问:
The uniform convergence of $\int_0^\infty \mathrm{e}^{-ax^2}x^{2n}\mathrm{d}x \quad w.r.t \ a$

For fixed $n\in \mathbb{N}$, study the uniform convergence of
$$\int_0^\infty \mathrm{e}^{-ax^2}x^{2n}\mathrm{d}x  \quad w.r.t \ a\in(0,\infty)$$.

Actually I want to compute the integral by differentiating
$$
\int_0^\infty \mathrm{e}^{-ax^2}\mathrm{d}x=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}},
$$
but we need to prove $$\int_0^\infty \mathrm{e}^{-ax^2}x^{2n+2}\mathrm{d}x $$
converges uniformly w.r.t $a>0$. Then the procedure is legal. The Weierstrass, Dirichlet and Abel criterion all seem do not work.

Appreciate any help!

答:
The convergence of the improper integral is uniform for $a$ in any interval $[A,\infty)$ where $A > 0$.

This follows from the Weierstrass M-test since $|e^{-ax^2}x^{2n}| \leqslant e^{-Ax^2}x^{2n}$ for all $a \geqslant A$ and for any fixed $n$ the improper integral $\int_0^\infty e^{-Ax^2}x^{2n} \, dx$ converges.

However, for any $c > 0$ and $a_c = 1/c^2 \in (0,\infty)$,

$$\sup_{a \in (0,\infty)}\left|\int_c^{\infty} e^{-ax^2}x^{2n} \, dx\right| \geqslant \sup_{a \in (0,\infty)}\int_c^{2c} e^{-ax^2}x^{2n} \, dx\geqslant \sup_{a \in (0,\infty)}c \cdot e^{-a (2c)^2} c^{2n}\geqslant  e^{-4a_c c^2} c^{2n+1}\\ = e^{-4}c^{2n+1} \underset{c \to \infty}\longrightarrow \infty$$

Thus, the improper integral fails to converge uniformly (with respect to $a$) on $(0,\infty)$.


问:
Example of a random variable s.t. first $2n$ moments exist and rest of the moments don’t exist
We have to construct a random variable $X$ such that its first $2n$ moments exist and rest of the moments don’t exist.

I cannot think of an example which satisfy this can, anybody help me to solve this? I tried various examples. Please help if anybody have knowledge about it.

答: The CDF $1-x^{-k}$ on $[1,\,\infty)$ is an example of a Pareto dstribution, with PDF $kx^{-k-1}$ on its support, so $\Bbb EX^n<\infty$ iff $0\le n<k$. For your purposes, $k=2n+1$ works. Or it would, if you'd said you want higher moments to be infinite. Since you want them to be nonexistent, make the distribution even so its PDF is $\tfrac12k|x|^{-k-1}$ on the support $\Bbb R\setminus(-1,\,1)$. If you'd prefer all of $\Bbb R$ as a support, with whatever smoothness conditions you might seek, I leave it to you to tweak this.

ContentLicense: CC BY-SA 4.0


问:
Karatzas&Shreve 5.4.33 - Equivalence of Three Local Martingales

I encoutered this problem while self-studying Karatzas&Shreve.I have worked out the equivalence relation between $\Lambda_t$ and $M_t$, which is setting $C_t = e^{-\alpha t} $ and $C_t = e^{\alpha t}$ respectively and using Fubini's. But I can't workout the equivalence between $M_t$(or $\Lambda_t$) and $N_t$.

Here are something I've tried:

(1) I tried to use Ito's Lemma since if we assume $M_t \in {\cal M}^{c,loc}$, then $f(X_t)$ is a semi-martingale. But the exponential term becomes a problem.

(2) If we assume $f(X_0) = 0$ (WLOG), then $f(X_t)\frac{d}{dt}\exp\left\{-\int_0^t \frac{{\cal A}f(X_s)}{f(X_s)}\right\} =\exp\left\{-\int_0^t \frac{{\cal A}f(X_s)}{f(X_s)}\right\}{\cal A}f(X_t)$ which looks slightly better not still going to nowhere.

(3) There is no obvious way to apply the hint.

So right now I am just trying random things to see if any of those leads to a spark. But so far none of then did. I am hoping someone could give me some pointers. Thanks in advance.

答:
Call the exponential $Y_t$, so that
$$
dY_t = -{\mathcal A f(X_t)\over f(X_t)}Y_t\,dt.
$$
Now use the "product rule", noting that the covariation between $f(X_t)$ and $Y_t$ vanishes because $Y_t$ is of finite variation:
$$
\eqalign{
d[f(X_t)Y_t)]
&= -\mathcal A f(X_t)Y_t\,dt+Y_tdf(X_t),\cr
&=-\mathcal A f(X_t)Y_t\,dt+Y_t\cdot(dM_t+\mathcal A f(X_t)\,dt)\cr
&=Y_t\,dM_t.\cr
}
$$


问:
Showing linear independence using matrices.

If $\mathbf{v}_1,...,\mathbf{v}_m \in F^n$ are written as rows of an $m\times n$ matrix $A$ and $B$ is the row-reduced echelon form of $A$, then $\{\mathbf{v}_1,...,\mathbf{v}_m\}$ is linearly independant if and only if $B$ has no all zero rows

I thought that a good way of showing this may be to show that linear dependence implies row of zeroes in reduced row echelon form and row of zeroes in reduced row echelon form implies linear dependence.

So I start from assuming that $B$ has a row of zeroes. I'm able to show that each row of $B$ is a linear combination of the rows of $A$ by considering the elementary row operations which reduce $A$ to $B$, and so since the rows of $A$ are $\mathbf{v}_1,...,\mathbf{v}_m$, I have that each row of $B$ is a linear combination of $\mathbf{v}_1,...,\mathbf{v}_m$.

So if $B$ has a row of zeroes, then we have that the zero vector is a linear combination of $\mathbf{v}_1,...,\mathbf{v}_m$. However, my issue is that I am unsure whether this linear combination could possibly be  $0\mathbf{v}_1+0\mathbf{v}_2+...+0\mathbf{v}_m = \mathbf{0}$ as in this case, having a row of zeroes wouldn't imply linear dependence of $\mathbf{v}_1,...,\mathbf{v}_m$. Also, I don't know how I would prove in the opposite direction: That if $\mathbf{v}_1,...,\mathbf{v}_m$ are linearly dependent then $B$ has a row of zeroes.

Can anyone help me with this? Is there a better way to approach this question?

答:
Nice problem :-)
This result is useful for to solve problem in linear algebra.  First, we will write an example of this statament amd then we will give a proof about this stament.

Example: Suppose that we need to analyze if the set $$S=\left\{\begin{pmatrix} 2 \\ 0 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 3 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 4\end{pmatrix}   \right\}$$
I think it's clear that $S$ is independent linearly, but we will use that statement

Informally, your result says that if by putting my vectors as rows of a matrix and then reducing by rows, no row is transformed into a row filled with zeros, then I can state that the set I took to build the rows was linearly independent .

Let $$A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{pmatrix}$$
using Gauss-Jordan's elimination we have $$A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4\end{pmatrix}\sim \cdots \sim \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}=B $$
since that no row in $B$ is filled with zeros, by the statement, we can see that the set is independent linearly.

Proof of the statament: Read here answer your question.


问:
Existence of additional independent RVs
Let $(\Omega, \mathcal{H},\mathbb{P})$ a probability space. Suppose $X : \Omega \to \mathbb{R}$ a random variable. Is it always possible to introduce an additional RV $Y: \Omega \to \mathbb{R}$ where $X,Y$ are independent and identically distributed?

This issue comes up in a proof in Cinlar's Probability and Stochastics, where essentially, we want to show that $Var(X)< \infty$ provided a certain property $P$ of $X$ holds. We first show it assuming $\mathbb{E}(X) =0 $, then show it for all integrable $X$ by supposing $Y$ is some RV iid with $X$ and then show that if we have $P(X)$, we also have $P(X-Y)$ (this is a consequence of how $P$ is defined and relies on the independence of $X$ and $Y$), and then $\mathbb{E}(X-Y) = 0$, so $2 Var(X) = Var(X-Y) < \infty$.

But there seems to be an issue in doing this: how do we know such a $Y$ exists? E.g. suppose that our probability space is $\Omega = (0,1), \mathcal{H} = \mathcal{B}_{(0,1)}, \mathbb{P} = \lambda$, where $\lambda$ is Lebesgue on $(0,1)$. Then let $X : \Omega \to \mathbb{R}$ be given by the identity. Then $\sigma(X)$, the $\sigma$-algebra generated by $X$, is $\mathcal{B}_{(0,1)}$. Thus the only $\sigma$-algebras independent of it are trivial, i.e., they only can have events with probability $0$ or $1$. So the question is, how do we justify the above procedure, i.e., how can we introduce new iid RVs?

Note the what Cinlar does is even worse, since he has some sequence $\{X_n\}$ (of independent RVs), and he wants to introduce a new sequence $\{Y_n\}$ independent of $\{X_n\}$ and so that $Y_j$ has the same distribution as $X_j$.

答:
Suppose $\Omega = \{1, 2 \}$, $\mathcal{F} = 2^{\Omega}$, $P(1) = P(2) = \frac12$.

Put $X(\omega) = \omega$. It's easy to see that this is counterexample.

Indeed, suppose that $Y$ has the same distribution and $X$ and $Y$ are independent. Then $P(X = 1, Y =1)$ must be equal to $\frac{1}{4}$, but there's no any event in $\mathcal{F}$ with such probability.

Title: Existence of additional independent RVs


问:
Solving coupled nonlinear ODEs
I am trying to find $z$ such that
$$\dot{z} = -1 + e^{-iz^*},$$
where $*$ denotes complex conjugate and the dots represent derivatives with respect to time. The time dependence of the dependent variables is suppressed for clarity of presentation. Letting $z=x+iy$,
$$\dot{x} = -1 -  e^{-y}\cos x, \quad \dot{y}=  e^{-y}\sin x,$$
together with the constraint that
$$A = -y+ e^{-y}\cos x$$
is a constant, i.e. $\dot{A}=0$.

The normal way I would proceed would be to eliminate $y$ in favor of $x$, which yields product log (ie Lambert W) functions, but the resulting ODE is complicated, and I cannot solve it in closed form. Note, if we take $z = i \log \zeta$, we can equate this to a problem involving a point vortex in a uniform stream, but I'm not sure this identification helps.

答:
$$\begin{cases}
\frac{dx}{dt}=-1-e^{-y}\cos(x) \\
\frac{dy}{dt}=e^{-y}\sin(x)
\end{cases}\quad\implies\quad
\frac{dy}{dx}=\frac{e^{-y}\sin(x)}{-1-e^{-y}\cos(x)}$$
$$\left(1+e^{-y}\cos(x) \right)dy+e^{-y}\sin(x)dx=0$$
$$dy+d\left(-e^{-y}\cos(x) \right)=0$$
$$y-e^{-y}\cos(x)=\text{constant}$$
This is equivalent to the so called "constrain" :
$$-y+e^{-y}\cos(x)=A$$
Thus this is not realy a constrain. This is the equation of trajectory corresponding to the solution of the original system of ODEs.

The above implicit equation can be explicitly written on various forms :
$$x(y)=\cos^{-1}\left( (y+A)e^y\right)$$
$$y(x)=-A+\text{W}\left(e^A\cos(x) \right)$$
W denotes the Lambert W function.

$$\frac{dy}{dt}=e^{-y}\sin(x)=\pm e^{-y}\sqrt{1-\cos^2(x)}=\pm e^{-y}\sqrt{1-(y+A)^2e^{2y}}$$
$$\frac{dy}{dt}=\pm\sqrt{e^{-2y}-(y+A)^2}$$
$$t=\pm\int \frac{dy}{\sqrt{e^{-2y}-(y+A)^2}}$$
This gives the analytic solution $t(y)$.

As far as I know, this integral cannot be expressed with a finite number of standard functions. A-fortiori don't expect the inverse function $y(t)$ on explicit form. Better use numerical calculus.

Title: Solving coupled nonlinear ODEs


问:
Find the absolute maximum and minimum of $f(x) = \left|\cos^2(x) - \frac{3}{4}\right|$ on $[0,\pi]$

Find the absolute maximum and minimum of $f(x) = \left|\cos^2(x) - \frac{3}{4}\right|$ on $[0,\pi]$

I have learnt that in order to find the absolute maximum, I need to check the value of the function at the:

sides of $[0,\pi ]$

in points that the function isn't differentiable.

in points that satisfy $f'(x)=0$

 How do I approach this question? I have no idea how to differentiate it with the absolute value, and how could I find the points that aren't differentiable in $[0,\pi ]$.
 any help is appreciated.

答:
Well, I will give you some (big) hints:

Notice that when $x\in\left[0,\frac{\pi}{6}\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\cos^2\left(x\right)-\frac{3}{4}$;

Notice that when $x\in\left[\frac{\pi}{6},\frac{5\pi}{6}\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\frac{3}{4}-\cos^2\left(x\right)$;

Notice that when $x\in\left[\frac{5\pi}{6},\pi\right]$ we have $\left|\cos^2\left(x\right)-\frac{3}{4}\right|=\cos^2\left(x\right)-\frac{3}{4}$;

Notice that:

$$\cos^2\left(x\right)=\frac{1+\cos\left(2x\right)}{2}\tag1$$

Notice that:

$$\frac{\text{d}}{\text{d}x}\left(\cos^2\left(x\right)-\frac{3}{4}\right)=\frac{1}{2}\cdot\left\{\frac{\text{d}}{\text{d}x}\left(1\right)+\frac{\text{d}}{\text{d}x}\left(\cos\left(2x\right)\right)\right\}-\frac{3}{4}\cdot\frac{\text{d}}{\text{d}x}\left(1\right)=$$
$$\frac{1}{2}\cdot\left\{0+\frac{\text{d}}{\text{d}x}\left(\cos\left(2x\right)\right)\right\}-\frac{3}{4}\cdot0=-\frac{1}{2}\cdot\sin\left(2x\right)\cdot\frac{\text{d}}{\text{d}x}\left(2x\right)=$$
$$-\frac{1}{2}\cdot\sin\left(2x\right)\cdot2\cdot\frac{\text{d}}{\text{d}x}\left(x\right)=-\frac{1}{2}\cdot\sin\left(2x\right)\cdot2\cdot1=-\sin\left(2x\right)\tag2$$

Notice that, likewise $(2)$ only with a flipped sign:

$$\frac{\text{d}}{\text{d}x}\left(\frac{3}{4}-\cos^2\left(x\right)\right)=\sin\left(2x\right)\tag3$$


问:
What is the rule used to calculate the terms of this expansion?
I did the following exercise : "Find Maclaurin series for $e^{-5x}$. "

My result was $e^{-5x} = (e^{-x})^5 = \left[\sum_{i=0}^{\infty}\frac{(-1)^nx^n}{n!} \right]^5$,   which I verified was correct with Wolfram Alpha.

In the book they had the result in expanded form like so : $e^{-5x} = 1 - 5x + \frac{5^2x^2}{2!} - \frac{5^3x^3}{3!} + \frac{5^4x^4}{4!} - ...$

I'm pretty sure this is some kind of application of the Multinomial Theorem or something?  I was just wondering, since there so obviously is a pattern here, what rule is applied here to find the first several terms of an infinite series raised to some power $k$?

Body:
Since $e^t=\sum_{i=0}^\infty \frac{t^i}{i!}$, just take $t=-5x$ to get
$$e^{-5x}=\sum_{i=0}^\infty \frac{(-5x)^i}{i!}=\sum_{i=0}^\infty \frac{(-1)^i 5^i x^i}{i!} = 1-5x+\frac{5^2}{2!}x^2-\frac{5^3}{3!}x^3+\frac{5^4}{4!} x^4+\ldots$$


问:
enumerability axiomas 1 implies 2 and vice versa
Be $X$ a topological space and $x \in X$. Prove that $x$ admits a fundamental system of enumerable neighbourhoods if, and only if, there is a local base in $x$ that is enumtable.

Definition. Be $X$ a topological space:

If each point of space $X$ has a fundamental system of enumerable neighbors, we say that it satisfies the first enumerability axiom.

When $X$ has an enumerable base we say that $X$ satisfies the second enumerability axiom.

When $X$ is separable (that is, it has an enumerable dense set) then we say that $X$ satisfies the third enumerability axiom.

(i) What this exercise says is that the first enumerability axiom implies the second enumerability axiom?
(ii) Can I consider the elements of this neighbourhood system to be the elements of the base, and vice versa? If so, then this demonstration is trivial.

答:
A space that satisfies your first axiom of enumerability is called a first countable space; one that satisfies your second axiom is called a second countable space.

It is not true that every first countable space is second countable: for instance, an uncountable space with the discrete topology is first countable but not second countable. Let $x\in X$; the exercise is to show that there is a countable fundamental system of nbhds of $x$ if and only if there is a countable local base at $x$.

Yes, if there is a countable local base $\mathscr{B}$ at $x$, then $\mathscr{B}$ is a countable fundamental system of nbhds of $x$. You have to work a little harder for the other direction. Let $\mathscr{N}$ be a countable fundamental system of nbhds of $x$. Then for each $N\in\mathscr{N}$ there is an open set $B_N$ such that $x\in B_N\subseteq N$. Let $\mathscr{B}=\{B_N:N\in\mathscr{N}\}$; $\mathscr{B}$ is countable, so to finish the proof you just have to show that $\mathscr{B}$ is a local base at $x$.


问:
Collapsing constant regions of path
Let $f: [0,1] \to X$ be some continuous map ($X$ some topological space).
Let introduce the equivalence relation on $[0,1]$
$y \sim_c z \iff $ $f$ constant on $[y,z]$ (or on $[z,y]$ if $z < y$)

Is the space $[0,1]/\sim_c$ with quotient topology homeomorphic to some interval [a,b]?

And for which topological spaces $X$ is this true?

答:
The statement is true iff $X$ is a $T_1$ space: See https://math.stackexchange.com/a/109333/58818 for a proof of the hard direction (that $T_1$ spaces have the desired property).

Let us show that if $X$ is not $T_1$, then there is a continuous function $f\colon I\to X$ such that the quotient $X/\!\sim_c$ is not homeomorphic to an interval. The idea is basically to find a sort of copy of the Sierpinski space within such $X$.

Since $X$ is not $T_1$, there are $a\neq b$ such that every neighbourhood of $b$ contains $a$.

Let $f\colon[0,1]\to X$ be given by
$$f(t)=\begin{cases}b&\text{ if }t\leq 1/2\\a&\text{ if }t>1/2\end{cases}$$

Then $f$ is continuous, and there are precisely two $\sim_c$-equivalence classes: $[0,1/2]$ and $(1/2,1]$. This means that the quotient $X/\!\sim_c$ has precisely two points, and is immediatelly not homeomorphic to an interval.

(In fact, the open sets in this quotient are only $X/\!\sim_c$, $\varnothing$ and $\left\{(1/2,1]\right\}$  (it is another copy of the Sierpinski space), and it is not even Hausdorff.)

Title: Collapsing constant regions of path


问:
Compactness of $K(S)$, if $K$ is an infinite-dimensional compact operator in Hilbert space

$H$ is infinite-dimensional Hilbert space. $K: H \rightarrow H$ is infinite-dimensional compact operator. Let $S$ be a unit sphere in H. The task is to proof that $K(S)$ couldn't be compact.

I know that:

Really we need to proof that $K[S]$ is not closed. If B is unit ball, $[B]$ - closure, then $K([B])$ is also closed(for compact operator in Hilbert space).

If $\mathrm{Ker} B$ is empty, i.e. $0 \notin K(S)$ then there is a sequence in $K(S)$ converging to $0$, so $K(S)$ isn't compact. But I don't understant how to proof noncompactness of $K(S)$ even if one of the basis vectors is mapped to $0$.

On the other side we can try to proof that if $K(S)$ is compact, then $\mathrm{Im}(K)=K(H)$ shuold have finite dimension. It is easy if $K(B)$ contains some ball(like a subspace in $K(H)$), but it's far from being true.

答:
Actually, if $K \colon X\to Y$ is a compact operator and $X$ is reflexive, then   $K(S)$ is    compact.  Notice that since $K$ is compact, it must be completely continuous, i.e., if $x_n \to x$ weakly in $X$, then $||K(x_n)-K(x)||\to 0$.
(This is a corollary of  Banach-Steinhaus's Theorem.)

To see that  $K(S) $ is compact,  let $ (K(x_n))_n $ be a sequence in $K(S)$. Since $X$ is reflexive, $S$ is weakly compact and thus (by Eberlein-Smulian's Theorem), we can find a weakly convergent subsequence,  $x_{n_k} \to x \in X$ weakly. By the complete continuity of $K$, it follows that $K x_{n_k} \to Kx \in K(S)  $ (so, it has a (norm) convergent subsequence).


问:
Generate a uniform distribution of orientations
I am rendering images of a 3D object and am trying to generate a uniform distribution of orientations for this object. I am unsure of how to do this, and considerations that are necessary to take into account.

Is a uniform distribution of orientations equivalent to a uniform distribution of rotations as apparent to a rendering camera? That is, if I generate uniform orientations, will it look uniform to the camera?

How do I generate a uniform distribution of orientations? Is it possible to generate uniform points on a d-sphere to get a set of rotations? If so, would this mean generating points on a 3D sphere or a 4D hypersphere?

答:
Simple in Mathematica:

```<code>Graphics3D[
   {Opacity[0.5], Yellow, Sphere[],
    Black, PointSize[0.02],
    Point /@ SpherePoints[200]}]
</code>```

Body:
Simple in Mathematica:

Graphics3D[
   {Opacity[0.5], Yellow, Sphere[],
    Black, PointSize[0.02],
    Point /@ SpherePoints[200]}]


问:
Find probability of discriminating company using Bayes theorem
After watching several tutorial video, I thought I finally understood the idea and tried few online quiz on Bayes theorem. I'm happy that I could solve some questions on my own but stuck with the following question :

Suppose that 5% of companies in a certain industry discriminate against Iowans. If a company discriminates, it will never hire someone from Iowa.
Suppose that 20 equally-qualified applicants apply for jobs at a company in this industry, and six are from Iowa. If this company hires four people from this set of applicants, but none are from Iowa, what is the probability that this company discriminates?

I could not relate the job applicant data with probability of discriminating company. Need some pointers to approach this prob. Any help is highly appreciated!! Thanks!

答:
Hint: Denote $D$ by the event that the company discriminates, and $E$ by the event that happened. Then $$P(D|E) = \frac{P(D)\cdot P(E|D)}{P(D)\cdot P(E|D) + P(\bar D) | P(E|\bar D)} $$

$P(E|D)=1$ as the company must only select non-Iowa people if it discriminates.

$P(E|\bar D) = \frac{14\choose 4}{20\choose 4} $ In this case, everyone is equally likely to be hired.


问:
Show the improper integral $\int_0^\infty \frac{\sin(x^3-ax)}{x}dx$ converges.
This is an exercise for the Dirichlet test for the convergence of improper integrals. This exercise is from the book called “A Course of Modern Analysis”. It does not specify what $a$, but I am thinking $a$ is a nonnegative number. I am not sure if this assumption is legit. Clearly, $\frac{1}{x}$ converges to 0 as $x \to \infty$. I am struggle to show $|\int_0^L \sin(x^3-ax)dx|$ for all $L$ is bounded. I tried to use u-sub but it does not make the problem easier. Any hint will be appreciated.

Body:
hint

$$\sin(x^3-ax)=\sin(x^3)\cos(ax)-\cos(x^3)\sin(ax)$$


问:
What does a horizontal bar over two vectors indicate?
I'm reading a physics paper dealing with angular mechanics. There is a vector G representing Force of gravity multiplied by a position vector C, all cross-producted with a Force vector. However, there is a horizontal bar (Vinculum?) over the vectors GC, as shown in the picture:
Angular Moment of Momentum Equation.
What does the bar represent? The only thing I could find elsewhere is that it is the complex conjugate, which makes no sense in this context.

答: Nevermind. My Physics teacher said it was the line formed between the two vectors. I am not sure what a non-geometric interpretation of it would be, but the solution was much simpler than I thought.</p>


问:
Continuous function by transfinite induction
Let   $A$  and  $B$ subsets of $\mathbb R$ such that both have cardinality $\mathfrak c.$ I want to construct a noncontact function $f\colon A\to B$  by transfinite induction such that $f$ is continuous function. I know how construct non continuous function by transfinite induction. This is just an question come to my mind I do not know how it can be done.

Any help will be appreciated greatly

答:
Not quite transfinite induction, but a complete answer nonetheless. If $A$ is disconnected, then you can write $A=A_1\amalg A_2$ for two non-empty subsets which are clopen in the subspace topology of $A$. Then, select two distinct elements $b_1,b_2\in B$; the function $$f:A\to B\\ f(a)=\begin{cases}b_1&\text{if }a\in A_1\\ b_2&\text{if }a\in A_2\end{cases}$$
is continuous.

If $A$ is connected, then $A$ is an interval, and it's all a matter of whether or not $B$ contains some interval with at least two points.
If it doesn't, then by intermediate value theorem all continuous functions $A\to B$ must be constant.

On the other hand, if without loss of generality $(b_1,b_2)\subseteq B$ for some $b_1<b_2$, then you can send $A\to (b_1,b_2)$ with one of the usual continuous functions from an interval into another (not necessarily onto). Say, for instance, $$f(a)=b_1+\frac{b_2-b_1}\pi\left(\frac\pi2+\arctan a\right)$$


答: Hint. Note that $$\left|\frac{\sin (x^3-ax)}{x}\right|\le \frac1x$$ for all $x>0.$

Title: Show the improper integral $\int_0^\infty \frac{\sin(x^3-ax)}{x}dx$ converges.

Body: This is an exercise for the Dirichlet test for the convergence of improper integrals. This exercise is from the book called “A Course of Modern Analysis”. It does not specify what $a$, but I am thinking $a$ is a nonnegative number. I am not sure if this assumption is legit. Clearly, $\frac{1}{x}$ converges to 0 as $x \to \infty$. I am struggle to show $|\int_0^L \sin(x^3-ax)dx|$ for all $L$ is bounded. I tried to use u-sub but it does not make the problem easier. Any hint will be appreciated.


答:
For any $x>\sqrt{\frac{a}{3}}$ we have that $g(x)=x^3-ax$ is an increasing function whose inverse function $h(x)$ behaves like $\sqrt[3]{x}$ for large values of $x$. $g(h(x))=x$ leads to $h'(x)=\frac{1}{g'(h(x))}=\frac{1}{3h(x)^2-a}$ and this gives

$$ \int_{\sqrt{a/3}}^{L}\sin(x^3-ax)\,dx = \int_{g(\sqrt{a/3})}^{g(L)}\frac{\sin(t)}{3h(t)^2-a}\,dt. $$
The last integral is convergent by Dirichlet's test/integration by parts since $\sin t$ has a bounded primitive and $\frac{1}{3h(t)^2-a}$ decreases to zero. As a consequence, $\int_{0}^{L}\sin(x^3-ax)\,dx$ is bounded.


问:
Having square roots in denominator in right angle trig
*apologies ahead of time, i am relativly new to this so I am unsure the proper keystrokes to do the fancy equations on here :| *

I can't remember if this is a general rule or not...

but when solving for $\sin$, $\cos$, $\tan$, etc., if you have a square root (imma abbreviate it as sqrt)in the denominator, are you supposed to rationalize it (multiplying the numerator and denominator by the sqrt value)?

for instance, It's been a while I took a math class, and I could've sworn one of my teachers expressed that you cannot have a sqrt in the denominator when giving your answers for right angle trig. But in this [Precalc crash course][1] on Youtube (it's the 5 hour one done by UNC@Chapel Hill posted by freeCodeCamp.org, time stamp 01:36:52 - 01:38:52), the professor left them as is.

ex. in the example equation she used, a right angle with a hypotenuse of $5$, and an opposite side of $2$, solve for sine, cosine, tangent, cosecant, secant, and cotangent.
knowing to solve for the last side, adjacent, you use pythagorean theorem, $a^2 + b^2 = c^2$, giving you adjacent = $\sqrt(21)$.

the professor said that $\tan = \frac{2}{\sqrt 21}$ and $\sec = \frac{5}{\sqrt 21}$.

isn't the answer though supposed to be $\frac{2 \sqrt 21}{21}$ for tangent? $\frac{5 \sqrt 21}{21}$ for secant?

help? clarifications?

edit:
y'all thx so much for the help and edits
[1]: https://www.youtube.com/watch?v=eI4an8aSsgw

答: It's only subjective. Most maths teachers find it ugly to put a square root at the denominator, but there is nothing false in doing it.</p>


答: Hint. Note that $$\left|\frac{\sin (x^3-ax)}{x}\right|\le \frac1x$$ for all $x>0.$ And it doesn't matter what $a$ is.


问:
Prove that in a sequence of chained subsets, the intersection is finite and nonempty
The heading is merely a simplified version. Currently, I am reading Understanding Analysis and working on the preliminaries. The question is:

If $A_1 \supseteq A_2 \supseteq A_3 \supseteq A_4\cdots$ are all finite, nonempty sets of real numbers, then the intersection $\bigcap_{n=1}^\infty A_n$ is finite and nonempty.

The book at this point has not formally defined finite. In addition, the only hint, in my opinion, offered by the book is the following question,

If $A_1 \supseteq A_2 \supseteq A_3 \supseteq A_4\cdots$ are all sets containing an infinite number of elements, then the intersection $\bigcap_{n=1}^\infty A_n$ is infinite as well.

With this question and an aforementioned example, I can solve this problem by defining the set $A_i = \{i,i+1,i+2\dots\}\subseteq N$ and a proof by contradiction.

However, when it comes to $A_i$ containing finite elements, I just don't now how to

Prove by definition

Understand the intuition behind can't find a counter example like the infinite version

答:
One way is to notice that a decreasing sequence of positive integers, in this case the cardinalities of the sets $A_k$, must eventually be constant. For $k\in\Bbb Z^+$ let $n_k=|A_k|$, the number of elements in $A_k$; $n_k$ is a positive integer. Let $N=\{n_k:k\in\Bbb Z^+\}$; $N$ is a non-empty set of positive integers, so it has a smallest element $m$. Let $\ell\in\Bbb Z^+$ be such that $n_\ell=m$.

$A_{\ell+1}\subseteq A_\ell$, so $n_{\ell+1}\le n_\ell=m$. But $m=\min N$, so $n_{\ell+1}\ge m$, and therefore $n_{\ell+1}=m$. Thus, $A_{\ell+1}\subseteq A_\ell$ and $|A_{\ell+1}|=|A_\ell|$ , so $A_{\ell+1}=A_\ell$. You can use this idea to prove by induction that $A_k=A_\ell$ for every $k\ge\ell$. Then you’re almost done. $A_k\supseteq A_\ell$ for $k=1,\ldots,\ell$, and $A_k=A_\ell$ for $k>\ell$, so

$$\bigcap_{k\ge 1}A_k=\bigcap_{k=1}^\ell A_k\cap\bigcap_{k>\ell}A_k=A_\ell\cap A_\ell=A_\ell\,.$$


答:
As $u \to 0$, $\sin(u)/u \to 1$, so in particular $$\lim_{x \to 0} \frac{\sin(x^3 - ax)}{x} = \left( \lim_{x \to 0} \frac{\sin(x^3 - ax)}{x^3 - ax} \right) \left( \lim_{x \to 0} \frac{x^3 - ax}{x} \right) = 1 \times -a = -a,$$

since $x^3 - ax \to 0$ as $x \to 0$. So the integrand is bounded in a neighborhood of $x = 0$ and, in fact, has a removable discontinuity at the endpoint $x = 0$.

As for convergence on the infinite end, looking at a graph, it's instantly clear that the integral is equal to an alternating series, with the positive terms being the areas above the curve, and the negative terms being the areas below the curve:

The mildly tedious part is verifying that the $n$th term of this series (the area between consecutive x-intercepts) is approaching $0$ as $x \to \infty$, but once you have that, you can just say that the improper integral converges by Alternating Series Test. This is a common way to handle improper integrals of functions that oscillate forever. The integrand doesn't have to approach $0$, it just has to be bounded: the same strategy can verify that, for instance, $$\int_0^\infty \cos(x^p) dx \text{ converges for any } p > 1.$$


问:
Differential Equation but by trial and error solution
Original question :

Find $f(x)$, $x\geq 0$, such that $$f'(x) =
 \frac{\sqrt{f(2x)-1}}{\sqrt{2}}$$

My work :

$\begin{align}
f'(x) &= \frac{\sqrt{f(2x)-1}}{\sqrt{2}}\\
f(2x) &= 2(f'(x))^2 + 1\tag{1.1}\\
f(x) &= 2\left(f'\left(\frac x2\right)\right)^2 + 1
\end{align}$

Suppose i want $f'(x)=x$ (a random guess).
Then $f(x)=\frac{x^2}{2} + C$ and
$$f(2x) = 2x^2 + C$$
So, we have $C=1$ from $(1.1)$

Thus, $f(x)=\frac{x^2}{2}+1$. But my friend said there's another solution involving exponential function idk maybe a hyperbolic? Since there's a similar property of hyperbolic function like this. But again, how do we know if there are no other solutions? How do you solve this?

Please help. Thanks

答:
Your friend is correct $f(x) = (e^x + e^{-x})/2$ is also a solution. As for the second part of you're question I'm not $100\%$ sure about the details but I think existence and uniqueness theorem for ODEs suggest that for many initial conditions $f(0) = y_0$ there will exist solutions. This makes we think that this problem might have infinitely many solutions however I don't know how many of these solutions will be globally defined.

EDIT: I thought about this more and what I said above doesn't make sense. However I think still think that more solutions should exist because we can substitute $f(x) = \sum_n c_n x^n$ and then solve for $c_n$ in the resulting equation.


问:
How to solve the equation $x^2-2y=z^2$?
Consider the following equation
$$x^2-2y=z^2,$$
according some theorems in my research, I found that the only integer solution of the equation such that $xy\neq 0$ is $$(x,y,z)=(2,2,0).$$ Now my question is: how to solve the equation (or what is the way or method)? Anyone can help me. Thanks in advance(
I tried to rewrite the equation as follows
\begin{align}
2y=x^2-z^2 \Rightarrow 2y=(x-z)(x+z)\Rightarrow y=\frac{(x-z)(x+z)}{2}
\end{align}
but it did not work).

答:
Write
$$
x^2-z^2=2y\tag 1
$$
Assume now that $d$ runs through all even divisors of $2y$. Then all  solutions  of (1) are (with given $y$):
$$
x=\frac{d}{2}+\frac{y}{d}\textrm{ and }z=-\frac{d}{2}+\frac{y}{d}\tag 2
$$

Title: How to solve the equation $x^2-2y=z^2$?


问:
How I can solve this differential equation ( i did transformed to u for make simply, then how i can do solve?
I transformed to du/dx - u/x = -x. then next how i can solve..? can you give any hint? original equation was this.

enter image description here

答:
$\frac{du}{dx} -\frac{u}{x} = -x$

By multiplying both sides by $1/x$, we get :

$\frac{1}{x} \frac{du}{dx} -\frac{u}{x^2}=-1$

Then we remark that $\frac{d(u/x)}{dx}=\frac{1}{x} \frac{du}{dx} - \frac{u}{x^2}$

So we get that $\frac{d(u/x)}{dx} = -1$.

We integrate : $\frac{u}{x} = -x+a$ for some $a$.

So $u(x) = -\frac{x^2}{2} +a x +c$ for some $c$.


问:
A simple finite field extension only has finitely many intermediate fields
Let $L/K$ be a finite and simple field extension, so $L = K(\alpha)$ and $[L:K] = n$. Show that $L/K$ only has finitely many intermediate fields $M$.

I'd like to know if my following proof is correct or how I could improve it:

Suppose $L = K(\alpha)$. For every intermediate field $M$ let $f_M \in M[X]$ be the minimal polynomial of $\alpha$ over $M$. Then for every $M, M'$ we have $M \subseteq M' \iff f_M \in M'[X]$. Now I wanna show that every $f_M$ uniquely defines the corresponding intermediate field $M$:

Let $M'$ be another intermediate field s.t $f_M \in M'[X]$. Then by the above we have $M \subseteq M'$ and by the definition of $f_M$ and $f_{M'}$ it follows that $\deg(f_{M'}) \leq \deg(f_M)$. Since every root of $f_{M'}$ in $M'$ is also a root of $f_M$ (I'm not sure about this one so in case this isn't necessarily true some clarification would be great!), we get $f_{M'} \mid f_M$. However, since both polynomials are irreducible and monic, we have $f_M = f_{M'}$ which gives us $M = M'$. Similarly we have that $f_M  \mid f_K$ in $L[X]$ for every $M$.

Now, since $L[X]$ is a UFD, we get:
$$f_K = \prod_{K \subseteq M \subseteq L}f_M.$$
Since this factorization is unique and all factors are monic and irreducible there are no units in the product so we get:
$$\deg(f_K) = \sum_{K \subseteq M \subseteq L}\deg(f_M) = n$$
which means there are only finitely many intermediate fields.

答:
I see a couple of problems with your proof. But we will manage to salvage the idea of your proof:

"Problem": The statement that $M\subset M'\Leftrightarrow f_M\in M'[X]$ needs a proof (unless you know this from somewhere). The implication "$\Rightarrow$" is immediate, the other direction is not. The problem is, that it could happen that $M\not\subset M'$ and $M'\not\subset M$.

Solution: If $f_M\in M'[X]$, then $f_M = f_{M\cap M'}$. Let $F:=M\cap M'$ Then we can conclude by a degree argument:
$$[L:K] =[L:F]\cdot[F:K]= [L:M]\cdot[M:F]\cdot[F:K]$$
But $[L:M]=[L:F]$ (since the minimal polynomials are the same) and hence $[M:F]=1$. Thus $M=F=M\cap M'$ and $M\subset M'$.

Problem: From your assumption that $f_M\in M'[X]$ you can't deduce that $f_M=f_{M'}$. As a counterexample look at any nontrivial extension $K(\alpha)/K$. $f_K\neq f_{K(\alpha)}$ since $f_{K(\alpha)}=X-\alpha$. The problem in your argument is the sentence: "However, since both polynomials are irreducible and monic...". Both are only irreducible as polynomials over their respective base fields.

Solution: Strike this paragraph. As you mention in the first paragraph: "Now I want to show that any $f_M$ uniquely determines the corresponding intermediate field $M$." Do just that! Assume that $f_M=f_{M'}$ and you can skip a couple of sentences in your proof.

Problem: Your claim $f_K = \prod_{K \subseteq M \subseteq L}f_M$ doesn't hold. I assume that again you are making the mistake of assuming that all of those $f_M$ are irreducible over $L[X]$? You can check that this is wrong with any non-trivial example. If $f_K = f_K\cdot g$, then $g=1$, so all the $f_M$ in your product would have to be units, thus constant).

Solution: You don't need this. It will suffice that in a UFD any element only has (up to associates) only finitely many divisors (why?). Any $f_M$ divides $f_K$, so there can only be finitely many such $f_M$. Thus there can only be finitely many intermediate fields.

Bonus: Why does any $f_M$ divide $f_K$? We know that $f_K\in M[X]$ and by definition (or a basic lemma, depending on your source) $f_M$ divides any polynomial $g\in M[X]$ that satisfies $g(\alpha)=0$. We know that $f_K(\alpha)=0$.


问:
What is the simplest way to do this combinatorics problem?
Question: Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to each other?

Here are the solutions:

In this case, we take the first case of all possible combinations: $\frac{8!}{2^4}$
But we see that we counted the possibilities with at least one pair together
so we have to subtract $\frac{7!}{2^3}\cdot \binom{4}{1}$ if we tie the two offending shirts together.

^This I don't get - why and what is being double-counted?

However, we see now that we double subtracted ones that have two pairs thus we must add the chance of two pairs so we must add $\frac{6!}{2^2}\cdot \binom{4}{2}$ and so on

Thus the final answer is equal to $864$

Is there a better way that doesn't involve BOTH adding AND subtracting?

答:
Because the problem involves rather small numbers, we can count by hand without using inclusion/exclusion.

Assume the first style of shirt that appears, left-to-right, is A, and that the next style is B, followed (disregarding any more A,B) by one of style C (so that the last style to appear in this sequence is D).  Any permutation of those four styles produces a distinct, but still valid, arrangement.  By counting the sequences with that fixed order of appearance, we will get a number that multiplied by $4!$ gives the answer to the original problem.

Accordingly the sequence resembles AB_C_D_ with the stipulation that underscores represent the combined locations of the four remaining shirts.  An underscore might correspond to zero or more of the four shirts, but with certain restrictions.  The first underscore cannot hold C or D styles, and the second underscore cannot hold a D style.

In consequence the final D has to go into the spaces represented by the last underscore, splitting it into two parts.  Furthermore we cannot leave the two D's adjacent, so something has to be put between them.  Let's represent this with an asterisk:

```<code>A B _ C _ D * D _
</code>```

Since something (rather than nothing) has to replace that asterisk, it makes sense to consider the seven cases of non-empty subsets $\mathscr S$ of the remaining styles A,B,C which might go there:

Case 1: $\mathscr S =$ {A,B,C}

If we put all the remaining shirts in the asterisk place (between the two D's), then the only choice we have is how to order them.  Thus this case leads to $3!=6$ possible outcomes.

Case 2: $\mathscr S =$ {A,B}

If we put the pair of shirts styled A,B in the asterisk place, we must choose the ordering of them and where else the last shirt, styled C, should go.  But the only valid place for that C is after the last D, so this case leads to $2!=2$ possible outcomes.

Case 3: $\mathscr S =$ {A,C}

If we put the pair of shirts styled A,C in the asterisk place, we must choose the ordering of them and where else the last shirt, styled B, should go.  As there are two valid places for that B, after the last D or between C and the first D, this case leads to $2! \times 2 = 4$ possible outcomes.

Case 4: $\mathscr S =$ {B,C}

If we put the pair of shirts styled B,C in the asterisk place, we must choose the ordering of them and where else the last shirt, styled A, should go. But that shirt A can go in three valid places (any of the three underscores will do), so this case leads to $2! \times 3 = 6$ possible outcomes.

Case 5: $\mathscr S =$ {A}

If we put shirt A in the asterisk place, then we must choose whether the remaining shirts B,C will be placed together or into separate underscore locations.

If together, then the first underscore is not valid (because the second C must go in the second or third underscore location), and if C goes into the second underscore, it would have to be preceded by B (either order works in the third underscore).  So that gives us $1+2 = 3$ possible outcomes.

If separately, then the first underscore is not valid (the two B's must not be adjacent) for either shirt, and the second underscore is not valid for the C shirt by itself.  So the only way to separate them is putting B into the second underscore and C into the third underscore, for only $1$ possible outcome.

Case 6: $\mathscr S =$ {B}

If we put shirt B in the asterisk place, then we must choose whether the remaining shirts A,C will be placed together or into separate underscore locations.

If together, then the first underscore is not valid (because the second C must go in the second or third underscore location), and if C goes into the second underscore, it would have to be preceded by A (either order works in the third underscore).  So that gives us $1+2 = 3$ possible outcomes.

If separately, then only the third underscore is valid for the C shirt by itself.  So if we separate them, we can choose to put A into the first or second underscore and C into the third underscore, for $2$ possible outcomes.

Case 7: $\mathscr S =$ {C}

If we put shirt C in the asterisk place, then we must choose whether the remaining shirts A,B will be placed together or into separate underscore locations.

If together, then in the first underscore location they would have to be ordered AB (to avoid adjacent B's),  but either order works in the second and third underscores.  So that gives us $1+2+2 = 5$ possible outcomes.

If separately, then only the second or third underscore is valid for the B shirt by itself, and then we can choose to put A in either of the two remaining underscore locations.  We get $2\times 2 = 4$ possible outcomes.

Finally we add up all the possible case outcomes:

$$ 6 + 2 + 4 + 6 + (3+1) + (3+2) + (5+4) = 36 $$

and multiply that by $4!$ to get the answer: $36 \times 24 = 864$.

Body:
Because the problem involves rather small numbers, we can count by hand without using inclusion/exclusion.

Assume the first style of shirt that appears, left-to-right, is A, and that the next style is B, followed (disregarding any more A,B) by one of style C (so that the last style to appear in this sequence is D).  Any permutation of those four styles produces a distinct, but still valid, arrangement.  By counting the sequences with that fixed order of appearance, we will get a number that multiplied by $4!$ gives the answer to the original problem.

Accordingly the sequence resembles AB_C_D_ with the stipulation that underscores represent the combined locations of the four remaining shirts.  An underscore might correspond to zero or more of the four shirts, but with certain restrictions.  The first underscore cannot hold C or D styles, and the second underscore cannot hold a D style.

In consequence the final D has to go into the spaces represented by the last underscore, splitting it into two parts.  Furthermore we cannot leave the two D's adjacent, so something has to be put between them.  Let's represent this with an asterisk:

A B _ C _ D * D _

Since something (rather than nothing) has to replace that asterisk, it makes sense to consider the seven cases of non-empty subsets $\mathscr S$ of the remaining styles A,B,C which might go there:

Case 1: $\mathscr S =$ {A,B,C}

If we put all the remaining shirts in the asterisk place (between the two D's), then the only choice we have is how to order them.  Thus this case leads to $3!=6$ possible outcomes.

Case 2: $\mathscr S =$ {A,B}

If we put the pair of shirts styled A,B in the asterisk place, we must choose the ordering of them and where else the last shirt, styled C, should go.  But the only valid place for that C is after the last D, so this case leads to $2!=2$ possible outcomes.

Case 3: $\mathscr S =$ {A,C}

If we put the pair of shirts styled A,C in the asterisk place, we must choose the ordering of them and where else the last shirt, styled B, should go.  As there are two valid places for that B, after the last D or between C and the first D, this case leads to $2! \times 2 = 4$ possible outcomes.

Case 4: $\mathscr S =$ {B,C}

If we put the pair of shirts styled B,C in the asterisk place, we must choose the ordering of them and where else the last shirt, styled A, should go. But that shirt A can go in three valid places (any of the three underscores will do), so this case leads to $2! \times 3 = 6$ possible outcomes.

Case 5: $\mathscr S =$ {A}

If we put shirt A in the asterisk place, then we must choose whether the remaining shirts B,C will be placed together or into separate underscore locations.

If together, then the first underscore is not valid (because the second C must go in the second or third underscore location), and if C goes into the second underscore, it would have to be preceded by B (either order works in the third underscore).  So that gives us $1+2 = 3$ possible outcomes.

If separately, then the first underscore is not valid (the two B's must not be adjacent) for either shirt, and the second underscore is not valid for the C shirt by itself.  So the only way to separate them is putting B into the second underscore and C into the third underscore, for only $1$ possible outcome.

Case 6: $\mathscr S =$ {B}

If we put shirt B in the asterisk place, then we must choose whether the remaining shirts A,C will be placed together or into separate underscore locations.

If together, then the first underscore is not valid (because the second C must go in the second or third underscore location), and if C goes into the second underscore, it would have to be preceded by A (either order works in the third underscore).  So that gives us $1+2 = 3$ possible outcomes.

If separately, then only the third underscore is valid for the C shirt by itself.  So if we separate them, we can choose to put A into the first or second underscore and C into the third underscore, for $2$ possible outcomes.

Case 7: $\mathscr S =$ {C}

If we put shirt C in the asterisk place, then we must choose whether the remaining shirts A,B will be placed together or into separate underscore locations.

If together, then in the first underscore location they would have to be ordered AB (to avoid adjacent B's),  but either order works in the second and third underscores.  So that gives us $1+2+2 = 5$ possible outcomes.

If separately, then only the second or third underscore is valid for the B shirt by itself, and then we can choose to put A in either of the two remaining underscore locations.  We get $2\times 2 = 4$ possible outcomes.

Finally we add up all the possible case outcomes:

$$ 6 + 2 + 4 + 6 + (3+1) + (3+2) + (5+4) = 36 $$

and multiply that by $4!$ to get the answer: $36 \times 24 = 864$.


问:
Prove $\lim\limits_{x\to0} xf(x) = 0$?
I am confused about the following proof on a textbook I'm reading.
Suppose $f:(0,b] \to \Bbb R$ is continuous, positive, and integrable on $(0,b]$. Suppose further that as $x \to 0$ from the right, $f(x)$ increases monotonically to $+\infty$. Prove that $\lim\limits_{x\to0} xf(x) = 0$.

It would be nice if I could write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$. Then for all $\epsilon > 0$ there exists a $\delta$ such that $0<x<\delta$ implies $0<f(x)x \le \int_a^{a+x} f(t)dt <\epsilon$. But this is not true since $f$ increases monotonically as $x\to0$. Instead of writing $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$, I should probably write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(a+x)dt = f(a+x)x$. But that way my proof does not work. Can someone give me a hint as to how to proceed?

EDIT: Thanks for the comments and I realized that the monotonicity of $f$ should be one of the conditions.

答:
Hint :
Try to see what you can do with $\int_x^{2x} f(t) dt$.

Title: Prove $\lim\limits_{x\to0} xf(x) = 0$?


问:
Matchings in a tree
Let $T$ be a tree.

Suppose that $T$ doesn’t have a perfect matching and let $M$ be a matching of $T$, $|M| = k$. Prove
that there exists a matching of the same cardinality $k$ which exposes at least a pendant vertex
of $T$.

My idea:

If $T$ has not a perfect matching it means that $\exists x$ in $V(T)$ and $x$ is exposed. I don't know how to use this in symmetric difference to find another matching $M_1$, $|M_1|=k$ which exposes at least a pendant vertex of $T$.

答:
First, here are some general ideas. In any graph, not just a tree, given a matching $M$ that is not perfect, we can try to find an augmenting path to improve the matching.

This is a path $v_0 v_1 v_2 \dots v_{2k+1}$ with the property that $v_1v_2, v_3 v_4, \dots, v_{2k-1} v_{2k}$ are edges of $M$, and $v_0$ and $v_{2k+1}$ are exposed. By deleting those $k$ edges of $M$, and adding the $k+1$ edges $v_0v_1, v_2v_3, \dots, v_{2k} v_{2k+1}$ instead, we get a bigger matching.

How might we find an augmenting path? Here is a procedure that may or may not work:

Start with any exposed vertex $v_0$.

Having found a vertex $v_{2i}$, if it has any exposed neighbor, let $v_{2i+1}$ be that neighbor, and stop.

Otherwise, if $v_{2i}$ has any neighbors whatsoever we haven't seen yet, let $v_{2i+1}$ be one of them, let $v_{2i+2}$ be the vertex it is matched to, and continue from $v_{2i+2}$.

Otherwise, if all of $v_{2i}$'s neighbors are already in the path, we give up. (Fancier algorithms do more here, but we don't need anything fancy for this problem.)

So let's say that $M$ is a matching in a tree $T$, and it leaves at least one vertex exposed. What can happen if we perform the procedure above on $T$?

It could succeed in finding a bigger matching $M'$ with $|M'|=|M|+1$. We get the new matching we want by deleting an edge from $M'$.

It could fail, stopping at a vertex $v_{2i}$ whose neighbors are all already on the path.

I claim that in the second case, $v_{2i}$ is a pendant vertex (prove this!) and we can use the path $v_0 v_1 v_2 \dots v_{2i}$ to make a new matching, of the same size as $M$, that leaves $v_{2i}$ exposed.

Title: Matchings in a tree


问:
Is $x$ an algebraic element over the field of rational functions $K(x)^p$?
Question: Is $x \in K(x)$ an algebraic element over the field $K(x)^p$?

Edit: Let $K$ be a field with char($K)=p>0$ and let $K(x)$ be the field of rational functions over $K$.

My attempt:
I basically tried to answer this by referring to:

Field $K (x)$ of rational functions over $K$, the element $x$ has no $p$th root.

Suppose to the contrary that $x$ is algebraic over $K(x)^p$, and so $x$ is a root of some $p$-degree polynomial such that;
$(\frac{f(x)}{g(x)})^p -x = 0$

$f(x)^p=g(x)^p * x$

Here we see the contradiction since the degrees of $f(x)^p= deg(f(x)*p)$ and $g(x)^p*x = \deg(g(x)*p+1)$.

I am totally lost on this, I've been using the 4th edition of Abstract Algebra by Beachy and there is hardly any mention of the field of rationals. Any hints and maybe suggestions on resources where I can find out more about the field of rationals would be greatly appreciated, thanks!

答:
$x$ is in fact algebraic over $K(x)^p$ (note the comments to the question, we only need that $x^p\in K(x)^p$. I think it might be confusing to you in which ring we're trying to find polynomials that have $x$ as a root. To get around this notational issue, let's call $F:=K(x)^p$.

Now $x$ is algebraic over $F$ if there is some polynomial $g\in F[Y]$ s.t. $g(x)=0$.
Let's look at the polynomial $g=Y^p-x^p$. We know that $x^p\in F$, so $g\in F[Y]$. Clearly also $g(x)=x^p-x^p=0$, so $x$ is algebraic over $F$.


问:
Solve three-variable system $x^2 - yz = 1,\>y^2 - xz = 2,\>z^2 - xy = 3$
I am going around in circles on this system

$$x^2 - yz = 1\\
y^2 - xz = 2\\
z^2 - xy = 3$$

I have tried a few things (below) but keep hitting a wall. I know that by adding the three equations we get:
$x^2 + y^2 + z^2 - yz - xz - xy =6$. And also that:

$(x-y)^2 + (x-z)^2 + (y-z)^2 = 2x^2+2y^2+2z^2-2xy-2xz-2yz=12$

I have tried the following:

$(x-y)(x+y+z)=(x^2-yz)-(y^2-xz)=-1$

$(x-z)(x+y+z)=(x^2-yz)-(z^2-xy)=-2$

$(y-z)(x+y+z)=(y^2-xz)-(z^2-xy)=-1$

So I think it follows that $x-y=y-z$ and therefore $x-z=2(x-y)$

Substituting in:

$(x-y)^2+[2(x-y)]^2+(x-y)^2=12$

$6(x-y)^2=12$

$x-y=\sqrt{2}$

Therefore: $y-z=\sqrt{2}$ and $x-z=2\sqrt{2}$

But when I rearrange and substitute into one of the original equations, the solutions I get for x, y and z don't actually work.

$y=x-\sqrt{2}$ and $z=x-2\sqrt{2}$

Substituting into the first equation:

$x^2-(x-\sqrt{2})(x-2\sqrt{2})=1$

$x=\frac{3\sqrt{2}}{2}$

Giving $y=\frac{\sqrt{2}}{2}, z=-\frac{\sqrt{2}}{2}$

Unfortunately, substituting these values into the three original equations only satisfies the second one. Can anyone please help?

答:
This type of the system is well known.  Its general form reads

$$x^2 - yz = a,\>\>\>y^2 - xz = b,\>\>\>z^2 - xy = c$$

and has the solutions given in the symmetric expression below

$$(x,y,z)=\pm \frac{(a^2-bc, b^2-ca, c^2 -ab)}{\sqrt{a^3+b^3+c^3-3abc}}
$$

So, substitute $a=1,\>b=2,\>c=3$ to obtain
$\pm(\frac{-5}{3 \sqrt{2}},\frac{1}{3 \sqrt{2}},\frac{7}{3
   \sqrt{2}})$.


问:
Prove that the function $ \zeta(x)=\sum^{\infty} \frac{1}{n^{x}}, x>1 $ is infinitely differentiable in $(1, \infty) . n=1$
Prove that if $a>1$ and $k \geq 1,$ then

a)  $\sum_{n=2}^{\infty} \frac{(\log n)^{k}}{n^{a}}<\infty$

b)  Prove that the function
$$
\zeta(x)=\sum^{\infty} \frac{1}{n^{x}}, x>1
$$
is infinitely differentiable in $(1, \infty) . n=1$
Hint: Use part (a) to prove part (b). You can use part (a) to prove (b), even if you do not know how to prove (a).

My try:

a)  As $\frac{(\log n)^{k}}{n^{a}}$ in monotonically decreasing because $\ln n$ grows slowly than $n^{a}$. Is there a rigorous way to prove it? I tried in this way that
$\frac{d}{dx}\left(\frac{(\log x)^{k}}{x^{a}}\right)=\frac{\frac{k}{x}(\log x)^{k-1}-a x^{a-1}}{x^{2 a}}=\frac{k(\log x)^{k-1}-a x^{a}}{x^{2 a+1}}$
and in this way inductively I was trying to see that $f(x)<f(y)$ whenn $x>y$ for some large $y$. But not convinced fully.

Thus we can use Cauchy's Condensation test, considering $f(n)=\frac{(\log n)^{k}}{n^{a}}$.

$\begin{aligned} 2^{n} f\left(2^{n}\right)=& \frac{2^{n}\left(\log 2^{n}\right)^{k}}{\left(2^{n}\right)^{a}}=\frac{\sin n^{k}(\log 2)^{k}}{2^{n(a-1)}} \\ &=\frac{n^{k} c}{2^{d n}} ; \text { where }\left(\log {2}\right)^{k}=c \text { finite constant }, d=a-1>0 \text { finite constant .}\end{aligned}$

Now, $\sum f(n)$ converges iff $\sum 2^{n} f\left(2^{n}\right)$ converges
iff $\sum \frac{n^{k}}{2^{d n}}$ converges.
By Ratio test the last series converges and hence the result.

For part b) I can see that $h(x)=\frac{1}{n^{x}}=n^{-x}$ then $h'(x)=-n^{-x}\log n$ and $h''(x)=n^{-x}(\log n)^2$ and so on. Then I can see that the series with term by term differentiation of $\zeta(x)$ is convergent. How to conclude my answer from there?

答:
Part (a): You can use the fact that $$\lim_{x \to \infty} \frac{(\log x)^k}{x^{(a-1)/2}} = 0$$ to show that $$\frac{(\log x)^k}{x^a} \leq \frac{1}{x^{(a+1)/2}},$$

and so $\sum_{n=1}^\infty \frac{(\log n)^k}{n^a}$ converges by direct comparison with $\sum_{n=1}^\infty \frac{1}{n^{(a+1)/2}}$, which is a $p$-series with $p = (a+1)/2 > 1$, hence convergent.

Part (b): Let $f_n(x) = n^{-x}$, then $\zeta(x) = \sum_{n=1}^\infty f_n(x)$ is a function series, so if the series $\sum_{n\geq 1} f_n(x)$ converges uniformly to $\zeta(x)$ on $[a, \infty)$ for any fixed $a > 1$, then it's legitimate to term-by-term differentiate it to get the derivative of the infinite sum. Since $a > 1$ is arbitrary here, that means $\zeta$ is infinitely differentiable on all of $(1, \infty)$ and its derivatives are gotten by term-by-term differentiation.


答:
HINT: If $M$ exposes some pendant vertex, we’re done, so suppose that $M$ covers every pendant vertex. Root $T$ at a non-pendant vertex $v$, so that the pendant vertices are the leaves of $T$. From each leaf there is a unique path to $v$, and every vertex of $T$ is on at least one of those paths, so there is a leaf $u$ such that the path $P$ from $u$ to $v$ contains an exposed vertex. Let $w$ be the first exposed vertex that we encounter when travelling $P$ from $u$ to $v$.

If the edges of $P$, listed in order starting at $u$, are $e_1,e_2,\ldots,e_n$, show that the odd-numbered edges of $P$ are in $M$, and the even-numbered edges are not in $M$. Conclude that $n$ is even. Show that if you form $M_1$ by deleting from $M$ the odd-numbered edges of $P$ and adding the even-numbered edges, $M_1$ will be a matching that exposes $u$ and has the same cardinality as $M$.

Title: Matchings in a tree

Body:
Let $T$ be a tree.

Suppose that $T$ doesn’t have a perfect matching and let $M$ be a matching of $T$, $|M| = k$. Prove
that there exists a matching of the same cardinality $k$ which exposes at least a pendant vertex
of $T$.

My idea:

If $T$ has not a perfect matching it means that $\exists x$ in $V(T)$ and $x$ is exposed. I don't know how to use this in symmetric difference to find another matching $M_1$, $|M_1|=k$ which exposes at least a pendant vertex of $T$.


问:
Obtaining and interpretation of the probability
Here is a part of the text of the game theory book. I got confused because of probability, which is here in a red frame. I understood that to obtain it, we multiply expression from the middle with probability $\frac{1}{n}$, but why we here should multiply? Are these probabilities independent? Also, I'm a bit lost with an interpretation of this "red" probability.

答:
this is incomplete, but as no one else has tried...
In  these  sort of game values, such as Shapley value, you get your value from the coalitions you are in. So if you are in a  coalition, you can leave and ruin the coalition. So the first set of equations, if there is a coalition of size s, in a population of n, what are the chances you are in that coalition.  Then 1/n shows up, because as the problem says, each size of coalition is equally likely,so each of those probabilities is multiplied by 1/n.</p>


答:
I’m assuming throughout that you mean for $K$ to have characteristic $p>0$. Perhaps you’re thrown by the possibility that $K$ is not perfect, in which case $\bigl(K(x)\bigr)^p$ is different from $K(x^p)$. Don’t worry, though: for our purposes, it doesn’t matter.

Let’s consider your field field $\mathscr L=\bigl(K(x)\bigr)^p$, in which there is an element $x^p$. I’ll call this element $t$. We note that there’s a field isomorphism $\varphi:K(x)\to\mathscr L$, by $\varphi(f)=f^p$. And the image of the element $x$ of $K(x)$ is $t\in\mathscr L$; just as $x$ has no $p$-th root in $K(x)$, so $t$ has no $p$-th root in $\mathscr L$. Thus the $\mathscr L$-polynomial $X^p-t$ is irreducible ($\dagger$). It has a root back up in $K(x)$, though, namely $x$. And there you are.

($\dagger$) I’ve used the fact that in a field $k$ of characteristic $p$, $X^p-b$  either has a root in $k$ or is $k$-irreducible.

Title: Is $x$ an algebraic element over the field of rational functions $K(x)^p$?

Body:
Question: Is $x \in K(x)$ an algebraic element over the field $K(x)^p$?

Edit: Let $K$ be a field with char($K)=p>0$ and let $K(x)$ be the field of rational functions over $K$.

My attempt:
I basically tried to answer this by referring to:

Field $K (x)$ of rational functions over $K$, the element $x$ has no $p$th root.

Suppose to the contrary that $x$ is algebraic over $K(x)^p$, and so $x$ is a root of some $p$-degree polynomial such that;
$(\frac{f(x)}{g(x)})^p -x = 0$

$f(x)^p=g(x)^p * x$

Here we see the contradiction since the degrees of $f(x)^p= deg(f(x)*p)$ and $g(x)^p*x = \deg(g(x)*p+1)$.

I am totally lost on this, I've been using the 4th edition of Abstract Algebra by Beachy and there is hardly any mention of the field of rationals. Any hints and maybe suggestions on resources where I can find out more about the field of rationals would be greatly appreciated, thanks!


问:
What books do you recommend on graph theory?
I am recently coming across a lot of questions relating to graph theory, one of them is the following:

Given a simple graph G with n vertices. One edge e of G is called edgy, if the vertices of G can be split up into to subtotals A and B such that there are at most 100 edges with one end in A and one end in B and one of these is e. Prove that G has at most 100n edgy edges.

I looked for graph theory books on google and these books came up:

https://www.google.com/search?client=firefox-b-d&q=graph+theory+book

I am looking for books containing graph theory used in IMO's, so that I can learn the theory required for such questions. Could you please suggest some books?

答: on the simple side, but I've always liked Oystein Ore, Graphs and their Uses, Anneli Lax New mathematical library.</p>


问:
Do I need to prove a function between two topological spaces $X, Y$ is onto to prove it induces a function on $Y$?
If there is a mapping $f: X\to Y$ between two topological spaces and $X$ has a topology $T_X$ then $Y$ will have a topology $T_f$ induced by $X$ and function $f$. But recently I got confused by something my teacher said while explaining identification functions:

"If I have a mapping from $X$ to $Y$ and $X$ has a topology and if I were able to prove the mapping was onto then I can say there is a topology $T_f$ induced on $Y$ by $T_X$ and $f$ and then we can complete our prove for identification function".

So my question is, why do I need the function to be onto in order to have an induced topology on $Y$? Now I know I need onto condition to prove identification function but I am asking for the induced topology itself why do i need onto condition for that?

答:
As mentioned in the comment, it is not necessary to have a surjective function to define a topology on the codomain of $f$, but it is sufficient to have one. Indeed, if you have a function $f : (X, \tau) \rightarrow Y$ that is onto, then it induces on $Y$ a topology that is called quotient topology, where the open sets are defined in the following way:

Def: A set $B \subseteq Y$ is said to be open in $Y$ if $f^{-1}(B)$ is
an open set in $X$.

Now you can see that if $f$ is not surjective, then you may have some $y \in Y$ for which $f^{-1}(\{y\})$ is the empty set, so open in $X$ by definition. Hence, these singletons result open in the topology induced by $f$ on $Y$ and you may want to avoid this (just as an example, if you assume more structure on the topology like to be $T_1$, then these sets would be both open and closed, hence the space would result always disconnected by these isolated points).


问:
Spectral analysis of a $d$-regular graph
Given $d$-regular graph $G=(V, E)$, the adjacency matrix $A$, and the normalized Laplacian $L=I-\dfrac{1}{d}A$, prove that for any vector $\vec{x}\in \mathbf{R^{|V|}}$:
$$ \vec{x}L\vec{x}^t=\dfrac{1}{d}\sum_{(u,v)\in E}(x_u-x_v)^2 $$

$u$ and $v$ are vertices, and $x_u$ and $x_v$ are values of the vector $\vec{x}$ corresponding to the vertices.

I don't know where to start.

答:
Say we have a graph with $N$ vertices whose connections are defined by the adjacency matrix $A$. Namely, vertex $i$ is connected to vertex $j$ if $A_{ij} =1$ and if $i$ and $j$ are not connected then $A_{ij}=0$. Let $D$ be the diagonal matrix defined as $D = \operatorname{diag}(\operatorname{deg}(A)_1, \operatorname{deg}(A)_2, \ldots,\operatorname{deg}(A)_N )$, where $\operatorname{deg}(A)_k$ is the degree of vertex $k$. By definition, this degree is $\operatorname{deg}(A)_k \equiv  \sum_{\ell=1}^{N} A_{k\ell}$. For a $d$-regular graph, we have $\deg(A)_k = d$ for all $k$, but we will work most generally here.

We define the Laplacian as $L \equiv D - A$. Defining the row vector $\textbf{x}^{T} = (x_1, x_2, \ldots, x_N)$, we then have
\begin{align}
\textbf{x} L \textbf{x}^{T} & = \sum_{i, j=1}^{N} x_{i} L_{ij} x_{j} = \sum_{i, j=1}^{N} x_{i} (\delta_{ij} \operatorname{deg}(A)_{j} - A_{ij}) x_{j}\\
& = \sum_{i, j=1}^{N} x_{i} \delta_{ij} \sum_{\ell=1}^NA_{j\ell}x_{j} - \sum_{i, j=1}^N x_{i}A_{ij} x_{j}\\
& = \sum_{i, \ell=1}^{N} x_{i}A_{i\ell}x_{i} - \sum_{i, \ell=1}^N x_{i}A_{i\ell} x_{\ell}\\
& = \sum_{i, \ell=1}^{N} A_{i\ell}(x_i^2 - x_i x_{\ell})\\
\end{align}
In the second equality, we used the Kronecker delta $\delta_{ij}$. In the first term of the fourth equality, we summed over the $j$ index and in the  second term of the fourth equality, we switched the dummy summation variable $j$ with $\ell$.

Now, because $A$ is the adjacency matrix for an unweighted graph, we have $A = A^T$. And thus $\sum_{i, \ell =1}^N A_{i\ell} x_i^2 = \sum_{i, \ell = 1}^N A_{i \ell} x_{\ell}^2$. Therefore, we have
\begin{align}
\textbf{x} L \textbf{x}^{T} & = \frac{1}{2} \sum_{i, \ell = 1}^NA_{i \ell}(x_i^2 + x_{\ell}^2 - 2 x_{i} x_{\ell})\\
& = \frac{1}{2} \sum_{i, \ell = 1}^NA_{i \ell}(x_i- x_{\ell})^2.
\end{align}
With $A$ the adjacency matrix, $A_{i\ell}$ is only non-zero when the connection $(i, \ell)$ is in the set of edges $E$. So we obtain

\begin{align}
\textbf{x} L \textbf{x}^{T}  = \frac{1}{2} \sum_{ (i, \ell) \in E}(x_i- x_{\ell})^2.
\end{align}


问:
How many orthogonal vectors to a given vector
Let's say I have a vector:

$$
v = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}.
$$

I was wondering how many vectors would be orthogonal to it. I noticed that these are some of the vectors orthogonal to it.
$$
\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix},
\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix},
\begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix},
\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix},
\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}.
$$

Obviously, there are more. Is there a principled way to count the number of orthogonal vectors to a given vector? Thanks!

答:
Ummm... an infinite number, of course:

Find a single vector orthogonal to yours, and rotate it by an arbitrary angle around your vector.

And if you also allow vectors of different magnitude, well then...


问:
About the type of numbers allowed by axioms and Nature
I have a question which has 4 different subcases or "avatars":

1) Has every "interesting" class of number been invented?
2) Has every "possible" class of number been invented?
3) Does Nature use every interesting class of number for something in the physical realm?
4) Does Nature use every possible class of number for something in the physical world?

Prime numbers, integer numbers, fractional numbers, irrational numbers, real numbers, complex numbers, quaternionic numbers, octonionic numbers, Cayley-Graves numbers, p-adic numbers, tropical numbers, surreal numbers, transfinite numbers, adelic numbers, grassmannian numbers, clifford numbers, Cayley-Dickson numbers, ternary numbers, ideal numbers, p-ary numbers, ... What else?

答: Nature has everything to do with mathematics. Mathematics is the natural science, inherent in the physical world, of quantity, with its amazing implications from set theory to category theory and far beyond. There are endless classes of integers alone. Just glance at the Bell Labs catalog of integer sequences. There is no reason to think that mathematical invention and discovery will ever reach an end as long as we exist. By the way, WE invent and discover what can exist in nature, not only in mathematics but in every area of our enterprise from agriculture and airplanes to Zeppelins and zoos. The idea that mathematics is not of the natural world was proposed by loafing weavers of invisible cloth long ago and is long long long overdue to be abandoned.</p>

ContentLicense: CC BY-SA 3.0


问:
Computation of $\int_0^1 x^{nx} dx$
I did this by using Feynman trick and differential equation. I guess there are a few functions presented as an integral, for which this differentiation trick does not hold true owing to the absence of derivative on some set; or owing to humans error - we all learn by making mistakes. Anyway, take a look at my solution (my attempt at solving the problem I post here!). We define

$$I(n) = \int_0^1 x^{nx}dx$$

Step 1. I take the derivative of the integral with respect to $n$, getting

$$\frac{dI}{dn} = \int_0^1 x^{nx} \cdot \ln x \cdot x \cdot dx$$

Step 2: Next, I notice that $(x^{nx})^\prime = n x^{nx} \left(\ln \left(x\right) +1\right)$ and using this I try to maintain the integral through the differential method:

$$\int_0^1 x^{nx} \ln \left(x\right) \cdot xdx = \frac{1}{n}\int_0^1 n \cdot x^{nx} \left(\ln \left(x\right) + 1 - 1 \right)xdx $$

So, I get

$$ \frac{dI}{dn} = \frac{1}{n} \int_0^1 n \cdot x^{nx} \left(\ln \left(x\right) + 1\right) xdx - \int_0^1 x^{nx} xdx$$

Note that I reduce $1/n \cdot n = 1$ in the second integral while in the other integral I keep it.

Step 3: I integrate the first integral by parts (you discerned my preparation for this during the previous step!); for that, I put $u = x$, then $du = dx$; and $dv = n\cdot x^{nx}\left(\ln \left(x\right) + 1\right)dx$, then $v = x^{nx}$. So, I have

$$\int_0^1 n \cdot x^{nx} \left(\ln \left(x\right) + 1\right) xdx =  x \cdot x^{nx} |_0^1 - \int_0^1 x^{nx} dx = 1 - I(n)$$

All steps before the following ones are considered (by me) to be correct and have thoroughly been verified; if my solution is incorrect, then I failed to solve my problem during the next problems!

Step 4: The second integral should also be integrated by parts; for that, I put $u = x$, then $du = dx$; and $dv = x^{nx}dx$. As an antiderivative of $x^{nx}$, I take $v = \int_0^x t^{nt}dt$. Thus,

$$\int_0^1 x^{nx} xdx = \left(x \int_0^x t^{nt} dt\right)|_0^1 - \int_0^1 \left(\int_0^x t^{nt}dt\right)dx$$

Note that

$$\left(x \int_0^x t^{nt} dt\right)|_0^1 = 1 \cdot \int_0^1 t^{nt}dt - 0 \cdot \int_0^0 t^{nt}dt = I(n)$$

Next, we try to switch the order of integration:

$$ \int_0^1 \int_0^x t^{nt} dt dx = \int_0^1 t^{nt}dt \int_0^t dx = \int_0^1 t^{nt} tdt$$

Then

$$ \int_0^1 x^{nx} x dx = \frac{1}{2} I(n)$$

Subsequently, the initial integral should satisfy the equation

$$ \frac{dI}{dn} = \frac{1}{n} - \frac{I}{n} - \frac{I}{2} = \frac{1}{n} - \frac{n+2}{2n}I(n)$$

Or, equivalently,

$$\frac{dI}{dn} + \frac{n + 2}{2n}I = \frac{1}{n}$$

which is the linear homogenous differential equation of the first order; in order to solve this, I used the method called Lagrange: in accordance with it, I had to solve the corresponding inhomogeneous equation $I^\prime + \varphi(n)I = 0$ and then vary the constant $C = C(n)$; let us dive in details: I had to solve the equation

$$ \frac{dI}{dn} + \frac{n+2}{2n}I = 0$$

or

$$\frac{dI}{I} + \frac{n+2}{2n}dn = 0$$

supposing that neither $I \ne 0$ and $n \ne 0$. Integrating yields:

$$ \ln I + \frac{n}{2} + \ln n = \ln C \Rightarrow I(n) = \frac{C}{n}e^{-\frac{n}{2}}$$

where

$$\int \frac{n + 2}{2n}dn = \int \left(\frac{1}{2}+\frac{1}{n}\right)dn =\frac{1}{n} + \ln n$$

See that $I = 0$ is a solution that might be produced from this one, so it is not an isolated solution and we did not lose it. Next, I vary $C = C(n)$. Then I find the derivative to be

$$\frac{dI}{dn} = \frac{d}{dn} \left(\frac{C(n)}{n}e^{-\frac{n}{2}}\right) = \frac{dC}{dn} \cdot \frac{1}{n} \cdot e^{-\frac{n}{2}} - C(n) \cdot \frac{n+2}{2n^2}e^{-\frac{n}{2}}$$

Putting $I$ and $dI/dn$ into the initial equation, I arrive at another

$$ \frac{dC}{dn} \cdot \frac{1}{n} \cdot e^{-\frac{n}{2}} - C(n) \cdot \frac{n+2}{2n}e^{-\frac{n}{2}} + \frac{n+2}{2n} \cdot \frac{C}{n}e^{-\frac{n}{2}} = \frac{1}{n}$$

or, equivalently,

$$ \frac{dC}{dn} \cdot \frac{1}{n} \cdot e^{-\frac{n}{2}} = \frac{1}{n} $$

From this,

$$C(n) =\int e^{\frac{n}{2}}dn = 2e^{\frac{n}{2}}+A$$

where A is another constant, this time, real constant. In our case, this constant is the specific value for which to be defined, I see that $I(1) = \int_0^1 x^x dx = \alpha$, then from

$$ I(n) = \frac{2}{n} + \frac{A}{n}e^{-\frac{n}{2}}$$

setting $n = 1$, I compute $A = \left(\alpha - 2\right)\sqrt{e}$

The terminal result is

$$ I(n) = \frac{2}{n} + \frac{\left(\alpha - 2\right)\sqrt{e}}{n}e^{-\frac{n}{2}} = \frac{2\sqrt{e^{n-1}}+\alpha -2}{n\sqrt{e^{n-1}}}$$

Besides which, $I(+\infty) = 0$. When $n = 2$ my formula gives the result which is approximately equal to that given by WolframAlpha programme; the difference is small but the error increases when $n$ becomes large, though the integral is still approaching zero at infinity and both formulae coincide.

Links:

The true calculation by Wolfram: https://www.wolframalpha.com/input/?i=%5Cint_0%5E1+x%5E%282x%29dx

Mine: https://www.wolframalpha.com/input/?i=%5Cfrac%7B2%5Csqrt%7Be%7D+%2B+%5Cint_0%5E1+x%5Ex+dx+-+2%7D%7B2%5Csqrt%7Be%7D%7D

答:
You messed up at this step: $$\int_0^1 \int_0^x t^{nt} dt dx = \int_0^1 t^{nt}dt \int_0^t dx = \int_0^1 t^{nt} tdt$$

Iterated integrals where the inner bounds depend on the outer ones can't be decoupled like that. Instead, it should be $$\int_0^1 \int_0^x t^{nt} dt dx = \int_{0}^{1}\int_{t}^{1}t^{nt}dxdt = \int_{0}^{1}t^{nt}\left(1-t\right)dt$$

which doesn't provide any information, as it just makes the integral "loop back" to $$\int_0^1t^{nt}tdt = \int_0^1t^{nt}tdt$$

The statement $$\int_0^1t^{nt}tdt = \frac{1}{2}I(n)$$ and everything after that is not correct then.

Title: Computation of $\int_0^1 x^{nx} dx$


答:
Notice that any vector of the form

$$\begin{pmatrix}x \\ y \\ -x-y\end{pmatrix}$$

will be orthogonal to

$$\begin{pmatrix}1 \\ 1 \\ 1\end{pmatrix}$$

Thus, there is an infinite number of such vectors, more precisely a continuum vectors spanning a two-dimensional subspace.

Title: How many orthogonal vectors to a given vector

Body:
Let's say I have a vector:

$$
v = \begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}.
$$

I was wondering how many vectors would be orthogonal to it. I noticed that these are some of the vectors orthogonal to it.
$$
\begin{bmatrix}-1 \\ 0 \\ 1\end{bmatrix},
\begin{bmatrix}1 \\ 0 \\ -1\end{bmatrix},
\begin{bmatrix}1 \\ -1 \\ 0\end{bmatrix},
\begin{bmatrix}-1 \\ 1 \\ 0\end{bmatrix},
\begin{bmatrix}-2 \\ 1 \\ 1\end{bmatrix}.
$$

Obviously, there are more. Is there a principled way to count the number of orthogonal vectors to a given vector? Thanks!


问:
$\mathrm{Hom}(R/I, R/J\otimes M)\cong ?$
Let $R$ be a Noetherian commutative ring, $I,J$ two ideal of $R$ and $M$ an $R-$module. Does anyone see the isomorphism $\mathrm{Hom}(R/I, R/J\otimes M)\cong \ldots$?
Thanks.</p>

答: $\mathrm{Hom}_R(R/I,N)\simeq (0:_NI)$, where $(0:_NI)=\{x\in N:Ix=0\}$. Since $R/J\otimes M\simeq M/JM$ one can take $N=M/JM$ and get $\mathrm{Hom}_R(R/I, R/J\otimes M)\simeq(JM:_MI)/JM$.</p>


答:
As written, this is not true. However, I am not sure if meant to include that $f(x)$ is a decreasing function as you mention later in your proof that it is a monic function. If this condition is not included, then the theorem is false. Consider the function

$$f(x)=\begin{cases}
      g_n(x) & \frac{1}{n^3}\leq x \leq \frac{1}{n^3+1}\\
      \frac{1}{\sqrt{x}}& \text{otherwise}
   \end{cases}$$

(for all $n\in\mathbb{N}$) where $g_n(x)$ is any continuous function such that

$g_n\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)=\frac{2n^3(n^3+1)}{2n^3+1}$

$g_n(x)$ is maximized at $x=\frac{2n^3+1}{2n^3(n^3+1)}$

$g_n\left(\frac{1}{n^3}\right)=\frac{1}{\sqrt{n^3}}$

$g_n\left(\frac{1}{n^3+1}\right)=\frac{1}{\sqrt{n^3+1}}$

$g_n(x)$ is minimized at $x=\frac{1}{\sqrt{n^3}}$

Note that these conditions imply $f(x)$ is continuous. Further, it is obvious that $f(x)>0$ and that $\lim_{x\to 0^+} f(x)=\infty$. For the final condition, note that

$$\int_{0}^b f(x)dx<\int_{0}^b\frac{1}{\sqrt{x}}dx+\sum_{n=1}^\infty \int_{\frac{1}{n^3}}^{\frac{1}{n^3+1}} g_n(x)dx$$

$$\leq \frac{2b^{3/2}}{3}+\sum_{n=1}^\infty\int_{\frac{1}{n^3}}^{\frac{1}{n^3+1}} \frac{2n^3(n^3+1)}{2n^3+1}dx=\frac{2b^{3/2}}{3}+\sum_{n=1}^\infty\frac{1}{n^6+n^3} \frac{2n^3(n^3+1)}{2n^3+1}<\infty$$

However, we know that

$$xf(x)\bigg|_{x=\frac{2n^3+1}{2n^3(n^3+1)}}=\frac{2n^3+1}{2n^3(n^3+1)}f\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)$$

$$=\frac{2n^3+1}{2n^3(n^3+1)}g_n\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)=\frac{2n^3+1}{2n^3(n^3+1)}\frac{2n^3(n^3+1)}{2n^3+1}=1$$

This implies $\lim_{x\to 0^{+}}xf(x)\neq 0$. However, I believe the original theorem would be true if you add the stipulation that $f(x)$ is decreasing.

Title: Prove $\lim\limits_{x\to0} xf(x) = 0$?

Body:
I am confused about the following proof on a textbook I'm reading.
Suppose $f:(0,b] \to \Bbb R$ is continuous, positive, and integrable on $(0,b]$. Suppose further that as $x \to 0$ from the right, $f(x)$ increases monotonically to $+\infty$. Prove that $\lim\limits_{x\to0} xf(x) = 0$.

It would be nice if I could write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$. Then for all $\epsilon > 0$ there exists a $\delta$ such that $0<x<\delta$ implies $0<f(x)x \le \int_a^{a+x} f(t)dt <\epsilon$. But this is not true since $f$ increases monotonically as $x\to0$. Instead of writing $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$, I should probably write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(a+x)dt = f(a+x)x$. But that way my proof does not work. Can someone give me a hint as to how to proceed?

EDIT: Thanks for the comments and I realized that the monotonicity of $f$ should be one of the conditions.


问:
Find $a$ so a quadratic expression describes a straight line (problem from a Swedish 12th grade ’Student Exam’ from 1931)

$x$ and $y$ are coordinates in an orthogonal coordinate system. If the constant $a$ is suitably chosen the equation
$$9x^2-3xy+ay^2+15x-16y-14=0$$
describe two straigth lines. Find the value of $a$ and the equations for the two lines.

I tried to complete the square (using Mathematica) but the resulting expression was not helpful’. Any idea where to start? TIA.

(N.B.: The problem is quoted from a Swedish 12th grade ’Student Exam’ from 1931. I find these problems rather fun to solve in the evenings.)

Body:
$$9x^2+(15-3y)x+ay^2-16y-14=0$$

$$\Delta = (15-3y)^2-36(ay^2-16y-14)=(9-36a)y^2+486y+729$$

Can you end it now?


问:
Separable topological space
Show that if $(X,\tau)$ is a separable topological space, then $\{x \in X : \{x\} \in \tau\}$ is an enumerable set.

I thought this way, if a topological space is separable, this implies that it has a dense and numerable set, that is, if this set is dense, then its closure is the topological space X, which has numberable points, finite or that has bijeção with the natural numbers.
Does this have anything to do with the elements of the topology being numberable? in particular discrete elements, formed by only one element ${x}$. In fact, how can I state that $\{x\}$ is in the topology?

答:
An uncountable space $X$ can have  a countable (numerable) dense subset $D$. For example, $X=\Bbb R$ and $D=\Bbb Q.$ And an extreme example: Let $X$ be any non-empty set; let $p\in X;$  let $\tau=\{\emptyset, \{p\}, X\};$ and let $D=\{p\}$.

Let $X^i$ be the set of isolated points of $X$. That is, $X^i=\{p\in X: \{p\}\in \tau \}. $

Let  $D$ be dense in $X.$ Then $D\cap U\ne\emptyset$ whenever $\emptyset \ne U\in \tau.$ So if $ p\in X^i$ then $\emptyset \ne  \{p\}\in \tau,$ so $\emptyset\ne D\cap \{p\},$ so $p\in D.$ Hence $$X^i\subseteq D.$$ Any subset of a countable set is countable. So if $D$ is countable then $X^i$ is countable.

Appendix. Let $E\subseteq X.$ The definition of $\overline E$ is $\bigcap \Bbb F_E$ where $\Bbb F_E=\{F: X\setminus F\in \tau \,\land \, F\supseteq E\}.$ An important fact is that if $U\in \tau$ and if $E\cap U$ is empty then $\overline E\cap U$ is empty. Because $X\setminus U\in \Bbb F_E,$ so $\overline E=\bigcap \Bbb F_E\subseteq X$ \ $U.$

Hence, another important fact: If $p\in U\in \tau$ and if $U\cap E$ is empty then $p\not\in\overline E.$ In particular, if $p\in X^i$ and if $p\not\in E$ then $E$ is not dense in $X,$ that is, $\overline E\ne X.$ Because, with $U=\{p\},$ we have $p\in U\in \tau$ and $E\cap U=\emptyset, $ so $\emptyset =\overline E\cap U=\overline E\cap \{p\},$ so $p\not\in\overline E.$

Title: Separable topological space


问:
For which values of $a$ the function $u(x,y)=x^2+x^4+axy+y^2$ is convex? concave?

Consider the function
$$
u(x,y)=x^2+x^4+axy+y^2
$$
For which values of $a$ the function $u(x,y)$ is convex? concave?

I know how to solve such kind of problems for a single variable function $f(x)$. There I would have to calculate a second derivative $f''_{xx}$ and find intervals where $f''_{xx}>0$ (convex) and $f''_{xx}<0$ (concave).

But what approach should I apply here? Is it $d^2u(x,y)$ instead of $f''_{xx}$ and checking if the matrix for $d^2u(x,y)$ is positive-definite (for convex) and negative-definite (for concave)?

答: For convexness, $d^2u(x,y)$ is positively semi-definite. Concaveness is opposite.


问:
Find all polynomials that satisfy a functional equation
I need help with this problem if anyone can contribute to solving this I would much appreciate it:

Find all polynomials $P$ that satisfy $P(1)=210$
and for every $x\in \mathbb{R}$: $$(x+10)P(2x)=(8x-32)P(x+6).$$

Body: $$P(x) = c (-8 + x) (-4 + x) (4 + x)$$


Title: Find all polynomials that satisfy a functional equation

Body:
I need help with this problem if anyone can contribute to solving this I would much appreciate it:

Find all polynomials $P$ that satisfy $P(1)=210$
and for every $x\in \mathbb{R}$: $$(x+10)P(2x)=(8x-32)P(x+6).$$

Body:
small hint

With $ x=-10 $, we get
$$P(-10+6)=P(-4)=0$$
and for $ x=4 $, it gives
$$P(8)=0$$

thus

$$P(x)=(x-8)(x+4)R(x)$$
with
$$R(1)=\frac{210}{-35}=-6$$


Title: Find $a$ so a quadratic expression describes a straight line (problem from a Swedish 12th grade ’Student Exam’ from 1931)

Body:
$x$ and $y$ are coordinates in an orthogonal coordinate system. If the constant $a$ is suitably chosen the equation
$$9x^2-3xy+ay^2+15x-16y-14=0$$
describe two straigth lines. Find the value of $a$ and the equations for the two lines.

I tried to complete the square (using Mathematica) but the resulting expression was not helpful’. Any idea where to start? TIA.

(N.B.: The problem is quoted from a Swedish 12th grade ’Student Exam’ from 1931. I find these problems rather fun to solve in the evenings.)

Body:
You can solve it by polynomial coefficient identification by doing

$$
(a_1 x+b_1 y + c_1)(a_2 x+b_2 y+c_2)=9 x^2 - 3 x y + a y^2 + 15 x - 16 y - 14
$$

resulting in

$$
\left\{
\begin{array}{l}
 a_1 a_2-9=0 \\
 a_1 b_2+a_2 b_1+3=0 \\
 a_1 c_2+a_2 c_1-15=0 \\
 b_1 b_2-a=0 \\
 b_1 c_2+b_2 c_1+16=0 \\
 c_1 c_2+14=0 \\
\end{array}
\right.
$$

solving for $a_1, b_1, c_1, a_2, b_2, a$ we get

$$
\left(3x+y+7\right)\left(3x-2y-2\right)c_2^2 = 0
$$


问:
Proving AoC(Axiom of Completeness) using NIP(Nested Interval Property)
NIP $\implies$ AoC.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Exercise 2.5.4 from Understanding Analysis by Abott.

The Bolzanno-Weierstrass theorem is extremely important, and so is the strategy employed in the proof. To gain some more experience with this technique, assume that the nested interval property is true and use it provide a proof of the Axiom of Completeness. To prevent the argument from being circular, assume also that $(1/2^n) \to 0$. (Why precisely is this last argument needed to avoid circularity?)

Does my proof check out? I hope my construction is rigorous.

Proof.

Claim. Every non-empty subset of real numbers that is bounded above has a least-upper bound.

Let $A$ be a subset of $\mathbf{R}$ bounded above by $M$. Bisect the interval $[-M,M]$ into two closed intervals $[-M,0]$ and $[0,M]$. Now, it must be that atleast one of these closed intervals contains both a point $a \in A$, and an upper bound for $A$. Select the half for which this is the case and label that interval $I_1$. Let $I_1 = [a_1,b_1]$ and let $s_1$ be an upper bound for $A$.

Bisect $I_1$ into two closed intervals of equal length and let $I_2$ be a half such that it contains both - a point in $A$ and an upper bound $s_2$ for $A$. In general, we construct a closed interval $I_k = [a_k,b_k]$ by taking a half of $I_{k-1}=[a_{k-1},b_{k-1}]$, containing a point in $A$ and an upper bound $s_k$ for $A$.

The set
\begin{align*}
 I_1 \supseteq I_2 \supseteq I_3 \supseteq \ldots
\end{align*}

forms a nested sequence of intervals. Also, $s_k \le s_{k-1} \le \ldots \le s_1$, so $s_k$ is a monotonically decreasing sequence. By the nested interval property, there exists atleast one point $x \in \mathbf{R}$, contained in every $I_k$. This provides us the candidate that we are looking for. It just remains to show that the sequence $(s_n) \to x$.

Pick an $\epsilon > 0$. By construction, the length of the interval $I_k = \frac{M}{2^{k-1}}$, which converges to $0$ as $k \to \infty$ (given). Thus, there exists $N \in \mathbf{N}$ such that, if $k \ge N$, the length of the interval $I_k$ can be made as small as we like; lesser than $\epsilon$. Since, $s_k$ and $x$ are both in the interval $I_k$, it follows that
\begin{align*}
 \absval{s_k - x} < \epsilon
\end{align*}

答:
There’s a problem with your argument at the very beginning: it’s possible that $A\cap[-M,M]=\varnothing$. You can avoid this problem by picking any $a_1\in A$, letting $b_1=M$, and starting with the interval $I_1=[a_1,b_1]$.

There is an even bigger problem when you choose one of the two subintervals. The righthand subinterval of $I_1$ contains the upper bound $M$ for $A$, so you could choose it to be $I_2$. Then when you bisect $I_2$, its righthand subinterval contains the upper bound $M$ for $A$, so you could choose it to be $I_3$. In this way you could end up choosing the righthand subinterval every time and getting $\bigcap_{k\ge 1}I_k=\{M\}$. And there’s no reason to think that $M$ is the least upper bound of $A$: for instance, it’s entirely possible that $A=[0,1]$ and $M=2$.

You need to make sure that you always choose a subinterval that contains points of $A$. If $L_k$ and $R_k$ are the left and right subintervals of $I_k$, let

$$I_{k+1}=\begin{cases}
R_k,&\text{if }A\cap R_k\ne\varnothing\\
L_k,&\text{otherwise.}
\end{cases}$$

(Why must you not choose $L_k$ when $A\cap R_k\ne\varnothing$?)

Now let $x$ be the unique point in $\bigcap_{k\ge 1}I_k$ and show that $x$ is the least upper bound of $A$. First show that it’ an upper bound for $A$, and then use the fact that $A\cap I_k\ne\varnothing$ for each $k\ge 1$ to show that it’s the limit of a sequence in $A$, so no smaller number can be an upper bound for $A$.


问:
Are 4-chromatic 3-connected unit distance graphs always rigid?
At braced heptagon we determined a graph was rigid. That graph is unit-distance, 3-connected, and 4-chromatic.  Here are some more graphs, all extremely constrained. I'm pretty sure they are all rigid graphs, but I'm not sure.

Are any of these graphs not rigid?
Are all unit-distance, 3-connected, 4-chromatic graphs rigid?

答:
I get:

the UnitDistance graphs above are rigid

the H-O-Fish graph is not infinetesimally rigid, but it might be finitely rigid

Further, using Mathematica 12.1.0

it doesn't seem to know the Exool... graphs

the embedding I get for the O-40 graph is not unit-distance

I may have to update my version of Mathematica


问:
Combination of arrangement and probability
Four guys and four girls are arranged in a row such that no two girls are together. What is the probability that any two of the four guys are together?</p>

答:
Use the laws for conditional probability: $ \displaystyle P(A | B) = \frac{P(A \cap B)}{P(B)} $.

In this problem, the total number of cases is simply $ 8! $ (which will eventually cancel out).

If no two girls are together, then there must be at least one guy between each of them. The last guy then has 5 possible positions - before the first girl, between first and second, and so forth. Hence, this has $ 4! \cdot 4! \cdot 5 $ possibilities.

Now, if no two guys are together, then the fourth guy has to be either before the first girl or after the last, so there are $ 4! \cdot 4! \cdot 2 $ possibilities.

Hence, the answer is $ \displaystyle \frac{\frac{4! \cdot 4! \cdot 2}{8!}}{\frac{4! \cdot 4! \cdot 5}{8!}} = \frac{2}{5} $.


问:
Application of Induction in the analyze of the convergence a sequence defined recursive.

Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad  n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.

My approach: Note that  $$a_{n +1}=\frac{1}{4-3a_{n}}, \quad n\geq 1$$
so, firstly  I would like to find $a_{n}$. Now, I was trying to find a  pattern but I can't find this
\begin{eqnarray*}
n=1 &\implies & a_{2}=\frac{1}{4-3a_{1}}=\frac{(3^{2-1}-1)-(3^{2-1}-3)a_{1}}{(2^{2})-(3^{2}-6)a_{1}}\\
n=2 & \implies & a_{3}=\frac{1}{4-3a_{2}}=\frac{1}{4-3 \left( \frac{1}{4-3a_{1}}\right)}=\frac{4-3a_{1}}{4(4-3a_{1})-3}\\
\vdots &\implies & \vdots \\
\end{eqnarray*}
If I know $a_{n}=a_{n}(a_{1})$, so I can analyze the denominator for the conclude when $a_{n}$ is not defined.

How can find $a_{n}$?

Also I know this problem was answered here. But I think, we can find an elementary solution using induction on $n$.

答:
Hint: If $a_1<1$, it is easy to see $a_n<1$ and then let $b_n=a_n-\frac13$.
If $a_1\in(1,\frac43)$, it is easy to see $a_n\in(1,\frac43)$ and then let $b_n=a_n-1$. You can do the rest.


答:
Update: Thanks Brian M. Scott for your insight.

I'll add the case where some $a_k=\frac 43$. Per Brian, we need to solve for the sequence $b_k$ such that $b_1=\frac 43$, $b_{k+1}=\frac{4b_k-1}{3b_k}$. This can be solved in a similar fashion, but easier because $b_1$ is given.

Note that
$$
b_{k+1} - 1 = \frac{b_k-1}{3b_k}$$$$
b_{k+1} - \frac 13 = \frac{b_k-\frac{1}{3}}{b_k}\tag 1
$$

From $(1)$ we conclude $b_k>\frac 13, \forall k$ via induction.

Then $\frac{b_{k+1}-1}{b_{k+1}-\frac 13} = \frac{1}{3} \frac{b_k-1}{b_k-\frac 13} \implies \frac{b_k-1}{b_k-\frac 13} = \frac{1}{3^{k-1}} \left( \frac{b_1 - 1}{b_1 - \frac 13}\right) = \frac{1}{3^k}$

Therefore $b_k = \frac{1 - \frac{1}{3^{k+1}}}{1-\frac{1}{3^l}} = \frac{3^{k+1} -1}{3^{k+1}-3}$ which is the same as Brian's results.

Original answer:

Since $1$ and $\frac 13$ are roots of the characteristic equation $x=\frac{1}{4-3x}$, we have

$$a_{n+1}-1 = \frac{3(a_n-1)}{4-3a_n}$$

$$a_{n+1}-\frac 13 = \frac{a_n-\frac 13}{4-3a_n}$$

So if no $a_n = \frac 13$ you have

$$\frac{a_{n+1}-1}{a_{n+1}-\frac 13} = 3 \frac{a_n-1}{a_n-\frac 13} = 3^n \frac{a_1-1}{a_1-\frac 13}$$

Of course you need to take care of the case where $a_1=\frac 13$.

Title: Application of Induction in the analyze of the convergence a sequence defined recursive.

Body:
Let $\left\{a_{n}\right\}$ be defined recursively by
$$
a_{n+1}=\frac{1}{4-3 a_{n}}, \quad  n \geq 1
$$
Determine for which $a_{1}$ the sequence converges and in case of convergence find its limit.

My approach: Note that  $$a_{n +1}=\frac{1}{4-3a_{n}}, \quad n\geq 1$$
so, firstly  I would like to find $a_{n}$. Now, I was trying to find a  pattern but I can't find this
\begin{eqnarray*}
n=1 &\implies & a_{2}=\frac{1}{4-3a_{1}}=\frac{(3^{2-1}-1)-(3^{2-1}-3)a_{1}}{(2^{2})-(3^{2}-6)a_{1}}\\
n=2 & \implies & a_{3}=\frac{1}{4-3a_{2}}=\frac{1}{4-3 \left( \frac{1}{4-3a_{1}}\right)}=\frac{4-3a_{1}}{4(4-3a_{1})-3}\\
\vdots &\implies & \vdots \\
\end{eqnarray*}
If I know $a_{n}=a_{n}(a_{1})$, so I can analyze the denominator for the conclude when $a_{n}$ is not defined.

How can find $a_{n}$?

Also I know this problem was answered here. But I think, we can find an elementary solution using induction on $n$.


问:
How I can solve this differential equation ( i did transformed to u for make simply, then how i can do solve?
I transformed to du/dx - u/x = -x. then next how i can solve..? can you give any hint? original equation was this.

enter image description here

答:
So we start from the equation :

$\frac{(x^2 + 2y^2)dx}{dy} = xy$.

We choose $u = x^2 + 2y^2$, so that we get :

$4\frac{dx}{du}u=x$ so $4\frac{dx}{x}=\frac{du}{u}$

So $4\ln x = \ln u +c$.

So $x^4 = c.u$

So $x^2 +2y^2 = x^4/c$.

So $y=\pm \sqrt{x^4/2c - x^2/2}.$
So $y = \pm x \sqrt{-1 + c_1 x^2}$.

Hope it helped.


问:
What is formal notation for rounding to the nearest hundredth?
For example, if =(X*Y), how would you denote formally that the  is to be rounded to the nearest hundredth in this operation?</p>

答:
I'm not avare of any notation, but you can of course, for your work, define one. For example, let's define:
$$ \mathrm{rd}_{100}(x) := 100\left\lfloor\frac{x}{100} + \frac 12\right\rfloor $$
Then, for example
$$ \mathrm{rd}_{100}(102) = 100\left\lfloor 1.02 + 0.5\right\rfloor = 100. $$


问:
Find the output $y(t)$ given the input and system response using Fourier Transform
I need to find the output $y(t)$ given the input $x(t)=\cos(t)$, and impulse response $h(t)=u(t)$ using the Fourier Transform.

I know that convolution in time domain is multiplication in the frequency domain, but since $\cos(t)$ has a Fourier Transform which involves the delta function, I am unable to figure out how to perform the multiplication in the frequency domain.

The answer in the time domain is $y(t)=\sin(t)$. I don't know how to get this. If someone could show the working, would be highly appreciated!

答:
From an engineering perspective, if you assume that

$$\delta(s-a) \cdot 0 = 0$$

and thus

$$\delta(s-a)\cdot\delta(s-b) = 0 \quad a \ne b$$

then

$$\begin{align*}y(t) &= h(t) \ast x(t)\\
\\
&= u(t)\ast \cos(t)\\
\\
&= \mathscr{F}^{-1}\left\{\mathscr{F}\left\{u(t)\right\}\cdot\mathscr{F}\left\{\cos(t)\right\}\right\}\\
\\
&= \mathscr{F}^{-1}\left\{\dfrac{1}{2}\left[\dfrac{1}{i\pi s}+\delta(s)\right]\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right]\right\}\\
\\
&= \mathscr{F}^{-1}\left\{\dfrac{1}{2\pi i s}\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right]\right\}\\
\\
&= \int_{-\infty}^\infty \dfrac{e^{2\pi i st}}{2\pi i s}\cdot\dfrac{1}{2}\left[\delta\left(s+\dfrac{1}{2\pi}\right)+\delta\left(s-\dfrac{1}{2\pi}\right)\right] ds\\
\\
&= \dfrac{1}{2}\left[\dfrac{e^{2\pi i \frac{-1}{2\pi}t}}{2 \pi i \frac{-1}{2\pi}}+ \dfrac{e^{2\pi i \frac{1}{2\pi}t}}{2 \pi i \frac{1}{2\pi}}\right] \\
\\
&= \dfrac{e^{it}-e^{-it}}{2i}\\
\\
y(t) &= \sin(t) \\
\end{align*}$$

Technically, instead of using a hand-waving $\delta(s-a)\delta(s-b) = 0 $ for $a\ne b$, I suppose I should have also done the two integrals with products of delta functions:

$$ \int_{-\infty}^{\infty} \dfrac{e^{2\pi i st}}{4} \delta(s)\delta\left(s+\dfrac{1}{2\pi}\right)ds $$

$$ \int_{-\infty}^{\infty} \dfrac{e^{2\pi i st}}{4} \delta(s)\delta\left(s-\dfrac{1}{2\pi}\right)ds $$

However, using the properties of the delta function, these will come out to be 0.


答:
Think of a plane orthogonal to a vector $(a,b,c)$ passing through $(x_0, y_0, z_0)$. This plane is given by the following:

$$
(−0)+(−0)+(−0)=0
$$

For an intuition read this answer. Then, since we can have a plane orthogonal to a given vector, and there's an infinite number of vectors living on a plane, we must have an infinite number of vectors orthogonal to a given vector.

Moreover, since we also need to specify a point on which the plane passes to, there are also infinite orthogonal planes, each of which have infinite vectors orthogonal to that given vector.


问:
Distribution of point-dependent function
I want to find the distribution function of $$(XY)^{-Z},$$ where $(X, Y, Z)$ are the coordinates of a point randomly selected from the cube $[0;1]^3$. I don't know how to solve it and how to start. Can you give me a hint, please?

答:
Assuming that $X,Y,Z$ are independent random variables, uniformly distributed over $[0,1]$, we may first define $W=XY$, which is obviously supported on $[0,1]$, and find the PDF of $W$. For any $k\in(0,1)$ we have

$$ \mathbb{P}[W<k]=\int_{0}^{1}\mathbb{P}[X<k/u]\,du =k-k\log k$$
hence the PDF of $W$ is $f_W(k)=\frac{d}{dk}(k-k\log k)=-\log(k)$. We apply the same approach to $W^{-Z}$, which is supported on $(1,+\infty)$. For any $k>1$

$$ \mathbb{P}[W^{-Z}<k]=\mathbb{P}[W^Z>\frac{1}{k}]=\int_{0}^{1}-\log(u)\mathbb{P}[Z<-\frac{\log k}{\log u}]\,du $$
$$ \mathbb{P}[W^{-Z}<k]=\int_{0}^{1/k}\log(k)\,du+\int_{1/k}^{1}-\log(u)\,du $$
$$ \mathbb{P}[W^{-Z}<k]=\frac{\log k}{k}+1-\frac{1+\log k}{k}=1-\frac{1}{k}$$
hence the PDF of $XY^{-Z}$ simply is $\frac{1}{k^2}$.


问:
Square integrable process with summable variations have SLLN
I'm trying to solve the following problem:

Let $X_1, X_2, \ldots$ be independent, square integrable random variables with $\sum_{n=1}^\infty \frac 1{n^2}\mathbf{Var}\left[X_n\right] < \infty$. Use the martingale convergence theorem to show the strong law of large numbers for $\left(X_n\right)_{n \in \mathbb N}$.

So I need to show $$\displaystyle \mathbf P \left[ \limsup_{n \to \infty} \left| \frac 1 n \tilde S_n\right|=0\right] = 1,$$
where $\tilde S_n = \sum_{i=1}^n \left(X_i - \mathbf E\left[X_i\right]\right)$. Since $X = (X_n)_{n \in \mathbb N}$ is independent, $\tilde S = \left(\tilde S_n\right)_{n \in \mathbb N_0}$ can be shown to be a square-integrable martingale with respect to the filtration $\mathbb F = \left(\mathcal F_n\right)_{n \in \mathbb N}$, $\mathcal F_n = \sigma(X_1, \ldots, X_n)$. If I can show that $\tilde S$ converges (which seems unlikely a priori), then it follows that $\frac 1 n \tilde S_n \to 0$.

I'm trying to use the following convergence theorem for square-integrable martingales:

Let $M$ be a square integrable martingale with square variation process $\langle M \rangle$. Then the following four statements are equivalent:

$\sup_{n \in \mathbb N} \mathbf E\left[M_n^2\right]<\infty$,

$\lim_{n \to \infty} \mathbf E\left[\langle M\rangle_n\right]<\infty,$

$M$ converges in $L^2$,

$M$ converges almost surely and in $L^2$.

The square variation process is
\begin{align*}
\left\langle \tilde S\right\rangle_n &= \sum_{i=1}^n \mathbf E\left[\left(\tilde S_i - \tilde S_{i-1}\right)^2\big| \mathcal F_{i-1}\right] = \sum_{i=1}^n \mathbf E\left[\left(X_i - \mathbf E\left[X_i\right]\right)^2\big| \mathcal F_{i-1}\right] = \sum_{i=1}^n \left( \mathbf E\left[E_i^2\right] - \mathbf E\left[X_i\right]^2\right) \\
&= \sum_{i=1}^n \mathbf{Var}\left[X_i\right].
\end{align*}
If I can show this is uniformly bounded, we're done, but it's not clear to me why it should be true. I'm sure it has something to do with the assumption that $\sum_{n=1}^\infty \frac 1{n^2}\mathbf{Var}\left[X_n\right] < \infty$, but in principle $\displaystyle \langle \tilde S \rangle$ could still be unbounded. Any thoughts? Is there another consequence of the martingale convergence theorem I'm overlooking?

答:
W.l.o.g. we may assume $EX_i = 0$. Consider $M_k = \sum_{i=1}^k \frac{X_i}i$. It's easy to see that the process $M_k$ is a square-integrble martingale w.r.t. $(\mathcal{F}_n)_{n \in \mathbb{N}}$.

$$ \langle M \rangle_{\infty} = \lim_k  \langle M \rangle_k= \lim_k \sum_{i=1}^k E( (M_i - M_{i-1})^2 | \mathcal{F}_{i-1})= \sum_{i=1}^{\infty} \frac{DX_i}{i^2} < \infty. $$

Hence $M_k$ converges a.s.

It's sufficient to apply Kronecker's lemma.

We know that
if $\sum_k a_k < \infty$, $b_k = k$, then $\lim_n \frac{\sum_{k=1}^n k a_k}{n} \to 0.$

Taking $a_k = \frac{X_k}k$ we get $\frac{\sum_{k=1}^n X_k}n \to 0$, q.e.d.