mm-77
==
Say curve Ked quadraticly by (from random sets of
(x,y)in space)parabola A that concave up for x= a to
bparabola B that concave down for x = b to cThe two parabolas
join smoothly at b with continuous ?st degreederivative.At b,
C changes concavity, which suddenly causes the tangent
vectorsto be ?mirrored against the curve. (Graphicly,
instead of beingalways above parabola A, now the tangents are
always below parabola B)This causes problem for me because in
3d, the Normal calculated fromthe Tangent suddenly become
mirrored against the curve. (from the TNBcoordinates view,
everything is now upside down after passing thechange in
concavity).Basicly I need an interpolation method that
prevents the TNB system
===
Well, you sound like you want to
suck Shaq off, and are jackin yourload over getting payton
and malone. I AM FROM hollywood unlike youhillbilly
types.Dont shoot your load too soon. You got the porno on
your webtvsystem I bet, so you cant even claim ignorance.
But claim something,cause your usenet post just sucked rats
ass.Perhaps you would like to suck Ronald Reagans
cock
===
ermmmthis post is very nice. only an artist can
appreciate this kind of work.hahahahahaha> Well, you sound
like you want to suck Shaq off, and are jackin your> load
over getting payton and malone. I AM FROM hollywood unlike
you> hillbilly types.>> Dont shoot your load too soon. You
got the porno on your webtv> system I bet, so you cant even
claim ignorance. But claim something,> cause your usenet post
just sucked rats ass.>> Perhaps you would like to suck Ronald
Reagans cock
===
> Well, you sound like you want to suck Shaq
off, and are jackin your> load over getting payton and
malone. I AM FROM hollywood unlike you> hillbilly types.> >
Dont shoot your load too soon. You got the porno on your
webtv> system I bet, so you cant even claim ignorance. But
claim something,> cause your usenet post just sucked rats
ass.> > Perhaps you would like to suck Ronald Reagans cock>
Dont you love it when someone makes a post like this but
leaves out all the attribution. AOL - still a bastion of
idiocy.-- Go Yankees!Go Lakers!!Go Rangers!Go Giants!! Fire
Fassle NOW!!!This is very upsetting to me (j/k, mostly) but I
just cannot seem to reallyUNDERSTAND why a number with an
irrational exponent exists. I believe itstrue, because
people much smarter than me say it is, but I dont see why.
Howcan 2 ^ pi, for example, be a number? What does 2 ^ pi
MEAN? 2 ^ 0.713 meansthe thousandth root of two ^ 713. But
if the exponent is irrational, isnt
itmeaningless?Its been
explained to me that since you can get closer and closer to
thenumber without reaching it (2^3.14, 2^3.141, 2^3.1415...)
it must exist sinceotherwise it would cause a hole in the
rationals...but how can an utterlymeaningless symbol like
2^pi be a hole? They see a hole, because it doesntEXIST and
they want it to exist. (Like I said, I DO believe them, I
just dontErin
===
Do you know calculus?Take a constant a,
and imagine an object moving on the real line sothat its
position x(t) at time t obeys x(t) = a x(t), with
startingvalue x(0)=1. That is, the velocity is proportional
to the coordinate.[Imagine a race-car driver starting at
marker 1 meter going at speed 1meter per second; when he
reaches marker 10 meters, he should be goingat speed 10
meters per second; when he reaches marker 20 meters, heshould
be goint at speed 20 meters per second; and so on. I used
a=1in this example.]Then the position at time t will be
exp(at). That is the MEANING ofexp(at). If we take the value
a so that x(1) = 2, that is exp(a) = 2,we call that value
a=ln(2). Using this value we have x(t) = exp(at) =2^t for
integer (or rational) t, so take this differential equation
asthe MEANING of 2^t for ALL values of t. If time pi makes
sense, then2^pi is the location of the race-car at that
time.Does that help? If you dont know calculus, perhaps you
will just haveto wait. Every calculus text has a section on
exponential growth anddecay.As an aside: Sometimes people
are confused by COMPLEX NUMBERS asexponents. Well, the same
differential equation method works toexplain them, too.-- G.
A. Edgar http://www.math.ohio-state.edu/~edgar/
===
For a
given group G does there always exist a group H such that
G=Aut(H)?
===
> For a given group G does there always exist a
group H such that G=Aut(H)?Not always. For example there is
no group H such that Aut(H) is acyclic group of order
3.Alireza Abdollahi
===
> Not always. For example there is no
group H such that Aut(H) is a> cyclic group of order 3.Can
you prove this?
===
>> For a given group G does there always
exist a group H such that G=Aut(H)?> > Not always. For
example there is no group H such that Aut(H) is a> cyclic
group of order 3.I guess if Aut(G) is of prime order pthen
G/Z(G) is of order p or 1, and so G is abelian.It is easy to
deal with _?ite_ abelian groups,but does the argument extend
to in?ite abelian groups?If an in?ite abelian group has only
a ?ite number of automorphismsis it necessarily
?itely-generated(in which case it would be easily dealt
with).Im sure this is answered at length in
Kaplanskys
book.-- Timothy Murphy tel: +353-86-233 6090
===
> >>> For a
given group G does there always exist a group H such that>>>
G=Aut(H)?>> >> Not always. For example there is no group H
such that Aut(H) is a>> cyclic group of order 3.> > I guess
if Aut(G) is of prime order p> then G/Z(G) is of order p or
1, and so G is abelian.> > It is easy to deal with _?ite_
abelian groups,> but does the argument extend to in?ite
abelian groups?> If an in?ite abelian group has only a ?ite
number of automorphisms> is it necessarily ?itely-generated>
(in which case it would be easily dealt with).Most abelian
groups have an obvious automorphismof order 2 :-)-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
===
> > >> > suppose H is a normal subgroup in G and
K is normal in H. Of course> > then K is a subgroup of G.> >
But is K normal in G? Proof or counterexample!> >> > This
problem occured to me some time ago, I dont have time to
?ure> > it out for myself these days so thanks for help!> >
May be that Ive overlooked something.> > But, as K is normal
in H it follows that> forall kin K ==> kin H> > The only
thing we gotta check now is whether> g^{-1}kg in G> but as
this holds for every h in H by hypothesis and -- by the
statement> above -- every k lies H it follows that K is also
normal in G.You have to check that g^{-1}kg is in K, not G.
The statement is not necessarily true. Consider S_4, the set
ofpermutations of four elements. Let G = A_4, the alternating
group ofS_4 (i.e. the even permutations). Let H = {1,
(12)(34), (13)(24),(14)(23)}, the Klein four-group. Then H is
normal in S_4 and hence inA_4.Let K = {1, (12)(34)}. Then K is
normal in H.However, K is not normal in A_4, since conjugating
(12)(34) by (123)gives(13)(24) which is not in K.Rob
===
In
Cantors Diagonal Argument reals are listed by their
decimalrepresentations. However, if the digits in a decimal
representationof a real are denoted d_1, d_2...d_n, then the
set of numbers that comprise the subscripts of d are
integers, and therefore are countably in?ite. Assuming the
validity of uncountablty in?ite, a countably in?ite set of
subscript numbers is not comprehensive; there must exist
decimal places greater than the nth place where nis an
integer. As the value of a given real is completed in
acountably in?ite number of digits, any remaining digits
must bezeros. If all reals have some point after which the
remaining digitsare zeros, diagonalization no longer
works.
===
What argument?> In Cantors Diagonal Argument reals
are listed by their decimal> representations. However, if the
digits in a decimal representation> of a real are denoted
d_1, d_2...d_n, then the set of numbers that> comprise the
subscripts of d are integers, and therefore are> countably
in?ite.OK> Assuming the validity of uncountablty in?ite,?>
a countably in?ite set of subscript numbers is not
comprehensive;comprehensive?> there must exist decimal places
greater than the nth place where n> is an integer.Hmmm.
This sounds like the dreaded there are in?itely many
integersdo there are in?ite integers argument.What do you
mean here?That for each positive integer n there is a decimal
place beyondthe n-th? (yes there is, for instance, the
(n+1)-th).or, that there is a decimal place beyond the n-th
for all positiveintegers n? (no, the decimal expansion is a
function from N to {0,..,9})> As the value of a given real is
completed in a> countably in?ite number of digits, any
remaining digits must be> zeros.Aha, you did mean the second
after all! But there are no remaining digits.> If all
reals have some point after which the remaining digits> are
zeros,They dont.> diagonalization no longer works.?-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of
Gentlemen
===
> In Cantors Diagonal Argument reals are listed
by their decimal> representations. However, if the digits in a
decimal representation> of a real are denoted d_1, d_2...d_n,
then the set of numbers that > comprise the subscripts of d
are integers, and therefore are > countably in?ite. Assuming
the validity of uncountablty in?ite, > a countably in?ite
set of subscript numbers is not comprehensive; > there must
exist decimal places greater than the nth place where n>
is an integer. As the value of a given real is completed in a>
countably in?ite number of digits, any remaining digits must
be> zeros. If all reals have some point after which the
remaining digits> are zeros, diagonalization no longer
works.A decimal representation consists of a countable number
of integers in the range 0..9. If you like you can think of a
decimal representation as a function N->{0,1,2,...9} where N
is the natural numbers.Its not clear what you mean when you
say that a countably in?ite set of subscript numbers is not
comprehensive. It is true that for any natural number n there
are decimal places past n. But each decimal place is indexed
by a natural number.
===
>In Cantors Diagonal Argument reals
are listed by their decimal>representations. However, if the
digits in a decimal representation>of a real are denoted d_1,
d_2...d_n, then the set of numbers that >comprise the
subscripts of d are integers, and therefore are >countably
in?ite. Assuming the validity of uncountablty in?ite, >a
countably in?ite set of subscript numbers is not
comprehensive; >there must exist decimal places greater than
the nth place where n>is an integer. Uh, right. Just like
if the idea of in?ite sets is valid then everyset must be
in?ite.Hmm.>As the value of a given real is completed in
a>countably in?ite number of digits, any remaining digits
must be>zeros. If all reals have some point after which the
remaining digits>are zeros, diagonalization no longer
works.************************David C. Ullrich
===
> In
Cantors Diagonal Argument reals are listed by their decimal>
representations. However, if the digits in a decimal
representation> of a real are denoted d_1, d_2...d_n, then
the set of numbers that > comprise the subscripts of d are
integers, and therefore are > countably in?ite. Assuming the
validity of uncountablty in?ite, > a countably in?ite set of
subscript numbers is not comprehensive; > there must exist
decimal places greater than the nth place where n> is an
integer. Your last statement is incorrect. There are
uncountably many realnumbers, but there also are uncountably
many decimal digit sequences d_1,d_2, ..., where each such
sequence is a countable collection of digits.Therefore, there
are plenty of digit strings to go around, with no needfor
uncountably-long strings to represent the real numbers.>As
the value of a given real is completed in a> countably
in?ite number of digits, any remaining digits must be>
zeros. If all reals have some point after which the remaining
digits> are zeros, diagonalization no longer works.Your
premise doesnt hold. There are no remaining digits.-- Dave
SeamanJudge Yohns mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>
===
> In Cantors Diagonal Argument reals are
listed by their decimal> representations. However, if the
digits in a decimal representation> of a real are denoted
d_1, d_2...d_n, then the set of numbers that > comprise the
subscripts of d are integers, and therefore are > countably
in?ite.I follow you up to here.> Assuming the validity of
uncountablty in?ite, > a countably in?ite set of subscript
numbers is not comprehensive; > there must exist decimal
places greater than the nth place where n> is an
integer. As the value of a given real is completed in a>
countably in?ite number of digits, any remaining digits must
be> zeros. ... a bit fuzzy but...>If all reals have some point
after which the remaining digits> are zeros, diagonalization
no longer works.This point does not exist. Moreover, there
are countably in?itesets such as the rationals for which the
above statement isnt trueeither.
===
> > >In Cantors Diagonal
Argument reals are listed by their decimal> >representations.
However, if the digits in a decimal representation> >of a
real are denoted d_1, d_2...d_n, then the set of numbers that
> >comprise the subscripts of d are integers, and therefore
are > >countably in?ite. Assuming the validity of
uncountablty in?ite, > >a countably in?ite set of subscript
numbers is not comprehensive; > >there must exist decimal
places greater than the nth place where n> >is an
integer. > > Uh, right. Just like if the idea of in?ite sets
is valid then every> set must be in?ite.> > Hmm.Well, a
number such as 1.0 _could_ be expressed in a ?ite number
ofplaces, but because of the validity of carrying out
calculations pastthe zero (in?itely so), this number is
thought to have an endlesstrail of zeros.I am asking: if
uncountably in?ite is valid, then in the same way we require
a (countably) in?ite number of decimal places even if the
_value_ of a number can be expressed in a ?ite number of
digits,why dont we require an uncountably in?ite number of
decimal places even though any numbers _value_ will be
expressed in a countably in?ite amount of places? (allow
calculations to continue as the new size permits....)dg> >As
the value of a given real is completed in a> >countably
in?ite number of digits, any remaining digits must be>
>zeros. If all reals have some point after which the
remaining digits> >are zeros, diagonalization no longer
works.> > ************************> > David C. Ullrich
===
I
have answered the following in response the OTHER place where
you posted the same comment. To post the same comment twice
under two different subject headings, presumably in the hope
that one of the postings will remain unanswered and thus
appear to be unanswerable, is very bad form.Since I have
already given an answer to the other identical posting that
you made, I wont bother here, except to point out that I
have made an extra remark at the end.>> >1. There exists two
point masses moving towards each other.>> >2. At time T they
collide at point P.>> >3. At time T two simultaneously paired
opposite forces of equal magnitude occur>> >at point P.>> >>
This is a contact force. On a side note, in Special
Relativity, all >> forces are contact forces.>> >> >4.
________ >> >Fill in the blanks.>> >> Point mass 1 exerts a
force on point mass 2, and point mass 2 exerts the >> paired
force on point mass 1. The motion of point mass 1 is affected
by >> the force exerted on it by point mass 2. Since the
force exerted on point >> mass 2 by point mass 1 is not on
point mass 1, then it has no effect on >> the motion of point
mass 1. While you continue to consider that there is >> no
effect of two paired contact forces on a system because they
balance, >> then you are not doing Newtonian Mechanics,
because you are making an >> assumption which is
contradictory to Newtonian Mechanics (speci?ally,>> it is
contradictory to Newtons Second Law of Motion).>You still
have it all wrong.>POINT mass A has mass Ma.>POINT mass B has
mass Mb.>At time t, a collision occurs at POINT p.>So at time
t we observe mass A and mass B existing simultaneously
at>point P. An equivalent observation would be that there
exists a SINGLE>POINT>mass C with mass Ma+Mb existing at
POINT P at time t.>Also at time t, we have the paired force
as predicted by Newtons 3rd>law. Given that there is a single
POINT mass at point p, then the>superposition principle
applies and the paired forces cancel each>other out as
predicted.>Do you see the problem yet?>> >> Newtons Three
Laws of Motion:>> >> 1. If a body has no forces acting on it,
then it either remains >> stationary or it moves
uniformly.>Empirical evidence implies otherwise.>> 2. The
time-derivative of the momentum of a body is equal to the >>
sum of the forces which are exerted on the body.>Contact
forces can never exist as they are always cancelled out at
the>point of contact.>> 3. Forces are paired in such a manner
that the forces in a pair >> are equal in magnitude and
opposite in direction. The two forces>> in a pair are caused
by the same mechanism. The same body >> experiences one of
the forces and exerts the other, so that if>> one force in a
pair is exerted on body A by body B, then the >> other force
in the pair is exerted on body B by body A.>The problem with
Newtons laws is that they do not de?e what a body>is and
isnt. Furthermore, Newtons laws do not prohibit the
existence>of POINT>masses and - as ive clearly shown - fails
to predict their behaviour>(unless logical fallacies
constitute strong evidence in academic>circles).As I
pointed out when I responded to your other ideration =
7.5058959(93) x 10^87 rad/s^2> 139) thermal transfer constant
= 6.2272531(21) x 10^88 kg/s^3-K > 140) current density
constant = 7.2263839(39) x 10^92 A/m^2 > 141) gravitational
density = 8.2089700(31) x 10^95 kg/m^3 > 142) energy density
= 7.3778543(28) x 10^112 kg/m-s^ 2 > 143) radiance constant =
3.2280937(18) x 10^119 kg/s^3-sr> 144) irradiance constant =
2.2118250(84) x 10^121 kg/s^3