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visits member for 1 year, 7 months
seen May 9 at 4:47

Second year undergraduate


Jul
2
awarded  Curious
May
17
awarded  Nice Question
Apr
11
revised How prove this result $\frac{x}{y}=\sqrt{\frac{\sqrt{5}+1}{2}}$
edited title
Apr
7
comment Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?
yes. expand the left side using the binomial formula
Apr
7
accepted Finding measure given by Riesz Representation Theorem
Apr
7
revised Finding measure given by Riesz Representation Theorem
added 11 characters in body
Apr
7
revised Finding measure given by Riesz Representation Theorem
added 298 characters in body
Apr
7
asked Finding measure given by Riesz Representation Theorem
Mar
9
awarded  Mortarboard
Mar
6
accepted $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$
Mar
6
awarded  Custodian
Mar
6
reviewed Approve suggested edit on Is minimizing 1/f(x,y) the same as maximizing f(x,y) if f(x,y) is linear?
Mar
6
comment $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$
Nice one, thanks!
Mar
6
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.
Mar
6
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
Mar
6
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.
Mar
5
answered $l^\infty(I)$ and $l^\infty(J)$ isometrically isomorphic with $|I| \not= |J|.$
Mar
3
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.
Mar
2
answered Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
Mar
2
answered Divergence for $p$ prime numbers and convergence for $m$ composite numbers