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Trying to learn math :)


Aug
8
answered Unique solution to a arbitrary non-linear system under monotonicity assumptions
Aug
8
comment Central manifold theorem => Stable/unstable manifold?
I don't think so, if the centre manifold is of dimension zero, then you have the common results of stable/unstable manifold theory. I must remark though that centre manifolds are very important as the dynamics there are given by the nonlinear terms of your system. Furthermore, centre manifolds are very useful, for example: 1) in singular perturbation problems, or the so called slow-fast systems; 2) In normal form theory; and related to it, 3) In reducing the dimension of the problem you are studying.
Aug
4
comment Stable manifold for bidimensional nonlinear dynamic system with complex eigenvalues
You are correct in your idea. Imagine a node. Every orbit is an invariant manifold. In fact every orbit is an invariant stable manifold. Their union form a 2 dimensional invariant stable manifold. The advantage of having real eigenvectors is that it is enough to study the generated spaces, in the sense that if you know what happens there, then you know what happens everywhere (in the linear case of course). For the complex case, recall that you can see a complex number in the real plane. So a 2d real system with complex eigs. can be seen in a 1d complex system (like passing to polar coords.)
Aug
2
answered Stable manifold for bidimensional nonlinear dynamic system with complex eigenvalues
Jul
30
awarded  Curious
Jul
29
comment Doubt with smooth extensions
I guess $\tilde\pi^{−1}(x)=(x,\pm\sqrt{-x})=(-y^2,\pm y)$ is the correct way of taking $\tilde\pi^{-1}$, but since I know which $y$ we are taking since the beginning, we could set $\tilde\pi^{-1}(y)=(-y^2,y)$.
Jul
29
revised Doubt with smooth extensions
added 1 character in body
Jul
29
asked Doubt with smooth extensions
Jul
27
accepted Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
Jul
27
comment Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
Thanks, I think this helps a lot. I'll accept the answer once I've checked everything carries on well in my project.
Jul
27
comment Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
@ellya indeed, as you say, that is how I am thinking of the operator $f$.
Jul
27
comment Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
This is indeed good for $g\in\mathbb R[x]$ as you say. What about $g\in\mathbb R[x_1,\ldots,x_n]$? ... is it possible to define $f_k(g)=\frac{1}{x_k}\int_0^{x_k}g(x_1,\ldots,x_{k-1},t,x_{k+1}\ldots,x_n)dt$ ?
Jul
27
comment Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
I've edited. I didn't know about formal integration, but I just googled it and I will check it, thank you.
Jul
27
revised Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
I have edited trying to make my thought clearer.
Jul
27
comment Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
I need to think on this as I do not understand very well your argument. I thought I was sure I was working on "maps from the space of polynomials onto itself". But it seems I am doing something wrong. Thanks.
Jul
27
comment Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
that is correct, I am assuming/thinking of $f$ as an operator that receives polynomials, not numbers.
Jul
27
comment Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
Thanks. Do you think there is some way to require that the function "acts on $x^q$ and not in factorisations"? Or something like that? Does it makes a difference that I am not defining $f(P^q)=P/(q+1)$ but requiring that in the particular case that the argument is something "exactly like" $x^q$ I then obtain $x^q/(q+1)$ ? I set $f(P)=P^2/\phi(P)$, this is well defined right?
Jul
27
asked Help with operator $f(x^q)=\frac{1}{q+1}x^q$.
Jul
25
accepted Help with function $f_r(x^q)=q^rx^{q-1}$
Jul
25
comment Help with function $f_r(x^q)=q^rx^{q-1}$
I'll use your first suggestion. Thanks a lot.