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Surb

 Discussion on question by George Orwe…

Imported from a comment discussion on math.stackexchange.com/q...
Surb
Sep 12, 2022 01:44
But this is obviously $0$, because if $X>9$, then $X<3$ will never happen...
 

 Discussion on answer by Surb: Using t…

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Surb
Aug 25, 2022 14:54
1) Yes it's something else. Namely, the volume of the $n-$ball of radius $R$ is $R^n$ times the volume of the unit $n-$ball.
Surb
Aug 25, 2022 14:54
1) As written in my OP, $\sigma $ is an element of $\partial B_1(0)$, but it's also the unit normal to the $n-$sphere. Also, $\mathrm d \sigma $ is an infinitesimal area of $\partial B_1(0)$. 2) Fubini theorem says that $$\int_{\mathbb R^n\times \mathbb R^m} f=\int_{\mathbb R^n}\int_{\mathbb R^m} f(x,y)\,\mathrm d y\,\mathrm d x=\int_{\mathbb R^m}\int_{\mathbb R^n}f(x,y)\,\mathrm d x\,\mathrm d y,$$ so, I used the first equality. 3) For your last question, why don't you try to prove it your self ;-)
Surb
Aug 25, 2022 14:54
$|B_R(0)|$ is the volume of $B_R(0)$. And by definition, $\int_A\,\mathrm d x=|A|$.@Brain
Surb
Aug 25, 2022 14:54
$g(x)\in \mathbb R$ (not $\mathbb R^n$... btw what would be the meaning of $g(x)f(x)$ if $f(x)\in\mathbb R^n$ and $g(x)\in \mathbb R^n$ ?) Notice that $x\cdot x\in \mathbb R$... that's why $g(x)\in \mathbb R$. And that's the reason why there is no $i^{th}$ component on $g$. @Brain
Surb
Aug 25, 2022 14:54
For your last comment, sorry, you are completely right, I should have said the unit $n-$sphere (and not the unit $n-$ball)... which is indeed the boundary of the unit $n-$ball... same for the $n-$sphere of radius $R$ (and not the $n-$ball of radius $R$). For your other comment, notice that $g(x)\in\mathbb R$ and $f(x)\in \mathbb R^n$ for all $x\in\mathbb R^n$. Therefore $F(x):=g(x)f(x)\in \mathbb R^n$ for all $x\in\mathbb R^n$, and thus indeed, $F:\mathbb R^n\to \mathbb R^n$. Let me know if something is still unclear. @Brain
Surb
Aug 25, 2022 14:54
The solution of your exercise ;-)
Surb
Aug 25, 2022 14:54
It more or less says (in an infinitesimal way) that the area of the $n-$ball of radius $R$ is $R^{n-1}$ times the area of the unit ball (i.e. $|\partial B_R(0)|=R^{n-1}|\partial B_1(0)|$). For exemple, if $n=2$, then $|\partial B_1(0)|=2\pi$ and indeed $2\pi R=|\partial B_R(0)|=R |\partial B_1(0)|$. For $n=3$, you can see that $|\partial B_1(0)|=\frac{4}{3}\pi$ and indeed, $|\partial B_R(0)|=\frac{4}{3}\pi R^2=R^2|\partial B_1(0)|$. However, the proof in dimension $n$ is a bit long (as your solution shows).
Surb
Aug 25, 2022 14:54
As I said, it implicitly prove that $\mathrm d \sigma _r=r^{n-1}\,\mathrm d \sigma $. As far as you have this formula, all details are included in my answer. @Brain
 

 Discussion on question by 100xln2: Ho…

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Surb
Aug 21, 2022 21:36
@SouravGhosh: "If any polynomial 𝑝(𝑥)∈𝐾[𝑥] with 𝑝(𝑇)=0 can be expressed as a product of distinct linear factors"... this will never ever happen... For all $T\in \mathcal L(V)$, there infinitely many polynomial $p(x)\in K[x]$ s.t. $p(T)=0$ and $p(x)$ can't be written as a product of linear factor...
Surb
Aug 21, 2022 21:36
@SouravGhosh: that's not true. Take $T$ being the identity. Then $T$ is diagonalizable, however, $P(x)=(x-1)(x^2+x+1)$ is s.t. $p(I)=0$ but you can't express $p$ as a product of linear factor.
Surb
Aug 21, 2022 21:36
your argument also work for linear map...
 

 Discussion on question by Seabourn: P…

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Surb
Jan 7, 2022 08:41
@XanderHenderson The set in my now deleted comment was containing the points you mentioned, for recall, I suggested to consider $H = \{t(x,y)\mid t\in[0,1], x,y\in \Bbb Q, \|(x,y)\|=2\} \cup \{(x,y)\mid x,y\in \Bbb R, \|(x,y)\|\leq 1\}$. I deleted the comment because this set is not open.
Surb
Jan 7, 2022 08:41
I think you are missing the assumption that $\cal H$ is an open set.
 

 Discussion on question by Jim Lars Sv…

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Surb
Dec 5, 2021 11:17
@DKNguyen You nail it! My guess is that by answering this question OP will, likely, provide all the ingredient to answer the original question.
 

 Discussion on question by F.V.: Prove…

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Surb
Nov 13, 2021 15:32
A better proof : since $x\mapsto x^n$ is convex, the inequality follows...
 

 2021 TeX - LaTeX Stack Exchange Moder…

Open discussion for tex.stackexchange.com/election/2
Surb
Nov 2, 2021 20:03
:)
Surb
Nov 2, 2021 20:03
yep
 

 Discussion on answer by Revenant_Evil…

Imported from a comment discussion on workplace.stackexchange....
Surb
Oct 30, 2021 06:07
"Diplomacy is the art of telling someone to go to Hell in such a way as to get them to look forward to the trip." Amazing quote!
 

 Discussion on question by CalculusLov…

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Surb
Aug 22, 2021 23:40
If you have that $\dim \ker(A-\lambda I)=1$ then $A$ is trigonalisable but not diagonalizable... Given that, it shouldn't be so hard to conclude...
Surb
Aug 22, 2021 23:40
what means $\gamma (\lambda )=1$ ?
 

 Discussion on question by questmath: …

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Surb
Mar 18, 2021 22:20
You have that for all $x,y\geq M$, $|f(x)-f(y)|<\varepsilon $ right ? And if $x,y\leq M$ and $|x-y|<\delta $, then $|f(x)-f(y)|<\varepsilon $. Right ? Then, if $x\leq M\leq y$ are s.t. $|x-y|<\delta $, then $|x-M|<\varepsilon $ and thus $|f(x)-f(M)|<\varepsilon $ and $|f(M)-f(y)|<\varepsilon $. @questmath
Surb
Mar 18, 2021 22:20
First, you must take $M$ s.t. $x\geq M$ implies $|f(x)-1|<\varepsilon /2$ (and not $x>M\implies |f(x)-1|<\varepsilon /2$. Now, if $x<M<y$, then $|f(x)-f(y)|\leq |f(x)-f(M)|+|f(M)-f(y)|\leq 2\varepsilon $ (I let you adapt things if you wish $|f(x)-f(y)|<\varepsilon $ instead of $2\varepsilon $) @questmath
Surb
Mar 18, 2021 22:20
$f$ is uniformly continuous on $[M,N]$. Therefore, there is $\delta >0$ s.t. for all $x,y\in [M,N]$, $|x-y|<\delta \implies |f(x)-f(y)|<\varepsilon $. What remains to prove it's if $x<M<y$, $y<M<x$,$x<N<y$ or $y<N<x$ are s.t. $|x-y|<\delta$, then $|f(x)-f(y)|<\varepsilon $. (which is rather straightforward) @questmath
Surb
Mar 18, 2021 22:20
@questmath: yes.
Surb
Mar 18, 2021 22:20
Your $\delta _2$ doesn't work. But you can use the fact that $f$ is uniformly continuous on $[M,N]$ to find a suitable $\delta $.
 

 Discussion on question by Studoku: Is…

Imported from a comment discussion on law.stackexchange.com/qu...
Surb
Nov 27, 2020 21:30
@Studoku "If they won they'd probably quit their job" interesting argument. What if it is a bad teacher mistreating children?
 

 Discussion on question by dwayneWhayn…

Imported from a comment discussion on workplace.stackexchange....
Surb
Oct 6, 2019 21:37
@Mars Amazing devilish game :D.
 

 Discussion on question by asmgx: Are …

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Surb
Sep 17, 2019 14:36
TIP: Look at the publication list of the blog's authors on google scholar.
 

 Discussion on question by John Colema…

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Surb
May 16, 2019 14:01
If you let them pass, you may allow them to have one more semester to increase their debt and fail eventually... Not sure this is helping them.
 

 Discussion between Theo Bendit and Surb

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Surb
Oct 23, 2018 12:30
is it known?
Surb
Oct 23, 2018 12:30
"if a bounded set has the property that every non-zero functional achieves its maximum uniquely, then is the set convex?"
I strongly believe that, in finite dimension, for each non-convex set, there exists a functional with 2 maxima.
Surb
Oct 23, 2018 11:50
Thank you very much
Surb
Oct 23, 2018 11:44
Interesting, I'll think about it. Could you recommend me a couple of papers on the problem?
Surb
Oct 23, 2018 11:38
(I'm going for a quick cigaret, I'll be back in 3min)
Surb
Oct 23, 2018 11:38
There exists U^*= U^*(S) in V^* such that if ||L|| has unique maximizer for all L in U^*, then the conjecture is verified for S
Surb
Oct 23, 2018 11:37
Something along the lines
Surb
Oct 23, 2018 11:37
"The metric projection is more akin to the set of points with optimal norm." This is what I had in mind
Surb
Oct 23, 2018 11:37
but, this is the point. In fact, the matrix norm above is in general NP-hard to compute.
However, with Perron-Frobenius theory you can find conditions on the norm and the operator under which the global maximizer is unique up to sign.
Surb
Oct 23, 2018 11:35
indeed
Surb
Oct 23, 2018 11:34
(this is a nonlinear eigenvalue problem)
Surb
Oct 23, 2018 11:34
where the exponents are take componentwise
Surb
Oct 23, 2018 11:33
(A^T(Ax)^(p-1))^(1/(q-1) = lambda x
Surb
Oct 23, 2018 11:33
If you write the critical point equation of this optimization problem you obtain something of the form
Surb
Oct 23, 2018 11:32
Consider the p norm ||.||_p on R^n and A \in R^{m x n}, then you wan to compute:
max_x ||Ax||_p / ||x||_q
Surb
Oct 23, 2018 11:31
the connection with your answer is the following:
Surb
Oct 23, 2018 11:31
to keep eigenvectors as rays and possibly change the eigenvalue on the ray if f is not homogeneous of degree 1
Surb
Oct 23, 2018 11:31
of some degree
Surb
Oct 23, 2018 11:31
and typically you'd assume f to be homogeneous
Surb
Oct 23, 2018 11:30
well it is just f(x) = lambda x