# froggie

 Nov 21 awarded Necromancer Oct 11 answered What is pluripotential theory? Oct 6 comment Show that $\dim H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(m)) = {n + m \choose n}$ if $m \geq 0$, and $0$ otherwise.@user99182: Sorry, I'm not an algebraic geometer, so I'm not comfortable answering that. But I will say your guess seems reasonable to me! Oct 6 reviewed Approve suggested edit on Solve n equation with n variables. Oct 6 answered Show that $\dim H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(m)) = {n + m \choose n}$ if $m \geq 0$, and $0$ otherwise. Sep 28 answered Show basin of attraction has only one connected component Jun 22 comment Finding operator norm@alans: Unless my computations are incorrect, I think $\|A\| = \pi$. Another way to see this is to use Young's inquality for convolutions. If $g(x) = \cos^2(x/2)$, then $(Af)(x) = (f*g)(x)$. By Young's inequality, $\|Af\|_2 \leq \|f\|_2\|g\|_1$. The $L^1$-norm of $g$ is $\|g\|_1 = \pi$, so $\|Af\|_2\leq \pi\|f\|_2$. Therefore $\|A\|\leq \pi$. Of course, we know it is exactly $\pi$ by looking at $A1$ as above. Jun 22 answered Finding operator norm Jun 18 reviewed Approve suggested edit on Proof needed for $\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$ Jun 18 comment Proof needed for $\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$@ˈjuː.zɚ79365: answer added. Also, if ever in the future you see a question left unanswered by me that you can answer, please feel free to add an answer yourself. I won't mind, seriously. Jun 18 answered Proof needed for $\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$ May 31 comment Are there mini-mandelbrots inside the julia set?@MarkMcClure: Very interesting, thank you! May 30 revised Are there mini-mandelbrots inside the julia set?edited tags May 30 answered Are there mini-mandelbrots inside the julia set? May 29 comment Using Nakayama's Lemma to prove isomorphism theorem for finitely generated free modules@rschwieb: Didn't know the attribution. Interesting answer you link to! May 29 answered Wirtinger derivative of composition of functions May 29 reviewed Approve suggested edit on Wirtinger derivative of composition of functions May 29 answered Using Nakayama's Lemma to prove isomorphism theorem for finitely generated free modules May 23 comment Divisibility of an expression involving the Möbius functionThanks for the comments! I haven't thought much about counting necklaces before, so I'll have to do some reading. Also, @GregMartin, that's a good rule of thumb to know. The dynamical interpretation I alluded to is essentially an inclusion-exclusion type argument, so it's probably a more natural interpretation. May 22 asked Divisibility of an expression involving the Möbius function