问:
find a matrix $A_{n \times n}$ which satisfy $A^{n}=0$ and $A^{n-1}\ne0$
I asking a question for my brother. He just started his algebra course not long ago and he got a question he is stuck on (he just started the course so all his knoweldge is based on matrices and their properties).

I need to find a Matrix $A_{2x2} \not=0$ which satisfies $A^2=0$.

Also, I need to find a Matrix $A_{3x3}$ which satisfies $A^3=0$ and $A^2\not=0$.

In the end I need to generalize the problem to $A_{n \times n}$ (find a matrix $A_{n \times n}$ which satisfy $A^{n}=0$ and $A^{n-1}\ne0$).

In the first question I wrote $A$ as

\begin{bmatrix} a & b \\ c & d \end{bmatrix}

after multiplying $A$ with itself I found out that every matrix $2x2$ in the form of

\begin{bmatrix} x & y \\ -\frac{x^2}{y} & -x \end{bmatrix}

when multiplied with itself equal zero.

In the second question I tried the same thing but it got to long to soon so I figured I am doing something wrong. also, the the equations I get when I am trying to find conditions for $A$ are to messy to deal with (9 in total).

I did manage to find some matrices that satisfies $A^3=0$ such as

\begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

but I still wish to know how should I find the conditions for it to happen.

About the third question, I guess I need to finish the second one to even begin thinking about the solution but I still can't see where should I start.

Body:
Try this:
$$A = \begin{bmatrix}
\color{red}0 & 1 & 1 & ... & 1 \\
0 & \color{red}0 & 1 & ... & 1 \\
... & ... & ... & ... & ...\\
0 & 0 & 0 & ... & 1 \\
0 & 0 & 0 & ... & \color{red}0 \end{bmatrix} $$

Just to feel what will happen:
$$A^2 = \begin{bmatrix}
0 & 1 & 1 & ... & 1 \\
0 & 0 & 1 & ... & 1 \\
... & ... & ... & ... & ...\\
0 & 0 & 0 & ... & 1 \\
0 & 0 & 0 & ... & 0 \end{bmatrix}\begin{bmatrix}
0 & 1 & 1 & ... & 1 \\
0 & 0 & 1 & ... & 1 \\
... & ... & ... & ... & ...\\
0 & 0 & 0 & ... & 1 \\
0 & 0 & 0 & ... & 0 \end{bmatrix} $$$$=
\begin{bmatrix}
\color{red}0 & \color{blue}0 & 1 & 2&  ... & n-1 \\
0 & \color{red}0 & \color{blue}0 & 1& ... & n-2 \\
... & ... & ...&... & ... & ...\\
0 & 0 & 0 & 0& ... & \color{blue}0 \\
0 & 0 & 0 & 0&... & \color{red}0 \end{bmatrix}$$
Each time you raise to another power, the upper diagonals will start to vanish consequently (actually, you can find an explicit formula for the $n^\text{th}$ power quite easily). Then, notice that $A^{n-1} \ne O$ but $A^n = O$.


问:
Find the cuvrvature and torsion of a space curve
$r(t)=t \cos(t) \vec e_1 + t \sin(t) \vec e_2 + (c-dt) \vec e_3$ is a space curve

Fint the torsion $\tau$ and the curvature $\kappa$

Attempt:

I know the formulas but it gets too complicated with standard formulas, can I directly use $\kappa = |r''(t)|$ or should I make it arc-length parameter first?

答:
$$\kappa =\frac{\sqrt{\left(x' y''-x'' y'\right)^2+\left(x'' z'-x' z''\right)^2+\left(y' z''-y'' z'\right)^2}}{\left(\left(x'\right)^2+\left(y'\right)^2+\left(z'\right)^2\right)^{3/2}}$$
If $\mathbf{r}(t)=(t\cos t;t\sin t;c-dt)$ then the curvature is

$$\kappa=\frac{\sqrt{d^2 \left(t^2+4\right)+\left(t^2+2\right)^2}}{\left(d^2+t^2+1\right)^{3/2}}$$

$\mathbf{r}'(t)=(\cos (t)-t \sin (t),\sin (t)+t \cos (t),-d)$

$\mathbf{r}''(t)=(-2 \sin (t)-t \cos (t),2 \cos (t)-t \sin (t),0)$

$\mathbf{r}'''(t)=(t \sin (t)-3 \cos (t),-3 \sin (t)-t \cos (t))$

The torsion is

$$\tau ={\frac {\det \left({\mathbf {r} ',\mathbf {r} '',\mathbf {r} '''}\right)}{\left\|{\mathbf {r} '\times \mathbf {r} ''}\right\|^{2}}}={\frac {\left({\mathbf {r} '\times \mathbf {r} ''}\right)\cdot \mathbf {r} '''}{\left\|{\mathbf {r} '\times \mathbf {r} ''}\right\|^{2}}}$$

$$\tau=-\frac{d \left(t^2+6\right)}{d^2 \left(t^2+4\right)+\left(t^2+2\right)^2}$$

ContentLicense: CC BY-SA 4.0


问:
Prove that every ideal of a regular ring is regular
Prove that every ideal of a regular ring is regular.

Let $R$ be a regular ring with unity, that is to say that each element $x \in R$ can be written as $x=xyx$ for all $y \in R$. And let $K$ be an ideal of $R$; we must see that $K$ is also regular.

Note: an ideal $K$ is a nonempty set in $R$, such that $x-y \in K$, y for all $x,y \in K$ also $xa \in K$ and $ax \in K$ for all $a \in R$

答:
There is an error in your characterization of regular rings. You write that $R$ is regular if each $x\in R$ can be written as $xyx$ "for all $y\in R$", but  of course you mean to say "for some $y\in R$". Incidentally, this is what is called a "Von Neumann regular" ring; there is a different meaning of "regular ring" in commutative algebra.

With this correction in mind, let $x\in K$. To show $K$ is Von Neumann regular, we need to find some $z\in K$ such that $x=xzx$.

Because $R$ is regular, there is some $y\in R$ such that $x=xyx$. This means that in fact $x=(xyx)yx=x(yxy)x$. Because $K$ is a two-sided ideal, and $x\in K$, $yxy\in K$, so letting $z=yxy$ gives the desired result.


问:
Simplify convoluted variables for partial solution?
Suppose one has

$ax' + by' = cx + dy,$ and I want to solve this in terms of an unknown relationship between $x$ and $y$ that will form a solvable differential equation.

As an example, if one had $x' + y' = x + y,$ one could suggest that if $u = x + y,$ then you'd arrive at
$u' = u$ and $u(t) = u_0e^t.$

Similarly, suppose $ax + by = u(t).$

I would then like to extrapolate the technique from that simpler case to this more complex case to achieve a similar result which I suspect is an exponential or pair of exponentials, possibly derived by assuming a change of coordinates related to $\sinh$ and $\cosh$.

How could I rewrite the right-hand sign in terms of $u$?

And if there is an answer to this, is there a generalization that makes sense in terms of Lie symmetries?

答:
Write your equation

$$ax' + by' = cx + dy, \tag{1}$$

under the form:

$$\forall t \in D, \ \ \ ax'(t)-cx(t)=-by'(t)+dy(t)$$

Therefore, the LHS and the RHS define a common function $f(t)$.

As a consequence, (1) equivalent to the system of 2 first order differential equations:

$$\begin{cases} \ \ \ x'(t)&-&cx(t)&=&f(t)\\
-by'(t)&+&dy(t)&=&f(t)\end{cases}\tag{2}$$

where $f$ can be considered in fact as an (almost) arbitrary function.

Of course, one could go a little further. Let us consider for example the case of the first equation in (2). Multiplying LHS and RHS by the integrating factor $a \exp(-\frac{c}{a}t)$, we can trasform it into

$$[ax(t) \exp(-\frac{c}{a}t)]'=\exp(-\frac{c}{a}t)f(t)$$

Integrating, we get:

$$x(t)=\frac{1}{a}\exp(\frac{c}{a}t)\int_0^T\exp(-\frac{c}{a}T)f(T)dT+C \tag{3}$$

with constant $C=x(0)$ (initial condition).

Of course, we have a similar expression for $y(t)$.

Conclusion: (2) (or even (3)) conveys the large degree of freedom we have for "solving" (1).

Edit: (3) can be written under a convolutive form:

$$x(t)=\frac{1}{a}\int_0^Te^{\alpha(t-T)}f(T)dT+C \ \ \tag{3}$$

As $y(t)$ can be expressed under the same form, we can conclude that the most general solution of (1) is:

$$\begin{cases}x(t)&=&A e^{\alpha t}\star f(t)+C\\y(t)&=&Be^{\beta t}\star f(t)+D\end{cases}$$

with fixed constants $A=\dfrac{1}{a}, B, \alpha, \beta$ (depending upon $a,b,c,d$) and arbitrary function $f$ and constants $C,D$.

Remark: we could as well have worked with Laplace Transform on (2)...


问:
Concept of Absolute value
I need to resolve this misconception quickly.
The question is very simple,
may I derive $|\sqrt y| = x-1$ from $y = (x-1)^2$, such that $x = |\sqrt y|+1$?

答:
From $y=(x-1)^2$ you get
$$\sqrt y=|x-1|\to x= 1\pm\sqrt y$$

Title: Concept of Absolute value


问:
How many options are there for 15 student to divide into 3 equal sized groups?
How many options are there for 15 students to divide into 3 equal-sized groups?

Now I know the solution is $\;\dfrac {15!}{5!5!5!3!}\;$ but I can't understand why.

Can anyone please enlighten me?

答:
There are $\,\binom{15}{5}\,$ ways to choose five students for group ONE,

Now, there are $\binom{10}{5}$ ways to choose five students for group TWO,

That leaves us with only five students: i.e., hence there is only $\,\binom 55 = 1\,$ way to choose the students for group THREE.

Multiplying these factors: $$\binom{15}{5} \times \binom{10}{5} \times \binom 55= \frac{15!}{10!\,5!}\times \frac{10!}{5!\,5!}\times 1 = \frac{15!}{5!\,5!\,5!}\tag{1}$$

Now we need to divide that total by $\,3!,\,$ since the labeling/ordering of the groups (group ONE, group TWO, group THREE) doesn't matter: since there are $3!$ ways of labeling the groups, we need divide the total given in $(1)$ by $\,3!$.

Hence, our final answer is:

$$\text{There are}\;\left(\frac{15!}{5!\,5!\,5!\,3!}\right)\;\text{ways to divide 15 students into three groups of equal size.}$$

ContentLicense: CC BY-SA 3.0


问:
Show that $F(s) = \log \zeta(s) + G(s)$ holds for $\text{Re }s>1$ and $F(s)$ does not have a meromorphic continuation for $\text{Re }s<1$.
Question

Let $F(s) = \sum_{p^k} p^{-ks}$, where the summation runs over all prime powers. Show that
$F(s) = \log \zeta(s) + G(s)$ holds for $\text{Re }s>1$, where $G(s)$ is analytic in the half-plane $\text{Re }s > 1/2$. Deduce from this that the function $F(s)$ does not have a meromorphic continuation for $\text{Re }s<1$.

My Attempt

I know that $$F(s) = \sum_{p} \sum_{k=1}^{\infty} \frac{1}{p^{ks}} = \sum_{p} \frac{p^{-s}}{1-p^{-s}}\, ,$$ and
$$\log \zeta(s) = \log \left(\prod_{p} \left(1-\frac{1}{p^s}\right)^{-1}\right) = \sum_{p} \log \left(1-\frac{1}{p^s}\right)^{-1} = \sum_{p} \sum_{k=1}^{\infty} \frac{1}{k\, p^{ks}}.$$

But how to proceed naturally, instead of reverse engineering as I am trying to do here?

答: $$G(s)= \sum_p \sum_{k\ge 2} p^{-sk} (1-1/k)$$ converges absolutely (bounded by $2\zeta(2\Re(s))$) thus is analytic for $\Re(s) > 1/2$ and $$F(s) = \log \zeta(s)+G(s)= -\log (s-1)+ \log ((s-1)\zeta(s))+G(s)$$ has a branch point at $s=1$


问:
Need help with logical expression involving existential and universal quantifier
P is the set of "all exams at university". E(x) denotes the predicate "x is easy". Translate the following sentence to formal logic. "Only one exam is easy." Solution:

I understand the first part, that there exists a test "x" for which the predicate "is easy" holds. But I don't understand the second part. I read it as such that for all y,z element of the set of exams, if y is easy and z is easy and there exist an exam x that is easy it follows that y = z.

答:
The first part says: "some exam is easy".

The second part says: "if there are two easy exams, they are equal".

See Uniqueness quantification.


问:
For the integral operator $(Af)(t) = \int_0^1K(t,s)f(s)ds$ in $L_2[0,1]$, prove that $\sum_{n,m=1}^{\infty} |a_{nm}|^2 = ||K||^2$
Let $K$ be a continuous function in $L_2[0,1]^2$. Let $A$ be the integral operator in $L_2[0,1]$:
$$(Af)(t) = \int_0^1K(t,s)f(s)ds$$

Since $L_2[0,1]$ is a separable hilbert space, then $A$ can be represented as a matrix $A = (a_{nm})$ in an orthonormal basis $\{e_n\}_{n=1}^{\infty}$.

I want to show that $\sum_{n,m=1}^{\infty} |a_{nm}|^2 = ||K||^2$, and it is allowed to assume that $\{e_n\}_{n=1}^{\infty}$ are continuous in order to avoid using Lebesgue integration.

However, I am having to understand how exactly I use the matrix representation with functions, so I'm not sure what to do.

Help would be appreciated.

答:
If $a_{nm}=(A(e_m),e_n)$, then

$$\sum_{n,m=1}^{\infty} |a_{nm}|^2=\sum_{n,m=1}^{\infty} |(A(e_m),e_n)|^2=\sum_{n,m=1}^{\infty} \left| \int_{0}^1 A(e_m)(t)\overline{e_n(t)}dt\right|^2=\sum_{n,m=1}^{\infty} \left|\int_{0}^1\int_{0}^1 K(t,s)e_m(s)\overline{e_n(t)}ds dt\right|^2$$

Now, using that $e_{nm}(t,s):=\overline{e_n(t)}e_m(s)$ is an orthonormal basis for $L^2([0,1]^2)$ we should have the last expression be equal to $\|K\|^2$, by parseval's identity. Here I am understanding that $\|K\|$ is the norm that comes from $L^2([0,1]^2)$.


答:
The appropriate translation "Only one exam in the university is easy" is

$$\exists x \in P\Big(E(x) \land \forall y\in P(E(y) \to (x=y))\Big).$$

The translation you posted says there does exist an easy exam, and for all exams, if they are both easy,  they are equal.

Title: Need help with logical expression involving existential and universal quantifier

Body:
P is the set of "all exams at university". E(x) denotes the predicate "x is easy". Translate the following sentence to formal logic. "Only one exam is easy." Solution:

I understand the first part, that there exists a test "x" for which the predicate "is easy" holds. But I don't understand the second part. I read it as such that for all y,z element of the set of exams, if y is easy and z is easy and there exist an exam x that is easy it follows that y = z.


答:
If you think about the first part in linear transformation terms, you want a map $A: \Bbb{R}^2 \to \Bbb{R}^2$ for which $A(A \vec{v}) = \vec{0}$ for any vector $\vec{v}$.

One option for such a transformation would be $$\langle x, y \rangle \mapsto \langle 0, x \rangle.$$ What would be the matrix of this transformation?

Body:
I asking a question for my brother. He just started his algebra course not long ago and he got a question he is stuck on (he just started the course so all his knoweldge is based on matrices and their properties).

I need to find a Matrix $A_{2x2} \not=0$ which satisfies $A^2=0$.

Also, I need to find a Matrix $A_{3x3}$ which satisfies $A^3=0$ and $A^2\not=0$.

In the end I need to generalize the problem to $A_{n \times n}$ (find a matrix $A_{n \times n}$ which satisfy $A^{n}=0$ and $A^{n-1}\ne0$).

In the first question I wrote $A$ as

\begin{bmatrix} a & b \\ c & d \end{bmatrix}

after multiplying $A$ with itself I found out that every matrix $2x2$ in the form of

\begin{bmatrix} x & y \\ -\frac{x^2}{y} & -x \end{bmatrix}

when multiplied with itself equal zero.

In the second question I tried the same thing but it got to long to soon so I figured I am doing something wrong. also, the the equations I get when I am trying to find conditions for $A$ are to messy to deal with (9 in total).

I did manage to find some matrices that satisfies $A^3=0$ such as

\begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}

but I still wish to know how should I find the conditions for it to happen.

About the third question, I guess I need to finish the second one to even begin thinking about the solution but I still can't see where should I start.


问:
Criteria for being a true martingale
Could you kindly list here all the criteria you know which guarantee that a continuous local martingale is in fact a true martingale? Which of these are valid for a general local martingale (non necessarily continuous)? Possible references to the listed results would be appreciated.

答:
Here is the other answer I was mentionning in my preceding post :

Tenth :

Given $X_t$ a continuous Itô diffusion starting at $x$ and taking values in $[l,\infty)$ with $l\in \mathbb{R}$. $X_t$ is moreover supposed to satisfy the following SDE :

$X_t=x+\int_{0}^{t}a(X_s)dB_s$

with :

- $B$ a standard Brownian Motion

- $a(x)^2>0$ for all $x\in (l,\infty)$

- $a^{-2}$ is locally integrable

This entails that $X$ has a weak and unique solution and that it is a local martingale (with absorbing boundary $l$).

Then the Second order equation $$\frac{1}{2}a^2(x)u''(x)=\alpha.u(x)$$ has two positve (linearly independent) unique solutions $\phi_{\alpha} $ and $\psi_{\alpha}$ up to a multiplicative constant ($\alpha$ is a fixed strictly positive constant).

Those solutions are respectively decreasing and increasing with boundary conditions in accordance with the boundary behaviour of $X$.

Under those assumptions $X$ is a martingale if and onyl if :
$$lim_{z\to +\infty}\psi_{\alpha}'(z)=+\infty$$

This is from Hulley and Palten "A Visual Criterion for Identifying Itô diffusions as Martingales or Strict Local Maritngale".

NB : The Criteria from George Lowther's blog is also given in the paper (with two references) and the connections are explicitly described.

Eleventh :

Another criteria with different tools and set of hypothesys (I would say that it is a little more specialised that the preceeding criteria) can be found in the paper by Mijatovic and Urusov "On the Martingale Property of Certain Local Martingale". I do not reproduce it here because it would be too long.

Regards

Title: Criteria for being a true martingale


问:
What is the math behind the "IQ Blox" puzzle?
One of my younger relatives showed me this curious puzzle (called "IQ Blox") while I was staying at their place. The premise is as follows: there's a leaflet included with the puzzle listing over a hundred "guides", which are initial positions of the white border bits ("walls") and colored pieces. From there you have to completely fill the board with the remaining pieces.

The leaflet claims there is only one solution for each guide. I was wondering if there's some specific math behind this puzzle. For example, if we represent it as a graph, there's this algorithm that, in combination with that algorithm, gives us all board configurations.

So, the question is:

How does one generate all the guides for a given set of pieces and walls, while also guaranteeing that there's one, and only one, solution? Or am I overthinking this, and it's simply bruteforced?

Thought, the search space seems quite large to me. A quick google search didn't reveal anything. Also, I'm not sure if this question would be more appropriate for Puzzles Stackexchange.

Thank you.

答:
Just eye-balling that puzzle, I really don't think brute-forcing this is much of a problem for a computer (or even a human!). The board is of limited size, there are a limited number of pieces, and all pieces are of size 4 or 5. Thus, given the placement of three wall pieces (and if you watch the video, you'll see that some puzzles place all 4 white pieces), the number of possible placements of each of the pieces is already quite limited, even if we take rotation and mirroring (i.e. flipping them over) into account.

Just to see this for yourself, just consider the puzzle with the three white walls pieces placed as given in that image, and think of where the yellow 'U' piece can go.  You'll find that there are really not a lot of options (also note that due to their symmetry, flipping the 'U' piece does not help, just as it doesn't help for the 'W' and 'T' pieces. And rotating the 'S' piece 180 degrees does not help either ...)

Moreover, after placing just one or two pieces, it will rule out possible positions for the remaining pieces even more quickly. Thus, I believe that even a completely dumb brute-force search algorithm will very quickly get to dead ends as it places the pieces, and should thus be able to cover the whole search space fairly quickly.

Of course, we can add some further intelligence to this algorithm as well.
For example, placing the 'U' piece in 4-9-15-21-16 creates an isolated 'island' of size 2: 5-10 ... and that of course means that this is not going to work given that all pieces are of size 4 or 5. Indeed, creating any island with size smaller than 4 means you have reached a dead end.

You could even check for islands of size 6 or 7, or any number that you cannot form by the size 4 and 5 pieces (e.g. 11 or 16 (yes, 16 also works since there are only three pieces of size 4)) but I believe a brute force algorithm that just checks for islands smaller than 4 will already be highly, highly efficient.

Indeed, in the puzzle as shown, the yellow U piece creates such islands in every of its already limited number of positions except for one. Thus, we can figure out the correct location of the U piece almost immediately, and from there, the rest will be very straightforward as well.

And as if that is not already enough, yet another thing that you can do is to try to figure out how chunks of puzzle pieces are forced (you effectively do some of this already when identifying islands). For example, we know that 5 and 10 will always have to belong to the same piece. Moreover, once you place the white walls, a bunch more of those connections are forced. And, once you place a puzzle piece, more connections are forced.

Consider, for example, what happens when we place the white walls as from the image shown on the website, and we place the green piece in 12-17-22-23-28:

Now, this is one of the few locations that the green piece can go without creating an island smaller than 4. However, immediately we can see that (because of the white walls) 2 and 7 need to be connected but (because of the green piece), 11, 6, and 1 need to be connected as well ... and 1 will need to be connected to 7 .. thus forcing a piece of size 5 ... of a shape that does not exist. So, we know that the green piece does not go here.

Adding this further bit of intelligence to the brute force search algorithm might make it even more efficient yet, though I am not sure, since merely checking for islands of size smaller than 4 might already make it very, very fast. Indeed, every bit of intelligence you add will slow down the basic loop, and so there is always this interesting trade-off between the number of loops you have to go through and how long it takes to pass through each loop (this is also why I think that checking for dead-end islands of size greater than 4 may not increase the efficiency either).

Anyway, I hope that at this point I have you convinced that a brute force algorithm should have no problem at all covering the whole search space. I truly think a brute force algorithm that checks for islands of size less than 4 will complete the search in a fraction of a second. But even without any further intelligence at all, I believe a totally dumb brute force algorithm will already be able to check for uniqueness in a fairly short time.

One final thought. Near the end of the video, you'll see that for some of the puzzles only one white piece is placed on the board. Now, having only one white piece would obviously increase the number of ways to place individual pieces significantly, and a brute force algorithm would slow down considerably. However, it actually confirms a suspicion I have: with this board and puzzle pieces, there are probably a fairly limited number of solutions to place all pieces in the board, period.

That is, even without any white wall pieces placed on the board at all, I bet that the number of solutions is fairly small (maybe a couple of hundred?). Thus, if you just brute force and get a complete list of all those completely unconstrained solutions (and you have to do this just once), then creating any new puzzle would be a snap: place some white pieces, and go through your already created list of unconstrained solutions to see if there is exactly one solution from that list that is compatible with the placed white pieces. This would probably take the time to verify that some set-up has a unique solution down to less than 1 microsecond.


问:
A confusion about the TREE function.
I know, from Kruskal's tree theorem, that the sequence of trees mentioned in the TREE function cannot be infinite. However, why can't the sequence of trees get arbitarily large, so that there is no infinite sequence, but there are arbitarily large sequences? As far as I know, Kruskal's theorem doesn't exclude such a possibility.</p>

答:
Note that the argument below is hinted at in the first couple sentences of this part of the wiki article on Kruskal's theorem, although it doesn't say explicitly why Konig's lemma applies.

Let's say that an $n$-TREE sequence is a finite sequence $(T_i)_{1\le i\le m}$ of rooted $\{1,...,n\}$-labelled trees where $T_i$ has at most $i$ vertices and for $1\le i<j\le m$ we have $T_i\not\le T_j$.

I've bolded the key part, the vertex counting condition; this is the crucial point which allows us to apply compactness in the form of Konig's lemma to get the desired result.

Specifically, note that the set of $n$-TREE sequences forms a tree $\mathfrak{T}_n$ itself, ordered by extension - so if $\langle T_1, T_2\rangle$ is an $n$-TREE sequence, then both $\langle T_1\rangle$ and $\langle T_1,T_2\rangle$ are nodes on $\mathfrak{T}_n$ and $\langle T_1,T_2\rangle$ is an immediate successor of $\langle T_1\rangle$ on $\mathfrak{T}_n$.

Now Kruskal's tree theorem says that $\mathfrak{T}_n$ has no infinite paths, which as you observe isn't quite what we want. But for each $i$ there are only finitely many $\{1,...,n\}$-labelled trees with at most $i$ vertices. So $\mathfrak{T}_n$ is finitely branching, and we can apply Konig's lemma to conclude that if it were infinite it would have an infinite path. This lets us get from the absence of infinite paths to the absence of arbitrarily long paths.

Title: A confusion about the TREE function.


问:
Conditions for quadratic variation
Good evening everyone. We are struggling showing the following:

we are given a filtered probability space (each filtration is complete) and $(B_t)_{t \geq 0}$ as a standard $\mathcal{F}_t$ Brownian Motion. In addition, we have the following process: $X = (B_t^2 - t)_{t \geq 0}$. We are asked to show that the quadratic variation (aka sharp brackets) of X is given by:
$$\langle X \rangle _ t = \int_0^t4 \cdot B_s^2 \ ds := A_t$$
We have seen a theorem that guarantees that $A_t$ is indeed the quadratic variation if the following four conditions are met:

X is a martingale;

$A_t$ is adapted, a.s. continuous, increasing and such that $A_0 = 0$;

X is squared integrable

$(X_t^2 - A_t)_{t \geq 0}$ is a martingale.

We showed those point in this way (and then get stuck in the last one):

X is integrable because sum of two integrable functions (namely $B_t^2$ and $-t$). Then we have to check:
$$E[X_t - X_s | \mathcal{F_s}] = 0 \\ E[X_t - X_s] = 0 \\ E[B_t^2 - t - B_s^2 + s] = 0$$
which is true by linearity of expectation and recalling that $E[B_t^2] = t \ \forall t \geq 0$

$A_t$ is adapted because it is integral of a continuous and adapted process (our $B_t^2$). It is a.s. continuous because integral of a.s. continuous function. It is increasing because $4 \cdot B_t^2 \geq 0 \ \forall t$ and basically we are just enlarging our interval of integration each subsequent step. $A_0 = 0$ trivially.

We checked that $E[X^2_t] < \infty$ which is true because we have: $$E[(B_t^2 - t)^2] = E[B_t^4 - 2B_t^2t + t^2] < \infty$$
because we know that $E[B_t^4] = 3t^2$

Is it quite right what we have done so far?

We are left with showing 4. Are there any hints on how to proceed? May Fubini be useful?

Thanks in advance and we hope this argument here will be useful for other users!!

Title: Conditions for quadratic variation

Body:
First calculate

\begin{align*}
X_{t}^{2} &= (B_{t}^{2}-t)^{2} \\
&= ([(B_{t}-B_{u})+B_{u}]^{2} - t)^{2} \\
&= (B_{t}-B_{u})^{4} + 4B_{u}(B_{t}-B_{u})^{3} + 6B_{u}^{2}(B_{t}-B_{u})^{2} + 4B_{u}^{3}(B_{t}-B_{u}) + B_{u}^{4} \\
&\quad - 2t[(B_{t}-B_{u})^{2} + 2B_{u}(B_{t}-B_{u}) + B_{u}^{2}] + t^{2}
\end{align*}
and $$X_{u}^{2} = B_{u}^{4}-2uB_{u}^{2}+u^{2}$$

before deducing that

\begin{align*}
\mathbb{E}_{u}[X_{t}^{2}] &= 3(t-u)^{2} + 6B_{u}^{2}(t-u) + B_{u}^{4} -2t(t-u) -2tB_{u}^{2} + t^{2} \\
&= B_{u}^{4}-2uB_{u}^{2}+u^{2} + 4(t-u)B_{u}^{2} + 2t^{2}-4ut+2u^{2} \\
&= X_{u}^{2} + 4(t-u)B_{u}^{2} + 2(t-u)^{2}.
\end{align*}

Next calculate

\begin{align*}
A_{t} &= \int_{0}^{t} 4B_{s}^{2}\, ds \\
&= A_{u} + \int_{u}^{t} 4B_{s}^{2}\, ds \\
&= A_{u} + 4\int_{u}^{t} (B_{s}-B_{u})^2 + 2B_{u}(B_{s}-B_{u}) + B_{u}^{2}\, ds,
\end{align*}

and take a conditional expectation to get

\begin{align*}
\mathbb{E}_{u}[A_{t}] &= A_{u} + 4 \int_{u}^{t} (s-u)+B_{u}^{2}\, ds \\
&= A_{u} + 4B_{u}^{2}(t-u) + 2(t^{2}-u^{2}) - 4u(t-u) \\
&= A_{u} + 4B_{u}^{2}(t-u) + 2(t-u)^{2}.
\end{align*}

Last but not least, take the difference to get

$$\mathbb{E}_{u}[X_{t}^{2}-A_{t}^{2}] = X_{u}^{2}-A_{u}.$$

NOTE: Obviously (for reference) It^o's lemma is a little faster...

\begin{align*}
d(X_{t}^{2}-A_{t}) &= 2X_{t}\, dX_{t} + d[X,X]_{t} - dA_{t} \\
&= 2X_{t}(2B_{t}\, dB_{t}) + 4B_{t}^{2}\, dt - 4B_{t}^{2}\, dt \\
&= 4X_{t}B_{t}\, dB_{t}.
\end{align*}


问:
On finding adjoint of transformation.
Let $V$ be an inner product space and $v,w\in V$ be fixed vectors. Define $T(u)=(u,v)w$.  How to find the adjoint mapping $T^*$?</p>

答:
From the definition of adjoint:
$(Tu, z) = (u, T^*z)$

So, for all $u, z \in V$:

$(Tu, z) = ( (u,v)w, z ) = (u,v) \cdot (w,z) = (u, \overline{(w,z)}v)
= (u, (z, w) v)$

So, $T^*(z) = (z,w)v$.

Title: On finding adjoint of transformation.


问:
Probability problem : why is this wrong
If you throw up 3 dice and want to calculate the probability of at least one six, my math book uses the complement rule as follows:

1- (5/6)^3

which is clear for me, but i tried it another way:

at least one six means either one, two, or three times a six, so i got this:

(3C1 * 15C2 + 3C2 * 15C1 + 3C3 )/ 18C3

So I noted there are 6 times 3, thus 18 elements with 3 dice. Picking 3 from 18 are all possible combinations. In the numerator I put the three possibilities, so one, two, or three times a 6.

But i compared the outcome and it wasnt equal to that of the book.

Can anyone explain why this is wrong?

Thanks in advance, also how can i write mathematical expressions on this site?

答:
$^{18}C_3$ or ${18 \choose 3}=816$ is not the number of equally likely ways of throwing $3$ dice.  Since each die can take six equally likely values, $6^3=216$ is.  Similarly with the rest of your calculation.

So the alternative calculation should be $$\frac{^3C_1\,5^2 + \, ^3C_2\,5   +  \,^3C_3}{6^3}$$
and this is the same as $1-\left(\frac 56\right)^3=\frac{91}{216}$

Title: Probability problem : why is this wrong


答:
A correct approach if you want to avoid the complement is
$$\frac{\sum_{k=1}^3 \binom{3}{k}5^{3-k}}{6^3} = \frac{\binom{3}{1}5^2+\binom{3}{2}5^1+\binom{3}{3}5^0}{6^3} = \frac{91}{216}$$

Title: Probability problem : why is this wrong

Body:
If you throw up 3 dice and want to calculate the probability of at least one six, my math book uses the complement rule as follows:

1- (5/6)^3

which is clear for me, but i tried it another way:

at least one six means either one, two, or three times a six, so i got this:

(3C1 * 15C2 + 3C2 * 15C1 + 3C3 )/ 18C3

So I noted there are 6 times 3, thus 18 elements with 3 dice. Picking 3 from 18 are all possible combinations. In the numerator I put the three possibilities, so one, two, or three times a 6.

But i compared the outcome and it wasnt equal to that of the book.

Can anyone explain why this is wrong?

Thanks in advance, also how can i write mathematical expressions on this site?


问:
Analytical solutions to system of inhomogeneous first order PDE
I am having trouble trying to find example or figure out how to solve systems of PDEs of this form.

$$ u_{t} + au_{x} = f(u,v,x) $$

$$ v_{t} + bv_{x} = g(u,v,x) $$

For cases where $a \geq 0$ and $b \geq 0$ and constant with $f(u,v,x)$ and $g(u,v,x)$ being linear.

For example:

$$ u_{t} + au_{x} = k(u - bv),$$
$$ v_{t} + bv_{x} = k(bv - u),$$

where $k > 0$, $b \geq 0$ and $c \geq 0$. Ideally I would like to have $b = b(x)$. Im using these equations to test a finite volume PDE code I am developing as part of my PhD work.

My question is do these equations have analytical solutions? If so how do I find them?

答:
No explicit solution formulas in general, but the theory does suggest a way to test your code. (The existence theory is on page 466 of Courant and Hilbert, Methods of Mathematical Physics Vol 2.) The first equation gives the derivative of $u$ in the direction $(a,1)$, that is,
$$ \frac{d}{dt}u(x_0+at,t) = ku(x_0+at,t)-kbv(x_0+at,t) $$
and a similar result using the second equation in the direction $(b,1)$.

So integrate
$$ ku(x_0+at,t)-kbv(x_0+at,t) $$
numerically, using the output of your code, from $t = 0$ to $t = T$
and you ought to get
$$ u(x_0+aT,T)-u(x_0,0)$$
pretty nearly, as a test of your code that does not require doing any
numerical derivatives.


问:
How to compute Nakayama functor explicitly?
I am reading the book Elements of representation theory of associative algebras, volume 1. I have some questions related to the Nakayama functor. On page 113-114 of the book, the Auslander-Reiten translation $\tau M$ of a module $M$ of the path algebra of the Kronecker quiver $Q=1\underset{\beta}{\overset{\alpha}{\Leftarrow}} 2$ is computed. The module $M$ is defined by $K \underset{0}{\overset{1}{\Leftarrow}} K$.

It is said that a minimal projective presentation of $M$ is given by
$$
0 \to P(1) \overset{p_1}{\to} P(2) \overset{p_2}{\to} M \to 0,
$$
where $$P(1)=S(1)= (K \underset{0}{\overset{0}{\Leftarrow}} 0), \quad P(2)= (K^2 \underset{\left( \begin{matrix}0 \\ 1\end{matrix} \right)}{\overset{\left( \begin{matrix}1 \\ 0\end{matrix} \right)}{\Leftarrow}} K).$$

We know that a basis of $K^2$ in $P(2)$ is $K\alpha\oplus K\beta$. The basis of $K$ in $P(1)$ is $K \varepsilon_1$. The map $p_1$ from $P(1)$ to $P(2)$ is
$$
\begin{matrix} K & \underset{0}{\overset{0}{\Leftarrow}} & 0 \\
f \downarrow & & \downarrow g \\
K^2 & \underset{\left( \begin{matrix} 0 \\ 1\end{matrix} \right)}{\overset{\left( \begin{matrix}1 \\ 0\end{matrix} \right)}{\Leftarrow}} & K
\end{matrix}
$$
Since the domain of $g$ is $0$, we must have $g=0$. How to compute the map $f$? Does $f$ send $\varepsilon_1$ to $\alpha$ or $\beta$ or $0$?

If we know the map $f$, then we know the map $p_1$. How to compute $\nu p_1$? Where $\nu=DHom_A(\cdot, A)$ is the Nakayama functor.

We know $I(1)$ and $I(2)$. If we know the map $\nu p_1: I(1) \to I(2)$, how to compute the kernel of $\nu p_1$? I know that the kernel of $\nu p_1$ must be of the form $K \underset{h_2}{\overset{h_1}{\Leftarrow}} K$. But I don't know how to show that $h_1=1$ and $h_2=0$. Thank you very much.

答:
For your first question, $f$ is given by $\epsilon_1\mapsto \beta$, or equivalently $p_1:\epsilon_1 x \mapsto \beta x$ for any $x\in kQ$.  You just need to stare at your diagram long enough to realise this...Note that the squares have to commute.  I usually prefer computing in (sort of) Loewy diagram form, as the algebra is basic and the field (so all simples are of 1 dimensional) is algebraically closed.  This technique which can be demonstrate as follows.  Turning your exact sequence into Loewy diagram:
$$
0 \to 1 \to
\left(\begin{array}{ccc} & 2 & \\  & \swarrow  \searrow & \\ 1 &  & 1 \end{array}\right) \to
\left(\begin{array}{c} 2  \\ {\scriptsize 1}\Downarrow {\scriptsize 0} \\ 1 \end{array}\right) \to 0
$$
Each number here denote the composition factor (simple module), so for example the middle $P(2)$, has composition factor $S(2)$ on top, and two copies of $S(1)$ in the bottom; the left hand side arrow correspond to multiplication by $\alpha$ on the right, the right hand side arrow correspond to multiplication by $\beta$.  So you see the top $S(2)$ of $P(2)$ is mapped to the top $S(2)$ of $M$, and the left hand side $S(1)$ in socle of $P(2)$ is mapped to $S(1)$ in the socle of $M$, because we need to arrows in the middle to match up (corresponding to the commuting square).  So now the thing in $P(2)$ that has not been mapped (i.e. the kernel) is the right hand side $S(1)$ in the socle.  So the map you want is to map $S(1)$ to the right hand side $S(1)$, i.e. given by $\epsilon_1 x\mapsto \beta x$.

Normally, you would need to compute $DHom_A(X,A)$ explicitly to see what the maps resulting maps look like.  Sometimes, we can just "clever guess", because it maps (ind) projective to (ind) injective "in a compatible way".  By the description of injective modules (this is in earlier proposition in the book), $\nu(S(1)\to P(2))= I(1) \to I(2) = D( \epsilon_1 A \to \epsilon_2 A) $ has Loewy diagram
$$
 \left(\begin{array}{ccc} 2 &  & 2\\  & \searrow  \swarrow & \\ & 1 & \end{array}\right) \to 2
$$
where left hand arrows corresponding to "multiplying $\alpha$ on the left, before taking $K$-linear dual $D$". "The compatibility" means that the map $\nu p_1$ is given by $D(x \beta \to x \epsilon_2)$, i.e. to map the right hand side composition factor $S(2)$ in the top of $I(1)$ to $I(2)=S(2)$.  Therefore the kernel, can be seen from the combinatorics easily, is given by "deleting the right hand side of the diagram of $P(2)$":
$$
\begin{array}{c} 2 \\ {\scriptsize 1}\Downarrow{\scriptsize 0} \\ 1 \end{array}
$$
so is just $M$ itself.


问:
A Markov chain having transitions probabilities being derived from a Poisson process
Here is a problem I am currently thinking about:

An airport shuttle makes tours to transport passengers from a train station to two possible terminals, $A$ and $B$. After carrying all of its passengers to their destination terminals, the shuttle returns to the train station to pick up the next passengers. The tour of the shuttle takes:

$-$12 minutes if the shuttle is carrying passengers headed to both terminals.

$-$10 minutes if the shuttle is carrying passengers headed to terminal $B$ but none to terminal $A$.

$-$8 minutes if the shuttle is not carrying any passengers headed to terminal $B$ (including if there are no passengers at all).

New passengers headed to terminal $A$ keep arriving at the train station with independent waiting times which follow $\text{Exp}(1/2)$ distribution, and new passengers headed to terminal $B$ arrive with independent waiting times which follow $\text{Exp}(1/6)$ distribution, independently also of each other. Always returning from its previous tour, the shuttle picks up all new passengers that have arrived meanwhile, and starts a new tour to carry them to their destination terminals.

(a) Construct a discrete time three state Markov chain describing the tours of the shuttle, such that the states indicate whether the tour has passengers headed to both terminals, has passengers headed to terminal $B$ but not to terminal $A$, or only has passengers headed to terminal $A$ (including the case of no passengers at all). Give the transition probability matrix of the Markov chain.

(b) In the long run, what is the average duration of the tours of the shuttle?

We're given that passengers arrive with independent waiting times, but what does this mean exactly? Do they mean to say that passengers arrive with independent interarrival times? Also, in  part (b) we're asked to evaluate the average duration of the tours in the shuttle. This is slightly confusing to me because the duration of the shuttle is fixed and not random. Any insight would be greatly appreciated.

答:
Here is my attempted solution. I am assuming the question means to say passengers arrive with independent interarrival times. Let state $I$ denote the event that shuttle contains passengers travelling to both terminals, state $II$ the event the the shuttle contains passengers travelling to terminal $B$ but not $A$, and state $III$ the event that the shuttle doesn't contain any passengers from terminal $B$. Let's first assume that the shuttle is in state $I$, meaning it's on a 12 minute trip containing passengers headed to both terminals.  Take $Y_A,Y_B$ as the random variables which count the number of passengers who arrive within this 12 minute period travelling to terminal $A$ and terminal $B$, respectively. Notice how $Y_A\sim \text{Poisson}\bigg(12\cdot \frac{1}{2}\bigg)$ whereas $Y_B\sim \text{Poisson}\bigg(12\cdot \frac{1}{6}\bigg)$. If $X_i\in\{I,II,III\}$ denotes the state of the shuttle on the $i^{\text{th}}$ trip, then $$P(X_{i+1}=I|X_i=I)=P(Y_A>0,Y_B>0)=(1-e^{-6})(1-e^{-2})$$ $$P(X_{i+1}=II|X_i=I)=P(Y_A=0,Y_B>0)=e^{-6}(1-e^{-2})$$ $$P(X_{i+1}=III|X_i=I)=P(Y_A>0,Y_B=0)+P(Y_A=0,Y_B=0)=e^{-2}$$ Using identical logic we arrive at a $3\times3$ transition probability matrix delineated below: $$\begin{equation*}
P =
\begin{bmatrix}
(1-e^{-6})(1-e^{-2}) & e^{-6}(1-e^{-2}) & e^{-2} \\
(1-e^{-5})(1-e^{-5/3}) & e^{-5}(1-e^{-5/3}) & e^{-5/3} \\
(1-e^{-4})(1-e^{-4/3}) & e^{-4}(1-e^{-4/3}) & e^{-4/3}
\end{bmatrix}
\end{equation*}$$ I think I am in good standing with part (a), but part (b) is still confusing. Would solving part (b) require us to find the steady$-$ state probabilities which would effectively reveal the the proportion of the time the shuttle is in state $I,II$ and $III$? Thank you for any insight.


问:
How do I interpret $\lim_{k \to \infty} \inf \{ s > 0 : \int_0^s g(x) \, dx > k\}$? (plus related question)
Let $t_k = \inf \{ s > 0 : \int_0^s g(x) \, dx > k\}$

Problem 1: How should I interpret $\lim_{k \to \infty} t_k$? Is it $\inf \{ s > 0 : \int_0^s g(x) \, dx = \infty\}$?

Problem 2: Assuming that $t = \inf \{ s > 0 : \int_0^s g(x) \, dx = \infty \}$, can we deduce that $\int_0^t g(x) \, dx < \infty$?

Attempt problem 2: Since $t = \inf \{ s > 0 : \int_0^s g(x) \, dx = \infty \}$, this means that for all $\varepsilon > 0$, we have that $\int_0^{t-\varepsilon} g(x) \, dx < \infty$, otherwise $t$ would not be the infimum. Since this holds for all $\varepsilon > 0$, we must have that $\int_0^t g(x) \, dx < \infty$ (I'm unsure if this holds in the limit as $\varepsilon \to 0$.)

答:
The quantity $t_k = \inf \{ s > 0 : \int_0^s g(x) \, dx > k\}$ depends on what number $k$ is. The expression $\lim_{k\to\infty} t_k$ should be interpreted as the limit of a function of $k,$ defined just as any limit is defined.

Now imagine, for example, theat $g(x) = e^{-x}.$ Then $\int_0^s g(x)\,ds = 1 - e^{-x} < 1.$ No matter how big $s$ gets, the integral will never be $+\infty.$ Therefore the set $\{ s > 0 : \int_0^s g(x) \, dx = \infty\}$ will be empty. The infimum of the empty set is $+\infty.$


问:
n-dimensional Euclidean space is separable
I'm self-teaching topology, and currently, I'm working on the topic of separable space. The following sentences are what I saw in Wikipedia:

Every compact metric space (or metrizable space) is separable.

Any topological space that is the union of a countable number of separable subspaces is separable. Together, these first two examples give a different proof that n-dimensional Euclidean space is separable.

The second statement makes me feel confused. How should I prove the union of a countable number of separable subspaces is separable, and why combining 1 and 2 indicates that n-dimensional Euclidean space is separable?

答:
To answer your second question first, note that the interval $I_k=[k,k+1]$ is compact for each $k\in\Bbb Z$. Let $\langle a_1,\ldots,a_n\rangle\in\Bbb Z^n$; then $\prod_{k=1}^nI_{a_k}$ is a compact metric space, so by (1) it is separable. Finally, $\Bbb Z^n$ is countable, and

$$\Bbb R^n=\bigcup\left\{\prod_{k=1}^nI_{a_k}:\langle a_1,\ldots,a_n\rangle\in\Bbb Z^n\right\}\,,$$

so (2) implies that $\Bbb R^n$ is separable.

To prove (2), let $X=\bigcup_{n\in\Bbb N}X_n$, where each $X_n$ is separable. This means that each $X_n$ has a countable dense subset $D_n$. Let $D=\bigcup_{n\in\Bbb N}D_n$; $D$ is the union of countably many countable sets, so $D$ is countable. Finally, let $U$ be any non-empty open subset of $X$; there is at least one $n\in\Bbb N$ such that $U\cap X_n\ne\varnothing$. And $U\cap X_n$ is a non-empty open subset of $X_n$, so $U\cap D_n\ne\varnothing$: $D_n$ is dense in $X_n$, so it intersects every non-empty open subset of $X_n$. But then $U\cap D\supseteq U\cap D_n\ne\varnothing$, so we’ve shown that every non-empty open subset of $X$ intersects $D$, i.e., that $D$ is dense in $X$. Thus, $X$ has a countable dense subset and is therefore separable.


问:
Particular holomorphic mappings of the Riemann sphere to itself
I am struggling to understand what prof. Penrose says in his "The Road to Reality" book (9.3):

There are many holomorphic mappings of the Riemann sphere to itself which send each hemisphere to itself, but which do not preserve the north or south poles (i.e. the points $z=0$ or $z=\infty$).These preserve the positive/negative frequency splitting but do not preserve the individual Fourier components $e^{-in\chi}$ or $e^{in\chi}$.

Then an exercise is proposed to show explicitely what these mappings are.

The first question I have in mind is whether that these holomorphic mappings of the sphere to itself necessarily correspond to Möbius transformations in the $\mathbb{C}$ plane.

Then, rather than the full solution, I am trying to find at least one example of a Möbius transformation that has the behaviour described; the only mapping that comes into my mind is a rotation through the "vertical" axis (hence a simple complex multiplication $az$ in terms of Möbius transformation) but this of course preserves the poles. I really cannot figure out how I can rotate or translate this sphere so that the north hemisphere still maps to the north, withouth keeping the north pole fixed. It really seems to me that prof. Penrose is meaning something else here, which I am not getting.

Last but not least, I am missing the connection between these holomorphic mappings and the fourier components of a function.

Any hint is welcome, thanks!

答:
If $h$ is analytic and non-zero at $e^{it}$ then (since $h(z)=\sum_k c_k (z-e^{it})^k$) so is $g(z)=1/\overline{h(\overline{z}^{\ -1})}$

If $h(e^{it}) = e^{i\theta}$ then $g(e^{it}) = e^{i\theta}$.

Thus $h$ is meromorphic $\Bbb{C\to C}$ and $|h(z)|=1$ on $|z|=1$ means that $g-h$ is analytic around $e^{it}$ and vanishes on $e^{ix},x-t\in (-\epsilon,\epsilon)$ ie. $g=h$.

Next since $h$ is analytic $\Bbb{P^1(C)\to P^1(C)}$ then it is a rational function $ h(z)=C\frac{\prod_j(z-a_j)}{\prod_i(z-b_j)}$, that $h$ sends each hemisphere to itself gives $|h(z)|=1$ on $|z|=1$ so that $h=g$ and hence $$h(z)=e^{i\theta}\prod_{j=1}^J \frac{z-a_j}{1-\overline{a_j}z}$$
Finally each $|a_j|<1$ since otherwise $h$ would have some zero on the $\infty$-hemisphere.


问:
Showing $\frac{f(a+2b)+f(a-2b)}{2f(a)}\leq\sqrt{1-\left(\frac{b}{a(1-a)}\right)^2}$ for $f(x)=\sqrt{x(1-x)}$ with some constraints
My question is from a inequality that is not proved (it is just implicitly mentioned I guess) in a book.

Specifically, let $a \in (0,1)$ and $b \in (0,1)$ with $a - 2b \geq 0$ and $a + 2b \leq 1$. Then with $f(x) := \sqrt{x(1-x)}$ we are asked to show that $$\frac{f(a+2b)+f(a-2b)}{2f(a)} \leq \sqrt{1-\left(\frac{b}{a(1-a)}\right)^2}$$

I am really stuck at this, the only related inequality that comes to mind is that $$\frac{\sqrt{1+2\alpha}+\sqrt{1-2\alpha}}{2} \leq \sqrt{1-\alpha^2},\quad \forall \alpha \in [0,\tfrac{1}{2}]$$

But I still cannot prove the desired inequality.

Any help is greatly appreciated!

答:
I will just bash it. Let's work on the square of the LHS first. You will actually find that:

$$f^2(a+2b) + f^2(a-2b) = 2a(1-a) - 8b^2 = 2f^2(a) - 8b^2$$
and
$$f(a+2b)f(a-2b) = \sqrt{(a^2-4b^2)((1-a)^2 - 4b^2)} = \sqrt{(f^2(a)+4b^2)^2 - 4b^2}$$
Therefore, the squared version of your inequality is:

$$LHS^2 = \dfrac{f^2(a) - 4b^2 + \sqrt{(f^2(a)+4b^2)^2- 4b^2}}{2f^2(a)}\leq 1 - \dfrac{b^2}{f^4(a)}=RHS^2$$

Now it looks like some nice substitution will simplify things:

$$\begin{cases}
f^2(a) = a(1-a) = x\in(0,\frac 14) \\
4b^2 = y\in (0,\frac 14)
\end{cases}.$$
Then, you have:
$$\dfrac{x-y+\sqrt{(x+y)^2-y}}{2x}\leq1 - \dfrac{y}{4x^2}\iff \dfrac{(x+y)^2-y}{4x^2}\leq \left(\frac 12-\frac{y}{4x^2}+\frac{y}{2x}\right)^2$$
and after a couple line of simplifying, you will get:
$$y(1-x)(yx-y+4x^2+2xy) + 4x^2y\geq 0$$
which is trivially true. There is no equality attained in the domain you specified, but $b =y= 0$ gives an equality.

P.S: I did try an naive AM-GM in the beginning, but that resulted in $a(1-a)\geq\frac 14$, which is only true when $a=\frac 12$, so some cute AM-GM is probably not gonna cut it.


问:
Notation to make a negative answer equal 0
So basically, I'm making a big equation, and one component of it is this

$$(n-5)+(n-7)$$

The n-value can be anything from 1 to far away. So, that obviously leads to some negative numbers. However, this part of the equation is only applicable whenever the n-value is greater than the subtracting numbers (respectively). So, is there some kind of notation I can use to signal that if the answer is a negative number, then it is to be dismissed, or just turned to 0? A kind of notation similar to the ceiling and floor functions, where a non-integer is turned to the closest integer (either up or down).

答:
The written notation would be
$$
X(n) = \left\{
    \begin{array}{ll}
        (n-5)+(n-7) & n \ge 6\\
        0 & n<6
    \end{array}
\right.
$$
In the comments were given notations that can be used in a programming language.


问:
Function $f$ not in $L^1$ but $\hat{f} \in L^1$
Can you have a function $f \notin L^1$ but its Fourier transform $\hat{f} \in L^1$? Ive been playing around with examples and I cant find one, but I also cant prove one doesn't exist.

答:
For
$$
f(x) = \frac{\sin x}{x} \not\in L^1(\mathbb{R})
$$
one has
$$
\hat{f}(\xi) = C \chi_{[-a,a]}(\xi) \in L^1(\mathbb{R})
$$
where $C$ and $a$ are some constants depending on the exact choice of definition of Fourier transform.


问:
Solving non-autonomous system of differential equations
Given the following 2x2 system of differential equations,

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y \end{cases}$

A solution is easy to fins by computing eigenvalues and eigenvectors of the associated matrix A

$A = \begin{bmatrix} 4 & -1 \\ 2 & 1 \end{bmatrix}$.

Therefore, knowing that eigenvalues are 2 and 3, and the related eigenvectors are $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ respectively, the solution is of the form

$Ae^{2t}\begin{bmatrix} 1 \\ 2 \end{bmatrix}+Be^{3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

What about if I make the initial system of equations non-autonomous? Resulting in something as follows

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y +f(t) \end{cases}$

where f(t) can be either a polynomial on t, an exponential ($e^{at}$) or a trigonometric function ($sin(at)$ or $cos(at)$).

What is now the procedure to follow this system of differential equations? For example, let $f(t)=t^2$.

How do I solve the following system of differential equations?

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y+t^2 \end{cases}$

答: You can express your non$-$autonomous  differential equation in the form $$\frac{d\vec{x}}{dt}-A\vec{x}=\vec{f}(t)$$ where $\vec{x}=\big(\begin{smallmatrix} x \\ y  \end{smallmatrix}\big)$, $A=\big(\begin{smallmatrix} 4 & -1\\ 2 & 1 \end{smallmatrix}\big)$, and $f(t)=\big(\begin{smallmatrix} 0 \\ t^2  \end{smallmatrix}\big)$. It turns out that $$\vec{x}=e^{At}\int e^{-At}f(t)dt$$ just as you might expect!


问:
Proof for triangle inequality for vectors
Generally,the length of the sum of two vectors is not equal to the sum of their lengths. To see this consider the vectors $u$ and $v$ as shown below.

By considering $u$ and $v$ as two sides of a triangle, we can see that the lengths of the third side is $\| u + v \|$ and we have $\| u + v \| \leq \|u\| + \|v\|$.  Under what circumstance equality occurs and how can one prove that?

答:
I always like to see and prove this visually:

One can find $A$,$B$ and $C$ points on the plane such that

\begin{align*}
\overrightarrow{AB}&={\overrightarrow{u}}\\ {\overrightarrow{BC}}&=\overrightarrow{v}\\
\overrightarrow{AC}&=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{u}+\overrightarrow{v}.
\end{align*}

Now fix points $A$ and $B$.

The questions that arise are the following: if we could move C to another position C', how should we move it such that:

the vectors $\overrightarrow{BC}$ and $\overrightarrow{BC'}$ have
the same length:
$\lvert\overrightarrow{BC'}\rvert=\lvert\overrightarrow{v}\rvert=\lvert\overrightarrow{BC}\rvert$?

the distance
$\lvert\overrightarrow{AC'}\rvert=\lvert{\overrightarrow{u}+\overrightarrow{v}}\rvert$ from $A$ to $C'$ is the largest possible?

what value would this largest distance take?

In other words, how should we rotate the vector $\overrightarrow{v}={\overrightarrow{BC}}$ around B such that the distance $\lvert\overrightarrow{u}+\overrightarrow{v}\rvert=\lvert\overrightarrow{AC}\rvert$ from $A$ to $C$ maximizes? What will be the max value?

The answer to $(1)$ are all the points $C'$ in the plane belonging to the circle with centre B and radius $\lvert{\overrightarrow{BC'}}\rvert=\lvert{\overrightarrow{v}}\rvert$.

Geometrically, the answer to $(2)$ seems to be the point $C'$ in the figure.

If this is the case, then the answer to (3) is obvious: $$\lvert{\overrightarrow{AC'}}\rvert=\lvert{\overrightarrow{AB}+\overrightarrow{BC'}}\rvert=\lvert{\overrightarrow{AB}}\rvert+\lvert{\overrightarrow{BC'}}\rvert
=\lvert{\overrightarrow{u}}\rvert+\lvert{\overrightarrow{v}}\rvert.$$
But why $C'$ in the figure is the right point?

We can use simple arguements from Euclidean Geometry to prove this arguement:
For the triangle $\triangle C'BC$ we have $\lvert{\overrightarrow{BC}}\rvert=\lvert{\overrightarrow{BC'}}\rvert$, hence the angles $\widehat{BC'C}=\widehat{BCC'}$ will be equal. We also have $\widehat{BCC'}\leq \widehat{ACC'}$, hence $\widehat{AC'C}=\widehat{BC'C}=\widehat{BCC'}\leq \widehat{ACC'}$. The last inequality holds for the triangle $\triangle ACC'$, and so the same inequality will hold for the segments that are opposite to these angles on that triangle:
$$\widehat{AC'C}\leq \widehat{ACC'}\implies \lvert{\overrightarrow{AC}}\rvert\leq \lvert{\overrightarrow{AC'}}\rvert.$$
Particularly, it holds that
\begin{align*}
\lvert{\overrightarrow{u}+\overrightarrow{v}}\rvert&=\lvert{\overrightarrow{AB}+\overrightarrow{BC} }\rvert\\
&=\lvert{\overrightarrow{AC}}\rvert\\
&\leq \lvert{\overrightarrow{AC'}}\rvert\\
&=\lvert{\overrightarrow{AB}}\rvert+\lvert{\overrightarrow{BC'}}\rvert\\
&=\lvert{\overrightarrow{AB}}\rvert+\lvert{\overrightarrow{BC}}\rvert\\
&=\lvert{\overrightarrow{u}}\rvert+\lvert{\overrightarrow{v}}\rvert
\end{align*}
which is exactly the triangle inequality and the reason that is called that way.


问:
Proof verification: inequality involving orthogonal projection
let $v \in V$ such that $W=[v]$. It is desired to show that
$$\|v-E(v)\| \leq \|v-w\|,\;\;\;\; \forall v \in V,\; and\;\;\forall w \in W$$
Where $E$ is the orthogonal projection of V in W.

Here's what I did: I simply used Cauchy-Schawrz inequality and the fact that $E(v)=w$. then,
$$\|v-w\|^2={\langle v-w, v-w \rangle} \leq \|v-w\|^2$$
but $E(v)=w$. therefore,
$${\langle v-E(v),v-E(v) \rangle} \leq \|v-w\|^2\\
\implies\|v-E(v)\|^2 \leq \|v-w\|^2 $$
Which lead us to the desired answer.
My question is: is it right? because since i assumed $E(v)=w$ wouldn't it imply that instead of an inequality sign I should have an "equal" sign? Any tips are welcome.

答:
No, it is not right. You did not prove that you have the inequality $\|v-E(v)\|\leqslant\|v-w\|$ for any $v\in V$ and any $w\in W$; you only proved that $\|v-E(v)\|\leqslant\|v-E(v)\|$ for any $v\in W$, which is trivial (you always have $x\leqslant x$). Besides, do you really need the Cauchy-Schwarz inequality in order to deduce that $\|v-w\|\leqslant\|v-w\|$?

You can prove it noting that\begin{align}\|v-w\|^2&=\|v-E(v)+E(v)-w\|^2\\&=\|v-E(v)\|^2+2\langle v-E(v),E(v)-w\rangle+\|E(v)-w\|^2\\&=\|v-E(v)\|^2+\|E(v)-w\|^2 \text{ ($v-E(v)$ and $E(v)-w$ are orthogonal)}\\&\geqslant\|v-E(v)\|^2.\end{align}


问:
Not linear Recursion
Solve recursion:$$a_{n+1}=a_n-a_n\cdot a_{n+1}$$. I tried transforming this recursion in form $$a_{n+1}=\frac{pa_n+q}{ra_n+s}$$ and then solve it as linear recursion. But not sure how to do it. Also not sure is there useful substitution for this.

答:
HINT: If you call
$b_n = \frac{1}{a_n}$
you get the recursion
$$b_{n+1} = b_n +1$$

Title: Not linear Recursion


问:
The path of complex structure.
$J = \left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)$. If $I(t)$ is a path in $\rm M(2n, \mathbb R)$ ($2n \times 2n $ real matrix) such that $I(0)=0$ and $I(t)J+JI(t)=0$, then for sufficiently small $t$ can we find a path $J(t) \in \rm M(2n, \mathbb R)$ such that (1) $J(0)=J$, (2) $J^2(t)+I^2(t)=-I_{2n}$, (3) $I(t)J(t)+J(t)I(t)=0$?</p>

答:
Post-bounty remarks: I am not experienced in complex structures and the like, but here are some ideas I had before I offered a bounty to this question.

Note that, if we add together the equations
\begin{align*}
J^2(t)+I^2(t)&=-I_{2n}\tag{a},\\
I(t)J(t)+J(t)I(t)&=0,\tag{b}
\end{align*}
we get
$$
(I(t)+J(t))^2=-I_{2n}.\tag{c}
$$
In other words, the original problem can be transformed into the existence problem of a square root $R(t)=I(t)+J(t)$ of $-I_{2n}$ such that the initial condition $R(0)=J$ and the equivalent form of $\text{(b)}$,
$$
I(t)R(t)+R(t)I(t) = 2I(t)^2,\tag{d}
$$
are satisfied.

Title: The path of complex structure.


问:
Prove the given surface has Gaussian curvature zero

If a regular surface is covered by a surface patch all of whose $u$- and $v$-coordinate curves are geodesics, then it has Gaussian curvature zero.

How can I show the above statement? Could you give any hint?

Edit: Using the following formula: Let $S$ be an oriented regular surface and $\sigma:U\subset\Bbb R^2\to V\subset S$ be a surface patch. Let $\gamma:I\to V$ be a regular curve which can be expressed as $\gamma(t) = \sigma(u(t),v(t))$ for some smooth functions $u,v:I\to\Bbb R$. Then $\gamma$ is geodesic if and only if $u''+(u')^2\Gamma_{11}^1+2(u'v')\Gamma_{12}^1+(v')^2\Gamma_{22}^1 = 0$ and $v''+(u')^2\Gamma_{11}^2+2(u'v')\Gamma_{12}^2+(v')^2\Gamma_{22}^2 = 0$.

From this, if we consider $\gamma$ as a coordinate curve (letting $u(t)=t$ and $v(t)=t$ resp.) then I get $\Gamma_{11}^1 =\Gamma_{11}^2 = \Gamma_{22}^1=\Gamma_{22}^2 = 0$. I don't know how to get further. I'm in introductory course in differential geometry by the way. So I want some explanation in elementary way.

The formula I wrote is in p.290 of this textbook.

答:
I believe this is false. Locally, on any surface whatsoever you can always find coordinates so that the $u$-curves and $v$-curves are geodesics. It is a standard result from differential equations that on a surface I can find local coordinates whose coordinate curves are tangent to two given linearly independent vector fields.

Choose a frame $X_0,Y_0$ (unit vectors) at $p_0$. Take the geodesics in those tangent directions. Call them $C$ and $\Gamma$. Take the tangent vectors to be $X$ and $Y$ respectively along those curves. Now choose $Y$ smoothly along $C$ and $X$ smoothly along $\Gamma$ extending these, maintaining linear independence. Now follow geodesics  from points of $C$ in the direction of $Y$ and follow geodesics from points of $\Gamma$ in the direction of $X$. Define the vector fields $X$ and $Y$ in the obvious way in a small neighborhood of $p_0$. We have constructed these so that they are tangent to geodesics at each point.


答:
$$\begin{cases} x' = 4x-y \\ {y'} = 2x+y +f(t) \end{cases}$$
Note that if you substract both DE you have:
$$x'-y' =2(x-y) -f(t) $$
$$u' -2u= -f(t) $$
$$(u(t)e^{-2t})'= -e^{-2t}f(t) $$
$$u(t)e^{-2t}= -\int e^{-2t}f(t) \,dt $$
Where $u(t)=x-y$.
$$x(t)=y(t)-e^{2t}\int e^{-2t}f(t) \,dt $$
Plug this in the second equation and solve.

Continuing from your result of integration in the comment:

$$u(t)=\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$
$$x-y=\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$
We have the first differential equation:
$$x'=4x-y$$
$$x'-4x=-x+\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$
$$x'-3x=\frac{1}{4}(2t^2+2t+1)+Ce^{2t}$$
$$(xe^{-3t})'=e^{-3t}\frac{1}{4}(2t^2+2t+1)+Ce^{-t}$$
Integrate. Don't forget the second constant of integration.

Title: Solving non-autonomous system of differential equations

Body:
Given the following 2x2 system of differential equations,

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y \end{cases}$

A solution is easy to fins by computing eigenvalues and eigenvectors of the associated matrix A

$A = \begin{bmatrix} 4 & -1 \\ 2 & 1 \end{bmatrix}$.

Therefore, knowing that eigenvalues are 2 and 3, and the related eigenvectors are $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ respectively, the solution is of the form

$Ae^{2t}\begin{bmatrix} 1 \\ 2 \end{bmatrix}+Be^{3t}\begin{bmatrix} 1 \\ 1 \end{bmatrix}$.

What about if I make the initial system of equations non-autonomous? Resulting in something as follows

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y +f(t) \end{cases}$

where f(t) can be either a polynomial on t, an exponential ($e^{at}$) or a trigonometric function ($sin(at)$ or $cos(at)$).

What is now the procedure to follow this system of differential equations? For example, let $f(t)=t^2$.

How do I solve the following system of differential equations?

$\begin{cases} \dot{x} = 4x-y \\ \dot{y} = 2x+y+t^2 \end{cases}$


问:
Continuous Function defined on compact set
Here's what I'm trying to prove:

let $K \subseteq \mathbb{R}^n$ be a compact nonempty set and let $f: K \to \mathbb{R}$ be a continuous function. Then:

$$\exists a,b \in K: \forall x \in K: f(a) \leq f(x) \leq f(b)$$

Proof Attempt:

We will just prove the existence of $b$ as a similar argument will apply for the existence of $a$. Suppose, for a contradiction, that:

$$\forall b \in K: \exists x \in K: f(x) > f(b)$$

Define a sequence $(x^{(k)})_{k \in \mathbb{N}}$ as follows:

Since $K \neq \varnothing$, take $x^{(1)}$ to be any element in $K$.

Once $x^{(k)}$ has been defined, define $x^{(k+1)}$ by picking the element of $K$ such that $f(x^{(k+1)}) > f(x^{(k)})$. We know that this exists by assumption.

Since $K$ is compact, the sequence above has a convergent subsequence $(x^{(k_m)})_{m \in \mathbb{N}}$ with limit $x$. However, by definition, the sequence $(f(x^{(k_m)}))_{m \in \mathbb{N}}$ cannot converge and this is impossible because $f$ is continuous and must map convergent sequences to convergent sequences. Hence, the desired $b$ must exist. $\Box$

Does the argument above work? If it doesn't, then why? How can I fix it?

答: No, it is not correct. You assert (without proof) that “the sequence $\left(f\left(x^{(k_m)}\right)\right)_{m \in \mathbb{N}}$ cannot converge”. That is not true. Suppose, for instance, that $K=[0,1]$ (with its usual topology), that $f\colon K\longrightarrow\Bbb R$ is defined by $f(x)=x$ and that $x^{(n)}=1-\frac1{n+1}$. Then the sequence $\left(x^{(n)}\right)_{n\in\Bbb N}$ converges, and so does the sequence $\left(f\bigl(x^{(n)}\bigr)\right)_{n\in\Bbb N}$.


问:
Five years ago, Tim was twice as old as Jim... Math Question
Five years ago, Tim was twice as old as Jim. In three years, Jim’s age will be $2/3$ that of Tim. What are the ages of Tim and Jim now?

答:
hint

Let $ T $ and $ J $ be the ages you want.

Five years ago, we had
$$T-5=2(J-5)$$

In three years, we will have,

$$(J+3)=\frac 23(T+3)$$

Now solve the system.

If you don't find that
$$T=21 \;\; and \;\; J=13$$
It means you made a mistake some minutes ago.


问:
Operators that improve strong convergence to norm convergence?
It is well known that if $K$ is a compact operator on a Hilbert space, and $T_n$ is a sequence of operators converging strongly to $T$, then $K T_n$ converges in norm to $K T$.

Question : are there other operators $K$ that also satisfy this property, but are not compact ?

Thanks

答:
The statement made by the OP that:

It is well known that if $K$ is a compact operator on a Hilbert space,
and $T_n$ is a sequence of operators converging strongly to $T$, then
$K T_n$ converges in norm to $K T$.

is being questioned here.  Nevertheless I have no qualms with the corresponding statement in which $K T_n$ and $KT$ are replaced by
$T_nK$ and $TK$, so I will read the present question with that modification.

My goal it thus to give a negative answer, namely to prove that if an operator $K$ has this property, then it is
necessarily compact.

$\newcommand{\one}{{\mathbb 1}}$
Lemma.  Let  $T$ be a bounded operator on a Hilbert space $H$  which is not compact.  Then there exists an
infinite dimensional subspace $K\subseteq H$, and a constant $c>0$, such that
$$
  \|T\xi \|\geq  c\|\xi \|, \quad \forall \xi \in  K.
  \tag 1
  $$

Proof. Let us assume first that $T$ is positive semi-definite.
For each $n\in \mathbb N$, let $\one_n$ be the characteristic function of the interval $[1/n, +\infty )$, and observe that
$$
  x\one_n(x) \ {\buildrel {n\to\infty}\over\longrightarrow}\ x,
  $$
uniformly on $[0,+\infty )$, so   it follows that
$$
  T\one_n(T) \to T,
  \tag  2
  $$
in norm.
Notice that the   $\one_n(T)$ form an increasing family of self-adjoint projections and we claim that they cannot
all  have finite rank.  This is because otherwise   $T\one_n(T)$ would also be finite rank, and then $T$ would be compact
by   (2).

Choosing any $n$ such that $\one_n(T)$ has infinite rank, let $K$ be the range of $\one_n(T)$, hence an infinite
dimensional subspace.   The proof will then be concluded once we show that
$$
  \|T\xi \|\geq  \frac1n   \|\xi \|, \quad \forall \xi \in  K.
  \tag 3
  $$
To prove this
notice that
$$
  x\one_n(x) \geq  \frac{\one_n(x)}n, \quad \forall x \in  [0,+\infty ),
  $$
so we have that
$$
  nT\one_n(T) \geq  \one_n(T).
  $$
For all $\xi $ in $K$,  a.k.a. the range of $\one_n(T)$, we then have that
$$
  \|\xi \|^2 =
  \langle \xi , \xi \rangle  =
  \langle \one_n(T)\xi , \xi \rangle  \leq
  n\langle T\one_n(T)\xi ,\xi \rangle  =
  n\langle T\xi ,\xi \rangle  \leq
  n\|T\xi \|\|\xi \|,
  $$
from where (3) follows.

Returning to the general case, in which $T$ is only assumed to be a bounded operator, let
$|T|=(T^*T)^{1/2}$, which is positive semi-definite.   Observing that
$$
  \|T(\xi)\| = \|\, |T|(\xi) \, \|,\quad \forall \xi \in  H,
  $$
we  see that any pair $(K,c)$ that works for $|T|$, also works for $T$.
QED

Theorem.    Let $T$ be an operator on a Hilbert space $H$ such that, whenever $\{S_n\}_n$ is a
sequence of bounded operators, strongly converging to zero, one has that $S_nT\to0$ in norm.  Then $T$ is compact.

Proof.  Assuming by contradiction that $T$ is not compact,  use the Lemma to find an infinite dimensional subspace $K\subseteq H$, and a
constant $c>0$,  such that  (1) holds.

Setting $L=T(K)$, is is easy to see that $T$ provides a bounded, linear, invertible map
$$
  \check T:K\to L,
  $$
with bounded inverse  and, in particular, that $L$ is a
closed  infinite dimensional subspace of $H$.

One may then easily  find a sequence of operators $\{S_n\}_n\subseteq B(L)$ which converges to zero
strongly, but not in norm.   Extending each $S_n$ to $H$ by setting it to be zero on $L^\perp$, it is clear that the
extended operators $\hat S_n$ also satisfy $\hat S_n\to0$ strongly, but not in norm.

By the assumption on $T$ we have that $\|\hat S_nT\|\to 0$, so
$$
  \|S_n\| =
  \|S_n\check T\check T^{-1}\|  \leq
  \|S_n\check T\| \|\check T^{-1}\| = $$$$ =
  \|\hat S_nT|_K\| \|\check T^{-1}\| \leq
  \|\hat S_nT\| \|\check T^{-1}\| \to 0
  $$
contradicting the choice of $S_n$.
QED


问:
Inverse function of a polynomial $x(x-1)(x+1) = 0 $
How to get the interval to get the inverse function of the polynomial $$x(x-1)(x+1) = f(x) $$, to show that the function is surjective.

The interval where the function is surjective is $x > 1$ and $x < -1$, how to get inverse on this interval ?

答:
Note that

$$x= y(y-1)(y+1)=y^3-y$$

is cubic in $y$. Use Cardano’s formula to obtain the inverse function below

$$y= \sqrt[3]{\frac x2+\sqrt{\frac{x^2}4-\frac1{27}}} +\sqrt[3]{\frac x2-\sqrt{\frac{x^2}4-\frac1{27}}}
$$
with $x\ne 0$.


答:
Let $t$ be Tim's current age and $j$ be Jim's current age. This is a problem with two equations and two unknowns. The most straightforward way to solve it is to first write out the two equations explicitly, and then solve for one of the variables so you get an equation of the form:

$$ x = \text{some expression not involving $x$} $$

Doing this will allow you to solve for the remaining variable.

Five years ago, Tim was twice as old as Jim.

This is a statement about the ratios of Tim's and Jim's ages five years ago, which we can express directly.

$$ (t-5) = 2 * (j - 5) $$
$$ t - 5 = 2j - 10 $$
$$ t = 2j - 5 $$

In three years Jim's age will be 2/3 that of Tim's.

$$ \frac{j+3}{t+3} = \frac{2}{3} $$
$$ 3j + 9 = 2t + 6 $$
$$ 3j - 2t = -3 $$

We can plug in our earlier equation $ t = 2j - 5 $.

$$ 3j - 2(2j - 5) = -3 $$
$$ 3j - 4j + 10 = -3 $$
$$ -j + 10 = -3 $$
$$ -j = -13 $$
$$ j = 13 $$

Which means $t = 2 * 13 - 5$ or $t = 21$

Let's check the cases

```<code>5 years ago
13-5    8
21-5   16

3 years from now
13+3   16
21+3   24
</code>```

This is indeed our solution.

Title: Five years ago, Tim was twice as old as Jim... Math Question

Body: Five years ago, Tim was twice as old as Jim. In three years, Jim’s age will be $2/3$ that of Tim. What are the ages of Tim and Jim now?

Body:
Let $t$ be Tim's current age and $j$ be Jim's current age. This is a problem with two equations and two unknowns. The most straightforward way to solve it is to first write out the two equations explicitly, and then solve for one of the variables so you get an equation of the form:

$$ x = \text{some expression not involving $x$} $$

Doing this will allow you to solve for the remaining variable.

Five years ago, Tim was twice as old as Jim.

This is a statement about the ratios of Tim's and Jim's ages five years ago, which we can express directly.

$$ (t-5) = 2 * (j - 5) $$
$$ t - 5 = 2j - 10 $$
$$ t = 2j - 5 $$

In three years Jim's age will be 2/3 that of Tim's.

$$ \frac{j+3}{t+3} = \frac{2}{3} $$
$$ 3j + 9 = 2t + 6 $$
$$ 3j - 2t = -3 $$

We can plug in our earlier equation $ t = 2j - 5 $.

$$ 3j - 2(2j - 5) = -3 $$
$$ 3j - 4j + 10 = -3 $$
$$ -j + 10 = -3 $$
$$ -j = -13 $$
$$ j = 13 $$

Which means $t = 2 * 13 - 5$ or $t = 21$

Let's check the cases

5 years ago
13-5    8
21-5   16

3 years from now
13+3   16
21+3   24

This is indeed our solution.


问:
Is a formula with no bound variables an atomic formula even if it contains a Quantifier?
For example my understanding is:

∃(P(x)) contains no bound variables even though we have a quantifier, it also always evaluates to either true or false.

Does this mean that it is an atomic formula?

Thanks

答:
No. An atomic formula has the form $R(t_1,\dots,t_n)$ or $t_1 = t_2$ (where $R$ is an $n$-ary relation symbol and $t_1,\dots,t_n$ are terms). Nothing else is an atomic formula.

Now a complex formula $\varphi$ may be logically equivalent to an atomic formula. But that does not make $\varphi$ itself atomic.


问:
What is the math behind the "IQ Blox" puzzle?
One of my younger relatives showed me this curious puzzle (called "IQ Blox") while I was staying at their place. The premise is as follows: there's a leaflet included with the puzzle listing over a hundred "guides", which are initial positions of the white border bits ("walls") and colored pieces. From there you have to completely fill the board with the remaining pieces.

The leaflet claims there is only one solution for each guide. I was wondering if there's some specific math behind this puzzle. For example, if we represent it as a graph, there's this algorithm that, in combination with that algorithm, gives us all board configurations.

So, the question is:

How does one generate all the guides for a given set of pieces and walls, while also guaranteeing that there's one, and only one, solution? Or am I overthinking this, and it's simply bruteforced?

Thought, the search space seems quite large to me. A quick google search didn't reveal anything. Also, I'm not sure if this question would be more appropriate for Puzzles Stackexchange.

Thank you.

答: At a glance it looks almost like a 2-dimensional partition puzzle.</p>


答: By Heine-Cantor theorem, the image of a compact set by a continuous function, $f: K \to \mathbb R$, is compact. Therefore, $f(K)$ is compact in $\mathbb R$ which is a closed set that can be include in an interval of the form $[a,b]$.  Moreover $f(K)$ admits lower and upper bounds which are in $f(K)$ because it is close.

Title: Continuous Function defined on compact set

Body:
Here's what I'm trying to prove:

let $K \subseteq \mathbb{R}^n$ be a compact nonempty set and let $f: K \to \mathbb{R}$ be a continuous function. Then:

$$\exists a,b \in K: \forall x \in K: f(a) \leq f(x) \leq f(b)$$

Proof Attempt:

We will just prove the existence of $b$ as a similar argument will apply for the existence of $a$. Suppose, for a contradiction, that:

$$\forall b \in K: \exists x \in K: f(x) > f(b)$$

Define a sequence $(x^{(k)})_{k \in \mathbb{N}}$ as follows:

Since $K \neq \varnothing$, take $x^{(1)}$ to be any element in $K$.

Once $x^{(k)}$ has been defined, define $x^{(k+1)}$ by picking the element of $K$ such that $f(x^{(k+1)}) > f(x^{(k)})$. We know that this exists by assumption.

Since $K$ is compact, the sequence above has a convergent subsequence $(x^{(k_m)})_{m \in \mathbb{N}}$ with limit $x$. However, by definition, the sequence $(f(x^{(k_m)}))_{m \in \mathbb{N}}$ cannot converge and this is impossible because $f$ is continuous and must map convergent sequences to convergent sequences. Hence, the desired $b$ must exist. $\Box$

Does the argument above work? If it doesn't, then why? How can I fix it?


问:
Rotation matrix between two frames
Frame A:

Frame B:

(x1 in frame A is perpendicular to the plane created by x2 and x3 and x3 in frame B is perpendicular to the plane created by x1 and x2)

I am trying to find the rotation matrix from A to B. I know that in order to do this, I need to write the unit vectors of B in terms of the coordinates of A so that:
$$R_{AB} = \begin{bmatrix} (R_{AB})_1 & (R_{AB})_2 & (R_{AB})_3\end{bmatrix}$$
How can I write the unit vectors of B in terms of coordinates of A with this little given info?

答:
I still find the description a bit vague. But if we take $e_1=(1,0,0)$, $e_2=(0,1,0)$, and $e_3=(0,0,1)$, then the new coordinate basis should be $e_1' = e_1$, $e_2' = (0,1/\sqrt2,1/\sqrt2)$, and $e_3'=(0,-1/\sqrt2,1/\sqrt2)$. Thus, the rotation matrix should be
$$R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/\sqrt2 & -1/\sqrt2 \\ 0 & 1/\sqrt2 & 1/\sqrt2\end{bmatrix}.$$

Title: Rotation matrix between two frames


问:
For vectors $0 \neq s,y \in \mathbb{R^n}$ with $(s,y)>0$ there exist only one positive definite symmetric matrix $Q$ such that $Qs = y$.
The statement above was given as an exercise to prove. I've done the existence part. The problem with it is that the part about uniqueness might be a typo. Is it unique?</p>

答:
This is clearly not true.

Take $s=y=e_1$ where $e_1$ is the first vector of the canonical basis then any diagonal matrix $Q$ with for diagonal $1, \lambda_2, \dots , \lambda_n$ where $\lambda_i \gt 0$ is symmetric positive definite and such that $Qs=y$.


问:
Can a process with a stochastic drift be a martingale?
In general, a process with a deterministic non-zero drift cannot be a martingale. What if the process has a stochastic drift, that has an expectation of zero? Could such a process ever be a martingale?

I am aware that the zero expectation of the stochastic drift is not a sufficient condition for the process to be a martingale, but is there a general mathematical theorem or a result that would prove that a process with any non-zero drift  (stochastic or deterministic) cannot be a martingale?

I provide a "counterexample" below (process with a zero-expectation stochastic drift, that is NOT a martingale).

Question: Could anyone either provide an example of a process with a stochastic drift that is a  martingale or alternatively reference a theorem or a result that shows that all processes with a drift can't be martingales?

Let: $$X_t:=\int_{h=0}^{h=t}W_hdh+\int_{h=0}^{h=t}1dW_h=\int_{h=0}^{h=t}W_hdh+W_t$$

We clearly have a zero-expectation drift term with:

$$\mathbb{E}\left[\int_{h=0}^{h=t}W_hdh+\right]=\int_{h=0}^{h=t}\mathbb{E}[W_h]dh=0$$

And in fact we have that:

$$\mathbb{E}[X_t|\mathcal{F}_0]=X_0$$

But, $\forall 0< s <t$:

$$\mathbb{E}[X_t | \mathcal{F}_s]=\mathbb{E}\left[\int_{h=0}^{h=s}W_hdh+\int_{h=s}^{h=t}W_hdh+W(t-s)+W(s) \mid \mathcal{F}_s\right]=\\=\int_{h=0}^{h=s}W_hdh+W_s+\mathbb{E}\left[\int_{h=s}^{h=t}W_hdh+W(t-s) \mid \mathcal{F}_s\right]=\\=X_s+\int_{h=s}^{h=t}\mathbb{E}[W_h|\mathcal{F}_s]dh=\\=X_s+W_s\int_{h=s}^{h=t}dh=\\=X_s+W_s(t-s)\neq X_s$$

So $X_t$ is not a martingale.

答:
I will prove it for an Ito process of the type
$$X_t = X_0 + \int_0^ta_udu + \int_0^tb_udW_u, \quad 0 \leq t \leq T. \tag{1}$$

We will need the following two theorems.

Theorem 1. Let M and N be local martingales. Then the sum $M+N$ is also a local martingale.

Theorem 2. Let $c \mathscr{M}_{0,loc}$ denote the space of continuous
local martingale that starts at zero. Let $M \in c \mathscr{M}_{0,loc}$ be of finite variation. Then $M \equiv 0$ a.s.

Answer: Let $X$ be a continuous local martingale of the type SDE(1). Then the drift must be zero.

Proof.
We know the process $\int_0^t b_udW_u, 0 \leq t \leq T,$ is a continuous local martingale. Then the process
$$A_t := \int_0^ta_udu= X_t - X_0  - \int_0^tb_udW_u, \quad 0 \leq t \leq T $$
is a sum of continous local martingales and, by Theorem 1, is also a continuous local martingale. Also, note that $A_0 = 0,$ so $A \in c \mathscr M_{0,loc}^2.$  But the process $A$ is of finite variation, so by Theorem 2, must be zero. This is the same as saying that $\int_0^ta_udu=0$ for all $0 \leq t \leq T.$ Since this integral is defined path by path, by elementary calculus we have that $a_t = 0$ for all $0 \leq t \leq T.$


问:
If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$
If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$

Is this true? This simple algebra should hold in order to finish my problem on Complex Analysis. Computing few numbers suggests that this is true, but I don't seem to be able to prove it.

Body: Hint: If $x$, $y\le1$ then $$0\le(1-x)(1-y)=1-x-y+xy$$


问:
Forcing Bezier Interpolation
I found this very informative site that discusses forcing bezier interpolation and the site gives formulae for calculating the control points so that the curve goes through a set of four points, y0, y1, y2 and y3, for t=0, 1/3, 2/3, 1 respectively.

The formulae given on this site are as follows:

```<code>P0 = y0
P1 = ( -5 * y0 + 18 * y1 -  9 * y2 + 2 y3 ) / 6
P2 = (  2 * y0 -  9 * y1 + 18 * y2 - 5 y3 ) / 6
P3 = y3
</code>```

What I am looking for is a more general form of the formulae for calculating P1 and P2. So instead of having fixed values of t of 1/3 and 2/3, rather to express them as a function of t1 and t2, where t1 is the value of t for point y1 and t2 is the value of t for y3. Can someone please help in deriving these more general formulae?

答:
A couple of comments (but too long for the comment box, I think).

First, if you try somewhat harder, you can actually get a planar cubic Bezier curve to interpolate six points, not just four. As you have realized, when interpolating four points, you have to make up parameter values for two of them. The standard choices are $\tfrac13$ and $\tfrac23$, but there's nothing magic about these -- you can use any values you like, and you'll still be able to solve the simultaneous equations and get a curve. But the shape of this curve is highly dependent on the choice you make, so it would be better if you didn't have to make it. By providing another two points, you can remove this troublesome choice. The details are described in this paper. It's probably more complicated than you need, right now, but maybe useful some day.

Secondly ... Peter Taylor's solution involves solving a system of linear equations (or, equivalently, inverting a 4x4 matrix). This isn't very difficult, but you can avoid it by using Lagrange polynomials. Once you have chosen parameter values for the two interior points, you can simply write down the equations of a cubic polynomial curve that passes through the 4 points. This is a standard technique that's explained in any decent numerical analysis text. There are some simple numerical examples on this Wikipedia page. Once you've done this, you might want to convert the curve from polynomial ("power basis") form to Bezier form. This is straightforward and very fast -- it just requires a matrix multiplication.

Peter's solution is simple and easy to implement, and it probably suits your needs. If you want to try either of the alternatives I suggested, ask again, and I can give you more details.

Title: Forcing Bezier Interpolation


答: Let $x = a^2+b^2, y = c^2+d^2 \implies x+y \le 1+xy \implies (x-1)(1-y) \le 0$, which is true since $ 0 \le x,y \le 1$.

Body:
If $a^2+b^2\leq 1$ and $c^2+d^2<1$, then $a^2+b^2+c^2+d^2\leq 1+a^2c^2+b^2d^2+b^2c^2+a^2d^2$

Is this true? This simple algebra should hold in order to finish my problem on Complex Analysis. Computing few numbers suggests that this is true, but I don't seem to be able to prove it.


答:
No, it's not unique. Consider
$$s = y = \begin{bmatrix} 1 \\ 1\end{bmatrix}$$
Then there are infinitely many positive semi-definite matrices $Q$ that have $s$ as a $1$-eigenvector, namely $Q s = s = y$. To construct these examples, just consider an eigendecomposition of a positive definite $Q$ which has $s$ as a $1$-eigenvector. Then the eigendecomposition is of the form
$S D S^\dagger$
where $s / \|s\|$ is a column of $S$ (which is unitary), the corresponding diagonal entry in $D$ is one, and the rest of the diagonal entries in $D$ are also positive. For example,
$$\left( \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \right) \begin{bmatrix} 1 & 0 \\ 0 & \mu \end{bmatrix} \left(\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\right)$$
satisfies the criterion for any positive $\mu$.

Title: For vectors $0 \neq s,y \in \mathbb{R^n}$ with $(s,y)>0$ there exist only one positive definite symmetric matrix $Q$ such that $Qs = y$.

Body: The statement above was given as an exercise to prove. I've done the existence part. The problem with it is that the part about uniqueness might be a typo. Is it unique?</p>


问:
$ABCD$ is a trapezium where $AB$ $||$ $CD$. Suppose $[\Delta BOC] = \frac{25}{9}$, and let $AB = b$, $CD = a$ where $a < b$.

$ABCD$ is a trapezium where $AB$ $||$ $CD$. Suppose $[\Delta BOC] = \frac{25}{9}$, and let $AB = b$, $CD = a$ where $a < b$. Find the value of $\frac{b}{a}$.

What I Tried: Here is a picture :-

I really have no idea on how to start this. I might proceed by giving variables to the side lengths and use similarity and area, but I am sure this will be length and quite complicated. There is as little information as possible, and I got no idea how to solve it.

Can anyone help me? Thank You.

答:
This is not an answer

As observed in the comments the trapezoid is not unique. See the picture below

$$
...
$$


问:
how to find the range for x for which the equation holds

how to find the value of x for which the equation holds?

答:
You have $\tan(\alpha) = x~$ and $~\cos(\beta) = \frac{1-x^2}{1+x^2}~$ with $2\alpha = \beta.$

Using the formula from my comment,

$$\tan(\beta) = \tan(2\alpha) = \frac{2x}{1-x^2} \implies
\sin(\beta) = \tan(\beta)\times\cos(\beta) = \frac{2x}{1+x^2}.$$

Now use the pythagorean theorem to obtain a constraint on $x$.

$$\sin^2(\beta) + \cos^2(\beta) = 1 \implies
\frac{4x^2}{1 + 2x^2 + x^4} + \frac{1 - 2x^2 + x^4}{1 + 2x^2 + x^4} = 1.$$

Clearly, this equation holds for any value of $x$.

Now contrast the question in your query's title with the question that you posed in your query.  Presumably, the intended answer is that the equation holds for any value of $x$.  Therefore, the allowable range of $x$ is any real number.

Addendum

Responding to the OP's latest comment.

First, consider $x = -1 \implies \alpha = (-\pi/4) \implies \beta = (-\pi/2) \implies 0 = \cos(\beta) = \frac{1 - [-1]^2}{1 + [-1]^2}.$

Therefore, the math clearly works.

However, if your book has adopted the arbitrary convention that the domain of the Arccosine function is $[0,\pi]$, then the sine of $\beta$ is not permitted to be negative.

Therefore, since $~\sin(\beta) = \frac{2x}{1+x^2},~$the only way to satisfy the arbitrary convention is to insist that $x$ be non-negative.


答:
This is the overkill approach.

Theorem: (Lagrange Inversion Theorem)
Suppose $z$ is defined as a function of $w$ by an equation of the form $$z=f(w),$$
where $f$ is analytic at a point $a$, and $f'(a)\ne 0$. Then we may solve for $w$ as
$$w=a+\sum_{n\ge1}\frac{q_n}{n!}(z-f(a))^n,\tag1$$
where $$q_n=\lim_{y\to a}\left[\left(\frac{\partial}{\partial y}\right)^{n-1}\left(\frac{y-a}{f(y)-f(a)}\right)^n\right].\tag 2$$

We may apply this theorem to the function $$z=f(w)=w(w+1)(w-1)=w^3-w.$$
Clearly $f$ is analytic. We can choose $a=0$ because $f'(0)=-1$. Thus
$$w=f^{-1}(z)=\sum_{n\ge1}\frac{q_n}{n!}z^n,$$
with $$q_n=\lim_{y\to 0}\left[\left(\frac{\partial}{\partial y}\right)^{n-1}\frac{1}{(y^2-1)^n}\right].$$
Then, recall that for  we have
$$\frac{1}{(1-q)^\alpha}=\sum_{k\ge0}\frac{(\alpha)_k}{k!}q^k,$$
where $(\alpha)_k=\Gamma(\alpha+k)/\Gamma(\alpha)$ and $\alpha\in\Bbb R$. Then,
$$\frac{1}{(y^2-1)^n}=\sum_{k\ge0}(-1)^n\frac{(n)_{k/2}}{(k/2)!}[2|k]y^k,$$
where
$$[2|k]=
\begin{cases}
1 & \text{if } 2\text{ divides }k\\
0 & \text{otherwise}
\end{cases}
= \frac{1+(-1)^k}{2}.
$$
Letting $p_n(y)=1/(y^2-1)^n$, we have the Taylor series of $p_n$ is
$$p_n(y)=\sum_{k\ge0}\frac{p_n^{(k)}(0)}{k!}y^k,$$
thus $$p_n^{(k)}(0)=(-1)^n\frac{(n)_{k/2}k!}{(k/2)!}[2|k].$$
And since $q_n=p_n^{(n-1)}(0)$, we have
$$q_n=(-1)^n\frac{(n)_{(n-1)/2}(n-1)!}{((n-1)/2)!}[2|n-1],$$
so that $$f^{-1}(z)=\sum_{n\ge1}(-1)^n\frac{(n)_{(n-1)/2}(n-1)!}{n!((n-1)/2)!}[2|n-1]z^n=\sum_{n\ge1}\frac{(-1)^n}{n}\frac{(n)_{(n-1)/2}}{((n-1)/2)!}[2|n-1]z^n.$$
This result is mess, so lets clean it up. Preforming the index shift $j=n-1$, we have
$$\begin{align}
f^{-1}(z)&=\sum_{j\ge0}\frac{(-1)^{j+1}}{j+1}\frac{(j+1)_{j/2}}{(j/2)!}[2|j]z^{j+1}\\
&=\sum_{j\ge0}\frac{(-1)^{2j+1}}{2j+1}\frac{(2j+1)_{j}}{(j)!}[2|2j]z^{2j+1}+\sum_{j\ge0}\frac{(-1)^{2j+2}}{2j+2}\frac{(2j+2)_{(2j+1)/2}}{((2j+1)/2)!}[2|2j+1]z^{2j+2}\\
&=-\sum_{j\ge0}\frac{(2j+1)_{j}}{(j)!}\frac{z^{2j+1}}{2j+1},\tag{3}
\end{align}$$
where we get $(3)$ from the fact that $[2|2j]=1$ and $[2|2j+1]=0$. Then using the definition of $(x)_n$, we have
$$f^{-1}(z)=-\sum_{n\ge0}\frac{\Gamma(3n+1)z^{2n+1}}{\Gamma(2n+2)\Gamma(n+1)}.$$
We are fortunate enough to express this as a hypergeometric series. To do so, write
$$t_n=\frac{\Gamma(3n+1)z^{2n+1}}{\Gamma(2n+2)\Gamma(n+1)}.$$
Then
$$\frac{t_{n+1}}{t_n}=\frac{(n+1/3)(n+2/3)}{n+3/2}\frac{27z^2/4}{n+1}, \qquad t_0=z.$$
Thus
$$f^{-1}(z)=-z\,_2F_1(\tfrac{1}{3},\tfrac{2}{3};\tfrac{3}{2};\tfrac{27}{4}z^2)\qquad |z|<\frac{2}{3\sqrt3}.$$

Edit: In case you're interested, here is some numerical verification of my answer.

Title: Inverse function of a polynomial $x(x-1)(x+1) = 0 $

Body:
How to get the interval to get the inverse function of the polynomial $$x(x-1)(x+1) = f(x) $$, to show that the function is surjective.

The interval where the function is surjective is $x > 1$ and $x < -1$, how to get inverse on this interval ?


问:
Does the limit $\lim\limits_{x \to\infty}x^2 \sin\frac{1}{x}$ not exist?
I used $u$ substitution for the limit $$\lim_{x \to\infty}x^2 \sin\frac{1}{x}$$ and got the limit does not exist by saying $u=\frac{1}{x}$. Is this correct and if so would that mean $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x^2}$ and $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x}$ also don't exist?</p>

答:
$$L=\lim_{x\to\infty}x^2\sin\left(\frac1x\right)$$
now we know that for very small $z$ that $\sin z\approx z$ and so:
$$\sin\left(\frac1x\right)\approx\frac1x$$
$$\therefore L=\lim_{x\to\infty}x^2.\frac1x\to\infty$$


问:
A question on weak convergence
Let $f_n(x)=1+\sin(n\pi x)$. I have to prove that $f_n^k$ converges weakly in $L^1([0,1])$ to
\begin{equation}
c_k:=\underset{a+2b=k}{\sum_{a,b\geq0}}\frac{k!}{a!b!b!}4^{-b}
\end{equation}
for $n\rightarrow+\infty$.

I cannot use Riemann Lebesgue lemma, so I don't know how to do this. Could somebody give me a hint?

答:
Hint. Write

\begin{align*}
f_n^k
&= \left( 1 + \frac{i}{2i}e^{n\pi ix} - \frac{1}{2i}e^{-n\pi ix}\right)^k \\
&= \sum_{\substack{a,b,c\geq 0 \\a+b+c=k}} \binom{k}{a,b,c} \left(\frac{1}{2i}e^{n\pi ix}\right)^b\left( -\frac{1}{2i}e^{-n\pi ix}\right)^c \\
&= \sum_{\substack{a,b,c\geq 0 \\a+b+c=k}} \binom{k}{a,b,c} \frac{1}{2^{b+c}i^{b-c}} e^{n\pi(b-c)ix}
\end{align*}

Title: A question on weak convergence


问:
How do I use MATLAB to solve an Euler Backward problem for a Differential Equation?
Is there a way to use MATLAB to solve an Euler Backward problem when the function I have is a differential equation?

My original problem was to simulate an electromagnetic and an electrical field for a certain interval in R^2 and in this interval an electron would enter with an initial speed in x, y direction. I then plotted the path the electron would take inside the field, but now I'm trying to approximate the path using Euler Backward.

Our differential equation in MATLAB looks like this:

```<code>dy(1)=y(3);
dy(2)=y(4);
dy(3)=y(4)+1;
dy(4)=-y(3)
</code>```

Where y=[x ; y ; u ; v] where u is the derivative of x and v is that of y.

答:
Given the vector-valued differential equation $\mathbf{y}'(t) = \mathbf{f}(\mathbf{y}(t))$, the Euler backward scheme is given by discretizing the time derivative and evaluating $\mathbf{f}$ at the value of $\mathbf{y}$ to be determined:

$$\frac{\mathbf{y}_{n+1} - \mathbf{y}_n}{h} = \mathbf{f}(\mathbf{y}_{n+1})$$

where $h$ is the time step and $t = nh$, so you have

$$\mathbf{y}_{n+1} - \mathbf{y}_n - h\mathbf{f}(\mathbf{y}_{n+1}) = 0$$

which is an implicit equation for $\mathbf{y}_{n+1}$. There are two interesting facts about the case you're trying to solve:

(1) the equation is basically linear, i.e. $\mathbf{f}(\mathbf{y}) = \mathbf{b} + A\mathbf{y}$ where $\mathbf{b}=(0,0,1,0)$ and $A$ is a matrix.

(2) the differential equation for $y_3$ and $y_4$ has no dependence on $y_1$ and $y_2$, implying that you can solve for $y_3$ and $y_4$ separately, and then calculate $y_1$ and $y_2$.

If we consider that $\mathbf{f}$ is linear, then we get the equation

$$\mathbf{y}_{n+1} - \mathbf{y}_n - hA\mathbf{y}_{n+1} - h\mathbf{b} = 0$$

which can be solved explicitly for $\mathbf{y}_{n+1}$, giving

$$\mathbf{y}_{n+1} = (I - hA)^{-1}(\mathbf{y}_n + h\mathbf{b})$$

where $I$ is the identity matrix. This requires finding the inverse of a 4 x 4 matrix which, whilst it is entirely possible using a computer, isn't necessarily desirable for stability reasons. It would be better to just consider the last two components of the equation, the $y_3$ and $y_4$ components. After manually inverting the 2 x 2 matrix you get

$$y_3^{n+1} = \frac{y_3^n + h y_4^n + h}{1+h^2}$$

$$y_4^{n+1} = \frac{y_4^n - hy_3^n}{1+h^2}$$

which you should easily be able to program in Matlab, and then use the values for $y_3$ and $y_4$ to calculate $y_2$ and $y_1$.

Note that I've given you a very specific implementation of backward Euler which is tailored for the problem at hand. A more general backward Euler solver would have to use a root-finding technique such as Newton-Raphson iteration to calculate $\mathbf{y}_{n+1}$ at each step.


问:
Sampling distribution for the mean of a large sample of salaries
Hello I am trying to solve this question. But, I can not find any approach, We know that avg is 600 and the question is asking for 500? and also if we sum the sd and avg we can not come near to the 1000£ on the second part. Could you please help me to solve this question?

Average income of nurses is 600£ and the standard deviation was 266£ . A sample of 1,000 nurses
was randomly chosen to verify if this sample would have a similar mean income as the entire population.
a) Probability that the mean income of the nurses sampled is less than 500£ .
b) Probability that the mean income of the nurses sampled is greater than 1000£ .

答:
Use the Central Limit Theorem: the sample mean $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ converges to
$$
\mathcal{N}(\mu, \sigma^2/n)
$$
in distribution, with $n = 1000$, $\mu = 600$ and $\sigma = 266$.

Therefore let
$$
f_\bar{X}(x) = \dfrac{1}{\sqrt{2\pi\sigma^2}} \cdot e^{\frac{-(x-\mu)^2}{2\sigma^2}}
$$
be the p.d.f. of the normal distribution given above, (a) is
$$
\mathbb{P}[\bar{X} \leq 500] = \int_{-\infty}^{500} f_\bar{X}(x) dx
$$

and (b) is simply
$$
\mathbb{P}[\bar{X} \geq 1000] = 1 - \mathbb{P}[\bar{X} \leq 1000] = 1 - \int_{-\infty}^{1000} f_\bar{X}(x) dx
$$


问:
Hard power series problem

Consider the differential equation $$(1+t)y''+2y=0$$
with the variabel coefficient $(1+t)$, with $t\in \mathbb{R}$.

Set $y(t)=\sum_{n=0}^{\infty}a_nt^n$. What are the first 4 terms in the associated power series?

An Attempt

We write
$$(1+t)y''=\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}$$
$$2y=\sum_{n=0}^{\infty}2a_nt^{n} $$

Combining these three gives us

$$\sum_{n=2}^{\infty}a_nn(n-1)t^{n-2}+ \sum_{n=2}^{\infty}a_nn(n-1)t^{n-1}+ \sum_{n=0}^{\infty}2a_nt^{n}$$

We now want all the powers to be $t^n$.
$$\sum_{n=0}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=0}^{\infty}2a_nt^{n} $$

We finally want the index of summation to all be the same.
$$2a_0+2a_2+ \sum_{n=1}^{\infty}a_{n+2}(n+2)(n+1)t^{n}+ \sum_{n=1}^{\infty}a_{n+1}(n+1)nt^{n}+ \sum_{n=1}^{\infty}2a_nt^{n}$$

Which simplifies to
$$2a_0+2a_2+ \sum_{n=1}^{\infty}\big[a_{n+2}(n+2)(n+1)+a_{n+1}n(n+1)+2a_n \big]t^n$$

Clearly, I'm not getting anywhere close to any of the answers given. I just can't find what I'm doing wrong. I have also tried with Maple, but that didn't give me any real result either.

I hope someone can help me out.

答:
Take $t=0$. You get from the ODE $y^{\prime \prime}(0)+y(0)=0$.

And answer b) is the only valid option providing that at least one is valid.

Title: Hard power series problem


问:
Looking for an example of a function with a Julia set of positive Lebesgue measure.
I am looking for an example of a function with a Julia set of positive Lebesgue measure.
I am certain these exist but cannot find a single example in the papers I have looked at.
The function does not have to be a quadratic, it can be of any form.

Thank you.

答:
There are many known classes of rational functions whose Julia set is the entire Riemann sphere.  For example, it is known that a post-critically finite rational function with no periodic critical points has the entire Riemann sphere as its Julia set.  (This is Corollary 16.5 of Milnor's Dynamics in One Complex Variable.)  Perhaps the simplest such example is the rational function
$$
f(z)=1-2z^{-2}.
$$
This has critical points $\{0,\infty\}$, with $f(0)=\infty$, $f(\infty)=1$, $f(1)=-1$, and $f(-1)=-1$.

Other Julia sets with positive Lebesgue measure are known.  For example, Curt McMullen proved in 1987 that any function of the form $f(z) = \sin(\alpha z+\beta)$ (where $\alpha\ne 0$) or the form $f(z)=\gamma e^z + \delta e^{-z}$ (with $\gamma,\delta\ne 0$) has a Julia set with positive Lebesgue measure.  See:

C. McMullen, Area and Hausdorff dimension of Julia sets of entire functions. Transactions of the American Mathematical Society 300.1 (1987): 329-342. Crossref.

In 2005, Xavier Buff and Arnaud Chéritat proved that there exist quadratic polynomials whose Julia sets have positive Lebesgue measure.  See:

X. Buff and A. Chéritat, Quadratic Julia sets with positive area. Annals of Mathematics (2012): 673-746. JSTOR. arXiv.


问:
How many options are there for 15 student to divide into 3 equal sized groups?
How many options are there for 15 students to divide into 3 equal-sized groups?

Now I know the solution is $\;\dfrac {15!}{5!5!5!3!}\;$ but I can't understand why.

Can anyone please enlighten me?

答:
I actually found a nice way of solving it (After reading the post on the comment):

Say i choose a person first in the line-up, Now that person have $\binom {14}4$ ways of choosing 4 other groups members, Then the next person in the line up has $\binom 94$ ways of choosing other groups members. Finally, the last member has $\binom 44=1$ ways of choosing his group members.

If we multiply all of the above: $\binom {14}4\times$$\binom 94$ and we get the exact same  result as stated. (Just written a bit differently.)


问:
Proving the uniqness of the minimizer
I consider der function $ F:\ \mathbb{R}^n\to \mathbb{R}^n  $ with $ F(x)=\|A\cdot x-b\|_2^2+\lambda \cdot \|x\|_2^2 $.

At first it is $$ \begin{align} F(x)&=\|A\cdot x-b\|_2^2+\lambda \cdot \|x\|_2^2\\&=(x^T\cdot A^T-b^T)\cdot (A\cdot x-b)+\lambda\cdot x^T\cdot x. \end{align}$$

Each component of $ \nabla F(x) $ is $$ \begin{align}\frac{\partial F}{\partial x_i}(x)&=e_i^T\cdot A^T\cdot (A\cdot x-b)+(x^T\cdot A^T-b^T)\cdot A\cdot e_i+\lambda\cdot e_i^T\cdot x+\lambda\cdot x^T\cdot e_i\\&=2\cdot (e_i^T\cdot A^T\cdot (A\cdot x-b)+\lambda \cdot e_i^T\cdot x),\quad i=1,...,n. \end{align}$$
So it is
$$ \nabla F(x)=2\cdot (A^T\cdot (A\cdot x-b)+\lambda\cdot x)=2\cdot ((A^T\cdot A+\lambda\cdot I_n)\cdot x-A^T\cdot b)\stackrel{!}{=}\textbf{0}\in \mathbb{R}^n. $$

Now I want to show that $$ (A^T\cdot A+\lambda\cdot I_n)\cdot x=A^T\cdot b $$

has an unique solution which means that $ A^T\cdot A+\lambda\cdot I_n $ invertible.

I tried two ways:

1.) $ A^T\cdot A $ is symmetric, so it can be
diagonalized it with orhtogonal transformation matrix $ S\in \mathcal{O}(n) $ such that $ A^T\cdot A+\lambda\cdot I_n=S\cdot D\cdot S^T+\lambda\cdot S\cdot S^T=S\cdot (D+\lambda\cdot I_n)\cdot S^T $.
We have $ \underbrace{\det(S)}_{\neq 0}\cdot \underbrace{\det(D+\lambda\cdot I_n)}_{=?} \cdot \underbrace{\det(S^T)}_{\neq 0}. $
The problem is now whether $ D+\lambda\cdot I_n $ is always invertible which I cannot see here.

2.) Suppose $ A^T\cdot A+\lambda\cdot I_n $ is not invertible. Then this matrix has the eigenvalue $ \mu=0 $. Let without loss of generality $ \omega $ be an eigenvalue of $ A^T\cdot A $ with an eigenvector $ v\neq 0 $ such that $ \mu=\omega+\lambda=0 $. We get
$ (A^T\cdot A+\lambda\cdot I_n)\cdot v=A^T\cdot A\cdot v+\lambda\cdot I_n\cdot v=(\omega+\lambda)\cdot v=\mu\cdot v=0\cdot v=0 $. In this case it is $ \lambda=-\omega $ and because of $ \lambda>0 $ the eigenvalue $
\omega $ is stricktly negative. Then $ A^T\cdot A $ is not positive definite.

I get stuck in both ideas and I cannot found annother way.

答:
Your second approach is correct: you show that if $A^\top A + \lambda I_n$ is not invertible, then $A^\top A$ has a strictly negative eigenvalue. But this is a contradiction since $A^\top A$ is positive semidefinite.

Explicitly: if $v$ is an eigenvector of $A^\top A$ with negative eigenvalue $\omega$, then $v^\top A^\top A v = \omega \|v\|^2 < 0$ while $v^\top A^\top A v = \|Av\|^2 \ge 0$, a contradiction.

Your first approach can also be salvaged. $D + \lambda I_n$ is a diagonal matrix, where the eigenvalues of $A^\top A$ appear in the diagonal entries of $D$. Since $A^\top A$ is positive semi-definite, the diagonal entries of $D$ are nonnegative. Since $\lambda$ is strictly positive, the diagonal entries of $D + \lambda I_n$ are also strictly positive, so $D + \lambda I_n$ is invertible.

Title: Proving the uniqness of the minimizer