问:
Prove that $Z_k := \int_{T_{2k+1}}^{T_{2k+2}}f^2(B_s) ~ds$ are independent and have same law
Consider $B=(B_t)_{t\geq 0}$ a real $\mathcal F_t$ - brownian motion starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider for $a,\ b \  \in \mathbb R, \ a T1 : B_t =b\}, \\ T_{2k+1} &= \inf\{t> T_{2k} : B_t =a\}, \quad T_{2k+2} = \inf\{t> T_{2k+1} : B_t =b\}.
\end{align}

I was interested in showing that the random variables $(Z_k, \ k \in \mathbb N)$ following defined are independent and have same distribution.

$$Z_k := \int_{T_{2k+1}}^{T_{2k+2}}f^2(B_s) ~ds$$ with $f\in \mathcal C ^0(\mathbb R,\mathbb R)$ and $f \not\equiv 0$.

I have tried to explore the strong Markov property but I had some difficult to deal with this approach and so to conclude it. However I believe being in the right way to solve it.

Could someone give me an advice on that problem, please? Thank's in advance.

Crossposted on overflow

答:
Hint: Can you show or is it clear to you that each $Z_k$ is distributed as
$$
\zeta=\int_0^\tau f^2(a+W_t)\mathrm dt,
$$
where $(W_t)$ is a standard Brownian motion starting from $W_0=0$ and $\tau$ is the first hitting time of $b-a$ by $W$, that is, $\tau=\inf\{t\geqslant0\mid W_t=b-a\}$?

Once you are done with this part, the independence property stems from the fact that each $Z_k$ is measurable with respect to
$$
\mathcal G_k=\sigma(B_{T_{2k+1}+t};0\leqslant t\leqslant T_{2k+2}-T_{2k+1}),
$$
and that the sigma-algebras $(\mathcal G_k)$ are independent. Another take on this is to consider
$$
\zeta^k_t=B_{T_{2k+1}+t}-a,
$$
and to show that the processes $(\zeta^k_t)_{0\leqslant t\leqslant T_{2k+2}-T_{2k+1}}$ are independent and distributed as $(\zeta_t)_{0\leqslant t\leqslant\tau}$.


问:
Video Lessons in Complex Analysis
Does anybody have some link for good video lessons of a complete course in Complex Analysis?

Grateful.

答: The complex analysis video series from professor Richard E. BORCHERDS's youtube channel is great.</p>

ContentLicense: CC BY-SA 3.0

Title: Video Lessons in Complex Analysis

ContentLicense: CC BY-SA 4.0


问:
Boyd & Vandenberghe, exercise 3.19(a) — I'm a bit confused
I was working through the exercises of chapter 3 from Boyd & Vandenberghe's Convex Optimization but I'm a bit confused by the solution for Ex 3.19(a). It says that:-

for $\alpha_1 \geq \alpha_2 \geq \alpha_3 \geq ... \geq \alpha_r \geq 0$ and $x_{[i]}$ denotes the $i^{th}$ largest component of $x$,

$f(x) = \sum_{i=1}^{r} \alpha_ix_{[i]} = \alpha_r(x_{[1]}+x_{[2]}+...+x_{[r]}) \ + \ (\alpha_{r-1}-\alpha_r)(x_{[1]}+x_{[2]}+...+x_{[r-1]}) \ + \ (\alpha_{r-2}-\alpha_{r-1})(x_{[1]}+x_{[2]}+...+x_{[r-2]}) \ + \ ... \ + \ (\alpha_1-\alpha_2)x_{[1]}$

I cant seem to understand how these two are equivalent. Any additional steps in between would be appreciated. Thank you.

答:
The right hand side of your equation is what's known as a telescoping sum, in which pairs of consecutive terms cancel. For instance, notice that adding the first two terms gives:

$$ a_r (x_{[1]} + \cdots + x_{[r]}) + (a_{r-1} + a_r)(x_{[1]} + \cdots + x_{[r-1]}) = a_r x_{[r]} + a_{r-1}(x_{[1]} + \cdots + x_{[r-1]}) $$

Then, adding the third term gives,

$$ a_r  x_{[r]} + a_{r-1}  x_{[r-1]} + a_{r-2}(x_{[1]} + \cdots + x_{[r-2]}) $$

And so on. Hence after adding all $r$ terms the sum collapses, and you are left with

$$ a_r  x_{[r]} + a_{r-1}  x_{[r-1]} + \cdots + a_1 x_{[1]} = \sum_{i=1}^r a_i x_{[i]}$$


问:
Find point nearest to the origin
Find the points on the curve $5x^2 - 6xy + 5y^2 = 4$ that are nearest the origin.

The first method I've tried is I've taken the derivative of the equation to optimize (Pythagorean Theorem) and also the function of the curve using implicit differentiation and plugged stuff in (the way I've done it so far). I got $y = x$ which I tried to plug into the function of the curve but didn't match the answers at the back of the textbook. I stared at it and figured I made a sign error somewhere since $-y = x$ worked but couldn't find it.

The second method I tried was solving for the function of the curve in terms of $y$ and tried both completing the square and using the quadratic formula to find $x$ which gave the same failed answer where $y$ subtracted to nonexistence when I plugged them into the function of the curve.

I don't know what lagrange multipliers are, which was the only solution on the web. I hope it could be done just with simple algebra (or calculus) since the textbook expects that (I think).

答:
You have to solve the following optimization problem:

Min $x^2+y^2$

Subject to:

$5x^2 -6xy+5y^2=4$

You could solve it using calculus as follows. Use the constraint to express $y$ in terms of $x$ and substitute that in the objective function. Then use the usual method of finding the optimal value by setting the $f'(x)=0$ and checking that $f''(x) >0$.

Title: Find point nearest to the origin


答:
Yes, because of the formula (valid for any $p\ne 1$)
$$\int\frac{\mathrm dx}{x^p}=-\frac 1{(p-1)x^{p-1}}\qquad({}+C),$$
so that
$$\int_0^p\frac{\mathrm dx}{x^p}=-\frac 1{(p-1)x^{p-1}}\biggr\vert_{0}^p$$
Observe that the function in the r.h.s. is defined for $x=0$ and is equal to $0$, since $p-1<0$.

Title: If $0 < p < 1$, then does $\int_{0}^{p}\frac{1}{x^p}dx =-\frac{1}{(p-1)(p^{p-1})}$.

Body: If $0 < p < 1$, then does $\int_{0}^{p}\frac{1}{x^p}dx =-\frac{1}{(p-1)(p^{p-1})}$ For example, if $p=0.5$ then $\int_{0}^{0.5}\frac{1}{x^{0.5}}dx =-\frac{1}{(0.5-1)({0.5}^{0.5-1})}$. I know $\int_{0}^{p}\frac{1}{x^p}dx = \frac{1}{1-p}$, but it's impossible to make this into the $-\frac{1}{(p-1)(p^{p-1})}$?


问:
Concave function proof
I have a problem with this proof. I know I should use the fact, that f(x) = $x^p$ is concave but I am not sure how to do it at all. Here is the task:

prove inequality $y^p - x^p \leq (y - a)^p - (x - a)^p$ provided that $0 \leq a \leq x \leq y$ and $0 \leq p \leq 1$.

What I have noticed is that $x - a < x < y - a < y$ and that f(x) = $x^p$ is concave but I don't know what to do next.

答:
Fix $h\geq 0$. Define $f:[0,\infty)\rightarrow \mathbb{R}$ by $f(t) = (t+h)^p - t^p$.
Clearly $f$ is continuous on $[0,\infty)$ and differentiable on $(0,\infty)$. Note that $f'(t) = p [ (t+h)^{p-1} - t^{p-1} ] \leq 0$, so $f$ is decreasing.

In particular, for $h = y-x$, we have that $f(x) \leq f(x-a)$ and the results follows.

Title: Concave function proof


问:
+Flattening the curve ...
Edited to correct errors identified below:

I have a cylinder with r = 1 and a central axis given by y = (√3)x/6. I think this is a cylinder intersecting the xz-plane at π/6. I am pretty sure that the intersection is an ellipse, in which case the formula for the intersection of the cylinder with the xz-plane would be (x^2)/(a^2) + (z^2)/(b^2) = 1, and in this case, I think a=2, b=1. Please confirm that I have not blundered this too badly.

So far; so good.

Suppose I were to "mark" my cylinder with the intersection and unroll it into a plane. What is the curve created by the mark?

Here's why I'm interested in this question. I have to cut a piece of stove pipe with an eight inch radius such that the stove pipe will make a thirty degree angle with the adjacent surface. I can unroll the stove pipe into a flat sheet, but I need to draw the cut line on it.

Thanks for the help,

Chris.

答:
Too long for a comment.

The cylinder equation is given by

$$
C(x,y,z) = \left(x+\frac{y}{2 \sqrt{3}}\right)^2-\frac{13}{12} \left(x^2+y^2+z^2-1\right)=0
$$

The intersection with the plane $x=0$ gives

$$
C_x(y,z) = \frac{12 y^2}{13}+z^2-1=0
$$

and with the plane $y=0$

$$
C_y(x,z) = \frac{x^2}{13}+z^2-1=0
$$

Title: +Flattening the curve ...


问:
Angle between the curves without differentiation
$y^2=4(x+1)$ and $x^2=4(y+1)$

Are the two curves, the answer online uses calculus and we have not been taught that.

can this be solved any other way?

Thanks in advance.

答:
Motivation:

Changing solution sets:

Suppose we have a quadratic:

$$ P(x) = ( x-a)(x-b)$$

Suppose I wanted another function for which the roots are square of the polynomial before, then I plug $ g(x) = \sqrt{x}$ into P:

$$ P(g) = ( \sqrt{x} -a ) ( \sqrt{x} - b) \tag{1}$$

So plugging in the inverse function of the desired result we want.

2. Rotations:

Using complex numbers we can find to rotate a number by clockwise ninty degrees we simply need to multiply by $-i$ i.e: if $ z= x+iy$ then rotated clockwise by ninty degree about origin is $-iz = -ix+y$. Hence under rotation of ninty about origin, we have the following map:

$$(x,y) \to (y,-x) \tag{2}$$

3. The actual answer:

Consider your parabola:
$y^2=4(x+1)$

If I wanted to rotate the whole curve by ninty counter clock, I plug in the coordinates with inverse action applied(2) (i.e: ninty clockwise) , this leads to:

$$(-x)^2 = 4(y+1)$$

This gives:

$$x^2 = 4(y+1)$$

Now, if our two curves have an intersection, this this point is uneffected by this rotation and hence it must be that the tangent of the original curve must be rotated by ninty degree. Hence angle between intersection is ninty.

The key idea here is knowing how solution sets of the equation change when you plug in things and rotation and work back your way to the observation about perp tangents


问:
Calculate surface normal and area for a non-planar quadrilateral
Given the four coordinates of the vertices, what is the best possible approximation to calculate surface area and outward normal for a quad?

I currently join the midpoints of the sides, thus dividing the quad into four triangles and a parallelogram, and find the normal to the plane of the parallelogram, but that doesn't give me correct results when the original four vertices aren't coplanar.

Is it possible to use isoparametric transformations for this purpose?  (This calculation is a part of a bigger FEM code)

Body:
Adrian is right, and Nominal Animals answer has the incorrect surface integral. The correct integral is
$$\begin{aligned}
A & = \int_0^1 \int_0^1 \left\lVert \vec{p}_u \times \vec{p}_v \right \rVert \, d u \, d v \\
~ & = \int_0^1 \int_0^1 \sqrt{ u^2 \, C_{uu} + 2 \, u \, v \, C_{uv} + v^2 \, C_{vv} + 2 \, u \, C_{u} + 2 \, v \, C_{v} + C_{0} } \, d u \, d v \\
\end{aligned}$$
where numbering the vertices such that $1$ and $4$ are diagonally opposite (as are $2$ and $3$),
$$\begin{aligned}
C_{uu} & = ( Y_\Delta Z_2 - Z_\Delta Y_2 )^2
         + ( Z_\Delta X_2 - X_\Delta Z_2 )^2
         + ( X_\Delta Y_2 - Y_\Delta X_2 )^2 \\
C_{vv} & = ( Y_\Delta Z_3 - Z_\Delta Y_3 )^2
         + ( Z_\Delta X_3 - X_\Delta Z_3 )^2
         + ( X_\Delta Y_3 - Y_\Delta X_3 )^2 \\
C_{uv} & = ( Y_\Delta Z_3 - Z_\Delta Y_3 )( Z_\Delta Y_2 - Y_\Delta Z_2 ) \\
~      & + ( Z_\Delta X_3 - X_\Delta Z_3 )( X_\Delta Z_2 - Z_\Delta X_2 ) \\
~      & + ( X_\Delta Y_3 - Y_\Delta X_3 )( Y_\Delta X_2 - X_\Delta Y_2 ) \\
C_{u}  & = ( Y_2 Z_\Delta - Z_2 Y_\Delta )( Y_2 Z_3 - Z_2 Y_3 ) \\
~      & + ( Z_2 X_\Delta - X_2 Z_\Delta )( Z_2 X_3 - X_2 Z_3 ) \\
~      & + ( X_2 Y_\Delta - Y_2 X_\Delta )( X_2 Y_3 - Y_2 X_3 ) \\
C_{v}  & = ( Y_\Delta Z_3 - Z_\Delta Y_3 )( Y_2 Z_3 - Z_2 Y_3 ) \\
~      & + ( Z_\Delta X_3 - X_\Delta Z_3 )( Z_2 X_3 - X_2 Z_3 ) \\
~      & + ( X_\Delta Y_3 - Y_\Delta X_3 )( X_2 Y_3 - Y_2 X_3 ) \\
C_{0}  & = ( Y_2 Z_3 - Z_2 Y_3 )^2
         + ( Z_2 X_3 - X_2 Z_3 )^2
         + ( X_2 Y_3 - Y_2 X_3 )^2  \\
\end{aligned}$$
where $C_{uu} \ge 0$, $C_{vv} \ge 0$, and $C_{0} \ge 0$; and
$$\begin{aligned}
X_\Delta & = x_4 - x_3 - x_2 + x_1 = (x_4 - x_3) - (x_2 - x_1) \\
X_2      & = x_2 - x_1 \\
X_3      & = x_3 - x_1 \\
Y_\Delta & = y_4 - y_3 - y_2 + y_1 = (y_4 - y_3) - (y_2 - y_1) \\
Y_2      & = y_2 - y_1 \\
Y_3      & = y_3 - y_1 \\
Z_\Delta & = z_4 - z_3 - z_2 + z_1 = (z_4 - z_3) - (z_2 - z_1) \\
Z_2      & = z_2 - z_1 \\
Z_3      & = z_3 - z_1 \\
\end{aligned}$$
but I am not sure if there is a feasible algebraic expression for the integral.  Numerically it is a simple matter, as the expression in the square root is numerically stable and just a bivariate quadratic function.

To calculate the cross sectional area in some direction $\hat{n}$, $\lVert\hat{n}\rVert = 1$, for example a flow in a specific direction through the surface, then
$$A_\hat{n} = \int_0^1 \int_0^1 \hat{n} \cdot \left( \vec{p}_u \times \vec{p}_v \right) \, d u \, d v$$
If $\hat{n} = (n_x, n_y, n_z)$, $n_x^2 + n_y^2 + n_z^2 = 1$ and $\hat{n}$ is in the same halfspace as the surface normal (i.e, $\hat{n} \cdot \left( \vec{p}_u \times \vec{p}_v \right) \ge 0$ for $0 \le u \le 1$, $0 \le v \le 1$, then
$$\begin{aligned}
A_\hat{n} & = \frac{n_x}{2} \bigl( (y_4 - y_1) (z_3 - z_2) - (z_4 - z_1) (y_3 - y_2) \bigr) \\
~         & + \frac{n_y}{2} \bigl( (z_4 - z_1) (x_3 - x_2) - (x_4 - x_1) (z_3 - z_2) \bigr) \\
~         & + \frac{n_z}{2} \bigl( (x_4 - x_1) (y_3 - y_2) - (y_4 - y_1) (x_3 - x_2) \bigr) \\
~         & = \frac{1}{2} \hat{n} \cdot \bigl( (\vec{p}_4 - \vec{p}_1) \times (\vec{p}_3 - \vec{p}_2) \bigr) \\
\end{aligned}$$
i.e. half the dot product between the unit flow direction vector and the cross product of the quadrilateral diagonals.  But do remember that this only applies if $\hat{n}$ is and the quadrilateral surface normal are in the same halfspace within the entire surface.


问:
How to find basis of the image of a matrix
I need to find the basis of the image of the linear transformation given by the following matrix:

$ \begin{pmatrix} 1 & 2 & 3\\ 0 & 1 & 3 \end{pmatrix} $

I've tried looking online, but have become more confused.

I thought that you would just find a maximal linearly independent subset of the column vectors and then that would be the basis of the image (in which case it would be $\begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and $\begin{pmatrix} 2 \\ 1 \end{pmatrix} $ or $\begin{pmatrix} 1 \\ 0 \end{pmatrix} $ and $\begin{pmatrix} 3 \\ 3 \end{pmatrix} $).

But I've heard that you have to first row reduce the matrix - why?

答:
$\begin{pmatrix}1 & 2 & 3 \\ 0 & 1 & 3 \end{pmatrix}\begin{pmatrix}x \\ y \\ z\end{pmatrix}= \begin{pmatrix} x+ 2y+ 3z \\ y+ 3z\end{pmatrix}= x\begin{pmatrix}1 \\ 0 \end{pmatrix}+ y\begin{pmatrix}2 \\ 1 \end{pmatrix}+ z\begin{pmatrix} 3 \\ 3 \end{pmatrix}$

Those are three vectors but since they are in $R^2$ they cannot be independent.  We can write $\begin{pmatrix}3 \\ 3 \end{pmatrix}$$= -3\begin{pmatrix}1 \\ 0 \end{pmatrix}+ 3\begin{pmatrix}2 \\ 1 \end{pmatrix}$.

So we can take $\begin{pmatrix}1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix}2 \\ 1 \end{pmatrix}$ as basis.  But in fact, since those span all of $R^2$ we could take any two independent vectors as basis, even $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix}0 \\ 1 \end{pmatrix}$.


问:
If $f'(t)$ exists then $\lim_{h,k\to0}\frac{f(t+h)-f(t-k)}{h+k}=f'(t)$
Consider $f'$ continuous...

I'm tried to prove it...with Taylor's expansion:

$f(t+h)=f(t)+hf'(t)+h\phi(t,h)$

$f(t-k)=f(t)-kf'(t)-k\phi(t,-k)$

Then:

$\frac{f(t+h)-f(t-k)}{h+k}=f'(t)+\frac{h}{h+k}\phi(t,h)+\frac{k}{h+k}\phi(t,-k)$

I don't know how I can prove $\lim_{h,k\to0}\frac{h}{h+k}\phi(t,h)=0$

答:
You went wrong way. It is simpler:

$$\frac{f(t+h)-f(t-k)}{h+k}=\frac{f(t+h)-f(t)+f(t)-f(t-k)}{h+k}=$$

$$h\frac{f(t+h)-f(t)}{h(h+k)}+k\frac{f(t)-f(t-k)}{k(h+k)}$$

And now

$$\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}=\lim_{k\to 0}\frac{f(t)-f(t-k)}{k}=f'(t)$$

making

$$\lim_{h,k\to 0} \left ( h \frac{f(t+h)-f(t)}{h(h+k)}+k\frac{f(t)-f(t-k)}{k(h+k)} \right ) =$$
$$f'(t)\lim_{h,k\to 0}\left ( \frac{h}{h+k}+\frac{k}{h+k} \right ) =f'(t)\lim_{h,k\to 0}\frac{h+k}{h+k}=f'(t)$$

Clarification:

To be totally precise, even for untrained eyes, this should be derived in the opposite direction.

If $\lim a$ exists and $\lim b$ exists then $\lim ab$ exists.

$$\lim_{h,k\to 0}\frac{h+k}{h+k}=1$$

$$\lim_{h\to 0}\frac{f(t+h)-f(t)}{h}=\lim_{k\to 0}\frac{f(t)-f(t-k)}{k}=f'(t)$$

So

$$\lim_{h,k\to 0}f'(t)\frac{h+k}{h+k}=f'(t)$$

$$\lim_{h,k\to 0} \left ( \frac{hf'(t)}{h(h+k)}+\frac{kf'(t)}{k(h+k)} \right ) = f'(t)$$

and now expand $f'(t)$ in two different but equivalent ways and there you go.

The entire trick here is dealing with expressions that have predefined limits.


问:
Equation for constructing the mate of the neckline and shoulder bevel

I have a following neckline construction figure.

ABC - a right-angles triangle;

Bold lines are known values but all others are unknown;

CD = L; AC = AD + DC;

A - it's a center point of S arc; AB = AD - radius of S;

The arc S touches the side AC at right angle;

L and S are known.

I have to find one from AB or BC or both.

Experimentally, I manage to find the required BC value for the given S and L, and I feel that such a system has at least one solution, because if the BC value is incorrect, the picture is distorted.

But unfortunately I can't build myself to find the correct equation. I ask for help.

答:
Let $\measuredangle A=\theta$ (in radians as usual), then we have the arc $S$ giving
$$
S=AB\cdot\theta
$$
i.e.
$$
AB=AD=S/\theta
$$
Now you know the side $AB=S/\theta$ and the hypotenuse $AC=L+S/\theta$ making the angle $\theta$, so
$$
\cos\theta=\frac{AB}{AC}=\frac{S}{L\theta+S}
$$
which you can solve numerically for $\theta$, then you have $AB=S/\theta$ and $BC=AC\sin\theta=(L+S/\theta)\sin\theta$.


问:
Number of ways to divide $n$ people into $2$ distinct groups, with at least $1$ member in each
This question was inspired by this programming problem.

The question is:
Find the number of ways to divide $n$ people into $2$ groups $A$ and $B$ such that each group has at least one member.

The accepted answer is:
$2^n -2$

I think the answer to this problem should have been:
$2 * {n \choose 2} * {2}^{n-2}$

Please correct me if i am wrong.

答:
It sounds like the reasoning behind your argument is as follows: since each group needs at least one person, use ${n\choose 2}$ to choose those two people, sort them into groups in one of $2$ ways, and then use $2^{n-2}$ to decide where the rest of the people should go. Here's the problem with that argument: the two people you chose at the beginning are not special, but are made special by your count. My faulty argument said to "choose those two people", but those two people chosen did not have to be anyone in particular, as long as we they were put into different groups.

For example, say $n=4$. One way to sort the groups would be to put $A$ and $B$ in group $1$, and $C$ and $D$ in group $2$. Your counting method would count this arrangement $4$ times. Your ${n\choose 2}$ could pick $A$ and $C$, then your method would put them in groups $1$ and $2$ respectively and distribute the rest of the people. However, it could also pick $A$ and $D$ and do the same, as well as for $B$ and $C$, and $B$ and $D$. This counts the same group multiple times because it treats the two people you chose at the beginning as being special, when they're not. That said, your argument does solve a similar, but different counting problem: given $n$ people, count the number of ways to divide them into two distinct groups, where each group has a leader.

To get the correct answer, ask yourself the following: how many ways are there to divide $n$ people into two groups? How many ways of doing that fail to satisfy the condition that each group is nonempty?


答: Each person can be assigned to either group, so 2 choices/person. That's $2^n$ choices. Two of them must be excluded as they would mean that either group will be empty. Hence the solution.

Title: Number of ways to divide $n$ people into $2$ distinct groups, with at least $1$ member in each

Body:
This question was inspired by this programming problem.

The question is:
Find the number of ways to divide $n$ people into $2$ groups $A$ and $B$ such that each group has at least one member.

The accepted answer is:
$2^n -2$

I think the answer to this problem should have been:
$2 * {n \choose 2} * {2}^{n-2}$

Please correct me if i am wrong.


问:
What is the price of the apples and how many were they brought?
The problem is as follows:

The cook said "I paid $12$ usd for the apples I bought" and they gave
me two apples because I claimed that the apples were too small, that
made me pay one dollar less per dozen. How many apples did the cook
bring?

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{16 apples}\\
2.&\textrm{18 apples}\\
3.&\textrm{12 apples}\\
4.&\textrm{15 apples}\\
\end{array}$

This problem is the typical word problem regarding a linear equation but I'm confused because I don't know the right intepretation.

The way how I put the equations is as follows.

The labels are $x$ the number of apples and $y$ the price of each apple.

Thus:

$12+2y=xy$

Because it mentions that a dozen price was one dollar less.

$y=12(y-1)$

The replacing this in the earlier equation yields.

$x=13$

This means that she ended up with $13$ plus the two additional apples that she had thus is $15$.

To me the most confusing part was to assume:

$y=12(y-1)$

I don't know if this is correct, thus I'd like this to be discussed in the answer please. The reason for which I am confused is that,

I think that the new prize of the apples (which ended up being used is one dollar less).

I don't know if this is the right answer, or if my math was correctly used. Hence I would like someone could answer in a more logical way because I'm confused.

答:
Let $c$=original cost per dozen in dollars

The cook paid $c-1$ so the number of apples (in dozens) that were bought was $\frac{12}{c-1}$

the number of dozens that would have been bought at the original price was $\frac{12}{c}$

the number from the two ways of buying should be the same

$\frac{12}{c-1} = \frac{12}{c} +\frac{2}{12}$

solving this gives $c=9$ and so 12/8 or 1.5 dozen apples, which is 18.


问:
GRE logic problem
So I am confused on the negation of the following:

Let $f,g$ be continuous on the reals.

Negate the statement:

For each $x \in \mathbb{R}$ there exists a $y \in \mathbb{R}$ such that if $f(x)>0$, then $g(y)>0$.

So the answer is

there exists an $x \in \mathbb{R}$ such that for each $y \in \mathbb{R}$  $f(x)>0$ and $g(y) \leq 0$.

Could someone explain this a bit more, that isn't the choice I selected and I am bothered I couldn't get the logic right here. Gotta be careful with quantifiers.

答:
$\neg$ for all = there exists.

$\neg$ there exists = for all.

Saying “if p then q” is the same as “$p \implies q.$”

The negation of $p \implies q$ is [$p$ and $\neg q$].

So $\neg(f(x) > 0 \implies g(y) > 0)$ is [$f(x) > 0$ and $g(y) \leq 0$]

Title: GRE logic problem


问:
Brownian motion recurrence theorems and Hausdorff Dimension
I need help with proving:

1.If $d>1$ then d-dimensional Brownian motion starting at $x$ has 0 probability to actually hit $y$.  Note that this is different from the usual notion of recurrence, where the brownian motion only has to get arbitrarily close.

2.If $d>2$ then d-dimensional Brownian motion $B_t$ satisfies $|B_t|\rightarrow \infty$ as $t \rightarrow \infty$ a.s.  For this one, I want a specific proof that my professor alluded to but did not give details on.  I already know that if you consider a sequence $S_n$ concentric spheres of radius $r^n$ with $r>1$, then if $x$ is of norm $r^N$ then the probability of a brownian motion starting at $x$ to hit $S_{N-1}$ before $S_{N+1}$ is $1/(r^{(d-2)}+1)<1/2$.  Now, he suggests that this be related to a random walk with drift.  I have trouble making this relationship precise.  Please assume that the Brownian motion is defined on some probability space.  It is okay to assume that it is everywhere continuous, instead of a.s. continuous, though the greater generality the better.  I think the strong Markov property has to be used here to elevate this proof idea from the status of being purely motivational.  I don't really need other proofs of 2.  There's a proof in Durrett that is legitimate as far as I can see.  I just want to know how this proof works.

3.Also, these questions have motivated me to ask questions about what paths of Brownian motions look like.  For some reason, people seem terribly interested in Hausdorff dimension.  If anybody can say a few words about why anybody should care about Hausdorff dimension, or how one usually goes about computing the Hausdorf dimension of a Brownian path, please let me know.

答:
This question is vastly too broad so let me answer part 1. Consider a Brownian motion starting at $x\ne0$, then one is interested in whether it hits $0$ or not. For every positive $r$, let $T_r$ denote the first hitting time of the sphere centered at $0$ with radius $r$. An idea is to compute $P_x(T_R\leqslant T_r)$ for every $r\lt|x|\lt R$ by solving explicitely a harmonic problem.

Then one can check that $\lim\limits_{r\to0}P_x(T_R\leqslant T_r)=1$, that is, $P_x(T_R\lt T_0)=1$, for every $R\gt|x|$. Now, $T_R\to+\infty$ almost surely when $R\to+\infty$ hence $P_x(T_0=+\infty)=1$. (Note that this is exactly the procedure suggested by your professor to solve part 2.)


问:
Proving the following shrinking lemma for binary covers
Prove a topological space is normal if and only if for each pair of open sets $U,V$ such that $U \cup V=X$, there exists open sets $U'$ and $V'$ such that $\overline U' \subset U$ and $\overline V' \subset V$ and $U' \cup V'=X$

Attempt:

$(\Rightarrow)$ Since $U \cup V=X \implies (X-U) \cap (X-V)=\varnothing$. $X-U$ and $X-V$ are disjoint closed sets. Then by normality of $X$, there are disjoint open sets $W,Z$ with $X-U \subset W$ and $X-V \subset Z$, and an equivalent definition of normality states there are open sets $W',Z'$ containing $X-U$ and $X-V$ with $\overline W' \subset W$ and $\overline Z' \subset Z$. Now I am unsure how to show that $W' \cup Z'=X$, which is what I need to show in the problem.

$(\Leftarrow)$ Suppose the second part of the statement holds. Let $A$ be closed in $X$ and $U$ be any open set containing $X$. Then the existence of an open set $U'$ containing $A$ such that $\overline U' \subset U$ implies $X$ is normal.

Comment:I am having difficulty understanding (why/if) $W' \cup Z'=X$ holds in the first direction. Also I am skeptical whether I started the second direction correctly by starting with a closed set, and an arbitrary open set containing this closed set. I also realize it may not be true that $U'$ contains the closed set $A$, which adds to my skepticism. Any ideas?

答:
In the first part of the proof you’re looking at the wrong pair of open sets. What you should do is let $U'=X\setminus\operatorname{cl}W'$ and $V'=X\setminus\operatorname{cl}Z'$. Certainly these are open sets, and

$$\begin{align*}
U'\cup V'&=(X\setminus\operatorname{cl}W')\cup(X\setminus\operatorname{cl}Z')\\
&=X\setminus(\operatorname{cl}W'\cap\operatorname{cl}Z')\\
&=X\,,
\end{align*}$$

since $\operatorname{cl}W'\cap\operatorname{cl}Z'\subseteq W\cap Z=\varnothing$.

In the second part you’ve omitted some crucial steps; in particular, you’ve not explained where $U'$ comes from. You need to let $V=X\setminus A$ and note that $U$ and $V$ are then open sets whose union is $X$; then you can apply your hypothesis to get open $U'$ and $V'$ such that $\operatorname{cl}U'\subseteq U$, $\operatorname{cl}V'\subseteq V$, and $U'\cup V'=X$. You need this in order to prove that $A\subseteq U'$.

Specifically, you know that $V'\subseteq V=X\setminus A$, so $V'\cap A=\varnothing$. On the other hand, $A\subseteq X=U'\cup V'$, so it must be the case that $A\subseteq U'$.


问:
How to express error function by confluent hypergeometric function
I am struggling to express error function by confluent hypergeometric function

Which I found on wikipedia: $\displaystyle \operatorname{erf}(x)=\frac{2x}{\sqrt \pi} \Phi \left(\frac{1}{2};\frac{3}{2};-x^2\right)$

I tried to Taylor expand it then integrate it which results in $\displaystyle \frac{2}{\sqrt \pi} \sum_{n=0}^{\infty}\frac {(-1)^n x^{2n+1}}{(2n+1)n!}$

From my book, this problem provided a hint $2n+1= \dfrac{\Gamma(3/2)}{\Gamma(1/2)}$, which this hints seems doesn't make any sense to me as the numerical value should be $\frac{1}{2}$

答:
Begin with
$$
\begin{aligned}
\operatorname{erf}(z)%
&=\frac{2}{\sqrt\pi}\sum_{k=0}^\infty\frac{(-1)^k z^{2k+1}}{k! (2k+1)}\\
&=\frac{2z}{\sqrt\pi}\sum_{k=0}^\infty\frac{(-z^2)^k}{k! (2k+1)}.
\end{aligned}
$$
Now,
$$
\begin{aligned}
\frac{1}{2k+1}=\frac{1}{2}\frac{1}{k+1/2}%
&=\frac{1}{2}\frac{\Gamma(k+1/2)}{\Gamma(k+3/2)}\\
&=\frac{1}{2}\frac{\Gamma(k+1/2)}{\Gamma(1/2)}\frac{\Gamma(1/2)}{\Gamma(3/2)}\frac{\Gamma(3/2)}{\Gamma(k+3/2)}\\
&=\frac{1}{2}\frac{(1/2)_k}{(3/2)_k}\frac{\Gamma(1/2)}{\Gamma(3/2)}\\
&=\frac{1}{2}\frac{(1/2)_k}{(3/2)_k}\frac{\Gamma(1/2)}{\frac{1}{2}\Gamma(1/2)}\\
&=\frac{(1/2)_k}{(3/2)_k}\\
\end{aligned}
$$
Substituting into our Taylor series then gives
$$
\begin{aligned}
\operatorname{erf}(z)%
&=\frac{2z}{\sqrt\pi}\sum_{k=0}^\infty\frac{(1/2)_k}{(3/2)_k}\frac{(-z^2)^k}{k!}\\
&=\frac{2z}{\sqrt\pi}{_1F_1}\left({1/2 \atop 3/2};-z^2\right),
\end{aligned}
$$
with the last equality being due to the definition of the generalized hypergeometric series.


问:
Let $\sigma, \tau \in S_n$. Prove $\sigma\tau \text{ is even} \iff \sigma \text{ and }\tau$ are both even or both odd.
I know that this has to be proven both ways. If $\sigma \tau$ is even then $\sigma$ and $\tau$ are both even or both odd, and if  $\sigma$ and $\tau$ are both even or both odd then $\sigma \tau$ is even.

If I let $\sigma \tau$ be even, then I know it can be written as a product of an even number of transpositions. Since $\sigma$ and $\tau$ are both in $S_n$, they must be disjoint. This means that the product of numbers of transpositions of $\sigma$ and $\tau$ will equal that of $\sigma \tau$.

If $\sigma$ is even, it must be the product of an even number of transpositions which is an odd number of permutations (though I'm not sure if I can just say that or if I need to prove it). If $\sigma$ is odd, it can be written as the product of an odd number of transpositions which is an even number of permutations.

The same can be said for $\tau$.

The product of two even numbers is even, the product of two odd numbers is even, but the product of an even and an odd number is odd. Therefore, for $\sigma \tau$ to be even, $\sigma$ and $\tau$ must be both even or both odd.

Does this make sense or am I missing something?

答:
Here is your first mistake:

Since $\sigma$ and $\tau$ are both in $S_n$, they must be disjoint.

You're on the right track for the proof overall though.


问:
Show that the set of polynomials with rational coefficients is countable.
Problem: Show that the set of polynomials with rational coefficients is countable.

Idea: We know that the set of rational numbers is denumerable. This implies that the set of rational numbers is countable. We also know that the degree that each polynomial can be is a natural number (I think, and I'm not sure how to word this.) Therefore, I think we can reason through this somehow, by showing that the plane $Q X N$ Is denumerable. I'm just not sure how to do this. Any ideas.

Note: this is for my introduction to proofs class study quide. So, I would prefer not to go to over the top.

答:
Write $\Bbb N_0=\Bbb N\cup\{0\}$

Theorem: Let $X\subset\Bbb C$ be a countable set. Then the set $$X[z]=\left\{\sum_{j=0}^{n}a_jz^j:n\in\Bbb N_0,a_j\in X\right\}$$
is countable.

Proof: Since $X$ is countable, so is $\mu(z)=\{az:a\in X\}$. Thus the sets
$$S_n(z)=\prod_{k=0}^{n}\mu(z^k)=\left\{(a_0,a_1z,a_2z^2,..,a_nz^n): (a_0,a_1,...,a_n)\in X^{n+1}\right\}$$
are countable for all $n\in\Bbb N_0$, because they are each a finite product of countable sets. Then for any $p=(p_1,...,p_n)\in\Bbb C^n$, write $\phi_n(p)=\sum_{k=1}^{n}p_k$. We then have that every set of degree-$n$ polynomials with coefficients in $X$, given by $$X_n[z]=\{\phi_n(p):p\in S_n(z)\},$$
is countable. And since $$X[z]=\bigcup_{n\in\Bbb N_0}X_n[z]$$
is a countable union of countable sets, it must also be countable. $\square$

Since $\Bbb Q$ is a countable subset of $\Bbb C$, we get that $\Bbb Q[x]$ is countable via the theorem.


问:
Any curve of constant width is a finite union of arcs of circles?
Any curve of constant width is a finite union of arcs of circles?</p>

答: No; there is no requirement that such a curve be composed of circular arcs.  See Must a curve of constant width be generated with an odd number of sides? for more information and examples.</p>


问:
Find point nearest to the origin
Find the points on the curve $5x^2 - 6xy + 5y^2 = 4$ that are nearest the origin.

The first method I've tried is I've taken the derivative of the equation to optimize (Pythagorean Theorem) and also the function of the curve using implicit differentiation and plugged stuff in (the way I've done it so far). I got $y = x$ which I tried to plug into the function of the curve but didn't match the answers at the back of the textbook. I stared at it and figured I made a sign error somewhere since $-y = x$ worked but couldn't find it.

The second method I tried was solving for the function of the curve in terms of $y$ and tried both completing the square and using the quadratic formula to find $x$ which gave the same failed answer where $y$ subtracted to nonexistence when I plugged them into the function of the curve.

I don't know what lagrange multipliers are, which was the only solution on the web. I hope it could be done just with simple algebra (or calculus) since the textbook expects that (I think).

答:
Hint: Using a change of coordinates $X=x+y$, $Y= x-y$, the problem becomes

$$ 5\frac{X^2+Y^2}{2} -6\frac{X^2-Y^2}{4} = 4,$$

or that

$$ X^2 + 4Y^2 = 4.$$

You should recognize this as an ellipse, with minor axis __ and major axis __.

Hence, the answer is __.

With regards to your differentiation solution:

$x^2 + y^2$ is minimized, when $2x + 2y \frac{dy}{dx} = 0$, or that $\frac{dy}{dx} = -\frac{y}{x}$.

In the original equation, we have $ 10x - 6y - 6x \frac{dy}{dx} + 10y \frac{dy}{dx}$, or that $\frac{dy}{dx} = \frac{6y-10x}{10y-6x}  $.

Hence, solving for $-\frac{y}{x} = \frac{dy}{dx} = \frac{6y-10x}{10y-6x}$, we get $-y(10y-6x)=x(6y-10x)$, or that $-10y^2 +6xy = 6xy - 10x^2$, so $10(x^2-y^2)=0$. This has solution set $x=y, x=-y$.

You then need to check the second order condition, to see which (if any) gives your a global minimum.

Title: Find point nearest to the origin


问:
prove a sequence diverges to infinity
I want to prove $$ \frac{2^n}{n^2} + (-1)^n \to \infty $$ I know that $ (-1)^n $ serves as a distractor here since it wouldn't affect the result as $ n \to \infty $. I try to come up with a sequence that's smaller than the one above and show that it diverges to infinity but fail to do so. I suppose I can utilize the idea $ \frac{x_{n+1}}{x_n} \to c $ such that $ c > 1 $ to show that $ x_n \to \infty $ but don't know where to start. Thank you for helping me.

Title: prove a sequence diverges to infinity

Body:
Let $$a_n=\frac{2^n}{n^2}$$

then

$$\lim_{n\to+\infty}\frac{a_{n+1}}{a_n}=2$$

So, for $ n\ge N$, (great enough) $$\frac{a_{n+1}}{a_n}>\frac 32$$

$$\frac{a_n}{a_{n-1}}>\frac 32$$
....
$$\frac{a_{N+1}}{a_N}>\frac 32$$
thus

$$a_n>(\frac 32)^{n-N-1}$$
hence
$$\lim_{n\to+\infty}a_n=+\infty$$

but

$$\frac{2^n}{n^2}+(-1)^n\ge a_n-1$$
done.


答:
$$ \frac{2^n}{n^2} + (-1)^n \geq \frac{2^n}{n^2} - 1  $$
so \begin{align*}
\lim_{n \rightarrow \infty} \frac{2^n}{n^2} + (-1)^n &\geq \lim_{n \rightarrow \infty} \left( \frac{2^n}{n^2} - 1 \right)  \\
    &= \lim_{n \rightarrow \infty} \left( \frac{2^n - n^2}{n^2} \right)  \\
    &\overset{L'H}{=} \lim_{n \rightarrow \infty} \left( \frac{\ln(2) 2^n - 2n}{2n} \right)  \\
    &\overset{L'H}{=} \lim_{n \rightarrow \infty} \left( \frac{\ln^2(2) 2^n - 2}{2} \right)  \\
    &= \lim_{n \rightarrow \infty} \left( \ln^2(2) 2^{n-1} - 1 \right)  \\
    &= \infty  \text{.}
\end{align*}

Title: prove a sequence diverges to infinity

Body: I want to prove $$ \frac{2^n}{n^2} + (-1)^n \to \infty $$ I know that $ (-1)^n $ serves as a distractor here since it wouldn't affect the result as $ n \to \infty $. I try to come up with a sequence that's smaller than the one above and show that it diverges to infinity but fail to do so. I suppose I can utilize the idea $ \frac{x_{n+1}}{x_n} \to c $ such that $ c > 1 $ to show that $ x_n \to \infty $ but don't know where to start. Thank you for helping me.


问:
Piecewise representation of $\int x^a \,dx$
I was reading about antiderivative of $\frac{1}{x}$ and they represented it as a piecewise function to include the negatives.

Like this:
$$  F(x) =
\begin{cases}
\ln (-x) + C_1, & x<0 \\
\ln (x) + C_2, &x>0
\end{cases}
$$

My question is about the antiderivative of $x^a$ and if you could represent it by including the peices from the antiderivative of $1/x$ along with the integration rule. Is this a correct way of represeting the $ \int x^a \,dx$ ?

$$  \int x^a \,dx =
\begin{cases}
\frac{x^{a+1}}{a+1} + C_1,  & a \neq -1 \\
\ln (-x) + C_2, & a = -1 \wedge (x<0) \\
\ln (x) + C_3, & a = -1 \wedge (x>0)
\end{cases}
$$

答:
Yes this is correct, I only have two remarks:

You should define $C_j$. (In your case probably real.)

You can simply write for $a=1$, $\log\lvert x\rvert+C$, where $C$  might take different values for positive or negative $x$.


问:
If $M$ is connected and $\int_M \omega = 0,$ then $\omega = 0$?
Let $M$ be a connected, orientable, and compact $(r+1)$-dimensional manifold and $\omega$ a $(r+1)$-differential form on $M$. If $\int_M \omega = 0$, then $\omega =0$?

I want to say that the answer is yes. The intuitive reason behind it is that if $\omega\neq 0$, since $M$ is connected then "it must be $\omega>0$ or $\omega<0"$, so we have $\int_M \omega \neq 0$. I've been trying to formalize this intuition by stating what means $\omega>0$ first. Since we can write $\omega(x) = a(u) du_1\wedge \cdots \wedge du_{r+1},x\in M,$ for $x = \varphi(u)$ and some chart $\varphi$, then $\omega>0$ would mean that $a(u)>0$ for every $x\in M$. But i'm struggling to use the fact that $M$ is connected to ensure that $a(u)>0$ should happen.

How can I proper formalize this intuition?

答:
This is not true. For instance, take $M$ to be a manifold (without boundary) and $\omega$ an exact form, so that $\omega=d\eta$. Then by Stokes' Theorem
$$
\int_{M}d\eta=\int_{\partial M}\eta=0.
$$
but being exact does not imply being zero. E.g. we can find $\eta$ so that $d\eta\ne 0$. For a simple example, take any nonconstant function $f$ on $S^1$, then $df\in \Omega^1(S^1)$ is exact but not zero.


问:
Number Theory Problem - finding a solution-forming function
$$\frac{n(n+1)}{2} = 4 q$$

What are all possible $f(k)$ and $g(k)$ such that for all positive integers $k$, $n=f(k)$ and $q=g(k)$ are integer solutions to the above.

答:
You have $n(n+1)=8q$. Thus, $8\mid n(n+1) \to 8\mid n$ or $8\mid n+1$ because $8=2^3$ and $\text{gdc}(n,n+1)=1$ for all $n$.

Case 1: $n=8k$

Then $8k(8k+1)=8q\to q=k(8k+1)$

Case 2: $n=8k-1$

Then $(8k-1)(8k-1+1)=8q\to q=(8k-1)k$

Finally, for all $k\in\mathbb{Z}$, $(n,q)=(8k,k(8k+1))$ and $(n,q)=(8k-1,(8k-1)k)$ are the solutions.


问:
To show that a sequence is bounded above by $3$.
Let $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2 x_n}$.

I would like to know if there is a way to see that this sequence is bounded above by 3 without induction.

I know how to use induction to show that it is in fact bounded by 2. I would like to not use induction though. Thanks for your time.

Body: OK. Let's give it a try. $x_{n} = \sqrt{2x_{n-1}}=2^{\frac{1}{2}}\cdot \sqrt{\sqrt{2x_{n-2}}}=2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}\cdot \sqrt[4]{x_{n-2}}= 2^{\frac{1}{2}+\frac{1}{4}}\cdot \sqrt[4]{\sqrt{2x_{n-3}}}= 2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}\cdot x_{n-3}^{\frac{1}{8}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots +\frac{1}{2^{n-1}}}\cdot x_1^{\frac{1}{2^{n-1}}}= 2^{\frac{1}{2}+\frac{1}{4}+\cdots + \frac{1}{2^n}}= 2^{1-\frac{1}{2^n}}< 2 < 3.$


问:
Find two vectors for basis R4
I have to find two vectors for a basis in $\mathbb{R}^{4}$and that basis needs to contain $ v = \begin{pmatrix} 1\\2\\-1\\0\end{pmatrix},u =\begin{pmatrix} 1\\0\\1\\3\end{pmatrix}$. Then the unitvectors $e2$ and $e4$ are two vectors for a basis together with $v$ and $u$, because these two vectors don't have a $1$ at the second and fourth spot, right?

Edit: $ e2 = \begin{pmatrix} 0 \\ 1 \\0 \\ 0\end{pmatrix}$ and $e4 = \begin{pmatrix} 0 \\ 0 \\0 \\ 1\end{pmatrix}$

答:
If I get your question correct, you want to extend the reduced basis $(\mathcal{B_2})$ of  $\mathbb{R}^{4}$. If yes, then $e_2$ and $e_4$ are not the vectors, with which $\mathcal{B_2}$ can be extended to $\mathcal{B_4}$.

If you are looking forward to extend $\mathcal{B_2}$, you may try to do the following: create a matrix $A_{4\times6}(\mathbb{R})$, such that its columns for sure span the whole $\mathbb{R^4}$:
$$ A = \left(\begin{array}{cc|cccc}
1 & 1 & 1 & 0 & 0 & 0\\
2 & 0 & 0 & 1 & 0 & 0\\
-1 & 1 & 0 & 0& 1 & 0\\
 0 & 3 & 0 & 0& 0 & 1\\
\end{array}\right) \xrightarrow[]{\text{RREF}} \left(\begin{array}{cc|cccc}
1 & 0 & 0 & 0 & -1 & \frac{1}{3}\\
0 & 1 & 0 & 0 & 0 & \frac{1}{3}\\
0 & 0 & 1 & 0 & 1 & -\frac{2}{3}\\
0 & 0 & 0 & 1 & 2 & -\frac{2}{3}\\
\end{array}\right)$$
Now, you have to pick the leading colums to the right of the dash in $RREF(A)$, i.e. the $3^{rd}$ and the $4^{th}$ columns of $RREF(A)$, and take corresponding columns in $A$ as the ones, which extend the basis $\mathcal{B_2}$ to $\mathcal{B_4}$ of $\mathbb{R^{4}}$. That is $\mathcal{B_4} = \mathcal{B_2} \cup \left\{e_1 = \begin{pmatrix}
1\\
0\\
0\\
0\\
\end{pmatrix}, e_2 = \begin{pmatrix}
0\\
1\\
0\\
0\\
\end{pmatrix}\right\}.$

If you need something else, then sorry for wasting your time :)

Title: Find two vectors for basis R4


问:
One point compactification of a topological space with the exclusion topology

Let $X$ a topological space and $\infty \in X$ with the topology  given by $\tau=\lbrace A \subset X \mid \infty \not \in A, X \setminus A \text{ finite } \rbrace $ Interpret $X$  as a one point compactification.

Let $M=X \setminus \lbrace \infty \rbrace$ and $m\in M$. Equipped with the topology $\tau=\lbrace U \subset M \mid X \setminus U \, \, \text{Is finite}, m\not \in U\rbrace $.

I claim that $M$ is locally compact and Hausdorff.

In case of can prove it I like see that $\overline{M}=X$ and hence $X$ was the compactification of $M$.

I have a hour trying to prove that $M$ is locally compact, but I don´t get it. Otherwise is easy see that $M$ is Hausdorff.

Any advice of how to construct $M$ or what things I forget consider for makes $X$ the compactification of $M$.

答:
The definition of $\tau$ is very poorly written. I would normally interpret it to mean that $\tau$ is the family of all subsets of $X$ that have finite complements and do not contain $\infty$, but that is not a topology on $X$, since it does not include the set $X$. Apparently the comma is being (mis)used to stand for or, and what is actually intended is that

$$\tau=\{A\subseteq X:\infty\notin A\text{ or }X\setminus A\text{ is finite}\}\,.$$

Now let $M=X\setminus\{\infty\}$. Then every $A\subseteq M$ satisfies the condition $\infty\notin A$, so every subset of $M$ is open in $X$. Now you should have no trouble showing that $M$ with the relative topology that it inherits from $X$ is locally compact and Hausdorff.

Your $\{U\subseteq M:X\setminus U\text{ is finite and }m\notin U\}$ is not a topology, since it does not include $M$, and is not the set of $A\cap M$ such that $A\in\tau$ no matter how the comma in the original definition of $\tau$ is interpreted.


答:
Let's go back to basics, where $M = [0,1]$.

I'm sure you can construct a non-zero function $f: M \to \Bbb R$ such that $\int_M f(x) \ \mathrm dx = 0$.

Title: If $M$ is connected and $\int_M \omega = 0,$ then $\omega = 0$?

Body:
Let $M$ be a connected, orientable, and compact $(r+1)$-dimensional manifold and $\omega$ a $(r+1)$-differential form on $M$. If $\int_M \omega = 0$, then $\omega =0$?

I want to say that the answer is yes. The intuitive reason behind it is that if $\omega\neq 0$, since $M$ is connected then "it must be $\omega>0$ or $\omega<0"$, so we have $\int_M \omega \neq 0$. I've been trying to formalize this intuition by stating what means $\omega>0$ first. Since we can write $\omega(x) = a(u) du_1\wedge \cdots \wedge du_{r+1},x\in M,$ for $x = \varphi(u)$ and some chart $\varphi$, then $\omega>0$ would mean that $a(u)>0$ for every $x\in M$. But i'm struggling to use the fact that $M$ is connected to ensure that $a(u)>0$ should happen.

How can I proper formalize this intuition?


问:
Average value of the function $f(x)=\cos^{2}x$ on the interval $[0, 2\pi]$
What exactly is the average value of the function $f(x)=\cos^{2}x$ on the interval $[0, 2\pi]$? Of course, we assume that the function is integrable on this interval, then we can use the average value theorem $\frac{1}{2\pi}\int_{0}^{2\pi}\cos^2xdx$. I'm not sure about this but is the final value $\frac{3\pi}{4}$? This seems reasonable enough as it is kind of an inflection point.

答:
Hints (especially good if you do not have trigonometric identities at hand):

Note that $\cos^2(x) = \cos(x) * \cos(x)$

Use integration by parts twice to get an expression of the form $\int_0^{2 \pi} \cos^2(x) \mathrm{d}x = \ldots$ by using $\cos^2(x) = 1-\sin^2(x)$.

The final value of the average is $1/2$.

Can you take it from here?

Alternatively, you can use the trigonometric identity that Hagen von Eitzen posted in the comments.


问:
Number of non-isomorphic graphs
A graph on 8 vertices, each has degree 5. I need to find the number of such graphs, which are non-isomorphic.
So, it's easy to show, that there are $|E| = \frac{8 \cdot 5}{2} = 20$ edges in such graphs, then it's easier to take the complement of such a graph into account (because if two graphs are isomorphic, then their complements are too). It has ${8 \choose 2} - 20 = 8$ edges. From this point I can see only 3 non-isomorphic graphs on 8 vertices with 8 edges: $C_8$, $C_7$ + 1 vertex connected to any other, $C_6$ + 2 vertices connected to any vertex/different vertices in the cycle. Am I missing anything? Or maybe my reasoning is wrong?

答:
In the complement, each vertex has degree $2$, so we have one of

$C_8$

$C_5+C_3$

$C_4+C_4$

I.e., we are looking for the number of ways to write $8$ as sum of (unordered) summands $\ge3$.

Title: Number of non-isomorphic graphs


问:
Probability of at least 1 person in Group A being in a group of X size with a general population
im trying to find a mathematical way to calculate the percentage chance that there is at least 1 Cheater in any given match chosen at random.

Game has 100 players per match

Total Players (including the Cheaters): 3000

there are 3 scenarios needing to be tested, a group of 500, 100, 25 Cheaters

i need to calculate the chance that there is at least 1 cheater in any random match, if someone could supply at least a formula and an example this would be great, thanks!

答:
I will use the following notation.

Population of Players: $P$

Number of Cheaters: $C$

Number of players in a match: $M$

The probability of at least one cheater in a match is $1$ minus the probability that there are no cheaters in the match.

$$1-\dfrac{\dbinom{P-C}{M}}{\dbinom{P}{M}}$$

In the scenarios you chose:

$$P=3000, M=100, C=500: \\ \\ 1-\dfrac{\dbinom{2500}{100}}{\dbinom{3000}{100}} \approx 100\%$$

$$P=3000, M=100, C=100: \\ \\ 1-\dfrac{\dbinom{2900}{100}}{\dbinom{3000}{100}} \approx 96.82\%$$

$$P=3000, M=100, C=25: \\ \\ 1-\dfrac{\dbinom{2975}{100}}{\dbinom{3000}{100}} \approx 57.3\%$$


问:
$X$ and $Y$ have a joint normal distribution with unity variance, zero mean, and correlation = $0.5$ -- what is $P(X > 2Y | X > 0)$?
I know how to find $P(X > 2Y | X > 0)$ for the case where $X$ and $Y$ are independent. I would use a graphical approach. It would come out to be $\frac{\frac{\pi}{2} - \arctan(0.5)}{\pi}$.

With a non-zero correlation, the bivariate distribution is no longer radially symmetric, and I believe for a correlation of $0.5$, it would stretch out the distribution along the $y = x$ line. How can I apply the graphical approach here?

答:
I will assume that the joint mean is $(\mu_X, \mu_Y) = (0,0)$, otherwise the question becomes much more difficult.  Then let

$$X = U, \quad Y = \frac{U + \sqrt{3} V}{2},$$

or equivalently

$$U = X, \quad V = \frac{2Y - X}{\sqrt{3}}.$$

Under this transformation, $(U, V)$ will be bivariate standard normal.  Then the desired probability becomes

$$\Pr[X > 2Y \mid X > 0] = \Pr[\sqrt{3} V < 0 \mid U > 0] = \frac{1}{2}$$ trivially.  The main computation is showing that $(U,V)$ has identity covariance matrix.


问:
Continuous of metric space
Definition : Let $(X, ρ), (Y, d)$ be metric spaces and $f: X \to Y$ some
function. We say that $f$ is continuous at $x_{0} \in X$ if for every sequence $(x_{n})$ in
$X$, from the fact that $\lim x_{n} = x_{0} $ it follows that $\lim f (x_{n}) = f (x_{0})$.
The function $f$ is continuous if it is continuous at every point $x_{0} \in X$

The Problem : Let $(X, d)$ be a metric space and $x_{0} \in X$. Show that the function $f: X \to \mathbb{R}$  given by
$f(x) = d (x, x_{0})$ is continuous.

Any idea how to use the definition in the task because I have no idea how to start

答: Let $y\in X$ be an arbitrary point, let $\{y_n\}\subseteq X$ be an arbitrary sequence which converges to $y$. Using the definition of $f$ and the triangle inequality we have for all $n\in \mathbb{N}$ that $|f(y_n)-f(y)|=|d(y_n, x_0)-d(y,x_0)|\leq d(y_n,y)$. Since $\{y_n\}$ converges to $y$ we have that $d(y_n,y)$ converges to zero, so $|f(y_n)-f(y)|$ converges to zero, so $f(y_n) $ converges to $f(y)$. Because the sequence is an arbitrary, we have a continuity of $f$ at $y$. Since $y$ is an arbitrary we have a continuity of $f$ on $X$.

Title: Continuous of metric space


问:
Unless in a Logic Problem
This is a problem from Kenneth H. Rosen's Discrete mathematics and its applications:

Translate the given statement into propositional logic using the propositions provided.

You cannot edit a protected Wikipedia entry unless you are an administrator.

Express your answer in terms of $e$:“You can edit a protected Wikipedia entry” and $a$: “You are an administrator.”

Based on the book, "$q$ unless $\neg p$" is logically equal to $p \rightarrow q$, which was hard to understand, but by accepting this, the solution would be $\neg a \rightarrow \neg e$ or equally $e \rightarrow a$. This means that "if you can edit a Wikipedia entry, then you are (necessarily) an administrator".

Comparing the two statements in bold, this is indeed true, but from my point of view, the converse statement is also true, which is:

"if you are an administrator, then you can (definitely) edit a Wikipedia entry."

In fact, the first answer I came up with was $a \rightarrow e$, and this was my logic; there may be others rather than administrators who can edit Wikipedia entries, but if the person is an administrator, there is no way he/she cannot edit them!

Now I understand that this is only what I imagined and is not based on the statement, but my question is why the biconditional statement is not acceptable? Is this a common natural language mistake which leads me to believe $e \leftrightarrow a$ should be the answer? And if so, then what is the point of solving these problems and using them in Logical Puzzles, while the natural language can make everything confusing?

答:
Typically 'not $P$ unless $Q$' translates to $\neg Q \to \neg P$ (or what is indeed the same: $P \to Q$) ... but there are certainly cases where we use this expression to also mean that $Q \to P$ ... and thus we would have the biconditional $P \leftrightarrow Q$.

This particular case may indeed be one of those cases where I think it makes sense to have a biconditional, as also pointed out by the commenter.

However, in general, it is typically a safe bet to go with just $P \to Q$. Consider this example:

'You will not pass the course unless you do all the homeworks'

For this, we can definitely say that as long as you do not do the course homeworks, you will not pass the course, i.e. $\neg H \to \neg P$ or $P \to H$: if we know that a student passes the course, then we know that they must have done all the homeworks, since this is apprently a necessary condition for passing the course.

However, just because a student does do all of the homeworks, that does not mean that they will now definitely pass the course: they could still fail all the quizzes, for example. So, we should not, in addition, also have that $H \to P$: doing all the homeworks may not be a sufficient condition to pass the course.

So this is why we typically only use a one-way conditional expressing the nessary condition, rather than using the biconditional. Indeed, even in this case, one could think of a scenario where someone is required to be an adminstrator in order to edit a protected Wikipedia page ... but other conditions need to be satisfied as well (e.g. One needs to have paid the Special Wikipedia Fee or recited the Secret Wikipedia Pledge)

Title: Unless in a Logic Problem


问:
Who first defined open sets in terms of neighborhoods?
One way to define a topology for a set (of "points") X is to first give a suitable definition of a neighborhood of $x$, for any point $x \in X$, and then to define an open set as any subset of $X$ that is a neighborhood of all of its elements.  A topology for $X$ is then defined as the family of all open subsets of $X$.

I would like to know who first proposed this particular approach to defining a topology.

(Note that another approach first defines open sets, and then defines neighborhoods in term of open sets.  Yet another approach takes the closure of a set as the fundamental definition.  Etc.)

EDIT: The motivation behind this question can be found in this other question.  In that question I give a verbatim rendition of a set of axioms defining neighborhoods and then defining open sets in terms of neighborhoods.  I argue that these axioms, as stated, are faulty.  Later (in my answer to the question) I show what I believe is the genesis of the error.  The problem lies in the subtle difference between the following two alternative versions of one of the axioms:

(B$3$) for each $x$, the set of neighborhoods of $x$ is closed under non-infinite intersections;

(B$3^{\prime}$) if $U_1, \dots, U_n$ are neighborhoods of $x$, then $U_1 \cap \dots \cap U_n$ is a neighborhood of $x$;

(Of course, one typically writes finite instead of the awkward non-infinite, but I've used the latter in (B$3$) to bring out the fact that the axiom applies to the empty intersection as well.  IOW, the axiom implies that, for each $x \in X$, the entire set $X$ is a neighborhood of $x$.)

In fact I've come across even more emphatic versions of (B$3^\prime$):

(B$3^{\prime\prime}$) if $U_1$ and $U_2$ are neighborhoods of $x$, then $U_1 \cap U_2$ is a neighborhood of $x$;

The problem with (B$3^\prime$) and (B$3^{\prime\prime}$) is that they fail to rule out the case in which some $x \in X$ has no neighborhoods at all, which in turn will imply that the entire space $X$ is not open.

My interest in this question is that I have found at least three separate mentions in the literature of the faulty set of axioms, one as recent as 1997.  It appears that this is a case of an error that has been propagated for decades, and I'm curious about its origin.  In fact, it's not hard to imagine a sequence of rewordings $\text{B}3 \to \text{B}3^\prime \to \text{B}3^{\prime\prime}$.

答:
Felix Hausdorff wrote 1914 in his book "Grundzüge der Mengenlehre":

(A) Jedem Punkt entspricht mindestens eine Umgebung $U_x$; jede Umgebung $U_x$ enthält den Punkt $x$.

  (B) Sind $U_x$, $V_x$ zwei Umgebungen desselben Punktes $x$, so gibt es eine Umgebung $W_x$, die Teilmenge von beiden ist ($W_x \subseteqq \mathfrak D(U_x, V_x)$).

  (C) Liegt der Punkt $y$ in $U_x$, so gibt es eine Umgebung $U_y$, die Teilmenge von $U_x$ ist ($U_y \subseteqq U_x$).

  (D) Für zwei verschiedene Punkte $x,y$ gibt es zwei Umgebungen $U_x, U_y$ ohne gemeinsamen Punkt ($\mathfrak D(U_x, U_y) = 0$).

My translation:

(A) Each point has at least one neighbourhood $U_x$; each neighbourhood $U_x$ contains $x$.

  (B) If $U_x$, $V_x$ are two neighbourhoods of the same point $x$, then there is a neighbourhood $W_x$, which is a subset of both ($W_x \subseteqq \mathfrak D(U_x, V_x)$).

  (C) If $y$ is a point of $U_x$, then there is a neighbourhood $U_y$, which is a subset of $U_x$ ($U_y \subseteqq U_x$).

  (D) Each two different points $x,y$ have neighbourhoods $U_x$, $U_y$ which do not have a common point ($\mathfrak D(U_x, U_y) = 0$).

As far as I know this is one the earliest definitions of the concept "topological space" (to be exact, of a Hausdorff space due to axiom (D)) and the first which uses neighbourhoods (note: $\mathfrak D$ denotes the intersection herein, it is the first letter of the german "Durchschnitt").


问:
Can every regular polygon be drawn on regular square grid?
Is it possible to draw every regular polygon with an even number of vertices on a regular square grid graph? I am lost with this, it works for all cases I've checked but does it work for all regular $n$-gons where $n$ is even?

答:
If you can do $2n$ then you can do $n$

But you cannot do $7$ even with a ruler and compass (and you can construct a grid with a ruler and compass) so you cannot do $14$ with a grid


问:
$a^x\equiv a^{x\bmod(m-1)}\pmod m$ for m prime.
Prove that $a^x\equiv a^{x\bmod(m-1)}\pmod m$ for m prime:

I attempted to prove the cyclicity using Fermat's little theorem, that $a^p\equiv a\pmod{p} $ but didn't succeed. Could you please explain to me how to prove it?

答:
$a^{p}\equiv a$ is the weak form of Fermat's Little Theorem.

Far stronger and much more relevant is the strong form that if $p\not \mid a$ then $a^{p-1} \equiv 1 \pmod p$.

And with that in mind the proof is immediate, isn't it?

$x \equiv y \pmod {p-1}$ if and only if $y = x + k(p-1)$ for some integer $k$ and so $a^y =a^{x + k(p-1)} = a^x \cdot a^{k(p-1)} \equiv a^x \cdot 1^k\equiv ax \pmod{p-1}$

Nit picking fine print:

  That is to say if $p\not \mid a$.

 If $p\mid a$ then $a^{anything} \equiv 0\pmod p$ so $a^x \equiv a^y$ whether or not $y\equiv x \pmod{p-1}$. .... Oh but $a^0 = 1$,.... so I should say $a^{anything\ but\ 0}\equiv 0 \pmod p$.

 Oh, I guess there is an exception.  If $p\mid a$ then $a\equiv 0\pmod p$ but $a^0 =1 \equiv pmod p$.  So we don't have $a^0 \equiv a^y \pmod p$ for any $y \ne 0$, even if $y \equiv 0 \pmod{p-1}$.


问:
An identity related to the second-order Eulerian numbers.
Recently, some of the remarkable properties of second-order
Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$  A340556 have been proved on MSE [  a ,
 b ,  c  ].

But there are also other notable identities to which these numbers lead.
For instance, we also noticed the following identity, which we haven't seen
elsewhere (reference?):

$$ \sum_{j=0}^{k} \binom{n-j}{n-k}
\left\langle\!\! \left\langle n\atop  j\right\rangle\!\! \right\rangle \,=\,
\sum_{j=0}^k (-1)^{j+k} \binom{n+k}{n+j} \left\{ n+j \atop  j\right \} \quad ( n \ge 0) $$

If we use the notation $ \operatorname{W}_{n, k} $ for these numbers,
we can also introduce the corresponding polynomials.

$$ \operatorname{W}_{n}(x) = \sum_{k=0}^n \operatorname{W}_{n, k} x^k \quad ( n \ge 0) $$

So far so routine, but then came the surprise:

$$ 3^n W_n\left(-\frac13\right) \, = \, 2^n \left\langle\!\!\left\langle - \frac{1}{2} \right\rangle\!\!\right\rangle_n  \quad ( n \ge 0) $$

On the right side are the numbers we recently asked about their
combinatorial significance!

Should I trust this strange equation?

答:
A sketch in three steps: First, we need a definition for $ \operatorname{W}_{n}(x)$. With this, we do not want to assume the
second-order Eulerian numbers. So we take the RHS of the identity.

The next and primary step is to show that in general

$$ \operatorname{W}_{n}(x) = (1 + x)^n
\left\langle \! \left\langle \frac{x}{1+x} \right\rangle \! \right\rangle _n. $$

Finally, plug in $ x = -\frac13 $, et voilà,
$$ \operatorname{W}_{n}\left( - \frac13 \right) = \left( \frac23 \right)^n
\left\langle \! \left\langle -\frac12 \right\rangle \! \right\rangle _n .$$

The choice of the notation $ \operatorname{W}_{n}(x) $ was not accidental.
We are speaking here about the Ward polynomials. Their coefficients are
the Ward numbers which be found at A269939.


问:
A function $f: \mathbb{R}^n \to \mathbb{R}^m$ is differentiable if and only if its components are differentiable
I have a function $f: \mathbb{R}^n \to \mathbb{R}^m$ and need to prove that it's differentiable if and only if its component functions are differentiable. The definition I'm working with is far different from what I'm used to, however. If $f$ is differentiable at some point $c$, then there exists a linear map $T$ such that
\begin{align*}
\lim\limits_{h \to 0} \frac{|f(c + h) - f(c) - Th|}{|h|} = 0.
\end{align*}
The definition I have seen before excludes the norm in the numerator, and it's easy to expand $f$ out in terms of its components and combine it with the length $\frac{1}{|h|}$. With this norm, I can't figure out how to proceed.

答: Your idea is fine, because $\text{something} \to 0$ iff $|\text{something}| \to 0$.


问:
You are tasked with putting the 27 identical dodgeballs away into 4 bins. This time, no bin can hold more than 7 balls
Question: After another gym class you are tasked with putting the 27 identical dodgeballs away into 4 bins. This time, no bin can hold more than 7 balls. How many ways can you clean up?
So I believe this is a problem of over counting, as I currently have

C(30,3) - [C(4,3) C(11,3) - C(4,4) C(3,3)]

But this is not that answer. Any help is appreciated, as I need to better understand this material.

答: If no bin can contain more than $7$ balls, and there are only $4$ bins, that means at most $28$ balls can be placed in the bins, and that this maximum is uniquely attained when each bin contains the maximum of $7$ balls.  Since we have only one less than this maximum number, it means that the only ways to place $27$ balls in the bins is to find the number of ways to remove one ball from the maximum configuration.  And since there are only $4$ ways to do this--in each case, picking the bin from which to remove a single ball--the answer is $4$.


答: Hint: if there were 28 balls instead of 27, each by would need to have $\frac{28}{4}=7$ balls in it, so there would only be one possible arrangement. Therefore, this problem is equivalent to asking how many ways you can remove 1 ball from one of the bins.

Title: You are tasked with putting the 27 identical dodgeballs away into 4 bins. This time, no bin can hold more than 7 balls

Body:
Question: After another gym class you are tasked with putting the 27 identical dodgeballs away into 4 bins. This time, no bin can hold more than 7 balls. How many ways can you clean up?
So I believe this is a problem of over counting, as I currently have

C(30,3) - [C(4,3) C(11,3) - C(4,4) C(3,3)]

But this is not that answer. Any help is appreciated, as I need to better understand this material.


问:
Why does $2$ ramify in $\mathbb{Z}[i]$ but not 3?
I know that we can write $(2)=(1+i)^2$, and from this question we have that $\mathbb{Z}[i]/(3)$ is a field. But perhaps I'm confused why this is so. In particular, doesn't the same computation in the selected answer to that question give us $\mathbb{Z}[i]/(2)\cong (\mathbb{Z}/2\mathbb{Z})[i]$, a field?

答: "$(\Bbb Z/2\Bbb Z)[i]$" is $(\Bbb Z/2\Bbb Z)[x]/(x^2+1) = (\Bbb Z/2\Bbb Z)[x]/((x+1)^2)$ which is not a field because e.g. $(x+1)(x+1) = 0$ in the quotient but $x+1 \ne 0$ in the quotient.


问:
Can I use the fact that {0} is a subring of every ring in this statements?
Prove:

a) A subring $S$ of a ring with unity $R$ does not have to be a ring with unity.

b) A subring $S$ of a ring $R$ without unity can contain a unity.

c) A subring $S$ of a ring $R$ can have a unity which is different than $1_R$

In b) and c), is it enough to say that {0} is a subring of every ring, and 0 is a unity in it?

Also, how to show a)?

答:
For part $a$, look at $2\mathbb{Z} \subseteq \mathbb{Z}$.

Your solution to parts $b$ and $c$ seem correct.


答:
a) You already know several subsets of $\Bbb{Z}$ which are closed under addition and multiplication and do not contain $1$.  They're kernels of $\Bbb{Z} \rightarrow \Bbb{Z}: z \mapsto [z \pmod{p}]$ for primes $p$.

b) and c)  If your definition of subring allows the trivial ring (which is usually the case), then the trivial ring, which you recite, is an example.

Title: Can I use the fact that {0} is a subring of every ring in this statements?

Body:
Prove:

a) A subring $S$ of a ring with unity $R$ does not have to be a ring with unity.

b) A subring $S$ of a ring $R$ without unity can contain a unity.

c) A subring $S$ of a ring $R$ can have a unity which is different than $1_R$

In b) and c), is it enough to say that {0} is a subring of every ring, and 0 is a unity in it?

Also, how to show a)?


问:
Definition of curvature and intuition
I am trying to derive the formula for curvature by finding the radius of best fitting circle. So let $$ \vec{r} \left(t \right)$$
be the parametrization of the curve in question. Let's define tangent lines at points $$ \vec{r} \left(t_0 \right) \ \mathrm{and} \ \vec{r} \left(t_0+\Delta t \right).$$
Then when we let $$ \Delta t \longrightarrow 0,$$
the intersection of normal lines should approach the center of the circle best fitting to point $$ \vec{r} (t_0).$$ And then we could find the radius of curvature R and define curvature as $$ \kappa= 1/R.$$

The problem is that computing this seems a bit harder than I thought. Most of my calculators can't do it. The reason why I use this geometric approach is that it sort of gives intuition to what curvature means. Instead of simply postulating the formula as definition of curvature this  has nice intuitive meaning.

So could someone help me derive the formula? I have seen some other derivations as well but those don't really consider the best fitting circles. And they mostly use something called arc length parameter which I am not familiar with.

答:
Very quickly since I have done it previously on this site: Suppose we have a curve $(x(t),y(t))$ on the plane.  The normal at parameter $t$ is
$$
x'(t)(x-x(t))+y'(t)(y-y(t))=0
$$
and similarly the normal at parameter $t+\Delta t$ is
$$
x'(t+\Delta t)(x-x(t+\Delta t))+y'(t+\Delta t)(y-y(t+\Delta t))=0
$$
Solving, we have
$$
x=\frac{\mbox{$x'(t)y'(t+\Delta t)x(t)+y'(t)y'(t+\Delta t)y(t)$}
-\mbox{$y'(t)x'(t+\Delta t)x(t+\Delta t)-y'(t)y'(t+\Delta t)y(t+\Delta t)$}}{x'(t)y'(t+\Delta t)-y'(t)x'(t+\Delta t)}
$$
and similarly for $y$.

Letting $\Delta t\to 0$, both numerator and denominator $\to 0$.  The first order term in denominator is $[x'(t)y''(t)-y'(t)x''(t)]\Delta t$ and the numerator is $[[x'(t)y''(t)-y'(t)x''(t)]x(t)-y'(t)[x'(t)^2+y'(t)^2]]\Delta t$, similarly for $y$.

So the center of curvature is
$$
\begin{bmatrix}x(t)\\y(t)\end{bmatrix}+\frac{x'(t)^2+y'(t)^2}{x'(t)y''(t)-y'(t)x''(t)}
\begin{bmatrix}-y'(t)\\x'(t)\end{bmatrix}
$$
and we can read off $R=\frac{(\dot{x}^2+\dot{y}^2)^{3/2}}{\dot x\ddot y-\ddot x\dot y}$.

Title: Definition of curvature and intuition


问:
Projection of vector spaces defined by Fourier expansion over all basis vectors
Let $V$ be a nontrivial subspace of vector space $U$ with a positive definite scalar product. $\{v_1,\ldots,v_k\}$ is a orthogonal basis for $V$. I am trying to show that a mapping $P:U\to U$ given by the Fourier expansion $$P(u)=\sum_{i=1}^k \frac{(v_i,u)}{(v_i,v_i)}v_i \hspace{1cm} \text{ for all } u\in U$$
is a projection of $U$ along $V^\perp$ onto $V$.

I see how the interior of this sum represents the projection of a vector $u$ onto $v$ but am not sure how this is along $v^\perp$ and moreso how to show this is a projection for the whole vector spaces?

答: Hint: It's enough to prove that $P(v)=v$ if $v\in V$ and $P(u)=0$ if $u\perp V$.


问:
"Principal uniform space" vs "discrete uniform space"?
Which terms are better for a uniform space such that the set of entourages is a principal filter?

"Principal uniform space" or "discrete uniform space"?

"Principal uniformity" or "discrete uniformity"?

答:
For any 2 sets $A,B$, let's define $$[\![A,B]\!]=\{L\subseteq B\mid
 A\subseteq L \}$$

Let  $X$ be a set. For each nonempty $A\subseteq X$, $[\![A,X]\!]$ is filter on $X$. Any filter on $X$ in this form, is called a principal filter.

Any uniformity $\mathcal D$ on a set $X$ of the form $[\![H,X^2]\!]$, where $H\subseteq X^2$, can be called a principal uniformity.

Assuming $\Delta_X\subseteq H$, where $\Delta_X$ is the diagonal, $[\![H,X^2]\!]$ is a uniformity on $X$ iff

$H=H^{-1}$.

$H=H\circ H$.

As examples:

$[\![X^2,X^2]\!]=\{X^2\}$ is the trivial uniformity. (trivial as applied to the topology (not as a filter)).

$[\![\Delta_X,X^2]\!]$ is the discrete uniformity (its topology is the discrete topology).

Note: Any filter of the form $[\![\{a\},X]\!]$ is called a trivial (ultra)filter.


答:
I am not sure what you want. You have essentially specified the answer to be a root of a certain polynomial equation.

If you want to express the solution in terms of radicals (i.e. $n$th roots) like $\phi = \frac{1+\sqrt5}2$ then I am afraid this is impossible, and the reason has to do with Galois theory, so you can ignore the following if you are unfamiliar with Galois theory.

But if you are, then e.g. the case where $m=7$ we have the polynomial $x^7 - x^6 - 1$ whose Galois group is $S_7$ (computed with PARI/GP) which is not solvable, so its roots cannot be expressed using radicals.

Title: Is there a general formula for the ratio limit of a(n) = a(n-1) + a(n-m).?

Body:
I am exploring the properties of a generalization of Fibonacci numbers and have been encountered an issue.

Specifically I am trying to find the limit of the ratio of the the sequence:

```<code>A(n) = A(n-1) + A(n-m)
A(n) = n+1 for 0 <= k < m
</code>```

when m = 2 this is the Fibonacci sequence.

It's clear that the solution is the real root of the polynomial:

x^m - x^(m-1) - 1 = 0

For n=1 the ratio is 2, n=2 is the golden ratio and n=3 is the supergolden ratio. But beyond that I haven't made much progress. I have written a script to generate the sequences and approximate the ratio which has given the following for the for 1 <= m <= 10:

```<code>1: 2.0: [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288]
2: 1.618033988749895: [1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946]
3: 1.465571231876768: [1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873]
4: 1.3802775690976141: [1, 2, 3, 4, 5, 7, 10, 14, 19, 26, 36, 50, 69, 95, 131, 181, 250, 345, 476, 657]
5: 1.324717957244746: [1, 2, 3, 4, 5, 6, 8, 11, 15, 20, 26, 34, 45, 60, 80, 106, 140, 185, 245, 325]
6: 1.2851990332453493: [1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 27, 34, 43, 55, 71, 92, 119, 153, 196]
7: 1.2554228710768465: [1, 2, 3, 4, 5, 6, 7, 8, 10, 13, 17, 22, 28, 35, 43, 53, 66, 83, 105, 133]
8: 1.2320546314285723: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 18, 23, 29, 36, 44, 53, 64, 78, 96]
9: 1.21314972305964: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 19, 24, 30, 37, 45, 54, 64, 76]
</code>```

Is there a known solution for this? Or a proof there isn't one?

Thank you


问:
K-isomorphic splitting fields in field with characteristic 2
The problem is as follows:

Let $K$ be a field of characteristic $2$, let $a$ be an element of $K$ which is not of the form $b^2 + b$ for any $b \in K$, let $f_a(X) = X^2 + X + a$, and let $L_a = K(\alpha)$ be a splitting field over $K$ of $f_a$, with $\alpha \in L_a$ a root of $f_a$.

Prove that if $a,a'$ are two elements of $K$ as above, then $L_a$ is $K$- isomorphic to $L_{a'}$ iff $a−a'$ is of the form $b^2 +b$ for some $b \in K$.

What I have so far:

$"\Leftarrow":$ If $a-a' = b^2+b$ for some $b \in K$, noting that $X^2+X+a = (X+\alpha)(X+\alpha+1)$ over $K(\alpha)$, we can write $X^2+X+a' = X^2+x+a+b^2+b = (X+\alpha+b)(X+\alpha+b+1)$, using the fact that the characteristic of $K$ equals $2.$ We have that $\alpha+b, \alpha+b+1 \in K(\alpha)$ and so $K(\alpha)$ is a splitting field of $f_{a'}$. Since $L_{a'}$ also is a splitting field of this polynomial, the two fields are $K$- isomorphic.

I can't seem to figure out the $"\Rightarrow"$ implication. I would be grateful for hints or solutions.

Thank you very much in advance!

答:
A roadmap to "$\Rightarrow$". Justify the following claims:

If $L_a\simeq_KL_{a'}$ then there exists an element $z\in L_{a'}\setminus K$ such that $z^2+z=a$.

If $\alpha'\in L_{a'}$ satisfies the equation $\alpha'^2+\alpha'=a'$ then there exist elements $k_0,k_1\in K$ such that $z=k_0+k_1\alpha'$.

$z^2+z=k_0^2+k_0+(k_1^2+k_1)\alpha'+k_1^2a'$. And this implies that we must have $k_1=1$.

$b=k_0$ works.

The idea is pretty much that an element $z=k_0+k_1\alpha\in L_a\setminus K$ has a minimal
polynomial (over $K$) of the form $x^2+x+m$ if and only if $k_1=1$. This is somewhat analogous to the more familiar case that if $n$ is a square-free integer, the element
$z=q_0+q_1\sqrt{n}\in\Bbb{Q}(\sqrt{n})\setminus\Bbb{Q}$ has a minimal polynomial of the form $x^2+q$ if and only if $q_0=0$. We want to "normalize" the quadratic minimal polynomials by adjusting the coefficient of the linear term. Basically because that gives us control of the Galois conjugate ($\overline{z}=z+1$ in the former case and $\overline{z}=-z$ in the latter case). The quirks of characteristic two lead us to this.
An analogue exists for other positive characteristics $p$. Look up Artin-Schreier extensions.


问:
Question About Laurent Series and Expanding about Infinity
So I am trying to find the asymptotic behavior of:

$$\left(1+\frac{x}{n}\right)^n$$

For very large $n$. I know I can use $z=\frac{1}{n}$ and obtain:

$$\left(1+xz\right)^\frac{1}{z}$$

And find a laurent series about $z=0$, but my question is how? How do I expand this about $z=0$? Mathematica gives the following result near $z=0$:

$$(1+xz)^\frac{1}{z}\approx e^x -\frac{1}{2}e^x (zx^2)+\frac{1}{24}z^2e^xx^3(3x+8)+...$$

Which is good enough since I only care about the those first two terms for the problem I'm working, but how do I get those terms? Can I get them from the bionomial theorem? I tried messing with that a bit but it didn't seem to lead me anywhere.

Body:
$$\ln \left(1+xz\right)^{1/z} = \frac1z \ln(1+xz) = \frac1z \left(xz - \frac12 (xz)^2 + O(z^3)\right) = x - \frac12 x^2z +O(z^2)$$
So:
$$(1+xz)^{1/z} = e^x e^{-\frac12x^2z+O(z^2)} = e^x \left( 1 - \frac12 x^2z + O(z^2) \right) = e^x - \frac12 e^x (zx^2) + O(z^2)$$


问:
A Quotient Representation of Singular Values of Symmetric Matrix
The question goes as:

Let $\sigma_1(A) \geq \sigma_2(A) \geq \cdots \geq \sigma_r(A)$ be all (non-zero) singular values of an order $n$ real matrix $A$. Prove:
\begin{align*}
\sigma_k(A) = \sup_{\substack{X \in \mathbb{R}^{k \times n}\\ X \neq 0}}\frac{\sigma_k(XA)}{\sigma_1(X)}, \quad
\sigma_{n + 1 - k}(A) = \inf_{\substack{X \in \mathbb{R}^{k \times n} \\ X \neq 0}}\frac{\sigma_1(XA)}{\sigma_k(X)},
\quad \forall k = 1, 2, \ldots, r.
\end{align*}

Take the first equality as example, by the definition of singular values and the representation of eigenvalues, I am able tho show that
$\sigma_k(A) \leq \sup_{X \in \mathbb{R}^{k \times n}}\frac{\sigma_k(XA)}{\sigma_1(X)}$. It seems difficult to obtain the reverse direction. In particular, I don't know how to use the condition that $A$ is symmetric.

答:
$\newcommand{\real}{\mathbb{R}}$
$\newcommand{\diag}{\mathrm{diag}}$

I will expand @J.Loreaux excellent comment to a more detailed answer for future references. Also, I will take @user8675309's suggestion to allow the singular values to be zero.

For any $1 \leq k \leq r$ and $X \in \real^{k \times n}$, suppose
$$X^TX = O\diag(\sigma_1^2(X), \ldots, \sigma_n^2(X))O^T.$$
is the spectral decomposition of $X^TX$, where $O$ is an order $n$ orthogonal matrix. Recall the notation: for two symmetric matrices $A$ and $B$, $A - B \geq 0$ means the matrix $A - B$ is positive semi-definite. It then follows that

\begin{align*}
  & \Delta_k := \sigma_1^2(X)A^TA - (XA)^T(XA) = \sigma_1^2(X)A^TA - A^TX^TXA  \\
= & A^T(\sigma_1^2(X)I_{(n)} - X^TX)A = A^TO\diag(0, \sigma_1^2(X) - \sigma_2^2(X), \ldots, \sigma_1^2(X) - \sigma_n^2(X))O^TA \geq 0.
\end{align*}

Together with $A^TX^TXA \geq 0$, by Courant-Fischer Theorem (see this link), we then have for every $1 \leq k \leq r$:
\begin{align*}
\lambda_k(\sigma_1^2(X)A^TA) = \lambda_k(A^TX^TXA + \Delta_k) \geq
\lambda_k(A^TX^TXA),
\end{align*}
which implies
\begin{align*}
\sigma_1(X)\sigma_k(A) \geq \sigma_k(XA).
\end{align*}
This shows
\begin{align*}
\sigma_k(A) \geq \sup_{\substack{X \in \real^{k \times n} \\ X \neq 0}}\frac{\sigma_k(XA)}{\sigma_1(X)}.
\end{align*}
The proof to the other inequality is analogous.


问:
Typo in Wikipedia article on Gaussian integral?
Relevant section of article

As of time of writing, the article says:

Suppose $A$ is a symmetric positive-definite (hence invertible) $n \times n$ precision matrix, which is the matrix inverse of the covariance matrix. Then,

$\int_{-\infty}^\infty \exp{\left(-\frac 1 2 \sum\limits_{i,j=1}^{n}A_{ij} x_i x_j \right)} \, d^nx =\int_{-\infty}^\infty \exp{\left(-\frac 1 2 x^{T} A x \right)} \, d^nx=\sqrt{\frac{(2\pi)^n}{\det A}} =\sqrt{\frac{1}{\det (A / 2\pi)}} =\sqrt{\det (2 \pi A^{-1})}$

where the integral is understood to be over $R^n$.

I am wondering if the three determinants on the far right should be $|\mathrm{det}A|,|\mathrm{det}(A/2\pi)|, |\mathrm{det}(2\pi A^{-1})|$ - in other words, have an absolute value sign. Reason for this is if the determinant is negative, then the three expressions are all imaginary, but I don't see why the Gaussian integral would a priori be imaginary or even how it can be imaginary in the first place, given that $A$ is positive-definite.

答:
$A$ is positive-definite, so $\det A > 0$, $\det(A/2\pi) > 0$, and $\det(2\pi A^{-1}) > 0$.

See Does a positive definite matrix have positive determinant for more on this.


问:
Design CFG for language
I am given the language:

$L = $ {${a^{n+m} b^{m+k}a^kb^n | m,n,k > 0 }$}

And i am asked to design a Context Free Grammar for this language. Can anyone help me in this one? Like if there is a step by step guide that i should follow? I am new and i don't know many things about CFG's. Please if you think that my question is too general please don't down-vote, please tell me where to go, like find a tutorial somewhere because i have seen a few in YouTube and i didn't find anything that would help me in this kind of CFG.  Thanks.

答:
I would start by grouping the expression into $a^n (a^m b^m) (b^k a^k) b^n$. This reduces the pattern to $a^n$ plus a string of the form $a^m b^m$ plus a string of the form $b^k a^k$ plus $b^n$.

You can specify a string of the form $a^m b^m$ by considering "adding $a$ to the left and $b$ to the right" as one step, and use the specification to be strings you can build using that step any times to the empty string. Explicitly:

$$T \to aTb ~|~ \epsilon$$

The other steps to building $L$ can be done similarly.

Title: Design CFG for language


问:
Natural isomorphisms between finite-dimensional vector spaces and their duals
There are several questions on this already, and my question arose after reading the explanation given in the following:

In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual?

We know a natural isomorphism $\eta:F \to G$ is defined as a natural transformation whose components are all isomorphisms in $\mathbf{D}$, where $F,G: \mathbf{C} \to \mathbf{D}$ are functors. What is the relation between isomorphisms in the category of vector spaces, which are just morphisms (and certainly not arrows between functors, it seems to me), and a "natural" isomorphism, i.e. the category-theoretic definition?

Also, in the answer to the quoted question, the author states that there is a natural isomorphism between a Euclidean vector space and its dual, and gives it as $V \mapsto V^*, v \mapsto \langle v,- \rangle$. What does this notation mean? And also, if there are natural isomorphisms between some finite-dimensional vector spaces, how can we "prove" that there are none in general, as the OP wants? It seems like the best we could do is a counterexample? Or am I missing a key distinction between Euclidean spaces and finite-dimensional spaces?

答:
$
\newcommand{\A}{\mathscr A}
\newcommand{\B}{\mathscr B}
\newcommand{\F}{\mathscr F}
\newcommand{\G}{\mathscr G}
\newcommand{\V}{\mathscr V}
\newcommand{\I}{\mathscr I}
\newcommand{\D}{\mathscr D}
$In order to better discuss this problem it is useful to have the precise categorical definition of a natural transformation in mind.

So, suppose that $\A$ and $\B$ are categories, and that $\F$ and $\G$ are functors from $\A$ to $\B$.

Definition.  A natural transformation from $\F$ to $\G$ is a correspondence that, to each object $X$ in $\A$,
assigns a morphism
$$
  \Phi_X: \F(X)\to \G(X),
  $$
satisfying the following crucial property:  if $X$ and $Y$ are objects in $\A$, and if  $f\in \text{Hom}(X, Y)$, then the
diagram
$\require{AMScd}$
\begin{CD}
  \mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\
  @V\mathscr{F}(f)VV @VV\mathscr{G}(f)V\\
  \mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y)
  \end{CD}
commutes.

This is, of course, presuming that both $\F$ and $\G$ are covariant functors.  If, on the other hand, both $\F$ and $\G$
are contravariant then the above diagram must be replaced by

\begin{CD}
\mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\
@A\mathscr{F}(f)AA @AA\mathscr{G}(f)A\\
\mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y)
\end{CD}

The question as to whether  there is a natural isomorphism from a vector space to its dual,  if phrased in the context
of category theory, should therefore be:

Question.  Considering the category $\V$ of vector spaces over a given field,  is there a natural transformation $\Phi$
from the identity functor
$$
  \I: \V\to \V,
  $$
to the dual functor
$$
  \D: \V\to \V,
  $$
namely the functor sending a vector space to its dual,  and such that $\Phi_X:X\to X^*$ is an isomorphism for every vector
space $X$?

The first major difficulty one encounters is that the identity functor $\I$ is covariant, while $\D$ is contravariant, so one
should first attempt to adapt the notion of a natural transformation for this situation.  Although this will soon be
understood  to be  a somewhat
foolish  pursuit, in the general case where our functor $\F$ above is covariant and $\G$, contravariant, we could ask
for the commutativity of the diagram
\begin{CD}
\mathscr{F}(X) @>{\Phi_X}>> \mathscr{G}(X)\\
@V\mathscr{F}(f)VV @AA\mathscr{G}(f)A\\
\mathscr{F}(Y) @>{\Phi_Y}>> \mathscr{G}(Y)
\end{CD}

Applying  this to the functors $\I$ and $\D$,  we would be asking for an a priori choice of isomorphism
$$
  \Phi_X:X\to X^*,
  $$
for every vector space $X$, such that

\begin{CD}
X @>{\Phi_X}>> X^*\\
@VTVV @AAT^*A\\
Y @>{\Phi_Y}>> Y^*
\end{CD}

commutes for every linear map $T$ from $X$ to $Y$.
However this is impossible for many reasons,  e.g. we may apply it twice,  once for $T$ and another time for $2T$,
leading up to
$$
  \Phi_X = T^*\Phi_YT = (2T)^*\Phi_Y(2T) = 4 T^*\Phi_YT = 4\Phi_X.
  $$

Of course this cannot hold!


问:
If $M$ is connected and $\int_M \omega = 0,$ then $\omega = 0$?
Let $M$ be a connected, orientable, and compact $(r+1)$-dimensional manifold and $\omega$ a $(r+1)$-differential form on $M$. If $\int_M \omega = 0$, then $\omega =0$?

I want to say that the answer is yes. The intuitive reason behind it is that if $\omega\neq 0$, since $M$ is connected then "it must be $\omega>0$ or $\omega<0"$, so we have $\int_M \omega \neq 0$. I've been trying to formalize this intuition by stating what means $\omega>0$ first. Since we can write $\omega(x) = a(u) du_1\wedge \cdots \wedge du_{r+1},x\in M,$ for $x = \varphi(u)$ and some chart $\varphi$, then $\omega>0$ would mean that $a(u)>0$ for every $x\in M$. But i'm struggling to use the fact that $M$ is connected to ensure that $a(u)>0$ should happen.

How can I proper formalize this intuition?

答:
Let $f:[0,2\pi]\to\mathbb{R}$, note that $M=[0,2\pi]$ connected, orientable and compact, and define $f(x)=\sin(x)$ a $0$-form, then

$$\int_M f(x) \,dx=0$$


问:
Markov Chain Transition Probability
When dealing with markov chains, say I am in state 0 on day 1, is the probability that I will be in state 0 on day 4 equal to the probability that I will be in state 0 on all of day 2, day 3 and day 4? I believe these are equivalent because to get from day 1 to day 4, you have to transition day 1-day 2 , day 2 - day 3 and day 3 - day 4.</p>

答: It is not necessary to be in the same states on days 2 and 3. For example, the states could be $0 \rightarrow 1 \rightarrow 1 \rightarrow 0$ or $0 \rightarrow 1 \rightarrow 2 \rightarrow 0$ and so on.</p>

Title: Markov Chain Transition Probability


问:
Order of elements of the group of $3 \times 3$ upper-triangular matrices over $\mathbb{Z}/p\mathbb{Z}$ with $1$'s in the diagonal

Suppose $p$ is an odd prime. Let $G$ be the group of $3 \times 3$ upper-triangular matrices over $\mathbb{Z}/p\mathbb{Z}$ with $1$'s down the diagonal. Show that every element of $G$ has order that divides $p$.

I know the objective is to show that for $g^{p} = 1$ for every $g \in G$, where $1$ is the identity matrix. However, I am stuck on this problem and don't know how to begin. Any help is appreciated.

答: You can prove by induction that$$(\forall n\in\Bbb N):\begin{bmatrix}1&a&c\\0&1&b\\0&0&1\end{bmatrix}^n=\begin{bmatrix}1&na&\frac{n(n-1)}2ab+nc\\0&1&nb\\0&0&1\end{bmatrix}.$$Therefore, if $p$ is an odd prime, you have$$\begin{bmatrix}1&a&c\\0&1&b\\0&0&1\end{bmatrix}^p=\operatorname{Id}_3.$$


问:
Find the essential spectrum $\sigma_e(T)$
Let $\mathcal{H}$ be a complex, separable, infinite-dimensional Hilbert space. Given an operator $T\in\mathcal{B(H)}$ and $\pi:\mathcal{B(H)}\rightarrow\mathcal{B(H)}/\mathcal{K(H)}$ the canonical quotient map, we define the essential spectrum of $T$ by $\sigma_e(T):=\sigma(\pi(T))$.

Let $\{e_n\}$ be the orthonormal basis for $\mathcal{H}$ and $\{\alpha_n\} \in \ell^{\infty}$.

Define $Te_n=\alpha_n e_n$.

Find the essential spectrum $\sigma_e(T)$ of the operator $T$.

What I've done:

I showed that $\sigma_e(T)\subseteq\sigma(T)$.

Also, I know that by Putnam-Schecheter, we have $\sigma(T)= \sigma_e \cup \Omega$, where $\Omega$ consists of some bounded components for the resolvent$\rho (\pi (T))$ and a sequence of isolated points in $\rho(\pi(T))$ converging to $\sigma_e(T)$.

Now I'm not sure how to continue. I think I need to use compactness at some point.

Any help is appreciated!

Thank you!

答:
I suggest you try to compute the ordinary spectrum first. You can do that by calculating $(T-z)^{-1}$ explicitly. Then for $z$ in the spectrum try to construct an operator $S$ such that $S(T-z)-1$ and $(T-z)S-1$ are compact (you can think about finite rank first; it may be handy in the full solution that compact operator may be approximated by finite rank operators).

There exists also a more direct way which requires additional knowledge. The essential spectrum can be related to the notion of a Fredholm operator and its index. In your case it is fairly easy to figure out when $T-z$ is Fredholm, provided that you know required definitions.


问:
Are there better estimates on Taylor remainders than Taylor's formula gives?
Given an $n+1$ times differentiable $f$, we have $f(x) = T_n(x) + R_n(x)$ with $$T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$$

and $$R_n(x) = f^{(n+1)}(z)\frac{(x-a)^{n+1}}{(n+1)!}$$

for some $z$ between $x$ and $a$. We can then take $M_n(x)$ to be the supremum of $|f^{(n+1)}(t)|$ on the interval between $a$ and $x$ to get the estimate

$$|R_n(x)| \leq M_n(x)\frac{|x-a|^{n+1}}{(n+1)!}.$$

In some cases we can use this to prove that for a given $x$, we have $\lim_{n\to \infty} R_n(x) = 0$, which gives convergence to $f(x)$ of the Taylor series for $f$ evaluated at $x$.

But there are cases where $x$ is in the interval of convergence for the Taylor series for $f$, but where $M_n(x)$ overshoots $|f^{(n+1)}(z)|$ badly, and grows fast enough that $M_n(x)\frac{|x-a|^{n+1}}{(n+1)!}$ does not converge, and thus this method could not be used to show convergence of the Taylor series for $f$ to $f$.

A specific example is the Maclaurin series $1+x+x^2+x^3+\dots$ for $f(x) = \frac{1}{1-x}$ evaluated at $0<x<1$. One can show that $M_n(x) = \frac{(n+1)!}{(1-x)^{n+2}}$, and hence the standard estimate gives
$$|R_n(x)| \leq M_n(x)\frac{|x-a|^{n+1}}{(n+1)!} = \frac{x^{n+1}}{(1-x)^{n+2}}.$$ The right hand side does not go to $0$ if $\tfrac{1}{2}\leq x < 1$, even though the series is geometric and hence clearly converges to $f$ for all of these $x$ values. How would one show that the Taylor series converged to $f$ here if we didn't happen to have a nice geometric series?

Are there general stronger estimates for the remainder $|R_n(x)|$ than the one given above?

答:
Let $f$ be a real-valued function defined on $X=(x_0-a,x_0+a)$ for some $x_0\in\mathbb{R}, a>0$. Let $f$ be $n+1$ times continuosly differentiable on X. Then Taylor remainder can be written in this way:
$R_n(x)=\frac{1}{n!} \int\limits_{x_0}^{x} (x-t)^n f^{(n+1)}(t)dt$ for all $x\in X$.
That form is called an integral form of Taylor remainder. It is used to prove that functions $(1+x)^c$ are real analytic on $(-1,1)$.


问:
Basics Question on Matrix Spaces: Selecting Basis
I am trying to understand the concept of a new vector space where vectors are instead matrices. In this example, I am looking at space M made of all 3 by 3 matrices. The text is as follows:

The dimension of M is 9; we must choose 9 numbers to specify an element of M. The space M is very similar to $\mathbb{R}^9$. A good choice of basis is:
$$
\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right], \ldots\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]
$$

The text goes on to specify: the subspace of symmetric 3 x 3 matrices S has dimension 6. When choosing an element of S we pick three numbers on the diagonal and three in the upper right, which tell us what must appear in the lower left of the matrix. One basis is the collection:
$$
\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right],\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]
$$

The text goes on to do the same for the subspace of all upper triangular matrices (also of dimension 6).  Lastly, it describes the intersection of symmetric and upper triangular to be the subspace of all 3 x 3 diagonal matrices.

Question: The basis and dimension of M, all 3 x 3 matrices, it makes sense that it would require 9 different matrices to span that space. It makes sense each one of the elements in the basis would be a matrix with 1 in a different position of the matrix.  However for the symmetric example, I don't see an easy way to come about the dimension or basis.  I am not clear how/why it's quickly apparent that the dimension of all symmetric matrices is 6. Is this just by intuition?

答:
Let $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}$ be a symmetric matrix.  Then symmetry forces the relations \begin{align*}
b &= d  \\
c &= g  \\
f &= h  \\
\end{align*}
So we are free to choose any $a$, $e$, and $i$ we like.  But a choice of any of the remaining entries in the matrix specifies the values of two entries in the matrix.  So of the six remaining values, we are only free to choose three of them.

The three choices on the diagonal and the three choices that specify pairs of entries gives only six choices.

This it not the first way I came to this answer (many years ago).  Take your thumb and cover up the "$d$", the "$g$", and the "$h$" since each of these is forced to be equal to an uncovered entry.  There are six uncovered entries -- you only get six free choices, then everything else is forced by symmetry.


答:
Let $N = \begin{pmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{pmatrix}$. Then you want to check that $(I + N)^p \equiv I \pmod p$.

So this follows from the binomial theorem and the fact that $N^3 = 0$.

Notice that this does not work for $p = 2$ because

$$(I + N)^2 = I + 2N + N^2 \equiv I + N^2 = I + \begin{pmatrix} 0 & 0 & ac \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$

Title: Order of elements of the group of $3 \times 3$ upper-triangular matrices over $\mathbb{Z}/p\mathbb{Z}$ with $1$'s in the diagonal

Body:
Suppose $p$ is an odd prime. Let $G$ be the group of $3 \times 3$ upper-triangular matrices over $\mathbb{Z}/p\mathbb{Z}$ with $1$'s down the diagonal. Show that every element of $G$ has order that divides $p$.

I know the objective is to show that for $g^{p} = 1$ for every $g \in G$, where $1$ is the identity matrix. However, I am stuck on this problem and don't know how to begin. Any help is appreciated.


答:
Let
$$
g = \begin{pmatrix}
1 & a & b \\
0 & 1 & c \\
0 & 0 & 1
\end{pmatrix}
$$
be an arbitrary matrix in your group $G$. Such a matrix $g$ is what is called unipotent: It is an addition of the identity matrix $I$ with a nilpotent matrix $N$; in this case
$$
N = \begin{pmatrix}
0 & a & b \\
0 & 0 & c \\
0 & 0 & 0
\end{pmatrix}.
$$
Note that $N$ is a nil-cube matrix, i.e., $N^3 = 0$. This lets you write (because $I$ and $N$ commute with each other)
$$
g^p = (I + N)^p = \sum_{k=0}^{p}\binom{p}{k}I^{p-k}N^{k} = I + pN + \binom{p}{2}N^2 = I.
$$


答:
No, they are not equivalent. To be in state 0 on day 4, you may be in state 1 on day 2, stae 2 on day 3 and jump to 0 on day 4. Any path which ends at 0 on day 4 will be good enough.

Being in that state all the time (not going anywhere) is one of the possibilities, but not necessarily the exclusive one.

If it is not possible to jump away from state 0, then the two are equivalent and happen with probability 1 (give that you are in state 0 on day 1 already).

Title: Markov Chain Transition Probability

Body: When dealing with markov chains, say I am in state 0 on day 1, is the probability that I will be in state 0 on day 4 equal to the probability that I will be in state 0 on all of day 2, day 3 and day 4? I believe these are equivalent because to get from day 1 to day 4, you have to transition day 1-day 2 , day 2 - day 3 and day 3 - day 4.</p>


问:
$a^x\equiv a^{x\bmod(m-1)}\pmod m$ for m prime.
Prove that $a^x\equiv a^{x\bmod(m-1)}\pmod m$ for m prime:

I attempted to prove the cyclicity using Fermat's little theorem, that $a^p\equiv a\pmod{p} $ but didn't succeed. Could you please explain to me how to prove it?

答: By Fermat's little theorem, $a^{p-1}\equiv 1\bmod p$, when $(a,p)=1$.  This means we can reduce the exponent $\bmod p-1$ in this case.  On the other hand when $a$ is a multiple of $p$, both sides are zero (or you get $0^0$ which is undefined).


问:
Proving the following shrinking lemma for binary covers
Prove a topological space is normal if and only if for each pair of open sets $U,V$ such that $U \cup V=X$, there exists open sets $U'$ and $V'$ such that $\overline U' \subset U$ and $\overline V' \subset V$ and $U' \cup V'=X$

Attempt:

$(\Rightarrow)$ Since $U \cup V=X \implies (X-U) \cap (X-V)=\varnothing$. $X-U$ and $X-V$ are disjoint closed sets. Then by normality of $X$, there are disjoint open sets $W,Z$ with $X-U \subset W$ and $X-V \subset Z$, and an equivalent definition of normality states there are open sets $W',Z'$ containing $X-U$ and $X-V$ with $\overline W' \subset W$ and $\overline Z' \subset Z$. Now I am unsure how to show that $W' \cup Z'=X$, which is what I need to show in the problem.

$(\Leftarrow)$ Suppose the second part of the statement holds. Let $A$ be closed in $X$ and $U$ be any open set containing $X$. Then the existence of an open set $U'$ containing $A$ such that $\overline U' \subset U$ implies $X$ is normal.

Comment:I am having difficulty understanding (why/if) $W' \cup Z'=X$ holds in the first direction. Also I am skeptical whether I started the second direction correctly by starting with a closed set, and an arbitrary open set containing this closed set. I also realize it may not be true that $U'$ contains the closed set $A$, which adds to my skepticism. Any ideas?

答:
I think applying "$X$ normal iff for all closed $A$ and open $O$ with $A \subseteq O$ we have $U$ open with $A \subseteq U \subseteq \overline{U} \subseteq O$" is more direct in this case:

Given the cover $\{U,V\}$, consider $V^\complement$, which is closed and a subset of $U$, so there is a $U'$ open such that $$V^\complement \subseteq U' \subseteq \overline{U'} \subseteq U$$ by normality. Then $\{U', V\}$ also covers $X$ (a point not in $V$ is in $U'$) and we do the same again: $U'^\complement$ is closed and a subset of $V$ so we have an open $V'$ such that $$U'^\complement \subseteq V' \subseteq \overline{V'} \subseteq V$$

and then $\{U',V'\}$ is the required cover as is now immediately clear. This procedure easily extends by induction to all finite open covers, of course.

The reverse (that having the refinements implies normality) is quite direct too:
let $A \subseteq O$ where $A$ is closed and $O$ is open, then $\{ A^\complement,O\}$ is an open cover of $X$ so we have refining $\{U',V'\}$ and $U' \subseteq \overline{U'} \subseteq A^\complement$ implies $A \subseteq \overline{U'}^\complement \subseteq U'^\complement$ and the latter closed set is a subset of $V' \subseteq O$ so that, defining $O'= \overline{U'}^\complement$ we have $A \subseteq O' \subseteq \overline{O'} \subseteq O$ showing normality.


问:
To show that a sequence is bounded above by $3$.
Let $x_1=\sqrt2$ and $x_{n+1}=\sqrt{2 x_n}$.

I would like to know if there is a way to see that this sequence is bounded above by 3 without induction.

I know how to use induction to show that it is in fact bounded by 2. I would like to not use induction though. Thanks for your time.

Body:
hint

$$x_{n+1}-3=\sqrt{2x_n}-3$$
$$=\frac{2x_n-9}{\sqrt{2x_n}+3}$$

$$=\frac{2(x_n-3)-3}{\sqrt{2x_n}+3}$$

$$<\frac{2(x_n-3)}{3}$$
$$<(\frac 23)^{n}(x_1-3)<0$$