580 reputation
19
bio website
location
age
visits member for 1 year, 6 months
seen 9 hours ago

Trying to learn math :)


2d
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
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
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
25
comment Help with function $f_r(x^q)=q^rx^{q-1}$
I'll use your first suggestion. Thanks a lot.
Jul
25
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.
Jul
25
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.
Jul
24
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 $q$.
Jul
24
comment Help with function $f_r(x^q)=q^rx^{q-1}$
I'll give a try on this idea :)
Jul
24
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.
Jul
24
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 $x$.
Jul
7
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?
Jul
1
comment problems based on connected bodies in dynamics
This looks like a standard physics course problem. I suggest you try and look and many similar problems available. Once you try and if you still have problems understanding its solution, you can come back and add your attempts. Then I think more people will be interested in helping you.
Jun
19
comment Orbit , trajectory, dynamical system
I would also say that the terms are "almost" equivalent. A perfect example is to imagine the "action of a matrix on a vector". You take a matrix $A$, a vector $x$, and then the set ${x,Ax,...,A^nx}$ is called the orbit of the action of $A$ through $x$. I guess in naive words a trajectory is mostly used when you can imagine an arrow flowing in some surface (in any dimension). Another difference is that trajectory gives you the idea of direction while when using orbit this is not really present... e.g. If A is invertible, the orbit of $A$ through $x$ is ${..., A^{-1}x,x,Ax,...}$
Apr
17
comment what are the equilibrium points of the following:
I belive you should try it. Equilibria means "zero velocity" . Just solve the equations $12-3xy-3x=0$ and $3xy-6y=0$. Then try to interprete the result.
Apr
8
comment Uniqueness of solution to differential equation
Fixed point theorem together with the assumption that the right hand side of the ODE is Lipschitz should be useful.