lentic catachresis
Reputation
5,228
Next privilege 10,000 Rep.
Access moderator tools
 Oct 26 revised Every $R$-module is free $\implies$ $R$ is a division ring edited to clarify Oct 26 answered Every $R$-module is free $\implies$ $R$ is a division ring Oct 26 accepted Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution? Oct 26 comment Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution? Thank you very much, it's crystal clear. In fact, after the first simple manipulation, the proof is of the general fact that a function with a constant limit in infinity must have vanishing derivative in infinity, which is geometrically obvious. I will accept, though, the other answer as it adresses the original question, I hope you will understand. Oct 23 accepted Particular case of Nakayama's lemma Oct 23 answered Particular case of Nakayama's lemma Oct 23 accepted Stability of autonomous linear systems of ODEs Oct 23 accepted Square roots of $-1$ in quaternion ring Oct 23 revised Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution? edited to clarify hypothesis, and add particular case I'm interested in Oct 23 suggested approved edit on Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution? Oct 23 comment Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution? @JuliánAguirre Would you please elaborate? I'm very interested in the autonomous case. Oct 23 asked Must a solution defined for $(t_0,+\infty)$ with bounded limit in $+\infty$ tend to a constant solution? Oct 22 revised Stability of autonomous linear systems of ODEs added 481 characters in body Oct 21 asked Stability of autonomous linear systems of ODEs Oct 18 awarded Yearling Oct 12 comment Particular case of Nakayama's lemma @Jack Ah, I see. Using elements as I was doing, it's $m=an_1\in M \Rightarrow m=a^2n_2\in M \Rightarrow \dots \Rightarrow m=a^kn_k=0$. Thank you. Oct 12 asked Particular case of Nakayama's lemma Oct 10 comment Isn't it cheating to consider an $\mathbb{R}^3$ vector as a “pure quaternion”? @Mariano: what is MU? Oct 1 comment Square roots of $-1$ in quaternion ring Ouch, I found my mistake. I must remember that the binomial theorem only works in commutative rings. Thank you! Oct 1 asked Square roots of $-1$ in quaternion ring