TITLE: Is $f(x):=(x-a)^n g(x)$ $n$ times differentiable at $a$, for any $g$ continuous at $a$?
QUESTION [1 upvotes]: This is a try to fill the holes in this previous exercise.
Let $f$ defined as $f(x):=(x-a)^n g(x)$, for some $g$ continuous in a neighborhood of $a$, we can say that $f$ is $n$ times differentiable at $a$?
If we define the quotient difference operator
$$\Delta^n_h f(x):=\frac1{h^n}\sum_{k=0}^n\binom{n}{k}f(x+(n-k)h)(-1)^k$$
then if $f$ is continuous in a neighborhood of $a$ and if Im not wrong we can write
$$\lim_{h\to 0}\Delta^n_h f(a)=f^{(n)}(a)$$
For the case of the title this reduces to
$$\begin{align}f^{(n)}(a)&=\lim_{h\to 0}\frac1{h^n}\sum_{k=0}^n\binom{n}{k}(-1)^k((n-k)h)^ng(a+(n-k)h)\\&=\lim_{h\to 0}\sum_{k=0}^n\binom{n}{k}(-1)^k(n-k)^ng(a+(n-k)h)\\&=\sum_{k=0}^n\binom{n}{k}(-1)^k(n-k)^ng(a)\end{align}$$
and for derivatives for $j\le n$ we have that $f^{(j)}(a)=0$.
It is this reasoning correct?

REPLY [3 votes]: The reasoning is not correct. The equality
$$\lim\limits_{h\rightarrow 0}\Delta_h^nf(a)=f^{(n)}(a)$$
is true, provided that the RHS exists. However, just because the limit on the LHS exist, it doesn't follow the RHS exists. Indeed, the statement in the question is false. This is because $f(x)$ can oscillate very rapidly in the neighbourhood of $a$, which would make the first derivative be discontinuous at $a$. To be precise, consider for example
$$g(x)=x\sin x^{-2}\text{ for }x\neq 0,\\g(0)=0$$ (check it's continuous) and $$f(x)=x^2g(x)=x^3\sin x^{-2}.$$
We then find
$$f'(x)=3x^2\sin x^{-2}-2\cos x^{-2},\\
f'(0)=0,$$
which is discontinuous at zero, hence $f$ is not twice differentiable.
For general $n$, you can similarly take $f(x)=x^{n+1}\sin x^{-n}$.


text:
\section{Conclusion}
This work presented a general framework for the multi-agent assortment optimization problem in sequential matching markets. While the related literature focuses on the MNL-Uniform model, we significantly expanded it to general choice models that satisfy the regularity axiom. Moreover, our approach allows us to consider the maximization of the expected revenue of the matching and the problem subject to cardinality constraints. When consumers are easy-to-match, we provide improved constant guarantees by using techniques from submodular optimization. On the other hand, when consumers follow a MNL model, our approach relies on LP relaxations and appropriate (randomized) rounding techniques.

Several research directions remain to be explored. We believe that a deeper analysis would give an improved guarantee for the problem studied in Section \ref{sec:mnl_unconstrained}. The computational study shows that simple assortment families, such as $V^N$, perform relatively well. Moreover, we think that the LP relaxation \eqref{eq:main_lp} can be used to obtain constant approximation guarantees for a combination of high- and low-value public scores $v_{ij}$, namely, when they are not restricted to be less than 1. Another possible research direction is to study other choice models whose system of choice probabilities can be described as linear inequalities. As conjectured in Section \ref{sec:further_challenges}, the Markov chain based model is a tentative option. Finally, we think that it would be interesting and promising from a practical perspective to look at models that allow a second round of matches, namely, remaining unmatched suppliers can be matched.

config: arxiv


TITLE: Quotient Space of Vector Spaces
QUESTION [1 upvotes]: I'm trying to prove that for subspaces of $V,U$ of a vector space $W$ it holds that
$
V=(U\cap V) \oplus V'
$
where $V'$ is isomorphic to the quotient space $V/(U\cap V)$. I think I got the intuition right but I'm not really sure how to show it. Can anyone help me?
And why is
$U+V=U+((U\cap V)\oplus V')=U\oplus V'$

REPLY [2 votes]: It is true without any hypothesis on dimensions.
Consider the short exact sequence:
$$0\to U\cap V\xrightarrow{\ i\ } V\xrightarrow{\ p\ } V/U\cap V\to 0$$
Let $\mathcal B=(e_i)_{i\in I}$ be a basis of the quotient space $V/U\cap V$. Consider $\mathcal B'=(e'_i)_{i\in I}$, with $p(e'_i)=e_i$. This is a set of linearly independent vectors of $V$, and mapping each $e_i$ to $e'_i$ defines a linear section $s$ of $p$.
Now for any $v\in V$, one checks $v-s(p(v))\in\ker p=U\cap V$. Setting $V'=s(V/U\cap V)$, we have proved that $\;V=(U\cap V)+V'$.
This sum is direct: indeed, if $u\in U\cap V$ lies in $V'$, i.e. if $u=s(p(v))$ for some $v\in V$, then $p(u)=0$ by definition, so, as $s$ is a section of $p$, $\;p(s(p(v))=p(v)=0$ and thus  $\;u=s(p(v))=0$.


TITLE: $|z-1|+|z-2|+|z-3|=6$ in the Argand Plane
QUESTION [4 upvotes]: In the argand plane  $$C:|z-1|+|z-2|+|z-3|=6 ~~~(1)$$ represents a bounded curve which is a rounded blob  mimicking an "ellipse".
Using the the inequality $$|Z_1+Z_2+Z_3| \le |Z_1|+|Z_2|+|Z_3|, ~~~~(2)$$ we can see that $$6=|z-1|+|z-2|+|z-3|\ge |3z-6| \implies |z-2| \le 2. ~~~(3)$$
The circle $C_1: |z-2|=2$ touches and remains outside  $C$.
Next, it will be interesting to use some other inequality or something to find the equation of a bounded curve $C_2$ which touches and remains inside $C$.  Eventually, $C$ will  touch $C_1$ and $C_2$ and it will be enclosed between them.
The question is to find the curve $C_2.$
Edit:
For $C_2$, the RMS-AM inequality gives:
$$\sqrt{\frac{|z-1|^2+|z-2|^2+|z-3|^2}{3}} \ge \frac{|z-1|+|z-2|+|z-3|}{3}=2~~~~(4)$$
The equality in above doesn't holds as $|z-1|=|z-2|=|z-3|$ cannot be met
$$\implies(x-2)^2+y^2 >\sqrt{\frac{10}{3}}=1.82574~~~~(5)$$
This circle $|z-2|=\sqrt{\frac{10}{3}}$ cannot be the required $C_2$ as  it cannot touch $C$ in (1).

REPLY [2 votes]: Let us consider the equivalent problem, with a $(-2,0)$ translation:
$$|z+1|+|z|+|z-1|=6\tag{1}$$
Let (C) be the corresponding curve (generalized elllipse). It is possible to give an intuitive graphical construction of (C) (see Fig. 2) that will naturaly involve the maximal circle. Morover, in this way, we will be able to provide parametric equations of (C) (see (4)).
Due to the invariance of (1) with respect to transformations $z \to -z$ and $z \to $\overline{z}$, it is enough to study (C) in the first quadrant.

Fig. 1: The thick blue curve is the part of (C) belonging to the first quadrant (passing in particular through point $(2,0))$. It has been plotted  using parametric equations (4). The other curves are level curves with equations $|z+1|+|z|+|z-1|=k$ for different values of $k$.
Let us introduce a parameter $m$ transforming (1) into the equivalent issue:
$$\begin{cases}\text{Circle }(C_m):&|z|&=&2m\\\text{Ellipse } (E_m):&|z+1|+|z-1|&=&6-2m\end{cases}\tag{2}$$
This is represented on Fig. 2:

Fig. 2: Circles $(C_m)$ and ellipses $(E_m)$ for values of $m \in [0,1]$ from red (small values of) $m$ to blue (values of $m$ close to $1$). Curve (C) in black. In fact $m \in [m_0/2,1]$ where $m_0=\sqrt{\tfrac{44}{3}}-2$... We have found back this critical value! The corresponding circle is thickened : it is the biggest circle that can be included into (C), and tangent to it (a single intersection solution).
Clearly, $m \in [0,3]$. In fact, the domain of variation of $m$ is much more narrow: this is easy to understand on Fig. 2: $m$ cannot be too small, because in this case the circle is itself too small to have an intersection with the ellipse.
The foci of $(E_m)$ are $z=-1$ and $z=1$ ; its semiaxes are:
$$a=3-m \ \ \text{and} \ \ b=\sqrt{a^2-1}.$$
A classical parametrization for the ellipse is:
$$\begin{cases}x&=&a \cos \theta\\y&=&b \sin \theta \end{cases} \ \iff \ \begin{cases}x^2&=&(3-m)^2 \cos^2 \theta\\y^2&=&((3-m)^2-1) \sin^2 \theta \end{cases}\tag{3}$$
If $(x,y)\in(C_m)\cap(E_m)$, we add the two right hand equations:
$$(2m)^2=(3-m)^2-\sin^2 \theta$$
For the first quadrant, where $\sin \theta \ge 0$ and $\cos \theta \ge 0$, we have a unique solution:
$$\sin \theta=\sqrt{(3-m)^2-4m^2}=\sqrt{-3(m-1)(m+3)}\tag{4}$$
meaning that the domain of variation of $m$ is included into $[0,1]$ (in fact, smaller).
We deduce from (3) and (4) the following parametric equations for the locus :
$$\begin{cases}x&=&(3-m)\sqrt{3m^2+6m-8}\\ y&=&\sqrt{-3(4-m)(2-m)(m-1)(m+3)}\end{cases}\tag{5}$$
(valid for the first quadrant as said upwards).
Remark: the critical value of the radius can be obtained (see legend of Fig. 2) by setting $x=0$ in (4), i.e. by solving $3m^2+6m-8=0$.


TITLE: Smooth morphism has smooth source
QUESTION [1 upvotes]: I am asking this question in order to understand the answer to this five years old question. I also added a comment there, but the user answering the question hasn't been active for years so I don't expect to get an answer there.
In the question mentioned above, it is claimed that a smooth morphism $X\to Y$ has smooth source, i.e. that $X$ is smooth. I was unable to find this statement anywhere and I have no idea at all about how to see this is true.
Any help is kindly appreciated.

REPLY [1 votes]: If $Y$ is smooth and $f:X \to Y$ is smooth then $X$ is smooth (this is Hartshorne III.10.1(c) for $Z$ a point). If $Y$ is singular then $X$ is singular as well (the simplest example is when $f$ is an isomorphism).


TITLE: Solve $\int_0^{\frac{\pi}{3}} \frac{\sin\theta + \sin\theta\tan^2\theta}{\sec^2\theta}d\theta$ without using integration by parts or substitution
QUESTION [0 upvotes]: I was required to solve $$\int_0^{\frac{\pi}{3}} \frac{\sin\theta + \sin\theta\tan^2\theta}{\sec^2\theta}d\theta$$
without using integration by parts or substitution. This is, I must use trigonometric identities to boil the expression down to identifiable antiderivatives and apply the fundamental theorem of calculus directly. Clearly
$$\frac{\sin\theta + \sin\theta\tan^2\theta}{\sec^2\theta} = \frac{\sin\theta}{\sec^2 \theta} + \sin\theta$$
The second term is easily integrable because $\int \sin x \space dx = -\cos x + C$. But I can't seem to transform $\frac{\sin\theta}{\sec^2 \theta}$ into an identifiable antiderivative. I'm sure there's some trigonometric identity or trick I'm missing, but after consulting Stewart's Calculus and the web I still did not find a possible solution.
Thanks in advance.
Ref: Problem 38, Section 5.4 in Stewart's Calculus: Early transcendentals, 7 ed.

REPLY [5 votes]: You may want to write the numerator as
$\sin\theta\color{blue}{(1+\tan^2\theta)}$
and compare the blue factor with $\sec^2\theta$ in the denominator. The integral is really a softball in disguise.


TITLE: Limiting operations within sets, or intervals.
QUESTION [0 upvotes]: I just have a question about using limits with union and intersection operations.
To prove the claim that the intersection of infinite collection of open sets may not be open (side question: is it possible for such an intersection to be open?), a common textbook example is goes something like: consider the intersection of sets defined by
$$G_n = (0, 1+ 1/n)$$
where it is clear that every set is open. From here, the argument from two of my sources pretty much goes as follows: the intersection of such sets is $(0, 1]$, which is not open. The end.
I understand that, and believe it. But I'm not sure how they applied the limiting operation with the intersection involved. The best argument I could come up with was just saying, well, $(0,1]$ is definitely contained in the intersection of all those sets and that for any $\epsilon_1 > 0$ in $(0, 1+ \epsilon_1)$ we can generate another set $(0, 1+ \epsilon_2)$ where $\epsilon_2 < \epsilon_1$ by blah blah... so that $1 + \epsilon_1 \notin (0, 1+ \epsilon_2) \cap  (0, 1+ \epsilon_1)$. I know that's true. But is that a good way of finding this intersection?

REPLY [2 votes]: They did not apply a limiting operation.
Addition is a binary operation, so when you want to add more than two things, you need to repeatedly add two things over and over until you work through them. This can only ever add finitely many things, so to do infinite sums you do a limiting operation.
Intersection, however, is an operation on arbitrary collections of sets. You don't repeatedly take pairwise intersections and then a limit — you just take the intersection of everything in one "step".
The definition is that $x \in \cap_{i \in I} X_i$  if and only if $x \in X_i$ for every $i \in I$. In the specific example:

$ x \in \bigcap_{n=1}^{\infty} \left(0, 1 + \frac{1}{n} \right) $
  if and only if, for every integer $n \geq 1$, you have $x \in \left(0, 1 + \frac{1}{n} \right)$

For $x > 1$, you can show that there exists an $n$ such that $x \notin \left(0, 1 + \frac{1}{n}\right)$, and thus you conclude that $x \notin \bigcap_{n=1}^{\infty} \left(0, 1 + \frac{1}{n} \right)$.


TITLE: Proving that three numbers are in A.P.
QUESTION [0 upvotes]: A number $a$ is the Arithmetic Mean(A.M.) between $b$ and $c$, $b$ is the Geometric Mean(G.M.) between $a$ and $c$. Prove that $$\frac{1}{a}, \frac{1}{c} and \frac{1}{b}$$ are in Arithmetic Progressions(A.P.).
I haven't been able to do much, but this is it :
As, $a$ is the A.M. between $b$ and $c$,
$$a = \frac{b+c}{2}$$ or, $$2a = b+c$$
And, $b$ is the G.M. between $a$ and $c$, so, $$b^2 = ac$$ or, $$b = \sqrt{ac}$$How do i proceed next?

REPLY [1 votes]: As $a$ is the Arithmetic mean, it lies between $b$ and $c$. But, as $b$ is the geometric mean it will lie between $a$ and $c$. This can only mean that $a=b$. And because $a=\frac{b+c}{2}$,  $a=b=c$. Clearly,$\frac{1}{a}= \frac{1}{c} = \frac{1}{b}$ and therefore, trivially, in AP.