mm-32

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The Lorentz transforms themselves are proof t' and x' cannot
possibly be just coordinates. Examination of their derivation
verifies their identity as

===

Opponents should first
actually find out what the content is, then think, then
request/submit-to arbitration by the appropriate neutral
mathematics authorities. Flaming the hard- working, selfless,
*.answers moderators evidences ignorance and atrocious
netiquette.Archive-name:
physics-faq/criticism/lorentz-intervals (SR) Lorentz t', x' =
Intervals (c) Eleaticus/Oren C. Webster
Thnktank@concentric.netof the
use of 'just coordinates' in any scientific formula.Defenders
of the Special Relativity faith are especiallyfond of telling
opponents of their space-time fairy talesthat they do not know
the difference between coordinatesand magnitudes. That may
often be so, but the fault lies ultimately with SR dogma. The
Lorentz-Einstein transformations cannot possibly be 'just
coordinates', which is the interpre-tation required to support
the many sideshow carnival actswith which the SR faithful
bedazzle the public, and establishtheir moral and intellectual
superiority.If I get in my car and drive steadily for a few
hours at 50 kilometers per hour, is 50t the distance I travel?
Of course not. The last time my hours-counting 'just
coord-inates' clock was set to zero was when Zeno first
reported one of his paradoxes to Parmenides. That was a long
time ago, so my t is not useful for such purposes unless you
also use my clock to established the starting time, perhaps
t0, and use the formula 50(t-t0) to calculate the distance. In
any case, my t is even then not 'just a coordinate' becauseit
always represents particular elapsed times that can beused in
the (t-t0) form to calculate perfectly good timeintervals
(elapsed times).Alternatively, I could (re)set my clock to
zero at the startof some meaningful time interval, in which
case my t shows a scientifically perfect current and/or end
time. In which case, the Lorentz-Einstein t'=(t-vx/cc)/g is a
function of an elapsed time interval (not 'just a coordinate')
and a time interval (-vx/cc; the interval amount the t' clock
is being screwed up at time t) and thus cannot be 'just a
coordinate' since neither of the independent variables is such
a 'just' thing. {Their meaning is shown below, step-by-step.]If
it takes me 50 minutes to cross the Interstate highway,was x/50
my velocity crossing it?Of course not. The origin of all my
axes is at the veryspot where Zeno first presented his first
paradox to Parmenides. That makes my x equal a couple of
thousands ofmiles, plus, and is not useful for such purposes
unless you establish the starting x value, perhaps x0, and use
theformula (x-x0)/50 to calculate my velocity. In any case,
even then my x is not 'just a coordinate' because it always
repesents particular distance intervalsthat can always be used
in the (x-x0) form for any and everyscientific
purose.Alternatively, I could move my x-axis origin to the
starting(zero) point of some meaningful distance, in which
case my x shows a scientifically perfect current and/or end
distance.In which case, the Lorentz-Einstein x'=(x-vt)/g is a
function of a current/ending distance interval (not 'just a
coordinate') and a distance interval (-vt; the interval amount
the x' axisis being screwed up at time t) and thus cannot be
'just a coordinate' since neither of the independent variables
is such a 'just' thing. {Their meaning is shown below,
step-by-step.]1. Introduction
with the obvious debunking of the use of 'just coordinates' in
any scientific formula. 3. The Lorentz-Einstein transforms. 4.
The 'just coordinates' argument. 5. Single-system,
little-purpose ambiguity. 6. Relating two coordinate
measures/systems. 7. Distances and moving coordinate axes. 8.
Time intervals. 9. Einstein's (1905) derivations. 10. A word
about intervals. 11. Intervals versus the Twins Paradox. 12.
SummarySpecial Relativity's
space-time circus is based onthe 'transformation' equations by
which it is believedone can relate a nominally 'stationary'
system's spaceand time coordinates to those of an inertially
(notaccelerating) moving other observer. That moving
observer's own physical body and coordinate system might have
been identical in size to those of the stationary observer
before the traveller began moving, but are 'seen' as very
different by the stationary observer when the relative
velocity of the two is great enough, a high percentage of the
velocity of light.Concerning ourselves - as is customary -
with justthe spatial coordinate axis that lies parallel tothe
direction of motion, and with time, Einsteinarrived at these
formulas that relate the movingsystem measures or coordinates
(x' and t') to thestationary system coordinates (x and t): x'
= (x - vt)/sqrt(1-vv/cc) (Eq 1x) t' = (t -
vx/cc)/sqrt(1-vv/cc) (Eq 1t)The v is for the two systems'
relative velocity as seen by the stationary observer, and is
positive if the dir-ection is toward higher values of x. By
concensus,the moving system x'-axis higher values also lie
inthat direction, and all axes parallel the other
system'scorresponding axis.We used vv to mean the square of v
but might use v^2for that purpose below. Similarly for
c.Because it is believed that no physical object canreach or
exceed c, the square-root term in bothdenominators is presumed
always less than one, which means that the formulas say both x'
and t' will tend tobe greater than x and t, respectively.
However,SRians call the x' result 'contraction' - which
meansshortening - and the t' result 'dilation' - whichmeans
increasing. The 'just
coordinates' argument is so patently ridiculousthat even
opponents have a hard time accepting just howsimple and
obvious its debunking can be, as shown in thissection.
However, further sections take a more arithmet-ical approach
that you'll maybe find more professorial.The 'just
coordinates' argument is that t is mot aduration, not a time
interval; it's just an arbitraryclock reading. But what if the
moving system observercomes speeding by while you make your
annual 'springforward' or 'fall back' change? The formula says
thatthe moving system clock's 'just coordinate' reading can be
calculated from yours: t' = (t - vx/cc)/sqrt(1-vv/cc) (Eq
1t)Imagine the moving system oberver's confusion if his clock
changes its reading while he's looking at it! If his clock
doesn't change when yours does, the formula is wrong; if it is
truly a 'just coordinates' formula. And then what happens if
you realize you were a day early and put your clock back to
what it had said previously?And suppose you are in NYC and
your twin in LA andboth are watching the moving observer.
You'll both beusing the same v because you are at rest wrt
(withrespect to) each other. You're on Eastern StandardTime
and your twin is on Pacific Standard Timemaybe. You have three
hours more on your clock than does your twin. On which 'just
coordinate' clock will the Lorentz transforms base the 'just
coordinate' time the moving system clock says? The formula
applies to both ofyour t-times: t' = (t - vx/cc)/sqrt(1-vv/cc)
(Eq 1t)Sure, the idea that you can change someone else'sclock
with no connection of any kind is really ridiculous, but Eqs
1x and 1t aren't MY equations. Are they yours? And we aren't
the ones to say x, t, x', and t' are just coordinates.If the
t' formula is actually either an elapsedtime formula, or the
basis of a t'/t ratio, thenthere is no implication that one
clock's readinghas anything to do with the other's. It can
only be rates of clock ticking, or how one time INTERVAL
compares to the other that the formula is
about.Since we're going to be
comparing measurements on twocoordinate systems in the next
section, let's go toour supply cabinet and get our yard-stick
(which weuse to measure things in inches) and our
meter-stick(which we use to measure things in
centimeters).Here, I'm getting mine. Oh! Oh!There's an ant on
mine, and he ... she ... sure ishanging on, right at the 3.5
inch mark of the yard-stick.Let's see if I can wave the stick
around enough thatshe'll let go. Nope.However, before I gave
up I waved the stick and theant 'all over the place.Always,
however, the ant was at the 3.5 mark on the yard-stick, and
always 3.5 away from the end of thestick, however far and
wide I have transported her.Neither of those 3.5 facts means
very much. Of thetwo, the distance aspect meant almost
nothing. Sothe distance was 3.5 from the end. So what?
Thatlength, distance, was not in use. And only maybethe ant
might have been concerned with just whatlocation, 'just
coordinate', on the stick she wasat.Just so with x and t.So,
is the 3.5 reading just a coordinate? Or adistance/length?
It's ambiguous in and of itself,and really makes no difference
what you say untilyou try to make use of the number. Hey, my
address is 5047 Newton Street. If youare looking for me and
you're at 4120 Newton, itis helpful information, because it
tells you whichdirection to go. Is that 'just coordinate'?
Where it really becomes useful, perhaps, is intelling you how
far away I am. That's not justa coordinate value, that's a
distance, length,interval.However, it is subtracting 4120 from
5047 that tells you which direction and how far. It is only
because both 5047 and 4120 are distances from the same point -
ANY same point - that the result means anything.My x - my
yardstick reading - is always a distanceor length; it is
impossible to be otherwise withan honest, competently designed
yardstick.Whether or not its reading is of good use in some
particular scientific formula depends on whether I put the
zero end of the yardstick at some usefulplace. As in the
introduction, we should eitherput it at the starting
location/end, or use tworeadings from it:
(x-x0).Taking care to not damage
our brave little ant, I placemy yard-stick onto the table, zero
end to the left, 36end to the right.Now I place the 'just
coordinate' meter-stick on the tablein the same orientation,
in a random location, and findthat the ant's coordinate on the
meter-stick is 51.The formula relating centimeters to inches is
cm=i*2.54but we want a formula similar to
x'=(x-vt)/sqrt(1-vv/cc).That would be c=i/.03937
approximately, but let's use x'for the meter-stick reading,
and x for the inch reading: x'=x/.3937. 3.5/.3937 = 8.89 Wait
a minute. It's not just science but definition that says
c=i/.3937=8.89, so something is wrong. 8.89is not 51.We
already knew that 51 cm was just an arbitrary coordinate.
Arbitrary not because that point isn't 51 cm from the zero end
of the meter-stick, but because the zero point was in an
arbitrary position.Let's put the meter-stick in a position
where it's zero point is at the yard-stick zero point.What is
the centimeter coordinate now? Hey. 8.89,just like the formula
says.The only way for a 'transform' like x'=x/g to work,
whatever g might be, is for both coordinate systemsto have
their zero points aligned, in which casesaying the two
measures are not intervals is pureidiocy.Noe that with both
zero points at the same positionboth x' and x are great
measures for scientificpurposes, in any and every case where
we were smartenough to put those zero points at a useful
location.There is one extension of x'=x/g that will let ususe
the meter-stick in arbitrary position. When the cm reading was
51, the zero point of theyard-stick read (51-8.89=) 42.11 cm.
If we call thatpoint x.z' we get x' = x.z' + x/.3937. = 42.11
+ 3.5/.3937 = 42.11 + 8.89 = 51.Obviously, in this formula
x/.3937 is the distancefrom the x' coordinate of the location
where x=0. An interval.Just as obviously, the fact that we now
have thecorrect formula for relating an x interval to
anarbitrary x' coordinate, does not mean that x'is anything
more than nonsense for use in anyscientific formula.Unless we
were smart enough to put the x zeropoint in a useful location,
and use (x'-x.z') inthe scientific formula. (x'-x.z') equals
the useful,Ratio Scale value x/.3937.So, we have discovered a
basic fact: a transformationformula like x'=x/g works only if
the two zero pointsof the coordinate systems coincide. That
makes it non-sense to say the two coodinates are only
coordinatesand not intervals. Both must be values that
representdistances from their respective zero points unless
youtake the proper steps to adjust for the discrepancy.Make
sure you understand that although the inclusionof x.z' made it
possible to correctly calculate x',the result is nonsense when
it comes to use of x'for general length/distance purposes; it
is x'-x.z' that is a useful number in such cases. It could
bethat we're measuring a sheet of paper with one endat x=0 and
the other at x=3.5; x'=51 is nonsense asa centimeter measure of
the paper.But, you say, the Lorentz transform contain a -vt
term.We discovered x'=x.z' +
x/g as the correct formulafor relating one coordinate to
another system's.But the Lorentz transform contains another
term, -vt/sqrt(1-vv/cc). What is it?Let's start with our x'=51
cm, x=3.5, x.z'=42.11 example.Every minute, let's move the
meter-stick one inch to ourright.At minute 0, the cm reading
was 51 cm.At minute 1, the cm reading is now 50 cm.At minute
2, the cm reading is now 49 cm.In this instance, v=1
inch/minute. And t was 0, 1, 2.What has happened is that we
have made our x.z' a lie,and increasingly so. -vt/.3937 is the
change in x.z'. x' = (x.z - vt/.3937) + x/.3937.Obviously,
vt/.3937 is not a coordinate; even most SRianswouldn't imagine
it was. It is an interval, the distanceover which the moving
system has moved since t=0.And, of course, x/.3937 is the
distance of our bravelittle ant from the point where x=0 and
the centimeterreading is x.z'-vt/.3937. Yes, every minute the
meter-stick moves to the right and the meter-stick
coordinateof the spot where x=0 gets less and less - and
eventuallynegative.Make sure you understand that every minute
the x' coordinate, because of -vt/g, becomes a better measure
of, say, the 3.5 paper we might be measuring with the
yard-stick, given that 51 was too big a number and-vt is
negative. That is, until the two origins coincide at x'=x=0,
and then it gets worse and worse.With -vt positive (because
v<0) the situation is different.With 51 and -vt positive, x'
just gets worse and worseover time.Quite obviously, the fact
that we now have thecorrect formula for relating an x interval
to anarbitrary x' coordinate even when the x' axis ismoving,
does not mean that x'is anything more than nonsense for use in
any scientific formula.Unless we were smart enough to put the x
zero point in a useful location, and use (x'-x.z'+vt/.3937)
inthe scientific formula. (x'-x.z'+vt/.3937) equals the
useful, Ratio Scale value
x/.3937.Instead of using our
sticks, let's get out two clocks.Mind you, we're not going to
deal with different clockrates here, just establish the same
basics as for distance.Your clock says 9:00 Eastern Standard
Time (EST) and we note that t=540 minutes when we put down the
clock.Blindly, let's turn the setting knob of your twin's
Pacific Standard Time clock and put it down before
us.According to what we see, EST's 540 minutes (9:00)
corre-sponds to PST's 14:30; t'=870.We know the formula
relating PST to EST is t' (pacific)= t (eastern) - 180
(minutes). Thus, it is not correct that the second clock can
have an arbitrary setting, because 870 <> 540-180. We know
that the two clocks are related by t' = t/1 since both are
using the same second, hour, etc units. But 870 (14:30 in
minutes) is not 540/1-180, so once again we know something is
wrong. However, t'=t.z' + t/1 works. EST midnight equals PST
0.0 (midnite) - 180, so t.z' = -180, and t' = -180 + 540/1 =
360.Since EST-180=PST, 9:00 EST is 6:00 PST = 360 minutes.We
see thus that like distance measures/coordinates, timeaxis
origins (zero points) must either be 'lined up' or adjusted
for. So, the Lorentz/Einstein t'=t/sqrt(1-vv/cc) must be the
moving system elapsed time interval since the time axes were
both at a common zero. There is no t.z' adjustment: t' = (t -
vx/cc)/sqrt(1-vv/cc) (Eq 1t)Make sure you understand that in
the clock case, if theEST is showing a good number for elapsed
time since thetravelling observer passed NYC, then the PST
clock issilliness. t.z' must be zero or must be taken out
oftime lapse calculations for the PST clock to be
usedintelligently, just as was true for x.z'.What is lacking
as yet for Lorentz t' is the -vx/cc term thatcorresponds to
the x' formula -vt term.Break it up into two parts: v/c and
x/c. v/c is a scaling factor that changes velocity from
whatever kind of unit you are using over to fractions of c.x/c
is distance divided by velocity, which is time. x/cis thus the
time interval since the two time axeshad a common zero point -
which they have to have in theLorentz transforms which do not
have the t.z' term welearned to use above.Thus,
(-vx/cc)/sqrt(1-vv/cc) is the interval amount the moving
system clock has been changed - since the common/adjusted time
- over and beyond the elapsed time intervalrepresented by
x/sqrt(1-vv/cc).We have discovered that the only way for t' to
be t/gis for t' and t to have a common zero point, just asfor
x' and x. It would be otherwise if the t' formulacontained an
adjustment t.z' under some name or other,but the necessity to
include such a term correlates100% with t' numbers that aren't
directly usable.As for x and x', our knowledge of how to setup
a properformula relating t and t' is of no use unless we
usethe knowledge in scientific formulas; (t'-t.z'+xv/gcc)gives
us the only directly useful value:
t/g.When we return to
Einstein's derivations of the transformformulas with a
well-focused eye, we find he was a wee bitconfused - or at
least self-contradictory.When he set up his (at first unknown)
tau=moving systemtime formulas, he created three particular
instances of tau.Tau.0 is the time at which light is emitted
at the movingorigin toward a mirror to the right that is
moving at rest wrt that moving origin and at a constant
distance from that origin. He lets the stationary time slot
have the value t,a constant, the stationary system starting
time.Tau.1 is the time at which the light is reflected. Helets
the stationary time be t+x'/(c-v); t is still aconstant and
x'/(c-v) is the time interval since t.Tau.2 is the time at
which the light gets (back) to themoving origin. The
stationary time value is put as t +x'/(c-v) + x'/(c+v); t is
still a constant and x'/(c-v)+ x'/(c+v) is the time interval
since t.On the thesis that the moving observer sees the time
tothe mirror as the same as the time back to the origin,he
sets .5[ tau.0 + tau.2 ] = tau.1.Tau.0 completely drops out of
the analysis and leavesno trace, and has no effect.Further, the
t you see in tau.0, tau.1, and tau.2 also completely drops out
with no trace and no effect, leaving us with exactly what
you'd get if you had explicilty said t' is an interval and so
is t.What doesn't drop out in the stationary time values
isx'/(c-v) and x'/(c+v), the time interval it takes forlight
to get to the fleeing mirror, and the time intervalit takes
for light to get back to the approaching origin.Thus, his
resultant t' formula is strictly based on time intervals in
the stationary system. Time intervals since some starting
time, yes, but time intervals.There is absolutely nothing in
the derived formulas that depends on arbitrary coordinates
like the constant t in the stationary time arguments.Let's
look at the x dimension; it is x'=x-vt [as x increasesby vt,
the effect over time is x'=(x+vt)-vt)], which Einstein
explicitly sets up as a constant stationary distance.He uses
that x' not just in the time interval parts of the stationary
time arguments, but also in the x (distance) stationary system
argument for the tau at the time light is reflected. x' can't
be the stationary system coordinate of the mirror at that
time. That value is x'+vt.x' is explicitly an interval,
distance. Thus, the whole tau derivation of the t' formula is
fully andexplicitly based on x' - a spatial
length/distance/interval -and the two time interals x'/(c-v)
and x'/(c+v).While we're at it, if the starting t is not zero,
his x'=x-vt formula is complete nonsense also. Given thatthere
was some L that was the mirror x-location and lengthwhen the
light is emitted, if t was already, say, 500, thenx'=L-vt
could have been a very negative
length.There are intervals, and
there are intervals.If we put our yard stick zero point at one
endof a piece of paper and read off the coordinateat the other
end of the paper, we have a goodmeasure of the paper's length,
a Ratio Scalemeasure. [Absolute temperature scales are
ratioscale.]If instead we put the one end of the paper at
theone inch mark (or the zero end of the stick oneinch 'into'
the length of the paper) we get measuresthat are one inch off
the true, ratio scale length.The two messed up measures are
still intervals,but they are Interval Scale measures.
[Householdtemperature scales are interval scale, which iswhy
your physics and chemistry professors won'tlet you use them
without first converting to theratio scale absolute
temperatures.)t'=t/g and x'=x/g represent ratio scale
measures,given that t and x were ratio scalae to start
with.t'=t.z'+t/g and t'=t/g-vx/gcc are both interval scale
measures, even given a good ratio scale tand a good ratio
scale x.x'=x.z'+x/g and x'=x/g-vt/g are both interval scale
measures, even given a good ratio scale xand a good ratio
scale t.Look for the (SR) Lorentz t', x' = degraded
measuresdocument soon at a newsgroup near
you.t'=(t-vx/cc)/g shows t'
being greater than t.The reason Special Relativity will not
allow theuse of its basic time equation in determining whatSR
has to say about the twins' ages, is that t' andx' are
supposedly just coordinates, and they say you have to take the
coordinate pairs (t',x') and (x,t)into consideration in both
the time and place the twins' separation started and the time
and place the twins reunited.Since t' and x' are actually both
intervals, notjust coordinates, the 'excuse' is spurious, and
is so even without use of the obvious (x_b-x_a) and(t_b-t_a)
usages.However, SR is right to be embarrassed by
theirtransformation formulas.Look for the (SR) Lorentz t', x'
= degraded measuresdocument at a newsgroup near you.
A. t'=t/g and x'=x/g can be
almost 'just coordinates' in the sense that the values
obtained may not be of much use except in the most primal and
useless way: how long and how far since/from the time/ place
they were zero. Even here, however, the zero points within
each of the two scale pairs (t',t) and (x'.x) must have been
lined up. If the zero points have been intelligently selected
(such as at the starting point and time of a trip) they can be
rationally used 'as is' in any valid sci- entific equation.B.
Even the interval scale t'=t.z' - xv/gcc + t/g and x'=x.z' -
vt/g + x/g are not 'just coordinates'. They can be used to
good effect by establishing the relevant starting times/points
and using (t'-t.z'+xv/gcc) and (x'-x.z'+vt/g), as the situation
may require.C. When you see vx/gcc or vt/g in use in any guise
with non-zero values, you know the resultant t' or x' is a
degraded, interval scale value.E-X: Anytime you do not see
what amounts to t.z' and xv/gcc in the time case, or x.z' and
vt/g in the distance case, you know that the t' and/or x' in
use are intervals. Period.Y: Either set your clock to zero at
the start of the relevant time interval, or use (t-t0), with
both being readings on the same clock. Either move your x-axis
origin to the starting end or point, or use (x-x0), with both
being readings on the same axis.Z: In _(SR) Lorentz t', x' =
Degraded (Interval) Scales_ we see that t' and x' satisfy the
mathematical tests for/of interval scales when -vt and -vx/cc
are not zero; thus, they must be intervals. When -vt and
-vx/cc are zero, t' and x' satisfy the much better
mathematical definition of ratio scales, and are thus not just
mere intervals, but (rescaled) good
ones.Eleaticus!---?---!---?---!---?---!---?---!---?---!---?---
!---?---!---?---!---?! Eleaticus Oren C. Webster
ThnkTank@concentric.net ?! Anything and everything that
requires or encourages systematic ?! examination of premises,
logic, and conclusions
?!---?---!---?---!---?---!---?---!---?---!---?---!---?---!---?
---!---?

===

Let f(x) be a function of one real variable and
let f*(x) be itsinverse.Suppose f(x)=f*(x) for all x. Is
anything known about the generalsolution set of this
equation?The only solution I can think of is f(x)=g*(a/g(x)).
Can you determine any other solutions?Eugene
Shuberthttp://www.everythingimportant.org

===

> Let f(x) be a
function of one real variable and let f*(x) be its> inverse. Suppose f(x)=f*(x) for all x. Is anything known about the
general> solution set of this equation?> The only solution I
can think of is f(x)=g*(a/g(x)). Can you > determine any other
solutions?0. What exactly are a and g above? What exactly
is a/g(X) when g(X) is 0 (as it must be for some X, in order
for g to have an inverse)?1. f(x)=f(-1)(x) -> f(f(x))=x. Unless
you have other conditions on f, beyond merely being a function
from R to R, that's about all you can say.2. E.g., suppose f
is defined for 0<=x, f(0)=0, and f is monotone decreasing.
Then f can be extended to x<0 in such a way that f(f) is the
identity function.

===

>Let f(x) be a function of one real
variable and let f*(x) be its>inverse.So I presume f is
supposed to be 1-1 and onto.>Suppose f(x)=f*(x) for all x. Is
anything known about the general>solution set of this
equation?This is equivalent to f(f(x)) = x.>The only solution
I can think of is f(x)=g*(a/g(x)). Can you >determine any
other solutions?Let A and B be disjoint sets of the same
cardinality, and h: A -> B1-1 onto. For a in A define f(a) =
h(a) and f(h(a)) = a, and for x not in A union B take f(x) =
x.If you want f to be continuous, it must have exactly one
fixed point p, A = (-infinity,p) and B = (p,infinity), and h
is a decreasing function from A to B. Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

> 2. E.g., suppose f is
defined for 0<=x, f(0)=0, and f is monotone > decreasing. Then
f can be extended to x<0 in such a way that> f(f) is the
identity function.but from f(f)=id it does not follow that f
is invertible because f maybounded so that its inverse does
not exist. But the algebraicdefinition of invertibility is
that left and right inverse exist,which is obviously the case
here. So do we have two differentdefinitions of invertibility
here?

===

> 2. E.g., suppose f is defined for 0<=x, f(0)=0,
and f is monotone >> decreasing. Then f can be extended to x<0
in such a way that>> f(f) is the identity function.> but from
f(f)=id it does not follow that f is invertible because f may>
bounded so that its inverse does not exist. But the algebraic>
definition of invertibility is that left and right inverse
exist,> which is obviously the case here. So do we have two
different> definitions of invertibility here?f is invertible
if there is a g with gf = id = fg. So if ff =id,f /is/
invertible (ff=id -> f not bounded). What is true (and maybe
what you meant?) was that f monotone decreasing only
impliesthat it's injective. For my construction to be correct,
I also needf to be surjective; this is assured if f is
continous and not boundedbelow.

===

> The only solution I can
think of is f(x)=g*(a/g(x)). Can you > determine any other
solutions?1. The term you seek is involution.2. A bit more
general (in terms of a function that actually can be written
down explicitely) would be a Moebius transform
f(x)=(ax+b)/(cx+d), the conditionto make this an involution
being left as an amusinghomework :-)-- Hauke Reddmann <:-EX8
For our chemistry workgroup,remove math from the addressFor
spamming, remove anything else

===

> f is invertible if there
is a g with gf = id = fg. So if ff =id,> f /is/ invertible
(ff=id -> f not bounded). What is true (and > maybe what you
meant?) was that f monotone decreasing only implies> that it's
injective. For my construction to be correct, I also need> f to
be surjective; this is assured if f is continous and not
bounded> below.Right, of course, fg=id implies surjectivity of
f.

===

> Let f(x) be a function of one real variable and let
f*(x) be its> inverse.> Suppose f(x)=f*(x) for all x.
Is anything known about the general> solution set of this
equation?> The only solution I can think of is
f(x)=g*(a/g(x)). Can you > determine any other solutions?>
0. What exactly are a and g above? What exactly is
a/g(X)> when g(X) is 0 (as it must be for some X, in order
for g to > have an inverse)?a is a constant. If g is an
invertible function with inverse g* thenthe function
f(x)=g*(a/g(x)) has the property that f(f(x)=x.For solutions
to f=f* my primary interest is in differentiablefunctions on
the interval (b, infinity). I was also thinking that f(x)could
equal f*(x) for all x. I now suppose that solutions for
thatcase are greatly limited. What's going on? I see that
f(x)=-x+a andf(x)=x is such that f(f(x)=x for all x but I
can't think of any othersolution.Please help.Eugene
Shuberthttp://www.everythingimportant.org

===

I believe that
the function f(x)=arcsinh(1/sinh(x)) is plenty explicit.
Eugene Shuberthttp://www.everythingimportant.org
boundary=----=_NextPart_000_0044_01C35F5C.03259350

===

--
-my question is wheather the equality <A,BxC>=<C,BxA>=<C,AxB>
is established or not.(where, A,B,C is a vector and <,>
means inner product, x means outer product)I look for this
in my linear algebra book, but there isn't about this.If this
is true, then please let me know why this is true.The reason
for wondering this, is following statement in my differential
geometry book.I read : k g(geodesic curvature)=<unit tangent
curve' , (unit tangent curve'' x unit normal vector)> =<unit
tangent curve'' , (unit normal vector x unit tangent curve'
)> =<unit normal vector , (unit tangent curve' x unit tangent
curve'')>thanks in advance...

===

> my question is wheather
the equality <A,BxC>=<C,BxA>=<C,AxB> is> established or not.
(where, A,B,C is a vector and <,> means inner> product, x
means outer product)If outer product means the vector
product in R^3, then thisis almost true. What is true is that
<A, B x C> = <C, A x B> = <C, A x B>but actually <C, B x A> =
- <C, A x B>.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen

===

>
this is originally informatic problem, but it's a math
challenge too.> I cant figure out what are the secret numbers
they want me to search> for such that the conversation can be
possible ( see the problem).>> Problem C> Secret Numbers>>
Input File: C.DAT> Program Source File: C.PAS or C.C or
C.CPP>> Two natural numbers a and b are chosen (1<a<b). Person
M is told the> multiple of a and b (a*b), and person S is told
the sum of a and b> (a+b). The discussion between M and S goes
like this:>> M: I do not know the numbers a and b.> S: I do not
know them either, but I knew you would not know them.> M: Now I
know the numbers!> S: Now I know them, too!>> Input: The input
is given in a text file. The input file contains> pairs of
natural numbers x and y (2<=x<y<=550), one pair per line. The>
input is guaranteed to be correct.>> Output: For each pair x,
y, find all pairs of a and b, such that> x<=a<b<=y and that
the given discussion is possible. Write these pairs> in a
single line, and finish that line with no more pairs. if
there> are a and b found in the given range, or write simply
no pairs. if> there are not. Separate the numbers of a pair
with a comma, terminate> each pair with a semi-colon, and
separate different pairs with a blank> after the semi-colon,
as shown in the example below.>> Example:> input> 2 10 <-no
pairs> 2 20 <-4,13; no more pairs>> this input is giving two
cases:> 1) x=2 and y=10 and it says no pairs for that that
range> 2) x=2 and y=20 and it returns the single possible pair
a=4 and b=13,> no more pairs in the range>> I tried all kinds
of tricks from crytography to understand what kind> of numbers
I am seeking for, but I failed.> Does any1 see the maths behind
those numbers?You do not need crytography (whatever that is)
tricks for this problem.What you do need, in the problem
definition, is an upper bound on a and b.Anyway.Suppose a=3
and b=4.M is told 12 and S is told 7.M deduces a=2, b=6or a=3,
b=4So M would say I do not know the numbers a and b.S
deducesa=2, b=5a=3, b=4So S would say, I do not know them
eitherBUT, if a=2 and b=5, M would have been told 10, and
would have been able todeduce a and b.SO S cannot say but I
knew you would not know them.S can only say that if all pairs
that add up to the sum give an ambiguousproduct to M.There is
no magic formula for this puzzle. Hope this helps.You may
like to look at the threadhttp://tinylink.com/?WZNSsZjEv2--
Clive Toothhttp://www.clivetooth.dk

===

What is the most
complex three dimensional combination of rotationsyou can get
before it can be described by a simpler combination
ofrotations? IE, how many different kinds of rotations and
rotationalaccelerations and torques, etc. can you all put
togethergyroscopically before you have to describe it in terms
of somethingsimpler?A better way of putting it is this: What is
the most complex rotationfunction or class of rotation
functions you can make? That is, whenyou have the most complex
rotation function or class of rotationfunctions, and additional
rotation function or class of rotationfunctions you add to it
does not increase its complexity.To envision this question,
imagine I have a gyroscope with dozens ofdifferent circles, or
chambers, and I set them in motion, one by one,and describe the
complex rotation by a number of mathematicalfunction. Exactly
how many times can I do this before I don't have todescribe it
by any more numbers of mathematical functions? Or is therea
limit at all?And now the physics part of the question: Is
there an angular energyand angular momentum four vector such
that we get an invariantquantity? Or do we count any energy
and momentum gotten from rotationsand spinning as just regular
energy and momentum? Meaning that thingsthat spin, even if
they're massless, can not go the speed of light ifthey have
any energy in their spinning, meaning they also cannot
havemomentum? Or if they do have momentum, they must have
energy, andcannot go the speed of light?I'm just wondering
this, because it seems to me that this means thatlight, which
is a number of photons, is NOT the fastest thing in
theuniverse then, since it has angular momentum, but no net
momentum inthe axises orthogonal, or normal, to its motion,
yet it still wouldhave energy, making it have nonzero mass,
and therefore not going atthe universal top speed.Or is the
spin something that doesn't follow these rules, IE,something
that is entirely nonphysical?(...Starblade Riven
Darksquall...)

===

In message
<4aa861fb.0308110143.3b189ff5@posting.google.com>, Starblade
>What is the most complex three dimensional combination of
rotations>you can get before it can be described by a simpler
combination of>rotations? IE, how many different kinds of
rotations and rotational>accelerations and torques, etc. can
you all put together>gyroscopically before you have to
describe it in terms of something>simpler?Addition of
infinitesimal rotations commutes. Angular rates of change are
limits involving infinitesimal rotations. So any number of
angular momenta, torques etc. can be summed according to the
normal rules of vector arithmetic.>>A better way of putting it
is this: What is the most complex rotation>function or class of
rotation functions you can make?A (pseudo) vector.>That is,
when>you have the most complex rotation function or class of
rotation>functions, and additional rotation function or class
of rotation>functions you add to it does not increase its
complexity.>>To envision this question, imagine I have a
gyroscope with dozens of>different circles, or chambers, and I
set them in motion, one by one,>and describe the complex
rotation by a number of mathematical>function. Exactly how
many times can I do this before I don't have to>describe it by
any more numbers of mathematical functions? Or is there>a limit
at all?No limit. Its angular momentum is a single vector
quantity.>>And now the physics part of the question: Is there
an angular energy>and angular momentum four vector such that
we get an invariant>quantity? Or do we count any energy and
momentum gotten from rotations>and spinning as just regular
energy and momentum? Meaning that things>that spin, even if
they're massless, can not go the speed of light if>they have
any energy in their spinning,Why not?>meaning they also cannot
have>momentum? Or if they do have momentum, they must have
energy, and>cannot go the speed of light?Why not?>>I'm just
wondering this, because it seems to me that this means
that>light, which is a number of photons, is NOT the fastest
thing in the>universe then, since it has angular momentum, but
no net momentum in>the axises orthogonal, or normal, to its
motion, yet it still would>have energy, making it have nonzero
mass,Why would having energy give it nonzero mass?Here's a
hint: E = mc^2 is not generally true.>and therefore not going
at>the universal top speed.>>Or is the spin something that
doesn't follow these rules, IE,>something that is entirely
nonphysical?>>(...Starblade Riven Darksquall...)Reading too
many fantasy novels is bad for you.-- Richard Herring

===

[snip
question asked several ways]> A better way of putting it is
this: What is the most complex rotation> function or class of
rotation functions you can make?The 3-D rotation group. IIRC,
it's SO(3), special, orthogonal,3 by 3 matrices.> To envision
this question, imagine I have a gyroscope with dozens of>
different circles, or chambers, and I set them in motion, one
by one,> and describe the complex rotation by a number of
mathematical> function. Exactly how many times can I do this
before I don't have to> describe it by any more numbers of
mathematical functions? Or is there> a limit at all?I'm sorry,
but you faded there. What does it mean for a gyroscopeto have
circles? What does it mean for a gyroscope to have
chambers?Are you talking about adding different gyrscopes with
differentrotating wheels?[vague, hazy question about angular
momentum and relativity snipped]Yes there is a relativistic
formulation of angular momentum.No, it does not mean that
light can't travel at c.Socks

===

Virgil refers to a proof of
Cantor of the uncountability of the reals. I believe he refers
to the nested sequences have real endpointsargument.Virgil
claims to not know the meaning of the words antidiagonal
orantisemitridiagonal, or rather, to have very little idea.
Thediagonal of the matrix is the sequence with the 'ith
element of thesequence being the element in the i'th row and
ccolumn of the matrix. The tridiagonal is the set of three
sequences consisting of one thatis having as its i'th element
the element in the i+1'th column andi'th row, the second being
the diagonal, and the third having the i'thelement being the
element in the i'th column and i+1'th row of thematrix. The
semitridiagonal is the set of the first and second orsecond
and third sequences of the tridiagonal. An antidiagonal is
asequence differing at each place from the diagonal.
Anantisemitridiagonal is a set of two sequences each differing
in eachplace from the two sequences of the
semitridiagonal.Virgil, the antisemitridiagonal argument is
about the 01 and 10subsequences.Virgil claims on behalf of the
status quo that an infinite set can notmap to another infinite
set that is not its powerset or a superset ofits powerset, in
that he claims that the rationals are not allowed tomap to the
powerset of the integers. I say powerset of integers isnot
powerset of rationals and powerset of integers is not
supersetof powerset of rationals, as well as powerset of
integers is propersubset of powerset of rationals. The
powerset proof has multiplelines of attack against it from
that an infinite set is not precludedfrom mapping to the
powerset of one of its proper subsets, thepowerset argument
fails to show otherwise, except that the mappingwould
invalidate it. The powerset argument only applies to a
giveinfinite set and its very own powerset, or that of a set
isomorphic toit in the way the evens are isomorphic to the
odds, in the naturals. I just say there is no immediate
injection from a set to its powerset.I think there are much
better ways to gauge the relative sizes ordensities of
infinite sets than cardinality, for example the value ofthe
asymptotic density of the subset of naturals in the naturals,
andthe various algebraic properties of the sequences that can
be said tocomprise the set.I say there are more reals than
rationals because there are infinitelymany proper supersets of
the rationals that are proper subsets of thereals. I say the
same for N and Z and E and N.I have a different context and
reason for claiming that not all theelements of Z+ are
elements of N, that being about how Z+ is a propersubset of
N.There's been some discussion on the list recently about
theprobability that a random real form a uniform distribution
over thereals is a rational. I think that the rationals and
irrationalsalternate on the real number line, and that the
probability of a realbeing a rational is one half. One half is
certainly positive. Theasymptotic density of the evens in the
naturals is one half, theprobability of arandom element from
a uniform distribution over the naturals beingan even number
is that value.Virgil, you're pretty funny. It's nice of you to
drop defense of thediagonal argument.Virgil, or anyone who
cares, please put forward Cantor's otherproof. If it's the
one I think it is, it applies to rationals as wellas reals,
nested sequences have rational endpoints.Ross

===

> Virgil
refers to a proof of Cantor of the uncountability of the
reals.> I believe he refers to the nested sequences have real
endpoints> argument.> Cantor's first proof of the
uncountability of the reals was indeed based on strictly
nested sequences of compact real intervals which do, of
necessity have real endpoints, but it is the intersection of
all of the intervals of such a sequence which is of interest,
since as Cantor constructed it (1) it cannot be empty and (2)
none of its points are in the listing from which it was
constructed, so that the listing must be incomplete.> Virgil
claims on behalf of the status quo that an infinite set can
not> map to another infinite set that is not its powerset or a
superset of> its powerset,Not at all, what I claim, and Contor
proved, is that no set can be mapped onto any set of cardinal
greater thanor equal to that of its power set. What Ross
claims I said would prevent some sets from being mapped onto
themselves. And while I do make mistakes, I did not make that
one. in that he claims that the rationals are not allowed to>
map to the powerset of the integers. Take any bijection from
the rationals to the integers, say f:Q -> I,then g(x) = {f(x)}
maps from the rationals to the power set of the integers, g Q
-> P(I), but not surjectively. I say powerset of integers is>
not powerset of rationals and powerset of integers is not
superset> of powerset of rationals, as well as powerset of
integers is proper> subset of powerset of rationals. Am I
supposed to deny any of this? What is the case which Ross does
not mention, however, is that those powersets are of equal
cardinality.> The powerset proof has multiple> lines of attack
against it from that an infinite set is not precluded> from
mapping to the powerset of one of its proper subsets, the>
powerset argument fails to show otherwise, except that the
mapping> would invalidate it. The powerset argument only
applies to a give> infinite set and its very own powerset, or
that of a set isomorphic to> it in the way the evens are
isomorphic to the odds, in the naturals.I cannot parse that
last sentece in any way that allows it to make sense. can
anyone else? > I just say there is no immediate injection from
a set to its powerset.I diagree. given set A and its powerset
P(A), then f: x -> {x} is immediately and trivially an
injection from A to P(A),unles Ross means by immediate
injection something which he has not explained here.> I
think there are much better ways to gauge the relative sizes
or> densities of infinite sets than cardinality, for example
the value of> the asymptotic density of the subset of naturals
in the naturals, and> the various algebraic properties of the
sequences that can be said to> comprise the set.Ross has been
here before. There areother ways of sizing sets, but
Cantor's has, in the long run, been the most fruitful.> I
say there are more reals than rationals because there are
infinitely> many proper supersets of the rationals that are
proper subsets of the> reals. I say the same for N and Z and E
and N.As long as Ross gives a clear definition of what he means
by more and as long as that definition is a partial order on
the class of sets, he has something that will work, at least
for him. But if Ross wants everyone else to give up what works
for them in favorof his own ordering, it will have to be a lot
more useful than Cantor's.> I have a different context and
reason for claiming that not all the> elements of Z+ are
elements of N, that being about how Z+ is a proper> subset of
N.If, by Z+, Ross means the positive integers, then for Z+ to
be a subset of N we also need N a subset of Z, and not all
constructions of number systems require this. On the other
hand, for every such construction there is a natural mapping
from N to Z which preserves additions and multiplications
and under which, Z+ is a subset of the image of N.> There's
been some discussion on the list recently about the>
probability that a random real form a uniform distribution
over the> reals is a rational. I think that the rationals and
irrationals> alternate on the real number line, and that the
probability of a real> being a rational is one half. In order
for Ross to support this allegation of alternation, Ross needs
to give a method for finding the next number after any given
number in the reals, or at least show that such a method is
possibleI await with interest Ross' futile attempts to do
so.One half is certainly positive. The> asymptotic density of
the evens in the naturals is one half, the> probability of a>
random element from a uniform distribution over the
naturals being> an even number is that value.There is no
probabilistic uniform distribution over the naturals
possible.> Virgil, you're pretty funny. It's nice of you to
drop defense of the> diagonal argument.The diagonal argument
is quite valid, but unnecessary, since other proofs exist. In
order to invalidate a theorem, one must either (1) find a
counterexample or (2) invalidate EVERY proof.Ross has (1)
found no counterexamples R's uncountbility.Ross has (2) not
invalidated any proof of R's uncountability.Therefore, R
remains uncoountable.> Virgil, or anyone who cares, please put
forward Cantor's other> proof. If it's the one I think it is,
it applies to rationals as well> as reals, nested sequences
have rational endpoints.Try Google. They are very good at
finding things.

===

So how many of you have designed your own
numerals? Sixteen shouldcover the most used bases (including
the cancerous one based on thefirst and third prime). Surely
you don't use the terribly designedArabic ones? I have my own
set brewing myself, just wanted to checkin on all of
yours.

===

Let f(n), g(n) and h(n) be three positive functions.
Prove or disprove:min{f(n), g(n), h(n)} = Omega(f(n) + g(n) +
h(n))There's a dispute that this can be proven whereas I think
it can't be. For three positive functions, I believe it can be
graphically shown that the minimum of either of these
functions will not (and cannot) have an asymptotic lower bound
of these functions altogether. I believe the following would be
correct:min{f(n), g(n), h(n)} = O(f(n) + g(n) + h(n))which
states that the minimun of the three positive functions will
have an asymptotic upper bound of a function I(n) such
that:I(n) = f(n) + g(n) + h(n)Does anyone have a different
view?-Andre

===

>Let f(n), g(n) and h(n) be three positive
functions. Prove or disprove:>min{f(n), g(n), h(n)} =
Omega(f(n) + g(n) + h(n))Of course not. What if f(n) = g(n) =
1 and h(n) = n?>There's a dispute that this can be proven
whereas I think it can't be. >For three positive functions, I
believe it can be graphically shown that >the minimum of
either of these functions will not (and cannot) have an
>asymptotic lower bound of these functions altogether.No. It
_can be_ true that min{f(n), g(n), h(n)} =
Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) =
Theta(h(n)).> I believe the >following would be
correct:>min{f(n), g(n), h(n)} = O(f(n) + g(n) + h(n))Yes, of
course, and the constant is 1.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

>>Let f(n), g(n) and h(n)
be three positive functions. Prove or disprove:>>min{f(n),
g(n), h(n)} = Omega(f(n) + g(n) + h(n))> Of course not.
What if f(n) = g(n) = 1 and h(n) = n?Yes, then the conjecture
would be false:min(f(n), g(n), h(n)) would be = 22 not greater
than C1(2+n), where C1 is some constant.>>There's a dispute
that this can be proven whereas I think it can't be. >>For
three positive functions, I believe it can be graphically
shown that >>the minimum of either of these functions will not
(and cannot) have an >>asymptotic lower bound of these
functions altogether.> No. It _can be_ true that min{f(n),
g(n), h(n)} = Omega(f(n)+g(n)+h(n)), > namely if f(n) =
Theta(g(n)) = Theta(h(n)).It still will not be true because
the minimum of f(n) = Theta(g(n)) = Theta(h(n)) will be f(n)
as it'll always be bounded by g(n) and h(n). In that case,
f(n) will still not be larger than Omega(f(n)+g(n)+h(n)).For
example, if f(n) = g(n) = 1 and h(n) = n:1 !=
Omega(2+n)-Andre>>I believe the >>following would be correct:>min{f(n), g(n), h(n)} = O(f(n) + g(n) + h(n))> Yes, of
course, and the constant is 1.> Robert Israel
israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel > University of British
Columbia > Vancouver, BC, Canada V6T 1Z2

===

>>Let f(n),
g(n) and h(n) be three positive functions. Prove or disprove:
>min{f(n), g(n), h(n)} = Omega(f(n) + g(n) + h(n))
>There's a dispute that this can be proven whereas I think
it can't be. >For three positive functions, I believe it can
be graphically shown that >the minimum of either of these
functions will not (and cannot) have an >asymptotic lower
bound of these functions altogether.>> No. It _can be_ true
that min{f(n), g(n), h(n)} = Omega(f(n)+g(n)+h(n)), >> namely
if f(n) = Theta(g(n)) = Theta(h(n)).>It still will not be true
because the minimum of f(n) = Theta(g(n)) = >Theta(h(n)) will
be f(n) as it'll always be bounded by g(n) and h(n). In >that
case, f(n) will still not be larger than
Omega(f(n)+g(n)+h(n)).min{f(n), g(n), h(n)} could be any of
the three, or sometimes one andsometimes another. Let's
consider f(n).If f(n) = Theta(g(n)) = Theta(h(n)), there are
constants C and D such thatf(n) >= C g(n) and f(n) >= D h(n).
So f(n) + g(n) + h(n) <= (1 + 1/C + 1/D) f(n), i.e.f(n) >=
1/(1 + 1/C + 1/D)(f(n) + g(n) + h(n))which says f(n) =
Omega(f(n) + g(n) + h(n)).Similarly, g(n) and h(n) are
Omega(f(n) + g(n) + h(n)), and so ismin(f(n), g(n), h(n)).>For
example, if f(n) = g(n) = 1 and h(n) = n:Then f(n) is not
Theta(h(n)).Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2

===

> min{f(n),
g(n), h(n)} could be any of the three, or sometimes one and>
sometimes another. > Let's consider f(n).> If f(n) =
Theta(g(n)) = Theta(h(n)), there are constants C and D such
that> f(n) >= C g(n) and f(n) >= D h(n). So True, but then
the following also holds for this: f(n) <= E g(n) and f(n) <=
F g(n), where E and F are constants. Shouldn't these
inequalities be held true?-Andre> f(n) + g(n) + h(n) <= (1 +
1/C + 1/D) f(n), i.e.> f(n) >= 1/(1 + 1/C + 1/D)(f(n) + g(n) +
h(n))> which says f(n) = Omega(f(n) + g(n) + h(n)).> Similarly,
g(n) and h(n) are Omega(f(n) + g(n) + h(n)), and so is>
min(f(n), g(n), h(n)).>>For example, if f(n) = g(n) = 1
and h(n) = n:> Then f(n) is not Theta(h(n)).> Robert
Israel israel@math.ubc.ca> Department of Mathematics
http://www.math.ubc.ca/~israel > University of British
Columbia > Vancouver, BC, Canada V6T 1Z2>

===

what I
actually meant was that if what you said was the case, then
the following would also hold:f(n) <= C2 g(n), f(n) <= D2 h(n)
where C2 and D2 are constants=>f(n) + g(n) + h(n) >= (1 + 1/C2
+ 1/D2) f(n), i.e.f(n) <= 1/(1 + 1/C2 +
1/D2)(f(n)+g(n)+h(n)and this would then mean that:min(f(n),
g(n), h(n)) = Theta(f(n) + g(n) + h(n))and in my understanding
not: min(f(n), g(n), h(n)) = Omega(f(n) + g(n) + h(n))The
reason being that when you choose f(n) = Theta(g(n)) =
Theta(h(n)), you're saying that f(n) is asymptotically tightly
bounded (upper and lower) by g(n) as well as h(n). However when
we say Omega(f(n) + g(n) + h(n)), it is my understanding that
this may or may not be true for Big-Oh(f(n) + g(n) + h(n)).
Please correct me if I'm wrong... can we really prove a
conjecture like that using Theta and then only proving for the
lower bounds (i.e. Omega) and leave out the upper
bounds?-Andre>> min{f(n), g(n), h(n)} could be any of
the three, or sometimes one and>> sometimes another. Let's
consider f(n).>> If f(n) = Theta(g(n)) = Theta(h(n)), there
are constants C and D such >> that>> f(n) >= C g(n) and f(n)
>= D h(n). So > True, but then the following also holds
for this: f(n) <= E g(n) and > f(n) <= F g(n), where E and F
are constants. Shouldn't these > inequalities be held true? -Andre>> f(n) + g(n) + h(n) <= (1 + 1/C + 1/D) f(n), i.e.>>
f(n) >= 1/(1 + 1/C + 1/D)(f(n) + g(n) + h(n))>> which says f(n)
= Omega(f(n) + g(n) + h(n)).>> Similarly, g(n) and h(n) are
Omega(f(n) + g(n) + h(n)), and so is>> min(f(n), g(n),
h(n)).> For example, if f(n) = g(n) = 1 and h(n) =
n:>> Then f(n) is not Theta(h(n)).>> Robert Israel
israel@math.ubc.ca>> Department of Mathematics
http://www.math.ubc.ca/~israel >> University of British
Columbia Vancouver, BC, Canada V6T 1Z2>

===

>> min{f(n),
g(n), h(n)} could be any of the three, or sometimes one and>>
sometimes another. >> Let's consider f(n).>> If f(n) =
Theta(g(n)) = Theta(h(n)), there are constants C and D such
that>> f(n) >= C g(n) and f(n) >= D h(n). So >>True, but then
the following also holds for this: f(n) <= E g(n) and >f(n) <=
F g(n), where E and F are constants. Shouldn't these
>inequalities be held true?Indeed. f(n) = Theta(g(n)) means C
g(n) <= f(n) <= E g(n) for some constants C and E, and
sufficiently large n. So? Why does thatbother you?Robert
Israel israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

>what I actually meant was
that if what you said was the case, then the >following would
also hold:>f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2
are constants>=>>f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n),
i.e.>f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n)>and this would
then mean that:>min(f(n), g(n), h(n)) = Theta(f(n) + g(n) +
h(n))>and in my understanding not: min(f(n), g(n), h(n)) =
Omega(f(n) + g(n) + >h(n))There is no contradiction between
these. If it's Theta, then it's alsoOmega.Let's back up a bit.
You said>For three positive functions, I believe it can be
graphically shown that >the minimum of either of these
functions will not (and cannot) have an >asymptotic lower
bound of these functions altogether.To which I replied> No. It
_can be_ true that min{f(n), g(n), h(n)} =
Omega(f(n)+g(n)+h(n)), > namely if f(n) = Theta(g(n)) =
Theta(h(n)).So my statement was that in this case, where f(n)
= Theta(g(n)) and f(n) = Theta(h(n)), min(f(n),g(n),h(n)) =
Omega(f(n)+g(n)+h(n))Now, as you noticed, more than that is
true in this case:min(f(n),g(n),h(n)) =
Theta(f(n)+g(n)+h(n))But that's fine: saying it's Theta(...)
means that it's both O(...) and Omega(...).Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

g(n), and h(n) can be any
positive functions.. by that definition f(n) = Theta(g(n)) =
Theta (h(n)) proves the conjecture true :)cheers,-Andre>what I actually meant was that if what you said was the
case, then the >>following would also hold:>>f(n) <= C2
g(n), f(n) <= D2 h(n) where C2 and D2 are constants>=>f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e.>>f(n)
<= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n)>>and this would then
mean that:>>min(f(n), g(n), h(n)) = Theta(f(n) + g(n) +
h(n))>>and in my understanding not: min(f(n), g(n), h(n))
= Omega(f(n) + g(n) + >>h(n))> There is no contradiction
between these. If it's Theta, then it's also> Omega.> Let's
back up a bit. You said>>For three positive functions, I
believe it can be graphically shown that >>the minimum of
either of these functions will not (and cannot) have an
>>asymptotic lower bound of these functions altogether.>
To which I replied>>No. It _can be_ true that min{f(n),
g(n), h(n)} = Omega(f(n)+g(n)+h(n)), >>namely if f(n) =
Theta(g(n)) = Theta(h(n)).> So my statement was that in
this case, where f(n) = Theta(g(n)) and > f(n) = Theta(h(n)),
> min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n))> Now, as you
noticed, more than that is true in this case:>
min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n))> But that's
fine: saying it's Theta(...) means that it's both O(...) and >
Omega(...).> Robert Israel israel@math.ubc.ca> Department of
Mathematics http://www.math.ubc.ca/~israel > University of
British Columbia > Vancouver, BC, Canada V6T 1Z2>

===

A
quickie:Prove or disprove f(n) = O(g(n)) implies
g(n)=Theta(f(n))cheers,Anderson> g(n), and h(n) can be any
positive functions.. by that definition f(n) = > Theta(g(n)) =
Theta (h(n)) proves the conjecture true :)> cheers,>
-Andre> what I actually meant was that if what you
said was the case, then > the following would also
hold:> f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and
D2 are constants> =>> f(n) + g(n) + h(n) >= (1 +
1/C2 + 1/D2) f(n), i.e.> f(n) <= 1/(1 + 1/C2 +
1/D2)(f(n)+g(n)+h(n)> and this would then mean
that:> min(f(n), g(n), h(n)) = Theta(f(n) + g(n) +
h(n))> and in my understanding not: min(f(n), g(n),
h(n)) = Omega(f(n) + > g(n) + h(n))>> There is no
contradiction between these. If it's Theta, then it's also>>
Omega.>> Let's back up a bit. You said> For three
positive functions, I believe it can be graphically shown >
that the minimum of either of these functions will not (and
cannot) > have an asymptotic lower bound of these functions
altogether.>> To which I replied> No. It _can be_
true that min{f(n), g(n), h(n)} = > Omega(f(n)+g(n)+h(n)),
namely if f(n) = Theta(g(n)) = Theta(h(n)).>> So my
statement was that in this case, where f(n) = Theta(g(n)) and
>> f(n) = Theta(h(n)), min(f(n),g(n),h(n)) =
Omega(f(n)+g(n)+h(n))>> Now, as you noticed, more than that
is true in this case:>> min(f(n),g(n),h(n)) =
Theta(f(n)+g(n)+h(n))>> But that's fine: saying it's
Theta(...) means that it's both O(...) >> and Omega(...).>>
Robert Israel israel@math.ubc.ca>> Department of Mathematics
http://www.math.ubc.ca/~israel >> University of British
Columbia Vancouver, BC, Canada V6T 1Z2>

===

> A quickie:>
Prove or disprove f(n) = O(g(n)) implies g(n)=Theta(f(n))This
can be clearly disproved:suppose f(n) = 1, g(n) = nthen 1 =
O(g(n)) can be true but not the other way round, i.e.:n !=
Theta(1) for any large value of n. This was quite obvious,
don't you think?-Andre> cheers,> Anderson>> f(n),
g(n), and h(n) can be any positive functions.. by that >>
definition f(n) = Theta(g(n)) = Theta (h(n)) proves the
conjecture >> true :)>> cheers,>> -Andre> what I
actually meant was that if what you said was the case, then
>> the following would also hold:>> f(n) <= C2
g(n), f(n) <= D2 h(n) where C2 and D2 are
constants>> => f(n) + g(n) + h(n) >= (1 +
1/C2 + 1/D2) f(n), i.e.>> f(n) <= 1/(1 + 1/C2 +
1/D2)(f(n)+g(n)+h(n)>> and this would then mean
that:>> min(f(n), g(n), h(n)) = Theta(f(n) +
g(n) + h(n))>> and in my understanding not:
min(f(n), g(n), h(n)) = Omega(f(n) + >> g(n) +
h(n))> There is no contradiction between these.
If it's Theta, then it's also> Omega.>> Let's back up a
bit. You said>> For three positive functions, I
believe it can be graphically shown >> that the minimum of
either of these functions will not (and cannot) >> have an
asymptotic lower bound of these functions
altogether.> To which I replied>> No. It
_can be_ true that min{f(n), g(n), h(n)} = >>
Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) =
Theta(h(n)).> So my statement was that in this
case, where f(n) = Theta(g(n)) and > f(n) = Theta(h(n)),
min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n))>> Now, as you
noticed, more than that is true in this case:>>
min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n))>> But that's
fine: saying it's Theta(...) means that it's both O(...) >
and Omega(...).>> Robert Israel israel@math.ubc.ca>
Department of Mathematics http://www.math.ubc.ca/~israel >
University of British Columbia Vancouver, BC, Canada V6T
1Z2>>

===

>> <deleted>> Consider, in the ring of
algebraic integers,>> P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 ->> 3(-1+mf^2 )x u^2 + u^3 f).>
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization>> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3)>> where w_1 w_2 w_3 = f, and> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),>> and at
m=0>> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),> so two of the b's must equal 0, which means>>
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)>> which is> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)> proving that w_1 w_2 must equal 1, if f is coprime to
3, which leaves> b_3 = 3.> Strictly speaking
you've proven that> w_1 w_2 b_3 = 3 and> w_1 w_2 w_3 = f>>
I already had that w_1 w_2 w_3 = f, and given that f is coprime
to 3,>> b_1 + b_2 + b_3 = 3, and knowing that b_1 = b_2 = 0,
b_3 = 3Where did this come from?When I do the factorization I
get:w_2w_3b_1 + w_1w_3b_2 + w_1w_2b_3 = 3And you know that 2
of the terms are zero. So say b_1=b_2=0So w_1w_2b_3 = 3Show
how you get your factorization b_1 + b_2 + b_3 ?Step by step,
not making prior assumptions about the w's>> which means w_1
w_2 = 1.>> Remember you have the coefficients from>> P(m) =
f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 ->> 3(-1+mf^2 )x u^2 + u^3
f)>> to use with> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3).> Essentially objections to how f^2
divides off now come down to> claiming that the w's are
functions of m, but consider that w_1 w_2 => 1, when m=0,
if f is coprime to 3.>> But that was an arbitrary
choice, so let f=3.>> Now w_1 w_2 = 3^{2/3} WITHOUT
REGARD TO m.>> Strictly speaking w_1 w_2 w_3 = 3 in this
case.>> Yup. And that is without regard to m, right? Or do you
disagree?Agree. Without regard to m.>> You've just chosen to
make different substitutions for w_1 w_2 in the two>
separate examples. Which is fine, it neither proves nor
disproves anything.>I'll stand by the above statement until
you can show your factorization.> Nope. All I did was switch
from f being coprime to 3 to f=3.>> That shows that the w's
are constant with regard to m.>> Remember, w_1 w_2 = 1, at
m=0, iff f is coprime to 3, but w_1 w_2 => 3^{2/3}, when f=3,
at m=0, and every other m, as they don't care about> m, so
they can't be functions of m. It's just not possible.>Does
anyone think they are functions of m?>> That is, the
w's are now all constant with regard to m and have the>
same value no matter what the value of m is.>>
Therefore, the factorization is>> P(m)/f^2 = (m^3 f^4
- 3m^2 f^2 + 3m) x^3 ->> 3(-1+mf^2 )x u^2 + u^3 f => (b_1 x + u)(b_2 x + u)(b_3 x + uf)>> where
you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are
not, which> shows a problem with the ring of algebraic
integers.>> I didn't ever suggest that the w's varied. I
was hoping to see you prove how> you could verify that a_3
was coprime to f at m== 0. Maybe next time?>> Well, if the
w's don't vary, then I can just check, as I do, with f>
coprime to 3, which gives me the factorization>> P(m)/f^2 =
(m^3 f^4 - 3m^2 f^2 + 3m) x^3 ->> 3(-1+mf^2 )x u^2 + u^3 f =>>
(b_1 x + u)(b_2 x + u)(b_3 x + uf)>> as the w's don't vary with
m, so I can just calculate them, right?>> Now then, if you
disagree with the values for the w's that I use> there, then
you must still be holding on to the notion that they are>
dependent on m.Nope. You can plug in any 2 values you want
into two of the w's and thencalculate the 3rd. Independent of
m.>> Oh, I guess that's a 3rd problem I have then. I
disagree with your assertion> that there's a problem with
the ring of algebraic integers. Perhaps when wesee> that
proof that a_3 must be coprime to f, we can consider this
assertion.>> Well a_3 = b_3, and you can just look at the
factorization above, with> the information that P(0)/f^2 is
coprime to f, to see that b_3 must> be.>> Now if you're still
trying to doubt that result, then you're still> holding on to
the notion of a variable dependency from the w's on m.>> You
probably gave the answer to why you're not following the math
by> showing you believe there's no problem with the ring of
algebraic> integers.Perhaps you just need to show me the math
of how you get tob_1 + b_2 + b_3> I've found the
Ring of Objects which includes the ring of algebraic>
integers, and does not have this problem, as the b's are all
included> in it.>> The Ring of Objects is the set
of all numbers where 1 is the only> member that is both a
unit, i.e. factor of 1, and an integer, where no>
non-unit member is a factor of any two integers that are
coprime.>> As I noted yesterday, there's a mistake with that
definition as I> forgot about -1, so it should be -1 and 1 are
the only members that> are bot ha unit and an integer. And no,
I didn't catch the mistake> myself as a poster pointed it out
yesterday.>> That definition and more is linked to from my
primary website>> http://groups.msn.com/AmateurMath> where you can also find information on my other math
research.>> Yup.> James HarrisPhil Nicholson

===

And now I
have to make a correction but luckily, it's not a big deal,but
maybe it'll clear out any remaining confusion. More below... >
<deleted>> Well, the argument which settles things is
*luckily* short and rather> direct, so I'll give it here.
Some may think it's exactly what> they've seen before, as
I've been posting it a lot of places, but I've> seen need
to put in minor corrections.>> Consider, in the ring of
algebraic integers,>> P(m) = f^2((m^3 f^4 - 3m^2 f^2 +
3m) x^3 ->> 3(-1+mf^2 )x u^2 + u^3 f).>
Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the
factorization>> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u
w_2)(b_3 x + u w_3)>> where w_1 w_2 w_3 = f, and> b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),>> and at
m=0>> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),> so two of the b's must equal 0, which means>>
P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)>> which is> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)> proving that w_1 w_2 must equal 1, if f is coprime to
3, which leaves> b_3 = 3.>> Strictly speaking
you've proven that> w_1 w_2 b_3 = 3 and> w_1 w_2 w_3 = f I already had that w_1 w_2 w_3 = f, and given that f is
coprime to 3,> b_1 + b_2 + b_3 = 3, and knowing that b_1 =
b_2 = 0, b_3 = 3> which means w_1 w_2 = 1.Yup, and I should
have remembered that further down. > Remember you have the
coefficients from> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3
-> 3(-1+mf^2 )x u^2 + u^3 f)> to use with> P(m)/f^2 =
(b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3).>
Essentially objections to how f^2 divides off now come down
to> claiming that the w's are functions of m, but consider
that w_1 w_2 => 1, when m=0, if f is coprime to 3.> But that was an arbitrary choice, so let f=3.>> Now
w_1 w_2 = 3^{2/3} WITHOUT REGARD TO m.> Strictly
speaking w_1 w_2 w_3 = 3 in this case.> Yup. And that is
without regard to m, right? Or do you disagree?Well I messed
up, as w_1 w_2 is coprime to 3, and does NOT equal3^{2/3},
which I should have known from P(0)/f^2 = u^2 (b_3 w_1 w_2 x +
u f) = u^2(3x + uf)as all that happens is that when f=3, that f
can now divide out, andit has to go through b_3. > You've
just chosen to make different substitutions for w_1 w_2 in the
two> separate examples. Which is fine, it neither proves nor
disproves anything.> Nope. All I did was switch from f being
coprime to 3 to f=3.> That shows that the w's are constant
with regard to m.> Remember, w_1 w_2 = 1, at m=0, iff f is
coprime to 3, but w_1 w_2 => 3^{2/3}, when f=3, at m=0, and
every other m, as they don't care about> m, so they can't be
functions of m. It's just not possible.Well I got the numbers
wrong but they are constant with regard to m.The easiest way
is just to look though at P(m)/3^3 = (m^3 3^3 - 3m^2(3) + m)
x^3 - (-1+mf^2 )x u^2 + u^3 = (b_1 x + u)(b_2 x + u)(b_3/3 x +
u).You might think that maybe the 3 goes some other way, but
look at P(m)/3^3 = (m^3 3^3 - 3m^2(3) + m) x^3 - (-1+mf^2 )x
u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)as you can
see that because 3 divides off from *each* factor, there'sonly
one way it can go, and it must be constant with regard to
m.It's interesting how it can *seem* like maybe f can go a lot
of wayswhen f doesn't equal 3, if you wish to try and force an
m dependency.> That is, the w's are now all constant with
regard to m and have the> same value no matter what the
value of m is.>> Therefore, the factorization is> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 ->>
3(-1+mf^2 )x u^2 + u^3 f =>> (b_1 x + u)(b_2 x +
u)(b_3 x + uf)>> where you'll notice that the b's are
algebraic integers with m=1,> f=sqrt(2), but that's a
special case as generally they are not, which> shows a
problem with the ring of algebraic integers.> I didn't
ever suggest that the w's varied. I was hoping to see you
prove how> you could verify that a_3 was coprime to f at
m== 0. Maybe next time?> Well, if the w's don't vary, then
I can just check, as I do, with f> coprime to 3, which gives me
the factorization> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3
-> 3(-1+mf^2 )x u^2 + u^3 f => (b_1 x + u)(b_2 x + u)(b_3
x + uf)> as the w's don't vary with m, so I can just
calculate them, right?Yup.> Now then, if you disagree with the
values for the w's that I use> there, then you must still be
holding on to the notion that they are> dependent on m.That's
going back to the poster Phil Nicholson who has been
soadamnant that m=0 is a special case, but that's really just
wishfulthinking on his part, as the math says otherwise.Here's
Phil's comments that follow.> Oh, I guess that's a 3rd
problem I have then. I disagree with your assertion> that
there's a problem with the ring of algebraic integers. Perhaps
when we see> that proof that a_3 must be coprime to f, we
can consider this assertion.> Well a_3 = b_3, and you can
just look at the factorization above, with> the information
that P(0)/f^2 is coprime to f, to see that b_3 must> be.Yup.>
Now if you're still trying to doubt that result, then you're
still> holding on to the notion of a variable dependency from
the w's on m.> You probably gave the answer to why you're
not following the math by> showing you believe there's no
problem with the ring of algebraic> integers.>> I've
found the Ring of Objects which includes the ring of algebraic integers, and does not have this problem, as the b's are
all included> in it.>> The Ring of Objects is the
set of all numbers where 1 is the only> member that is
both a unit, i.e. factor of 1, and an integer, where no>
non-unit member is a factor of any two integers that are
coprime.> As I noted yesterday, there's a mistake with that
definition as I> forgot about -1, so it should be -1 and 1 are
the only members that> are bot ha unit and an integer. And no,
I didn't catch the mistake> myself as a poster pointed it out
yesterday.> That definition and more is linked to from
my primary website>>
http://groups.msn.com/AmateurMath>> where you can also
find information on my other math research.> Yup.Cool
beans.James Harris

===

>> And now I have to make a correction
but luckily, it's not a big deal,> but maybe it'll clear out
any remaining confusion. More below...Let's see...>>
<deleted> Well, the argument which settles things is
*luckily* short and rather> direct, so I'll give it
here. Some may think it's exactly what> they've seen
before, as I've been posting it a lot of places, but I've seen need to put in minor corrections.>>
Consider, in the ring of algebraic integers,>>
P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 ->>
3(-1+mf^2 )x u^2 + u^3 f).> Now using b_1,
b_2, b_3, w_1, w_2, and w_3, I have the factorization> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u
w_3)>> where w_1 w_2 w_3 = f, and>>
b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),>> and at
m=0>> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf),> so two of the b's must equal 0, which means> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)>>
which is>> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) =
u^2(3x + uf)>> proving that w_1 w_2 must equal 1,
if f is coprime to 3, which leaves> b_3 = 3.>> Strictly speaking you've proven that> w_1 w_2 b_3 =
3 and> w_1 w_2 w_3 = f>> I already had that w_1 w_2 w_3
= f, and given that f is coprime to 3,>> b_1 + b_2 + b_3 =
3, and knowing that b_1 = b_2 = 0, b_3 = 3The correct
factorization isb_1w_2w_3 + b_2w_1w_3 + b_3 w_1 w_2 = 3Yours
is correct if w_1=w_2=w_3 = 1But w_1w_2w_3 = f.Hence a
contradiction.>> which means w_1 w_2 = 1.>> Yup, and I
should have remembered that further down.>> Remember you
have the coefficients from>> P(m) = f^2((m^3 f^4 - 3m^2
f^2 + 3m) x^3 ->> 3(-1+mf^2 )x u^2 + u^3 f)>> to use
with> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x
+ u w_3).> Essentially objections to how f^2
divides off now come down to> claiming that the w's are
functions of m, but consider that w_1 w_2 => 1, when
m=0, if f is coprime to 3.>> But that was an
arbitrary choice, so let f=3.>> Now w_1 w_2 =
3^{2/3} WITHOUT REGARD TO m.>> Strictly speaking w_1
w_2 w_3 = 3 in this case.>> Yup. And that is without
regard to m, right? Or do you disagree?>> Well I messed up, as
w_1 w_2 is coprime to 3, and does NOT equal> 3^{2/3}, which I
should have known from>> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f)
= u^2(3x + uf)>> as all that happens is that when f=3, that f
can now divide out, and> it has to go through b_3.>
You've just chosen to make different substitutions for w_1 w_2
in the two> separate examples. Which is fine, it neither
proves nor disprovesanything.>> Nope. All I did was switch
from f being coprime to 3 to f=3.>> That shows that the w's
are constant with regard to m.>> Remember, w_1 w_2 = 1, at
m=0, iff f is coprime to 3, but w_1 w_2 => 3^{2/3}, when
f=3, at m=0, and every other m, as they don't care about> m,
so they can't be functions of m. It's just not possible.>> Well
I got the numbers wrong but they are constant with regard to
m.>> The easiest way is just to look though at>> P(m)/3^3 =
(m^3 3^3 - 3m^2(3) + m) x^3 ->> (-1+mf^2 )x u^2 + u^3 =>> (b_1
x + u)(b_2 x + u)(b_3/3 x + u).> You might think that maybe
the 3 goes some other way, but look at>> P(m)/3^3 = (m^3 3^3 -
3m^2(3) + m) x^3 ->> (-1+mf^2 )x u^2 + u^3 =>> (a_1/3 x +
u)(a_2/3 x + u)(a_3/3 x + u)>> as you can see that because 3
divides off from *each* factor, there's> only one way it can
go, and it must be constant with regard to m.>> It's
interesting how it can *seem* like maybe f can go a lot of
ways> when f doesn't equal 3, if you wish to try and force an
m dependency.>> That is, the w's are now all constant
with regard to m and have the> same value no matter what
the value of m is.>> Therefore, the factorization
is>> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 -> 3(-1+mf^2 )x u^2 + u^3 f =>> (b_1 x +
u)(b_2 x + u)(b_3 x + uf)>> where you'll notice
that the b's are algebraic integers with m=1,>
f=sqrt(2), but that's a special case as generally they are
not, which> shows a problem with the ring of algebraic
integers.>> I didn't ever suggest that the w's varied.
I was hoping to see you provehow> you could verify that
a_3 was coprime to f at m== 0. Maybe next time?>> Well,
if the w's don't vary, then I can just check, as I do, with f coprime to 3, which gives me the factorization>>
P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 ->> 3(-1+mf^2 )x
u^2 + u^3 f =>> (b_1 x + u)(b_2 x + u)(b_3 x + uf)>>
as the w's don't vary with m, so I can just calculate them,
right?>> Yup.>> Now then, if you disagree with the values
for the w's that I use> there, then you must still be
holding on to the notion that they are> dependent on m.>>
That's going back to the poster Phil Nicholson who has been
so> adamnant that m=0 is a special case, but that's really
just wishful> thinking on his part, as the math says
otherwise.>> Here's Phil's comments that follow.>> Oh, I
guess that's a 3rd problem I have then. I disagree with
yourassertion> that there's a problem with the ring of
algebraic integers. Perhaps whenwe see> that proof that
a_3 must be coprime to f, we can consider this assertion.> Well a_3 = b_3, and you can just look at the factorization
above, with> the information that P(0)/f^2 is coprime to f,
to see that b_3 must> be.>> Yup.Yup.It's well proven, many
times that a_3 = 3 when m = 0. So a_3 is coprime to fwhen f
is coprime to 3, when m = 0.You just need to extend this proof
to show it's behaviour when m<>0.You have further proven I
think that a_3 is not coprime to f, when f=3 for anym.So
you've only got that last bit to go. Prove that a_3 is (or is
not) coprime tof when f coprime to 3, and m<>0.Phil
Nicholson.>> Now if you're still trying to doubt that
result, then you're still> holding on to the notion of a
variable dependency from the w's on m.>> You probably gave
the answer to why you're not following the math by> showing
you believe there's no problem with the ring of algebraic>
integers.> I've found the Ring of Objects
which includes the ring of algebraic> integers, and does
not have this problem, as the b's are all included> in
it.>> The Ring of Objects is the set of all
numbers where 1 is the only> member that is both a unit,
i.e. factor of 1, and an integer, where no> non-unit
member is a factor of any two integers that are coprime.>>
As I noted yesterday, there's a mistake with that definition as
I> forgot about -1, so it should be -1 and 1 are the only
members that> are bot ha unit and an integer. And no, I
didn't catch the mistake> myself as a poster pointed it out
yesterday.>> That definition and more is linked to
from my primary website>>
http://groups.msn.com/AmateurMath>> where you can
also find information on my other math research.>> Yup.>>
Cool beans.> James Harris

===

>> And now I have to make a
correction but luckily, it's not a big deal,> but maybe it'll
clear out any remaining confusion. More below...> Let's see...>
<deleted> so two of the b's must equal 0,
which means>> P(0)/f^2 = w_1 w_2 u^2 (b_3 x +
u w_3)>> which is>> P(0)/f^2 =
u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)>>
proving that w_1 w_2 must equal 1, if f is coprime to 3, which
leaves> b_3 = 3.> Strictly
speaking you've proven that> w_1 w_2 b_3 = 3 and>
w_1 w_2 w_3 = f>> I already had that w_1 w_2 w_3 = f,
and given that f is coprime to 3,>> b_1 + b_2 + b_3 =
3, and knowing that b_1 = b_2 = 0, b_3 = 3> The correct
factorization is> b_1w_2w_3 + b_2w_1w_3 + b_3 w_1 w_2 = 3>
Yours is correct if w_1=w_2=w_3 = 1> But w_1w_2w_3 = f.> Hence
a contradiction.That is correct. What I gave was wrong. The
correct one gives b_3 w_1 w_2 = 3, at m=0.<deleted>
Well a_3 = b_3, and you can just look at the factorization
above, with> the information that P(0)/f^2 is coprime to
f, to see that b_3 must> be.>> Yup.> Yup.> It's well
proven, many times that a_3 = 3 when m = 0. So a_3 is coprime
to f> when f is coprime to 3, when m = 0.> You just need to
extend this proof to show it's behaviour when m<>0.> You have
further proven I think that a_3 is not coprime to f, when f=3
for any> m.> So you've only got that last bit to go. Prove
that a_3 is (or is not) coprime to> f when f coprime to 3,
and m<>0.> Phil Nicholson.Ok, so you accept that f^2 divides
off as a *constant* with respect tom, when f=3. However, you're
now apparently still considering thepossibility that it divides
off as a function of m, if f does notequal 3.Is that
correct?And now it seems you think I need to prove that there
isn't thisreally weird math function that can switch from not
being a functionof m to being a function of m dependent on
whether or not it iscoprime to 3.Is that a correct
assessment?James Harris

===

> Ok, so you accept that f^2
divides off as a *constant* with respect to> m, when f=3.
However, you're now apparently still considering the>
possibility that it divides off as a function of m, if f does
not> equal 3.>> Is that correct?>> And now it seems you think
I need to prove that there isn't this> really weird math
function that can switch from not being a function> of m to
being a function of m dependent on whether or not it is>
coprime to 3.>> Is that a correct assessment?No not at all a
correct assessment.Let v = -1 + mf^2The a's you refer to are
the 3 solutions to.a^3 + 3va^2 - (v^3 + 1) = 0Correct?The
second term is always a multiple of 3, regardless of m.When
f=3 the third term is a multiple of 3, regardless of
m.Therefore for the equation to exist, a is also a multiple of
3 when f=3.When m=0, we have v=-1a^3 -3a^2 = 0 so one of the
a's is equal to 3, independent of f.You claim that one of the
solutions to the equation, call it a_3, is coprime toffor all
other values of m and f. (At least when m and f are relatively
coprime)If you have a proof for this, you're invited to
present it here.The cubic is not that weird. Oh and a hint,
you can prove a_3 is coprime to fif a_1/f and a_2/f are
algebraic integers. So your proof will need to addressthe
situation when a_2/f and a_1/f are not algebraic
integers.Until you do that your work has a clear gap in it. So
your conclusions have novalue until you fill this gap.>
James Harris

===

> Ok, so you accept that f^2 divides off as
a *constant* with respect to> m, when f=3. However, you're
now apparently still considering the> possibility that it
divides off as a function of m, if f does not> equal 3.> Is that correct?>> And now it seems you think I need to
prove that there isn't this> really weird math function that
can switch from not being a function> of m to being a
function of m dependent on whether or not it is> coprime to
3.>> Is that a correct assessment?> No not at all a
correct assessment.Actually it has to be, which I'll show in a
bit. > Let v = -1 + mf^2> The a's you refer to are the 3
solutions to.> a^3 + 3va^2 - (v^3 + 1) = 0> Correct?Yup.>
The second term is always a multiple of 3, regardless of m.>
When f=3 the third term is a multiple of 3, regardless of m.>
Therefore for the equation to exist, a is also a multiple of 3
when f=3.Yup.> When m=0, we have v=-1> a^3 -3a^2 = 0 so one of
the a's is equal to 3, independent of f.Yup.> You claim that
one of the solutions to the equation, call it a_3, is coprime
to> f> for all other values of m and f. (At least when m and f
are relatively coprime)> If you have a proof for this,
you're invited to present it here.What you didn't show is the
full system which defines the a's. Theexpressions that show
that system are P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 -
3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x +
uf).And as you've pointed out, > a^3 + 3va^2 - (v^3 + 1) = 0is
also true as an equation within the system.You *have* to
consider the entire system, rather than try to pullpieces out
of it, and claim a single piece is the entire thing.Now then,
P(m) is a consideration of what I call the uber-polynomial
f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3
f)where the focus is on it as a polynomial with respect to
m.That's just one of many possible looks.Now the expression is
complicated enough that it pays to use m=0 toget a look at the
constant term with respect to m, so I set m=0, andget P(0) =
u^2 f^2(3x + uf)which tells me that with the factorization
P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)the factor f^2
divides off through only *two* of the a's, which is afact that
no one can dispute.What you now need though to try and disagree
with the conclusion thatthey divide off that way for all m is
the belief that they divide offdifferently *dependent* on
m.However, looking at the constant term of the polynomial
P(m), rememberthe uber-polynomial is considered to be a
polynomial with respect tom, which is just one of many looks,
does NOT constrain m, which is animportant point.Consider P(x)
= x^2 + 2x + 1, and P(0) = 1. Does that constrain x?Why would
you think that my use of m=0, with P(m), would actually puta
constraint on the system? I'm merely using m=0 to pull out
theconstant term with respect to m.Why do I want to look at
it?Because now I can check P(m)/f^2, by again looking at *its*
constantterm.When I do so, I find that the constant term is
P(0)/f^2 = u^2(3x + uf)so I know that f^2 divided off from it,
leaving it now coprime to f,as long as x and 3 are coprime to
f.But P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)and the
*only* way to get P(0)/f^2 is for f to divide out through
onlytwo of the a's, so that I have P(m)/f^2 = (a_1/f x +
u)(a_2/f x + u)(a_3 x + uf)as from before I know that two of
the a's equal 0, when m=0, so I cansee that the constant term
P(0)/f^2 = u^2(3x + uf) corresponds to
thatfactorization.Dispute with that fact require the belief
that f^2 divides off with adependency on m.Readers might find
that hard to understand, but remember the*constant* term of
any polynomial is that part which doesn't have thekey
variable.So the constant term for P(x) = x^2 + 2x + 1, doesn't
have x in it.Since I can check the constant term for P(m), by
setting m=0, forposters to dispute with the conclusion
gathered, they must be claimingthat f^2 divides off in some
way with the variable m in the factorsof f^2 that divide
off.Remember my use of m=0 at P(m) gives me the *constant*
term, just likewith P(x) = x^2 + 2x + 1, x=0, gives me P(0) =
1, which is distinctivein that it does not have a dependency
on x.So the refusal by Phil Nicholson to admit that he's
asserting adependency on m, is clearly against mathematical
logic.But consider this exchange from earlier in the post,
with my commentsstarting:> And now it seems you think I need
to prove that there isn't this> really weird math function
that can switch from not being a function> of m to being a
function of m dependent on whether or not it is> coprime to
3.>> Is that a correct assessment?> No not at all a
correct assessment.Here the poster is fighting the
*mathematical* requirements of his ownassertions!!!> The cubic
is not that weird. Oh and a hint, you can prove a_3 is coprime
to f> if a_1/f and a_2/f are algebraic integers. So your
proof will need to address> the situation when a_2/f and
a_1/f are not algebraic integers.You continue to try and
reduce the expressions P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m)
x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3
x + uf)to a simpler system.Here you've tried to reduce
everything down to the expression> a^3 + 3va^2 - (v^3 + 1) =
0presumably because it's something that you understand
readily.However, the system cannot be so reduced.> Until you
do that your work has a clear gap in it. So your conclusions
have no> value until you fill this gap.Repeatedly you make
claims without being able to base them onmathematical
logic.Repeatedly you try to ignore equations or claim they're
more limitedthan they are.Why won't you at least admit the m
dependency your position wouldrequire on how f^2 divides
off?Repeatedly you've claimed you're not asserting such a
dependency,which goes against the math.James Harris

===

>Lots deleted this time.>I wouldn't mind
doing a sanity check, then restate my problem if you
don't>> mind.>Fine.>>We have>>P(m) = f^2((m^3
f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)= m^3 f^6
x^3 - 3m^2 f^4 x^3 + 3mf^2 x^3 +3xu^2 f^2 -3mf^4 xu^2 + u^3
f^3This is a polynomial with four variables. The coefficients
are 1, -3, 3, 3, -3, 1. If you wish to view any of m,f,x,u as
a number, then it changes the terms of the polynomial and the
coefficients. It is *still* a polynomial. Please stop talking
about uber-polynomials as they simply don't exist.> What the
math sees is an uber-polynomial, not a polynomial, as a>
polynomial has constant coefficients, while with the
uber-polynomial> you have all those variables.Perhaps I
shouldn't respond to this, but it's just one example of how
James doesn't understand the terminology, or the importance of
clear definitions. As defined, P(m) is a function from
polynomials to polynomials, but that's unimportant. What's
been bothering me for a while is that James doesn't appear to
know what a polynomial *is*.-- Will Twentyman

===

>Your
assumption that the field of algebraic numbers is required,
which>you've repeated several times, but have been unable to
prove.>As long as we're mindful that the b's are not
always algebraic integers.> I have said so myself, and in
fact that is why there's a problem with> the ring of algebraic
integers.> It may seem esoteric to readers on sci.physics and
sci.skeptic, but> mathematics requires zero errors, and what
I've managed to show with> some fascinatingly basic algebra is
an error created by that> definition of algebraic integers as
roots of monic polynomials with> integer coefficients.> That
definition leaves gaps by not including certain numbers that>
should be included which leads to fascinating contradiction
like that> ***in the ring of algebraic integers*** you can
have abc = 5, where a,> b and c are coprime to 5.Can you
provide even *one* example of such an a, b, c? Not something
abstract, but actual numbers?> That coprime just means
they don't share non-unit factors, i.e. not> factors of 1,
with 5, but they multiply together to give 5, and a, b> and c
are each algebraic integers.> Mathematicians missed this
little thing for over a hundred years, but> I can prove
there's a problem in the ring with a short argument using>
basic algebra, which comes at the end of this post.> -- Will
Twentyman

===

> Barbier's theorem is that all curves of constant
width of width w > have the same perimeter, pi * w.>
http://www.cut-the-knot.org/ctk/Barbier.shtml> gives a proof
without calculus. Is there a simple proof that uses calculus? http://mathworld.wolfram.com/CurveofConstantWidth.htmlAs the
width rotates around through 180 degrees = pi radians, the
locus of the two ends of the width is exactly the curve of
constant width. So consider what amount of perimeter is
traced out by the two ends as the width rotates through an
infinitesimal angle d theta. Consider the two widths, one at
angle theta, and the other at angle theta + d theta. These two
widths meet at some point along the widths, lets say the point
of intersection divides one of the widths into pieces of
length a and w-a. Not the piece of curve traced as the width
goes from theta to theta + d theta is perpendicular, the
lengths of these pieces of curve is a a d theta and (w-a) d
theta respectively. So adding all these pieces together for
all theta, we getperimeter = int_0^pi a d theta + int_0^pi
(w-a) a thetawhich is pi*w.-- Stephen
Montgomery-Smithstephen@math.missouri.eduhttp://
www.math.missouri.edu/~stephen

===

let p(x)=x^2 + ax + b in
R[x]if factor ring R[x]/(p(x)) is field ,find basis of field
R[x]/(p(x)) on Ri think
that basis is {1,x}but in solution paper,deg p(x) =2a exist
such that p(a)=0thus {1,a} is
basis-what do you think about it
?

===

>let p(x)=x^2 + ax + b in R[x]>>if factor ring R[x]/(p(x))
is field ,>>find basis of field R[x]/(p(x)) on
R>>i think that basis is
{1,x}>>but in solution paper,>>deg p(x) =2>>a exist such that
p(a)=0>>thus {1,a} is
basis>>->what do you think about
it ?_If_ basis of field R[x]/(p(x)) on R meansa basis for
R[x]/(p(x)), regarded as a vectorspace over R then I think
you're right andthe solution paper is obviously wrong
because{1, a} is not independent.There are at least three
possibilities:(i) I'm wrong(ii) the solution paper is
wrong(iii) my assumption that basis of field R[x]/(p(x)) on
R means a basis for R[x]/(p(x)), regarded as a vectorspace
over R is wrong, the phrase actually meanssomething else
here.************************David C. Ullrich

===

Let me revise
that a little bit:>let p(x)=x^2 + ax + b in R[x]First we
should note that there are too many a'sin the notation -
let's change p to p(x)=x^2 + Ax + B.>>if factor ring
R[x]/(p(x)) is field ,>>find basis of field R[x]/(p(x)) on
R>>i think that
basis is {1,x}>>but in solution paper,>>deg p(x) =2>>a
exist such that p(a)=0>>thus {1,a} is
basis>>->>what do you think
about it ?>>_If_ basis of field R[x]/(p(x)) on R means>a
basis for R[x]/(p(x)), regarded as a vector>space over R then
I think you're right and>the solution paper is obviously wrong
because>{1, a} is not independent.>>There are at least three
possibilities:>>(i) I'm wrong>>(ii) the solution paper is
wrong>>(iii) my assumption that basis of field R[x]/(p(x)) on
R >means a basis for R[x]/(p(x)), regarded as a vector>space
over R is wrong, the phrase actually means>something else
here.It's true that {1, x} is a basis (or rather {1 + <p>, x +
<p>}is a basis, where <p> is the ideal generated by p.)But when
I read the statement a exist such that p(a)=0thus {1,a} is
basis I was assuming that a was an elementof R, which is
obviously impossible. I suspect you didn'ttell us exactly what
the solution paper says: If a is anelement of some _extension_
of R and p(a) = 0 then{1, a} is a basis as
well.>************************>>David C.
Ullrich************************David C.
Ullrich

===

oh...sorry.solution paper
isdeg p(x) =2for a sucht that
p(a)=0{1,a} is R[x]/(p(x)).what
do you think about it??sorry, i have infirm
english~

===

oh...sorry.solution paper
isdeg p(x) =2for a sucht that
p(a)=0{1,a} is basis of
R[x]/(p(x)).what do you think
about it??sorry, i have infirm english~

===

> let p(x)=x^2 + ax
+ b in R[x]> but in solution paper,> deg p(x) =2> a
exist such that p(a)=0> thus {1,a} is basisI think the
author is using the letter a in two differentsenses (as a
coefficient and as a zero of the polynomial p).This is very
dodgy practice.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen

===

sorry, i am foolp(x) = x^2 + Ax +B
===
>> Can anyone provide me with the citation of Cantor's
paper where he>> 1/2>> 1/3, 2/3>> 1/4, 2/4,
3/4>> 1/5, 2/5, 3/5, 4/5>> 1/6, 2/6 ...>> ...>So the number of rationals between 0 and 1 is the square of
the number>of integers between 0 and infinity?>> Don't
be silly; it's not the square, it's the triangular number.>>
Shift second column up one, third column up two, etc.> So
there, you've a square.Alright, it's kind of a triangle. But
aren't there some redundantnumbers? Namely, 2/4 = 1/2. Doesn't
this kind of skew the results? Ifthese were points in the x-y
plane, then if you can draw a straightline between two
equivilant numbers, then any point also on that samenumber is
also equivilant. How do you elimiate equivilant numbers insuch
a manner?Also, a square number divided by two and a triangular
number areessentially the same for large enough numbers, at
leastproportionately. That is to say, if you take a triangular
number anddivided it by a square number, then at large enough
numbers you'll getcloser and closer to 1/2 until you 'get to
infinity'... or supposedlyget to infinity, but it's not
possible so I put it in quotes.(...Starblade Riven
Darksquall...)

===

>>So the number of rationals between 0
and 1 is the square of the number>>of integers between 0
and infinity?>> Don't be silly; it's not the square, it's
the triangular number.>> Shift second column up one, third
column up two, etc.>> So there, you've a square.> Alright,
it's kind of a triangle. But aren't there some redundant>
numbers? Namely, 2/4 = 1/2. Doesn't this kind of skew the
results? If> these were points in the x-y plane, then if you
can draw a straight> line between two equivilant numbers, then
any point also on that same> number is also equivilant. How do
you elimiate equivilant numbers in> such a manner?It should be
clear that every rational number appears at least once (infact,
infinitely many times) in the triangle. This is all that's
needed,because of the Cantor-Bernstein theorem. If you have an
injection from Qinto N (which the triangular map provides; just
take the smallestavailable n for a given q), and you also have
an injection from N into Q(the inclusion map), then the C-B
theorem guarantees that there is abijection between Q and N.>
Also, a square number divided by two and a triangular number
are> essentially the same for large enough numbers, at least>
proportionately. That is to say, if you take a triangular
number and> divided it by a square number, then at large
enough numbers you'll get> closer and closer to 1/2 until you
'get to infinity'... or supposedly> get to infinity, but it's
not possible so I put it in quotes.The remark about triangular
numbers was a joke.-- Dave SeamanJudge Yohn's mistakes revealed
in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>

===

How do you define Computational
Mathematics and what is it consistedof(sciences) ? This might
sound a stupid question but I am deadserious.I am interested in
Math for computers in general but that is so HUGE.I mean, it's
all about Algebra, algorithms, etc. etc.Can anyone help me by
leading me to the correct direction?

===

>> My strategy is
quite straightforward, > You've just assumed what you need
to prove.I find that to be the most straightforward method of
proof.Infinitely efficient. Much faster than dividing both
sides of an equationby 0, for example.Phil

===

[snip]Earth to
James...Please post one, count 'em: *one*, number which
belongs in the ring of algebraic integers but whichsomehow
got 'left out'. You have repeatedly claimed that the ring of
algebraic integers is incompletebecause it fails to include
numbers which belong in it. With all due contempt I believe
you have someobligation to produce at least one of them -- a
number, mind you, not an ambiguous expression consistingof
ill-defined symbols.Name one such number, if you can. If not,
stop claiming that such numbers exist. (In your
breathlesslyanticipated reply, please do not include
terminology like: amazing, fascinating, and odd.
Theseexpose your mental state, not the underlying mathematics
of the subject at hand.)--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

===

> [.snip.]>theory:>> THEOREM: If r is a root of a non-monic
polynomial> with integer coefficients, ***irreducible***
over> the rationals, the r cannot be an algebraic
integer.>>and conclude that c1 cannot be an
algebraic integer.> You must also include the
hypothesis that the polynomial is> primitive. Since
nonzero constants are units in Q[x], they are not>
considered nontrivial factors, so the hypothesis must be
explicitly> included.>>That is correct. And it is
true that c1 cannot be an algebraic>>integer.>>That
has not been under debate.> Which is why I said I was
nitpicking: pointing out a minor error>> that is well
understood.>>It is true that c1 cannot be an algebraic
integer.Which was not what I addressed. I included the stuff
before it toprovide context, and then pointed out that a
hypothesis was missingfrom the Theorem quoted.What exactly is
your problem with that? Are you asserting that theTheorem was
correct as written? Are you asserting that the theorem iswrong
as corrected according to the mistake I pointed out?Or are you
simply committing the fallacy of Argumentum ad
logicam?http://www.infidels.org/news/atheism/logic.html#
logicam [.snip.]>> 65x^3 - 12x + 1 = (a_1 x + 1)(a_2 x +
1)(a_3 x + 1)>>any a's exist, within the ring of
algebraic integers, such that>>sqrt(5) is a factor of them
in that ring.> This is rather confused. You have a not
at the beginning, a whether>> or not after that, and a
qualifier any for the a's. It's pretty>> close to nonsense.
What you are really saying, presumably, is:> Given a1,
a2, a3 algebraic integers such that> 65x^3 - 12x + 1 =
(a_1 x + 1)(a_2 x + 1)(a_3 x + 1)> [as a polynomial
identity], then none of a_1,a_2,a_3 are multiples>> (in the
ring of algebraic integers) of sqrt(5).>>It is true that
neither a_1, a_2, nor a_3 has sqrt(5) as a factor>***in the
ring of algebraic integers***.>> This is also true, and has
been established.>>Yup as I've stated.As I've ->PROVEN<- (and
was proven by Bengt, and Dale, and Dik, andNora, and David,
and many others), and as you DENIED, mightily, formonths on
end. But thanks for finally coming around, even if now youare
trying to claim credit for a result you do not understand
andcannot prove.>>The problem is that neither a_1, a_2, nor
a_3 have ANY non-unit>>factors in common with 5 in the ring
of algebraic integers.> And that's false. I am pretty sure
that Dale produced explicit common>> factors; but in any case,
your claim here is certainly false, since>> their product is
not coprime to 65.>>I've proven it true that neither a_1, a_2
nor a_3 have ANY non-unit>factors in common with 5 in the ring
of algebraic integers.This is paraphrased from the
posthttp://groups.google.com/groups?selm=3F148963.6000800%
40farir.comTake the polynomial 65x^3 - 12x + 1, and factor it
as65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1)with a1, a2, a3
(necessarily) algebraic integers (in fact, minus theroots of
x^3 - 12x^2 + 65). Let z be any root of that polynomial.
Thatis, z will be either -a1, -a2, or -a3. It is trivial that
any commonfactor between z and 5 will be a common factor of
the corresponding ai.Take the following three polynomials q(x)
= 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37
x - 104Then any product of q(z), r(z), s(z) and integers will
be an algebraicinteger, necessarily.We have that q(z)*r(z) =
64 z^4 - 640 z^3 - 1536 z^2 + 4160 z + 8325Note that (64 x +
128)*(x^3 - 12 x^2 + 65) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160
x + 8320so (65z + 128)(z^3 - 12z^2 + 65) = 64z^4 - 650z^3 -
1536z^2 +4160z+8320.But z^3-12z^2+65 = 0. So 64z^4 - 650z^3 -
1536z^2 +4160z+8320 = 0.Therefore, q(z)*r(z) = 64 z^4 - 640
z^3 - 1536 z^2 + 4160 z + 8325 = (64 z^4 - 640 z^3 - 1536 z^2
+ 4160 z + 8320) + 5 = 5.Likewise, take r(z)*s(z) = 32 z^4 -
312 z^3 - 864 z^2 + 2081 z + 4680This time, note that (32 x +
72)*(x^3 - 12 x^2 + 65) = 32 x^4 - 312 x^3 - 864 x^2 + 2080 x
+ 4680So (32z+72)*(z^3-12z^2+65) = 32z^4 - 312z^3 - 864z^2 +
2080z + 4680.But since z^3-12z^2+65 is equal to 0, it follows
that32z^4 - 312z^3 - 864z^2 + 2080z + 4680=0.
Therefore,r(z)*s(z) = 32z^4 - 312z^3 - 864z^2 + 2081z + 4680 =
(32z^4 - 312z^3 - 864z^2 + 2080z + 4680) + z = z.Therefore,
r(z), which is an algebraic integer, is a common factor of5
and of z. It only remains to show that it is not a unit.The
claim is that r(z) is a root of f(x) = x^3 - 969 x^2 + 315 x
+5. If this is so, then r(z) is not a unit in the ring of all
algebraicintegers, since this is a monic irreducible
polynomial with integercoefficients whose constant term is
neither 1 nor -1.As Dale noted, we havef(r(z)) = (r(z))^3 -
969 (r(z))^2 + 315 (r(z)) + 5 = (8 z^2 - 4 z - 45)^3 - 969 (8
z^2 - 4 z - 45)^2 + 315 (8 z^2 - 4 z - 45) + 5 = 512 z^6 - 768
z^5 - 70272 z^4 + 70592 z^3 + 731136 z^2 - 374400 z -
2067520.Letting w(z) = 512 z^3 + 5376 z^2 - 5760 z - 31808we
have that p(z)*w(z) = 512 z^6 - 768 z^5 - 70272 z^4 + 70592
z^3 + 731136 z^2 - 374400 z - 2067520and therefore, f(r(z)) =
p(z)*w(z). But we know that p(z)=0, sof(r(z))=0. This proves
that r(z) is not a unit, and yet is a commonfactor of z and
5.Letting z be -a1, -a2, and -a3, in turn, you obtain common
factors ofEACH of a1, a2, a3 with 5 in the ring of all
algebraic integers, whichare not units.>> Lemma. Let R be the
ring of all algebraic integers, and let a, b, c be>> any
elements of R. If a and b are coprime to c, then a*b is
coprime to>> c.> Proof. We use the characterization of
coprime valid for commutative>> rings with 1: a and b are
coprime in R if and only if there exist x>> and y in R such
that ax+by = 1. >>By that definitionThat property is
EQUIVALENT to being coprime in the algebraicintegers, which is
EQUIVALENT to having no common factors other than units.> only
*one* of the a's is coprime to 5, but none of>them has a
factor in common with 5 either.Nonsense. By that definition,
NONE of the a's are coprime to 5, andALL have common factors
with 5 which are not units. See the explicitcalculations
above.>The ring of algebraic integers is really screwed
up.Something is, but it is not the ring of algebraic
integers.>For those who don't understand, consider that in the
ring of evens,>which does not have 1, you can't use that
definition of coprime that>Arturo Magidin gives, though it is,
interestingly enough, true that in>fact 2 is coprime to 6 in
the ring of evens because 2(3) = 6, and 3 is>not in the
ring.The ring 2Z is not a ring with unit, so the
characterization givenabove does not apply. As it happens, 2
and 6 are coprime in the ringin question, but NOT for the
incoherent reason you give. They arecoprime because the ideal
generated by 2 is maximal, contains theideal generated by 6,
and is not prime.>However, rather than use dueling definitions
or argue about>definitions I can simply switch to saying that 2
does not share>non-unit factors in the ring of evens with 6.But
it does not matter: above I have reproduced EXPLICIT
commonfactors of each of a1, a2, and a3 with 5 in the ring of
algebraicintegers, and none of them are units. Your claim is
as dead as italways was. [.snip.]>> So, assume you were
correct and neither a_1, a_2, nor a_3 have ANY>> non-unit
factors in common with 5 in the ring of algebraic>> integers.
Then, by the lemma, neither does a1*a_2; and applying the>>
lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65,
which>> clearly has 5 as a nonunit common factor with 5. This
contradicts the>> assumption that none of a_1, a_2, a_3 have
common non-unit factors>> with 5 in the ring of algebraic
integers. Therefore, your assertion is>> false.>>Well by your
definition of coprimeIt's not my definition. It's a
property which is equivalent to theSTANDARD definition (which
is NOT has no common factors other thanunits, as you have
been told numerous times), and which in turn isequivalent, IN
THE RING OF ALL ALGEBRAIC INTEGERS, to the propertythat they
have no common factors.> NONE of the a's have a factor
in>common with 5, in the ring of algebraic integers, and you
cannot prove>that any of them do.Weird. Because above I
exhibited explicit factors. >Now if you don't want to call
that coprime fine. It doesn't change>the situation.No, it
does not. You're wrong. [.snip.]You continue to argue that a
singular case implies a general case, andto argue that a
numeric identity implies a polynomial one. Both arecolossal
errors.

===


===
====================================

===


===
======Why do you take so much trouble to expose such
a reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A man's capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does most
of all. He has made at least ten publications, full of figures
few readers can critize. A great many people are staggered to
this extend, that they imagine there must be the indefinite
something in the mysterious all this. They are brought to
the point of suspicion that the mathematicians ought not to
treat all this with such undisguised contempt, at least. --
A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de
Morgan

===


===
====================================

===


===
=====Arturo Magidinmagidin@math.berkeley.edu===>>
Take your favorite quadratic extension of Q. Let p be a
rational prime> which splits in the ring of integers of K,
(p)=PQ, with P<>Q. One can> then invert elements which are in
P but not in Q to obtain a ring> larger than the ring of
integers of K, which still satisfies the> condition that the
only rational integers which are units are 1 and -1. That's
the point of the example I posted.> One can do a symmetric
argument and do the same with elements in Q> but not in P.
Then one can consider the subrings of the ring of all>
algebraic numbers which include the algebraic integers, and
EITHER the> elements inverted from P-Q, OR the elements
inverted from Q-P, but> noth both. These rings will each,
separately satisfy your conditions,> and be incomparable
(neither one is contained in the other).> But the problem is
that by choosing the elements carefully, one> obtains elements
with the property that any ring which contains BOTH> those
inverses will necessarily contain 1/p. For example, choose a>
quadratic extension which is a PID, and let (p)=(a)(b), with
(a)<>(b)> primes. Then take R[1/a], and R[1/b], where R is the
ring of all> algebraic integers. Each of them satisfies the
property you want, but> any ring which contains BOTH 1/a and
1/b will necessarily contain 1/p,> which should not be allowed
in The Object Ring.That's rather obvious. To wit: any ring
that contains a non-integralalgebraic number w along with all
its conjugates necessarily containsa non-integral rational
number, since the ring then contains all thecoefficients of
the minimal polynomial of w, at least one of whichis a
non-integral rational number. Thus once one allows
adjunctionsof non-integral algebraic numbers there are
restrictions on whatlater extensions are possible if such
extensions are required to notcontain any non-integral
rational number. Contrast this with integralextensions, where
there are no restrictions whatsoever on just whichalgebraic
integers may be adjoined -- integral extensions never changea
nonunit in the base ring into a unit in the extension. That is
oneof the key properties of integral extensions . It is of
course an obvious consequence of lying-over for an integral
extension R < Tsince: c nonunit => cR < P for a max ideal P,
which is lain over byprime Q in T which necessarily contains
the extension cT, so cT < 1,so c stays nonunit in T (one can
also easily prove this directly bymassaging the the minimal
poly of c). So for algebraic integers theproperty of being
unit or nonunit is absolute -- independent of theambient field
containing the algebraic integer. Any enlargement of the
algebraic integers will fail to satisfy these crucial
properties(among others). In fact one can easily show that
integral extensionsmay be characterized in terms of the
lying-over (LO) property andthe incomparable (INC) property,
e.g. consider the followingTHEOREM For commutative rings R < T
the following are equivalent(1) R < T is an integral
extension(2) A < B has INC and LO for all A,B with R < A < B <
T(3) A < A[u] has INC and LO for all A,B with R < A < T, all u
in T-Bill Dubuque

===

>> Take your favorite quadratic
extension of Q. Let p be a rational prime>> which splits in
the ring of integers of K, (p)=PQ, with P<>Q. One can>> then
invert elements which are in P but not in Q to obtain a ring>>
larger than the ring of integers of K, which still satisfies
the>> condition that the only rational integers which are
units are 1 and -1. >>That's the point of the example I
posted.Yes; I believe I mentioned that the example was
suggested by the postquoted (yours); at least the previous
version, which contained myincorrect followup to the example,
did. Sorry if I wasn't clearer ingiving credit where it is
due.

===


===
======================================

===


===
===Why do you take so much trouble to expose such a
reasoner as Mr. Smith? I answer as a deceased friend of mine
used to answer on like occasions - A man's capacity is no
measure of his power to do mischief. Mr. Smith has untiring
energy, which does something; self-evident honesty of
conviction, which does more; and a long purse, which does most
of all. He has made at least ten publications, full of figures
few readers can critize. A great many people are staggered to
this extend, that they imagine there must be the indefinite
something in the mysterious all this. They are brought to
the point of suspicion that the mathematicians ought not to
treat all this with such undisguised contempt, at least. --
A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de
Morgan

===


===
====================================

===


===
=====Arturo
Magidinmagidin@math.berkeley.edu

===

function f:R -> Rf(x) =
(x^2)*[1/x] if x /= 0f(x) = 0 if x=0show, continuous on
x=0([a] gauss notation, ex
[2.4]=2)if 0<x<1if x=q/p (
p<=q)natural number k,r exist such that p=qk+r (0<= r
<q)[1/x]=[p/q]=[(qk+r)/q]=kx ->0 => k -> infinitythus, limf(x)
= lim(x^2)[1/x] if x->0+=lim (q^2)k / (q^2)(k^2)+2qkr+(r^2) = 0
if k->infinitysimilarlyif -1>x>0 ~~~~lim f(x) =0 if x->0-thus
left limit = right limit = function valuetherefore, continuous
on 1-this
solved course is possible ?? impossible??i think that proved
method was rational approach.i doubt that this satisfy all
approach??my question is that this proved course is correct ?
or incorrect?thank to sir.

===

> function f:R -> R> f(x) =
(x^2)*[1/x] if x /= 0> f(x) = 0 if x=0> show, continuous
on x=0> ([a] gauss notation, ex [2.4]=2) if 0<x<1> if x=q/p (
p<=q)[...]> this solved course is possible ?? impossible??>
i think that proved method was rational approach.> i doubt
that this satisfy all approach??> my question is that this
proved course is correct ? or incorrect?You are close to a
solution in the case where f: Q->Q,where Q is the set of all
rational numbers.However, since f:R->R and R is larger than Q,
itturns out that the proof doesn't work for thecase where the
domain is R.I would suggest that you plot the graph of[1/x]
for 0<x<1 and secondly for-1<x<0 . Can you find a simple
expression in x,g(x), such that | [1/x] | <= |g(x)| for anyx
such that 0<x<1? |x| is absolute value of x.That seems like a
good starting point...David Bernier

===

> function f:R -> R>
f(x) = (x^2)*[1/x] if x /= 0> f(x) = 0 if x=0>> show,
continuous on x=0> ([a] gauss notation, ex [2.4]=2)> 1/x - 1 <
[1/x] <= 1/x> if 0<x<1> if x=q/p ( p<=q)>Why is x rational?>
natural number k,r exist such that p=qk+r (0<= r <q)>>
[1/x]=[p/q]=[(qk+r)/q]=k>> x ->0 => k -> infinity>> thus,
limf(x) = lim(x^2)[1/x] if x->0+>> =lim (q^2)k /
(q^2)(k^2)+2qkr+(r^2) = 0 if k->infinity>> similarly> if
-1>x>0 ~~~~lim f(x) =0 if x->0->I've doubt that x < 0 is as
easy.> thus left limit = right limit = function value>
therefore, continuous on 1>I'm not convinced.

===

The following
exercise is in a book on one complex variable (byKrantz), in
the section on evaluating real integrals by integratingover a
contour:Calculate the integral from 0 to infinity of 1/(x^3 +
x + 1). I don't know which contour to use. Any
suggestions?Mike

===

>The following exercise is in a book on
one complex variable (by>Krantz), in the section on evaluating
real integrals by integrating>over a contour:>>Calculate the
integral from 0 to infinity of 1/(x^3 + x + 1). >>I don't know
which contour to use. Any suggestions?This is one where you
need to do that thing where you stickin a log(z), integrate
around that keyhole-shaped thing andwatch the log(z) drop out
in the end.>Mike************************David C.
UllrichX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer:
YesX-Coriate: admin@interspeed.co.nzX-Ecrate:
tanandtanlawyers.comX-Punge: Micro$oft

===

at 10:34 PM, Leon
<Leon@nospam.net> said:>Both of these theorems would be
referred to as the T theorem as if>there was a unique theorem.
But these are not the same sentences.They are logically
equivalent. You will find cases where the same word is defined
differently, orwhere the same name is used for different
theorems. But this isn't oneof them.-- Shmuel (Seymour J.)
Metz, SysProg and JOATReply to domain Patriot dot net user
shmuel+news to contact me. Do not replyto
spamtrap@library.lspace.org

===

>>But I suspect the conjecture
may imply Bertrand's postulate, i.e. if >>A_{1,1} >= 2,
A_{1,j} = 1 otherwise, A_{i,j+1} = |A_{i,j} - A_{i+1,j}|,
>>and A_{i,1} is strictly increasing, then A_{i+1,1} < 2
A_{i,1}>2 3 5 7 9 15 33>1 2 2 2 6 18>1 0 0 4 12>1 0 4 8>1 4
4>1 0 ^^^ That 1 would be a 3. (Maybe I didn't say it
explicitly enough, butthe equation A_{i,j+1} = |A_{i,j} -
A_{i+1,j}| is supposed to include the case i=1).Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2

===

> Take the difference
sequence of the primes, and iteratively take the> difference
sequence of the previous sequence, but make sure you take>
all differences in absolute value.> 2 3 5 7 11 13 15 17 19 .
. . .> 1 2 2 4 2 2 2 2 . . . .> 1 0 2 2 0 0 0 . . . .> 1 2 0 2
0 0 . . . .> 1 2 2 2 0 . . . .> The claim is that the
leftmost diagonal is always 1 (except the initial 2).> This
seems to be some sort of delicate statement> Not all that
delicate! As we shall see. It's almost worthless.> about
the distribution of primes. Has anyone seen this claim?>
Yes, in some popularization or other; and I didn't think much
of it then,> and don't now. It struck me as having been made
by someone with little> real number sense. (But in that, it is
similar to Goldbach's.)> I don't want to rain on anyone's
parade, (well actually I don't really> mind doing that as you
probably know), but this is a parade that should> have been
rained out before it left the marshalling yard.> The
conjecture will apply to almost any sequence of numbers all
of> the same parity, except for a first one of different
parity, that doesn't> grow too fast. And fast is loose -
basically, size doubling is the limit.> In a sense, the
algorithm and initial conditions have been chosen quite>
cunningly; the absolute value of the differences ensures that
any local> variations (as in the primes) get smoothed out
fairly quickly, and the> parity conditions ensure that there
will always be a one at the very> left, provided there isn't
any sudden fast growth or decay anywhere.> As examples, here
is a sequence beginning with primes 2,3,5 and continuing> to
grow as fast as possible consistent with the conjecture.> 2
3 5 9 17 33 65 129 ... essentially just doublings.> 1 2 4 8 16
32 64 ...> 1 2 4 8 16 32 ... not really very exciting, is it? And if some of the terms are lowered, it has no effect so
long as> the subsequent terms are lowered accordingly (within
VERY easy limits!),> as the absolute differences will smooth
out local variations, as I say.> As a second example, we go
to the other extreme from primes.> (Contrary to what someone
was hoping about composite numbers.)> Here is a sequence of
very round numbers according to a criterion> which I have
devised but will keep hidden for now.> 1 2 4 8 12 20 24 36
48 60 90 120 180 200 240 300 ...> 1 2 4 4 8 4 12 12 12 30 30
60 20 40 60 ...> 1 2 0 4 4 8 0 0 18 0 30 40 20 20 ...> 1 2 4 0
4 8 0 18 18 30 10 20 0 ...> 1 2 4 4 4 8 18 0 12 20 10 20 ...> 1
2 0 0 4 10 18 12 8 10 10 ...> 1 2 0 2 6 8 6 4 2 0> 1 2 2 4 2 2
2 2 2> 1 0 2 2 0 0 0 0> 1 2 2 2 0 0 2> 1 2 0 2 0 2> As you
can see, they still satisfy the conjecture.> (OK I admit I had
to work a little to make it so, but not a lot.)> Finally, I
was going to make an example with the primes, but each one>
plus-or-minus one, more or less at random. But it was far too
easy.> So here is an example that shows the slow growth of
the primes isn't> necessary either; each term is 1.5 times the
previous one, rounded> to the nearest odd number. It grows
geometrically fast, though not> close to the doubling
limit.> 2 3 5 7 11 17 25 37 55 83 125 187 ...> 1 2 2 4 6 8
12 18 28 42 62 ...> 1 0 2 2 2 4 6 10 14 20 ...> 1 2 0 0 2 2 4
4 6 ...> Again, you can plus-or-minus two to all the terms
to introduce more> local randomness, but provided you don't
damage the very low ones> too badly, it won't make any
difference.> Well, ICBW, but AFAICS the conjecture is a
crock.>
-
-----> Bill Taylor
W.Taylor@math.canterbury.ac.nz>
-
-----> If we knew what we were doing it wouldn't
be called research>
-
-----Bill,As I am sure you have gathered, I have
very little number sense also,I will be the first to admit
it.Now for the sake of me possibly learning something here I
propose thisone question.which uses all the primes just twice)
when you force an error (reversetwo adjoining primes) down
about 75 terms into the sequence the 4distinct left diagonal
deltas are consistent with their patterns untilabout 105 terms
(rows). At this point the integrity of any patternsare
lost.This is understandable, but as the 4 delta diagonals
continue to about130 rows the integrity of these patterns
return intact and willprobably continue until---->oo baring
any other errors in the sequence! Why?Yes, I agree there is a
parity issue with the even prime, but with mysequence I
believe something else is happening. Can you duplicate this
with another sequence that has 4 distinct leftdiagonal delta
patterns ---->oo?Sorry, that was two questions.  If you can
create another sequence, with 4 distinct left diagonaldelta
patterns and causing an error 75 or more terms into the
sequenceand somewhat duplicating what happens in my sequence
where the errorcauses a breakup of the 4 patterns for 25 of
more rows and thenreturning to the original 4 patterns.Then I
will say your argument has merit. A non-response to this post
I will consider as no one can create thisnew sequence with the
above criteria.I believe this to be an interesting
challenge.Dan

===

here, there are quite a lot and I'll read
them tonight.But I just thought I'd post a follow-on to my
earlier remarks.I should have done this years ago when I first
became underwhelmed bythe conjecture as applied to primes, but
I didn't, so I'm grateful to Danfor bringing up this topic, so
I can now do a little more and then pass onthe baton to someone
else.So I'm going to look at the set of admissable sequences
under the givencriterion, with a view to seeing if we can find
out how remarkable it IS to find that the primes are among
them; and you know I suspect - not very.Actually, I'm not
going to look at a sequence, but rather a sequence ofof finite
sequences.... yuk....ANYWAY.Let's look at how sequences can
begin, and maintain the delta-1's property.For Sloane-like
standardisation, I take only sequences of even numbers
thatbegin with a 1, so that all its deltas might be of the
same sort.So any sequence must begin 1. There is just 1 of
these. :)It can continue 1 0 or 1 2 - 2 initial subsequences
of length 2.These may continue 1 0 0 1 0 2 1 2 0 1 2 2 1 2 4 -
so 5 3-sequences.These may continue...1000 1002 1020 1022 1024
1026 1200 1202 1204 1220 1222 1224 1240 1242 1244 1246* 1248
.There are 17 of these 4-sequences. I did these by hand so
there may bemistakes, but I'm pretty confident of those up to
here. I also did a quicklook ahead at how many 5-sequences
there were, which is not indeed verydifficult when you have
the previous lot written down in front of you;so computering
them will be fairly easy too.Anyway, those 17 4-sequences can
be followed by, respectively,2 5 3 4 5 8 3 4 7 4 3 6 5 4 5 5 9
different continuations,giving a total of 82 admissable
5-sequences. [*]BTW - there is oneparticularly interesting
case:- 1246 is the ONLY 4-sequence that cannot becontinued
with a 0! This sort of thing will become much commoner
later, OC.(For example, one finds in going from length 5 to
6, there appear cases where the next admissable entries form
a non-contiguous range of even numbers!)I stopped there, and
leave the next many cases to someone with more
efficientprogramming skills than me!The sequence of counts
itself is an intriguing sequence...1 2 5 17 82 ... and has a
smell of factorial growth about it! I wonder if it already
appears in Sloane?But just how common this makes admissable
sequences among those of the samelength and beginning with 1
but otherwise even, is still open to definition.I suppose we
would have to limit it to sequences whose every term was
boundedby the appropriate power of 2, as given by the
maximum sequence 1 2 4 8...These universes of finite
sequences are indeed factorial in number!And the admissable
sequences as a proportion of their universe does not(as one
might thus expect) plummet at all fast - they are surprisingly
common,as it happens. Here are the counts:- size of # oflength
universe admissables 1 1 1 100% 2 2 2 100% 3 6 5 83% 4 24 17
71% 5 120 82 65% 6 720 ??? 7 5040 ? ...and I eagerly await a
great mountain of figures! 8 .
.

===


===
=====================Another snippet.There is
OC the delta way to reduce an (n+1)sequence to an
n-sequence,which can be reversed in many different ways. (It's
a many-one function.)But there is also a very gross way in
which an n-sequence can be extendedto an (n+1)-sequence,
namely just by adjoining another entry, as I've beendoing
above. (The reverse, truncation, is also a many-one function
OC.)So one can thus make up an intriguing directed graph, or
rather two such,a trivial one obtained by adjoining admissable
entries, and a more trickyone obtained by differencing a
sequence to a one-shorter one. One canthus also superimpose
these two to get a two-sorted digraph of all admissablefinite
sequences, with 2 sorts of arcs, by adjoining & delta-ing
respectively.The bi-digraph so obtained has some very
intriguing near-patterns, but notquite perfect enough to be
general rules. It would also be a fun study though.However,
none of the above has any very direct relevance to the
INFINITE admissable sequences (such as primes?)! :-)Well then;
there
y'go.---
---Bill Taylor
W.Taylor@math.canterbury.ac.nz--
----Education is a
process of telling a carefully chosen sequence of lies in
whichthe amount of deliberate deception gradually tends
towards zero. - John
Baez---
---

===

|>>I am iterating a certain
function, and I keep all the resultant values|>>in an array:
e:=array(0..499);|>>I need a local proc(e,n) which takes as
input the array of entries, its|>>actual size (actual size may
be less than 500) and heuristically tries|>>to determine if the
entries are close to a k-cycle.On second thought, why not just
do it this way?> findperiod:= proc(e, n, eps) local i,k; for k
from 1 to n-1 do for i from 0 to n-k-1 while abs(e[i] - e[i+k])
< eps do od: if i = n-k then RETURN(k) fi; od; 0 end;This
should work with complex data, and with all versions of
Maple.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2

===

[snip]> |>>I
need a local proc(e,n) which takes as input the array of
entries, its> |>>actual size (actual size may be less than
500) and heuristically tries> |>>to determine if the entries
are close to a k-cycle.> On second thought, why not just do
it this way?> findperiod:= proc(e, n, eps)> local i,k;>
for k from 1 to n-1 do> for i from 0 to n-k-1 while abs(e[i] -
e[i+k]) < eps do od:> if i = n-k then RETURN(k) fi;> od;> 0>
end;> This should work with complex data, and with all
versions of Maple.Hmm, I am getting some strange results with
your findperiod. Let's lookat the code.First, the code I am
using for the iteration (f(z)=c^z, and I amiterating using
z=1):# c: complex base# z: first complex exponent# n:
iteration count# m: cycle count# o: show cyclic component
count> iterf:=proc(c,z,n,m,o,eps)local
i,j,k,l,e;i:=evalf(z);e:=array(0..499);for k from 0 to n-1 do
i:=evalf(c^i); e[k]:=i; if abs(evalf(Re(i)))>1e5 then
printf(`real component out of bounds`); break; fi; if
abs(evalf(Im(i)))>1e5 then printf(`imaginary component out of
bounds`); break; fi; if abs(evalf(Re(i)))<1e-5 then
j:=evalf(Im(i)); #if Re close to zero, zap Re i:=evalf(j*I);
e[k]:=i; #update array entry fi; if abs(evalf(Im(i)))<1e-5
then j:=evalf(Re(i)); #if Im close to zero, zap Im
i:=evalf(j); e[k]:=i; #update array entry fi; if (k+o) mod m =
0 then print(i); fi; od;findperiod(e,n,eps); #call Robert's
code after iteration is done.end:>findperiod:= proc(e, n, eps)
local i,k,f; for k from 1 to n-1 do for i from 0 to n-k-1 while
abs(e[i] - e[i+k]) < eps do od: if i = n-k then RETURN(k); fi;
od; 0;end:>eps:=1e-2; eps := .1> iterf(3-3*I,1,10,1,0,eps);
#works correctly! 3. - 3. I6.642369647 - 2.875670132
I-1540.034993 - 80.14662298 I01.3. - 3. I 6.642369647 -
2.875670132 I -1540.034993 - 80.14662298 I01.5But:>
t:=(n,k)->exp(2*k*Pi*I/n);>eps:=1e-1;>
iterf(t(3,2),1,33,1,0,eps);[...snip results]32While:>
iterf(t(3,2),1,30,1,0,eps);> iterf(t(3,2),1,31,1,0,eps);>
iterf(t(3,2),1,32,1,0,eps);All return 0 (cannot find a
cycle)Now, 3-3*I is a true 5-cycle, whereas the 2nd, 3rd root
of unity t(3,2)appears to be a pseudo 17-cycle, which
coalsesces into a 1-cycle. YourfindCycle returns 32, for
eps=0.1 and n > 32 and 0 for eps < 0.1 or n <32.Is k=32
correct in this case? It appears it should rather be k=17,
atleast before the values eventually coalesce into the
1-cycle.> Robert Israel israel@math.ubc.ca> Department of
Mathematics http://www.math.ubc.ca/~israel> University of
British Columbia> Vancouver, BC, Canada V6T 1Z2--
Ioannishttp://users.forthnet.gr/ath/jgal/_____________________
______________________Eventually, _everything_ is
understandable.

===

>[snip]>While:>>
iterf(t(3,2),1,30,1,0,eps);>> iterf(t(3,2),1,31,1,0,eps);>>
iterf(t(3,2),1,32,1,0,eps);>>All return 0 (cannot find a
cycle)I'm not sure what's going wrong, but it looks to me like
for example iterf(t(3,2),1,30,1,0,eps) doesn't assign values to
e[k]for k >= 30.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2

===

> I am
iterating a certain function, and I keep all the resultant
values> in an array: e:=array(0..499);> I need a local
proc(e,n) which takes as input the array of entries, its>
actual size (actual size may be less than 500) and
heuristically tries> to determine if the entries are close to
a k-cycle.> I tried to figure out a quick and dirty solution
to this, but after a> while it became obvious that the problem
is rather difficult.> For starters one needs to make several
assumptions. One needs to have an> eps, which determines how
close the entries need to be to be considered> a match for
the cycle and I need the proc to tell the user in case> there
is no match.> There are other difficulties also: Cases where
there are accidental> matches, but where the rest of the
entries do not comprise a cycle.> It looks to me that in the
worst case, the algorithm seems to be O(n^2),> since I need to
check all entries against all previous ones for matches.>
I'd appreciate if anyone can help a bit with this.> -->
Ioannis> http://users.forthnet.gr/ath/jgal/>
___________________________________________> Eventually,
_everything_ is understandable.You have options here
...restart;with(networks):MUTE:=(proc() NULL end):findperiod:=
proc(e, n, eps) local i,k,store,j; j:=1; for k from 1 to n-1 do
for i from k+1 to n do if abs(e[i] - e[k]) < eps then
store[j]:=[i,k];j:=j+1; break;fi od od; store; end:#Remove
the break in findperiod if you want all the
loopsRANDPOINT:=subs(R1=rand(360),R2=rand(1024),unapply(('proc
(alpha,r)evalf(r*(cos(alpha)+I*sin(alpha)))end')('R1()'/180*Pi
,'R2'()/1024*m+delta),delta,m)):print(`Setting up random point
set
:`);assign('Px'=mul(x-RANDPOINT(1,1),i=1..8)),assign('Pxf'=
unapply(Px,x)),print(`Morfingpoly set
up`):assign('fx'=RootOf(Px+x0,x)),assign('fx0f'=(evalf@RootOf)
(Px,x)),assign('pf'=unapply(subs(subs([(proc(P,x,x0,p,xn)
locali;seq((D@@i)(p)(xn)=unapply(diff(P,x$i),x)(x0),i=1..
degree(P,x))end)(collect(Pxf(x),x),x,fx0f,p,fx0)],[seq(convert
(diff(g(x0),x0$i),D)=subs(RootOf(p(_Z)+x0)=fx0,collect((diff)(
RootOf(p(x)+x0,x),x0$i),D)),i=1..8)]),x0=0,g(0)=fx0f,convert(
series(g(x),x=x0,9),polynom)),x)),print(`Morfing
function`=eval(pf));assign('PS'=[seq((evalf@pf)(RANDPOINT(0,1)
),i=1..100)]),plots[complexplot](PS,style=point,title=`
Pointset`);print(`Random point set
ready`);assign('sol'=findperiod(PS,nops(PS),0.001)),print(`
Loops
located`);print(`Loopcount`=nops(map(op,convert(sol,set)))),
assign('CS=CS'),MUTE(new('CS'),addvertex(map(op,convert(sol,
set)),'CS'),map(proc(L)addedge({op}(L),'CS'); end,((proc(L)
local
i;[seq([L[i-1],L[i]],i=2..nops(L)),(op@map)(op,[entries](sol))
]end)@sort@[op]@map)(op,convert(sol,set)))),print(plots[
display](draw(CS),color=red));Chris

===

>> Now, 3-3*I is a
true 5-cycle, whereas the 2nd, 3rd root of unity t(3,2)>
appears to be a pseudo 17-cycle, which coalsesces into a
1-cycle.It looks rather like a 16-cycle. That explains number
32 - two cycles.In Maple 9, one can use FFT for finding the
cycle length. Here is theprocedure,C:=proc(V,h) local
VR,VC,a,b,c;VR,VC:=map(Re,V),map(Im,V):a:=DiscreteTransforms:-
FourierTransform(VR);b:=DiscreteTransforms:-FourierTransform(
VC);# print(plots[listplot](map(abs,convert(a,list))));#
print(plots[listplot](map(abs,convert(b,list))));c:=proc(L)
local k; for k from 2 to op([2,2],L)-1 do
ifabs(L[k])>abs(L[k-1]) and abs(L[k])>=abs(L[k+1]) and
abs(L[k])>h thenreturn round(op([2,2],L)/(k-1))fi od; 1
end;ilcm(c(a),c(b))end:V should be a complex Array and h is
the minimal height of peaks. I'll useh=1.5 in most of the
examples below.Example:V:=Array(1..500,x->x mod 20 + (x mod
7)*I,datatype=complex[8]):C(V,1.5); 140Uncommenting prints in
the procedure C, one can see the spectrograms of theFFT of
real and imaginary parts of V.Now, I'll use the following
iteration procedure,it:=proc(c,n) local
V,i;V:=Array(1..n,datatype=complex[8]);V[1]:=evalf[20](c);for
i from 2 to n do V[i]:=evalf[20](c^V[i-1]) od;V end:For your
first example,V:=it(3-3*I,10):C(V,1.5); 5Correct! Now, the
example with
t(3,2),t:=(n,k)->exp(2*k*Pi*I/n);V:=it(t(3,2),500):C(V,1.5);
16Finally, Robert Israel's example,V:=it(15.384972737405246873
+ 0.42943156081631657205*I,100):C(V,1000); 4Alec
Mihailovshttp://webpages.shepherd.edu/amihailo/

===

> I am
iterating a certain function, and I keep all the resultant
values> in an array: e:=array(0..499);>> I need a local
proc(e,n) which takes as input the array of entries, its>
actual size (actual size may be less than 500) and
heuristically tries> to determine if the entries are close to
a k-cycle.Is k fixed? Then k should be a parameter also. Or do
you want to checkif there are cycles for any k?> I tried to
figure out a quick and dirty solution to this, but after a>
while it became obvious that the problem is rather
difficult.>> For starters one needs to make several
assumptions. One needs to have an> eps, which determines how
close the entries need to be to be considered> a match for
the cycle and I need the proc to tell the user in case> there
is no match.>> There are other difficulties also: Cases where
there are accidental> matches, but where the rest of the
entries do not comprise a cycle.>> It looks to me that in the
worst case, the algorithm seems to be O(n^2),> since I need to
check all entries against all previous ones for matches.Just
sort the list. Then matches will be next to each other.
That'sO(n*log(n)). If the list is complex numbers, sorting
them by magnitudeshould be sufficient if it is unlikely that
list contains numbers whosemagnitudes differ by less than
epsilon when the numbers themselves arefuther than epsilon
apart.> I'd appreciate if anyone can help a bit with this.Is
the list tail-recursive? That is, does each entry depend only
on theprevious entry: x[n+1]:= f(x[n]). For example, a
sequence generated byNewton's method would be like this. Then
any single repeat is enough tomake a cycle.The average case
complexity may be much better than the worse case. Ifthere are
in fact no entries are even close to being a cycle, this can
bequickly determined.Here's a first pass:CycleCheck:=
proc(e,n,eps) local L,i; L:= sort( [seq]([e[i],i], i= 0..n-1)
,(a,b)-> abs(a[1]) < abs(b[1]) ); for i to n-1 do if
abs(L[i][1]-L[i+1][1]) < eps then print(`Probable match at
positions`); return L[i][2], L[i+1][2] fi od; print(`No
match`)end proc;

===

> Just sort the list. Then matches will be
next to each other. That's> O(n*log(n)). If the list is complex
numbers, sorting them by magnitude> should be sufficient if it
is unlikely that list contains numbers whose> magnitudes
differ by less than epsilon when the numbers themselves are>
futher than epsilon apart.My answer was written days ago,
before seeing the other answers. Forsome reason, there was a
big delay in posting it. The subsequentanswers supercede my
answer.

===

The Primary Universal Base Unit Values1) mass
(hc/G)^1/2 = 5.4563026(39) x 10^-8 kg2) length (hG/c^3)^1/2 =
4.0507625(38) x 10^-35 m3) time (hG/c^5)^1/2 = 1.3511889(41) x
10^-43 s4) current e/(hG/c^5)^1/2 = 1.1857530(90) x 10^24 A5)
temperature b/(hG/c^3)^1/2 = 3.5518626(82) x 10^32 K6)
substance M/(hc/G)^1/2 = 1.6605387(31) x 10^-27 kmol7)
intensity (hG/c^5)^1/2/sr = 1.9720204(06) x 10^-45 cd8)
Einstein solid angle = 6.8517999(97) x 10^1 sr9) Planck plane
angle = 1.3703599(97) x 10^2 radThe Primary Universal Physical
Constants001) irradiance constant i. = 4.5211591(52) x 10-122
s^3/kg002) radiance constant i.= 3.0978078(26) x 10-120
s^3-sr/kg003) radiant volume constant = 1.3554094(15) x
10^-113 m-s^2/kg004) measurement volume = 6.6467654(65) x
10^-104 m^3005) graviton volume constant = 1.2181812(31) x
10^-96 m^3/kg006) luminous efficacy = 3.7229937(53) x 10^-96
cd-sr-s^3/kg-m^2007) electric current volume = 1.3838190(49) x
10^-93 m^2/A008) luminous energy = 1.8257115(55) x 10^-86
cd-sr-s009) electric charge volume = 4.1485851(42) x 10^-85
m^3/A-s010) molar volume = 4.0027765(33) x 10^-77 m^3/kmol011)
moment of inertia = 8.9530708(38) x 10^-77 kg-m^2012) graviton
fluidity = 1.0031235(26) x 10^-70 m-s/kg013) measurement area
= 1.6408677(14) x 10^-69 m^2014) electric moment =
6.4900363(91) x 10^-54 A-s-m015) Euclid capacitance =
5.234567901... x 10^-48 A^2-s^4/kg-m^2016) magnetic moment =
1.9456639(62) x 10^-45 A-m^2017) luminous intensity =
1.9720204(06) x 10^-45 cd018) graviton frequency i. =
1.3511889(41) x 10^-43 s019) luminous flux = 1.3511889(41) x
10^-43 cd-sr020) graviton moment = 2.2102186(33) x 10^-42
kg-m021) inductance constant = 3.4877980(18) x 10^-39
kg-m^2/A^2-s^2022) absorption-emission = 2.4763819(58) x
10^-36 s/kg023) graviton wavelength = 4.0507625(38) x 10^-35
m024) Planck constant = 6.6260687(65) x 10^-34 kg-m^2/s025)
relative expansion = 2.8154241(58) x 10^-33 /K026) electric
resistivity = 1.0456155(41) x 10^-30 kg-m^3/A^2-s^3027)
unified substance = 1.6605387(31) x 10^-27 kmol028) kinematic
viscosity = 1.2143880(58) x 10^-26 m^2/s029) inverse electric
current = 8.4334589(42) x 10^-25 /A030) Boltzmann constant=
1.3806502(93) x 10^-23 kg-m^2/s^2-K031) thermal resistance =
9.7866124(96) x 10^-21 s^3-K/kg-m^2032) graviton molality =
3.0433405(93) x 10^-20 kmol/kg033) elementary charge =
1.6021764(62) x 10^-19 A-s034) primary radiation =
5.9552136(16) x 10^-17 kg-m^4/s^3035) specific heat =
2.5303770(42) x 10^-16 m^2/s^2-K036) magnetic flux q. =
2.0678336(42) x 10^-15 kg-m^2-sr/A-s^2-rad037) magnetic flux =
4.1356672(77) x 10^-15 kg-m^2/A-s^2038) electric permittivity =
1.2922425(96) x 10^-13 A^2-s^4/kg-m^3039) magnetic exposure =
2.9363775(58) x 10^-12 A-s/kg040) electric constant =
8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3041) magnetic pole
strength = 4.8032041(96) x 10^-11 A-m042) Newton constant =
6.6723641(43) x 10^-11 m^3/kg-s^2043) density of states =
2.0392015(07) x 10^-10 s^2/kg-m^2044) S-B primary constant =
1.3897143(30) x 10^-9 kg/s^3-K^4045) radiant distribution
constant = 3.335640952... x 10^-9 s/m046) graviton mass
constant = 5.4563026(39) x 10^-8 kg047) molar Planck constant
= 3.9903126(87) x 10^-7 kg-m^2/s-kmol048) magnetic constant =
1.256637061... x 10^-6 kg-m/A^2-s^2-sr049) electric
conductance = 3.8740458(43) x 10^-5 A^2-s^3/kg-m^2050)
conductance q. = 7.7480916(72) x 10^-5
A^2-s^3-rad/kg-m^2-sr051) magnetic permeability =
8.6102258(15) x 10^-5 kg-m/A^2-s^2052) fine-structure constant
= 7.2973525(36) x 10^-3 /rad053) second radiation constant =
1.4387752(29) x 10^-2 m-K054) dielectric constant =
1.4594705(05) x 10^-2 /sr055) gravitational momentum =
1.6357583(80) x 10^1 kg-m/s056) relative permeability =
6.8517999(97) x 10^1 sr057) inverse fine-structure =
1.3703599(97) x 10^2 rad058) impedance of vacuum = 3.767303134
x 10^2 kg-m^2/A^2-s^3-sr059) molar gas constant = 8.3144720(88)
x 10^3 kg-m^2/s^2-kmol-K060) spin angle constant =
9.3894326(23) x 10^3 sr-rad061) i. conductance q. =
1.2906403(83) x 10^4 kg-m^2-sr/A^2-s^3-rad062) von Klitzing
constant = 2.5812807(61) x 10^4 kg-m^2/A^2-s^3063) inverse
gravitational mass = 1.8327429(14) x 10^7 /kg064) Faraday
constant = 9.6485341(30) x 10^7 A-s/kmol065) speed of light in
vacuum = 2.99792458 x 10^8 m/s066) graviton energy constant =
4.9038802(52) x 10^9 kg-m^2/s^2067) Josephson primary =
2.4179894(88) x 10^14 A-s^2/kg-m^2068) Josephson q. =
4.8359789(67) x 10^14 A-s^2-rad/kg-m^2-sr069) electric
displacement = 3.9552465(66) x 10^15 A-s/m070) absorbed dose =
8.987551787 x 10^16 m^2/s^2071) luminous density =
2.7467669(26) x 10^17 cd-sr-s/m^3072) gravity displacement =
4.4930470(15) x 10^18 kg-s/m^2073) molar mass constant =
3.2858629(17) x 10^19 kg/kmol074) magnetic potential =
1.0209601(87) x 10^20 kg-m/A-s^2075) thermal conductance =
1.0218040(21) x 10^20 kg-m^2/s^3-K076) electric current
constant = 1.1857530(90) x 10^24 A077) luminance constant =
1.2018155(94) x 10^24 cd/m^2078) luminous flux density =
8.2346000(82) x 10^25 cd-sr/m^2079) Avogadro constant =
6.0221419(79) x 10^26 /kmol080) gravitational field =
1.3469816(08) x 10^27 kg/m081) electric potential =
3.0607616(38) x 10^28 kg-m^2/A-s^3082) electric conductivity =
9.5637446(19) x 10^29 A^2-s^3/kg-m^3083) Celcius temperature =
3.5518626(82) x 10^32 K084) graviton wave number =
2.4686709(99) x 10^34 /m085) mass flow rate constant =
4.0381492(72) x 10^35 kg/s086) molar energy = 2.9531863(13) x
10^36 kg-m^2/s^2-kmol087) surface concentration =
1.0119881(80) x 10^42 kmol/m^2088) graviton frequency =
7.4008894(66) x 10^42 /s089) superforce constant =
1.2106066(96) x 10^44 kg-m/s^2090) luminous intensity i. =
5.0709414(41) x 10^44 /cd091) angular velocity = 1.0141882(87)
x 10^45 rad/s092) electric flux density = 9.7642024(91) x 10^49
A-s/m^2093) radiant intensity = 5.2968673(52) x 10^50
kg-m^2/s^3-sr094) graviton field strength = 2.2187308(44) x
10^51 m/s^2095) superpower constant = 3.6293075(70) x 10^52
kg-m^2/s^3096) magnetic flux density = 2.5204148(03) x 10^54
kg/A-s^2097) thermal conductivity = 2.5224979(53) x 10^54
kg-m/s^3-K098) magnetic field strength = 2.9272342(65) x 10^58
A/m099) absorbed dose rate = 6.6515877(33) x 10^59 m^2/s^3100)
graviton surface density = 3.3252544(33) x 10^61 kg/m^2101)
electric field strength = 7.5560134(90) x 10^62 kg-m/A-s^3102)
measurement area i. = 6.0943364(99) x 10^68 /m^2103) dynamic
viscosity = 9.9688619(97) x 10^69 kg/m-s104) molar
concentration = 2.4982658(70) x 10^76 kmol/m^3105) surface
tension constant= 2.9885896(41) x 10^78 kg/s^2106) electric
charge density = 2.4104603(51) x 10^84 A-s/m^3107) angular
acceleration = 7.5058954(07) x 10^87 rad/s^2108) thermal
transfer = 6.2272175(40) x 10^88 kg/s^3-K109) electric current
density = 7.2263783(36) x 10^92 A/m^2110) luminous efficacy i.
= 2.6860104(16) x 10^95 kg-m^2/cd-sr-s^3111) graviton density
constant = 8.2089591(81) x 10^95 kg/m^3112) measurement
density = 1.5044911(77) x 10^103 /m^3113) radiant density
constant = 7.3778445(74) x 10^112 kg/m-s^2114) radiance
constant = 3.2280892(03) x 10^119 kg/s^3-sr115) irradiance
constant = 2.2118221(59) x 10^121 kg/s^3The Primary Universal
Gravitation EquationsE=(hc^5/G)^1/24.9038802 x 10^9 kg-m^2/s^2
= [(6.6260687 x 10^-34 kg-m^2/s)(2.4216061 x 10^42
m^5/s^5)/(6.6723641 x 10^-11 m^3/kg-s^2)]^1/24.9038802 x 10^9
kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2[rsu 3.9 x
10^-8]E=c^5/Gv4.9038802 x 10^9 kg-m^2/s^2 = (2.4216061 x 10^42
m^5/s^5)/(6.6723641 x 10^-11 m^3/kg-s^2)(7.4008894 x 10^42
/s)4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9
kg-m^2/s^2[rsu 3.9 x 10^-8]E=hc^3/Gm4.9038802 x 10^9
kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s)(2.6944002 x 10^25
m^3/s^3)/(6.6723641 x 10^-11 m^3/kg-s^2)(5.4563026 x 10^-8
kg)4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9
kg-m^2/s^2[rsu 3.9 x 10^-8]E=h(Gd)^1/24.9038802 x 10^9
kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s)[(6.6723641 x 10^-11
m^3/kg-s^2)(8.2089591 x 10^95 kg/m^3)]^1/24.9038802 x 10^9
kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2[rsu 3.9 x
10^-8]G=c^3/mv6.6723641 x 10^-11 m^3/kg-s^2 = (2.6944002 x
10^25 m^3/s^3)/(5.4563026 x 10^-8 kg)(7.4008894 x 10^42
/s)6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11
m^3/kg-s^2[rsu 3.9 x 10^-8]G =wc^4/E6.6723641 x 10^-11
m^3/kg-s^2 = (4.0507625 x 10^-35 m)(8.0776087 x 10^33
m^4/s^4)/(4.9038802 x 10^9 kg-m^2/s^2)6.6723641 x 10^-11
m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2[rsu 3.9 x
10^-8]G=c^4/F6.6723641 x 10^-11 m^3/kg-s^2 = (8.0776087 x
10^33 m^4/s^4)/(1.2106066 x 10^44 kg-m/s^2)6.6723641 x 10^-11
m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2[rsu 3.9 x
10^-8]G=c^5/P6.6723641 x 10^-11 m^3/kg-s^2 = (2.4216061 x
10^42 m^5/s^5)/(3.6293075 x 10^52 kg-m^2/s^3)6.6723641 x
10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2[rsu 3.9 x
10^-8]E is graviton energy constant, h is Planck constant,c is
speed of light in vacuum, G is Newton constant,v is graviton
frequency, m is graviton mass constant,d is graviton density
constant, w is graviton wavelength,F is superforce constant, P
is superpower constant,e is elementary charge, b is second
radiation constant,M is molar mass constant, i. is inverse, q.
is quantumhttp://physics.nist.gov/cuu/Constants/[UPDATED 1998
CODATA-NIST VALUES]

===

So what is your point and what has this
to do with mathematics?This is a math group. Go talk to the
physickers.>The Primary Universal Base Unit Values>>1) mass
(hc/G)^1/2 = 5.4563026(39) x 10^-8 kg>2) length (hG/c^3)^1/2 =
4.0507625(38) x 10^-35 m>3) time (hG/c^5)^1/2 = 1.3511889(41) x
10^-43 s>4) current e/(hG/c^5)^1/2 = 1.1857530(90) x 10^24 A>5)
temperature b/(hG/c^3)^1/2 = 3.5518626(82) x 10^32 K>6)
substance M/(hc/G)^1/2 = 1.6605387(31) x 10^-27 kmol>7)
intensity (hG/c^5)^1/2/sr = 1.9720204(06) x 10^-45 cd>8)
Einstein solid angle = 6.8517999(97) x 10^1 sr>9) Planck plane
angle = 1.3703599(97) x 10^2 rad>>The Primary Universal
Physical Constants>>001) irradiance constant i. =
4.5211591(52) x 10-122 s^3/kg>002) radiance constant i.=
3.0978078(26) x 10-120 s^3-sr/kg>003) radiant volume constant
= 1.3554094(15) x 10^-113 m-s^2/kg>004) measurement volume =
6.6467654(65) x 10^-104 m^3>005) graviton volume constant =
1.2181812(31) x 10^-96 m^3/kg>006) luminous efficacy =
3.7229937(53) x 10^-96 cd-sr-s^3/kg-m^2>007) electric current
volume = 1.3838190(49) x 10^-93 m^2/A>008) luminous energy =
1.8257115(55) x 10^-86 cd-sr-s>009) electric charge volume =
4.1485851(42) x 10^-85 m^3/A-s>010) molar volume =
4.0027765(33) x 10^-77 m^3/kmol>011) moment of inertia =
8.9530708(38) x 10^-77 kg-m^2>012) graviton fluidity =
1.0031235(26) x 10^-70 m-s/kg>013) measurement area =
1.6408677(14) x 10^-69 m^2>014) electric moment =
6.4900363(91) x 10^-54 A-s-m>015) Euclid capacitance =
5.234567901... x 10^-48 A^2-s^4/kg-m^2>016) magnetic moment =
1.9456639(62) x 10^-45 A-m^2>017) luminous intensity =
1.972020religion.> FOR THERE IS A NICE PLACE IN HELL FOR
IDIOTS LIKE YOU!Tsk tsk, do you really talk with your mom with
that mouth? Like I said.. don't talk about Hell, which
according to you, was given (haha) by the Indo-Aryas
(double haha)... maybe what you call heaven might turn out to
be Hell? - Just a thought..-Andre>>-Andre