distribute the 49 among the factors howeveryou want - for
example, as (7*x - 7*a)*(7*x - 7*b)*(x - c),but then note
that neither 7*a nor 7*b can equal 7:remember, a, b, and c
are nonunits. It's easy to show no other distribution works
either. Bottom line: there is no factorization that looks
likewhat you have above. This is a dead end.> I think few of
you> would have a problem realizing that only two of the c's
can have 7 as> a factor with that example!!!Now then, you
have as a zero of the factorization x = -7/c_1, so let> x=
-7/c_1, so you have49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) =
0which is2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.Please tell me
that example is simple enough that now you'll quit> ?hting
me on my work.> Obviously the roots cannot be algebraic
integers. This is justanother way of proving what I showed
above, that is, there isno factorization of the form you
started with. And so what???> My guess is that polynomial
irreducible over Q, but I haven't checked> it since I
recently deleted off my math software because I had to>
reinstall Windows XP.> It's trivial to check. The only
possible roots are 2, -2, 1, and -1.None work.> Now then,
that example should cause certain posters to post>
retractions, or they can try and run.> Ridiculous. You
blundered on this one. Best admit it and move on.> Some of
you might want to get legal advice.> You're going to sue us
for pointing out your mistake? Nora B. James Harris>
http://mathforpro?.blogspot.com/
===
>[...]Now then, that
example should cause certain posters to post>retractions, or
they can try and run.Some of you might want to get legal
advice.Oops. Sounding like a raving lunatic _again_. You
knowsome people are starting to wonder about that...>James
Harris>http://mathforpro?.blogspot.com/David C. Ullrich
===
>
I posted an example earlier today to try and help those of you
who've> been so thoroughly confused by posters like Arturo
Magidin, Nora> Baron and Dik Winter to see how it all works,
and it turns out that> if you were smart enough you should
have noticed something.Consider(c_1 x + 7)(c_2 x + 7)( c_3 x
+ 2) = > 49(x^3 + 5x^2 + 3x + 2)where the c's *should* be
algebraic integers. I think few of you> would have a problem
realizing that only two of the c's can have 7 as> a factor
with that example!!!Now then, you have as a zero of the
factorization x = -7/c_1, so let> x= -7/c_1, so you
have49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0which
is2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.Please tell me that
example is simple enough that now you'll quit> ?hting me on
my work.Now what exactly is that supposed to show? Looks like
you are completely incapable of writing down one coherent
thought. You may want to check
outhttp://www.mentalhealth.com/rx/p23-pe07.htmlwhich won't
explain the maths, but it will explain the real problem you
are having.
===
> I posted an example earlier today to try and
help those of you who've> been so thoroughly confused by
posters like Arturo Magidin, Nora> Baron and Dik Winter to
see how it all works, and it turns out that> if you were
smart enough you should have noticed something.Consider(c_1 x
+ 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)where
the c's *should* be algebraic integers. Why?
===
In sci.math,
James
Harris<jstevh@msn.com><3c65f87.0311081551.261b9a0b@
posting.google.com>:> I posted an example earlier today to
try and help those of you who've> been so thoroughly confused
by posters like Arturo Magidin, Nora> Baron and Dik Winter to
see how it all works, and it turns out that> if you were
smart enough you should have noticed something.Consider(c_1 x
+ 7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)where
the c's *should* be algebraic integers.Sorry, they're
not.If
P(x) = 49*(x^3 + 5*x^2 + 3*x + 2) = 49 * (x - x_1) * (x -
x_2) * (x - x_3)for some x_i, then for some permutation of
these x_i,c_1 = -7/x_1c_2 = -7/x_2c_3 = -2/x_3(note that x_1
* x_2 * x_3 = -2 so c_1 * c_2 * c_3 = 49, as required)If one
substitutes x = -7/c, one getsc^3 * P(-7/c)/49 = 2*c^3 -
21*c^2 + 245*c - 343which is irreducible. Sorry James.If one
substitutes x = -2/c, c_3 satis?sc^3*P(-2/c)/98 = c^3 -
3*c^2 + 10*c - 4so we get one out of three.[rest snipped]--
#191, ewill3@earthlink.netIt's still legal to go
.sigless.
===
>I posted an example earlier today to try and
help those of you who've>been so thoroughly confused by
posters like Arturo Magidin, Nora>Baron and Dik Winter to see
how it all works, and it turns out that>if you were smart
enough you should have noticed something.Consider(c_1 x +
7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)where
the c's *should* be algebraic integers. I think few of
you>would have a problem realizing that only two of the c's
can have 7 as>a factor with that example!!!Let A denote the
ring of all algebraic integers and let r1, r2, r3denote the
zeros of f(x) = x^3 + 5x^2 + 3x + 2. Since f is monic and its
coef?ients are rational integers, itszeros are in A, but none
of the zeros is a unit in A since f isirreducible over Z and
its constant coef?ient is not a unit.The r's are not all
coprime to 2 in A since 2 dividesr1 * r2 * r3 = -2 in A.If
exactly two of the r's, say r1 and r2, were coprime to 2 in
Athen 2 would divide r3 in A, so r3 / 2 = -1 / (r1 * r2)
would be inA and each of -r1 * r2, r1 * r2, r1 and r2 would
be a unit in A.If exactly one of the r's, say r1, was coprime
to 2 in A then 2would divide r2 * r3 in A, so (r2 * r3) / 2 =
-1 / r1 would be in Aand r1 would be a unit in A.Hence, in
the factorization f(x) = (x - r1)(x - r2)(x - r3), noneof the
r's is coprime to 2 in A.The polynomial 49 * f(x) is not
primitive and there is an in?ityof ways in which its content
may be distributed over its factors.If c1, c2, c3 are any
algebraic integers whose product is 49 thenthere is a
factorization of 49 * f(x) in A[x] as49 * f(x) = (c1 * x - c1
* r1)(c2 * x - c2 * r2)(c3 * x - c3 * r3),and, since it has
the same zeros as f, every factorization of 49 * f(x) in A[x]
has this form. Since none of the r's is coprime to 2 in A,
none of the c * r's iscoprime to 2 in A. In particular, none
of them can be equal to -7.The supposed example cannot
exist.I can see no way out. You will have this problem in any
ring inwhich 2 and 7 are coprime.Now then, you have as a zero
of the factorization x = -7/c_1, so let>x= -7/c_1, so you
have49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0which
is2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.Please tell me that
example is simple enough that now you'll quit>?hting me on
my work.My guess is that polynomial irreducible over Q, but I
haven't checked>it since I recently deleted off my math
software because I had to>reinstall Windows XP.Now then, that
example should cause certain posters to post>retractions, or
they can try and run.Some of you might want to get legal
advice.>James Harris>http://mathforpro?.blogspot.com/John
Roberts-Jones
===
>I posted an example earlier today to try
and help those of you who've>been so thoroughly confused by
posters like Arturo Magidin, Nora>Baron and Dik Winter to see
how it all works, and it turns out that>if you were smart
enough you should have noticed something.Consider(c_1 x +
7)(c_2 x + 7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)where
the c's *should* be algebraic integers. I think few of
you>would have a problem realizing that only two of the c's
can have 7 as>a factor with that example!!!Let A denote the
ring of all algebraic integers and let r1, r2, r3> denote the
zeros of f(x) = x^3 + 5x^2 + 3x + 2.> Since f is monic and
its coef?ients are rational integers, its> zeros are in A,
but none of the zeros is a unit in A since f is> irreducible
over Z and its constant coef?ient is not a unit.The r's are
not all coprime to 2 in A since 2 divides> r1 * r2 * r3 = -2
in A.If exactly two of the r's, say r1 and r2, were coprime
to 2 in A> then 2 would divide r3 in A, so r3 / 2 = -1 / (r1
* r2) would be in> A and each of -r1 * r2, r1 * r2, r1 and r2
would be a unit in A.If exactly one of the r's, say r1, was
coprime to 2 in A then 2> would divide r2 * r3 in A, so (r2 *
r3) / 2 = -1 / r1 would be in A> and r1 would be a unit in
A.Hence, in the factorization f(x) = (x - r1)(x - r2)(x -
r3), none> of the r's is coprime to 2 in A.The polynomial 49
* f(x) is not primitive and there is an in?ity> of ways in
which its content may be distributed over its factors.> If
c1, c2, c3 are any algebraic integers whose product is 49
then> there is a factorization of 49 * f(x) in A[x] as> 49 *
f(x) = (c1 * x - c1 * r1)(c2 * x - c2 * r2)(c3 * x - c3 *
r3),> and, since it has the same zeros as f, every
factorization of > 49 * f(x) in A[x] has this form. Since
none of the r's is coprime to 2 in A, none of the c *
r's is>
coprime to 2 in A. In particular, none of them can be equal to
-7.> The supposed example cannot exist.Now that's funny. You
must mean that the c's can't be algebraicintegers as,
of
course, they can be algebraic numbers, right?I mean, you are
NOT trying to say that the c's just can't exist atall,
right?
> I can see no way out. You will have this problem in any ring
in> which 2 and 7 are coprime.So I guess you're saying that if
2 and 7 are coprime in a ring thenthe c's, can't be in
that
ring. You're wrong, and it's easy to
prove.I'll use a
separate thread though, for my own reasons.James Harris
===
>>
>>I posted an example earlier today to try and help those of
you who've>>been so thoroughly confused by posters like
Arturo Magidin, Nora>>Baron and Dik Winter to see how it all
works, and it turns out that>>if you were smart enough you
should have noticed something.>>Consider>>(c_1 x + 7)(c_2 x +
7)( c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)>>where the c's
*should* be algebraic integers. I think few of you>>would
have a problem realizing that only two of the c's can have 7
as>>a factor with that example!!!> Let A denote the ring
of all algebraic integers and let r1, r2, r3>> denote the
zeros of f(x) = x^3 + 5x^2 + 3x + 2.> Since f is monic
and its coef?ients are rational integers, its>> zeros are in
A, but none of the zeros is a unit in A since f is>>
irreducible over Z and its constant coef?ient is not a
unit.> The r's are not all coprime to 2 in A since 2
divides>> r1 * r2 * r3 = -2 in A.> If exactly two of the
r's, say r1 and r2, were coprime to 2 in A>> then 2 would
divide r3 in A, so r3 / 2 = -1 / (r1 * r2) would be in>> A
and each of -r1 * r2, r1 * r2, r1 and r2 would be a unit in
A.> If exactly one of the r's, say r1, was coprime to 2
in A then 2>> would divide r2 * r3 in A, so (r2 * r3) / 2 =
-1 / r1 would be in A>> and r1 would be a unit in A.>
Hence, in the factorization f(x) = (x - r1)(x - r2)(x - r3),
none>> of the r's is coprime to 2 in A.> The polynomial
49 * f(x) is not primitive and there is an in?ity>> of ways
in which its content may be distributed over its factors.>>
If c1, c2, c3 are any algebraic integers whose product is 49
then>> there is a factorization of 49 * f(x) in A[x] as>> 49
* f(x) = (c1 * x - c1 * r1)(c2 * x - c2 * r2)(c3 * x - c3 *
r3),>> and, since it has the same zeros as f, every
factorization of >> 49 * f(x) in A[x] has this form. >
Since none of the r's is coprime to 2 in A, none of the c *
r's is>> coprime to 2 in A. In particular, none of them can
be equal to -7.>> The supposed example cannot exist.Now
that's funny. You must mean that the c's
can't be
algebraic>integers as, of course, they can be algebraic
numbers, right?I mean, you are NOT trying to say that the c's
just can't exist at>all, right?We seem to be agreed that there
is no factorization in the givenform of 49 * f(x) in A[x]. If
we instead work in the ring of polynomials over the
algebraicnumbers, then of course49 * f(x) = (c1 * x + 7)(c2 *
x + 7)(c3 * x + 2),where c1 = -7 / r1, c2 = -7 / r2 and c3 =
-2 / r3 = r1 * r2.Only c3 is an algebraic integer.I hope you
are not suggesting that c1 and c2 should be algebraicintegers
on the grounds that they may be written as the ratio oftwo
algebraic integers. >> I can see no way out. You will have
this problem in any ring in>> which 2 and 7 are coprime.So I
guess you're saying that if 2 and 7 are coprime in a ring
then>the c's, can't be in that ring. Yes, if 2 and 7
are not
units.> You're wrong, and it's easy to
prove.I'll use a
separate thread though, for my own reasons.I look forward to
seeing it.>James HarrisJohn Roberts-Jones
===
I posted an
example earlier today to try and help those of you
who've>been so thoroughly confused by posters like Arturo
Magidin, Nora>Baron and Dik Winter to see how it all works,
and it turns out that>if you were smart enough you should
have noticed something.Consider(c_1 x + 7)(c_2 x + 7)( c_3 x
+ 2) = 49(x^3 + 5x^2 + 3x + 2)where the c's *should* be
algebraic integers. I think few of you>would have a problem
realizing that only two of the c's can have 7 as>a factor
with that example!!! Let A denote the ring of all algebraic
integers and let r1, r2, r3> denote the zeros of f(x) = x^3 +
5x^2 + 3x + 2. Since f is monic and its coef?ients are
rational integers, its> zeros are in A, but none of the zeros
is a unit in A since f is> irreducible over Z and its constant
coef?ient is not a unit. The r's are not all coprime to 2 in
A since 2 divides> r1 * r2 * r3 = -2 in A. If exactly two of
the r's, say r1 and r2, were coprime to 2 in A> then 2 would
divide r3 in A, so r3 / 2 = -1 / (r1 * r2) would be in> A and
each of -r1 * r2, r1 * r2, r1 and r2 would be a unit in A. If
exactly one of the r's, say r1, was coprime to 2 in A then 2>
would divide r2 * r3 in A, so (r2 * r3) / 2 = -1 / r1 would be
in A> and r1 would be a unit in A. Hence, in the factorization
f(x) = (x - r1)(x - r2)(x - r3), none> of the r's is coprime
to 2 in A. The polynomial 49 * f(x) is not primitive and
there is an in?ity> of ways in which its content may be
distributed over its factors.> If c1, c2, c3 are any
algebraic integers whose product is 49 then> there is a
factorization of 49 * f(x) in A[x] as> 49 * f(x) = (c1 * x -
c1 * r1)(c2 * x - c2 * r2)(c3 * x - c3 * r3),> and, since it
has the same zeros as f, every factorization of> 49 * f(x) in
A[x] has this form. Since none of the r's is coprime to 2 in
A, none of the c * r's is> coprime to 2 in A. In particular,
none of them can be equal to -7.> The supposed example cannot
exist. Now that's funny. You must mean that the c's
can't be
algebraic> integers as, of course, they can be algebraic
numbers, right? I mean, you are NOT trying to say that the
c's just can't exist at> all, right? I can see no way
out.
You will have this problem in any ring in> which 2 and 7 are
coprime. So I guess you're saying that if 2 and 7 are coprime
in a ring then> the c's, can't be in that ring.
You're wrong,
and it's easy to prove. I'll use a separate thread
though,
for my own reasons.>Can't get out of the argument again, eh?
MAYBE if you took a few years andlearned some math, you'd see
that we're not lying like you think we are. I'dthink
if you
did this 7 years ago, you could almost have a Ph.D by now.
Butas I said before, you obviously don't care about the math,
you just want thefame and money. I do have a challenge for
you; go an entire thread withoutany childish insults.David
Moran James Harris
===
I can see no way out. You will have
this problem in any ring in>which 2 and 7 are coprime.I
should have said ... in which 2 and 7 are coprime
non-units.John Roberts-Jones
===
My guess is that polynomial
irreducible over Q, but I haven't checked> it since I
recently deleted off my math software because I had to>
reinstall Windows XP. Now then, that example should cause
certain posters to post> retractions, or they can try and
run. Some of you might want to get legal advice.> James
Harris> http://mathforpro?.blogspot.com/psychopathology:
narcissism---------------------------------------------------
-----------------------------A pattern of traits and
behaviours which signify infatuation and obsessionwith one's
self to the exclusion of all others and the egotistic
andruthless pursuit of one's grati?ation, dominance and
ambition.
===
>>My guess is that polynomial irreducible over
Q, but I haven't checked>>it since I recently deleted off my
math software because I had to>>reinstall Windows XP.>>Now
then, that example should cause certain posters to
post>>retractions, or they can try and run.>>Some of you
might want to get legal advice.>>James
Harris>>http://mathforpro?.blogspot.com/> psychopathology:
narcissism---------------------------------------------------
-------------------------> ----A pattern of traits and
behaviours which signify infatuation and obsession> with
one's self to the exclusion of all others and the egotistic
and> ruthless pursuit of one's grati?ation, dominance and
ambition.I think you've nailed it.Gib
===
Dear
Flip,http://malignantsel?ripod.com/npdglance.htmlThere
is a lot more here:http://malignantsel?ripod.com/Take
care.Sam Vaknin
===
Dear Flip,>
http://malignantsel?ripod.com/npdglance.html There is a
lot more here: http://malignantsel?ripod.com/ Take care.
Sam Vakninyou have excellent writing skills.I wonder if you
could answer a couple of questions regarding JSH's NPD?1. Is
it bad to feed the troll?2. Does feeding the troll only serve
to increase his high from all of theattention, particularly
when the comments are against his point of view?3. How can he
be helped?4. What can we do to subdue the troll?
===
> I wonder
if you could answer a couple of questions regarding JSH's
NPD?1. Is it bad to feed the troll?2. Does feeding the troll
only serve to increase his high from all of the> attention,
particularly when the comments are against his point of
view?3. How can he be helped?4. What can we do to subdue the
troll?> You may take a look at
http://www.mentalhealth.com/dis/p20-pe07.htmlwhich discusses
diagnosis and treatment. Treatment on sci.math is just plain
impossible. Ignoring him completely is probably the best we
could do for his mental health. To be able to treat him, you
would have to be a person of the highest possible authority
(and if you read his opinion of Andrew Wiles, you will see
that this person will be hard to ?d), and that person would
have to not contradict him (which would only make him feel
threatened and strengthen his illness), but gently lead him
to a realistic outlook on the world. Forget it, you won't ?d
anyone who could do that.
===
I will not go into any speci?s
in my response. Please do not forget that only a quali?d
mental health diagnostician- can determine whether someone
suffers from NPD and this, followinglengthy tests and
personal interviews.Still, these may be of
interest:http://malignantsel?ripod.com/journal67.
htmlhttp://malignantsel?ripod.com/faq4.htmlTake
care.Sam
===
> I posted an example earlier today to try and
help those of you who've> been so thoroughly confused by
posters like Arturo Magidin, Nora> Baron and Dik Winter to
see how it all works, and it turns out that> if you were
smart enough you should have noticed something. Consider (c_1
x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) where
the c's *should* be algebraic integers. I think few of you>
would have a problem realizing that only two of the c's can
have 7 as> a factor with that example!!! Now then, you have
as a zero of the factorization x = -7/c_1, so let> x= -7/c_1,
so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which
is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that
example is simple enough that now you'll quit> ?hting me on
my work. My guess is that polynomial irreducible over Q, but
I haven't checked> it since I recently deleted off my math
software because I had to> reinstall Windows XP. Now then,
that example should cause certain posters to post>
retractions, or they can try and run. Some of you might want
to get legal advice.> James Harris>
http://mathforpro?.blogspot.com/James, I thought you were
going back to lancing blisters and emptyingbedpans. What
happened?Lurch
===
> I posted an example earlier today to try
and help those of you who've> been so thoroughly confused by
posters like Arturo Magidin, Nora> Baron and Dik Winter to
see how it all works, and it turns out that> if you were
smart enough you should have noticed something. Consider (c_1
x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) where
the c's *should* be algebraic integers. I think few of you>
would have a problem realizing that only two of the c's can
have 7 as> a factor with that example!!! Now then, you have
as a zero of the factorization x = -7/c_1, so let> x= -7/c_1,
so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which
is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that
example is simple enough that now you'll quit> ?hting me on
my work. My guess is that polynomial irreducible over Q, but
I haven't checked> it since I recently deleted off my math
software because I had to> reinstall Windows XP. Now then,
that example should cause certain posters to post>
retractions, or they can try and run. Some of you might want
to get legal advice.>Legal advice? For what? Maybe if you
took some math courses, you'd have anidea of what everyone is
talking about and that none of us are lying.David Moran James
Harris> http://mathforpro?.blogspot.com/
===
> I posted an
example earlier today to try and help those of you who've>
been so thoroughly confused by posters like Arturo Magidin,
Nora> Baron and Dik Winter to see how it all works, and it
turns out that> if you were smart enough you should have
noticed something.>Hey James. Nora has the patience of an
absolute saint. Youshould be honored that she spends so much
time on herthoughtful reasoned replies to you. But no. You'd
ratheravoid answering questions, avoid any responseto the
points brought up, and avoid avoiding calling peopledog shit.
You are pitiful.
===
those of us who try to kill?e him.As
you've no doubt noticed though Robin, his posts (to
sci.mathanyway) do have an Achilles heel - a four-digit year
at the endof the title.If your newsreader supports wildcard
titles in kill rules, allone needs is a ?ter on titles of
the form /(19|20)dd$/(in Perl wildcard
syntax).-----------------------------------------------------
----------------------John R Ramsden
(jr@adslate.com)---------------------------------------------
------------------------------Eternity is a long time,
especially towards the end. Woody Allen
===
>Indeed, algebraic
topology seems to be an extremely powerful tool: Yes!>S^2 is
simply connected, Yes!>so every continuous image of it also
isNo! Take a balloon (S^2), de?t to a disk, roll it up
like a crepeor something to get an interval, and then join
the ends togetherto get a circle, which is not simply
connected.Sort of at the basis of the theory of covering
spaces is theobservation that the continuous image of a
simply-connected space isnot necessarily simply
connected.dave
===
>>so every continuous image of it also is
simply connected.> No! Take a balloon (S^2), de?t to a
disk, roll it up like a crepe> or something to get an
interval, and then join the ends together> to get a circle,
which is not simply connected.What do you mean with roll it
up like a crepe? And joining the endstogether is no
continuous operation!?Sort of at the basis of the theory of
covering spaces is the> observation that the continuous image
of a simply-connected space is> not necessarily simply
connected.> Yes, I see that I was wrong. But given a simply
connected top. space X and atop. space Y together with an
epimorphism f:X->Y, can we say that Y issimply connected? My
idea was that any continuous curve s:[0,1]->Y can bepulled
back to a continuous curve r:[0,1]->X such that f(r)=s so
that wecan prove that Y has to be simply connected. Can we
turn this idea into aproof? According to your post, the
answer seems to be no. So can you givean example where this
pull-back does not work?In the present case, we would have
X=S^2, Y=R^2{(0,0)} and f any continuousmap X->Y. Of course
it need not be an epimorphism, so the conclusion waswrong
even if the proposition mentioned above was correct.I'll
think about it further, but this is it for now.Tobias--
===
>
so every continuous image of it also is simply connected.>>
No! Take a balloon (S^2), de?t to a disk, roll it up
like a crepe>> or something to get an interval, and then join
the ends together>> to get a circle, which is not simply
connected.> What do you mean with roll it up like a crepe?
Sorry, too many math for poets lectures. That really just
means to takea projection to an interval. Everything to this
point is just theprojection pi : R^3 --> R^1, restricted to
S^2. (That's continuous.)>And joining the ends together is no
continuous operation!?Sure it is; it's the map f : t |--> (
cos(2pi t), sin(2pi t) ).Perhaps you're thinking of
continuous maps _with continuous inverses_,which this isn't.
If that's what you mean by continuous image, thenthe right
phrase is homeomorphic image, and then yes, thehomeomorphic
image of a simply-connected space is also simply-connected.>>
Sort of at the basis of the theory of covering spaces is the>>
observation that the continuous image of a simply-connected
space is>> not necessarily simply connected.>Yes, I see that
I was wrong. But given a simply connected top. space X and a>
top. space Y together with an epimorphism f:X->Y, can we say
that Y is> simply connected? No, that is still exactly the
setting for covering spaces. The usualexample is the same f I
gave above, where X is the real line andY is the circle.> My
idea was that any continuous curve s:[0,1]->Y can be> pulled
back to a continuous curve r:[0,1]->X such that f(r)=sThat's
called the path-lifting property; it's a consequence of f
being a covering map. (It's also possessed by other maps,
such as ?rations.)> so that we can prove that Y has to be
simply connected.No! First of all, the fact that you can lift
paths does not mean thatyou can lift loops; that is, if s is a
loop in Y, then it's alsoa path, and so it lifts to a path r
if f is a covering (say); butthis lifted path r need not be a
loop, i.e., there's no reason youmust have r(0)=r(1). If you
can't see this with formulas, draw apicture of a helix (a
spring) X, casting a circular shadow. The shadow, Yis a
circle and the projection is the map f. Draw a loop around Y,
andthen lift it back to X and you'll see you don't get
a
loop.Also you want to be a little careful : the fact that you
can liftpaths does not by itself guarantee that you can lift
other maps suchas homotopies (maps from [0,1]x[0,1] into Y).
If you _are_ in thecontext of covering spaces, there is a
pretty characterization ofwhat can be lifted: a map r : Z -->
Y induces a homomorphismr* : pi_1(Z) --> pi_1(Y), from which
you can show that a _necessary_condition for liftability is
that the subgroup r*(pi_1(Z)) mustbe contained in the image
f*(pi_1(X)). If f is a covering, thenthis condition is also
suf?ient. In particular, any map from acontractible space Z
into a space Y will lift to a map from Zinto any cover X of
Y.I know you never said you interested only in covering maps,
but thesethings put your confusion into stark relief. You
might want to thinkabout some of the classical examples of
these. Start with anysimply-connected space X and a discrete
group G of self-homeomorphismsof X and then let Y be the
quotient X/G . You could take(X,G) to be (R, Z), (R^2, Z^2),
(S^2, Z/2Z), (complex half-plane, modular group), (S^3,
binary icosahedral group), etc.dave
===
> Sorry, too many math
for poets lectures. That really just means to take> a
projection to an interval. Everything to this point is just
the> projection pi : R^3 --> R^1, restricted to S^2. (That's
continuous.) Sure it is; it's the map f : t |--> ( cos(2pi
t), sin(2pi t) ).> Ah, now your example is clear. Nice!>
Perhaps you're thinking of continuous maps _with continuous
inverses_,> which this isn't. If that's what you mean
by
continuous image, then> the right phrase is homeomorphic
image, and then yes, the> homeomorphic image of a
simply-connected space is also simply-connected.>> Yes, I
see that I was wrong. But given a simply connected top. space
X>> and a top. space Y together with an epimorphism f:X->Y,
can we say that Y>> is simply connected?>In the meantime, I
could ?ure this out. A morphism (here: a continuousmap)
f:X->Y is called an epimorphism iff there is a morphism
g:Y->X suchthat fg=id_Y. Using this de?ition, and any closed
curve s:[0,1]->Y, wecan easily lift it to gs:[0,1]->X and
still it is closed: gs(0)=gs(1).Because X is simply
connected, there is continuous c:[0,1]^2->X withc(0,t)=gs(t)
and c(1,t)=x for some x in X. Applying f gives
usfc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t) and fc(1,t)=f(x). So
Y is simplyconnected.But what does it mean for f to be an
epimorphism? fg=id_Y is equivalent to:for every S subset Y we
have g-(f-(S))=S (where the - sign denotespre-images; this is
just set-theoretic). Because of continuity reasons,
anecessary condition isS open <=> f-(S) open (*)For
surjective f with the property (*), you can take any
set-theoreticg:Y->X with fg=id_Y to prove the other direction
of the following theorem:f is an epimorphism in the category
of topological spaces, iff it issurjective and has property
(*).This is equivalent to that f be a quotient map. So being
an epimorphism ispretty special, so I agree with you that
continuous images ofsimply-connected spaces need not be
simply-connected.Tobias--
===
>> Yes, I see that I was wrong.
But given a simply connected top. space X>> and a top. space Y
together with an epimorphism f:X->Y, can we say that Y>> is
simply connected? In the meantime, I could ?ure this out. A
morphism (here: a continuous> map) f:X->Y is called an
epimorphism iff there is a morphism g:Y->X such> that
fg=id_Y. No - that's _not_ the usual (category-theoretic)
de?ition of epimorphism. Instead, f: X --> Y is an
epimorphism iff whenever g_1,g_2 : Y --> Z are morphisms such
that g_1 o f =g_2 o f, it follows that g_1 = g_2. (Your notion
might be called a split epimorphism ... )> Using this
de?ition, and any closed curve s:[0,1]->Y, we> can easily
lift it to gs:[0,1]->X and still it is closed: gs(0)=gs(1).>
Because X is simply connected, there is continuous
c:[0,1]^2->X with> c(0,t)=gs(t) and c(1,t)=x for some x in X.
Applying f gives us> fc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t)
and fc(1,t)=f(x). So Y is simply> connected. In the context
of covering spaces (where Dave Rusin was explaininf stuff),
the only covering maps f: X --> Y with X simply connected and
f an epimorphism in your sense are _homeomorphisms_ ... so Y
_would_ be simply connected in this (fairly uninteresting)
case.> But what does it mean for f to be an epimorphism?
fg=id_Y is equivalent to:> for every S subset Y we have
g-(f-(S))=S (where the - sign denotes> pre-images; this is
just set-theoretic). Because of continuity reasons, a>
necessary condition isS open <=> f-(S) open (*)For surjective
f with the property (*), you can take any set-theoretic>
g:Y->X with fg=id_Y to prove the other direction of the
following theorem:f is an epimorphism in the category of
topological spaces, iff it is> surjective and has property
(*).> This is equivalent to that f be a quotient map. ... Yep
...> So being an epimorphism is> pretty special, so I agree
with you that continuous images of> simply-connected spaces
need not be simply-connected.Tobias
===
No - that's _not_ the
usual (category-theoretic) de?ition> of epimorphism. Instead,
f: X --> Y is an epimorphism iff> whenever g_1,g_2 : Y --> Z
are morphisms such that g_1 o f =g_2 o f,> it follows that
g_1 = g_2. (Your notion might be called a split epimorphism
... )> Sorry for this one, the only reference I have for
category theory is Lang'sAlgebra, and he de?es epimorphism
and monomorphism only for abeliancategories.My notion is much
stronger; it implies the other, but if we take a set X andlet
X_1 be X with the discrete topology, X_2 with the indiscrete
topology,and de?e f:X_1->X_2 as the identity, then f is
continuous. It is anepimorphism, but does not have a right
inverse. In the context of covering spaces (where Dave Rusin
was explaininf> stuff), the only covering maps f: X --> Y
with X simply connected and> f an epimorphism in your sense
are _homeomorphisms_ ... so Y _would_> be simply connected in
this (fairly uninteresting) case.> Maybe, but since I am not
familiar with covering maps, I did not considerthis case.--
Tobias FritzStudent of Mathematics and PhysicsUniversity of
Heidelberg, Germany
===
>Does anyone know if there is any
potential one-way function>which is associative other than
modular exponentiation?How is this associative? Is (a^b)^c =
a^(b^c)? I think not!(And therefor I am not?)>That is, given
a function f(x,y) (which is easy to compute),>- given z and
x, it is dif?ult to ?d y such that z = f(x,y)>- f(f(x, y1),
y2) = f(f(x, y2), y1)That doesn't look like the associative
law, either.Setting the words aside, it sort of looks like
what you want occursin any group: given a group element x and
an integer n we computethe power x^n in the group; certainly
(x^n)^m = (x^m)^n, which iswhat you appear to request in your
last line. And yes, it's oftendif?ult to take an element in a
cyclic group and decide what power ofthe generator it is.Your
initial example is of this type, where the group is (Z/kZ)^*
, themultiplicative group mod k (where k is any ?ed
integer). Anotherexample commonly used for one-way functions
is to use the group ofpoints in an elliptic curve mod
p.dave
===
As most here are already aware (but just in case
some aren't),> if> you shift the digits of a number to the
left or right one time then it's> like multiplying or
dividing the number by its radix. So if you perform a> binary
left shift on say: 00011011 (27 base 10) you get: 00110110 (54
base 10 or 27 * 2) So you can multiply a number by any power
of two very quickly (for a> computer, register shifts are
much faster than mult/div. instructions) by> shifting. What
about numbers that are not powers of two? They can bedone> by
distributing the operation across numbers that are powers of
two andsum> to the number you're trying to multiply by. Say
you want to multiply by> 800. 800 is not a power of two, but
32, 256, and 512 are(32+256+512=800).> So: n*800 =
(n*32)+(n*256)+(n*512) = (n<<5)+(n<<8)+(n<<9) The same is
true for division. Binary right shifting is the same as>
dividing by two. I can't seem to work out the algebra though.
Cansomeone> tell me how (or is it possible) to do division by
a general integer using> shifts?> -- Best wishes,> AllenYou
could try not working binary but convert to the divisor as
radixinstead, and then reconvert to binary.Does this
help?
===
>> Ok, I've probably asked this question before, but
I've forgot the answer...>> What's the derivate of
f(x)=x^x
and f(x)=x^(x^(x-1)) ?>> I guess asking for the integral of
these functions would be foolish?>x^x = e^(x*log(x))>Now use
the chain and product rules.Another method, more general, is
to differentiate withrespect to each x as if all the others
are constant.So the derivative of x^a is a*x^(a-1); setting
a=x, x^x,and the derivative of b^x is b^x*log(b), and this
yieldsx^x*log(x). Add the two.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue University
<bok1tb$4i8g@odds.stat.purdue.edu>
===
> What's the derivate
of f(x)=x^x and f(x)=x^(x^(x-1)) ?>x^x = e^(x*log(x))>Now use
the chain and product rules. Another method, more general, is
to differentiate with> respect to each x as if all the others
are constant. So the derivative of x^a is a*x^(a-1); setting
a=x, x^x,> and the derivative of b^x is b^x*log(b), and this
yields> x^x*log(x). Add the two.>Hm, the chain rule for
functions of two variables. df(u,v)/dx = f_u(u,v) du/dx +
f_v(u,v) dv/dxThus df(x,x)/dx = f_x1(x,x) + f_x2(x,x)where
_x1 is with respect to the ?st x.
===
>Ok, I've probably
asked this question before, but I've forgot the
answer...>What's the derivate of f(x)=x^x and
f(x)=x^(x^(x-1)) ?Logarithmic differentiation.Doug
===
>Ok,
I've probably asked this question before, but I've
forgot the
answer...>What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1))
?In general, (u^v)' = v u^{v-1} u' + u^v ln(u)
v', wherever
it's de?ed.The answer can be derived from this.
===
>Ok, I've
probably asked this question before, but I've forgot the
answer...>What's the derivate of f(x)=x^x and
f(x)=x^(x^(x-1)) ?Think u^v = exp(v*ln(u)), with standard
rules for derivatives.
===
> Ok, I've probably asked this
question before, but I've forgot theanswer...> What's
the
derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ?> I guess asking
for the integral of these functions would be foolish?Use
logarithmicdrivative:f(x) = x^x
===
> Ln(f(x)) = x*Ln(x)

===
>f'(x)/f(x) = 1*Ln(x) + x *(1/x) = 1 + Ln(x) ===>f'(x)
=
x^x(1 + Ln(x))For the other, aply that twice.-- Ignacio
Larrosa Ca.96estroA Coru.96a
(Espa.96a)ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Some people
may think that this test is irrelevant for admissions
committeesfor pure math, but there were some interesting
questions that were asked onthe exam (it was held today and I
took it)...For instance,1. Hom(Z/2Z x Z/2Z, S_3) is isomorphic
to what?2. Let f(x) = x^2 + Bx + C, let B and C be independent
random variablesuniformly distributed on [0,1]. What is the
probability that the roots off(x) are distinct and real?There
were other interesting ones as well, but I can't remember
them...Theseare pretty easy if you're not under pressure to
do them in less than average3 minutes each (or maybe you
think 30 seconds under pressure issuf?ient...but give
undergraduates some slack :-) )It's good to see that the test
requires more than memorization and somethinking...needless to
say I got the 1st one wrong and the 2nd one right...In any
case, I don't think I did as well as I would have liked, and
hopethat doesn't impact my chances at some of the top schools
in representationtheory and algebra in general (princeton,
harvard, berkeley), but I haveheard that this test isn't
looked at very much, it's more recommendationsand extra
stuff/research...only I have heard that if you do poorly it
raisesa red ?ut I'm con?ent I don't have to worry
about
that)Just a post if anyone's interested,Mike
===
[GRE
questions]>1. Hom(Z/2Z x Z/2Z, S_3) is isomorphic to what?>2.
Let f(x) = x^2 + Bx + C, let B and C be independent random
variables>uniformly distributed on [0,1]. What is the
probability that the roots of>f(x) are distinct and
real?After 10 seconds of thought: 2:1 odds against.I ?st
learned to count at the poker table, at the age of 1 or so,
goingA, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. :)
===
> [GRE
questions]>1. Hom(Z/2Z x Z/2Z, S_3) is isomorphic to what?>2.
Let f(x) = x^2 + Bx + C, let B and C be independent random
variables>uniformly distributed on [0,1]. What is the
probability that the rootsof>f(x) are distinct and real?
After 10 seconds of thought: 2:1 odds against.>That seems to
be incorrect. Explanation? We look for the probability
thatB^2/4 > C...We can immediatelyput an upper bound on the
probability at 1/4 since if B = 1, then C must beless than or
equal to 1/4...> I ?st learned to count at the poker table,
at the age of 1 or so, going> A, 2, 3, 4, 5, 6, 7, 8, 9, 10,
J, Q, K. :)
===
>And believe me, if you don't speak any
French, the >waiters in France won't bring you soup. > Have
you considered the possibility that the reason waiters don't >
bring you soup might *not* be that you don't speak French?I
think that when I order soup in France (in English), I would
get soup.As the French have the word soupe which is some form
of soup. E.g.bouillabaisse is a soupe. But in that case it
will be the maincourse.A better example would be ordering
cold water in Italy. Do not besurprised when you get hot
water.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025
jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
>Message-id:
<Ho27Kq.6uz@cwi.nl>And believe me, if you don't speak any
French, the >waiters in France won't bring you soup.> Have
you considered the possibility that the reason waiters don't>
bring you soup might *not* be that you don't speak French?I
think that when I order soup in France (in English), I would
get soup.>As the French have the word soupe which is some
form of soup. E.g.>bouillabaisse is a soupe. But in that case
it will be the main>course.A better example would be ordering
cold water in Italy. Do not be>surprised when you get hot
water.I ordered two lattes at an AutoGrill in Italy, and got
exactly (and only) whatI ordered...two cups of milk.Marvin
Sebournosugeography@aol.com >dik t. winter, cwi, kruislaan
413, 1098 sj amsterdam, nederland,>+31205924131>home:
bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
>And believe me, if you don't
speak any French, the>waiters in France won't bring you
soup.Have you considered the possibility that the reason
waiters don't> bring you soup might *not* be that you
don't
speak French?This is unthinkable if you have ever been to
Quebec or France. The onlyway I could get a cup of coffee in
Quebec City was for my wife to orderit for me. :-)Chuck-- ...
The times have been, That, when the brains were out, the man
would die. ... Macbeth Chuck Simmons
chrlsim@earthlink.net
===
|> Have you considered the
possibility that the reason waiters don't|> bring you soup
might *not* be that you don't speak French?||This is
unthinkable if you have ever been to Quebec or France. The
only|way I could get a cup of coffee in Quebec City was for
my wife to order|it for me. :-)Since this is a thread about
social problems.... ;-)I think it varies depending on the
part of the city. I went on a trip toQuebec city with
friends, mostly to the old city (inside the wallsonce set up
to defend the city), which gets quite a bit of tourism.We
were there to ring bells in the cathedral there.Some of us
knew a bit of French, and tried using it, but thewaiters and
waitresses would always use English with us.I think sometimes
the ?st person to order was someone whodidn't know any
French, and simply ordered in English, andthe net result was
much the same.The Vancouver Sun once ran a piece on language.
They arrangedfor two people to ask where the nearest
laundromat was, oneasking in French in Vancouver, and the
other asking in Englishin Quebec. They got a much higher rate
of success in Quebec.ObMath anecdote. In Littlewood's
Miscellany, he describes alecture given by someone from a
region where the ?u' soundsin Duke and duck were swapped
over. He made repeatedreferences to Fuchs in his talk, but
pronounced the other way....Afterward someone pointed this
out, and he said yes, he'd realizedthis was not the usual
pronunciation, but he said it like thatbecause Ladies were
present.Keith Ramsay
===
if f is differentialbe on (0,
in?ite)and lim [f(x) +f'(x)] = L (x->in?ite)show that lim
f(x) = L (x->in?ite) and lim f'(x) = 0
(x->in?ite)------------------------------------it seems to
be trivial. but i no
touch.....oops........help.....me......please......
===
if f
is differentiable on (0, in?ite)and lim [f(x) + f'(x)] = L
(x->in?ite)show that lim f(x) = L (x->in?ite) and lim f'(x)
= 0 (x->in?ite)Note: this classic problem is well-known due
to the factthat it appeared in Hardy's classic calculus
textbook witha solution more awkward than the elegant
L'Hopital approach.For further details and references on the
L'Hospital-based approach see my prior post [1], and
[2].-Bill Dubuque[1]
http://google.com/groups?threadm=y8zognpn1wa.fsf%
40berne.ai.mit.edu[2] Martin D. Landau; William R. Jones. A
Hardy Old Problem.Math. Magazine 56 (1983) 230-232.
===
0 = L
- L = lim f+f' - lim f = lim f'However when f e^x ->
k; then f
= fe^x e^-x -> 0 f e^-x = f e^x e^-2x -> 0L = lim f+f' - lim
2f = lim f'-f0 = lim f = lim f e^-x / e^-x = (f' -
f)e^-x /
-e^-x = lim f-f' = -L0 = lim f+f' - lim f = lim
f'So if lim f
= k and lim f' exists, then lim(x->oo) f+f' = k + lim
f' k =
lim(x->oo) f = k + lim f'; lim f' = 0If lim
f' doesn't exist,
then for n >= 2 f_n(x) = sin(x^n)/x -> 0 (f_n)'(x) = nx^(n-1)
cos (x^n)/x - (sin x^n)/x^2 -> oscillation----
===
> if f
is differentiable on (0, in?ite)> and lim [f(x) + f'(x)]
= L (x->in?ite)> show that lim f(x) = L (x->in?ite) and
lim f'(x) = 0
(x->in?ite)http://google.com/groups?threadm=y8zognpn1wa.fsf%
40berne.ai.mit.edu>|>| f e^x (f + f') e^x>| lim f + f'
= L =>
lim f = lim ----- = lim ------------ = L>| x->oo x->oo x->oo
e^x x->oo e^x Provided f e^x -> +-oo, in which case> 0 = L -
L = lim f+f' - lim f = lim f'No proviso is needed.
This form
of L'Hopital's rule needs onlythat the denominator is
in?ite, not also the numerator [1][2].So there's no need to
treat this case specially as you do below.-Bill Dubuque[1] A.
E. Taylor, L'Hospital's Rule Amer. Math. Monthly, Vol.
59, No.
1 (Jan., 1952), pp.
20-24.http://links.jstor.org/sici?sici=0002-9890(195201)59:3%
3C20%3E[2] A. M. Ostrowski, Note on the Bernoulli-L'Hospital
Rule Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp.
239-242.http://links.jstor.org/sici?sici=0002-9890(197604)83:
3%3C239%3E> However when f e^x -> k; then> f = fe^x e^-x ->
0> f e^-x = f e^x e^-2x -> 0L = lim f+f' - lim 2f = lim
f'-f0
= lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim
f-f'
= -L> 0 = lim f+f' - lim f = lim f'So if lim f = k and
lim f'
exists, then lim(x->oo) f+f' = k + lim f'> k =
lim(x->oo) f =
k + lim f'; lim f' = 0If lim f'
doesn't exist, then for n >=
2> f_n(x) = sin(x^n)/x -> 0> (f_n)'(x) = nx^(n-1) cos (x^n)/x
- (sin x^n)/x^2 -> oscillation
===
> if f is differentialbe on
(0, in?ite) and lim [f(x) +f'(x)] = L (x->in?ite) show that
lim f(x) = L (x->in?ite) and lim f'(x) = 0
(x->in?ite)>Let's
suppose L=0 (substitute f by f-L).We must prove that if
f(x)+f'(x)->0 (x->inf), then f(x)->0
(x->inf)f(x)+f'(x)=o(1)
(x->inf)multiply by e^(x) :e^(x)[f(x)+f'(x)]=o(e(x))
(x->inf)integrate from 0 to t : (the derivative of e^(x)f(x)
is e^(x)[f(x)+f'(x)])e^(t)f(t)-f(0)=o(e(t)) (t->inf)Divide by
e^t :f(t)=o(1) which means that f(t)->0 (t->inf)(excuse my
english)
===
> if f is differentialbe on (0, in?ite)and lim
[f(x) +f'(x)] = L (x->in?ite)show that lim f(x) = L
(x->in?ite) and lim f'(x) = 0 (x->in?ite)Well. Perhaps
L'Hopital's Rule has some use, after all.--Ron Bruck
===
X,Y
is topology space.function f : X -> Yany B <= Y (symbol <= is
inclusion relation),cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy
(symbol cls = closure)show that f is
continuous.--------------------------------um......i
think......i use that any closed A <= Y, f^(-1) (A) is closed
set. <=> f : contii would show that f^(-1) cls(B) is closed
set.cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1)
(cls(B))thus f^(-1) cls(B) is closed.it is
right??please......advice....for me....thank you
===
>i would
show that f^(-1) cls(B) is closed set. >cls{f^(-1) cls(B)} <=
f^(-1) cls(cls(B)) = f^(-1) (cls(B)) >thus f^(-1) cls(B) is
closed. >it is right??Almost.You forgot to start with f^-1(cl
B) subset cl f^-1(cl B)hypothesis and closed B ==> f^-1(B)
closed f^-1(B) subset cl f^-1(B) subset f^-1(cl B) = f^-1(B)
f^-1(B) = cl f^-1(B)Extend exercise unto theoremf continuous
iff for all A, f(cl A) subset cl f(A) iff for all A, cl
f^-1(A) subset f^-1(cl A) iff for all A, bd f^-1(A) subset
f^-1(bd A)----
===
bok3np$ocd$1@news.hananet.net...> X,Y is
topology space. function f : X -> Y any B <= Y (symbol <= is
inclusion relation), cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy
(symbol cls = closure) show that f is continuous.>Take B a
closed subset of Y; you have cl(f^(-1)(B)) C f^(-1)(cl(B))
=f^(-1)(B).Hence cl(f^(-1)(B)) = f^(-1)(B), and f^(-1)(B) is
closed.--Julien Santini,Etudiant en licence au CMI de
Ch.89teau-Gombert,FRANCE.
===
Julien Santini
<santini.julien@wanadoo.fr> grava .88 la saucisse et au
marteau:> Julien Santini,> Etudiant en licence au CMI de
Ch.89teau-Gombert,> FRANCE.Est-ce vraiment n.8ecessaire? :)--
Nicolas
===
Nicolas Le Roux <nicolas@bisounours.net> scribbled
the following:> Julien Santini <santini.julien@wanadoo.fr>
grava .88 la saucisse et au marteau:>> Julien Santini,>>
Etudiant en licence au CMI de Ch.89teau-Gombert,>> FRANCE.>
Est-ce vraiment n.8ecessaire? :)Que est-ce que est vraiment
n.8ecessaire?-- /-- Joona Palaste (palaste@cc.helsinki.?
------------- Finland ----------
http://www.helsinki.?~palaste --------------------- rules!
--------/Ice cream sales somehow cause drownings: both happen
in summer. - Antti Voipio & Arto Wikla
===
> X,Y is topology
space.function f : X -> Yany B <= Y (symbol <= is inclusion
relation),cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy (symbol
cls = closure)show that f is
continuous.--------------------------------um......i
think......i use that any closed A <= Y, f^(-1) (A) is closed
set. <=> f : contii would show that f^(-1) cls(B) is closed
set.cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1)
(cls(B))thus f^(-1) cls(B) is closed.it is right??No. You
said that you wanted to proved that a certain condition
impliesthat f is continuous, but what you did prove was the
reverseimplication.Jose Carlos Santos
===
Can the direct
isogeny be found between these two elliptic curves?It is
known that they are in a set of 8 isogenous curves, but
whatis the exact transformation that takes x <--> X, i.e.
points on (1)to points on (2)?(1) y^2 = (x + 366)*(x^2 -
366*x + 2625489)(2) Y^2 = (X - 1551)*(X^2 + 1551*X +
497214)Both curves have conductor 210.More generally, how
might one ?d the isogeny betweeny^2 = x^3+ax+b and Y^2 =
X^3+AX+B, when the curves havethe same conductor? (a,b,A,B
are known)
===
> There's more to my work than just arguing on
Usenet, so I'd like to> point out that my paper Advanced
Polynomial Factorization is slated> to be published:> The
Mega Foundation is an organization of high IQ people, and I'm
glad> to be associated with them. To learn further about the
organization> you can use Google, or see: Why a group like
the Mega Foundation?>
http://www.ultrahiq.org/Mega/WhyMega.htm>Mr. Harris,to me,
this seems equivalent to the copyright you had on the web
page youcreated for your famed Proof of FLT.Now, you are once
again hell-bent on crying, cursing, cheating and trying todo
anything to propagate your ?lies.You also have some sort
of greatness disorder that makes you think you areabove every
person you come in contact with. IIRC, this is a
NarcissisticPersonality Disorder.Here is excerpt from the web
site: http://www.suite101.com/welcome.cfm/npdNarcissists
attract abuse. Haughty, exploitative, demanding,
insensitive,and quarrelsome - they tend to draw opprobrium
and provoke anger and evenhatred. Sorely lacking in
interpersonal skills, devoid of empathy, andsteeped in
irksome grandiose fantasies - they invariably fail to
mitigatethe irritation and revolt that they induce in
others.Here is a clinical
de?ition:http://www.behavenet.com/capsules/disorders/
narcissisticpd.htm***Diagnostic criteria for 301.81
Narcissistic Personality Disorder (cautionarystatement)A
pervasive pattern of grandiosity (in fantasy or behavior),
need foradmiration, and lack of empathy, beginning by early
adulthood and present ina variety of contexts, as indicated
by ?e (or more) of the following:(1) has a grandiose sense
of self-importance (e.g., exaggerates achievementsand
talents, expects to be recognized as superior without
commensurateachievements)(2) is preoccupied with fantasies of
unlimited success, power, brilliance,beauty, or ideal love(3)
believes that he or she is special and unique and can only
beunderstood by, or should associate with, other special or
high-status people(or institutions)(4) requires excessive
admiration(5) has a sense of entitlement, i.e., unreasonable
expectations ofespecially favorable treatment or automatic
compliance with his or herexpectations(6) is interpersonally
exploitative, i.e., takes advantage of others toachieve his
or her own ends(7) lacks empathy: is unwilling to recognize
or identify with the feelingsand needs of others(8) is often
envious of others or believes that others are envious of him
orher(9) shows arrogant, haughty behaviors or
attitudesReprinted with permission from the Diagnostic and
Statistical Manual ofMental Disorders, fourth Edition.
Copyright 1994 American PsychiatricAssociation***Perhaps
instead of spewing your useless nonsense, you should study
thisdisorder, go see a doctor and then drug and drink
yourself to death.Think about, eh, let's wack you!
===
... >
1. I de?e: > w1(x) = gcd(5 a1(x) + 7, 49) > Ok,
w_1(x) = 7, which works. > Where do you see w1(x) = 7? That
can only be true when you *know* that > (5 a1(x) + 7) is
divisible by 7 in the algebraic integers, otherwise the >
greatest common divisor can *not* be 7. > Which is the
problem with the ring of algebraic integers.Nope, there is
*not yet* a problem. I have de?ed w1(x) *as I de?ed it*.You
are immediately seeing that that w1(x) equals 7, which is
blatantlyfalse as the de?ition of gcd *does not warrant
that*. > At this moment the only thing we know about w1(0)
= 7. For some > values of x w1(x) can even be larger than 7.
> That's nonsense, easily proven by the factorization.That
is *not* nonsense. > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5
b_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22Yup,
this is the polynomial. Note that when x = 7y + 5 it can
bedivided by additional factors 7. So when x is of that form,
it is*not possible* that the gcd's deliver 7, 7 and 1. For
instance,when x = 5 the value is 2^3 * 7 * 13 * 23 * 2213.
See that 7 inthe factors? > where no other factorization
works as long as 7 is not a factor of 22. > That's because
I've isolated the factors of 22, which is obvious by >
inspection. > So you think that (7/w_1(x)) (7/w_2(x))
(22/w_3(x)) = 22 makes a > difference Dik Winter? Not as long
as 7 is not a factor of 22, it > doesn't.*Why* do you think
that 22/w3(x) must *also* be an algebraic integer?That is
just plain stupid. You only need that (5 b3(x)/w3(x) +
22/w3(x))is an algebraic integer. > Understand Dik Winter?
You do understand factors, right?Yes, but I seriously doubt
your understanding. > So you know where I got (7/w_1(x))
(7/w_2(x)) (22/w_3(x)) = 22 from, > right?Yes, that is true
with my factorisation too. > Now adding in more variables,
like sticking in w's may look like a > good way to hide the
truth Dik Winter, but it's childish. > Besides, I'm
bored
now that I've shown that math society is so stupid > as to
try and ?ht such a basic result. Now the only question is
how > to start yanking funds so that they can be redirected
to real > research.You would not know a basic result when you
saw one.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025
jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
> ...>
Pray answer the following quesions:> Ok.> 1. I de?e:>
w1(x) = gcd(5 a1(x) + 7, 49)> Ok, w_1(x) = 7, which works.
Where do you see w1(x) = 7? That can only be true when you
*know* that> (5 a1(x) + 7) is divisible by 7 in the algebraic
integers, otherwise the> greatest common divisor can *not* be
7. Which is the problem with the ring of algebraic integers.
At this moment the only thing we know about w1(0) = 7. For
some> values of x w1(x) can even be larger than 7. That's
nonsense, easily proven by the factorization. (5 a_1(x)/7 +
1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 -
360 x + 22 where no other factorization works as long as 7 is
not a factor of 22.That factorization has only been
demonstrated for the case x = 0. It has not been proven for
?x'generally. Since you *know* this, your presentation is
deceptive and dishonest.--There are two things you must never
attempt to prove: the unprovable -- and the
obvious.--Democracy: The triumph of popularity over
principle.--http://www.crbond.com
===
... > Oh yeah, easy
mistake, but yours Dik Winter is MUCH more embarrassing. >
After all the factorization is > (5 a_1(x) + 7)(5 a_2(x) +
7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x +
22) > where you want to divide through by w_1(x) w_2(x)
w_3(x) = 49, so > doing that gives > (5 a_1(x)/w_1(x) +
7/w_1(x))(5 a_2(x)/w_2(x) + 7/w_2(x))(5 > b_3(x)/w_3(x) +
22/w_3(x)) = 300125 x^3 - 18375 x^2 - 360 x + 22 > and
notice that necessarily, if you multiply everything out and >
simplify, you have > (7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22 >
> which leaves you shit out of luck Dik Winter.Yes, the
product of those three terms is 22. What is the problem?You
have (7/7)(7/7)(22/1) = 22. Looks pretty similar to me.-- dik
t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam,
nederland; http://www.cwi.nl/~dik/
===
... > Oh yeah, easy
mistake, but yours Dik Winter is MUCH more embarrassing. >
After all the factorization is > (5 a_1(x) + 7)(5 a_2(x) +
7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22)
> where you want to divide through by w_1(x) w_2(x) w_3(x) =
49, so > doing that gives > (5 a_1(x)/w_1(x) + 7/w_1(x))(5
a_2(x)/w_2(x) + 7/w_2(x))(5 > b_3(x)/w_3(x) + 22/w_3(x)) =
300125 x^3 - 18375 x^2 - 360 x + 22 > and notice that
necessarily, if you multiply everything out and > simplify,
you have > (7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22 > which
leaves you shit out of luck Dik Winter. > Why are you so
hateful toward people who make mistakes? Everyone makes >
mistakes, at least he admits when he is wrong. By the way,
your use of > swearing makes you look very immature.The
strangest about this is that there is no mistake. The product
indeedcomes to 22. When you do the multiplication James
suggest you come at22, just what was needed. So I really do
not understand what James isdoing here.-- dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home:
bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
>Oh yeah, easy mistake, but yours
Dik Winter is MUCH more embarrassing.Much more embarrassing
than starting a new thread every time youcan't talk your way
out of a corner?DougX-Cise:
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===
>1. Why is there a cross product in the
de?ition of torque T = r x>F ?Because you're doing Physics
in 3 dimensions, you can mapskew-symmetric tensors into
vectors. For more than 3 dimensions, youhave to explicitly
treat the torque as a skew symmetric tensor or a2-form.>2.
How is torque a tensor? Because the de?ition of cross
product makes it transform like one.If you don't want to
treat it as a tensor or a 2-form, than you needto treat it as
a vector and pay close attention to your weights.Either way,
you need to be careful to distinguish covariant indicesfrom
contravariant.-- Shmuel (Seymour J.) Metz, SysProg and
JOATnot reply to spamtrap@library.lspace.org
===
> Q( a x b )
= 1/(detQ) ( Qa x Qb )> where Q is the matrix of the change
of basis, but I'm not quite seeing> how that matches the
A' =
Q^{-1} A Q form.As you say, under the change of orthonormal
basis, the vectors a and b> go to a' = Qa and b' = Qb.
In
component form,a_i = (sum over m) Q_(im) a_m and b_i = (sum
over n) Q_(in) b_nI've used different summation indices in
order that the substitutions> below makes sense.Because both
the new and old bases are orthonormal, Q is an orthogonal>
matrix, i.e., Q^T = Q^(-1), i.e if M = Q^(-1), M_(ij) =
Q_(ji). Thus,> det Q = +/- 1. If in addition, right-handed
orthonormal bases are> taken to right-handed orthonormal
bases, then det Q = 1, and Q is called> a special orthogonal
matrix. The set of all special orthogonal matrices> forms the
group SO(3), where the notation is fairly self-explanatory ->
S for special (det +1), O for orthogonal (inverse =
transpose), 3 for> 3 by 3.De?e matrix A by A_(ij) = a_i b_j
- a_j b_i. Under a transformation,A'_ij = a'_i
b'_j - a'_j
b'_i = (sum over m,n) [Q_(im) a_m Q_(jn) b_n - Q_(jn) a_n
Q_(im) b_m] = (sum over m,n) [Q_(im) a_m b_n Q_(jn) - Q_(im)
a_n b_m Q_(jn)] = (sum over m,n) [Q_(im) (a_m b_n - a_n b_m)
Q_jn] = (sum over m,n) [Q_(im) A_(mn) Q^(-1)_nj]A = Q A
Q^(-1)This is a little different than A = Q A Q^(-1) because,
the way Feynman> de?es things, a'_i = (sum over n)
Q^(-1)_(in) a_n. It's a good> exercise to show this.sure what
you were meaning to show here, if you will note the last 4
linesof your posting.that is inconsistent with how I speci?d
the change of basis, it shouldhave been A' = Q A Q^{-1} and
that is what you have.I also note that Feynman's matrix
object T_ij = x_i F_j - x_j F_i doesindeed transform as T' =
Q T Q^{-1}, as you have shown. I am still mysti?d as to what
real value there is in taking a niceaxial vector object like
Torque and making the anti-symmetric matrix out of it...
<2202379a.0310270854.d34b089@posting.google.com>
<po1spvg49r7qug87h6r4qoa8tvrbhimgd9@4ax.com>
<2202379a.0310281028.2884dbfa@posting.google.com>
<09iupvkca3n0ur4e7bcvroeviu9isfbbit@4ax.com>
<2202379a.0310310905.20a78472@posting.google.com>
<77seqvotf0ul8f8nmds64ilpesolg1dhhm@4ax.com>
<2202379a.0311050918.1ff64462@posting.google.com>
<3fa9ddbf$4$fuzhry+tra$mr2ice@news.patriot.net>
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===
at 11:25 AM, dynamics@vianet.on.ca (Ken S.
Tucker) said:>But the spectra itself is formed by photons,No.
The term continuous spectra is not the same as[1] the usage
ofthe term spectrum in optics. An observable in QM is a
Hermitianoperator, and such an operator has a characteristic
that is known inthe literature as its spectrum.Read any good
book on Functional Analysis for a explanation. Or
readPrinciples of Quantum Mechanics (P.A.M. Dirac) for an
explanationfrom a Physics perspective.[1] But the spectrum in
optics is related to the spectrum of the Hamiltonian.-- Shmuel
(Seymour J.) Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
>>But the spectra itself is
formed by photons,>No. The term continuous spectra is not the
same as[1] the usage of>the term spectrum in optics. An
observable in QM is a Hermitian>operator, and such an
operator has a characteristic that is known in>the literature
as its spectrum.Actually, the term spectrum is generic to
linear algebras, and is notspeci? to operator algebras. If A
is an algebra over a ?ld F, thespectrum of an element a of A
would be spec(a) = { f in F: (a-f) has no two-sided inverse
in A }. <3fad7a93$1$fuzhry+tra$mr2ice@news.patriot.net>
<boml4a$4gn$1@uwm.edu>X-Cise:
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===
at 12:11 AM, whopkins@alpha2.csd.uwm.edu (Mark)
said:>Actually, the term spectrum is generic to linear
algebras, and is>not speci? to operator algebras. Correct,
although I don't know whether the more general meaning isever
used in QM. I'm certainly not aware of the ?ld[1] being
otherthan R, C or H.[1] If you use the de?ition that
requires commutativity, then omitH.-- Shmuel (Seymour J.)
Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
>But the spectra itself is
formed by photons,>Ken >>No. The term continuous spectra is
not the same as[1] the usage of>>the term spectrum in optics.
An observable in QM is a Hermitian>>operator, and such an
operator has a characteristic that is known in>>the
literature as its spectrum.>>Seymour>Actually, the term
spectrum is generic to linear algebras, and is not>speci? to
operator algebras. If A is an algebra over a ?ld F,
the>spectrum of an element a of A would be>spec(a) = { f in
F: (a-f) has no two-sided inverse in A }.>MarkThis is
amusing, see this thread has been cross postedto sci-math and
sci-physics, so there are different meanings to words. Anyway
I'm the guy who is talkingabout physics, but I'll try
to make
it interesting for thehigher class posters in sci-math, who
may disagree.An invariant has no units, speci?ally invariant
massor energy. Two invariant masses can be equated ona balance
without resort to units. I argue the invariant mass/energy can
only changeby small but ?ite amounts, hence discontinuously.
Let E be the invariant then dE =0 as it has no continuous
function and derivative, although E =/=constant.Transactions
are quantized to the nearest penny,thus d(money) =0.For
mathematical masochists, one might suggesta Fourier series to
form a digital pulse (penny),however, the wavelengths required
exceed thesize of the penny to form the square pulse, ((SUM i,
(n=1+2*i) 1/n sin n*w*t)).So a wave mechanical solution is an
approximation.but is DOA mathematically.Returning to physics,
I maintain we cannotin?itely divide the two masses on the
balance,nor can we in?itesmally vary either mass.Therefore
dE =0 is a Law of Physics and describes how invariant energy
varies. This might look like a peculiar law but compare this
to the GR geodesical equation of motion, given by the
absolute derivative DU^u/ds =0.(Absolute acceleration does
not exist)This equation is very successful. Now 4-momentum is
E*U^u so a quantum-mechanical generally relativistic geodesic
is,DE^u/ds =0 .(absolute force does not exist)This equation
rules out any force like f^u (or f_u).So it sets all the
components of Lorentz force to zero to describe how a charge
moves geodesically.That's a powerful constraint, and
*appears* to beveri?d so far.One might call DE^u =0 a uni?d
?ld theory,but I think it's only a requirement. The
metricthat satis?s this equation is the ?ld, and I suspect
the uni?d ?ld equation will turn outto be G_uv=kT_uv.
Einstein has suggested (in his later years) using
anti-symmetrical metricsto go from G_uv = kT_uv to DE^u=0, to
unify.Ken S. Tucker
<2202379a.0311052323.77555cbc@posting.google.com>X-Cise:
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===
at 11:23 PM, dynamics@vianet.on.ca (Ken S.
Tucker) said:>As I understand, the indices forming the
relatively>de?ed tensor K are meaningless after the
de?ition,>but I have some uncertainty.I'm not sure what
you're trying to say, but change to polarcoordinates and
observe what happens.>The relative tensor K_uv is actually a
4th rank>(counting weight) like the Riemann-Christoffel
>Curvature tensor. The weight has nothing to do with the
rank.>It is very evident the Kronecker delta (d^u_v) >serves
as an orthogonal metric,No it doesn't.>relativity of K_uv?--
Shmuel (Seymour J.) Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
Mr.
Harris,Psychopathology------------------------Narcissism:A
pattern of traits and behaviors which signify infatuation and
obsessionwith one's self to the exclusion of all others and
the egotistic andruthless pursuit of one's grati?ation,
dominance and ambition.Perhaps research in to your delusional
?s of grandeur would be a moreappropriate vocation for
you?Still pissing out your research?
===
>> Similarly,
Cramer's proof of the Levy-Cramer theorem that>> the sum of
two independent random variables is not normal>> unless they
both are is probably the only easy proof, but>> it likewise
obscures the ideas. Any time characteristic>> functions are
used to prove a probability theorem, the>> concepts are not
even present.>Do you happen to know of a probabilistic proof
of the Cramr-Lvy >theorem?There are entropy proofs. Any
published?-- A.
===
> If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x
+ 2) = 49(x^3 + 5x^2 + 3x + 2) with the c's algebraic
integers, I think few of you would have a> problem realizing
that only two of the c's have 7 as a factor. But, of course,
you're looking at *functions* of x, as you have f_1(x) = c_1
x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write
it as> (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2
+ 3x + 2). Notice that dividing both sides by 49 gives
(f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x
+ 2 as long as you're in a ring where 7 is not a factor of
22. I want to emphasize that point as notice there's only
*one* way to> divide through by 49 if 7 is not a factor of
22. Usually you can *see* the other factors of 7, but I want
you to> abstract, and generalize. Please pay careful
attention to that example. You may see people who reply
claiming that the word polynomial has> some signi?ance, as
if it's a mystical thing which refutes basic> logic, so if
something isn't polynomial it no longer behaves>
logically.All functions that I know of, all of them behave
logically when examinedfrom the view point of
calculus.EXAMPLE:Describe the graph of f(x)=sin(x) without
graphing.Well, I know that a derivative measures change and I
also know that when theinstantaneous change is 0, there is
some sort of a critical point(determined by the 2nd
derivative).Let's say we want to describe the graph of f(x)
in words. Begin by ?dingf'(x).f'(x)=cos(x)We know that
a
function has a critical point at a point where the
derivativeis 0.cos(x)=0 at x=pi/2 and 3pi/2Ok, we know where
the critical points are, but we don't know what happens
atthose points. So take the 2nd derivative.
f''(x)=-sin(x)When x=pi/2,
f''(x)=-1. When x=3pi/2
f''(x)=1Therefore I can now analyze my results. When
x=pi/2,
the 2nd derivative isnegative, which means it's decreasing so
pi/2 must be a maximum. Whenx=3pi/2, the 2nd derivative is
positive so 3pi/2 must be a minimum.In fact, if I graph this,
my predictions are correct.My point is that all elementary
functions that I know of behave logically ifyou examine it
from a calculus viewpoint, which may not be a bad idea foryou
to do.David Moran Now then, in my advanced factorization work,
I just use functions of x> that are a lot more complicated
than f_1(x) = c_1 x, and unfortunately> there are people who
can use an unfamiliar leap in complexity to> confuse others.
Some of you have learned various advanced math topics, now
imagine if> in your classrooms there were some hecklers who
continually hollered> out at your teacher, or otherwise
disrupted the class? What if when there were dif?ult
concepts those hecklers would try to> confuse everyone as
they sought to discredit the mathematics? If you ?d that
hard to imagine, imagine me in your class with you>
questioning the professor and calling him names. How much
would you have learned? I need those of you interested in
mathematics to focus on the basics,> so that you can
understand the advanced.> James Harris>
http://mathforpro?.blogspot.com/
===
You may see people who
reply claiming that the word polynomial has> some
signi?ance, as if it's a mystical thing which refutes basic>
logic, so if something isn't polynomial it no longer behaves>
logically.> No both polynomial and non-polynomial functions
behave logically.It is just that polynomials have some
properties that arenot true in general.Let P(x) by a
polynomial divisible by p, and let g_1(x) andg_2(x) be
polynomials such that P(x) = g_1(x) * g_2(x)Then p divides at
least one of g_1(x) and g_2(x).If g_1(x) and g_2(x) are not
polynomials, this is no longer true.Counterexample.Let P(x) =
15-3x. Then P(x) is divisible by 3.Let g_1(x)=4-sqrt(1+3x) and
g_2(x)=4+sqrt(1+3x).Then P(x) = g1(x)*g2(x)list a few valuesP(
0) = 15, g_1( 0) = 3, g_2( 0) = 5P( 1) = 12, g_1( 1) = 2, g_2(
1) = 6P( 5) = 0, g_1( 5) = 0. g_2( 5) = 8P( 8) = -9, g_1( 8) =
-1. g_2( 8) = 9P(16) = -33, g_1(16) = -3. g_2(16) = 11Note how
the factor of 3 switches back and forthbetween g_1 and g_2
depending on the value of x.Thus neither g_1(x) nor g_2(x) is
divisible by3 for all x. In particular the fact that g_1(0)is
divisible by 3 does not mean that g_1(1) isdivisible by 3. -
William Hughes
===
> If you saw > (c_1 x + 7)(c_2 x + 7)(
c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2) > with the c's
algebraic integers, I think few of you would have a > problem
realizing that only two of the c's have 7 as a factor.How do
you know that this is possible with algebraic integer c1 to
c3?I do not think this is a valid factorisation in the
algebraic integers.Could you provide the monic polynomial
with integer coef?ients ofwhich the c's are the roots? More
to the point, x^3 + 5x^3 + 3x + 2factors as (x - r1)(x -
r2)(x - r3) in the algebraic integers. Wherethe r's are the
roots of the polynomial. We can multiply the factorsby 7/r1,
7/r2 and 2/r3 to get your factorisation. So we get: (7/r1 x +
7)(7/r2 x + 7)(2/r3 x + 2).So c1 = 7/r1, c2 = 7/r2, c3 =
2/r3.Of these only c3 is an algebraic integer (r3 is a
divisor of 2).Neither 7/r1, nor 7/r2 are algebraic integers.
Unless you claimthat r1 and r2 are units (they are divisors
of 2). But they arenot, because the constant term of a
polynomial (see the correctusage here, James) must be 1 for a
root to be a unit.-- dik t. winter, cwi, kruislaan 413, 1098
sj amsterdam, nederland, +31205924131home: bovenover 215,
1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
===
In
sci.math, James
Harris<jstevh@msn.com><3c65f87.0311080557.3c8f97f1@
posting.google.com>:> If you saw(c_1 x + 7)(c_2 x + 7)( c_3 x
+ 2) = > 49(x^3 + 5x^2 + 3x + 2)with the c's algebraic
integers, I think few of you would have a> problem realizing
that only two of the c's have 7 as a factor.Not clear at all
without knowing precisely what the roots are.Since two of the
roots are complex in this case I might havea chance of not
using trig, but it would take some work.I can tell you that
the roots, as reported by Pari/GP,
areapproximately:-4.424076847035404679966063971-
0.2879615764822976600169680144 -
0.6075770270241740255965589263*I-
0.2879615764822976600169680144 +
0.6075770270241740255965589263*Ibut this doesn't help much as
algebraic units are provably densearound (0 + 0i) -- sqrt(n+1)
- sqrt(n) is a unit.But, of course, you're looking at
*functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x,
and f_3(x) = c_3 x, so I could also write it as> (f_1(x) +
7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2).Notice
that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 +
1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2as long as you're in a
ring where 7 is not a factor of 22.Only an issue modulo 3 or
5, AFAICT.... :-)I can posit a simpler example. Let's assume
for gigglesthe following equation, similar to yours:(d_1 * x
+ 7) * (d_2 * x + 7) * (d_3 * x + 2) = 49 * (x^3 + 2)and
carry through your logic. I know the roots to this equation.
:-)We know thatx^3 + 2 = (x + 2^(1/3)) *(x + 2^(1/3) * (-1/2
+ sqrt(3) * I/2)) *(x + 2^(1/3) * (-1/2 - sqrt(3) *
I/2))(GP/Pari veri?s this nicely, as it turns
out).Therefore, one can pick d_1 = 7 / (2^(1/3)),d_2 = 7 /
(2^(1/3) * (-1/2 + sqrt(3) * I/2)),d_3 = 2 / (2^(1/3) * (-1/2
- sqrt(3) * I/2)).Since -1/2 + sqrt(3) * I/2 and -1/2 -
sqrt(3) * I/2 are bothunits (as is easily veri?d by
multiplying them together),they factor little in the
analysis. I can just as easilypositd_1' = d_2' = 7 /
(2^(1/3))d_3' = 2 / (2^(1/3)) = 2^(2/3)Now d_3' turns
out to
be an algebraic integer, solving theequation y^3 - 4 = 0.
(d_3 solves this equation too.)d_1' ? Well, lessee; the most
obvious equation is2 * y^3 - 343 = 0which is clearly not
going to lead to anything resembling analgebraic integer.
Since d_1' requires a cube root no equationof lesser degree
will do.But I may need those roots to your equation, and not
justapproximations, either. If one positsN(x) = x^3 + 5*x^2 +
3*x + 2then, if one substitutes x = y-5/3, one can rewrite
this as:N(y-5/3) = y^3 - 16/3 * x + 169/27Now substitutey = w
+ 16/(9*w)(Vi.8fta's substitution, with p = -16/3) and
getN(w+16/(9*w)-5/3) = (729*w^6 + 4563*w^3 +
4096)/(729*w^3)Surprise: w^3 = (-4563
sqrt(4563^2-4*729*4096)) / (2 * 729)= (-4563 sqrt(8877033))
/ (2 * 729) , by our old friend,the quadratic formula.Since
8877033 = 3^9*11*41 and 4563 = 3^3 * 13^2, we can divideby
3^3/3^3 (or sqrt(3^6)/3^3), resulting in w^3 = (-169
sqrt(12177)) / (2 * 27)orw^3 = (-169 3*sqrt(1353)) / (2 *
27)I'm going to take the + sign; the - sign will yield
slightlyidentical results, as it turns out.Because GP/Pari
does odd (but somewhat logical)things with (-1)^3, I'm going
to take the sign out now.The results can be written:x1 = -5/3
- 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) - 16/(3*((169 -
3*sqrt(1353)) / 2)^(1/3))x2 = -5/3 - 1/3 * ((169 -
3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) -
16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) * (-1/2 + sqrt(3) *
I/2)x3 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) *
(-1/2 + sqrt(3) * I/2) - 16/(3*((169 - 3*sqrt(1353)) /
2)^(1/3)) * (-1/2 - sqrt(3) * I/2)It's worth noting that(169
- 3*sqrt(1353)) * (169 + 3*sqrt(1353)) = 16384 = 2^14,so one
can rewrite the above asx1 = -5/3 - 1/3 * ((169 -
3*sqrt(1353)) / 2)^(1/3) - 1/3 *
((169+3*sqrt(1353))/2)^(1/3)x2 = -5/3 - 1/3 * ((169 -
3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) - 1/3 *
((169+3*sqrt(1353))/2)^(1/3) * (-1/2 + sqrt(3) * I/2)x3 =
-5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 +
sqrt(3) * I/2) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) * (-1/2 -
sqrt(3) * I/2)If you look carefully you'll see that using the
alternate solutionw^3 = (-169 - 3*sqrt(1353)) / (2 * 27)will
merely swap x2 and x3 in the ?al list.In this form, ?uring
out whether xi is an algebraic integer is aLost Cause(tm),
although we know the answer is yes since theyall solve the
equation x^3 + 5*x^2 + 3*x + 2 = 0, as one canverify if one
multiplies(x-x1)*(x-x2)*(x-x3)in GP/Pari.However, there's a
different route, making the exactroots entirely irrelevant.
(Sorry, guys -- but it *was*an interesting solution using
Vi.8fta's substitution. :-) )We are hypothesizing(c_1*x +
7)(c_2*x + 7)( c_3*x + 2) = 49(x^3 + 5 x^2 + 3x + 2).for some
algebraic numbers ci. This hypothesis turns out to bequite
valid, as it turns out -- but what are the ci?Since we
knowx^3 + 5x^2 + 3x + 2 = (x-x1)*(x-x2)*(x-x3)we can deduce
that c1 = -7/x1, c2 = -7/x2, and c3 = -2/x3, for somerotation
of x1, x2, x3. Are these algebraic integers?We note that, if
xi solvesx^3 + 5 x^2 + 3x + 2 = 0,then 1/xi has to solve2*z^3
+ 3*z^2 + 5*z + 1 = 0,7/xi has to solve2*z^3 + 3*7*z^2 +
5*7^2*z + 7^3 = 0,and -7/xi has to solve2*z^3 - 3*7*z^2 +
5*7^2*z - 7^3 = 0.-7/xi is clearly *not* an algebraic integer
(as the aboveis not a reducible polynomial over
Q).Fortunately, -2/xi is:2*z^3 - 3*2*z^2 + 5*2^2*z - 2^3 =
0orz^3 - 3*z^2 + 10*z - 4 = 0Well, 1 out of 3 ain't bad.This
counter-proof generalizes quite nicely for most cubic
polynomialsx^3 + a_2 * x^2 + a_1 * x + a_0with non-unit a_0
not divisible by 7.I'm afraid you may be out of luck, James.
:-)[rest snipped]-- #191, ewill3@earthlink.netIt's still
legal to go .sigless.
===
>If you saw(c_1 x + 7)(c_2 x + 7)(
c_3 x + 2) = > 49(x^3 + 5x^2 + 3x + 2)with the c's
algebraic integers, I think few of you would have a>problem
realizing that only two of the c's have 7 as a factor.>No -
the problem here is that this polynomial doesnot factor at
all in the algebraic integers in the form you have shown. Its
factorization is necessarily of the form 49*(x - r1)*(x -
r2)*(x - r3),where r1, r2, and r3 are nonunit *algebraic
integers* whichare the roots of x^3 + 5*x^2 + 3*x + 2 = 0,
and are such that -r1*r2*r3 = 2.Knowing this, you can
distribute the 49 among the factors in many different ways.
However, *none* of thoseways will end up having constant
terms of 7, 7, and 2.Why? Because 7 and 2 are coprime in the
algebraic integers. It is not possible to write 2, which you
wantas the constant term of the third factor, as a product of
a nonunit factor of 7 and a nonunit factor of 2.Andrzej.>But,
of course, you're looking at *functions* of x, as you have
f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I
could also write it as>(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2)
= 49(x^3 + 5 x^2 + 3x + 2).Notice that dividing both sides by
49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5
x^2 + 3x + 2as long as you're in a ring where 7 is not a
factor of 22.I want to emphasize that point as notice there's
only *one* way to>divide through by 49 if 7 is not a factor of
22.Usually you can *see* the other factors of 7, but I want
you to>abstract, and generalize.Please pay careful attention
to that example.You may see people who reply claiming that
the word polynomial has>some signi?ance, as if it's a
mystical thing which refutes basic>logic, so if something
isn't polynomial it no longer behaves>logically.Now then, in
my advanced factorization work, I just use functions of
x>that are a lot more complicated than f_1(x) = c_1 x, and
unfortunately>there are people who can use an unfamiliar leap
in complexity to>confuse others.Some of you have learned
various advanced math topics, now imagine if>in your
classrooms there were some hecklers who continually
hollered>out at your teacher, or otherwise disrupted the
class?What if when there were dif?ult concepts those
hecklers would try to>confuse everyone as they sought to
discredit the mathematics?If you ?d that hard to imagine,
imagine me in your class with you>questioning the professor
and calling him names.How much would you have learned?I need
those of you interested in mathematics to focus on the
basics,>so that you can understand the advanced.>James
Harris>http://mathforpro?.blogspot.com/
===
Is there a simple
proof, or counterexample, to the propositionthat if we
consider the expansion of the fractional partof some
(irrational) number x in some base, and know that it is
normal,then it follows that the expansion of 1/x is also
normal?I'd appreciate a pointer to any literature in this
area.Rob Shaw
===
>Is there a simple proof, or counterexample,
to the proposition>that if we consider the expansion of the
fractional part>of some (irrational) number x in some base,
and know that it is normal,>then it follows that the
expansion of 1/x is also normal?I'd appreciate a pointer to
any literature in this area.>What do you mean by normally
distributed digits? The normal distribution is a continuous
one.Nor would it make sense to speak of x - ?) being
normally distributed, since the normal dsitrubtion is
supported by the entire real line.-- Stephen J. Herschkorn
herschko@rutcor.rutgers.edu
===
>Is there a simple proof, or
counterexample, to the proposition>that if we consider the
expansion of the fractional part>of some (irrational) number
x in some base, and know that it is normal,>then it follows
that the expansion of 1/x is also normal?I'd appreciate a
pointer to any literature in this area. What do you mean by
normally distributed digits? The normal > distribution is a
continuous one.Nor would it make sense to speak of x -
?) being normally > distributed, since the normal
dsitrubtion is supported by the entire > real line.A real
number is said to be normal (to the base 10) if for all n
every string of digits of length n appears with frequency
10^(-n) in its decimal expansion. This has nothing to do with
the (Gaussian) normal distribution that you are thinking of.
As others have noted, it seems unlikely that x normal implies
1/x normal, but I don't know what's in the literature.
Let y
be normal to base 2, and let x be the number you get by
making believe y is a decimal expansion. Then x isn't normal
- it has only the digits 0 and 1 - but I'd expect 1/x to be
normal (though I wouldn't be able to prove it).--
===
Is
there a simple proof, or counterexample, to the
proposition>that if we consider the expansion of the
fractional part>of some (irrational) number x in some base,
and know that it is normal,>then it follows that the
expansion of 1/x is also normal?I seriously doubt it. Here's
an indication of why:Let's talk about base 2. Start with x =
.01010101... .Of course x is not normal, but it does contain
the rightnumber of 0's and 1's; it's weakly
normal in base 2,
ina sense. But x = 1/3, so 1/x = three = 11.000... , which
isfar from normal.I wouldn't be surprised if you could
actually constructa counterexample along these lines - you'd
need to verifya few things...>I'd appreciate a pointer to any
literature in this area.>Rob ShawDavid C. Ullrich
===
Is there
a simple proof, or counterexample, to the proposition>that if
we consider the expansion of the fractional part>of some
(irrational) number x in some base, and know that it is
normal,>then it follows that the expansion of 1/x is also
normal?I seriously doubt it. Here's an indication of
why:Let's talk about base 2. Start with x = .01010101... .>
Of course x is not normal, but it does contain the right>
number of 0's and 1's; it's weakly normal
in base 2, in> a
sense. But x = 1/3, so 1/x = three = 11.000... , which is>
far from normal.You're in the land of rationals. Rationals
and normality don't mix.(Take that out of context,
man!)Phil-- Unpatched IE vulnerability: window.open search
injectionDescription: cross-domain scripting,
cookie/data/identity theft, command executionExploit:
http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm

===
>>Is there a simple proof, or counterexample, to the
proposition>>that if we consider the expansion of the
fractional part>>of some (irrational) number x in some base,
and know that it is normal,>>then it follows that the
expansion of 1/x is also normal?> I seriously doubt it.
Here's an indication of why:> Let's talk about base
2.
Start with x = .01010101... .>> Of course x is not normal,
but it does contain the right>> number of 0's and 1's;
it's
weakly normal in base 2, in>> a sense. But x = 1/3, so 1/x =
three = 11.000... , which is>> far from normal.You're in the
land of rationals. >Rationals and normality don't mix.(Take
that out of context, man!)Well, that's amusing enough to make
it seem possible that youwere just trying to be funny. In case
you were also trying tomake a serious point:I certainly didn't
mean that this was a counterexample, orsomething that would
lead to a counterexample with no newideas required. But what
I had in mind was something_vaguely_ like this: Suppose we
could ?d x_n such thateach n-bit sequence in the binary
expansion of x_n occurswith the right frequency, but 1/x_n
has a terminatingbinary expansion. Then let x consist of a
long stretchof the bits of x_1, followed by a much longer
stretchof the bits of x_2, etc. For suitable values of
longand longer x will be normal, but it doesn't seem
sounlikely that 1/x would turn out to be abnormal.>Phil--
>Unpatched IE vulnerability: window.open search
injection>Description: cross-domain scripting,
cookie/data/identity > theft, command execution>Exploit:
http://safecenter.net/liudieyu/WsFakeSrc/
WsFakeSrc-MyPage.htmDavid C. Ullrich
===
What is the re?n
in a cylindrical mirror x^2+y^2 = a^2 of astraight line y+b =
0 as seen from a viewpoint (0,-c,h) ( a < b << c)? Is it a
catenary ? If not, what is it?
===
In 3D, y+b=0 is a plane.
Right? Are you looking for its re?n, orre?n of a
line?> What is the re?n in a cylindrical mirror x^2+y^2
= a^2 of a> straight line y+b = 0 as seen from a viewpoint
(0,-c,h) ( a < b << c)> ? Is it a catenary ? If not, what is
it?
===
> In 3D, y+b=0 is a plane. Right? Are you looking for
its re?n, or> re?n of a line?> What is the
re?n in a cylindrical mirror x^2+y^2 = a^2 of a>
straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a <
b << c)> ? Is it a catenary ? If not, what is it?It's probably
safe to assume, since this is a line on the ?f themall,
that the equation is y+b = 0 and z = 0. Now that it's been a
fewdays, can I give a hint? The symmetry of the problem --
whatcoordinate system should you be using?
===
When is the
homework due?> What is the re?n in a cylindrical mirror
x^2+y^2 = a^2 of a> straight line y+b = 0 as seen from a
viewpoint (0,-c,h) ( a < b << c)> ? Is it a catenary ? If
not, what is it?
===
> When is the homework due?:) this is my
curiosity about curvarture of lines when the virtualimage is
distorted in re?n.I observed clear images of square
lines of ?osaic in a polishedcylindrical ?ot in
Twelve Oak Mall, Detroit. What is commonlyknown is the
optics: image-object distances are connected by
standardformula (u-f)(v-f)=f^2 where f=a/2 or focal length,
u,v are object &image distances etc. in re?n. I did not
do much of image / raytracing in 3D before .I believed that
cos(slope angle) in the x-z projection is
inverselyproportional to z-coordinate, as it is for a
catenary z=A cosh(x/A).I asked this question expecting that
there could be elementarygeometrical operations with straight
lines, cylinders, re?n &c.My imagination surrounded vague
thoughts about geodesic deviation, butI right now I would
thank in advance for any ensuing pointers
andclari?ations.
===
What is the re?n in a cylindrical
mirror x^2+y^2 = a^2 of a> straight line y+b = 0 as seen from
a viewpoint (0,-c,h) ( a < b << c)> ? Is it a catenary ? If
not, what is it?What have you done so far?
===
Why don't you
guys go to sci.numerology?> a) E N G E L E> 5 14 7 5 12 5 =
48 48+ Dad 21 6 23 48+ Mom 4 11 23 188 Paulette 8 8 59
220/145 902> Paulette 100 Linda 40 Engele 48 Paulette was
born on the 8th day of the 8th month, her name begins with>
the 16th (8+8th) letter of the alphabet. Her given initials
add togetherfor> 28. Her last two names differ in value by 8
and add together for 88. Her> full name adds to 188. She has
19 letters (8th prime). Her names add tothe> 75th, 28th and
33rd non-primes, together for 136, corresponding to Numbers>
19 (8th prime), prettier as the numbers 1 through to 16 (8+8)
add together> for 136. * * * * * * *> b) This family stopped
to provide stats. 200 Mathew 13 11 66 317/48> Mathew 70
William 79 Shaw 51 195 Jeanine 7 10 77 280/85> Jeanine 58
Marie 46 Taylor 91 131 Aidan 7 5 03 127/238> Aidan Michael 51
Shaw 51 * * * * * * *> c) 97+ Dad 3 3 47 62/303 218 Ambrose 27
3 84 87/279> Ambrose 73 James 48 Pritchard 97 Ambrose was born
on the 27th, the vowels in his given names add to 27. Ambrose
was bon in 84 (7 times the 7th non-prime), he was born on
theprime).> He has 21 (7+7+7) letters, it's the number of
chapters in Bible Book 7.is> Bible Book 59 (the 17th prime).
The vowels in his ?st name add to the 21> (7+7+7) chapters
of Bible Book 7. He has 7 vowels in all, they add to
37chapter> 7). Dad and Ambrose were born on days of the year
adding to 149, there are149> verses in Bible Book 48 while
his middle name adds to 48. Ambrose was born on day 87
(29+29+29), his ?st name adds to the 73> adds to 218
(109+109, or twice the 29th prime). Daryl Shawn Kabatoff> Box
7134 Saskatoon Sask.> S7K 4J1
===
This guy is totally out of
touch with reality. So whatwould be the harm in simply
ignoring him from thisday forward and forever, and permitting
him to retirein the triumphant knowledge (as far as he's
concerned)that he is absolutely and totally correct in
everythinghe's ever said? He would be quite happy, and then
youall could have a lot more fun communicating with
realmathematicians. You all must know by now that youare
wasting your time. I mean the ones of you whoare responding
in a thoughtful reasoned way to James.The others, who are
simply calling him a horse's ass,should join another group,
maybe themutualasskicking.math group.
===
> This guy is
totally out of touch with reality. So what> would be the harm
in simply ignoring him from this> day forward and forever, and
permitting him to retire> in the triumphant knowledge (as far
as he's concerned)> that he is absolutely and totally correct
in everything> he's ever said?he wouldn't retire. he
is a
world-class crackpot and troll.have you seen his picture?> He
would be quite happy, and then you> all could have a lot more
fun communicating with real> mathematicians. You all must
know by now that you> are wasting your time. I mean the ones
of you who> are responding in a thoughtful reasoned way to
James.> The others, who are simply calling him a horse's
ass,> should join another group, maybe the>
mutualasskicking.math group.
===
> This guy is totally out of
touch with reality. So what> would be the harm in simply
ignoring him from this> day forward and forever, and
permitting him to retire> in the triumphant knowledge (as far
as he's concerned)> that he is absolutely and totally correct
in everything> he's ever said? He would be quite happy, and
then you> all could have a lot more fun communicating with
real> mathematicians. You all must know by now that you> are
wasting your time. I mean the ones of you who> are responding
in a thoughtful reasoned way to James.> The others, who are
simply calling him a horse's ass,> should join another group,
maybe the> mutualasskicking.math group.irresistable. You
yourself have added yet another JSH-related post to this
newsgroup.
===
>>This guy is totally out of touch with
reality. So what>>would be the harm in simply ignoring him
from this>>day forward and forever, and permitting him to
retire>>in the triumphant knowledge (as far as he's
concerned)>>that he is absolutely and totally correct in
everything>>he's ever said? He would be quite happy, and then
you>>all could have a lot more fun communicating with
real>>mathematicians. You all must know by now that you>>are
wasting your time. I mean the ones of you who>>are responding
in a thoughtful reasoned way to James.>>The others, who are
simply calling him a horse's ass,>>should join another group,
maybe the>>mutualasskicking.math group.> irresistable. You
yourself have added yet another JSH-related post to > this
newsgroup.And you another ... and me too.Gib
===
Francis
Harrington> This guy is totally out of touch with reality. So
what> would be the harm in simply ignoring him from this> day
forward and forever, and permitting him to retire> in the
triumphant knowledge (as far as he's concerned)> that he is
absolutely and totally correct in everything> he's ever said?
He would be quite happy, and then you> all could have a lot
more fun communicating with real> mathematicians. You all
must know by now that you> are wasting your time. I mean the
ones of you who> are responding in a thoughtful reasoned way
to James.> The others, who are simply calling him a horse's
ass,> should join another group, maybe the>
mutualasskicking.math group.Talking to Harris _about
mathematics_ is a waste of time, yes. After eightyears of
claiming to have a proof of FLT, he doesn't know an
algebraicinteger is, or what a ring extension is. Conclusion?
He doesn't _care_whether his statements are true or false.Even
if he had no audience at sci.math, would he stop posting? It
costs himnothing to cross-post to other rooms from which he
would still get his arsekicked.The JSH thing is tedious most
of the time, and morbid all the time, butstill good for the
occasional laugh. Try Google with these keywords: guffaw
idiot polynomialLH
===
I'm having a hard time with the
following problem:Let f be a continuous function on [1,oo)
such thatf^4 is Lebesgue integrable. Prove that if a > 4 then
|integral of f(x^a)| < oo wherethe integral is also over
[1,oo).Also, give an example where the integral of f(x^4) is
in?ite.Any hints or input on this would be greatly
appreciated.Hugh
===
> I'm having a hard time with the
following problem:Let f be a continuous function on [1,oo)
such that> f^4 is Lebesgue integrable. Prove that if a > 4
then |integral of f(x^a)| < oo where> the integral is also
over [1,oo).Also, give an example where the integral of
f(x^4) is in?ite.Any hints or input on this would be greatly
appreciated.Can you write the integral of f(x^a)
differently?Do you know Hoelders inequality?HTH,
===
> I'm
having a hard time with the following problem:Let f be a
continuous function on [1,oo) such that> f^4 is Lebesgue
integrable. Prove that if a > 4 then |integral of f(x^a)| <
oo where> the integral is also over [1,oo).Also, give an
example where the integral of f(x^4) is in?ite.I'll give you
a hint for a = 5: consider a change of variable inint_1^infty
f(x^5)^4 x^4 dx,then think Holder.--Ron Bruck
===
> William
Elliot <marsh@xx.com> ha scritto nel messaggio>Is there a
fast way to determine X and Y both integer such that:>X^2 -
Y^2 = A>Of course the answer to your question is
yes.>>Indeed, you phrased your question incorrectly.>>Factor
x^2 - y^2 = (x-y)(x+y) = a>>Thus x-y and x+y are factors of
a>>For each factorization of a = nm you have>>x-y = n; x+y =
m>>solve for x and y. Beware,>>if the solution turns out not
be an integer, discard it.>>Do this for all factorizations of
a.>>Collect the solutions you ?d.>>To make sure you
understand the process>>solve x^2 - y^2 = 12 for integers,
x,y.>>First list all the ways of factoring 12.>>Let's see
what you try.>>What are your answers?>>x^2 - y^2 = 66 has no
solution. Why?bacause it has tre factors.Think of the form of
the factorization of the leftside: x^2 - y^2 = (x - y)(x +
y)Now, let A = x - y, and B = x + y. What can you sayabout
the relationship between A and B? What sorts ofnumbers can
occur as A and B? For instance, you canprobably see that B -
A = 2y. What does that say aboutA and B (thought of together,
not just one at a time)?Having decided what numbers A and B
that can appear inthis factorization, now look at all the
factorizationsof 66. What sorts of numbers do you get from
thosefactorizations?If this isn't enough of a hint,
you're
trying too hard.The answer is really pretty
simple.ThanxmarcoDale.
===
> Saying that one random variable
is> equal to another one means that the value of the ?st is,
with > probability 1, equal to the value of the second.Huh? I
think you're understating. Saying that one random variable>
is equal to another is stronger than what you're stating --
it means> that *every instance* of the value of the ?st
variable is equal to> the value of the second one. (yes, what
I just said implies that> the probability is one -- but
probability one does not imply what I> said).Please correct
me if I'm mistakenI won't insist on this point. I
think it
depends on your point of view. E.g. in analysis L^1 functions
are not really functions, but rather equivalence classes for
equality almost everywhere.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Here is an important equation
I am encountering.>b <= x * ( (1-x)^a ) Inequality, not
equation.I accept.Solve for x in terms of a and b. Unlikely
to have a closed-form solution in general.>Okay, for my
purposes, a is an even integer and b is a ratioanl number
less> than one.> I am looking for a solution in the range
(0,1).Particularly, I am interested in the smallest value of
x which will satisfy> the inequality. If you can't get me the
smallest value, can you get me> something as small as possible
in a closed form solution. Right now the> smallest I have is
the place where x * ( (1-x)^a ) is maximised.... that place
being x_0 = 1/(1+a). I assume b < f(x_0) = a^a/(1+a)^(1+a),
where f(x) = x (1-x)^a. It seems that
f'''(x) > 0 for 0 < x <
1/(1+a), so if x_1 is any point with 0 <= x_1 <= x_0 with
f(x_1) < b, f(x) > f(x_1) + f'(x_1) (x-x_1) + 1/2
f''(x_1) (x
- x_1)^2.So if x_1 is not itself a solution, solve the
quadratic f(x_1) + f'(x_1) (x - x_1) + 1/2
f''(x_1) (x -
x_1)^2 = b;any solution x_2 of this with x_1 < x_2 < x_0 will
have f(x_2) > b.In particular, with x_1 = 0 you get x_2 = (1 -
sqrt(1-4ab))/(2a) ifb<1/(4a).Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
>but the Greeks started it.>
Just about every thing the Greeks did was rediscovered by
others> as it was needed in their time. Much of Archemedies
work was for War > efforts as they were needed. Sure they
have being ?st name > recognition but played no more role
than India and other countries > to thought today.I have
heard that Euclid's Elements is the book with more editions>
published than any other except the Bible. But wasn't his
school in> Alexandria? So perhaps he should be assigned to
Egypt (even if he was> Greek by culture and ancestry).And in
fact the MacTutor site does assign Euclid to Egypt,
althoughthere's no reliable information about his birthplace
or ancestry.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
===
>>but the
Greeks started it.>> Just about every thing the Greeks did
was rediscovered by others>> as it was needed in their time.
Much of Archemedies work was for War >> efforts as they were
needed. Sure they have being ?st name >> recognition but
played no more role than India and other countries >> to
thought today.>> I have heard that Euclid's Elements is the
book with more editions>> published than any other except the
Bible. But wasn't his school in>> Alexandria? So perhaps he
should be assigned to Egypt (even if he was>> Greek by
culture and ancestry).>And in fact the MacTutor site does
assign Euclid to Egypt, although>there's no reliable
information about his birthplace or ancestry.The Greeks in
Alexandria seem to have been mainly of Greekancestry, or to
be from those of other groups, mainly notEgyptians, who
joined the Greeks. The Ptolemies made noattempt to submerge
the other cultures, and the Greeks weresuf?iently xenophobic
(barbarian was the Greek for thosewho did not speak Greek,
basically) that they made no attemptto do so. When the
Seleucids did, they ran into open revolts.-- This address is
for information only. I do not claim that these viewsare
those of the Statistics Department or of Purdue
University.Herman Rubin, Department of Statistics, Purdue
University
===
>and the Greeks were>suf?iently xenophobic
(barbarian was the Greek for those>who did not speak Greek,
basically) Use of the term barbarian might suggest xenophobia
in a modernEnglish speaker, but Herodotus for example seems to
use it in aquite neutral sense to mean non-Greeks.-- Richard--
FreeBSD rules!
===
> 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--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/
===
>I was just
trying to see who in fact borned the most mathematicians. >My
SWAG list was not intended to be complete but just a start.
Ok, I >blew it after France at least according to MacTutor
link below. >However I was close with Italy, Germany and
former Russia but totally >underestimated England and the
U.S. and way over estimated India and>Greece. This is a
quantative analysis. The next step is qualitative>analysis
which is why I asked you guys the question.I doubt that
anybody would claim that being in the MacTutor list is all
that signi?ant as an indication of mathematical importance:
it's mainly a question of which people somebody has bothered
to writea biography of. >>There sure have been a lot of
important French mathematicians;>>Fourier, Galois,
d'Alembert...>Fermat - but his last theorem was published
after his death. >So during his life time it did not play a
major role in mathematical>developement.Fermat's importance
in mathematics does not rest on his Last Theorem.He made many
contributions in various areas. Although he didn't
publishmuch, he corresponded with many important
mathematicians, so he wasquite in?al in his lifetime.>
The same can be said of Da Vinci and other Greats
whose>certain works were suppressed during their
lifetime.Although da Vinci did study a lot of mathematics,
I'm not aware ofany really signi?ant contributions he made
to mathematics. >>but the Greeks started it.>Just about every
thing the Greeks did was rediscovered by others>as it was
needed in their time. Much of Archemedies work was for War
>efforts as they were needed. Sure they have being ?st name
>recognition but played no more role than India and other
countries >to thought today.This is just silliness. The main
contribution of the Greeks wasto discover the idea of
mathematical proof. Nearly all Western mathematicians until
very recent times got their ?st taste ofmathematics from
studying Euclid.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
===
> at 10:26
AM, rge11x@netscape.net (robert egri) said:>Please forgive me
my ignorance, I missed a few classes in college>thanks to some
heavily hopped Czechish Pilsner, but could you>elaborate on
this list especially regarding Czechoslovakia
and>Taiwan?Others have addressed some of the other countries.
My recollection is> that Hamilton was from Ireland and that
Chern was from Taiwan.Shiingshen Chern. Born 26 Oct 1911 in
Chia-hsing (or Jiaxing),Chekiang province(now Zhejiang),
China.You, too, need some well-hopped Pilsner to refresh your
memory ...
===
> There are lots of metrics. In fact, if d is
any bounded> metric, and there are lots of bounded metrics
equivalent> to any one, take D(f,g) = int d(f(x),g(x)) dP(x).
This> is equivalent to the topology you
speci?d.Noel.
===
-Steven> This question may be too simple,
but it is important to me.> In 7D space, I have two hyper
planes:> (*1):> (x1 x2 x3 x4 x5 x6 x7)*(a1 a2 a3 a4 a5 a6
a7)T = 0> (*2):> (x1 x2 x3 x4 x5 x6 x7)*(b1 b2 b3 b4 b5 b6
b7)T = 0> where T is the transpose sign, * is the matrix
multiplication sign.> Now I ?st get the
subspace(sub-hyperplane) S1 which is the> intersection of
(*1) and (*2). Then I combine the subspace S1 and a> vector>
v1 = (a1 a2 a3 a4 a5 a6 a7)T to form a hyper plane (*3), so
that this> new hyperplane is the span of S1 and v1, and S1,
v1 are on the> hyperplane (*3). The questions what is the
formula of this new> hyperplane?> Let u be a vector such that
the induced hyperplane x*(u^T)=0 (like in (*1)> and (*2))
contains S1. Then u is a linear combination of a and b.>
Proof: Every linear combination of a and b clearly has this
property, this> gives a 2D vector space. On the other hand,
the orthogonal complement of S!> (that is the set of all such
u) only is 2D because S1 is 5D and 2+5=7. qed.So we make the
ansatz u=s*a+t*b with unknown coef?ients s and t. Because>
v1 also lies in the hyperplane (*3) we have v1*(u^T)=0 which
gives a linear> condition on s and t. The result is 1D
solution space for s and t. It is 1D> because every multiple
of any solution u also is a solution. Just pick any> value
where not both s and t are zero and calculate u.
===
>Is it
possible to convert a problem where we are trying to maximize
a>quantity (pro?) into a problem where we are trying to
minimize a>(different) quantity (loss)?> Maximizing f is the
same as minimizing -f. Or perhaps what you have in> mind is
the equivalence in Linear Programming between solving the
primal> problem (a max problem) and the dual problem (a min
problem).Yes. The equivalence in Linear programming was what
I was thinking> about.> If yes, can we do the transform in
linear>time? In particular, is it possible to transform a
Network ?oblem where we try to maximize the ?tween
the source and sink>into a shortest path problem where we are
trying to ?d the shortest>between two points (say source and
sink)?> Maximum ? equivalent to minimal cut. If it's on
a planar graph,> and the source and sink are on the same
face, then this is equivalent> to a shortest-path problem in
the dual graph.> Here, I want to know if the Maximum ? a
network can be> formulated as a Linear programming problem.
Also, can you please> elaborate as to why the Maximal ?
equivalent to the> shortest-path problem _only_ for planar
graphs? Certainly maximum-?ith source s and sink t) is a
linear programming problem:maximize fsubject to sum_j (x_{sj}
- x_{js}) = f sum_j (x_{tj) - x_{jt)) = -f sum_j (x_{ij) -
x_{ji}) = 0 for all other i 0 <= x_{ij} <= c_{ij} for all i,j
where c_{ij} is the capacity of the arc i -> j, and x_{ij} the
? that arc (make x_{ij} = 0 if there is no arc i -> j).
Robert Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
>Is it possible to convert a
problem where we are trying to maximize a>quantity (pro?)
into a problem where we are trying to minimize a>(different)
quantity (loss)?> Maximizing f is the same as minimizing -f.
Or perhaps what you have in> mind is the equivalence in
Linear Programming between solving the primal> problem (a max
problem) and the dual problem (a min problem).> Yes. The
equivalence in Linear programming was what I was thinking>
about.> If yes, can we do the transform in linear>time? In
particular, is it possible to transform a Network ?oblem
where we try to maximize the ?tween the source and
sink>into a shortest path problem where we are trying to ?d
the shortest>between two points (say source and sink)?>
Maximum ? equivalent to minimal cut. If it's on a planar
graph,> and the source and sink are on the same face, then
this is equivalent> to a shortest-path problem in the dual
graph.> Here, I want to know if the Maximum ? a network
can be> formulated as a Linear programming problem. Also, can
you please> elaborate as to why the Maximal ? equivalent
to the> shortest-path problem _only_ for planar graphs?
Certainly maximum-?ith source s and sink t) is a linear
programming problem:maximize f> subject to> sum_j (x_{sj} -
x_{js}) = f> sum_j (x_{tj) - x_{jt)) = -f> sum_j (x_{ij) -
x_{ji}) = 0 for all other i> 0 <= x_{ij} <= c_{ij} for all
i,j> where c_{ij} is the capacity of the arc i -> j, and
x_{ij} the ? > that arc (make x_{ij} = 0 if there is no
arc i -> j). > Pradip
===
What conditions should be satis?d
so that a singular solution envelope for F(x,y,c) = 0 exists
for parametric variation of c ?It is known that singular
solution should satisfy the above F anddel F/ del c = 0 ;
perhaps it is a condition using second orderpartial
derivative.In the crank - connecting rod example, we have a
crank of radius Rcentered on origin, angle c with x-axis. The
end [R cos(c), R sin(c)]connects to a point on x-axis
(reciprocating piston) with connectingrod length L ;It is
found that an envelope exists in c limits + - ArcTan(L/R),
andthere is no envelope outside these limits.
===
I learned
all of my linear algebra over the reals or complexes,so this
seemingly simply problem has baf?.X is a vector space
over the ?ld F_2 = {0,1} of?ite dimension. K is a subspace
and K^perp is itsorthogonal complement,K^perp := {x in X with
<x,k>=0 for all k in K}.The equation I want to prove isdim K +
dim (K^perp) = dim X.I've tried on my own, and noticed many
discouraging things.Normally (in R^n, say), we have that K
and K^perp onlyintersect at zero. But this is no longer
necessarily true.I don't think that you can extend any
orthonormal basis ofa subspace to an orthonormal basis of the
entire space. Forone thing, Gram-Schmidt certainly doesn't
work any more.Many of these troubles probably stem from the
fact that thereare many nonzero vectors v for which <v,v>=0.
-Tyler
===
I think I understand now. The equalitydim(Null(A))
+ dim(Range(A)) = dim(X)is still true,where A is a linear
transformation in the ?ite-dimensional space X. I say still
true becauseI was not certain that it would hold for
vectorspaces over arbitrary ?lds. (I reviewed theproof to
con?m it still works.)We want to check that dim(F) +
dim(F^perp) = dim(X).Choose a basis of F, and let A have
these elementsas rows. Then Null(A) = F^perp; and the row
rankof A is clearly dim(F). Since row rank = col rank=
dim(Range(A)), we are done.
===
in message
<43b20b30.0311091614.90ffa88@posting.google.com>:> I think I
understand now. The equality dim(Null(A)) + dim(Range(A)) =
dim(X) is still true,> where A is a linear transformation in
the ?ite-> dimensional space X. I say still true because> I
was not certain that it would hold for vector> spaces over
arbitrary ?lds. (I reviewed the> proof to con?m it still
works.) We want to check that dim(F) + dim(F^perp) = dim(X).>
Choose a basis of F, and let A have these elements> as rows.
Then Null(A) = F^perp; and the row rank> of A is clearly
dim(F). Since row rank = col rank> = dim(Range(A)), we are
done.You can't assume Null(A) = F^perp without knowing more
aboutthe inner product (sesquilinear form) you're using. As
othershave pointed out, you really need to take into account
whetherthe form is degenerate. For example, if X has a
non-zero radicalR then dim(R) + dim(R^perp) = dim(R) + dim(X)
> dim(X).-- Jim Heckman
===
>I learned all of my linear algebra
over the reals or complexes,>so this seemingly simply problem
has baf?.X is a vector space over the ?ld F_2 = {0,1}
of>?ite dimension. K is a subspace and K^perp is
its>orthogonal complement,What is the inner product for the
vector space, though? There arecanonical inner products in
R^n and in C^n, but I don't know if thereis one for vector
spaces over F_2...>K^perp := {x in X with <x,k>=0 for all k
in K}.>The equation I want to prove isdim K + dim (K^perp) =
dim X.>I've tried on my own, and noticed many discouraging
things.>Normally (in R^n, say), we have that K and K^perp
only>intersect at zero. But this is no longer necessarily
true.>I don't think that you can extend any orthonormal basis
of>a subspace to an orthonormal basis of the entire space.
For>one thing, Gram-Schmidt certainly doesn't work any
more.>Many of these troubles probably stem from the fact that
there>are many nonzero vectors v for which <v,v>=0.What is
your de?ition of inner product? What are the axioms
yourequire of an inner product? In some books, an inner
product isrequired to be positive de?ite (i.e., <v,v>=0, and
<v,v>=0 if andonly if v =
0)...
===
=======================================

===
========It's not denial. I'm just very selective
about
what I accept as reality. --- Calvin (Calvin and
Hobbes)
===
======================================

===
=========Arturo Magidinmagidin@math.berkeley.edu===>X
is a vector space over the ?ld F_2 = {0,1} of>?ite
dimension. K is a subspace and K^perp is its>orthogonal
complement,>K^perp := {x in X with <x,k>=0 for all k in
K}.>The equation I want to prove isdim K + dim (K^perp) = dim
X.As you noted, inner products don't always behave likethe
standard one in R^n. But this result still holds if,for
example, <u,v> is still de?ed as the dot product.In that
case, you can describe K^perp as the solutionto a system of
linear equations. Do you know how to computethe dimension of
a nullspace (kernel)?Note that the result is _false_ if for
example <u,v>=0 forall u and v. You'll need to look to see
what propertiesyou're assuming about the inner product before
you try toprove something about it! (Nondegeneracy, for
starters.)dave
===
> I learned all of my linear algebra over
the reals or complexes,> so this seemingly simply problem has
baf?.X is a vector space over the ?ld F_2 = {0,1} of>
?ite dimension. K is a subspace and K^perp is its>
orthogonal complement,> K^perp := {x in X with <x,k>=0 for
all k in K}.> The equation I want to prove isdim K + dim
(K^perp) = dim X.I've tried on my own, and noticed many
discouraging things.> Normally (in R^n, say), we have that K
and K^perp only> intersect at zero. But this is no longer
necessarily true.> I don't think that you can extend any
orthonormal basis of> a subspace to an orthonormal basis of
the entire space. For> one thing, Gram-Schmidt certainly
doesn't work any more.> Many of these troubles probably stem
from the fact that there> are many nonzero vectors v for
which <v,v>=0. -TylerIt is best to think of K^perp not as a
subspace of X but as a subspaceof the dual space X*. (Now via
your pairing <x,y>, it is true that X*is identi?d with X, but
that just confuses things.)-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
> I learned all of
my linear algebra over the reals or complexes,> so this
seemingly simply problem has baf?.X is a vector space
over the ?ld F_2 = {0,1} of> ?ite dimension. K is a
subspace and K^perp is its> orthogonal complement,> K^perp :=
{x in X with <x,k>=0 for all k in K}.> The equation I want to
prove isdim K + dim (K^perp) = dim X.What do you mean by
<x,k>? You have not de?ed it.Jose Carlos Santos
===
Sorry.Fix
a basis of X (suppose dim X = m). Then let
thecoordinates/coef?ients of x and k bex1...xm and k1...km
respectively. Now we cande?e <x,k>:=x1k1 + x2k2 + ... +
xmkm,in a manner analogous to the standard innerproduct. It
is linear and symmetric, but doesnot really give a
norm.
===
>Sorry.Fix a basis of X (suppose dim X = m). Then
let the>coordinates/coef?ients of x and k be>x1...xm and
k1...km respectively. Now we can>de?e <x,k>:=x1k1 + x2k2 +
... + xmkm,>in a manner analogous to the standard
inner>product. It is linear and symmetric, but does>not
really give a norm.So? Not all inner products are positive
de?ite.
===
=====================================

===
==========It's not denial. I'm just very selective
about what I accept as reality. --- Calvin (Calvin and
Hobbes)
===
======================================

===
=========Arturo
Magidinmagidin@math.berkeley.edu
===
>Sorry.Fix a basis of X
(suppose dim X = m). Then let the>coordinates/coef?ients of
x and k be>x1...xm and k1...km respectively. Now we can>de?e
<x,k>:=x1k1 + x2k2 + ... + xmkm,>in a manner analogous to the
standard inner>product. It is linear and symmetric, but
does>not really give a norm.So? Not all inner products are
positive de?ite.This is, of course, depending on your
de?ition of inner product.The one usually given in linear
algebra and hilbert spaces insistsupon it.Take the usual
de?ition of inner product for a vector space over thereal or
complex ?ld. For vectors x,y,z and scalars a,b. We require1)
linear <ax+by,z> = a<x,z>+b<y,z>2) (hermitian) symmetric
<x,y> = conj<y,x>3) positive de?ite <x,x>= 0 and <x,x> iff x
= 0.However, you are right, some, physics in Lorentz space and
this topicof ?ite ?lds, allow a broader de?ition. Imho we
should havegiven it a different name since not all the
properties of the positivede?ite inner product (PD IP) carry
over to a non-PD IP.For example, for ?ite ?lds,
Gramm-Schmidt fails to generate a setof orthogonal vectors
from a set of indendent ones.-- Johan KULLSTAM
<kullstj-nn@comcast.net> sysengr
===
> Well, I discovered
something rather very odd (at least to me) today.> Let n be a
positive integer, and let s(n) denote the sum of all the>
divisors of n. So, for example, s(6) = 1+2+3+6 = 12. It turns
out there is a class of numbers with the property that a) n is
a perfect square> b) s(n) is a perfect squareThis one is in
Sloane's online database of offbeat sequences, numberA008847.
Go tohttp://www.research.att.com/~njas/sequences/and enter1,
9, 20, 180, 1306and click Search. The search will cough up
more terms of the table, andsome references in the lit.Larry