问:
Confusion about Suprema Properties and Spivak's Proof of the Intermediate Value Theorem
If it's possible, I'm wondering if someone can clarify the following for me as part of the proof of the Intermediate Value Theorem by Spivak in the 4th edition of his Calculus (proof and auxiliary theorem given at the bottom of the post). The line I am focusing on is "there is some number $x_0$ in $A$ which satisfies $\alpha−\delta<x_0<\alpha$ (because otherwise $\alpha$ would not be the least upper bound of $A$)." In particular, I am not entirely convinced that I am internalizing the part in brackets.

My question is, why does the fact that $x_0<\alpha$ (where $\alpha$ is of course the supremum of the set, which exists as established earlier in the proof) necessarily require that $x_0$ is in $A$?

Just for example, $a-1 < \alpha$, but $a-1$ is not in $A$. Now this is a cheeky example, since it can (or so it seems to me) easily be established that we can find a $\delta$ so that this $x_0$ is between $a$ and $\alpha$. But even then, I feel like I am missing something. Why must this $x_0$ be in $A$ based on the definition of suprema? The proof, and the part I am confused about, intuitively make sense to me based on my hazy characterization of the supremum of a set as having the property that any number less than it (and greater than some other element in the set - here $a$ for example is known to be in the set and $x_0$ is greater than $a$) should be in the set, but that seems to me to roughly rest on the notion that the set contains all numbers (again, and unfortunately, roughly speaking) between that number known to be in the set and the supremum.

This question is admittedly closely related to this one, but I thought it was sufficiently different because it seems I am more confused about properties of suprema than about the proof as a whole.

Theorem 7-1 (Intermediate Value Theorem):
If $f$ is continuous on $[a,b]$ and $f(a) < 0 < f(b)$, then there is some number $z$ in $[a, b]$ such that $f(x)=0$.

Proof: Define the set $A$ as follows:

$$A=\{x : a \le x \le b, \text{ and } f \text{ is negative on the interval } [a,x]\}.$$
Clearly $A \neq \varnothing$, since $a$ is in $A$; in fact, there is some $\delta>0$ such that $A$ contains all points $x$ satisfying $a \le x < a + \delta$; this follows from Problem 6-16, since $f$ is continuous on $[a,b]$ and $f(a)<0$. Similarly, $b$ is an upper bound for $A$ and, in fact, there is a $\delta > 0$ such that all points $x$ satisfying $b - \delta < x \le b$ are upper bounds for $A$; this also follows from Problem 6-16, since $f(b)>0$.

From these remarks, it follows that $A$ has a least upper bound $\alpha$ and that $a < \alpha < b$. We now wish to show that $f(\alpha) = 0$, by eliminating the possibilities $f(\alpha) < 0$ and $f(\alpha) > 0$.

Suppose first that $f(\alpha)<0$. By Theorem 6-3, there is a $\delta>0$ such that $f(x)<0$ for $\alpha - \delta< x < \alpha + \delta$. Now there is some number $x_0$ in $A$ which satisfies $\alpha - \delta< x_0 < \alpha$ (because otherwise, $\alpha$ would not be the least upper bound of $A$).

This means that $f$ is negative on the whole interval $[a,x_0]$. But if $x_1$ is a number between $\alpha$ and $\alpha+\delta$, then $f$ is also negative on the whole interval $[x_0,x_1]$. Therefore $f$ is negative on the interval $[a,x_1]$, so $x_1$ is in $A$. But this contradicts the fact that $\alpha$ is an upper bound for $A$; our original assumption that $f(\alpha)<0$ must be false.

In doing the proof, Spivak uses Theorem 6-3, given below:

Theorem 6-3:
Suppose $f$ is continuous at $a$, and $f(a)>0$. Then $f(x)>0$ for all $x$ in some interval containing $a$; more precisely, there is a number $\delta > 0$ such that $f(x)>0$ for all $x$ satisfying $|x−a|<\delta$. Similarly, if $f(a)<0$, then there is a number $\delta>0$ such that $f(x)<0$ for all $x$ satisfying $|x−a|<\delta$.

答:
This question is probably no longer active but...

Another way of seeing why all $x< \alpha$ in the interval must be in $A$ is as follows:

We've defined the set A as:

$$A=\{x : a \le x \le b, \text{ and } f \text{ is negative on the interval } [a,x]\}.$$

Note especially the requirement "$f$ is negative on the interval $[a,x]$".

Suppose there exists some $x_0$ in $[a,\alpha)$ such that $f(x_0)$ is not negative.

This $x_0$ will not be in $A$. Furthermore, no $x > x_0$ can be in $A$, because the "interruption" at $x_0$ means any intervals that contain $x_0$ will not be negative over the whole interval $[a,x]$.

Therefore, $x_0$ is an upper bound of $A$. But $x_0$ is less than $\alpha$, a contradiction!

Because of the way we've defined $A$, any $x$ not in $A$ instantly becomes an upper bound on $A$.

Every point in the interval less than $\sup A$ must be in $A$.

This is not true of all bound sets. It's a consequence of the way this particular set is defined.


问:
Evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n F_{2n}}{9^n},$ where $F_{2n}$ is the nth Fibonacci number
Evaluate $\sum_{n=0}^{\infty} \frac{(-1)^n F_{2n}}{9^n},$ where $F_{n}$ is the $n^{th}$ fibnocci number

My idea:

if $a(n)$ is an eventually monotone decreasing sequence, converging to 0, here $a(n) = \frac{ F_{2n}}{9^n},$

but how do we find sum of this series, thank you

答:
Using https://artofproblemsolving.com/wiki/index.php/Binet%27s_Formula

The summation, $$=(-1/9)^n\frac{a^{2n}-b^{2n}}{a-b}=\dfrac{\sum(-a^2/9)^n-\sum(-b^2/9)^n}{a-b}$$

where $a,b$ are the roots of $$t^2-t-1=0$$

Now both are https://en.m.wikipedia.org/wiki/Geometric_progression#Infinite_geometric_series

ContentLicense: CC BY-SA 4.0


问:
Jacobian for polar decomposition
Let $f:\mathrm{Mat}_n(\mathbb{C})\to\mathbb{C}$ be some function and let us suppose we want to make a change of variables in the integral $$ \int_{A\in \mathrm{Mat}_n(\mathbb{C})}f(A)\mathrm{d}{A} $$ from $A$ to $|A| U$, i.e., the polar decomposition of $A$, where $|A|\equiv\sqrt{A^\ast A}$ and $U$ is the unique partial isometry with kernel equal to that of $A$ (there is a theorem saying it exists).

What is the Jacobian matrix of the transformation $A \mapsto (|A|, U$)? I.e., what is $J$ such that the following equation holds:
$$ \int_{A\in \mathrm{Mat}_n(\mathbb{C})}f(A)\mathrm{d}{A} = \int_{P\geq0,U^\ast U\,\mathrm{idempotent}}f(P U)|\det(J(P,U))|\mathrm{d}{P}\mathrm{d}{U}$$

I tried to calcualte it but I'm not getting anything simple. In particular, I've written $A = A_R + i A_I$ with $A_R = \frac{1}{2}(A+A^\ast); A_I = \frac{1}{2i}(A-A^\ast)$ so that $A$ is parametrized by two self-adjoint matrices. In turn, we may write $|A| = \exp(H_1) ; U = \exp(i H_2)$ for two self-adjoint matrices $H_1,H_2$ (assuming for a moment that $A$ is invertible so that $U$ is actually unitary). Hence we want to calculate the Jacobian of the transformation $(H_1,H_2)\mapsto(A_R,A_I)$ from $\mathrm{Herm}_n(\mathbb{C})^2\to \mathrm{Herm}_n(\mathbb{C})^2$.

This, however, starts to get ugly, with the differential of the exponential map for example being given by functional calculus of the adjoint super operator (https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map) and having to use the determinant of a block matrix formula.

Is there an easier way out?

Possible solution:

In Edelman's PhD thesis there are given Jacobians to get from a matrix A to its LQ decomposition, and from its LQ decomposition to its Cholesky decomposition (Theorem 3.1). This possibly solves the problem as follows:

\begin{align} \int_{A\in \mathrm{Mat}_n(\mathbb{C})}f(A)\mathrm{d}{A} &= \int_{L\text{ lower triangular},\,U\text{ unitary}} f(LU)\prod_{i=1}^{n}L_{ii}^{2n-2i+1}\mathrm{d}{L}\mathrm{d}{U} \\ &=2^{-n}\int_{P\geq0,\,U\text{ unitary}} f(\sqrt{P}U)\mathrm{d}{P}\mathrm{d}{U}\\&=2^{-n}\int_{P\geq0,\,U\text{ unitary}} f(PU)|\det(P\otimes I+I\otimes P^\ast)|^2\mathrm{d}{P}\mathrm{d}{U}\\&=2^{-n}\int_{P\geq0,\,U\text{ unitary}} f(PU)\prod_{1\leq i,j\leq n}(\lambda_i(P)+\lambda_j(P))^2\mathrm{d}{P}\mathrm{d}{U}\end{align}

with the usual abuse of notation that $\mathrm{d}{L}$ integrates only over the non-zero elements of $L$, $\mathrm{d}{U}$ is the volume element within the unitary group, and $\mathrm{d}{P}$ the volume element on self-adjoint matrices (so only $n$ real and $\frac{1}{2}n(n-1)$ complex matrix elements). $\lambda_j(P)$ is the $j$th eigenvalue of the matrix $P$.

Remaining question: Why is the LQ-decomposition change of variables valid for complex matrices? A complex unitary $n\times n$ matrix is $n^2$ real parameters, whereas a lower triangular matrix is $n(n+1)$ real parameters. On the other hand, a complex matrix is $2n^2$ real parameters, so there seem to be $n$ real parameters too many in this decomposition? (This is not a problem if matrices have real entries). Note that for the Cholesky decomposition this is not an issue since then the lower triangular matrix has positive entries on its diagonal.

Could it be possible to make an LQ decomposition for complex matrices where the lower triangular has positive entries on the diagonal? Is this what Edelman is referring to?

Unfortunately, precisely for the complex LQ decomposition he does not give a reference nor a proof.

答:
This is a little late, but for anyone who might still be interested in his Jacobian, see:

Cadet A. Polar coordinates in Rnp; application to the computation of the Wishart and beta laws. Sankhya: The Indian Journal of Statistics Series A. 1996;58(1):101–114; https://www.jstor.org/stable/25051086

Also, see the paper below for Jacobians of other matrix decompositions and extensions to  singular matrices:

José A.Diaz-Garcia, Graciela Gonzalez-Farias, Singular random matrix decompositions: Jacobians, Journal of Multivariate Analysis, Volume 93, Issue 2, April 2005, Pages 296-312; https://doi.org/10.1016/j.jmva.2004.03.002

Title: Jacobian for polar decomposition


问:
A completely regular topological space which is $T_0$ but not $T_1$?
The question pretty much says it all.  I need a completely regular (the definition not requiring $T_1$) topological space which is $T_0$ but not $T_1$.  I've sifted through Counterexamples in Topology but I just can't find one.  I've also tried google and searching this site but I'm not coming up with anything.

It seems the counterexample should be simple but everything I try doesn't work (like Sierpinski topology, open extension topology, topology on $\mathbb{R}$ whose open sets are $(a,+\infty)$, etc.).

EDIT: By a completely regular topological space I mean a topological space $(X,\tau)$ so that for every $x\in X$ and closed $\mathscr{C}\subseteq X$ such that $x\not\in\mathscr{C}$ there exists a continuous
function $f:X\to[0,1]$ so that $f(x)=0$ and $f(\mathscr{C})=\{1\}$.

By a $T_0$ space I mean a topological space in which any two distinct points are topologically distinguishable.

By a $T_1$ space I mean a space in which for any two distinct points
both of them are contained in neighborhoods which do not contain the other.

I was thinking completely regular + $T_0$ did not imply $T_1$ since the definitions of Tychonoff/$T_{3\frac{1}{2}}$ spaces I've seen require completely regular and $T_1$ rather than completely regular and $T_0$.

EDIT EDIT: Turns out no such counterexample exists.

答:
If $X$ is completely regular (closed set and point can be separated by a continuous real-valued function), then it is also regular. Let $x\in X$ be a point and $A\subset X$ be a closed set not containing $x$, then there is a continuous function $f:X\to\Bbb R$ with $f(x)=0$, $f(A)=1$, so $f^{-1}([0,\frac12))$ and $f^{-1}((\frac12,1])$ are disjoint open neighborhoods of $x$ and $A$, respectively.

To show that a regular $T_0$-space $X$ is $T_2$, let $x,y\in X$. There is a closed set $A$ containing one point, say $y$, but not the other. By regularity, there are disjoint open sets $U\ni x$ and $V\supseteq A\ni y$. So $X$ is Hausdorff.

ContentLicense: CC BY-SA 3.0


问:
How to figure out this Polar Equation to find out domain of Θ values to get the entire graph without repeating?
I understand that the period for the graph 4 + 2sin(2Θ/3) is 3π. The way I got that was by 2π/b. b is 2/3. Therefore I did 2π/(2/3) and got 3π for the period. Can someone guide me to solve part b? I need to get the domain values of Θ to get the whole graph without repeating any part of the graph. I posted a picture of the graph below and part b in case I didn't explain it clearly.

答:
For the graph to repeat, $r$ needs to repeat, as well as the angle

The angle repeats when $\theta$ changes by $2\pi$

$r$ repeats when $\theta$ changes by $3\pi$ as you have found. So you need the $\theta$ that is the smallest multiple of both $2\pi$ and $3\pi$, which is $6\pi$. So $6\pi$ is the answer.

You can also see that the graph makes 3 loops around the origin hinting that it repeats after $3\times2\pi$


问:
$P_n$ is a.s. to P, $Q_n$ is L1 to Q. $P(P_n \neq Q_n) \leq 1/n$. Is $P=Q$ a.s.?
$\left\{P_{n}\right\}_{n=1,2, \ldots}$ and $\left\{Q_{n}\right\}_{n=1,2, \ldots}$ are two sequences of r.v.s. For every $n$ , $\mathbb{P}\left(P_{n} \neq Q_{n}\right) \leq 1 / n . P_{n} \stackrel{\text {a.s.}}{\rightarrow} P$ and $Q_{n} \stackrel{L^{1}}{\rightarrow} Q$. Can we conclude that $P=Q$ a.s.?

Is it right?

答:
Let $\epsilon > 0$ be given. Note that
$$\mathbb P(|P-Q| > \epsilon) \leq \mathbb P(|P - P_n| > \epsilon/3) + \mathbb P(|P_n - Q_n| > \epsilon /3) + \mathbb P(|Q_n - Q| > \epsilon/3).$$
Letting $n \to \infty$, the right-hand side tends to $0$ (because both a.s. and $L^1$ convergence imply convergence in probability). This implies that $P=Q$ a.s.


问:
Interpreting a question about finite combinations of arithmetic operations
"Let $S$ be the set of numbers obtained from $\sqrt{2}$ and $\sqrt{3}$ by finitely many arithmetic operations ($+$, $-$, $\times$, $\div$). Show that $S$ is countable."

I can't figure out the definition of $S$. Is it true that $\{\sqrt{2}+\sqrt{3}, \sqrt{2}+\sqrt{3}\times\sqrt{2}, \sqrt{2} \div \sqrt{3}+\sqrt{3}\times\sqrt{2} \}\subseteq S$?

答: Yes, all of those would be in $S$.  Here's a way you could define $S$ more precisely.  Recursively define the set $S_n$ of numbers obtained from $\sqrt{2}$ and $\sqrt{3}$ by $n$ operations as follows.  Let $S_0=\{\sqrt{2},\sqrt{3}\}$.  Given $S_n$, define $S_{n+1}$ as the set of all numbers of the form $a+b,a-b,a\times b,$ or $a\div b$ for $a,b\in S_n$.  Then, define $S=\bigcup_{n\in\mathbb{N}}S_n$.


问:
Prove $2^{\lceil \lg (n-1) \rceil} \ge n-1$
The inequality is from "Introduction to algorithm" of CLRS, section 25.1. What I tried so far is to express $n-1 = e^{\ln(n-1)}$; consider the case when $n-1 = 2^m$ for some integer m; assumed for the purpose of contradiction that the opposite is true. Could someone help?

Thanks.

Update: the log in the question is log of base 2, not e.

答:
Note that $2^{\lg(n-1)} = n-1$ for all $n > 1$. (Here, we use the base $2$ logarithm $\lg$ instead of $\ln$, as using the natural logarithm would lead the inequality being false for all sufficiently large $n$. This usage is probably what the problem implied.) Then, since $\lceil x\rceil \geq x$ for all $x$, and $2^{x}$ is an increasing function for all $x$, we have:

$$2^{\lceil\lg(n-1)\rceil}\geq 2^{\lg(n-1)} = n-1\ \blacksquare$$


问:
Naturality of the lie bracket.
I heard the term natural transformation for functors. Does the naturality of the lie bracket has something to do with that?

Naturality of the lie bracket : $F_*[V_1,V_2]=[F_*V_1,F_*V_2]$ where the star denotes the pushforward and $V_1$, $V_2$ are vector fields.

答:
Yeah, these ideas can be expressed in the language of category theory.
For instance, the fact that the star is a subscript (instead of a superscript) on $F_*$ indicates that there is a covariant functor (instead of a contravariant functor) behind the scenes. To be precise, the naturality you seek involves the category Diff with smooth manifolds as objects and diffeomorphisms as morphisms, as well as the category Vect$_\mathbf{R}$ with real vector spaces as objects and linear maps as morphisms.

In this language, $\mathfrak{X}$ is a functor from Diff to Vect$_\mathbf{R}$ sending diffeomorphisms $F\colon M\to N$ of smooth manifolds to linear maps $F_*\colon\mathfrak{X}(M)\to\mathfrak{X}(N)$, where $\mathfrak{X}(M)$ denotes the space of smooth vector fields on $M$. That is, $\mathfrak{X}\colon F\mapsto F_*$.

Similarly, we may define a functor $\mathfrak{X}\times\mathfrak{X}$ from Diff to Vect$_\mathbf{R}$ by sending $F$ to $F_*\times F_*$, where
$(F_*\times F_*)\colon\mathfrak{X}(M)\times\mathfrak{X}(M)\to\mathfrak{X}(N)\times\mathfrak{X}(N)$ is defined as one may expect — namely by sending $(X,Y)$ to $(F_*X,F_*Y)$.

The Lie bracket $[\cdot,\cdot]\colon\mathfrak{X}\times\mathfrak{X}\to\mathfrak{X}$ is then a natural transformation from the functor $\mathfrak{X}\times\mathfrak{X}$ to the functor $\mathfrak{X}$, as the identity $F_*[X,Y]=[F_*X,F_*Y]$ is precisely the statement that the diagram

$$\require{AMScd}\begin{CD}\mathfrak{X}(M)\times \mathfrak{X}(M)     @>F_*\times F_*>>  \mathfrak{X}(N)\times \mathfrak{X}(N)\\@VV[\cdot,\cdot]V        @VV[\cdot,\cdot]V\\\mathfrak{X}(M)     @>F_*>>  \mathfrak{X}(N)\end{CD}$$

commutes.

Title: Naturality of the lie bracket.


问:
Totally decomposable vectors as quotient of monomorphisms by SL
Consider vector spaces $V$ and $W$ of dimensions $m$ and $n\geq m$, respectively.
Both the special linear group $\operatorname{SL}(V)$ as well as the general linear group $\operatorname{GL}(V)$ act smoothly and freely from the right on the smooth manifold of monomorphism $\operatorname{Mono}(V;W)$ by precomposition. Denote the space of totally decomposable $m$-vectors in $W$ by $\bigwedge^m_sW$.

Fix a basis $e_1,\ldots,e_m$ of $V$ and define the surjective map
$$\pi\colon \operatorname{Mono}(V;W)\to \bigwedge\nolimits^m_sW\setminus\{0\}, A\mapsto Ae_1 \wedge \cdots \wedge Ae_m.$$
I can show that the map
$\Phi\colon \operatorname{Mono}(V;W)/\operatorname{SL}(V) \to \bigwedge^m_sW\setminus\{0\}$ defined by $\Phi([A]_{\operatorname{SL}}) = \pi(A)$ is well-defined and bijective.
Analoguously, I can show that $\Psi\colon \operatorname{Mono}(V;W)/\operatorname{GL}(V) \to \bigwedge^m_sW\setminus\{0\}$ defined by $\Phi([A]_{\operatorname{GL}}) = \{t \pi(A)\colon t\in\mathbb{R}^{\times}\}$ is well-defined and bijective.
Hence the following diagram commutes:

My main issue is whether $\Phi$ and $\Psi$ are diffeomorphisms. To even start to show this each space should be a smooth manifold.
I know that by the Plücker embedding $\mathbb{P}(\bigwedge^m_sW\setminus\{0\})$ is the Grassmannian and thus a compact smooth manifold of dimension $m(n-m)$.
The quotient on the bottom right also should have dimension $nm-m^2=m(n-m)$ but I'm unsure how to show that it's a smooth manifold. If $\operatorname{GL}$ acted properly then the quotient manifold theorem would be my friend. Is this the case?

The quotient on the middle right should have dimension $nm-(m^2-1)=m(n-m)+1$. The totally decomposable $m$-vectors are a projective variety of "non-projective" dimension $m(n-m)+1$. So these arguments at least indicate to me, thankfully, that my bijectivity proofs are not bogus.
I'm unsure if the spaces on the middle level are actually manifolds. On this level, also, the map $\pi$ does depend on the chosen basis whereas (an analogous map) on the bottom level does not. Maybe another indication that the bottom-level compactification makes things "better"?

What "regularity" of $\Phi$ could I even expect? I want to show that the totally decomposable vectors are isomorphic to the middle right quotient in an appropriate category.

答:
Let $U$ be the tautological (rank-$m$) vector bundle over the Grassmannian $\mathrm{Gr}(m,W)$ and let
$$
P := \mathrm{Iso}(V,U)
$$
be the corresponding principal $\mathrm{GL}(V)$-bundle over $\mathrm{Gr}(m,W)$. Then
$$
P \cong \mathrm{Mono}(V;W).
$$
Indeed, the map from left to right is given by taking an isomorphism $V \to U$ to the composition $V \to U \to W$, where the second arrow is the tautological embedding, and it is easy to see this map is an isomorphism and, moreover, it is $\mathrm{GL}(V)$-equivariant. Taking quotients by $\mathrm{GL}(V)$ and $\mathrm{SL}(V)$, you can obtain the two isomorphism you are interested in.


问:
Use the pumping lemma to show that the language is not context-free
I want to show that this language is not context-free:

I struggle with the pumping lemma for CFL and i would like to know how to solve this problem. Thank you

(skipping the classic introductory text of the proof)

During my proof i chose the word $z = a^pb^pc^{p-1}$ ( correct me if this isn't a good choice)

The word z can be cut as $sv^iuw^it$

Let's take i = 2

From there i know that we have to explore all cases:

Case 1: v and w are only made of a's so we have $a^{p+|vw|}b^pc^{p-1}$ so this isn't in the language because $|vw| >= 1$

Case 2: it's the same thinking but v w are only made of b's

Case 3: v and w are only made of c's, if $i=2$ then we have $v^2$and $w^2$ so $|vw|=4$ so we have $a^pb^pc^{p-1+|vw|}$ = $a^pb^pc^{p-1+4}$ which mean here that it won't be in the language since we need $m>=n$

From there ( if everything yet makes sens) i am not too sure on how to continue this problem, could anyone help out ? Thanks a lot.

答:
The tricky cases are when there is an $r\le\frac{p}2$ such that $v=a^r$ and $w=b^r$, because then if you make $i>1$, $sv^iuw^it$ is still in $L$. However, the pumping lemma says that $sv^iuw^it\in L$ for all $i\ge 0$, so you can pump down to $i=0$, so that $sv^iuw^it=a^{p-r}b^{p-r}c^{p-1}$. If $r>1$, this word is not in $L$, because $p-r<p-1$. Unfortunately, if $r=1$ the word is $a^{p-1}b^{p-1}c^{p-1}$, which is in $L$. The problem here is with your choice of $z$: you should have let $z=a^pb^pc^p$. This is in $L$, since the definition of $L$ requires only that $m\ge n$, not that $m>n$, and now everything works, both the cases that you’d already considered and the one that I just did.

The remaining possibilities are:

$v=b^r$ for some $r\ge 1$ and $w=c^s$ for some $s\ge 1$, in which case pumping up or down will give you unequal numbers of $a$s and $b$s and hence a word not in $L$;

$v=a^rb^s$ for some $r,s\ge 1$, in which case $w$ cannot contain any $c$s (because $|vuw|\le p$), so pumping down will drop the number of $a$s below the number of $c$s and give you a word not in $L$; and

$v=b^rc^s$ for some $r,s\ge 1$, in which case $w$ contains only $c$s, and pumping up or down gives you unequal numbers of $a$s and $b$s and hence a word not in $L$.

In every case, therefore, we can pump $z$ to get a word not in $L$, and the pumping lemma then implies that $L$ is not context free.


问:
Find the probability that a 4 will appear on the second dice.
Two dice are rolled. If the sum of the two dice is 7 or less than 7. Find the probability that a 4 will appear on the second dice.

I am trying to do this exercise. But my result is different from the one provided by the book. In the book it appears that the answer is $1/4$. This is what I have done. It's okay?

Let $P(A)$ be The probability that the sum is equal to 7 or less than $7$ and let $P(B)$ be the probability that a 4 appears on the second die, then

$P(A)=\frac{21}{36}=\frac{7}{12}$

$P(B)=\frac{6}{36}=\frac{1}{6}$

So $P(A\cap B)=\frac{3}{36}=\frac{1}{12}$

Thus $\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{12}}{\frac{7}{12}}=\frac{1}{7}$

If I try $P(A|B)$ and I get  $1/2$

答: Using your table there are $21$ rolls $7$ or less of which $3$, ($(1,4),(2,4),(3,4)$) have $4$ as the second die.  Net is $\frac{1}{7}$.


问:
Proving that a 5-regular graph with ten vertices is non planar
How can we prove that a 5-regular graph with ten vertices is non planar? Do we use $E \leq 3V-6$?</p>

答: Hint:  What is a regular graph? How many edges are there? You need the handshaking lemma.</p>


答:
Your working is correct.

With one of the dice being $1$, you can have max sum of $7$ with total of $6 \times 2 - 1 = 11$ possibilities. (Numbers can interchange places on dice and subtract one as we did count $(1, 1)$ twice.)

Now with $2$, there are $10$ possibilities (but subtract 2 possibilities for $(1,2)$ which is covered and one for $(2, 2)$). Now you have $7$ possibilities.

With $3$, $3$ possibilities.

No point in checking with $4$, as all the possible cases summing to $7$ or less have been covered already.

So total of $21$ possibilities to get sum of $7$ or less on two dice.

$4$ on the second dice with sum being less than equal to $7$ - $(1, 4), (2, 4), (3, 4) \,$ - 3 possibilities.

So $\frac{1}{7}$ is correct.

Body:
Two dice are rolled. If the sum of the two dice is 7 or less than 7. Find the probability that a 4 will appear on the second dice.

I am trying to do this exercise. But my result is different from the one provided by the book. In the book it appears that the answer is $1/4$. This is what I have done. It's okay?

Let $P(A)$ be The probability that the sum is equal to 7 or less than $7$ and let $P(B)$ be the probability that a 4 appears on the second die, then

$P(A)=\frac{21}{36}=\frac{7}{12}$

$P(B)=\frac{6}{36}=\frac{1}{6}$

So $P(A\cap B)=\frac{3}{36}=\frac{1}{12}$

Thus $\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{12}}{\frac{7}{12}}=\frac{1}{7}$

If I try $P(A|B)$ and I get  $1/2$


问:
Why is $\frac{(x-3)}{0} = \frac{(y+2)}{0} = \frac{(z+5)}{11}$ a valid equation for a line in space when division by zero is undefined?
Oh yes, I do know division by zero is not defined.
But look at this problem:

Find the Cartesian equation of the line that passes through $(3,-2,-5)$ and $(3,-2,6)$.

And the answer is conveniently stated to be:
$$\frac{(x-3)}{0} = \frac{(y+2)}{0} = \frac{(z+5)}{11}$$

What gives us the liberty to divide by zero?

Insert a value of $x$; $2$ for example. $2$ divided by zero is infinity. But $\frac{(z+5)}{11}$ will give us a real value!

Please help me clear this silly confusion.

答:
The more precise formula for the line that passes through $(a,b,c)$ and $(d,e,f)$ is that it is given by the three equations
\begin{align*}
(x-a)(e-b)&=(y-b)(d-a)\\
(x-a)(f-c)&=(z-c)(d-a)\\
(y-b)(f-c)&=(z-c)(e-b).
\end{align*}
When $d-a$ and $e-b$ and $f-c$ are all nonzero, these can be combined into the single pair of equations
$$
\frac{x-a}{d-a} = \frac{y-b}{e-b} = \frac{z-c}{f-c}.
$$
However, that convenience comes at a cost of loss of generality, as you've noted. Nevertheless, the formula above does give the correct answer here:
\begin{align*}
(x-3)((-2)-(-2))&=(y-(-2))(3-3)\\
(x-3)(6-(-5))&=(z-(-5))(3-3)\\
(y-(-2))(6-(-5))&=(z-(-5))((-2)-(-2)),
\end{align*}
which simplifies to the correct equations for this line:
\begin{align*}
0&=0 \\
x&=3 \\
y&=-2.
\end{align*}


问:
Finding the Equation of Parabola

Write the equation in the form $y=a(x-h)^{2}+k$ with zeros -4 and 8, and an optimal value of 18.

I'm not sure what "optimal value" means first of all- I think it means that the maximum value has a y-value of 18. What I've done so far:

$y=a(x+4)(x-8)$. Then to calculate the x-value of the vertex: $\frac{-4+8}{2}=2$ and then you substitute $x=2$ into the original equation to get the y-value of the vertex, which is:

$y=(2+4)(2-8)\implies y=-36$. Then $y=a(x-2)^{2}-36$, and then since we know a point on the line, I subbed in $(-4,0)$, which means that $0=a(-4-2)^{2}-36\implies 0=36a-36\implies a=1.$ So what I'm getting is $y=-(x-2)^{2}-36$

Title: Finding the Equation of Parabola

Body:
Use factored form: $y=a(x+s)(x+t)$

$y=a(x+4)(x-8)$

'Optimal Value'= $y$ value of $(x,y)$

$18$ = $y$ value of $(x,y)$

$x$ value of $(x,y)= (s+t)/2
                = (-4+8)/2
                = 4/2
                = 2$

Therefore $(x,y) = (2,18)$

Substitute $(2,18)$ into $y=a(x+4)(x-8)$

$18=a(2+4)(2-8)$

$18=a(6)(-6)$

$18=a(-36)$

$-36a=18$

$a=18/-36$

$a=1/-2$

Therefore the equation of the parabola is $y= 1/-2(x+4)(x-8)$


问:
Set of infinite tuples is closed, bounded, non-compact

Let $\mathbb{R}^{\infty}$ be the set of all "infinite-tuples" $x=(x_1,x_2,\ldots)$ of real numbers that end in an infinite string of $0$'s. Define an inner product on $\mathbb{R}^{\infty}$ by the rule $(x,y)=\sum x_iy_i$. Let $\|x-y\|$ be the corresponding metric on $\mathbb{R}^{\infty}$. Define $e_i=(0,\ldots,0,1,0,\ldots,0,\ldots)$, where $1$ appears in the $i$th place. Let $X$ be the set of all the points $e_i$. Show that $X$ is closed, bounded, and non-compact.

$X$ is bounded: Clear, since $\|e_i\|=1$ for all $i$.

$X$ is closed: Suppose that $y=(y_1,y_2,\ldots)$ is a limit point of $X$, and let $y_n=y_{n+1}=\ldots=0$. Then for $i\geq n$, we have $$\|e_i-y\| = \sqrt{\sum_j ((e_i)_j-y_j)^2}\geq \sqrt{((e_i)_i-y_i)^2} = 1.$$ Let $d=\min(\|e_1-y\|,\|e_2-y\|,\ldots,\|e_{n-1}-y\|,0.5)$. If $d>0$, the ball of radius $d/2$ centered at $y$ doesn't contain any element of $X$, and so $y$ is not a limit point. So $d=0$, and $y=e_i$ for some $i$. Hence $X$ is closed.

$X$ is not compact: We must show that there exists an open collection covering $\{A_{\alpha}\}$ of $X$ for which no finite subcollection covers $X$. What collection might we choose here?

答:
Assume that $X$ is compact.

Any element of $X$ is an isolated point of $X$.

So, any function $f:X\to\mathbb{R}$ is a continuous function.

So, any function $f:X\to\mathbb{R}$ takes a maximum value on $X$.

$g:X\ni\mathbf{e_i}\to i\in\mathbb{R}$ is a continuous function but obviously $g$ doesn't take a maximum value on $X$.

This is a contradiction.

So, $X$ is not compact.


问:
Good estimator of $p$ knowing $\operatorname{Bin}(2,p)$
We consider the statistical sample ${1,0,2,1,1,0,1,0,0}$ and we assume that the values of this sample come from a binomial distribution $\operatorname{Bin}(2, p)$.

How can we estimate the value of $p$?

For me, an intuitive estimator is $\hat{p} = \frac{1}{n}\sum_{i=1}^n X_i$. So $\hat{p}$ should be just $\frac{2}{3}$. Where am I wrong?

答:
If $X_1,X_2,\dots,X_n$ are binominal $(2,p)$, and $Y:=\sum_{i=1}^{n}X_i$ then $Y$ is binomial $(2n,p)$.

Since $E[Y]=2np$ and you observed $Y=6$ in your sample, the maximum likelihood estimator will arise from solving $$2n\hat p=6$$

The solution is $\hat p=\frac{6}{18}$, or $\frac{1}{3}$.

As for the proposed answer of $2/3$, just look at your sample. Most of the numbers are closer to $0$ than $2$ right?

Another way to approach is that if you take the estimators for each of your individual sample points, you will get $\frac{1}{2}, 0, 1, \frac{1}{2}, \frac{1}{2}, 0, \frac{1}{2}, 0, 0$ (just dividing by two). The average of these numbers is $\frac{1}{3}$.


问:
Does having a zero eigenvalue preclude a matrix from being indefinite?
If a $3\times3$ matrix has a positive eigenvalue, a negative eigenvalue, and a zero eigenvalue, is it then, by definition, indefinite? I think so, since the matrix has both a positive and a negative eigenvalue. However, my optimisation lecture notes categorically claims that as a long as there is any zero eigenvalue, the stationary-point test fails.

I am not including the details of the stationary-point test as it is tangential to the substance of my question.

答:
If the matrix is $2\times 2$ then having a zero eigenvalue stops it being indefinite, though not in an interesting way.

A $2\times 2$ matrix has only two eigenvalues, so if one is zero it is not possible for one to be positive and one to be negative.


问:
Basis of a subspace of a matrix
Given
$$ A =
    \begin{pmatrix}
    -1 & -1 & 0 \\
    0 & 1 & 3 \\
   0 & 0 & -2
    \end{pmatrix}
$$
consider the subspace $W$ of $\Bbb{R}^{3}$ given by $W= \{x \in \Bbb{R}^{3}: Ax = −2x\}$. Find a basis for W

答:
The characteristic polynomial is $\rm{det}(A-xI)=(x-1)(x+1)(x+2)$.  The eigenvalues are thus $1,-1$ and $-2$, each with geometric multiplicity one.  Thus the dimension is one.

So we need $A\vec v=-2\vec v$.  Then we can compute the kernel of $A+2I$.  We get $\begin{pmatrix}1&-1&0\\0&3&3\\0&0&0\end{pmatrix}\to\begin{pmatrix}1&-1&0\\0&1&1\\0&0&0\end{pmatrix}$.  Thus by back-substituting, we get $(1,1,-1)$ as a basis.

Title: Basis of a subspace of a matrix


答:
Let me start with an analogy. There are a number of ways of finding the equation of a line on the plane given two points.

For example, given the points $(a,b)$ and $(c,d)$, there’s the point/slope formula, where we first calculate the slope $m$ (which is $m=\frac{d-b}{c-a}$) and then the equation is given by
$$ y-b = m(x-a)$$
(or $y-d = m(x-c)$).

Or you can use the 2-point formula, which says that the line has equation
$$(c-a)(y-b) = (d-b)(x-a).\qquad\qquad(1)$$
This is sometimes “simplified” to $y-b = \frac{d-b}{c-a}(x-a)$
so that you can see the connection with the point/slope formula.

Or there is the “intercepts” formula, where you try to figure out the values of the points $(0,h)$ and $(l,0)$ the line goes through, and use those.

But the thing is that not all methods work for all lines: they all work for lines that have nonzero slope; and all but the last one work for lines that have zero slope. But the only one that will give you the right answer when the line is vertical is $(1)$: if $a=c$, you have a vertical line and the equation is of the form $x=a$, which you can get from equation (1) since you will get $0(y-b)=(d-b)(x-a)$, which simplifies to $(d-b)(x-a)=0$, which is equivalent (since $d\neq b$) to $x-a=0$. The line is described simply by saying what the $x$ coordinate is, $x=a$, without any specification on $y$.

The same thing is happening with your points. Because both points have the same $x$-coordinate and the same $y$-coordinate, your line is parallel to the $z$-axis, and so it is described by simply saying the constant value of $x$ and the constant value of $y$, and not mentioning $z$. The equation you are trying to use can only be used for lines that do not lie on planes parallel to the coordinate planes, which is not the case here. It’s like trying to use the point-slope formula for a vertical line. The line is specified by the equations $x=3$, $y=-2$.

(Or, to use another analogy, it’s like trying to use the quadratic formula to solve an equation of degree $1$, say $0x^2 + 2x-4=0$, and then complaining that the formula is telling you to divide by $0$... wrong tool for the problem.)

Title: Why is $\frac{(x-3)}{0} = \frac{(y+2)}{0} = \frac{(z+5)}{11}$ a valid equation for a line in space when division by zero is undefined?

Body:
Oh yes, I do know division by zero is not defined.
But look at this problem:

Find the Cartesian equation of the line that passes through $(3,-2,-5)$ and $(3,-2,6)$.

And the answer is conveniently stated to be:
$$\frac{(x-3)}{0} = \frac{(y+2)}{0} = \frac{(z+5)}{11}$$

What gives us the liberty to divide by zero?

Insert a value of $x$; $2$ for example. $2$ divided by zero is infinity. But $\frac{(z+5)}{11}$ will give us a real value!

Please help me clear this silly confusion.


问:
Example of null tensor product

Does there exists a commutative noetherian ring $R$ with unit which has two ideals $I_1$ and $I_2$ such that $I_1\otimes_R I_2=0$ and $I_1I_2\neq 0$?

I've tryed to use a quotient of the multivariable polynomial ring but i can't find an example. Any suggestion?

答: There is an onto homomorphism $I_1\otimes_R I_2\to I_1I_2$ for $I_1,I_2$ ideals of $R$.

Title: Example of null tensor product


问:
Does $f(x)=f(1/x) \forall x$ put any restrictions on the derivative of $f$?
Consider a functions with the property that $f(x)=f(1/x) \forall x \not = 0$ and $f$ continuous.

Does this tell us anything about the derivatives of $f$?

I feel like it should, since if $f$ is increasing on $(2,3)$ then $f$ is decreasing on $(\frac{1}{3}.\frac12)$

But I feel like this (and potentially other properties) should be easy to show via application of chain rule/product rule.

(Another property that I have a hunch may be true is that there is some relationship about concavity/convexity to the left and right of $x=1$)

I recall, however, that $f(2)=g(2)$ does not mean that $f'(2)=g'(2)$ so I'm not whether I can even take the derivative of both sides to use in proofs.

答:
Let $g$ be any even function defined on $\Bbb R$.

Now let $f$ be defined as $f(x)=g(\log x)$. Then $f(1/x)=g(\log(1/x))=g(-\log x)
=g(\log x)=f(x)$.

So $f$ will be differentiable, continuous, etc., as the original $g$ was.


问:
Optimal Rule for Rank Loss
In a binary classification setting, the classification risk of a classifier $h$ is defined by
$$
R(h) = \mathbb{P}(Y \neq h(X)),
$$
where $(X,Y) \sim P$. It is well known that the classifier that minimizes $R(h)$ over all measurable functions is the Bayes classifier,
$$
h^*(x) = \mathbb{1}\{ \mathbb{E}[Y|X=x] \ge 1/2\}.
$$

Another example is in the regression setting, where the `regression function' $f(x) = \mathbb{E}[Y|X=x]$ minimizes mean squared error $\mathbb{E}[(Y- f(X))^2]$.

I am interested in the 'ranking' risk
$$
\rho(h) := \mathbb{P}( (Y - Y')(h(X)-h(X')) < 0),
$$
where $(X,Y), (X',Y') \sim P$ are independent draws. The $Y$ need not be binary here. Could anyone point me to a derivation of the optimal ranking function $h$ that minimizes $\rho(h)$?

答:
This goes the same way as the proof for classification as well as for pairwise ranking (where $h: \mathcal{X} \times \mathcal{X} \rightarrow \mathbb{R}$):

Note that the risk can be written as:

$$\mathbb{P}(Y-Y' \neq h(X) - h(X')) = \mathbb{E} 1(Z=1)1(h(X)<h(X') + 1(Z=-1)1(h(X)>h(X')$$

where $Z = Y-Y'$.

Now, $\mathbb{E} 1(Z=1)1(h(X)<h(X')) = \mathbb{E}[\mathbb{P}(Y = 1, Y' =0|X, X')1(h(X)<h(X'))]$ by conditioning on $X,X'$ and by definition of $Z$.

Applying independence: $\mathbb{P}(Y = 1, Y' =0|X, X') = \mathbb{P}(Y = 1|X)\mathbb{P}(Y' = 0|X') = \eta(X)(1-\eta(X'))$ where $\eta$ is the Bayes classifier: $\eta(x) = \mathbb{P}(Y=1|X=x)$.

The same work for the second terms gives: $\mathbb E [1(Z=-1)1(h(X)>h(X'))] = \mathbb E[\eta(X')(1-\eta(X))1(h(X)>h(X')]$

Putting it together we see that the best $h$ picks the term that is minimal of $\eta(X')(1-\eta(X))$ and $\eta(X)(1-\eta(X'))$ based on whichever of $h(X)$ or $h(X')$ is larger. This can be done with the Bayes classifier $\eta$.

EDIT: For general $Y$ we follow a similar approach, but of course, $\eta(x)$ is no longer useful for us.

$$\mathbb P(1(Y>Y') \neq 1(h(X)>h(X'))) = \mathbb P (Y>Y', h(X)<h(X')) + \mathbb P (Y<Y', h(X)>h(X'))$$

Looking at just the first term:

$$\mathbb P (Y>Y', h(X)<h(X')) = \mathbb E[ \mathbb P(Y>Y'|X,X')1(h(X)<h(X')))]$$

and now the second term:

$$\mathbb P (Y<Y', h(X)>h(X')) = \mathbb E[ \mathbb P(Y<Y'|X,X')1(h(X)>h(X')))]$$

So pointwise for each $X, X'$, the best $h$ picks out the minimum of $P(Y<Y'|X,X')$ and $P(Y>Y'|X,X')$ given only the relative values of $h(X)$ and $h(X')$. By symmetry, the best we can do (in expectation) is with $h(x) = \mathbb E[P(Y>Y'|X, X') | X=x]$ where we marginalize over $Y', X'$. This is optimal in the same way that conditional expectation is optimal (in the mean square sense).

Title: Optimal Rule for Rank Loss


问:
$P_n$ is a.s. to P, $Q_n$ is L1 to Q. $P(P_n \neq Q_n) \leq 1/n$. Is $P=Q$ a.s.?
$\left\{P_{n}\right\}_{n=1,2, \ldots}$ and $\left\{Q_{n}\right\}_{n=1,2, \ldots}$ are two sequences of r.v.s. For every $n$ , $\mathbb{P}\left(P_{n} \neq Q_{n}\right) \leq 1 / n . P_{n} \stackrel{\text {a.s.}}{\rightarrow} P$ and $Q_{n} \stackrel{L^{1}}{\rightarrow} Q$. Can we conclude that $P=Q$ a.s.?

Is it right?

答: Convergence in probability and convergence in $L^{1}$ both imply a.s. convergence for a subsequence. Since $\mathbb P(|P_n-Q_n| > \epsilon) \leq \frac  1n$ it follows that $P_n-Q_n \to 0 $ in probability. Choose a sub-sequence of $(P_n-Q_n)$ which tends to $0$ (so that $Q_n \to P$ as. along this subsequence) and then choose  a further subsequence along which $Q_n \to Q$ a.s.. It is now clear that $P=Q$ a.s..


问:
Solution of IVP $y'=y^2, y(0) =0$.
Consider the ordinary differential equation $$y'=y^2, y(0) =0$$

Clearly $y=0$ is a solution of it. Can i say that this is only solution? I tried as follows

Choose rectangle $|x|\leq a, |y|\leq b$, then $|f(x, y) |\leq b^2$ . Now using existence theorem we have $$h=min\{a, \frac{b}{b^2}\}
=\frac{1}{b}\to\infty$$ So by Existence theorem we can say that $y=0$ is a global solution of given differential equation . But I don't know about uniquness of solution. Did I correctly apply Existence theorem? Please suggest. Thank you.

答:
Suppose there exist a different solution $y\ne 0$. Then divide both sides of the equation by $y^2$:$$\frac{y'}{y^2}=1$$
Integrating, you get $$-\frac 1y=x+C$$or $$y=-\frac 1{x+C}$$But this can't verify the condition $y(0)=0$ for any finite $C$.

Title: Solution of IVP $y'=y^2, y(0) =0$.


问:
Integrating the Difference between Sine and its Maclaurin Series
The integral in question:

Let $n$ be a nonnegative integer.

$$
\int_0^\infty
\frac{1}{x^{2n+3}}
\left (
\sin x -
\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!}
\right )
\mathrm{d}x
$$

I was given this problem to try for fun from a professor about two months ago and haven't made a dent in it.
If possible, I'd like someone to show how they get to their answer, but please keep in mind that I haven't learned contour integration or complex analysis, but know of a few tricks like Feynman's.

What I've tried:

Writing out the terms to integrate by segments, using the Cosine Maclaurin series and substituting (pi/2 -x) to get the Sine series in a different way, integration by parts (2n+3) times or such to simplify, and a lot more I'm probably forgetting.

I know this site is full of integral-calculating gods, so if this piques your interest please have an attempt and share below.

Body:
Hoping that you enjoy hypergeometric functions
$$I_n=\int_0^\infty
\frac{1}{x^{2n+3}}
\left (
\sin x -
\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!}
\right )
\,dx$$

$$\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!}=\frac{(-1)^n x^{2 n+3} \,
   _1F_2\left(1;n+2,n+\frac{5}{2};-\frac{x^2}{4}\right)}{(2 n+3)!}+\sin (x)$$

$$I_n=\frac{(-1)^{n+1}}{\Gamma (2 n+4)}\int_0^\infty \, _1F_2\left(1;n+2,n+\frac{5}{2};-\frac{x^2}{4}\right)\,dx$$
$$I_n=(-1)^{n+1}\,\frac{   \pi}{2\,\Gamma (2 n+3)}$$


问:
New to equipotent sets. How to do this proof
Prove that if (A-B)$\approx$(B-A)then B $\approx $A

2.21 Theorem If $f : A\mapsto B $is a bijective function, then $f^{-1}$ : B → A is a bijective function

Attempted Proof
I know there is bijective function.
Let $f:(A-B)\mapsto(B-A). $

Then by 2.21 $f^{-1}:(B-A)\mapsto(A-B)$ is a function.

Then ?
Help

答:
Take a look at this venn diagram:

Your goal is to construct a function $f:A\to B$ which is a bijection. To do this, you can map the points in the overlap $A\cap B$ onto themselves.

Next the points in $A-B$ (the white part on the left) need to be mapped over to $B-A$ (the white part on the right)... but didn't you already assume there was a map (maybe called $g$) that can do this? Hope this helps.


答:
The equation you got as an answer is just an - IMO - unfortunate different way of giving the vector equation of the line in parameter form.

The line in parameter form is:

$$\begin{pmatrix} x\\ y\\z \end{pmatrix} = \begin{pmatrix} 3\\ -2\\-5 \end{pmatrix}+ t\begin{pmatrix} 3 -3\\ -2-(-2)\\6-(-5) \end{pmatrix} = \begin{pmatrix} 3\\ -2\\-5 \end{pmatrix}+t\begin{pmatrix} \color{blue}{0}\\  \color{green}{0}\\ \color{orange}{11} \end{pmatrix} $$

Rearranging this gives

$$\begin{pmatrix} x-3\\ y+2\\z+5 \end{pmatrix} = t \begin{pmatrix} \color{blue}{0}\\  \color{green}{0}\\ \color{orange}{11} \end{pmatrix} $$

If all entries of the direction vector were different from zero, you could solve this vector equation coordinatewise for $t$ which then gives your form of the equation. Now, some people are so attracted to this form that they extend it formally also to direction vectors with zeros in some coordinates. This then leads to

$$\frac{x-3}{\color{blue}{0}} = \frac{y+2}{\color{green}{0}} =\frac{z+5}{\color{orange}{11}}$$


问:
Manually Finding Values of Inverse Trigonometric Functions
I'm trying to solve (for $x$) some problems such as $\arctan(0)=x$, $\arcsin(-\frac{\sqrt{3}}{{2}})=x$, etc.

What is the best way to go about this? So far, I have been trying to solve the problems intuitively (e.g. I ask myself what value of sine will give me $-\frac{\sqrt{3}}{{2}}$?), maybe drawing a triangle to help. Is there a better way to solve these problems?

答: A combination of unit circle and "famous triangles" is perhaps the best strategy. Have drawing of the three famous triangles, $30,60,90$,  with sides $1,2,\sqrt 3$, and $45,45,90$, with sides $1,1,\sqrt 2$  and $0,90,90$ with sides $0,1,1$. These will give you the reference angle you are looking for. Then for finding $z$ in $\arctan(a)=z,\arccos(a)=z,\arcsin(a)=z$ locate the axis on the unit circle and apply the definition to find the right quadrant for $z$. For sine the axis is the vertical $y$ axis and final angle is in fourth or first quadrant. For cosine the axis the horizontal $x$ axis and the angle is on first or second quadrant. For tangent the axis is $x=1$ line and the answer is on first or fourth quadrant.</p>


问:
Does $A(x, m) = 1 - A(m, x)$ imply some symmetry in partial derivatives?
This is a self-made problem. I have a function $A$ of variables $x, m \geq 0$. A satisfies the following symmetry:

$$A(x, m) = 1 - A(m, x)$$

Is there an equation that relates the partial derivative of $A$ wrt the first argument to the partial derivative of $A$ with respect to the second argument? It seems like there should be something straightforward but I'm not sure how to deduce it.

Thank you!

答:
I suggest using the definition of partial derivative to have a clearer point of view on the situation. The partial derivative of $A$ with respect to the first variable at the point $(x_0,m_0)$ is defined by :

$$\partial_1 A(x_0,m_0)=\lim_{h\to0} \frac{A(x_0+h,m_0)-A(x_0,m_0)}{h}.$$

Using the symmetry relation, we have :

$$\lim_{h\to0} \frac{(1-A(m_0,x_0+h))-(1-A(m_0,x_0))}{h}=\lim_{h\to0} \frac{-A(m_0,x_0+h)+A(m_0,x_0)}{h}.$$

The last term is equal, also by the definition partial derivative, to $- \partial_2 A(m_0,x_0)$.

Thus :

$$\partial_1 A(x_0,m_0)=- \partial_2 A(m_0,x_0)$$


答:
All you need to do is differentiate both sides of the symmetry equation-

$$(\partial_1 A)(x,m) = \frac d{dx}[A(x,m)] = \frac d{dx}[1-A(m,x)] = - (\partial_2 A)(m,x)$$

or did you want something more?

Title: Does $A(x, m) = 1 - A(m, x)$ imply some symmetry in partial derivatives?

Body:
This is a self-made problem. I have a function $A$ of variables $x, m \geq 0$. A satisfies the following symmetry:

$$A(x, m) = 1 - A(m, x)$$

Is there an equation that relates the partial derivative of $A$ wrt the first argument to the partial derivative of $A$ with respect to the second argument? It seems like there should be something straightforward but I'm not sure how to deduce it.

Thank you!


问:
GCSE Vector Question Being Pulled For Being 'Too Hard'
So apparently this was lined up as the last question on a GCSE paper but is being pulled because it was deemed 'too hard'. I personally have studied it for over an hour and my feeling is not that it's just too hard, rather that it's unsolvable without more information. Or am I missing something obvious?

Here is the full question:

All I managed to glean are the following expressions:

$\vec{CX} = \frac{m}{m+1}(a+b)$

$\vec{XE} = \frac{1}{m+1}(a+b)$

$\vec{FX} = \frac{n}{n+1}(2a-\frac{1}{2}b)$

$\vec{XM} = \frac{1}{n+1}(2a-\frac{1}{2}b)$

I've no idea how one would find values for $m$ (the scalar ratio of $\vec{CX}$ to $\vec{XE}$) or $n$ without being given the vectors $a$ and $b$.

答:
By the Ratio Theorem $\left(\text{or using }\,\vec{EX}=\vec{EM}+\frac1{n+1}\vec{MF}\right),$
$$\vec{EX}=\frac{-1}{n+1}\left(2\mathbf{a}+\frac{n}2 \mathbf{b}\right).$$

Since $CXE$ is a straight line, for some real $\lambda,$
$$\begin{aligned}\vec{EX}&=\lambda\vec{EC}\\&=-\lambda\left(\mathbf{a}+\mathbf{b}\right).\end{aligned}$$

Equating the above expressions for $\vec{EX}:$
$$\frac{2}{n+1}\mathbf{a}+\frac{n}{2(n+1)} \mathbf{b}=\lambda\mathbf{a}+\lambda\mathbf{b}.$$

Since $\mathbf{a}$ and $\mathbf{b}$ are not collinear,
$$\frac{2}{n+1}=\lambda=\frac{n}{2(n+1)}\\n=4.$$


答:
If $f$ is differentiable at $x$, then by the chain rule, we get
$$f'(x) = -\frac{f'(1 / x)}{x^2}.$$

Title: Does $f(x)=f(1/x) \forall x$ put any restrictions on the derivative of $f$?

Body:
Consider a functions with the property that $f(x)=f(1/x) \forall x \not = 0$ and $f$ continuous.

Does this tell us anything about the derivatives of $f$?

I feel like it should, since if $f$ is increasing on $(2,3)$ then $f$ is decreasing on $(\frac{1}{3}.\frac12)$

But I feel like this (and potentially other properties) should be easy to show via application of chain rule/product rule.

(Another property that I have a hunch may be true is that there is some relationship about concavity/convexity to the left and right of $x=1$)

I recall, however, that $f(2)=g(2)$ does not mean that $f'(2)=g'(2)$ so I'm not whether I can even take the derivative of both sides to use in proofs.


问:
Question about required inequality for proof that Hilbert Transform is weak-(1,1)
I'm reading a couple of proofs that $H$, the Hilbert transform, is weak-$(1,1)$ and have struggled to justify an inequality that all the proofs mention can be proved without justification.

Given
$$
b(x) = \sum_j b_j(x)
$$
then they use the fact that
$$ |Hb(x)| \leq \sum_j |Hb_j(x)| \text{ almost everywhere}.$$

Interestingly, both proofs mention that this is immediate if the above sum is finite and can be proved otherwise using the fact that $\sum_j Hb_j$ converges to $Hb$ in $L^2$.  Indeed the finite case is clear, and the infinite case seems intuitive to me, but I'm having trouble using norm convergence to show the inequality!  Here's what I can do.

By the linearity of the transform we have that $H(\sum_{j=1}^n b_j(x)) = \sum_{j=1}^n Hb_j(x)$.  As stated above we can immediately have
$$
\left| \sum_{j=1}^n Hb_j(x) \right| \leq \sum_{j=1}^n |Hb_j(x)|,
$$
and so by the $L^2$ convergence we have
$$
\int |Hb(x)|^2dx \leq \int \sum_{j=1}^n |Hb_j(x)|^2 = \sum_{j=1}^n \int |Hb_j(x)|^2.
$$
This is true for all $n$, thus we have
$$
\int |Hb(x)|^2dx \leq \sum_j \int |Hb_j(x)|^2 = \int \sum_j |Hb_j(x)|^2.
$$

However I now seem to have reached an impasse, as there seems no reason to me why this inequality of integrals would imply an inequality of integrands.  I can't help but feel I've gone awry somewhere, or went in the wrong direction.  Any clarity is much appreciated!

答:
Here's an approach that still uses integrals, but in a slightly different way.

Suppose to the contrary that
$$| Hb(x) |
\leq
\sum_j |Hb_j (x)|
$$
does not hold almost everywhere. Then we can find a set $E$ of finite positive measure where the opposite (strict) inequality holds. Therefore the integral
$$
\int_E
\bigl(
|Hb(x)| - \sum_j |Hb_j (x)|
\bigr) \, dx
$$
has a positive value, call it $A$.

Now using Hölder's inequality, the fact that the series $\sum_j Hb_j$ converges to $Hb$ in $L^2$ implies that the series $\sum_j \chi_E Hb_j$ converges to $\chi_E Hb$ in $L^1$, where $\chi_E$ denotes the characteristic function of $E$.

On the other hand, for any $N$ we have
\begin{align*}
\left\|
\chi_E Hb - \sum_{j=1}^{N} \chi_E Hb_j
\right\|_{1}
&= \int
\biggl|
\chi_E(x) Hb(x)
- \sum_{j=1}^{N} \chi_E(x) Hb_j(x)
\biggr|
\, dx \\
&\geq \int_E
\biggl(
| Hb(x) |
- \sum_{j=1}^{N} | Hb_j(x) |
\biggr)
\, dx \\
&\geq \int_E
\biggl(
| Hb(x) |
- \sum_j | Hb_j(x) |
\biggr)
\, dx \\
&= A.
\end{align*}
This contradicts the $L^1$ convergence, implying the result.


问:
Poisson process of alternating sources, pairing products of the two of them
I have a problem I've been thinking for a while, but it's confusing me quite a lot.

There are two independen machines, A and B. A produces a product at a rate of 2/minute, and B does it at a rate of 3/minute, both of them following a Poisson process.

The products of both machines get into a line where a person packs them by pairs as they arrive.

I need to find the mean and the variance of the time I have to wait until there is a package formed by a product of A and a product of B.

I don't really know where to start, because I can't even decide what the "success" is. I know that i can get the mean of both the machines, but how can i be sure i'm alternating between the two machines?

Thank you very much beforehand.

答:
This is a magic property of Poisson processes.

Let $\mu = 2$ and $\lambda = 3$ be the rates of your two machines

Now I want to know the probability that I get nothing out of either machine in the first $t$ minutes.

I get nothing out of machine $A$ with probaiblity $e^{-\mu t}$ and nothing out of machine $B$ with probability $e^{-\lambda t}$.

So as the machines are independent I get nothing out of the either machine with probability $e^{-(\mu+\lambda)t}$.

So the two machines working together act like a Poisson machine producing at a rate of $5$ per minute.

You should be able to recognise the distribution of the waiting time above and calculate the mean.

Now lets try a similar approach for the question you actually asked so that I can pretend that was a warm up exercise and not just idiot boy here not reading the question.

We already know that the first product comes out of the machine after an exponential waiting time mean $\frac 15$ and variance $\frac 1{25}$.
So the amount of time I have to wait before I get two is the sum of two independent exponentials, hence has mean $\frac 25$ and variance $\frac 2{25}$.

So how many pairs do I need to see before I get a package with one from each?
For a fixed time $t$ the probability I've seen one product from $A$ and none from $B$  is $e^{-(\lambda+\mu)} \frac \mu{0!\cdot 1!}$ and the probability I've seen one from $B$ and none from $A$ is $e^{-(\lambda+\mu)} \frac \lambda{0!\cdot 1!}$ so conditioned on the event only one product has been seen the probability it came from $A$ is $\frac{\mu}{\mu+\lambda}$ or $\frac 25$.

Now the output of the machines is independent and memoryless so the sequence of products is an iid sequence.  Therefore the probability the firs pair gives a proper package is $2\cdot \frac 25\cdot\frac 35 = \frac{12}{25}$.

Let $N$ be the number of packages we make before we get a good one. (So the $N$th package is good) We have
$$P(N = n) = \left(\frac {13}{25}\right)^{n-1}\left(\frac {12}{25}\right)$$

Now the packages arrive at independent intervals of length $T_i$ each of which has $E(T) = \frac 15$ and $E(T^2) = \frac 2{25}$ so with total Waiting time
$$T = \sum_{i=1}^N T_i$$
So if I know $N$ I can write
$$E(T|N) = NE(T_i) = \frac N5$$and
$$E(T^2|N) = \left(\sum_{i=1}^N T_i\right)^2 = \sum_{i=1}^N E(T_i)^2 + 2\sum_{i\neq j} E(T_i)E(T_j)$$ $$= NE(T_i^2) + (N^2-N)E(T_i)^2 = \frac{N^2 + N}{25}$$

You get to take it from here. You know the distribution of $N$ so you can work out  $E(T) = \frac{E(N)}5$ and $E(T^2) = \frac{E(N^2+N)}{25}$ So you should be able to get the mean and variance.


问:
Expected value in a joint distribution
Let $X$ and $Y$ be continuous random variables with joint density function
$$ f(x,y)=0.25$$ for $$0 \leq x \leq 2$$ and $$x-2 \leq y \leq x$$
What's $$E[X^3Y]$$

The solution is $$ 0.25  \int_{0}^{2} \! \int_{x-2}^{x} x^3yf(x,y) \,dy\,dx$$

Can I switch to $dxdy$?

$$ 0.25  \int_{0}^{2} \! \int_{y}^{2} x^3yf(x,y) \,dx\,dy$$+$$ 0.25  \int_{0}^{2} \! \int_{-2}^{y+2} x^3yf(x,y) \,dx\,dy$$

The result is different.

答: When you switch the order of integration you get $\int_{-2}^{2} \int_{y\vee 0}^{y+2} x^{3}yf(x,y)\ dxdy$ where $y \vee 0 $ is the maximum  of $y$ and $0$.  Now see if you get the same answer. [Split the intergal as $\int_0^{2}\int_y^{y+2} x^{3}y\ dxdy +\int_{-2}^{0}\int_0^{y+2} x^{3}y\ dxdy $].

Title: Expected value in a joint distribution


问:
Integrate: $\int_0^\infty \frac{\sin^2(x)}{x^2}dx$
I am trying to integrate $\displaystyle \int_0^\infty \frac{\sin^2(x)}{x^2} dx$  by method of contour. I am considering the following contour but I am not being able to. Also I am not sure if it's right approach.

$$\int_\Gamma f(z) dz + 2 \int_0^\infty f(z) dz + \int_\gamma f(z)dz = 0$$
The first integral tends to zero as $R \to \infty $, but letting $\epsilon \to 0$, for the last integral I am getting.
$$\int_{\pi }^0 \frac{\sin^2({\epsilon e^{i\theta}})}{\epsilon^2 e^{i2\theta }} i \epsilon e^{i\theta}d\theta = 0$$

ADDED:: Taking the above contour, we do not have any pole inside the contour.
$$\int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz + \int_\Gamma \frac{1 - e^{i2z}}{2z^2} dz + \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz = 0 \hspace{1 cm }(1)$$

$\displaystyle \int_\Gamma \frac{1 - e^{i2z}}{2z^2} dz \to 0$ due to Jordan Lemma. To evaluate $\displaystyle \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz $ let the radius of the small semi circle be $\epsilon \to 0 $.

$$\lim_{\epsilon \to 0}\int_\pi^0 \frac{1 - e^{i2\epsilon e^{i\theta}}}{2\epsilon ^2 e^{i2\theta}} i \epsilon e^{i\theta}d\theta =\lim_{\epsilon \to 0} \int_\pi^0 \frac{1 - 1 - 2i\epsilon e^{i\theta} + O(\epsilon^2)}{2\epsilon e^{i\theta}} i  = \lim_{\epsilon \to 0} \int_\pi^0 1+  O(\epsilon) d\theta = -\pi \hspace{1 cm }(2)$$

From $(2)$, $(1)$ reduces to
$$\int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz - \pi = 0 \implies \Re \int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz = \int_{-\infty}^{\infty}\frac{\sin(x)^2}{x^2}dx = \pi$$

Including singularity at $z = 0$, we will have that small inner circle on lower plane.
$$\int_{-\infty}^\infty \frac{1 - e^{i2z}}{2z^2}dz + \int_\Gamma \frac{1 - e^{i2z}}{2z^2} dz + \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz = 2 \pi i \text{Residue}[f(z), z = 0] = 2\pi \hspace{1 cm }(3)$$

As for $\displaystyle \int_\gamma \frac{1 - e^{i2z}}{2z^2} dz = \lim_{\epsilon \to 0}\int_{-\pi}^0 \frac{1 - e^{i2\epsilon e^{i\theta}}}{2\epsilon ^2 e^{i2\theta}} i \epsilon e^{i\theta}d\theta = \pi \hspace{1 cm }(4)$

From $(3)$ and $(4)$ we get the same result.

答:
What you try to do won't work since your function is (almost) analytic insde the path you take and thus won't help you to evaluate the real integral.

Let us try the following:

$$\cos 2x=1-2\sin^2x\implies \sin^2x=\frac{1-\cos2x}{2} \;\text{define}\;\;f(z):=\frac{1-e^{2iz}}{2z^2}:$$

$$\text{Res}_{z=0}(f)=\lim_{z\to 0}\,zf(z)=\lim_{z\to 0}\frac{1-e^{2iz}}{2z}\stackrel{\text{l'Hospital}}=-i$$

Question: The above implies $\,z=0\,$ is a simple pole...why is this so and not a double one?

Taking your contour, taking the limits and etc. and using the lemma and, specially, its corollary in the 2nd. answer here , we get after comparing real and imaginary parts

$$\int\limits_{-\infty}^\infty\frac{\sin^2x}{x^2}dx=\pi\;\;\ldots\ldots$$


问:
Is there a math symbol meaning "If" or "when"?
To introduce a logic premise normally people use the word If, or Iff (if and only if). Is there a math symbol with the same meaning?

答: You could say that $\implies$ (implies) means if, so, $A \implies B$ would mean "If $A$ then $B$", and $A\iff B$ would mean iff, so it would mean "$A \implies B \wedge B \implies A$" (or "If $A$ then $B$" and "If $B$ then $A$")


问:
How to prove recursive sequence converges? - $a_{n+1}=1/(a_n+1)$
I have a recursive sequence defined by $$a_{n+1}=\frac{1}{a_n+1}$$, where $a_0=7$.

The sequence is bounded, but it's not monotonic. So how does one show, that the sequence converges?
I already found it's limit under the assumption, that it is converging. Which is $a=\frac{\sqrt5-1}{2}$.

Note, that I'm not allowed to use epsilon proofs or proof by induction.

答:
You could even go beyond the problem of the convergence.

This is a first-order rational difference equation (have a look here). If you look at my answer here for a more general case.

Let $m=0$, $x=1$, $y=1$ that is to say $p=-1$ and $q=1$ you should find
$$a_n=\frac{K \left(\frac{1}{2} \left(1+\sqrt{5}\right)\right)^n+\left(\frac{1}{2}
   \left(1-\sqrt{5}\right)\right)^n}{K \left(\frac{1}{2}
   \left(\sqrt{5}-1\right)\right)^{-n-1}+\left(\frac{1}{2}
   \left(1-\sqrt{5}\right)\right)^{n+1}}$$ and the condition $a_0=7$ leads to $K=\frac{27-7 \sqrt{5}}{22}$ and the limit you already found.

This generates the sequence
$$\left\{7,\frac{1}{8},\frac{8}{9},\frac{9}{17},\frac{17}{26},\frac{26}{43},\frac
   {43}{69},\frac{69}{112},\frac{112}{181},\frac{181}{293},\frac{293}{474},\frac
   {474}{767},\frac{767}{1241},\frac{1241}{2008},\cdots\right\}$$


问:
Why is this linear algebra proof correct? (dimension of the kernel and the image)
I was asked to prove the following theorem about vector spaces:

Let $K$ be a field and let $V,W$ be vector spaces over $K$. If there is a linear transformation $T:V\to W$ such that $\mathrm{Ker}(T)$ and $\mathrm{Im}(T)$ are both finite dimensional, then $V$ is finite dimensional as well.

I tried to solve but I eventually gave up and looked up the solution. Turns out that I could only find variations of this particular proof (a variation of the proof can be found on page 12 of this document). However, I'm having trouble understaing why it is correct, as the proof in the answer starts with the assumption that $\mathrm{Ker}(T)$ has a basis $\{u_1,\dots,u_n\}$, and then a finite spanning set of $V$ is constructed from there. I think that makes sense, but what if $T$ is injective? Then we would have $\mathrm{Ker}(T)=\{0_{V}\}$, so that the only basis of $\mathrm{Ker}(T)$ is the empty set, that has no vectors in it, so the proof wouldn't make sense, right?

I know this may be a silly question but I'm really having a bad time with this problem, any help would be really appreciated. :)

答: Essentially the same proof works when  $ker(T)=\{0_V\}$. Let $\{Tw_1,Tw_2,..,Tw_k\}$ be  a basis for the image of $T$. Take any $x \in V$. Then $Tx$ is a linear combination of the basis elements, say $Tx=\sum c_iTw_i$. But then $T(x-\sum a_iw_i)=0$ so $x=\sum a_iw_i$ We have proved that $V$ is spanned by $w_1,w_2,...,w_k$ so $V$ is finite dimensional.


问:
Let's consider non-decreasing sequence of integer not negative numbers and $d_1 \geq d_2 \geq \cdots \geq d_n \geq 0$
Prove that such sequence is a graph sequence  of some graph $G$ which does not contain loops iff $d_1 \leq d_2 + d_3 + \cdots + d_n$.

IMPORTANT: by condition $d_1 + d_2 + \cdots + d_n = 2k$ for some integer $k$

I was trying to solve this one by assuming that $d_1 > d_2 + d_3 + \cdots + d_n$. I was planning on coming to contradiction. But I haven't managed to.

All ideas for solution will be much appreciated!

答:
To prove the "only if" direction, note that each edge coming out of the vertex of degree $d_1$, also has an endpoint which is one of
the vertices of degree $d_2,d_3,\cdots,d_n$.


问:
An open cover that is not locally finite

I could not understand why example $13.4$ is not locally finite.

Can you give me an explanation please.

答:
This open cover is very far from being locally finite.  In fact, no real number has a neighbourhood which intersects only finitely many $U_{r,n}$.

For example, if $U$ is any (open) neighbourhood of $0$, then clearly $0 \in U \cap U_{0,n}$ (and so $U \cap U_{0,n} \neq \emptyset$) for all $n \in \mathbb{Z}^+$ (and there are infinitely many of these).


问:
Proving $\gcd(4k+2,k^2+k) = 2$ by direct proof
I came across this problem in a textbook (not an assignment) and was interested in how to solve it.

I imagine $\gcd(4k+2,k^2+k)$ can easily be proved to be an even integer because:

$4k+2 = 2(2k+1)$, which would be even because its $2$ times and integer, and

$k^2+k$ can be proved to be even by using $2$ cases, one where $k$ is even and is equal to $2n$ and the other for an odd $k$, where we can set it equal to $2n+1$, where $n \in \mathbb{Z}$.

From here, if we let $\gcd(4k+2,k^2+k)=x$, we can say if $x$ is the $\gcd$ of $(4k+2,k^2+k)$ then it must divide any linear combination of the 2. I'm stuck on this part, because there's so many directions this could go.

I'm thinking $x|(k^2+5k+2)$ or $x|(4k^2-2)$, but I can't find a way to prove the gcd isn't some even integer greater than $2$. Any insight is appreciated!

答: HINT: If a prime $p$ divides both $2k+1$ and $k(k+1)$, then either $p$ divides $k$ or $p$ divides $k+1$. Show that the first leads to an immediate contradiction and the second leads to the conclusion that $p$ divides $k$.


答:
Let us consider an arbitrary (not necessarily commutative) field $K$, two left $K$-vector spaces $V$ and $V'$ and a $K$-linear map $f \in \mathrm{Hom}_{\operatorname{\mathit{K}-\mathbf{Mod}}}(V, V')$ such that $\mathrm{Ker}f$ and $\mathrm{Im}f$ are both finite dimensional. Fix an arbitrary supplementary subspace $U$ of the kernel $\mathrm{Ker}f$ in $V$ (any subspace admits a supplementary by virtue of elementary vector space theory). It is easy to show that the linear map $g\colon={}_{\mathrm{Im}f|}f_{|U}\in \mathrm{Hom}_{\operatorname{\mathit{K}-\mathbf{Mod}}}(U, {\mathrm{Im}f})$ obtained by restricting $f$ is an isomorphism, whence $\mathrm{dim}_KU=\mathrm{dim}_K\mathrm{Im}f=\mathrm{rk}_K f$. Indeed, we have $\mathrm{Ker}g=\mathrm{Ker}f \cap U=\{0_V\}$, meaning that $g$ is injective and $\mathrm{Im}f=f\left[U+\mathrm{Ker}f\right]=f[U]+f[\mathrm{Ker}f]=f[U]=g[U]=\mathrm{Im}g$, meaning that $g$ is also surjective.

As $U$ and the kernel are supplementary, it follows that $\mathrm{dim}_K V=\mathrm{dim}_K \mathrm{Ker}f+\mathrm{dim}_K U$. Since the sum of two finite cardinals (which are nothing else than natural numbers in the axiomatic system I adhere to) is also finite, we gather that $\mathrm{dim}_K V \in \mathbb{N}$, in other words $V$ is finite dimensional.

The above argument is actually part of the standard proof of the notorious rank-nullity relation, valid for any vector space morphism $h \colon V \to V'$ and according to which $\mathrm{dim}_K V=\mathrm{dim}_K \mathrm{Ker}h+\mathrm{dim}_K \mathrm{Im}h$.

Title: Why is this linear algebra proof correct? (dimension of the kernel and the image)

Body:
I was asked to prove the following theorem about vector spaces:

Let $K$ be a field and let $V,W$ be vector spaces over $K$. If there is a linear transformation $T:V\to W$ such that $\mathrm{Ker}(T)$ and $\mathrm{Im}(T)$ are both finite dimensional, then $V$ is finite dimensional as well.

I tried to solve but I eventually gave up and looked up the solution. Turns out that I could only find variations of this particular proof (a variation of the proof can be found on page 12 of this document). However, I'm having trouble understaing why it is correct, as the proof in the answer starts with the assumption that $\mathrm{Ker}(T)$ has a basis $\{u_1,\dots,u_n\}$, and then a finite spanning set of $V$ is constructed from there. I think that makes sense, but what if $T$ is injective? Then we would have $\mathrm{Ker}(T)=\{0_{V}\}$, so that the only basis of $\mathrm{Ker}(T)$ is the empty set, that has no vectors in it, so the proof wouldn't make sense, right?

I know this may be a silly question but I'm really having a bad time with this problem, any help would be really appreciated. :)


答:
For the forward (only if) direction, , let $G$ be a graph that does not contain any loops and has degree sequence $d_1,d_2,\ldots,d_n$. Any edge incident on the vertex of degree $d_1$ must also be incident on one of the vertices of degree $d_2,\ldots,d_n$. That is, the number of edges incident on this vertex is upper-bounded by the sum of the remaining degrees, which is exactly what we want.

For the backward (if) direction, consider the sequence $d_1-d_n,d_2,\ldots,d_{n-1}$. If $d_1-d_n$ is the largest element in this sequence, then we clearly have $$d_1-d_n\leq d_2+\cdots+d_{n-1}.$$
On the other hand, if $d_2$ is the largest element, then because $d_1\geq d_2$ and $d_{n-1}\geq d_n$, we have $$(d_1-d_n)+d_3+\cdots+d_{n-1}\geq d_2 + (d_{n-1}-d_n)\geq d_2.$$

Try using induction to complete the argument from here: construct a graph with this degree sequence, add a new vertex $v_n$, and add $d_n$ edges from the vertex with degree $d_1-d_n$ to $v_n$.

Title: Let's consider non-decreasing sequence of integer not negative numbers and $d_1 \geq d_2 \geq \cdots \geq d_n \geq 0$

Body:
Prove that such sequence is a graph sequence  of some graph $G$ which does not contain loops iff $d_1 \leq d_2 + d_3 + \cdots + d_n$.

IMPORTANT: by condition $d_1 + d_2 + \cdots + d_n = 2k$ for some integer $k$

I was trying to solve this one by assuming that $d_1 > d_2 + d_3 + \cdots + d_n$. I was planning on coming to contradiction. But I haven't managed to.

All ideas for solution will be much appreciated!


答:
"B if A" is "if A, then B".

Thus, we may symbolize it with:

$A \to B$.

Also "B when A" is "if A, then B".

Title: Is there a math symbol meaning "If" or "when"?

Body: To introduce a logic premise normally people use the word If, or Iff (if and only if). Is there a math symbol with the same meaning?


问:
Let $C(0,1)$ be the real linear space of all real functions continuous on ...

How a operator can be shown to be skew symmetric. Here the space is not finite dimensional.

答:
By definition we have:

\begin{eqnarray*}( Tf,g)&=&\int_0^1\left(\int_0^xf(t)dt\right) g(x)dx
\\&=&\int\int_{0\leq t\leq x \leq 1}f(t) g(x)dt\,dx
\\
\end{eqnarray*}

Similarly we have:

\begin{eqnarray*}( Tg,f)&=&\int_0^1\left(\int_0^tg(x)dx\right) f(t)dt
\\&=&\int\int_{0\leq x\leq t \leq 1}f(t) g(x)dt\,dx
\\
\end{eqnarray*}

Adding we get:

\begin{eqnarray*}( Tf,g)+(Tg,f)&=&\int\int_{0\leq t\leq x \leq 1}f(t) g(x)dt\,dx
+\int\int_{0\leq x\leq t \leq 1}f(t) g(x)dt\,dx\\&=&\int_0^1 \int_0^1 f(t) g(x)dt\,dx\\&=&\int_0^1 0 dx =0,
\end{eqnarray*}
as required.


问:
Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $
I have a trouble with a integral:
Using this Laplace trasform equation:
$$\begin{align}
 \int_0^\infty F(u)g(u) \, du  & = \int_0^\infty f(u)G(u) \, du \\[6pt]
 L[f(t)] & = F(s) \\[6pt]
 L[g(t)] & = G(s)
 \end{align}$$

Applying to compute this integral:

$$I = \int_0^\infty \frac{\sin^4 x}{x^3} \, dx $$

答:
Here is how we compute the LT of $\sin^4{t}$ directly.  As mentioned in @Mhenni's solution, write as $\sin^4{t} = (e^{i t}-e^{-i t})^4/(2 i)^4$

$$F(u) = \frac{1}{16} \int_0^{\infty} dt \, (e^{i 4 t} - 4 e^{i 2 t} + 6 - 4 e^{-i 2 t} + e^{-i 4 t}) e^{-u t}$$

Evaluate separately and combine judiciously:

$$F(u) = \frac18 \left [ \frac{u}{u^2+16} - 4 \frac{u}{u^2+4} + \frac{3}{u}\right]$$

As also mentioned above, $g(u) = u^2/2$ (This is a direct look-up in the table, or it may be evaluated using residue theory.)  Now, when we multiply this through, it looks like the integral will be divergent.  However, we may cancel out the $u$ piece by using the fact that $u^2/(u^2+a^2) = (1/a^2) (1-1/(u^2+a^2))$.  We then get

$$\begin{align}F(u) g(u) &= \frac{1}{16} \frac{u^3}{u^2+16} - \frac14 \frac{u^3}{u^2+4} + \frac{3}{16} u \\ &= \frac{u}{16} - \frac{u}{u^2+16} - \frac{u}{4} + \frac{u}{u^2+4}+\frac{3}{16}u\\ &= \frac{u}{u^2+4} - \frac{u}{u^2+16} \end{align}$$

To do the integral, note that we need to be careful because, taken individually, the integrals diverge - but the divergences cancel.  Thus write

$$\begin{align}\int_0^{\infty} du \, F(u) g(u) &= \lim_{R \to \infty} \left (\int_0^R du\, \frac{u}{u^2+4} - \int_0^R du\, \frac{u}{u^2+16}  \right ) \\ &= \frac12 \lim_{R \to \infty}\left [ \log{(R^2+4)} - \log{4} - \log{(R^2+16)} + \log{16}\right]\\ &= \frac12 [\log{2^4} - \log{2^2} ] \\ &= \log{2}\end{align}$$


问:
Closed form of $I = \int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$
I need help with this integral:

$$I = \int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$$

Where $κ,λ>0$.

Neither Mathematica nor Maple could find a closed form for this integral.

Let $G$ follow a $\Gamma\left(\kappa+1,\lambda\right)$ distribution, i.e. its density can be written as $f_G\left(x\right)=\mathbb{I}_{\mathbb{R}_+^\ast}\left(x\right)\ x^\kappa\ e^{-\ \frac{x}{\lambda}}\frac{1}{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}$.

For fixed values of κ and λ, I can use Monte-Carlo to simulate : $$I=\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}\int_{0}^{+\infty}{f_G\left(t\right)\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}=\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}\ \mathbb{E}\left(\sin^2{\left(\frac{G\pi}{2\kappa\lambda}\right)}\right)$$

But I would actually rather have a closed form. Any help or insight will be very much appreciated.

PS : although fluent in english, I mostly study math in french...

Edit : following another user's suggestion, I know have the following :

$$I=\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$$

$$I=\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\frac{1-\cos{\left(\frac{\pi t}{\kappa\lambda}\right)}}{2}dt}$$

$$I=\frac{1}{2}\left(\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}dt}-\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\cos{\left(\frac{\pi t}{\kappa\lambda}\right)}dt}\right)$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\left(e^{i\frac{\pi t}{\kappa\lambda}}+e^{-i\frac{\pi t}{\kappa\lambda}}\right)dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}+i\frac{\pi t}{\kappa\lambda}}\ dt}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}\ -\ i\frac{\pi t}{\kappa\lambda}}\ dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{\left(\kappa-i\pi\right)t}{\kappa\lambda}\ }dt}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{\left(\kappa+i\pi\right)t}{\kappa\lambda}\ }dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{\left(\frac{\kappa\lambda}{\kappa-i\pi}u\right)^\kappa e^{-u\ }\frac{\kappa\lambda}{\kappa-i\pi}du}-\frac{1}{4}\int_{0}^{+\infty}{\left(\frac{\kappa\lambda}{\kappa+i\pi}u\right)^\kappa e^{-u\ }\frac{\kappa\lambda}{\kappa-i\pi}du}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa-i\pi}\right)^{\kappa+1}\int_{0}^{+\infty}{u^\kappa e^{-u\ }du}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa+i\pi}\right)^{\kappa+1}\int_{0}^{+\infty}{u^\kappa e^{-u\ }dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa+i\pi}\right)^{1+\kappa}\Gamma\left(1+\kappa\right)-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa-i\pi}\right)^{1+\kappa}\Gamma\left(1+\kappa\right)$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{4}\left(2-\left(\frac{\kappa}{\kappa+i\pi}\right)^{1+\kappa}-\left(\frac{\kappa}{\kappa-i\pi}\right)^{1+\kappa}\right)$$

Not sure what to do about the complex numbers I get in the end though...

答:
I do not agree with the statement "NeitherMathematica nor Maple could find a closed form for this integral"

Mathematica find a quite simple expression of the antiderivative in terms of the gamma function which can simplify to
$$f(t)=\frac{1}{4} t^{\kappa +1} \left(E_{-\kappa }\left(\frac{t (\kappa -i \pi )}{\kappa  \lambda
   }\right)+E_{-\kappa }\left(\frac{t (\kappa +i \pi )}{\kappa  \lambda
   }\right)-2 E_{-\kappa }\left(\frac{t}{\lambda
   }\right)\right)$$ where appear the exponential integral function. The same for the definite integral but here we face the problem of your other question.

Using the formulation in terms of the expoential integral,
there is no problem when $t\to \infty$ since the result is just $0$. Where the problem starts to be unpleasant is when I try to evaluate $f(0)$.

Now, all my congratulations for your work !


问:
Elementary controls question
I'm new to differential equations as well as to controls so I'd like to show my thought process for a problem.

Let

$$
\begin{align}
\dot x &= z \\
\dot z &= -\frac{d}{dx}V_\alpha(x)
\end{align}
$$
where $V_\alpha(x) = x^4 - \alpha x^2$

We want to show that $(z,x) = (0,0)$ is an equilibrium point or all $-1 \leq \alpha \leq 1$

I started with substituting in our equation and evaluating what $\dot z$ is.

$$
\begin{align}
-\frac{d}{dx} (x^4 - \alpha x^2) &= -(4x^3 - 2\alpha x) \\
&= -2(2x^2 - \alpha x)
\end{align}
$$

I know that I'm supposed to show that at $(z,x)=(0,0)$ I'm supposed to see that when $\alpha$ is on the interval $[-1,1]$ that I have a real component but $z$ isn't in the equation...

Another aspect that I'm familiar with but still not comfortable is that I can put the equations in matrix form but this doesn't make sense to me in this context since it's just one variable.

I know I'm probably pretty far off the mark but some guidance would be well appreciated.

答:
For equilibrium all derivatives must be zero. So you have to find $x$, $z$ such that

$$
\begin{align}
\dot{z} &= 0 \\
\dot{x} &= 0
\end{align}
$$

You have the system

$$
\begin{align}
\dot{z} &= z \\
\dot{x} &= - \frac{d}{d x} V_{\alpha}(x)
\end{align}
$$

and $V_{\alpha}(x) = x^4 - \alpha x^2$. Your derivative is wrong, because $\frac{d}{d x} V_{\alpha}(x) = 4 x^3 - 2 \alpha x$ and so your system is

$$
\begin{align}
\dot{z} &= z \\
\dot{x} &= -4 x^3 + 2 \alpha x
\end{align}
$$

First look at the first equation: You get zero derivative only when

$$
z = 0 \tag{1}
$$

Now look at the second equation: You want $-4 x^3 + 2 \alpha x = 0$. Factor out $x$ you get:

$$
x(\alpha - 2 x^2) = 0 \tag{2}
$$

This is true if $x = 0$ for all $\alpha$ so you have a equilibrium at $(z,x)=(0,0)$.

For this equilibrium the value of $\alpha$ does not matter. However you have two more equilibrium points, one at $(z,x) = (0,\frac{\sqrt{2}\sqrt{\alpha}}{2})$ and one at $(z,x) = (0,-\frac{\sqrt{2}\sqrt{\alpha}}{2})$ because there $(1)$ and $(2)$ are also zero.

Some more notes:

You mentioned a "real component", but that is not really related to finding equilibrium points.

You don't need $z$ in the equation because you have a second equation where $z$ appears.

Usually you put only linear systems in matrix form. Then you can find the equilibrium points easily by solving a system of linear equations (see this answer for some more information). However your system is nonlinear so the matrix form doesn't really help.

You don't have only one variable, you have two variables: $z$ and $x$.

Title: Elementary controls question


问:
Proof of Cauchy Integral Formula using homotopy of curves
This is from Stein and Shakarchi.

Give another proof of the Cauchy integral formula

$f(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta-z}d\zeta$

using homotopy of curves.

(Hint: Deform the circle $C$ to a small circle centered at $z$, and
note that the quotient $\frac{f(\zeta)-f(z)}{\zeta-z}$ is bounded)

I found other posts about this problem but I couldn't find the answer to my question.
I found that the problem reduces to the problem of showing that $\frac{1}{2\pi i}\int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta$ approaches to zero as $r$ goes to zero. I think this is where I need to use the theorem of homotopy in the book (integrals over holomorphic function on two path-homotopic curves is same). But since end points of different circles are not exactly the same, they are not path-homotopic. How can I modify the theorem to apply it in the problem?

答: The integrals along $a$ and $b$ cancel as they are the same line in opposite directions.  Thus the integral around $C_r$ is the same as the integral along the path $aC_rb$, which is homotopic to the path $C_1$.


答:
Let $C$ be circle centered at $z_0$ and let $C_r$ be a circle centered at $z_0$ contained within $C$. We consider

$$\frac{1}{2\pi i} \int_{C} \frac{f(\zeta)}{\zeta-z_0}d\zeta = \frac{1}{2\pi i} \int_{C} \frac{f(\zeta)-f(z_0)}{\zeta-z_0}d\zeta + \frac{f(z_0)}{2\pi i} \int_{C} \frac{1}{\zeta-z_0}d\zeta$$

Now, observe that $C$ can be continously deformed into $C_r$. Thus, we have

$$\frac{1}{2\pi i} \int_{C} \frac{f(\zeta)}{\zeta-z_0}d\zeta=\frac{1}{2\pi i} \int_{C_r} \frac{f(\zeta)-f(z_0)}{\zeta-z_0}d\zeta + \frac{f(z_0)}{2\pi i} \int_{C_r} \frac{1}{\zeta-z_0}d\zeta$$

Now, we prove that the first integral approaches zero as $r \rightarrow 0$:

$$\bigg| \int_{C_r} \frac{f(\zeta)-f(z_0)}{\zeta-z_0}d\zeta \bigg| \leq \int_{C_r} \bigg|\frac{f(\zeta)-f(z_0)}{\zeta-z_0} \bigg| d \zeta \leq M \ell(C_r)=2\pi Mr \rightarrow 0$$

when $r \rightarrow 0$. The last integral is equal to $f(z_0)$, which proves the theorem.

Title: Proof of Cauchy Integral Formula using homotopy of curves

Body:
This is from Stein and Shakarchi.

Give another proof of the Cauchy integral formula

$f(z)=\frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta-z}d\zeta$

using homotopy of curves.

(Hint: Deform the circle $C$ to a small circle centered at $z$, and
note that the quotient $\frac{f(\zeta)-f(z)}{\zeta-z}$ is bounded)

I found other posts about this problem but I couldn't find the answer to my question.
I found that the problem reduces to the problem of showing that $\frac{1}{2\pi i}\int_{C_r} \frac{f(\zeta)}{\zeta-z}d\zeta$ approaches to zero as $r$ goes to zero. I think this is where I need to use the theorem of homotopy in the book (integrals over holomorphic function on two path-homotopic curves is same). But since end points of different circles are not exactly the same, they are not path-homotopic. How can I modify the theorem to apply it in the problem?


问:
Why is $2\mathbf{\hat{k}}\times \mathbf{\hat{j}} = -2\mathbf{\hat{i}}$?
Why is   $2\mathbf{\hat{k}}\times \mathbf{\hat{j}} = -2\mathbf{\hat{i}}$?

How do we do it?

Please tell how to solve it.

答: $$\vec {2k} \times \vec j = 2(\vec {k} \times \vec j)=2(-\vec{i})=-2\vec i$$


问:
Prove that $f$ can have at most finitely many zeros.

Let $f : \Bbb C \longrightarrow \Bbb C$ be an entire function such that $$\lim\limits_{z \to 0} \left \lvert f \left (\dfrac {1} {z} \right ) \right \rvert = \infty.$$ Then show that $f$ can have at most finitely many zeros.

What I can observe is that if $f$ has infinitely many zeros and zero set of $f$ is bounded then by Bolzano-Weierstrass theorem it has a limit point. Then by identity theorem it follows that $f \equiv 0$ on $\Bbb C.$ But then the above limit becomes $0,$ a contradiction. Hence if the zero set of $f$ has infinite cardinality then the set has to be unbounded. Now how do I proceed? Any help in this regard will be appreciated.

Thanks in advance.

答: There exists $\delta >0 $ such that $|z| <\delta $ implies $|f(\frac  1 z)| >1$. In particular $|z| >\frac 1  {\delta} $ implies $f(z) \neq 0$. So all the zeros are in $\{z: |z| \leq \frac1 {\delta}\}$. If there are infinitely many zeros in this compact set there would be a limit point for the zeros which leads to a  contradiction. .


问:
Closed Graph of a function and the continuous inverse implies the image is closed.
Let $X$ and $Y$ be Banach spaces, and let $A \subseteq X$ be a subspace.

Suppose $T: A \rightarrow Y$ is a injective linear map with a closed graph.

Let $B=T(A)$.

Show if the inverse map $T_{^{-1}} : B \rightarrow X$ is continuous, then $B$ is closed.

Attempt:

Suppose $(y_n) \subset B$ and $y_n \rightarrow y$. Then we want to show $y \in B$.

By the Closed Graph Theorem, we have that $T$ is continuous.

We also know that the restrictions map $A \rightarrow T(A)$ is a bijection.

From here I'm not sure how to continue.

Thanks in advance!

答:
Suppose $y_n$ is in $B$ convergent to some $y$. It follows that $y_n$ is a Cauchy sequence, since $T_{-1}$ is continuous it is Lipschitz continuous you have that $x_n:=T_{-1}(y_n)$ is also Cauchy, as such it admits a limit $x\in X$.

So $x_n\in A$ is a convergent sequence in $X$ with $T(x_n)=y_n$ also convergent. Since $T$ has closed graph you have that the limits $x\in A$ and $y\in B$.


问:
A question regarding moment generating function
The moment generating function of a real-valued random variable $X$ on a probability space $(\Omega,\mathscr A,P)$ is defined as $G_{X}(t)=E[e^{tx}]$ $(t \in \mathbb{R})$. Prove that $G_{\alpha X+\beta}(t)=e^{\beta t}G_{X}(\alpha t)$ for $\alpha, \beta \in \mathbb{R}$.

Attempt

$G_{X}(t)=E[e^{tx}]$, then

$$G_{\alpha X+\beta}(t)=E[e^{t(\alpha x+\beta)}]=E[e^{\alpha xt}e^{\beta t}]=E[e^{\alpha xt}]E[e^{\beta t}]=e^{\beta t}E[e^{\alpha xt}]=e^{\beta t}G_{X}(\alpha t).$$

Show that $G_{X+Y}=G_{X} \cdot G_{Y}$.

Attempt

$$G_{X+Y}(t)=E[e^{t(x+y)}]=E[e^{xt}e^{yt}]=E[e^{xt}]E[e^{yt}]=G_{X}(t)\cdot G_{Y}(t).$$

Questions

Are my attempts correct?

In general, does $E[XY]=E[X]\cdot E[Y]$? If yes, what conditions do the random variables $X, Y$ have to satisfy for $E[XY]=E[X]\cdot E[Y]$ to be true?

Suppose we have a random variable $X$, and $g$ is not a function of $x$, then does it follow that $E[g]=g$ for all $g$?

Thank you for your time.

答:
Yes.

$E[XY]=E[X]E[Y]$ means that $X$ and $Y$ are orthogonal. Independence of $X$ and $Y$ is a sufficient (but not necessary) condition for this equality to be true.

Yes, if the expectation is with respect to $X$.


问:
Proving Riordan's identity that $\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1$
In a book titled  "Advances in Problem Solving," authored by Sailesh Shirali, Riordan's identity is mentioned which can also be written as
$$S=\sum_{k=1}^{n} {n-1 \choose k-1} \frac{k!}{n^k}=1~~~~(1)$$
Here, we prove it by the integral representation of $(j+1)!$ as:
$$S=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{(j+1)!}{n^{j+1}}=\sum_{j=0}^{n-1} {n-1 \choose j} \frac{1}{n^{j+1}} \int_{0}^{\infty} x^{j+1} e^{-x} dx$$
$$\implies S=\frac{1}{n^{n}}\int_{0}^{\infty} x ~(n+x)^{n-1} e^{-x} dx=n^{-n}\int_{0}^{\infty} [(x+n)^n-n(x+n)^{n-1}]~ e^{-x} dx$$ $$S=-n^{-n}\left. (x+n)^n~ e^{-x}\right|_{0}^{\infty}=1.$$

It will be interesting to see other approaches for proving (1).

答:
We shall count the number of length $n+1$ sequences $(x_1,\ldots,x_{n+1})$ over a set of $n$ elements in two different ways.

The first is to just independently choose each $x_i$, yielding $n^{n+1}$ sequences.

Alternatively, let $x_{k+1}$ be the first repeated element in the sequence (a repetition must occur because the length of the sequence is greater than $n$). The first $k$ elements of the sequence can be chosen in $k!\binom{n}{k}$ ways (they must all be distinct), and $x_{k+1}$ can be chosen in exactly $k$ ways (choosing which of the first $k$ elements is repeated). The elements at the remaining $n-k$ positions can then be chosen in $n^{n-k}$ ways (there is no restriction on them). Multiplying these together and summing over $k$, we get

$$n^{n+1}=\sum_{k=0}^nk!\binom{n}{k}\cdot k\cdot n^{n-k}$$
$$1 = \sum_{k=0}^n \frac{k}{n}\binom{n}{k}\cdot\frac{k!}{n^k}$$
$$1=\sum_{k=1}^n\binom{n-1}{k-1}\frac{k!}{n^k}$$
which is exactly what we want.


问:
Is my method wrong for solving this integration question?
I wish to solve this question
$$\int_{0}^{1} \int_{0}^{1} \frac{y}{1+xy}\ dx\ dy$$

I take $1 + xy = t$ i.e. $y\ dx = dy$

New Limits = $1$ to $1 + y$.

Expression becomes $\int -1/2\sqrt{t}$.

Still getting wrong answer after second integration :(

Body: The inside integral is $\int_1^{1+y}\frac  1 t dt=\ln t|_1^{1+y}=\ln (1+y)$ and the answer is $\int_0^{1} \ln (1+y)dy=[(1+y)\ln (1+y)-(1+y)]|_0^{1}=2\ln 2 -1$.


问:
BODMAS Rule ("Of" vs "Division")?
What if the precedence of "of" in an expression ?

Consider this expression :-

(2/3) of (4/5) / (6/5) <== so first division will happen first of "of" will be applied ?

Title: BODMAS Rule ("Of" vs "Division")?

Body:
It will be $\frac{2}{3}$ of $(\frac{\frac{4}{5}}{\frac{6}{5}})$

$$\frac{2}{3}\times (\frac{\frac{4}{5}}{\frac{6}{5}})$$


问:
Newbie vector spaces question
So browsing the tasks our prof gave us to test our skills before the June finals, I've encountered something like this:

"Prove that the kernel and image are subspaces of the space V: $\ker(f) < V, \operatorname{im}(f) < V$, where $<$ means a subspace."

Is it just me or there's something wrong with the problem? I mean: I was rewriting the tasks from the blackboard by hand so I may have made a mistake but is a problem like this solvable or I messed up and should rather look for the task description from someone else? Cause for the time being, I don't see anything to prove here since we don't know what V is, right?

答: Implicit in the question is that $f:V\rightarrow V$ is a linear transformation from $V$ to $V$.  What you need to do in this problem is show that the kernel and image are not only subsets of $V$, but actually linear subspaces.  That is, you need to prove that they are closed under scalar multiplication and vector addition.</p>

Title: Newbie vector spaces question


问:
Limit of infinite product $ \lim_{x\to 1^-} \prod_{n=0}^{\infty } \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}$
How do we calculate this limit:

$$ \lim_{x\to 1^-}  \prod_{n=0}^{\infty } \left(\frac{1+x^{n+1}}{1+x^n}\right)^{x^n}$$

I tried to take log and convert the resulting sum of logs into an integral but couldn't proceed after that. The answer is given as $\frac{2}{e}$ which probably means that taking log is the first step. Any help is appreciated.

Edit:

This is given as the solution but I couldn't understand where step 3(they express the limit between two inequalities) comes from.

答:
With $t_n=\ln(1+x^n)$, i.e. $x^n=e^{t_n}-1$, we have
$$\sum^\infty_{n=0}x^n\,(\ln(1+x^{n+1})-\ln(1+x^n))=-\sum^\infty_{n=0}(e^{t_n}-1)\,(t_n-t_{n+1}).$$ Since $t_0=\ln 2$ and $t_n\to0$ as $n\to\infty$, the sum on the RHS is an integral sum for $$\int^{\ln2}_0(e^t-1)\,dt$$ and since $0<t_n-t_{n+1}\le t_0-t_1=\ln2-\ln(1+x)\to0$ as $x\to1^{-}$,
$$\lim_{x\to1^{-}}\sum^\infty_{n=0}x^n\,(\ln(1+x^{n+1})-\ln(1+x^n))=-\int^{\ln2}_0(e^t-1)\,dt=\ln2-1.$$
EDIT: Here, we have an infinite integral sum, but this sum is the integral of a step function with value $e^{t_n}-1$ in $(t_{n+1}, t_n)$, and this function converges to $e^t-1$ monotonically, because of the monotony of $e^t-1$. So convergence is ensured in this case, too.


问:
Proving/disproving a given set $A$ is open in $\Bbb R^2$ with the usual metric
$
\newcommand{\R}{\mathbb{R}}
$
Let $(\R^2, d)$, where $d$ denotes the usual Euclidean distance, be a metric space. A set $A$ is defined as follows:

$$A = \left\{ (x,y) \in \R^2 \; \middle| \; e^{2x} + \left( \frac y 5 \right)^4 < 1 \right\}$$

I think the set should be open in this metric space.

To prove it, is it correct to show the boundary set of $A$ is a subset of $\R^2\setminus A$?

Let $P_0(x_0,y_0)$ be an arbitrary element belongs to the boundary set.

There exists a converging sequence in $A$ that converge to $P_0$.

Also, there exists a converging sequence in the complement of $A$ that is also converging to $P_0$.

Then we have for all $n>N$ for some $N$,

$$e^{2x_n} + \left( \frac {y_n} 5 \right)^4 \ge 1$$

And since the sequence when $n \to \infty$ is $P_0 (x_0,y_0)$.

So after taking the limit we have

$$e^{2x_0} + \left( \frac {y_0} 5 \right)^4 \ge 1$$

Thus $P_0$ is a element of $\R^2\setminus A$.

Does my proof make sense?

答:
There's another approach:
To answer your main question, what do you know of inverse images of open sets, of continuous functions?

Is the given function continuous?
And is (-∞,1) open in R?


问:
Gaussian Elimination, Question Check.
I'm going through my practice problems, and just want to know if I am doing this right:

$$ 2x_1 - 3x_2 = -2$$
$$ 2x_1 + x_2 = 1$$
$$ 3x_1 + 2x_2 = 1$$

And this is my solution:
$$
\begin{align}
\begin{bmatrix}
2 & 3 & -2\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
2 & 1 & 1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & -1 & -1\\
3 & 2 & 1
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & -1 & -1\\
0 & -1 & -2
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & -1 & -2
\end{bmatrix}
\end{align}
$$

$$
\begin{align}
\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & 0 & -1
\end{bmatrix}
\end{align}
$$

Can anyone tell me if I did this right, or if I did a mistake where? Thanks.

答:
first  consider $2x_1 - 3x_2 = -2$ and $2x_1 + x_2 = 1$ your solution is $x_1=\frac{1}{8} , x_2=\frac{3}{4}$

now consider :
$2x_1 + x_2 = 1$ and $3x_1 + 2x_2 = 1$ your solution is $x_1=1,x_2=-1$

so your system does not an answer

and 3 th row of  your last matrix \begin{align}
$\begin{bmatrix}
1 & -3/2 & -1\\
0 & 1 & 1\\
0 & 0 & -1
\end{bmatrix}
\end{align}$
say that your matrix  has not answer too

($ 0x_1+0x_2=1 $ is impossible)

Title: Gaussian Elimination, Question Check.


问:
System of infinite equations
While working I encountered the following system of equations:

$\binom{0}{0} x_0+\binom{1}{0} x_1+\binom{2}{0} x_2+\binom{3}{0} x_3+...=\alpha x_0$

$\binom{1}{1} x_1+\binom{2}{1} x_2+\binom{3}{1} x_3+...=\alpha x_1$

$\binom{2}{2} x_2+\binom{3}{2} x_3+...=\alpha x_2$

$\binom{3}{3} x_3+...=\alpha x_3$

$\vdots$

where $\alpha$ is a fixed real number. I want to find the values of $x_1,$ $x_2,$ $\ldots$ in terms of $x_0.$ The commonly used substitution method is not working in this case. Binomial coefficient is not allowing us to use the substitution method. Is there anyway to find the values of $x_1,$ $x_2,$ $\ldots$? Please help.

答:
@Jean Marie, you have explained this so beautifully. Thank you for refering the paper by Edelman and Strang. I wonder when we consider $q$-binomial coefficient in place of ordinary binomial coefficient will the result will be analogues that is

$x_k=\frac{1}{G_k}x_0$

where $G_k=\sum_{j=0}^{k}\binom{k}{j}_q$.

Please notice that $G_k=2^k$ when $q\to1$.

Title: System of infinite equations


问:
Differentiation on tangent on a point

$$\text{tangent of } e^{x^2} + \ln x + e^{x\ln x}$$

How to find the tangent of this curve at $(1, e+1)$ and differentiate it?

答:
Let $y=e^{x^2}+\ln x+e^{x\ln x}$.

Using the fact that $e^{\ln x}=x$, we have
$$y=e^{x^2}+\ln x+x^x$$

Let us suppose for simplicity, $u=e^{x^2}$, $v=\ln x$ and $w=x^x$.
$$y=u+v+w$$
$$\implies \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}\quad (*)$$

Differentiating $u$

To differentiate $u=e^{x^2}$, we let $t=x^2$.
By Chain Rule,
$$\frac{du}{dx}=\frac{du}{dt}\cdot \frac{dt}{dx}$$
$$\frac{du}{dx}=\frac{d}{dt}(e^t)\cdot \frac{d}{dx}(x^2)$$
$$\frac{du}{dx}=2xe^{x^2}\quad (1)$$

Differentiating $v$ is trivial as we know $(\ln x)'=1/x$. So,
$$\frac{dv}{dx}=\frac{1}{x}\quad (2)$$

Differentiating $w$

For this, take $\ln$ of both sides
$$\ln w =x\ln x$$
Differentiate both sides w.r.t. $x$,
$$\frac{1}{w}\cdot\frac{dw}{dx}=x\cdot\frac{1}{x}+\ln x\cdot 1$$
$$\frac{1}{x^x}\cdot \frac{dw}{dx}=1+\ln x$$
$$\frac{dw}{dx}=x^x(1+\ln x)\quad (3)$$

Using $(1)$, $(2)$ and $(3)$ in $(*)$, we get
$$\frac{dy}{dx}=2xe^{x^2}+\frac{1}{x}+x^x(1+\ln x)$$

Thus the slope of $y$ at the point $(1,e+1)$ is obtained by putting $x=1$,
$$\left[\frac{dy}{dx}\right]_{x=1}=2e+1+1=2e+2$$

Thus, the tangent has slope $m=2e+2$ and lies on $(x_0,y_0)=(1, e+1)$. By point-slope form,
$$y-y_0=m(x-x_0)$$
So, the equation of required tangent is
$$y-e-1=(2e+2)(x-1)$$
$$(2e+2)x-y-(e-1)=0$$

Hope this helps :)

Title: Differentiation on tangent on a point


问:
What is the distribution of $ZX$ when $Z$ is standard normal and $X$ is binary random variable, and $Z$, $X$ are independent?
Let $Z \sim \mathcal{N}(0,1)$ and $X$ be binary random variable taking values in $\{0,1\}$ with some arbitrary pmf. Also, let $Z$ and $X$ are independent. I am trying to find the distribution of $ZX$.

Here is my attempt:

Let $S=ZX$. Then,

$f_S(s) = f_{ZX}(zx)=\sum_{x} f_{Z|X}(z|x)P(X=x)$

But since $Z$ and $X$ are independent, this can be written as

$f_S(s) = \sum_{x} f_{Z}(z)P(X=x) = f_Z(s) \sum_x P(X=x) = f_Z(z)$.

I am not sure if I am making some blunder here, but somehow I don't find this quite correct. Any help is much appreciated.

答:
$ZX$ does not even have a density since $P(ZX=0) >0$.  You have to work with distribution functions.
$P(ZX \leq x)=P(X=0)+P(X=1)P(Z \leq x)=1-p+p\Phi(x)$ if $x \geq 0$ and $P(ZX \leq x)=p\Phi (x)$ if $x < 0$ where $p=P(X=1)$.


问:
If you throw a dice $10$ times, what is the probability that there is at least one $1$ and at least one $6$?

If you throw a dice $10$ times, what is the probability that there is
at least one $1$ and at least one 6?

My Idea:

An analogous question one could ask oneself is: Suppose, you have a combination lock with $n$ different cogwheels. For each cogwheel one has $6$ possible numbers.

So in total we have $6^n$ ways.

If we negate "at least one 1 and at least one 6", we get "no $1$ or no $6$ or neither a $1$ or a $6$" Wouldn't this just be:$$1-\frac{5^n+5^n+4^n}{6^n}$$?

答:
So you're right there are $5^n$ possibilities when you take one out and there are $4^n$ possibilities when you take both out. However there is some overlap. Some of those $5^n$ cases with "no $1$" also have "no $6$", which means you're over counting. Essentially you're counting "no $1$ and no $6$" three times, so you need to take two out, which should give:

$$
1 - \frac{2\cdot5^n - 4^n}{6^n}
$$

There's a theorem for this in probability that's slipping my mind right now, but I'll look for it.

Edit: Yep here is: the addition law of probability. In the second part you're trying to find the probability that there are "no $1$'s" or "no $6$'s":

$$
P(A\cup B) = P(A) + P(B) - P(A \cap B)
$$

In this case $P(A) = P(B) = \frac{5^n}{6^n}$ and the probability of neither (i.e. the intersection) is $P(A\cap B) = \frac{4^n}{6^n}$ which gives what we deduced above.


问:
How to calculate the expectation of the Wishart distribution?
Let $X \sim \mathcal{W}_p(V,\nu)$ follow a central Wishart distribution with scale matrix $V$ and $\nu$ degrees of freedom.
Its p.d.f. is given by:
$$
\frac{|\mathbf{X}|^{(\nu-p-1)/2} e^{-\operatorname{tr}(\mathbf{V}^{-1}\mathbf{X})/2}}{2^\frac{\nu p}{2}|{\mathbf V}|^{\nu/2}\Gamma_p(\frac \nu 2)}
$$
Its expectation is given by:
$$
E[X]=\nu V
$$
How do we actually calculate this expected value? What is the general procedure for matrix valued distributions?</p>

答:
$X$ follows a wishart distribution, $X \sim W_{p}(\Sigma,v)$, if
\begin{equation}
 X = \sum_{i=1}^{v}Z_{i}Z_{i}^{T}
\end{equation}
where $Z_i$ are independent identically p-dimensional normally distributed such that $Z_{i} \sim N_{p}(0,\Sigma)$, and $\Sigma$ is the covariance matrix of $Z_{i}$.

The expectation of a random matrix A is defined as
$\mathbb{E}(A)_{i,j} = \mathbb{E}(A_{i,j})$, i.e it's the expectation of each of its elements. This definition is compatible with random vectors.

Now
\begin{align*}
 \mathbb{E}(X) &= \mathbb{E} \left( \sum_{i=1}^{v}Z_{i}Z_{i}^{T} \right)
\\&= \sum_{i=1}^{v} \mathbb{E}(Z_{i}Z_{i}^{T})
\\ &= \sum_{i=1}^{v} \mathbb{E}(Z_{1}Z_{1}^{T})
\\ &= v\mathbb{E}(Z_{1}Z_{1}^{T})
\end{align*}

Where the equality follows from the linearity of the expectation and that $Z_{i}$ are identically distributed.

But for any random vector $Y$,
$Cov(Y) = \mathbb{E}(YY^{T}) - \mathbb{E}(Y)\mathbb{E}(Y^{T})$. The mean of $Z_{1}$ is zero and hence
$\mathbb{E}(Z_{1}Z_{1}^{T}) = Cov(Z_{1}) = \Sigma$.

Therefore
\begin{equation}
 \mathbb{E}(X) = v\Sigma.
\end{equation}


答:
$$\begin{align}
P[\text{at least one } 1 \cap \text{at least one } 6] &= 1-P[\text{no }1 \cup \text{no }6] \\
\\
&= 1 - \left( P[\text{no }1] + P[\text{no }6] - P[\text{no }1 \cap \text{no }6]  \right) \\
\\
&= 1 - \frac{5^n+5^n-4^n}{6^n}
\end{align}$$

Title: If you throw a dice $10$ times, what is the probability that there is at least one $1$ and at least one $6$?

Body:
If you throw a dice $10$ times, what is the probability that there is
at least one $1$ and at least one 6?

My Idea:

An analogous question one could ask oneself is: Suppose, you have a combination lock with $n$ different cogwheels. For each cogwheel one has $6$ possible numbers.

So in total we have $6^n$ ways.

If we negate "at least one 1 and at least one 6", we get "no $1$ or no $6$ or neither a $1$ or a $6$" Wouldn't this just be:$$1-\frac{5^n+5^n+4^n}{6^n}$$?