问:
What is the probability that at least one red ball is drawn when three balls are drawn from a bag with $5$ red and $5$ yellow balls?
I am having a tough time trying to understand the concept of probability and figuring out which is the right way to solve this problem. So I would really appreciate a lot if you can help me with this question

Problem:  There are 5 red balls and 5 yellow balls. If we make 3 draws from this set, without replacing the ball drawn in each set, what is the probability that in the end we have drawn at least one red ball.

Method 1:

Calculate the probability of not drawing any red ball in all three draws and subtract that from 1.

The probability of drawing 3 yellow balls in succession : 5/10 * 4/9 * 3/8

So then the probability of drawing at least one red ball is 1-(5/10 * 4/9 * 3/8) = 11/12

Method 2:
Create the possible color combinations that the 3 draws can produce. I can come up with only 8 such combinations:

RRR, YYY, YRR, YRY, YYR, RRY, RYR, RYY

So based on these combinations I see that there is only 1 combination that has no red ball. So isthe probability of drawing a red ball then 7/8?

Which is the right answer and the right way of solving this problem.

答: The first is correct. In your second method you haven’t taken into account the fact that the eight different outcomes have different probabilities.</p>


问:
Show that each eigenvalue of matrix $M_Z \Sigma M_Z$ is less than the corresponding one of $M_X \Sigma M_X$
Let $Z:=[X \hspace{0.2cm} Y]$ be an $n\times k$ partitioned real matrix with rank $k$ and

$$M_X:=I_n-X(X'X)^{-1}X'$$

$$M_Z:=I_n-Z(Z'Z)^{-1}Z'$$

If $\Sigma$ is an $n\times n$ positive semi-definite matrix, how can I show that each eigenvalue of $M_Z \Sigma M_Z$ is less than the corresponding eigenvalue of $M_X \Sigma M_X$? (ordering the eigenvalues of each matrix in decreasing order)?

EDIT: I made the following progress:

Let $\Sigma^{1/2}$ denote the square root of $\Sigma$. If $A$ and $B$ are two square $n\times n$ matrices, then the characteristic polynomials of $AB$ and $BA$ coincide. Therefore the eigenvalues of $M_Z \Sigma M_Z$ and $M_X \Sigma M_X$ are identical to those of $\Sigma^{1/2} M_Z \Sigma^{1/2}$ and $\Sigma^{1/2} M_X \Sigma^{1/2}$ respectively (here I used the idempotence of $M_Z$ and $M_X$).

Now we have the following decomposition:

$$M_Z=M_X-P$$

(see here) where $P$ is a projection matrix. Plugging in we get

$$A=B+C$$

where $A:=\Sigma^{1/2} M_X \Sigma^{1/2},B:=\Sigma^{1/2} M_Z \Sigma^{1/2}$ and $C:=\Sigma^{1/2} P \Sigma^{1/2}$ are all positive semi-definite matrices.

From Weyl's inequalities we have

$$\lambda_i(A)\geq \lambda_i(B)+\lambda_n (C)\geq \lambda_i(B)  \hspace{1cm} i=1,\dots,n$$

which is what we wanted.

Is this ok?

答:
Note that $M_X$ and $M_Z$ are each orthogonal projections onto a subspaces of $\Bbb R^n$, and the image of $M_Z$ (the left nullspace of $Z$) is a subspace of the image of $M_X$ (the left nullspace of $X$).

Now, let $U$ denote a matrix whose columns form an orthonormal basis for the image of $M_Z$. Let $V$ be such that the columns of $[U \ \  V]$ form a basis of the image of $M_X$. We have
$$
M_Z = UU^T, \quad M_X = \pmatrix{U & V}\pmatrix{U&V}^T = UU^T + VV^T.
$$
Now, we note that whenever both products are defined, $AB$ and $BA$ have the same non-zero eigenvalues. Thus, the matrices
$$
M_Z(\Sigma M_Z), \quad \Sigma M_ZM_Z = \Sigma M_Z = (\Sigma U)U^T, \quad U^T \Sigma U=: P_Z
$$
will each have the same non-zero eigenvalues. Similarly, the matrices
$$
M_X \Sigma M_X, \quad \Sigma M_X = \Sigma \pmatrix{U & V} \pmatrix{U & V}^T, \\
\pmatrix{U & V}^T \Sigma \pmatrix{U & V} = \pmatrix{U^T\Sigma U & U^T \Sigma V\\ V^T \Sigma U & V^T \Sigma V} =: P_X
$$
will have the same non-zero eigenvalues.

However, note that $P_Z$ is a submatrix of $P_X$, which means that we can apply the Cauchy interlacing theorem. Thus, if we order the eigenvalues in decreasing order, we have
$$
\lambda_i(P_X) \geq \lambda_i(P_Z) \geq \lambda_{i + \ell}(X)
$$
where $\ell$ denotes the number of columns in $Y$.

ContentLicense: CC BY-SA 4.0


问:
Functions that are metrics on the x-axis but not metrics on R^2 as a whole?
Are there functions that can be defined on $\mathbb{R}^2$, are metrics on the whole x-axis (i.e, the set of all points whose second coordinates are $0$), but not metrics on $\mathbb{R}^2$ as a whole? I am especially wondering about the existence of the ones that break the triangle inequality.

More generally, are there functions that can be defined on $\mathbb{R}^2$, but are metrics only on a few straight lines or some other subsets of $\mathbb{R}^2$, but not on $\mathbb{R}^2$ as a whole?

Thanks!

Edit:

I guess some piecewise functions may fit the description. For example, we let $ d = |x - y| $ only on $[-1, 1] \times [-1, 1]$, but let $ d = (x - y) ^ 2 $ elsewhere on $\mathbb{R}^2$. However, I was looking for "non-piecewise" functions.

答: You can take $d\bigl((x,y),(z,t)\bigr)=|x-z|+(y-t)^2$. The restriction to $\{(x,0)\mid x\in\Bbb R\}$ is the usual distance there. But\begin{align}d\bigl((0,0),(0,1)\bigr)+d\bigl((0,1),(0,2)\bigr)&=2\\&<4\\&=d\bigl((0,0),(0,2)\bigr).\end{align}


问:
$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$
Someone please help me to solve this integral
$$\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}\,dx$$
I have tried to use the formula
$$ab^2=\frac{(a+b)^3+(a-b)^3+2a^2}{6}$$
to reduce the original integral to three integrals
$$\frac{1}{6}\int_0^1\frac{\ln^3(x+x^2)}{1-x}\,dx+\frac{1}{6}\int_0^1\frac{\ln^3(\frac{x}{1+x})}{1-x}\,dx+\frac{1}{3}\int_0^1\frac{\ln^3x}{1-x}\,dx$$
But I couldn't figure out how to solve these.
Another way is to use series representation of $$\ln(x), \quad \ln(1+x),\quad \frac{1}{1-x}.$$
But I don't know how to solve the multiple sums.
Please help me out.

Body:
$
\displaystyle\int_0^1\frac{\ln^2(x)\ln(1+x)}{1-x}dx=\displaystyle\int_0^1\displaystyle\int_0^1\frac{x\ln^2(x)}{(1+xy)(1-x)}dxdy$

$=\displaystyle\int_0^1\frac{1}{1+y}\left(\displaystyle\int_0^1\frac{\ln^2(x)}{1-x}-\frac{\ln^2(x)}{1+yx}dx\right)dy$

$=\displaystyle\int_0^1\frac{1}{1+y}\left(2\zeta(3)-\displaystyle\sum_{n=1}^{\infty}(-y)^{n-1} \displaystyle\int_0^1x^{n-1}\ln^2(x)dx\right)dy$

$=2\zeta(3)\ln(2)-2\displaystyle\int_0^1\frac{1}{1+y} \displaystyle\sum_{n=1}^{\infty}\frac{(-y)^{n-1}}{n^3}dy=2\zeta(3)\ln(2)+2\displaystyle\int_0^1\frac{\operatorname{Li}_3(-y)\ }{y(1+y)}dy$

$=2\zeta(3)\ln(2)+2\displaystyle\int_0^1\frac{\operatorname{Li}_3(-y)\ }{y}-\frac{\operatorname{Li}_3(-y)\ }{1+y}dy$

$=2\zeta(3)\ln(2)+\frac{7}{4}\zeta(4)-2\left(\ln(1+y)\operatorname{Li}_3(-y)\ \right)_0^1+\displaystyle\int_0^1\frac{\operatorname{Li}_2(-y)\ \ln(1+y)}{y}dy$

$=2\zeta(3)\ln(2)+\frac{7}{4}\zeta(4)+\frac{3\ln(2)\zeta(3)}{2}-2\displaystyle\int_0^1 \operatorname{Li}_2(-y)(-\operatorname{Li}_2^{'}(-y))dy$

$=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}-2\left(-\frac{(\operatorname{Li}_2(-y))^2}{2}\right)_0^1=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}+(\operatorname{Li}_2(-1))^2$

$=\frac{7}{4}\zeta(4)+\frac{7\zeta(3)\ln(2)}{2}+\left(-\frac{\pi^2}{12}\right)^2$


问:
Action of $\pi_1(G)$ on $\pi_n(G)$ is trivial for a topological group $G$, i.e. $G$ is a n-simple space.
My question is a follow up of this question Bijection between the free homotopy classes $[S^{n},X]$ and the orbit space $\pi_n/\pi_1$.

For a topological group $G$, there is a natural action of $\pi_1(G)$ on $\pi_n(G)$. I want to show that this action is trivial i.e. each orbit under this action is a singleton.

For $\gamma:[0,1]\rightarrow G$ s.t. $\gamma(0)=\gamma(1)=g$ and $f:\mathbb{D}^n\rightarrow G$ s.t. $f(\delta \mathbb{D}^n)=g$, I have been able to show that there is a based homotopy between the $\gamma_*f$ and $f.\gamma$ where the first one is the normal action of $\pi_1(G)$ on $\pi_n(G)$, and the second one is between the product $(f.\gamma)(r,\theta)=f(r,\theta)\gamma(r)$ where $(r,\theta)$ is the generalized polar coordinates of $\mathbb{D}^n$ (n-dimensional disk).

I don't know how to proceed further.

Any help is appreciated.

答:
Your "product action" can be written as $(f\cdot \gamma)(x) = f(x) \cdot \gamma(\lVert x \rVert)$. I therefore think the basepoint of $G$ should be the neutral element $e \in G$ in order that $(f\cdot \gamma)(S^{n-1}) = f(S^{n-1})$, i.e. that $f \cdot \gamma$ and $f$ map $S^{n-1}$ to the same basepoint.

Now define
$$H : D^n \times I \to G, H(x,t) = f(x) \cdot \gamma(t\lVert x \rVert +1-t). $$
Then all $H_t$ have the property that $H_t(S^{n-1}) = f(S^{n-1})$. Moreover $H_0 = f$ and $H_1 = f \cdot \gamma$. This shows that the action is trivial.


问:
Set-Builder Notation: "$A$ is a set a of natural numbers that contains $n_0$ and contains $k+1$ whenever it contains $k$"
As the title suggests, I wondered if there was a way to encode a set with the following properties using set-builder notation:

$A$ is a set a of natural numbers that contains $n_0$ and contains $k+1$ whenever it contains $k$

At first I thought that I could invoke the Axiom of Separation/Specification:

$A=\{x \in \mathbb N | x = n_0 \lor x-1 \in A\}$

but I have become wary of this attempt because of the self-referencing that takes place in second portion of the disjunction... $x-1 \in A$. My understanding is that this is not allowed.

For the sake of argument, assume that $n_0 \gt 1$. It therefore seems as though this proper subset of $\mathbb N$ cannot be created using the Axiom of Separation/Specification (at least as written above).

Interestingly, one can prove that a set adhering to the properties "contains $n_0$ and contains $k+1$ whenever it contains $k$" necessarily contains all elements of $\mathbb N \geq n_0$. At this point, one could certainly write $A$ in set-builder notation (in a way that allows the Axiom of Separation/Specification to be invoked) as follows:

$A=\{x \in \mathbb N | x \geq n_0\}$.

At quick glance, the following seems to be true:

$\forall x \in A (x = n_0 \lor  x-1 \in A) \leftrightarrow \forall x \in A (x \geq n_0)$ (Edit: This statement actually depends on $A$...i.e. it can be true, but it can also be false)

So I guess my question is:

Why can one FOL representation of the properties of $A$ be "accepted" by the Axiom of Separation/Specification but another semantically equivalent FOL representation of the properties of $A$ be "rejected"

答:
The reason why $A=\{x\in \mathbb{N}:x\ge n_0\}$ is an acceptable definition while $$A=\{x\in \mathbb{N}:x=n_0\vee (x>0\wedge x-1\in A)\}$$
is not is just that with self-reference in general, it's not clear there will be exactly one (if any) set satisfying the definition.  For example, consider $A=\{x\in \mathbb{N}:x\notin A\}$. If $A$ exists, is $0\in A$? is $0\notin A$?

So why can we sometimes get around this?  Well, any definition needs to do two things:

it needs to have something that satisfies it, and

it needs to have only one thing satisfy it.

So any definition of a set $A$ shouldn't include $A$ because it often won't uniquely describe anything. Sometimes it can, but it's not obvious. $\{x\in \mathbb{N}:\varphi(x)\}$, on the other hand, always uniquely describes something because it's merely talking about properties of elements of $\mathbb{N}$, not the resulting set we (in some sense) haven't formed yet.  So this is why the axiom is phrased the way it is.

The property you describe isn't a definition because (2) isn't true, the $A$ you want isn't unique: $\mathbb{N}$ also has $n_0\in\mathbb{N}$ with $k\in\mathbb{N}\rightarrow k+1\in\mathbb{N}$.  But the fact that something satisfies this property is a big deal, because that means (1) works for many things, and to get (2) we just need to pick out one of them.

For you, the key point that makes this work is that the property you've stated is a closure property: you want $n_0\in A$, and $A$ to be closed under the operation $k\mapsto k+1$.  In this way, you want to define $A$ to be the least (i.e. $\subseteq$-least) set with with that property.  Because we know some such sets exist, we can pick out the least one just by considering
$$A=\bigcap\{X\subseteq\mathbb{N}:X\text{ has the property}\}\text{.}$$
In general, we can do this with closure properties so long as we have a starting point: we should be able to generate $A$ just by taking our starting point and continually applying the operations to all of its elements infinitely many times until we don't need to anymore.


答:
Your first method is correct.  Let's correct your second method.

There are
$$\binom{10}{3}$$
ways to select three of the ten balls.

The number of selections with at least one red ball is
$$\binom{5}{3} + \binom{5}{2}\binom{5}{1} + \binom{5}{1}\binom{5}{2}$$
where the first term counts the number of ways of selecting three red balls, the second term counts the number of ways of selecting two red and one yellow ball, and the third term counts the number of ways of selecting one red and two yellow balls.  Observe that
$$\frac{\dbinom{5}{3} + \dbinom{5}{2}\dbinom{5}{1} + \dbinom{5}{1}\dbinom{5}{2}}{\dbinom{10}{3}} = \frac{11}{12}$$
The problem with your second method is that the eight events are not equally likely.

Title: What is the probability that at least one red ball is drawn when three balls are drawn from a bag with $5$ red and $5$ yellow balls?

Body:
I am having a tough time trying to understand the concept of probability and figuring out which is the right way to solve this problem. So I would really appreciate a lot if you can help me with this question

Problem:  There are 5 red balls and 5 yellow balls. If we make 3 draws from this set, without replacing the ball drawn in each set, what is the probability that in the end we have drawn at least one red ball.

Method 1:

Calculate the probability of not drawing any red ball in all three draws and subtract that from 1.

The probability of drawing 3 yellow balls in succession : 5/10 * 4/9 * 3/8

So then the probability of drawing at least one red ball is 1-(5/10 * 4/9 * 3/8) = 11/12

Method 2:
Create the possible color combinations that the 3 draws can produce. I can come up with only 8 such combinations:

RRR, YYY, YRR, YRY, YYR, RRY, RYR, RYY

So based on these combinations I see that there is only 1 combination that has no red ball. So isthe probability of drawing a red ball then 7/8?

Which is the right answer and the right way of solving this problem.


问:
Decidability of bi-cartesian closed categories
This is actually in reference to the question posed here
https://stackoverflow.com/posts/66285948/edit
but is more appropriate as a question to be posed on a non-coding site.

I provide a partial answer, but not the full answer, relaying the full question here as an unresolved question. Perhaps others can take the baton.

The question is: what decision procedures are there, for bi-cartesian closed categories? (A secondary question is: what normal order reduction strategies are there for them?)

The following observations will help illustrate why the initial object is being excluded in the reference cited in the original link. Things start to collapse when initial objects are combined with exponentials; or terminal objects with co-exponentials. This collapse becomes complete when both exponentials and co-exponentials are present; in which case, the decideability problem becomes trivial, reducing to provability in a logic, while the "normal form" reduction problem becomes more difficult - ironically.

In a category, an initial object $0$, terminal object $1$, product $A,B ↦ A×B$, co-product $A,B ↦ A+B$, exponential $A,B ↦ B^A$ and co-exponential $A,B ↦ A_B$ (if we may be excused in denoting it like so) correspond respectively to the constants & connectives in logic $⊥$ (false), $⊤$ (true), $A,B ↦ A∧B$ (and), $A,B ↦ A∨B$ (or), $A,B ↦ A⊃B$ (only if) and $A,B ↦ A⊂B$ (unless) (again: if we may be excused in denoting it this way).

This correspondence is made more precise in the following way: if the pre-order $A ≤ B ⇔ ∃f: A → B$ is defined on objects, and if objects are identified up to equivalence ($A ≡ B$ if and only if $A ≤ B$ and $B ≤ A$), then $⊤$, $⊥$, $∧$ and $∨$ are the lattice top, bottom, meet and join respectively, while $⊃$ and $⊂$ are respectively the implication and co-implication operators respectively of a Heyting and co-Heyting lattice.

If all of the following $\{ ∧, ⊃, ⊥ \}$ are present together then so is $⊤$, and then the category practically reduces to a logic. More precisely: it has a negation connective, defined up to equivalence by $¬A ≡ A ⊃ ⊥$ and the full subcategory of the regular objects (i.e. those objects $A$ for which $A$ and $¬¬A$ are equivalent) is a partial order and, in fact, reduces to a Boolean algebra (thus making all such equivalences as also isomorphisms).

That is, if $A$, $B$ are regular objects and $f: A → B$, and $f': A → B$, then $f = f'$; and such $f$ exists if and only if $A ≤ B$. Since $¬¬¬A$ and $¬A$ are isomorphic, then the regular objects comprise all the negations and the negative objects $¬A$ form a subcategory that is actually a Boolean logic.

A categorical algebra that includes either of the full sets of connectives $\{ ∧, ⊃, ⊥ \}$ or $\{ ∨, ⊂, ⊤ \}$ therefore contains a categorical algebra for Boolean algebra, itself, as a subalgebra. One such algebra was, in fact, given by Spencer-Brown in the "Laws of Form" though he did not recognize that it was a categorical algebra; since category theory was virtually unknown at the time and categorical algebras probably weren't even known of at all, at that time. Spencer-Brown actually defined a groupoid for Boolean algebra, not just a category, since his arrows were all reversible and were actually isomorphisms.

It can be described equivalently as a braided monoidal category, whose objects are strings, with the product operator $x,y ↦ xy$ is denoted by juxtaposition, the terminal object by the empty string $λ$. The category is endowed with an extra operator (denoted by circling the operand) which could be denoted here $x ↦ ❨x❩$. The isomorphisms he defined were Position $❨❨x❩x❩ ↔ λ$, Transposition $❨❨xz❩❨yz❩❩ ↔ ❨❨x❩❨y❩❩z$ and an implied isomorphism $xy ↔ yx$ for commutativity - which is already part of the definition for braided monoidal categories. It is coherent (only one morphism per arrow), and a decision procedure was given on Spencer-Brown's Laws of Form. The functor $x ↦ ❨x❩$ is more properly denoted an "anti-functor" since it maps the category to its reverse.

If Transposition is replaced by morphisms for Number $xx ↔ x$, Order $❨❨x❩❩ ↔ x$ and Ortho $❨xy❩❨y❩ ↔ ❨y❩$, then this suffices to characterize orthocomplemented lattices. I don't know how the procedure specified by Spencer-Brown can be adapted to this generalization. Nor am I even sure if it collapses to a coherent category.

The partial "collapse" can be seen directly as follows, where we'll consider the combination $\{ ∧, ⊃, ⊥ \}$ that combines exponentials and initial objects: Cartesian closed categories with initial objects.

Each connective can be defined equivalently as a functor on the category that comes equipped with a natural bijection that provides the categorical algebra for the connective and proof rules for the corresponding logic.

For the following, adopt the following conventions. For a category $$, let $||$ denote its objects and for $A, B ∈ ||$, let $(A,B)$ denote the set of morphisms for the arrow $A → B$. Let $$ denote the unit category with $|| = \{0\}$ and $(0,0) = \{1_0\}$.

For a category $$, let $^+$ denote the reversal of category $$; where $|^+| = ||$ and for $A, B ∈ |^+|$,
$$^+(A,B) = \{ f ¦ A ← B: f \} = \{ f ¦ f: B → A \} = (B,A);$$
composition defined in $^+$ as the reversal of composition in $$. Finally, for categories $$ and $$, let $×$ denote the direct their product; where $|×| = ||×||$, and for $(A,A'), (B,B') ∈ |×|$,
$$×((A,A'),(B,B')) = (A,B) × (A',B'),$$
with composition defined componentwise.

If $$ is the category in question, then the functors include some or all of $⊤, ⊥:  → $; $∧, ∨: × → $; $⊃:^+× → $; $⊂:×^+ → $; and the natural bijections corresponding to each one are given, respectively, by
$$C ∈ |^+| ↦ (C,⊤) ⇔ (0,0),$$
$$(A,B,C) ∈ |××^+| ↦ (C,A∧B) ⇔ (C,A)×(C,B),$$
$$(A,B,C) ∈ |^+××^+| ↦ (C,A⊃B) ⇔ (C∧A,B),$$
$$D ∈ || ↦ (⊤,D) ⇔ (0,0),$$
$$(A,B,D) ∈ |××| ↦ (A∨B,D) ⇔ (A,D)×(B,D),$$
$$(A,B,D) ∈ |^+××| ↦ (A⊂B,D) ⇔ (A,B∨D).$$

In detail, for $∧$, $⊃$, $⊤$ and $⊥$, we can write them as
$$(f: C → A, g: C → B ⇔ h: C → A∧B) ⇒ (<f,g> = h ⇔ ⊤h = f, ⊥h = g),$$
$$(f: C∧A → B ⇔ h: C → A⊃B) ⇒ (⋀f = h ⇔ ⋁h = f),$$
$$f: C → ⊤ ⇔ f = <>_C,$$
$$g: ⊥ → D ⇔ g = []_D,$$
(where I'm operator-overloading by also using $⊤$ and $⊥$ as prefix operators to denote the components of the reverse direction of the natural bijection for $∧$).

The last two conditions are where the "collapse" starts to take place.

In the cases of $⊤$ and $⊥$, "naturality" of the bijections is given by the conditions just stated above, while for $∧$, it is given equivalently by the conditions that
$$⊤(h∘m) = ⊤h∘m, ⊥(h∘m) = ⊥h∘m,$$
where $m: C' → C$. So, we could just as well write $⊤h = ⊤1_{A∧B}∘h$ and $⊥h = ⊥1_{A∧B}∘h$ (taking $1_{A∧B}$ and $h$, respectively, in place of $h$ and $m$).

First, $⊥∧A ≤ B$ is always true and there is a unique $f: ⊥∧A → B$, since
$$f: ⊥∧A → B ⇔ ⋀f: ⊥ → A⊃B ⇔ ⋀f = []_{A⊃B} ⇔ f = ⋁[]_{A⊃B}.$$

Second, by this uniqueness, since
$$⊤1_{⊥∧A}: ⊥∧A → ⊥, []_{⊥∧A}: ⊥ → ⊥∧A,$$
and thus
$$[]_{⊥∧A}∘⊤1_{⊥∧A}: ⊥∧A → ⊥∧A;$$
and since
$$1_{⊥∧A}: ⊥∧A → ⊥∧A,$$
then it follows that
$$[]_{⊥∧A}∘⊤1_{⊥∧A} = ⋁[]_{A⊃(⊥∧A)} = 1_{⊥∧A}.$$

Third, since
$$⊥1_{⊥∧A}: ⊥∧A → A; []_{⊥∧A}: ⊥ → ⊥∧A,$$
then
$$⊥1_{⊥∧A}∘[]_{⊥∧A}: ⊥ → A$$
thus
$$⊥1_{⊥∧A}∘[]_{⊥∧A} = []_A.$$

Fourth, suppose $A ≤ ⊥$ and that $f: A → ⊥$. Then
$$⊤1_{⊥∧A}∘<f,1_A> = ⊤(1_{⊥∧A}∘<f,1_A>) = ⊤<f,1_A> = f.$$
$$⊥1_{⊥∧A}∘<f,1_A> = ⊥(1_{⊥∧A}∘<f,1_A>) = ⊥<f,1_A> = 1_A.$$
Thus
$$[]_A∘f = ⊥1_{⊥∧A}∘[]_{⊥∧A}∘⊤1_{⊥∧A}∘<f,1_A> = ⊥1_{⊥∧A}∘1_{⊥∧A}∘<f,1_A> = ⊥1_{⊥∧A}∘<f,1_A> = 1_A,$$
while $f∘[]_A: ⊥ → ⊥$ and $1_⊥: ⊥ → ⊥$ therefore $f∘[]_A = []_⊥ = 1_⊥$.

Therefore, $f = {([]_A)}^{-1}$, thus making the morphism $f: A → ⊥$ is unique.

When $A ≤ ¬B ≡ B ⊃ ⊥$, then there is a unique $f: A → ¬B$, since
$$f: A → ¬B = B ⊃ ⊥ ⇔ ⋁f: A∧B → ⊥ ⇔ ⋁f = {([]_{A∧B})}^{-1} ⇔ f = ⋀{([]_{A∧B})}^{-1}.$$

It is therefore also possible to define $⊤ ≡ ⊥ ⊃ ⊥$, thereby making the morphisms $f: A → ⊤$ unique, since
$$f: A → ⊤ = ⊥ ⊃ ⊥ ⇔ ⋁f: A∧⊥ → ⊥ ⇔ ⋁f = {[]_{A∧⊥}}^{-1} = ⊥1_{A∧⊥} ⇔ f = ⋀⊥1_{A∧⊥},$$
and we could therefore define $<>_A ≡  ⋀⊥1_{A∧⊥} = {[]_{A∧⊥}}^{-1}$. So, it is indeed a Cartesian closed category, as I alluded to above.

答:
The Wikipedia page on Cartesian closed categories has a subsection on bi-Cartesian closed categories which asserts that (1) the equational algebra cannot be axiomatized by a finite set of equational identities and (2) the word problem is still open. It cites the following reference, which should be worth a look

Remarks on Isomorphism for Typed Lambda Calculus with Coproducts
https://www.dicosmo.org/Papers/lics02.pdf

It is dated (2001) so it may be superseded by Ryan Wisnesky's link. If so, then the Wikipedia page needs to be updated. (Any volunteers, Ryan? :) )

A comment on "finite axiomatizability" - take this with a grain of salt because of an important detail that it leaves out. Kleene algebras, for instance, were proven by Conway in the 1960's to not be finitely axiomatizable by equational identities and this prematurely put a stop on the whole algebraization programme for formal language and automata theory ... on much the same way that Minsky's "no go" result put a premature stop on neural net research in the 1960's and 1970's. One of our colleagues revived the whole algebraization programme in the mid 1990's by showing that Kleene algebras are finitely axiomatizable - in terms of a small set of equational identities and one conditional equational identity (a least fixed point property).

Since then, the whole field of has taken off culminating in (among other things) the successful lifting of the algebraic framework of regular expressions to context-free expressions, as we've brought to completion the early attempt by Chomsky and Schuetzenberger in the early 1960's to algebraize the "type 2" tier in the Chomsky hierarchy, grandfathering in the older early 1970's formalism that was based on a fixed-point calculus formalism.

I expect that a similar result will emerge for bi-Cartesian closed categories and that one might be able to provide a small finite set of equational identities with a few conditional equational identities as an axiomatization.

As for embedding the "regular" objects of a bi-Cartesian closed category into Spencer-Brown's Laws of Form categorical algebra, the latter needs to be extended first, since it is (after all) only a groupoid, so it can't handle arrows that are not reversible. I conjecture that it suffices to just include the arrows for the terminal object, and that this should be enough to produce the other irreversible arrows.

Since we already have an extension of the Curry-Howard correspondence to classical logic - involving continuations - then it would be interesting to see how this can be related to Spencer-Brown's categorical algebra ... and also interesting to see whether and how it can be hybridized with a categorical algebra for bi-Cartesian closed categories.


问:
Example of a group which is abelian and has finite (except the $e$) and infinite order elements.

Exercise 7: Show that the elements of finite order in an abelian
  group $G$ form a subgroup of $G$

I just solved this exercise but I can't find example of a group which is abelian and has finite (except the $e$) and infinite order elements. Are there any well known such groups?

For not abelian group with other properties group of matrixes I know.

答:
The multiplicative group of units $U$ from $(\mathbb{C},+,\cdot)$ is an abelian group. Of the real numbers in $U$, $-1$ has order $2$, and besides $e=1$, all other real numbers will have infinite order. Also, $i \in U$ and $i$ has order 4.

$-1$ and $i$ are both examples of $2^{nd}$ and $4^{th}$ roots of unity, respectively. That is, they are solutions to $z^2=1$ and $z^4=1$, respectively.

In general, an $n^{th}$ root of unity in $\mathbb{C}$ is a complex number $z$ satisfying $z^n=1$. For every $n \in \mathbb{N}$ there are $n$ distinct complex numbers which satisfy $z^n=1$, $z=1$ being a solution for every $n \in \mathbb{N}$.

ContentLicense: CC BY-SA 3.0


问:
Z confidence interval
$W_1, \ldots, W_5$ be iid $Normal(3.1, 5.2^2)$.  What is the probability that $\left(\frac{\bar W - 3.1}{\hat \sigma / \sqrt{5}} \geq 2.1\right)$?

I am not sure how to start. Is this equivalent to the probability that $Z \leq 2.1$?

答:
You're close, but it's actually $1-P[t \le 2.1]$.

The distribution of $\bar{W} = \frac{1}{5}\sum_i^5 {W_i}$ is

$$ \bar{W} \sim \mathscr{N} \left( 3.1, \frac{5.2^2}{5} \right) . $$

Normally, $ (\bar{W} - 3.1)/\left( \frac{5.2^2}{\sqrt{5}}\right) $ would be the standardization of $\bar{W}$ into a standard normal $Z \sim \mathscr{N}(0,1)$. Because of the use of $\hat{\sigma}$ in the question, this is presumably coming from a sample of $5$ observations of $W$, and we are using a $t$-distribution with $4$ degrees of freedom, not a standard normal.

So, we can rewrite the probability we want in terms of $t$:

$$ P \left[ \frac{\bar{W} - 3.1}{\frac{\sigma}{\sqrt{5}}} \ge 2.1 \right] $$
$$ = 1 - P \left[ \frac{\bar{W} - 3.1}{\frac{\sigma}{\sqrt{5}}} \le 2.1 \right] $$
$$ = 1 - P \left[ t \le 2.1 \right] . $$
$$ \approx 0.052 .$$

Title: Z confidence interval


问:
Let A, B and C bounded sets of $\mathbb{R}$...
Let $A$, $B$ and $C$ bounded sets of $\mathbb{R}$. Define $D_H(A,B)$ as

$$D_H(A,B)=\max\Bigg\{\underset{x\in A, y \in B}{\inf}|x-y|, \underset{x\in A, y \in B}{\sup}|x-y|\Bigg\}$$

Prove that: $$D_H(A,B)\leq D_H(A,C)+D_H(B,C)$$

This is and old exercise from the begin of the course I'm studying, but until today I am not able to prove this. I only need a hint. I'm stucked for a few hours in this exercise. In the moment this exercise was presented we were only allowed to use the definition of $\sup$ and $\inf$ of a set, the completeness of $\mathbb{R}$, stuff like that, nothing more elaborated than that, like sequences, for example.

I tryed to use the triangle inequality, but I felt like I had to separate in many cases and it didn't look right.

答:
Claim #1: $D_{H}(A,B)=\sup_{a\in A,b\in B}|a-b|$

Proof of Claim #1: Pick any $x\in A$ and $y \in B$. Then $$\inf_{a\in A,b\in B}|a-b|\leq |x-y| \leq \sup_{a\in A,b\in B}|a-b|$$ This implies $$\inf_{a\in A,b\in B}|a-b| \leq \sup_{a\in A,b\in B}|a-b|$$ and so $$D_{H}(A,B)=\sup_{a\in A,b\in B}|a-b|$$

Claim #2: For any fixed $c\in C$ we have $$\sup_{a\in A,b\in B}|a-b| \leq \sup_{a\in A}|a-c| + \sup_{b\in B}|b-c|$$

Proof of Claim #2: Choose any $x\in A$ and $y\in B$. Then $$|x-y|\leq |x-c|+|y-c|\leq \sup_{a\in A,b\in B}\Big[ |a-c|+|b-c|\Big] $$This shows $\sup_{a\in A,b\in B} \Big[ |a-c|+|b-c|\Big]$ is an upper bound of the set $$\Big\{|x-y|:x\in A,y\in B\Big\}$$ so by the definition of $\sup$ we get $$\sup_{a\in A,b\in B}|a-b| \leq \sup_{a\in A,b\in B}\Big[ |a-c|+|b-c|\Big]$$ and using this we have $$ \sup_{a\in A,b\in B}\Big[ |a-c|+|b-c|\Big]=\sup_{a\in A}|a-c| + \sup_{b\in B}|b-c|$$ So finally, $$\sup_{a\in A,b\in B}|a-b| \leq \sup_{a\in A}|a-c| + \sup_{b\in B}|b-c|$$

Claim #3: For any fixed $c\in C$ we have $$\sup_{a\in A}|a-c|\leq \sup_{a\in A,z\in C}|a-z|$$ $$\sup_{b\in B}|b-c| \leq \sup_{b\in B,z\in C}|b-z|$$

Proof of Claim #3: Choose $x\in A$ arbitrarily. Then $$|x-c|\leq \sup_{a\in A,z\in C}|a-z|$$ This shows $\sup_{a\in A,z\in C}|a-z|$ is an upper bound of the set $$\Big\{|x-c|:x\in A\Big\}$$ so using definition of $\sup$ we have $$\sup_{a\in A}|a-c|\leq \sup_{a\in A,z\in C}|a-z|$$ An identical argument also proves $$\sup_{b\in B}|b-c| \leq \sup_{b\in B,z\in C}|b-z|$$

Claim #4: $D_{H}(A,B)\leq D_H(A,C)+D_H(B,C)$

Proof of Claim #4: Combine all three previous claims to get the result.


问:
Solving a cubic equation on an interval
$d(t)=-3t^3+45t^2+600t$ where $0\leq t \leq 10$

What is the value of $t$ if $d(t) = 3750$?

Practice question from my textbook I'm having a little trouble with for some reason. Would like to have the steps clarified In case this kind of question shows up in a future test. Thanks.

答:
This is mainly an extended comment on the answers that have already been posted, but it may be of help in approaching problems of this type, where you can assume the problem poser has carefully constructed things to have a nice answer.

As the other answers observe, the cubic simplifies initially to

$$t^3-15t^2-200t+1250=0$$

which you can then attack by hoping it has an integer root. The key observation to make is that since the coefficients $15$, $200$, and $1250$ are all divisible by $5$, then $t^3$, if it's an integer, must also be divisible by $5$. So if we let $t=5u$, the equation becomes $125u^3-15\cdot25u^2-1000u+1250=0$, which simplifies to

$$u^3-3u^2-8u+10=0$$

The smaller coefficients here make this a much easier cubic to think about, and it might even jump right out at you that $u=1$ is a root, since $3+8=1+10$. This leads to the solution $t=5u=5$.

Title: Solving a cubic equation on an interval


问:
How is the sum of simplices defined?
I have recently started to learn about simplices, simplicial complexes and simplicial homology. I understand that for some set of points $\{ p_0, \ldots, p_k \} \subset \mathbb{R}^n$ a simplex $\sigma$ is the set

$$\sigma \enspace \equiv \enspace < \; p_0, \ldots, p_k \; > \enspace := \enspace \Big\{ \; x \in \mathbb{R}^n \; \Big | \; x = \sum_{i = 0}^k \lambda_i p_i \, , \; \lambda_i > 0 \, , \; \sum_{i = 0}^k \lambda_i = 1 \; \Big\}$$

Now, the author of my script started adding and substracting simplices, e.g. when defining the boundary of $\sigma$ he writes

$$\partial_k \sigma \enspace = \enspace < \; p_1, \ldots, p_k \; > - < \; p_0, p_2, \ldots, p_k \; > + \ldots + < \; p_0, \ldots, p_{k-1} \; >$$

However, he did not define what the operations $+$ and $-$ mean in this context. I also seem to have trouble finding a rigorous definition of those operations for simplices on the web. Can anyone explain?

答:
The short answer is: the maps $\partial$ are defined on the free abelian groups generated by $n$-simplices, hence it makes sense to speak about sums of simplices in these groups.

I'll try to give a down to earth answer, for which we will need to detour first. Consider the set $\mathbb{Z}[X]$ of integer polynomials. An element there is an expression
$$
a_0 + a_1 X + \ldots + a_nX^n
$$
where each $a_i$ is an integer, and $X$ is an indeterminate. Even though we might be quite familiar with thinking of polynomials in this way, the terms 'expression' and 'indeterminate' are informal. A rigorous definition of $\mathbb{Z}[X]$ could be the set of sequences of integers that are eventually zero,

$$
\mathbb{Z}[X] = \{(a_n)_{n \geq 0} \in \mathbb{Z}^{\mathbb{N}_0} : \text{there exists $k \in \mathbb{N}_0$ such that $a_n = 0$ for all $k \geq n$}\}. \tag{1}
$$

Given integers $a_1,\ldots,a_n\in \mathbb{Z}$, we then define

$$
a_0 + a_1 X + \ldots + a_nX^n := (a_0,a_1,\ldots,a_n,0,0,\ldots) \tag{2}.
$$

All the operations on polynomials can be defined in terms of sequences as in $(1)$ and are compatible with the usual notation $(2)$. This is a way to formally capture the idea of having some unknown element $X$ that can be 'scaled' by integers, multiplied by itself, and such that we can make sense of the sums of expressions like this.

In the same spirit, suppose you have a set $X$ and you want to have another set $M$ such that:

$X$ is, in some sense, contained in $M$,

we can make sense of multiplying $x \in X$ by an integer $k \in \mathbb{Z}$,

we can give meaning to an expression like $17x+28y+3z$ for some $x,y,z \in X$.

A way to do this is to define the free abelian group $\mathbb{Z}^{(X)}$ with basis $X$. This concept is closely related to bases of vector spaces. The definition (technically, one definition, there are 'many' equivalent ones) is as follows: as a set $\mathbb{Z}^{(X)}$ consists of functions $f \colon X \to \mathbb{Z}$ such that $f(x) \neq 0$ for finitely many $x \in X$. In other words, the elements of $\mathbb{Z}^{(X)}$ are finitely supported functions $\mathbb{Z} \to X$. We can make sense of sum and multiplication of elements here in the same way we do to define a vector space structure on $\mathbb{R}^X$. Namely:

the sum of $f,g \in \mathbb{Z}^{(X)}$ is the function $(f+g)(x) := f(x)+g(x)$.

if $k$ is an integer and $f \in \mathbb{Z}^{(X)}$, we define $(k\cdot f)(x) := kf(x)$.

You can check that these operations define once again elements of $\mathbb{Z}^{(X)}$.

Now, let's go back to the analogy with polynomials; in particular, to the relation between definition $(1)$ and notation $(2)$. Even though formally the free abelian group is defined as finitely supported functions, we want to think of it as "$\mathbb{Z}$-linear combinations of $X$". To do this, we write
$$
x := \chi_{\{x\}}, \quad \chi_{\{x\}}(y) = \begin{cases}1 &\text{if $x=y$}\\
0 &\text{otherwise}\end{cases}
$$
You can check that, following this notation, every element of $\mathbb{Z}^{(X)}$ can be written as a finite sum
$$
a_1 x_1 + \ldots +a_n x_n
$$
for some $x_i \in X$ and integers $a_i$. Moreover, two such expressions are equal if the elements of $X$ appearing are the same, and the coefficients accompanying each element coincide.

In this case, we consider $\mathbb{Z}^{(C_n)}$ with $C_n$ the set of $n$-simplices. In $\mathbb{Z}^{(C_n)}$ it makes perfect sense to speak of $3 \sigma + 22 \tau$ for some pair of simplices $\sigma,\tau$ or any expression of this sort; likewise the maps $\partial_k$ are well defined.

I hope this gives a little bit more context to the first sentence of this answer.

Title: How is the sum of simplices defined?


问:
A Stirling number identity representing the second-order Eulerian numbers.
Graham, Knuth, and Patashnik give in
CMath
a thorough introduction to the Stirling numbers.
On table 250 and table 251, they compile two pages of
Stirling number identities.

Of course, there are many more such identities. We give here one
which is not included and which we could not find anywhere else
(which means little), but Knuth would perhaps like.
They can be proved with the techniques from CMath. Maybe someone
enjoys showing us the proof?

As in CMath, the Stirling cycle numbers are unsigned.
{n, k} and [n, k] are defined for 0 <= k <= n, and for this
range the following identity holds:

$$ \sum \limits_{j\,=\,0}^{k} (-1)^{k-j}\binom{2n+1}{k-j}
{n+j \brace j}
= \sum \limits_{j\,=\,0}^{n-k} (-1)^j \binom{2n+1}{j}
{n+m \brack m} $$

For better readability we use the shortcut $ m = n-k-j+1$.

References to the OEIS are A132393, A048993, and A340556.

答:
We seek to show that with $0\le k\le n$ the following identity holds:

$$\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j} =
\sum_{j=0}^{n-k} (-1)^j {2n+1\choose j} {2n-k-j+1\brack n-k-j+1}.$$

We will start with the LHS. The chapter 6.2 on Eulerian Numbers
of Concrete Mathematics by Knuth et al. proposes the formula

$${n\brace m} = (-1)^{n-m+1} \frac{n!}{(m-1)!} \sigma_{n-m}(-m)$$

where $\sigma_n(x)$ is a Stirling polynomial and we have the identity

$$\left(\frac{1}{z} \log\frac{1}{1-z}\right)^x
= x \sum_{n\ge 0} \sigma_n(x+n) z^n.$$

We get

$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^x
= x \sigma_{n-m}(x+n-m)$$

and hence

$$[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}
= -n \sigma_{n-m}(-m)$$

which implies that for $n\ge m\ge 1$

$$\bbox[5px,border:2px solid #00A000]{
{n\brace m} = (-1)^{n-m} \frac{(n-1)!}{(m-1)!}
[z^{n-m}] \left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n}.}$$

This gives for the LHS

$$\sum_{j=1}^k (-1)^{k-j} {2n+1\choose k-j}
(-1)^n \frac{(n+j-1)!}{(j-1)!} [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-j}
\\ = (-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
[w^{k-1}] (1+w)^{2n+1}
\\ \times \sum_{j=1}^k {n+j-1\choose n} (-1)^{j-1} w^{j-1}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-j+1}.$$

Now the coefficient extractor in $w$ enforces the upper limit of the
sum and we may extend $j$ to infinity, getting

$$(-1)^{n-k+1} n! [z^n]
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{-n-1}
[w^{k-1}] (1+w)^{2n+1}
\frac{1}{(1+w/(\frac{1}{z} \log\frac{1}{1-z}))^{n+1}}
\\ = (-1)^{n-k+1} n! [z^n]
[w^{k-1}] (1+w)^{2n+1}
\frac{1}{(w+\frac{1}{z} \log\frac{1}{1-z})^{n+1}}.$$

Continuing,

$$ (-1)^{n-k+1} n! [z^n]
[w^{n+k}] (1+w)^{2n+1}
\frac{1}{(1+\frac{1}{w} \frac{1}{z} \log\frac{1}{1-z})^{n+1}}
\\ = (-1)^{n-k+1} n! [z^n]
[w^{n+k}] (1+w)^{2n+1}
\sum_{q\ge 0} {n+q\choose n} (-1)^q \frac{1}{w^q}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^q
\\ = (-1)^{n-k+1} n! [z^n]
\sum_{j=n+k}^{2n+1} {2n+1\choose j}
{n+j-(n+k)\choose n} (-1)^{j-(n+k)}
\\ \times
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j-(n+k)}
\\ = (-1)^{n-k+1} n! [z^n]
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j}
\left(\frac{1}{z} \log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} n!
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j} [z^{n+j}]
\left(\log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} n!
\sum_{j=0}^{n-k+1} {2n+1\choose j+n+k}
{n+j\choose n} (-1)^{j}
\\ \times \frac{j!}{(n+j)!} \times (n+j)! [z^{n+j}]
\frac{1}{j!} \left(\log\frac{1}{1-z}\right)^{j}
\\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1}
{2n+1\choose j+n+k} (-1)^j {n+j\brack j}
\\ = (-1)^{n-k+1} \sum_{j=0}^{n-k+1}
{2n+1\choose 2n-j+1} (-1)^{n-k-j+1} {2n-k-j+1\brack n-k-j+1}
\\ = \sum_{j=0}^{n-k+1}
{2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$

The Stirling number is zero for $j=n-k+1$ and we get at last

$$\sum_{j=0}^{n-k}
{2n+1\choose j} (-1)^j {2n-k-j+1\brack n-k-j+1}.$$

This is the RHS and we have the claim.


问:
maximal antichain
I don't understand the definition of Jech (set theory) for "maximal antichain". Let $B$ a boolean algebra and $A$ a subalgebra of $B$.
$W\subseteq A^+$ is a maximal antichain if $\sum W=1$ and $W$ antichain. As $A^+\subseteq B^+$ and $1\in A^+\cap B^+$, $W$ is also a maximal antichain in $B$ (is there a mistake here ?). But it is saying that $W$ need not be maximal in $B$ .... why ?
Thanks </p>

答:
It is easy to check that $W$ is an antichain in $B$ as well. So we will have to find a way to violate $\sum W = 1$ in $B$. The preservation of finite suprema means we will need to reach for an infinite $W$. An example is below.

Consider $B = \mathcal P(\Bbb N)$, and let $A$ be the subalgebra of finite subsets of $\Bbb N$ not containing zero and their complements:

$$A = \{S \subseteq \Bbb N: \text{$S$ finite and $0 \notin S$, or $S$ infinite and $0 \in S$}\}$$

Then the set of singletons of positive integers $W = \{\{n\}: n > 0\}$ is clearly an antichain in $A$, and in $A$, $\sum W = \Bbb N$, while in $B$, $\sum W = \Bbb N_{>0}$.

Title: maximal antichain


问:
Random variable with finite moments $E(X^k)=k!$
Suppose that $X$ is a random variable with all of its moments finite, given by $ \mathbb{E}(X^k) = k! , k \in \mathbb{N}$. If $X$  has a moment generating function, find the distribution of $X$.

I'm lost in here. If someone could give me a hint of how to solve it, I appreciate any help. Thanks in advance.

答:
Consider the moment generating function
$$ E(e^{tX}) = \sum_{x=0}^\infty e^{tx} P(X=x) = \sum_{n=0}^\infty \frac{t^n}{n!} E(X^n) = \sum_{n=0}^\infty t^n = \frac{1}{1-t}$$
provided that $|t|<1$. This looks like the MGF for the Exponential distribution with parameter 1.


问:
How to prove a function is a regular Borel measure on $\mathbb{R}$
We have a function $w: \mathbb{Q} \rightarrow [0, \infty]$ which obeys the following property:
$$
\sum_{r \in \mathbb{Q} \cap (-b, +b)} w(r) < \infty, \ \ \ \ \ \ 0 < b < \infty
$$

for $E \subseteq \mathbb{R}$, define the function $\mu(E) = 0$, if $E = \emptyset$ and
$$\mu(E) = \sum_{r \in \mathbb{Q} \cap E} w(r), \ \ \ \ \ \ E \neq \emptyset.$$

How do I prove that $\mu$ is a regular Borel measure on $\mathbb{R}$ ?

I tried to relate the measure defined here with the Lebesgue-Stiletjes measure on $\mathbb{R}$, then from that conclude somehow that the function defined above is regular.

P.S: a regular Borel measure satisfies the following:

if $A \subseteq \mathbb{R}$ is measurable and given $\epsilon > 0$, then there is an open set $G \subseteq \mathbb{R}$, such that:

$A \subseteq G$.

$\mu\left( G \setminus A \right) < \epsilon$.

This is a homework question, I am not asking for solutions, just for hints.

Thank you,

Regards.

答:
Note that $\mu(E)=\sum_{r \in \Bbb{Q}}w(r)\delta_q(E)$ is defined on the powerset of $\Bbb{R}$.

Here $\delta_q$ denotes the Dirac measure.

It is true that if $\mu_n$ is a sequence of measures on a sigma algebra then $\sum_n\mu_n$ is also a measure.(A good exercise for you).From this, by taking an enumeration of the rationals,you can deduce that $\mu$ is a measure,

Since it is defined on the powerset then it is defined on every Borel set.

Now from hypothesis,the Borel measure $\mu$ is locally finite on $\Bbb{R}$,thus regular.

A reference for that can be found at: http://fourier.math.uoc.gr/~papadim/

In the measure theory lecture notes(go to course notes),on page 64,theorem 5.7

Theorem 5.7 is a  general theorem for regularity of Borel measures on a topological space.

It would be quite instructive to read its proof ,becuase uses techniques that you will encounter often in measure theory.


问:
Prove that $2^{1/3}$ is irrational.
Proof. Assume by way of contradiction that $\sqrt[3] 2$ is rational. Then $\sqrt[3] 2$ can be written as $a/b$ where $ a,b\in\mathbb{N},b\neq0$. Then $(a^3/b^3 = 2) = (a^3 = 2b^3)$. Thus $2|a^3$ and by the lemma, $2|a$. Then $a=2x $ where $x\in\mathbb{Z} $. Substituting $a=2x$ into $a^3 = 2b^3$ we get $8x^3 = 2b^3$, simplifying we get $4x^3 = b^3$. Which implies $4|b^3$ and then by the lemma, $4|b$. This is a contradiction because if a is a multiple of $2$ and $b$ is multiple of $4$, then they are not in lowest terms. Hence $\sqrt[3] 2$ is irrational.

I think I am good up til the $2|a^3$ part. I don't know if I can say that, "by the lemma $2|a$". Any tips or help is much appreciated!

答:
Since $2|a^3$, then $2|a$ because $2$ is prime (Euclid's Lemma).

Later on when you say $4|b^3$, you can argue that this implies $2|b^3$.

By the same argument as before, $2|b$ (as mentioned in the comments). This contradicts that $\text{gcd}(a,b)=1$ (you will need to state this as well in your proof because the fraction $\dfrac{a}{b}$ must be reduced to lowest terms).

Title: Prove that $2^{1/3}$ is irrational.


问:
Prove that the given set is a finite set.
$A = \{(a,b) \in N^2 | (2+a)(2-b) \ge 2(a-b)\}$. Prove that A is a finite set.

I'm having problem with proving A is a finite set. So far,

Let $x, y \in A$. Then,
\begin{equation*}
\begin{aligned}
    (2+x)(2-y) &\ge 2(x-y)\\
    4-2y+2x-xy &\ge 2x-2y\\
    4-xy &\ge 0\\
    xy &\le 4
\end{aligned}
\end{equation*}

This is what I have, but I'm not sure that showing $xy \le 4$ is enough to show that we have finite set. The reason why I thought this was enough because x and y are natural numbers. I searched other proofs and saw someone used contradiction to prove this is a finite set by assuming A is a infinite set. Do you guys think I need contradiction? or is it enough to prove A is a finite set?

答:
Of course it's fine to show directly it's a finite set if that is what you're trying to prove! Your proof is fine. If we write it out very explicitly: you have shown shown for an arbitrary element $(x,y) \in A$, we have $xy \leq 4$. Since we know that $x$ and $y$ are natural numbers, this then implies that ($x=1 \text{ and } y=4$) or ($x=2 \text{ and } y=2$) or ($x=4 \text{ and } y=1)$ etc. There will only be a finite number of terms here (you can write them all out explicility). Hence you've shown whenever $(x,y)$ is any element of $A$, it must be either $(1,4)$ or $(2,2)$ or $(4,1)$ or one of these other finitely many solutions.

I.e. you have proven that $A\subset \{(1,1),(1,2),(1,3),(1,4),(2,2),(4,1),(3,1),(2,1)\}$ (in fact we have exact equality by the contrapositive statement that not being one of these pairs means you are not in $A$). So obviously $A$ is a finite set!

Also, I can't see a way to prove this by contradiction that is more straightforward than what you've done. What you've done proves that the number of solutions for the equation, and hence the elements of $A$, is a finite number. You could say that if $A$ were infinite, then $A$ would have to contain a solution that is not one of the possible ones. But this is entirely pointless, as the result is already proved.


问:
A Calculus Question on onto functions with a specified range.
The following question was from a mock test of a competitive exam.

Suppose $f:\mathbb{R} \to [-8,8]$ is an onto function and $f(x) = \dfrac{bx}{(a-3)x^3 + x^2 + 4}$ where $a,b \in \mathbb{R}^+$. If the set of all values of $m$ for which the equation $f(x) = mx$ has three distinct real solutions in the open interval $(p,q)$, then find the value of $a+b+p+q$.

I tried the problem and obtained an answer. According to me, it is $43$. I will post the answer later. My solution was quite involved and long. I want to know if there is a quick solution to this problem.

Thank you.

答:
Since $f:\mathbb{R} \to [-8,8]$,

$\max f(x)=\dfrac{bx}{(a-3)x^3+x^2+4}=8$ , we need to have $a=3$ so as to make Polynomial have real roots. Here $x=+2, b=32$ .

And also $\min f(x) \dfrac{bx}{(a-3)x^3+x^2+4}=-8$, you get $x=-2$ and $b=32$

$f(x)=mx$ gives a straight line with $m$ as a slope. Now the function $f(x)=\dfrac{bx}{x^2+4}$ needs to have slope that is equal to $m$. So, $f'(x)=-\dfrac{32(x^2-4)}{(x^2+4)^2}=m$, when $x= \pm 2 \implies m=0$ and when $x=0$ you get $m=8$, therefore $m \in [0,8]$ claiming the roots to be $(2,-2,0)$


问:
proving $\deg(fg) = \deg(f) + \deg(g)$ in a $\mathbb Z$-graded integral domain.
In a $\mathbb Z$-graded integral domain $B = \bigoplus_{i \in \mathbb Z} B_i$ my definition for the degree function is as follows:

Given $f \in B,$ define $\deg : B \rightarrow \mathbb Z \cup \{-\infty \}$ by $$\deg(f) = \mathrm{maxSupp}{(f)} \text{ if } f \neq 0 \text{ and } \deg(0) = -\infty $$

My definition for the  $\mathrm{Supp}$ of $f$ is as follows $\{i \in \mathbb Z| f_i \neq 0\}.$

Here is the statement I want to prove:

$\deg(fg) = \deg(f) + \deg(g).$

Here is my trial:

Let $f = f_0 + f_1 + \dots +f_k \neq 0$ where $f_i \in B_i, 0 \leq i \leq k, f_k \neq 0, \deg(f) = k$ and let $g = g_0 + g_1 + \dots +g_l \neq 0$ where $g_j \in B_j, 0 \leq j \leq l, g_l \neq 0, \deg(g) = l.$

Now, $\deg(fg) = \deg ((\sum_{i = 0}^k f_i)(\sum_{j = 0}^l g_j)) = \deg (\sum_{n=0}^{l+k} \sum_{i + j = n, i \geq 0, j \geq 0} f_i g_i)$ and $f_lg_k \neq 0 \implies \deg (fg) = l + k = \deg(f) +  \deg(g)$ this is because we are in an integral domain.

If either $f$ or $g$ is $0,$ then $fg = 0$ and so $\deg(fg) = -\infty = \deg(f) + \deg(g) = -\infty$

But I think that my trial has 2 mistakes which are:

1- I took $i,j$ starting from $0.$ even though according to the support I defined above they should be in $\mathbb Z.$ How can I correct the proof to include $i,j$ as integers?

2- I think I used that I have $\mathbb Z$-graded integral domain $B$ in the wrong place. Could anyone help me in correcting this please?

答:
You are correct about $(1)$, this does not cover all cases. I don't think $(2)$ is right though, you are using that $B$ is a domain and $\mathbb{Z}$-graded correctly. You just have to adapt the proof to the general case.

To do that, prove (or use) that
$$
\left(\sum_{i=m}^ka_i\right) \cdot \left(\sum_{j=n}^l b_j\right) = \sum_{s = m+n }^{k+l}\left(\sum_{i+j=s} a_ib_j\right)
$$

for any $m \leq k, n \leq l$.


问:
Neighbor fractions problem
I have posted an exercise below about basic proof writing with fractions and I would appreciate if you could check my work/comment on the logical structure of my proofs.

Problem

Fractions $\frac{a}{b}$,$\frac{c}{d}$ are called neighbor fractions if their difference $\frac{ad-bc}{bd}$ has numerator $±1$, that is, $ad-bc=±1$.

Prove that:

(a) in this case neither fractions can be simplified.

(b) if $\frac{a}{b}$,$\frac{c}{d}$ are neighboring fractions, then $\frac{a+c}{b+d}$ is between them and is a neighboring fraction for both $\frac{a}{b}$ and $\frac{c}{d}$.

(c) no fraction $\frac{e}{f}$ with positive integer $e$ and $f$ such that $f<b+d$ is between $\frac{a}{b}$ and $\frac{c}{d}$.

Proof of (a):

Assume that $\frac{a}{b}$ can be simplified. Then there exist $k,p,q\not= 0\inℤ$ s.t. $a=kp$ and $b=kq$. Substituting in for $a$ and $b$ we obtain, $kpd-kqc=±1$. Factoring out $k$ we get $k(pd-qc)=±1$. Now,  $k\not=0$, and the product with $k\not=0$, so $(pd-qc)\not=0$. Therefore, k=$±1/(pd-qc)$ with $(pd-qc)\in ℤ$. So $k\notinℤ$, a contradiction. An analogous argument shows for $\frac{c}{d}$.      $\blacksquare$

Proof of (b):

Without loss of generality, assume $\frac{a}{b}<\frac{c}{d}$. It follows that $ad<bc$. Adding $dc$ to both sides we get $ad+dc<bc+dc$, and then factoring out $d$ from LHS and $c$ from RHS we get $d(a+c)<c(b+d)$ and so $\frac{a+c}{b+d}<\frac{c}{d}$. Likewise, adding $ab$ to both sides we get $ad+ab<bc+ab$, and then factoring out $a$ from LHS and $b$ from RHS we get $a(b+d)<b(c+a)$ and so $\frac{a}{b}<\frac{a+c}{b+d}$.

By the definition of neighboring fractions, we need to show $a(b+d)-b(a+c)=±1$ and $d(a+c)-c(b+d)=±1$. Because $\frac{a}{b}$,$\frac{c}{d}$ are neighboring fractions, we get $ad-bc=±1$. It follows that $-bc=±1-ad$ and $ad=±1+bc$.

$a(b+d)-b(a+c)=ab+ad-ba-bc=ad-bc=±1+bc-bc=±1$

$d(a+c)-c(b+d)=da+dc-cb-cd=da-cb=da±1-ad=±1$ $\blacksquare$

Proof of (c):

How would you show part (c)?

答:
Your proof of (a): I'll ignore the typos for now, and focus on the maths. You need to assume $k\neq \pm 1$ as well (otherwise, you're not simplifying at all). Then, (WLOG that $ad-bc = 1$ and $k=1$) the contradiction comes from that $k$ is forced to be an integer between $0$ and $1$.

Your proof of (b): Both parts look fine.

For part c: Are you assuming that $\frac{a}{b}$ and $\frac{c}{d}$ are neighbor fractions? Otherwise, there are counterexamples such as $\frac12 < \frac35 < \frac34$, for example.

Good work on writing this up. You've asked a good question.

Title: Neighbor fractions problem


问:
Use the epsilon-delta definition to prove
Use $\epsilon$ - $\delta$ to prove that $\lim_{x\rightarrow0} \frac {\sqrt{4+x}-2} {x} = \frac{1}{4}$

Given $\epsilon > 0$, we want to find a $\delta$ such that if $|x-0| < \delta$, $\left|\frac{\sqrt{4+x}-2}{x} - \frac{1}{4}\right| < \epsilon$.
I multiplied by $\frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}$ to rationalize the numerator and combined the fractions to get $\left|\frac{2-\sqrt{4+x}}{4(\sqrt{4+x}+2)}\right|$, but I am not sure where to go from here to relate it to $\delta$.

答:
Hint:

The denominator is never smaller than $8$ (for $\ x\ge-4\ $, which must be the case for it to be well-defined).

And\begin{align} |2-\sqrt{4+x}|<t&\iff 2-t<\sqrt{4+x}<2+t\\
  &\Leftarrow |x| <2t+t^2\ \&\ t<\frac{1}{2}\end{align}


问:
Enumerate partitions of identical objects
I have a problem concerning enumerating partitions of a set of identical objects. I know, now someone is going to talk about Stirling number of second kind, but I'm quite sure this is not the answer.

I will describe the problem with an example. Suppose to have a set of $n$ identical objects. The problem is to enumerate and to generate all the partitions obtained with $k$ subsets, like in the following example.

Let $n=5$, $k=3$. Define $B_i$, with $i=1,\dots,k$ the set of the partition. Hence, I can partition the set of $n$ objects by constructing the following partitions:

$|B_1|=3$, $|B_2|=1$, $|B_3|=1$.

$|B_1|=2$, $|B_2|=2$, $|B_3|=1$.

Notice that: $(i)$ must result $B_i\neq \emptyset$; $(ii)$ partitions like, for instance, $|B_1|=1$, $|B_2|=3$, $|B_3|=1$ or $|B_1|=1$, $|B_2|=1$, $|B_3|=3$ must not be considered, since are in practice already considered in the first partition (i.e. $|B_1|=3$, $|B_2|=1$, $|B_3|=1$).

Since objects are all identical, I only express the sets cardinality and I doesn't care about the specific object allocated to a set of the partition.

The problem is particular useful in the formulation of algorithms for assignment problems.

答:
Don Knuth gives an algorithm for generating all partitions of $n$ into $k$ parts in Algorithm H on p. $2$ of pre-fascicle 3B of The Art of Computer Programming, which is a draft of Sections 7.2.1.4–5. This is available as a gzipped PostScript file on Knuth's website and as a low-quality PDF elsewhere.

Here's the algorithm (in my words; Knuth gives it in pseudocode):

Denote the parts by $a_1$ to $a_k$. Start out with $n-k+1,1,\dotsc,1$. Generate partitions by moving $1$ from $a_1$ to $a_2$ as long as this doesn't make $a_2$ greater than $a_1$. Then look for the first $j$ with $a_j\lt a_1-1$. If there isn't one, you're done. If there is, increment $a_j$ by $1$, set all parts between $a_1$ and $a_j$ to $a_j$ and choose $a_1$ to make the sum come out right; then go back to generating partitions.


问:
Is this the right notion of "relatively arithmetical"?
Say I want to tease out an arithmetical property from the ether: I get it's $\Pi^0_n$ $(\Sigma^0_n)$, and I get some indices for programs that compute structures (their (atomic) diagrams) with that property. So lets say I want to check for a certain property among those guys, and I manage to show this is $\Pi^0_m$ $(\Sigma^0_m)$ relative to this set of indices I was talking about.

Is this new property $\Pi^0_{m + n}$ $(\Sigma^0_{m+n})$ in some absolute sense? If there is already some terminology for this (probably yes) please let me know. Also if necessary please smuggle in the word complete where suitable.

答:
Yes, everything is basically additive. However, there's a slight catch:

Suppose $A$ is $\Sigma^0_1$ and $B$ has the form $$\{x: \exists y\varphi(x,y)\},$$ where $\varphi$ is a formula with only bounded quantifiers in which $A$ gets used as a unary predicate symbol. Now there are two plausible guesses for the complexity of $B$, namely $\Sigma^0_1$ (since everything is existential) or $\Sigma^0_2$ (since there are two quantifiers involved).

The answer depends on how the unary predicate corresponding to $A$ gets used in $\varphi$. If it occurs only positively, then $B$ is $\Sigma^0_1$; otherwise, however, $B$ is merely guaranteed to be $\Sigma^0_2$. Indeed you can show that the following are equivalent for a set $X\subseteq\mathbb{N}$:

$X$ is $\Sigma^0_2$.

$X$ is c.e. relative to some co-c.e. oracle.

$X$ is c.e. relative to some c.e. oracle.

The point is that in a Turing reduction, both positive and negative information about the oracle is used (contrast this with enumeration reductions).

This indicates a dangerous ambiguity with phrases like "$\Sigma^0_m$-over-$\Sigma^0_k$" - we need to be careful about whether the inner predicate occurs positively or both positively and negatively. From what you've described, the latter is the correct interpretation. We then have pretty much what one should expect, e.g. in this sense $\Sigma^0_m$-over-$\Sigma^0_n$ is $\Sigma^0_m$-over-$\Pi^0_n$ is $\Sigma^0_{m+n}$. But in other contexts we might want to pay attention to finer restrictions and have "positivity" conditions involved in the way the inner predicate gets used.

However, basically everything else gets incredibly messy. For example:

Suppose $X$ is a high c.e. set, $X\le_m Y$, and $Y$ is "$\Pi^0_1$-complete-over-$X$." Then $X$ is not $\Sigma^0_1$ complete but $Y$ is $\Pi^0_2$-complete.

That "$X\le_mY$" bit is important: we can have e.g. a c.e. set $X$ and a set $Y$ such that $Y$ is "c.e.-over-$X$" but $Y$ doesn't compute $X$!

So folding in e.g. completeness questions makes things very fiddly.


问:
Integrating $\log(-ix)\exp(-ix)/x^2$
I would like to compute a few integrals like
$$\int_{-\infty}^\infty\frac{\log(-ix)\exp(-ix)}{x^2}\,dx$$
To be clear, here the path of integration is really $z = \epsilon i + x$, so that it avoids the singularity at $x=0$, and the branch cut of $\log$. This expression is "well-behaved" in several ways: it falls off faster than $1/x^{3/2}$ on the real line, so it converges decently quickly; in fact, falls off quickly everywhere in the lower half-plane of $z$. Exponentially quickly, thanks to the $\exp$ term. The function is holomorphic everywhere except the branch cut at $x\le 0$, so I can deform it pretty easily.

The $\log \exp / x^k$ form seems impossible to find an antiderivative for. So this seems like something that should be doable using Cauchy residue theorem and related tricks. The function grows quickly on the upper half plane, so I can't just deform it "up and away" and show that it's zero. One can easily reduce it to a keyhole contour around the branch cut, but neither the "along the cut" nor the "around the pole" term are zero, they both depend on the radius of the keyhole, and neither one seems solvable analytically.

Anyone have tricks to tackle this?

答: Assuming $\epsilon>0$, the given integral, after the substitution $-ix=z$, is $$i\int_{\epsilon-i\infty}^{\epsilon+i\infty}z^{-2}e^z\log z\,dz=-if'(2),\qquad f(s)=\int_{\epsilon-i\infty}^{\epsilon+i\infty}z^{-s}e^z\,dz=\frac{2\pi i}{\Gamma(s)}$$ (the last equality is basically Hankel's integral). The final result is then $\color{blue}{-2\pi(1-\gamma)}$.

Title: Integrating $\log(-ix)\exp(-ix)/x^2$


问:
What does the notation $\mathbb{Z}_q[X]/(X^n+1)$ mean and how is multiplication defined there?
I recently encountered the notation which is referred to as ring (in the description of a cryptography scheme, Dilithium, here, page 4):

[...] The key generation algorithm generates a $k \times l$ matrix $\bf{A}$ each of whose entries is a polynomial in the ring $R_q = \mathbb{Z}_q/(X^n+1)$

I am not entirely sure how to read this notation:

Is this simply the polynomials of maximum degree where all coefficients are in $Z_q$?

How is the multiplication of two polynomials defined in this ring?

Does the matrix $\bf{A}$ now consist of $l$ polynomials or $k \times l$ polynomials?

答:
$q$ is prime and $\mathbb{Z}_q=\mathbb{Z}/q\mathbb{Z}$ is the field with $q$ elements, $\mathbb{Z}_q[X]$ is the ring of polynomials with coefficients in $\mathbb{Z}_q$.

At first you can think to $R=\mathbb{Z}_q[X]/(X^n+1)$  as $\mathbb{Z}_q[X]$ quotiented by the relation $X^n=-1$. It means that in $\sum_{j=0}^J a_j X^j$ you can replace $X^j$ by $(-1)^{\lfloor j/n \rfloor} X^{j-n \lfloor j/n \rfloor}$, this gives for each polynomial a unique representative of degree $<n$.

Two such polynomials are added and multiplied as in $\mathbb{Z}_q[X]$, except that two elements of $R$ are equal iff they have the same representative of degree $<n$.

To clarify what is happening you should look first at $$\Bbb{C}\cong\Bbb{R}[X]/(X^2+1)$$
In $\Bbb{R}[X]$ quotiented by the relation $X^2=-1$ every element has a representative of the form $a+bX$ where $X$ is a square root of $-1$, so this is the complex numbers.

Finally $(X^2+1)$ means the ideal of $\Bbb{R}[X]$ generated by $X^2+1$, and the quotient of a ring $S$ by an ideal $I$ has a natural ring structure: its elements are subsets of $S$ of the form $u+I$ and $(u+I)+(v+I) =(u+v)+I$,$ (u+I)(v+I)=uv+I$.


问:
Smoothly connecting PDEs with finite differences

A PDE with non-smooth inhomogeneity

Let $\mathcal{L}$ be a second-order, linear, elliptic differential operator acting on $\mathcal{C}^2([0,2]^2)$.

I'm numerically solving the inhomogeneous PDE
\begin{align*}
\mathcal{L}u(x,y)+(x-1)^+=0,
\end{align*}
where $(\cdot)^+$ denotes the positive part.

Put differently, I solve two PDEs which need to be connected at $x=1$:
\begin{align*}
\begin{cases}
\mathcal{L}u(x,y)=0 & \text{for }(x,y)\in[0,1]\times[0,2], \\
\mathcal{L}u(x,y)+x-1=0& \text{for }(x,y)\in(1,2]\times[0,2].
\end{cases}
\end{align*}

Approximating  all partial derivatives by central differences, I get the nine-point stencil
\begin{align*}
c_1 u_{i-1,j-1} + c_2 u_{i,j-1} + c_3 u_{i+1,j-1} + c_4 u_{i-1,j} + c_5 u_{i,j} &\\
  + c_6 u_{i+1,j}+ c_7 u_{i-1,j+1} + c_8 u_{i,j+1} + c_9 u_{i+1,j+1} + (x_i-1)^+ &=0.
\end{align*}
Thus, $u$ is the solution to a system of linear equations.

Problem

Plotting the solution $u$ for a fixed $y$, it all looks fine and perfect. However, a plot of $\frac{\partial u}{\partial x}$ as a function of $x$ shows that the derivative is not smooth at $x=1$. The above FD scheme works fine with value-matching (the solution $u$ is perfectly continuous) but struggles with smooth-pasting at $x=1$ (the derivative is not smooth).

Question

How do I ensure smooth-pasting with a finite difference scheme at $x=1$?

Some of my failed attempts include

Impose that forward and backward differences at $x=1$ equal each other (= 2nd order central difference is zero).

Use higher order approximations around $x=1$ such as $u_{xx}\approx \frac{-u(-2h)+16u(-h)-30u(0)+16u(h)-u(2h)}{12h^2}$ and $u_{x}\approx \frac{u(-2h)-8u(-h)+8u(h)-u(2h)}{12h}$ and central differences everywhere else.

Approximating partial derivatives using points only from one side of $x=1$ (i.e. only using either $u(0), u(h), u(2h)$ or instead $u(0),u(-h),u(-2h)$).

Imposing that $u_x\approx\frac{u_{i+1,j}-u_{i-1,j}}{2h}$ equals an average of partial derivatives at $1+h$ and $1-h$.

Note: This problem arises as part of a larger system of free boundary problems. Thus, it's necessary to solve the PDE numerically. This question is also posted here.

答:
Let's say your solution $u$ is continuous at $x=1$, as are $u_x$ and $u_{xx}$, but not $u_{xxx}$. I think that's the situation you have. Using the standard templates centered at $x=1$ won't give you the accuracy you might expect, so you need to adjust.

One way to explore this is to consider the function
\begin{equation}
f(z) =
\begin{cases}
a_0 + a_1z + a_2z^2 + f_3z^3 + f_4z^4 + \cdots &\text{ for } z < 0, \\
a_0 + a_1z + a_2z^2 + g_3z^3 + g_4z^4 + \cdots &\text{ for } z > 0.
\end{cases}
\end{equation}
I've shifted from $x=1$ to $z=0$ just to make the typing a bit easier.
It's pretty smooth and you can argue $f$, $f'$, and $f''$ are continuous from the left and right. The standard $(1, -2, 1)/h^2$ template, which usually gives an $O(h^2)$ estimate for the second derivative now yields only $O(h)$. Give it a try. And the template $(-\tfrac{1}{12}, \tfrac{16}{12}, -\tfrac{30}{12}, \tfrac{16}{12}, -\tfrac{1}{12})/h^2$, which you'd expect to deliver $O(h^3)$, doesn't, it's also $O(h)$.

Instead, you'll need to respect the differences between the two regions. One way to do this is with a template that does exactly that. Try $(-\tfrac{1}{4}, 2, -\tfrac{7}{2}, 2, -\tfrac{1}{4})/h^2$. You'll see it's $O(h^2)$.

For most of your grid the standard template works fine and is precisely what you want, but it's the loss of higher derivatives at $x=1$ that tells you closer attention is required there.


问:
Giving a 1-hour talk to highschool math club: any topic suggestion?
I've been invited (by my kid) to give a one hour talk to her highschool math club.  Last year (right before the pandemic hit) I did two such talks on probability, and they loved it.  I'm looking for topic suggestions.

Logistics: There will be about $10-15$ kids total, from grades $9-12$.  These kids are mostly "A" students, and some of them have skipped a math grade, but they're nowhere near Olympiad standard.  The talk will be on March $1$st, $2021$, and will be on zoom.  I will be making slides this weekend.

I don't want this to become an "educational" lecture, so I'm looking for fun examples to motivate certain math areas.  E.g. last year my talks on probability culminated with Polya Urn and Buffon's Noodle.  Also, I'd like everything (well, $95\%$) to be fully explainable within the one hour, i.e. I don't want to just spew facts and then tell them "just trust me".

My own math skill tends toward discrete: comp-sci, combinatorics, etc.  However I'm open to any idea - esp. if it has worked for you before! - as long as I can learn it during one weekend.  :)

Some ideas:

modulo arithmetic esp. finite fields: Perfect shuffle is fun.  Would cryptography (e.g. RSA) be too difficult?  What about codes e.g. Hamming?  Any other fun examples of modulo arithmetic?

non-Euclidean geometry: especially how a POINT in elliptic geometry is actually $2$ "ordinary points".  Can the alternates to the parallel postulate (and some consequence e.g. angle sum in a triangle) be explained sufficiently at highschool level in an hour?

finite geometry: Personally I find Finite projective planes very beautiful.  Do they (or finite affine planes) have any application?  (Besides the kiddie game Spot It?)

algorithm: This is actually my work area, but I don't know what I can cover in an hour.  Maybe the $O(N \log N)$ lower bound for sorting?  Some of these kids don't even have coding experience... :(

graph theory: Edge coloring of a complete graph is one of my favorite pictorial proofs.  Eulerian tour.  Euler characteristic (of a planar graph) is related.  What else?  Shortest path would require discussion of algorithm.  Hall's marriage theorem is surprising and neat but I don't think I can prove it from scratch in an hour.

combinatorics: Start with ${N \choose m}$ and then stars-and-bars for sure.  What are your favorite (elementary) examples?  Can Burnside's lemma be proved from scratch (at highschool level, without group theory) in an hour?

Comments on above and/or any other suggestion most welcome!

(Apologies if I'm tagging too widely...)

答:
Have you considered game theory (e.g. Nim or Nim-like games)? Will students have the chance to interact and play a few games against each other?

Since you say it's a "math club", are these kids AMC 10/12 level?

Among your suggestions, I find combinatorics the most promising. There are lots of angles of attack: 1. story-problem/story-proof, 2. algebraic proof, 3. generating functions.

Aw man you're making me miss my high-school times. Can you record this and put it on the internet?


问:
Extending a local homeomorphism to a global one on a topological manifold $M$.
Suppose I have a topological $n$-manifold $M$ and a homeomorphisms $\varphi : U \to \mathbb{B}^n$ where $\mathbb{B}^n$ is the open unit-ball in $\mathbb{R}^n$ and $U \subseteq M$ is some open set.. Suppose that I have a homeomorphism $\xi : D^n \to D^n$ such that $\xi$ is fixed on the boundary of $D^n$. Then note that $\psi = \varphi^{-1} \circ \xi|_{\mathbb{B}^n} \circ \varphi : U \to U$ is a homeomorphism. Is it then possible to extend $\psi$ to a homeomorphism $\widetilde{\psi} : M \to M$? If not, are there any other conditions we need on $\varphi$ to ensure this is the case?

Similarly when is it possible to extend a homeomorphism $\zeta : \operatorname{Int}(M) \to \operatorname{Int}(M)$ to all of $M$?

答:
Assume $U$ is tamely embedded with boundary $\mathbb{S}^{n-1}$.  Here $M$ is assumed to have no boundary.  If you let $\xi$ be the identity then we are just talking about when $\phi$ extends to an orientation-preserving homeomorphism on the boundary.  If there is such an extension, then you can take a collar neighborhood $N$ around $\mathbb{S}^{n-1}$ and apply the annulus theorem + Schoenflies to construct a homeomorphism on all of $M$ by letting $f \equiv \text{id}_M$ on $N^c$ and homotoping to the identity in the annulus $N$ (which is possible via the degree theory of sphere mappings).  Note that homotopic sphere homeomorphisms are homotopic via homeomorphisms.

EDIT: Actually, there's a bit more to it; I forgot about the degree $-1$ case.  If you have your homeomorphic extension to the boundary of $U$ but it's the wrong orientation, there may still be a homeomorphism on $M$.  For example, reflecting the open disc in the plane, viewed as a subset of the sphere; the whole sphere can be flipped.  Assuming $M$ is connected, then any two open balls with tame boundaries can be carried onto one-another via a homeomorphism of $M$, so you want to have the boundary extension + orientation-preserving OR homeo$(M)$ contains an element which flips the orientation on a tame $(n-1)$-sphere which bounds an open ball.

But in general this is not always possible, even in the plane: For example, take a homeomorphism of $\mathbb{D}^2$ to itself that preserves the distance of any point to zero, but which rotates faster and faster unboundedly as you approach $\mathbb{S}^1$.  Then as you approach the unit circle via a straight line from the center its trajectory will be dense in $\mathbb{S}^1$ so won't have an extension.  You can do similar things in any dimension larger than $1$; in the even dimensions use similar rotations, and in odd dimensions you can "kneed the dough back and forth" more and more as you approach the boundary - maybe someone can give an explicit map in the odd dimensions which is easier to write down than what I'm talking about.

If your sphere isn't tamely embedded then even having the extension to the boundary won't be strong enough to imply what you want, unfortunately.  Extension of homeomorphisms isn't a 'solved problem' in topological manifolds of dimension higher than $2$ - there won't be a simple answer, only partial and technical results which may only hold in specific dimensions.

If the boundary of $U$ self-intersects in $M$ but is 'otherwise tame' in some nebulous sense then I have no idea.  There would be all sorts of cases depending on how it intersected with itself, probably.


问:
Flipping Coins, Expected Value Interpretation
Say we have the following problem:

We are flipping some coin. The probability of it landing on heads is $p$. Is the expected number of coin flips we need to get the expected number of heads to be 1 different from the expected number of coin flips we need to get one head?

P.S. when I say are they the same I mean are they equivalent questions, not are the values of both the same (which they are).

I would think these are the same but I don't know how to describe why.

Thanks in advance :)

答:
Let $X\sim Ber(p)$, where $P(X=1)=p$, and $P(X=0)=1-p$.

You are asking whether

$$E(E(X=1))=E(X=1).$$

This is true because $E(X=1)$ is a constant (in fact, $p$), so you can take it out of the $E(\cdot)$ on the left hand side.

What do you mean when you say "are those equivalent questions"?


答:
Conway's rational tangles, continued fractions, and the "greedy" Euclidean algorithm!

Here are some Math Circles notes by Tom Davis on the topic. You're going to want to have two pieces of thick rope, at least 10 feet in length, and hopefully have at least 4 volunteers to hold the ends for a very fun interactive activity.

Pictured: construction of the tangle
$$
-4 + \frac{1}{2+\frac{1}{3}} = -\frac{25}{7}
$$

Title: Giving a 1-hour talk to highschool math club: any topic suggestion?

Body:
I've been invited (by my kid) to give a one hour talk to her highschool math club.  Last year (right before the pandemic hit) I did two such talks on probability, and they loved it.  I'm looking for topic suggestions.

Logistics: There will be about $10-15$ kids total, from grades $9-12$.  These kids are mostly "A" students, and some of them have skipped a math grade, but they're nowhere near Olympiad standard.  The talk will be on March $1$st, $2021$, and will be on zoom.  I will be making slides this weekend.

I don't want this to become an "educational" lecture, so I'm looking for fun examples to motivate certain math areas.  E.g. last year my talks on probability culminated with Polya Urn and Buffon's Noodle.  Also, I'd like everything (well, $95\%$) to be fully explainable within the one hour, i.e. I don't want to just spew facts and then tell them "just trust me".

My own math skill tends toward discrete: comp-sci, combinatorics, etc.  However I'm open to any idea - esp. if it has worked for you before! - as long as I can learn it during one weekend.  :)

Some ideas:

modulo arithmetic esp. finite fields: Perfect shuffle is fun.  Would cryptography (e.g. RSA) be too difficult?  What about codes e.g. Hamming?  Any other fun examples of modulo arithmetic?

non-Euclidean geometry: especially how a POINT in elliptic geometry is actually $2$ "ordinary points".  Can the alternates to the parallel postulate (and some consequence e.g. angle sum in a triangle) be explained sufficiently at highschool level in an hour?

finite geometry: Personally I find Finite projective planes very beautiful.  Do they (or finite affine planes) have any application?  (Besides the kiddie game Spot It?)

algorithm: This is actually my work area, but I don't know what I can cover in an hour.  Maybe the $O(N \log N)$ lower bound for sorting?  Some of these kids don't even have coding experience... :(

graph theory: Edge coloring of a complete graph is one of my favorite pictorial proofs.  Eulerian tour.  Euler characteristic (of a planar graph) is related.  What else?  Shortest path would require discussion of algorithm.  Hall's marriage theorem is surprising and neat but I don't think I can prove it from scratch in an hour.

combinatorics: Start with ${N \choose m}$ and then stars-and-bars for sure.  What are your favorite (elementary) examples?  Can Burnside's lemma be proved from scratch (at highschool level, without group theory) in an hour?

Comments on above and/or any other suggestion most welcome!

(Apologies if I'm tagging too widely...)


问:
Proving $\frac 1 4 < \left( \int_a^b [f']^2 \ dx\right)^{1/2}\left( \int_a^b x^2[f(x)]^2 \ dx\right)^{1/2}$
In chapter 6 of Rudin's Principles of Mathematical Analysis, problem 15, he writes

Suppose $f$ is a real, continuously differentiable function on $[a,b], f(a) = f(b) = 0$, and

$$\int_a^b f^2(x)\ dx = 1.$$

Prove that

$$ \int_a^b x f(x) f'(x) \ dx = -\frac 1 2 $$

and that

$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx > \frac 1 4. $$

I've shown everything up to demonstrating

$$\int_a^b[f'(x)]^2 \ dx \cdot \int_a^b x^2f^2(x)\ dx \ge \frac 1 4 $$

However, I can't figure out how to eliminate the case where equality holds.  I've seen several other resources say that equality would imply

$$ f'(x) = \lambda xf(x) $$

but I can't figure out why that would be implied.

Resources I've already looked at: How to show the inequality is strict?

Baby Rudin Chapter 6, Problem 15 : Strict inequality

Prob. 15, Chap. 6, in Baby Rudin: If $f$ is a real, continuously differentiable function on $[a, b]$, . . .

Proving a strict inequality (Application of Hölder's Inequality)

Show that if $f(a)=f(b)=0$ and $\int_a^b [f(x)]^2dx=1$, then $\int_a^b [f'(x)]^2dx\cdot\int_a^b [xf(x)]^2dx\gt \frac14$

答:
Use Cauchy Schwartz inequality, $p(x)$ and $q(x)$ are continuous in $[a,b]$, then
$$\left(\int_{a}^{b} p(x) q(x) dx \right)^{2} \le \int_{a}^{b} p^2(x) dx \int_{a}^{b} q^2(x) dx,$$
by choosing $p(x)=f'(x), q(x)=x f(x).$ Equality holds when $p(x)=t q(x)$.

Here the equality will hold if $$txf(x)=f'(x) \implies f(x)=C e^{tx^2/2}.$$
Using $f(a)=0=f(b)$, we get $C=0$. so the equality will hold for the trivial  case of $f(x)=0.$


问:
Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$
I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits_{0}^{1} \frac{\log(1-u^2+u^4)}{1+u^2}\mathrm{d}u.$$But now I am stuck again.

答:
$$I=\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}dx\overset{t=1/x}=\int\limits_{1}^{\infty} \frac{\log(1-t+t^2)}{\sqrt{t}(1+t)}dt-\int\limits_{1}^{\infty} \frac{\log(t^2)}{\sqrt{t}(1+t)}dt$$

$$I=\frac{1}{2}\bigl(\int\limits_{0}^{\infty} \frac{\log(1-t+t^2)}{\sqrt{t}(1+t)}dt-\int\limits_{1}^{\infty} \frac{\log(t^2)}{\sqrt{t}(1+t)}dt\bigr)\overset{x=\sqrt{t}}=$$$$=\frac{1}{2}\int\limits_{-\infty}^{\infty} \frac{\log(1-x^2+x^4)}{1+x^2}dx-4\int\limits_{1}^{\infty} \frac{\log(x)}{1+x^2}dx$$

Branch points of logarithm: $e^{\frac{\pi{i}}{6}}, e^{\frac{5\pi{i}}{6}}, e^{\frac{-\pi{i}}{6}}, e^{\frac{-5\pi{i}}{6}}$

For the first couple of roots we close the contour in the bottom half of the complex plane, for the second - in the upper-half.

$$I=\frac{2\pi{i}}{2*2i}\log\Bigl((-i-e^{\frac{\pi{i}}{6}})(-i-e^{\frac{5\pi{i}}{6}})(i-e^{\frac{-\pi{i}}{6}})(i-e^{\frac{-5\pi{i}}{6}})\Bigr)-4G$$
$$I=\frac{\pi}{2}\log\biggl(4(1+\sin{\frac{5\pi}{6}})(1+\sin{\frac{\pi}{6}})\biggr)-4G=\frac{\pi}{2}\log(3^2)-4G=\pi\log3-4G$$


问:
Eigenvalues of outer product matrix of two N-dimensional vectors
I have a vector $\textbf{a}=(a_1, a_2, ....)$, and the outer product $M_{ij}=a_i a_j$. What are the eigenvalues of this matrix? and what can you say about the co-ordinate system in which $M$ is diagonal? I have proved that the only eigenvalue of the matrix is the norm of the vector squared, and that one of the eigenvectors is $\textbf{a}$ itself.

$$Mu=a a^{T}u=\lambda u\\\ \implies a^{T}(aa^{T})u=a^{T}\lambda u \implies a^{T}a (a^{T}u)=\lambda a^{T}u \implies ||a||^2=\lambda$$

Also it is obvious that $a$ is the eigenvector of $aa^{T}$, which implies $M= \begin{pmatrix} ||a||^2 &0 &0 &..... \\0 & ||a||^2\\.\\.\\\\. \end{pmatrix}$

Is this correct? What are all the eigenvectors of this the outer product? and what intuitively happens in the transformed basis when M is diagonal?

答:
Say $b \perp a$ then $a^{\text{T}}b=0$ and therefore $Mb=aa^{\text{T}}b=0$. So, just find $n-1$ independent vectors that are orthogonal to $a$ and you have $n-1$ new eigenvectors of $M$, all with eigenvalue $0$.

So the spectrum of your matrix $M$ is $(\|a\|,0,\ldots,0)$. The transformed basis is just what happens when you realign your axes to correspond with the eigenvectors. That's when you see that $M$ projects on one direction, the direction of $a$.


问:
Deck of Cards Probability: One card missing(randomly), Probability it's a jack?
The question is:

A card is missing from a deck of 52 cards. You draw a card, What's the
probability that it's a jack?

I been struggling with probability and would need help understanding this question. What I got so far is that I need to consider two scenerios.

A: Jack is one of the missing cards.

B: Jack is not one of the missing cards.

P(A) = 4/52

P(B) = 48/52

To continue I feel like I might be on the right direction but I could be getting my probabilities wrong and that's what is confusing me and I don't know what do after. Any suggestion are welcomed. Thank you.

答:
Use the events, $A:$ the missing card is a Jack, $B:$ the missing card is not a Jack, and $C:$ a Jack is drawn.

You have $\mathsf P(A)=4/52$ and $\mathsf P(B)=48/52$

The Law of Total Probability says: $\mathsf P(C)=\mathsf P(A)\,\mathsf P(C\mid A)+\mathsf P(B)\,\mathsf P(C\mid B)$

$\mathsf P(C\mid A)$ is the probability for drawing a Jack when three are left among the fifty-one remaining cards.

$\mathsf P(C\mid B)$ is the probability for drawing a Jack when all four are among the fifty-one remaining cards.

$~\\~\\~$

What do you notice about the result?   Is this surprising?


问:
$\mathbb{Q}$ is not a Baire space?
The condition of 'X is a Baire space': (cite from Topology by Munkres)

Given any countable collection $\{ A_n \}$ of closed sets of X each of which has empty interior in X, their union $\bigcup A_n$ also has empty interior in X.

So for rational numbers $\mathbb{Q}$ with the usual topology of $\mathbb{R}$:

(1) countable collection $\{ Q_i \}$ : one can index all rational numbers with  positive integers, so let $Q_i = \{q_i\}$ be
one-point set.

(2) $Q_i$ is closed, since $\mathbb{R}- Q_i= (-\infty, q_i) \cup (q_i, \infty)$ is open in $\mathbb{R}$.

(3) $Q_i$ has empty interior, since $Q_i$ contains no open set of $\mathbb{R}$.

(4) $\mathbb{Q}=\bigcup Q_i$ has empty interior, since $\mathbb{Q}$ contains no open set of $\mathbb{R}$.

Thus, $Q_i$ is a Baire space. What's wrong here?

答: $\Bbb Q$ doesn't have an empty interior.  The relevant topology is the topology on $\Bbb Q$.  That topology is the topology inherited from the usual topology on $\Bbb R$, so a basis for the open sets of the topology are the rational points of any open interval.

Title: $\mathbb{Q}$ is not a Baire space?


问:
Every preorder is a topological space
Let $(X,\leq)$ be a preorder, i.e. $\leq$ is a reflexive and transitive binary relation on $X$. I want to show that $\leq$ induces a topological structure on $X$. Hence I need to specify when a subset $A$ of $X$ must be considered open. My personal guess is that $A$ is open iff it is contained in its interior, where the interior $A^°$ of $A$ is defined as follows:

$$A^°:=\{s\in X: \forall f\in X (s\leq f\rightarrow f\in A\}$$

Do you think this makes sense? If not, which could be the right definition of an open?

A "dual" definition for the interior of $A$ could be

$$A^°:=\{s\in X: \forall p\in X (p\leq s\rightarrow p\in A\}$$

Do the two definitions work in order to get a topological space? Do they define the same topology on $X$ (if they define any?)

In the same spirit, we could define a topology by mean of closed subsets. I say that a subset $C$ of $X$ is closed when the closure of $C$ is contained in $C$, and i define the closure of $\overline{C}$ of $C$ as follows:
$$\overline{C}:=\{s\in X:\exists f\in X(s\leq f\wedge f\in C\}$$
or, dually, as
$$\overline{C}:=\{s\in X: \exists p\in X(p\leq s\wedge p\in C\}$$
The question is the same: do this define a topology on $X$? If so, what kind of topology is this? Has it some evident description, is it the obviuous one, or natural in some sense?

答:
If you have a simple order, then the following approach by Munkres is a nice one:

The following will be the basis elements of the topology, and the generate the topology using this basis.

All intervals of the form (a,b) where $a,b \in X$.

All intervals $(a_{0},b)$ where $a_{0}$ is the smallest element (if any) of the space $X$.

All intervals $(a,b_{0})$ where $b_{0}$ is the largest element (if any) of the space $X$.

It is easily seen that this forms a basis of a topology.

This is one way to define a topology on an ordered set. For example, the usual topology on $\mathbb{R}$ is the one obtained by this method.

Title: Every preorder is a topological space


答:
My own math skill tends toward discrete: comp-sci, combinatorics, etc.
However I'm open to any idea - esp. if it has worked for you before! -
as long as I can learn it during one weekend.  :)

Some ideas:

modulo arithmetic esp. finite fields: Perfect shuffle is fun.  Would
cryptography (e.g. RSA) be too difficult?  What about codes e.g.
Hamming?  Any other fun examples of modulo arithmetic?

...

algorithm: This is actually my work area, but I don't know what I can cover in an hour.  Maybe the $O(N \log N)$ lower bound for
sorting?  Some of these kids don't even have coding experience... :(

You did probability before.  Even without cryptography, I think it's fascinating that one can pretend to have probability using a pseudorandom number generating algorithm.  The simplest algorithms are very simple, but still depend on basic results in modulo arithmetic, finite fields, etc.  You could teach some basic statistical tests on PRNG quality if you want.

You can get what you need from Knuth's TAOCP vol. 2, ch. 3, or maybe Kneusel's Random Numbers and Computers (too shallow?) or Niederreiter & Winterhof (too deep?).

Another idea: Fourier analysis.  Nice graphics. Sum cool curves, infinite series, integrals--or stick with discrete cases.  You can even transform Boolean functions.


答:
You are correct so far. There are two cases to consider and they are disjoint: the missing card is or is not a jack. If the jack is missing, then there are only $3$ possible jacks to draw from the remaining $51$-card deck. Similarly, if the missing card is not a jack, then there are $4$ possible jacks to draw from the remaining $51$-card deck.

Remember how conditional probability works: $P(A \cap B) = P(B) \, P(A \mid B)$.

Try to calculate this yourself, but click below to reveal a spoiler when you're ready.

 $$\begin{align*}\def\j{\text{draw jack}}\def\jm{\text{jack missing}}\def\njm{\text{jack not missing}}P(\j) &= P(\j \cap \jm) + P(\j \cap \njm) \\&= P(\jm) \, P(\j \mid \jm) \\&\qquad\qquad {}+ P(\njm) \, P(\j \mid \njm) \\&= \frac{4}{52}\frac{3}{51} + \frac{48}{52}\frac{4}{51} \\&= \frac{1}{13}\frac{3}{51} + \frac{12}{13}\frac{4}{51} \\&= \frac{51}{13 \cdot 51} \\&= \frac{1}{13} \end{align*}$$

Title: Deck of Cards Probability: One card missing(randomly), Probability it's a jack?

Body:
The question is:

A card is missing from a deck of 52 cards. You draw a card, What's the
probability that it's a jack?

I been struggling with probability and would need help understanding this question. What I got so far is that I need to consider two scenerios.

A: Jack is one of the missing cards.

B: Jack is not one of the missing cards.

P(A) = 4/52

P(B) = 48/52

To continue I feel like I might be on the right direction but I could be getting my probabilities wrong and that's what is confusing me and I don't know what do after. Any suggestion are welcomed. Thank you.


问:
Partitioning a union of convex shapes
A union of two axes-parallel rectangles can always be partitioned into at most $3$ pairwise-disjoint axes-parallel rectangles:

Is there an analogous lemma for convex figures? I.e, is there a number $k$ such that a union of two convex figures can always be partitioned into at most $k$ pairwise-disjoint convex figures?

答:
As Michael Burr mentions in a comment, this is not bounded with unrestricted convex figures, because we can take two regular $n$-gons rotated at an offset from one another:

I thought I would illustrate a proof that this takes at least $n+1$ convex pieces, since it was not originally clear to me how to rigorously show that the number of pieces in such a construction is unbounded. ($n+1$ is the best possible, by taking one of the original polygons and the $n$ leftover triangles.)

Consider which of the $2n$ "outer" vertices lie in the same convex piece. Clearly, no two such vertices which are adjacent can lie in the same convex piece. We wish to allocate them into pieces so that the diagonals of different pieces don't intersect.

Lemma: Given $k\ge 2$ points in cyclic order, say that a good allocation is a way of grouping the points together so that no two adjacent points are in the same piece, and given points $p_1,p_2$ in piece $A$ and $q_1,q_2$ in piece $B$, the lines $p_1p_2$ and $q_1q_2$ do not intersect. Then any good allocation uses at least $\left\lfloor \frac k2\right\rfloor +1$ pieces.

For $k=2n$, this will give us the desired result.

Proof: By induction on $k$. Obviously, it is true when $k=2$. Now, suppose it holds up to $k$, and consider a good allocation of $k+1$ points. If no two points are in the same piece, we have $k+1\ge\left\lfloor \frac k2\right\rfloor +1$ pieces.

Otherwise, suppose two points $P$ and $Q$ are in the same piece. Let there be $a$ and $b$ points on either side of the line $PQ$, with $a+b=k-1$. By assumption, $P$ and $Q$ are not adjacent, so $a,b\ge 1$. But then we can think of $PQ$ as a single new "point", $R$, and apply the lemma to the $a+1$ points on one side and the $b+1$ points on the other. We conclude that there are at least $\left\lfloor \frac {a+1}2\right\rfloor +\left\lfloor \frac {b+1}2\right\rfloor +2$ pieces, except we have to subtract $1$ to avoid overcounting the piece containing $PQ$. But

$$\left\lfloor \frac {a+1}2\right\rfloor +\left\lfloor \frac {b+1}2\right\rfloor +1 \ge \frac a2+\frac b2+1 = \frac{k+1}2$$

So there are at least $\left\lceil\frac{k+1}2\right\rceil$ pieces. But for integer $k$, $\left\lceil\frac{k+1}2\right\rceil=\left\lfloor \frac k2\right\rfloor +1$, so the proof is complete.

Title: Partitioning a union of convex shapes


答: Trying to enter into the mind of the OP I think she wants to mimic the usual euclidean topology on $\mathbb{R}$: so you can define a basis for your topology given by $ \{ (a,b):a,b\in X \}$. You have to check that it satisfies the condition for being a basis and if it's true or not that order preserving endomorphism are continuous with this topology on both domain and codomain.</p>

Title: Every preorder is a topological space

Body:
Let $(X,\leq)$ be a preorder, i.e. $\leq$ is a reflexive and transitive binary relation on $X$. I want to show that $\leq$ induces a topological structure on $X$. Hence I need to specify when a subset $A$ of $X$ must be considered open. My personal guess is that $A$ is open iff it is contained in its interior, where the interior $A^°$ of $A$ is defined as follows:

$$A^°:=\{s\in X: \forall f\in X (s\leq f\rightarrow f\in A\}$$

Do you think this makes sense? If not, which could be the right definition of an open?

A "dual" definition for the interior of $A$ could be

$$A^°:=\{s\in X: \forall p\in X (p\leq s\rightarrow p\in A\}$$

Do the two definitions work in order to get a topological space? Do they define the same topology on $X$ (if they define any?)

In the same spirit, we could define a topology by mean of closed subsets. I say that a subset $C$ of $X$ is closed when the closure of $C$ is contained in $C$, and i define the closure of $\overline{C}$ of $C$ as follows:
$$\overline{C}:=\{s\in X:\exists f\in X(s\leq f\wedge f\in C\}$$
or, dually, as
$$\overline{C}:=\{s\in X: \exists p\in X(p\leq s\wedge p\in C\}$$
The question is the same: do this define a topology on $X$? If so, what kind of topology is this? Has it some evident description, is it the obviuous one, or natural in some sense?


问:
Continuous Image of a Closed Interval is Closed.
$f:A\to \mathbb{R}$ is a continuous function, where $A$ is a closed set which is also an interval. Then $f(A)$ is also a closed set as well as an interval.

Is the statement true?

$A$ may not be bounded, that is $A$ can be of the form $[a,\infty)$

If I take $A:=(-\infty,\infty)=\mathbb{R}$ then this is not true. Because there are functions from $\mathbb{R}$ whose image is $(0,1)$.

So If $A$ is a proper closed subset of $\mathbb{R}$, is the statement true?

答: Yes, $f$ attains its maximal $K$ and minimal $T$ on closed interval $A$, and then use intermediate value theorem to conclude that $f$ attains  every $t \in [T,K]$ on $A$.


问:
What is the use of caligraphic characters in mathematics?
The only one I am familiar with is $\mathcal{L}$ for Laplace transforms, but I have now encountered other symbols such as $\mathcal{F}$ and $\mathcal{M}$. I'm curious: does each calligraphic letter have a specific use case, or are they just alternatives to the standard $f$ and $F$ types (and other letters)?

答:
So far I've seen:

$\mathcal{A}, \mathcal{B}$ for bases in linear algebra or function classes

$\mathcal{C}, \mathcal{D}$ for integration domains in the complex plane

$\mathcal{E}$ looks like a "capital epsilon ($\epsilon$)" so it might be used together with other capital letters (e.g. matrices) like $f(A+\mathcal{E})$

$\mathcal{F}, \mathcal{G}, \mathcal{H}$ for function classes

$\mathcal{I}, \mathcal{J}, \mathcal{K}$ for index sets (often uncountable)

$\mathcal{L}$ for linear functionals (linear map that takes in a function)

$\mathcal{M}$ (not sure)

$\mathcal{N}$ for normal distribution

$\mathcal{O}$ for big-O notation

$\mathcal{P}$ for power set

$\mathcal{Q}$ (not sure)

$\mathcal{R}$ for Rademacher complexity, or response function when solving DEs using Fourier transform

$\mathcal{S}, \mathcal{T}, \mathcal{U}, \mathcal{V}$ transformations (often linear)

$\mathcal{X}, \mathcal{Y}, \mathcal{Z}$ domains of random variables or random samples


问:
Composition of Integrable Function and Strictly Increasing Continuous Function
I have been struggling to prove that if $f$ is integrable on $B$ and $g$ is continuous and strictly increasing on $A$ such that $Img \subset B$, then $f\circ g$ is integrable on $A$.

I know that this is true, for example, in the case when $f$ is integrable on $B=[0,1]$ and $g=x^n$ with $A=[0,1]$, then $f\circ g=f(x^n)$ is integrable on $A$. This is easily shown by Lebesgue criterion for Riemann integrability, because if $f$ is integrable on $[0,1]$, then it is discontinuous on the subset of $[0,1]$ of measure zero. Then since $g:[0,1] \rightarrow [0,1]: x \mapsto x^n$ is continuous bijection, the set of discontinuities of $f \circ g$ has measure zero. So, $f \circ g$ is integrable on $[0,1]$ by Lebesgue criterion again.

Is my argument above for the special case example correct? If yes, can one generalize it somehow?

答:
I have been struggling to prove that if $f$ is integrable on $B$ and $g$ is continuous and strictly increasing on $A$ such that $Img \subset B$, then $f\circ g$ is integrable on $A$.

I don't think this is generally true.

Consider this counterexample:

Let $f$ be the characteristic function on the ternary cantor set $C$. Since $C$ is measure 0 and $f$ is continuous at points outside of $C$, $f$ is riemann integrable. Let $g$ be an ambient homeomorphism of $\mathbb{R}$ that carries the fat cantor set $F$ to $C$. We can do this since every cantor space on $\mathbb{R}$ are ambiently homeomorphic. Since $g$ is a continuous bijection on $\mathbb{R}$ (it's a homeomorphism of $\mathbb{R}$ to itself), it must be strictly monotone. However, $f \circ g$ restricted to $[0,1]$ is not riemann integrable since it's discontinuity set is $F$, which is not measure 0 by construction.


答:
Let $A = [0, \infty)$ and $f(x) = \arctan(x)$. Then $f(A) = [0, \pi/2)$, which is not a closed interval.

The result is true if $A$ is a compact (closed and bounded) interval, in which case $f(A)$ is also a closed and bounded interval.

Title: Continuous Image of a Closed Interval is Closed.

Body:
$f:A\to \mathbb{R}$ is a continuous function, where $A$ is a closed set which is also an interval. Then $f(A)$ is also a closed set as well as an interval.

Is the statement true?

$A$ may not be bounded, that is $A$ can be of the form $[a,\infty)$

If I take $A:=(-\infty,\infty)=\mathbb{R}$ then this is not true. Because there are functions from $\mathbb{R}$ whose image is $(0,1)$.

So If $A$ is a proper closed subset of $\mathbb{R}$, is the statement true?


问:
Vector perpendicular to 3d vector?
Given a unit vector $v_1 \in \mathbb{R}^3 : |v_1| = 1$, let $v_1 = (x_1,y_1,z_1)$

What's the simplest way to find another unit vector $v_2 = (x_2, y_2, z_2)$ such that they are perpendicular (ie $v_1 \cdotp v_2 = 0$) ?

Notice the constraints are:

\begin{align}
    x_1x_2 + y_1y_2 + z_1z_2 &= 0,\\
    {x_1}^2 + {y_1}^2 + {z_1}^2 &= 1,\\
    {x_2}^2 + {y_2}^2 + {z_2}^2 &= 1.
\end{align}

What is a solution for $x_2$, $y_2$, $z_2$ in terms of $x_1$, $y_1$, $z_1$ that is easy to express and calculate?

(Yes there is an infinite family of such vectors $v_2$ for a given $v_1$, I just want one, any one)

答:
Pick any vector ${\bf v}_2 \neq {\bf v}_1$, then compute ${\bf v}_1 \times {\bf v}_2$, and normalize the result.

(Oh... I just saw @JMoravitz stated this in a comment.  Upvote his comment, not my "answer.")

Title: Vector perpendicular to 3d vector?


答: I've always been a fan of the coin problem. It's basically a generalization of the question, "what is the largest score in American football that cannot be achieved (say, ignoring safeties)?" It's got a built-in hook with the story of the "McNugget numbers". It very naturally motivates $\gcd(a_1, \ldots, a_n)$ and you can discuss existence of the Frobenius number. As I recall there's a nice graphical proof of the formula for two coins. The topic typically uses some basic number theory like Bezout's lemma, but if anything that's a feature and not a bug.