TITLE: A question about Hatcher exercise 2.1.23
QUESTION [2 upvotes]: I'm trying to solve a problem on barycentric subdivision. The problem deals with any delta complex in general, so I can't find a way to formulate some argument at all...I can't even see how to express any delta complex in general and apply the barycentric subdivision. Could anyone show me how to solve this problem?

REPLY [0 votes]: I'll first give a brief excursus about the different ways to look at a delta complex

A general Delta complex $X$ can be described combinatorially as a sequence of sets $X_0,X_1,\dots$ together with maps $d_i:X_n\to X_{n-1}$ for any $i\in[n]=\{0<\dots<n\}$ such that
  $$ d_j d_i = d_i d_{j+1}, \qquad \text{whenever }\; i\le j $$ whenever $j\ge i$. The elements of $X_n$ are called $n$-simplices of $X$, and the maps $d_i$ are the face maps of $X$.
  Note that any monotone injection $f:[k]\to[n]$ can be written uniquely as a composite
  $$ f=\delta_{i_{n-k}} \dots \delta_{i_1}, \qquad i_1<\dots<i_{n-k} $$
  where $\delta_i:[m-1]\to[m]$ denotes the monotone injection which omits the value $i$, for any $m$. The indices in the decomposition of $f$ are then simply those elements of $[n]$ which are not assumed by $f$. Such an $f$ can be thought of as representing the face of $\Delta^n$ which is spanned by the basis vectors $e_{f(0)},\dots, e_{f(k)}$. If we denote the restriction of $\sigma$ to this face by $B_f(\sigma)$, then we see that
  $$
B_f B_g (\sigma) = B_{gf}(\sigma) \quad \text{for any monotone injections $f:[k]\to[l]$ and $g:[l]\to[n]$ }
$$
  This property is actually equivalent to the rule $(1)$, where $d_i=B_{\delta_i}$.

A delta complex $X$ has a realization $|X|$ which is constructed as follows: For any $n$-simplex $\sigma$ take a standard-$n$-simplex $\Delta^n_\sigma$. Now identify its $i$-th subsimplex with the standard-$(n-1)$-simplex $\Delta^{n-1}$ corresponding to the face $d_i\sigma$ using the linear inclusion $\Delta^{n-1} \to \Delta^n$ which preserves the ordering of the vertices. For example, you may have the delta complex $X_0=\{v\} \leftleftarrows X_1=\{e\}$, where the two arrows represent $d_0$ and $d_1$. Then you have a line $\Delta^1$ and a point $\Delta^0$, and both endpoints of the line are glued to that point, so we get a circle.
The barycentric subdivision $\mathrm{Bd}(X)$ has as $n$-simplices all sequences
$$
(\sigma, f_1, \dots, f_n), \quad \sigma\in X_m,\, f_i:[k_i]\hookrightarrow[k_{i-1}],\, 0\le k_n<\dots<k_{0}=m
$$
This can be thought of as taking a simplex $\sigma$ and then choosing faces of $\sigma$ such that each face belongs to the previously chosen face. The face map $d_i$ sends $(\sigma, f_n, \dots, f_1)$ to
$$
\begin{cases}
(B_{f_1}\sigma, f_2, \dots, f_n)  &\text{if } i=0 \\
(\sigma, f_1, \dots, f_{i-1}, f_{i}f_{i+1},\dots, f_n)  &\text{if } 0<i \\
(\sigma,f_1,\dots, f_{n-1})  &\text{if } i=n
\end{cases}
$$
Now can you show that if $X$ has no self-identifications (meaning $B_f\sigma\ne B_g\sigma$ whenever $f\ne g$), then this $\mathrm{Bd}(X)$ produces a simplicial complex?


TITLE: Approximation of a two-variable function by tensor products
QUESTION [2 upvotes]: Let $X$ and $Y$ be compact metric spaces and $f: X \times Y \to \mathbb{R}$ be a continuous function.
We know that, for every $n \in \mathbb{N}$, by the Stone-Weierstrass theorem, there exist $k_n \in \mathbb{N}$ and continuous functions $f^{(1,n)}_i$, $i \in \{1, \ldots, k_n\}$, on $X$ and continuous functions $f^{(2,n)}_i$, $i \in \{1, \ldots, k_n\}$, on $Y$, such that

$$ \sup_{x \in X, y \in Y } \bigg| f(x,y) - \sum_{i=1}^{k_n} f^{(1,n)}_i(x) f^{(2,n)}_i(y) \bigg| < \frac{1}{n} .$$

Is it possible to choose these functions such that

$$ \sup_{n \in \mathbb{N}}   k_n < + \infty, $$
$$   \sup_{n \in \mathbb{N}}  \sup_{i \in \{1, \ldots, k_n\}}\sup_{x \in X} \Big| f^{(1,n)}_i(x) \Big| < +\infty, \quad \quad \sup_{n \in \mathbb{N}}
 \sup_{i \in \{1, \ldots, k_n\}}\sup_{y \in Y} \Big| f^{(2,n)}_i(y) \Big| < +\infty \quad ? $$

REPLY [3 votes]: Instead of the function $f$ consider the associated function $F: X \to C(Y)$, which is continuous. Let $\varepsilon>0$, and for each $x_0$ take a neighbourhood $U(x_0)$ where $\|F(x)-F(x_0)\|<\varepsilon$, from this cover of $X$ take a finite subcover $U(x_1),\dots,U(x_n)$ and a partition of unity $f_1,\dots,f_n$ subordinate to this subcover. Write $g_k=F(x_k)$ and $g(x,y)= \sum_{k=1}^n f_k(x) g_k(y)$. Then $\|f-g\|\le3\varepsilon$, all $f_k$ satisfy $0\le f_k\le 1$ and of course $\sum f_k = 1$. Also $\|g_k\|\le \|f\|$. So the bounds on $f_k$ and $g_k$ and the sum condition on the $f_k$ are ok.
However, in general one cannot have the sum condition $\sup_y \sum |g_k(y)| < \infty$ as well. If so, one would have an estimate for the ``weak $\ell_2$-norm'' $\sup_y (\sum |g_k(y)|^2)^{1/2}$ as well  and likewise for $\sup_x [\dots]$. In terms of tensor norms this means that the injective tensor norm (corresponding to the sup-norm of $f$) is equivalent to Grothendieck's Hilbertian norm $w_2$, defined on the tensor product of two Banach spaces as follows: If $\|(x_j)\|_2^w := \sup_{x^*\in B_{E^*}} \|(x^*(x_j))\|_{\ell_2}$, put $\|u\|_{w_2}=\inf \|(x_j)\|_2^w\|(y_j)\|_2^w$, where the infimum is taken over all representations of the tensor $u\in E\otimes F$ as $u=\sum x_k \otimes y_k$. The punchline here is that for spaces of continuous functions, the Hilbertian norm is equivalent to the projective norm by Grothendieck's inequality. Hence the injective and the projective tensor norms would be equivalent on spaces of continuous functions, which they aren't.


TITLE: Is $\mathbb{F}_q(x)$ a perfect field?
QUESTION [0 upvotes]: It is well known that the finite fields $\mathbb{F}_q$ are perfect i.e. every finite extension of $\mathbb{F}_q$ is separable. I am wondering if a pure transcendental extension $\mathbb{F}_q(x) \supseteq \mathbb{F}_q$ remains perfect? Can someone tell me if this is true and where can I find a proof? Thanks in advance!

REPLY [2 votes]: A purely transcendental extension $F(x)$ is never perfect when $F$ is a field of characteristic $p \neq 0$; the polynomial $f(t) = t^p - x$ is an irreducible over $F(x)$, but has repeated roots.
One way to see this is that if we let $F(y)$ be another purely transcendental extension, then $F(x) \cong F(y^p)$. Since $F(y)$ is generated over $F(x)$ by a root of $f(t)$ and $\deg(f) = [F(y) : F(x)]$ (which, for example, can be seen by looking at formal Laurent series), we see that $f$ is irreducible over $F(x)$. And over $F(y)$, we have the factorization
$$ f(t) = t^p - x = (t - y)^p $$
so $f$ has repeated roots.


TITLE: Quotient of Dedekind rings by a power of a prime ideal
QUESTION [1 upvotes]: Let $ A $ be a Dedekind ring, let $ P$ be a nonzero prime ideal of $ A $. What can we say about $ P^n $ or $ A/P^n $ in general? For example, is it true that $ A/P^n $ is a vector space over $ A/P $ of dimension $ n $?

REPLY [1 votes]: No, it's not true since this ring, as an $A$-module, is not killed by $P$ as an $A$-module (as a counter-example, consider $\mathbf Z/p^n\mathbf Z$: is it a $\mathbf Z/p\mathbf Z$-vectorspace.
The nearest assertion which is true is that, for instance $P^{n-1}/P^n$ is an $A/P$-vector space.
The only other thing I can see that can be said  about $A/P^n$ is that it's a local (noetherian) ring of Krull dimension $0$.


TITLE: Continuity of f with supremum metric - Lipschitz condition
QUESTION [1 upvotes]: Let $$ X=C([0,1]) $$ $$ f: [0,1] \rightarrow {R}  $$ with supremum metric:
$$  d(f,g)=\sup_{x \in [0,1]}|f(x)-g(x)| $$
where $$ f,g \in X
$$
 Check if  $$F: X \rightarrow X $$ $$ F(f)=\arctan{(f(x))} $$ is continous
My attempt:
I want to say that this function fulfils Lipschitz contidion.
I have:
$$ d(F(f), G(g))= \sup_{x \in [0,1]}|\arctan{f(x)}-\arctan{g(x)}|$$
This is the point where I am stuck. I wanted to say that
$$ \sup_{x \in [0,1]}|\arctan{f(x)}-\arctan{g(x)}|\leqslant \sup_{x \in [0,1]}|f(x)-g(x)|$$
Then it would be everything, but I'm not sure that this is true. Can you help me with this?

REPLY [1 votes]: Actually,
Use Lagrange inequality giving :
$$ \exists c\in]0,\frac{\pi}{2}[, \forall(p,q)\in]0,\frac{\pi}{2}[ , (\arctan(p) -\arctan(q)) \leq |p-q|*\arctan'(c) $$


TITLE: Proof of limit to infinity using epsilon-N method
QUESTION [0 upvotes]: I wanted to prove this: $\lim_{x\rightarrow c} f(x)g(x)= \infty$ when $\lim_{x\rightarrow c} f(x) = L (L>0)$ and $\lim_{x\rightarrow c} g(x) = \infty$ by using epsilon-N method.
I used two inequalities derived by using epsilon-delta and epsilon-N method on two limit conditions. I thought that by fixing L as $L>\epsilon>0$, I could make the form of $f(x)g(x)>K (K>0)$ using inequalities. Well I cannot derive that form, my result was always $K<0$. How can I prove this statement? Thank you.

REPLY [0 votes]: Let $M>0$. Since $f(x)\to L$ id $x\to c$, there is $\delta_1 >0$ s.t. $$f(x)\geq \frac{L}{2}>0,$$
whenever $|x-c|<\delta_1 $. Moreover, since $g(x)\to \infty $, when $x\to c$, there is $0<\delta <\delta _1$ s.t. $$g(x)\geq \frac{2M}{L},$$
if $|x-c|<\delta $. In particular, if $|x-c|<\delta $, we have $$f(x)g(x)\geq M,$$
and thus $f(x)g(x)\to \infty $ whenever $x\to c$.


TITLE: Conformally Map Region between tangent circles to Disk
QUESTION [2 upvotes]: Suppose we are given two circles, one inside the other, that are tangent at a point $z_0$. I'm trying to map the region between these circles to the unit disc, and my thought process is the following:
I feel like we can map $z_0$ to $\infty$, but I'm not really sure about this. If it works, then I get a strip in the complex plane, and I know how to handle strips via rotation, translation, logarithm, power, and then I'm in the upper half-plane (if I've thought this through somewhat properly). My problem really lies in what point $z_0$ can go to, because I thought the point symmetric to $z_0$ (which is $z_0$ in this case) had to be mapped to $0$. Is this the right idea, and if what other details should I make sure I have? Thanks!

REPLY [2 votes]: As it turns out, writing down an explicit formula is very long, but describing the general process is easy (and is similar to Chris C's, although I still feel his solution has a hole). Take $z_0$ to be the point of tangency, $z_1$ to be the antipodal point on the inner circle, and $z_2$ to be the antipodal point on the outer circle (in particular, these points are on the boundaries of the circles). Assuming we're given equations for the circles, we can explicitly calculate these antipodal points. From here, rotate (if necessary) so that these points are all on the real axis and translate so that $z_2$ is at $0$. Now, given the equations of the circles, we can ensure $z_0$ and $z_1$ lie on the positive real axis. From here, we can use a Moebius transformation to send $z_2\mapsto 0$ (so $0$ is fixed), $z_1\mapsto 1$, and $z_0\mapsto\infty$. Since angles are preserved and this maps the circles to lines, this maps the region to a vertical strip in the complex plane of width $1$. Now rotate this by $\pi/2$, dilate it by $\pi$, and apply the exponential map. This brings us to the upper half-plane, and now we can use the standard map to go the unit disc.


TITLE: Product rule for matrix-valued functions and Differentiability of matrix multiplication
QUESTION [5 upvotes]: Let $F, G: \mathbb{R}^{n \text{ x } n} \to \mathbb{R}^{n \text{ x } n}$ be two differentiable functions. Defining $(F \cdot\ G)(A):=F(A)G(A)$ using the usual matrix multiplication, prove that $F\cdot\ G$ is differentiable and express $(F\cdot\ G)'$ in terms of $F$, $F'$, $G$, $G'$.

What I thought of doing was to write $A=[a_{ij}]_{i, j \in {1, ..., n}}$ and express $F$ as a matrix $[f_{ij}]_{i, j \in {1, ..., n}}$ of differentiable functions $f_{ij}:\mathbb{R} \to \mathbb{R}$ (same idea for $G$), and I imagine $(F\cdot\ G)'$ will look something like $F'\cdot\ G+F\cdot\ G'$, but I don't know how to formalize it.
[obs.: the matrix norm considered here is $||A||=\sup_{|v|=1}|A(v)|$, where $v \in \mathbb{R^n}$ and $|\cdot\ |$ is the usual euclidean norm]

REPLY [2 votes]: $E = \mathbb{R}^{n \times n}$ is a Banach space (which, with the given norm, is isomorphic to the space of continuous linear maps $\mathscr{L}(\mathbb{R}^n; \mathbb{R}^n)$).
The usual matrix multiplication $\mu \colon E \times E \to E$, $(B, C) \mapsto BC$ is bilinear; and it is continuous, because $\|BC\| \leqslant \|B\|\|C\|$ for all $B, C \in E$. Continuity follows from e.g. J. Dieudonne, Foundations of Modern Analysis (1969), proposition (5.5.1).
By e.g. Dieudonne's proposition (8.1.4), $\mu$ is differentiable at every point $(B, C)$ in $E \times E$, and its derivative at $(B, C)$ is the continuous linear mapping
\begin{equation}
\tag{1}\label{eq:mu}
\mu'(B, C) \colon E \times E \to E, \ (U, V) \mapsto BV + UC.
\end{equation}
By e.g. Dieudonne's proposition (8.1.5), the function
$$
H \colon E \to E \times E, \ A \mapsto (F(A), G(A))
$$
is differentiable at every point $A \in E$, and - under the natural linear isometry between $\mathscr{L}(E; E \times E)$ and $\mathscr{L}(E; E) \times \mathscr{L}(E; E)$, where both products have the $\sup$ norm - the derivative at $A$ is identified with the continuous linear mapping
$$
H'(A) = (F'(A), G'(A)).
$$
(I had better now spell that out in more detail - especially as I haven't properly studied differential calculus in Banach spaces! It took some brass nerve to post this answer. Partly it's in order to learn some of this stuff myself, but it is also to show that an answer requires virtually no calculation.)
The natural linear isometry in question sends $L : E \to E \times E$ to $(\operatorname{pr}_1 \circ L, \operatorname{pr}_2 \circ L)$, where $\operatorname{pr}_i$ ($i = 1, 2$) are the projection mappings $E \times E \to E$. So we have
\begin{equation}
\tag{2}\label{eq:H}
H'(A) \colon E \to E \times E, \ T \mapsto (F'(A)(T), G'(A)(T)).
\end{equation}
By the Chain Rule (which is Dieudonne's proposition (8.1.2)),
\begin{align*}
(F \cdot G)'(A) & = (\mu \circ H)'(A) \\ & = \mu'(H(A)) \circ H'(A) \\ & = \mu'(F(A), G(A)) \circ H'(A).
\end{align*}
Therefore, by \eqref{eq:mu} and \eqref{eq:H},
\begin{align*}
(F \cdot G)'(A)(T) & = \mu'(F(A), G(A))(H'(A)(T)) \\ & = \mu'(F(A), G(A))(F'(A)(T), G'(A)(T)) \\ & = \boxed{F(A)G'(A)(T) + F'(A)(T)G(A)}
\end{align*}
- which agrees with rmdmc89's answer.
By the same argument, the result holds generally for differentiable functions $F: W \to \mathbb{R}^{n \times p}$, $G: W \to \mathbb{R}^{m \times n}$, where $W$ is an open subset of a Banach space and $m, n, p$ are positive integers.


TITLE: What is the physics behind control/restriction valves?
QUESTION [2 upvotes]: I was recently looking at some closing characteristics of valves (the relationship between the opening area of a valve and its stroke) and couldn't wrap my head around something.

Why exactly the flow rate of an incompressible fluid decreases as you decrease the valve opening? What is the physics behind it?
If the fact that the fluid flowing is incompressible implies that different cross-sections of a tube have the same flow rate, what am I missing here?
I appreciate any insights, thanks in advance.

REPLY [1 votes]: A local restriction (valve) in a fluid conduit (pipe) causes a pressure drop (in engineering terms: 'head loss'). Since as volume throughput through the pipe depends on total pressure drop over the length of the pipe, this 'head loss' causes the volume throughput to be reduced accordingly.


TITLE: Intersection between orthogonal complement of a subspace and a set
QUESTION [15 upvotes]: Consider the normed vector space $E=\mathbb{R}^n$. Define
$ P=\{x \in \mathbb{R}^n: x_i \geq 0, \forall i \}$.
Let $M$ be a subspace such that $M \cap P = \{0\}$. I want to see that $M^\perp \cap {\rm Int}(P) \neq \emptyset $. This seems obvious geometrically, any idea how a short proof would look like?

REPLY [2 votes]: This is an elementary proof I discovered using a geometric version of Hahn-Banach. Hope you like it. Thanks to all of you who contributed to the discussion. :)
Suppose $M^\perp \cap \rm{Int}P=\emptyset$. Then since both sets are convex and $\rm{Int}P$ is open, it follows that there exists $0 \neq f \in E^*$ that separates them. That is, there exists $\alpha \in \mathbb{R}$ such that
$f(x) \leq \alpha \leq f(y) \text{ for all } x\in M^\perp, y \in \rm{Int} P. $
By Reisz's Lemma $\exists x^* \in E$, $x^*\neq0$, such that $f(x)=(x,x^*)$. Now it is easy to see that since $M^\perp$ is a subspace, the first inequality implies that $(x,x^*)=0$ for all $x \in M^\perp$, ie $x^* \in (M^\perp)^\perp=M$ (since $E$ is finite dimensional).
Therefore, from the second inequality we have $(y,x^*) \geq 0$ for all $y\in \rm{Int}P$. Finally, one only needs to note that $e_i \in \overline{\rm{Int}P}$ for all $i=1,\cdots,n$. The continuity of the inner product now implies that $(e_i,x^*) \geq 0$ for all $i$. I.e., $x^* \in P \cap M$ and $x^* \neq 0$, a contradiction.
$\square$


TITLE: What are the connections (if any) between "kernel" in "kernel density estimation," "kernel of a matrix," and "kernel method"?
QUESTION [2 upvotes]: I'm getting my understanding of kernel density estimation from pages 6-7 of this PDF.
If there are conceptual relationships between the "kernels" in each of these topics, I'd like to understand them.

REPLY [0 votes]: Kernel density estimation and Kernel methods(referring to interpolation or regression by using kernels) employ the same use of kernel. The kernel of a matrix is unrelated.
In the first two problems the kernel acts a basis function centered at a data points $k(x, x_i)$. The idea is that if we take linear combinations of these basis functions we can describe other functions $f(x) = \sum_i w_ik(x,x_i)$
In kernel density estimation we approximate a pdf as the average of a bunch of these basis functions. Typically people use radial basis functions $k(x,x_i) = \exp(-||x-x_i||^2/h)$
To get a good fit we need to tune the length scale parameters $h$. For general curve fitting with basis functions we do basically the same process but typically do a weighted average.


TITLE: Free objects in several constructs.
QUESTION [0 upvotes]: Let C be one of these constructs (in the sense that this is a concrete category over Set): Alg$(\Omega)$ or Set$^{op}$. Set$^{op}$ is a concrete category through a contravariant power-set functor. And I’m trying to find free objects in each of these categories.
How to use the definition of a free object to deduce them in these cases?
A free object over a set $X$ is an C-object $A$ such that there exists a universal arrow over $X$.
A universal arrow over a set $X$ is a structured arrow $X\overset{u}\longrightarrow |A|$ with domain $X$ that has the following universal property: for each structured arrow $X\overset{f}\longrightarrow |B|$ with domain $X$ there exists a unique C-morphism $f’:A\longrightarrow B$ such that $X\overset{u}\longrightarrow |A|\overset{f’}\longrightarrow |B|=X\overset{f}\longrightarrow |B|$.

REPLY [0 votes]: Using the definition you provide, a free object of $\mathbf{Set}^{\mathrm{op}}$ over a set $X$ is a set $A$ together with a function $u : X \to \mathcal{P}(A)$ such that, for all sets $B$ and functions $f : X \to \mathcal{P}(B)$, there is a unique function $f' : B \to A$ such that
$$f = X \xrightarrow{u} \mathcal{P}(A) \xrightarrow{(f')^{-1}} \mathcal{P}(B)$$
where ${f'}^{-1}$ is the function sending each $U \subseteq A$ to its preimage ${f'}^{-1}[U]$ under $f'$.

This is a bit technical, so let's consider the case when $X=1=\{0\}$ is a singleton set.
In this case, specifying a free object of $\mathbf{Set}^{\mathrm{op}}$ over $X$ is then equivalent to specifying a set $A$ and a subset $U \subseteq A$ such that, for all subsets $V \subseteq B$, there is a unique function $f' : B \to A$ such that ${f'}^{-1}[U] = V$.
This looks like a characteristic function. So let's define $A=2=\{0,1\}$ and $U=\{1\}$. Given a subset $V \subseteq B$, there is indeed a unique function $f' : B \to 2$ such that ${f'}^{-1}(\{1\})=V$, namely the characteristic function $\chi_V : B \to 2$.
So when $X=1$, the free object on $X$ is $2$ equipped with the map $1 \to \mathcal{P}(2)$ which picks out the singleton subset $\{1\}$.

Since $2 \cong \mathcal{P}(1)$ suggests that, in general you should take $A=\mathcal{P}(X)$ and take $u : X \to \mathcal{P}(\mathcal{P}(X))$ to be "the only thing it can be": define $u$ by
$$u(x) = \mathcal{F}_x := \{ U \subseteq X \mid x \in U \}$$
Given $f : X \to \mathcal{P}(B)$, define $f' : B \to \mathcal{P}(X)$ by
$$f'(b) = f^{-1}(\{b\}) = \{ x \in X \mid b \in f(x) \}$$
Then
$$\begin{align*}
f'^{-1}[u(x)]
&= {f'}^{-1}[\mathcal{F}_x] \\
&= \{ b \in B \mid f'(b) \in \mathcal{F}_x \} \\
&= \{ b \in B \mid \{ y \in X \mid b \in f(y) \} \in \{ U \subseteq X \mid x \in U \} \} \\
&= \{ b \in B \mid x \in \{ y \in X \mid b \in f(y) \} \} \\
&= \{ b \in B \mid b \in f(x) \}\\
&= f(x)
\end{align*}$$
I'll leave it up to you to check uniqueness.

I imagine the argument for $\mathbf{Alg}(\Omega)$ will take on a similar form. My general advice would be to look at a special case when $X$ is some simple, easy-to-understand set, such as the empty set or a singleton.


TITLE: Marginal PMF from joint discrete PMF, y=0,1, .. n ; x=0,1, ...
QUESTION [0 upvotes]: Studying for actuarial P1 exam, onto conditional discrete distributions.
See question-at-hand from textbook at this link>>1<<:

Problem 43.6
Let $X$ and $Y$ be discrete random variables with joint probability function
$$p_{X,Y}(x,y) = \begin{cases}\frac{n!\,{y}^x\,{(p\,\mathsf e^{-1})}^y\,{(1-p)}^{n-y}}{y!\,(n-y)!\,x!} & :y\in\{0,1,\cdots, n\}; x\in\{0,1,\cdots\} \\[1ex] 0 & : \textsf{otherwise}\end{cases}$$
(a) Find $p_Y(y)$.
(b) Find the conditional probability distribution of $X$, given $Y=y$.
Are $X$ and $Y$ independent?   Justify your answer.

I don't know how to approach (a).  Sometimes with a discrete joint pmf you can find the complete marginal pmfs by constructing a table and summing over the rows/columns, but that does not seem to be an option here, and I do not know any other option to find a marginal pmf of a discrete distribution.
Any help would be appreciated.
Thanks,
Alex

REPLY [1 votes]: Hint: You can rewrite your joint pmf as:
$${p}_{_{X,Y}}(x,y) = \binom{n}{y} p^y (1-p)^{n-y}\cdot \frac{y^x \mathsf e^{-y}}{x!} \quad\big[y\in\{0,\ldots, n\}, x\in\{0,\ldots\infty\}\big]$$
Does anything look familiar about that?

Thanks for responding. RHS consists of a geometric random variable with parameters (n, y) multiplied by a poisson random variable with parameter y. -- AlexanderWright

Almost.   $Y$ has a Binomial distribution, and $X$ has a conditionally Poisson distribution for a given $Y$.
This suggests that $Y$ is the count of successes in a series of $n$ iid Bernoulli trials (of success rate $p$) and that count determines the arrival rate of Poisson events, the count of which is then $X$.
$$Y\;\sim\; \mathcal{Bin}(n, p) \\ X\mid Y\;\sim\;\mathcal{Pois}(Y)$$

The condition that must be met for two random variables X and Y to be independent is that $p_{_{X,Y}}(x,y) = p_{_X}(x) \cdot p_{_Y}(y)$, correct? What I see here is a $p_{_X}(x) \cdot p_{_Y}(y)$, although it just so happens that $p_{_X}(x)$ has parameter $y$. Is this just a convenient case where a dependent joint distribution can be separated into two neat marginal distributions? Any input appreciated

No, it's not that case at all.   The very point is that, because of the $y^x$ factor, the joint distribution is not separable into independent marginal distributions.   Although we can separate into two factors which are identifiably distribution functions, one is clearly conditional.
Thus we have : $p_{_{X,Y}}(x,y) = p_{_Y}(y)\cdot p_{_{X\mid Y}}(x\mid y)$.
The factor, $p_{_{X\mid Y}}(x\mid y)$, sometimes written as $p_{X\mid Y=y}(x)$, is the conditional probability mass function of $X$ given $Y = y$.

The marginal for $X$ would be:
$$p_{_X}(x) = \frac 1{x!} \sum_{y=0}^n \binom{n}{y} {(p\mathsf e^{-1})}^y(1-p)^{n-y}y^x$$
This would be ...difficult... to evaluate.   Fortunately it's unnecessary to do so, since clearly the probability mass of $X$ varies under conditions of given $Y$.   Thus : dependence.


TITLE: Finding all solutions to $\cos \theta = −\frac{1}{\sqrt{2}}$ in a given interval
QUESTION [0 upvotes]: Find all solutions between $-2\pi$ and $2\pi$ of the following equation:
  $$\cos \theta = −\frac{1}{\sqrt{2}}$$
  Give all answers in radians.

Now I have drawn the graph between $−2\pi$ and $2\pi$.
Am I correct thinking that $\cos \theta = 2\pi - \theta$?
I am completely at a loss of where to draw the values on the graph to work this out.

REPLY [0 votes]: By symmetry, $\cos\theta = \cos(-\theta)$.

Moreover, since two coterminal angles have the same cosine,
\begin{align*}
\cos(\theta + 2n\pi) & = \cos\theta, n \in \mathbb{Z}\\
\cos(-\theta + 2n\pi) & = \cos(-\theta) = \cos\theta, n \in \mathbb{Z}
\end{align*}
where $\mathbb{Z}$ is the set of integers.
One solution for the equation
$$\cos\theta = -\frac{1}{\sqrt{2}}$$
is
$$\theta = \arccos\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}$$
Hence, the general solution is
\begin{align*}
\theta & = \frac{3\pi}{4} + 2m\pi, m \in \mathbb{Z} & \theta = -\frac{3\pi}{4} + 2n\pi, n \in \mathbb{Z}
\end{align*}
The requirement that $-2\pi \leq \theta \leq 2\pi$ implies that $m = 0, -1$ and that $n = 0, 1$, which yields the solutions
$$\theta = \pm \frac{3\pi}{4}, \pm \frac{5\pi}{4}$$


TITLE: Prove $(H \times 1)(1\times K)= H\times K$ where $H,K$ are groups.
QUESTION [0 upvotes]: Prove $(H \times 1)(1\times K)= H\times K$ where $H,K$ are groups.
Suppose $x=ab,a\in H\times 1,b\in 1\times K$
Then $x=(h,1)(1,k)$ where $h\in H,k\in K$
Hence $x=(h,k)\in H\times K$
Let $(h,k)\in H\times K$
Then $(h,k)=(h,1)(1,k)\in (H\times 1)(1\times K)$
What I want to ask is $(h,1)(1,k)=(h,k)$ always true and is there any difference between $(H\times 1)(1\times K)$ and $(H\times 1)\times(1\times K)$?

REPLY [0 votes]: Multiplication in $H\times K$ is given by
$$(h_1,k_1)(h_2,k_2)=(h_1h_2,k_1k_2)$$
so, yes, $(h,1)(1,k)=(h1,1k)=(h,k)$.
Now, elements of $H\times K=(H\times 1)(1\times K)$ are pairs $(h,k)$ with $h\in H$ and $k\in K$. Elements of $(H\times 1)\times(1\times K)$ are pairs of pairs $((h,1),(1,k))$ with $h\in H$ and $k\in K$. Of course, there is an obvious isomorphism $$H\times K\to (H\times 1)\times (1\times K)$$ so this difference is superficial.


TITLE: Thermodynamics : Doubt in First Law
QUESTION [1 upvotes]: I just started my study of thermodynamics. I am having problem in uderstanding a concept.
Suppose I drop an ice-cube and a rock of cube shape. Which one will reach the ground first? I know ratio of Work to Heat equals Joules Constant. Therefore any change in potential energy will result in heat which in return will melt ice cube thus mass is changing. Will internal energy of ice cube change till it reaches the ground? If mass of ice cube is changing what will be the ratio of kinectic energy of ice cube to that of rock.

REPLY [0 votes]: The falling process can be quite complex when we come to answer your with one word: stone or water.
When an object falls, it will be subjected to the following two forces: gravity force and air drag force. If the air drag force can be ignored, the gravity force acts on the object converting the potential energy to a kinetic energy. The answer can easily be that they reach the ground at the same time.
When there is viscous force, the object shape and its surface viscous force play an important role (e.g. a feature vs. a stone of the same mass). The viscous force is a function of velocity. If they have the same shape, then they reaches the ground at the same time.
Viscous force does work that converts kinetic energy to heat. This heat will heat up air in vicinity and ice. It can melt the ice cube. The water can change shape easily. With the shape change, the ice will have less drag and reach the ground first.
Viscous force can also shear off some of the water, whereas it would not tear off any material of the stone. This will make the ice smaller and less drag. Ice will hit the ground first.
The ice may completely evaporate before it reaches ground. In this case, a rock will hit the ground first.
Now, back to the first law, there are components of potential energy, kinetic energy and internal energy. But there is also an important component: work. Nothing can be ignored and all stone should be turned.


text:
\section{Provable Satisfiability of Weak Learning Condition using Over-parameterized Models}\label{sec:wlc-by-overparameterized-model}

In this section, we show that there exists a model that for any w.l.c.\ parameters $\alpha$ and $\beta$, we can find a weak learner which \textit{probably} satisfies the w.l.c.\ using the gradient descent algorithm.
To ensure the w.l.c., the set of transformation functions $\mathcal{B}$ must be sufficiently large so that it can approximate all possible values of the negative gradient $-\nabla \widehat{\mathcal{L}}$ can take.
We can accomplish it by leveraging the universal approximation property of MLPs, similarly to graph isomorphism networks (GIN)~\cite{xu2018how}, but for a different purpose.
We adopt the recent studies that proved the global convergence of over-parameterized MLPs trained by a tractable algorithm (e.g.,~\cite{pmlr-v97-arora19a,du2018gradient}).

Let $R\in \mathbb{N}_+$.
For the parameter $\Theta = (\theta_{ri}) \in \mathbb{R}^{R \times N}$, we consider an MLP with a single hidden layer $f_\Theta: \mathbb{R}^C\to \mathbb{R}$ defined by
\begin{equation*}
    f_{\Theta}(\bm{x}) := \frac{1}{\sqrt{R}} \sum_{r=1}^R a_{r} \sigma (\theta_{r\cdot}^\top \bm{x}),
\end{equation*}
where $a_{r}\in \{-1, 1\}$ and $\sigma$ is the ReLU activation function: $\sigma(x) := x\vee 0$ (we apply ReLU in an element-wise manner for a vector input).
We define the set of transformation functions $\mathcal{B}:= \{(f_\Theta, \ldots, f_\Theta) \mid \Theta \in \mathbb{R}^{R\times N} \}$.
At the $t$-th iteration, given a gradient $\nabla\widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})$, we initialize the model with $a_{r}\overset{\text{i.i.d.}}{\sim}\mathrm{Unif}(\{-1, 1\})$ and $\theta_{ri} \overset{\text{i.i.d.}}{\sim} \mathcal{N}(0, I)$ independently and train it with the gradient descent to optimize $\Theta$ by minimizing the mean squared error between the output of the model and the properly normalized negative gradient.
Using the result of~\cite{du2018gradient}, we obtain the following guarantee.
\begin{proposition}\label{prop:wlc_by_overparameterized_model}
    Suppose Algorithm \ref{alg:train} finds $g^{(s)} \in \mathcal{G}^{(s)}$ ($s\in [t]$) such that $X^{(t)} = g^{(t)} \circ \cdots \circ g^{(1)}(X) = \begin{bmatrix}\bm{x}_1 & \cdots & \bm{x}_N \end{bmatrix}^\top\in \mathbb{R}^{N\times C}$ ($\bm{x}_i\in \mathbb{R}^C$) satisfies the conditions that $\bm{x}_i\not =0$ for all $i\in [N]$ and $\bm{x}_{i} \nparallel \bm{x}_{j}$ for all $i\not = j \in [N]$.
    Let $\delta, \alpha, \beta > 0$.
    Then, there exists $R = O(N^6\delta^{-3})$ such that for all $\nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})$, with a probability of at least $1-\delta$, the gradient descent algorithm finds $b^{(t)}\in \mathcal{B}^{(t)}$ such that the $t$-th weak learner $f^{(t)}$ satisfies the $(\alpha, \beta, -\nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)}))$-w.l.c.
\end{proposition}

\begin{proof}
    We define $H^\infty\in \mathbb{R}^{N\times N}$ by $H^\infty_{ij} := \mathbb{E}_{\bm{w}\sim \mathcal{N}(0, I)}[\bm{x}_i^\top \bm{x}_j \bm{1}\{\bm{w}^\top \bm{x}_i\geq 0\} \bm{1}\{\bm{w}^\top \bm{x}_j \geq 0\} ]$.
    Let $\lambda_0$ be the lowest eigenvalue of $H^\infty$.
    Under the assumption, we know $\lambda_0 > 0$ by Theorem 3.1 of~\cite{du2018gradient}.
    We train the parameter $\Theta_{\cdot c\cdot}$ using the dataset $((\bm{x}_1, -\alpha[\widehat{\mathcal{L}}(\widehat{Y}^{(t-1)}]_{1}), \ldots, (\bm{x}_N, -\alpha[\widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})]_{N})$.
    We denote the parameter of the MLP at the $k$-th iteration of the gradient descent by $\Theta^{(k)}$.
    We denote the output of the model $f_{\Theta^{(k)}}$ by $\bm{u}^{(k)} := (f_{\Theta^{(k)}}(\bm{x}_1), \ldots, f_{\Theta^{(k)}}(\bm{x}_N))^{\top}$.
    By~\cite[Theorem 4.1]{du2018gradient}, with probability $1-\delta$, we have
    \begin{equation*}
        \|\bm{u}^{(k)} + \alpha\nabla\widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})\|_2
        \leq \left(1 - \frac{\eta\lambda_0}{2}\right)^k
        \|\bm{u}^{(k)} + \alpha\nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)}) \|_2,
    \end{equation*}
    where $\eta = O\left(\frac{\lambda_0}{N^2}\right)$.
    Set
    \begin{equation*}
        k := \log\left(\frac{\beta \|\nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})\|_{2}}{\|\bm{u}^{(0)} + \alpha\nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})\|_2\vee 1}\right)
        \left(\log \left(1 - \frac{\eta\lambda_0}{2}\right)\right)^{-1}.
    \end{equation*}
    Then, with probability $1-\delta$, we have
    \begin{equation*}
        \|\bm{u}^{(k)} + \alpha \nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})\|_{2} \leq \beta \|\nabla \widehat{\mathcal{L}}(\widehat{Y}^{(t-1)})\|_{2},
    \end{equation*}
    which means $\bm{u}^{(k)}$ satisfies $(\alpha, \beta, -\nabla\widehat{\mathcal{L}}(\widehat{Y}^{(t-1)}))$-w.l.c.
\end{proof}

\begin{remark}
    Reference \cite{du2018gradient} assumed that any feature vector $\bm{x}$ of the training data satisfies $\|\bm{x}\| = 1$.
    However, as commented in~\cite{du2018gradient}, we can loosen this condition as follows: there exists $c_{low}, c_{high} > 0$ such that any feature vector $\bm{x}$ satisfies $c_{low} \leq \|\bm{x}\|\leq c_{high}$.
\end{remark}
Although this instantiation provably satisfies the w.l.c.\ with high probability, its model complexity is extremely large because it has as many as $O(N^6)$ parameters.
As we saw in Section~\ref{sec:discussion}, such a large model complexity is inevitable as long as the gradient can take arbitrary values.

config: arxiv


text:
\begin{definition}[Definition:Consequence Function with Probability]
Let $G$ be a [[Definition:Game|game]].
Let $P$ be a [[Definition:Player|player]] of $G$.
Let $A$ be the [[Definition:Set|set]] of [[Definition:Move|moves]] available to $P$.
Let $C$ be the [[Definition:Set|set]] of [[Definition:Consequence|consequences]] of those [[Definition:Move|moves]].
Let the [[Definition:Consequence|consequences]] of those [[Definition:Move|moves]] be affected by a [[Definition:Random Variable|random variable]] on a [[Definition:Probability Space|probability space]] $\Omega$ whose [[Definition:Realization of Random Variable|realization]] is not known to the [[Definition:Player|players]] before they make their [[Definition:Move|moves]].
A '''consequence function''' for $P$ is a [[Definition:Mapping|mapping]] from $A \times \Omega$ to $C$:
:$g: A \times \Omega \to C$
interpreted as that $g \left({a, \omega}\right)$ is the [[Definition:Consequence|consequence]] when the [[Definition:Move|move]] is $a \in A$ and the [[Definition:Realization of Random Variable|realization]] $\omega$ of the [[Definition:Random Variable|random variable]] is $\omega \in \Omega$.
\end{definition}

config: wiki


TITLE: Matrix proof by induction
QUESTION [2 upvotes]: Problem
Let $A_n\in M\left(\mathbb{R}\right)$ such that
$$A_n=\begin{pmatrix}1+x_1&x_2&...&x_n\\ x_1&1+x_2&...&x_n\\ ...&...&...&...\\ x_1&x_2&...&1+x_n\end{pmatrix}$$
Demonstrate that $\det(A_n)=1+x_1+x_2+...+x_n$
Thoughts
I think that the best way is by induction.
Starting with a inductive base
let say $n=2$
$$A_2=\begin{pmatrix}1+x_1&x_2\\ x_1&1+x_2\end{pmatrix}$$
$$\det(A_2)=1+x_2+x_1$$
Its true for that case
Now the inductive hypothesis would be for $n=k$
$$\det(A_k)=\begin{pmatrix}1+x_1&x_2&...&x_k\\ x_1&1+x_2&...&x_k\\ ...&...&...&...\\ x&x_2&...&1+x_k\end{pmatrix}=1+x_1+x_2+...+x_k$$
Now we need to prove that $n=k+1$
$$\det(A_n)=\begin{pmatrix}1+x_1&x_2&...&x_k&x_{k+1}\\ x_1&1+x_2&...&x_k&x_{k+1}\\ x_1&x_2&...&...&....\\ ...&...&...&1+x_k&...\\ x_1&x_2&...&x_k&1+_{k+1}\end{pmatrix}=1+x_1+x_2+...+x_{k+1}$$
Then of that I'm unsure about how to proceed, I tried something like representing them by their cofactors but didn't let me anywhere. Any help would be appreciated.

REPLY [0 votes]: If you subtract the second line of $A_n$ from the first one, the determinant remains the same and the entries of the first line become are $1,-1,0,0,\ldots,0$. Now, if you subtract the third line from the second one, the entries of the second line become $0,1,-1,0,\ldots,0$. And so on. So$$\det A_n=\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\x_1&x_2&x_3&x_4&\ldots&x_{n-1}&1+x_n\end{vmatrix}.\tag1$$Now, if you add the first line times $-x_1$ to the last line, the $RHS$ of $(1)$ becomes$$\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\0&x_1+x_2&x_3&x_4&\ldots&x_{n-1}&1+x_n\end{vmatrix}.\tag2$$And now, if you add the second lines times $-x_1-x_2$ to the last line, the RHS of $(2)$ becomes$$\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\0&0&x_1+x_2+x_3&x_4&\ldots&x_{n-1}&1+x_n\end{vmatrix},$$and so on… In the end, you will get$$\begin{vmatrix}1&-1&0&0&\ldots&0&0\\0&1&-1&0&\ldots&0&0\\0&0&1&-1&\ldots&0&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&0&\ldots&1&-1\\0&0&0&0&\ldots&0&1+x_1+\cdots+x_n\end{vmatrix}=1+x_1+\cdots+x_n.$$


TITLE: small tip on If $a_n>0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=l$ prove that $\lim a_n^{\frac{1}{n}}=l$
QUESTION [1 upvotes]: I have the following problem:
If $a_n>0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=l$ prove that $\lim_{n\to\infty} a_n^{\frac{1}{n}}=l$.
Now I know I have no work to show but I really don't even know how to start. I need a small tip on how to get going with this problem and not a whole solution.
Thanks in advance!

REPLY [1 votes]: Let $v_{1}=u_{1}, v_{2}=\frac{u_{2}}{u_1}, .....,v_{n}=\frac{u_{n}}{u_{n-1}},.....$
Then $v_{n}> 0$  for all $n\in \mathbb{N}$ and $\lim v_{n}=l>0.$
This imply $\lim \log v_{n}=\log l.$ Then by cauchy theorem,
$\lim \frac{\log{v_{1}}+\log{v_{2}}+......+\log{v_{n}}}{n}=\log l.$
i.e $\lim \log (v_{1}. v_{2}....v_{n})^{1/n}=\log l$.
i.e $ \lim (v_{1}. v_{2}....v_{n})^{1/n}=l.$
i.e $\lim u_{n}^{1/n}=l$.


TITLE: Showing that a function $G : [0,1) \times [0,1) \rightarrow [0,1)$ is not a surjection
QUESTION [4 upvotes]: While proving $|\Bbb C|=|\Bbb R|$ in Just and Weese's Set Theory book, they construct a function $G : [0,1) \times [0,1) \rightarrow [0,1)$ where $\langle x,y\rangle \in [0,1) \times [0,1)$ such that $x = +.x_0x_1x_2...$ and $y=+.y_0y_1y_2...$ and $G(x,y) = +.x_0y_0x_1y_1x_2y_2...$
It is clear to me why this function is one-to-one, however he makes the comment that it does not map $[0,1) \times [0,1)$ onto $[0,1)$. But since each digit of $z\in G$ is free to be any number 0 through 9 without restriction, I am led to the conclusion that G is onto. My guess at a solution would be to create some function that alters $z$ in such a way that it cannot be written as $+.x_0y_0x_1y_1x_2y_2...$  however I cannot find such a function.

REPLY [2 votes]: You must first fix some convention as to what do you do with numbers admitting two different decimal representations, otherwise the functions $G$ is not defined. Whatever choice you make, consider $x=0.1919191919191919191919....$. It can't be in the image of the function $G$ since the only way to interleave two decimal representations to obtain $x$ is by $G(0.11111....,0.99999.....)$. However, that pair is not in the domain of the function.


text:
{\bf Problem.} Jeff decides to play with a Magic 8 Ball. Each time he asks it a question, it has a 2/5 chance of giving him a positive answer. If he asks it 5 questions, what is the probability that it gives him exactly 2 positive answers?
{\bf Level.} Level 4
{\bf Type.} Counting & Probability
{\bf Solution.} There are $\binom{5}{2}=10$ ways to choose 2 of 5 questions to give a positive response to. For each choice, there is a probability of $\left( \frac{2}{5} \right)^2 \left( \frac{3}{5} \right)^3$ that those 2 questions will yield a positive answer and the other 3 will not. Therefore, the total probability that exactly 2 of questions give a positve answer is $10 \left( \frac{2}{5} \right)^2 \left( \frac{3}{5} \right)^3 = \boxed{\frac{216}{625}}$.

set_name: MATH