问:
How to show $Var(T_n)=\infty$?
This question is from George Casella Statistical Inference textbook question 10.5.

For the mean $\bar{X_n}$ of n iid normal observations with $EX=\mu$ and $VarX=\sigma^2$, if we take $T_n=\sqrt{n}/\bar{X_n}$

then show $Var(T_n)=\infty$.

The solution is :

I don't quite understand this solution. Please see my process is correct or not. First, I need to know the pdf of $\bar{X_n}$, which is $n(\mu, \sigma^2/n)$. Hence,

$$E(T_n^2)=\int_{-\infty}^{\infty}\frac{n}{x^2}*pdf=\int_{-\infty}^{\infty}\frac{n}{x^2}\frac{\sqrt{n}}{\sqrt{2\pi\sigma^2}}exp(-\frac{n(x-\mu)^2}{2\sigma^2})$$.

Is my above step correct?

答:
Let's first show that $\mathsf E T_n^2=\infty$. Letting $T_n=\sqrt n/\bar X_n$ with $\bar X_n\sim\mathcal N(\mu,\sigma^2/n)$ we have
$$
\begin{aligned}
\mathsf ET_n^2
&=\int_{-\infty}^\infty\frac{n}{x^2}f_{\bar X_n}(x)\,\mathrm dx\\
&=\underbrace{\int_{-\infty}^0\frac{n}{x^2}f_{\bar X_n}(x)\,\mathrm dx}_{>0}+\int_0^1\frac{n}{x^2}f_{\bar X_n}(x)\,\mathrm dx+\underbrace{\int_1^\infty\frac{n}{x^2}f_{\bar X_n}(x)\,\mathrm dx}_{>0}\\
&>\int_0^1\frac{n}{x^2}f_{\bar X_n}(x)\,\mathrm dx\\
&=\frac{n^{3/2}}{\sqrt{2\pi}\sigma}\int_0^1\frac{1}{x^2}\underbrace{\exp\left(-\tfrac{(x-\mu)^2}{2\sigma^2/n}\right)}_{\leq 1}\,\mathrm dx\\
&\geq\frac{n^{3/2}}{\sqrt{2\pi}\sigma}\int_0^1\frac{1}{x^2}\,\mathrm dx\\
&=\infty,
\end{aligned}
$$
which completes the proof.

Now comes a bit of controversy: Does this prove $\mathsf{Var}T_n=\infty$? NO!!!

Let me be clear that this has nothing to do with Casella's proof of $\mathsf{Var}T_n=\infty$ simply being incomplete. Instead, the issue is that $\mathsf{Var}T_n$ is not infinite but rather undefined. Let me explain:

Working directly from the definition of variance we have
$$
\mathsf{Var}T_n=\mathsf E T_n^2-(\mathsf ET_n)^2.
$$
It has already been shown that $\mathsf ET_n^2=\infty$ so the fact that $\mathsf{Var}T_n$ is undefined stems from $\mathsf ET_n$ being undefined. Why is this? Working from the definition of expected value for continuous random variables we write
$$
\mathsf ET_n=\int_{-\infty}^\infty\frac{\sqrt n}{x}f_{\bar X_n}(x)\,\mathrm dx=\int_{-\infty}^0\frac{\sqrt n}{x}f_{\bar X_n}(x)\,\mathrm dx+\int_0^\infty\frac{\sqrt n}{x}f_{\bar X_n}(x)\,\mathrm dx=-\infty+\infty,
$$
which is indeterminant. This fact is due to $\lim_{x\to 0^-}x^{-1}=-\infty$ and $\lim_{x\to 0^+}x^{-1}=\infty$ so the pole in the integrand at $x=0$ is what is causing problems. If we're clever, we may try to sneak up on the pole using limits as to cancel out the infinities in our problem. To do this we let $a,b,\epsilon>0$ and write
$$
\mathsf ET_n(a,b,\epsilon)=\int_{-\infty}^{-a\epsilon}\frac{\sqrt n}{x}f_{\bar X_n}(x)\,\mathrm dx+\int_{b\epsilon}^\infty\frac{\sqrt n}{x}f_{\bar X_n}(x)\,\mathrm dx
$$
and claim $\mathsf ET_n=\lim_{\epsilon\to 0^+}\mathsf ET_n(a,b,\epsilon)$. If there is any hope of $\mathsf ET_n$ being well-defined then $\lim_{\epsilon\to 0^+}\mathsf ET_n(a,b,\epsilon)$ should at least give the same value for any choice of $a,b$ (see disclaimer at end of question). Let's now do some numerical work. For the sake of simplicity let $\mu=1$, $\sigma=1$, $n=1$. Using Mathematica we numerically evaluate $\mathsf ET_n(a,b,\epsilon)$ for increasingly small values of $\epsilon$ to find:
$$
\left(
\begin{array}{cccc}
\epsilon & \mathsf ET_n(1,1,\epsilon) &\mathsf ET_n(2,1,\epsilon) &\mathsf ET_n(3,1,\epsilon)\\
 10^{-1} & 0.676438 & 0.820142 & 0.894529 \\
 10^{-2} & 0.719939 & 0.885241 & 0.980932 \\
 10^{-3} & 0.724295 & 0.891774 & 0.989643 \\
 10^{-4} & 0.72473 & 0.892427 & 0.990514 \\
 10^{-5} & 0.724774 & 0.892493 & 0.990601 \\
 10^{-6} & 0.724778 & 0.892499 & 0.99061 \\
 10^{-7} & 0.724778 & 0.8925 & 0.99061 \\
 10^{-8} & 0.724778 & 0.8925 & 0.99061 \\
\end{array}
\right)
$$
And so we see that for different choices of $a,b$, $\lim_{\epsilon\to 0^+}\mathsf ET_n(a,b,\epsilon)$ converges to different (yet finite) values. In fact, using this approach gives a different answer for every choice of $a,b$! We thus conclude that $\mathsf ET_n$ and $\mathsf{Var}T_n$ are undefined.

Small disclaimer: Even if $\mathsf ET_n=\lim_{\epsilon\to 0^+}\mathsf ET_n(a,b,\epsilon)$ did yield the same solution for all $a,b$ this would still not be enough to prove that $\mathsf ET_n$ exists. We would instead need to show that
$$
\lim_{\delta\to 0^+}\lim_{\epsilon\to 0^+}\int_{-\infty}^{-\delta}\frac{\sqrt n}{x}f_{\bar X_n}(x)\,\mathrm dx+\int_{\epsilon}^\infty\frac{\sqrt n}{x}f_{\bar X_n}(x)\,\mathrm dx
$$
converges (which it doesn't). The purpose of this example was to show how we can make the integral representation of $\mathsf ET_n$ "equal" to any value we want by changing $a,b$.

Title: How to show $Var(T_n)=\infty$?


问:
Unique Periodic orbit
So I have the following ODE:

$\ddot{x}-(1-x^{2}-\dot{x}^{2})\dot{x}+x=0$

Now I have converted this ODE to polar coordinates which resulted in the following:

$\dot{x}=rcos(\theta)$ and $x=rsin(\theta)$

I am trying to figure out what the unique period orbit $\gamma$ is as I want to find the flow on this orbit. However, I am not sure how to use this system of polar cooridnates to quantify the orbit. Any ideas?

答: Note that if $x^2 + \dot{x}^2 = 1$ the ODE says $\ddot{x} + x = 0$, and a solution of this is $x = \cos(t+c)$ which does have $x^2 + \dot{x}^2 = 1$.

ContentLicense: CC BY-SA 4.0

Title: Unique Periodic orbit


问:
conditional expectation in markov chain proof
In Durrett's probability theory and examples, when proving chapman-kolmogorov's equation, we do the following:

\begin{align*}
P_x(X_{n+m} = z) &= E_x[P_x(X_{m+n} = z \mid \mathcal{F}_m)]\\
&= E_x [ P_{X_m} (X_n = z)] \quad (\text{by Markov Property})\\
&= \sum_y P_x(X_m = y) P_y(X_n = z)
\end{align*}

To get the first and second equality, I think what he did was
\begin{align*}
P_x(X_{n+m} = z) &= E_x[1_{\{X_{m+n} = z\}}]\\
&= E_x[1_{\{X_{n} = z\}} \circ \theta_m]\\
&= E_x[E_x[1_{\{X_{n} = z\}} \circ \theta_m \mid \mathcal{F}_m]] \quad  (?)\\
\end{align*}
where $\theta_m$ is a shift operator.

How can you take the conditional expectation on $\mathcal{F}_m$ when we don't know whether $1_{\{X_{n} = z\}} \circ \theta_m \in \mathcal{F}_m$?

答: We do not need that  $1_{\{X_{n} = z\}} \circ \theta_m$ is $\mathcal F_m$-measurable (and it does not hold in general). What is used is rather that $\mathbb E\left[X\right]=\mathbb E\left[\mathbb E\left[X\mid\mathcal G\right]\right]$ for any $\sigma$-algebra $\mathcal G$ and any integrable random variable $X$. And this holds without assuming that $X$ is $\mathcal G$-measurable.


问:
The Picard Number of a Complete Intersection
I don't remember where but I've read that the Picard number of a complete intersection of dimension at least 3 is $1$. Why?

In Dimension $2$ case the Picard number of $P^1\times P^1$ is $2$, by the Segre Embedding it is a hypersurface in $P^3$. But anything else indicated that the picard number of a complete intersection of dimension at least 3 is $1$ ? Thanks!

答:
The Picard number of projective space is $1$. Moreover, there's a version of the Lefschetz hyperplane theorem for Picard groups: if $X$ is a smooth projective variety of dimension $\geq 4$ over a field of characteristic zero and $D \subset X$ is an ample effective divisor, then the restriction map $\operatorname{Pic}(X) \to \operatorname{Pic}(D)$ is an isomorphism. (For reference, see example 3.1.25 in Lazarsfeld's Positivity in Algebraic Geometry I.)

So, for any smooth complete intersection of dimension $\geq 3$ over a field of characteristic zero, you can apply the above argument inductively to conclude that the Picard group is isomorphic to that of the ambient projective space, and thus is isomorphic to $\mathbb{Z}$. I'm not sure what happens in the case of singular complete intersections (actually, I suppose you can handle some singularities with the above inductive method, but all the singularities have to come "at the last step", i.e., it has to be given by an ample divisor in a 4-dimensional smooth complete intersection) or over a field of positive characteristic.


问:
Proving that for all numbers $a, b\in\mathbb{R}$, $\min \{a, b\} \le (a+b)/2$
Can anyone help me solve the two cases that are derived from this problem?</p>

答:
If $a=b$ you get:
$$min\{a,a\}=\frac{1}{2}2a=a$$
If $a>b$,
$$min\{a,b\}=b$$
So
$$a+b\gt 2b$$
and $a-b\gt0$. Because $a\gt b$ this is true.</p>

ContentLicense: CC BY-SA 3.0


问:
Interesting problem involving gradient
Let $f$ be a differentiable function such that $\nabla f(1,1)=(1,-3)$ and $\nabla f(1,0)=(-3,3)$. Define the function, $$h(x,y)=f(x^2,x^2-y^2).$$ Find, $||\nabla h(1,1)||$.

My attempt:

We have that, $f_x(1,1)=1$, $f_y(1,1)=-3$, $f_x(1,0)=-3$ and $f_y(1,0)=3$. How can I relate the derivatives of $h$ and $f$? I tried to use the chain rule but I couldn't find the value of $||\nabla h(1,1)||$.

答:
By the chain rule we have

$$h_x(x,y)=f_x(x^2,x^2-y^2) \cdot 2x+f_y(x^2,x^2-y^2) \cdot 2x$$

and

$$h_y(x,y)=f_x(x^2,x^2-y^2) \cdot 0+f_y(x^2,x^2-y^2) \cdot (-2y).$$

Now compute $\nabla h(1,1)$ and then $||\nabla h(1,1)||.$

Title: Interesting problem involving gradient


答:
You have $h(x, y) = f(u(x), v(x, y))$ with $u(x) = x^2$ and $v(x, y) = x^2 - y^2$.

The chain rules tells you that

$\frac{\partial h}{\partial x}(x, y) = \frac{\partial f}{\partial x}(u(x), v(x, y))u'(x) + \frac{\partial f}{\partial y}(u(x), v(x, y)) \frac{\partial v}{\partial x}(x, y)$

and

$\frac{\partial h}{\partial y}(x, y) = \frac{\partial f}{\partial y}(u(x), v(x, y)) \frac{\partial v}{\partial y}(x, y)$.

You should be able to find these expressions, then substitute $x = y = 1$.

Title: Interesting problem involving gradient

Body:
Let $f$ be a differentiable function such that $\nabla f(1,1)=(1,-3)$ and $\nabla f(1,0)=(-3,3)$. Define the function, $$h(x,y)=f(x^2,x^2-y^2).$$ Find, $||\nabla h(1,1)||$.

My attempt:

We have that, $f_x(1,1)=1$, $f_y(1,1)=-3$, $f_x(1,0)=-3$ and $f_y(1,0)=3$. How can I relate the derivatives of $h$ and $f$? I tried to use the chain rule but I couldn't find the value of $||\nabla h(1,1)||$.


问:
Sketch the curve $\frac{ax + b}{cx + d}$
I am working through a pure maths book out of interest. I am stuck on this problem:

If $a$, $b$, $c$ and $d$ are positive, sketch the curve $y = \frac{ax + b}{cx + d}$ when

i) $ad - bc \lt 0$

ii)  $ad - bc = 0$

iii)  $ad - bc \gt 0$

I can work out the basics. Thus

$x$-intercept:  $f(x) = 0 \implies \frac{ax + b}{cx + d} = 0 \implies x = \frac{-b}{a}$

so intercept is $[\frac{-b}{a}, 0]$

$y$-intercept: $f(0) = \frac{b}{d}$ at point $[0,\frac{b}{d}]$

horizontal asymptote: $\frac{a}{c}$

vertical asymptote: $cx + d = 0 \implies x = \frac{-d}{c}$

I can also see that $\frac{ax + b}{cx + d} = \frac{a}{c}\left(1 + \frac{\frac{b}{a}-\frac{d}{c}}{x + \frac{d}{c}}\right)$

but I cannot put all this together to see how one could sketch the graphs except in one case: where $ad -bc = 0$ because then $\frac{b}{a}-\frac{d}{c} = 0$ and the graph will be a horizontal line equal to $\frac{a}{c}$

I think not knowing the actual magnitudes makes it difficult for me to visualise.

答:
Assuming $c \not=0$, you can write $$y = \frac ac + \frac{bc-ad}{c^2(x+\frac dc)}$$

so $y$ is not defined when $x=-\frac dc$ and

if $ad=bc$ you have $y$ otherwise being constant $\frac ac$

if $ad > bc$ you have a hyperbola with horizontal asymptote $y=\frac ac$ and vertical asymptote $x=-\frac dc$ using the second and fourth quadrants

if $ad < bc$ you have a hyperbola with horizontal asymptote at $y=\frac ac$ and vertical asymptote at $x=-\frac dc$ using the first and third quadrants

Meanwhile if $c=0$ and $d \not=0$, you have the straight line $y=\frac ad x+\frac bd$, with the slope affected by the sign of $\frac ad$, which is equivalent to the sign of $ad-bc$


问:
Basis of a basis
I'm having troubles to understand the concept of coordinates in Linear Algebra.

Let me give an example:

Consider the following basis of $\mathbb R^2$:

$S_1=\{u_1=(1,-2),u_2=(3,-4)\}$ and $S_2=\{v_1=(1,3),v_2=(3,8)\}$

Let $w=(2,3)$ be a vector with coordinates in $S_1$, then $w=2u_1+3u_2=2(1,-2)+3(3,-4)=(11,-16)$.

When I tried to found the coordinates of $w$ in $S_2$, I found the following problem:

Which basis $(11,-16)$ belongs to? I suppose the same of $u_1$ and $u_2$, but which basis $u_1$ and $u_2$ belongs to? and if I discover the basis of $u_1$ and $u_2$, what's the basis of the basis of $u_1$ and $u_2$?

I found an infinite recurrence problem and I was stuck there.

Maybe I'm seeing things more complicated than it is, but it seems that there is a deeper and philosophical question inside of this doubt, I couldn't see what a coordinate really is.

I would be very grateful if anyone help me with this doubt.

答:
When starting out, it's often easy to confuse the basis coordinates and the elements of $\Bbb R^2$ because they are all written as ordered pairs with parentheses.

Let's try to keep them separate by using $\langle a ,b\rangle$ for coordinates in $S_1$, $[a,b]$ for coordinates in $S_2$ and just $(a,b)$ for elements of $\Bbb R^2$. This should help you envision "which coordinates belong to which basis," and help you keep those separate from the elements of $\Bbb R^2$.

Then $\langle2,3\rangle$ is the coordinates for the vector $2(1,-2)+3(3,-4)=(11,-16)$

To find the coordinates $[a,b]$ for $(11,-16)$, you'll have to solve the following equation:

$$a(1,3)+b(3,8)=(11,-16)$$

Comment: what others have said about interpreting $(a,b)$ as coordinates of "the standard basis" is correct, but I don't find it immediately relevant. I think it's pedagogically more helpful to emphasize their identity as elements of the vector space, rather than yet another set of coordinates.

We need to get you to recognize that the coordinates of a point in a basis are the coefficients you need to manufacture that vector using the basis :)

Title: Basis of a basis


问:
$Av>0, v\ge 0$, $Aw<0, w\ge 0$ for a square matrix $A$ of certain type
Consider an $n\times n$ matrix $A = B - D$, where $B$ has positive coefficients and $D$ is diagonal with positive coefficients.

I want to prove that for such $A$ there don't exist vectors $v, w \in\mathbb{R_{\ge 0}^n}$ such that $Av$ has strictly positive coefficients and $Aw$ has strictly negative coefficients.

Geometric interpretation: hyperplanes $x_i = \pm 1$, $x_i=0$ define $2^n$ cubes. Consider a cube $C$, pick points $P_1, P_2, ..., P_n$ from inside each neighbour of $C$ (one point from each). Let $O$ be the origin. Then convex hull of $OP_1P_2...P_n$ intersects either $C$ or the cube symmetric to $C$ trivially (the intersection is $O$).

I proved the claim for $n=2$, and drew a picture in Geogebra for $n=3$, which made me believe that the fact should be true.

Can it be proven using Farkas' lemma? Seems like very similar setting, but the lemma involves $A^T$.

If not, then what to do? I see that it is enough to prove that the plane $P_1P_2...P_n$ doesn't intersect one of $C$, $-C$, because then I could separate $C$, $-C$ by drawing parallel plane through $O$, but this is where I'm stuck.

答:
Here's a solution due to Pavel Myaktinov.

We work by induction on n. We will show the step for $n=3$, it obviously generalises for $n>3$.

If $A$ has a non-negative row, then obviously there is no vector $v$ such that $v\ge 0$, $Av<0$. So, the diagonal entries of $A$ are negative. We may scale any row by a non-negative number, so, WLOG the diagonal entries of $A$ are $-1$.

Then we have a system of inequalities (where $\vee$ stands either for $>$ or for $<$ and all letters are positive)

$$    \begin{cases}
      x \vee ay + cz\\
      y \vee cx + dz \\
      z \vee ex + fy
    \end{cases}\,.
 $$
If this system has a solution, then the following system does, too:

$$    \begin{cases}
      y \vee c(ay+cz) + dz \\
      z \vee e(ay+cz) + fy
    \end{cases}\,.
 $$

Note that the coefficient of $z$ in the first inequality is positive and so is the coefficient of $y$  in the second inequality. That means that the new matrix $2\times 2$ is again of the form $B - D$ and we've succesfully reduced the case $n=3$ to the case $n=2$.


问:
Persistence barcodes given a sequence of abstract homology groups
For a sequence of nested complexes $K_1\subset K_2 \subset K_3$, I have calculated the first homology groups at each level,

$$
H_1(K_1)=\left\cong \mathbb{Z}^2\\
H_1(K_2)=\left\cong \mathbb{Z}^3\\
H_1(K_3)=\left/\left\cong\left\cong\mathbb{Z}^2\\
$$
where $a$, $b$, and $c$ are 1-simplexes and the homology groups are constructed with integer coefficients from these generating elements.

In this scenario, how do I go about drawing the barcode for $H_1$? I know the rank of $H_1$ should match the number of generating elements, so it should be $2,3,2$ in this case.

For the first two groups ($H_1(K_1)$ and $H_1(K_2)$) I could imagine taking one bar for the element $a$ and one bar for $b$, that are both present at all stages. However, the quotient $H_1(K_3)$ is also isomorphic to $\left$, so using this strategy can lead to at least two very different barcodes (see image below)?

This represents whether $c$ is used as a generating element in $H_1(K_3)$ or not. The rank (number of bars) is correct for both barcodes. Anyway, it seems I am wrong on how to construct barcodes since I expected a single correct answer. How should one go about this?

答:
When looking at filtered modules $\mathbb{V} = \{V_t\}$, it's useful to consider the ranks over an interval in addition to the ranks just at single time points, i.e., look at $$\beta^{s,t} = \operatorname{rank}(V_s \to V_t)$$ not just when $s = t$, but also for $s < t$.

In your example, consider the homomorphism $H_1(K_1) \to H_1(K_3)$, which is an isomorphism and has rank 2.  So the barcode should contain two full bars spanning from beginning to end.  This is consistent with your first answer but not the second.

In general, this is encapsulated in the so-called elder rule: when two classes become homologous, the older one survives and the death should be paired with the class that was born more recently.


问:
canonical one-form
The canonical one-form is defined here:
http://books.google.nl/books?id=uycWAu1yY2gC&lpg=PA128&dq=canonical%20one%20form%20hamiltonian&pg=PA128#v=onepage&q&f=false

My problem is this:
It states that if $(x_1,\dots x_n)$ are local coordinates in $N$,
a 1-form $\alpha\in T^*_x N$ is represented by $\alpha = \Sigma^n_{j=1} y_jdx_j$
It then goes on to define a special 1-form $\lambda$ on $T^*N$ by
$\lambda = \Sigma^n_{j=1} y_jdx_j$.

This to me looks the same as $\alpha$ and as a 1-form on $N$ and
not on $T^*N$. What am I missing here?

答:
This sort of thing can be a bit confusing!

A local one-form on $N$ is given by $\sum_j y_j(x) dx_j$; this gives coordinates $(y_1,\ldots,y_n)$ on the fibres of $T^*N$, so we have local coordinates $(x_1,\ldots,x_n,y_1,\ldots,y_n)$ on the total space of $T^*N$.  Therefore a local one-form on $T^*N$ is $\sum_j\big(\alpha_j(x,y) dx_j + \beta_j(x,y) dy_j\big)$.

The canonical one-form on $T^*N$ is given by $\alpha_j = y_j,~ \beta_j = 0$.  In words, it has no components 'pointing along the fibre', and its transverse components are given by the point we're sitting at in the fibre.

Title: canonical one-form


问:
Find multiple tangents for parabola
I have a parabola $x = 0.5y^2 - 4y + 3$. I need to find a tangent(s) that go throught point $(2,-1)$. I have tried:

$y - (-1) = m(x-2)$

$4y = 0.5y^2 - x + 3$

$y = (0.5y^2)/4 - x/4 + 3/4$

$m(x-2) - 1 = (0.5y^2)/4 - x/4 + 3/4$

But I'm pretty sure that's incorrect, could I have some advise on the problem? Thanks a lot

答:
Can you see where these equations come from?

$$
\begin{cases}
x= 2 + m(y+1) &\qquad&(1)\\
m = -4+y &&(2)\\
x = \frac 12 y^2 -4y +3 &&(3)\\
\end{cases}
$$

The tangency point belongs to both the parabola and the tangent line (equations 1 and 3), and the slope of the tangent line is the derivative at the tangency point (equation 2).

This way you find two solutions, for $y=-1\pm\sqrt{11}$, with slopes $-5 \pm \sqrt{11}$.

Title: Find multiple tangents for parabola


问:
Comparison Test Proof (Hammack's Book of Proof, Third Edition, Exercise 13.8.2)
I am currently going through Hammack's Book of Proof (3rd edition), and have been stuck on the following exercise for the past two weeks.

Prove the comparison test: Suppose $\Sigma a_k$ and $\Sigma b_k$ are series. If $0 \le a_k \le b_k$ for each $k$, and $\Sigma b_k$ converges, then $\Sigma a_k$ converges. Also, if $0 \le b_k \le a_k$ for each $k$, and $\Sigma b_k$ diverges, then $\Sigma a_k$ diverges.

Now, the reader is instructed to prove this using "Definition 13.7 (and Definition 13.5, as needed)" from the book which goes as follows.

Definition 13.7

A series $\sum_{k=1}^{\infty} a_k$ converges to a real number $S$ if its sequence of partial sums $\{s_n\}$ converges to $S$. In this case we say $\sum_{k=1}^{\infty} a_k = S$.

We say $\sum_{k=1}^{\infty} a_k$ diverges if the sequence $\{s_n\}$ diverges. In this case $\sum_{k=1}^{\infty} a_k$ does not make sense as a sum or does not sum to a finite number.

Definition 13.5

A sequence $\{a_n\}$ converges to a number $L \in \mathbb{R}$ provided that for any $\epsilon > 0$ there is an $N > \mathbb{N}$ for which $n > N$ implies $|a_n - L| < \epsilon$.

If $\{a_n\}$ converges to $L$, we denote this state of affair as $\lim_{n \to \infty} a_n = L$

If $\{a_n\}$ does not converge to any number $L$, then we say it diverges.

I am currently focusing on the first statement. Also, that statement intuitively makes sense to me: if the series $\Sigma a_k$ has only non-negative terms, then it will either converge or diverge to infinity. However, since $0 \le a_k \le b_k$, it follows that $\Sigma a_k \le \Sigma b_k$ and thus, if $\Sigma a_k$ did diverge to infinity we would have a contradiction. I am able to formalize this as follows.

Proof. For the sake of contradiction, suppose $\Sigma a_k$ diverges to infinity. That is, for every $L \in \mathbb{R}$ there exists a number $N' \in \mathbb{N}$ for which $n \ge N'$ implies $A_n > L$ where $A_n$ is the $n$-th partial sum of $\Sigma a_k$. Also, because $\Sigma b_k$ converges to some number $B \in \mathbb{R}$, we have that for every $\epsilon > 0$, there exists a number $N'' \in \mathbb{N}$ for which $n > N''$ implies $|B_n - B| < \epsilon$ where $B_n$ is the $n$-th partial sum of $\Sigma b_k$. Further, notice that if $0 \le a_k \le b_k$ for each $k$, then $A_n = (a_1 + a_2 + \dots + a_n) \le (b_1 + b_2 + \dots + b_n) = B_n$.

Now, take $\epsilon > 0$ and $L = B - \epsilon$. Also, let $N = \max(N', N'')$. Then, if $n > N$ we have $A_n > B - \epsilon$ and $|B_n - B| < \epsilon$, or $B_n < \epsilon + B$. Thus we have $B_n < \epsilon + B < A_n$ and $A_n \le B_n$, a contradiction. Therefore, $\Sigma a_k$ must converge.
$$\tag*{$\blacksquare$}$$

The problem I have is that although divergence to infinity implies divergence, divergence does not imply divergence to infinity. Therefore, for my proof to be complete, I would need to show that if $\Sigma a_k$ diverges then it must diverge to infinity. However, I am unable to formalize this idea. I have looked it up online and stumbled accross proofs that uses Cauchy sequences or the monotone convergence theorem, but nothing that uses only the definitions above. I also tried a direct proof without any luck.

Any pointers or help would be greatly appreciated.

答:
I finally managed to complete my proof by showing that if $\Sigma a_k$ diverges, then it must diverge to infinity.

Proof. Suppose the series $\Sigma a_k$ diverges. Then, for all $L \in \mathbb{R}$, there exists a $\epsilon > 0$ such that for all $N \in \mathbb{N}$, there is a number $n>N$ for which $|A_n - L| \ge \epsilon$ where $A_n$ is the $n$-th partial sum of $\Sigma a_k$. Also suppose that $a_k \ge 0$ for all $k$. We must show that this implies that $\Sigma a_k$ diverges to infinity, or, in other terms, that for all $L > 0$, there is a $N \in \mathbb{N}$ for which $n > N$ implies $A_n > L$.

Now, from $|A_n - L| \ge \epsilon$ we have $(A_n - L \ge \epsilon) \lor (A_n - L \le - \epsilon)$.

Case 1.
Suppose $A_n - L \ge \epsilon$. Thus, we have a number $\epsilon > 0$ such that for all $N \in \mathbb{N}$ there is a $n > N$ for which $A_n \ge L+ \epsilon$. However, from $a_k \ge 0$ we have that $A_{n+1} \ge A_n$ for all $n$. Therefore, $n' > n$ implies $A_{n'} \ge A_n \ge \epsilon + L$. That is, for all numbers $L+\epsilon = L' > 0$, we have a $N \in \mathbb{N}$ for which $n > N$ implies $A_n > L'$. That is, by definition, $\Sigma a_k$ diverges to infinity.

Case 2.
Suppose $A_n - L \le -\epsilon$. Now, from the definition of divergence, we also have a $N' > n$ for which there is a number $n' > N'$ such that $|A_{n'} - L| \ge \epsilon$, or $(A_{n'} - L \ge \epsilon) \lor (A_{n'} - L \le -\epsilon)$. From case 1 above we know that $A_{n'} - L \ge \epsilon$ implies convergence to infinity. Now, for $A_{n'} - L \le -\epsilon$, taking the difference of $A_{n'}$ and $A_n$ we get $A_{n'} - A_{n} \le 0$. That is, $(a_{n+1} + a_{n+2} + \dots + a_{n'}) \le 0$ and so it follow that $a_{n'} = 0$ for all $n' > N' > n$. Then, for any $\epsilon > 0$, we have a number $n \in \mathbb{N}$ such that  $n' > n$ implies $|A_{n'} - A_n| = 0 < \epsilon$. Therefore, by definition, we have that $\Sigma a_k$ converges, a contradiction.

The above two cases show that if $\Sigma a_k$ diverges, then it must diverge to infinity. $$\tag*{$\blacksquare$}$$


问:
A characterisation of semi-continuity

Show that function $J$ is semi-continuous at point $v\in U$ iff $\forall \epsilon >0$ there exists $\delta >0$ such that $\forall u\in U\cap L_\delta (v)$, $J(u)>J(v)-\epsilon $.

$L_\delta (v)$ is open ball with center in $v$ and radius $\delta$.

Definition of semi-continuity is "Function $J: U\rightarrow R, U\subset R^n$ is semi-continuous from under at point $v\in U$ if $\lim\limits_{k \to \infty} \inf J(u_k)\geq J(v)$ for every sequence $\{u_k\}_{k\in N}$ that converges to $v$".

Please, can anyone help me to show this.

答:
"If" part: Suppose that the "$\varepsilon/\delta$" condition holds, and let $u_k$ be a sequence as in definition of semicontinuity. Fix $\varepsilon$, and take $\delta$ as in the condition. Then for some $n_0$, for $k > n_0$ we have $u_k \in L_\delta(u)$. Hence $J(u_k) > J(v) - \varepsilon$. Passing to $\liminf_k J(u_k) \geq J(v) - \varepsilon$. Since $\varepsilon > 0$ was arbitrary, you get $\liminf_k J(u_k) \geq J(v)$

"Only if" part:
 Suppose that $J$ is semicontinuous from below. If the "$\varepsilon/\delta$" condition was false, then for some $\varepsilon > 0$ we would be able to find $u_k$ with $|u_k - v| < \frac{1}{k}$ but $J(u_k) \leq J(v) - \varepsilon$ (else, $\delta = 1/k$ would work). Passing to the limit we find $\liminf_k J(u_k) \leq J(v) - \varepsilon$, contradicting semicontinuity.

Title: A characterisation of semi-continuity


问:
Any subgroup $H$ of $G$ contains a normal subgroup of $G$ that contains all normal subgroups of $G$ contained in $H$
As in the title, I need to prove the following statement: Let $G$ be a group and $H$ a subgroup. Then $H$ contains a normal subgroup of $G$, say $N$, such that all normal subgroups of $G$ that are contained in $H$ are also contained in $N$.

My thought is to take the union $R$ of all the normal subgroups of $G$ that are contained in $H$, and let $N$ be the normal subgroup of $G$ generated by $R$, and then prove that $N$ is contained in $H$. But I do not know how to proceed. Any help is appreciated.

答:
Let $\{N_\lambda\}_{\lambda\in\Lambda}$ be the family of all normal subgroups of $G$ contained in $H$. The family is nonempty, since $1\subseteq H$ is a normal subgroup of $G$. I claim $N:=\sum_{\lambda\in\Lambda}N_\lambda$, the group generated by all the $N_\lambda$'s, satisfied the conditions given. First, since $N_\lambda\subseteq H$ for each $\lambda\in\Lambda$, clearly we also have $N\subseteq H$. Moreover, every normal subgroup of $G$ contained in $H$ is one of the $N_\lambda$, which is by definition contained in $N$.

Thus, all that is left to show is that $N$ is a normal subgroup of $G$. But this is clear, since an element of $N$ is of the form $h_1h_2\dots h_n$, where $h_n\in N_{\lambda_n}$, and so for each $g\in G$, we have $g(h_1h_2\dots h_n)g^{-1}=(gh_1g^{-1})(gh_2g^{-1})\dots(gh_ng^{-1})\in N$. (This argument also shows uniqueness.)

Alternatively, you can just let $N=\bigcap_{g\in G}gHg^{-1}$, which is easily checked to satisfy all the necessary properties.


问:
The height of a tangential trapezoid
Let $ABCD$ be a tangential trapezoid, touching the circle $k(O;r)$. Show that the height of the trapezoid $h=2r$.

Let $M$ and $N$ be the points on  $AB$ and $DC$, respectively, at which the circle is tangent $\Rightarrow OM\perp AB, ON\perp CD$ (The radius of a circle is perpendicular to the tangent line through its endpoint on the circle's circumference.) What can we do next?

答:
Proving angle $MON = 180^{o}$ should complete the proof, as that would mean the line $MN$ is the diameter of the circle.
In order to prove this draw a horizontal line passing through the $O$ which touches the side $AD$ at a point say $E$. Angle $MOE$ is $90^o$ (can be proved by using the fact that the line is horizontal and sum of angles of a quadrilateral is $360^o$), Similarly Angle $NOE$ is $90^o$. Thus Angle $MON$ is $180^o$

Title: The height of a tangential trapezoid


问:
Verifying a set theory statement within a proof
Let $\dots E_N \subset E_2 \subset E_1$, $E = \cap_{i = 1}^{\infty} E_i$ and $E_i$ decreases to $E$. $E$ is a measurable set, and $m(E_k) < \infty$ for some $k$.

Show $$E_1 = E \cup \cup_{j = 1}^{\infty} G_j$$

where $G_j = E_j - E_{j + 1}$ and $G_j$ are disjoint.

We were asked to verify this statement within a proof in my measure theory class, but I am unfortunately quite weak when it comes to set theory. I never took a course in it, and I heard my Discrete Math professor didn't really provide me with sufficient tools. So I'm trying to fill in what I can as problems come up. Can anyone help me with this problem? Thanks!

答: Every $x \in E \cup \cup_{j=1}^{\infty} G_j$ is clearly in $E_1$.  Now suppose that $x \in E_1$.  If $x$ is in every $E_i$ then $x$ is in $E$ and thus in $E \cup \cup_{j=1}^{\infty} G_j$.  If $x$ is not in every $E_i$ then let $N$ be the largest $N$ such that $x \in E_N$.  Note then that $x$ is not in $E_{N+1}$.  Such an $N$ exists since $x$ is not in all $E_{i}$ and the sequence is decreasing with respect to set containment.  Hence $x$ is in $G_{N}$ and thus in $E \cup \cup_{j=1}^{\infty} G_j$.


问:
Why is a function defined to have only a single value at a point?
I'm aware people have asked why a function cannot have more than one value before, on this site and on others, but I found none of the answers satisfactory, which mostly seemed to amount to "because it's defined that way", and did not discuss the reason for this definition. I do not understand the rationale behind this definition, it just seems to me a strange constraint on what otherwise was a very general notion of a "mathematical transform".

An argument could be raised that, under the accepted definition, certain properties we can conveniently say are true for all functions, won't be true for all functions under the alternative definition. For example,
$$\forall x,y.x=y\implies f(x)=f(y) \text{ for any function }f$$
But consider the converse of this same statement,
$$\forall x,y. f(x)=f(y)\implies x=y\text{ for any }\textbf{injective}\text{ function }f$$

We're already used to saying that some properties hold only for certain kinds of functions; it isn't too far a stretch to then assign a category to the functions which can produce only a single value, and say the first statement holds true for all functions of that category.

Is there then a logical reason for this special constraint or is it just this way due to convention or historical reasons?

答:
There are many "transforms" from one set to another, if you think of a transform as just a set of pairs $(x,y)$.

One of the most useful transforms is the one that corresponds to the intuitive idea that you have some kind of a "rule" that tells you an element of the codomain whenever you're given an element of the domain. The cost of a load of coal depends on the number of tons you buy. The distance you travel at a fixed speed depends on how long you drive. The position of a satellite in  orbit around the Earth depends on [some set of inputs].  In each case the input determines just a single answer. It's that idea that mathematicians have codified in the abstract definition of a function.

If you changed the definition of a function to include "multi-valued functions" no actual mathematical facts would change. But many of the most common and useful statements about functions would have to start

If $f$ is a function [that is single valued] then ...

which would complicate exposition.

The accepted definition is not

a strange constraint on what otherwise was a very general notion of
a "mathematical transform".

It's the choice we've made precisely because it makes dealing with the most common situation easiest.


问:
Integral curves associated to the movement of a pendulum
The equation of a pendulum is given by $\ddot{\theta}=-\sin(\theta)$ so we write this as a system
$$\begin{cases} \theta'=\dot\theta\\
(\dot\theta)'=\ddot\theta=-\sin(\theta)\end{cases}$$
The vector field associated to this system is $\vec{F}(\theta,\dot\theta)=(\dot\theta,-\sin\theta)$
and the integral curves are given by $$\dot\theta^2=2\cos(\theta)+c\;\;\;\; (*)$$ I want to understand how to obtain the last equation $(*)$.

My undestanding is that if we have a curve $\gamma(t)=(\theta(t),\dot\theta(t))$ then we need the vector $\gamma'(t)$ to be colinear to the vector field $\vec{F}(\theta,\dot\theta)=(\dot\theta,-\sin\theta)$ but I can't see how write this into equations that lead to $(*)$.

答:
Multiply the first equation by $\dot \theta$ to obtain
\begin{align}
\dot \theta \ddot \theta & = -\dot \theta \sin \theta \\
\frac{1}{2} \frac{d}{dt} \dot \theta^2 & = \frac{d}{dt} \cos \theta \\
\frac{d}{dt} \left(\frac{1}{2} \dot \theta^2 - \cos \theta \right) & = 0 \\
\frac{1}{2}\dot \theta^2 - \cos \theta & = E,
\end{align}
where $E$ is a constant of integration.


答:
We have that if $x\in E\cup \bigcup_{i=1}^{\infty} G_i$ then since $E\subseteq E_1$ and $G_i=E_i\setminus E_{i+1}\subseteq E_i\subseteq E_1$ we have $x\in E_1$. So $E\cup \bigcup_{i=1}^{\infty} G_i\subseteq E_1$$

if instead $x\in E_1$ then there are two cases. If $x\in E$ then $x\in E\cup \bigcup_{i=1}^{\infty} G_i$. If $x\notin E$ then $x\notin \bigcap_{i=1}^\infty E_i$ but $x\in E_1$. Since the $E_i$ are a downwards chain under inclusion or rather $E_1\supseteq E_2\supseteq\cdots$ there must be a smallest $n$ such that $x\notin E_n$ if there wasn't a smallest $n$ then $x$ would be in all of the $E_i$ and therefore would be in the intersection. Being the smallest $x\in E_{n-1}$ so $x\in E_{n-1}\setminus E_n=G_{n-1}$ therefore in either case if $x\in E_1 $ then $x\in E\cup \bigcup_{i=1}^{\infty} G_i$. So $E_1 \subseteq  E\cup \bigcup_{i=1}^{\infty} G_i$.

This means that the two sets are subsets of each other and thus must be equal

Title: Verifying a set theory statement within a proof

Body:
Let $\dots E_N \subset E_2 \subset E_1$, $E = \cap_{i = 1}^{\infty} E_i$ and $E_i$ decreases to $E$. $E$ is a measurable set, and $m(E_k) < \infty$ for some $k$.

Show $$E_1 = E \cup \cup_{j = 1}^{\infty} G_j$$

where $G_j = E_j - E_{j + 1}$ and $G_j$ are disjoint.

We were asked to verify this statement within a proof in my measure theory class, but I am unfortunately quite weak when it comes to set theory. I never took a course in it, and I heard my Discrete Math professor didn't really provide me with sufficient tools. So I'm trying to fill in what I can as problems come up. Can anyone help me with this problem? Thanks!


问:
Is there a connected $n$-sheeted covering space of a closed orientable surface for any positive integer $n$?
Let $S$ be a closed orientable surface. For any positive integer $n$, is there a connected $n$-sheeted covering space of $S$? This is certainly not true if $S=S^2$ because $S^2$ is simply-connected. The result is true for $S=T^2$, since $\pi_1(T^2)=\Bbb Z^2$ has an index $n$ subgroup for each $n$. However I cannot handle the other cases. I know that this is equivalent to finding an index $n$ subgroup of $\pi_1(S)$ but the $\pi_1$ for the other surfaces are quite complicated, I think. Any hints?

答:
Since the first homology is the abelianization of the fundamental group, and $H_1(S)=\mathbb{Z}^{2g}$, where $g$ is the genus, we see that we have a surjective composition $\pi_1(S) \rightarrow H_1(S) \rightarrow \mathbb{Z}$, where this last map is just projection onto the first coordinate.

Since taking the preimage of a subgroup $H \leq G$ under a surjective map preserves index, we deduce that $\pi_1(S)$ contains an index n subgroup given by the preimage of $n\mathbb{Z}$ under the above composite. As you mention, covering space theory implies we have the covering you ask for.


问:
What is a strictly positive definite covariation?
We have martingales $(M^1,...,M^d)$. What does it mean if the covariation $(\langle M^j,M^i\rangle)$ is strictly positive definite?

Does someone know a reference?

What does it mean if I integrate with respect to this positive covariation $\langle M^i,M^j\rangle$ ?

答:
Right from the start, that the (quadratic)-covariation is positive definite has notting to do with the given martingales $(M^1,...,M^d)$.

The covariation is a symmetric bilinear form. And for symmetric bilinear form beeing positive definite is a well know definition. But for beeing definite you can't just consider the covariation on the set of semimartingales (the biggest set for which the covariation is defined, as far as I know), as all finite variation processes have zero quadratic variation and thus the zero-value is not unique. But if you consider the equivalence class $V$ in which all processes are identified, iff their difference has zero quadratic variation, then the zero-value is unique. This means for all $X\in V$ $\langle X,X\rangle\geq0$ and equality iff $X$ is the zero-value in $V$.

On this space $V$ the covariation is a positive definite symmetric bilinear from, which means it is a scalar product. Thus $V$ equiped with $\langle \cdot,\cdot\rangle$ is a inner product space.


问:
Two estimations in analytic number theory
I want to show the following two estimation:
$$
\sum_{p|n}\frac{\log p}{p}=O(\log\log n)
$$
And
$$
\sum_{p|n}\frac1p=O(\log\log\log n)
$$

First assume $$n=p_1^{\alpha_1}...p_s^{\alpha_s}$$ then the first can be rewritten as:
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}
$$

I think i should bulid some connections between s and n, and could i do the following?
$$
\sum_{i=1}^s\frac{\log p_i}{p_i}=O(\sum_{i=1}^s\frac{1}{i})
$$

If this is true, why?

答:
The only ingredient needed is the elementary (Chebysev estimates) $Ak\log k < q_k < Bk\log k, k \ge 2$ where $q_1=2, q_2=3,...$ is the sequence of primes. This implies:

$\frac{1}{q_k} < \frac{A_1}{k \log k}, k \ge 2$ and $\frac{\log q_k}{q_k} < \frac{A_2}{k}, k \ge 1$.

Also $\frac{\log x}{x}$ is decreasing for $x \ge 3$ so if $p \ge q_k, \frac{\log p}{p} < \frac{A_2}{k}, k \ge 2$

If $n=p_1^{\alpha_1}...p_s^{\alpha_s}$ (in increasing order of the primes) we have:

$\log n=\sum_{k=1}^s\alpha_k \log p_k \ge s\log 2$ since $\alpha_k \ge 1, \log p_k \ge \log 2$, hence :

$s \le C\log n$

Then using that $p_k \ge q_k$ we get that:

$\sum_{p|n}\frac{\log p}{p} \le \sum_{k=1}^{C\log n}\frac{A_2}{k} \le A_3  \log \log n$

$\sum_{p|n}\frac{1}{p} \le \sum_{k=2}^{C\log n}\frac{A_1}{k\log k} \le A_4 \log \log  \log n$, so we are done!


问:
Laplace Transformations
can someone kindly help me with these few questions? :)

Find $L{e^tf(t)}$ in terms of $f*(s)$ and state a range of $s$ which this is defined.

I couldn't figure this out. You use the definition but i then get $e^t(1-s)f(t)$ and then i don't know what to do....

Using the Convolution Theorem, find the function f(t) satisfying the equation

$$f(t) = \int_0^t e^u f(t-u)\mathrm du + e^t$$

I know how to take the LT of both sides. I'm not sure how to figure out the LT of the integral though. Ive for f(s) = something + 1/(s-1) at the moment.

答:
Here is a detailed solution to the modified -- substantially more challenging -- problem (see the OP's comments below the previous answer; in particular, it is stated there that this is not homework).

To find $f:[0,\infty) \to \mathbb{R}$ satisfying the equation
$$
f(t) = \int_0^t {u e^u f(t - u)\,du}  + e^t, \;\; t \geq 0,
$$
begin, as in the previous answer, by writing
$$
\hat f(s) = \hat \varphi (s) + \frac{1}{{s - 1}},\;\; s > 1,
$$
where this time $\hat \varphi$ is the Laplace transform of the convolution
$$
(f*te^t )(t) = \int_0^t {ue^u f(t - u)\,du} \,\bigg(= \int_0^t {f(u)(t-u)e^{t - u}\,du}\bigg).
$$
By the convolution theorem,
$$
\hat \varphi(s) = \hat f(s) \frac{1}{{(s - 1)^2 }}.
$$
It is worth noting that the term $1/(s-1)^2$ can be derived as follows, recalling that an exponential random variable with density function $\lambda e^{-\lambda t}$, $t \geq 0$, has mean equal to $1/\lambda$ (here $\lambda = s -1 > 0$):
$$
\int_0^\infty  {e^{ - st} te^t \,dt}  = \frac{1}{{s - 1}}\int_0^\infty  {t(s - 1)e^{ - (s - 1)t} \,dt}  = \frac{1}{{(s - 1)^2 }}.
$$
Solving for $\hat f(s)$ (using the above expression for $\hat \varphi(s)$) gives
$$
\hat f(s) = \frac{{s - 1}}{{s^2  - 2s}} = \frac{{(s - 2) + 1}}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{{s(s - 2)}} = \frac{1}{s} + \frac{1}{s}\frac{1}{{s - 2}}.
$$
Assuming that $s > 2$, it follows by inversion (and the Convolution Theorem) that
$$
f(t) = 1 + (1 * e^{2t})(t),\;\; t \geq 0.
$$
(Indeed, note that $\int_0^\infty  {e^{ - st} 1\,dt}  = \frac{1}{s}$ and $\int_0^\infty  {e^{ - st} e^{2t} \,dt}  = \frac{1}{{s - 2}}$.)
Finally, from
$$
(1 * e^{2t})(t) = \int_0^t {e^{2u} 1\,du} \,\bigg( = \int_0^t {1e^{2(t - u)} \,du} \bigg) ,
$$
it follows that
$$
f(t) = 1 + \frac{{e^{2t}  - 1}}{2} = \frac{{e^{2t}  + 1}}{2},\;\; t \geq 0.
$$
Indeed, this $f$ satisfies the original equation; that is, as one can easily verify, it holds
$$
\frac{{e^{2t}  + 1}}{2} = \int_0^t {ue^u \frac{{e^{2(t - u)}  + 1}}{2}\,du}  + e^t .
$$

EDIT (in response to the OP's comment below). While inverting $1/s$ gives $1$ and inverting $1/(s-2)$ gives $e^{2t}$, inverting $1/(s(s-2))$ does not give the product of $e^{2t}$ and $1$; rather, by the Convolution Theorem, it gives the convolution of $e^{2t}$ and $1$. Since $(1 * e^{2t})(t) = \frac{{e^{2t}  - 1}}{2}\,( = \int_0^t {e^{2u} \,du} )$,
$$
f(t) = 1 + \frac{{e^{2t}  - 1}}{2} = \frac{{e{}^{2t} + 1}}{2}
$$
(which was verified by substitution into the original equation).
However, as the OP observed, the solution can be obtained more elementarily by splitting $\hat f(s)$ into partial fractions. Specifically,
$$
\hat f(s) = \frac{{s - 1}}{{s^2  - 2s}} = \frac{1}{{2s}} + \frac{1}{{2(s - 2)}},
$$
from which it follows, by inversion, that
$$
f(t) = \frac{1}{2} + \frac{1}{2}e^{2t}  = \frac{{e^{2t}  + 1}}{2}.
$$

Title: Laplace Transformations


问:
Continuity of bilinear form
Let $X$ be a Banach space and $b(\cdot,\cdot):[0,C] \times X \to \mathbb{R}$ be a form that is continuous wrt. the first argument, linear wrt. the second argument and satisfies
$$b(t,x) \leq K\lVert x \rVert_X$$
a boundedness condition for all $x$ and all $t \in [0,C]$.
Suppose $x_h \to x$ in $X$ as $h \to 0.$
Does
$$b(t+h, x_h) \to b(t,x)$$ as $h \to 0?$
How do I prove this?</p>

答:
Fix $\epsilon >0$.

If $b$ is continuous wrt the first argument, then for any $x\in X$ and $\epsilon>0$, $\exists \delta>0$ such that $|h|<\delta$ implies $|b(t+h,x)-b(t,x)|\leq \dfrac{\varepsilon}{2}$.

Now, since $x_h\rightarrow x$, it means that there exists also $\eta>0$ such that $|h|<\eta\Rightarrow \|x_h-x\|<\dfrac{\varepsilon}{2C}$.

When $|h|<\min(\delta,\eta)$, $$\begin{align*}|b(t+h,x_h)-b(t,x)| & =|b(t+h,x_h+x-x)-b(t,x)| \\ &=|b(t+h,x_h-x)+b(t+h,x)-b(t,x)|\end{align*}$$
by linear w.r.t the second argument, so
$$\begin{align*}|b(t+h,x_h)-b(t,x)| & \leq |b(t+h,x_h-x)|+|b(t+h,x)-b(t,x)| \\ & \leq |b(t+h,x_h-x)|+\dfrac{\varepsilon}{2} \\ & \leq C\|x_h-x\| + \dfrac{\varepsilon}{2} \\ & \leq \varepsilon\end{align*}$$

Title: Continuity of bilinear form


问:
Prove $ \sum_{k=1}^{3n-1} (-1)^k\cos^n(\frac{k\pi}{3n}) =1 $ for any positive integer $n$
Prove $ \sum_{k=1}^{3n-1} (-1)^k\cos^n(\frac{k\pi}{3n}) =1 $ for any positive integer $n$

I tried to convert $\cos(x)$ into $\frac{e^{ix}+e^{-ix}}{2}$ and plugged in, and then expanded it using binomial resulted in very long exponential terms

$$\sum_{k=1}^{3n-1} e^\frac{4ik\pi}{3n}+{n \choose 1}e^\frac{(4n-2)ik\pi}{3n}+{n \choose 2}e^\frac{(4n-3)ik\pi}{3n}+...+{n \choose 1}e^\frac{-(4n-2)ik\pi}{3n}+e^\frac{-4ik\pi}{3n}$$
then i'm stuck on this step. can anyone give my ideas or show me a full solution? Thanks.

答:
First of all, this sum equals $-1$, not $1$.

As you have already noticed:
\begin{align}
\sum_{k=1}^{3n-1} (-1)^k \cos\left(\frac{k\pi}{3n}\right)^n &= \sum_{k=1}^{3n-1} (-1)^k\left(\frac 12\left(\exp\left(i\frac{k\pi}{3n}\right)+\exp\left(-i\frac{k\pi}{3n}\right)\right)\right)^n \\
&= 2^{-n} \sum_{k=1}^{3n-1} (-1)^k \sum_{j=0}^n \binom nj \exp\left(i\frac{(2j-n)k\pi}{3n}\right)
\end{align}

Since $\exp(kx)=\exp(x)^k$, one gets a sum of truncated geometric series. Since $0<|(j+n)/(3n)|<1$ and thus $\exp\left(2\pi i\frac{(j+n)}{3n}\right)\neq 1$ for $0\le j\le n$, this can be rewritten as follows:
\begin{align}
2^{-n} \sum_{k=1}^{3n-1} (-1)^k \sum_{j=0}^n \binom nj \exp\left(i\frac{(2j-n)k\pi}{3n}\right) &= 2^{-n} \sum_{j=0}^n \binom nj \sum_{k=1}^{3n-1} \left(-\exp\left(i\frac{(2j-n)\pi}{3n}\right)\right)^k \\
&= 2^{-n} \sum_{j=0}^n \binom nj \sum_{k=1}^{3n-1} \exp\left(2\pi i\cdot\frac{j+n}{3n}\right)^k \\
&= 2^{-n} \sum_{j=0}^n \binom nj \frac{\exp\left(2\pi i\cdot\frac{j+n}{3n}\right)-\exp\left(2\pi i\cdot\frac{j+n}{3n}\right)^{3n}}{1-\exp\left(2\pi i\cdot\frac{j+n}{3n}\right)}\\
&= 2^{-n} \sum_{j=0}^n \binom nj \cdot(-1) \\
&= -1
\end{align}


问:
Alternating series of powers of Cosine-squared
I am looking for a nice(er) proof of the following identity:
$$\sum_{k=1}^{n-1} (-1)^k \cos^{2j}\frac{\pi k}{2n} = -\frac{1}{2}, ~ 0<j<n,~j,n\in \mathbb{Z}^+.$$
I have been able to prove it using the following strategy. Express the cosine in terms of the complex exponential, obtaining the summand $(-1)^k\left(1/2+e^{i\pi k/n}/4+e^{-i\pi k/n}/4\right)^j$. Use the trinomial expansion to expand the bracket, then sum over $k$ which is now a geometric series. In particular, I find
$$-\sum_{p+q\leq j} {j\choose p~q} \frac{1}{2^{j-p-q}}\frac{1}{4^p}\frac{1}{4^q}
\frac{(-1)^{n+p-q}+e^{i(p-q)\pi/n}}{1+e^{i(p-q)\pi/n}}=-\sum_{p+q\leq j} {j\choose p~q} \frac{1}{2^{j-p-q}}\frac{1}{4^p}\frac{1}{4^q}\times\begin{cases}1, ~ n+p-q=\text{even}\\0, ~ n+p-q=\text{odd}\end{cases},$$
where the odd case follows as the result would be imaginary. The trinomial then reconstitutes to $$-(1/2+1/4+1/4)^j/2 \pm (1/2-1/4-1/4)^j/2=-1/2,$$ where the choice of sign depends on the parity of $n$.

Is there some deeper interpretation of this identity, and/or a slicker way to prove it? There is some clear overlap with Sum of Fourth Powers of cosine series has closed form solution., epecially the answer given there by Claude Leibovici.

答:
I would do a series of simplifications, to reduce to sums-of-roots-of-unity. First of all, $$\sum_{k=1}^{n-1}(-1)^k\cos^{2j}\frac{\pi k}{2n}=S\implies\sum_{k=0}^{2n-1}(-1)^k\cos^{2j}\frac{\pi k}{2n}=2S+1$$ (this is $\sum_{k=0}^{2n-1}a_k$ with $a_k=a_{2n-k}$, $a_n=0$ and $a_0=1$). Next, to get rid of $(-1)^k$, use $$\sum_{k=0}^{2n-1}(-1)^k a_k=2\sum_{k=0}^{n-1}a_{2k}-\sum_{k=0}^{2n-1}a_k;$$ thus, $2S+1=2S_{n,j}-S_{2n,j}$ where $S_{n,j}=\sum_{k=0}^{n-1}\cos^{2j}(k\pi/n)$.

Now introduce the $n$-th root of unity $\omega=\exp(2i\pi/n)$. By the binomial formula, $$2^{2j}S_{n,j}=\sum_{k=0}^{n-1}\omega^{-kj}(1+\omega^k)^{2j}=\sum_{r=0}^{2j}\binom{2j}{r}\sum_{k=0}^{n-1}\omega^{k(r-j)}.$$ The inner sum is $0$ unless $n\mid(r-j)$. If $0<j<n$, the latter holds if and only if $r=j$.

Finally, we get $S_{n,j}=\frac{n}{2^{2j}}\binom{2j}{j}$ for $0<j<n$, and $2S+1=0$, hence $S=-1/2$.


问:
How to find the value of the difference between two determinants divided by $(b-a)$ and $(b-c)$?
The problem is as follows.

Let $B$ to be:

$B=\frac{\left|\begin{matrix}(a-b)&-2\\0&(b-c)\end{matrix}\right|-\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|}{(b-a)(b-c)}$

After solving for $B$ which of the following expressions match the answer?.

The choices given by my workbook are as follows:

$\begin{array}{ll}
1.&c-a-1\\
2.&3b-a+1\\
3.&3c-a-1\\
4.&3c-3a-1\\
\end{array}$

I'm not sure exactly how to understand the official solution from my workbook as it is difficult to get.

So the lines from below is the solution that is proposed by my book, could someone help me please to understand what did the author meant?

$B=\frac{\left|\begin{matrix}(a-b)&-2\\0&(b-c)\end{matrix}\right|-\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-3\left|\begin{matrix}1&1&1\\a&b&c\\bc&ac&ab\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-\frac{3}{abc}\left|\begin{matrix}a&b&c\\a^2&b^2&c^2\\abc&abc&abc\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-\frac{3}{abc}\left|\begin{matrix}abc&abc&abc\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-3\left|\begin{matrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|}{(b-a)(b-c)}$

Relying on Vandermonde's determinant:

$B=\frac{(a-b)(b-c)-3(b-a)(c-b)(c-a)}{(b-a)(b-c)}$

$B=\frac{(b-a)(b-c)[(-1)+3(c-a)]}{(b-a)(b-c)}$

$B=3c-3a-1$

Therefore the answer is choice 4.

That's where it ends the explanation of the official answer. But to me the most disturbing or confusing things from such conclusion is how on earth does these factorizations are justified?

I mean:

$\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|=3\left|\begin{matrix}1&1&1\\a&b&c\\bc&ac&ab\end{matrix}\right|$

$\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|=\frac{3}{abc}\left|\begin{matrix}a&b&c\\a^2&b^2&c^2\\abc&abc&abc\end{matrix}\right|$

$\frac{3}{abc}\left|\begin{matrix}a&b&c\\a^2&b^2&c^2\\abc&abc&abc\end{matrix}\right|=\frac{3}{abc}\left|\begin{matrix}abc&abc&abc\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|$

$\frac{3}{abc}\left|\begin{matrix}abc&abc&abc\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|=3\left|\begin{matrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|$

I'm confused exactly why do these manipulations work. Thus I beg someone could explain to me these

I do remember that when you factor out a term from a matrix or a determinant this must affect all rows:

For example:

$\left|\begin{matrix}2&4\\6&8\end{matrix}\right|=2\left|\begin{matrix}1&2\\3&4\end{matrix}\right|$

But in the earlier cases it does seem to only affect the first row, but is this valid?.

Please I require someone could help me to address these doubts because I'm just totally confused.

Thus it would really help me a lot if someone could help me to understand why you can also swap the rows and the determinant seem unaffected.

答: You are wrong in your belief that $\det(\lambda A)=\lambda\det(A)$ when $A$ is a $n\times n$ matrix. Actually, when you multiply one row (or line) by $\lambda$, the determinant of the new matrix is the determinant of the original one time $\lambda$, and that's what is used in that solution.


问:
How to separate real and imaginary parts of $\arccos(x+iy)$?
I know how to separate the real and imaginary parts of $\ln z$ and $\arctan z$:
$$\begin{align}\ln(x+iy)&=\frac{1}{2}\ln(x^2+y^2)+i\arctan\frac{y}{x}\\
\arctan(x+iy)&=\frac{1}{2}\arctan\frac{2x}{1-x^2-y^2}+\frac{i}{2}\operatorname{artanh}\frac{2y}{1+x^2+y^2}\end{align}$$
and I know how to prove both of these; in particular for the second one we can make use of the fact that $\arctan z=-i\operatorname{artanh}(iz)$. I would now like to know how to separate the real and imaginary parts of $\arccos(x+iy)$; I've thought of using the identity $\arccos(z)=\operatorname{arcosh}(iz)$, but I don't know how to use this, as the argument of the $\ln$ becomes very complicated.

Thank you very much for your help.

答:
Let
$$
\arccos(x+iy)=u+iv.\tag1
$$
Then
$$\begin{align}
x+iy&=\cos(u+iv)=\cos u\cosh v-i\sin u\sinh v\\
\\
&\implies
\begin{cases}
x=\hphantom-\! \cos u\,\!\cosh v\\
y=-\sin u\,\sinh v
\end{cases}.
\end{align}$$

Eliminating $u$ one obtains:
$$\begin{align}
&\left(\frac x{\cosh v}\right)^2+\left(\frac y{\sinh v}\right)^2=1\\
&\implies\cosh^4v-(x^2+y^2+1)\cosh^2v+x^2=0\\
&\implies\cosh^2v=\frac12\left[x^2+y^2+1+\sqrt{(x^2+y^2+1)^2-4x^2}\right]\tag{$*$}\\
&\implies\cosh^2v=\frac14\left[\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right]^2\\
&\implies\cosh v=\frac12\left[\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}\right],\tag2
\end{align}$$
and from this:
$$
\cos u=\frac{x}{\cosh v}=\frac12\left[\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}\right].\tag3
$$

Resolving $(2)$ and $(3)$ and plugging the obtained values of $u$ and $v$ into $(1)$ gives rise to the final expression:
$$\begin{align}
\arccos(x+iy)&=\arccos\frac{\sqrt{(x+1)^2+y^2}-\sqrt{(x-1)^2+y^2}}2\\
&+i\operatorname{arccosh}\frac{\sqrt{(x+1)^2+y^2}+\sqrt{(x-1)^2+y^2}}2.\tag4
\end{align}$$

Note added:

In implication $(*)$ the possibility
$$
\cosh^2v=\frac12\left[x^2+y^2+1-\sqrt{(x^2+y^2+1)^2-4x^2}\right]
$$
was disregarded since for $y\ne0$ it would imply $\cosh^2v<1$, which is impossible for real $v$.


问:
Why is convexity necessary in monotonic binary relations?
First, let's start with the definition of a monotone binary relation:

Definition: Let $C \subset \mathbb{R}^n$ be nonempty and convex. A binary relation $\succeq$ on $C^m$ is monotonic if for any $(x_1, \dots, x_m)$ and $(y_1, \dots, y_m)$ in $C^m$ if $(x_i, \dots, x_i) \succeq (y_i, \dots, y_i)$ for $i = 1, \dots, m$ imply $(x_1,\dots, x_m) \succeq (y_1, ... y_m)$

My question: Why does $C$ need to be convex? Couldn't we just choose any non-empty subset of $\mathbb{R}^n$?

My guess: Because we eventually want to assume continuity, then any convex combination between $x$ and $y$ must be in $C$, hence $C$ must be convex.

答:
As mentioned in the comments, convexity on $C$ is assumed to use the fact that any convex subset of $\mathbb{R}^n$ is connected. (proof can be found anywhere really, but in Math Exchange, we have this post.)

Connectedness is used in the proof of the following proposition:

Proposition: Let $\succeq$ be a complete, monotonic, transitive and continuous binary relation on $C^m$. For any $(x_1, \dots, x_m) \in C^m$, there is $z \in C$, such that:

$$(x_1, \dots, x_m) \sim (z,\dots, z)$$


问:
Probability theory : Event definition and Subset
In Probability theory, we define $Event$ like this

"Subset of Sample Space"

For Example, When we play Coin-toss game (just one-flip), then the sample space would be constructed:

$\{\{H\},\{T\}\}$

And there are just only two events in textbook explanation

1.Head  $\{H\}$

2.Tail  $\{T\}$

However, when we think about a 'subset', there is also $\{\{H\},\{T\}\}$!!

Then except 2, there are also the other events : $\{\{H\},\{T\}\}$, $\{ø\}$
but why the theory doesn't consider just above two as events..?? In definition, $\{\{H\},\{T\}\}$, $\{ø\}$ are also subset of sample space.

答:
Generally, an event is defined as e subset of the sample space $\Omega$, while a result is defined as an element of $\Omega$. In your case, $H$ is a result, while $\{H\}$ and $\emptyset$ are events.

Either your book has a different definition, which is unlikely, or this part of the book is just wrong.


答:
I don't know if you are using your book's notation, but usually we would say that the sample space (for flipping a coin) is $\{H,T\}$ (Note that $H$ and $T$ are presented as individual elements, not as sets). The elements of the sample space are ${\underline {\rm outcomes}}$ of the experiment. So this experiment has two outcomes: $H$ and $T$.

Events, as you noted, are subsets of the sample space. So the events in this situation are
$\emptyset, \{H\}, \{T\},$ and $\{H,T\}$.

By the way, what you wrote: $\{\emptyset\}$, is not an event.

Title: Probability theory : Event definition and Subset

Body:
In Probability theory, we define $Event$ like this

"Subset of Sample Space"

For Example, When we play Coin-toss game (just one-flip), then the sample space would be constructed:

$\{\{H\},\{T\}\}$

And there are just only two events in textbook explanation

1.Head  $\{H\}$

2.Tail  $\{T\}$

However, when we think about a 'subset', there is also $\{\{H\},\{T\}\}$!!

Then except 2, there are also the other events : $\{\{H\},\{T\}\}$, $\{ø\}$
but why the theory doesn't consider just above two as events..?? In definition, $\{\{H\},\{T\}\}$, $\{ø\}$ are also subset of sample space.


问:
Partial derivatives of homogenous functions.
Consider a homogenous function $f(x, y)$ of degree $r$. That is,

$f(tx, ty) = t^{r}f(x,y)$.

In all textbooks,

$\frac{t{\partial f(tx, ty)}}{\partial x} = \frac{t^{r}{\partial f(x, y)}}{\partial x}$.

I think it should be written as
$\frac{t{\partial f(tx, ty)}}{\partial (tx)} = \frac{t^{r}{\partial f(x, y)}}{\partial x}$

because of the chain rule, because $f(tx, ty)$ is first a function of $tx$ and $ty$ (and only 'ultimately' of $x$ and $y$) and then $tx$ and $ty$ are functions of $x$ and $y$ respectively. Please help.

答:
Consider a function $f$ that is homogeneous of degree $r$, i.e.,

$f(tx, ty) = t^rf(x, y)$ ......$(1)$

Differentiate both sides of equation (1) with respect to $t$. Then by Chain Rule:

$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}x + \frac{\partial f(tx, ty)}{\partial(ty)}y = rt^{r-1}f(x, y)
\end{equation}$ ......$(2)$

Differentiate both sides of equation $(1)$ with respect to $x$. Then by chain rule:

$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}t = t^r\frac{\partial f(x, y)}{\partial(x)}
\end{equation}$

$\implies$$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)} = t^{r-1}\frac{\partial f(x, y)}{\partial x}
\end{equation}$ ......$(3)$

Similarly,

$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(ty)} = t^{r-1}\frac{\partial f(x, y)}{\partial y}
\end{equation}$ ......$(4)$

Put $(3)$ and $(4)$ in $(2)$ to get the result:

$\begin{equation}
\frac{\partial f(x, y)}{\partial x}x + \frac{\partial f(x, y)}{\partial y}y = rf(x, y)
\end{equation}$


问:
Is there some special algorithm to minimize polynomial functions?
I have an optimization problem of the form

$$\begin{array}{ll} \underset{\textbf{x} \in \Bbb R^4}{\text{minimize}} & f(\textbf{x})\\ \text{subject to} & \textbf{x} \in \Omega\end{array}$$

where $f$ is a degree-$8$ multivariate polynomial function,  and $\Omega$ is a $4$-dimensional box (i.e., the Cartesian product of $4$ closed intervals). I would like to solve it in the most efficient way.

答:
I would look into global optimization based on interval analysis: https://github.com/JuliaIntervals/IntervalOptimisation.jl

It's not necessary the fastest, but it gets you the global minimum.


答:
Determinant is a sum of products where no two multipliers in a product come from the same row or column. For example, for 3×3 matrix:

$$
\det\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32} \\ - a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}
$$

See how in each of the term, we have all 3 indices (1,2,3) in the first place (row) and in the second place (column).

That means, that if we multiply the row by the same number $k$, exactly one element in each product will be affected:

$$
\det\begin{vmatrix}k a_{11}& ka_{12}&ka_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix} = (ka_{11})a_{22}a_{33}+(ka_{12})a_{23}a_{31}+(ka_{13})a_{21}a_{32} \\- (ka_{13})a_{22}a_{31}-(ka_{12})a_{21}a_{33}-(ka_{11})a_{23}a_{32} = \\
k\left(a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32} - a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}\right) = \\
k \det\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}
$$

The same is true for the columns.

Regarding the manipulations:

the first row was divided by 3

the first column was multiplied by $a$, the second by $b$, the third by $c$

we switched rows 2 and 3 and then 1 and 2. Each switch brought a $-1$ coefficient. Two of them resulted in the same value as before.

the first row was divided by $abc$

Title: How to find the value of the difference between two determinants divided by $(b-a)$ and $(b-c)$?

Body:
The problem is as follows.

Let $B$ to be:

$B=\frac{\left|\begin{matrix}(a-b)&-2\\0&(b-c)\end{matrix}\right|-\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|}{(b-a)(b-c)}$

After solving for $B$ which of the following expressions match the answer?.

The choices given by my workbook are as follows:

$\begin{array}{ll}
1.&c-a-1\\
2.&3b-a+1\\
3.&3c-a-1\\
4.&3c-3a-1\\
\end{array}$

I'm not sure exactly how to understand the official solution from my workbook as it is difficult to get.

So the lines from below is the solution that is proposed by my book, could someone help me please to understand what did the author meant?

$B=\frac{\left|\begin{matrix}(a-b)&-2\\0&(b-c)\end{matrix}\right|-\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-3\left|\begin{matrix}1&1&1\\a&b&c\\bc&ac&ab\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-\frac{3}{abc}\left|\begin{matrix}a&b&c\\a^2&b^2&c^2\\abc&abc&abc\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-\frac{3}{abc}\left|\begin{matrix}abc&abc&abc\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|}{(b-a)(b-c)}$

$B=\frac{(a-b)(b-c)-3\left|\begin{matrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|}{(b-a)(b-c)}$

Relying on Vandermonde's determinant:

$B=\frac{(a-b)(b-c)-3(b-a)(c-b)(c-a)}{(b-a)(b-c)}$

$B=\frac{(b-a)(b-c)[(-1)+3(c-a)]}{(b-a)(b-c)}$

$B=3c-3a-1$

Therefore the answer is choice 4.

That's where it ends the explanation of the official answer. But to me the most disturbing or confusing things from such conclusion is how on earth does these factorizations are justified?

I mean:

$\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|=3\left|\begin{matrix}1&1&1\\a&b&c\\bc&ac&ab\end{matrix}\right|$

$\left|\begin{matrix}3&3&3\\a&b&c\\bc&ac&ab\end{matrix}\right|=\frac{3}{abc}\left|\begin{matrix}a&b&c\\a^2&b^2&c^2\\abc&abc&abc\end{matrix}\right|$

$\frac{3}{abc}\left|\begin{matrix}a&b&c\\a^2&b^2&c^2\\abc&abc&abc\end{matrix}\right|=\frac{3}{abc}\left|\begin{matrix}abc&abc&abc\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|$

$\frac{3}{abc}\left|\begin{matrix}abc&abc&abc\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|=3\left|\begin{matrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{matrix}\right|$

I'm confused exactly why do these manipulations work. Thus I beg someone could explain to me these

I do remember that when you factor out a term from a matrix or a determinant this must affect all rows:

For example:

$\left|\begin{matrix}2&4\\6&8\end{matrix}\right|=2\left|\begin{matrix}1&2\\3&4\end{matrix}\right|$

But in the earlier cases it does seem to only affect the first row, but is this valid?.

Please I require someone could help me to address these doubts because I'm just totally confused.

Thus it would really help me a lot if someone could help me to understand why you can also swap the rows and the determinant seem unaffected.


问:
Second Order Differential Equation - Two Irregular points
Consider the following second order differential equation

$$
z^2 \psi''(z)+(z a+b) \psi'(z)+(b z+c) \psi(z) = 0
$$

where $a,b$ and $c$ are arbitrary real numbers. The equation has two irregular singular points at $z = 0$ and $z = \infty$. Mathematica is unable to solve this problem. Of course if $b = 0$ then the solution is trivially given by $z$ to some power. But otherwise, I'm not sure how to attack this equation. Any help is greatly appreciated!

答:
Motivated by this, you can build an asymptotic expansion with $b \ll 1$, with $\psi = \sum_j \psi_j b^j$. We have
\begin{align}
\sum_{j=0} b^j (z^2 \psi_j'' + a z \psi_j' + c \psi_j) + b^{j+1} (\psi_j' + z \psi_j) & = 0 \\
z^2 \psi_0'' + a z\psi_0' + c \psi_0 + \sum_{j=1} b^j (z^2 \psi_j'' + a z \psi_j' + c \psi_j + \psi_{j-1}' + z \psi_{j-1}) & = 0
\end{align}
Thus, we need to solve the following systems iteratively:
\begin{align}
& \mathcal O(1): \quad L \psi_0:=z^2 \psi_0'' + a z\psi_0' + c \psi_0 = 0\\
& \mathcal O(b^j): \quad L \psi_j = -(\psi_{j-1}' + z \psi_{j-1}), \quad j \ge 1.
\end{align}
You can build up your solution with the number of terms desired. If $b$ is not small, then this will not work well.


问:
Intersection of nonempty compact sets are nonempty
Question:
If ${Sn}$ is a sequence of nonempty compact sets with $S_{n+1} ⊂ S_{n}$ for
every $n$, show that ${\bigcap}_{n=1}^∞
S_{n} \neq ∅$.

My Attempt
Choose $S_{k}$ from sequence ${Sn}$. Assume ${\bigcap}_{n=1}^∞ S_{n} = ∅$.

Such that no points inside $S_{k}$ are in unions of all the other $S_{n}$
Set $T_{\alpha} = S_{\alpha}^{c}$.

$S_{k}$ is contained in the open cover formed by several $T_{\alpha}$

Since $S_{k}$ is compact, there exists a finite subcover of several $T_{\alpha}$ that covers $S_{k}$.

This means $S_{k} ⊂ T_{\alpha_1}{\bigcap}T_{\alpha_2}{\bigcap}T_{\alpha_3}{\bigcap}....{\bigcap}T_{\alpha_n}$

This also implies $S_{k}{\bigcap}S_{\alpha_1}{\bigcap}S_{\alpha_2}{\bigcap}S_{\alpha_3}{\bigcap}....{\bigcap}S_{\alpha_n} = ∅$

However, the condition says, $S_{n+1} ⊂ S_{n}$ which means $S_{k}{\bigcap}S_{\alpha_1}{\bigcap}S_{\alpha_2}{\bigcap}S_{\alpha_3}{\bigcap}....{\bigcap}S_{\alpha_n} = ∅$ is not possible, thus making contradiction.

Thus ${\bigcap}_{n=1}^∞
S_{n} \neq ∅$.

I'm not sure whether this would be the legitimate proof for this question, especially the last part about the contradiction. Could anyone verify or suggest a better method to address this question?

Thanks in advance!

答:
This is actually not true in general you need that the the compact sets are also closed.
A simple counter example is the reals with the topology that has all sets of the form $(x,\infty)$ Any set of the form $[y,\infty)$ is going to be compact but it's not closed since the only closed sets are of the form $(-\infty,x]$. So taking the sequence of sets $([n,\infty) )_{n\in\mathbb{N}}$ then the intersection is empty even thought they are all compact. In a $T_2$ space like the real line with its standard topology all compact sets are closed. Your statement is true in this case.

Edit: I see that there are no general topology tags on your question. Just ignore this part if you haven't taken any introductory courses on point-set topology. The following proof works for the Real numbers because of the Heine-Borel Theorem which tells us compact sets are closed.

Proof: Given a sequence $S_n$ of decreasing non empty closed compact sets we can assume by taking a subsequence that it's strictly decreasing. assume that $\bigcap_{n} S_n=\emptyset$ then one defines an sequence of open sets as the complements  $A_n=S_n^c$ for $n \geq 2$ Using Demorgan $\bigcup A_n=\bigcup S_n^c=(\bigcap S_n)^c=X$ So it's an open cover of the entire topological space $X$ so it must also cover $S_1$ so there must be a finite open subcover of $S_1$ since it's compact. But $A_n\subseteq A_{n+1}$ for all $n$ so if there is a finite subsequence there must be an $n$ such that $A_n$ covers $S_1$ Since $S_n\subseteq S_1$ and $A_n=S_n^c$ we get  $S_n=S_1\cap S_n=S_1\setminus S_n^c =S_1\setminus A_n=\emptyset$. But this contradicts the fact that all of the $S_n$ were non empty so the intersection can't be non empty.


问:
validity of proof $\pi$ is irrational from Euler's product formula
Many articles and blog posts (and some textbooks) mention in passing that Euler's product formula can be used to prove that $\pi$ is irrational.

$$ \zeta(s) = \sum_n \frac{1}{n^s} = \prod_n \left( \frac{1}{1-p^{-s}} \right) $$

Taking $s=2$ gives us the Basel proble, and so:

$$ \frac{\pi^2}{6} =\prod_n \left( \frac{1}{1-p^{-2}} \right) $$

Since we know that there are infinite primes, and that each factor in the product is finite, we can say that $\pi$ can't be written as a product of finite rationals.

Is this correct?

This is the explanation given in several sources, but my concern is that we also need to show the infinite product can't be reduced, for example by "telescoping".

I would appreciate replies that were suitable for readers without university maths training.

答: No. Infinitely many positive rationals can have a positive rational product. For example, $\prod_{k=1}^\infty\frac{1-(k+2)^{-2}}{1-(k+1)^{-2}}$ telescopes to $\tfrac43$, which is an example of the concern you raise.


问:
How to check if a subset is compact and a k-dimensional submanifold?
I'm given the following subset S of $\mathbb{R}^4$: $$x^4 + y^4 + z^4 + t^4 = 1$$
$$x + y + z + t = 1$$

Now, given these two, how do I show this is a compact set? I've looked at a number of examples but I'm very far behind on the knowledge required for this so I'm really not sure what to do. If anyone could provide an outline of what I need to do/use to show this, it would be super helpful!

I also needed to show that this set was a 2-dimensional submanifold of $\mathbb{R}^4$. I did this by calculating the gradient of both equations and equating them with a constant, and I got 2 sets of points, but they don't satisfy both the equations. I'm not exactly sure what to do next to show that it's a submanifold so any help in that regard would be really useful as well! :(

Thanks in advance!!

答: For the Euclidean space $E=\mathbb{R}^4,$ a subset $A$ is compact iff it is closed and bounded. The first equation (of unit sphere in $E$) gives that your intersection is bounded. The second equation is of a plane in $E$ and so is closed. A closed subset of a bounded set is closed and bounded. Therefore...


问:
If we're given characteristic and minimal polynomial of a linear transformation, how can we find all its possible Jordan forms?
Determine all possible Jordan forms of a linear transformation with characteristic polynomial $(x−2)^4(x−3)^3$ and minimal polynomial $(x−2)^2(x−3)^2$.

答: Hint: The exponent of $(x - \lambda)$ in the characteristic polynomial is the sum of the sizes of the Jordan blocks associated with $\lambda$. The exponent of $(x - \lambda)$ in the minimal polynomial is the maximum among the sizes of the Jordan blocks associated with $\lambda$.


问:
How to prove the Vandermonde's determinant for a $3\times 3$ matrix when rows and columns have been swapped?
The problem is as follows:

The following determinant is named after french mathematician Alexandre-Théophile Vandermonde who lived in the late 18th century.

Prove this determinant

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=(b-a)(c-b)(c-a)$

What I've attempted to do was to use this well known identity on matrices:

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=1\left|\begin{matrix}b&c\\b^2&c^2\end{matrix}\right|-1\left|\begin{matrix}a&c\\a^2&c^2\end{matrix}\right|+1\left|\begin{matrix}a&b\\a^2&b^2\end{matrix}\right|$

Then working in the right side of this equation I'm getting:

(Here I'm using the notation of capital $V$ for Vandermonde's determinant)

$V=(bc^2-cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$

Factoring the common terms I'm getting

$V=bc(c-b)-ac(c-a)+ab(b-a)$

But now what? I'm stuck, what else should it be done here? Can someone help me here on how to prove this determinant? Could it be that my method wasn't that right? Please help.

答:
Subtract the first column from the second column and factor out a common term from the second column.

$\left|\begin{matrix}
1 & 1 & 1\\
a & b & c\\
a^2 & b^2 & c^2
\end{matrix}\right|=\left|\begin{matrix}
1 & 0 & 1\\
a & b-a & c\\
a^2 & b^2-a^2 & c^2
\end{matrix}\right|=(b-a)\left|\begin{matrix}
1 & 0 & 1\\
a & 1 & c\\
a^2 & b+a & c^2
\end{matrix}\right|$

Now do that again.

$(b-a)\left|\begin{matrix}
1 & 0 & 1\\
a & 1 & c\\
a^2 & b+a & c^2
\end{matrix}\right|=
(b-a)\left|\begin{matrix}
1 & 0 & 0\\
a & 1 & c-a\\
a^2 & b+a & c^2-a^2
\end{matrix}\right|=
(b-a)(c-a)\left|\begin{matrix}
1 & 0 & 0\\
a & 1 & 1\\
a^2 & b+a & a+c
\end{matrix}\right|$

Last, subtract the second column from the third column

$=
(b-a)(c-a)\left|\begin{matrix}
1 & 0 & 0\\
a & 1 & 0\\
a^2 & b+a & c-b
\end{matrix}\right|$

That final discriminant is obviously $c-b$, and you're done.


问:
Lebesgue–Stieltjes integral from 0 to $\infty$ on $\mathbb{R}^+$
In the Stochastic analysis course we encountered the following integral

$\int_0^\infty H^2_sd[M,M]_s$,

where $H_s$ is a predictable process,  $M_s$ is a uniformly integrable martingale in $L^2$, $[M,M]_s$ is the quadratic variation.

We were working on the measure space $(\mathbb{R}^+ \times \Omega, \mathcal{B}(\mathbb{R}^+) \otimes \mathcal{F}) $. As far as I understand, we consider the integral $w$-wise ($\omega \in \Omega$) as a Lebesgue–Stieltjes integral, with a stochastic process being a function $s \mapsto X_s(\omega), \ X_s(w): \mathbb{R}^+ \rightarrow \mathbb{R}$ for any $\omega$.

The question is: how the integral is defined (taking into account we are in $\mathbb{R}^+$ and the integral is to $\infty$)?

My guess was: we define it via a limit as $s \rightarrow \infty$ of the integral from 0 to $s$. However, my professor declined it (saying "there is no problem here as with a Riemann integral").

答:
Let $\omega\in\Omega$ be fixed and let $F\!:\mathbb{R}_+\to \mathbb{R}_+$ be defined as $F(s)=[M,M]_s(\omega)$. Then $F$ is a non-negative, right-continuous and non-decreasing function with $F(0)=0$, and hence
$$
\mu_F((a,b])=F(b)-F(a),\quad0\leq a<b<\infty
$$
defines a Borel-measure on $(\mathbb{R}_+,\mathcal{B}(\mathbb{R}_+))$. If $g\!:\mathbb{R}_+\to \mathbb{R}_+$ is a measurable function, then integrating with respect to $F$ is just ordinary Lebesgue-integration with respect to $\mu_F$, i.e.
$$
\int_0^\infty g(s)\,\mathrm dF(s):=\int_{\mathbb{R}_+}g(s)\mu_F(\mathrm ds).\tag{1}
$$
Now, the integral
$$
\int_0^\infty H_s^2\mathrm d[M,M]_s
$$
is just the $\omega$-wise construction of the integral in $(1)$.</p>


问:
Find multiple tangents for parabola
I have a parabola $x = 0.5y^2 - 4y + 3$. I need to find a tangent(s) that go throught point $(2,-1)$. I have tried:

$y - (-1) = m(x-2)$

$4y = 0.5y^2 - x + 3$

$y = (0.5y^2)/4 - x/4 + 3/4$

$m(x-2) - 1 = (0.5y^2)/4 - x/4 + 3/4$

But I'm pretty sure that's incorrect, could I have some advise on the problem? Thanks a lot

答:
Equation of parabola is $x = 0.5y^2 - 4y + 3$. You can write tangent points to the parabola as $(0.5t^2 - 4t + 3, t)$.

Slope at the tangent point $ = \displaystyle \frac{dy}{dx} = \frac{1}{t-4}$

So the tangent lines are given by,

$\displaystyle y-t = \frac{1}{t-4} (x - 0.5t^2 + 4t - 3) $

As the tangents go through point $(2, -1)$

$\displaystyle -1-t = \frac{1}{t-4} (2 - 0.5t^2 + 4t - 3) $

Solving, $t = -1 \pm \sqrt{11}$

So equation of tangent lines can be written as,

$ \displaystyle y + 1 = \frac{1}{-5 \pm \sqrt{11}} (x - 2)$

Title: Find multiple tangents for parabola


问:
Two vectors linearly dependent on a third vector?
I know how to check if a vector or a matrix is linearly dependent or independent , but how do I apply it on this problem?

Let V1 , V2 , V3 be vectors
How do I prove that the vector V3 = ( 2, 5, -5) is linearly dependent on V1 = ( 1,-2,3) and V2 = (4,1,1) ?

Will it be enough or correct if I solved the equation:
α1V1 + α2V2 - V3 = 0
and proved it has a solution?

答:
Yes, that will work.

Or you can check that$$\begin{vmatrix}2&1&4\\5&-2&1\\-5&3&1\end{vmatrix}=0.$$It follows from this that there are numbers $\alpha$, $\beta$ and $\gamma$ such that $\alpha v_3+\beta v_1+\gamma v_2=0$. If $\alpha=0$, then $v_1$ and $v_2$ would be linearly dependent, and therefore one of them would be the other one times a scalar. But that's clearly not the case. So, $\alpha\ne0$ and therefore$$v_3=-\frac\beta\alpha v_1-\frac\gamma\alpha v_2.$$


问:
Converting second order ODE to polar coordinates
I have the following ODE:

$\ddot{x}-(1-x^{2}-\dot{x}^{2})\dot{x}+x=0$

Now I want to convert this to polar coordinates so that I have an equation for $\dot{\theta}$ and $\dot{r}$. The first thing I have tried is reducing the order to first order by introducing:

$u=x$ and $v=\dot{x}$

But this then leads to a very complicated formula due to the $\dot{x}^{2}$ term. So I am not sure how to then simplfiy that to get the polar ODE's I am looking for. Would appreciate some help with how to simplify such a complex system or perhaps another approach to making turning this ODE into polar coordinates.

Thanks a lot!

答:
Following your first step:
\begin{align}
\dot u & = v \\
\dot v & = (1 - u^2 - v^2)v - u
\end{align}
Let $u = r\cos \theta$ and $v = r \sin \theta$, then note that $r^2 = u^2 + v^2$ so $r \dot r = u \dot u + v \dot v$ and $\tan \theta = v/u$ and so $\sec^2 \theta \dot \theta = \frac{u \dot v - v \dot u }{u^2}$. I will do the $r$ equation, you can plug in to find one for $\dot \theta$:
$$ \dot r = (1-r^2)r \sin^2 \theta $$


问:
Are there any named probability distribution of $x$ proportional to $e^{-\alpha x}(1-e^{-x})^{\beta -1}$?
Are there any named probability distribution of $x$ like:

$$
Pr(x)=\frac{\Gamma(\alpha+\beta)} {\Gamma(\alpha)\Gamma(\beta)} e^{-\alpha x}(1-e^{-x})^{\beta -1}
$$
where $x\in(0,\infty), \alpha>0,\beta >0$. In particular I would like to have an analytical expression of the mean and variance of $Pr(x)$ in terms of $\alpha$ and $\beta$.

Note that the transformation $y(x)=e^{-x}$ yields a beta distribution with parameters $\alpha$ and $\beta$.

答:
This is, unsurprisingly, called a beta exponential distribution (although I'm not sure it's a widely used distribution, so this name may not be as canonical as one might hope). Nadarajah and Kotz define this distribution given parameters $(\lambda,a,b)$ as having probability density function
$$f(x)=\frac{\lambda}{B(a,b)}e^{-b\lambda x}\left(1-e^{-\lambda x}\right)^{a-1},$$
so your distribution has parameters $(1,\beta,\alpha)$. They show that, if $X$ is a random variable following this distribution,
$$\mathbb E[X^n]=\frac{(-1)^n}{\lambda^n B(a,b)}\frac{\partial^n}{\partial p^n}B(a,1+p-a)\bigg|_{p=a+b-1},$$
which allows you to calculate the mean and variance. The linked article gives many more calculations of standard properties of this distribution, which may be of interest.


答:
To solve this you can approach with basic methodology:

Consider the equation $c_1(1,-2,3)+c_2(4,1,1)=(2,5,-5).$ Then you will have system of linear equations:

$c_1+4c_2=2, -2c_1+c_2=5, 3c_1+c_2=-5$.

Finding that whether $v_3$ is linearly dependent on $v_1$ and $v_2$ is equivalent to finding the solution of the above system.

A trivial computation shows that

$c_1=-2, c_2=1.$

Thus, the vector $v_3$ is linearly dependent on $v_1$ and $v_2$.

Title: Two vectors linearly dependent on a third vector?

Body:
I know how to check if a vector or a matrix is linearly dependent or independent , but how do I apply it on this problem?

Let V1 , V2 , V3 be vectors
How do I prove that the vector V3 = ( 2, 5, -5) is linearly dependent on V1 = ( 1,-2,3) and V2 = (4,1,1) ?

Will it be enough or correct if I solved the equation:
α1V1 + α2V2 - V3 = 0
and proved it has a solution?


问:
Brownian motion with 2 absorbing boundaries
Consider the stochastic process
\begin{equation*}
X_t = \mu t + \sigma W_t
\end{equation*}
where $W_t$ is the standard Brownian motion. Suppose that $X_0 = x \in (0,1)$ and that $0$ and $1$ are absorbing boundaries. I would like to know how to tackle the following questions:

What is the probability that $X_t$ will be absorbed at the $x=1$ boundary? i.e. What is the probability that $X_t$ hits $x = 1$ without ever hitting $x = 0$.

What is the probability that $X_t$ will never be absorbed?

What is the probability that $X_t$ will remain unabsorbed after $t > T$?

My Thoughts

I know that if there is only one boundary at $x = 1$ then the probability of hitting $x = 1$ at some time $t > T$ is given by
\begin{equation}
Prob(T(1) > T) = \int_{-\infty}^{1-x}e^{-\frac{1}{2}\frac{(s-\mu T)^2}{\sigma^2 T}}ds - e^{\frac{2\mu(1-x)}{\sigma^2}}\int_{-\infty}^{x-1}e^{-\frac{1}{2}\frac{(s-\mu T)^2}{\sigma^2 T}}ds.
\end{equation}
The formula can be derived using the reflection principle and some change of measure. We can see that the probability approaches zero as $T\rightarrow \infty$. Therefore, in the one boundary case, $X_t$ would be absorbed a.s.. However, I'm not sure how to get a similar formula if we also have a boundary at $x = 0$ since I don't think the reflection principle technique will apply. However, adding a boundary at $x = 0$ will not change to answer to 2, that is there is a zero probability that $X_t$ will never be absorbed.

答:
Though I guess that this already has been answered on MSE somewhere I will present what I know about it.

First let me make some notations. For a process $Y$ let

$T_a^Y := \inf \{t:Y_t =a\}$ and $T^Y_{a,b} = \min (T^Y_a , T^Y_b)$

For symplicity let $\sigma = 1, \mu \geq 0$.

First of all the answer to 2. is clear by $\Bbb P (T^X_{0,1} < \infty)\geq \Bbb P (T_{1}^X < \infty) \geq \Bbb P (T_1^W < \infty) =1$. The last inequality follows by the fact that the drift is positive. That the last probability is $1$ is well-known from the theory of Brownian motion.

At this point let me just note that parts of the problem can be reformulated as a two-sided so called first passage time problem over affine linear time dependent boundaries, i.e. for a (continuous) function $b: [0,\infty ) \to \Bbb R$ denote the hitting time (or first passage time) as $\tau_b^Y := \inf\{ t: Y_t = b(t) \}$. Then for $b_+(t) = 1 - \mu t$ and $b_-(t) = -\mu t$, we have that

$$\Bbb P (T^X_{0,1} > t) = \Bbb P (\tau^W_{b_+ , b_-} > t)$$

I found that A modification of the sequential probability ratio test to reduce the sample size by Anderson (1960) seems to cover this situation.

(Old text of this part: As far as I know for affine linear functions $b_i (t) = \alpha_i + \beta_i t$ hitting probabilties $\Bbb P (\tau_{b_1 ,b_2}^W > t)$ have only been established yet for $\alpha_1 < \alpha_2$ and $\beta_1 < 0 < \beta_2$ (for example see Some conditional crossing results of Brownian motion over a piecewise-linear boundary by Mario Abundo (2002) in Statistics & Probability Letters 58), which does not cover your case. )

But your problem as special case is even simpler. By the Girsanov theorem $(X_s)_{s\in [0,t]}$ is a Brownian motion under the measure $\Bbb Q_t =L_t d \Bbb P$ on $\mathcal F_t = \sigma (X_s , s\in [0,t])$, where $L_t = \exp (-\mu W_t - \frac 1 2 \mu^2 t )= \exp (-\mu X_t  + \frac 1 2 \mu^2 t)$.

For the purpose of addressing 2. let $A^X = \{T^X_{0,1} > t\} \in \mathcal F_t$ which depends in some particular way on the path $(X_s)_{s\in[0,t]}$.
Speaking in densities we can make the following informal computations (which can be made rigorous)

$$\Bbb P(X_t = z , A^X) = \Bbb E_{\Bbb P} [ 1_{A^X}1_{\{X_t = z\}}] = \Bbb E_{\Bbb P} [ 1_{A^X}1_{\{X_t = z\}} \frac 1 {L_t} L_t]\\ = \Bbb E_{\Bbb Q_t} [ 1_{A^X}1_{\{X_t = z\}} \exp (\mu X_t  - \frac 1 2 \mu^2 t)] = \Bbb E_{\Bbb P} [ 1_{A^W}1_{\{W_t = z\}} \exp (\mu W_t  - \frac 1 2 \mu^2 t)]\\ = \Bbb E_{\Bbb P} [ 1_{A^W}1_{\{W_t = z\}} \exp (\mu z  - \frac 1 2 \mu^2 t)] = \Bbb P (W_t = z , A^W)  \exp (\mu z  - \frac 1 2 \mu^2 t)$$

where by $A^W = \{T^W_{0,1} > t\}$. Thus the event $A^W$ depends on the path $(W_s)_{s\in[0,t]}$ in the same way $A^X$ does on $(X_s)_{s\in[0,t]}$.

Now the advantage is: The density
$$\Bbb P ( W_t = z , T^W_{0,1} > t)$$
for $z\in (0,1)$ is known and there are essentially two represantations (one is good for large $t$, the other for small $t$) (see for example Brownian motion and classical potential theory by Port and Stone at Proposition 8.2 in Chapter 2 or Brownian motion and Stochastic Calculus by Karatzas and Shreve at Proposition 8.10 in Chapter 2 ).

Thus we have
$$\Bbb P (T^X_{0,1} > t) = \int_0^1 \Bbb P(X_t = z , T^x_{0,1} > t) dz = \int_0^1 \Bbb P (W_t = z , T^W_{0,1} > t)  \exp (\mu z  - \frac 1 2 \mu^2 t) dz$$

which can be computed further.

In fact one could try to address 1. in the same way: By setting $A^X = \{T^X_1 < T^X_0 , T^X_{0,1} \leq t\}$ we get
$$\Bbb P (T^X_1 < T^X_0) = \lim_{t\to \infty} \Bbb P (T^X_1 < T^X_0 , T^X_{0,1} \leq t)\\= \lim_{t\to \infty} \int_{\Bbb R} \Bbb P (W_t = z,T^W_1 < T^W_0 , T^W_{0,1} \leq t) \exp (\mu z - \frac 1 2 \mu^2 t) d z \\= \lim_{t\to \infty} \int_{\Bbb R} \Bbb P (W_t = z,W_{T^W_{0,1}} =1, T^W_{0,1} \leq t) \exp (\mu z - \frac 1 2 \mu^2 t) d z$$

By the strong Markov property

$$\Bbb P (W_t = z,W_{T^W_{0,1}} =1, T^W_{0,1} \leq t) = \int_0^t \Bbb P (W_{t-s} = z | W_0 =1) p(s)ds$$

where $p(s) = -\frac{d}{ds} \Bbb P (T^W_{0,1} > s)$ is the density of $T^W_{0,1}$.


问:
Gallery of unlabelled trees with n vertices
Can anyone point me to a gallery (printed or online) of unlabelled trees, sorted according to their order (i.e., number of vertices)?  That is, for each order n in oeis.org/A000055 (up to maybe n=11 or so) I'd like to see a visualization of all the trees.

I found Gary Fredericks' online database Small Simple Graphs but there are a few problems which make it unsuitable for my purposes:  1) It's for graphs in general as opposed to trees.  (Though maybe there is some way of using the filters in this tool to suppress the graphs with cycles?)  2) It works only for n up to 9.  3) It shows only five graphs at a time, though I'd like to see all the trees for a given n on the same web page (or printed page).

Alternatively can someone point me to some code which can automatically generate images of such trees, preferably in a vector format?  Then I could arrange them myself into a single web page or LaTeX document.

答: Frank Harary's Graph Theory has all the trees from p = 1 to p=10 displayed on pp. 233-234; the 10-vertex trees take up the whole page 234.


问:
Coproduct in ${\bf Set}$ of the same object
In a category $\mathfrak{C}$ (for instance ${\bf Set}$) I want to compute the coproduct of the objects $X$ and $X$. Since they are not disjoint, I can consider these sets as two disjoint copies of the same sets and, next, compute their coproduct. Now, using the definition of coproduct, it is immediate to verify that $(X,1)$, where $1$ denotes the identity morphism. So, there must be an isomorphism between $X$ and the coproduct of $X$ with itself. Which is such an isomorphism?

答:
No, it's not true unless $X$ is the initial object.

The coproduct of a singleton set $X$ with itself will contain $2$ elements, which is not isomorphic to $X$


问:
Improper integral related to Gamma function
I found the following integral which seems an extension of the gamma function
$$
\int_{0}^{\infty}\left[1 -
\mathrm{e}^{-\left(\large ux^{a} + vx^{b}\right)}\right]
x^{-1 - c}\,\mathrm{d}x,
$$
where $u,v,a,b,c$ are all positive constant such that $c \in \left(0,1\right)$ and $a > b > c$.

In the case of $v = 0$, I evaluated the integral thanks to the change of variable $y = x^{a}$ in terms of the Gamma function, but, when both $u$ and $v$ are strictly positive, I can not make the exponent linear in $x$ when $a \ne b$. I also try to use the series representation of
$1 - \mathrm{e}^{-\left(ux^{a} + vx^{b}\right)}$ but it does not work. Do you have any suggestions or know some noteworthy extensions of the Gamma function that can help me?

Thanks a lot!

P.S.
I'm new to "mathematics", so I also appreciate suggestions on how to ask questions effectively here :)

答:
Integration by parts results in a linear combination of two integrals of the form
\begin{align*}
\int_0^\infty x^{d-1}e^{-ux^a-vx^b}\,dx&=\sum_{n=0}^\infty\frac{(-v)^n}{n!}\int_0^\infty x^{bn+d-1}e^{-ux^a}\,dx\\&=\frac1{u^{d/a}}\sum_{n=0}^\infty\frac1{n!}\left(\frac{-v}{u^{b/a}}\right)^n\Gamma\left(\frac{bn+d}{a}\right).
\end{align*}
This series converges, but there's no general closed form for it (other than just a name).


问:
Is $f'(x)$ Riemann integrable if it is bounded
The fact that $f(x)$ is differetiable dosen't guarantee the integrability of $f'(x)$ on $[a,b]$ beacuse $f'(x)$ could be unbounded, which violates the assumption of Riemann integral. However, what's the case if $f'(x)$ is bounded? Since $f'(x)$ doesn't have any discontinuity point of the first kind and $f'(x)$ is bounded, $f'(x)$ has only discontinuity points as $\sin \frac1x$ dose near $x=0$ if there are any.

Could $f'(x)$ has uncountably many discontinuity points of that kind? Otherwise, $f'(x)$ should be integrable on $[a,b]$. I think $f'(x)$ is integrable if it is bounded, could someone please give me a hand here? Or is there a counterexample?

Thanks in advance!

答: Yes, there are differentiable functions $f$ such that $f'$ is bounded but not Riemann-integrable. The standard example is Volterra's function.


问:
Solving a non-linear optimisation problem with min type constraint.
The problem is $\min_{\boldsymbol{\mu}}f(\boldsymbol{\mu})$ given that
$\min|\mu_i-\mu_j|\geq\delta>0$, where $\boldsymbol{\mu} = (\mu_1,\ldots,\mu_n)^{\rm T}$. I am having difficulty to convert $\min|\mu_i-\mu_j|\geq\delta>0$ to a set of linear constraints so that I can get the solution (at least numerically). Any help will be appreciated. Thanks in advance.

答:
If the $\min|\mu_ - \mu_|$ were in the objective function, you could write a smooth reformulation by introducing additional continuous variables. However, since it is a constraint, I guess you need to introduce binary variables to model the disjunction.

Unless you can order your $\mu$s by increasing order?


问:
If a form is invertible then $\forall x, \exists y$ such that $B(x,y) \neq 0$
The question is the following:

If a matrix that represents a form $B:E \times E \rightarrow K$ is invertable, show that, $\forall x_0 \in E, \exists y_0$ such that $B(x_0,y_0) \neq 0$. $E$ is an Euclidean space and K is a field.

I thought about using the fact that the invertibility of $B$ means the eigenvalues are nonzero, but couldn't really put a statement in place.

答: If the matrix $A$ is in the basis $(e_i)$, let $B_0(e_i,e_j)=\delta_{ij}$ be the scalar product, $B(x,y)=B_0(A(x),y)$, if $A$ is invertible $A(x)\neq 0 $ if $x\neq 0$ and  there exists $y$ such that $B_0(A(x),y)\neq 0$ since $B_0$ is non degenerated, write for example $A(x)=u_1e_1+..+u_ne_n$ there exists $i$ such that $u_i\neq 0$, $B(A(x),e_i)\neq 0$.


问:
category theory - what is the difference between morphism and arrow?
The Arrow Category confuses my understanding between morphism and arrow.
I can understand a morphism is a mapping that preserves the context structure.

is Arrow just an alias of morphism or to some-extent mapping ?

```<code>Axioms:
1. for each arrow f, target * source (f) = source(f),
</code>```

isn't source of f just the data ?, then target of a data is just meaningless.

答:
In my understanding, the Arrow Theory is not meant to be a category, so you should not understand "data" as the objects of some category. It is rather another way of expressing the same information that is given by a category. Category theory and Arrow theory are two points of view of the same theory, and the point of the Arrow theory's point of view is that it is much clearer that what category theory focus on are morphisms, and not objects (category theory is about relationship between objects and not the objects themselves).

About the distinction between arrow and morphism, at least in a categorical context, these terms are usually used interchangeably.


问:
differential equation (proof)
Consider on $(0,+\infty)$ the differential equation
\begin{align}
\label{eq}
x^2y''(x)+xy'(x)+y(x)=10\ . \quad\quad(1)
\end{align}
For a twice derivable function $y:(0,+\infty)\rightarrow\mathbb{R}$, define $u:\mathbb{R}\rightarrow\mathbb{R}$ as
\begin{align}
u(t)=y\left(e^t\right).
\end{align}
Show that $y$ solves $(1)$ if and only if the corresponding $u$ satisfies
\begin{align}
u''(t)+u(t)=10\ .
\end{align}

Do I just replace $y$ with $u$ in the equation and then try to solve it?

答:
$$x^2y''(x)+xy'(x)+y(x)=10 \quad\quad(1)$$
This is Cauchy-Euler's differential equation. Change variable $x=e^t$:
$$y'_x=\dfrac {dy}{dt}\dfrac {dt}{dx}=e^{-t}y'_t$$
$$y''_x=(-e^{-t}y'_t+e^{-t}y''_t)e^{-t}$$
$$y''_x=(y''_t-y'_t)e^{-2t}$$
The equation becomes:
$$-y'_t+y''_t+y'_t+y_t=10$$
$$y''_t++y_t=10$$

Title: differential equation (proof)


问:
Expressing trigonometric expression in multiple angles using complex methods.
How would you express $\cos^3x\sin^3x$ in terms of $\sin(6x)$ and $\sin(2x)$ via complex methods?

答:
I'm unsure what you mean by "via complex methods". But this is a relatively simple conversion, by usual methods.

\begin{eqnarray} \cos^3(x)\sin^3(x) &=& \bigl(\cos(x)\sin(x)\bigr)^3
\\ &=& \Biggl(\frac{1}{2}\sin(2x)\Biggr)^3
\\ &=& \frac{1}{8}\sin^3(2x)
\\ &=& \frac{1}{8}\Biggl(\frac{1}{4}\biggl(3\sin(2x) - \sin(6x)\biggr)\Biggr)
\\ &=& \frac{1}{32}\Biggl(3\sin(2x) - \sin(6x)\Biggr)\end{eqnarray}


问:
Fano planes probability
I have to find the threshold probability for containing a Fano plane. Since its a 3-hypergraph with 7 vertices and 7 hyperedges I was following the same approach as in:

https://math.stackexchange.com/a/4073729/874698

I have a small doubt here: since the Fano plane has a particular shape, thus, when calculating the variance, I believe we should get only the following cases

(i) no two Fano planes share a hyperedge

(ii) two Fano planes share one hyperedge

(iii) two Fano planes share all 7 hyperedges, i.e. they are the same.

Essentially, I think it is not possible for two Fano planes to share two or more hyperedges. Is it correct? Or its shape can be distorted so as to only keep the essential structure? Thanks in advance.

答:
A quick way to check this theory is to ask: how many hyperedges are left in the Fano plane if we delete one vertex? (Doesn't matter which one; the Fano plane is very symmetric.)

We still have $4$ edges.

So it's possible for two Fano planes to intersect and share $6$ of their vertices and $4$ of their edges.

What if we delete $2$ vertices? You should be able to convince yourself with a similar diagram that $2$ edges are always left. (Again, this happens no matter which two vertices you delete - there's a high degree of symmetry.) So it's possible for two Fano planes to intersect and share $5$ of their vertices and $2$ of their edges.

If we delete $3$ vertices, we are left with either $0$ or $1$ edges. So it seems like we should consider the case where two Fano planes share $4$ of their vertices and $1$ of their edges. But no - at that point, if the Fano planes share only $1$ edge, we might as well assume that they share only the $3$ vertices in that edge.

Finally, if two Fano planes share fewer than $3$ vertices, they can't share any edges - so the dominant term of this form comes from Fano planes that share no vertices.

You might ask: what's the general strategy for making sure we don't miss all the cases?

Well, we don't have to get all the cases. We just have to make sure that we consider the leading cases: the most efficient ways to intersect. So far, we've found intersections with

$0$ common vertices, $0$ common edges

$3$ common vertices, $1$ common edge

$5$ common vertices, $2$ common edges

$6$ common vertices, $4$ common edges

$7$ common vertices, $7$ common edges

and we can prove that any other situation is dominated by one of these: it either has the same number of common vertices and fewer common edges, or the same number of common edges but more common vertices, or is worse on both counts. That's how we tell, because sharing $x$ common vertices and $y$ common edges gives an $O(n^{2\cdot7 - x} p^{2\cdot7-y})$ term, and $p \to 0$ as $n \to \infty$.

More advanced technique is possible. Once we know the threshold $p = n^{-\alpha}$, we can compare terms by their behavior at that threshold! This comes down to comparing the density of a graph or hypergraph (the density is $\frac{\text{# edges}}{\text{# vertices}}$) to the density of its subgraphs. As long as you don't find any subgraphs with higher density than the entire graph, the variance method should work.

Title: Fano planes probability


问:
$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$ by substitution
I have to solve this integral $$ \int \frac{1+\sin 4x}{(\sin x -\cos x) \cdot \cos x}\, dx$$
It seems that the most convenient way to operate is doing the substitution $ \tan x = t$.
Then after some passages the integral becomes:
$$  \int \frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2}\, dx  $$

Now I'm trying the coefficient A,B,C,D,E and F such as:

$$\frac{t^4-4t^3+2t^2+4t+1}{(t-1)(t^2+1)^2} = \frac{A}{t-1}+\frac{2B(t-1)+C}{t^2+1}+ \frac {d}{dx} \frac{D(t-1)^2+E(t-1)+F}{(t-1)(t^2+1)}$$

Perhaps I made mistakes in this last passage because I find some coefficients that don't lead to the result suggested by the book.

Body:
$$\frac{{{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1}{\left( t-1\right)  {{\left( {{t}^{2}}+1\right) }^{2}}}=-\frac{4}{{{t}^{2}}+1}+\frac{4-4 t}{{{\left( {{t}^{2}}+1\right) }^{2}}}+\frac{1}{t-1}$$

Details:
$$\frac{{{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1}{\left( t-1\right)  {{\left( {{t}^{2}}+1\right) }^{2}}}=\frac{B t+C}{{{t}^{2}}+1}+\frac{D t+E}{{{\left( {{t}^{2}}+1\right) }^{2}}}+\frac{A}{t-1}$$

$$\frac{{{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1}{{{t}^{5}}-{{t}^{4}}+2 {{t}^{3}}-2 {{t}^{2}}+t-1}=\frac{\left( B+A\right)  {{t}^{4}}+\left( C-B\right)  {{t}^{3}}+\left( D-C+B+2 A\right)  {{t}^{2}}+\left( E-D+C-B\right)  t-E-C+A}{{{t}^{5}}-{{t}^{4}}+2 {{t}^{3}}-2 {{t}^{2}}+t-1}$$

$$\left( B+A\right)  {{t}^{4}}+\left( C-B\right)  {{t}^{3}}+\left( D-C+B+2 A\right)  {{t}^{2}}+\left( E-D+C-B\right)  t-E-C+A={{t}^{4}}-4 {{t}^{3}}+2 {{t}^{2}}+4 t+1$$
Solve system
$$-E-C+A=1\operatorname{,}E-D+C-B=4\operatorname{,}D-C+B+2 A=2\operatorname{,}C-B=-4\operatorname{,}B+A=1$$
We get
$$D=-4\operatorname{,}B=0\operatorname{,}E=4\operatorname{,}C=-4\operatorname{,}A=1$$