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Jul
29
comment spectral radius
Yep. I suggest you write it up as an answer and accept it. I am too lazy (it's bedtime in my corner of the world). And we don't like unanswered questions lying around.
Jul
29
comment $\liminf_n a_n = \inf_n a_n$ if $a_n \ge a_m$ when $n\mid m$
Yeah, your best bet would be an exercise where the object is to learn how $\liminf$ works.
Jul
29
answered If a given # is $70$% of $X$. How do you determine what $X$ is?
Jul
29
comment spectral radius
What is the spectral radius of $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$? Oh, and for the triangle inequality, try $x=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $y=\left(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right)$!
Jul
29
comment $\liminf_n a_n = \inf_n a_n$ if $a_n \ge a_m$ when $n\mid m$
I took the liberty of editing the question title. I was unable to make sense of it after reading it several times, mainly because I was reading it as one implication, where the left side made no sense. Hope that's ok. – I suspect a reference will be hard to find; as you say, the result is very easy, and I imagine it is rarely used.
Jul
29
revised $\liminf_n a_n = \inf_n a_n$ if $a_n \ge a_m$ when $n\mid m$
Original question title was too confusing
Jul
28
comment How to find $\int_0^1 \int_x^1 \arctan(\frac{y}{x})dxdy$?
What kind of calculus course (or text) will even pose such a problem without teaching about different orders of integration first? That surprises me.
Jul
28
comment How to find $\int_0^1 \int_x^1 \arctan(\frac{y}{x})dxdy$?
Have you tried inverting the order of the integrals?
Jul
25
comment Functions with rank $n$.
Wait, I forgot a subtle point. You must rule out the minimum being on the boundary of the ball. At a minimum on the boundary, you can't know that $Dg=0$. But considering the value of $\|f(0)\|$ shows that this case cannot happen.
Jul
25
comment Functions with rank $n$.
You have the right idea, but taking norms/absolute value isn1t right, for the norm of a matrix product (or a matrix-vector product) is usually not the product of the norms. Instead, go back to $Dg=0$ (yes, that must hold at a minimum) and notice that the invertibility of $Df$ implies $f=0$ at the minimum. And then you're done.
Jul
25
comment Functions with rank $n$.
No, $g$ is a scalar function, so its derivative should be a vector (i.e., the gradient), not a matrix. Write $g=\sum_{k=1}^n f_k^2$, where $f=(f_1,\ldots,f_n)$ and then take partial derivatives. Express the result as a vector; you should get a product of $Df$ and a vector. (Or just take partial derivatives of the scalar product $f\cdot f$ directly, if the resulting algebra doesn't intimidate you.)
Jul
25
answered Functions with rank $n$.
Jul
24
comment Is there any commonality between Math induction and Logic induction?
And to the OP: No, the two concepts are quite different. Apart from the name there is a certain superficial similarity, but that seems to dissipate once you look closer.
Jul
24
comment Is there any commonality between Math induction and Logic induction?
@GitGud The induction mentioned in the video is certainly not reasoning by mathematical probability, although I suppose Bayesian thinking could aid such induction. He's really talking about inductive reasoning in the philosophical sense of arguing from special cases to the general.
Jul
18
answered How is an $n$ sphere in $m$ dimensions defined?
Jul
8
comment Definition of $C^k$ boundary
@Lost1 That's right.
Jul
7
awarded  Enlightened
Jul
7
awarded  Nice Answer
Jun
27
comment an inequality for multiplication of cubic numbers
Try to put $a_i=2$ for all $i$ and see where that gets you …
Jun
26
comment Prove that $\det\left[A^{T}B-B^{T}A\right]=\det[A+B]\cdot\det\left[A-B\right]$
@HenningMakholm Good point.