问:
$\epsilon$-regular pair of vertex sets definition
Introducing Szemeredi's Regularity Lemma requires the definition of  $\epsilon$-regularity, which I cannot get good intuition for, because the official intuition clashes with, well let me explain.

Definition: for a fixed $\epsilon>0$ we say that the pair of vertex sets $(A,B)$ (which are subsets of the two vertex sets of a bipartite graph in most situations) is $\epsilon$-regular if for any subsets $A'\subset A, B'\subset B$ with $|A'|\ge\epsilon|A|$ and $|B'|\ge\epsilon|B|$ we have $$|d(A,B)-d(A',B')|\le\epsilon$$
where $d(A,B):=\frac{\#\text{edges between }A\text{ and }B}{|A||B|}$ is called the edge density (and this last definition happens to make perfect sense).

My issue with the way $\epsilon$-regularity is defined is with the condition $|A'|\ge\epsilon|A|$ and $|B'|\ge\epsilon|B|$.

$\epsilon$-regularity is supposed to capture the randomness of a graph: taking subsets won't result in a dramatical change in edge density.

But to me it would make way more sense to have $|A'|\ge(1-\epsilon)|A|$ and $|B'|\ge(1-\epsilon)|B|$ as a condition instead. According to the original definition, the smaller the $\epsilon$ the closer the edge densities are.

I guess in my version of the definition it is easier to be $\epsilon$-regular and it is actually always implied by the original $\epsilon$-regularity if $\epsilon<1/2$, because if if the densities are close for any sets of size $\ge \epsilon |A|$ and $\ge \epsilon|B|$ they will be close for any sets that are bigger than these bounds (just by the transitive property of $\ge$.

My confusion came from the fact that being close in size will result to being close in edge density, but the original definition wants more than that property (which  maybe is even true for graphs that are far from being random).

答:
When $A'$ and $B'$ are close in size to $A$ and $B$, such as the condition $|A'| \ge (1-\epsilon)|A|$ and $|B'| \ge (1-\epsilon)|B|$, we expect $d(A',B')$ to be close to $d(A,B)$, no matter what the graph structure is. So your version of the definition is far too weak.

For example, when $|A'| \ge (1-\epsilon)|A|$ and $|B'| \ge (1-\epsilon)|B|$, we can write $$|A'||B'|d(A',B') \le |A||B|d(A,B) \le |A'||B'| d(A',B') + 2\epsilon |A| |B|$$ just by counting edges. The first inequality says "there are at least as many edges between $A$ and $B$ as between $A'$ and $B'$." The second inequality says "we get an upper bound on the edges between $A$ and $B$ by assuming all the edges not between $A'$ and $B'$ are there." Anyway, dividing by $|A||B|$, we get
$$
   (1-2\epsilon + \epsilon^2) d(A',B') \le d(A,B) \le (1+\epsilon^2) d(A',B')
$$
which is a form of saying "$d(A',B')$ is close to $d(A,B)$". It follows that $|d(A',B') - d(A,B)| \le 2\epsilon$. This doesn't quite imply your definition, but it's halfway there.

We don't want the $\epsilon$-regularity condition to almost hold for any pair $(A,B)$! We want the statement "$(A,B)$ is an $\epsilon$-regular pair" to actually mean something. So we ask for as much as possible - because the regularity lemma says we can.

You can imagine a more general $(\epsilon,\delta)$-regularity that says $$|d(A,B) - d(A',B')| \le \epsilon$$ whenever $|A'| \ge \delta |A|$ and $|B'| \ge \delta |B|$. But for both $\epsilon$ and $\delta$, the smaller they are, the better, and the regularity lemma says it's possible to make both of them as small as we want. So we just set them equal to simplify the statement.

Finally, we want the regularity condition to be useful. When you start using the regularity lemma, you'll see that the actual form is what we need to find large substructures in our graph.


问:
Proof of injective functions
Let $f:\mathbb{N}\to\mathbb{N}$ and $N_k=\{1,2,3,...,k\}$ for all $k\in\mathbb{N}$.

I have to prove two statements. If $f_{\big|N_k}$ is injective function for all $k\in\mathbb{N}$, then $f$ is injective function(one to one) and second if $f[N_k]=N_k$ for all $k\in\mathbb{N}$, then $f$ is identity function.

Note that $f_{\big|N_k}$ is restricted domain of function and $f[N_k]=N_k$ is image of function.

I prove the first statement, let $x_1,x_2\in\mathbb{N}$ and assume that $f(x_1)=f(x_2)$. If $x_1,x_2\in{N_k}$, then we are done but if not ?
Next, the second statement I have to show that $f(x)=x$ for all $x\in\mathbb{N}$ but how to show that ?

Tanks for someone to help me.

答:
Consider two natural numbers not equal to each other.  Label the smaller one $j$ and the larger one $k$.  $j< k$ so $j, k \in \mathbb N_k$.  $f|_{\mathbb N_k}$ is injective so $f(j) \ne f(k)$.  But $j$ and $k$ were arbitrary so for any $j,k \in \mathbb N$ if $j\ne k$ then $f(j) \ne f(k)$ and $f$ is injective.

Let $n \in \mathbb N$.  $n\in \mathbb N_n$ so $f(n) \in \mathbb N_n$ so $f(n) \le n$.

If $f(n) < n$ then let $f(n) = k < n$.  But $f(\mathbb N_k) = \mathbb N_k$ so there is an $m \in \mathbb N_k$ so that $f(m) = k = f(n)$.  But by 1) $f$ is injective so $m = n$.  But $n \not \in \mathbb N_k$.  So $f(n) \not < n$ and $f(n)\le n$ and so $f(n) = n$.

ContentLicense: CC BY-SA 4.0

Title: Proof of injective functions


问:
Preserving structures
Category theory abstracts the notion of the preservation of structure by means of morphisms. Is there a description of what it means to preserve structure of different types of mathematical structures (topological spaces, modules, rings, etc.)? </p>

答:
Typically, structures on a set $A$ are relations, that is subsets $R_A$ of some $A^n$. In that case $f\colon A\to B$ preserves the structure if $(a_1,\ldots,a_n)\in R_A$ implies $(f(a_1),\ldots,f(a_n))\in R_B$.
An example of this are groups where we can consider the group law as ternary relation $\{(a,b,c)\in G^3\mid a\circ b=c\}$.
Of course this also works with infinitary relations.

Topological spaces are a bit different, as here the structure on $A$ is a subset $T_A$ of $\mathcal P(A)$, which introduces some contravariance. As you know, $f\colon A\to B$ is continuous if the inverse images of open $B$-sets are open $A$-sets. That is, $f\colon A\to B$ preserves the stucture if the induced map $f^{-1}\colon \mathcal P(B)\to\mathcal P(A)$ behaves as above, i.e. $U\in T_B$ implies $f^{-1}(U)\in T_A$.

Of course, all these can be combined and even more complex structures can be considered.

ContentLicense: CC BY-SA 3.0

Title: Preserving structures


问:
Let $f\in\mathcal{L}^1(\mathbb{X},\Sigma,\mu)$ Prove the transformation $A\to\int_Afd\mu$ for $A\in\Sigma$ is sigma additive.

Let $f\in\mathcal{L}^1(\mathbb{X},\Sigma,\mu)$ Prove the transformation $A\to\int_Afd\mu$ for $A\in\Sigma$ is sigma additive.

My try is the following

Defining the transformation  $\nu(A)=\int_A fd\mu$ and setting $f(A)_-=\max\{-\nu(A),0\},f(A)_+=\max\{\nu(A),0\}$.

$f(A)=f_+(A)-f(A)_-$ and $f(A)_-,f(A)_+$ are non-negative functions.

Assuming that $A=\biguplus_{i=1}^{\infty}A_i$ .

We know that $\int_A|f|=\int_A f_++\int_Af_-<\infty$ and therefore each integral is finite.

$\nu(A)=\nu(\biguplus A_i)=\int_{\biguplus(A_i)}f=\int_{\biguplus(A_i)}f_+-f_-\underset{(**)}{=}\int_{\biguplus(A_i)}f_+-\int_{\biguplus(A_i)}f_-\underset{(*)}{=}\sum_{i=1}^{\infty}\int_{A_i}f_+-\sum_{i=1}^{\infty}\int_{A_i}f_-=\sum_{i=1}^{\infty}\int_{A_i}f_+-\int_{A_i}f_-=\sum_{i=1}^{\infty}\int_{A_i}f_+-f_-=\sum\nu(A_i)$

(*) I've seen before the same statment for non-negative functions.

(**) is true because$\int_{A_i}f_+,\int_{A_i}f_-<\infty$

答:
The step (*) in your question is because of the monotone convergence theorem. The sequence of functions $1_{\cup_{i=1}^n A_i}f_+$ and $1_{\cup_{i=1}^n A_i}f_-$ are monotone and $\lim_{n \rightarrow \infty}1_{\cup_{i=1}^n A_i}f_+ =1_A f_+, $ so by the monotone convergence theorem, you can "take the limit outside of the integral".

\begin{align}
\nu(A)d\mu=\int_{A}f d\mu &=\int_{A}f_+ -f_- d\mu\underset{(**)}{=}\int_{A}f_+ d\mu -\int_{A}f_- d\mu =\int_{\Omega} 1_{A}f_+-\int_{\Omega}1_{A}f_- \\
&=\int_{\Omega} \lim_{n \rightarrow \infty} 1_{\cup_{i=1}^{n} A_i}f_+-\int_{\Omega} \lim_{n \rightarrow \infty} 1_{\cup_{i=1}^{n} A_i}f_- \\
&=\sum_{i=1}^{\infty} \int_{\Omega} 1_{A_i}f_+ \sum_{i=1}^{\infty} -\int_{\Omega}1_{A_i}f_-=...
\end{align}


问:
Find number of solutions for equation: $~x+y+z=n~$ where $~x,~y,~z~$ are non-negative whole numbers and $~x\le y\le z~$.

Find number of solutions for equation: $~x+y+z=n~$ where $~x,~y,~z~$ are non-negative whole numbers and $~x\le y\le z~$.

First I used substitution $~y=x+k,~ z=y+k~$ where $~k\ge 0~$(that is $y=x+k, z=x+2k$). Then after I plug that into equation I got, $~3x+3k=n~$. Then after using generating functions I get $~(1-t^3)^{-2}~$ or $~\sum\limits_{n\ge 0}{n+1\choose n} t^{3n}~$. Is this correct?

Body:
The number $S(n)$ of solutions are
$$S(n)=\frac{1}{72} \left(6 n^2+36 n+9 (-1)^n+16 \cos \left(\frac{2 \pi  n}{3}\right)+47\right)$$
and its generating function is
$$F(x)=\frac{1}{(1-x)^3 (x+1) \left(x^2+x+1\right)}=\frac{1}{(1-x) \left(1-x^2\right) \left(1-x^3\right)}$$
first values, starting from $n=0$,  are
$$0,1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, 24, 27, 30, 33, 37, 40,\ldots$$


问:
Zariski and Euclidean topologies
I've shown that every open set in the Zariski topology is open in the Euclidean topology, but I wonder why they are not equivalent. I'm searching for an open set in the Euclidean topology that is not open in Zariski.</p>

答: Hint:  Consider the affine line $\mathbb{R}$, and let $U = \mathbb{R}\setminus \mathbb{Z}$.

Title: Zariski and Euclidean topologies


问:
Is the shortest distance between a point and a curve always normal to the curve?
Suppose we have some point $P$ and some differentiable function $f$. Then let $P_C$ be a point on $f$ such that $|P-P_C|\le |P-f(t)|$ for any $t$. In other words, $P_C$ is one of the points closest to $P$ and lies on $f$. Intuitively, it seems to me that the line normal to $f$ at $P_C$ will always intersect $P$. However, I don't know of a proof for this. Is my intuition correct? Is there a proof of this statement? Is this true for all dimensions? If it's not true for all differentiable functions, is there a subset of functions for which it is true?

答:
Since there seems to be some misconceptions here, let me elaborate a bit on my comment. If $M$ is a $k$-dimensional submanifold (a subset that is locally smoothly equivalent to a ball in $\Bbb R^k$) of $\Bbb R^n$, then it has a $k$-dimensional tangent space at the point $p\in M$, often denoted $T_pM$. Orthogonal to this $k$-dimensional subspace is the $(n-k)$-dimensional space called the normal space of $M$ at $p$, often denoted $N_pM$. Given a point $Q\in\Bbb R^n$, if the point $p\in M$ is closest to (or, indeed, farthest from) $Q$, then essentially the calculation that @QuangHoang did will establish that the vector $Q-p$ is orthogonal to $T_pM$, i.e., is an element of the normal space $N_pM$. (You can think of this intuitively as an infinitesimal Pythagorean Theorem.)

The proof, as I suggested, is to parametrize a neighborhood $U$ of $p\in M$ by a ball $B$ in $\Bbb R^k$ by $\varphi\colon B\to U\subset\Bbb R^n$. Let's assume $\varphi(0)=p$.  Now consider the function
$$F\colon B\to\Bbb R, \quad F(s) = \|\varphi(s)-Q\|^2.$$
If $F$ has a (local) extreme point at $0$, then the chain rule tells us that for any vector $v\in\Bbb R^k$,
$$DF(0)v = 2 (\varphi(0)-Q)\cdot D\varphi(0)v = 0,$$
which says that $\varphi(0)-Q = p-Q$ is orthogonal to the image of $D\varphi(0)$. That image is the tangent space $T_pM$.


问:
How to do integral $\int\frac1{x^3+x+1}dx$?
I've been stuck by this integral. I only knew that roots of the cubic equation are complex.
Please help me with this. $$\int\frac{dx}{x^3+x+1}$$

答:
$x^3+x+1$ has one real and a pair of complex roots. The real root is $r= -0.6823$, given analytically by the Cardano’s formula. Decompose the integrand in terms of $r$
\begin{align}
\frac1{x^3+x+1}&= \frac1{(x-r)(x^2+rx-1/r)}
=\frac r{2r +3}\left(-\frac1{x-r}+\frac{x+2r}{x^2+rx-1/r}\right)
\end{align}

and then integrate to obtain
$$\int \frac{dx}{x^3+x+1}
=\frac r{2r +3}\left(-\ln\frac{x-r}{\sqrt{ x^2+rx-1/r}}
+\frac{ 3r \tan^{-1}\frac{2x+r}{\sqrt{1-3/r} } }{\sqrt{1-3/r}}\right)
$$
where $r^3=-r-1$ is recognized in above derivation.


问:
Isomorphism between two irreducible root systems is conformal
Let $R$ and $R'$ be irreducible root systems in the real inner product spaces $E$ and $E'$. Prove that $R$ and $R'$  are isomorphic iff there exists a scalar $\lambda \in \mathbb{R}$ and a vector space isomorphism $\varphi: $ $E \to E'$ such that $\varphi(R)=R'$ and $(\varphi(u),\varphi(v))=\lambda(u,v)\text{ for all }u,v \in E$.
(Introduction to Lie algebra Erdmann Karin- Mark Wildon, Exercise 11.15, page 124.)

I just can prove the "if" part and I get stuck with the "only if part". I highly appreciate who can give me some ideas.

Thank in advance

答:
Two root systems $R$ and $R’$ are said to be isomorphic if there is a vector space isomorphism $\phi : E \to E’$ such that:

(1) $\phi(R) = R’$

(2) for any two roots $\alpha$ and $\beta \in R$, $\langle \alpha, \beta \rangle = \langle \phi(\alpha), \phi(\beta) \rangle$.

Now, for the only if direction, you only need to prove that $\lambda(u,v) = (\phi(u), \phi(v))$.
Notice that if $u,v \in R$ this follows easily from condition (2), by definition of $\langle u,v \rangle$ and $\langle \phi(u), \phi(v) \rangle$.


问:
Set Theory without Powerset
I'm looking for references for a set theory which does not include the axiom of powerset but includes axioms allowing taking cartesian product. Please refer me to such if you know about any.</p>

答:
Elaborating on the comments:

First of all, even within the classical $\mathsf{ZFC}$-style context, powerset-free theories play an important role. The example I know most about is $\mathsf{KP}$, which basically consists of set theory without choice or powerset, and with separation and replacement dramatically restricted to "simple" formulas. An ordinal $\alpha$ such that $L_\alpha\models\mathsf{KP}$ is called admissible, and admissible ordinals occur not just in set theory but also in computability theory and proof theory. There is also "$\mathsf{ZFC}$ without powerset" (the details of which are more complicated than one might guess); this theory doesn't have the same significance outside set theory that $\mathsf{KP}$ does, but within set theory it too is quite important. Basically, one way we analyze a given "big" set theory is by looking at how "small" pieces of its models behave. Such truncated models are especially interesting when they satisfy a "strong-but-not-too-strong" theory, since then they occur frequently but still have some decent closure properties. Both $\mathsf{KP}$ and $\mathsf{ZFC}^-$ (not to be confused with $\mathsf{ZFC-}$, per the above-linked article) are such a happy-medium theories in the right context, but there are many others.

Outside the classical context there are additional examples but to my knowledge they don't enjoy nearly the same prominence as the $\mathsf{ZFC}$ fragments mentioned above. The SEP article on alternative set theories mentions two, pocket set theory (which Asaf Karagila brought up above) and an alternative due to Vopenka. Each has the property that there is a maximal cardinality. Note that this alone does not negate powerset - consider e.g. $\mathsf{NFU}$ or positive set theories, which in fact have a universal set.

Title: Set Theory without Powerset


答:
I interpret the question as follows: What does it mean for a map to preserve structure (and therefore to be called a morphism)?

Very roughly, there are two sorts of structures: Algebraic structure and topological structure.

A) Algebraic structure on a set is given by a set of operations of certain arities such that certain equations hold. A map preserves this algebraic structure if it commutes with the operations in the obvious sense. Examples are group homomorphisms between groups, ring homomorphisms between rings, and isometries between metric spaces (viewing the metric as a binary real-valued operation).

B) A topological structure is given by a set of "admissible" subsets satisfying certain properties (often closure properties). Examples are topological spaces, measurable spaces, partial orders (consider the down-sets) and filters. A map preserves this structure iff the preimage of an admissible subset is admissible. In the above examples we get continuous maps, measurable maps, increasing maps and filter maps.

We can also mix algebraic and topological structures. There are lots of interesting and important examples, but let me only mention three of them:

1) Topological groups and Lie groups. A morphism of topological (Lie) groups is a continuous (smooth) group homomorphism.

2) Schemes, or more generally ringed spaces. A morphism of ringed spaces $(X,\mathcal{O}_X) \to (Y,\mathcal{O}_Y)$ is a continuous map $f : X \to Y$ (topological part) together with a homomorphism of sheaves of rings $f^\# : \mathcal{O}_Y \to f_* \mathcal{O}_X$ (algebraic part).

3) C*-algebras, more generally Banach algebras. A morphism of C*-algebras is a continuous algebra homomorphism which also preserves the involution. Actually it turns out that we get continuity for free, and the norm is always $\leq 1$.

The most general notions of "structure" and "homomorphism" are developed and studied within model theory.

Title: Preserving structures

Body: Category theory abstracts the notion of the preservation of structure by means of morphisms. Is there a description of what it means to preserve structure of different types of mathematical structures (topological spaces, modules, rings, etc.)? </p>


问:
The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is
I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$

$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$

$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$

From here I am not able to process further to think of the way that will give me the solution.

答:
It's much easier if we restrict variables to be non-negative.

So we have $y=z-20, x=z+w-20, z\ge 0, w \ge 0$

$2x+y=99 \implies 2(z+w-20)+5(z-20)=99 \implies 7z+2w=239$

Then $z\equiv 1 \pmod 2 \implies z=2m+1, m \ge 0$

$2w\equiv 1 \pmod 7  \implies w\equiv 4 \pmod 7 \implies w=7n+4, n\ge 0$

Finally, $7(2m+1)+2(7n+4)=239 \implies m+n=16$

So there are 17 solutions.


问:
orthogonality exercise in Hilbert space
Let X a Hilbert space, S is a closed vector subspace of X. Suppose that $(x_n)$ is a Cauchy sequence in X. Prove the following statements:

The sequence $(P_S(x_n))$ is a Cauchy sequence in S.

$x_n - P_S(x_n) \in S^{\perp}$

答:
My answer is the following (I'm not sure if my answer is correct):

I have no answer.

If $x_n - P_S(x_n) \in S^{\perp}$, then it is true that:

$(x_n-P_S(x_n))P_S(x_n) \in S^{\perp}$

$(P_S(x_n)+P_S^{\perp}(x_n)-P_S(x_n))P_S(x_n) \in S^{\perp}$

$P_S^{\perp}(x_n)P_S(x_n)=0$

Then it is proved that $x_n - P_S(x_n) \in S^{\perp}$

Title: orthogonality exercise in Hilbert space


问:
Technique for finding cardinalites in combinatorics
In combinatorics I see the following technique used a lot:

Let A and B be sets.

We try to find a function $f:A→B$ such that all the fibers of the elements of $B$ under $f$ have the same cardinality $\kappa$.

If we manage to do that, we conclude that $|A|=\kappa|B|$.

Example:

Find the number of subsets of $\{1,2,3,4\}$ whose cardinality is $2$.

Let A be the set of ordered pairs that have two distinct elements from $\{1,2,3,4\}$.

Let B be the set of subsets of $\{1,2,3,4\}$ whose cardinality is $2$.

Let $f:A→B$, $f(〈x,y〉)=\{x,y\}$.

The fiber of each $\{x,y\}∈B$ under $f$ has exactly $2$ elements: $〈x,y〉$ and $〈y,x〉$.

From that, we conclude that $|A|=2|B|$, and we know that
$|A|=4⋅3=12$, so we get $|B|=6$.

My questions:

Is there a name for this technique? Is there a name for this type of function?

答: The method is called overcounting. I sometimes hear $f$ described as a $\kappa$-to-one function, in analogy with a bijection being one-to-one.


问:
Is there any such NO-answer example for Betweenness (TOP) problem?

Problem Statement

The input to a betweenness problem is a collection of ordered triples of items. The items listed in these triples should be placed into a total order, with the property that for each of the given triples, the middle item in the triple appears in the output somewhere between the other two items. The items of each triple are not required to be consecutive in the output

There are two versions of the problem defined as follows

The Decision version of the betweenness problem in which an algorithm must decide whether or not there exists a valid solution?

The Normal (Hard) version of the betweenness problem in which an algorithm must provide a total ordering which satisfies the conditions or says NO when there no solution can be found.

in the original published paper, where Opatrný (in 1979) showed that its decision problem is NP-Complete, it has been defined precisely as a Total ordering problem

definition. The Total Ordering Problem (TOP) is the following: given a finite set
S and a set of ordered triples R $\subseteq$ SxSxS determine whether there exists a total
ordering of S such that for any element (a,, b, c) in R either a < b, b < c or c < b, b < a
(we say that such an ordering satisfies R). A solution for TOP would be an algorithm
which takes an instance (S, R) of TOP and outputs true if and only if there exists a total
ordering of S satisfying R.

Question

Is there a proposed nontrivial example (for intuition) in which there will be NO answer for the betweenness decision problem?

means there is no total ordering of $S$ exists that satisfying R

My works

I came up with this example, but the bi-orientation nature of answers makes it possible to find the answer.

$(a,b,c) \in R$ and $(X, c, b) \in R$ for $X \ne a$
has solution of abcX

I also guess, if a circle appears in the representation graph of R simulation there will be no total ordering, but I'm not sure I'm right or how to prove it.
by graph representation of R, I mean a->b->c if $(a,b,c) \in R$ for all $(ai,bi,ci) \in R$

updated

As @amrsa suggested in the comments there are possibilities for trivial approval or disapproval of the statement. followings are roughly explained

if $R = \{(x,y,z), (a,b,c)\}$ then there is a total ordering of e.g. xyzabc.

if $R=\{(a,b,c), (b,c,a)\}$ then there will not be any ordering.

the second statement may not be the only condition for the nonexistence of total order, cause if it was, we can remove all of them in polynomial time and do the topological sort and the exact answer (I guess).

wikipedia

答:
Reichenbach, H. pointed out that R is not totally orderable when it
includes the ordered triples

$(A_1, A_2, A_3), (A_1, A_2, A_4), (A_4, A_2, A_3)$

for each clause, if you fix its orientation you face a contradiction.

for example.
let fix reverse order for first clause meaning $(A_3, A_2, A_1)$
so the second clause must rotate to $(A_4, A_2, A_1)$ to be satisfiable beside first clause, till now we have $A_3 < A_2$ and $A_4 < A_2$ but the third caluse says $A_4 < A_2 <  A_3$ or $A_3 < A_2 < A_4$ which both states $\bot$ means our first guess was wrong for every other it reachse $\bot$ the same way.

Actually, if we assume any clauses in R to be fixed orientated (meaning we know they will not be rotated), the representation graph must be a DAG.

Reichenbach, H. (1956). The direction of time. Berkeley, Los Angeles: University of California Press.


问:
Differential forms and determinants
2-forms are defined as
$du^{j} \wedge du^{k}(v,w) = v^{j}w^{k}-v^{k}w^{j} = \begin{vmatrix} du^{j}(v) & du^{j}(w)  \\ du^{k}(v) & du^{k}(w) \end{vmatrix}$

But what if I have two concret 1-forms in $R^{3}$? For example $(2dx-3dy+dz)\wedge (dx+2dy-dz)$ which gives

$(2dx-3dy+dz)\wedge (dx+2dy-dz)=-7dy \wedge dx +3dz \wedge dx - dy \wedge dz= 7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz$ I know this is the same as the vector product between $(2,-3,1)^{T}$ and $(1,2,-1)^{T}$.

What is the relationship with the determinant? Because when I calculate the wedge product between two 1-forms in $R^{2}$ then I get the value of the determinant $(2dx+4dx)\wedge (3dx+9dy) = -18 dx\wedge dy +12 dx \wedge dy = 6 dx \wedge dy$, but for 1-forms in $R^{3}$ I get the vector product. And is  the interpretation right that parts of the area spanned by the 2-form above (the vector product) is that 7 get's projected onto the $xy$ plane, 3 onto $zx$ and  1 onto $yz$? Or is it in this case another coordinate system with planes $dxdy$, $dzdx$ and $dydz$? And since all the differential forms are functions of vectors, what happens when they come into the picture? Because doesn't $7 dx \wedge dy + 3 dz \wedge dx + dy \wedge dz$ looks like this?

$7 \begin{vmatrix} dx(v) & dx(w)  \\ dy(v) & dy(w) \end{vmatrix} + 3 \begin{vmatrix} dz(v) & dz(w)  \\ dx(v) & dx(w) \end{vmatrix} + \begin{vmatrix} dy(v) & dy(w)  \\ dz(v) & dz(w) \end{vmatrix}$
For some vectors $v,w \in R^{3}$?

答:
This is explained very nicely in Hubbard and Hubbard's book Vector Calculus, Linear Algebra, and Differential Forms, page 560 (Example 6.1.2) in the fourth edition.

Let's focus on a particular example: how the 2-form $dx\wedge dy$ on $\mathbb{R}^3$ is related to determinants. Well, given two three-dimensional vectors $\mathbf{u}=[u_1, u_2, u_3]^t$ and $\mathbf{v}=[v_1, v_2, v_3]^t$, we have

$$dx\wedge dy(\mathbf{u},\mathbf{v})=dx(\mathbf{u})dy(\mathbf{v})-dy(\mathbf{u})dx(\mathbf{v})=u_1v_2-u_2v_1=\det\begin{pmatrix}u_1 & v_1\\u_2 & v_2\end{pmatrix}$$

where the final matrix is the matrix we get by stacking $\mathbf{u}$ and $\mathbf{v}$ into a $3\times 2$ matrix and then selecting the rows corresponding to the $x$ and $y$ coordinates. We get a square matrix we can take the determinant of by selecting the appropriate rows.

More generally, consider a $k$-form $d_{i_1}\wedge\cdots\wedge d_{i_k}$ on $\mathbb{R}^n$. The value of this form on a $k$-tuple of $n$-dimensional vectors can be found first by stacking the $k$ vectors into an $n\times k$ matrix, then selecting the $k$ rows corresponding to the indices $i_1, \ldots, i_k$, and finally taking the determinant of the resulting $k\times k$ matrix.

Title: Differential forms and determinants


问:
Find angle $x$ in the isosceles $\triangle ABC$.
Find angle $x$ in the isosceles $\triangle ABC$.

Can anyone help me with this problem?

I was stuck. I made my findings in this diagram but I didn't know how can I spot the angle that's needed.

Can anyone solve it?

答:
Here is a solution using trigonometry.

Using law of sines,

In $\triangle CDF, \displaystyle \frac{CF}{\sin (130^0-x)} = \frac{DF}{\sin20^0} \,$ (as $\angle BDC = 130^0$)

In $\triangle BDF, \displaystyle \frac{BF}{\sin x} = \frac{DF}{\sin10^0}$

As $BF = CF$, we have

$\displaystyle \frac{\sin (130^0-x)}{\sin20^0} = \frac{\sin x}{\sin10^0}$

$\displaystyle \frac{\sin (50^0+x)}{\sin20^0} = \frac{\sin x}{\sin10^0}$

$\displaystyle \frac{1}{2}\sin (50^0+x)= \sin x \cos10^0$

$\displaystyle \sin 30^0 \sin (50^0+x)= \sin x \sin80^0$

Now using the identity $2 \sin A \sin B = \cos (A-B) - \cos (A+B)$, we have

$\cos (20^0 + x) - \cos (80^0 + x) = \cos (80^0 -x) - \cos (80^0 + x)$

$\implies 2x = 60^0, x = 30^0$


问:
is there exist complete language in R department?
I tried to proof if there is exist a complete language in R department. My main idea was to show that there is exist complete language in RE department and conclude from that that there is exist a complete language in R because R⊆RE.

I got stucked in this stage, hope someone can help me out.

Full definition for complete lang in R deparment:
For given lang department R , lets say language L1 is complete in R if L1⊆R and for each L2∈R , L2≤L1

答:
Completeness only makes sense with respect to some reducibility. For example, the phrase "$\mathsf{NP}$-complete" refers to polynomial-time many-one reductions, whereas when we say that the Halting Problem is a complete r.e. set the relevant notion is arbitrary-time many-one reductions. And there are even reducibility notions which aren't of "many-one flavor" at all - e.g. Turing reducibility, which while less relevant in complexity theory is the main reducibility notion in computability theory.

That said, your question does have a strong negative answer. According to every reducibility notion I know of, either every set in $\mathsf{REC}$ is complete for $\mathsf{REC}$ with respect to that reducibility or there is no complete set in $\mathsf{REC}$ for that reducibility. In particular, Turing reducibility (trivially) has the first behavior, and every resource-bounded reducibility (e.g. polytime $m$-reducibility) has the second (because we can always "pad out" a given language in a computable way to make it more complicated with respect to a resource-bounded reducibility notion).

Note that this is in contrast with the class $\mathsf{RE}$. While the same argument as for $\mathsf{REC}$ shows that there is no $\mathsf{RE}$-complete language with respect to any resource-bounded reducibility, the situation is nontrivial for (resource-unbounded) many-one reducibility: the Halting Problem is $\mathsf{RE}$-complete, but the degree structure of r.e. sets modulo many-one reducibility (or even the coarser notion of Turing reducibility) is extremely rich. In general, $\mathsf{RE}$ is much better behaved than $\mathsf{REC}$, and this is one of the key foundational observations of computability theory; for example, we can enumerate the $\mathsf{RE}$ languages in an $\mathsf{RE}$ way but we can't enumerate the $\mathsf{REC}$ languages in a $\mathsf{REC}$ (or even $\mathsf{RE}$) way.


问:
How to derive the sum of an arithmetic sequence?
I'm attempting to derive a formula for the sum of all elements of an arithmetic series, given the first term, the limiting term (the number that no number in the sequence is higher than), and the difference between each term; however, I am unable to find one that works. Here is what I have so far:

Let $a_0$ be the first term, $a_n$ be the last term, and $x$ be the difference between each term.

If we have the sequence $a_0, a_0 + x, a_0 + 2x ... a_n$, then we can add the first and last term, the second and second last term, etc., to quickly find the sum based on the number of terms. Thus, the sum of these terms is $n\frac{(a_n + a_0)}{2}$, where $n$ is the total number of terms.

The number of terms must be the number of times the first term was increased by $x$ plus one (to account for the first term), and so $n = \frac{(a_n - a_0)}{x} + 1$.

Thus, the sum is equal to $(\frac{(a_n - a_0)}{x} + 1)\frac{(a_n + a_0)}{2}$.

However, I am unable to integrate the limiting term in the place of $a_n$; any ideas for how to make this work?

In case my definition of a limiting term is ambiguous; an example would be if there was a set $3, 6, 9$; I'd like to be able to replace $a_n$ (which is $9$ in this case) with any number above $9$, and below $12$, and still get the same answer.

答: Essentially, you want to "round" $a_n$ down to the greatest number which can be expressed as $a_0 + kx$ for some integer $x$. One way to do so is to replace $a_n$ by $(a_0+x\lfloor\frac{a_n-a_0}{x}\rfloor)$ in your summation formula.</p>


问:
Local Homeomorphism with constant fiber is a covering map
Let $f:X\to Y$ be a local homeomorphism between Hausdorff spaces such that $f$ is surjective and $f^{-1}(y)$ is compact $\forall y\in Y$. If $|f^{-1}(y_1)|=|f^{-1}(y_2)|,\forall y_1,y_2\in Y$ then show that $f$ is a covering map.

Attempt: I know $f^{-1}(y)$ is finite as it has a finite cover $\cup_{i=1}^nU_i\supseteq f^{-1}(y)$ with each $U_i$ open such that $f_{U_i}$ is a homeomorphism. I considered $V_i\subseteq U_i$ with $V_i\cap V_j=\emptyset$ but the problem is the preimage of $\cap_{i=1}^nf(V_i)$ is not necessarily in $\cup_{i=1}^nU_i$ otherwise I would have found an evenly covered neighborhood of $y$. I need to show that $f$ is closed so I can use the set $(\cap_{i=1}^nf(V_i))\backslash (f(X\backslash \cup_{i=1}^nU_i))$ instead or consider another approach by using the constant preimage assumption which I do not see how it fits there...

答: It is much simpler than I thought it seems: the sets $V_i$ are the disjoint restrictions of the $U_i$ as defined in my question above. Suppose that $|f^{-1}(y)|=n,\forall y\in Y$. The problem is that taking the set $V=\cup_{i=1}^nf(V_i)$, I want to show that $f^{-1}(V)\subseteq \cup_{i=1}^nV_i$ so I can use the local 1-1 property and conclude that the preimages are exactly the open sets $f^{-1}(V)\cap U_i$. This is straghtforward however: if $y_0\in f^{-1}(V)$ then there exist $v_1\in V_1,...,v_n\in V_n: f(v_i)=y_0$ as we simply have $v_i=f_{V_i}^{-1}(y_0)$. But $|f^{-1}(y_0)|=n \implies f^{-1}(y_0)=\{v_1,...,v_n\}$ and thus $f^{-1}(y_0)\subseteq \cup_{i=1}^nV_i, \forall y_0\in Y$. The result now follows immediately.


问:
equivalent properties of cardinals

Let $A\neq \emptyset$ be a set. Show that the following are equivalent:

$|\mathbb{N}| = \aleph_0\leq |A|$

$|A| = n|A|$ for any $n\geq 2\in \mathbb{N}$

$|A| = \aleph_0 |A|$

I want to show $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1).$

$(1)\Rightarrow (2)$. Suppose $(1)$ holds. Then there is an injection $f : \mathbb{N}\to A.$ It suffices to show that $|A| = |A| + |A| = |A\times 2| = |A\times \{0\}\cup A\times \{1\}|$, where $2 = \{0,1\},$ since an inductive argument can be applied to give the required result. So we need to define a bijection $f : A \to A\times 2.$ Let $\mathcal{F} := \{(X,f) : X\subseteq A, f : X\to X\times 2\text{ is a bijection}\}.$ Observe that $\mathcal{F}$ is nonempty because $A$ contains a denumerable subset $X$, and thus there exists a bijection $f : X\to X\times 2.$ Order the elements of $\mathcal{F}$ by extension. That is, let $(X_1, f_1) \leq (X_2, f_2)$ if $X_1\subseteq X_2$ and $f_2\vert_{X_1} = f_1$. Let $C = \{(X_i, f_i)\}_{i\in I}$  be a chain in $\mathcal{F}.$ Let $X = \cup_{i\in I} X_i$ and $f : X\to X\times 2$ be defined so that $f(x) = f_i(x),$ where $i\in I$ is so that $x\in X_i.$ Then note that $(X,f)$ is an upper bound for $C,$ and is in $\mathcal{F}.$ So by Zorn's Lemma, $\mathcal{F}$ has a maximal element, say $(Y, g).$ Suppose $|Y| < |A|.$ Then $|A\backslash Y| = |A| - |Y| = |A|.$ So $A\backslash Y$ has a denumerable subset $Z.$ Since $Z$ is denumerable, there is a bijection $h : Z\to Z\times 2.$ Then let $i : Y\cup Z\to (Y\cup Z)\times 2, i(w) = \begin{cases}g(w),&\text{ if $w \in Y$}\\
h(w),&\text{ otherwise }\end{cases}$
Then $(Y,g) < (Y\cup Z, i),$ a contradiction. So $|Y| = |A|.$ Hence $|A| = |Y| = |Y\times 2| = |Y| + |Y| = |A| + |A|,$ as required.

$(2)\Rightarrow (3).$ One can find a bijection between $\{1,\cdots, n\}\times A$ and $\{1,\cdots, n+1\}\times A$ for any $n\geq 1$, since $n|A| = (n+1)|A|$ (indeed $n|A| = |A|$ so $(n+1)|A| = n|A| + |A| = |A| + |A| = |A|$). Hence, one can find a bijection between $\{1\}\times A$ and $\mathbb{N}\times A,$ which implies $|A| = |\{1\}\times A| = |\mathbb{N}\times A| = \aleph_0 |A|.$

$(3)\Rightarrow (1).$ If $|A| = \aleph_0 |A|,$ then there is a bijection $f : A \to \mathbb{N}\times A.$ Fix $a \in A$ and let $h : \mathbb{N}\to A, h(b) = f^{-1} (b,a).$ Observe that $h$ is injective because if $b\neq d,$ then since $f^{-1}$ is injective, $h(b) = f^{-1}(b,a) \neq f^{-1}(d, a) = h(d).$ Thus, $\aleph_0 \leq |A|.$

Is this incorrect? Especially $(2)\Rightarrow (3)$ and $(3)\Rightarrow (1).$ Also, WHY is $f$ well-defined in $(1)$? If I choose $i\in I$ randomly, isn't it possible that $x\neq y $ but $x\in X_i$ and $y \in X_j, i\neq j$ and $f_i(x) = f_j(y)$?

答:
(1)$\implies$(2): The idea is very nice that we could in theory stop at any subset of cardinality $|A|$, however, I think, when you stated $|Y|<|A|\implies |A\setminus Y|=|A|$, you implicitly used the statement to be proved: $|Y|+|Y|=|Y|$
for infinite $|Y|$.

Instead, I think, we should prove that $Y=A$: suppose $f:Y\to Y\times 2$ is a bijection and $x\in A\setminus Y$, then we can shift one of the columns in a denumerable subset $Y_1$ of $Y$ in order to obtain a bijection $\,Y\cup\{x\}\,\longrightarrow \,(Y\cup\{x\})\times 2$.

(2)$\implies$(3): It's not correct.

By $2|A|=|A|$, we can cut $A$ into half: there are disjoint $A_0$ and $B_0$ such that
$|A_0|=|B_0|=|A|$ and $A=A_0\cup B_0$. This can be done with $B_0$ again, yielding $A_1$ and $B_1$, and then so on, cut $B_i$ to $A_{i+1},\ B_{i+1}$.

Fix bijections $h_n:A\to A_n$, then
map $(n,a)\in\Bbb N\times A$ to $h_n(a)$. This is a bijection to $\bigcup_nA_n\subseteq A$.

(3)$\implies$(1): It's correct.

Title: equivalent properties of cardinals


答:
First point

The projection $P_S$ onto $S$ is a contracting map. Therefore
$$\Vert P_S(x) - P_S(y) \Vert \le \Vert x-y\Vert$$ for all $x,y \in X$. This implies that $\{P_S(x_n)\}$ is a Cauchy sequence if $\{x_n\}$ is Cauchy.

Second point

By definition of a projection, we have $\langle P_S(x) , y \rangle = \langle x , P_S(y) \rangle$ for all $x,y \in X$.

As for $y \in S$, $P_S(y) =y$, we get that for all $y \in S$:

$$\langle P_S(x_n) - x_n, y \rangle = 0$$

which proves that $x_n - P_S(x_n) \in S^{\perp}$.

Title: orthogonality exercise in Hilbert space

Body:
Let X a Hilbert space, S is a closed vector subspace of X. Suppose that $(x_n)$ is a Cauchy sequence in X. Prove the following statements:

The sequence $(P_S(x_n))$ is a Cauchy sequence in S.

$x_n - P_S(x_n) \in S^{\perp}$


问:
$y=\sin^2x+\sin x+1$. Find the range of $y$
The range of $\sin x$ is $[-1,1]$ for $x \in \mathbb{R}$.

答:
If you use calculus, then it won't be much time consuming as you just need to differentiate and apply Maxima-minima rule.
But if you're not allowed to use calculus, then the best method I would suggest is the COMPLETING SQUARE METHOD .
See, you'll have to convert the trigonometric terms into a perfect square somehow.
Here, the equation may be converted to-

Sin²x+Sinx+1
= Sin²x+2(1/2)(sinx)+(1/2)²-(1/2)²+1
=(Sinx+1/2)²+(3/4)

Now we have only one parameter i.e (sinx+1/2)
So, for maximum, put the maximum value of (sinx+1/2) there, which is 3/2 (at X=π/2 radians)

So the maximum value of the expression is :

(3/2)²+3/4=12/4

Similarly, for minimum value, put the minimum value of |sinx+1/2| , which is equal to zero.

Here I used the modulus as we will have to square the term when we will put it in y....... Thus, the negative value will also become positive, but we want the minimum value, so we will put the minimum absolute value there (if an even power is associated in the expression).
Thus the minimum value becomes:
(0)²+3/4=3/4
Thus the range of y is [3/4,3].

Now you see, this method won't help you always but I can assure that this works most of the time.
So it's your liberty to use the calculus method (which works always) or this method.


问:
Does adding polynomial terms in linear regression models not violate multicollinearity assumptions?
Linear regression has couple of assumptions. One of them is independence of feature vectors or they should not be correlated. If we add a polynomial features to feature vector, does it not create correlated feature?

Also, could anyone share good data set link for performing Regression splines/smoothing splines/piecewise polynomial regression?

答:
Usually, you don't assume independence of features, and not even that they are pairwise uncorrelated, rather that they are not too much correlated to the extent that your model is unstable.

Adding powers of some features may or may not cause problems. That is, $x^2$ and $x$ are not linearly dependent (i.e., correlated), hence no violation and no danger of perfect multicollinearity. However, clearly it does not imply zero correlation as well,i.e., $cov (X, X^2)$ maybe non zero. Namely,  you may add noise (variance) to your model and harm the stability of the regression coefficients if the $x^2$ does not help to "explain" the conditional variablity of $Y$.


答:
We identify the key angle $\angle AFB = 100^{\circ}$.

Construct an equilateral $\triangle HBF$ as shown. Join $HA$.

$\angle HFB = 60 \Rightarrow \angle HFA = 40 = \angle HBA \Rightarrow HBFA$ is cyclic.

Then $\angle HAB = \angle HFB = 60 \Rightarrow H$ lies on line $AC$.

Now $\angle HGB = \angle AGB = 50 = \angle HBG \Rightarrow HG=HB=HF$

Thus $H$ is center of circle through $B,F,G$.

Hence $\angle BGF = \frac{1}{2}\angle BHF=30^{\circ}$.

Body:
Find angle $x$ in the isosceles $\triangle ABC$.

Can anyone help me with this problem?

I was stuck. I made my findings in this diagram but I didn't know how can I spot the angle that's needed.

Can anyone solve it?


问:
In the area of sphere, which angle with $0\leq \alpha<2\pi$ and which angle with $0\leq \beta < \pi$?
I am lacking the skill of visualizing the problem, picture here, to decide the right intervals. The way I do it currently is to try things but the technique fails with anything more complicated to 2D. So how do you know which angle correspond to which angle range?

[update]

Let's break this into parts. Suppose there are two nodes, one $N$ in the north pole and one $S$ in the south pole. The distance $N-S$ is $\pi$. The distance $N-N$ is either $0$ or $2\pi$. Now, please, break this problem with nodes and graph-theoretically. See the paths are clearly different, the ending points vary: one is even and one is odd (trivially even if $N-N=0$ length accepted).

答: If you extend it to a globe, $d\theta$ represents the interval between two meridians of longitude, so should go from $0$ to $2\pi$.  $d\phi$ represents the interval between two parallels of latitude, so should go over a range of $\pi$.  On the globe, it is from $\frac{-\pi}{2}$ to $\frac{\pi}{2}$, but usually we measure from the pole and range over $0$ to $\pi$</p>


问:
Factorized Conditional Expectation w.r.t. a smaller domain
Let $X$ and $Y$ be real-valued random variables with joint density denoted by $f:=f^{(X,Y)}$ for sake of brevity.
Prove that
$$
E[X \mid \{Y=y\}] := \frac{1}{f^Y\!(y)} \int_\mathbb{R} x \cdot f(x,y) \, \lambda(dx)
$$
for all $y\in\mathbb{R}$ with $f^Y\!(y)>0$ is a factorized conditional $E[ X \mid Y = \cdot \,]$ with domain $\{f^Y > 0\}$.

I cannot understand the concept of Factorized Conditional Expectation. Here, the exercise asks to show that via the given formula a factorized conditional expectation is defined, not on the entire real numbers but on the smaller domain $\{f^Y>0\}$, where the formula actually makes sense.

How can I do that?

答:
I guess that by "factorized conditional expectation" of $X$ given $Y$ you mean a Borel measurable function $h:\Bbb R\to\Bbb R$ such that $h(Y)$ is integrable and  a version of $E[X\mid\sigma\{Y\}]$. The latter means that
$$
E[h(Y)\cdot 1_B(Y)] = E[X\cdot 1_B(Y)]\qquad(\dagger)
$$
for each Borel subset $B$ of $\Bbb R$. One is  then is inclined to interpret $h(y)$ as $E[X\mid Y=y]$.

Assuming $X$ is integrable, the numerator in your expression is well defined for $\lambda$-a.e. $y$; more precisely, if $y\in G:=\{y:\int_{\Bbb R} |x|f(x,y)\lambda(dx)<\infty\}$. The ratio itself is well defined if, in addition, $f_Y(y)>0$. Define $H:=G\cap\{f_Y>0\}$, noting that $P[Y\in H]=1$, and then
$$
h(y):=\cases{{\int_{\Bbb R} x\cdot f(x,y)\lambda(dx)\over f_Y(y)},& $y\in H$;\cr
0,& $y\notin H$.\cr}
$$
Now computing as did @Furrer, it's straightforward to check that ($\dagger$) holds with this choice of $h$.


问:
Fake proof regarding continuity.

So to my understanding the fallacy in this argument is that it's circular. Since he didn't declare an ϵ at the beginning of the proof, his choice of ϵ is dependant on δ, which in turn is dependant on ϵ. We were taught to always begin our proofs with "Let ϵ>0 ... ", or "We choose ϵ=[blank/fill in later]>0 ... " and the fact that he doesn't raised my suspicion. But I can't TRULY point out his mistake, and my thought regarding the fallacy is purely intuitive, and the fact that everything else looks fine really makes me wonder if there is something that I am missing..

答:
The catch is that when it states that "there exists an $ \epsilon_0 > 0 $", this $ \epsilon_0 $ depends on $ x_1 $, it is $ \epsilon_0(x_1) $. But $ x_1 $ was chosen based on $ \delta_1 $ which in turn was defined based on the choice of $ \epsilon_0 $. So the first $ \epsilon_0 $ and the second are not the same positive number!

The second $ \epsilon_0 $ is equal to $ \epsilon_0 = \epsilon_0 (x_1(\delta_1(\epsilon_0))) $

Title: Fake proof regarding continuity.


问:
$\alpha$-convexity of a function - formal definition and examples (related to a particular homework)
I'm trying to solve an exercise that asks if a function $f: \mathbb{R}^2 \to \mathbb{R}$ is $\alpha$-convex. The function in question is defined by: $f(x,y):=x^2 + 2x + xy + y^2 - 4y + 3.$

I know the for convexity, we need to calculate the Hessian of $f$ and see if it's non-negative definite (and positive-definite for strict convexity), but upon an internet search, I could locate anything useful for when a function $f$ is $\alpha$-convex. So I'd appreciate a formal definition and the solution to the exercise above.

P.S. My guess would be if a function $f$ is $\alpha$-convex, $\alpha > 0,$ then $Hess(f) - \alpha I$ is non-negative definite, but I'm not sure!

答:
Yes. A function $f$ is $\alpha$-convex, if and only if $\nabla^2f(x) - \alpha I \succeq 0 \ \forall x$.

We can easily deduce that this $\alpha$ correspond to the minimum possible eigen value of the Hessian.

Hint: $\nabla^2f(x) - \alpha I \succeq 0 \ \implies y^T (\nabla^2f(x) - \alpha I) y = y^T\nabla^2f(x)y - \alpha \geq0$. Can you find the $y$ for which we get the minimum value?

In your case,
$$
\lambda_{min} = 1
$$

Hence, the function $f$ is $1$-convex.


问:
Evenly distributing $n^2$ ball between $n$ pockets by chosing $2$ of them at a time.
We have $n^2$ balls distributed among $n$ pockets (some of them may be empty). We can choose any two pockets of which sum of balls is even and then distribute the balls equally among these two pockets. We can repeat it as many times as we wish. For every natural, even number $n$ such that $2 ≤ n ≤ 10$ decide if by using this method we can always distribute the balls so that at the end every bag contains n balls, no matter how many balls there are in each pocket at the beginning.

I think and I am informed it is true that we can do it for every n number of pockets which is a power of two. I suppose, we split them in two until we are left with $n/2$ pairs of pockets. Then we make a move on each pocket and make our way out merging the subsets of pairs. For $n=4$ we have two pairs which we first want to even out and then make the move on the first elements from the each pair an then on the second elements from each pair. However the number of ball combined in the selected pockets has to be even. That feels to be the case as if $n$ is a power of two, $n^2$ is even so we have an even number of odd numbers of balls. How would a proper proof of that look? How to prove we cannot do it if $n$ is not a power of two?

答:
Let's do it for n=6 and n=10

For n=6 If you take it as $({7,7,7,7,4,4})$ or ${5,5,5,5,8,8}$ or ${3,3,3,3,12,12}$ or ${1,1,1,1,16,16}$ or ${9,9,9,9,0,0}$. Here the sum is 36 but you can't make it into ${6,6,6,6,6,6}$.

For n=10 If you take it as ${1,1,1,1,1,1,1,1,46,46}$ or ${9,9,9,9,9,9,9,9,14,14}$ or something of the form {o,o,o,o,o,o,o,o,e,e} where o is odd and e is even

So basically doesn't work for n=6 or n=10 or any n of the form 4k+2 for k>0.

It does work for $n=2^k$

It doesn't work for odd numbers also.

Am I making sense?


问:
Simplifying an algorithm to evaluate $\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n(x_i\&x_j)(x_j|x_k)$ for non-negative integers
Let $x_1, x_2, \ldots, x_n$ sequence of non-negative integers. I need to calculate $$\sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}(x_i \& x_j)(x_j | x_k)$$
Here "$\&$" stands for bitwise AND, and "$|$" stands for bitwise OR

I know how to write an algorithm that runs ~ $O(n^3)$. How can you get rid of the triple amount?

答:
I recommend to take a look at these two

https://codeforces.com/contest/1466/problem/E

https://codeforces.com/blog/entry/86126/


问:
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$?
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.

答: Since $\sin x$ is concave for $x\in[0,\,\pi/2]$, if $x\in[0,\,1]$ then $\sin x\ge x\sin1\implies1-\cos x\ge x^2\tfrac12\sin1$. The $x^2$ coefficient approximates $0.42$. The generalization $x\in[-1,\,1]$ follows from $1-\cos x,\,cx^2$ being even.


问:
Proper notation of average with a subscript
When notating a mean, does the bar extend over a subscript as well? Which of these is correct?

$$\bar S_T$$

or

$$\overline {S_T}$$

答: After going through some textbooks, I've consistently found that the first notation $\bar S_T$ is the correct one to use.


问:
What should actually be demonstrated in this exercise about left and right derivatives?
I am redoing my math education after 35 years in order to be up to date with my children's curriculum (I usually have a nanosecond to give an answer because of magical dad knowledge). My older son is in high school and they are going through derivatives.

One of the exercises I was preparing for him (this is beyond the homework, he is in an "advanced" track and I want to show him some more interesting problems) I stumbled upon the following problem. It is a translation from French so feel free to edit in order to improve the wording

We define

$$
\begin{align}
f: \mathbb{R} & \rightarrow \mathbb{R} \\
x & \mapsto \begin{cases}
      \text{$x^2 - 1$ if $x < 0$} \\
      \text{$x^2 + 1$ if $x \geq 0$}
    \end{cases}
\end{align}
$$

Demonstrate that, for $a=0$, $f$ is derivable on the right but not on the left.

Both $x^2 - 1$ and $x^2 + 1$ are derivable in $\mathbb{R}$ but $f$ is defined for $0$ as $x^2 + 1$. So technically the derivative in $0$ will be $2 \times 0 = 0$

It is however derivable on $0^+$ ($2x$) and on $0^-$ ($2x$).

The only idea I have is that by taking the definition of the left derivative I have
$$ \lim_{h \rightarrow 0^-} \frac{f(x+h) - f(x)}{h} $$
which I think is not defined because the limit approaching on the left does not exist (as the function is undefined there). But I think I must be missing something because this is too obvious.

答:
You are on the right track, but as I note in a comment, you are somewhat oversimplifying.

We do want to use the definitions of left/right derivative, so let's do that.

$$\lim\limits_{h\to0^-}\frac{f(x+h)-f(x)}h=\lim\limits_{h\to0^-}\frac{(0+h)^2-1-(0^2+1)}h=\lim\limits_{h\to0^-}\frac{h^2-2}h=\infty$$

$$\lim\limits_{h\to0^+}\frac{f(x+h)-f(x)}h=\lim\limits_{h\to0^+}\frac{(0+h)^2+1-(0^2+1)}h=\lim\limits_{h\to0^+}h=0$$

There is no such thing as an infinite derivative, so this shows that only the right derivative is well defined at $0$.


问:
If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$
Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded.
Show that $f'(x)\to0$ as $x\to\infty$.
(This is inspired by a comment/answer to a different question)</p>

答:
Let $|f''|\le 2M$ on $\mathbb R$ for some $M>0$. By Taylor's expansion, for every $x\in\mathbb R$ and every $\delta>0$, there exists $y\in[x,x+\delta]$, such that
$$f(x+\delta)=f(x)+f'(x)\delta+\frac{1}{2}f''(y)\delta^2.$$
It follows that
$$|f(x+\delta)-f(x)-f'(x)\delta|\le M\delta^2.\tag{1}$$
Since $\lim_{x\to\infty}f(x)=0$, fixing $\delta>0$ and letting $x\to\infty$ in $(1)$, we have
$$\limsup_{x\to\infty}|f'(x)|\le M\delta.$$
Since $\delta>0$ is arbitrary, the conclusion follows.</p>


答:
We work on $[0,1]$.

Let's note $f:x\mapsto 1-\cos x - \frac{x^2}{3}$.

Then $f':x\mapsto \sin x - \frac{2}{3}x$ and $f'':x\mapsto \cos x - \frac{2}{3}$.

$f''(x)\geqslant 0 \Leftrightarrow \cos x \geqslant 2/3 \Leftrightarrow \cos x \leqslant a$ with $a=\text{arccos} 2/3$ which is around $0.84$ rad.

So $f'$ is increasing on $[0,a]$ and decreasing on $[a,1]$. And we have $f'(0)=0$ and $f'(1)>0$ so $f'>0$.

So $f$ is increasing, and since $f(0) = 0$, the function is always positive. This gives the answer.

Body: How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.


问:
Prove that polynomial $1+x+x^2$ of $\mathbb{R}[x]$ is irreducible over $\mathbb{R}$. Do not use the Eisenstein criterion.
Prove that $1+x+x^2$ is irreducible over $\mathbb{R}[x]$. Do not use the Eisenstein criterion.

I don’t know where to start. Could someone help me?

答:
For all $x\in\mathbb R$, $x^2+x+1=\left(x+\frac12\right)^2+\frac34>0$, so $x^2+x+1$ has no roots in $\mathbb R$.

A quadratic polynomial with no roots is irreducible.


问:
Indicator functions and vector spaces
The following is an exercise I have encountered in my linear algebra exercise book (I will not provide the reference as it is an Italian one). The solution was attached to it; still, as it is the first time I encounter the indicator function, I am unsure about its understanding.

Consider the vector space $ℝ^{[0,1]}$ and its subset $S=\{1_{[0,2^{-n}]} : n \in \mathbb N_0\}$, where $1_A$ is the indicator function of $A$.
Let $W = \operatorname{span}(S)$ and consider the function $g := 1_{(\frac12, 1]}$.

Question: Does $g$ belong to $W$?

Solution: Yes, indeed $g$ can be rewritten as $$g = 1_{[0,1]} - 1_{[0,\frac12]}$$

My reasoning: Since we have to rewrite $g$ as an element of $W$, the lower end of the interval where the indicator function is defined must be equal to $0$. To do so, we exploit the fact that the interval $[0,1]$ can be decomposed as $$[0,1] = [0,\tfrac12] \cup (\tfrac12,1].$$

Yet, I think I have an issue with the full understanding of how an indicator function works. For example, in our case, when is $g(x)$ equal to 1? When $x∈[0,1]$ or when $x∈[0,\tfrac12]$?

Thank you for your attention.

答:
If $X$ is a set, and $S$ a subset of $X$, the indicator function $\mathbf 1_S : X \to \mathbb R$ is $1$ in $S$ and $0$ elsewhere, that is, for each $x \in X$ we have
$$\mathbf 1_S(x) = \begin{cases}
1 & \textrm{if } x \in S, \\[0.3mm]
0 & \textrm{if } x \notin S. \end{cases}$$

Now, if $A$ and $B$ are two subsets of $X$, the indicator function $\mathbf 1_{A  \cup B}$ is the same as $\mathbf 1_A+\mathbf 1_B-\mathbf 1_{A \cap B}$. This is because for $x \in A \cup B$ we have three cases: $(1)$ $x \in A$ and $x \notin B$; $(2)$ $x \in B$ and $x \notin A$; $(3)$ $x \in A \cap B$. In the first case we have $$\mathbf 1_{A  \cup B}(x) = 1 = 1+0-0 = \mathbf 1_A(x)+\mathbf 1_B(x)-\mathbf 1_{A \cap B}(x),$$
in the second case we have
$$\mathbf 1_{A  \cup B}(x) = 1 = 0+1-0 = \mathbf 1_A(x)+\mathbf 1_B(x)-\mathbf 1_{A \cap B}(x),$$ and in the third case we have
$$\mathbf 1_{A  \cup B}(x) = 1 = 1+1-1 = \mathbf 1_A(x)+\mathbf 1_B(x)-\mathbf 1_{A \cap B}(x).$$
Also, if $x \notin A \cup B$ then $\mathbf 1_A(x)+\mathbf 1_B(x)-\mathbf 1_{A \cap B}(x) = 0+0-0 = 0 = \mathbf 1_{A  \cup B}(x)$.

Thus $\mathbf 1_{A  \cup B} = \mathbf 1_A+\mathbf 1_B-\mathbf 1_{A \cap B}$, and in particular, if $A$ and $B$ are disjoint ($A \cap B = \varnothing$) we have that $\mathbf 1_{A  \cup B} = \mathbf 1_A+\mathbf 1_B$. Hence, in your problem $X = [0,1]$ and $[0,1] = [0,\tfrac12] \cup (\tfrac12,1]$ tells us that $$\mathbf 1_{[0,1]} = \mathbf 1_{[0,\frac12]} + \mathbf 1_{(\frac12,1]} \implies 1_{(\frac12,1]} = \mathbf 1_{[0,1]} - \mathbf 1_{[0,\frac12]}.$$

Title: Indicator functions and vector spaces


问:
Sitting younger, older alternately.
In a group of eight friends, everyone is of different age. Everyone decided to take a seat by an eight-person round table. The youngest of them sat at seat 1. Later they sat: older, younger, older, younger, etc. [seat 2 is older than seat 1, seat 3 is younger than seat 2, ..., seat 1 is younger than seat 8] How many ways can they sit?

Actually, I feel this is the hardest of all the problems I have posted recently.

My first fhought was that you have to split them evenly into the "younger" and "older" groups, but you can find solutions that satisfy the requirements but don't follow this rule.

What is true though, is you cannot the currently youngest person as the "older" one or currently olders person as the "younger" one.
Since it is an archival local math contest, I know the answer is supposed to be 272.

答:
Once the youngest has sat, the other seven people form a line of alternating age comparisons:
$$1B<C>D<E>F<G>1$$
It is well-known that the number of ways to arrange the other seven people in such a line is given by the up/down number $A_7=272$. The Wikipedia page linked describes a simple recurrence for these numbers.

Title: Sitting younger, older alternately.

Body:
Once the youngest has sat, the other seven people form a line of alternating age comparisons:
$$1<A>B<C>D<E>F<G>1$$
It is well-known that the number of ways to arrange the other seven people in such a line is given by the up/down number $A_7=272$. The Wikipedia page linked describes a simple recurrence for these numbers.


答:
To determine whether or not $f'(c)$ exists, one can see if the following is true:

$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c} = \lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}.$$

If either the LHS or RHS to the above equation does not exist, or the LHS and RHS exists but have different values, then $f'(c)$ does not exist.

For your piecewise function, let's determine whether or not the derivative of $f$ at $c = 0$ exists.

$\lim_{x\to 0^+}\left(\frac{f(x)-f(0)}{x-0}\right) = \lim_{x\to 0^+}\left(\frac{x^2+1 - 1}{x}\right) = \lim_{x\to 0^+}\left(x\right) = 0.$

However,

$\lim_{x\to 0^-}\left(\frac{f(x)-f(0)}{x-0}\right) =  \lim_{x\to 0^-}\left(\frac{x^2-1 - 1}{x}\right) = \lim_{x\to 0^-}\left(\frac{x^2-2}{x}\right) = \lim_{x\to 0^-}\left(x - \frac{2}{x}\right) \to +\infty,$

or rather it's more proper to say that the latter limit does not exist.

Since the second limit does not exist, $f'(0)$ does not exist.

Title: What should actually be demonstrated in this exercise about left and right derivatives?

Body:
I am redoing my math education after 35 years in order to be up to date with my children's curriculum (I usually have a nanosecond to give an answer because of magical dad knowledge). My older son is in high school and they are going through derivatives.

One of the exercises I was preparing for him (this is beyond the homework, he is in an "advanced" track and I want to show him some more interesting problems) I stumbled upon the following problem. It is a translation from French so feel free to edit in order to improve the wording

We define

$$
\begin{align}
f: \mathbb{R} & \rightarrow \mathbb{R} \\
x & \mapsto \begin{cases}
      \text{$x^2 - 1$ if $x < 0$} \\
      \text{$x^2 + 1$ if $x \geq 0$}
    \end{cases}
\end{align}
$$

Demonstrate that, for $a=0$, $f$ is derivable on the right but not on the left.

Both $x^2 - 1$ and $x^2 + 1$ are derivable in $\mathbb{R}$ but $f$ is defined for $0$ as $x^2 + 1$. So technically the derivative in $0$ will be $2 \times 0 = 0$

It is however derivable on $0^+$ ($2x$) and on $0^-$ ($2x$).

The only idea I have is that by taking the definition of the left derivative I have
$$ \lim_{h \rightarrow 0^-} \frac{f(x+h) - f(x)}{h} $$
which I think is not defined because the limit approaching on the left does not exist (as the function is undefined there). But I think I must be missing something because this is too obvious.


问:
How do I simplify this sum?
What technique should I use for simplifying

$$\sum_{k=1}^\infty \frac{k-1}{k(k+1)(k+2)} ?$$

答:
Observe that
$$k-1 = k + (k+1) - (k+2).$$
Therefore,
\begin{align}\sum_{k=1}^\infty\frac{k-1}{k(k+1)(k+2)} &= \sum_{k=1}^\infty\frac{1}{(k+1)(k+2)}+\sum_{k=1}^\infty\frac{1}{k(k+2)}-\sum_{k=1}^\infty\frac{1}{k(k+1)} \\ &= \sum_{k=1}^\infty\left(\frac{1}{k+1}-\frac{1}{k+2}\right)+\frac{1}{2}\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+2}\right)-\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+1}\right)\end{align}
Can you take it from here?

Title: How do I simplify this sum?


问:
Prove that $A^2=\text{tr}(A)A$ where $A$ is $n\times n$ matrix of rank $1$
Let $A$ be a matrix of $n\times n$ with rank $1$. I need to prove that $A^2=\text{tr}(A)A$, without using eigenvalues.

答:
A matrix of rank $1$ $A$ can be written as $UV^T$ with $U,V\in \mathbb{K}^n$ two column vectors.

So $AA = UV^T\cdot UV^T = U(V^TU)V^T=U(\text{tr}A)V^T = (\text{tr} A) A$.


问:
2 dimensional symmetric random walk
Let $\left\{\left(X_{n}, Y_{n}\right)\right\}_{n=0}^{\infty}$ be a 2 -dimensional symmetric random walk. Namely, this is a Markov chain where $\left(X_{n+1}, Y_{n+1}\right)$ takes one of the following 4 values with equal probability:
$$\left(X_{n}+1, Y_{n}\right),\left(X_{n}-1,Y_{n}\right),\left(X_{n}, Y_{n}+1\right),\left(X_{n}, Y_{n}-1\right)$$
Suppose that $X_{0}=Y_{0}=0$.

Define $T:=\inf \left\{n \geq 0: \max \left(\left|X_{n}\right|,\left|Y_{n}\right|\right)=3\right\} .$
I want to find the value of $\mathbb{E}[T]$ and $\mathbb{P}\left(X_{T}=3, Y_{T}=0\right)$.
Thanks for your help!

答:
Let $k_{ij} = \mathbb E[T\mid X_0 = i, Y_0=j]$. So we want $k_{00}$. Noting the clear symmetry in both axes, we have the equations
\begin{align*}
k_{00}&=1+k_{10} \\
k_{10}&=1+\frac{1}{4}k_{00}+\frac{1}{4}k_{20}+\frac{1}{2}k_{11} \\
k_{20}&=1+\frac12k_{12}+\frac14k_{10} \\
k_{11}&=1+\frac12(k_{12}+k_{10}) \\
k_{12}&=1+\frac14(k_{11}+k_{20}+k_{22})\\
k_{22}&=1+\frac12k_{12}
\end{align*}
If you crunch through this system, you will get $\boxed{k_{00}=\tfrac{135}{13}}$.

For the second part, let $p_{ij}$ be the probability, conditional on $X_0=i$, $Y_0=j$, that $(X_T,Y_T)$ is one of $(\pm3,0),(0,\pm3)$. So we want $p_{00}/4$. Similar to above, we can get the system
\begin{align*}
p_{00}&=p_{10} \\
p_{10}&=\frac{1}{4}p_{00}+\frac{1}{4}p_{20}+\frac{1}{2}p_{11} \\
p_{20}&=\frac14+\frac12p_{12}+\frac14p_{10} \\
p_{11}&=\frac12(p_{12}+p_{10}) \\
p_{12}&=\frac14(p_{11}+p_{20}+p_{22})\\
p_{22}&=\frac12p_{12}
\end{align*}
Solving gives $p_{00}=\frac{4}{13}$, so $\boxed{\mathbb P(X_T=3,Y_T=0)=\tfrac{1}{13}}$.

In case you're interested, similar computations to the above gives $\mathbb P(X_T=3, Y_T=1)=\frac{3}{52}$ and $\mathbb P(X_T=3,Y_T=2)=\frac{3}{104}$.

Another aside: if you have a particular aversion to simultaneous equations, you might like to try some kind of martingale trick. This is especially appealing when you realise that the stopping time can be redefined as $T=\inf\left\{n\geq0\mid X_n^2+Y_n^2\geq9\right\}$. Why is this nice? Well, you can show that $M_n=X_n^2+Y_n^2-n$ is martingale, so by the optional stopping theorem,
$$\mathbb E[M_T]=\mathbb E[M_0]=0\implies\mathbb E[T]=\mathbb E\left[X_T^2+Y_T^2\right].$$
We know that $X_T^2+Y_T^2\in\{9,10,13\}$. So we certainly have the bounds $9<\mathbb E[T]<13$. Unfortunately, I don't think we can glean much more information than this from martingale magic.

Title: 2 dimensional symmetric random walk


问:
Confused about Tensor Product of R-modules
In Tu's book on Differential Geometry he first defines $Free(V\times W)$ as:

$$\sum r_i(v_i, w_i), r_i \in R, (v_i, w_i) \in V \times W$$
where the sum is finite.

The way I understand it is the above construction is of formal combinations and forgets the actual structure of the modules. In other words, if $v_1+v_2 = v_3$, it isn't true that in $Free(V\times W)$ that $(v_1, 0) + (v_2, 0) = (v_3, 0)$

Now to form the tensor product we quotient by the submodule, $S$ spanned by elements of the form:
$$ (v_1 + v_2, w) - (v_1, w) - (v_2, w)\\
 (v,w_1 + w_2) - (v, w_1) - (v, w_2)\\
 (rv,w) - r(v,w)\\
 (v, rw) - r(v,w)$$
Then we have a map from the product to the tensor product,
$$(v,w) \rightarrow v\otimes w$$

However, if $v_3 = v_1 + v_2$, then I can't show that $$v_3\otimes w = v_1 \otimes w + v_2 \otimes w$$
which should be the case if $\otimes$ is a module homomorphism bilinear map.

答:
Since $V \otimes W := \operatorname{Free}(V \times W)/S$ and $\otimes : V \times W \to V \otimes W$ is defined by $$(v,w) \mapsto v \otimes w := (v,w)+S,$$ the condition $(v_1 + v_2, w) - ((v_1, w) + (v_2, w)) \in S$ tells us that $$(v_1 + v_2, w)+S = ((v_1, w) + (v_2, w))+S = ((v_1,w)+S) + ((v_2,w)+S)$$
which is the same as $(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w$. Also observe that the other relations which define $S$ gives us \begin{align} v \otimes (w_1+w_2) &= v \otimes w_1 + v \otimes w_2, \\ (rv) \otimes w &= r(v \otimes w), \\ v \otimes (rw) &= r(v \otimes w).\end{align}

Recall that if $M$ is an $R$-module and $S$ is a submodule of $M$, the quotient $M/S$ is defined by $M/\!\sim$, where $$(\forall u_1,u_2 \in U) \quad u_1 \sim u_2 \iff u_1-u_2 \in S.$$ In this case, the equivalence class of $m \in M$ is given by $m+S := \{m+s : s \in S\}$ (hence $m+S = m'+S \iff m-m' \in S$), and we define an $R$-module structure in $M/S$ by $$(\forall m,m' \in M)(\forall r \in R) \quad r(m+S)+(m'+S) := (rm+m')+S.$$


问:
The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is
I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$

$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$

$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$

From here I am not able to process further to think of the way that will give me the solution.

答: This is a Linear Diophantine equation. As the first step we need to check whether the greatest common divisor (GCD) of 2 and 5 divides 99 (it does). Once we have done that we need to find a way to express the GCD of 2 and 5 as $2x_0+5y_0 = 1$, where 1 is their GCD and $x_0$, $y_0$ are integers. $x_0$ and $y_0$ can be found using Euclidean Algorithm, in this case $x_0=-2$ and $y_0=1$. Now we multiply both the sides of the equation by 99 to get $99=2(-198)+5(99)$. Consequently, we set $x_0=-198$ and $y_0=99$. Voila! You have found the answer! Oh, not yet. We need to fix few things. We have to keep the constraints in mind. Here comes the general solution to our equation, $x=x_0+5t$ and $y=y_0-2t$, here (x,y) is our general solution. We apply the given inequalities on x and y to get the following range for t (remember that t is an integer), $43\leq t \leq 59$. So the number of possible integral values of t is $59-43+1=17$, and corresponding to each value of t we have a different solution, so there are a total of 17 solutions.


问:
Solving Recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n\geq 1, a_0=2$?
How can I solve recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n \geq1 $ and $a_0 = 2$?

答:
general sequence is $a_n= (3^n)a_0 +n \times 5\times 3^n$
substitute $a(n-1)$ for $a(n-2)$ and check the pattern you can easily figure it out.


问:
Show that $dX_t=\frac{X_t}{1-t}dt+dW_t$ can be written as $X_t=(1-t)\int_{0}^{t}\frac{1}{1-s}dW_s$
I have been reading the following link on Brownian Bridge definition, and came across the following statement (bullet point 9 in the link above):

Suppose $W_t$ is a standard Brownian motion, define $X_1=1$, then, for $h \in [0,1]$, the process $X_t$ is a Brownian Bridge:

$$X_t=(1-t)\int_{h=0}^{h=t}\frac{1}{1-h}dW_h \tag{1}$$

I can actually understand the proof of this statement presented in the link above and do not have an issue with the claim that $X_t$ above is a Brownian Bridge. However, the author then goes on to state that:

"In differential form, the above can be written as:"

$$dX_t=\frac{X_t}{1-t}dt+dW_t \tag{2}$$

I actually am unable to connect the differential form with the equation (1) given for $X_t$.

When I rewrite the differential form in the "long-hand" notation, I get ($X_0:=0$):

$$X_t=\int_{h=0}^{h=t}\frac{X_h}{1-h}dh+\int_{h=0}^{h=t}dW_h=\int_{h=0}^{h=t}\frac{X_h}{1-h}dh+W_t$$

The above is clearly not the same as the former definition of $X_t$ given in equation (1). I am thinking there might be some Ito's lemma application to a smartly defined function $F(X_t,t)$, that I haven't been able to figure out (I've tried playing around with variants of $F(X_t,t):=X_te^t$, but to no avail).

Is there a way to "solve" the differential equation (2) into (1), or has the author made a typo?

Edit: having read the answer linked in the comment below, and in the spirit of my own answer on notation to another question here, I have attempted to rewrite the linked answer using the long hand notation (because I struggle with interpreting some of the steps of the short-hand notation answer):

I am still getting a wrong answer. Could you please help me spot where I am going wrong?.

The "trick" in the linked aswer seems to be applying Ito's lemma to a function $F(W_t,t):=\frac{W_t}{1-t}$. The derivatives are:

$$\frac{\partial F}{\partial t}=\frac{-W_t}{(1-t)^2}, \frac{\partial F}{\partial W_t}=\frac{1}{1-t}, \frac{\partial^2 F}{\partial t^2}=0$$

We also have that:

$$W_t=W_0+\int_{h=0}^{h=t}a(W_h,h)_{=0}dh+\int_{h=0}^{h=t}b(W_h,h)_{=1}dW_h$$ so that:

$$\frac{W_t}{1-t}=\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial W}*a(W_h,h)_{=0}+\frac{\partial^2 F}{\partial W^2}_{=0}*b(W_h,h)\right)dh+\int_{h=0}^{h=t}\frac{\partial F}{\partial W}b(W_h,h)_{=1}dW_h=\\=\int_{h=0}^{h=t}\left(\frac{-W_h}{(1-h)^2}\right)dh+\int_{h=0}^{h=t}\left(\frac{1}{1-h}\right)dW_h$$

Multiplying through by $1-t$ then gives:

$$W_t=(1-t)\int_{h=0}^{h=t}\left(\frac{-W_h}{(1-h)^2}\right)dh+X_t$$

Therefore we have:

$$X_t=(1-t)\int_{h=0}^{h=t}\left(\frac{W_h}{(1-h)^2}\right)dh+W_t$$

Focusing on the term $(1-t)\int_{h=0}^{h=t}\left(\frac{W_h}{(1-h)^2}\right)dh$, we can write:

$$\int_{h=0}^{h=t}\left((1-t)\int_{s=0}^{s=h}\frac{1}{1-h}dW_s\right)\frac{1}{1-h}dh$$

Note that the term in the bracket above, i.e. $\left((1-t)\int_{s=0}^{s=h}\frac{1}{1-h}dW_s\right)$ in fact does not equal to $X_h$ (as defined in equation (1)), so in fact we do not have that:

$$X_t=\int_{h=0}^{h=t}\frac{X_h}{1-h}dh+W_t$$

答:
Let $Y_{t} = \int_{0}^{t} \frac{1}{1-s}\, dW_{s}$. Next take a look at

$$X_{t} = (1-t) \int_{0}^{t} \frac{1}{1-s}\, dW_{s} = (1-t)Y_{t}$$

and differentiate using It^o's lemma

\begin{align*}
dX_{t} &= -Y_{t}\, dt + (1-t)\, dY_{t} + d[ 1-t, Y_{t} ] \\
&= - Y_{t}\, dt + (1-t)\cdot \frac{1}{1-t}\, dW_{t} \\
&= -\frac{X_{t}}{1-t}\, dt + dW_{t}
\end{align*}

and so there is indeed a typo.

If you want to solve

$$dX_{t} =  \frac{X_{t}}{1-t}\, dt + dW_{t},$$

then (as in ODEs) use integrating factor

$$\mu(t) = e^{-\int_{0}^{t} \frac{1}{1-s}\, ds } = 1-t$$

to solve SDE

\begin{align*}
d \left (X_{t}(1-t) \right) = (1-t)\, dX_{t} - X_{t}\, dt =: (1-t)\, dW_{t}
\end{align*}

for solution

\begin{align*}
X_{t} = \frac{1}{1-t}X_{0} + \frac{1}{1-t} \int_{0}^{t} (1-s)\, dW_{s}.
\end{align*}

Note of caution: you should not apply It^o's lemma to solve the SDE. This only works in the case that it admits a strong solution (cf. Oksendal, Chapter 5).


问:
Is there a way to get an approximation of $\det( \sigma_s^2A+B)^{1/N}$ in the GLRT?
Both $A_{N \times N}$ and $B_{N \times N}$ are symmetric Toeplitz matrix, $\sigma_s^2$  is a constant.

In detection theory, the computation of $\det(\sigma_s^2A+B)^{1/N}$ arises in the computation of the GLRT (sometimes referred to as sphericity test). In particular, $A_{N \times N}$, $B_{N \times N}$ and $\sigma_s^2$ represent the covariance of received signal, the noise covariance and the received signal power, respectively.

I need to find how $\det(\sigma_s^2A+B)^{1/N}$ varies as $\sigma_s^2$ varies. An approximated expression or bound could work.

Using the Minkowski determinant inequality i found a bound $$\det(\sigma_s^2A+B)^{1/N}\ge \sigma_s^2\det(A)^{1/N}+\det(B)^{1/N}$$ However this inequality yields a not sufficiently tight approximation.

A tight bound or (better an approximation of the above determinant) can be found, maybe under specific conditions on the matrix A and B?

答:
Here is a full computation for the derivative of
$$
f(\epsilon) = \det(B + \epsilon A)^{1/N}.
$$
Let $g(\epsilon) = \det(I + \epsilon B^{-1}A)$, so that $f(\epsilon) = \det(B)^{1/N}g(\epsilon)^{1/N}$. By the chain rule, we have
$$
f'(\epsilon) = \frac 1N \det(B)^{1/N} g(\epsilon)^{1/N - 1} \cdot g'(\epsilon)
\\ = \frac 1N \det(B)^{1/N}\det(I + \epsilon B^{-1}A)^{1/N - 1} \cdot \operatorname{tr}(B^{-1}A).
$$
Plugging in $\epsilon = 0$ yields $f'(0) = \frac 1N \det(B)^{1/N}\operatorname{tr}(B^{-1}A)$, which leads to the Taylor expansion
$$
f(\sigma_s^2) = \det(B)^{1/N} + \frac 1N \det(B)^{1/N}\operatorname{trace}(B^{-1}A)\sigma_s^2 + O(\sigma_s^4)
$$
that I mentioned in a comment on this question.


问:
Expectation for arriving earlier and later
I was thinking to get these using the first and second moment by $E(X)=\frac{(b+a)}{2} $ and $E(X^2)=\frac{b^3-a^3}{3b-3a}$ for $a=2,b=8$, but I don't think this is the solution. I have found some sort of solution with order statistics and Beta distributions, but I would like to solve it without it, in a simpler way.

答:
Hint (almost solution):

Put $\xi$ = arrival time of customer A, $\eta$ = arrival time of customer B. Then the time of first (earlier) arrival is $\zeta = \min(\xi, \eta)$. Thus we know $P(\zeta > x) = P( \xi > x, \eta >x) = P( \xi > x)P(\eta >x)$. For nonnegative r.v. $\zeta$ we have:
$$E \zeta = \int_{0}^{\infty} P( \zeta>x) dx.$$
Hence we know $E \zeta $.

Put $\bar{\zeta}$ = time of second arrival $= \max(\xi, \eta)$, then $\bar{\zeta} + \zeta =   \min(\xi, \eta) + \max(\xi, \eta) = \xi + \eta$. As $E\xi= E\eta = \frac{b+a}{2}$ we get that
$$E\bar{\zeta} + E\zeta = E(\xi + \eta) = b+a$$ and thus we know $E\bar{\zeta} $.


答:
Easy can you solve it by calculating the expectation of  min and Max

(b)

The distribution of $Z=max(X,Y)$ is

$$F_Z(z)=\Big(\frac{z-2}{6}\Big)^2$$

Derivating you get $Z$ density with expectation

$$E(Z)=\int_2^8 z\frac{z-2}{18}dz=6 pm$$

Similar result for (a): find the distribution of $U=min(X,Y)$ then the expectation

Result: 4 pm

The CDF of an uniform is

$$F_X(x)=\frac{x-a}{b-a}=\frac{x-2}{8-2}$$

The CDF of the Max is

$$F_Z(z)=P(Z\leq z)=P(X\leq z; Y\leq z)=P(X\leq z)P(Y\leq z)=\Big(\frac{z-2}{8-2}\Big)^2$$

The CDF of min is

$$P(U>u)=P(X>u,Y>u)=P(X>u)P(Y>u)=\Big(1-\frac{u-2}{6}\Big)^2$$

Thus

$$F_U(u)=P(U\leq u)=1-\Big(\frac{8-u}{6}\Big)^2$$

Derivating you get the density with expectation

$$E(U)=\int_2^8 u\Big(\frac{8-u}{18}\Big)du=4$$

Title: Expectation for arriving earlier and later

Body: I was thinking to get these using the first and second moment by $E(X)=\frac{(b+a)}{2} $ and $E(X^2)=\frac{b^3-a^3}{3b-3a}$ for $a=2,b=8$, but I don't think this is the solution. I have found some sort of solution with order statistics and Beta distributions, but I would like to solve it without it, in a simpler way.


问:
Existence of a special pentagonal tiling
Is there a pentagonal tiling composed of only one shape of pentagon so that each pentagon touches exactly 5 other pentagons?
Two pentagons are in touch if they share at least one common point.

Few examples:

Here each pentagon touches exactly 6 other pentagons:

Here each pentagon touches exactly 7 other pentagons:

Here each pentagon touches exactly 8 other pentagons:

Here each yellow pentagon touches 5 other pentagons, each pink pentagon touches 6 other pentagons and each blue pentagon touches 7 other pentagons:

I just can not find any tiling that there are only 5 touches per pentagon.

Maybe it is trivial question but I am not very familiar with tiling theory.

答: If such a tiling existed, it would necessarily have to be edge-to-edge with $3$ pentagons at every vertex. The tiling's skeleton (the graph formed by the vertices and edges of contact) would therefore be a $3$-regular infinite planar graph with each face being of size $5$. But by Euler's formula, this should be a dodecahedron, which is not a tiling. Thus no such tiling exists.


问:
Find a second degree polynomial that goes through 3 points
I am having trouble calculating the quadratic curve $f(x)$ that goes through 3 points;
for example a curve that goes through $A(1,3), B(-1,-5), and C(-2,12)$.
I can only guess that the curve is upwards and that I may create the system:
$$
y_1 = ax^2_1 + bx_1 + c\\
y_2 = ax^2_2 + bx_2 + c\\
y_3 = ax^2_3 + bx_3 + c
$$
assuming that the points are in the format $A(x,y)$ and from this point what do I do?
Do I build a matrix and use the Gaussian eliminations?

EDIT: I also know that $f(4) = 120$

Body:
For the three points
P1=(x1,y1),
P2=(x2,y2),
P3=(x3,y3),

use the coefficients a, b and c,

a = y1/(x1*x1 - x1*x2 - x1*x3 + x2*x3) + y2/(-x1*x2 + x1*x3 + x2*x2 - x2*x3) + y3/(x1*x2 - x1*x3 - x2*x3 + x3*x3)

b = - (x2*y1)/(x1*x1 - x1*x2 - x1*x3 + x2*x3) - (x3*y1)/(x1*x1 - x1*x2 - x1*x3 + x2*x3) - (x1*y2)/(-x1*x2 + x1*x3 + x2*x2 - x2*x3) - (x3*y2)/(-x1*x2 + x1*x3 + x2*x2 - x2*x3) - (x1*y3)/(x1*x2 - x1*x3 - x2*x3 + x3*x3) - (x2*y3)/(x1*x2 - x1*x3 - x2*x3 + x3*x3)

c = (x2*x3*y1)/(x1*x1 - x1*x2 - x1*x3 + x2*x3) + (x1*x3*y2)/(-x1*x2 + x1*x3 + x2*x2 - x2*x3) + (x1*x2*y3)/(x1*x2 - x1*x3 - x2*x3 + x3*x3)

for the polynomial
y = a*x² + b*x + c


问:
Maclaurin expansion for sech$(x)$
I am a bit unsure where I have gone wrong in working this out.

Sech$(x)=2/(e^x+e^{-x}).$

Maclaurin expansions:

$e^x = 1+ x +  x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 -  x^4/24;$

so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$  (I think this is where I have gone wrong.)

The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).

How would I work this out?

Title: Maclaurin expansion for sech$(x)$

Body:
From your expansions of $e^x$ and $e^{-x}$, we have $e^x+e^{-x}=2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots$,

so  sech$( x)=\dfrac2{e^x+e^{-x}}=\dfrac{2}{2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots}=\dfrac1{1+\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots}$

$=1-\left(\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots\right)+\left(\dfrac{x^2}{2}+\dfrac{x^4}{4!}+\cdots\right)^2\cdots$

$=1-\dfrac{x^2}2-\dfrac{x^4}{24}+\dfrac{x^4}{4}\cdots=1-\dfrac{x^2}2+\dfrac{5x^4}{24}\cdots$


问:
Pick two random people in the world, what's the probability of them knowing each other?
I have one question popped in my head out of nowhere at 3AM.

If we pick two random people in the world, what's the probability of them knowing each other?

You can assume that there are 7 billion people in the world and the idea of Six Degrees of Separation is relevant, or maybe you can assume one person has a circle of friends up to 100 people. You can provide any other assumptions to make it simple.

Please help answer my 3AM thought so I can sleep peacefully.

答:
All you need to know is the average number of people the average person knows, plus the total population of the world. (By "knows" here, I am assuming a reciprocal relationship, discounting the fact that everyone "knows" lots of celebrities who don't know them from Adam.) If $K$ represents this average, and $P$ represents total population, then the answer is $K/(P-1)$.

To elaborate, just a bit, we have

$$K={1\over P}\sum_{i=1}^Pk_i$$

where $k_i$ is the (precise) number of people known to person $i$.  When you pick two people at random, the first person will be person $i$ with probability $1/P$, and the second person will be known to person $i$ with probability $k_i/(P-1)$, and thus the probability the two people will know one another is $\sum_{i=1}^P{1\over P}{k_i\over(P-1)}=K/(P-1)$.

As a practical matter, the probability is pretty small, probably well under one in a million.  Even if you restrict just to the U.S., the probability is likely to be around one in half a million: It has been estimated that the average American knows about $600$ people.

Note, the six-degrees-of-separation phenomenon and the Birthday Paradox play no role here. They would have role if we were assembling (at random) some medium-sized group of people and asking, for example, for the probability that some pair of them are connected by a chain entirely within the group.


问:
Let $H$ be a normal subgroup of the group $G$. Show that if the factor group $G/H$ is finite, and $H$ is finite, then $G$ must be finite.
Let $H$ be a normal subgroup of the group $G$. Show that if the factor group $G/H$ is finite, and $H$ is finite, then $G$ must be finite.

My attempt:

Since $G/H$ and $H$ are finite sets, we know that there are finite left cosets of $H$ in $G$, i.e., $|gH|=|H|< \infty,$ for each $g \in G$

We have that $G$ is the union of all these cosets of $H$ in $G$, and since we have finite cosets, $G$ must be finite.

Is this right?

答:
Yes, your proof is fine.

Alternatively, note that $[G:H]=\lvert G/H\rvert$ and $$| G|=[G:H]\lvert H\rvert.$$


问:
Why does $f(x)=f(x+1)$ imply $f(x) \equiv$ constant if $f$ is polynomial?
A polynomial function $P: \mathbb{R} \mapsto \mathbb{R}$ has the following property:
$$\forall x\in\mathbb{R} \space \bigg(P(x)=P(x+1)\bigg)$$
Why does this imply that $P \equiv \text{constant}$?

EDIT: When posting the question for the first time I forgot to mention that we are talking about a polynomial function.

答:
No.

For example,
$$
f(x)=\sin(2\pi x).
$$


答: Since $\operatorname{sech}x$ is even, write $\operatorname{sech}x=\sum_{j=0}a_j(x^2)^j$ so$$1=\sum_{j,\,k\ge0}\frac{a_j}{(2k)!}(x^2)^{j+k}\implies\delta_{n0}=\sum_{j=0}^n\frac{a_j}{(2n-2j)!}.$$Now get the first few $a_j$ by solving simultaneous equations.

Title: Maclaurin expansion for sech$(x)$

Body:
I am a bit unsure where I have gone wrong in working this out.

Sech$(x)=2/(e^x+e^{-x}).$

Maclaurin expansions:

$e^x = 1+ x +  x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 -  x^4/24;$

so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$  (I think this is where I have gone wrong.)

The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).

How would I work this out?


答:
It doesn't. Every function with period $1$ has this property.

For example, if $f(x) = \sin 2 \pi x$, then $f(x+1) = \sin (2\pi x +2\pi) = \sin 2\pi x = f(x)$ for all $x$.

Title: Why does $f(x)=f(x+1)$ imply $f(x) \equiv$ constant if $f$ is polynomial?

Body:
A polynomial function $P: \mathbb{R} \mapsto \mathbb{R}$ has the following property:
$$\forall x\in\mathbb{R} \space \bigg(P(x)=P(x+1)\bigg)$$
Why does this imply that $P \equiv \text{constant}$?

EDIT: When posting the question for the first time I forgot to mention that we are talking about a polynomial function.


问:
Rational Expression Question. (Word problem)
Joe got a mark of $\dfrac{44}{50}$ on one test and $\dfrac{32}{x}$ on another test. If the average mark on the two tests was 80%, what value was the second test out of?

My revised attempt:

Still confused, is this correct? Seems odd that the answer I am getting is a repeating decimal:

$$\dfrac 12 \left(\dfrac{44}{50}+\dfrac{32}{x}\right)=.8$$

$$\dfrac{44}{50}+\dfrac{32}{x}=\dfrac{.8}{.5}$$

$$\dfrac{32}{x}=\bigg(\dfrac{8}{5}-\dfrac{44}{50}\bigg)x$$

$$32=\dfrac{8}{5}x-\dfrac{44}{50}x$$

$$50\bigg(32=\dfrac{8}{5}x-\dfrac{44}{50}x\bigg)$$

$$1600=80x-44x$$

$$\dfrac{1600}{36} = \dfrac{36x}{36}$$

$$\boxed{x=44.4}$$

答: The average is half the sum.  So your first should be $\dfrac 12 \left(\dfrac{44}{50}+\dfrac{32}{x}\right)=.8$</p>


Title: Solving Recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n\geq 1, a_0=2$?

Body: How can I solve recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n \geq1 $ and $a_0 = 2$?

Body:
Hint: $$a_{n+1} - 3a_n = 5 \cdot 3^{n+1} = 3 (a_n-3a_{n-1})$$

Can you start from here?

 $$a_{n+1} - 6a_n + 9a_{n-1} = 0 \implies \frac{a_{n+1}}{3^{n+1}} - 2\frac{a_n}{3^n} + \frac{a_{n-1}}{3^{n-1}}=0$$ Let $a_n = 3^n b_n$ then $$b_{n+1} - 2b_n  + b_{n-1} = 0 \implies b_{n+1} - b_{n} = b_n-b_{n-1} = \cdots = b_1 - b_0 = 7 - 2 =5$$ Therefore $b_n=b_0+5n=5n+2, a_n = 3^n (5n+2). \blacksquare$


答:
Note that the OP has clarified that they are assuming $f$ is a polynomial.

This follows from the standard result that every nonconstant polynomial has only finitely many zeroes: just consider the polynomial $Q(x)=P(x)-P(0)$. We'll have $Q(z)=0$ for every integer $z$, so $Q$ - and hence $P$ itself - must be constant.

Incidentally, the standard result above may feel obvious at first but is surprisingly delicate; you should go through the proof in careful detail.

Alternatively, it's easy to show that if $P$ is a nonconstant polynomial then $\lim_{x\rightarrow\infty}P(x)$ is either $+\infty$ or $-\infty$ depending on the sign of the leading coefficient of $P$. In particular, no nonconstant polynomial is bounded. But every continuous periodic function is bounded, as a consequence e.g. of the Extreme Value Theorem. (This argument was also mentioned by lulu in a comment above, as I was typing this.)

(It's easy to miss the application of EVT here, so let me unpack it. Periodicity gives us that $ran(P)=\{P(x): x\in [0,p]\}$ where $p$ is the period of $P$. Now we need to show that $\{P(x): x\in [0,p]\}$ is bounded. This is where continuity comes in: since $P$ is continuous and $[0,p]$ is a closed and bounded interval, $P$ has a maximum and minimum value on $[0,p]$. Note that we can have discontinuous periodic functions which are unbounded: consider e.g. the extension $\alpha$ of $\tan$ gotten by setting $\alpha(x)=17$ whenever $\tan(x)$ is undefined.)


答:
Since $\text{rank}(A) = 1$, it has eigenvalue $0$ with multiplicity of $n-1$ and the other eigenvalue $\text{tr}A$ (because the sum of eigenvalues is equal to the trace).

Therefore, $$p(x) = (x-\text{tr}A)x^{n-1}$$ where $p(x)$ is the characteristic polynomial of $A$. Now, observe that the minimal polynomial of $A$ divides $(x-\text{tr}A)x$, and thus, $$(A - \text{tr}A)A = 0 \implies\boxed{A^2 = \text{tr}A \cdot A}$$

Title: Prove that $A^2=\text{tr}(A)A$ where $A$ is $n\times n$ matrix of rank $1$

Body: Let $A$ be a matrix of $n\times n$ with rank $1$. I need to prove that $A^2=\text{tr}(A)A$, without using eigenvalues.


问:
What is $\int x^re^xdx$?
Is there any simple way to get integral of $e^{x}x^{r}, r \in \mathbb{R}$?

Basically I want to solve this:

$$\displaystyle \int \frac{e^t(4t^2+1)}{2t \sqrt{t}}dt$$

so I will appreciate any help in both of the above problems.

答:
Hint: Break up the integral into two integrals
$$
\int \frac{e^t(4t^2 + 1)}{2t \sqrt{t}} dt = \int 2e^t \sqrt{t} dt + \int \frac{e^t}{2t^{3/2}}dt
$$

Then use integration by parts on one of them until you have a constant multiple of the other, and see what happens.

Note: In general, we can't find $\int e^x x^r dx$. The definite integral $\int_0^\infty x^r e^x dx$ is (almost) the gamma function.

Title: What is $\int x^re^xdx$?


答: Suppose there exists nonconstant polynomial $P$. Then it has at least one root and since it is periodic it has infinitely roots. But such polynomial is constant polynomial $P(x)=0$ for all $x$. A contradiction.


问:
$n$ vertex graph with more edges than its Turan number: $e(G) = t_r(n) + 1$
I have been struggling with this problem for quite some time now and I cannot think of a way to proceed with either part:

Suppose that G is a graph with $n > r + 1$ vertices and $t_r(n) + 1$ edges:

(a) Prove that for every $p$ with $r + 1 < p <= n$ there is a subgraph $H$ of $G$ with $|H| = p$ and $e(H) \geq t_r(p) + 1$.

(b) Prove that $G$ contains two copies of $K_{r+1}$ with exactly $r$ common vertices.

For part a), I have so far tried to use induction on $n$. For the base case with $n = r+2$ we have only $H = G$ and $n = p$ to check, which is clearly true. For the $n$th case I think we want to find a vertex $v$ in $G$ of degree $\delta(T_r(n))$, so that $H = G-v$ satisfies the induction hypothesis and gives all of the required subgraphs.

This is where I am stuck, I cannot seem to find anything close to this required vertex. As $e(G) \geq t_r(n) + 1$ we know $K_{r+1} \leq G$, so there is a vertex $v$ with $d(v) \geq r$, which is not enough as $(q-1)r \leq \delta(T_r(v)) \leq qr-1$.

Perhaps useful I thought would also be the fact that if $|G| = n, e(G) = t_r(n), K_{r+1} \nleq G$ then $\delta(G) \leq \delta(T_r(v)) \leq \Delta(T_r(v)) \leq \Delta(G)$, this would give us a large enough vertex to delete, but we need $G$ to not contain $K_{r+1}$.

Likewise I did not get very far in the second part of the question, apart from the fact that a $K_{r+1}$ would exist in G in that case, but I cannot see how to get a second copy.

I noticed that someone has asked this identical problem a few years ago, but unfortunately, that one does not have any responses - which is why I have posted again as I do not have enough reputation to start a bounty without losing some rights on this site.

答:
a) We use induction on p. For a  base case of $p=n$, we are given that such a subgraph exists (since the entire graph is itself a subgraph satisfying the required conditions). Now assume that such a subgraph, $H$, exists on $l+1$ vertices. We note the identity: $t_r(l)=t_r(l+1)-\delta (T_r(l+1))$.

What we are looking for is a vertex in the subgraph $H$ which we can remove so that $H-v$ is a subgraph of G on $l$ vertices with sufficiently many edges.

Of course, using the identity, if there does exist a vertex, $v$, of degree less than or equal to $\delta (T_r(l+1))$, then we can just remove this vertex, and $H-v$ has $e(H-v)\geq t_r(l+1)+1-d(v)\geq t_r(l)+1$.

Otherwise, if no such vertex exists, then all vertices have degree greater than this. Now consider removing any vertex from the graph. So there are $l$ vertices, each with degree greater than $\delta (T_r(l+1))-1$ (the -1 comes from the fact each vertex could be adjacent to the removed vertex), and $\delta (T_r(l+1))-1=l+1-\lceil \frac{l+1}{r} \rceil -1 = l-\lceil \frac{l+1}{r} \rceil$, therefore $e(H-v) \geq \frac{l(l+1-\lceil \frac{l+1}{r} \rceil)}{2}$, which we can use as an upper bound for $t_r(l)+1$ by thinking about the structure of $T_r(n)$ (although care is needed for all the cases).

b) Once we have part a) this part should be fairly simple. Apply the lemma for p=r+2, that is, we have that there is a subgraph of G, H, on r+2 vertices with at least $t_r(r+2)+1$ edges. Intuitively what we should be thinking about is the fact that for n 'not much larger' than r, $T_r(n)$ is 'nearly' a complete graph. In fact, $T_r(r+2)$ is a complete r-partite graph with r-2 parts of size 1, and 2 parts of size 2 (from the definition of $T_r(n)$, so it is only 2 edges short of $K_{r+2}$, those 2 edges being, for each of the parts of size 2, the edge inbetween the 2 vertices. Therefore our subgraph H is either $K_{r+2}$ or $K_{r+2}-e$ for any edge (since we have $t_r(n)+1$ edges in H). Clearly if $H=K_{r+2}$, taking any two distinct sets of r+1 vertices, they both form a copy $K_{r+1}$ with r common vertices. In the case $H=K_{r+2}-e$, let the two endvertices of e be in distinct subsets of r+1 vertices, and, since e is the only missing egde, the two subset of r+1 vertices form two copies of $K_{r+1}$ with r common vertices.


问:
Indicator functions and vector spaces
The following is an exercise I have encountered in my linear algebra exercise book (I will not provide the reference as it is an Italian one). The solution was attached to it; still, as it is the first time I encounter the indicator function, I am unsure about its understanding.

Consider the vector space $ℝ^{[0,1]}$ and its subset $S=\{1_{[0,2^{-n}]} : n \in \mathbb N_0\}$, where $1_A$ is the indicator function of $A$.
Let $W = \operatorname{span}(S)$ and consider the function $g := 1_{(\frac12, 1]}$.

Question: Does $g$ belong to $W$?

Solution: Yes, indeed $g$ can be rewritten as $$g = 1_{[0,1]} - 1_{[0,\frac12]}$$

My reasoning: Since we have to rewrite $g$ as an element of $W$, the lower end of the interval where the indicator function is defined must be equal to $0$. To do so, we exploit the fact that the interval $[0,1]$ can be decomposed as $$[0,1] = [0,\tfrac12] \cup (\tfrac12,1].$$

Yet, I think I have an issue with the full understanding of how an indicator function works. For example, in our case, when is $g(x)$ equal to 1? When $x∈[0,1]$ or when $x∈[0,\tfrac12]$?

Thank you for your attention.

答:
As you are interested in how an indicator function works. I think the following clarification may help.

Definition. Let $A$ be a subset of $[0,1]$, then
the indicator function is defined as follows
\begin{align}
\mathbf 1_A(x)=
\begin{cases}
1, \text{ if } x\in A\\
0, \text{ if } x\notin A
\end{cases}
\end{align}

Back to your problem,
\begin{align}
g(x)=\mathbf 1_{(0.5,1]}(x)=
\begin{cases}
1, \text{ if } x\in (0.5,1]\\
0, \text{ if } x\in [0,0.5]
\end{cases}
\end{align}

We now prove that $g=\mathbf 1_{[0,1]}-\mathbf 1_{[0,0.5]}$. Indeed,

If $x\in (0.5,1]$, then $\mathbf 1_{[0,1]}(x)-\mathbf 1_{[0,0.5]}(x)=1-0=1=g(x)$.

If $x\in [0,0.5]$, then $\mathbf 1_{[0,1]}(x)-\mathbf 1_{[0,0.5]}(x)=1-1=0=g(x)$.

My comments.

Regarding the notation $\mathbb R^{[0,1]}$. It is well known that $\mathbb R^n$ is the vector space in which each vector has $n$ components. Similarly, $\mathbb R^{[0,1]}$ is the vector space in which each vector has "$[0,1]$ components", in other words, $\mathbb R^{[0,1]}$ is the space of function over $[0,1]$.

Regarding your resoning. To prove that $g\in span(S)$, you should rewrite $g$ as a combination of elements in $S$. Notice that $S$ contains the set of the form $[0,2^{-n}]$, thus you should use
\begin{align}
[0,1]\setminus [0,0.5]=(0.5,1]
\end{align}
and then we have
\begin{align}
\mathbf 1_{[0,1]}-\mathbf 1_{[0,0.5]}=\mathbf 1_{(0.5,1]}=g.
\end{align}

Title: Indicator functions and vector spaces


问:
How do I show $T:V \rightarrow W$ there are bases $B$ of $V$ and $B'$ of $W$ such that $A_{T,B,B'} \in S$?
Let $V$ and $W$ be the real vector spaces, and let $T: V \rightarrow W$ be a linear map. If the dimension of $V$ and $W$ are 3 and 5, respectively, then for any bases $B$ of $V$ and $B'$ of $W$, we can represent $T$ by a $5 \times 3$ matrix $A_{T,B,B'}$.

Find a set $S$ of $5 \times 3 $ matrices as small as possible such that for any $T:V \rightarrow W$ there are bases $B$ of $V$ and $B'$ of $W$ such that $A_{T,B,B'} \in S$.

Can anyone suggest some direction, how do I find the set $S$?

答:
Hint: Gaussian elimination. Any operation you perform on rows and columns is allowed since you can change the basis on $V$ and on $W$. The question then becomes, if you can do operations on both rows and columns, what are the possible end results.

Second hint: Rank is relevant here.

Hope that helps,


问:
Bus arrival density function
```<code>Fred lives in Blissville, where buses always arrive exactly on time, with the time
between successive buses fixed at 10 minutes. Having lost his watch, he arrives at
the bus stop at a random time (assume that buses run 24 hours a day, and that the
time that Fred arrives is uniformly random on a particular day).

 **What is the distribution of how long Fred has to wait for the next bus? What is
the average time that Fred has to wait, and what is the density function (PDF) or his waiting time till the next bus?
</code>```

So the answer is:
$E[X] = 5$
and the PDF is $\frac{1}{10}$.

Can you explain me why?
Why do I need to get the average between 0 and 10 for the $E[X]$ and why the pdf is $\frac{1}{10}$?

I mean I want to understand the inside of the calculation.. the lambda here is 1/10 but other than that.. I know that PDF is not just the lambda so how can be it 1/10?

Thanks.

答:
The buses arrive always on time every 10' thus the distribution is uniform in $[0;10]$
The density of a continuous uniform is

$$f(x)=\frac{1}{b-a}=\frac{1}{10-0}=\frac{1}{10}$$

Its expectation ( the average time he has to wait) is

$$E(X)=\frac{a+b}{2}=\frac{0+10}{2}=5$$

Title: Bus arrival density function


问:
What does mean $df/dg$ (derivative of a function with relation to another)?
I have on a textbook a system of two ODEs as follow

$$\dfrac{df}{dx}=af(x)g(x)\\
\dfrac{dg}{dx}=-bg(x)$$

So, the author studies $df/dg$ writing $\dfrac{df}{dg}=-\dfrac{a}{b}f(x)$.

I understand the arithmetic, but I am not sure of mathemathical/geometric meaning.

Thank you in advance for light and maybe some lemmas or theorems of reference that it is valuable (are when is valuable).

答:
I have ranted numerous times on this site about that horrendous notation. It really means nothing and can be very misleading. I would suggest you write
$$\frac{df}{dx} = af(x)g(x) = -\frac ab f(x)\frac{dg}{dx},$$
and then divide by $f(x)$ to get
$$\frac1{f(x)}\frac{df}{dx} = -\frac ab \frac{dg}{dx}.$$
Now integrate both sides (with respect to $x$) and you get
$$\ln |f(x)| = -\frac ab g(x) + c.$$
[You could write the previous equation as $\frac{df}f = -\frac ab dg$, and then you get where your textbook should end up.]


答:
Suppose $P$ has degree $D\ge1$

Differentiate $D-1$ times.

$$P(x)=P(x+1) \implies \frac{d^{D-1} P(x)}{dx^{D-1}}=\frac{d^{D-1} P(x+1)}{dx^{D-1}} $$

Both sides are linear polynomials. Let $P^{(D-1)}(x) = ax+b $. Then, $$ax+b = a(x+1)+ b \implies a=0$$

That’s a contradiction. This means D must be zero.


答:
Let $ n\in\mathbb{N} $, we have for any $ k\in\mathbb{N} $ :$$ P\left(k+1\right)-P\left(k\right)=0\Longrightarrow\sum_{k=0}^{n-1}{\left(P\left(k+1\right)-P\left(k\right)\right)}=0\iff P\left(n\right)=P\left(0\right) $$

Since $ n $ is arbitrary, $ \left(\forall n\in\mathbb{N}\right),\ P\left(n\right)=P\left(0\right) $.

Consider a polynomial function $ Q:x\mapsto P\left(x\right)-P\left(0\right) $.

$ Q $ has an infinite number of zeros, In fact $ \mathbb{N}\subset Q^{-1}{\left(\left\lbrace 0\right\rbrace\right)} $.

Thus : $$ Q=0 $$

Which means : $$ \left(\forall x\in\mathbb{R}\right),\ P\left(x\right)=P\left(0\right) $$