mm-98
===
Invariant - something that does not vary. YesCovariant
- if one changes the other changes.>Contravariant - if one
increases the other decreases.> Simple but am I close?> No,
not even in the right ball park.> Co/contra variant is about
how vectors and tensors transform between> coordinates.Are
the following covariant:sin(pi/2 - x) = cos(x) tan(pi/2 - x)
= cot(x) sec(pi/2 - x) = csc(x) <40095f1f_2@127.0.0.1>
===
Invariant - something that does not vary.>Yes> Covariant -
if one changes the other changes.>> Contravariant - if one
increases the other decreases.>> Simple but am I close?>No,
not even in the right ball park.>Co/contra variant is about
how vectors and tensors transform between>coordinates. Are
the following covariant: sin(pi/2 - x) = cos(x)> tan(pi/2 -
x) = cot(x)> sec(pi/2 - x) = csc(x)>No no. Those equations
are identities. Theyre arent even
invariants.Constants are
invariants. The length of a line segment over
rigidtransformations is an invariant. The end points of a
line seqment arentinvariant over rigid transformations. 0
and 1 are invariants oversquaring. 0 and -1 are invariants
over cubing.conclusion.
===
> Co/contra variant is about how
vectors and tensors transform between>>
coordinates.examples?
===
>Invariant - something that does not
vary.>> Yes>>Covariant - if one changes the other
changes.>>Contravariant - if one increases the other
decreases.>>Simple but am I close?>> No, not even in the
right ball park.>> Co/contra variant is about how vectors and
tensors transform between>> coordinates.> Are the following
covariant:> sin(pi/2 - x) = cos(x)> tan(pi/2 - x) = cot(x)>
sec(pi/2 - x) = csc(x)> No no. Those equations are
identities.how to interpret it.>Theyre arent even
invariants.> Constants are invariants. The length of a line
segment over rigid> transformations is an invariant. The end
points of a line seqment arent> invariant over rigid
transformations.Ok. I understand that.>0 and 1 are invariants
over> squaring.0*0 = 01*1 = 1>0 and -1 are invariants over
cubing.0*0*0 = 0-1*-1*-1 = -1
===
>Invariant - something that
does not vary.>> Yes>>Covariant - if one changes the other
changes.>>Contravariant - if one increases the other
decreases.>>Simple but am I close?>> No, not even in the
right ball park.>> Co/contra variant is about how vectors and
tensors transform between>> coordinates.> Are the following
covariant:> sin(pi/2 - x) = cos(x)> tan(pi/2 - x) = cot(x)>
sec(pi/2 - x) = csc(x)> No no. Those equations are
identities. Theyre arent even invariants.> Constants
are
invariants. The length of a line segment over rigid>
transformations is an invariant. The end points of a line
seqment arent> invariant over rigid transformations. 0 and 1
are invariants over> squaring. 0 and -1 are invariants over
cubing.
conclusion.http://mathforum.org/library/drmath/view/54027.
html <40095f1f_2@127.0.0.1>
<20040117220343.M35539@agora.rdrop.com>
<400aa7b6_2@127.0.0.1>
===
>>> Covariant - if one changes the
other changes.>>> Contravariant - if one increases the other
decreases.>> Are the following covariant:>> sin(pi/2 - x) =
cos(x)>> tan(pi/2 - x) = cot(x)>> sec(pi/2 - x) = csc(x)>No
no. http://mathforum.org/library/drmath/view/54027.html>Bah,
read it carefully. He made unique to him slang phrase co
variantwhich means something about sin & cos, tan & cot, sec
& csc.
===
[.snip.]>The theorem correctly states that, using
these standard de?itions and >>in ring of algebraic
integers, or any other ring with identity, if ?f >>divides
?a*b and is coprime to ?a then ?f
divides ?b.I agree with
you, but as Bill Dubuque has pointed out many times,>coprime
does have other ?standard meanings (though the one you
use>is, as far as I am aware, the ?gold standard in
algebraic number>theory). We know that James uses coprime to
mean any common>divisors are units; in the ring of algebraic
integers the two notions>are equivalent, but that is a hard
fact to establish. It is, of>course, provable; but it ->is<-
hard to do so. > Its funny that so many posted as if
something were so obvious only to> have Magidin come in and
claim that its not.> Since you are now arguing about whether
the result is obviousor not, I assume that you agree that the
result is true. [1]Where were we...There are two very similar
statements. The ?st, Lemma 1, is false.The second, Lemma 1
is true. The only difference is thatin Lemma 1, s and t are
allowed to vary with m. It is, however,interesting to note
that Lemma 1 does hold if g_1 and g_2are polynomials. The
importance is that James assumes thatLemma 1 holds for
non-polynomial factors [2]. This, not surprisinglyleads to a
contradiction. Lemma 1: Let P(m) be a polynomial with
coef?ients in A, the algebraic integers. Let each coef?ient
be divisible (in A) by the algebraic integer f. Let P(m) =
g_1(m) * g_2(m), with g_1(m) and g_2(m) functions from A to
A. Then there exist s and t in A such that s*t=f; and for all
m in A, s divides g_1(m) and t divides g_2(m).Lemma 1: Let
P(m) be a polynomial with coef?ients in A, the algebraic
integers. Let each coef?ient be divisible (in A) by the
algebraic integer f. Let P(m) = g_1(m) * g_2(m), with g_1(m)
and g_2(m) functions from A to A. Then there exist s(m) and
t(m) in A such that s(m)*t(m)=f; and for all m in A, s(m)
divides g_1(m) and t(m) divides g_2(m). - William Hughes[1]
Actually, no one has claimed that this speci? result (the
equivalence between two de?itions of coprime in the
algebraic integers) is obvious. However, the result is very
well known to James. When I stated that the issue hascome up
at least once before I was being ironic. The issuehas come up
many times.[2] The fact that the ?st thing that so many
people say after looking at James work is why do you use
results at 0 to make claims for divisiblity not at 0 does not
seem to bother James at all. (I think he suspects a
conspiracy.) James has not offered any proof of his claims;
he considers them obvious. He has given a number of examples,
all of them involving polynomials.
===
Well, you can extract
bezier curves out of edges of a patch, but you canjust as
well extract them anywhere on the surface (well, not
exactlyanywhere) to produce isoparms. Bezier patch is de?ed
as a tensor productof two bezier curves and that produces 16
control points.Googling for tensor product and bezier patch
should yield promising results.> hi all, I assume that a
bezier patch has 16 control points and 4 corners control>
points are attached onto the surface. It implies that there
are 4 bezier> curves on the edges of the patch. My question
is how can the middle 4> control points change the shape of
the surface?> newbie...>hi guys,2^(x+1)+2^(3-x)=17logging
gives(x+1)lg2+(3-x)lg2=lg17expanding
giveslg2x+lg2+3lg2-lg2x=lg17but lg2x-lg2x=0.....goinn
mad...graphed it so i know x=3 but cant get
therenumerically.....please help i know im missing something
simpleshaun
===
<20040118120121.28470.00000202@mb-m15.aol.com>:
>2^(x+1)+2^(3-x)=17logging gives(x+1)lg2+(3-x)lg2=lg17No.
log(3+7) =/= log(3) + log(7)-- Stan Brown, Oak Road Systems,
Cortland County, New York, USA http://OakRoadSystems.com/For
many Americans the danger of tyranny lies not in government
butin employers or insurance companies or HMOs, from which we
needgovernment to protect us. To say that any worker is free
to escapean oppressive employer by getting another job is
about as realisticas to say that any citizen is free to
escape an oppressive govern-ment by emigrating. -- Steven
Weinberg, in /Facing Up/ (2001)
===
Hey to all, I need to
solve the following equation, but not using trial anderrorany
ideas?(1.2)^n > 1.6 + 0.4nI guess I fergot my basicsDoug
===
I
dont think this will help you (ProductLog). If n is assumed
to beReal...equ = 1.2^n > 1.6 + 0.4*nIncrease its
precission...equ = Rationalize[equ](6/5)^n > 8/5 +
(2*n)/5List @@ Reduce[equ, Reals]n < (-4*Log[5] + 4*Log[6] +
ProductLog[(3125*Log[5] -3125*Log[6])/2592])/(Log[5] -
Log[6])andn > (-4*Log[5] + 4*Log[6] + ProductLog[-1,
(3125*Log[5] -3125*Log[6])/2592])/(Log[5] -
Log[6])Numerically...N[%]{n < -2.3801395955782705, n >
9.073517983127845}-- Dana> Hey to all, I need to solve the
following equation, but not using trialand> error any ideas?
(1.2)^n > 1.6 + 0.4n I guess I fergot my basics Doug
===
> Hey
to all, I need to solve the following equation, but not using
trial and> error> any ideas?> (1.2)^n > 1.6 + 0.4n> I
guess I fergot my basics> Doug> Firstly, it is not an
equation unless you use > to mean is equal to.it should be
called an inequality.Secondly, such equations or
inequalities, with unknowns as exponents as well as in linear
expressions, generally do not have exact solutions, but must
be found by approximation methods.
===
>Hey to all, I need to
solve the following equation, but not using trialand>error>
any ideas?> (1.2)^n > 1.6 + 0.4n> I guess I fergot my basics>
Doug> Firstly, it is not an equation unless you use > to
mean is equal to. it should be called an inequality.
Secondly, such equations or inequalities, with unknowns as
exponents as> well as in linear expressions, generally do not
have exact solutions,> but must be found by approximation
methods.One such method that I would use for this is the
iteration method.1. Rearrange to give an equation with n= on
one side in terms of n on theother. E.g. n=((1.2)^n -
1.6)/0.42. Now add iterative subscripts. I usually use n for
the subscripts - butsince this is the variable, well use p
instead. n(p+1) = ((1.2)^n(p) - 1.6)/0.43. Now set n(1) to a
reasonable guess (say 1) and ?d n(2) using thisformula. Now
feed this into the equation again to give n(3), n(4), ....
etc.The equation converges to a value of about -2.38.This
will only give you an approximation though - unless of course
you couldrepeat the process to ?d n(in?ity).Also, Virgil was
right - this is an inequality! This method is just givingone
value for which the two sides are equal! If you want the
other, justrearrange the original equation to give n= in
terms of the other n.Hope this helpsVassagoP.S. I know about
initial conditions affecting convergence etc. but haveskipped
over this for the sake of simplicity. If anyone else feels
they wantto expand on the issue - feel free.
===
For n in
[-2.38014;9.07352] is the equation wrong...PS: I dont really
like cross-postingMax> Hey to all, I need to solve the
following equation, but not using trialand> error any ideas?
(1.2)^n > 1.6 + 0.4n I guess I fergot my basics Doug
===
> For
n in [-2.38014;9.07352] is the equation wrong...> PS: I
dont really like cross-posting> Max> >Hey to all, I need
to solve the following equation, but not using trial>
and>error> any ideas?> (1.2)^n > 1.6 + 0.4n> I guess I fergot
my basics> Doug>Cross posting beats multiple postings all
hollow.It isnt an equation.I reqally dont like top
posting.
===
Use misc.test -- thats what its there for.--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.comThoroughness. I always tell my
students, but they areconstitutionally averse to painstaking
work. -- Emma Thompson, in /Wit/ (2000)