# lyj

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 Sep30 awarded Explainer Sep24 awarded Autobiographer Jul2 awarded Curious May17 awarded Nice Question Apr11 revised How prove this result $\frac{x}{y}=\sqrt{\frac{\sqrt{5}+1}{2}}$ edited title Apr7 comment Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$? yes. expand the left side using the binomial formula Apr7 accepted Finding measure given by Riesz Representation Theorem Apr7 revised Finding measure given by Riesz Representation Theorem added 11 characters in body Apr7 revised Finding measure given by Riesz Representation Theorem added 298 characters in body Apr7 asked Finding measure given by Riesz Representation Theorem Mar9 awarded Mortarboard Mar6 accepted $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$ Mar6 awarded Custodian Mar6 reviewed Approve Is minimizing 1/f(x,y) the same as maximizing f(x,y) if f(x,y) is linear? Mar6 comment $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$ Nice one, thanks! Mar6 comment $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$ In retrospect, I spent too much time thinking about how to control where characteristic functions went, and it makes sense that that didn't work, since those are not the extreme points on the unit ball. Mar6 comment $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$ That's true, but after seeing this solution, I don't feel as if I just didn't know enough to do the problem, since the notion of extreme point is very intuitive and is just a sort of generalization of vertices of a (convex) polytope Mar6 comment Divergence for $p$ prime numbers and convergence for $m$ composite numbers Unfortunately no. I've heard from a friend that this problem was actually given as homework for a class at cambridge (?), but obviously not many people got it. Mar5 answered $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$ Mar3 comment Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. what you've done is equivalent to saying $n\mid 0,$ which is not really anything. see @Álvaro's hint.