Mac Dual Monitor without apple menu bar and Prevent Second monitor to go black while one is fullscreen, is possible?

I am trying to find the $L$-smoothness constant of the following function (logistic regression cost function) in order to run gradient descent with an appropriate stepsize.

The function is given as $f(x)=-frac{1}{m} sum_{i=1}^mleft(y_i log left(sleft(a_i^{top} xright)right)+left(1-y_iright) log left(1-sleft(a_i^{top} xright)right)right)+frac{gamma}{2}|x|^2$ where $a_i in mathbb{R}^n, y_i in{0,1}$,$s(z)=frac{1}{1+exp (-z)}$ is the sigmoid function.

The gradient is given as
$nabla f(x)=frac{1}{m} sum_{i=1}^m a_ileft(sleft(a_i^{top} xright)-y_iright)+gamma x $.

My ideas was that the smoothness constant $L$ has to be bigger than all the eigenvalues of the hermitian of the given function, this follows from the fact that if $f$ is $L$-smooth, $g(x)=frac{L}{2} x^T x-f(x)$ is a convex function and therefore the hessian has to be positive semi-definite.
The second-order partial derivatives of $f$ are given as

$ frac{partial^2 }{partial x_k partial x_j}f(x)=frac{1}{m} sum_{i=1}^ms(a_i^{top} x)left(1-s(a_i^{top} x)right)[a_i]_k[a_i]_j+gammadelta_{ij} $

from the following github post (https://github.com/ymalitsky/adaptive_GD/blob/master/logistic_regression.ipynb) i know that $ L=frac{1}{4} lambda_{max }left(A^{top} Aright)+gamma$ , where $lambda_{max }$ denotes the largest eigenvalue, which seems good since i figured out that $s(a_i^{top} x)left(1-s(a_i^{top} x)right)leq frac{1}{4}$ for all $x$.

But i am not able to fit everything together. I would appreciate any help.

A topological space is compact if and only if every ultrafilter is convergent.

While I was reading the proof of the one Side of theorem above, there is something I could not understand. Following is the proof of of the one side of the theorem.

Let $X$ be compact and assume that $mathcal{F}$ is the ultrafilter on $X$ without a limit point. Then for each $xin X$, there exists an open neighborhood $U_{x}$ of it such that each $U_{x}$ does not contain any member of $mathcal{F}$. Since $mathcal{U}={U_{x} : xin X}$ is an open cover of $X$, there exists a finite subfamily ${U_{x_{i}}: i=1,2,…,n}$ of $mathcal{U}$ such that $X=bigcup_{i=1}^{n} U_{x_{i}}$. Let $Ainmathcal{F}$ be fixed. Then $A=(Acap U_{x_{1}})cap (Acap U_{x_{2}})…(Acap U_{x_{n}})inmathcal{F}$ and thus there exists an $iin{1,2,…,n}$ such that the subset $Acap U_{x_{i}}$ is in $mathcal{F}$ which is a contradiction.

The thing that I could not understand, why there exists $iin{1,2,…,n}$ such that $Acap U_{x_{i}}$ must be in $mathcal{F}$? If you clarify this, it would highly be appreciated. Thank you.

In this paper (Equation 2.6 and 2.7) the author seems to suggest that one can represent the $text{GL}^+(4,mathbf R)$ group using the product of two exponentials: $exp (epsilon cdot T) exp (u cdot J)$, where $T$ are the generators of shears and dilation, and $J$ are the generators of Lorentz transformations.

My take on the subject is that since $T$ and $J$ do not commute, one cannot write $G$ as a product of these two exponentials. One must instead write $G=exp ( epsilon cdot T + u cdot J )$. It appears to me the author is wrong.

Is the author correct, or am I?

How can I represent $text{GL}^+(2,mathbf R)$ as the matrix product $G=TH$ where $H=text{SO}(2)$?

What is the pushout in the category of commutative unital $C^{ast}$-algebras? Is it the tensor product? Is it the same as in the category of noncommutative unital $C^{ast}$-algebras?

Suppose I have a polynomial that is given a form
$$
f(x)=x^n – a_{n-1}x^{n-1} – ldots – a_1x – 1
$$

where each $a_k$ can be either $0,1$.

I’ve tried a bunch of examples and found that the maximum real root seems to be between $1,2$, but as for specifics of a polynomial of this structure I am not aware.

Using IVT, we can see pretty simply that $f(1)leq0$ and $f(2)> 0$ so there has to be a root on this interval, but thats a pretty wide range was wondering if this was previously studied

Could anyone please clarify to me what active particles, in particular active deformable particles are? I have never heard of them, and I am quite curious

Let $X$ be a smooth projective variety over an algebraically closed field $k$ of characteristic zero.

Is there a description of $operatorname{Aut}(operatorname{Coh}(X))$, i.e. the autoequivalences of the category $operatorname{Coh}(X)$?

Clearly it contains $operatorname{Aut}(X)ltimesoperatorname{Pic}(X)$ as a subgroup.

$$int_{n+1}^{infty}f(x)dx le R_n le int_{n}^{infty}f(x)dx$$

What does this actually mean?

Let’s use n=5.

The $R_5$ is the error of the partial sum $S_5$

That error is less than the sum of the remaining terms from 5 to $infty$ ?

Why is that?

Also, why is it greater than sum of the remaining terms from 6 to $infty$ ?