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visits member for 2 years, 5 months
seen Dec 12 at 10:14

Dec
12
awarded  Caucus
Aug
12
awarded  Critic
Jul
26
awarded  Yearling
Jul
26
answered $\forall v\in V:(Tv,v)=0\implies T^{\star}=-T$
Jul
26
revised True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$
deleted 2 characters in body
Jul
26
comment How can it be proved that the geometric mean function is concave?
Did you try applying $ln()$ to the arithmetic mean function ?
Jul
26
comment True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$
What you have written are just implications and not equivalences. You can't say that $$||v+v ||=||v||+||v||$$ implies that $$||Tv+v ||=||Tv||+||v||$$
Jul
26
revised True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$
added 133 characters in body
Jul
26
answered True or false: If $||Tv+v ||=||Tv||+||v||$, then $1$ is eigenvalue of $T$
Jul
26
revised find the non constant polynomial so that P(x) δ_1' = δ_1'
Added latex delimiters
Jul
26
suggested approved edit on find the non constant polynomial so that P(x) δ_1' = δ_1'
Jul
25
revised Duality and the Minimax Theorem
added 335 characters in body
Jul
25
revised Duality and the Minimax Theorem
added 335 characters in body
Jul
25
revised Duality and the Minimax Theorem
added 335 characters in body
Jul
25
revised Duality and the Minimax Theorem
added 335 characters in body
Jul
25
answered Duality and the Minimax Theorem
Jul
25
comment Duality and the Minimax Theorem
$v$ is not a constant. The variables of the problem are ($x_1$,$x_2$,...$x_{m_1}$,$v$). What's $A$?
Jul
22
comment How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$
@egreg, sorry I've written the answer before noticing that you added yours. Your answer was sufficient and complet
Jul
22
revised How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$
added 498 characters in body
Jul
22
comment How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$
Hey Simar, if you don't mind but $\sqrt(x^2) = |x|$ not $x$