# user58533

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Trying to learn math :)

 2d comment Every solution of the system is attracted to the center manifold **tex text is the extension. Aug24 comment Every solution of the system is attracted to the center manifold I made it on Inkscape together with an extension called text ext, which allows you to "write Latex". Aug8 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. Aug4 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.) Jul29 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 $\tilde\pi^{-1}(y)=(-y^2,y)$. Jul27 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. Jul27 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 $f$. Jul27 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 $f_k(g)=\frac{1}{x_k}\int_0^{x_k}g(x_1,\ldots,x_{k-1},t,x_{k+1}\ldots,x_n)dt$ ? Jul27 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. Jul27 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. Jul27 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. Jul27 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? Jul25 comment Help with function $f_r(x^q)=q^rx^{q-1}$ I'll use your first suggestion. Thanks a lot. Jul25 comment Help with function $f_r(x^q)=q^rx^{q-1}$ This is indeed useful. No problem on taking polynomials without constant term. I am just not quite sure if the division by $X$ may carry some problems elsewhere. I'll continue working with your idea and once I check everything goes fine I'll accept your answer. Thanks. Jul25 comment Help with function $f_r(x^q)=q^rx^{q-1}$ I get it now, I think it is better to call it operator. Jul24 comment Help with function $f_r(x^q)=q^rx^{q-1}$ Uhm so, $x^q$ is arbitrary, so I see it as what you say in the end, as a family labeled by r. If I could use the $q$ parameter I think the problem becomes somewhat trivial. The thing is that the function (functional????) $f_r$ should not depend on $q$. Jul24 comment Help with function $f_r(x^q)=q^rx^{q-1}$ I'll give a try on this idea :) Jul24 comment Help with function $f_r(x^q)=q^rx^{q-1}$ @Semiclassical not really, I hope/wish it exists, it'll help me in a bigger purpose. Jul24 comment Help with function $f_r(x^q)=q^rx^{q-1}$ @N.S. uhm $f_r$ has as arguments as I understand monomials $x^q$, derivative is taken with respect to $x$. Jul7 comment For all topological conjugation $$h: \Delta_1 \rightarrow \Delta_2$$ we have to $h(\omega(p))=\omega(h(p))$, for all $p \in \Delta_1$ As far as my understanding goes, when $h$ is a homeomorphism, what you wrote is the definition of topological conjugacy, isn't it?