argh
what is D
what is the directional derivative of something in the direction of (1,0,0) ?
f is a function R^n -> R and Df is also a map R^n -> R
at the point
Df(v) = directional derivative of f in the direction v
i defined D as i meant it above
Yeah, Df(x)(v)
i gave the direction
Ok
what is the directional derivative of f in the direction c'(0) = (1,0,0) ?
Sorry
it's @f/@x^1, hey, D is just @/@x^1
@f/@x^1
Ok, that shows that D = @/@x^1 but how is D = (1,0,0)
in other words the curve c begets a tangent vector c'(0) and this corresponds to the operator 'directional derivative in the direction c'(0)'
drfurious, i didn't say D = (1,0,0)
c'(0) = (1,0,0)
--> _Virginia (mIRC@ppp220219.fx.ro) has joined #Mathematics
in other words i gave an isomorphism between the two common definitions of tangent spaces
with (1,0,0) -> @/@x^1
<-- _Virginia (mIRC@ppp220219.fx.ro) has left #Mathematics
(a_1, ..., a_n) <-> sum a_i @/@x^i...
yes
<->
I see
<-- Decoy has quit (Dead socket)
Not =
--> Decoy (hidden-use@adsl-63-195-149-246.dsl.scrm01.pacbell.net) has joined #mathematics
<-- drumzz has quit ()
So dc/dt = dx^i/dt @/@x^i or dc/dt <-> dx^i/dt @/@x^i ?
so if c is a curve dc/dt at a point = (c_1', ..., c_n') at that point = dc_i/dt @/@x^i
i would say =
dc/dt is a tangent vector, it can be viewed in either 'view' of tangent vectors
But dc/dt = (dx^1/dt,dx^2/dt,dx^3/dt) = dx^i/dt e_i
dc^i/dt i would say first
and second of all, no
So if you say equal, you are saying e_1 = (1,0,0) = @/@x^i
dc/dt is the tangent vector defined by c
dc/dt is not the rate of change of c wrt to t?
meaning you can view it either as a differential operator or as a plain old vector
even tho this distinction is prolly semantic
drfurious, it's that too
in our case, we are talking about tangent vectors, so it's a tangent vector
The time rate of c wrt t is the same as a differential operator?
err
but i prolly wouldn't have problems writing = even between the different notations as long as the isomorphism has been established
The rate of change of c wrt t is the same as a differential operator?
i mean it's like (x_1, ..., x_n) <-> x_i e_i
drf, yes, that's what we just established
Hmm
What I saw that we established is that "rate of change of c wrt t" <-> "a differential operator"
But <-> is not the same as =
--> muto (afish@1Cust9.tnt12.sjc4.da.uu.net) has joined #mathematics
no
that's not what you say
saw
you saw that the two ways of viewing tangent vectors were <->
dc/dt is just a tangent vector
it can thus be viewed in either realm
and second of all, the distinction between <-> and = is semantic, once the isomorphism has been established
in fact, the differential operator view is much nicer
so anyways... dc/dt at a point = dc^i /dt @/@x^i is i think the relation you were looking for
It is straightforward to show that dc/dt = (dc^1/dt,dc^2/dt,dc^3/dt)
yes, and the directional derivative in the direction v is just (@/@x^1, ., @/@x^3) . v
dc/dt = (dc^1/dt,dc^2/dt,dc^3/dt) = dc^i/dt e_i = dc^i/dt @/@x^i ==> e_1 = @/@x^1 = (1,0,0)
yes
So @/@x^1 = (1,0,0)
yes
again, directional derivative in the direction (1,0,0) is @/@x^1
DrFurious] Ok, that shows that D = @/@x^1 but how is D = (1,0,0) drfurious, i didn't say D = (1,0,0)
i didn't
but once an isomorphism is established i can replace <-> with = as freely as i want to
But D = @/@x^1 and @/@x^1 = (1,0,0) so D = @/@x^1
if you feel uncomfortable, take every = as <->
err
But D = @/@x^1 and @/@x^1 = (1,0,0) so D = (1,0,0)
and you must know the difference between 'i didn't say...' and 'it isn't true that...'
--> reuben (r@r154-104-dsl.sea.lightrealm.net) has joined #mathematics
Ok
--- Hurwitz gives channel operator status to reuben
Now I'm confused even worse :)
In everything you said, I can replace <-> with =?
sure
Or vice versa
or = with <-> if you want
Ugh
That is nasty
the distinction is entirely semantics, but if it's helpful to you...
So I understand that @/@x^1 <-> (1,0,0)
ok
and thus dc/dt <-> dc^i/dt @/@x^i
And I understand that dc/dt <-> dc^i/dt @/@x^i
But I want to know what exactly dc/dt equals
then replace <-> with =
it depends on which view you want to take
That is nasty
they are all equivalent
no it's not
isormorphic is not = and I will never write = to mean isomorphic
the only possible nasty thing is the abuse of notation...
that's exactly the point, it is an abuse of notation
eric, this is getting silly. (SxS)xS is merely isomorphic to S^3, yet most mathematicians write (SxS)xS = S^3 without incident
using dc/dt to mean different notions, but they do that for a reason, the calculus 'rate of change' is easier to understand, and the derivation is the most useful, but is less so
drf, R^2 x R is only isomorphic to R^3
at some point you have to bite the bullet and say "okay those are equal"
if you can't tell them apart, is there really a point to trying to distinguish them with notation
Exatly, so I would never write (S x S) x S = S^3 or R^2 x R = R^3
then you are overly pedantic
but there are many arguments (I don't have any offhand, but there are) where preserving the difference between (SxS)xS and S^3 simply adds unnecessary, confusing, useless, and altogether harmful work
I think you'd have a hard time finding a textbook that has (S x S) x S = S^3
and you should not ask what 'something equals' without picking one of the several equivalent notions of that
drf, if S = R, that's all over the place
I can find plenty of textbooks with worse abuses of notation than that
you are making it harder for yourself by trying to distinguish with notation what are indistinguishable objects
I do not want to take S^3 = functions from 3 to S, then SxS = ordered pairs {{a},{a,b}} of elements of S, (SxS)xS = ordered pairs of SxS and S, and work back from one to the other, to one to the other, every single time I want to switch
You could even simply define S x S x S := {(a,b,c) | a,b,c in S}
drf, that is yet another nonequal isomorphic definition
you prove the correspondence once, and then you use it. Freely. Transparently. As often as needed.
Huh? I though S x S := {(a,b) | a,b in S}
drfurious, obviously if you want to make such a distinction, and you can, then you can't ask what 'dc/dt' 'equals' until you specify what notion of dc/dt you wish to use, since mathematicians consider equivalent things, well uh, equivalent
(SxS)xS = {((a,b),c)}
and you're not even being honest here, since (a,b) = {{a},{a,b}}
I guarantee you that if you fully resolve every detail, this will get out of control fast
(SxS)xS isn't equal to Sx(SxS) or SxSxS or S^3, pedantically speaking
nevertheless they are naturally isomorphic, so you prove the isomorphism, and then use it
freely
"naturally" is important.
I thought S^n := S x .... x S (n copies)
DrF, S^n = set of functions from n to S
S^2 = set of functions from 2 to S
where 2 = {0,1}
or S^n = S x (S^((n-1)
S x S = ordered pairs of elements of S
where an ordered pair is (a,b) = {{a},{a,b}}
another isomorphic defn :)
an ordered triple is (a,b,c) = {{a},{a,b},{a,b,c}}. This concept is again nonequal to S^3 = functions from 3 = {0,1,2} to S
do you see my point? All these are nonequal, yet people move between them freely
Galois I see the point you are trying to make, but I don't know if I'll swallow these examples. There seems there should be some ='s in all of that.
you want me to write down what all of these are?
S^3 = functions from 3 to S. 3 = {0,1,2}. A function is a subset of {0,1,2} x S.
so it's a set of elements of the form {{x},{x,s}} where x=0 or x=1 or x=2
that's S^3
let's look at SxSxS
an element of this is {{a},{a,b},{a,b,c}}
quite distinct from a function in the previous case
If you have S^3 != S x S x S, then it could be that the definitions were not proper
now let's look at (SxS)xS
something in SxS is {{a},{a,b}}
cross that with S, you get {{{{a},{a,b}}}, {{{a},{a,b}},c}}
Sx(SxS) is yet again different
I know (S x S) x S != S x (S x S)
all four of S^3, SxSxS, (SxS)xS, and Sx(SxS) are different sets. I have in fact described all four.
drf, ok, let's take the proper defn, the tangent space is differentials, and let's defined dc/dt at a point = dc^i/dt @/@x^i
But I think S^3 should = S x S x S or else the definitions are simply screwed up
heh
drf, there are two competing definitions which I have described in full detail
each is equally valid, but they are not equal
nevertheless, people do use both.
True, but whose to say the definitions aren't screwed? :)
the tangent vector situation is exactly identical
the derivations definition, and the M/M^2 definition, and the actual vector definition for manifolds in euclidean space
>galois< hehe, you wanna argue that he isn't irritating now? :)
three definitions. Nonequal. All isomorphic, and canonically so.
People move between them without pause.
People move between them without pause once they understand all three and how each is related
ok then
yes!
and my point is, do this first!
then worry about the chain rule
FIRST be able to move between them without pause
THEN try proving theorems
I'm not saying you can't move between them, I am just not confortable doing so and there should be a way to do things with all ='s without having to resort to different notions
not the other way around
don't try to prove a theorem the hard way deliberately as an excuse to avoid having to get comfortable with the definitions
I just think I should be able to pick a notion and stick with it and everything should follow as equalities
I guarantee you it is easier the other way around. 1) First get comfortable with the equivalences. 2) prove the theorem using those equivalences.
3) if at that point, and only at that point, you want to go back and do it the hard way, then do so
but at least at that point you know what you're doing
drf, that is a nice fantasy, but it doesn't play out in real life
the whole reason there are five different definitions is that they are all useful
Not even for this basic stuff?
drf, and you sometimes can, sometimes the different notions are actually useful, sometimes it's helpful to use one definition to prove Theorem A, and an equivalent one to prove Theorem B
even for a theorem as silly as the chain rule, you want all of them available
eric?
Yeah?
--> Althea- (althea@dialup-63.214.198.252.Dial1.Philadelphia1.Level3.net) has joined #mathematics
thea wanted to say hi to you :)
'lo peeps
Althus!!
Erikah!!! :)
You in Phillie?!
ayup :)
Wow :)
hmm
cincinatti???
How's life?!!
You've got any news?
's ok :) :) how have you been??
No, I'm in Chicago
I choose to define S^3 as functions from 3 to S. This definition is useful, though I do not defend that point here. But an element of S^3 under this definition is a set {(0,a),(1,b),(2,c)}
Dunno why it says cincinnatti :)
keeping in mind (x,y) = {{x},{x,y}}
{(0,a),(1,b),(2,c)} is naturally isomorphic to, but not equal to, (a,b,c)
Eric, i 'graduated' on monday :)
Whoa! Congrats! :)
Anything else?
yur, except i've still got 2 summer courses AND i didn't go to my own graduation!
Have a kid or anything? :)
(didn't tell my parents either) :)
no, david didn't want any :P
* Althea- *ducks*
hehe :)
David. Get with the program! :)
hmm
So you need to fly back to england this summer to finish up?
--> decimal (guh@user-vcauvk8.dsl.mindspring.com) has joined #mathematics
i'm picking up something... althea + galois?
nooo...i'm finishing up in filthy-delphia, then going back to england
or different david?
--- Evariste gives channel operator status to decimal
david=galois
yus
Althus wishes :)
She goes for the brainy type
yur, which would explain why i had hots for *you* once :P
hehe yeah right
so, a tangent vector is 1) a derivation on germs of functions, 2) a functional on M_x/M_x^2, 3) a choice of tangent vector of a chosen curve in a coordinate chart
You just loved me for my body
mathgirls = evil
<-- jrt has quit (Read error: 54 (Connection reset by peer))
ALthus isn't a mathgirl, shes a math groupie
Or should I say, math aid
:)
math gropie more like :P
hehe :)
Keep your hands to yourself young lady!
Can't leave you alone for one minute :)
*laugh*
that's not what i meant :P
but i suppose i brought it upon myself with my comment :P
Galois I'm familiar with 1) and 3) but not 2)
*grope*
it's fairly easy to go between 1) and 2)
Maybe, if I knew what M_x was :)
M_x = functions which are 0 at the point x
the only difference between 1) and 2) is that 2) requires the function to have zero value at x
this is easily arranged, by subtracting off a constant
I guess going between 1) and 3) is not so easy for me
for functions which are 0 at the point x, leibniz rule is particularly simple: (fg)' = f g' + g f', but f=g=0 at the point, so (fg)' = 0
so it just says the ideal (M_x)^2 maps to 0
(
geocities.com/siliconvalley/park) (
geocities.com/siliconvalley)