问:
Bounding the sum of absolute values of an orthogonal matrix's entries
The Problem: Let $A \in O(n)$, the set of $n \times n$ matrices which satisfy $A^t = A^{-1}$. Denote the entries of $A$ by $a_{i,j}$. I wish to show the following bounds:

$$n \le \sum_{i=1}^n \sum_{j=1}^n \left| a_{i,j} \right| \le n \sqrt{n!}$$

Context: This comes up as part of a homework assignment, so I would prefer only nudges in the right direction. I have proved a couple of similar identities for orthogonal matrices $A$:

$$\sum_{i=1}^n \sum_{j=1}^n a_{i,j}^2 = n \qquad \left| \sum_{i=1}^n \sum_{j=1}^n a_{i,j} \right| \le n$$

The first was fairly easy since the sum in question is essentially $\langle A,A \rangle_F$ where $\langle \cdot, \cdot \rangle_F$ denotes the Frobenius inner product, which can also be given by $\langle B,C \rangle_F = \operatorname{trace}(B^t C)$ for real matrices. Using that $A^t = A^{-1}$ made that easy.

The second followed in part from the first, alongside using Cauchy-Schwarz. You can rewrite the sum in terms of the sum of the column vectors instead and apply Cauchy-Schwarz on the dot product with the all-ones vector. You get two sums in the end, one of which is precisely that in part (a).

Granted those are mostly the high points. However, I'm not really sure how to prove either of the desired bounds for the problem I stated at the outset. The triangle inequality by itself isn't enough to get the lower bound of $n$, and I can't quite figure out a framing of the problem or an inner product to use to make use of Cauchy-Schwarz for either bound. I've also struggled to find any other case of someone asking this or proving this elsewhere online, MSE included.

Does anyone have any ideas to lead me in the right direction?

Update: Proving the upper bound is actually quite easy, thanks to a nudge from copper.hat. Define $\mathcal O$ to be the matrix consisting of all ones, and define $\widetilde A$ to be the matrix such that $\widetilde a_{i,j} = |a_{i,j}|$. Then we see that

$$\sum_{i=1}^n \sum_{j=1}^n \left| a_{i,j} \right| = \langle \widetilde A, \mathcal O \rangle_F$$

From here, we just square both sides, apply Cauchy-Schwarz and one of the already-proven identities, and with a little algebra easily show

$$\sum_{i=1}^n \sum_{j=1}^n \left| a_{i,j} \right| \le n \sqrt n$$

This is actually a tighter bound than $n \sqrt{n!}$ (as $n \le n!$ for all $n \in \Bbb Z^+$, trivially).  So this somewhat makes me wonder what method was originally intended.

Still not sure on the lower bound though... But thanks for your insight copper.hat!

答:
$
\newcommand{\ipf}[2]{ \Big \langle #1, #2 \Big \rangle_F}
\newcommand{\ds}{ \sum_{i=1}^n \sum_{j=1}^n}
\newcommand{\AA}{\widetilde{A}}
\newcommand{\aa}{\widetilde{a}}
\newcommand{\OO}{\mathcal{O}}
$
Posting an answer to this since copper.hat and Gerry Myerson were gracious enough to fill me in on enough details to move forward.

The Upper Bound:

Define $\AA$ to be the matrix that is only absolute values of $A$'s entries, i.e. $\aa_{i,j} = |a_{i,j}|$. Denote by $\OO$ the all-ones matrix. Obviously, we can then see that

$$\ds |a_{i,j}| = \ds |a_{i,j}| \cdot 1 = \ipf \AA \OO$$

We can then square both sides and apply Cauchy-Schwarz:

$$\left( \ds |a_{i,j}| \right)^2 = \ipf \AA \OO^2 \le \ipf \AA \AA \ipf \OO \OO$$

The former is clearly given by

$$\ipf \AA \AA = \ds |a_{i,j}||a_{i,j}| = \ds |a_{i,j}|^2 = \ds a_{i,j}^2 = n$$

from one of the equalities I already proved. Meanwhile,

$$\ipf \OO \OO = \ds 1 = n^2$$

more trivially. Therefore, we've seen that

$$\left( \ds |a_{i,j}| \right)^2 \le n \cdot n^2 = n^3 \implies  \ds |a_{i,j}| \le n^{3/2} = n \sqrt n$$

This is a tighter bound than originally desired, but works since $n \le n!$ $\forall n \in \Bbb Z^+$ trivially.

The Lower Bound:

Recall: the columns and rows of orthogonal matrices form orthonormal bases of $\Bbb R^n$. In particular, that will mean that the $j$th column (or row, but I prefer thinking in columns) of $A \in O(n)$, for any $j$ we like, will have unit length under the Euclidean norm, i.e.

$$\sqrt{ \sum_{i=1}^n a_{i,j}^2 } = 1 \implies \sum_{i=1}^n a_{i,j}^2 = 1$$

Note that we're summing entirely nonnegative quantities. Therefore, between the above and the fact that this applies for all $j$, we have

$$a_{i,j}^2 \in [0,1] \implies a_{i,j} \in [-1,1]$$

for all $i,j \in \{1,2,\cdots,n\}$. We also note that $x^2 \le x$ on $[0,1]$ and $x^2 \le -x$ on $[-1,1]$; or, more compactly, $x^2 \le |x|$ on $[-1,1]$. Therefore, now that we know $a_{i,j} \in [-1,1]$ $\forall i, j$, then

$$a_{i,j}^2 \le |a_{i,j}|$$

If we sum over both sides over the full range of $i,j$, then the inequality is maintained (as we're adding only nonnegative terms, all of which fulfill that relation) and we have

$$\ds a_{i,j}^2 \le \ds |a_{i,j}|$$

Note, however: the right-hand sum is precisely the one we want a bound on. Meanwhile, the left-hand one is one for which I already know an exact value: as stated in the question body itself, it is $n$. Thus,

$$n\le \ds |a_{i,j}|$$

giving the desired lower bound.


问:
What are the mean and stdev for a dataset that is approx. lognormally or Weibull-distributed?
Given are particle sizes and corresponding percentages of the total volume. Data:

```<code>    size    vol%
    0.594   0.03
    0.675   0.11
    0.872   0.25
    0.991   0.34
    1.28    0.55
    1.45    0.65
    1.88    0.89
    2.13    1.04
    2.75    1.44
    3.12    1.66
    4.03    2.16
    4.58    2.43
    5.92    3.03
    6.72    3.35
    8.68    4.05
    9.86    4.41
    12.7    5.09
    14.5    5.38
    18.7    5.7
    21.2    5.67
    27.4    5.15
    31.1    4.66
    40.1    3.38
    45.6    2.68
    58.9    1.42
    66.9    0.92
    86.4    0.18
    98.1    0.03
</code>```

Plot using Excel:

Reading up on particle size distribution on WP, I found that this kind of data usually follow lognormal or Weibull-distributions. So I followed some YT tutorials on checking whether that is the case and arrived at this:

So it's not perfect (2nd is Weibull) but I want to follow through with it if possible. However, when I extract the $\mu$ and $\sigma$ from the equation of the regression line, I don't get the original distribution (plot using WA):

Is my approach/are $\mu$ and $\sigma$ correct?

Could I just take $\frac{1}{n}\sum_1^n x\cdot f(x)$ for $\mu$, and from that then calculate $\sigma$ as I would for a discrete distribution using $\sqrt{(\frac{1}{n}\sum_1^n (\mu-x)^2)}$?

答:
It's difficult to say whether you are doing the estimation correctly based on what you have provided.  Lognormal model fitting is very sensitive to outliers.  The usual approach for model fitting is to transform the data to the log scale, then fit a normal distribution to the transformed data.  Then the values for $\mu$ and $\sigma$ are then used to calculate the lognormal mean and variance.  If $Y = e^X$ is lognormal where $X$ is normal with mean $\mu$ and variance $\sigma^2$, then

$$\operatorname{E}[Y] = e^{\mu + \sigma^2/2}, \quad \operatorname{Var}[Y] = (e^{\sigma^2} - 1)e^{2 \mu + \sigma^2}.$$

The fit based on a small amount of data is likely to be very poor.  Based on the regression plot you have included, I would say that the data is not lognormally distributed.  If you include a Weibull plot, I would be able to see whether the Weibull fit is actually better.

ContentLicense: CC BY-SA 4.0


问:
The properties of convex function on the closed unit interval $[0,1]$.
Consider a continuous and convex function $F(x):[0,1]\longrightarrow\mathbb{R}$. I am wondering if

$F(x)$ is continuously differentiable in $[0,1]$

$F(x)$ is of bounded variation in $[0,1]$

$F(x)$ is absolute continuous in $[0,1]$.

The second one is correct, due to this post Proving a convex function is of bounded variation.

However, the remaining two became mysterious to me. Royden's chapter 6 answers them if we have an open interval.

Corollary 17: Let $\varphi$ be a convex function on $(a,b)$. Then $\varphi$ is Lipschitz, and therefore absolutely continuous on each closed, bounded subinterval $[c,d]$ and $(a,b)$

Theorem 18: Let $\varphi$ be a convex function on $(a,b)$. Then $\varphi$ is differentiable except at a countable number of points.

By the Theorem 18, it is hard to believe that $F(x)$ will become differentiable in $[0,1]$. But I cannot find a counterexample. That is, a convex function that is continuous on $[0,1]$ but is not differentiable.

The Corollary 17 gives us pretty nice result, but seems like it does not apply to the closed interval. Is it possible to say that if we have $F(x)$ on $[0,1]$ is convex, then it will be convex on $(-\epsilon, 1+\epsilon)$? and then we can use Corollary 17 to conclude that it is absolutely continuous on $[0,1]\subset (-\epsilon, 1+\epsilon)$.

Thank you!

答:
Given real numbers $a<b$, let us show that a continuous and convex function $F(x):[a,b]\longrightarrow\mathbb{R}$ is absolutely continuous. Since $F$ is a continuous function on a compact set $[a,b]$ it attains its minimum at some point $c\in [a,b]$. Convexity of $F$ implies that $F$ is non-increasing on $[a,c]$ and non-decreasing on $[c,b]$. So it suffices to
consider a case when $F$ is monotonic on $[a,b]$.

Let $\varepsilon>0$ be any number. Since the function $F$ is continuous at $a$ and $b$, there exist $0<\delta'<|b-a|$ such that if $x,y\in [a,b]$ and $|x-a|\le\delta’$, $|y-b|\le\delta’$ then $|F(a)-F(x)|\le\varepsilon/3$ and $|F(b)-F(y)|\le\varepsilon/3$. The monotonicity of $F$ implies that for any family $(x_n,y_n)$ of disjoint open intervals contained in $[a,a+\delta’]\cup [b-\delta’,b]$ we have $\sum |F(y_n)-F(x_n)|\le 2\varepsilon/3$.

By Corollary 17, $F$ is absolutely continuous on $(a+\delta’, b-\delta’)$, so there exists a real number $\delta\le \delta’$ such that for any finite family $(x_n,y_n)$ of disjoint open intervals contained in $(a+\delta’, b-\delta’)$ of the total length at most $\delta$ we have $\sum |F(y_n)-F(x_n)|\le \varepsilon/3$.

The above easily implies that any finite family $(x_n,y_n)$ of disjoint open intervals contained in $[a, b]$ of the total length at most $\delta$ we have $\sum |F(y_n)-F(x_n)|\le \varepsilon$.


问:
Retract of $F_2$ onto $[F_2,F_2]$
I'm trying to show that the commutator subgroup  $[F_2,F_2]$ is not a retract of $F_2$. I was trying to do the proof by contradiction so I assume there is a retract $f: F_2 \to [F_2,F_2]$, then I know this is surjective and then I tried to apply abelianization to get the map $f^{ab}: \mathbb{Z}^2 \to {1}$ but this doesn't give me any contradictions since this map is surjective. Then I was thinking to apply the universal property of quotient groups somehow because I know $F_2/ [F_2,F_2] = \mathbb{Z}^2$ and that is a surjective map but I can't find anything that gives me a contradiction. I'm not sure where to go from here.

答: We can make the previous argument more precise. The group $[F_2,F_2]$ is a free group. It is freely generated by elements of the form $a^n b^m [a,b]b^{-m}a^{-n}$, for $n, m\in \bf Z$, as it is the fundmantal group of the Cayley graph of $\bf Z^2$ viewed as a quotient $F_2/[F_2,F_2]$, so it is not finitely generated.

Title: Retract of $F_2$ onto $[F_2,F_2]$


问:
Give some example of linear functional?
Give  some example  of linear functional ?

My attempt : I know  the definition of linear functional

linear  functional  on $V$ is a  function $ f : V \to F$ such that

$f(a_1 v_1 + a_2v_2)=a_1f(v_1) + a_2f(v_1)$ for all $ a_1,a_2 \in F$ and all $v_1,v_2 \in V$

Here   i don't  know how to find the example of linear functional

But i can find the example of linear transformation  take  $f : \mathbb{R}^3 \to \mathbb{R}^3$ defined

$f(x,y,z)= (x+y,y+z,z+x)$

答:
An example that is close to the example you have of a linear transformation:

$$f(x,y,z)=x+y$$

This is a linear functional on $\mathbb{R}^3$ or, more generally, $F^3$ for any field $F$.

A much more interesting example of a linear functional is this: take as your vector space any space of nice functions on the interval $[0,1]$, for example the space of continuous functions or the space of polynomials or (if you prefer a finite dimensional space) the space of polynomials of degree at most $20$. Now you have a linear functional

$$\phi(f)=\int_0^1 f(x)\,dx$$.

Title: Give some example of linear functional?


问:
Asymptote for $y=\frac {x^2}{kx+1}$
We want to find the asymptote of
$$y=\frac {x^2}{kx+1}$$
as $x\to \infty$.
By synthetic division or binomial expansion, we arrive at $$y=\frac 1k \left(x-\frac 1k\right)$$
which is the correct answer.

However we could also have divided top and bottom by $x$ to get:
$$y=\frac x{k+\frac 1x}$$
and conclude that as $x\to\infty$, $\frac 1x \to 0$, hence $y\to \dfrac kx$ which is close but not the right answer. How do we reconcile this method with the ones mentioned earlier?

Thanks.

答:
Your second analysis is not accurate. You cannot just remove $1/x$ because it goes to zero. You need to find the second term in the expansion at infinity, in this case the constant term. You can do
$$
y=x(k+1/x)^{-1}=\frac x{k}(1+1/kx)^{-1}=\frac x{k}(1-1/kx+\dots)=\frac 1{k}(x-1/k)+\dots
$$

Title: Asymptote for $y=\frac {x^2}{kx+1}$


问:
Does anyone know of a much faster way to solve this polynomial question?
A monic cubic $p(x) $ is divided by $(x^2 +x +1)$ and the remainder is $(2x+3)$. When $p(x)$ is divided by $x(x+3)$, the remainder is $5(x+1)$. Find $p(x)$

So the way I solved it was through using long division - as you can imagine very lengthy, and not very efficient in exam situations. I then equated the remainders with the given remainders. However, in an exam situation, would there be an even faster way? I am open to suggestions.

答: here is a trick: as the polynomial has leading coefficient 1 and degree 3 we have $$p(x)=(x+K)(x^2+x+1)+2x+3$$  Now use the fact that $$p(x)=x(x+3)g(x)+5(x+1)$$ $$p(0)=5(0+1)=5$$ $$K=2$$


问:
Random walk on $\mathbb{Z}^3$ crosses $XY$ plane infinitely often
I need to show that the simple symmetric random walk on $\mathbb{Z}^3$ crosses the $XY$ plane infinitely often. The hint given is to consider the movement relative to the $XY$ plane.

I know that $\mathbb{P}[X_1=(1,0,0)]=...=\mathbb{P}[X_1=(0,0,-1)]=1/6$. I also know that the plane would be denoted by $\{(x,y,0): x,y \in \mathbb{R}\}$. In order to show it crosses it infinitely often, I can try proving that $G(i,j)=\mathbb{E}_i(N(j))=\sum_{n=1}^{\infty} \mathbb{E}_i (\mathbb{1}_{\{j\}}(X_n))$, which is the expected number of times that the Markov Chain visits state $j$ starting from state $i$, goes to infinity.

However, I do not seem to understand how to proceed further. Any help/source would be appreciated!

答:
Hint: consider what happens to the $z$-coordinate during the random walk. It either:

increases by $1$ (probability $1/6$)

stays the same (probability $2/3$)

decreases by $1$ (probability $1/6$)

Showing that the simple symmetric random walk on $\Bbb{Z}^3$ crosses the $xy$-plane infinitely often, then, is equivalent to showing that the random walk on $\Bbb{Z}$ that takes steps of $\pm 1$ with probability $1/6$ each, and otherwise stays in place with probability $2/3$, crosses the point $0$ infinitely often. Can you finish from here?


问:
Help In Understanding Concepts of Double Integral
I would be grateful if anyone could help me understand a concept in double integrals.

I'm attaching a screenshot from a MV Calc textbook. I'm interested in problems 35 and 36.

I have been taught (I thought) that when you have two "z = " curves, you should solve for their intersection and this will give you the level curve on the xy plane.

I have the answer key to these problems. In problem 36, this is exactly what is done. You get a line, y = 1, which forms the upper bound of the region.

But on problem 35, this was NOT the process done. Instead, the region was created only by the two "y = " functions. When you find the intersection of the two planes, I get a line, y = -8/3 + x. But this apparently does not have any impact on the region.

Hmmm.....

I thought you needed to find a level curve of the "z = " function on the xy plane. Indeed, problem 36 did this. Why didn't problem 35?

Here are the photos:

答:
The planes can be rewritten as $z_1=2-x-y$ and $z_2=2x+2y+10$. Then

$$z_2-z_1=3x+3y+8\,,$$

so if $-1\le x,y\le 1$, then $z_2-z_1\ge-3-3+8=2$. If $\langle x,y,z\rangle$ is in side the vertical cylinder between $y=1-x^2$ and $y=x^2-1$, clearly $-1\le x,y\le 1$, so the $z_2$ plane lies entirely above the $z_1$ plane within that cylinder: the intersection of the planes is outside the cylinder and has no effect on the volume in question, which can therefore be obtained by integrating $z_2-z_1$ over the region in the $xy$-plane bounded by the two parabolas.


问:
Quadratic expression with any matrix has quadratic expression with symmetric matrix equivalent
Given $x \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ show that there exists a symmetric matrix $B \in \mathbb{R}^{n \times n}$  (which means $B^T = B$) for all A such that:
$$x^TBx=x^TAx$$

This statement totally makes sense to me when I try it with different examples but not sure how to formally prove it.

答: You just have to balance the coefficients of the quadratic form given by $A$. Just take $b_{ij}=(a_{ij}+a_{ji})/2$.


问:
Ordinary or Partial differential equation
Pic of the equation

$$\frac{\partial y}{\partial x}+\frac{\partial y}{\partial x}=0$$
I saw this in an imageboard site and it bothers me. OP said it's a valid equation, if so then $y$ is constant. But in that format, doesn't it violates the rule for partial derivative, where it should have a one dependent variable and two or more independent variables. So it lingers me to say that it may be an ODE.

答:
Probably there is a typo in the equation
$$\frac{\partial y}{\partial x}+\frac{\partial y}{\partial x}=0$$
If the equation was a correct PDE at least two variables would be involved for example $x$ and $t$. The PDE would be :
$$2\frac{\partial y(x,t)}{\partial x}=0$$
Writting $y(x,t)$ makes clear that a second variable exists and makes clear what is its name.

Then the general solution is not $y=$constant but is :
$$y(x,t)=f(t)$$
where $f$ is an arbitrary function. The case $y=f(t)=$constant is not the general solution but is only a particular solution.

However all the above is without interest. Most probably the PDE without typo is :
$$\frac{\partial y}{\partial x}+\frac{\partial y}{\partial t}=0$$
and the general solution would be $\quad y=F(x-t)\quad$ where $F$ is an arbitrary function (to be determined according to some boundary condition).


问:
How to find if particle is moving in clockwise direction or not?
I've been given a vector equation in the form of r(t), how do i find if the particle is going in clockwise direction or in anticlockwise direction?</p>

答: I assume your question is about a motion on the plane. You can just look at $r'(t)\times r''(t)$. This vector is perpendicular to the plane. Assuming you choose a right framework in 3D, it is parallel to your third unit vector $\mathbf{k}$. If its third coordinate is positive, the motion is counterclockwise, otherwise it is clockwise. This is because $r''(t)$ always points to the side where the center of curvature is located.


问:
Evaluate $\iint_{S}xz\,dy\,dz$ where $S=\{(x,y,z)\mid x\geq0, y\geq0, z\geq0,2x+2y+z=2\}$
Problem

Let $S=\{(x,y,z)\mid x\geq0, y\geq0, z\geq0,2x+2y+z=2\}$, and $a=(2,2,1)$ be a normal vector oriented outside. Evaluate the surface integral $$\iint_{S}xz\,dy\,dz$$

I am having a hard time interpreting this question since every surface integral I have ever seen was in a form $\iint_{S}f(x,y,z)\,dS$. I don't even know what this integral even means.

If it was in a form $\iint_{S}f(x,y,z)\,dS$ then I can calculate using the method written here, but I am not even sure what is the first step.

What would be the first step to solve this question? Any help is appreciated.

By the way, the normal vector $a$ may be unnecessary to the question. If it is not required to solve it, you can ignore it.

答:
All it means is you have a scalar field $f(x,y,z) = xz$ instead of a vector field.

Your surface is a plane given by $2x+2y+z=2 \,$ where $(x\geq0, y\geq0, z\geq0)$.

Given the integral is wrt $y, z$, rewrite your plane in the form $g(y,z) = x = 1 - y - \frac{z}{2} \,$.

$\displaystyle \frac{\partial g}{\partial y} = - 1, \frac{\partial g}{\partial z} = - \frac{1}{2}$

Based on your parametrization, your projected triangle in YZ plane $(x = 0)$ gives you upper bound of $y$ as $(1-\frac{z}{2})$ and upper bound of $z$ as $2$.

Your integral becomes $\displaystyle \iint_S (1 - y - \frac{z}{2}) z \, \sqrt{1 + (\frac{\partial g}{\partial y})^2 + (\frac{\partial g}{\partial z})^2} \, dy \, dz$


答:
Let $\gamma_i\in C[a,b]$ and $V=C^n [a,b]$ and $F:V\to\Bbb{R}$.Then,
$$F(v)=\int_a^b [\gamma_0(x)v(x)+\gamma_1(x)h'(x)+...+\gamma_nh^{n}(x)]dx$$ is a linear functional.

Title: Give some example of linear functional?

Body:
Give  some example  of linear functional ?

My attempt : I know  the definition of linear functional

linear  functional  on $V$ is a  function $ f : V \to F$ such that

$f(a_1 v_1 + a_2v_2)=a_1f(v_1) + a_2f(v_1)$ for all $ a_1,a_2 \in F$ and all $v_1,v_2 \in V$

Here   i don't  know how to find the example of linear functional

But i can find the example of linear transformation  take  $f : \mathbb{R}^3 \to \mathbb{R}^3$ defined

$f(x,y,z)= (x+y,y+z,z+x)$


问:
Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$
If $a,b,c \in R$ find the minimum value of $$E=\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$$
My attempt:
Let $x=a-b,y=b-c,z=c-a$

So $x+y+z=0$

We need to minimize $$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{z}\right)^{4}+\left(\frac{z}{x}\right)^{4}$$
$\implies$
$$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{x+y}\right)^{4}+\left(\frac{x+y}{x}\right)^4$$
Letting $\frac{x}{y}=t$ we get:
$$E=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)^4}{t^4}$$
Now i tried using derivatives, but very tedious.

Body:
The minimum of $E = E(t) = t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}, \ t \ne 0, -1$ admits a closed form:
$$E_{\min} = \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}}
+ \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}.$$
Note: $E_{\min} \approx 12.5037895937134016209875697987512$.

Proof: Denote
$$Q = \frac{1}{6}\sqrt[3]{98972 + 3396\sqrt{849}}
+ \frac{1}{6}\sqrt[3]{98972 - 3396\sqrt{849}} + \frac{7}{3}.$$
We have
$$Q^3-7Q^2+3Q-898 = 0, \quad Q-7 > 0, \quad Q^2 - 7Q + 3 > 0. \tag{1}$$

We have
\begin{align}
&[E^2 + (Q-7)E + Q^2-7Q+3](E - Q)\\
=\ & E^3 - 7E^2 + 3E - Q^3 + 7Q^2 - 3Q\\
=\ & E^3 - 7E^2 + 3E - 898\\
=\ & \left(t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}\right)^3 - 7\left(t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}\right)^2\\
&\qquad  +
3\left(t^4 + \frac{1}{(t+1)^4} + \frac{(t+1)^4}{t^4}\right) - 898\\
=\ & \frac{1}{(t+1)^{12}t^{12}}f(t)[g(t)]^2\\
\ge\ & 0 \tag{2}
\end{align}
where
\begin{align}
f(t) &= t^{18} + 6 t^{17} + 19 t^{16} + 40 t^{15} + 57 t^{14} + 60 t^{13} + 93 t^{12} + 308 t^{11}\\
 &\quad + 901 t^{10} + 1832 t^9 + 2650 t^8 + 2836 t^7 + 2316 t^6 + 1466 t^5\\
 &\quad + 717 t^4 + 264 t^3 + 70 t^2 + 12 t + 1, \\
g(t) &= t^9 + 3t^8 + t^7 - 3t^6 + t^5 + 2t^4 - 10t^3 - 13t^2 - 6t - 1,
\end{align}
and it is not difficult to prove that $f(t) > 0$ for all $t$ in $\mathbb{R}$.

Since $E^2 + (Q-7)E + Q^2-7Q+3 > 0$, from (2), we have $E \ge Q$.
Also, $g(t) = 0$ has at least one real root, and thus the minimum of $E$ is exactly equal to $Q$. We are done.


答:
You're misinterpreting what it means to find the asymptote of a function. It's NOT the same as finding the limit of the function.

And in your second attempt, you apparently set out to find the limit of the function, $\lim\limits_{x\to\infty}y(x)$ — which, as I said, is not the same thing, and which you didn't do correctly. Let me reiterate: once you said that you want to see what happens to $y$ as $x\to\infty$, you're talking about $\lim\limits_{x\to\infty}y(x)$. But if you're finding this limit with respect to $x$, you can't leave some of the $x$'s still in the "answer". As $x\to\infty$, not only does $1/x\to0$ in the denominator, but also $x\to\infty$ in the numerator; and as a result we find that
$$\lim_{x\to\infty}y(x)=\lim_{x\to\infty}\frac{x}{k+\frac{1}{x}}=\frac{\infty}{k+0}=\infty,$$
which is certainly true, but not particularly useful.

Finding an asymptote for a function $y(x)$ means that we need to find a linear function $ax+b$ such that $\lim\limits_{x\to\infty}[y(x)-(ax+b)]=0$. Your second approach doesn't address this question.

Title: Asymptote for $y=\frac {x^2}{kx+1}$

Body:
We want to find the asymptote of
$$y=\frac {x^2}{kx+1}$$
as $x\to \infty$.
By synthetic division or binomial expansion, we arrive at $$y=\frac 1k \left(x-\frac 1k\right)$$
which is the correct answer.

However we could also have divided top and bottom by $x$ to get:
$$y=\frac x{k+\frac 1x}$$
and conclude that as $x\to\infty$, $\frac 1x \to 0$, hence $y\to \dfrac kx$ which is close but not the right answer. How do we reconcile this method with the ones mentioned earlier?

Thanks.


答:
You only need to consider the $z$-axis.

Imagine the random walk only crosses the plane a finite number of times. Then all remaining walks would have either more ups than downs, or vice versa.

This gives Dyck paths, and the probability of a Dyck path over all paths tends to zero.

Title: Random walk on $\mathbb{Z}^3$ crosses $XY$ plane infinitely often

Body:
I need to show that the simple symmetric random walk on $\mathbb{Z}^3$ crosses the $XY$ plane infinitely often. The hint given is to consider the movement relative to the $XY$ plane.

I know that $\mathbb{P}[X_1=(1,0,0)]=...=\mathbb{P}[X_1=(0,0,-1)]=1/6$. I also know that the plane would be denoted by $\{(x,y,0): x,y \in \mathbb{R}\}$. In order to show it crosses it infinitely often, I can try proving that $G(i,j)=\mathbb{E}_i(N(j))=\sum_{n=1}^{\infty} \mathbb{E}_i (\mathbb{1}_{\{j\}}(X_n))$, which is the expected number of times that the Markov Chain visits state $j$ starting from state $i$, goes to infinity.

However, I do not seem to understand how to proceed further. Any help/source would be appreciated!


问:
volume between two surfaces
In my textbook they solve this problem like this, Find volume between two surfaces $\quad$  $x^2+y^2-1\leq z \leq 2+\frac{2}{3}(x^2+y^2), \ \ z\geq0$

They have called the first region $D_1=x^2+y^2-1=0$, and second region $D_2=x^2+y^2=9$ and then they solve this problem with two integral
$$\iint _{D_1}   \left(2+\frac{2}{3}(x^2+y^2)-0\right) dxdy + \iint_{D_2} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right)dxdy$$

My question is that why they counted the first integral? and why they divided the region into two parts?

Thanks in advance

答: Projection on $Oxy$ of intersection of paraboloids  $z=x^2+y^2-1$ and $z= 2+\frac{2}{3}(x^2+y^2)$  is $x^2+y^2=9$, so first integral is volume of cylinder with projection $x^2+y^2=1$ and "roof" $z= 2+\frac{2}{3}(x^2+y^2)$. Second integral is difference between two paraboloids, but, they should take it over $D_2 \setminus D_1$.

Title: volume between two surfaces


答:
Note that $A = \dfrac 12(A + A^T) + \dfrac 12(A - A^T)$.

Since $x^TAx$ is a scalar, then $x^TAx = (x^TAx)^T = x^TA^Tx$

Let $B =  \dfrac 12(A + A^T)$. Then

\begin{align}
   x^TAx
      &= \dfrac 12x^T(A + A^T)x + \dfrac 12x^T(A - A^T)x \\
      &= x^TBx + \dfrac 12x^TAx - \dfrac 12x^TA^Tx \\
      &= x^TBx + \dfrac 12x^TA^Tx - \dfrac 12x^TA^Tx \\
      &= x^TBx
\end{align}

Title: Quadratic expression with any matrix has quadratic expression with symmetric matrix equivalent

Body:
Given $x \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ show that there exists a symmetric matrix $B \in \mathbb{R}^{n \times n}$  (which means $B^T = B$) for all A such that:
$$x^TBx=x^TAx$$

This statement totally makes sense to me when I try it with different examples but not sure how to formally prove it.


问:
For the probability function p(n) which is recursive, how to obtain the maximum n under the condition p(n)≤0.1?
here is the recursive function

$p(1) = 0$

$p(n+1) = (1-10^{-8}n)p(n)+10^{-8}n$

I want to get the maximum n that makes $p(n)≤0.1$

I tried to get the solution of the function but it failed. It seems possible to use something like Chernoff Bounds,Chevyshev’s Inequality,Markov’s Inequality to calculate the bound of n, but I don't know how to start.

The original question is to choose 8 numbers from $0-9$ arbitrarily, arrange and combine them in order. Find the largest n that after repeating n times, probability of duplicates in n numbers$≤0.1$

(I want to find largest n as possible,under the premise of computability, without computer calculation. The answer may not be the number exactly at the critical point, but the number close to the critical point in the calculable range)

Can anyone help me?

答:
Using a computer, the maximum value of $n$ for which $p(n) \le 0.1$ is $$p(4590) \approx 0.0999628 \\
p(4591) \approx 0.100004.$$

Let $c = 10^{-8}$.  Note that the given recursion may be written as $$1 - p(n+1) = (1 - cn)(1 - p(n)).$$

This suggests defining an auxiliary sequence $b(n) = \log (1 - p(n))$, hence
$$\log(1 - p(n+1)) = \log (1 - cn) + \log(1 - p(n))$$ or $$b(n+1) = \log(1 - cn) + b(n)$$ and we get the telescoping sum $$b(n) = \sum_{k=0}^n \log(1 - cn) = \log \prod_{k=0}^n (1 - ck),$$ consequently $$p(n) = 1 - \prod_{k=1}^{n-1} (1 - ck).$$


问:
Conditional expectation stable by passing into the limit in $L^{1}$?
Does the conditional expectation converges when the sigma-algebre converges ?
e.g:
$$Y \in L^{1}, (X_{n})_{n \ge 1} \xrightarrow[n \to \infty]{L^{1}}  X_{\infty}
 \Rightarrow \mathbb{E}[Y | X_{n}] \xrightarrow[n \to \infty]{L^{1}} E[Y | X_{\infty}]$$ ?

My intuition is to say Yes: but I don't know how to properly prove it.

答:
There are statements of the type "the conditional expectation converges if the $\sigma$-algebras converge" (e.g. Lévy's upward theorem). The implication which you stated is, however, in general false.

Consider $(0,1)$ with Lebesgue measure and $Y:=1_{(0,1/2)}$. Set $X_n=0$ if $n$ is even and $X_n = \frac{1}{n} 1_{(0,1/2)}$ if $n$ is even. Clearly, $X_n \to 0$ in $L^1$. Moreover, $\mathbb{E}(Y \mid X_n) = \mathbb{E}(Y)=\frac{1}{2}$ if $n$ is even. If $n$ is odd, then

$$\mathbb{E}(Y \mid X_n) = 1_{(0,1/2)} c_1 + 1_{[1/2,1]} c_2$$

where $$c_1 := 2\mathbb{E}(Y\cdot 1_{(0,1/2)}) \qquad c_2 = 2 \mathbb{E}(Y\cdot 1_{[1/2,1)}).$$
By definition of $Y$, we have $c_1 = 1$ and $c_2=0$; thus, $\mathbb{E}(Y \mid X_n) =  1_{(0,1/2)}$. Consequently, we have shown that

$$\mathbb{E}(Y \mid X_n) = \begin{cases} \frac{1}{2}, & \text{$n$ is even} \\ 1_{(0,1/2)}, & \text{$n$ is odd}, \end{cases}$$

and this sequence clearly fails to converge as $n \to \infty$.


问:
Conceptual doubt in finding the range of values in this question
The question is to find The set of values of $\lambda$ such that $\tan^2\theta+\sec\theta=\lambda$

By forming a quadratic in sec and using the quadratic equation I get
$$\sec\theta=\frac{-1 \pm \sqrt(5+4\lambda)}{2}$$

Now applying range conditions for sec I get two inequalities

$\lambda\geq1$ and $\lambda\geq-1$

But the answer is only $\lambda\geq1$

I'm not able do understand why lambda can't have values in between -1 and 1 as it is also a solution.

Thanks in advance.

答:
The range of values of $\theta$ in your problem is probably restricted to $(-\pi/2,\pi/2)$. Otherwise for example if $\theta=\pi$ your function is equal to $-1$. Assuming that, the values of the secant are in $[1,\infty)$. In terms of the secant, your LHS is
$$
\sec^2\theta+\sec\theta-1
$$
a quadratic function of $\sec\theta$. This quadratic function $t\to g(t)= t^2+t-1$ is increasing on $(-1/2,+\infty)$. Therefore the image of $[1,\infty)$ is $[g(1),\infty)=[1,\infty)$.


Title: volume between two surfaces

Body:
In my textbook they solve this problem like this, Find volume between two surfaces $\quad$  $x^2+y^2-1\leq z \leq 2+\frac{2}{3}(x^2+y^2), \ \ z\geq0$

They have called the first region $D_1=x^2+y^2-1=0$, and second region $D_2=x^2+y^2=9$ and then they solve this problem with two integral
$$\iint _{D_1}   \left(2+\frac{2}{3}(x^2+y^2)-0\right) dxdy + \iint_{D_2} \left(2+\frac{2}{3}(x^2+y^2) -(x^2+y^2-1)\right)dxdy$$

My question is that why they counted the first integral? and why they divided the region into two parts?

Thanks in advance

Body: $$\int\limits_{r=0}^3 2 \pi r (( 2 + 2 r^2/3) - \max[0,(r^2 - 1)])\ dr = 13 \pi$$


问:
How to find a single forumla for a series
I am putting pricing strategy for a product

This product for 1 user is for $100

for 2 user is for $190

for 3 user is for $271

and so on

there for the equation will be

1 user = 100

2 users = $100 + 100\times0.9$

3 users = $100 + 100\times0.9 + 100\times0.9^2$

4 users = $100 + 100\times0.9 + 100\times0.9^2 +  100\times0.9^3$

..

There for for $N$ users the price will be

$N$ users = $100 (1 + 0.9 + 0.9^2 + 0.9^3 + ... + 0.9^N)$

How to put this in a single formula so I do not need to calculate all $N-1$ users first?

答:
You can solve it with the geometric formula. Therefore, you get
$$ \sum_{k=1}^{n} 100 \cdot 0.9^{k-1} = 100\cdot \frac{1-0.9^{k}}{1-0.9} $$


问:
On the proof of the Stochastic Exponential by Protter (Thm 37 of Ch 2)
This is a theorem on the stochastic exponential from Protter's Stochastic Integration and Differential Equations. I have seen some questions about this theorem and the book from the site, but not on these questions which I believe are not so easy to understand as written in the proof. I would greatly appreciate it if anyone who has read through this text or know the subject well could enlighten me on these questions.

Here, $X$ is cadlag, adapted, semimartingale and $\Delta X_s = X_s - X_{s-}$.

We define the quadratic covariation process $[X,Y]$ of two semimartingales $X$ and $Y$ as the unique adapted cadlag process with paths of finite variation on compacts that makes $XY - [X,Y]$ a local martingale and can be computed by $$[X,Y] = X_0 Y_0 + \lim_{n \to \infty} \sum_i (X^{T_{i+1}^n} -X^{T_i^n})(Y^{T_{i+1}^n} -Y^{T_i^n})$$ where convergence is uniform on compacts in probability for a sequence of random partitions $\sigma_n$ tending to identity and $\sigma_n$ is the sequence $0=T_0^n \le T_1^n \le \cdots T_{k_n}^n$ with $T_i^n$ stopping times. Tending to identity means that $\lim_n \sup_k T_k^n = \infty$ and $\Vert \sigma_n \Vert = \sup_k |T_{k+1}^n - T_k^n| \to 0$ a.s.

Moreover, $[X,Y]^c$ denotes the path by path continuous part of $[X,Y]$.

And a semimartingale is called a pure jump if $[X,X]^c = 0$.

I have attached the theorem and proof at the bottom.

Questions:
It is written in the proof that clearly the product $$\Pi_{s \le t} (1+\Delta X_s) \exp(-\Delta X_s)$$ is cadlag, adapted. But why is this infinite product cadlag, adapted? Adapted I think follows since this is just a limit of adapted processes, but I cannot see how to show that this is right continuous with left hand limits.

Second, the proof shows that since $X$ has cadlag paths, it has only a finite number of $s$ such that $|\Delta X_s| \ge 1/2$ on each compact interval for each fixed $\omega$, and so it suffices to show that $$V_t = \Pi_{0 < s \le t} (1+\Delta X_s 1_{|\Delta X_s| <1/2}) \exp (-\Delta X_s 1_{|\Delta X_s | <1/2})$$ converges and is of finite variation.

Convergence is clear since we need only multiply finite terms to get $Z_t$. But how do we get that $Z_t$ is also of finite variation if $V_t$ is?

In the second part of the proof uses Ito's formula (or the Change of Variables) by setting $K_t = X_t - \frac{1}{2}[X,X]_t^c$ and $S_t = \Pi_{s \le t} (1+\Delta X_s) \exp(-\Delta X_s)$, and $f(x,y) = ye^x$ and set $Z_t = f(K_t, S_t).$

In going to the second equality in (*), it is claimed that $S$ is a pure jump process and so $[K,S]^c = [S,S]^c = 0$. But I cannot prove how to show that $S$ is a pure jump process, i.e. its quadratic variation has no continuous part, and why does this give $[K,S]^c=0$?

In fact, I cannot see where he uses $[K,S]^c = 0$ since the only thing that changes in the two forms is that $\frac{1}{2}\int_{0+}^t Z_{s-} d[K,K]_s^c$ becomes $\frac{1}{2} \int_{0+}^t Z_{s-} d[X,X]_s^c$, so it seems like we only need that $[K,K]_s^c = [X,X]_s^c$. Where does $[K,S]^c$ come into play here?

Finally, it is claimed that $Z_s = Z_{s-} (1+ \Delta X_s)$. Why does this hold?

I would greatly appreciate some help with these questions.

答:
I will present a more detailed argument of the cadlag-adapted property of $\prod_{s \leq t} (1+\Delta X_s)\exp(-\Delta X_s)$.

For adaptedness, fix a time point $T$ and consider the above term at $T$. Note that for each $s \leq T$, the process $(1+\Delta X_s)\exp(-\Delta X_s)$ is $\mathcal F_T$ measurable, because $X_s$ and $X_{s^-}$ are both measurable and continuous functions of adapted processes are adapted.

Now, we want to show that the infinite product is adapted : for this, we use an exercise from chapter $1$ of the same book, that states that a cadlag function on a compact interval will , for any $\epsilon >0$ have only finitely many jumps of size greater than $\epsilon$.

With this in mind, we define for $\epsilon > 0$ the process $X^{T}_{\epsilon}=\prod_{s \leq T : |\Delta X_s| > \epsilon} (1 + \Delta X_s)\exp(-\Delta X_s)$. Then $X^T = \lim_{\epsilon \to 0} X^T_{\epsilon}$.

Why is $X_{\epsilon}^T$ measurable? Let me go in detail here : we'll define $X^T_{\epsilon , n}$ random variables now in a logical way. First throw out the set on which cadlag is violated. On the rest, for each $\omega$ in this good set, there is a finite set $\{k_{1,\omega},...,k_{m_{\omega},\omega}\}$ (sorted ascending) depending on $\omega$ such that the set of jumps is bigger than $\epsilon$. $X^T_{\epsilon,n}(\omega)$ is the product of $(1+\Delta X_s)\exp(-\Delta X_s)$ over the first $n$ elements in the jump set.

Now clearly $X^T_{\epsilon,n}$ is $\mathcal F_T$ measurable, and then we can take $n \to \infty$ to get $X^T_{\epsilon}$ being measurable, and finally $\epsilon \to 0$ to finish it.

The cadlag was easier and I had got things wrong. We have the process $M_t = \prod_{s \leq t} (1+\Delta X_s)\exp(- \Delta X_s)$ which we want to show is cadlag. I don't know why I was trying to show each term was cadlag, because it most certainly is not.

In fact, we just go by definition. Let $t_n \downarrow t$. Then we know that $|\frac{M_{t_n}}{M_t}| = \prod_{t < s \leq t_n} (1+\Delta X_s)\exp(- \Delta X_s)$. First , consider for each $\epsilon > 0$ the product $Y^{\epsilon}_n = \prod_{t < s \leq t_n : |\Delta X_s| > \epsilon} (1+\Delta X_s)\exp(- \Delta X_s)$. This product is finite for each $\omega$, so clearly as $n \to \infty$ this product goes to $1$ for each $\omega$, that is $\lim_{n \to \infty} Y^n_{\epsilon} = 1$ a.s. so $\lim_{\epsilon \to 0} \lim_{n \to \infty} Y^{\epsilon}_n = 1$ a.s. Now, we have $\lim_{n \to \infty} |\frac{M_{t_n}}{M_t}| = \lim_{n \to \infty}\lim_{\epsilon \to 0} Y^{\epsilon}_n$.

Note that for each $x$ we have $(1+x)e^{-x} \leq 1$. Therefore, increasing the number of terms in the product defining $M_t$ decreases it. This leads to monotonicity in $Y_{\epsilon}^n$ : as $\epsilon \to 0$ the number of terms increases, so $Y_{\epsilon}^n$ increases. Similarly, as $n$ increases the number of terms decreases so the product increases. Finally, using the usual limit exchange theorem for monotone sequences, it follows that both double limits exist and are equal.

I want you to see if you can argue left limit existence similarly.

We are trying to show that $Z_s$ is a semimartingale. The product of two semimartingales is a semimartingale, therefore we just need to show that the two processes $\exp(X_t - \frac 12[X,X]_t^c)$, and the infinite product discussed above, are semimartingales.

The fact that the former is a semimartingale follows from the fact that $X_t - \frac 12[X,X]_t^c$ is a semimartingale, and $\exp$ is a $C^2$ function so the Ito formula applies.

That the latter is a semimartingale, and furthermore a quadratic pure jump one, is proved using theorem $26$, for which we only need to show that the infinite product is of finite variation on compacts.

To show that the infinite product is of finite variation on compacts, the argument breaks the infinite product into two parts , one of which consists of only finitely many terms a.s. and the other is $V_t$.

The finite part is of finite variation, because that is a monotone bounded process (as $s$ goes up , the number of terms increases and each term is $\leq 1$, with boundedness following from the finiteness of terms). That $V_t$ is of finite variation is explained in the text. The product of two processes of finite variation is of finite variation as well, so this follows.

The confusion is that $V_t$ being of finite variation implies that $Z_t$ is. This is not true : once $V_t$ is of finite variation, then its product with that finite part is a semimartingale (which is the entire jump part), and the rest follows from the fact that a product of semimartingales is one too. That , and Theorem $26$ is why we do this entire maneuver.

Then for the Ito formula we write $f(K_t,S_t) = e^{K_t}S_t$ where $K$ and $S$ are the two parts as specified. As mentioned earlier, theorem $26$ tells us that $S$ is a quadratic pure jump process i.e. $[S,S]^c = 0$.

Now, look at the proof of theorem $28$, which tells you that if $[X,X]^c = 0$ for a semimartingale $X$, then $[X,Y]^c = 0$ for any semimartingale $Y$. Thus, since $[S,S]^c  = 0$ we get $[K,S]^c = 0$.

Next, we look at the integration steps. Note that  we want $Z_0 = 1$ so that's why we subtract $1$.

The first equality $Z_t - 1 = ...$ follows from the Ito multidimensional rule (Theorem $33$), with $f$, $K_t$ and $S_t$ as given. The key point is that in the multidimensional rule we have the cross variations $d[X^i,X^j]^c_s$, and since $[K,S]^c = [S,S]^c = 0$(This is the only point in the entire computation that $[K,S]^c = 0$ is used) we know that these particular terms vanish. That is why there's only one integral with a $[K,K]^c_s$ term. The rest of the terms are straightforward from the formula, and usual partial differentiation rules.

For the second equality, nothing changes except expressing $dK_s$ and $d[K,K]^c_s$ in terms of $dX_s$ and $d[X,X]^c_s$, each of which are straightforward applications of the Ito Formula, along with the fact that $d[X,[X,X]^c] = 0$ and $d[[X,X]^c,[X,X]^c] = 0$. (Which is proved in the edit).

Now, the fact that $Z_{s} = Z_{s^-}(1+\Delta X_s)$ : note that $Z_{s^-} = e^{K_{t^-}} S_{t^-}$, and therefore $\frac{Z_s}{Z_s^-} = e^{K_t - K_{t^-}}(\frac{S_t}{S_{t^-}}) = e^{\Delta K_t}\frac{S_t}{S_{t^-}}$ (The ratios can be taken as the definitions indicate a.s. positivity of the stochastic processes involved).

Now, $\Delta K_t = \Delta X_t -  \frac 12 \Delta [X,X]^c_t = \Delta X_t$ since the other term is zero by continuity of the stochastic process. Note that $S_{t^-} = \prod_{0 < s \boxed{\color{blue}{<}} t} (1+ \Delta X_s)\exp(- \Delta X_s)$, so the ratio $\frac{S_t}{S_{t^-}}$ is just the term for $t$, which is $(1+ \Delta X_t)\exp(- \Delta X_t)$.

So the answer is just $e^{\Delta X_t} (1+\Delta X_t)\exp(-\Delta X_t) = (1 + \Delta X_t)$, as desired.

In a similar way, it is clear that $Z_{s^-}\Delta X_s = Z_{S^-} \Delta K_s$ (from definition of $K$). The rest follows from simplification.

It is possible that there are still doubts, so kindly clarify. It seems that most of your confusions were clarified by referring to previous theorems, maybe the author should have done this more often.

Furthermore, to obtain more intuition behind the Doleans-Dade exponential, I suggest you look at the case where $X$ is a continuous semimartingale (for which there's no $S_t$ so just $e^{K_t}$ would have been $Z_t$) which will help you understand the continuous situation better. The exponential martingale that comes out of the Doleans-Dade exponential is key in the Girsanov theorem.

EDIT : The explanation that $[X,[X,X]^c] = 0$ follows from the fact that $[X,X]^c$ is continuous, adapted and has finite variation on compacts(being continuous), hence is a pure jump semimartingale. It follows from Theorem $28$ that :
$$
[X,[X,X]^c]_t = \sum_{0 < s \leq t} \Delta X_s \Delta [X,X]^c_t = 0
$$

because the process $[X,X]^c_t$ doesn't jump, it is a continuous process so that $\Delta$ term is zero (Note : this feels like a contradiction : how can a process be "pure jump" and continuous? Turns out this is confusing but true. See George Lowther's blog post here). Similar stuff follows for $[[X,X]^c,[X,X]^c] = 0$.


答:
I assume you are dealing with two dimensional motion.

You're given a vector equation in the form $\displaystyle \vec{r(t)} = x(t)\mathbb i + y(t)\mathbb j$

Letting $\displaystyle |\vec{r(t)}| = r$ and $\displaystyle \theta$ be the argument of the particle's position vector measured in a counterclockwise sense from the origin, you can determine the polar equation as follows:

$\displaystyle x(t) = x = r\cos\theta$

and $\displaystyle y(t) = y = r\sin\theta$

Dividing one by the other, you get $\displaystyle \tan\theta = \frac yx$

Differentiating implicitly with respect to time $\displaystyle t$,

$\displaystyle \sec^2\theta \ \dot\theta = \frac{x\dot y - y\dot x}{x^2}$

Noting that $\displaystyle \sec^2\theta = 1 + \tan^2 \theta$,

$\displaystyle (1 + \frac {y^2}{x^2}) \dot \theta = \frac{x\dot y - y\dot x}{x^2}$

$\displaystyle \dot \theta = \frac{x\dot y - y\dot x}{x^2} \cdot \frac{x^2}{x^2+y^2} = \frac{x\dot y - y\dot x}{r^2}$

As the denominator is always non-negative, you only need to compute the sign of the numerator. If it's positive, the particle is moving counterclockwise, and vice versa.

Title: How to find if particle is moving in clockwise direction or not?

Body: I've been given a vector equation in the form of r(t), how do i find if the particle is going in clockwise direction or in anticlockwise direction?</p>


问:
Issues with regularity: $Y$ is $T_1$ iff there is regular space $X$ s.t. all continuous function from $X$ to $Y$ is constant.
I am interested in the following proposition:

$Y$ is $T_1$ iff there is regular space $X$ s.t. all continuous function from $X$ to $Y$ is constant.

The translation by Martin Sleziak of the paper Wann sind alle stetigen Abbildungen in Y konstant? written by Horst Herrlich contains the following construction:

Definition of a space $Q$. First we start with some given space $Y$.

The spaces $R_i$ for $i=1,2$ and points $r_i\in R_i$ are constructed in such way that every continuous map from $R_i$ to $Y$ is constant on some neighborhood of $R_i$.

A space $T=R_1\times R_2\setminus \{(r_1,r_2)\}$. This space has the property, that for every continuous map $f$ from $T$ to $Y$ there exist neighborhoods $U_i$ of $r_i$ such that $f$ is constant on $U_1\times U_2 - \{(r_1; r_2)\}$.

We take countably many homeomorphic copies $T\times\{n\}$ of the space $T$ for each integer $n$. We add two new points $a$, $b$ with local neighborhood bases $\{\bigcup T_m; m\ge n\}\cup \{a\}\subseteq B$ and $\{\bigcup T_m; m\le n\}\cup \{b\}\subseteq B$, $\color{red}{\text{then we get a regular space } R}$.

I have issues with the red sentence. It seems to me that every neighbourhood of $a$ is sure to intersect $T_n$ for some $n\in \Bbb Z$. In particular, if the closed set $A$ has nonempty intersections with each $T_n$, then it is impossible to separate $A$ and $a$. Am I missing something obvious?

答:
(Just to kick it out of the unanswered queue)

Thanks to @BrianM.Scott comment:

The problem does not arise: the definition of the topology ensures that a closed set that meets each $T_n$ already contains both $$ and $$, so of course it cannot be separated from them. A set that meets $T_n$for infinitely many positive integers $$ has $$ in its closure, and one that meets $T_n$ for infinitely many negative integers $n$ has $$ in its closure.


答:
The sum of the geometric series $1+r+r^2+\ldots+r^N$ is given by $\frac{1-r^{N+1}}{1-r}$ if $r\not=1$ (and by $N+1$ if $r=1$), Thus in your case you may calculate $100\frac{1-0.9^{N+1}}{1-0.9
}=1000({1-0.9^{N+1}})
$.

Comment: As mentioned in a comment below $\frac1{0.1}=10$ and not $100$.

Title: How to find a single forumla for a series

Body:
I am putting pricing strategy for a product

This product for 1 user is for $100

for 2 user is for $190

for 3 user is for $271

and so on

there for the equation will be

1 user = 100

2 users = $100 + 100\times0.9$

3 users = $100 + 100\times0.9 + 100\times0.9^2$

4 users = $100 + 100\times0.9 + 100\times0.9^2 +  100\times0.9^3$

..

There for for $N$ users the price will be

$N$ users = $100 (1 + 0.9 + 0.9^2 + 0.9^3 + ... + 0.9^N)$

How to put this in a single formula so I do not need to calculate all $N-1$ users first?


问:
What about this $\lim_{x \to \infty}\frac{3x+4}{\sqrt[5]{x^9+3x^4+1}}$?
When I saw this limit, I didn't even try to solve it by an algebraic method. I thought about the assyntotic concept.

In the example,

$$\frac{3x+4}{\sqrt[5]{x^9+3x^4+1}}\sim \frac{3x}{\sqrt[5]{x^9}}(x \to \infty)$$

So,

$$\lim_{x \to \infty}\frac{3x}{\sqrt[5]{x^9}}=3\lim_{x \to \infty}\frac{x}{x^{\frac{9}{5}}}=3\lim_{x \to \infty}\frac{1}{x^{\frac{4}{5}}}=0$$

And then the original limit is equal zero.

Is my thought correct? Would the algebraic method less difficult as I thought?
Thanks.

答:
Yes. Both your line of thought and final answer are correct. If you want to be a bit more rigorous, you could proceed as below:

For $x>0$, we have $$x^9 < x^9 + 3x^4 + 1 \implies \sqrt[5]{x^9} < \sqrt[5]{x^9 + 3x^4+1} \implies 0 < \dfrac{3x+4}{\sqrt[5]{x^9 + 3x^4+1}} < \dfrac{3x+4}{\sqrt[5]{x^9}}$$
Now use the squeeze theorem or the sandwich theorem, to conclude what you want.

ContentLicense: CC BY-SA 3.0


答:
It is accepted convention that anticlockwise rotation is positive.

In the oval/loop below with above accepted convention (around pole/origin at O) the rotation

$ A\to B$ is positive and  $ B\to A$ is negative.

and they are reversed if convention is made positive for clockwise rotation.

So an arbitrary sense of rotation is inconclusive regarding its sign.


问:
Homogenous PDE, changing of variable
I have an PDE $\dfrac{df}{d \xi}-\xi\dfrac{d x_1}{d \xi}=0$ homogeneous for $\xi$, where $f:\mathbb{R}^n\to\mathbb{R}$ is function of $x_1,\cdots,x_n:\mathbb{R}\to\mathbb{R}$, which in turn are functions of $\xi$ (so, $\dfrac{df}{d \xi}$ is a total derivative).

I also have $\xi=y/z\in\mathbb{R}$, where $y,z\in\mathbb{R}$, and one said that once the PDE is homogeneous in $\xi$ I have $\dfrac{df}{d y}-\xi\dfrac{d x_1}{d y}=0$.

I missed this gap. I imagine that I can multiply the first ODE by $\dfrac{d\xi}{d y}=\dfrac{1}{z}$, so

$$\dfrac{df}{d \xi}\dfrac{d\xi}{d y}-\xi\dfrac{d x_1}{d \xi}\dfrac{d\xi}{d y}=\sum_{i=1}^n \dfrac{\partial f}{\partial x_i}\dfrac{d x_i}{d\xi}\dfrac{d\xi}{d y}-\xi\dfrac{d x_1}{d y}=\sum_{i=1}^n \dfrac{\partial f}{\partial x_i}\dfrac{d x_i}{d y}-\xi\dfrac{d x_1}{d y}=\dfrac{d f}{d y}-\xi\dfrac{d x_1}{d y}=0.$$

But I am not sure. Indeed, I'd like to know how state this only from the fact of homogeneous.

Thanks so much.

Related:
Equality between two total derivatives

答:
Assume that $f:\textbf{R}^n\rightarrow\textbf{R}$ and $f=f(x_1,x_2,\ldots,x_n)$ is a function of $n$ variables. By saying that $x_i=x_i(\xi)$, then $C:\overline{x}=\{x_1(\xi),x_2(\xi),\ldots,x_n(\xi)\}$, $\xi\in\textbf{R}$, then $C$ is one dimentional object in $\textbf{R}^n$ and  hence $C$ is a curve of $\textbf{R}^n$. Then
$$
\frac{df}{d\xi}=\sum^{n}_{k=1}\frac{\partial f}{\partial x_k}\frac{dx_k}{d\xi}
$$
is the derivative of $f$ allong $C$ (or total derivative of $f$ allong the curve $C$). You also have the equation:
$$
\frac{df}{d\xi}-\xi\frac{dx_1}{d\xi}=0\Leftrightarrow \sum^{n}_{k=1}\frac{\partial f}{\partial x_k}\frac{dx_k}{d\xi}=\xi\frac{dx_1}{d\xi} \tag 1
$$
If $\xi=u y$, then $\frac{d\xi}{dy}=u$. Hence
$$
\frac{df}{d\xi}-\xi\frac{dx_1}{d\xi}=0\Leftrightarrow \frac{df}{dy}\frac{dy}{d\xi}-\xi\frac{dx_1}{dy}\frac{dy}{d\xi}=0\Leftrightarrow \frac{df}{dy}\frac{1}{u}-\xi\frac{dx_1}{dy}\frac{1}{u}=0\Leftrightarrow
$$
$$
\frac{df}{dy}-\xi\frac{dx_1}{dy}=0.\tag 2
$$
This answer your first question about the change of variables.

About the homogenicity

However if $f$ is homogeneous function then we have even more

If the function $f$ is homogeneous of degree $\lambda$. Then setting $x_i=uy_i$ in equation (1) we have, (knowing that $f(x_1,x_2,\ldots,x_n)$ and $(x_1,x_2,\ldots,x_n)\rightarrow x_1$ are homogeneous i.e.   $f(uy_1,uy_2,\ldots,uy_n)=u^{\lambda}f(y_1,y_2,\ldots,y_n)$ and $(ux_1)=ux_1$ of degree 1):
$$
\sum^{n}_{k=1}\frac{\partial f}{\partial x_k}(uy_1,uy_2,\ldots,uy_n)\left(u\frac{dy_k}{d\xi}\right)-\xi\left(u\frac{dy_1}{d\xi}\right)=0\Leftrightarrow
$$
$$
u^{\lambda-1}\sum^{n}_{k=1}\frac{\partial f}{\partial y_k}(y_1,y_2,\ldots,y_n)\left(u\frac{dy_k}{d\xi}\right)-\xi u\frac{dy_1}{d\xi}=0
$$
$$
u^{\lambda-1}\sum^{n}_{k=1}\frac{\partial f}{\partial y_k}(y_1,y_2,\ldots,y_n)\frac{dy_k}{d\xi}-\xi \frac{dy_1}{d\xi}=0.\tag 3
$$
(That is because when $f(x_1,x_2,\ldots ,x_n)$ is homogeneous of degree $\lambda$, then $\frac{\partial f}{\partial x_{j}}$ is homogeneous of degree $\lambda-1$ i.e. $\frac{\partial f}{\partial x_j}(uy_1,uy_2,\ldots,uy_j,\ldots,uy_n)=u^{\lambda-1}\frac{\partial f}{\partial x_j}(y_1,y_2,\ldots,y_n)$). Hence when $\lambda=1$, then (3) becomes:
$$
\sum^{n}_{k=1}\frac{\partial f}{\partial y_k}\frac{dy_k}{d\xi}-\xi\frac{dy_1}{d\xi}=0.\tag 4
$$
Hence if $f=f(x_1,x_2,\ldots,x_n)$ is homogeneous of degree 1, then equation (1) is homogeneous PDE (invariant under any transformation of variables of the form $x_i=uy_i$, $i=1,2,\ldots,n$).

Title: Homogenous PDE, changing of variable


问:
a question about $L^p$ space and Holder inequality
let $(X,A,\mu)$ a measurable space.$1<p, q, s<\infty$

a.Prove that if $f\in L^p$ and $g\in L^q$ such that $1/p+1/q=1/s$ then $fg\in L^s$ and $\|fg\|_s\leq \|f\|_p\|g\|_q$.

b.Prove that if $f\in L^p$ , $g\in L^q$ and $h\in L^s$ such that $1/p+1/q+1/s=1$ then $fgh\in L^1(\mu)$ and $\|fgh\|_1\leq \|f\|_p\|g\|_q\|h\|_s$

My trial:

a. $\|fg\|_s= (\int_X|fg|^s)^{1/s}=(\int_X|fg|^s)^{1/p+1/q}= (\int_X |fg|^s)^{1/p} (\int_X |fg|^s)^{1/q}\leq$ (p,q>s) $(\int_X |fg|^p)^{1/p}  (\int_X |fg|^q)^{1/q}= \|f\|_p\|g\|_q$ and since $\|f\|_p , \|g\|_q < \infty$ then $\|fg\|_s$ is finite too.so $fg \in L^s$

b. Denote $1/p+1/q=1/p'$ so by part a we get $fg\in L^{p'}$ and: $\|fg\|_{p'} \leq \|f\|_p \|g\|_q$, then we have $1/p'+1/s=1$ therefore by holder ineq1uality we get
$\|fgh\|_1\leq \|fg\|_{p'} \|h\|_s \leq \|f\|_p\|g\|_q\|h\|_s$

Is what I solved right?
Glad to her any notifications.

答:
If I understand correctly, in part (a),

you have used that $p>s$ implies that
$$ \int_X |F|^s \le \int_X |F|^p$$
(and similarly for $q>s$). This is not true. For example consider $X=[1,\infty)$ with the usual Lebesgue measure, and $f(x) = 1/x$. Then $f\in L^2(X)$ but not $L^1(X)$, so $$\infty=\int_1^\infty\frac1{x} dx \not\le \int_1^\infty \frac1{x^2}dx<\infty. $$

(a counterexample also in $L^1$ is  $f(x)=\mathbf 1_{[1,N]}/x$, for some $N$ very large; $\mathbf 1_A$ is the indicator of $A$.)

You then further went on to write $\|fg\|_p \|fg\|_q = \|f\|_p \|g\|_q$. This is also not true, and can be seen by scaling considerations: if you multiply $f$ by $2$, the left hand side is multiplied by $4$, but the right hand side only by $2$. So the equality is false.

Instead, here's a big hint: try use the normal Holder's inequality with the exponents $1=\frac1{p/s}+\frac1{q/s}$.

Some other minor points: you should write $fg\in L^s$, not $\|fg\|_s\in L^s$; and $p'$ is normally reserved for the Hölder conjugate; so in part (b), perhaps use $s'$ instead. (The proof for (b) is correct.)


问:
Prove that $3x+2y-2xy-2\geq0$ when $x,y\in[0.5,1)$?
I am looking for the proof that $3x+2y-2xy-2\geq0$ when $x,y\in[0.5,1)$. This seems to be true when plotted out. Is there a simple and elegant way to show this? Thanks.

答: Rewrite the LHS of inequality as $(3x+2y)(1-x)+3x^2-2$ , now see if you want to minimise this expression, $y$ will take it's minimum value as $1-x$ is positive, so minimum occurs at $y=0.5$ hence you are left with $3x+1-x-2 \geq 0 $ which is nothing but $x \geq 0.5$, which is true.


问:
What type of Hypergeometric series is this?
I am trying to find a closed form for the series

$$ \sum^\infty_{n=0} \frac{1}{n!} \frac{1}{n+1}(-z)^n {}_2F_2\left(m,n+1;\frac{1}{2},n+2; b z\right)$$

$m$ is a nonzero positive integer, and $b$, $z$ are positive real numbers. I to rewrite the sum as

$$ \sum^\infty_{n=0} \sum^\infty_{q=0}
 \frac{1}{n!} \frac{1}{q!}
 \frac{(m)_q}{(\frac{1}{2})_q}
\frac{(1)_{q+n}}{(2)_{q+n}}
(-z)^n (b z)^q$$

any idea what type of multi-variable hypergeometric function is the last equation?

答:
Here is a suggestion, not a complete answer.

Since $n+q$ seems prevalent,
let $k = n+q$.

$\begin{align}
\sum^\infty_{n=0} \sum^\infty_{q=0}
 \frac{1}{n!} \frac{1}{q!}
 \frac{(m)_q}{(\frac{1}{2})_q}
\frac{(1)_{q+n}}{(2)_{q+n}}
(-z)^n (b z)^q
&=\sum^\infty_{n=0} \sum^\infty_{q=0}
 \frac{1}{n!} \frac{1}{q!}
 \frac{(m)_q}{(\frac{1}{2})_q}
\frac{(1)_{q+n}}{(2)_{q+n}}
z^{n+q}
(-1)^n b^q\\
&=\sum^\infty_{k=0} \sum^k_{q=0}
 \frac{1}{(k-q)!} \frac{1}{q!}
 \frac{(m)_q}{(\frac{1}{2})_q}
\frac{(1)_{k}}{(2)_{k}}
z^{k}
(-1)^{k-q} b^q\\
&=\sum^\infty_{k=0}
\frac{(-1)^k(1)_{k}}{(2)_{k}}
z^{k}
\sum^k_{q=0}
 \frac{1}{(k-q)!} \frac{1}{q!}
 \frac{(m)_q}{(\frac{1}{2})_q}
(-1)^{q} b^q\\
\end{align}
$

The next step would be to do something
with the inner sum,
but I'll stop here,
since I'm not much of a hypergeometric function
expert.


问:
Finding specific interval
The question is to find $[a,b]$ such that $$\forall x,y : x\in [a,\frac {a+b}{2}],y\in[\frac{a+b}{2},b]\Rightarrow x+y,xy \in [a,b]$$
I tried like below
$$a \leq x \leq\frac {a+b}{2}\\
\frac {a+b}{2} \leq y \leq b\\
\to \frac {3a+b}{2} \leq x+y \leq \frac {a+3b}{2}\\\to
a\leq \frac {3a+b}{2}\leq \frac {a+3b}{2}\leq b$$ this all get $a\leq b $

and $$a.\frac {a+b}{2} \leq xy \leq b.\frac {a+b}{2}\\
a\leq a.\frac {a+b}{2} \leq b.\frac {a+b}{2}\leq b$$ for a>o will be $$2a\leq a^2+ab \to 2\leq a+b\\
b.\frac {a+b}{2} \leq b \to a+b\leq 2$$
this mean $a+b=2$

first :Am I on right track?

second: I got stuck on this,because It revert to trivially $a<b$

Please guide me to find the solution? or give an Idea to take over

thanks in advance

Title: Finding specific interval

Body:
$$a \le \frac{3a+b}2 \iff 0 \le a+b$$

$$\frac{a+3b}{2}\le b \iff a + b \le 0$$

Hence we have $a+b=0$.

$$a=-b$$

$|x| \le b$, $|y| \le b$, we want $|xy| \le b$

$$b^2 \le b$$

$$b^2-b \le 0$$

$$b(b-1) \le 0$$

Hence,
$$0 \le b \le 1, a = -b$$


问:
Functional derivative of the square of a functional
How should I compute the functional derivative of the following functional:

$$
F(p(x)) = \left[\int\cos(x)p(x)dx - \int\cos(y)q(y)dy\right]^2
$$

I know that the functional derivative $\frac{\delta F}{\delta p(x)}$ is defined as

$$
\left[\frac{d}{d\epsilon}F(p +\epsilon\phi)\right]_{\epsilon=0} = \int \frac{\delta F}{\delta p(x)}\phi(x)dx
$$

but I am not sure how the chain rule is defined here in the case of a function of a functional, i.e., what's the functional derivative of $F = g(H(p(x)))$ where $g$ is a differentiable function and $H$ is a functional.

答:
The problem you found in the application of the formula for functional derivatives above is due the fact that it is not correct, since it is valid only for particular classes of functionals: the general definition of functional derivative is exactly the right side of the one you tried to use, i.e.
$$
\frac{\delta F}{\delta p} [p](\phi)\triangleq \left[\frac{\mathrm{d}}{\mathrm{d}\epsilon}F(p +\epsilon\phi)\right]_{\epsilon=0} = \lim_{\epsilon \to 0} \frac{F(p +\epsilon\phi) -F(p)}{\epsilon}\label{1}\tag{1}
$$
Said that, applying the definition to the quadratic functional under analysis we get
$$
\frac{\delta F}{\delta p} [p](\phi) = 2\left[\int\cos(x)p(x)\mathrm{d}x - \int\cos(y)q(y)\mathrm{d}y\right] \int\cos(x)\phi(x)\mathrm{d}x
$$
Edit following the comment of the Asker.

In order to clarify some of the doubts pointed out in the comment, I expanded a bit definition \eqref{1} above to analyze in a clearer way its meaning: it states that, in order to calculate the functional derivative of a functional $F$ at a "point" $p$, you should calculate the ordinary first order derivative of the functional applied to the variation $p+\epsilon \phi$, considered as a function of the real variable $\epsilon$. This implies that we should proceed in the calculations by using all the ordinary rules for differentiation including the chain rules: thus, when the functional $F$ is given by the composition of a ($C^\infty$-real valued) function of a real variable and a (real valued) functional $H$, i.e.
$$
F(p)=g\circ H(P)= g\big(H(p)\big),
$$
we have that
$$
\begin{split}
\frac{\delta F}{\delta p} [p](\phi) &= \left[\frac{\mathrm{d}}{\mathrm{d}\epsilon}g\big( H(p +\epsilon\phi)\big)\right]_{\epsilon=0} = \lim_{\epsilon \to 0} \frac{g\big( H(p +\epsilon\phi)\big) -g\big( H(p) \big)}{\epsilon}\\
& = \lim_{\epsilon \to 0} \frac{g\big( H(p +\epsilon\phi)\big) -g\big( H(p)\big)}{ {H(p +\epsilon\phi) -H(p)}}\cdot\frac{H(p +\epsilon\phi) - H(p)}{\epsilon}\\
&=\left. \frac{\mathrm{d}g(x)}{\mathrm{d} x}\right|_{x=H(p)}\cdot \left[\frac{\mathrm{d}}{\mathrm{d}\epsilon}H(p +\epsilon\phi)\right]_{\epsilon=0}\\
&=\left. \frac{\mathrm{d}g(x)}{\mathrm{d} x}\right|_{x=H(p)}\cdot \frac{\delta H}{\delta p} [p](\phi)
\end{split}
$$
which is basically an application (and the basic proof in disguise) of the chain rule formula.

Notes

An analysis of the general definition of functional derivatives is offered in this Q&A, where you can have a look in order to have more information and some references on the concept.

Regarding references: almost any book in the calculus of variation includes several examples of calculations of functional derivatives. However, they mostly (if not exclusively) deal with functional of integral type, which are not of the same kind of the functional studied here. Therefore my advice is to have a look at this, this, this and this Q&A which deal with functionals of non integral type. And, apart from consult the references given in the cited Q&A if needed, I advice to look at the first volume [1] of the monograph by Mariano Giaquinta and Stefan Hildebrandt, in my opinion a wonderful reference.

[1] Giaquinta, Mariano; Hildebrandt, Stefan (1996), Calculus of Variations I. The Lagrangian Formalism, Grundlehren der Mathematischen Wissenschaften, 310 (1st ed.), Berlin: Springer–Verlag, pp. xxix+475, ISBN 3-540-50625-X, MR 1368401, Zbl 0853.49001.


问:
nth Fibonacci number divided by 2 to the n
Is there any way to calculate
$\sum_{n=1}^{\infty} \frac{F_{2n}}{2^{2n}}$
where $F_n$ is the nth Fibonacci number, $F_1=0$, $F_2=1$, $F_n = F_{n-1} + F_{n-2}$

答:
The generating function of $F_{2n}$ (as defined here) is
$$\frac{x-x^2}{x^2-3x+1}=\sum_{n=1}^\infty F_{2n}x^n$$
Substitute $x=\frac14$ to get the sum as
$$\sum_{n=1}^\infty\frac{F_{2n}}{2^{2n}}=\frac{1/4-1/16}{1/16-3/4+1}=\frac35$$


问:
How to calculate return time from a transition probability matrix?

Consider the transition probability matrix with state space $\{0,1,2,...5\}$ as follows
$$P= \begin{bmatrix} 1/2 & 1/2 & 0 & 0 & 0 & 0 &\\
1/3 & 2/3   & 0 & 0 & 0 & 0\\
0 & 0 & 1/8 & 0  & 7/8 & 0\\
1/4 & 1/4 & 0 & 0 & 1/4 & 1/4\\
0 & 0 & 3/4 & 0 & 1/4 & 0\\
0 & 1/5 & 0 & 1/5 & 2/5 & 1/5\end{bmatrix}$$
a) Let $C=\{0,1\}$. Then find $\rho_{3}(C)$ i.e., absorption probability to the set $C$ starting from $3$.

b) Find $m_0=\mathbb{E}_0(T_0)$, where $\mathbb{E}_i(T_i) := \mathbb{E}(T_i \mid X_0=i)$

Here is what I have thought of.
I know that $\{0,1\}$ is a recurrent irreducible closed set and $\{3,5\}$ is another irreducible closed set. Hence starting from $3$, it would be impossible to go to any other state except $5$ and so the absorption probability of  $\rho_3(C)$ must be $0$. Coming to the second part, I am not sure how to exactly do it? We haven't explicitly learnt how to calculate hitting time, and so I don't know how to start calculating the conditional expectation using this. I also read that $\mathbb{E}_i [T_i]=m_i=\frac{1}{\pi_i}$ where $\pi_i$ is the stationary distribution. But $P$ isn't an irreducible Markov Chain so it doesnt have a unique stationary distribution. So I don't know how to proceed using that formula either. I am unaware of any other formula to calculate return time, so any help would be appreciated!

答:
Your understanding of absorption probailities is slightly incorrect. Here is a formula for absorprtion probabilities-
$$\rho_C(x)= \sum_{y \in C} P(x,y) + \sum_{y \in \mathcal{S}_{T}} P(x,y) \rho_C(y)$$

Given that $C=\{0,1\}$ and $\mathcal{S}_T = \{3,5\}$, we have

\begin{align}
    \rho_C(3) &=P(3,0)+P(3,1)+P(3,3)\rho_C(3) + P(3,5) \rho_C(5)\nonumber\\
    &= 1/4 + 1/4 + 1/4 \rho_C(5)\nonumber\\
    \implies &\rho_c(3)- 1/4 \rho_C(5) = 1/2
\end{align}

and

\begin{align}
    \rho_C(5) &=P(5,0)+P(5,1)+P(5,3)\rho_C(3) + P(5,5) \rho_C(5)\nonumber\\
    &= 1/5 + 1/5 \rho_C(3) + 1/5 \rho_C(5)\nonumber\\
    \implies & 4/5\rho_C(5)- 1/5 \rho_C(3) = 1/5
\end{align}

Then, solving the two equations, we get $\rho_C(3)=3/5$ and $\rho_C(5)=2/5$.

For $m_0$, consider the irreducible closed classes in this M.C. We have $\pi P= \pi \implies P^T \pi^T= \pi^T$. Thus,

$$ \left[\begin{array}{cc}
1/2 & 1/3 \\
1/2 & 2/3
\end{array}\right]
\left[\begin{array}{c}
\pi_0 \\
\pi_1 \\
\end{array}\right]
=
\left[\begin{array}{c}
\pi_0 \\
\pi_1 \\
\end{array}\right]
$$
Now we have the equations
$$1/2 \pi_0 + 1/3\pi_1= \pi_0 \Rightarrow -1/2 \pi_0+ 1/3 \pi_1=0$$
$$ \pi_0 +\pi_1=0$$
Solving this yields $\pi(0) = 2/5$ and $\pi(1) = 3/5$.


问:
Force from right angle to the opposite side

$ABC$ is a triangle, right angled at $A$. The resultant of the forces acting along $AB$, $AC$ with magnitudes $\frac1{AB}$ and $\frac1{AC}$ respectively is the force along $AD$, where $D$ is the foot of the perpendicular from $A$ onto $BC$. The magnitude of the resultant is
(1)$\frac{{AB}^2+{AC}^2}{(AB)^2(AC)^2}$
(2) $\frac{(AB)(AC)}{AB+AC}$
(3) $\frac1{AB}+\frac1{AC}$
(4) $\frac1{AD}?$

I understand that the forces given are $AB$ unit vector and $AC$ unit vector. So, by parallelogram law of vectors, $AD$ would be in the direction of $AB+AC$?

Also, by triangle law, I get that $BC=AC-AB$. And, $BC\cdot AD=0$. How to proceed from here?

答:
Note that it looks like the question is using AB to refer to the length of the segment AB of the triangle, and the same with AC.

If we label $\vec{i}$ as a unit vector along AB, and $\vec{j}$ as a unit vector along AC (where $\vec{i}$ and $\vec{j}$ are perpendicular to each other),

then our resultant force is just

$\vec{f} = (\frac{1}{AB})\vec{i} + (\frac{1}{AC})\vec{j}$

We just use the pythagorean theorem to get the magnitude of the resultant

$|\vec{f}| = \sqrt{(\frac{1}{AB})^2+(\frac{1}{AC})^2}$

Now we can use a theorem from geometry

$\frac{1}{AB^2}+\frac{1}{AC^2} = \frac{1}{AD^2}$

This theorem is described in the altitudes section here:
https://en.wikipedia.org/wiki/Right_triangle#:~:text=The%20Pythagorean%20theorem%20states%20that,meet%20at%20a%20right%20angle).

So we finally get

$|\vec{f}|=\frac{1}{AD}$


问:
How to understand C(X)'' = bounded Borel measurable functions?
Let $X$ be a compact metric space and $C(X)=\{ f:X\rightarrow \mathbb{R} \ | \ \ f \ continuous\}$ with the uniform norm. It is a separable Banach space.

1) I'm aware of the fact that $C(X)^*$, the space of continuous linear functionals $C(X)\rightarrow \mathbb{R}$ coincide with $M(X)$, the space of signed regular borel measures on $X$.

This intuitively makes sense because $\mu\in M(X)$ can be naturally be seen as an evaluation map $f\mapsto \mathbb{R}$ by means of integration.

2) I'm also aware that $(C(X)^*)^*$, the dual of the dual, coincide with the set of bounded Borel measurable functions $F:X\rightarrow\mathbb{R}$ (source: P. Lax, Functional Analysis).

UPDATE Landscape provided a counterexample in the comments.
My source was: P. Lax "Functional Analysis" 2002, Page 82, Theorem 14(ii).
I guess this might be a mistake, perhaps fixed in some errata somewhere.

New question:
C.f. Landscape's example, let $g_A$ be the extension (given by Hahn-Banach) of $f_A$ to $C([0,1])^{**}$. Looking at $g_A$ as a function $X\rightarrow\mathbb{R}$, $g_A$ must satisfy by construction the equation $\int_{[0,1]}g_A \ d \ \delta_x= 1$ if $x\in A$ and $0$ otherwise. This seems to imply that $g_A$ is the characteristic function of $A$. Hence not Borel if $A$ is not Borel.

Thus, $g_A$ as a function $(X\rightarrow\mathbb{R})$ is uniquely determined by the construction starting from $f_A$. But is it $g_A$ as an element of $C([0,1])^{**}$ uniquely determined? If so, it looks to me we would have a reasonably well defined notion of integration for non-measurable sets.

Thanks!

Screen shot of Lax's theorem 14 on page 82:

答:
Let $X$ be a compact metric space.  Let $\mathcal B$ be the collection of Borel sets on $X$.  Banach space $C(X)$ is the set of all (necessarily bounded) continuous real-valued functions $f : X \to \mathbb R$, with norm
$$
\|f\|_\infty = \max\{|f(x)| : x \in X\}
$$

Banach space $M(X)$ is the set of all countably-additive signed ($\mathbb R$-valued) measures on the sigma-algebra $\mathcal B$.  The norm is the total variation: for $\mu \in M(X)$, write $\mu = \mu^+ - \mu^{-}$ where $\mu^+$ and $\mu^-$ are positive measures, singular to each other, and let
$$
\|\mu\|_1 = \mu^+(X)+\mu^-(X)
$$

The pairing $M(X) \times C(X) \to \mathbb R$ defined by
$$
\langle\mu,f\rangle = \int_X f \;d\mu
$$
identifies $C(X)^\ast$ isometrically with $M(X)$ in these norms.

Another description of $M(X)$ may be obtained as follows.  Let $\{\tau_i : i \in I\}$ be a maximal (under inclusion) family of mutually singular probability measures on $(X,\mathcal B)$.  (Such a family exists by Zorn's Lemma.  But it is not unique.)  Identify $M(X)$ isometrically with the $l^1$-direct sum of the family of all the $L^1(\tau_i)$ spaces:
$$
M(X) \approx \left(\bigoplus_{i\in I} L^1(X,\mathcal B,\tau_i)\right)_{1}
$$

Once we describe $M(X)$ in this way, we get the corresponding description of the dual as the $l^\infty$-sum of the spaces $L^\infty(\tau_i)$.
$$
C(X)^{\ast\ast}\approx \left(\bigoplus_{i\in I} L^\infty(X,\mathcal B,\tau_i)\right)_{\infty}
$$


问:
Show that $\lim_{n\to\infty}\int_0^1|f_n(x)-1|\,dx = 1$ if $\lVert f_n\rVert_{1} = 2$ and $\lim_{n\to\infty}f_n(x) = 1$
Suppose that $\{f_n\}_{n=1}^\infty$, $f_{n}:[0,1]\to\mathbb{R}$, is a sequence of measurable functions such that for every $x\in [0,1]$
$$\lim_{n\to\infty}f_n(x)=1$$
and for every $n$: $$\int_{0}^{1}|f_{n}(x)|\,dx = 2.$$
Prove that
$$\lim_{n\to\infty}\int_{0}^{1}|f_{n}(x) - 1|\,dx = 1$$

What I tried/have: Such a sequence is clearly possible since we can take $f_n = 1+n\chi_{(0,\frac{1} {n})}$. The proposition is also trivial if $f_n\geq 1$. I also managed to get, using the reverse triangle inequality that
$$\liminf_n \int_0^1|f_n(x) - 1|\,dx\geq \liminf_n\int_0^1|f_n(x)|\,dx - 1 = 1$$
but I'm quite stuck on proving that the $\limsup$ is less than or equal to 1. Since it is not possible to find a dominating function (eg. consider the sequence I stated before) and there is no clear way to construct a monotone sequence, I feel like I need to use Fatou's Lemma somehow. Could someone please give me some tips on how I might show that the limit superior is at most $1$? Or maybe, if there is an easier approach, how I would directly compute the limit?

答:
It is sufficient to show $\limsup_n \int |f_n-1| \le 1$.

Let $\epsilon >0$ then Egorov gives some measurable $B \subset [0,1]$ such that $m B < \epsilon$ and $f_n \to 1$ uniformly on $B^c$.

Then $\int |f_n-1| = \int_B|f_n-1|+ \int_{B^c}|f_n-1| \le \int_B|f_n| + mB + \int_{B^c}|f_n-1|$ so
$\limsup_n \int |f_n-1| \le \limsup_n \int_B|f_n| + \epsilon$.

Since $\int |f_n| = \int_B |f_n|+ \int_{B^c}|f_n| = 2$, we see that
$\lim_n \int_B |f_n| = 2-m B^c$, that is
$\limsup_n \int |f_n-1| \le 2-m B^c + \epsilon \le 1+2 \epsilon$. Since $\epsilon>0$
was arbitrary we have the desired result.


问:
How to find domain of definition of solution?
For the given initial-value problem:
$$\dfrac{dy}{dt}=\dfrac{1}{\left(y+2\right)^{2}}, \quad y(0)=1$$
we are asked to solve it then to state the domain of definition of the solution. So first of all I separated the variables and applied the initial condition and obtained:
$$y(t)=\left(3t+27\right)^{1/3}-2$$

but now I can't figure out why does the solution exist only when $t>-9$? I am having trouble understanding also what does this have to do with $\dfrac{dy}{dt}$ not being defined at $y=-2$. Why are we letting
$$3t+27 >0 $$

what if $t=-10$? how does this make $y$ undefined?

答:
$$\dfrac{dy}{dt}=\dfrac{1}{\left(y+2\right)^{2}}$$
$$t=\int (y+2)^2dy=\frac13 (y+2)^3 +c$$
Condition : $t(1)=0=\frac13 (1+2)^3 +c \quad\implies\quad c=-9 $
$$t=\frac13 (y+2)^3-9$$
For $y$ and $t$ reals this implies $t>-9$.

If the problem is from Physic the speed $y'(t)$ is infinite at $(t=-9\:,\:y=-2)$ .  This excludes the branch $y\leq -2$.


问:
Shortest orthogonal integer vectors and their cross product

Question:

  Given an arbitrary non-zero vector $v \in \mathbb{Z}^3$ with $\gcd(v) = 1$, there exist linear independent non-zero vectors $f, s \in \mathbb{Z}^3$ orthogonal to $v$ and minimal with respect to the Euclidean norm. Their cross product must be a scalar multiple of $v$, say $f \times s = \lambda v$. Obviously $\lambda \in \mathbb{Z}$, but is $\lambda = \pm 1$?

With help from Rodrigo de Azevedo I made progress.

This answer provably provides two basis vectors spanning the lattice of vectors orthogonal to $v$.

Then an algorithm due to Lagrange finds the shortest two basis vectors $f$ and $s$, given any two basis vectors $b$ and $c$ :

if $\lVert b \rVert < \lVert c \rVert$:

  $\quad$ $b, c := c, b$

  while $\lVert c \rVert < \lVert b \rVert$:

  $\quad$ $b, c := c, b - \lfloor \frac{\langle\,b,c\,\rangle }{{\lVert c \rVert}^2}\, \rceil c$

  return $b, c$

Here $\lVert \cdot \rVert$ denotes the Euclidean norm, $\lfloor \cdot \rceil$ the nearest integer function and $\langle\,\cdot, \cdot\,\rangle$ the dot product.

On my computer checking 10000 random vectors takes roughly 5 minutes. So far I haven't found a counterexample.

Motivation:

Consider the homogeneous diophantine equation
$$v_1x_1+v_2x_2+v_3x_3=0$$
over $\mathbb{Z}$ with $v$ given and $\gcd(v)=1$.

A version of Siegel's lemma states, that there exist two non-trivial linear independent solutions $f, s$, such that
$$\max(|f_i|) \max(|s_i|) \leq \lVert v \rVert. $$
Since
$$|\lambda| \lVert v \rVert = \lVert f \times s \rVert \leq  \lVert f \rVert \lVert s \rVert < 3 \max(|f_i|) \max(|s_i|) \leq 3 \lVert v \rVert,$$
we have $|\lambda| \leq 2$. I was hoping for either an example with $|\lambda| = 2$ or an intuitive explanation why $|\lambda|$ always equals $1$.

答:
As an extension to the Rodrigo de Azevedo's answer, it must be mentioned that the $b_1$ and $b_2$ vectors, selected as $(-v_2,v_1,0)$ and $(-v_3,0,v_1)$, do not generally produce the right results by just using the lattice reduction. To generate correct source vectors for that purpose, it is needed to first find out 2 integers $p$ and $q$ so that $p v_2+q v_3 = GCD(v_2,v_3) v_1$. This can be made by applying GCD method, and produces two different solutions $(p_1,q_1)$ and $(p_2,q_2)$, when $|p_i|<v_3$ and $|q_i|<v_2$. In fact, if $p_1 > 0$ and $q_1 < 0$, then $p_2 = p_1 - v_3 < 0$ and $q_2 = q_1 + v_2 > 0$. Then you calculate new $b_1' = (p_1 b_1+q_1 b_2)/v_1$ and $b_2' = (p_2 b_1+q_2 b_2)/v_1$. The division is possible because all the corresponding values are multiples of $v_1$. The lattice reduction may now be performed using $b_1'$ and $b_2'$, and assumedly produces the right results. (Note that in the practical example (1,2,3) this extra step was avoided by using $v_1$ = 1).

Let us consider a general case $(v_1,v_2,v_3) = (aef,bdf,cde)$, where $a$, $b$, $c$, $d$, $e$, and $f$ are positive integers having no pairwise common factors. We note that $GCD(v_2,v_3) = d$, $GCD(v_1,v_3) = e$, and $GCD(v_1,v_2) = f$, but $GCD(v_1,v_2,v_3) = 1$. We get $b_1 = (bd,-ae,0)$ and $b_2 = (cd,0,-af)$, where we have divided the values with their common factors $f$ and $e$, respectively. Now we find out $p$ and $q$ so that $pb + qc = a$, where $0 < p < c$ and $0 < |q| < b$. The candidates for shortest orthogonals are then $b_1'= (pb_1+qb_2)/a = (d,-pe,-qf)$ and $b_2'=((p-c)b_1+(q+b)b_2)/a = (d,-(p-c)e,-(q+b)f)$. Both are orthogonal to $(v_1,v_2,v_3)$ and $b_1' × b_2'  = (ef(pb+qc),bdf,cde) = (aef,bdf,cde) = (v_1,v_2,v_3)$. This candidate pair thus fulfills the condition $λ=±1$.

Note that by exchanging the roles of $v_1$, $v_2$, and $v_3$ we (may) get more candidate pairs, which may (or may not) be even shorter, but still fulfilling the condition $λ=±1$.


问:
Borel sigma algebras of subsets of $\mathbb{R}$
I'm trying to prove a result that seems to be used routinely, but the proof is skipped over.

Let $\mathbb{R}$ have the Euclidean metric, and this defines a topology. Let $A \subset \mathbb{R}$ be equipped with the subspace topology. Then,

$\mathcal{B}(A) = \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$

If $A \in \mathcal{B}(\mathbb{R})$, then $\mathcal{B}(A) = \{ B \in \mathcal{B}(\mathbb{R}) : B \subset A\}$.

The proof for the $\subset$ direction of 1. is simple; note that $\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$ is a sigma algebra, which contains all the open sets of $A$. There are also other questions on stackexchange which address this direction of the proof. However, I am stuck on the $\supset$ direction. Does anyone have a nice proof for this?

(2. is an immediate consequence of 1., as far as I can see)

答:
This question permits a very elegant and general solution, which I would like to try and present. We begin by introducing a very elementary and equally important construction. For arbitrary map $f \colon A \to B$ and subsets $X \subseteq A$ respectively $Y \subseteq B$ we denote by $f[X]$ respectively $f^{-1}[Y]$ the direct respectively inverse images through $f$ of the subsets in question. We proceed to introduce the direct and inverse image maps as follows:
$$\begin{align}
\hat{f} \colon \mathscr{P}(A) &\to \mathscr{P}(B)\\
\hat{f}(X)&=f[X]\\
\check{f} \colon \mathscr{P}(B) &\to \mathscr{P}(A)\\
\check{f}(Y)&=f^{-1}[Y].
\end{align}$$
We also introduce another highly important notion: given fixed set $A$, we say $\mathscr{K}$ is a closure system on $A$ if $A \in \mathscr{K} \subseteq \mathscr{P}(A)$ such that for every nonempty $\varnothing \neq \mathscr{M} \subseteq \mathscr{K}$ we have that $\bigcap \mathscr{M} \in \mathscr{K}$. In natural language, $\mathscr{K}$ is a collection of subsets of $A$ containing $A$ in particular and closed with respect to arbitrary intersections. Given arbitrary subset $X \subseteq A$ we introduce the notation:
$$[X]_{\mathscr{K}}\colon=\bigcap_{\substack{M \in \mathscr{K}\\M \supseteq X}}M$$
for the $\mathscr{K}$-substructure generated by $X$, which is easily seen to be the minimum (with respect to inclusion) of the subset $\{M \in \mathscr{K} \mid M \supseteq X\}$.

We proceed to adopt the formal (and usual) definition according to which $\mathscr{M}$ is a $\sigma$-algebra on $A$ if $\varnothing \neq \mathscr{M} \subseteq \mathscr{P}(A)$ satisfies the axioms:
$$\begin{align}
&\left(\forall X \right)\left(X \in \mathscr{M}^{\mathbb{N}} \Rightarrow \bigcup_{n \in \mathbb{N}} X_n \in \mathscr{M}\right)\\
&(\forall X)(X \in \mathscr{M} \Rightarrow A \setminus X \in \mathscr{M}).
\end{align}$$
For fixed set $A$ we denote by $\mathscr{Alg}^{\sigma}(A)$ the collection of all $\sigma$-algebras on $A$ and remark that it is a closure system on the powerset $\mathscr{P}(A)$.

Given an arbitrary map $f \colon A \to B$ we signal the following two important constructions:

Given a $\sigma$-algebra $\mathscr{M} \in \mathscr{Alg}^{\sigma}(A)$ it holds that $\mathscr{N}\colon=\check{f}^{-1}\left[\mathscr{M}\right] \in \mathscr{Alg}^{\sigma}(B)$. We call this $\sigma$-algebra on $B$ the direct image of $\mathscr{M}$ through $f$ and denote it by $f^{*}\left[\mathscr{M}\right]$. We also remark that it is by definition constituted of those $Y \subseteq B$ such that $f^{-1}[Y] \in \mathscr{M}$.

Given a $\sigma$-algebra $\mathscr{N} \in \mathscr{Alg}^{\sigma}(B)$ it holds that $\mathscr{M}\colon=\check{f}\left[\mathscr{N}\right] \in \mathscr{Alg}^{\sigma}(A)$. We call this $\sigma$-algebra on $A$ the inverse image of $\mathscr{N}$ through $f$ and denote it by $f_{*}\left[\mathscr{N}\right]$.

We make the important observation that given any two $\sigma$-algebras $\mathscr{M} \in \mathscr{Alg}^{\sigma}(A)$ and $\mathscr{N} \in \mathscr{Alg}^{\sigma}(B)$ we have the following dualisation relation: $f^{*}\left[\mathscr{M}\right] \supseteq \mathscr{N} \Leftrightarrow \mathscr{M} \supseteq f_{*}\left[\mathscr{N}\right]$.

Let us also note that in the particular case when $B \in \mathscr{M} \in \mathscr{Alg}^{\sigma}(A)$, if we denote by $\mathrm{i}^B_A \colon B \to A$ the respective inclusion map we have the following simple description: $\left(\mathrm{i}^B_A\right)_{*}\left[\mathscr{M}\right]=\{X \in \mathscr{M} \mid X \subseteq B\}$.

With these preparations in place, we can now state a very general:

Proposition. Let $f \colon A \to B$ be an arbitrary map and $\mathscr{Y} \subseteq \mathscr{P}(B)$ be an arbitrary collection of subsets of $B$. Then the relation $f_{*}\left[\left[\mathscr{Y}\right]_{\mathscr{Alg}^{\sigma}(B)}\right]=\left[\check{f}\left[\mathscr{Y}\right]\right]_{\mathscr{Alg}^{\sigma}(A)}$ holds.

Proof. For ease of notation let us write $\mathscr{X}\colon=\check{f}\left[\mathscr{Y}\right] \subseteq \mathscr{P}(A)$, $\mathscr{N}\colon=\left[\mathscr{Y}\right]_{\mathscr{Alg}^{\sigma}(B)}$, $\mathscr{M}\colon=\left[\mathscr{X}\right]_{\mathscr{Alg}^{\sigma}(A)}$ respectively $\mathscr{M'}\colon=f_{*}\left[\mathscr{N}\right]$. Our objective thus becomes to show that $\mathscr{M}=\mathscr{M'}$.

On the one hand we have by definition that $\mathscr{Y} \subseteq \mathscr{N}$ whence $\mathscr{X}=\check{f}\left[\mathscr{Y}\right] \subseteq \check{f}\left[\mathscr{N}\right]=\mathscr{M'}$. Since $\mathscr{M'}$ is a $\sigma$-algebra on $A$ including $\mathscr{X}$ we infer that $\mathscr{M} \subseteq \mathscr{M'}$.

On the other hand, since  by construction $\mathscr{M} \supseteq \mathscr{X}=\check{f}\left[\mathscr{Y}\right]$ we have $f^{*}\left[\mathscr{M}\right]=\check{f}^{-1}\left[\mathscr{M}\right] \supseteq \check{f}^{-1}\left[\check{f}\left[\mathscr{Y}\right]\right] \supseteq \mathscr{Y}$. Since $f^{*}\left[\mathscr{M}\right]$ is a $\sigma$-algebra on $B$ including $\mathscr{Y}$ we infer that $f^{*}\left[\mathscr{M}\right] \supseteq \mathscr{N}$ which by virtue of the dualisation relation mentioned above leads to $\mathscr{M} \supseteq \mathscr{M'}$, thus bringing our argument to an end. $\Box$

Given a fixed topology $\mathscr{T}$ on a set $X$ let us abbreviate by $\mathscr{B}\left(\mathscr{T}\right)\colon=\left[\mathscr{T}\right]_{\mathscr{Alg}^{\sigma}(X)}$ the Borel $\sigma$-algebra associated to the space $(X, \mathscr{T})$. If we adopt notations similar to the ones above for initial and final topologies induced by arbitrary maps, the general proposition above applies in the following particular form to Borel $\sigma$-algebras:

Corollary.  Let $(Y, \mathscr{T})$ be an arbitrary space and $f \colon X \to Y$ an arbitrary map. We then have the relation $f_{*}\left[\mathscr{B}\left(\mathscr{T}\right)\right]=\mathscr{B}\left(f_{*}\left[\mathscr{T}\right]\right)$.

The result you are interested in is a special case of the above corollary, when the map $f$ is an inclusion map.


问:
How does Cantor's diagonal argument work?
I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the wikipedia article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):

\begin{align}
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.

My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:

\begin{align}
s_0 &= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
s_4 &= (0,1,1,\mathbf{1},\dots)\\
s_4 &= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\
s_5 &= (1,0,0,0,\dots)\\
s_6 &= (1,0,0,1,\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:

For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,

$$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$

...snip...

Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.

Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.

I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:

$$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 & \mathrm{if\ } s_{m,m} = 1\\1 & \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$

This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because... it's turtles all the way down!?  Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.

Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".

What am I missing?

答:
The idea, in a nutshell, is to assume by contradiction that the reals are countable. Due to that you can write them as $a_i$ for $i\in\mathbb N$. This list contains all the real numbers.

By taking the number produced by the diagonal argument you ensure at the $n$-th step that you are not any of $a_1,\ldots,a_n$. When the induction "ends" you have produced a real number which is none of the $a_i,\ i\in\mathbb N$.

This could only mean one thing: the enumeration did not capture all the real numbers like you assumed it to.

Since the enumeration was not specific, but arbitrary, we deduce that any countable enumeration of the reals will not cover the entire real numbers. This, by definition, means that the real numbers are uncountable.

Addendum:

After reading more carefully most of this thread (on its comments, and so on) I think that I should add a few words about proof by contradiction - which is a very common method in mathematics.

"Mathematics is a science of deductions", said my linear algebra prof. on the first day of my B.Sc.

"We assume A,B,C and deduce D,E,F", he continued.

The important thing about the ability to deduce things, is to remain consistent. In simple words it means not to be able to prove both something and its negation. Why is this important? Because it is simple to prove anything from contradiction, in fact according to this xkcd it is even possible to derive phone numbers (although I never managed to do that myself).

Proof by contradiction exploits the assumption that we use a consistent axiomatic system. If we assumed one thing, and derived from it contradiction then our assumption was false.

Since something is either true or false, if it is false then its negation is true.

So what did I do in the proof above? I started by assuming mathematics as we know it is consistent, then added the assumption "The reals are countable". If they are countable, then by definition we can list them as above.

The diagonal argument shows that regardless to how you are going to list them, countably many indices is not enough, and for every list we can easily manufacture a real number not present on it. From this we deduce that there are no countable lists containing all the real numbers. This is to say that the reals are not countable, which contradicts our assumption that they are countable.

I should add that some of the proofs by contradiction do not actually require the contrapositive assumption. In this case, for example, I could have said "given an arbitrary countable list of real numbers we can procure a real number not on the list" and deduce from that that the real numbers are uncountable. Sometimes, however, contradiction is slightly more essential for the proof.


问:
Prove that $1-2(1-y)(1-x)\geq0$ when $x\in[0.5,1)$ and $y\in[0,0.5)$?
I am looking for the proof that $f(x,y)=1-2(1-y)(1-x)\geq0$ when $x\in[0.5,1)$ and $y\in[0,0.5)$. This seems to be true when plotted out.

Using the second partial derivative test does not seem to work as the critical points do not land within the restricted boundaries.

Is there a simple and elegant way to show this?

Thanks.

答: You can reduce the inequality to $1/2\geq(1-x)(1-y)$ and then the bounds for $x$ and $y$ make the result immediate.


问:
Solve equation $xy^2-(3x^2-4x+1)y+x^3-2x^2+x=0$ , in $\mathbb R$
Solve equation $xy^2-(3x^2-4x+1)y+x^3-2x^2+x=0$ , in $\mathbb R$

This question was asked by another user but unfortunately was deleted when I was typing the answer. I did not want my time wasted, so I asked it again with new condition.

Solution:

Let $y=tx$ we get:

$(t^2-3t+1)x^3+2(2t-1)x^2+(1-t)x=0$

$x=0$

$(t^2-3t+1)x^2+2(2t-1)x+(1-t)=0$

$\Delta'=(2t-1)^2+(t-1)(t^2-3t+1)=k^2$

$\Delta'=t^3=k^2$

If $t=a^2$ then $t^3=(a^3)^2=k^2$

For $a∈\mathbb N$ we can have infinite solutions;

$(t, k)=(1, 1), (4, 8), (9, 27), \cdot\cdot\cdot$

$(t, k)=1, 1$  ⇒$(x, y)=(0, 0),(-2, -2) $

$(t, k)=(4, 8)$ ⇒ $(x, y)=(3, 12), (-\frac 15, -\frac 45)$

$(t, k) =(9, 27)$ ⇒ $(x, y)=(\frac 45, \frac{36}5),(\frac{-2}
{11}, -\frac{18}{11})$

Any alternative solution?

Here is a plot by wolfram

答:
Not sure why you put $y=ax$, since this is clearly a quadratic equation in $y$: $$xy^2-(3x^2-4x+1)y+x^3-2x^2+x=0$$

Its discriminant is $$D=(3x-1)^2(x-1)^2-4x^2(x-1)^2=(x-1)^3(5x-1)$$

So $y$ is defined for $x\in \left(-\infty,{1\over 5}\right]\cup [1,\infty)$ and $x\ne 0$: $$y={(x-1)\Big(3x-1\pm \sqrt{(x-1)(5x-1)}\Big)\over 2x}$$


答:
Since $x>0$ you can reduce this to showing $1-2(1-y)(1-x) \ge 0$.

It is immediate that $(1-y)(1-x) \le 1 \cdot {1 \over 2}$, so
$1-2(1-y)(1-x) \ge 1-2 \cdot 1 \cdot {1 \over 2} \ge 0$.

Title: Prove that $1-2(1-y)(1-x)\geq0$ when $x\in[0.5,1)$ and $y\in[0,0.5)$?

Body:
I am looking for the proof that $f(x,y)=1-2(1-y)(1-x)\geq0$ when $x\in[0.5,1)$ and $y\in[0,0.5)$. This seems to be true when plotted out.

Using the second partial derivative test does not seem to work as the critical points do not land within the restricted boundaries.

Is there a simple and elegant way to show this?

Thanks.


问:
discrete math - equivalence relation
I know that equivalence relations must be reflexive, symmetric and transitive. But how do I obtain the sets of equivalence relation from a specific relation?

Below is the question:
Let S be {1,2,3}. How many binary relations R on S are there such that
(i) R is reflexive?
(ii) R is symmetric?
(iii) R is an equivalence relation?

Attached is the solution.

For part (i) and (ii), how does the 3^2 -3 come about? I understand the 2 is the number of choices of whether the relation is symmetric or not symmetric.

For part (iii), how does the answer come up with the equivalence relations? I observed that the sets of relations are all possible relations in S = {1,2,3}.

I am really confused by the counting of number of reflexive/symmetric/transitive relations :(

答:
There is a natural correspondence between relations on $n$ objects and $n×n$ binary matrices. A reflexive relation is a matrix with a diagonal of all ones and a symmetric relation is a symmetric matrix.

$3^2-3$ is the number of off-diagonal entries of the matrix for the $3$-element case. If the relation is only constrained to be reflexive, they may be any value whatsoever. If the relation is only constrained to be symmetric, the two halves of off-diagonal entries (split by the diagonal) must be equal – one half may be specified arbitrarily, then the second half is fixed.

Equivalence relations on $n$ elements are counted by the Bell numbers, being equivalent to partitions of the $n$-element set (the smaller sets being equivalence classes). For three elements, the count of five relations is easy to perform manually.

Title: discrete math - equivalence relation


问:
Laurent Expansion of a given function
$$f(z)=\frac{z^2-2}{z^3}\cos\left(\frac{1}{z+1}\right)$$ I suppose $f(z)$ has an essential singularity at $z=-1$.
I found the Laurent expansion to be $$\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\frac{(-1)^{n}[2.C(m+2,m)-1]}{(2n)!}(z+1)^{m-2n}$$
Is this right? If wrong then what will be the Laurent expansion of $f(z)=\frac{z^2-2}{z^3}\cos\left(\frac{1}{z+1}\right)$ ?

答: Yes, $f$ has an essential singularity at $-1$. If you compute the Taylor series of $\frac{z^2-2}{z^3}$ centered at $-1$, which turns out to be $\sum_{n=0}^\infty(n^2+3n+1)(z+1)^n$. And you have$$\cos\left(\frac1{z+1}\right)=1-\frac1{2!(z+1)^2}+\frac1{4!(z+1)^4}-\cdots$$Therefore\begin{align}f(z)&=\frac{z^2-2}{z^3}\cos\left(\frac1{z+1}\right)\\&=\left(1+5(z+1)+11(z+1)^2+\cdots\right)\left(1-\frac1{2!(z+1)^2}+\frac1{4!(z+1)^4}-\cdots\right)\end{align}and you can use this to comput the Laurent series of $f$ centered at $-1$, although I don't think that there is a nice closed expression for it. For instance, the constant term will be$$1-\frac{11}{2!}+\frac{29}{4!}-\cdots$$whereas the coefiicient of $(z+1)^{-1}$ (that is, the residue of $f$ at $-1$) will be$$-\frac5{2!}+\frac{19}{4!}-\frac{41}{6!}+\cdots$$

Title: Laurent Expansion of a given function


问:
Laurent Series and residue of $\frac{z}{(z-1)(z-3)}$ around z = 3
As mentionned in the title, I'd like to get the function's Laurent series and after its residue, I have tried to separate the two denominators to get a partial fraction but I still have a z at numerator that's bothering me ...

So far : $$
        \frac{z}{(z-1)(z-3)} = \frac{-z}{2(z-1)} + \frac{3z}{(z-3)}
$$

I can obviously try $\omega = z-3$ and change my variable but I don't feel it'll will work ... Because I will still have this z on top of my fraction ...

Could you help me ?
Thank you very much ;)

Body:
$$\frac1{z-1}=\frac1{z-3+2}=\frac12\frac1{1+\frac{z-3}2}=\frac12\left(1-\frac{z-3}2+\frac{(z-3)^2}4-\ldots\right)=$$

$$=\frac12-\frac{z-3}4+\ldots$$

So

$$\frac z{(z-1)(z-3)}=\frac12\left(\frac3{z-3}-\frac1{z-1}\right)=\frac32\frac1{z-3}-\frac14+\frac{z-3}8+\ldots$$

so the residue is $\,3/2\,$ .


问:
can i write this $\int_{\mathbb{R}^2} \psi(\textbf{x}) \, d^{2}\textbf{x} = 2\pi \int_{0}^{\infty} \rho \, \phi(\rho)\, d\rho $?
If the function $\psi$ is isotropic, (i.e.  $\psi(\textbf{x}) = \phi(|\textbf{x}|)$, where $\phi \in L^{1}(\mathbb{R})$, then can i write
\begin{equation}
 \label{eq:2.1}
 \int_{\mathbb{R}^2} \psi(\textbf{x}) \, d^{2}\textbf{x} = 2\pi \int_{0}^{\infty} \rho \, \phi(\rho)\, d\rho ,
\end{equation}
where $\rho=|\textbf{x}|$ is the norm of $\textbf{x}$ in $\mathbb{R}^{2}?$

答:
You can view it as a substation,
$$\iint_{\mathbb R^2}\psi(x,y)dxdy$$
Let $x=r\cos(\theta)$, $y=r\sin(\theta)$.(i.e. polar coordinates) for which the jacobian is known to be $r$.
$$\int_0^{2\pi}\int^\infty_{-\infty} \phi(r^2)r~dr~d\theta$$

$$2\int_0^{2\pi}\int^\infty_0 \phi(r^2)r~dr~d\theta$$

now, doing simple substitution $r^2=\rho$
$$\int_0^{2\pi}\int^\infty_0 \phi(\rho)d\rho d\theta$$
Because the integral inside isn't dependent on $\theta$ then it is equal to
$$2\pi\int^\infty_0 \phi(\rho)d\rho$$

3D

$$\iiint_{\mathbb R^3}\psi(x,y,z)dxdydz$$
using spherical coordinates:
$$\int^{2\pi}_0\int^{\pi}_0\int^\infty_0\phi(r^2)r\sin(\theta)drd\theta d\alpha$$
$$2\pi\int_0^\pi sin(\theta)d\theta \int_0^\infty\phi(r^2)rdr$$
$$4\pi\int^\infty_0 \phi(\rho)d\rho$$
I'm unsure of the existence of a general solution.


问:
Nature of generalized integral
I am trying to find the nature of this generalized integral

$$\int_1^\infty \frac{x^2\sin x}{e^x \ln x}dx$$

I tried absolute convergence then comparison but didn't work since I found $\frac{x^2}{e^x \ln x}$

答:
Hint

Break the integral into
$$\int_1^\infty \frac{x^2sin x}{e^x ln x}dx=\int_1^e \frac{x^2sin x}{e^x ln x}dx+\int_e^\infty \frac{x^2sin x}{e^x ln x}dx$$
The second integral is convergent compared to $x^2e^{-x}$ and for the first one, the function is $\sim {1^2\times \sin 1\over e^1\ln x}$. Now what are the natures of
$$
\int_1^e{dx\over \ln x}
$$
and
$$
\int_e^\infty{x^2e^{-x}dx}
$$?

Title: Nature of generalized integral


问:
Prove that $3x+2y-2xy-2\geq0$ when $x,y\in[0.5,1)$?
I am looking for the proof that $3x+2y-2xy-2\geq0$ when $x,y\in[0.5,1)$. This seems to be true when plotted out. Is there a simple and elegant way to show this? Thanks.

答:
The inequality is equivalent to

$$
1\geq (3-2y)(1-x)
$$

Since both $(3-2y)$ and $(1-x)$ are positive in the range, we just need to prove the inequality is valid when both $x$ and $y$ are mininum.