# user58533

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Trying to learn math :)

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 18h accepted Help with function $f_r(x^q)=q^rx^{q-1}$ 18h comment Help with function $f_r(x^q)=q^rx^{q-1}$ I'll use your first suggestion. Thanks a lot. 23h 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. 23h 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. 1d 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 $q$. 1d comment Help with function $f_r(x^q)=q^rx^{q-1}$ I'll give a try on this idea :) 1d 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. 1d 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 $x$. 1d asked Help with function $f_r(x^q)=q^rx^{q-1}$ 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? Jul1 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. Jun19 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 ${..., A^{-1}x,x,Ax,...}$ Jun9 answered Is $\gamma(t) = \left( A \cos(\sqrt{a} t),B \cos \! \left( \sqrt{b} t \right) \right)$ dense in the rectangle $[- A,A] \times [- B,B]$? Apr30 answered derive an equation for this mass spring damper Apr19 answered near identity change of coordinates Apr17 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. Apr17 revised Find values of the parameters in Predator prey model added 869 characters in body Apr11 answered Help Finding Saddle Connections of a System Apr11 revised 2nd order linear differential equation added 117 characters in body Apr11 answered 2nd order linear differential equation