问:
Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$

Let $p,q,r$ denote sides $QR, PR, PQ$ of $\Delta PQR$ respectively. Then prove that $p\cos^2 (R/2) +r \cos^2 (P/2) = \frac{p+q+r}{2}$

Can I get a hint so that I can get started? I tried using the standard half angle formula, but they don’t work.

答:
The standard half-angle formula $\cos^2 \frac x2 = \frac{1+\cos x}2$ is helpful if combined with the law of cosines.

Using above half-angle formula, the equation to be shown is equivalent with

$$p\cos R+ r \cos P = q$$

Now, just replace $\cos R = \frac{q^2+p^2-r^2}{2pq}$ and $\cos P = \frac{q^2+r^2-p^2}{2qr}$:
\begin{eqnarray*} p\cos R+ r \cos P
& = & \frac 1{2q}\left(q^2+p^2-r^2+q^2+r^2-p^2\right)\\
& = & q
\end{eqnarray*}


问:
On a sphere, does a rhumb line always exist between two points? If so, how to calculate it?
We can calculate the bearing between two points (described by lat, lon with an elevation of 0 for simplicity) on Earth by using the Haversine formula, which yields an initial heading and a final heading. When this line is followed the distance is minimal.

But is there such a thing as a "straight line" which is not distance minimal that can be cast between any two points such that the initial heading is the same as the final heading always? How can this bearing be calculated if so? (As noted in comments, such a path is called a rhumb line or loxodrome.)

Example: If point A is some point that lies on the equator of Earth, and B is a point 200 miles to the west of point A, you could simply get from A to B by going 200 miles on a 270 heading. You could also follow a great circle path which would result in less distance but would involve an initial and final bearing which change as a function of your position on the line. Does this straight line exist for any two points on a sphere?

I am fairly sure the answer is yes, but I'd like to know how to calculate this if I am correct. What would this line be called? I've heard it described at LOS (line of sight) I think, but unsure of if that is the same idea.

答:
I expected that this question was a duplicate, but I couldn't locate another instance of it.

But is there such a thing as a "straight line" which is not distance minimal that can be cast between any two points such that the initial heading is the same as the final heading always? ... What would this line be called?

What you're describing is a rhumb line, also called a loxodrome or a path of constant heading, namely, a path on the earth's surface such that the angle between the direction of travel and the meridian (longitude line) is constant. You can produce equations for a rhumb line by parameterizing the sphere and imposing the constant-angle condition, which leaves a single functional parameter that controls how fast you move along the path; to do this it's convenient to use latitude-longitude coordinates.

How can this bearing be calculated if so?

Carrying out the above procedure (and ignoring the poles) shows that the points on the loxodrome

through the point with latitude $\phi_0$ and longitude $\theta_0$ and

making a constant angle $\alpha$ with the meridians

are characterized by the relation

$$\phantom{(\ast)} \qquad (\theta - \theta_0) \cot \alpha = \operatorname{gd}^{-1} \phi - \operatorname{gd}^{-1} \phi_0 , \qquad (\ast)$$

where $\theta$ denotes longitude and $\phi$ latitude, and where $\operatorname{gd}^{-1}$ is the inverse of the Gudermannian function: $$\operatorname{gd}^{-1} = \operatorname{artanh} \circ \sin .$$

In particular, if we fix a destination point $(\theta, \phi) = (\theta_1, \phi_1)$ and let $\alpha$ vary, we can solve for the heading $\alpha$ that will take us from $(\theta_0, \phi_0)$ to $(\theta_1, \phi_1)$. We split into two cases:

If $\phi_0 = \phi_1$, i.e., the points lie on the same parallel, then $\alpha = \pm \frac{\pi}{2}$, that is, the only headings that take you between the points are due west and due east.

On the other hand, if $\phi_0 \neq \phi_1$, then since $\operatorname{gd}^{-1}$ is strictly increasing on the set $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ of (non-polar) latitudes, $\operatorname{gd}^{-1}(\phi_0) \neq \operatorname{gd}^{-1}(\phi_1)$, and rearranging gives $$\alpha = \arctan \frac{\theta_1 - \theta_0}{\operatorname{gd}^{-1} \phi - \operatorname{gd}^{-1} \phi_0} .$$ Notice that adding any integer multiple of $2 \pi$ to, say, $\theta_1$ gives another representation $(\theta_1 + 2 \pi k, \phi_1)$ of the destination point and (since $\arctan$ is strictly increasing), giving an infinitely many headings that take you from the departure point to the destination point,
$$\boxed{\alpha = \arctan \frac{\theta_1 - \theta_0 + 2 \pi k}{\operatorname{gd}^{-1} \phi - \operatorname{gd}^{-1} \phi_0} , \qquad k \in \Bbb Z} .$$ Difference choices of $k$ correspond to rhumb lines that wrap around the sphere different numbers of times.

Does this straight line exist for any two points on a sphere?

The above computation shows that it does, at least provided neither of the points is a pole, and generically that there are infinitely many rhumb lines connecting two points; you can check readily that you can always find a rhumb line passing through a pole and a given second point, and the only headings that take you between the two are $\alpha = 0, \pi$.

Remark Like you say, rhumb lines are typically not distance-minimizing; in fact, among all rhumb lines, only arcs of the equator and of longitude lines are. Nor are they straight in a naive sense: A typical rhumb line curves away from a geodesic ("straight") path. To see this, imagine you're standing a meter from the North Pole facing east. If you walk straight ahead, you'll walk along a great circle that will pass near the South Pole. But the constant-heading path is just a small circle around the North Pole that bends to your left. This curving behavior is also what you'd expect a priori: Anything worth calling "straight lines" ought to be a feature of the geometry itself and in particular independent of the choice of coordinates, but different choices of poles give different rhumb lines.

On the other hand, one can look for a map projection that has the property that rhumb lines on the Earth correspond to straight lines on the map. But equation $(\ast)$ shows that, for any given rhumb line, $\theta$ and $\operatorname{gd}^{-1}(\phi)$ are affinely related, so the projection $$(\theta, \phi) \mapsto (\theta, \operatorname{gd}^{-1}(\phi))$$ has that property. One can check that this projection is conformal (that is, it preserves angles between vectors) and moreover that it is essentially the only projection that has these two properties; this is the classic Mercator projection.

ContentLicense: CC BY-SA 4.0


问:
Proving f'(x) is measurable on I if d is lebesgue integrable on I and f'(x) exists almost everywhere
This question is from Tom Apostol Mathematical analysis and I am struck on how to prove this.

Question: If f is lebesgue measurable on an open interval I and if f'(x) exists almost everywhere on I , prove that f' is measurable on I.

On I there exists {$s_n (x)$} of step functions  such that {$s_n (x)$} $\to f(x)$ .But I need to show that that there exists {$t_n(x)$ } of step functions such that {$t_n(x)$ } $\to f'(x)$. But I am not getting any ideas on how should I use existence of f'(x) for it/

Can you please help me with it?

答: $f'(x)=\lim \frac {f(x+\frac 1n) -f(x)} {1/n}$ a.e. and limit of a sequence of measurable functions is  measurable.


问:
Define an algebra $\Sigma$ over the set $S=\{(A,A)(A,B),(B,A),(B,A)\}$
Like the title says, I need to Define an algebra $\Sigma$ over the set $S=\{(A,A),(A,B),(B,A),(B,B)\}$

I think this is trivial, we can take $\mathscr P(S)$ as the largest algebra $\Sigma$ over $S$.

The second question is, how can I calculate $Pr(\sigma_i)$, We can take $Pr$ to be the usual probability measure.

答:
First, yes, $\mathscr{P}(S)$ is an algebra over $S$.

Then define $Pr$ on the singleton sets of $\mathscr{P}(S)$ however you want so that they (1) are non-negative and (2) sum to $1$.  This may be done arbitrarily, as there is no "usual" probability on an arbitrary set.  The non-singleton sets are given probabilities by
$$Pr(A) = \sum_{i \in A} Pr(\{i\})$$


问:
Evaluating $\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$
I have been working on the following integral:
$$\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$$
where $t$ is any nonnegative real number. Would anyone be able to provide a hint or provide a solution on how such an integral should be approached? I have a hunch it involves differentiation under the integral sign.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
{\expo{-tx^{2}} - \expo{-x^{2}} \over x}\,\dd x} =
\\[5mm] = &\
-\int_{0}^{\infty}\ln\pars{x}\expo{-tx^{2}}\pars{-2tx}\dd x +
\int_{0}^{\infty}\ln\pars{x}\expo{-x^{2}}\pars{-2x}\dd x
\\[5mm] = &\
\int_{0}^{\infty}\ln\pars{x^{1/2} \over t^{1/2}}
\expo{-x}\,\dd x -
{1 \over 2}\int_{0}^{\infty}\ln\pars{x}\expo{-x}\dd x =
\bbx{-\,{1 \over 2}\,\ln\pars{t}} \\ &
\end{align}


问:
Topology for convergent sequences
Let $(X,\tau)$ be a topological space, and consider the family $\mathcal{F}$ of the topologies over $X$ such that the convergent sequences for each $\gamma \in \mathcal{F}$ are the same as the convergent sequences for $\tau$, with the same limits. That is,
$$
  \mathcal{F} = \{\gamma \;|\; x_n \xrightarrow{\gamma} x \Leftrightarrow x_n \xrightarrow{\tau} x\}
$$

It is easy to see that the topology $\tau_M$ generated by the topologies in $\mathcal{F}$ is such that $\tau_M \in \mathcal{F}$. The $\tau_M$ is the finest topology with the same convergent sequences as $\tau$, converging to the same point.

But is it true that the topology
$$
  \tau_m = \bigcap_{\gamma \in \mathcal{F}} \gamma
$$
is itself in $\mathcal{F}$?

In other words, does it follow that the convergent sequences for $\tau_m$ and their limits are the same as the convergent sequences and their limits in $\tau$? Are there any nice counterexamples?

In other words again, do we have a weakest topology such that the convergent sequences and their limits are the same as the ones for $\tau$?

Edit: Fixed according to @joriki's comment.

答:
Taking the intersection can introduce new convergent sequences.

Let $p$ be a point not in $\omega$. Let $X=\{p\}\cup\omega$. For each free ultrafilter $\mathscr{U}$ on $\omega$ define a topology $\tau_{\mathscr{U}}$ on $X$ as follows. Points of $\omega$ are isolated, and the basic open nbhds of $p$ in $\tau_{\mathscr{U}}$ are the sets of the form $\{p\}\cup U$ such that $U\in\mathscr{U}$. The only convergent sequences of $\langle X,\tau_{\mathscr{U}}\rangle$ are the trivial ones. Let $$\tau=\bigcap_{\mathscr{U}\in\beta\omega\setminus\omega}\tau_{\mathscr{U}}\;.$$ A set $W\subseteq X$ belongs to $\tau$ iff $p\notin W$, or $W\in\mathscr{U}$ for all $\mathscr{U}\in\beta\omega\setminus\omega$. But the only subsets of $\omega$ that belong to all free ultrafilters on $\omega$ are the cofinite sets, and It follows that

$$\tau=\wp(\omega)\cup\{W\subseteq X:p\in W\text{ and }|\omega\setminus W|<\omega\}\;.$$

That is, points of $\omega$ are isolated in $\langle X,\tau\rangle$, and the basic open nbhds of $p$ are the sets of the form $\{p\}\cup(\omega\setminus F)$ such that $F$ is finite. But then $\langle k:k\in\omega\rangle$ is a sequence in $X$ that $\tau$-converges to $p$.

Now let $\tau'$ be the intersection of all topologies on $X$ having only the trivial convergent sequences. Clearly $\tau'\subseteq\tau$, so $\langle k:k\in\omega\rangle$ converges to $p$ in $\langle X,\tau'\rangle$.

ContentLicense: CC BY-SA 3.0

Title: Topology for convergent sequences


答:
Using differentiation under the integral sign makes it quite easy: let
$f(t) = \int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$, then
$$f'(t) = \int_0^\infty \frac{\partial}{\partial t}\left(\frac{e^{-tx^2}-e^{-x^2}}{x}\right)dx = \int_0^\infty \frac{-x^2 e^{-tx^2}}{x} dx = -\int_0^\infty x e^{-tx^2} dx = \left[\sqrt{t}x = s, dx = \frac{ds}{\sqrt{t}}\right] =
-\frac{1}{t}\int_0^\infty s e^{-s^2} ds = -\frac{1}{2t} \int_0^\infty e^{-s^2} d(s^2) =
-\frac{1}{2t} e^{-s^2} \vert_0^{+\infty} = -\frac{1}{2t}
$$
So, $f'(t) = -\frac{1}{2t}$, $f(1) = 0$ (easy to see), so $f(t) = -\frac{1}{2} \ln t$.

Title: Evaluating $\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$

Body:
I have been working on the following integral:
$$\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$$
where $t$ is any nonnegative real number. Would anyone be able to provide a hint or provide a solution on how such an integral should be approached? I have a hunch it involves differentiation under the integral sign.


答:
Without using Frullani integral's, we have
$$\int \frac{e^{-kt^2}}x \,dx=\frac 12 {\text{Ei}\left(-t x^2\right)}$$ where appears the exponential integral function (we could also use the gamma function).

If $a>0$, then
$$\int_a^\infty \frac{e^{-kt^2}}x \,dx=\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)+\log
   \left(a^2\right)\right)$$
$$\int_a^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx=\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)-\Gamma
   \left(0,a^2\right)\right)$$ Using Taylor series
$$\frac{1}{2} \left(\Gamma \left(0,a^2 t\right)-\Gamma
   \left(0,a^2\right)\right)=-\frac{\log (t)}{2}+\frac{1}{2} a^2 (t-1)+\frac{1}{8} a^4
   \left(1-t^2\right)+O\left(a^6\right)$$


问:
Real Analysis Theorem Question
This is a Theorem covered in my textbook:

Let $n \geq 2$ be an integer. Suppose that $b_{1}, b_{2}, \ldots , b_{n}$ are positive real numbers that are not all equal. If $b_{1}b_{2}\cdots b_{n} = 1$, then prove that $b_{1} + b_{2} + \ldots + b_{n} > n.$

I know that mathematical induction will be used and why it was used. I also fully comprehend the base case and will just include the ind. hypothesis/step.

Suppose the result is valid for some positive integer $p \geq 2$. Let $b_{1}, b_{2}, \ldots ,b_{p}, b_{p+1}$ be positive real numbers that are not all equal and satisfy $b_{1}b_{2}\cdots b_{p}b_{p+1} = 1$. Without loss of generality, we may assume that the numbers are in increasing order: $b_{1} \leq b_{2} \leq \ldots \leq b_{p} \leq b_{p+1}$.

This is where I am stuck:

By the assumptions on these numbers, we must have $b_{1} < 1 < b_{p+1}$.

My question is: why? I tried proving this with the given information (from above) but had little to no success. I can intuitively see why this makes sense, but cannot show this step.

Any help would be greatly appreciated!

答: Since they are not all equal, there is $i,j \in \{1, \ldots, p+1\}$ with $i< j$, such that $b_i \neq b_j$ and as your sequence $\{b_1, \ldots, b_{p+1}\}$ is ordered you must have $b_1 \le b_i < b_j \le b_{p+1}$. Moreover, if $b_1 \ge 1$, then $b_{p+1} b_1 > 1$ which contradicts your induction hypothesis. The same argument holds for $b_{p+1} \le 1$.

Title: Real Analysis Theorem Question


问:
Ross's Probability 10th edition Chapter 2, question 22
I am currently self-studying Sheldon Ross' Intro to Probability and I am stuck on this problem. I would appreciate any hints on how to begin with:

Each of 52 people are given a deck of cards, which they are asked to shuffle independent of each other. What is the probability that

the order of the cards in each shuffled deck is unique?

There is exactly one card that occupies the same position in the shuffled decks received from all 52 persons?

all cards occupy the same position in all the shuffled decks?

I appreciate any hints!

Edit: For 1. my strategy is to say that or each shuffle, there are $$52!$$ possible orderings, so there are $52!$ unique orderings where no card is fixed, $51!$ ordering where 1 card is held in the same place, and $(52-i)!$ orderings where we fix $i$ of the cards in the same place, so, there are $$\prod_{i=0}^{51}(52-i)!$$ possible unique orderings, out of $(52!)^{52}$ possible orderings, so the probability that the order of the shuffled cards is unique is
$$\frac{\prod_{i=0}^{51}(52-i)!}{(52!)^{52}}=\frac{52!}{52!}\frac{51!}{52!}\dots\frac{1!}{52!}=1\frac{1}{52}\frac{1}{52\cdot 51}\cdots\frac{1}{52\cdot 51\cdots (52-i+1)}\cdots\frac{1}{52!}$$
Which is $$\prod_{i=1}^{51}\frac{1}{52\cdots{(52-i+1)}}$$. This is the answer in the back of the book but I am unsure about the reasoning

For part b), I'm thinking, there are 52 choices for the card that occupies the same position over all decks, and there are 52 choices for the position in where it should be. Then, just like in the previous one, there are $51!$ ways to order the rest of the cards, $50!$ if we keep an additional card fixed and so on. Thus, there are $$52(52)(51!)(50!)\cdots(3!)(2!)$$ possible orderings where the one card is in the same position in all decks so the probability is

$$\frac{52(52)(51!)(50!)\cdots(3!)(2!)}{(52!)^{52}}$$

答:
As suggested by @Ted Shifrin, I shall solve a more tractable problem of $4$ packs of $4$ cards distributed to $4$ people.

Each pack can be shuffled in $4!$ ways, and each of the $4$ persons can receive any one shuffle, so total ways of distribution $= (4!)^{4} = D,\;say$

(1) Permuting the total shuffles between the $4$ people, $\;Pr\;= \dfrac{^{4!}P_{4}}{D}$

(2) Choose "special" card and place in line in $4^2$ ways, $3!$ shuffles of remaining $3$ cards need to be distributed to $4$ people, thus $Pr\; = \dfrac{{4}^2\times ^{3!}P_{4}}{D}$

(3) Each person has the same shuffle from $4!\;$ thus $\dfrac{4!}{D}$

You should be able to easily transmute it to the required scale.


问:
Cartan subalgebras of classical Lie algebras are diagonal matrices
I want to prove that Cartan subalgebras of $A_l, B_l, C_l, D_l$ consist of the respective diagonal matrices.

As for $A_l$: let $\mathfrak{h}$ the set of diagonal matrices of $\mathfrak{sl}(l+1)$. We have that $\mathfrak{h}$ is of course toral; to prove that it is maximal, I showed that if $A \in \mathfrak{sl}(l+1)$ and $A$ commutes with every element of $\mathfrak{h}$, then $A$ must be diagonal. (In particular I chose $A = E_{ij}$ from the basis of $\mathfrak{sl}(l+1)$, computed $[A, H]$ with $H \in \mathfrak{h}$ and $H$ having $1$ at $i$-th line and $-1$ at the $j$-th line and found that $[A, H] \neq 0$). However, I noticed that this concludes the proof for every other classical Lie algebra, since every diagonal matrix in $\mathfrak{sp}(2l), \mathfrak{so}(2l), \mathfrak{so}(2l+1)$ must have trace zero. Is this sufficient to complete the proof or if $\mathfrak{h}$ is the set of diagonal matrices in each classical algebra, do I have to check that $[A, \mathfrak{h}] \neq 0$ for each $A$ in the basis of those algebras? I think this is not necessary since every element of those bases is a linear combination of elementary matrices. Am I missing something?

Also, what about uniqueness of the Cartan subalgebra in these cases? Since it consists of diagonal matrices, could we argue that it is unique?

答:
To expand on comments:

As Dietrich Burde points out, of the various characterisations of Cartan subalgebras in semisimple Lie algebras (cf. 1, 2), the one that likely works best here is that of a maximal toral subalgebra (toral = abelian and consisting of $ad$-semisimple elements).

Then, we should agree on what Lie algebras we even mean here: Namely, the split simple forms of types $A_n, B_n, C_n, D_n$ (over a field $K$, let's assume characteristic zero to be safe). For $A_n$, everybody knows that it is $\mathfrak{sl}_{n+1}(K)$, the $n+1 \times n+1$-matrices over $K$ with trace $0$. As per 3, the others are:

For type $B_n$,

$$\mathfrak{so}(2n+1) := \{M \in M_{2n+1}(K): M^TS+SM=0\} \text{, where } S=\begin{pmatrix}
O&I_n&O\\
I_n&O&O\\
O&O&1
\end{pmatrix}
$$

$$=\{\left(\begin{matrix}A&B&-f^T\\C&-A^T&-e^T\\e&f&0\end{matrix}\right):  A,B,C\in M_n(K), B=-B^T, C=-C^T , e, f \in M_{1\times n}(K)\}.$$

For type $D_n$,

$$\mathfrak{so}(2n) := \{M \in M_{2n}(K): M^TS+SM=0\} \text{, where }
S=\begin{pmatrix}
O&I_n\\
I_n&O
\end{pmatrix}
$$

$$=\{\left(\begin{matrix}A&B\\C&-A^T\end{matrix}\right):  A,B,C\in M_n(K), B=-B^T, C=-C^T \}.$$

For type $C_n$,

$$\mathfrak{sp}(2n) := \{M \in M_{2n}(K): M^TS+SM=0\} \text{, where }
S=\begin{pmatrix}
O&I_n\\
-I_n&O
\end{pmatrix}
$$

$$=\{\left(\begin{matrix}A&B\\C&-A^T\end{matrix}\right):  A,B,C\in M_n(K), B=B^T, C=C^T \}.$$

Now what you want to show is that in each of these algebras, those diagonal matrices which sit inside it form a CSA. (Note that these are never all diagonal matrices. E.g. in all of them, the trace of every matrix has to be $0$. Further, e.g. $\pmatrix{1&0&0&0&0\\0&-1&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0} \notin \mathfrak{so}(5)$ etc.)

It is nearly trivial that diagonal matrices describe $ad$-semisimple elements. It is well known that they commute with each other. Leaves to show the maximality. Indeed, one can show (seemingly) more, namely, that in each case, the diagonal matrices inside the given algebra are their own centraliser.

To show that, I would recommend first looking at the case of $\mathfrak{sl}_n$. The crucial step is learning to "see" centralisers of individual diagonal matrices. E.g. you should be able to see with a glance that

the centraliser of $\pmatrix{3&0&0&0&0\\0&0&0&0&0\\0&0&-3&0&0\\0&0&0&0&0\\0&0&0&0&0} \in \mathfrak{sl}_5$ is $\pmatrix{*&0&0&0&0\\0&*&0&*&0\\0&0&*&0&0\\0&*&0&*&*\\0&0&0&*&*}$, or that

the centraliser of $\pmatrix{1&0&0&0&0\\0&1&0&0&0\\0&0&-1&0&0\\0&0&0&-1&0\\0&0&0&0&0} \in \mathfrak{sl}_5$ is $\pmatrix{*&*&0&0&0\\*&*&0&0&0\\0&0&*&*&0\\0&0&*&*&0\\0&0&0&0&*}$, or that

the centraliser of $\pmatrix{17&0&0&0&0\\0&-17&0&0&0\\0&0&-17&0&0\\0&0&0&17&0\\0&0&0&0&0} \in \mathfrak{sl}_5$ is $\pmatrix{*&0&0&*&0\\0&*&*&0&0\\0&*&*&0&0\\*&0&0&*&0\\0&0&0&0&*}$.

All these things can be shown with direct matrix computations, but it is strongly recommended to instead learn to think of them in terms of root space decompositions, cf. 4. This allows you to quickly grasp answers to questions like 5.

Once you have that, you should be able to quickly see that indeed in all cases, the subalgebra made up of diagonal matrices (that is, those diagonal matrices that actually lie in the respective Lie algebra) are self-centralising. (Use that the centraliser of a subalgebra is the intersection of the centralisers of all its elements.)

Then in hindsight, you can actually even use something like

the centraliser of $\pmatrix{1&0&0&0&0\\0&2&0&0&0\\0&0&-1&0&0\\0&0&0&-2&0\\0&0&0&0&0} \in \mathfrak{sl}_5$ is $\pmatrix{*&0&0&0&0\\0&*&0&0&0\\0&0&*&0&0\\0&0&0&*&0\\0&0&0&0&*}$ (and hence so is its centraliser in $\mathfrak{so}(5)$, where there are just a few restrictions on the asterisks anyway), and that kind of thinking should lead you right into the usefulness of regular elements (which this one is, and the other ones above were not), and give you a feeling connected to your previous question.

As final words of warning, as noted in a comment, a) these algebras, even when written exactly as matrices as above, contain many other Cartan subalgebras which are not given by diagonal matrices; and b) note that at least over certain base fields, one can alternatively represent the Lie algebras by matrices which do not even contain non-trivial diagonal ones. E.g. $\mathfrak{so}_3(\mathbb C)$ is isomorphic to $\{\pmatrix{0&a&b\\-a&0&c\\-b&-c&0}: a,b,c \in \mathbb C\}$. There's infinitely many Cartan subalgebras in there, but they're hard to see ...


问:
Prove that $x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}\ge x_n+2n$

St.Petersburg 1999 :Let $x_0>x_1>\dots >x_n$ be real numbers. Prove that $$ x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}\ge x_n+2n $$

I tried Titu's inequality.

So we get $$ x_0+\frac{1}{x_0-x_1}+\frac{1}{x_1-x_2}+\dots+\frac{1}{x_{n-1}-x_n}= x_0+\frac{1^2}{x_0-x_1}+\frac{1^2}{x_1-x_2}+\dots+\frac{1^2}{x_{n-1}-x_n} \ge x_0 +\frac{n^2}{x_0-x_n}.$$

So we need to show that $$x_0 +\frac{n^2}{x_0-x_n}\ge x_n+2n .$$

Now note that $$\frac{n^2}{x_0-x_n}=\frac{n^2}{x_0}+\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}$$

So we need to show that $$x_0 +\frac{n^2}{x_0}+\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}\ge x_n+2n .$$

Note that by AM-GM $$x_0 +\frac{n^2}{x_0}\ge 2n$$.

So it's enough to show that $\frac{x_n \cdot n^2}{x_0\cdot(x_0- x_n)}\ge x_n$ .

Enough to show that $$\frac{n^2}{x_0\cdot(x_0- x_n)}\ge 1.$$

Which I am not able to. Any hints? Thanks in advance!!

答:
Note that your last inequality need not be true, esp for $ x_0 > n$, so unfortunately that's futile.

The reason being that in the previous inequality, equality holds iff $ x_0 =n$. And when $ x_0 > n$, we have some leeway to work with, that is needed for your last inequality.

Instead, do you see that you can prove this, similar to what Albus stated?

So we need to show that $$x_0 +\frac{n^2}{x_0-x_n}\ge x_n+2n .$$


问:
Edge percolation on $\mathbb{Z}^2$: probability that two neighbouring vertices are connected?
I'm considering edge percolation on $\mathbb{Z}^2$ with parameter $p$, so that edges are present with probability $p$.

Is it known how to express the probability $P(p)$ that $(0,0)$ is in the same connected component as $(1,0)$ as an explicit function of $p$?

A very crude bound is
$$
p \leqslant P(p) \leqslant 1-(1-p)^4,
$$
but I'm sure people have derived better bounds, or even an exact expression.

Simulations suggest that $P(1/2)$ is close to $3/4$.

Thanks.

Edit:

The following is an approximation of the graph of the function $p\mapsto P(p)$ for $p=0,.01,\ldots,.82$.

答:
Indeed $P(1/2)=3/4$. With probability $1/2$, the edge $e$ joining the vertices is present, and they're connected. With probability $1/2$, the edge $e$ is not present, and then either the vertices are connected by a path not including $e$, or they're in separate clusters, and in the latter case, since there's no percolation in the critical phase, there's a closed path in the dual lattice that's composed of duals of absent edges, surrounds one of the vertices and includes the edge $e'$ dual to $e$. Removing $e'$ from such a path yields a path joining the endpoints of $e'$ that doesn't include $e'$. Thus these two situations (the endpoints of $e$ joined by a path and the endpoints of $e'$ joined by a path) are dual, and since the square lattice is self-dual, they must have the same probability. As exactly one of them obtains, that probability must be $1/2$. Thus $P(1/2)=1/2+1/2\cdot1/2=3/4$.

We can extend this to uncover a hidden symmetry in $P(p)$. Let $R(p)$ denote the probability that the vertices are connected by a path not including $e$; then $P(p)=p+(1-p)R(p)$. By the above duality argument, $R(p)=(P(p)-p)/(1-p)$ exhibits the symmetry $R(p)=1-R(1-p)$, which is antisymmetry about the point $R(1/2)=1/2$. Here's a plot of $R(p)$ that confirms this symmetry:

(Right-click on the image and open in a new tab/window to see the full resolution.)

The plot was obtained using this code, which associates each edge of a finite square lattice (with periodic boundary conditions) with a threshold uniformly distributed on $[0,1]$, then adds the edges in order of increasing thresholds and updates a histogram for $P'(p)$ when adjacent sites become connected. The red, green, blue and magenta lines show calculations on lattices of dimensions $5$, $10$, $20$ and $50$, respectively. The plot seems to be nearly converged up to its graphical resolution at dimension $50$. $10000$ runs were averaged for each lattice size, and the coincidence of the results for different sizes indicates that the statistical errors are at or below the plot's graphical resolution.

We can also obtain some analytical results using inclusion–exclusion. For small $p$, short paths between the vertices dominate, so we can derive a power expansion of $R(p)$ at $p=0$. We only have to consider self-avoiding paths, since other paths always contain a shorter path as a subpath. There are two paths of length $3$ connecting the vertices, which are disjoint from each other, so we arrive at $R(p)=2p^3+O(p^4)$ without having to deal with overlap between paths. There are six paths of length $5$, and merging them with one of the paths of length $3$ adds at least one more edge, so we can even get up to $5$th order without overlaps, $R(p)=2p^3+6p^5+O(p^6)$. There are six different unions of paths of length $3$ and $5$ that have $6$ edges, and one such union of the two paths of length $3$, so up to $6$th order inclusion–exclusion yields $R(p)=2p^3+6p^5-7p^6+O(p^7)$.

After this it becomes hard to avoid errors, so we'd better get some more help from our electronic friends. Here's code that finds all self-avoiding paths between the vertices and performs the complete inclusion–exclusion for all paths and unions containing up to $14$ edges. The result is

$$
R(p)=2p^3+6p^5-7p^6+28p^7-54p^8+140p^9-297p^{10}\\+690p^{11}-1534p^{12}+3710p^{13}-9285p^{14}+O(p^{15})\;,
$$

and thus

$$
P(p)=p+2p^3-2p^4+6p^5-13p^6+35p^7-82p^8+194p^9\\-437p^{10}+987p^{11}-2224p^{12}+5244p^{13}-12995p^{14}+O(p^{15})\;.
$$

Each of the self-avoiding paths corresponds to a self-avoiding loop obtained by adding the edge $e$. The number of self-avoiding loops of length $l$ is conjectured to be asymptotic to $B\mu^ll^{-\beta}$ with $\mu\approx2.63816$ (see Analytic Combinatorics, p. 365). The initial coefficients above suggest that inclusion–exclusion doesn't affect the exponential growth rate $\mu$, so we can expect the power expansion to converge up to $p=1/\mu\approx0.379052$. This is confirmed by a comparison of the numerical results with the expansions of $R(p)$ up to $5$th order and up to $13$th order:

The expansion up to $5$th order (in green) is already a very good approximation up to $p\approx1/\mu$, and around and beyond $1/\mu$ the expansion up to $13$th order (in blue) is actually worse. A detailed plot of values around $p=0.25$ indicates that the higher-order terms aren't simply wrong and do improve the approximation in the interior of the interval of convergence:

To summarize, for $p\lesssim0.4$ for practical purposes $R(p)$ is well approximated by

$$
R(p)\approx2p^3+6p^5\;,
$$

and thus

$$
P(p)\approx p+(1-p)(2p^3+6p^5)\;.
$$

(Note that this is derived manually above, so it doesn't rely on the correctness of the linked code.)


答:
Indirectly if we assume that $b_1 < 1 < b_{p+1}$ is NOT true, there are two cases:

$b_{p+1} \le 1$, which implies that $0 < b_1 \le \ldots \le b_{p+1} < 1$ (one of the inequalities being strict), implying $\prod b_i < 1$, or

$1 \le b_1$ which similarly would imply $\prod b_i > 1$.

And both cases lead to a contradiction with $\prod b_i = 1$.

Title: Real Analysis Theorem Question

Body:
This is a Theorem covered in my textbook:

Let $n \geq 2$ be an integer. Suppose that $b_{1}, b_{2}, \ldots , b_{n}$ are positive real numbers that are not all equal. If $b_{1}b_{2}\cdots b_{n} = 1$, then prove that $b_{1} + b_{2} + \ldots + b_{n} > n.$

I know that mathematical induction will be used and why it was used. I also fully comprehend the base case and will just include the ind. hypothesis/step.

Suppose the result is valid for some positive integer $p \geq 2$. Let $b_{1}, b_{2}, \ldots ,b_{p}, b_{p+1}$ be positive real numbers that are not all equal and satisfy $b_{1}b_{2}\cdots b_{p}b_{p+1} = 1$. Without loss of generality, we may assume that the numbers are in increasing order: $b_{1} \leq b_{2} \leq \ldots \leq b_{p} \leq b_{p+1}$.

This is where I am stuck:

By the assumptions on these numbers, we must have $b_{1} < 1 < b_{p+1}$.

My question is: why? I tried proving this with the given information (from above) but had little to no success. I can intuitively see why this makes sense, but cannot show this step.

Any help would be greatly appreciated!


问:
Finding the intersection of two "half" hyperbolas with equations of the form $\sqrt{(x-x_1)^2+(y-y_1)^2}-\sqrt{(x-x_2)^2+(y-y_2)^2}=C$
I have two hyperbola equations that need to find an interaction,
I tried to use the GeoGebra to draw the graph, but I cannot implement this using C++ as I am not able to convert the two equations like
$x = \text{??}$ or $y = \text{??}$.

Is there any software or anyway I can make these two equations to be the format like: $x = \text{??}$ and $y = \text{??}$ ?
For your convenience, I will paste the equation here:

$$\begin{align}
\sqrt{(x-x_1)^2+(y-y_1)^2}-\sqrt{(x-x_2)^2+(y-y_2)^2}=C_1 \\[6pt]
\sqrt{(x-x_1)^2+(y-y_1)^2}-\sqrt{(x-x_3)^2+(y-y_3)^2}=C_2
\end{align}$$

Thank you so much for your help, appreciated

答:
There is an analytical solution to your problem (it is almost half a mile long).

Instead, take the first equation and square it. This gives a quadratic equation in $y(x)$ (two roots). Plug in the second equation, repeat squaring to have a single equation in $x$ and ... have fun !


问:
Can it be proven that a product of factors where two or more factors have non-integer exponentiation is a irrational number?
I'm reading a course in Discrete Mathematics and exploring different ways things can be proven.
One classic question seems to be prove, for instance, that ${\sqrt{30}}$ is a irrational number.

By proof through contradiction and going through the motion we do following:
${\sqrt{30}} = \frac{a}{b}$ where $a$, $b$ are integers, $b\ne0$ and $gcd(a,b)=1$.
We end up with $2*3*5*b^2=a^2$

Now one can look at substitution and see that we end up with a contradiction.

But, a different argument at this point could be:
Since all factors of $a^2$ must be of power $2n$, where $n$ is integer $\ge 1$, $a=(q^n_1*q^n_2*..) \Rightarrow a^2=(q^{2n}_1*q^{2n}_2*..)$. Since we have $2*3*5*b^2=a^2$ that would mean that if $a^2$ have these factors then they would have exponential of $^{2n+1}$ after multiplication with $b^2$, if $a^2$ doesn't have these factors they would have exponential of $^1$ after multiplication.

Taking the square root of $a^2$ would result in these factors have a exponential that are non-integer. Can we prove that for any combination of factors where the exponentials are fractions that the product is irrational?

答:
Welcome to MSE!

$\sqrt 4$ is an integer, not irrational.
In your argument, $4b^2=a^2$, so $2^2b^2=a^2$ and there is no way to find a contradiction.

So if the number is a perfect square, like 4, 9, 16, the proof of contraction fails as it should.


Body: take $u=x^2$ to get $$ \int_{0}^{\infty}\dfrac{e^{-tu}-e^{-u}}{2u}du = \dfrac{-\ln(u)}{2} $$


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: In Group theory i would say "Structure theorem for finitely generated abelian groups"

Title: Landmarks of subjects of mathematics


问:
Prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$
I am trying to prove that $d_1\left({x,y}\right) = \max\{\left|{x_j-y_j}\right|:j=1,2,...,k\}$ is a metric on $\mathbb{R}^k$. So far I know that $\text{dist}\left({x,y}\right) \leq d_1\left({x,y}\right)$ where $\text{dist}\left({x,y}\right)$ is the standard distance function in k-dimensional Euclidian space. I also know that $d_1\left({x,y}\right) \leq \text{dist}\left({x,y}\right) \leq \sqrt{k}\cdot d_1\left({x,y}\right)$. I have already proven that $d_1\left({x,y}\right) \geq 0$, that $d_1\left({x,y}\right) = 0 \iff x=y$, and that $d_1\left({x,y}\right) = d_1\left({x,y}\right)$, but I am stuck on the triangle inequality property of this particular metric. Any hints or suggestions would be greatly appreciated. </p>

答:
Hint: For any particular $j$, we have $|x_j-z_j|\le |x_j-y_j|+|y_j-z_j|$.

Note also that for example $|x_j-y_j|\le \max|x_i-y_i|$.


问:
Quotient space: $\mathbb{R}^3$ by a line.
I am trying to understand, geometrically, a plot of the quotient of $\mathbb{R}^3$ quotiented by some line $L$, through the origin. The Wikipedia article states;

Similarly, the quotient space for $\mathbb{R}^3$ by a line through the origin can again be represented as the set of all co-parallel lines, or alternatively be represented as the vector space consisting of a plane which only intersects the line at the origin.

$L$ is a one-dimensional subspace, so this quotient space must surely be two-dimensional. However, a set of lines, as in the case of quotienting $\mathbb{R}^2$ by a line $L$, is one-dimensional. I'm having trouble both understanding this and visualizing the result.

As another matter, I don't want to just take for granted that the equivalence classes are sets of "co-parallel" lines, but my attempt to prove this by picking an arbitrary point $(x,y,z)$ in $\mathbb{R}^3$ and considering the set of $(a,b,c)$  such that $(x,y,z) - (a,b,c) \in L$ didn't produce a workable result (in fact, it produced a plane, rather than a line).

答:
Consider a line $L$ passing throw origin, as a vector subspace of $\mathbb{R}^3$,

so the quotient space will be the space of equivalent relations that is geometrically a plan perpendicular to $L$, which is a $2$-dimensional vector subspace in $\mathbb{R}^3$.


问:
Asymptotic stability for nonautonomous system with unique solution
I'm concerned with the system $\frac{dy}{dt} = \frac{1}{yt}$, of course restricting so that $y$ and $t$ are real and cannot be $0$. I've been able to come up with a form for the unique solution to this equation at any point $(t_0, y_0)$. I want to know if all of these unique solutions are stable/asymptotically stable.

The unqiue particular solution at $(t_0, y_0)$ I've come up with is $\varphi(t, t_0, y_0) = \sqrt{y_0^2 + t_0^{-2}-t^{-2}}$. Based on the definition then, wouldn't $\phi(t)$ and $\varphi(t,t_0,y_0)$ have the exact same form, since this solution is exactly that solution where plugging in $t_0$ yields $y_0$? Based on this I believe that all of these solution are in fact stable, and they must be asymptotically stable as well, since in the limit these two solutions will eventually equal. Am I thinking about this wrong?

答:
Consider the case when the solution is positive. The case of a negative solution can be considered similarly.
We have to estimate
$$
|\phi(t)-\varphi(t, t_0, y_0)|=\left|\sqrt{\phi(t_0)^2 + t_0^{-2}-t^{-2}}-\sqrt{y_0^2 + t_0^{-2}-t^{-2}}\right|
$$
using the fact that  $|\phi(t_0) - y_0| < \delta$. First notice that
$$
|\phi(t)^2-\varphi(t, t_0, y_0)^2|=\left|\phi(t_0)^2 + t_0^{-2}-t^{-2}-y_0^2 - t_0^{-2}+t^{-2}\right|=|\phi(t_0)^2-y_0^2|
$$
$$
=|\phi(t_0)-y_0|\cdot|\phi(t_0)+y_0|<\delta |\phi(t_0)+y_0|.
$$
Since the solution $\phi(t)$ is given, we know the value of $\phi(t_0)$; however, we don't know what is $y_0$ exactly. But $|\phi(t_0) - y_0| < \delta$, thus,
$$
|\phi(t_0)+y_0|=|y_0-\phi(t_0)+2\phi(t_0)|\le2|\phi(t_0)|+\delta=2 \phi(t_0)+\delta.
$$
Hence
$$
|\phi(t)^2-\varphi(t, t_0, y_0)^2|<2\delta \phi(t_0)+\delta^2.
$$
Then notice that
$$
\phi(t)+\varphi(t, t_0, y_0)=
\sqrt{\phi(t_0)^2 + t_0^{-2}-t^{-2}}+\sqrt{y_0^2 + t_0^{-2}-t^{-2}}
$$
$$
>\sqrt{\phi(t_0)^2 }+\sqrt{y_0^2}>|\phi(t_0)|=\phi(t_0),
$$
which implies
$$
|\phi(t)-\varphi(t, t_0, y_0)|=\frac{|\phi(t)^2-\varphi(t, t_0, y_0)^2|}
{\phi(t)+\varphi(t, t_0, y_0)}<\frac{2\delta \phi(t_0)+\delta^2}{\phi(t_0)}
=2\delta +\frac{\delta^2}{\phi(t_0)}.
$$
Now let
$$
\epsilon=2\delta +\frac{\delta^2}{\phi(t_0)}.
$$
Expressing
$$
\delta = \phi(t_0)\left(\sqrt{1+\frac{\epsilon}{\phi(t_0)}}-1\right)
$$
we obtain the $\delta$ from the definition. Hence, stability is proven.

The solution $\phi(t)$ is not asymptotically stable since
$$
\lim_{t\to+\infty} \phi(t)= \lim_{t\to+\infty} \sqrt{\phi(t_0)^2 + t_0^{-2}-t^{-2}}=
\sqrt{\phi(t_0)^2 + t_0^{-2}},
$$
but
$$
\lim_{t\to+\infty} \varphi(t, t_0, y_0)= \lim_{t\to+\infty} \sqrt{y_0^2 + t_0^{-2}-t^{-2}}=
\sqrt{y_0^2 + t_0^{-2}},
$$
i.e.
$$
\lim_{t\to+\infty} \varphi(t, t_0, y_0)\ne \lim_{t\to+\infty} \phi(t).
$$


问:
Example of an unbounded convex set with no recession direction?
Is it true that every unbounded convex set in $R^n$ has a recession direction? I think there is a counterexample, because I have not assumed that the convex set is closed.

I have not been able to come up with a counterexample though. Does anyone know if this is true?

答:
In a finite dimensional space, any unbounded convex has a recession direction, should it be closed or not.

That may not be the case for infinite dimensional spaces though.

See here for details.


问:
General integral for partial differential equation
I want to find a general integral for $xz_x + yz_y = z^2 + 1$, where $z = z(x,y)$ is implicitly a function of two variables. I'm not sure how to go about this.

答:
Elaborating on my comment, the complete Lagrange-Charpit equations read

$$  \frac{\mathrm{d}x}{x} = \frac{\mathrm{d}y}{y} = \frac{\mathrm{d}z}{1+z^2} = \frac{\mathrm{d}p}{(2z-1)p} = \frac{\mathrm{d}q}{(2z-1)q}, $$

with $p=z_x$ and $q=z_y$. As in @Aryadeva's answer, from the first equality we get the characteristic $x/y=c_1$. If you then go with the equation formed with the 2nd and 3rd fraction, you obtain their solution. Alternatively, the 1st and 3rd would give you $z = \tan\left[ \log x + f(x/y) \right]$.

In order to obtain a complete integral you would need to use the 3rd and either 4th or 5th equation together with the fact that

$$\mathrm{d} z = p \mathrm{d}x + q \mathrm{d}y. \quad(1)$$

Let me use for instance 4th and 5th fractions to get

\begin{equation}
\frac{\mathrm{d}p}{(2z-1)p} = \frac{\mathrm{d}q}{(2z-1)q} \implies \frac{\mathrm{d}p}{p} = \frac{\mathrm{d}q}{q} \implies q = a_1 p,
\end{equation}

where $a_1$ is a constant of integration. Let us now use the 3rd and 4th fractions to arrive at

$$ \frac{2z-1}{z^2+1} \mathrm{d}z = \mathrm{d}z \implies p = a_2 (1+z^2)\mathrm{e}^{-\operatorname{atan}{z}},$$

with $a_2$ another constant of integration. Combine this into $(1)$ to finally get at

$$ \mathrm{d}z = p (\mathrm{d}x + a_1 \mathrm{d}y) = a_2 (1+z^2)\mathrm{e}^{-\operatorname{atan}{z}} (\mathrm{d}x + a_1 \mathrm{d}y), $$
or alternatively

$$  \frac{\mathrm{e}^{\operatorname{atan}{z}}}{z^2+1} \mathrm{d}z = \mathrm{d}\left( \mathrm{e}^{\operatorname{atan}{z}}  \right) = \mathrm{d}(A x + B y),$$ where I have renamed the constants $a_2=A$ and $a_1a_2=B$. The result in my comment follows upon integration and putting the new constant of integration as a function of $c_1 = x/y$, i.e.,

$$z = \tan\left[ \log(Ax+By) + f(x/y) \right]$$


问:
Bounded sequence in Hilbert space contains weak convergent subsequence
In Hilbert space $H$, $\{x_n\}$ is a bounded sequence then it has a weak convergent subsequence.

Is there any short proof? Thanks a lot.

答: Suppose $M$ bounds the sequence. Then, if we think of $H$ as sitting inside $H^{**}$, then for any $T \in H^\ast$ with $\|T\| \leq 1$, we have $\|x_n(T)\| = \|Tx_n\| \leq \|x_n\| \leq M$, so the operator norms of the $x_n$ thought of as operators on $H^\ast$ are bounded by $M$. Apply Banach-Alaoglu. (To the unit ball of $H^{\ast \ast}$.)</p>


问:
Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.

Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.

What I have tried: Put $x = \frac{1}{9}$. We have, $\sqrt[3]3\big(4x^\frac{1}{3} - 2x^\frac{1}{3} +  x^\frac{1}{3})$
$$\rightarrow \sqrt[3]3\big(x^\frac{1}{3}(4-2+1)\big)$$
$$\rightarrow \sqrt[3]3 \times \frac{1}{3\sqrt[3]x}$$
$$\rightarrow \frac{3}{27x} = \frac{1}{9x}$$
Putting back the value of $x$ gives the answer to be $1$.

However, in my text the answer is given to be $\sqrt[3]2 + 1$ , so where did I go wrong?

Can anyone help me?


问:
Find the Region for Change of Variables in a Double Integral Problem
Let $D$ be the region surrounded by the quadrilateral with vertices $O(0,0) , A(2,1), B(s,t)$ and $C(1,3)$ with $s$ and $t$ constants.

1- Using the change of variables $(x,y)=(au+bc+cuv, du +ev+fuv)$ Find the constants $a, b, c, d$ such as $E:0\le u \le 1 ; 0\le v\le 1$ corresponds to $D$.

2- Calculate $\iint_D xy \,dx\,dy$

I am stuck at the first question and I cannot seem to find $a, b, c$ and  $d$.

I tried to write the region $D$ such as

$$
\left\{
\begin{array}{c}
0\le x\le 1 ; \frac x2\le y\le 3x \\
1\le x\le 2 ; \frac x2\le y\le \frac{t-3}{s-1}(x-1)+3 \\
2\le x\le s ; \frac{t-1}{s-2}(x-2)+1\le y\le  \frac{t-3}{s-1}(x-1)+3 \\
\end{array}
\right.
$$

Then I tried to substitute with $u$ and $v$ but I was not able to find the constants.

Any hint will be very much appreciated

答:
The vertices of $E$ are $(0,0), (0,1), (1,0), (1,1).$  The suggested change of variables is $$(x,y)=F(u,v)=(au+bc+cuv, du +ev+fuv)$$  Along the edges of $E$, one of the variables $u,v$ is constant, so if we look at the coordinate functions of $F$, that is, $$F_1(u,v)=au+bc+cuv\\F_2(u,v)=du +ev+fuv$$
they are simply linear functions along the edges, because the $uv$ term will disappear.  So, it appears that $F$ will map the edges of $E$ to the edges of $D$.

We have $$\begin{align}
F(0,0)&=(0,0)\\
F(0,1)&=(bc,ev)\\
F(1,0)&=(a+bc,d+ev)\\
F(1,1)&=(a+b+c,d+e+f)
\end{align}$$

Now, let's guess that $$\begin{align}
(bc, ev)&= (2,1)\\
(a+bc,d+ev)&=(1,3)\\
(a+b+c,d+e+f)&=(s,t)
\end{align}$$

We must have $a=-1$, $d=2$, $bc=2$, $ev=1$.  The last equation gives $$s=-1+b+\frac2b\\t=2+e+\frac1e$$ and we can solve for $b$ abd $e$ in terms of $s$ and $t$.

There's no guarantee that this will work, though I'd be surprised if it doesn't.  Once you get $F$, check that it works.  If it doesn't, and you're sure it doesn't, go back and make a different guess about which vertex maps to which vertex, and see if that works.

Good luck.


问:
How to find the image of this linear transformation?
Let $T\colon\mathbb{R}^3\to\mathbb{R}^3$ be a linear transformation such that $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2x+2y-z\\2x+5y-2z\\4x+4y-2z\end{pmatrix}$. I am asked to find the image of $T$.

So, let $p=\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix}$ be in the image of $T$.
$$
T\begin{pmatrix}x\\y\\z\end{pmatrix}=p \implies \begin{pmatrix}2x+2y-z\\2x+5y-2z\\4x+4y-2z\end{pmatrix}=p \implies \begin{matrix}2x+2y-z=p_1\\2x+5y-2z=p_2\\4x+4y-2z=p_3\end{matrix}
$$
Clearly, $p_3=4x+4y-2z=2p_1\implies p_3=2p_1$. Now, I think if I can show that $p_1$ and $p_2$ does not depend on each other I can prove that the plane $p_3=2p_1$ is the image of $T$.

I do not know how to proceed from here.

答:
Note that
$$T\begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}2&2&-1\\2&5&-2\\4&4&-2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}\qquad\forall\:\begin{pmatrix}x\\y\\z\end{pmatrix}\in\mathbb R^3$$

So image of $T$ is the colspan of this matrix (which is of rank $2$). Can you proceed now?


答:
You can write
$$T\left(\begin{array}{c} x\\ y\\ z\end{array}\right)=\left(\begin{array}{rrr} 2 & 2 & -1\\ 2 & 5 & -2\\ 4 & 4 & -2\end{array}\right)\left(\begin{array}{c} x\\ y\\ z\end{array}\right).$$
Take a basis of $\mathbb{R}^3$. For example, the canonical basis $B=\{\vec u_1(1,0,0),\vec u_2(0,1,0),\vec u_3(0,0,1)\}$, and compute:
$$T\left(\begin{array}{c} 1\\ 0\\ 0\end{array}\right)=\left(\begin{array}{c} 2\\ 2\\ 4\end{array}\right),\quad T\left(\begin{array}{c} 0\\ 1\\ 0\end{array}\right)=\left(\begin{array}{c} 2\\ 5\\ 4\end{array}\right),\quad T\left(\begin{array}{c} 0\\ 0\\ 1\end{array}\right)=\left(\begin{array}{c} -1\\ -2\\ -2\end{array}\right).$$
As every vector in $T(\mathbb{R}^3)$ can be written as a linear combination of $T(\vec u_1)=(2,2,4)$, $T(\vec u_2)=(2,5,4)$ and $T(\vec u_3)=(-1,-2,-2)$, you know that $T(\mathbb{R}^3)$ is the linear span of these vectors:
$$T(\mathbb{R}^3)=\big\langle(2,2,4),(2,5,4),(-1,-2,-2)\big\rangle.$$
However, the above set is not a basis of $T(\mathbb{R}^3)$ because the vectors are dependents (you can check that with Gauss or the determinant). You can remove any one of the vectors and then you'll have a basis of the image.

Title: How to find the image of this linear transformation?

Body:
Let $T\colon\mathbb{R}^3\to\mathbb{R}^3$ be a linear transformation such that $T\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2x+2y-z\\2x+5y-2z\\4x+4y-2z\end{pmatrix}$. I am asked to find the image of $T$.

So, let $p=\begin{pmatrix}p_1\\p_2\\p_3\end{pmatrix}$ be in the image of $T$.
$$
T\begin{pmatrix}x\\y\\z\end{pmatrix}=p \implies \begin{pmatrix}2x+2y-z\\2x+5y-2z\\4x+4y-2z\end{pmatrix}=p \implies \begin{matrix}2x+2y-z=p_1\\2x+5y-2z=p_2\\4x+4y-2z=p_3\end{matrix}
$$
Clearly, $p_3=4x+4y-2z=2p_1\implies p_3=2p_1$. Now, I think if I can show that $p_1$ and $p_2$ does not depend on each other I can prove that the plane $p_3=2p_1$ is the image of $T$.

I do not know how to proceed from here.


答:
We can assume that the line $L$ goes in the $x$-direction, so it is generated by $(1,0,0)$. This does not change the outcome, since this can be achieved by a simple linear transformation.

Then you want to know what the space $\mathbb R^3$ looks like, when you impose the equivalence relation $v \sim w$ iff $v-w \in L$. You can also phrase this relation like: $v \sim w$ iff $v = w + (a, 0, 0)$ for some $a \in \mathbb R$. This means that the coset of $ w $ (the set of elements that are equivalent to $w$) is a line, parallel to $L$, through $w$. This is what wikipedia means when they talk about co-parallel lines: you partition the space into lines that are parallel to $ L $.

For each coset we can pick a unique representative $ w = (0, a, b) $ with $ (a, b) \in \mathbb R^2 $ (this unique representative is just the intersection of the co-parallel line with the plane $x=0$). The quotient space is the space of the cosets, is the space of the representatives, is $ \mathbb R^2 $.

Does this make sense?

Title: Quotient space: $\mathbb{R}^3$ by a line.

Body:
I am trying to understand, geometrically, a plot of the quotient of $\mathbb{R}^3$ quotiented by some line $L$, through the origin. The Wikipedia article states;

Similarly, the quotient space for $\mathbb{R}^3$ by a line through the origin can again be represented as the set of all co-parallel lines, or alternatively be represented as the vector space consisting of a plane which only intersects the line at the origin.

$L$ is a one-dimensional subspace, so this quotient space must surely be two-dimensional. However, a set of lines, as in the case of quotienting $\mathbb{R}^2$ by a line $L$, is one-dimensional. I'm having trouble both understanding this and visualizing the result.

As another matter, I don't want to just take for granted that the equivalence classes are sets of "co-parallel" lines, but my attempt to prove this by picking an arbitrary point $(x,y,z)$ in $\mathbb{R}^3$ and considering the set of $(a,b,c)$  such that $(x,y,z) - (a,b,c) \in L$ didn't produce a workable result (in fact, it produced a plane, rather than a line).


问:
How to plot cycloid curve?
I have having trouble plotting the curve. It only plots half of the curve, what may be missing here?

https://en.wikipedia.org/wiki/Cycloid#Equations

x = r arccos (1 - y/r) - sqrt (2ry - y^2)

Plot https://www.geogebra.org/calculator/nvujgvgw

答:
The problem is that the form of equation you have entered into geogebra is treated by geogebra as a function. Therefore, for the sqrt() function it only returns the positive value.

To get geogebra to plot the full curve you will have to enter the parametric equation given in Wikipedia using the geogebra curve() command.
For example, with $r$ set to 5,
$$Curve(5 (t-sin(t)),5 (1-cos(t)),t,0,4 π)  $$

This should do two cycles.

Title: How to plot cycloid curve?


问:
Evaluating $\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$
I have been working on the following integral:
$$\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx$$
where $t$ is any nonnegative real number. Would anyone be able to provide a hint or provide a solution on how such an integral should be approached? I have a hunch it involves differentiation under the integral sign.

答:
If $f'(x)$ is continuous and the integral exists then
$$\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(0)-f(\infty)] \ln(b/a);~ a, b>0~~~~(1),$$
it is called Frullani's integral, see
https://en.wikipedia.org/wiki/Frullani_integral

Proof of Frullani's theorem

$$I=\int_0^\infty \frac{e^{-tx^2}-e^{-x^2}}{x}dx=\int_{0}^{\infty} \frac{e^{(\sqrt{t}x)^2}-e^{(x)^2}}{x}dx=-\frac{1}{2}\ln(t)~~~~(2)$$
as $a=\sqrt{t}$ and $b=1$.


问:
Central limit theorem for exponential random variable
Consider the following problem:

Suppose that $(e_n)_n$ are independent exponentially distributed random variable with $E[e_n]=\mu_n.$ If $$\lim_n\max_{1 \leq k \leq n} \frac{\mu_k}{\sum_{r=1}^n\mu_r}=0.$$
then $$\sum_{k=1}^n(e_k-\mu_k)/\sqrt{\sum_{k=1}^n\mu^2_k} \implies N(0;1).$$

It's obvious that we need to prove Lindeberg condition, so in order to prove it, we need the assumption $\lim_n\max_{1 \leq k \leq n} \frac{\mu_k}{\sum_{r=1}^n\mu^2_r}=0$ and not the above.

Is it possible the there is a typo in this exercise? Is there any counterexample to disprove it?

Note that this question was asked before here, but it's not confirmed.

答:
Suppose $ \{X_n,n\ge 1\}$ are exponentially distributed random variables with $ \mathsf{E}[X_n]=\mu_n $. Then
\begin{equation*}
 \lim_{n\to\infty}\frac{\max\limits_{1\le i\le n}\mu_i}{\sum\limits_{j=1}^n \mu_j}=0 \tag{1}
\end{equation*}
is not sufficient for the following CLT
\begin{equation*}
 \frac{\sum\limits_{i=1}^n (X_i-\mu_i)}{\sqrt{\sum\limits_{i=1}^{n}\mu_j^2}}
 \Rightarrow N(0,1). \tag{2}
\end{equation*}
For example, $\mu_n=\frac1n$. In this case,

\begin{equation*}
 \max\limits_{1\le i\le n}\mu_i=1 , \qquad
 \lim_{n\to\infty}\sum_{j=1}^{n}\mu_j=\infty,
\end{equation*}
hence (1) holds.

On the other hand, since $ \mathsf{Var}[X_n]=\mu_n^2=\frac1{n^2} $,
\begin{equation*}
 c^2=\sum_{j=1}^{\infty} \mathsf{Var}[X_j]=\sum_{j=1}^{\infty}\frac1{j^2}\Big(=\frac{\pi^2}{6}\Big)<\infty.
\end{equation*}
Using Kolmogorov Convergence Criterion(Please refer to the book
"S. I. Resnick, A Probability Path, Birkhäuser Boston, 2005." Th.7.3.3, p212.
) to get both
\begin{equation*}
 S_1=\sum_{j=1}^{\infty}(X_j-\mu_j),\quad  S_2=\sum_{j=2}^{\infty}(X_j-\mu_j)
\end{equation*}
converge almost surely.
Mean while $(X_1-\mu_1)$, $S_2  $ are independent mutually,  hence $ S_1=(X_1-\mu_1)+S_2 $
couldn't be Gaussian distributed and
\begin{equation*}
 \lim_{n\to\infty}\frac{\sum\limits_{i=1}^n (X_i-\mu_i)}{\sqrt{\sum\limits_{i=1}^{n}\mu_j^2}}
 =\frac{S_1}{c}\stackrel{d}{\ne}N(0,1).
\end{equation*}
This means that (1) is not sufficient for (2)


问:
Let $A,\,B$ be subsets of the set $X$. Prove one of de Morgan's Laws of Set Theory
Let $A,\,B$ be subsets of the set $X$. Prove one of de Morgan's Laws of set theory:$$X\setminus(A\cap B)=(X\setminus A)\cup(X\setminus B)$$

答:
We have:

$x\in X\setminus (A\cup B)\Leftrightarrow x\in X\wedge (x\not\in A\cup B)\Leftrightarrow x\in X\wedge (x\not\in A\vee x\not \in B) \Leftrightarrow (x\in X\wedge x\not\in A)\vee (x\in X\wedge x\not\in B) \Leftrightarrow x\in (X\setminus A)\cup (X\setminus B)$.

Please note the use of some laws from propositional calculus, like distributivity and De Morgan's law.


问:
Find the probability of the outcome which has highest chance of occurring.
A biased die has numbers $1, 2, 3, 4, 5, 6$. The probability of obtaining one of the numbers is greater
than ${1\over 6}$, whereas the probability of obtaining a number opposite to it is less than ${1\over 6}$ . The remaining four numbers each have a probability of ${1\over 6}$ of being obtained. Given that any two opposite faces add up to $7$. When two such dice are rolled, probability of obtaining a sum $7$ is ${47\over 288}$. If the number which has the highest probability of being obtained has probability ${p\over q}$, for ($p, q$) = $1$, find $p + q$.

Well I know that $P(E)$ = ${No.\;of\;Favourable\;Outcomes\over Total\;Outcomes}$ but I am confused as to how to find the Probability in this case. Because the formula which I wrote, I guess, is applicable only when all the possible outcomes have equal chances of occurring which is not happening in this case.

答:
Call a number "good" if the probability that it comes up is $\frac14$, and "bad" otherwise.  Let $P$ be the bad number whose probability is $p$ and let $Q$ be the other bad number.

The first die lands on a good number with probability $\frac16$ and then the number we need on the other die is also good, so the probability of rolling $7$ with two good numbers is $\frac46\cdot\frac16=\frac19.$

If the first die lands on $P$ (probability $p$) then we need $Q$ on the other die (probability $q$) so the probability of this event is $pq$.  Similarly, if the first dies shows $Q$, we need $P$ on the second die, and the probability is again $pq$.

This accounts for all the probabilities, so the probability of rolling $7$ is $$\frac19+2pq= {47\over 288}$$

Take it from here.


问:
Definition of a p-value
I'm studying all statistics by Wasserman and on page 156 he defines p-values:

10.11 Deﬁnition. Suppose that for every $\alpha \in (0,1)$ we have a size $\alpha$ test with rejection region $R_{\alpha}$. Then,

$$\text{p-value} = \inf\{\alpha: T(X^n)\in R _{\alpha}\}$$

That is, the p-value is the smallest level at which we can reject
$H_0$.

I've already seen some answers here to this definition in Wasserman book, but my question is more basic. I want to know what is this exponent $n$. What does $X^n$ mean?

答: This is an unfortunate use of poorly-chosen notation.  Presumably, $X^n$ is intended to indicate a sample of size $n$; i.e., $X^n = (X_1, X_2, \ldots, X_n)$, from which a test statistic defined by some function $T$ of the sample is calculated.  This notation is probably inherited from expressions like $\mathbb R^n$.  But for obvious reasons, one should probably avoid writing things this way.

Title: Definition of a p-value


答:
Let $P(A)$ be the event of getting the number 'A' on the dice

$2P(1)P(6) + 2P(2)P(5) + 2P(3)P(4) = \dfrac{47}{288}$

Let $P$(biased number) $= x$

Therefore,

$2x(\frac{1}{3} - x) + \dfrac{1}{18} + \dfrac{1}{18} = \dfrac{47}{288}$

Title: Find the probability of the outcome which has highest chance of occurring.

Body:
A biased die has numbers $1, 2, 3, 4, 5, 6$. The probability of obtaining one of the numbers is greater
than ${1\over 6}$, whereas the probability of obtaining a number opposite to it is less than ${1\over 6}$ . The remaining four numbers each have a probability of ${1\over 6}$ of being obtained. Given that any two opposite faces add up to $7$. When two such dice are rolled, probability of obtaining a sum $7$ is ${47\over 288}$. If the number which has the highest probability of being obtained has probability ${p\over q}$, for ($p, q$) = $1$, find $p + q$.

Well I know that $P(E)$ = ${No.\;of\;Favourable\;Outcomes\over Total\;Outcomes}$ but I am confused as to how to find the Probability in this case. Because the formula which I wrote, I guess, is applicable only when all the possible outcomes have equal chances of occurring which is not happening in this case.


问:
Determinants of free modules
This should be easy, but I'm stuck.

Let $A$ be a DVR and $B$ a finite algebra over $A$ that is free as an $A$-module. For $b \in B$ one can define the norm of $b$ as the determinant of multiplication by $b$ on $B$. Suppose that there is a basis $f_1, \ldots, f_n$ of $B$ over $A$ and $n$ elements $a_i \in A$ such that $a_1f_1, \ldots, a_n f_n$ is a basis of the free $A$-module $bB$. Then, one should show that $N(b) = a_1 \cdots a_n$. How can this be done?

Regards,
Heidar

答:
Multiplication by $b$ has image $bB$ and sends the basis $\{f_1,\dots,f_n\}$ to $\{bf_1,\dots,bf_n\}$. This is also an $A$-basis of $bB$ (why?). Now write $bf_i=\sum_{j=1}^na_{ij}a_jf_j$, where $A=(a_{ij})$ is an invertible matrix. Then $N(b)=a_1\cdots a_n\det A$. (Note that since $A$ is invertible, then $\det A$ is an invertible element of $A$.)

Edit. As noted @Jiangwei Xue in his comment to the question this is the maximum we can get.

Title: Determinants of free modules


答:
You may obtain the sum $7$ with $1+6,2+5,3+4,4+3,5+2,6+1$. Among these sums, two are biased and the other $4$ have probability $\frac{1}{6}\times\frac{1}{6}$. The other two pairs have faces with probabilities $\frac16+x$ and $\frac16-x$ for some positive $x$.

Hence, the probability of obtaining the sum $7$ is

$$\frac{4}{36}+2(\frac{1}{6}+x)(\frac{1}{6}-x)=\frac{47}{288}$$

$$\frac{1}{36}-x^2=\frac12(\frac{47}{288}-\frac{4}{36})=\frac{5}{192}$$

$$x^2=\frac{1}{36}-\frac{5}{192}=\frac{1}{576}=(\frac{1}{24})^2$$

Hence, the highest probability is $\frac16+\frac1{24}=\frac{5}{24}$ and $p+q=5+24=29$.


问:
Invariant factors of order 27000
I am doing some practice problems to prepare for the exam and want to be sure I understand correct:

(1) Find invariant factors of abelian group $\mathbb{Z}_6 \times \mathbb{Z}_{100} \times \mathbb{Z}_{45}$

My answer is: $3^2*2^2*5^2$ and $3*2*5$

(2) Let $p$ be a prime. Let $A$ be an abelian group of order $p^4$ that contains an element of order $p^2$. State all possibilities for the elementary divisors:

My answer is: 2 possibilities: $p^2, p^2$ and $p^2, p, p$

答:
Your answer for (1) is correct.

For (2), note that we can list all abelian groups of order $p^4$ by looking at the partitions of $4$, which are
$$4,\ 3+1,\ 2+2,\ 2+1+1,\ 1+1+1+1$$
One should also note that if $G$ is a cyclic group with $|G| = n$, for each divisor $d$ of $n$, there exists a unique subgroup of $G$ with order $d$. Also, since that subgroup is also cyclic, its generators must have order $d$ as well. Can you revisit your answer for (2) now?

Title: Invariant factors of order 27000


问:
Given that $X_1,X_2,...,X_i$ are random variables, find maximum likelihood estimator of $a$
$X_1,X_2,X_3,...X_i,$ are random variables and

$$f(x,a)= 0.5e^{-|x-a|}$$

with
$$-∞ < x < +∞$$
$$-∞ < a < +∞$$

Find maximum likelihood estimator of $a$.

I solved this problem as follows

$$L(a) = \prod_{i=1}^{n} 0.5e^{-|xi-a|}$$

$$L(a) = 0.5^n \prod_{i=1}^{n} e^{-|x-a|}$$

$$LL(a) = -n ln2 + \sum-|x-a|$$

But when I take this derivative, I do not achieve anything.

答:
If I asked you for the value of $x$ for which the function $$f(x) = 1 - |x - 2|$$ attains its global maximum, would you try to differentiate $f$ to find it?

So, why would you expect that taking the derivative of the log-likelihood will yield the desired solution?  You need to understand that finding the extrema of a function is not always about differentiation.  Such is the case here:  the sum of absolute deviations is not something you can differentiate and get a critical point.  Instead, think of what it means to look for $a$ that minimizes $$\sum_{k=1}^n |x_k - a|$$ for some set of numbers $(x_1, x_2, \ldots, x_n)$.  For instance, what if $n = 4$ and $(x_1, x_2, x_3, x_4) = (1, 3, 7, 11)$?  What value of $a$ would you pick?  Does a unique value of $a$ exist that minimizes such a sum?


答:
X1,X2,X3 are random variables and...

First well copy the text of your exercise...say $(X_1,X_2,X_3)$ is a simple random sample from the following density....

Then, as suggested, the function $\hat{a}$ that minimize the quantity  $\Sigma_i |X_i-a|$ is the Sample median: $X_{(2)}$

The proof is very simple:

Let's start with the definition of $E|X-a|$; without loss of generality we can do the brainstoming for a continuous rv

$$E|X-a|=\psi(a)=\int_{a}^{\infty}(x-a)f(x)dx+\int_{-\infty}^{a}(a-x)f(x)dx$$

$$\psi'(a)=-\int_{a}^{\infty}f(x)dx+\int_{-\infty}^{a}f(x)dx$$

(using Leibnitz integral rule)

That is

$$-[1-F_X(a)]+F_X(a)$$

Setting it $=0$ you get

$$F_X(a)=1/2$$

Thus

$$a=Me$$

The results can be found here, as your density is a known one: the Laplace distribution

Title: Given that $X_1,X_2,...,X_i$ are random variables, find maximum likelihood estimator of $a$

Body:
$X_1,X_2,X_3,...X_i,$ are random variables and

$$f(x,a)= 0.5e^{-|x-a|}$$

with
$$-∞ < x < +∞$$
$$-∞ < a < +∞$$

Find maximum likelihood estimator of $a$.

I solved this problem as follows

$$L(a) = \prod_{i=1}^{n} 0.5e^{-|xi-a|}$$

$$L(a) = 0.5^n \prod_{i=1}^{n} e^{-|x-a|}$$

$$LL(a) = -n ln2 + \sum-|x-a|$$

But when I take this derivative, I do not achieve anything.


问:
Find the value of $ A + B + C + D + E. $
KTOM Indonesia, October 2020

$\sin(5^{\circ})\sin(55^{\circ})\sin(65^{\circ})=\displaystyle\frac{A\sqrt{B}+C\sqrt{D}}{E}$

where $ B $ and $ D $ are two natural numbers that are not divisible by the square of any prime, $ A $ and $ C $ are integers, and $ E $ is a natural number. Find the value of $ A + B + C + D + E. $

答:
Use this identity-
$$sin(\theta)\cdot sin(60-\theta)\cdot sin(60+\theta)=\frac{1}{4}sin(3\theta)$$


问:
Is the transformation $T$ orthogonally diagonalisable?
Let $T: P_3 \rightarrow P_3$ be the transformation $T(f)=5 \operatorname{proj}_v(f) -2f$ where $V$ is a subspace spanned by the vectors $1 + 2t+t^2$ and $t +t^2+ 3t^3$.

The inner product on the vector space $P_{3}$ is defined as:  $\textit{f} \cdot \textit{g} =\int_{-1}^1 fg \,dt$

a) determine a matrix representation (D) for the transofmration S relative to a basis $\beta$ of eigenvalues

b) is the transformation $T$ orthogonally diagonalisable?

I solved a) by choosing a basis $\beta=\{f_1, f_2, g_1, g_2\}$ where $f_1, f_2$ form a basis for $V$ and $g_1, g_2$ form a basis for the orthogonal complement of $V.$ However, I don't know how to reason on b). The solution says that it is orthogonally diagonalisable because you can choose a ON-basis for $V$ and an ON-basis for the orthogonal complement of $V$, thus forming an ON-basis of eigenvectors for $P_3$.

How do I know that I can form an ON-basis for the two subspaces and why is that sufficent to show that $T$ is orthogonally diagonalisable? Can't an ON basis be formed for all subspaces using the Gram-Schmidt process, thus you can find an ON-basis for all eigenspaces, which then would imply that all matrices are orthogonally diagonalisable which is not the case. Any help would be greatly appreciated!

The answer to a) is
$$D= \begin{bmatrix}
3 & 0 & 0 & 0\\
0 & 3 & 0 & 0\\
0 & 0 & -2 & 0\\
0 & 0 & 0 & -2\\
\end{bmatrix}$$

答: If you apply Gramm-Schmidt to $\{f_1,f_2\}$, you will get a set $\{h_1,h_2\}$ of two orthonormal elements of $V$. Besides, $T(h_1)=3h_1$ and $T(h_2)=3h_2$. And if you apply Gramm-Schmidt to $\{g_1,g_2\}$, you will get a set $\{h_3,h_4\}$ of other two orthonormal elements of $V$. Besides, $T(h_3)=-2h_3$, $T(h_4)=-2h_4$, and, since $h_1,h_2\in V$ and $h_3,h_4\in V^\perp$, the set $\{h_1,h_2,h_3,h_4\}$ is also orthonormal. Finaly, the matrix of $T$ with respect to the basis $\{h_1,h_2,h_3,h_4\}$ as the one that you got.


问:
What does this notation mean? ( $X=\inf\{ x : U \le F(x) \}$ )
What does ( $X= \inf\{ x : U \le F(x) \}$ ) mean?

I know the concept of inf, found here

I found the notation in the title at the end of the chapter on the Inverse Transform Sampling Method where $F(x)$ is the CDF cumulative density function of X and $U$ a random number generate with Uniform distribution $[0,1]$

答:
$\newcommand{\R}{\mathbb R}$
Not sure how much you know about probability, but random numbers like $U$ are formally defined as a function from a probability space $(\Omega,\mathcal F,P)$ to $(\R,\mathcal B(\R)$. You can see more about this here.

Here the probability space is $(\R,\mathcal B(\R))$ and $X$ is a random variable from $\R\to\R$, formally $X(\omega)=\inf\{x:U(\omega)\le F(x)\}$ for $\omega\in\R$.


问:
Show ker$(\varphi)=\langle x^2-x\rangle$
I have $\varphi : \mathbb{Q}[x]\to\mathbb{Q}^2$ with $\varphi (f)=(f(0),f(1))$ being a ringhomomorphism. $\mathbb{Q}^2$ is a ring equipped with the operations $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)\cdot (c,d)=(a\cdot c,b\cdot d)$, with $(0,0)$ and $(1,1)$ being their respective neutral elements.

I want to first show that ker$(\varphi)\subseteq \langle x^2-x\rangle $. Doesn't every element in ker$(\varphi)$ have the form $f=f(2)x^2+f(3)x^3+\dots +f(n)x^n$? For the inclusion to hold, I would like to have no remainder after division of $f$ by $x^2-x$. But this is not true for $f=x^2$?

EDIT: I think I'm being confused by this part in my textbook:

答:
$\langle x^2-x\rangle \subset \ker \varphi$ is obvious because $\varphi(x^2-x) = (0^2-0,1^2-1) = (0,0)$.

Conversely, $f \in  \ker \varphi$  means that $(f(0),f(1)) = (0,0)$, i.e. $f(0) = 0$ and $f(1) = 0$. Thus $f$ has zeros at $0$ and $1$. Zero at $0$ means that $f$ has the form $f(x) = f_0(x)\cdot (x-0)$. We have $f_0(1) = f(1) = 0$, thus $f_0(x) = f_1(x) \cdot (x-1)$. Therefore $f(x) = f_1(x)\cdot(x^2-x)$, i.e. $f \in \langle x^2-x\rangle$.


问:
Good approximations for $\int_{-\infty}^0 \bigg(\tanh(ax+b)\tanh(cx+d) - 1\bigg)~\mathrm{d}x$
I have an integral

$$I(a,b,c,d) = \int_{-\infty}^0 \bigg(\tanh(ax+b)\tanh(cx+d) - 1\bigg)~\mathrm{d}x$$

which I need to approximate analytically (to use it in further steps within a machine learning approach) with $a,c\in\mathbb{R}_+$ and $b,d\in\mathbb{R}$. What would be the best approach to obtain an anaytic expression approximating $I(a,b,c,d)$?

答:
For $a=c$, I get an answer

$$I(a,b,c,d) = -\frac{1}{a}(\coth(d - b) (d + \log(\cosh(d)) + \log(1 - \tanh(b))))$$

unless I screwed up? I used the definition of $\tanh(\cdot)$, rewrote it in terms of $p=\exp(ax)$ and did partial fraction decomposition.


问:
Prove that $\frac{b-a}{b}< \ln(\frac{b}{a})<\frac{b-a}{a}$
So I have hardships trying to prove that $\frac{b-a}{b}< \ln(\frac{b}{a})<\frac{b-a}{a}$, where $\, 0<a \left(\frac{b}{a}\right)^b $ and $\mathrm{e}^{b-a}> \left(\frac{b}{a}\right)^a $.

It might be wrong, but seems right to me. Anyways, I don't feel like I'm closer to the solution knowing these. Could you help me out, please?

答:
We are given $0<a<b$. Hence, $\frac{a}{b-a} < \frac{b}{b-a}$.

Now suppose, $f(x) = \ln{x}$. Suppose also that $c$ lies between $\frac{a}{b-a}$ and $\frac{b}{b-a}$. Hence, if we now apply the Mean Value Theorem here, we will get $$f^\prime(c) = \frac{f(\frac{b}{b-a})-f(\frac{a}{b-a})}{\frac{b}{b-a}-\frac{a}{b-a}}$$ $$= \frac{\ln{(\frac{b}{b-a})-\ln(\frac{a}{b-a})}}{\frac{b}{b-a}-\frac{a}{b-a}}$$ Or $$f^\prime(c)=\ln{\frac{b}{a}}$$ Since $f(x) = \ln{x}\tag{1}\label{eq1}$ we have $f^\prime(x) = \frac{1}{x}$. Therefore, $$f^\prime(c)=\frac{1}{c}\tag{2}\label{eq2}$$ From (1) and (2), we get $$c=\frac{1}{\ln{\frac{b}{a}}}\tag{3}\label{eq3}$$ Since, $c$ lies between $\frac{a}{b-a}$ and $\frac{b}{b-a}$, we have $$\frac{a}{b-a} < c < \frac{b}{b-a}$$ Or $$\frac{b-a}{b}<\frac{1}{c}<\frac{b-a}{a}$$ Or, from equation (3) $$\frac{b-a}{b}<\ln{\frac{b}{a}}<\frac{b-a}{a}$$


问:
How can we make it easy to solve quadratics without the Quadratic Formula?
For example ,

$$5x^2 - 7x-2=0$$

Then, we only need to find is when
$$5x^2-7x=2$$
$$x(5x-7)=2$$
Since that when we get $2-2=0$.

Is there any way we can find them easily without solving quadratics using formula?

I am just getting very enthusiastic in this topic.

答:
Unfortunately, your strategy does not help that much because it is very difficult to solve
$$
x(5x-7)=2
$$
without simply returning to the original equation. This is because there is a $2$ on the RHS. The reason we want to be able to write a quadratic in the form
$$
(x-p)(x-q)=0
$$
is so that we can exploit the zero-product property: if $a \times b=0$, then $a=0$ or $b=0$. Once we have written an equation in the form $(x-p)(x-q)=0$, we can still instantly deduce that $x=p$ or $x=q$. This is something we can't do when we have something like
$$
x(5x-7)=2 \, .
$$

Instead, I would suggest dividing through by $5$, which yields
$$
x^2 - \frac{7}{5}x - \frac{2}{5} = 0 \, .
$$
Then, add $2/5$ to both sides:
$$
x^2 - \frac{7}{5}x = \frac{2}{5} \, \tag{*}\label{*} .
$$
The LHS is almost a perfect square. Note that
$$
\left(x-\frac{7}{10}\right)^2=x^2-\frac{7}{5}x+\frac{49}{100} \, .
$$
If we subtract $49/100$ from this equation, we get
$$
\left(x-\frac{7}{10}\right)^2-\frac{49}{100}=x^2-\frac{7}{5}x \, .
$$
Hence, the equation $\eqref{*}$ can be rewritten as
$$
\left(x-\frac{7}{10}\right)^2-\frac{49}{100}=\frac{2}{5} \, .
$$
Add $49/100$ to both sides:
$$
\left(x-\frac{7}{10}\right)^2 = \frac{2}{5}+\frac{49}{100} = \frac{89}{100} \, .
$$
Take the square root of both sides:
$$
x-\frac{7}{10} = \pm\sqrt{\frac{89}{100}} \, .
$$
Add $7/10$ to both sides:
$$
x = \frac{7}{10} \pm \sqrt{\frac{89}{100}} \, ,
$$
and we are done. This method of solving quadratics is known as completing the square. Unlike factorisation, it can be used to solve any quadratic. In fact, the quadratic formula comes from completing the square on the general quadratic equation
$$
ax^2 + bx + c = 0 \, .
$$
Please let me know if you have any questions.


问:
Is this an exact differential or not?
I have the 1-form
$$dz=2xy\, dx+(x^{2}+2y)\, dy$$
And I want to integrate it from $(x_{1},y_{1})$, to $(x_{2},y_{2})$.

If I'm not drunk, checking mixed partials, I find that $dz$ is an exact differential. BUT, when I want to calculate explicitly the integral

$$\int_{\sigma}dz$$
where $\sigma$ is

i) $(x_{1},y_{1})\to(x_{2},y_{1})\to(x_{2},y_{2})$

ii) $(x_{1},y_{1})\to(x_{1},y_{2})\to(x_{2},y_{2})$

I find that the integral has a different value if I take the trajectory ii) instead of i). Why this happens?

Calculating the integral for i)

$\int dz=\int_{x_{1}}^{x_{2}}2xy_{1}\, dx+\int_{y_{1}}^{y_{2}}(x_{2}^{2}+2y)\, dy=y_{1}(x_{2}^{2}-x_{1}^{2})+x_{2}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}$

Calculating the integral for ii)

$\int dz=\int_{y_{1}}^{y_{2}}(x_{1}^{2}+2y)\, dy+\int_{x_{1}}^{x_{2}}2xy_{2}\, dx=x_{1}^{2}(y_{2}-y_{1})+y_{2}^{2}-y_{1}^{2}+y_{2}(x_{2}^{2}-x_{1}^{2})$

答:
Actually $dz$ is an exact differential:
$$z = x^2y + y^2. $$
And integral does not depend on path.
I think you lost something when you integrate. I just can't understand how do you integrate on these path.</p>

OwnerDisplayName: Anuar

Title: Is this an exact differential or not?


问:
Why is $\mathbb{Z}[i\sqrt5]$ isomorphic to $\mathbb{Z}[X]/(X^2+5)$?

Why is $\mathbb{Z}[i\sqrt5]$ isomorphic to $\mathbb{Z}[X]/(X^2+5)$ ?

$\mathbb{Z}[i\sqrt5]$={$a+ib\sqrt5,a,b\in \mathbb{Z}$}

Also can $\mathbb{Z}[i\sqrt5]$ be seen as a ring of polynomial in $"i\sqrt5"$ or is it just a notation looking like the polynomial ring notation but with another meaning ?

答:
Define

$$\phi: \Bbb Z[x]\to\Bbb Z[\sqrt5\,i]\;,\;\;\;\phi(f(x)):=f(\sqrt5\,i)$$

It is not hard to show this is a ring homomorphism, and

$$\ker\phi=\{f\in\Bbb Z[x]\;|\;f(\sqrt5\,i)=0\;$$

All you need now is to prove that $\;f(\sqrt5\,i)=0\iff (x^2+5)\,\mid f(x)\;$ , which means $\;f(\sqrt5\,i)=0\iff f(x)=(x^2+5)g(x)\;$ , for some $\;g(x)\in\Bbb Z[x]\;$ .

The above also explains why in $\;\Bbb Z[\sqrt5\,i]=\Bbb Z[\sqrt{-5}]\;$ we get an expression with only linear polynomials in $\;\sqrt{-5}\;$ ....


问:
Elevator stops in building
Bill is working in the 38th floor of an 100-floors building. This building has a strange elevator which only has 2 buttons: the green one which takes you to the next floor every time you press it (and if you are at the last floor, it takes you to the 1st) and a red button which takes you to a random floor every time you press it. Once you enter the elevator, the first stop is selected at random, without pressing any button. Bill wants to eventually arrive at his office at the 38th floor. How is he going to do it, with the least number of steps?

I am thinking of a very simplistic approach: Since the first stop is decided at random, depending on the number of the floor where the elevator stops, Bill will decide which button to use: If he is near the 38th and before it, he will keep pressing the green button until he reaches the 38th. If it stops in a floor $>38$, he will press the red button until the elevator stops at some floor $<38$, but I assume the problem involves the expected number of stops etc probabilities, which I am not very familiar with!
Any help is really appreciated!

答:
Since @MiloBrandt gave us a perfect and complete answer, I'm gonna give a partial answer for a simpler problem that I hope someone will find interesting.

Since we are considering an elevator (and not a circular bus) I'm gonna consider the same problem under the hypothesys that we always start from the floor $0$ (the lowest floor of the palace) and instead of evaluating the lowest number of pushes I'm gonna put the focus on three different strategies and evaluate which is "the best" to gain some time but with the result to reach always the 38th floor.

As I said, we can distinguish 3 different strategies, of 3 different players.
The first one is an "old man" strategy: "keep pushing the green botton", you are gonna take 38 pushes but you are sure to arrive at destination.

The "kid" strategy: "keep pushing the red botton and hope". Eventually they are going to reach the 38th floor but with a probability of 1/100 of getting the exact floor at every push. Since this is a better strategy than the old man's one only if the number of pushes is lower that 38, we can conclude that "on average" only 37 times out of 100 they will use less pushes than the old man, so this is not the best strategy "on average".

Then we can identify a "middle" strategy: push the red button. If you get to a floor that is between the first floor and the 38th then push the green botton until you get to the 38th floor.

Even in this case, to beat the old man's strategy, we have to use less than 38 pushes.
At the firts push we can get a floor $n$ such that $2\le n \le38$. If this did not happened, then we have lost one chance. Than we need to get a floor such that $3\le n \le 38$ and so on..

On every step we need to reach at least a floor higher than the prewious.
If I did not made any error the probability this does not happens is: $63/100 \cdot 64/100 \cdot 65/100 \dots = 0.00029\dots$

In conclusion: the middle strategy is better than the old man strategy quite everytime and kids should grow up to understand that there exists a better strategy than theirs!

Title: Elevator stops in building


问:
A question on two polynomial ideals generated by the same polynomials have the same generator
Here is the statement I want to prove:

Let $K$ be a subfield of $F$, and $f,g\in{K[x]}$. Let $M_K=fK[x]+gK[x]$ and $M_F=fF[x]+gF[x]$. Show that $M_K$ and $M_F$ have the same monic generator.

My argument is the following: Suppose that $M_K=dK[x]$, then d is the unique and monic generator of $M_K$. Suppose $q\in{M_K}$, then there exists $q_1,q_2\in{K[x]}$ such that $q=fq_1+gq_2$ and there exists $q_3\in{K[x]}$ such that $q=dq_3$. So, $dq_3=fq_1+gq_2$. Also, since $q_1,q_2,q_3\in{K[x]}$, we have $q_1,q_2,q_3\in{F[x]}$. Thus, $fq_1+gq_2\in{M_F}$, i.e, $dq_3\in{M_F}$. I claim that since $q_3$ is an arbitrary polynomial in $F[x]$, it follows that $M_F=dF[x]$. Hence, d is also the unique and monic generator of $M_F$.

It seems logical but since I started with an assumption q belonging to $M_K$, I could not be sure. Is a converse implication needed, or is the idea completely wrong?

答:
Let $d$ be the monic greatest common divisor of $f$ and $g$. So, $M_K=dK[x]$. Since $d\in{M_K}$, there exists $q_1,q_2\in{K[x]}$ such that $d=fq_1+gq_2$.

Now let $r$ be an arbitrary polynomial over the field $F$. Consider $dr=(fq_1+gq_2)r$, so $dr=fq_1r+gq_2r$, where $q_1r$ and $q_2r$ are polynomials over $F$. By the definition of the ideal $M_F$, we have $dr\in{M_F}$. We showed that multiplication of any polynomial over $F$ with $d$ belongs to $M_F$. Hence, $M_F=dF[x]$.


问:
What's the most elementary way to solve this trigonometric problem?

A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.

This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.

Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).

Thanks in advance.

答:
Notice that by Pythagoras we have:
\begin{align}
y^2+1&= \left\vert BC\right\vert^2\\
(2+y)^2+1&= \left\vert AC\right\vert ^2
\end{align}
Taking roots and adding both equations yield:
$$\sqrt{y^2+1}+\sqrt{(2+y)^2+1}=\left\vert BC\right\vert+\left\vert AC\right\vert=4$$
Now we can solve this equation to get the value of $y$.


答:
I will give a simple solution using ellipse which is rather more easier.

Since the thread is tied to $A$ and $B$, we deduce that $A$ and $B$ are the focii of the ellipse. If we rotate the ellipse in the counter clockwise direction and make it horizontal, then we can use the equation for standard ellipse which is $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$ where $AD$ now becomes the $x$ axis, $DC$ the positive $y$ axis, with origin being the midpoint of $AB$. So difference between focii is given to be $2ae=2$ and length of string $=2a=4$ which implies $a=2$ and $e=\dfrac{1}{2}$. So, $b^2=a^2(1-e^2)=4(1-0.25)=3$. So, the equation of ellipse becomes, $$\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$$ Putting $y=1$, we get $x=\pm \sqrt{\dfrac{8}{3}}$. So, $BD=\sqrt{\dfrac{8}3}-1$.

Title: What's the most elementary way to solve this trigonometric problem?

Body:
A bead is threaded onto a light, inextensible string of length $4m$. One end of the string is fixed to a point, $A$, on a (vertical) wall. The other end of the string is attached to a point $B$ on the wall exactly $2m$ directly below $A.$ The bead is held in place so that it is at a distance of $1m$ from the wall, such that the string is taut and the plane the string is in is perpendicular to the plane the wall is in. Find the two possible vertical components of the displacement from $B$ to the bead.

This is a question I came up with, and I thought there should be some relatively simple trigonometric methods to get to the answer. I tried lots of stuff but nothing seemed to work.

Of course, we could find the equation of the ellipse that corresponds to the locus of points the bead could be when the string is taut, and then find the two values of $y$ when $x = 1$. But I'm looking for more elementary methods involving only trigonometry and Pythagoras. This is because I want the answer to be aimed at secondary school students who know elementary trigonometry only (Pythagoras, addition angle formulae, R addition formulae etc).

Thanks in advance.


问:
Weak solution operator of Elliptic PDE is lower bounded or not?
I am interested in the operator $T: v\mapsto u$ where $u$ is the weak solution of the following well-studied Elliptic PDE
\begin{align}
-\text{div} (A\nabla u) &=v \text{ in } \Omega,\\
u & =0 \text{ on } \partial \Omega.
\end{align}
Here, for each $x\in \Omega$, $A(x)$ is a symmetric matrix such that, exists $c>0$, $y^TA(x)y>c||y||^2_2$ for all vector $y$.

It is well known that the operator $T$ is upper bounded (continuous). I am wondering if it is lower bounded or not, i.e. does there exist a constant $\gamma>0$ such that
\begin{align}
||T v||\geq \gamma ||v|| \text{ for all } v.
\end{align}

I know that an operator is lower bounded iff it is injective and has closed range (see Theorem 2.5 here). However, since I am not familiar with PDE, I don't know where to begin. Could any one give me some clues to prove it? Thank you!

答:
No, this operator is not lower bounded in your sense. If it were, we would have
$$
\gamma \int_\Omega |\nabla u|^2\leq c\gamma\int_\Omega A\nabla u\cdot\nabla u=c\gamma\int_\Omega vu\leq c\gamma\|u\|_2\|v\|_2\leq c\|u\|_2^2
$$
for $u\in C_c^\infty(\Omega)$. To see that this is not true, take any $u\in L^2(\Omega)\setminus H^1(\Omega)$ and a sequence $(u_n)$ in $C_c^\infty(\Omega)$ such that $u_n\to u$ in $L^2$.


问:
$U(f,K,\epsilon)=\{g:g\in L$ and $|f(x)-g(x)|<\epsilon$ for all $x\in K\}$, is Let $C=\{U(f,K,\epsilon)\}$. Is $T=\{\cup B, B\in C\}$ a topology.
Let $L$ be the set of all continuous real-valued functions defined on $\mathbb{R}$. For every compact subset $K$ of $\mathbb{R}$, $f\in L$, and positive number $\epsilon >0$ let

$U(f,K,\epsilon)=\{g:g\in L$ and $|f(x)-g(x)|<\epsilon$ for all $x\in K\}$

Let $C=\{U(f,K,\epsilon): K$ is compact subset of $\mathbb{R}$, $f\in L, \epsilon >0\}$. Further, let $T=\{\cup B, B \text{ is a subcollection of } C\}$. Is $T$ a topology for $L$?

What I tried:

I think it's a topology.

$\emptyset\in T$, since $\emptyset$ is the union of empty subcollection of $L$.

$L\in T$, since $f(x)=0$ is continuous, and every continuous function on a compact set is bounded, so for every continuous function $g$, exists $\epsilon >0$, s.t. $|g(x)-0|<\epsilon$ on a compact set.

Arbitrary union in $T$: since by defniton of $T$, $\cup\cup B=\cup B\in T$.

But I don't know how to show that $\cap^n_1(\cup B)\in T$.

（I don't quite sure whether $f$ in $C$ is fixed or not, if $f$ is fixed, then this question seem to be easy.)

答:
We assume that in the definition of the family $C$, $ f$  is not fixed. We claim that $C$ is a basis of a topology on $L$. To show this, for each $f\in L$ and each elements $U(f_1,K_1,\epsilon_1),\dots, U(f_n,K_n,\epsilon_n)$ containing $f$ there exists a compact subset $K$ of $\Bbb R$ and $\delta>0$ such that $U(f,K, delta)\subset U(f_i,K_i,\epsilon_i)$ for each $1\le i\le n$.
For each such $i$, a function $g_i(x)=|f_i(x)-f(x)|$ is a continuous function on a compact set $K_i$, so it attains it maximum $m_i$ at some point $x_i$ of $K_i$. Since $g_i(x_i)<\epsilon_i$, there exists $\delta_i>0$ such that $m_i+\delta_i<\epsilon_i$. Put $K=K_1\cup\dots K_n$ and $\delta=\min\{\delta_1,\dots, \delta_n\}$. Let $g\in (f,K, \delta)$ be any function. Pick any  $1\le i\le n$ and any $x\in K_i$. Then
$$|g(x)-f_i(x)|\le |g(x)-f(x)|+|f(x)-f_i(x)|<\delta+m_i<\delta_i+m_i<\epsilon_i,$$
so $g\in U(f_i,K_i,\epsilon_i)$.


问:
Show that there do not exist any distinct natural numbers a,b,c,d such that

Show that there do not exist any distinct natural numbers a,b,c,d such that $a^3+b^3=c^3+d^3$ and $a+b=c+d$.

答:
Mark $a+b=x$ then $b=x-a$ and $d=x-c$. Notice that $x\ne 0$. No we have: $$a^3+(x-a)^3 = c^3+(x-c)^3$$ and thus $$-3x^2a+3xa^2 = -3x^2c+3xc^2$$

so \begin{align}xa-a^2= xc-c^2& \implies x(a-c) = (a-c)(a+c)\\ &\implies x=a+c \\& \implies a+c=a+b \\&\implies c=b\end{align}

A contradiction.


问:
Maclaurin series for $(\sin(x^3))^{1/3} $
I need to find Maclaurin series for $(\sin(x^3))^{1/3} $

My attempt:

I found first derivative. $f '(x) = \frac{(x^2 \cdot  \cos x^3)}{(\sin x^3)^{2/3}}$. Hence, $f'(0)$ does not exists and there is no Maclaurin series for the function above. Nevertheless, this is incorrect answer.

By the way, I saw the same question on this forum, but the answer doesn't solve my problem

答:
In order to find the Maclaurin series of $(\sin(x^3))^{1/3}$ at $x=0$ up to to $x^{13}$, it is better not to compute it by using the definition.
Instead, I suggest to start with the series of $\sin(x^3)$,

$$\sin(x^3)=x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}+O(x^{21}).$$
It follows that
$$(\sin(x^3))^{1/3}=\left(x^3-\frac{x^9}{3!}+\frac{x^{15}}{5!}+O(x^{21})\right)^{1/3}=
x\left(1-\frac{x^6}{3!}+\frac{x^{12}}{5!}+O(x^{18})\right)^{1/3}.$$
Now use the expansion $(1+t)^{1/3}=1+\frac{t}{3}-\frac{t^2}{9}+O(t^3)$ and finish the job.

P.S. It turns out that the first derivative at $0$ exists and
$$f'(0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{(\sin(x^3))^{1/3}}{x}=1.$$


问:
Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$
Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$
The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$
In this case the remainder should be $$\operatorname {f}(1) = 1^3+1 = 2$$
so the remainder should be 2 for any value of $x$

But for $x=3$
$$f(3) = 3^3+1$$
when this is divided by $x-1$ which is $3-1$ which equals to $2$, $$\frac {3^3+1}{3-1}$$ $$= \frac {27+1}{2}$$ $$= \frac {28}{2}$$ $$= 14, remainder = 0 ≠ 2$$
Please explain me why this happened

答:
Evaluating the polynomial remainder theorem at $\,x =\,$ some number does yield a valid congruence but the result may not be the canonical remainder because leastness need not be preserved. Let's see how the proof actually $\rm\color{#c00}{breaks \ down}$ after evaluation at $\,x = b$.

$\begin{align}
f(x) = (x\!-\!a)q(x) + f(a) &\Rightarrow f(x)\equiv  f(a)\!\!\!\pmod{\!x\!-\!a} \Rightarrow f(x)\,\bmod x\!-\!a = f(a)\\[.3em]
{\rm so}\ \,f(b) =\, (b\!-\!a)\,q(b) + f(a) &\Rightarrow f(b)\equiv  f(a)\!\!\pmod{\!b\!-\!a} \,\color{#c00}{\not\Rightarrow}f(b)\ \,\bmod b\!-\!a = f(a)\\[.3em]
{\rm e.g.}\,\
x^3\!+\!1 = (x\!-\!1)q(x) + 2 &\Rightarrow x^3\!+\!1\,\equiv\,  2\!\pmod{\!x\!-\!1} \Rightarrow x^3\!+\!1\bmod x\!-\!1 = 2\\[.3em]
{\rm so}\ \ \ \ \ 3^3\!+\!1 =\:\! (3\!-\!1)q(3) + 2 &\Rightarrow\, 3^3\!+\!1\,\equiv  \,2\!\pmod{\!3\!-\!1} \color{#c00}{\not\Rightarrow} 3^3\!+\!1\bmod\, 3\!-\!1 = 2\\
\end{align}$

Obviously we can remedy this by doing a $\rm\color{#0a0}{final}$ modular reduction to ensure we obtain the least number congruent to $\,f(b),\,$ i.e. the remainder, e.g. above  $\,3^3\!+\!1\bmod 2 = 2\color{#0a0}{\bmod 2} = 0,\,$ and $\,f(a)\bmod b\!-\!a = f(b)\color{#0a0}{\bmod b\!-\!a}.\,$ This is simply the remainder form of the congruence  $\,f(a)\equiv f(b)\pmod{ b\!-\!a}.\,$  The inference is done more naturally in congruence language:
$\!\bmod b\!-\!a\!:\,\ a\equiv b\Rightarrow f(a)\equiv f(b)\,$ for any polynomial $f(x)$ with integer coef's, by the Polynomial Congruence Rule.


问:
Transport equation solution clarification and alternative
Consider the following problem:

Problem 1. Write down an explicit formula for a function $u$ solving the initial-value problem
$$\left\{\begin{array}{rcl}
u_t+b\cdot Du+cu=0 & \text{on} & \mathbb{R}^n\times(0,\infty),\\
u= g & \text{on} & \mathbb{R}^n\times\{t=0\}.
\end{array}\right.$$
Here $c\in\mathbb{R}$ and $b\in\mathbb{R}^n$ are constants.

Sol: Fix $x$ and $t$, and consider $z(s):=u(x+bs,t+s)$.
Then
\begin{align*}
\dot z & =b\cdot Du+u_t\\
& =-cu(x+bs,t+s)\\
& =-cz(s)
\end{align*}
Therefore, $z(s)=De^{-cs}$, for some constant $D$. We can solve for $D$ by letting $s=-t$. Then,
\begin{align*}
z(-t) & =u(x-bt,0)\\
& =g(x-bt)\\
& =De^{ct}
\end{align*}
i.e. $D=g(x-bt)e^{-ct}$. Thus, $u(x+bs,t+s)=g(x-bt)e^{-c(t+s)}$ and so when $x=0$, we get
$$u(x,t)=g(x-bt)e^{-ct}.$$

In this solution to an exercise in Evans, Why is $z$ chosen as such?

Also, what would the solution look like using Fourier transform? Is there a nice way to do it?

答:
The curves of the form $\gamma(s)=(x+bs,t+s)$ are the characteristic curves of this equation. You can find the general theory of characteristic curves for first order PDEs on the Evans, or on other books on PDEs.

$z$ is $u$ when valued on one of this curves. The motivation for defining the characteristic curves this way (in this case) is that the PDE
$$\frac{\partial u}{\partial t}+b\cdot\nabla u+ cu =0$$
is equivalent to say that the directional derivative of $u$ in direction of the $n+1$-dimensional vector $(b,1)$ is equal to $-cu$.
So for $u$ being a solution of the PDE is equivalent that all its restrictions on the flow lines of this vector field solve a certain ODE. The flow lines of this vector field are the characteristic curves.

You can do the same for every linear first order PDEs.


问:
Find eigenvalues without having the matrix.
Let $\{v_1,v_2,v_3,v_4\} $ be a vector basis of $\mathbb{R}^4$ and $A$ a constant matrix of $\mathbb{R}^{4\times 4}$ so that:
$$A v_1=-2v_1,\hspace{0.5cm} Av_2=-v_1,\hspace{0.5cm} Av_3=3v_4,\hspace{0.5cm}Av_4=-3v_3$$
Can I find the eigenvalues of the matrix A?
I know that $\lambda_1=-2$ is a trivial eigenvalue but I don't know how to calculate the others.

答:
You actually know the representation of your matrix related to your basis. How do you represent linear application as matrix?
Then you can calculate your characteristic polynomial and work from there if you want a standard way to solve this.</p>


问:
Is there a faster way to calculate a few diagonal elements of the inverse of a huge symmetric positive definite matrix?
I asked this on SO first, but decided to move the math part of my question here.

Consider a $p \times p$ symmetric and positive definite matrix $\bf A$ (p=70000, i.e. $\bf A$ is roughly 40 GB using 8-byte doubles). We want to calculate the first three diagonal elements of the inverse matrix $({\bf A^{-1}})_{11}$, $({\bf A^{-1}})_{22}$ and $({\bf A^{-1}})_{33}$.

I have found this paper by James R. Bunch who seems to solve this exact problem without calculating the full inverse $\bf A^{-1}$. If I understand it correctly he first calculates the Cholesky decomposition, i.e. the upper triangular matrix $\bf R$ which satisfies $\bf A=R^T R$, which needs $\frac16p^2+\frac12p^2-\frac23p$   floating point operations (multiplications/divisions) using the LINPACK function SPOFA. He then proceeds to calculate individual diagonal elements of the inverse $({\bf A^{-1}})_{ii}$ using an expression which exploits the sparsity of ${\bf R}^T{\bf y}={\bf e}_j$   and which requires $\frac12(p-i)^2+\frac52(p-i)+2$   floating point operations. (I don't understand the full details of this, so I can't currently sum it up correctly).

The paper is based on LINPACK; it isn't cited by anyone, so it seems nobody cared for the last 23 years? After reading this, I'm wondering whether this is still the best way of doing things, or whether a modern LAPACK-based approach could avoid the Cholesky decomposition?

In short, is there a quicker way to calculate those diagonal elements of the inverse?

答:
To get the first column of $A^{-1}$, solve $A \, X_1 \, = \, e_1 \, = \, [1 0 0 ...]$;

to get the first two, solve $\qquad \qquad A \, [X_1 X_2] \, = \, [e_1 e_2]$;

and you have $X_{11} \, X_{22} ...\ $
Or am I missing something ?

(There's a wide range of solvers for various sparse matrices, direct and iterative;
they vary a lot in robustness, number of papers, etc.
For benchmarking, sparsesolvers.py
on gist.github
is a uniform wrapper around a dozen or so solvers in scipy.)


答:
Your mistake is in believing that the derivative does not exist, because you tried to obtain it by symbolic differentiation. But back to the true definition,

$$f'(0)=\lim_{x\to0}\frac{(\sin(x^3))^{1/3}}{x}=\lim_{t\to0}\frac{(\sin(t))^{1/3}}{t^{1/3}}=\left(\lim_{t\to0}\frac{\sin(t)}t\right)^{1/3}=1$$ (by continuity of the cube and cubic root functions.)

Note that this value coincides with the limit of the symbolic expression,

$$\lim_{x\to0}\frac{x^2\cos(x^3)}{\sin(x^3)^{2/3}}=\lim_{t\to0}\frac{t^{2/3}}{\sin(t)^{2/3}}=1$$ though this is not a proof.

Body:
I need to find Maclaurin series for $(\sin(x^3))^{1/3} $

My attempt:

I found first derivative. $f '(x) = \frac{(x^2 \cdot  \cos x^3)}{(\sin x^3)^{2/3}}$. Hence, $f'(0)$ does not exists and there is no Maclaurin series for the function above. Nevertheless, this is incorrect answer.

By the way, I saw the same question on this forum, but the answer doesn't solve my problem