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Apr
11
comment Min of concave symmetric function on a convex set
I was just guessing about $(1/3,1/3,1/3)$, but after doing the calculations I see that it's an extreme point. In addition to the points you mentioned $(0,1/2,1/2)$ is also one of the extreme points. Since the maximum occurs at $(1/3,1/3,1/3)$, your minimum point is either $(0,1/2,1/2)$ or $(1/4,1/4,1/2)$.
Apr
10
comment Min of concave symmetric function on a convex set
For $N=3$ you draw the set in 3D and vertices of the set you get would be the extreme points. I think (0,0.5,0.5) is one of them, but (1/3,1/3,1/3) is not.
Apr
10
comment Min of concave symmetric function on a convex set
You are basically right about the minimum being attained at an extreme point. The minimizer will be a point on the boundary of $C$ that cannot be written as a convex combination of the other points in $C$. I don't think without knowing $\varphi$ you can make any better statement. Actually, I think you can cook up different $\varphi$s that have minimum at different extreme points.
Mar
22
accepted Find a projection of a $k$-simplex with minimal “radius”
Mar
14
comment Show that if $\displaystyle\int_0^1f(x)dx=a$, then $\displaystyle\int_0^1\sqrt{f(x)}dx\ge a^{2/3}$
The first inequality is just the Hölder inequality. In general, $$\Vert g\Vert _{L_2} ^2\leq \Vert g\Vert_{L_p}\cdot \Vert g\Vert_{L_q},$$ with $\frac{1}{p}+\frac{1}{q}=1$ which holds particularly for $p=1$ and $q=\infty$.
Mar
14
answered Show that if $\displaystyle\int_0^1f(x)dx=a$, then $\displaystyle\int_0^1\sqrt{f(x)}dx\ge a^{2/3}$
Mar
8
comment $A'B=I \rightarrow B'B \geq (A'A)^{-1}$
For any matrix $Q$, $Q′Q$ is positive semi-definite. You see it immediately when you write down the definition.
Mar
8
answered $A'B=I \rightarrow B'B \geq (A'A)^{-1}$
Mar
7
comment A question regarding a prefix code
I don't think your argument is convincing. The probability of any codeword of length $n$ being drawn is $\sigma^{-n}$. You need to be more rigorous. I would suggest to look at $k_n$ the number of codewords of length $n$ and upper bound it using the fact that the codewords are prefix-free.
Mar
7
comment A question regarding a prefix code
Where did you use the fact that the codes are prefix-free? Where do those probabilities come from?
Mar
6
answered Special Matrix 2-norm and F-norm Inequalities
Mar
5
comment KKT conditions for nonsmooth convex problems
Just to clarify, the KKT conditions are necessary and sufficient only when some constraint qualifications hold.
Mar
5
answered Eigenvalues of a submatrix
Mar
3
revised $ABCD$ right angle trapezoid
Typo corrected
Mar
3
comment $ABCD$ right angle trapezoid
@user126154 Sorry, I meant $BO/DO$ rather than $BO/EO$. You get the idea. I corrected the error.
Mar
3
answered $ABCD$ right angle trapezoid
Mar
3
comment Is this sequence $x_{n+1}=1+\frac{a-1}{1+x_n},$ with $a>1, x_1>0$, convergent or divergent?
For the first few members of the sequence write the upper and lower bounds that you can get starting from $u_1=0<x_1<\infty$. The pattern is that $x_n$ is bounded between a $u_n$ and $a/u_n$. Then derive the relation between $u_n$ and $u_{n+1}$ from the recurrence formula for $x_n$.
Mar
2
revised Is this sequence $x_{n+1}=1+\frac{a-1}{1+x_n},$ with $a>1, x_1>0$, convergent or divergent?
deleted 32 characters in body; deleted 3 characters in body
Mar
2
answered Is this sequence $x_{n+1}=1+\frac{a-1}{1+x_n},$ with $a>1, x_1>0$, convergent or divergent?
Mar
1
comment Condition number vs. reconstruction error
I see. However, still if you look at the plot you see that most of your matrices have condition number around 3. You can't expect much variation in the performance there. Perhaps that range of condition number is good enough for the available numerical precision. As I said, to see the effect of condition number you need to carefully draw matrices whose condition numbers cover a wider range (let's say from 1 to 1000).