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23:00
then I claim its a field
not sure about that or anything
anyway
$D:F(\Bbb R)\to F(\Bbb R):x\mapsto x'$ is a surjection
@Alizter Isn't $\mathbf{Hom}(\Bbb R, \Bbb R)$ in general a field?
Then I consider $F(\Bbb R)/\Bbb R$
@BalarkaSen I have no clue
i.e., functions from R to R?
no, of course it's not.
nor is F(R) one.
$f(x) = x^2$ is differentiable. The inverse $\sqrt{x}$ is not a function
@BalarkaSen Who said function composition was an operation?
@Alizter your operation is composition right?
oh darn
23:04
addition and multiplication
my stuff works up to here
Ah. then $\mathbf{Hom}(\Bbb R, \Bbb R)$ is a field too.
We could probabalyly generalise later
i.e., just functions. no differnetiability.
Anyway considering the cosets
@Alizter OK, continue.
23:05
?
Bascially meaning that $f+c=f+a$ even $c\ne a$
with me? @BalarkaSen
$f+c=f+a$
@BalarkaSen For $a,c\in\Bbb R$
this is getting confusing. continue from the beginning, @Alizter
in a formalized way
also assuming that $\Bbb R \subset F(\Bbb R)$
@BalarkaSen I will need to decipher the rest
Ill formalise for tomorrow
23:07
what were you doing? F(R) \to F(R) right? what's the map?
differentiation
f(x) goes to f'(x)?
@Alizter So that F(R) of yours is functions from R to R with infinite differentiability, not just differentiability?
23:09
No.
those are functions
then $f \mapsto f'$ is not defined.
however in a way such that $\Bbb R$ is a subset
consider a function f which is differentiable but not twice differentiable.
f' isn't in F(R)
so f \to f' is not defined.
yeah there are so many holes in this
@BalarkaSen I would stick to polynomial functinos
so you need F(R) to be elts in Hom(R, R) and infinitly differentiable.
@Alizter Then you just have R[X]
And no field.
23:11
@BalarkaSen Told you its a bunch of crap
why not let F(R) to be functions which are infinitely differentiable?
Or just consider something completely different
A limit for the nth derivative of a funciton :)
ok, so what next? you consider maps from F(R) to F(R). then?
(F(R) is now redefined with infinite differentiability)
$\displaystyle f^{(n)}(x)=\lim_{\Delta x\to0}\frac1{\Delta x^n}\sum^n_{k=0}\binom{n}{k}(-1)^{n-k}f(x+k\Delta x)$
that is one of my nicer findings
proven and all
yeah, that should be obvious
23:14
@BalarkaSen But yeah I think I was trying to set D up as an automorphism
around a field
then extend it with some fnuctions or something
GALOIS
D means differentiation?
stuff
@BalarkaSen Yes I am lazy
ok so F(R) \to F(R) is an automorphism defined as f \to Df.
what next?
it is an automorphism.
Pretend F(R) is the least set such that it is a field
uh oh. not quite.
23:16
then I wanted to study extensions
it's the differential operator.
indeed you are kinda getting at differential extensions here.
keep it up and you'll get to differential galois theory quickly.
in particular some interesting galois groups that arise from D over the polynomials extended with logarithms or some stuff liek that
@Alizter oh?
how do you define field extensions here?
just the usual notion?
but what about the differential?
yes
logarithmic polynomials?
so you are extending something by logarithms.
23:18
Yeah that would be ncie
right it makes sense.
maybe extending rational functions would make more sense
@Alizter you have the field of rational functions over R.
@Alizter exactly what i was trying to tell
R(X) be your field. you are extending by logarithms. maybe call it R(X, log)? can you preserve differentials?
think about it. you'll make much more sense if you can develop this idea.
in fact, i'd ask you to forget about R. think about C.
Difficult
well, C has much more structure.
especially in the sense of analytic structure.
23:25
Well one preservation would be the solution of solving $f'+1/x=f$?
smashes head on table
I MUST SLEEP @BalarkaSen
YOU TOO
is it late there?
heya @blue
heya
@Alizter you can define the differentials as d(log(x)) = dx/x. Simple enough, no?
1/x is in your F(R).
i think i really need sleep now. bu-byes.

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