问:
Orthogonal Trajectories to the solutions of $\frac{dy}{dt} = y$
By using the properties of the dot product, I get as answer
$y_{ortho} = \frac{1}{C}e^{-t}$ , if  $y = Ce^{t}$ and $C \neq 0$.

If a try to use the inner product of function $\int fg$, I don't get $0$ as I would expect.

What is wrong with this answer?

Thanks in advance.

答:
In the original DE $y'=y$
,substitute  $y' \to -\dfrac 1 {y'}$:
$$-\dfrac 1 {y'}=y$$
$$y'y=-1$$
$$\int ydy=-\int dt$$
The equation of the orthogonals curves are:
$$y^2+2t=C$$


问:
What is the "Mean curvature vector" in $\mathbb{R}^4$?
I need to understand the terminology "Mean curvature vector" in $\mathbb{R}^4$. How can we understand this terminology ? For example, what is "mean curvature vector" of a plane in $\mathbb{R}^4$, of a 2-dimensional sphere in $\mathbb{R}^4$, 2-dimensional cylinder in a hypersurface?

答:
Let $\vec u\in\mathbb R^2$ be coordinates for a variable point $\vec x\in\mathbb R^4$ on the surface. A basis for the tangent plane at $\vec x$ is given by $\vec e_k=\partial\vec x/\partial u_k$ (for $k=1,2$), which has a corresponding reciprocal basis $\vec e^k$ given by

$$\vec e^1=\frac{(\vec e_2\cdot\vec e_2)\vec e_1-(\vec e_1\cdot\vec e_2)\vec e_2}{(\vec e_1\cdot\vec e_1)(\vec e_2\cdot\vec e_2)-(\vec e_1\cdot\vec e_2)^2},$$

$$\vec e^2=\frac{(\vec e_1\cdot\vec e_1)\vec e_2-(\vec e_1\cdot\vec e_2)\vec e_1}{(\vec e_1\cdot\vec e_1)(\vec e_2\cdot\vec e_2)-(\vec e_1\cdot\vec e_2)^2}.$$

(Verify that $\vec e^k\cdot\vec e_l=\delta^k_l$.) Define the projection $P_\parallel$ onto the tangent plane, $\vec v\mapsto P_\parallel(\vec v)=\sum_k(\vec v\cdot\vec e^k)\vec e_k$, and the rejection $P_\perp$ from the tangent plane, $\vec v\mapsto P_\perp(\vec v)=\vec v-P_\parallel(\vec v)$. Then one possible definition of the mean curvature vector is

$$2\vec H=\sum_{k,l}\left(\vec e^k\cdot\vec e^l\right)P_\perp\left(\frac{\partial\vec e_k}{\partial u_l}\right)=\sum_{k,l}\left(\vec e^k\cdot\vec e^l\right)P_\perp\left(\frac{\partial\partial\vec x}{\partial u_l\partial u_k}\right).$$

It can be shown that this is independent of the choice of coordinates. The factor of $2$, or the dimension of the manifold, may or may not be included as a matter of convention.

The mean curvature vector, or any type of curvature, of a plane is $0$.

For a 2D sphere with radius $a$ and centre $0$ in 4D, one parametrization is

$$\vec x=(a\cos u_1\cos u_2,\;a\sin u_1\cos u_2,\;a\sin u_2,\;0)=a\,\hat x$$

(the hat indicating a unit vector). I won't go through the calculations; the result is

$$2\vec H=-\frac2a\hat x.$$

ContentLicense: CC BY-SA 4.0


问:
Is it possible to prove from the definition of big $O$ that $5n^3+7n+1$ is $O(n^3)$?
Is it possible to prove from the definition of big O that $5n^3+7n+1$ is $O(n^3)$? Can this be generalised to any case where you have to (and what is the procedure for  working it out?) I guess the main question is there a specific term for this kind of procedure?

from the workings I've been given in class given $T\le O(f(n))$ if $n=1$ then $5n^3+7n+1 = 5^3+7+1 (23)$. Although I'm not sure how that relates to proving the above.

Thanks,

P.L.G.

答:
Big O notation is fundamentally (sort of) a measure of growth rate. So the growth rate of a polynomial $f(x)$ is going to be bounded by its leading term. Since the coefficients on the leading term are constants, they are irrelevant to the growth rate, and thus removed.

Wikipedia has a good example of this, applied to $f(x)=6x^4-2x^3+5$, and goes through, step by step, the process used to derive this. I'm not going to post it here, but it's pretty straightforward. Conceptually, it's the above statement.

ContentLicense: CC BY-SA 3.0


问:
MLE of a linear combination of random variables
Let $x_1$ and $x_2$ be observed random variables. Let $y$ be defined as:
\begin{align}
y = \theta_1 x_1 + \theta_2 x_2
\label{1}
\end{align}
with $\theta_1$ and $\theta_2$ real-valued coefficients. I have prior knowledge that $y \sim N(0,\sigma^2)$, i.e. that $y$ is Gaussian with zero mean value and variance $\sigma^2$. How do I define a maximum likelihood estimator for $\theta_1$ and $\theta_2$, assuming $\sigma$ known?

SOLUTION ATTEMPT
\begin{align}
\log{L(\theta_1, \theta_2|y)} = \log{\prod_{i=1}^{N} p(y_i|\theta_1,\theta_2)} = -\frac{1}{2} N \log{(2 \pi \sigma^2)} -\frac{1}{2} \frac{1}{\sigma^2} \sum_{i=1}^{N} (\theta_1 x_{1,i}+\theta_2 x_{2,i})^2
\label{2}
\end{align}
Clearly the best solution is $\theta_1 = \theta_2 = 0$, but I expect a solution depending on $x_1$, $x_2$ and $\sigma$.

答:
If you want to solve for MLE normally, you need the joint distribution of $Y,X_1$ and $X_2$. However, since we know nothing about the distribution of $X_1$ and $X_2$, we can't obtain the exact joint distribution or the likelihood function. The one you wrote is not the likelihood function.

If $X_1$ and $X_2$ are independent, at least we can write it as
$$
f(x_1,x_2,y|\theta_1,\theta_2) = f_Y(y|x_1,x_2,\theta_1,\theta_2)f_{X_1}(x_1)f_{X_2}(x_2)
$$
Fortunately,
$$
EY = \theta_1EX_1 +\theta_2EX_2 = 0\\
Var(Y) = \theta_1^2Var(X_1) + \theta_2^2Var(X_2)=\sigma^2
$$
which must be true regardless of the distributions of $X_1$ and $X_2$ (If they are independent). Therefore, we can solve for $\theta_1$ and $\theta_2$ in terms of $EX_1,EX_2,Var(X_1)$ and $Var(X_2)$.

Now due to the invariance property of MLEs, the MLE for $\theta_1$ and $\theta_2$ are obtained by replacing $EX_1,EX_2,Var(X_1)$ and $Var(X_2)$ by their respective MLEs.

Edit: I have to correct my statement. You can still solve for MLE the normal way, since $f_{X_1}$ and $f_{X_2}$ do not depend on $\theta_1$ and $\theta_2$. However, if you just differentiate w.r.t to $\theta_1$ or $\theta_2$, you will get the same equations twice. To get the exact solution, we need the additional constraint, which is $Var(Y) = \sigma^2$. This method is equivalent to the one above, as we are implicitly computing the MLEs of $\theta_1$ and $\theta_2$ by computing the MLEs of $EY$ and $Var(Y)$.


问:
Finding a tetrahedron in a unit sphere
I am trying to solve an exercise where I have to find a tetrahedron in a unit sphere. My first thought and i think it is not a bad idea is to let one point be at $(0,1,0)$ (assume this is $A$) and then find a plane

$y = -d$

so that

$|\overline{AB}|$ = $|\overline{BC}|$

using a new point $O = (0, d, 0)$ which will be the centroid of the triangle $BCD$. Now i know that the points $B, C, D$ will be on the sphere so that the equation

$x^2 + z^2 = 1 - d^2$

holds true.
Furthermore, I know that

$2 * sin(60) * \overline{OB} = \overline{BC}$.

I feel like i have collected all the parts of the puzzle, but i still cannot solve it. Any suggestions?

EDIT: Since somone mentioned it in the comments, it should be a regular tetrahedron

答:
Plane $y = d$ will cut the sphere $x^2 + y^2 + z^2 = 1$ such that you get a circle

$x^2 + z^2 = 1 - d^2 \,$ (so radius is $\sqrt{1-d^2}$) as you rightly found.

Now this circle is circumcircle of the base of our tetrahedron (equilateral triangle). So the side of the equilateral triangle is $\sqrt{3(1-d^2)} \,$.

Now the distance from the vertex $P (0,1,0)$ to the center of the circle at $y = d$ will be $(1-d)$.

So the distance from the vertex $P$ to any of the three vertices on the base will be

$ = \sqrt{1-d^2 + (1-d)^2} = \sqrt{2 - 2d} \,$ (by Pythagoras)

Equating both as the tetrahedron is regular $\sqrt{2 - 2d} = \sqrt{3(1-d^2)}$

Solving it, you get value of $d = 1, -\frac{1}{3}$.

Now we take $d = - \frac{1}{3}$ and find $3$ points on the circle $x^2 + z^2 = \frac{8}{9}$ at $y = -\frac{1}{3}$ such that it is an equilateral triangle. As it is symmetric around $y$ axis, you will have multiple such triples. Pick one for example where $x = 0$ and then the rest of the two points.

Title: Finding a tetrahedron in a unit sphere


问:
Example of a non-continuous functional on C[0,1] with L2-Norm
do You know some simple example of a non-continuous functional on $C[0,1]$ with the $L^2$-Norm? Surely we can build such example using the Hamel basis but I ask myself if there is some simpler example. On $C^1[0,1]$ we could simply take differentiation. But on $C[0,1]$?
Thanks in advance.

答:
If you take $\phi(f)=f(1)$, this is unbounded. Define
$$
f_n(t)=\begin{cases}
0,&\ 0\leq t\leq 1-\tfrac1{n^2}\\[0.3cm]
nt+\tfrac1n-n,&\ t>1-\tfrac1{n^2}
\end{cases}
$$
Then $\phi(f_n)=n$, while
$$
\|f_n\|_2^2=\int_{1-\tfrac1n}^1(nt+\tfrac1n-n)\,dt=\tfrac1{3n^4}.
$$


问:
How to calculate with uncertainties in this example?
One exercise says to drop something from a certain height and measure the time it takes for it to hit the floor and then calculate the gravitational acceleration, but can you show me how to calculate with the uncertainties I got?

The times i got was 0.20, 0,18s, 0,13s.. ten times. So $\overline{t}=\frac{0.20 + 0.18..}{10}$ and $\Delta t = \frac{max-min}{2}$ so the uncertainty in time is $(0,203 \pm0,09)s$ and the height is $(0,715\pm 0,05)m$. (Im not sure about the number of significant digits?)

To calculate the gravitational acceleration we have $s=V_0t + \frac{1}{2}at^2 \Rightarrow a =\frac{2s}{t^2}$. So is

$$a=\frac{2\cdot (0,715\pm0,05)m}{(0,203\pm0,09)s^2}$$

correct? And how do I calculate that? I appreciate your time and help.

答:
You measured $n=10$ times, lets call them $t_i$ with $ i=1,\dots,10$.
The mean time is given by
$$
\bar{t}:= \frac{1}{n} \sum_{i=1}^n t_i = \frac{0.20+0.19+0.13+\dots}{10}s \,.
$$
The uncertainty on the mean time is given by
$$
\sigma_{\bar{t}} = \sqrt{\frac{1}{n(n-1)} \sum_{i=1}^n (\bar{t}-t_i)^2} \,.
$$

The height was measred to be $h=0.715m$ and its uncertainty $\sigma_h=0.05m$.

Now we have to compute gravitational accerleration
$$
a=\frac{2h}{\bar{t}^2} = \frac{2\cdot 0.715m}{\bar{t}^2} \,.
$$
Its uncertainty can be calculated via error propagation
$$
a=a(h,t) \implies \sigma_a
= \sqrt{\left(\frac{\partial a}{\partial h} \sigma_h\right)^2
+\left(\frac{\partial a}{\partial t} \sigma_{\bar{t}}\right)^2}
= \sqrt{\left(\frac{2}{\bar{t}^2} \sigma_h\right)^2
+\left(\frac{4h}{\bar{t}^3} \sigma_{\bar{t}}\right)^2} \,.
$$


问:
Example of Schur decomposition

Find the Schur decomposition for the matrix $$ A=\begin{bmatrix}1&1\\
-2&-1\end{bmatrix}$$

I have attempted to do as the question asks via finding the eigenvectors ($u_1=(-1-i,2)^T$ corresponding to $\lambda_1=i$ and $u_2=(-1+i,2)$ corresponding to $\lambda_2=-i$), using Gram-Schmidt and then make the obtained vectors orthonormal and finally obtain $U=[v_1\ v_2],$ where $v_1$ and $v_2$ are the orthonomal vectors obtained. However in my attempts of this I am running into issues (complex vectors for example when solution shouldn't have complex entries according to the MATLAB decomposition). I would appreciate if anyone could help.

答:
Recall that in the proof of Schur decomposition, we used induction. Generally speaking, we will do it recursively to find a Schur decomposition of some square matrix.

For this one, you have obtained the eigenvalues and the corresponding eigenvectors. A nice job. Then you can start with any eigenvalue, say $\lambda_1=i$. The corresponding eigenvector is $u_1=[-1-i,2]^T$. Nonetheless, you have to normalize it:
\begin{align}
||u_1||&=\sqrt{|-1-i|^2+|2|^2}=\sqrt{2+4}=\sqrt{6}. \\
v_1&=\frac{u_1}{||u_1||}=\frac{1}{\sqrt{6}}[-1-i,2]^T.
\end{align}
Here $|c|$ is the modular of a complex number $c$. Next, you will have to find a unitary matrix $U_1=[u_{11}\ u_{12}]$ where $u_{11}=v_1$. In general, this unitary matrix $U_1$ is not unique. Anyway, here we can easily find one: $$U_1=\frac{1}{\sqrt{6}}\begin{bmatrix} -1-i & 2 \\ 2 & 1-i \end{bmatrix}.$$
Now we can see that $$U_1^*AU_1=\frac{1}{6}\begin{bmatrix} -1+i & 2 \\ 2 & 1+i \end{bmatrix}\begin{bmatrix} 1 & 1 \\ -2 & -1 \end{bmatrix}\begin{bmatrix} -1-i & 2 \\ 2 & 1-i \end{bmatrix}=\begin{bmatrix} i & -2+i \\ 0 & -i \end{bmatrix}.$$It seems that we are done now. But for matrices of order at least $3$, we will have a matrix of the form: $$\begin{bmatrix} \lambda_1 & * \\ 0 & A_{(n-1)\times(n-1)} \end{bmatrix}$$Then we apply almost the same step to the block $A_{(n-1)\times(n-1)}$.

As for the MATLAB output, according to the manual, if a matrix has real and complex eigenvalues, you have to set the flag to be complex to get a triangular matrix. In addition, the Schur decomposition of a matrix is not unique, so the MATLAB output might be different from what we obtained here.

Title: Example of Schur decomposition


问:
Finite atomic measure space can be a countable disjoint union of atoms
An atom A in a measure space is a measurable set with positive measure such that every measurable subset of A has the same measure of A or 0.

An atomic measure is a measure space such that in every measurable set there is an atom.

Let's say that I have a positive atomic measure space $(X, \Sigma, \mu)$ such that $\mu(X)<\infty$.
I want to prove that X is a countable disjoint union of atoms and a set with zero measure.

My attempt:

X is a measurable set therefore exist an atom $A_1 \subseteq X$, if $\mu(X/A_1) = 0$ we are done, if not we define $A_2$ as the sub atom of $X/A_1$ and we precede this way for every n. What I need to prove is that $\mu(X/\bigcup_{n=1}^{\infty}A_n) = 0$.

答:
Am not sure if you can prove that $\mu\bigl(X\setminus \bigcup_{n=1}^{\infty}A_n\bigr)=0$ since you dont have any condition in your selection of the $A_n$'s, but this can be done if you strenghthen your recursive construction of the $A_n$'s.

The above argument is an example of an "exhaustion argument", we will "exhaust" all the of the atoms of $X$.

Let $\mathcal{A_1}=\{A\in \Sigma:\, A\subseteq A,\ A\ \text{is an  atom}\}$ and $\alpha_1=\sup_{A\in \mathcal{A_1}}\mu(A)>0$. Then, we find an atom $A_1\subseteq X$ such that $\mu(A_1)\geq 2^{-1}\alpha_1$ (this is our condition). As you said, if for every $B\subseteq X\setminus A_1$ we have $\mu(B)=0$ then we write $X=A_1\cup B$ and we are done. Suppose now, that $X$ cannot be written as finite disjoint union of atoms and a set of zero measure. Then continuing as before, recursively we find a sequence $A_n$ of atoms such that

$1)$ $\mu(A_{n+1})\geq 2^{-1}\alpha_{n+1}$

$2)$ $\alpha_{n+1}=\sup_{A\in \mathcal{A_{n+1}}}\mu(A)$

$3)$ $\mathcal{A_{n+1}}=\{A\in \Sigma:\, A\subseteq X\setminus(A_1\cup...\cup A_{n}),\,\ $A$\, \text{is an atom}\}$

Now, if $A=\bigcup_{n=1}^{\infty}A_n$ we will show that $\mu(X\setminus A)=0$. Since, the $A_n$'s are disjoint by $(1)$ we have
$$\mu(A)=\sum_{n=1}^{\infty}\mu(A_n)\geq \sum_{n=1}^{\infty}\frac{\alpha_n}{2}$$
Now, $\mu$ being finite implies that $\alpha_n\to 0$ as $n\to \infty$. Suppose now that $X\setminus A$ has positive measure. Then, $X\setminus A$ would contain an atom, say $B$. But $B\subseteq X\setminus A$ implies that $B\subseteq X\setminus (A_1\cup ...\cup A_{n})$ for every $n$. So, since $B$ is an atom it follows that $B\in \mathcal{A_{n+1}}$. Hence, by the definition of the $\alpha_n's$ we must have $\mu(B)\leq \alpha_{n+1}$ for every $n$. So, $B$ must have zero measure, which contradicts the fact that $B$ is an atom and must have a positive measure.


问:
USAMO problem solution
I had asked for a hint here USAMO problem hint. i had tried induction once but i thought it won't work so i left it, but after seeing @lulu 's comment, i decided to give it a go again. Please see if my solution is correct.

[USAMO 2003] Prove that for every positive integer n there exists an n-digit number divisible by $5^n$ all of whose digits are odd.

MY SOLUTION: So first thing, i checked some small cases and figured we could generate the number with (n+1) digits satisfying the property by adding a number to it's front, ie. adding b 10$^n$ to the number with n digits.

We'll proceed with induction, let P(n) mean there exists an n-digit number divisible by $5^n$ all of whose digits are odd.

P(1) is true as 5|5.

Let P(k) be true, ie. let 5$^k$ | $a_ka_{k-1}...a_1$  with $a_i$  $\neq$ 2l for i $\in$ {1,2...k} .

I'll attempt to prove that by adding $ b \cdot 10^k $  with $ b \in {1,3,5,7,9} $ . we can have a number that is divisible by $5^{k+1}$.

So we want 5$^{k+1}$ $|$ $ b \cdot 10^k $ +  $a_ka_{k-1}...a_1$ . --> eq.1

Let $a_ka_{k-1}...a_1$ = $5^km  $

So from eq. 1, inputting $a_ka_{k-1}...a_1$ = $5^km  $ , we will get

5$^{k+1}$ $|$ $ b \cdot 10^k $ +  $5^k$m, then dividing throught by $5^k$ , we need

5 $|$ $2^k \cdot b + m$

as  $ b \in {1,3,5,7,9} $ , $\equiv$ 0,1,2,3,4 (mod 5)

So now m $\equiv$ 0,1,2,3,4 (mod 5) , let m $\equiv$ r (mod 5),

We need $2^k \cdot b + r  =0 (mod 5)$

now,$2^k \equiv$ 1,2,3,4 (mod 5)

so painstakingly going through each and every case of possible values of $2^k$ and m (mod 5) (there are 16 cases), we prove that we can find a $ b \in {1,3,5,7,9} $ such that 5 $|$ $2^k \cdot b + m$ .

This is the first time i have written so much in latex, so i am sorry if there is any mistake.

If you were a grader, out of 7, how many points would you give me?

答:
I don't think math.se answers can address how they'd mark it, but I can advise on a cleaner way to write answers, because your ideas are right but they could do with algebraic clarity, and clarity regarding modulo arithmetic. (If you find yourself claiming that if we do something we eventually get a particular outcome, try to state this as an existence theorem that's either obvious, well-known or proven within your work.) To wit:

We claim some sequence $a_n$ of $n$-digit numbers in base $10$, all
digits odd, satisfies $5^n|a_n,\,10^n|a_{n+1}-a_n$. In particular
write $a_n=5^nb_n,\,a_{n+1}=a_n+10^nc_n$, so $b_1=1$ (because $a_1=5$)
and$$5^{n+1}b_{n+1}=a_{n+1}=c_n10^n+5^nb_n\iff5b_{n+1}=c_n2^n+b_n,$$so
it suffices to choose $c_n\in\{1,\,3,\,5,\,7,\,9\}$ with
$5|c_n2^n+b_n$. This choice is possible because these $5$ choices of
the $c_n$ each achieve a different residue class modulo $5$ (because
$5\nmid k2^n$ for $k\in\{2,\,4,\,6,\,8\}$), and exactly one achieves
$5|c_n2^n+b_n$.

Title: USAMO problem solution


问:
ABCD is a cyclic quadrilateral whose two diagonals are perpendicular.
ABCD is a cyclic quadrilateral whose two diagonals are perpendicular. If $R$ is the radius of the circumcircle, prove that:

$AB$$2$ + $BC$$2$ + $CD$$2$ + $DA$$2$ = $8R$$2$

If the centre of the circle is O, I tried drawing radii perpendicular to each of the sides. That way, we get 3 pairs of congruent right triangles.

答:
Lemma. If $ABCD$ is a cyclic orthodiagonal quadrilateral centered at $M$, then $$\angle AMB+\angle CMD=\angle BMC+\angle DMA=180^\circ$$

Here you have a "proof without words"

Thus, we have that $DA$ and $BC$ form a right triangle with a diameter, and hence $DA^2+BC^2=4R^2$. We similarly obtain $AB^2+CD^2=4R^2$...


问:
No-monotone sequence
A no-monotone sequence can have limit?

Can I consider as no-monotone sequence $a_n=\frac{(-1)^n}{n}$?

答:
The sequence is not monotone for if
$n$ is even then $n + 1$ is odd and $\frac{(- 1)^{n + 1}}{n + 1} < 0 <
\frac{(- 1)^n}{n}$ and if $n$ is odd then $n + 1$ is even so
that $\frac{(- 1)^n}{n} < 0 < \frac{(- 1)^{n + 1}}{n + 1}$. Further if $n$ is
even we have $\left| \frac{(- 1)^n}{n} \right| = \left| \frac{1}{n} \right| =
\frac{1}{n}$ and if $n$ is odd we have $\left| \frac{(- 1)^n}{n} \right| =
\left| \frac{- 1}{n} \right| = \left| \frac{1}{n} \right| = \frac{1}{n}$. Now
for $0 < \varepsilon$ there exists a $N \in \mathbb{N}$ such that $0 <
\frac{1}{N} < \varepsilon$ hence if $n \geqslant N$ we have that $\left|
\frac{(- 1)^n}{n} - 0 \right| = \left| \frac{(- 1)^n}{n} \right| = \frac{1}{n}
\leqslant \frac{1}{N} < \varepsilon$ proving that $\underset{n - <
\infty}{\lim} \frac{(- 1)^n}{n} = 0$. So the answer is yes.

Title: No-monotone sequence


问:
Sketch of proof concerning a conjecture of Vasile Cirtoaje : $(1-x)^{(2x)^k}+x^{(2(1-x))^k}\leq 1$ with $k\geq1$ and $0<x\leq \frac{1}{2}$

Claim

Prove that [1]:

Let $0<x\leq 0.5$ and $k\geq 1$ a real number then we have :

$$h(k)=(1-x)^{(2x)^k}+x^{(2(1-x))^k}\leq 1$$

Sketch of proof :

The case $k\geq 1+\alpha$ :

We define $\alpha$ as :

$$\lim_{x\to 0.5^-}=\frac{\ln\Big(2(2^{2x}x^2(1-x)2)\Big)}{\ln(2x)}=\frac{\ln\Big(2(1-(2^{2x}x^2(1-x)2))\Big)}{\ln(2(1-x))}=\alpha=1.69315\cdots$$

We have :

$$x^{2(1-x)}\leq2^{2x}(1-x)2x^2$$

And :

$$(1-x)^{2x}\leq 1-2^{2x}(1-x)2x^2$$

For a sketch of proof see Refinements of the inequality $f(x)=x^{2(1-x)}+(1-x)^{2x}\leq 1$ for $0<x<0.5$

So we have :

$$x^{(2(1-x))^k}\leq f(x)^{(2(1-x))^{k-1}}=(2^{2x}(1-x)2x^2)^{(2(1-x))^{k-1}}$$

And :

$$(1-x)^{(2x)^k}\leq (1-2^{2x}(1-x)2x^2)^{(2x)^{k-1}}$$

Now we want $y\geq 1$ a real number :

$$(2x)^{k-1}\geq (2f(x))^{y}$$
And
$$(2(1-x))^{k-1}\geq (2(1-f(x)))^{y}$$

Solving this we have

$$\frac{\ln\Big(2(1-(2^{2x}x^2(1-x)2))\Big)}{\ln(2(1-x))}\leq \frac{k-1}{y}\leq \frac{\ln\Big(2(2^{2x}x^2(1-x)2)\Big)}{\ln(2x)}\quad (1)$$

The constraint $(1)$ is honoured if :$$\alpha=\frac{k-1}{y}$$

So we have :

$$x^{(2(1-x))^k}\leq (f(x))^{(2(1-f(x))^{\frac{k-1}{\alpha}} }\quad (2)$$

And :

$$(1-x)^{(2x)^k}\leq (1-f(x))^{(2f(x))^{\frac{k-1}{\alpha}}}\quad (3)$$

Adding $(2)$ and $(3)$ we have :

$$h(k)\leq (1-f(x))^{(2f(x))^{\frac{k-1}{\alpha}}}+(f(x))^{(2(1-f(x))^{\frac{k-1}{\alpha}}}$$

We can repeat the reasoning $p$ times ($p\geq 1$ a natural number) and the last exponent becomes :

$$1\leq \frac{k-1}{\alpha^p}\leq \alpha$$

It's true because we can always choose $k$ such that :

$$\alpha^p\leq k \leq \alpha^{p+1}$$

So we reduce this case to the case $1\leq k\leq 1+\alpha $

Case $k\leq 1+\alpha $

Using Bernoulli's inequality and derivatives we have $1\leq k\leq 2$ and $0<x\leq \frac{1}{4}$

$$(1-x)^{(2x)^k}\leq 1-x(2x)^k$$
And
$$x^{(2(1-x))^k}\leq x(2x)^k$$

In the same idea we have $2\leq k\leq 1+\alpha$ and $0<x\leq \frac{1}{3}$

$$(1-x)^{(2x)^k}\leq 1-x(2x)^k$$
And
$$x^{(2(1-x))^k}\leq x(2x)^k$$

Now  to finalize the sketch of proof we have :

$h(k)$ is decreasing for $1\leq k\leq 2$ and $ \frac{1}{4}\leq x\leq 0.5\quad (4)$

$h(k)$ is decreasing for $2\leq k\leq 1+\alpha$ and $ \frac{1}{3}\leq x\leq 0.5\quad (5)$

Questions :

It is good and coherent ?

How to show $(4)$ and $(5)$?

Thanks in advance !

Regards Max

[1] Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions",
The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137.
https://eudml.org/doc/223938

https://link.springer.com/article/10.1186/1029-242X-2013-468

答:
I prefer to do this this analytical process.

Look at the function with lower k:

I mathematica it is often sensible to look at the function sequence on the borders of the interval of interest. In this case it is possible to state this values for all k. $f(0)=1$ and $f(0.5)=0$. Inbetween the function can either be bigger or smaller than $1$. The functions are polynoms in x, so monomials, but with mixed order. So they are steady and differentiable on the interval and the boundary. There positive as well as negative. But both terms are positive on the interval for all k $1>0.5>x>0$.

There must be for each k a global unique minimum or maximum. This property decides whether the function are bigger or smaller than $1$ in the interval $(0,0.5)$.

The first derivative looks even as nice as the function:

This few derivative show the beauty and complexity of the derivatives in full.

$$(1 - x)^(2^k x^
   k) (-((2^k x^k)/(1 - x)) + 2^k k x^(-1 + k) Log[1 - x]) +
 x^(2^k (1 - x)^k) ((2^k (1 - x)^k)/x - 2^k k (1 - x)^(-1 + k) Log[x])$$

Because this already has the logarithm function combined with the polynomials it is not solvable in a closed form.

With numerical methods it is easy to find the zeros of the functions sequence. It starts with

$${0.216454, 0.281732, 0.326587, 0.357129, 0.378871, 0.395017, \
0.407435, 0.417264, 0.425226}$$

What we need for a prove is only that there is one for shure for all k.

All these function value at the extrema are smaller than $1$.

$${0.990665, 0.973494, 0.960633, 0.951346, 0.944465, 0.939202, \
0.935062, 0.931727, 0.928985}$$

This is in the start already a continuously falling sequence. Since these are polynomial this will continue for all k. At that is the prove with the methods of numerics and analysis.

$$\lim_{k\rightarrow\infty}f(x_{Extrema,k})=0$$

with

$$\lim_{k\rightarrow\infty}x_{Extrema,k}=0.5$$

Therefore the

$(1−)^{(2)^}+^{(2(1−))^}≤1$ with $≥1$ and $0<≤\frac{1}{2}$ and $k$ natural

ist true.

$\lim_{k\rightarrow\infty}(1−)^{(2)^}+^{(2(1−))^}=1$ if $\log(1 - x) < -\log(2)$ $and$ $\log(x) > -\log(2)]$

So the function sequence converges pointwise towards 1, but remains the extrema which is proven as a global unique minimum but the minimun converges to 0.5. The limit therefore is constant $1$.

These can be interpreted as a sum of two sequences. The first one is the same as the second one if 1-x=t and 0.5<t<1. So the first part increases with k to 1 and the second to 0 with k. Like the functions sequence $x^k$ converges to 0 on $(0,1)$ converges this functions sequence to $1$. $x^k$ converges pointwise and global on $(0,1)$. The function sequence (1−)^{(2)^}+^{(2(1−))^} just converges pointwise on $(0,0.5)$. The global minimum on $(0,0.5)$ converges to $0.5$ and $1$ from below.


问:
Surface integral with unit sphere

Let S be the surface defined by $$z^2+y^2+x^2=1.$$  Compute the
surface integral $$\int_S -z^2-y^2-x^2dS.$$

How can I approach this? I've taken a look at multiple examples, but I don't think I've understood them properly. Thank you in advance!

Title: Surface integral with unit sphere

Body:
On $S$, $x^2+y^2+z^2=1$, thus,
$$\int_S-(x^2+y^2+z^2)=-\int_S 1$$
Where $\int_S 1$ is simply the area of the unit sphere, $8\pi^2$.
$$\int_S -z^2-x^2-y^2=-8\pi^2$$


问:
How can I solve this recurrence?
I have a weird recurrence relation and don't know how to solve it:
$$a_n = pa_{n-1} + qa_{n+1} + cb_n$$
$$b_n = p'b_{n+1} + q'a_n$$
$$a_0 = 1$$
$p,q,c,p',q' \in [0,1]$ and $p+q+c=1,p'+q'=1$.

Thanks for help.

答:
Use "generatingfunctionology" techniques. Define $A(z) = \sum_{n \ge 0} a_n z^n$ and similarly $B(z) = \sum_{n \ge 0} b_n z^n$. Shift indices:
$$
\begin{align*}
a_{n + 1} &= p a_n + q a_{n + 2} + c b_{n + 1} \\
b_n       &= p b_{n + 1} + q a_n
\end{align*}
$$
Multiply by $z^n$, add over $n \ge 0$ to get:
$$
\begin{align*}
\frac{A(z) - a_0}{z^2}
  &= p A(z) + q \cdot \frac{A(z) - a_0 - a_1 z}{z^2}
       + c \cdot \frac{B(z) - b_0}{z} \\
B(z)
  &= p \cdot \frac{B(z) - b_0}{z} + q A(z)
\end{align*}
$$
Given $a_0$, $a_1$, $b_0$ this is a linear system of equations in $A(z)$ and $B(z)$. Solve it, expand in partial fractions and you are set.</p>

Title: How can I solve this recurrence?


问:
Quotient space of $C[0,1]$
Construct the quotient space of $C[0,1]$ with the subspace
$$
M:=\left\{f \in C[0,1]: f\left(\frac{1}{2} \pm \frac{1}{2^n}\right)=0, n\in \Bbb N\right\}.
$$
I really don't know how to solve it, I would appreciate a hint or example to help me understand it.

答:
You have a sequence $\{x_n\}$ and $M=\{f\in C[0,1]:\ f(x_n)=0,\ n\in\mathbb N\}$.  So two functions will be equal in the quotient if they agree on all $x_n$. Thus a class will be defined by its values in the set  $\{x_n\}$. Being bounded, it looks like we can identify the quotient with $\ell^\infty(\mathbb N)$. But there is an added factor, which is that $\{x_n\}$ as given in the question has an accumulation point, $t=1/2$. So the values $f(x_n)$ converge to $f(1/2)$ since $f$ is continuous, and then the candidate for the quotient is $c$, the space of convergent sequences.

Now, let's do it formally. Define $\pi:C[0,1]/M\to c$ by $\pi(f+M)=\{f(x_n)\}_n$.

Well defined: if $f-g\in M$, then $f(x_n)=g(x_n)$ for all $n$. And, as $x_n\to 1/2$ and $f$ is continuous, $f(x_n)\to f(1/2)$, so $\pi(f+M)$ is convergent.

Linearity is obvious, as $\pi$ is an evaluation.

Injective: if $f(x_n)=0$ for all $n$, then $f\in M$

Surjective: given $y\in c$, we can construct $f$ as linear segments joining the points $(x_n,y_n)$. As $x_n\to1/2$, we define $f(1/2)=\lim_nx_n$.

Title: Quotient space of $C[0,1]$


答:
This question actually allows a nice quick solution.
Since our integrand can be written as $-(x^2+y^2+z^2)$, and this equals $-1$ on $S$, we can rewrite the integral as $-\int_{S}dS$ which is minus the surface area of the unit sphere and so our integral equals $-4\pi$.

More generally, we would look to parametrise the surface $S$ in terms of some function $\mathbf{r}(u,v)$, and then substitute $dS=|\mathbf{r}_u \wedge \mathbf{r}_v|dudv$. In this particular case, spherical polars would be a natural choice.

Title: Surface integral with unit sphere

Body:
Let S be the surface defined by $$z^2+y^2+x^2=1.$$  Compute the
surface integral $$\int_S -z^2-y^2-x^2dS.$$

How can I approach this? I've taken a look at multiple examples, but I don't think I've understood them properly. Thank you in advance!


问:
Commutator ideal of $\mathfrak{gl}(V)$ for $V$ infinite-dimensional
It is well-known that the commutator ideal of $\mathfrak{gl}(V)$ is $\mathfrak{sl}(V)$, the Lie algebra of traceless endomorphisms, if $V$ is a finite-dimensional vector space. I am now wondering what is known about $[\mathfrak{gl}(V), \mathfrak{gl}(V)]$ if the vector space $V$ is infinite-dimensional. I am in particular wondering about the the codimension of $[\mathfrak{gl}(V), \mathfrak{gl}(V)]$ in $\mathfrak{gl}(V)$.

答:
Motivated by YCor’s answer I came up with the following argumentation, which shows that every element of $\mathfrak{gl}(V)$ is a commutator.

The vector space $V$ is isomorphic to the direct sum $V^{\oplus \mathbb{N}_{\geq 1}}$.
We may therefore assume that $V = W^{\oplus \mathbb{N}_{\geq 1}}$ for some vector space $W$.
We can thus represent every endomorphism $f$ of $V$ as a matrix
$$
  f
  \equiv
  \begingroup
  \renewcommand{\arraystretch}{1.3}
  \begin{pmatrix}
    f_{11}  & f_{12}  & f_{13}  & f_{14}  & \cdots  \\
    f_{21}  & f_{22}  & f_{23}  & f_{24}  & \cdots  \\
    f_{31}  & f_{32}  & f_{33}  & f_{34}  & \cdots  \\
    f_{41}  & f_{42}  & f_{43}  & f_{44}  & \cdots  \\
    \vdots  & \vdots  & \vdots  & \vdots  & \ddots
  \end{pmatrix} \,.
  \endgroup
$$
Each matrix entry $f_{ij}$ is an endomorphism of $W$, namely the composite $\pi_i \circ f \circ \iota_j$, where $\iota_j$ the inclusion from $W$ into the $j$-th direct summand of $V$, and $\pi_i$ the projection from $V$ onto its $i$-th direct summand $W$.
This matrix is pointwise column-finite, in the sense that for every element $w$ of $W$ and every column index $j$ the images $f_{ij}(w)$ vanish for all but finitely many row indices $i$.
Suppose conversely that we have given a family of endomorphisms $g_{ij}$ of $W$ with $i, j \geq 1$ such that the resulting matrix is pointwise column-finite.
Then this matrix describes an endomorphism $g$ of $V$.

Let $s$ be the shift endomorphism of $V$ given by
$$
  s( (w_1, w_2, w_3, \dotsc) )
  =
  ( w_2, w_3, w_4, \dotsc )
$$
for all $(w_1, w_2, w_3, \dotsc) \in V$.
As a matrix, this endomorphism is given by
$$
  s
  \equiv
  \begin{pmatrix}
    0 & 1 &   &   &         \\
      & 0 & 1 &   &         \\
      &   & 0 & 1 &         \\
      &   &   & 0 & \ddots  \\
      &   &   &   & \ddots
  \end{pmatrix}
$$
We therefore have
\begin{align*}
  [s, f]
  &=
  s f - f s
  \\[1em]
  &\equiv
  \begin{pmatrix}
    f_{21}  & f_{22}  & f_{23}  & f_{24}  & \cdots  \\
    f_{31}  & f_{32}  & f_{33}  & f_{34}  & \cdots  \\
    f_{41}  & f_{42}  & f_{43}  & f_{44}  & \cdots  \\
    f_{51}  & f_{52}  & f_{53}  & f_{54}  & \cdots  \\
    \vdots  & \vdots  & \vdots  & \vdots  & \ddots
  \end{pmatrix}
  -
  \begin{pmatrix}
    0       & f_{11}  & f_{12}  & f_{13}  & \cdots  \\
    0       & f_{21}  & f_{22}  & f_{23}  & \cdots  \\
    0       & f_{31}  & f_{32}  & f_{33}  & \cdots  \\
    0       & f_{41}  & f_{42}  & f_{43}  & \cdots  \\
    \vdots  & \vdots  & \vdots  & \vdots  & \ddots
  \end{pmatrix}
  \\[1em]
  &=
  \begingroup
  \renewcommand{\arraystretch}{1.5}
  \begin{pmatrix}
    f_{21}  & f_{22} - f_{11} & f_{23} - f_{12} & f_{24} - f_{13} & \cdots  \\
    f_{31}  & f_{32} - f_{21} & f_{33} - f_{22} & f_{34} - f_{23} & \cdots  \\
    f_{41}  & f_{42} - f_{31} & f_{43} - f_{32} & f_{44} - f_{33} & \cdots  \\
    f_{51}  & f_{52} - f_{41} & f_{53} - f_{42} & f_{54} - f_{43} & \cdots  \\
    \vdots  & \vdots          & \vdots          & \vdots          & \ddots
  \end{pmatrix} \,.
  \endgroup
\end{align*}
Let now $g$ be an endomorphism of $V$.
We can use the above explicit calculation of the commutator $[s, f]$ to construct an endomorphism $f$ with $[s, f] = g$.

Indeed:
Starting with the first column we choose the entry $f_{11}$ arbitrary and $f_{i1}$ as $g_{i-1,1}$ for every $i \geq 2$.
Once the entries $f_{ij}$ are constucted for all $i \geq 1$ and $j = 1, \dotsc, k$, we choose the entries $f_{1,k+1}$ arbitrary and the entry $f_{i,k+1}$ as $g_{i-1, k+1} + f_{i-1, k}$ for every $i \geq 2$.
The matrix described by these entries $f_{ij}$ is pointwise column-finite because the matrix associated to $g$ is pointwise column-finite.
It follows that the entries $f_{ij}$ describe an endomorphism $f$ of $V$.
By construction, this endomorphism satisfies $[s, f] = g$.

We have thus shown that every endomorphism of $V$ is a commutator (namely a commutator of the form $[s, f]$ for some endomorphism $f$ of $V$).


问:
What is the maximum number of T-shaped polyominos (shown below) that can be put into a 6x6 grid without any overlaps? The blocks can be rotated.

I just drew the figure and manually tried the question but I am wondering is there a way to do this problem via permutations and combinations.

PS: I got answer as 7.

答:
As noted above it's possible to tile the 6x6 square with 8 T-tetrominoes, with 4 squares left over. FWIW, there's a simple tiling of the 4 x 4 square, so there are trivial tilings of any 4n x 4m rectangle.

Here's a symmetrical 6×6 solution I found by hand:

(JPEG version)

I hacked my code from my answer to Tiling of a $9\times 7$ rectangle to solve this new problem. Due to the nature of the hack, I'm not totally confident that I've found all the solutions, but I managed to find 184, which include rotations and reflections. So the true answer is something higher than 184 / 8 = 23.

Anyway, here's a slow-motion GIF anim of those results. I suppose it would look nicer if the 4 empty squares were all the same colour, and a different colour to the tetrominoes, but I didn't feel up to hacking that part of the program too.


问:
Prove that the language of only constant symbols is complete
Take the language made up of only constant symbols and no function or relation symbols $L = \{c_1\ne c_2, c_2 \neq c_3, c_1 \neq c_3, ...  \}$. I want to show that $L$ is a complete theory.

The way I think this should be done is to first show that $L$ has countably many non-isomorphic countable models and then secondly show that there is a countable model of $L$ into which all countable models can be elementary embedded. And then lastly we can conclude that $L$ is complete. Can somebody just write down how the first part is done?

答:
This should really be a comment, but it's too long:

What you describe is not the easiest way to answer the question. Here are a couple approaches which will go quicker:

Isomorphisms of countable reducts. Suppose $\mathcal{M,N}$ are countable models of $L$ and $\varphi$ is a sentence in the language of $L$; we'll show that $\mathcal{M}\models\varphi\iff\mathcal{N}\models\varphi$. Since $\varphi$ is a single sentence, it only involves finitely many constant symbols $c_1,...,c_n$. Let $\mathcal{M}',\mathcal{N}'$ be the reducts of $\mathcal{M},\mathcal{N}$ respectively to the language consisting of just these finitely many constant symbols. We have $\mathcal{M}'\models\varphi\iff\mathcal{M}\models\varphi$ and similarly for $\mathcal{N}'$ and $\mathcal{N}$. With a bit of thought, you can show that $\mathcal{M}'\cong\mathcal{N}'$ and this finishes the problem.

This works for models of $L$ of the same cardinality, not just countable ones, but it might be helpful to restrict attention to countable models when they suffice.

Uncountable categoricity. Rather than look at the countable models, think about the models of size (say) $\aleph_1$. It's not hard to show that there is exactly one model up to isomorphism of cardinality $\aleph_1$, and by the Lowenheim-Skolem theorem this means that $L$ is complete.

Note that both approaches above require understanding what models of $L$, or $L$'s finite fragments, look like. So broadly speaking this is still running into the concern you have in the OP. But I think each of the approaches above, being meaningfully simpler, will help demystify things.

On that note, here are a couple hints to get started:

Suppose $\mathcal{M}\models L$. Then $\mathcal{M}$ has a particular distinguished countable subset, namely the set $\{c_i^\mathcal{M}: i\in\mathbb{N}\}$ of elements named by constant symbols. Can you show that this does not have to be all of $\mathcal{M}$?

After doing so, you have that any model of $\mathcal{M}$ splits into two pieces, the piece above and everything else; what properties of these pieces determine $\mathcal{M}$ up to isomorphism?


问:
Erdös theorem which gives a weaker lower bound for R(k,k)
I  have proved the theorem of Erdös that gives a lower bound for $R(k,k)$ and it is exponential in k. I have used probabilistic method to find this lower bound: $R(k,k) > (1+o(1))\frac {k}{(e√2)} √(2^k )$.

My question is: How can we construct a family of graphs that show this : $R(k,k) ≥ (k-1)^2$

I know that this is a weaker lower bound, but how to prove it?

I checked this link: Lower bound for the Ramsey number $r(k,k)$ but still I am missing some details to understand the hint that is given in this link.

If someone could help me, I would appreciate it :)

答:
Suppose that $G$ is the union of $k-1$ disjoint $(k-1)$-cliques, so that $\overline{G}$ is the complete $(k-1)$-partite graph $K_{k-1,k-1,\dots,k-1}$.

Another way to say this is that $V(G) = V(\overline{G})$ is partitioned into $k-1$ sets $S_1, S_2, \dots, S_{k-1}$ of size $k-1$. Two vertices are adjacent in $G$ if they're in the same set, and they're adjacent in $\overline{G}$ if they're in different sets.

Now suppose that we pick any $k$ distinct vertices.

They cannot all be in the same set, because each set only has size $k-1$, and we are picking $k>k-1$ vertices. So at least two of them are in different sets - and those two vertices are not adjacent in $G$. Therefore $G$ does not have a clique of size $k$.

By the pigeonhole principle, since we have $k$ vertices and each of them is in one of the $k-1$ sets $S_1, S_2, \dots, S_{k-1}$, two vertices must be in the same set. Those two vertices are not adjacent in $\overline{G}$. Therefore $\overline{G}$ does not have a clique of size $k$.


问:
Let $X$ be a normed space and $M$ a finite dimension subspace, $M\neq X$. Prove that $\exists x\in X$ such that $||x||=d(x,M)=1$. And prove...
So, i'm facing this problem

Let $X$ be a normed space and $M$ a finite dimension subspace of $X$, $M\neq X$.

$a)$ Prove that $\exists x\in X$ such that $\left\lVert x\right\rVert=d(x,M)=1$.

$b)$ Prove that if $dim(X)=\infty$ then $\exists \{x_n\}$ in the unit sphere of $X$ such that $\left\lVert x_n-x_m\right\rVert\geq 1$, $\forall n,m\in \mathbb{N}$, $n\neq m$

What i've tried so far: Let $x \in X$ with $\left\lVert x \right\rVert =1$, since $M$ has finite dimension, then $\exists y \in M \ : \ d(x,M)=d(x,y)= \left\lVert y-x \right\rVert$. And is easy to see that $d(y,y-\frac{y-x}{\left\lVert y-x \right\rVert})=1$ and $$\left\lVert y-\frac{y-x}{\left\lVert y-x \right\rVert} \right\rVert =\frac{1}{\left\lVert y-x \right\rVert}\left\lVert y\left\lVert y-x \right\rVert -y+x\right\rVert \geq  \frac{d(x,M)}{\left\lVert y-x \right\rVert}=\frac{\left\lVert y-x \right\rVert}{\left\lVert y-x \right\rVert}=1$$
where i used that $y\left\lVert y-x \right\rVert -y\in M$. So i found an element of $X$, $z:=y-\frac{y-x}{\left\lVert y-x \right\rVert}$, $d(z,M)=1$ and $\left\lVert z \right\rVert\geq 1$. But i think i can't prove that $\left\lVert z \right\rVert=1$ this way

Any help would be much appreciated.

答:
For $(a)$ take $z=\frac{1}{d}\cdot (x-y)$ where $d=d(x,M)$. Then, $||z||=1$. And, for $y'\in M$ we have
\begin{align}
||z-y'||&=\bigl|\bigl|\frac{1}{d}(x-y)-y'\bigl|\bigl|\\
&=\frac{1}{d}\bigl|\bigl|x-y-dy'\bigr|\bigr|\\
&\geq \frac{d}{d}=1
\end{align}

Hence, $d(z,M)\geq 1$. For the other inequality,
$$d(z,M)\leq ||z-0||=||z||=1$$

Therefore, $d(z,M)=1=||z||$.  Now, for $(b)$, pick $x_1$ on the unit sphere. And let $M_1=span(x_1)$. Since, $X$ is infinite dimensional, $M_1\neq X$. Hence, there is some $x\notin M_1$. Since, $M_1$ is a finite dimensional subspace by $(a)$ there is some $x_2$ on the unit sphere such that $d(x_2,M)=1=||x_2||$. Now, let $M_2=span(x_1,x_2)$ and repeat the same argument as before. Continuing this way, recursively, we construct a sequence $x_n$ such that $||x_n-x_m||\geq 1$ for every $n \neq m$.


问:
Bundle Automorphisms, Structure Groups and Gauge Groups
I am trying to get my head around the mathematical foundations of gauge theory and wanted to check that I am correct in thinking the following is true.

If $E$ is a $G$-principle bundle over $M$ then we call $G$ the structure group of $E$ and it is the group of transition functions at any point $x \in M$.

The group of bundle automorphisms is precisely $G$ at any point $x \in M$.

We call the group of bundle automorphisms the group of (global) gauge transformations.

We call $G$ the gauge group.

We call the group of transition functions on some neighbourhood $U$, the group of (local) gauge transformations. It is the group of bundle automorphisms over $U$.

I feel that I may be mixing up global and local notions here a bit, particularly with regard to automorphisms and transition functions. Is there a better way to phrase the interaction between the group of transition functions and the automorphism group?

I'd appreciate mathematical precision, since the hand-wavy arguments in physics books are precisely what confuses me on these points! Many thanks in advance!

答: This is not really an answer. It seems that I can't add a comment (low reputation). Just wanted to correct a small (but crucial) inaccuracy in Henry T. Horton's fine answer: in general $\mathscr{G}$ can be identified with $C^\infty(M,G)$ only when $G$ is Abelian or the principal bundle $E$ is trivializable. See e.g. Husemoller, Fibre bundles, Proposition 1.7 p. 81.</p>


问:
Showing that in a given set of 7 distinct integers that there must be at least one pair in the set whose sum or difference is a multiple of 10.
Show that, given any set of seven distinct integers, there must be at least one pair in the set whose sum
or difference is a multiple of 10.

I've been thinking a lot on how to show that this is true, but I've not managed to come up with anything yet. If anyone could offer their viewpoint I'd appreciate it.

答:
Modulo $10$, numbers are $0, (1,-1), (2,-2), (3,-3), (4,-4), 5$.

If six numbers are all distinct modulo $10$

seventh is either identical to one of previous in which case difference is divisible by $10$.

seventh is negative to one of previous, then sum is divisible by $10$.


问:
What does the following logical expression $\phi_{M,S,k}$ mean?
I recently started my logic course. In my logic textbook I have the following logical expression and the exercise is to determine what it means:

Let $M$ be a finite set, $S \subseteq2^M$ and $k \in \mathbb{N}$

$$\phi_{M,S,k} := \bigwedge_{M'\subseteq M, |M'|=k} \left(\bigwedge_{i \in M'} X_i \land \bigwedge_{j \in M \backslash M'} \lnot X_j \land \bigwedge_{X \in S} \bigvee_{l \in X}X_l \right)$$

答: If the outer $\bigwedge$ were instead $\bigvee$, the expression would be asserting the existence of a $k$-subset that covers $S$.  If you interpret $X_i$ as a binary variable that indicates whether element $i\in M$ is selected, this proposition would be that the set covering problem of minimizing $\sum_{i\in M} X_i$ subject to $\sum_{i\in A} X_i \ge 1$ for all $A \in S$ has upper bound $k$.


问:
Differentiating a boundary condition at infinity
A typical boundary condition for an initial boundary value problem is
$$ \lim_{x\rightarrow\infty} T(x,t) = T_\infty.$$

For example, this might be the temperature at the end of a very long rod.  Under what conditions is it equivalent to instead enforce the boundary condition
$$ \lim_{x\rightarrow\infty} \frac{\partial T}{\partial x} = 0$$

or, perhaps worded differently, is it ever the case, given $T\rightarrow T_\infty$ as $x\rightarrow\infty$, that $\partial T/\partial x\ne0$ as $x\rightarrow\infty$?  The alternative boundary condition makes sense physically, but does there exist a formal mathematical treatment of this?

答:
As gerw wrote, in general neither condition implies the other. However, in the context of PDE you may have extra information about the function which allows to pass from one condition to the other.

Fact. If $T\to 0$ and $T''$ is bounded, then $T'\to 0$.

I remember the following proof from  Littlewood's Miscellany. Draw the graph of $T'$ (in red):

If $T'$ is not small somewhere, then one can fit a triangle (black) under the graph, so that the height of the triangle is not small, and the slope of its legs is not large (bounded by $|T''|$). Therefore, the area of this triangle is not small, contradicting the assumption $T\to 0$. $\Box$

This fact could be applicable to solutions of second-order equations, which give a way to control $T''$.


问:
Infinite series of metric spaces is metric space
Suppose $(X,d) $ is a complete metric space with $U_1,U_2,...$ nonempty open subsets, with none equal to $X.$ Let $U=
    \bigcap_{n=1}^{\infty } U_n \neq \emptyset$ and define $d_n $ on
$U_n $ as $$d_{n}(x,y) =\text{min} (D_{n} (x,y),1)$$ where $$D_{n}
=d(x,y)+\lvert \frac{1}{d(x,U_n^c) } - \frac{1}{d(y,U_n^c)}
\rvert. $$ Define $$D(x,y)=\sum_{n=1}^{\infty } \frac{1}{2^n} d_{j}
(x,y). $$

I have a question about showing that it's a metric under the
condition that $D(x,y)=0 \iff x=y.$ I'm trying to prove it in the
$\implies $ direction. Since $\frac{1}{2^n} $ is a geometric
series and $d_n(x,y)$  is less than or equal to 1, we know that the series converges.
This implies that at some point, $\frac{1}{2^{n} } d_n(x,y)$ goes
to 0, but this is only possible if $d_n(x,y)=0,$ which means that
$x=y.$ Is this correct?

I also want to show that if $\{ x_{n}\}  $ is a Cauchy sequence in $(U,D),$ then it is a Cauchy sequence in each $(U_n,D_n).$ My
attempt is as follows. Since $\{ x_{n}\}  $ is Cauchy in $U,$ we
have for any $\epsilon >0,\exists N,\forall n,m \geq N,$
$$D(x_n,x_{m} )=\sum_{n=1}^{\infty } \frac{d_n(x_n,x_m)}{2^n}< \epsilon.  $$
We can rewrite this as
$$\sum_{k=1}^{n-1 }
\frac{d_k(x_n,x_m)}{2^k}+\frac{d_n(x_n,x_m)}{2^n}+\sum_{k=n+1}^{\infty
} \frac{d_k(x_n,x_m)}{2^k}< \epsilon $$
$$\implies d_n(x_n,x_m)<(\epsilon-\sum_{k=1}^{n-1 }
\frac{d_k(x_n,x_m)}{2^k} -\sum_{k=n+1}^{\infty
} \frac{d_k(x_n,x_m)}{2^k})2^n$$
Set $$\epsilon = \frac{\epsilon}{2^n}+ \sum_{k=1}^{n-1 }
\frac{d_k(x_n,x_m)}{2^k} +\sum_{k=n+1}^{\infty
} \frac{d_k(x_n,x_m)}{2^k}>0. $$
After substituting and canceling terms we get $d_n(x_n,x_{m} )<
\epsilon .$ Since $\epsilon $ is arbitrary, $D_n(x_n,x_m)<\epsilon
.$ Thus $\{ x_{n}\}  $ is Cauchy in each $(U_n,D_n).$ Is
this argument correct?

I want to show that $(U,D)$ is complete. Suppose $\{ x_{n}\}  $ is a
Cauchy sequence in $(U,D)$ that converges to $L.$
If $L \in U^c,$ then
$$ \forall \epsilon >0 \lim_{n \to \infty} d(x_n,U^c) \leq \lim_{n
  \to \infty} d(x_n,L) < \epsilon.$$ By the contrapositive, we have if
$$ \exists \epsilon >0 \lim_{n \to \infty} d(x_n,U^c) > \lim_{n \to
  \infty} d(x_n,L) \geq \epsilon ,$$
then $L \in U.$
So we want to show that there is an $\epsilon >0$ such
that
$$\lim_{n \to \infty} \frac{1}{d(x_n,U^c)} \leq \frac{1}{\epsilon
} .$$
Since
$$\frac{1}{d(x_n,U^c)} \to \frac{1}{d(L,U^c)},$$
we have that $\frac{1}{d(x_n,U^c)}$ is bounded. I'm not sure how to
incorporate any other details about $D(x,y)$ other than the $d(x_n,U^c)$ term.

答:
Your reasoning in $(1)$ isn’t at all clear. I would simply prove the contrapositive. If $x,y\in U$ and $x\ne y$, then $d(x,y)>0$, so $D_n(x,y)\ge d(x,y)>0$, and therefore $d_n(x,y)>0$. Clearly this implies that $D(x,y)>0$.

In $(2)$ you can’t set

$$\epsilon=\frac{\epsilon}{2^n}+\sum_{k=1}^{n-1}\frac{d_k(x_n,x_m)}{2^k}+\sum_{k=n+1}^\infty\frac{d_k(x_n,x_m)}{2^k}\,.$$

First, it makes no sense: it’s like saying ‘let $x=x+1$’. You need to use a different name for the new value. Next, you’re trying to show that $d_k(x_n,x_m)$ is less than the original $\epsilon$, not some new $\epsilon$. Finally, this sentence makes no sense:

Since $\epsilon$ is arbitrary, $D_n(x_n,x_m)<\epsilon$.

Unless $D_n(x_n,x_m)=0$, it is not less than $\epsilon$ for arbitrary $\epsilon$. And you probably mean $D_k$, not $D_n$. You appear to have made a similar error in the displayed line

$$D(x_n,x_{m} )=\sum_{n=1}^{\infty } \frac{d_n(x_n,x_m)}{2^n}< \epsilon\,:$$

the index of summation has to be different from the subscripts on $x$.

What you need to do here is let $\epsilon>0$ and $k\in\Bbb Z^+$. Since $\langle x_n:n\in\Bbb Z^+\rangle$ is $D$-Cauchy, there is an $n_0\in\Bbb Z^+$ such that $D(x_n,x_m)<\frac{\epsilon}{2^k}$ whenever $n,m\ge n_0$. But then for all $n,m\ge n_0$ we have

$$\frac{d_k(x_n,x_m)}{2^k}\le D(x_n,x_m)<\frac{\epsilon}{2^k}$$

and therefore $d_k(x_n,x_m)<\epsilon$. Thus, $\langle x_n:n\in\Bbb Z^+\rangle$ is $d_k$-Cauchy and therefore $D_k$-Cauchy. (You should have no trouble proving that a sequence is $d_k$-Cauchy if and only if it is $D_k$-Cauchy.)

The that you start $(3)$ makes no sense. To show that $\langle U,D\rangle$ is complete, you need to start with an arbitrary $D$-Cauchy sequence in $U$ and show that it converges; you cannot simply assume that it converges.

HINT: The sequence $\langle x_n:n\in\Bbb Z^+\rangle$ is $d_k$-Cauchy for each $k\in\Bbb Z^+$, and each $\langle U_k,d_k\rangle$ is complete, so for each $k\in\Bbb Z^+$ there is a $y_k\in U_k$ such that $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $y_k$ in $\langle U_k,d_k\rangle$. Show that there is a $y\in U$ such that $y_k=y$ for each $k\in\Bbb Z^+$, and show that $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $y$ in the metric $D$.


问:
The rate of change in Newton's Law of Cooling
I know how to solve the following problem but I have a confusion concerning the sentence in bold.

A metal bar is taken from an inside room and dropped into a large container of boiling water.The initial temperature of the bar was 20° C. If it is known that the temperature of the the bar increases 2°C in 1 second, How long will it take the bar to reach 98° C?

The solution of the problem is as following

The differential equation is

$\frac{dT}{dt}=k(T_m-T)$

Its solution is

$T=Ce^{-kt}+T_m$

Substitute by $T_m=100$ and by the initial condition $T(0)=20$ in the solution of the DE

$T(0)=100+C$

Hence $C=-80$

Hence $T=100-80e^{-kt}$

To get the constant k, we should substitute by the condition $T(1)=22$ in the solution
$22=100-80e^{-k}$

Hence $k=0.02532$

Then, we can substitute by $T=98$ in the solution to get the required time.

My question is:

Can we deal with the information "the temperature of the the bar increases 2° in 1 second" that this is the rate of change at $T=0$? I mean:

$\frac{dT}{dt}\big|_{t=0}=2$

Hence, we can substitute in the DE

$\frac{dT}{dt}\big|_{t=0}=k(100-T(0))$

$2=k(100-20)$

Hence

$k=0.025$

We got finally approximately the same value for the constant k.

Note:

I think it is not allowed to deal with the given info as a rate because we divided the increase in temperature by the duration which is not exactly the derivative or the rate; we should divide by a duration that is infinitely small.
Also, I think that if the given info is the rate, it should be written in the form  2°c/sec.

I felt confused because when I deal with the given info as a rate, it gives me approximately the same solution.

答:
Your first solution is correct.  In the second, you are treating the mixing up instantaneous rate of change and average rate of change.  The information given, that the temperature of the the bar increases 2°C in 1 second, is the average rate of change, $$\frac{T(1) - T(0)}{1} = T(1) - T(0) = 2,$$ not the instantaneous rate of change, $\frac{dT}{dt}(0)$.  Remember that as the temperature changes, so does the rate at which it changes.

That said, remember that we can use the derivative at a point to give us a linear approximation of our function at a point.  For a function $f$, you may have seen this written as $$f(x) \approx f(a) + f'(a)(x - a).$$  For your function $T$, then, we could write $$T(1) \approx T(0) + T'(0)(1 - 0)$$ or after rearranging, $$T(1) - T(0) \approx T'(0)$$ which gives you $$2 \approx \frac{dT}{dt}(0).$$  So, essentially what you did in the second solution, is use an approximation for $\frac{dT}{dt}(0)$, which gave you an approximation of the solution.


问:
Convergence of (-1)^n * sqrt(n)/n series
What I am having problems with is the following series.

$$\sum_{n=1}^\infty (-1)^n\frac{\sqrt{n}}{n+907}$$

So far I have tried Abel's test and while It is apparent that the series $$(-1)^n \frac{1}{n+907}$$ converges, I couldn't prove that $\sqrt{n}$ is bounded above.

The next thing I tried was Leibniz test for alternating series and while I could prove that
$$\lim_{n \to\infty}\frac{1}{n+907}=0$$ I couldn't prove that $\frac{1}{n+907}$ is monotonically decreasing.

Up to now I have run out of ideas about what to try. Note that I cannot test for absolute convergence, I must do it by another method.

答:
The function $\frac{\sqrt x}{x+907}$ is monotonically decreasing – the numerator of derivative (the denominator is always positive, so we can ignore it) is
$$(x+907)/(2\sqrt x)-\sqrt x=\frac{907}{2\sqrt x}-\frac12\sqrt x$$
and since $\sqrt x$ grows faster than $1/\sqrt x$, this derivative is negative at sufficiently large $x$. Thus, by the alternating series test, the series converges.


问:
Which proportions of areas can be formed by circles in the plane?
Consider two circles in the plane, one red and one blue with a purple intersection.

$\hskip2.5in$

The proportion of the area in the plane that is occupied by each color can be seen as a point on the 3-simplex if we say that each triple is (Red Area, Blue Area, Purple Area). The illustration above would correspond to some point close to (0.14, 0.84, 0.02). All points on the 3-simplex can be realized by circles in this way. If we think of this as a function, $f_2 : X^2 \rightarrow \triangle^3$ (where $X$ is some parameterization of circles), then we would say that it's surjective.

If we now add a green circle to the plane we would be able to form 7 different colors.

$\hskip2.5in$

And now we get a new function, $f_3 : X^3 \rightarrow \triangle^7$. The difference is that this function isn't surjective. For example: consider a configuration where we only see yellow, purple and cyan, all having an area of $1/3$ each.

If we let $F_3$ be the fraction of points on the simplex that are in the image of $f_3$ then we can ask ourselves: what is the value of $F_3$?

This can of course be generalized to more circles with $f_n : X^n \rightarrow \triangle^{2^n-1}$ and $F_n$ defined similarly.

In that case we can easily prove that $F_n \leq F_{n-1}$ by considering the restrictions on all circles but one and then adding the last one. We can also prove the stronger statement that
$$
    F_n \leq \min(\{F_p^{\lfloor n/p \rfloor} \mid 1 < p < n\})
$$
This can be proved by splitting the circles into groups of size $p$ and using the fact that if one of the groups can't be formed then the whole configuration can't be formed. A corollary of this is that
$$
(\exists n.  F_n \neq 1) \implies \left(\lim_{n \rightarrow \infty} F_n=0\right)
$$
But this argument doesn't construct the value of any $F_n$.
So my questions are

What is the value of $F_3$?

What about for higher $n$?

If we can't find exact values, are there at least better bounds?

答:
Observe that $X$ has real dimension $3$, since a circle in the plane is described by the coordinates of its center and its radius. So $X^n$ should be thought of as a $3n$-dimensional space - indeed, we can parameterize it by $3n$ real numbers in a natural way as described above, and when we do so the areas of each of the $2^n-1$ possible regions will be given by a function of these $3n$ parameters which is given by a piecewise composition of "nice" functions (e.g. "if these three circles are in the following type of configuration and all other circles are sufficiently far away, the area is given by the lune formed by $A$ and $B$ minus the intersection with $C$" and so on). In particular, the areas of these regions are not given by any sort of pathological space-filling curve functions of the $3n$ parameters.

This means that the image of $X^n$ under $f_n$ in $\Delta^{2^n-1}$ has dimension at most $3n$, while the latter space has dimension $2^n-2$.

In fact, for all circle arrangements $x\in X^n$, $f_n(x)$ is invariant under any combination of translation, rotation, and scaling of $x$, so we can effectively quotient out by a $4$-dimensional space of symmetries preserving $f_n$. This brings our dimension down to $3n-4$.

But $3n-4< 2^n-2$ for all $n>2$, so in fact we can conclude that $F_n=0$ for every $n>2$.


答:
$\sqrt n$ is evidently not bounded, therefore trying to prove that it is bounded is a waste of time.

$\frac1{n+907}$ is monotonically decreasing in quite an easy way, however that fact is completely irrelevant for the purpose of using Leibniz criterion: that criterion states quite clearly that $\sum_{k=0}^\infty (-1)^k a_k$ converges if $a_k$ decreases to $0$.

However, what is true is that $\frac{\sqrt n}{n+907}$ is positive, it converges to $0$ and it is eventually decreasing, meaning that there is some $M$ such that, for all $n>m>M$, $\frac{\sqrt m}{m+907}>\frac{\sqrt n}{n+907}$. Since you have this, you can write the series as $$\sum_{n=M+1}^\infty (-1)^n\frac{\sqrt n}{n+907}=\sum_{k=1}^M(-1)^k\frac{\sqrt k}{k+907}+\sum_{n=M+1}^\infty (-1)^n\frac{\sqrt n}{n+907}$$

and prove convergence by applying Leibniz to the second sum.

Personally, the way I like to prove that $\frac{\sqrt{n}}{n+907}$ is eventually decreasing is by evaluating the sign of the following derivative $$\frac{d}{dx}\left[\frac{\sqrt x}{x+907}\right]=\frac{-\frac12\sqrt x+\frac{907}{\sqrt x}}{(x+907)^2}$$

which is eventually $<0$.

Title: Convergence of (-1)^n * sqrt(n)/n series

Body:
What I am having problems with is the following series.

$$\sum_{n=1}^\infty (-1)^n\frac{\sqrt{n}}{n+907}$$

So far I have tried Abel's test and while It is apparent that the series $$(-1)^n \frac{1}{n+907}$$ converges, I couldn't prove that $\sqrt{n}$ is bounded above.

The next thing I tried was Leibniz test for alternating series and while I could prove that
$$\lim_{n \to\infty}\frac{1}{n+907}=0$$ I couldn't prove that $\frac{1}{n+907}$ is monotonically decreasing.

Up to now I have run out of ideas about what to try. Note that I cannot test for absolute convergence, I must do it by another method.


问:
How to prove that $\sum\pi_i = \sum\frac{1}{E_iT_i} = 1$ in an irreducible Markov chain with stationary distribution $\pi$?
In Durrett's book Chapter 5 Theorem 4.6, It said that if p is irreducible and has stationary distribution, then $\pi_i=\frac{1}{E_iT_i}$. Where $T_i$ is the first time the markov process returns back to state i. $E_iT_i$ is the expected number of times needed to return back to state i given the initial state is $i$.

Also in the proof of theorem 4.7 said, if p is ireducible and i is positive recurrent, then, $\pi_j=\frac{\sum_{n=0}^{\infty}P_i(X_n=j, T_i \gt n)}{E_iT_i}$ is a stationary distribution.

Does that mean $\frac{\sum_{n=0}^{\infty}P_i(X_n=j, T_i \gt n)}{E_iT_i} = \frac{1}{E_jT_j}$? How can we prove that?

And also how can we prove that $\sum\frac{1}{E_jT_j} = 1$ in an irreducible Markov chain with stationary distribution $\pi$?

答:
What text are you referring to, exactly? Probability: Theory and Examples Fifth Edition by Rick Durrett? The book proves these theorems in detail, so what part of the proof are you having trouble with?

The theorem you are asking about is numbered 5.5.11, not 4.7 - perhaps you have an old edition of the text? The claim that if $p$ is irreducible and $i$ positive recurrent, then
$$
\pi_j = \frac{\sum_{n=0}^\infty \mathbb P(X_n=j, T_i > n\mid X_0=i)}{\mathbb E[T_i\mid X_0=i]}
$$
defines a stationary distribution is stated as part of theorem 5.5.12.

As for the statement in the title of the question, if $\pi$ is a stationary distribution then by definition it must sum to one. The other equality follows from Theorem 5.5.7 in the book, which states:

Let $i$ be a recurrent state, and let $T=\inf\{n\geqslant 1: X_n=i\}$. Then
$$
 \mu_i(j) = \mathbb E\left[\sum_{n=0}^{T-1} \mathsf 1_{\{X_n=j\}}\right] = \sum_{n=0}^\infty \mathbb P(X_n=j,T>n\mid X_0=i)
 $$
defines a stationary measure.

In the text, the author uses Fubini's theorem to show the above. The equality $\pi_i = \frac1{\mathbb E[T_i\mid X_0=i]}$ follows from summing $\mu_i$ over all states and using Fubini's theorem to show that the sum is in fact equal to $\mathbb E[T_i\mid X_0=i]$. Finally, the result that an irreducible and positive recurrent Markov chain has a unique stationary measure up to constant multiples is used. This is Theorem 5.5.9 in the text, and the proof requires a bit of finesse. From this it follows that $\pi_i = \frac{\mu_i(i)}{\mathbb E[T_i\mid X_0=i]}$, and since by definition $\mu_i(i)=1$ , we have $\pi_i = \frac{1}{\mathbb E[T_i\mid X_0=i]}$. Since this equality is true for each state, it remains true when summing over all states, and hence
$$
\sum_i \pi_i = \sum_i \frac{1}{\mathbb E[T_i\mid X_0=i]},
$$
as desired.


问:
ABCD is a cyclic quadrilateral whose two diagonals are perpendicular.
ABCD is a cyclic quadrilateral whose two diagonals are perpendicular. If $R$ is the radius of the circumcircle, prove that:

$AB$$2$ + $BC$$2$ + $CD$$2$ + $DA$$2$ = $8R$$2$

If the centre of the circle is O, I tried drawing radii perpendicular to each of the sides. That way, we get 3 pairs of congruent right triangles.

答:
Here's my solution.

Use this construction

to prove that $AB = 2R\cos(CAD)$.

By the law of sines we have $CD = 2R\sin(CAD)$, so $AB^2 + CD^2 = 4R^2$.


问:
Given $T_1 = 0$ and for $i > 1$: $T_i = 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j$ Prove via induction that $T_i = \sum\limits_{j=1}^{i-1} \frac{1}{j}$

Given $T_1 = 0$ and for $i \in \mathbb{N}, i > 1$:
\begin{align*}
  T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j \\
\end{align*}
Manually computing terms $2,3,4$:
\begin{align*}
  T_2 &= 1 \\
  T_3 &= 1 + \frac{1}{2} \\
  T_4 &= 1 + \frac{1}{3} \left(1 + 1 + \frac{1}{2} \right) = 1 + \frac{1}{2} + \frac{1}{3} \\
\end{align*}

Prove by induction that this is:

\begin{align*}
  T_i &= \sum\limits_{j=1}^{i-1} \frac{1}{j} \\
\end{align*}

The given equation is clearly satisfied for terms $1,2,3,4$. We need to show the inductive step. Assuming it's true for all $j < i$, show that it is also true for $i$.

\begin{align*}
  T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} \sum\limits_{k=1}^{j-1} \frac{1}{k} \\
  T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-2} \frac{i-1-j}{j} \\
\end{align*}

Body:
$$\begin{align*}
T_{i+1}&=1+\frac1i\sum_{j=1}^iT_j\\
&=1+\frac1i\sum_{j=1}^i\sum_{k=1}^{j-1}\frac1k\\
&=1+\frac1i\sum_{k=1}^{i-1}\sum_{j=k+1}^i\frac1k\\
&=1+\frac1i\sum_{k=1}^{i-1}\frac1k\sum_{j=k+1}^i1\\
&=1+\sum_{k=1}^{i-1}\frac{i-k}{ik}\\
&=1+\sum_{k=1}^{i-1}\left(\frac1k-\frac1i\right)\\
&=1+\sum_{k=1}^{i-1}\frac1k-\frac{i-1}i\\
&=1+\sum_{k=1}^{i-1}\frac1k-1+\frac1i\\
&=\sum_{k=1}^i\frac1k
\end{align*}$$


问:
Limit of a function f and limit of 1/f
I have a question I've been pondering for a while,  but did not succeed at proving it. It is as follows:
if a function $f(x)$ satisfies that $\lim_{x\to a}f(x) = L$, then is it true that $\lim_{x\to a}\frac{1}{f(x)} = \frac{1}{L}$, with $L\neq 0$?

I've tried to prove it with the $epsilon-delta$ definition but I couldn't reach any conclusion. Can somebody help me please.

答:
Here is an $\varepsilon-\delta$ proof.

Let $\varepsilon > 0$. Apply the definition of $\lim_{x \rightarrow a} f(x)=L$ with $$\varepsilon' = \min \left\lbrace \frac{|L|}{2}, \varepsilon \frac{L^2}{2} \right\rbrace$$

which is $>0$ because $L \neq 0$. You get that there exists $\alpha > 0$ such that for every $x \in [a-\alpha, a+\alpha]$, one has $|f(x)-L|\leq \varepsilon'$, which means that for every $x \in [a-\alpha, a+\alpha]$, one has
$$|f(x)-L|\leq \frac{|L|}{2} \quad \text{and} \quad |f(x)-L|\leq \varepsilon \frac{L^2}{2} \quad \quad  (*)$$

In particular, the first inequality gives that for every $x \in [a-\alpha, a+\alpha]$, $$|f(x)| \geq \frac{|L|}{2} \quad \quad (**)$$
which implies that $f$ does not vanish on $[a-\alpha, a+\alpha]$. You can then write that for every $x \in [a-\alpha, a+\alpha]$, $$\left|\frac{1}{f(x)}-\frac{1}{L} \right| = \frac{|f(x)-L|}{|f(x)|\times |L|}$$ so using both inequalities $(*)$ and $(**)$, you have $$\left|\frac{1}{f(x)}-\frac{1}{L} \right|\leq \frac{\varepsilon \frac{L^2}{2} }{\frac{|L|}{2} \times |L|} = \varepsilon$$

and you are done.

Title: Limit of a function f and limit of 1/f


问:
Definition of algebra generated by an operator.
Suppose T is an operator (defined on a Hilbert space, say).  Does the "algebra generated by $T$" have a commonly understood meaning?  I couldn't find a satisfactory answer after some internet searching. It seems to me that there are two ways one could define the algebra $A$ generated by $T$ (without including additional structure on $A$, like closure), namely $A=\{p(T): p$ is a polynomial}, or $A=\{p(T): p$ is a polynomial with $p(0)=0$}. Both definitions contain only the operators which are linear combinations of powers of $T$. In the first definition, we allow for a power of $0$, so that the identity operator belongs to the algebra. Is this the standard definition, or is it something else?

答:
The algebra generated by $T$ is usually defined to be the intersection of all algebras containing $T$. If you work out this definition, you will find that the result is one of the algebras you mention, depending on whether you require algebras to have a unit.

More genreally, if $A$ is an algebra and $S \subseteq A$ is a subset, then the subalgebra generated by $S$ is the intersection of all subalgebras $B \subseteq A$ containing $S$. Again, you have to specify whether or not you require all algebras to be unital. (On this point, different authors use different conventions.)


问:
Should $\mathbb{N}$ contain $0$?
This is a classical question, that has led to many a heated argument:

Should the symbol $\mathbb{N}$ stand for $0,1,2,3,\dots$ or $1,2,3,\dots$?

It is immediately obvious that the question is not quite well posed. This convention, as many others, are not carved in stone, and there is nothing to prevent mathematician $A$ define $\mathbb{N}$ to be the positive integers including $0$, and mathematician $B$ to define $\mathbb{N}$ to be the positive integers excluding $0$. It does not seem that one definition is accepted widely enough for it to be "the right definition", and even if this was the case, the fashion might change in the future.

I am, however, hoping that there might be a semi-mathematical reason to prefer one notion over the other. For example, I have spent much of my mathematical life believing that $0 \in \mathbb{N}$ because: 1) morally, $\mathbb{N}$ is the cardinalities of finite sets 2) the empty set is a set with $0$ elements. However, recently I realised that this reasoning applies to $\omega$ rather than $\mathbb{N}$, and - much to my horror - I saw $\omega$ and $\mathbb{N}$ used side by side with the only distinction being that $0 \in \omega$ while $0 \not \in \mathbb{N}$. For another example, $\mathbb{N}$ seems to be a much nicer semigroup if $0 \not \in \mathbb{N}$ (and in any case, adding $0$ to a semigroup is a more natural operation than removing it), which would an argument for taking $0 \not \in \mathbb{N}$.

The arguments mentioned above are, of course, rather weak, but perhaps just enough to tip the scale. In any case, this is the general type of argument I am looking for.

Question: Does there exists a convincing argument for deciding if $0 \in \mathbb{N}$ ?

(I consider it quite possible that the answer is negative because in some context one convention is preferable, and in other context the other one. The problem could be dismissed by using $\mathbb{Z}_+$ (or even $\mathbb{Z}_{>0}$ and $\mathbb{Z}_{\geq 0}$) to avoid confusion, but note that in some context one definitely does not want to do this. I would be interested in an argument that is universal in the sense that it makes overall mathematical landscape more elegant, and does not spoil any detail too much. I do not hope that the argument would be convincing to every mathematician, especially one working in a very specific and narrow area.)

答: My 2¢: Besides the in my opinion other good arguments why $\mathbb{N}$ should contain $0$, I argue for a purely notational one: since $\mathbb{Z}_+$ is a perfectly good name for the set $\{ 1, 2, 3, \ldots \}$ there is no need for a second name for this set. Hence $0 \in \mathbb{N}$.</p>

Title: Should $\mathbb{N}$ contain $0$?


问:
Derivative of $\frac{x/2}{\sin(x/2)}$ is bounded in $[0,\pi]$
Given the function $g(x)=\frac{d}{dx}\left(\frac{x/2}{\sin(x/2)}\right)$, I want to prove it is bounded in $[0,\pi]$.

For me, it is immediate that it is bounded in $(0,\pi]$, as $\frac{x/2}{\sin(x/2)}$ is continuous in $(0,\pi]$. Please, correct me if this needs to be proved with more detail.

Then, how I prove it is bounded in 0? Is it useful to compute the limit of $\frac{x/2}{\sin(x/2)}$ in $0$?. Many thanks in advance!

Body:
$$g(x)=f'(x)=\frac{\frac{1}{2}\sin\left(\frac{x}{2}\right)-\frac{x}{4}\cos\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}$$
Simplify
$$f'(x)=\frac{\frac{1}{4}\left(2\sin\left(\frac{x}{2}\right)-x\cos\left(\frac{x}{2}\right)\right)}{\sin^2\left(\frac{x}{2}\right)}= \frac{2\sin\left(\frac{x}{2}\right)-x\cos\left(\frac{x}{2}\right)}{4\sin^2\left(\frac{x}{2}\right)}\le\frac{1}{2}$$
From the graph below


问:
Orthogonal Trajectories to the solutions of $\frac{dy}{dt} = y$
By using the properties of the dot product, I get as answer
$y_{ortho} = \frac{1}{C}e^{-t}$ , if  $y = Ce^{t}$ and $C \neq 0$.

If a try to use the inner product of function $\int fg$, I don't get $0$ as I would expect.

What is wrong with this answer?

Thanks in advance.

答:
Given ODE
$$ \frac{dy}{dt}= y $$

has solutions $$ y=C_1e^t$$

Its reciprocal $$ \frac {1}{y}=e^{-t}/C_1$$

is unrelated to the discussion. (Negative reciprocal of slope does not lead to the reciprocal of the function itself).

Orthogonal trajectories to given ODE are obtained by using negative reciprocal $ y'\to \dfrac{-1}{y'}$, so

$$ \frac{dy}{dt}=\frac{-1}{y} $$
has solutions $$ y=\sqrt{C_2- 2 t}.$$


问:
Vector calculus "identities"
Do these "identities" even make sense? $F$ is a 3D vector field. For the second equation, the LHS is a vector but the RHS is a scalar.
$$\nabla\cdot\Delta F=\Delta(\nabla\cdot F),\,\nabla\times\Delta F=\Delta(\nabla\times F)$$

答: The first equation means $\partial_i(\partial_j\partial_j F_i)=\partial_j\partial_j(\partial_i F_i)$. The second means $\epsilon_{ijk}\partial_j(\partial_l\partial_l F_k)=\partial_l\partial_l(\epsilon_{ijk}\partial_j F_k)$.

Title: Vector calculus "identities"


问:
Finding a tetrahedron in a unit sphere
I am trying to solve an exercise where I have to find a tetrahedron in a unit sphere. My first thought and i think it is not a bad idea is to let one point be at $(0,1,0)$ (assume this is $A$) and then find a plane

$y = -d$

so that

$|\overline{AB}|$ = $|\overline{BC}|$

using a new point $O = (0, d, 0)$ which will be the centroid of the triangle $BCD$. Now i know that the points $B, C, D$ will be on the sphere so that the equation

$x^2 + z^2 = 1 - d^2$

holds true.
Furthermore, I know that

$2 * sin(60) * \overline{OB} = \overline{BC}$.

I feel like i have collected all the parts of the puzzle, but i still cannot solve it. Any suggestions?

EDIT: Since somone mentioned it in the comments, it should be a regular tetrahedron

答:
You have better work in spherical coordinates
for instance with the mathematical convention
$$
\left\{ \matrix{
  x = r\sin \varphi \cos \theta  \hfill \cr
  y = r\sin \varphi \sin \theta  \hfill \cr
  z = r\cos \varphi  \hfill \cr}  \right.
$$
polar angle $\varphi$, azimuthal angle $\theta$, and $r = 1$ for the unitary sphere.

Take a vertex in the pole $(0,0,1)$.

The other three vertices will be, by symmetry, on azimuthal planes at $120^{\circ}$ from each other.

So $\theta = 0, \pm 2/3 \pi$.

For the angle $\varphi$ consider that it is the same for all the three vertices at the base,
and that their position vectors, summed up shall give a null resultant with the polar vertex.

So they are at $z= -1/3$ or $\cos \varphi = -1/3$, i.e. $\sin \varphi =2 \sqrt{2}/3$.

And you have all the elements to fix the other three vertices.

Title: Finding a tetrahedron in a unit sphere


问:
Simplifying $F(\sin^{-1}\sqrt{2/(2-p)},1-p^2/4)$ (for a minimal surface)
The tetragonal distortion of the Schwarz P minimal surface has a Weierstrass–Enneper parametrisation of $g(w)=w,f(w)=(w^8+\lambda w^4+1)^{-1/2}$ and a Bonnet angle of $\pi/2$ where $\lambda<-2$. This surface can be decomposed into "catenoids" between square rings:

I have worked out that the aspect ratio (height over base side) of this catenoid is $\sqrt{-ps}\frac ZX$ where
$$p=\lambda+\sqrt{\lambda^2-4},s=\sqrt{2-\lambda}$$
$$X=K\left(\frac12+\frac1s\right)$$
$$Z=F\left(\sin^{-1}\sqrt{\frac2{2-p}},1-\frac{p^2}4\right)$$
(Here the last argument of elliptic integrals is the parameter $m$.)

I am unable to simplify $Z$ to a complete elliptic integral – I think it can be done, since the formula for the aspect ratio of the unit cell of the conjugate T (tetragonal D) surface only involves complete integrals. Can simplification on $Z$ be done?

In particular, can the Landen and Gauss transformations be applied to $Z$ to reduce it?

答:
In Carlson symmetric form:
$$Z=F\left(\sin^{-1}\sqrt{\frac2{2-p}},1-\frac{p^2}4\right)=\sqrt{\frac2{2-p}}R_F\left(\frac p{p-2},-\frac p2,1\right)$$
Now apply a descending Landen transformation (DLMF 19.22.18, .23) to get a zero as one argument, then scale to make another argument one:
$$=\sqrt{\frac2{2-p}}R_F\left(0,\frac{4p}{p-2},\frac{2-p}2\right)=\frac2{2-p}R_F\left(0,\frac{-8p}{(2-p)^2},1\right)$$
$R_F$ may now be converted back into a complete Legendre elliptic integral. Its parameter is quite complicated in $p$ alone, but substituting its definition $p=\lambda+\sqrt{\lambda^2-4}$ gives a surprisingly simple result:
$$\boxed{Z=\frac2{2-p}K\left(1+\frac{8p}{(2-p)^2}\right)=\frac2{2-p}K\left(\frac{\lambda+2}{\lambda-2}\right)}$$

$K(p^2/4)$ where $p$ has the same meaning appears in the T surface's aspect ratio formula. By the same approach as detailed above, but using an ascending Gauss transformation, I was able to simplify it to $\frac2{2-p}K\left(\frac4{2-\lambda}\right)$.


答:
One one speaks about things generated by a set (linear subspaces, subgroups, subrings, and so on) there is usually a category in the background and then the generated object is supposed to be the smallest "acceptable" thing containing the set of generators.

If the background category is that of

rings, you would only take polynomials without constant term,

unital rings, then take all polynomials,

C$^*$-algebras, then take (non commutative) polynomials in $T$ and $T^*$, with or without constant term, and close it.

Title: Definition of algebra generated by an operator.

Body: Suppose T is an operator (defined on a Hilbert space, say).  Does the "algebra generated by $T$" have a commonly understood meaning?  I couldn't find a satisfactory answer after some internet searching. It seems to me that there are two ways one could define the algebra $A$ generated by $T$ (without including additional structure on $A$, like closure), namely $A=\{p(T): p$ is a polynomial}, or $A=\{p(T): p$ is a polynomial with $p(0)=0$}. Both definitions contain only the operators which are linear combinations of powers of $T$. In the first definition, we allow for a power of $0$, so that the identity operator belongs to the algebra. Is this the standard definition, or is it something else?


问:
Remainder when $20^{15} + 16^{18}$ is divided by 17
What is the reminder, when $20^{15} + 16^{18}$  is divided by 17.
I'm asking the similar question because I have little confusions in MOD.
If you use mod then please elaborate that for beginner.
Thanks in advance.</p>

答: consider this over $\mathbb{Z}/17\mathbb{Z}$. Addition and multiplication are well defined, so we can without harm say $20^{15} + 16^{18} \equiv 3^{15} + (-1)^{18}$ mod 17. You want to show what it is equivalent to mod 17. To do this you would just crunch it out using Fermats litle theorem and stuff like that.</p>


问:
First-order model theory problem
Are the following structures isomorphic?

a) $(\mathbb{R};=;+)$ and $(\mathbb{Q};=;+)$;

b) $(\mathbb{Z}, <)$ and $(\mathbb{Z},>)$;

c) $(\mathbb{N}, <)$ and $(\mathbb{N}, >)$;

答:
Your intuition is somewhat correct:

a) No. An isomorphism would require a bijection of sets, but the cardinalities of $\mathbb{Q}$ and $\mathbb{R}$ are different (the former is countable, the later is not).

b) Yes. Let $f: \mathbb{Z} \to \mathbb{Z}$ be the map $f(x)=-x$. This is clearly a bijection of sets, and also an embedding of structures as:
$x < y$ iff $ -x > -y$, i.e. $(x,y) \in   <$ iff $ (f(x),f(y)) \in   >$.

c) No (assuming equality is a logical symbol). Call the first structure $N_1$, and the second $N_2$. Suppose that they were isomorphic. Then they would satisfy the same sentences. But $N_1 \models \exists x \forall y (x \neq y \implies (x<y)$ (take $x=0$), and so $N_2 \models \exists x \forall y (x \neq y \implies (x>y)$. This is a contradiction as $\mathbb{N}$ has no upper bound, so the two structures can't be isomorphic.

Title: First-order model theory problem


问:
An example separating the amalgamation property from a weaker variant
Let's consider an equational theory $T$ in an algebraic language language and the following statement (S): for every formula $\phi, \psi$, if $T \vdash \phi \to \psi$, then there exists a formula $\chi$ such that $T \vdash \phi \to \chi \land \chi \to \psi$.  Here,   $\phi$, $\psi$, $\chi$ are existential, universal, and quantifier-free formulas, respectively.  One can show that (S) holds if and only if $T$ has the amalgamation property.

Now let's consider a similar statement (S'), which is just like (S) but $\phi, \psi, \psi$ are restricted to positive formulas.  Basich (1974) showed that (S') corresponds to a weaker variant of the amalgamation property sometimes called the flat amalgamation property.  Here, only diagrams of the form $B \leftarrow A \rightarrow A * F(X)$ are required to be amalgamated, where $*$ denotes the so-called free product, and $F(X)$ is the free $T$-algebra generated by $X$.

People's interest in both of these prima facie distinct principles suggest that they are actually different.  I cannot, however, find an equational theory $T$ satisfying (S') but not (S).  If I understand correctly, every variety of normal modal algebras satisfying (S') also satisfies (S), and these varieties are the only ones I can apply what little intuition I have regarding these things.

What $T$, if at all, satisfies (S') but not (S)?

答:
The theory of rings (when I say "ring", I mean "commutative ring with unity") is an example, which I believe motivates the terminology "flat amalgamation property".

For any ring $A$ and any set $X$, the free product (coproduct) $A * F(X)$ is $A\otimes_{\mathbb{Z}} \mathbb{Z}[X] \cong A[X]$, the polynomial ring in variables $X$ with coefficients from $A$. Now for any embedding of rings $f\colon A\to B$, we get a canonical embedding of polynomial rings $A[X]\to B[X]$ (by applying $f$ to the coefficients), so $B[X]$ witnesses the flat amalgamation property.

The fancy way to say this is that the ring map $\mathbb{Z}\to \mathbb{Z}[X]$ is a flat morphism of rings, so taking the tensor product with $\mathbb{Z}[X]$ preserves injectivity. In particular, if $A\to B$ is injective, so is $A\otimes_ \mathbb{Z} \mathbb{Z}[X] \to B\otimes_\mathbb{Z} \mathbb{Z}[X]$,  which is isomorphic to the canonical map $A[X]\to B[X]$.

On the other hand, the theory of rings does not have the amalgamation property (for embeddings). If there is a witness $D$ to the amalgamation property for embeddings $f\colon A\to B$ and $g\colon A\to C$, then this induces a map from the tensor product $B\otimes_A C$ (which is the pushout of $B$ and $C$ over $A$) to $D$. Since the maps $B\to D$ and $C\to D$ are injective, the maps $B\to B\otimes_A C$ and $C\to B\otimes_A C$ must also be injective. But this is not always the case - see here for an example where the map $A\to B\otimes_A C$ is not even injective.

Edit: If the example linked to above looks mysterious, here is a logician's point of view on what's happening there. Let $\varphi(a,b)$ be the formula $\exists x\, (x^2 = 0\land xb = a\land a\neq 0)$, and let $\psi(a,b)$ be the formula $\exists y\, (ya = b)$.

These two formulas are inconsistent, since together they imply $a = xb = xya = x^2yb = 0$ and $a\neq 0$. So the existential formula $\varphi(a,b)$ implies the universal formula $\lnot \psi(a,b)$. But there is no quantifier-free formula $\chi$ such that $\varphi(a,b)\rightarrow \chi(a,b)$ and $\chi(a,b)\rightarrow \lnot \psi(a,b)$.

This can be proven by finding a ring $A$ with distinguished elements $a$ and $b$ and extensions $A\subseteq B$ and $A\subseteq C$ such that $B\models \varphi(a,b)$ and $C\models \psi(a,b)$. If we had an interpolating $\chi(a,b)$, it would be true in $B$, hence true in $A$, and hence true in $C$ (since it is quantifier-free), but then $\lnot\psi(a,b)$ would be true in $C$, contradiction.

Well, the idea is to  fix a field $k$ to work over and "freely adjoin" witnesses subject to the constraints: take $A = k[a,b]$, $B = A[x]/(x^2,a-bx)$, and $C = A[y]/(b-ay)$. But then this needs to be adjusted a bit to make the maps from $A$ to $B$ and $C$ injective, e.g. $a = bx$ and $x^2=0$ implies $a^2 = b^2x^2 = 0$, so we need to make $a^2 = 0$ in $A$. Continuing along these lines (and maybe making things more symmetrical for aesthetic reasons), we arrive at the example in the link.


问:
Why "Stable" in Stable vector bundle and Stable homotopy theory?
Why/what/how does stable mean in the

Stable vector bundle?

Stable homotopy theory?

Of course, it is described in the Wiki texts in a formal way. But I wonder what are the intuitions and concepts behind both stables in the text.

Another way to phrase it may be that when we remove stable, what makes the differences for vector bundle and homotopy theory?

答:
Any good answer to this would be very long, so perhaps some people can answer with alternative perspectives.

Essentially stability removes "pathologies" that only happen in low dimensions. For example, there is only one tangent bundle to a manifold, but many normal bundles. However, if we make the codimension of our embedding large enough all normal bundles become isomorphic, so stabilization here allows us to talk about "the" stable normal bundle. There are other reasons to care about stable bundles, for example, most of our characteristic classes care only about the stable isomorphism type. And perhaps the most important reason is that stable bundles (or the slightly different virtual bundles), form a group rather than a monoid.

When it comes to homotopy theory, stable homotopy theory tends to be more algebraic than homotopy theory. This can be seen at the level of spaces. There is no natural group structure on $[X,Y]$ but if $X$ is a suspension, then this naturally has a group structure. Moreover, if $X$ is a double suspension, this is even nicer than a group, it is an abelian group.

One really concrete way of seeing the algebraic difference between spaces and spectra is through the following example: Quillen proved the remarkable result that (subject to a small connectivity assumption) spaces up to rational equivalence have the exact same homotopy theory as rational differential graded commutative algebras. The proof of this is rather involved.

Of course we can ask what spectra up to rational equivalence are? The answer is just rational chain complexes, and it is very easy to see. It follows from the fact that rationally every spectrum is equivalent to the generalized Eilenberg-MacLane spectrum with the same homotopy groups.

To me, chain complexes are entirely algebraic, but products on a chain complex can encode a lot of geometry (i.e. cup products on singular cochain complexes or de Rham wedge products, etc.)

Now what "pathology" is this a result of? I haven't thought about this in a while, but I think it is due to the fact that in the stable world the rational Eilenberg-MacLane space is equivalent to the rational Moore space (the rational sphere). However, unstably they are not equivalent. This is due to an unstable obstruction in the cohomology of the even dimensional Eilenberg-MacLane space, the cup square, so passing to the stable world this vanishes.

This will imply that Postnikov towers for rational spectra are all trivial, but for rational spaces they could have nontrivial k invariants. Of course, there is so much more that can be said, so I hope others, who assuredly know a lot more stable homotopy theory than me, can contribute


问:
Setting up linear system
Setup a linear system and solve to get the equation of a plane that contains points $x = (4,0, −2)$, $y = (2,3, -1)$ and $z = (0,0,1)$.
How do we set up an equation like do we use transformation matrix?

答:
General equation of a plane is $$ax+by+cz=d$$
Plug the coordinates and get the system
$$\begin{cases}
4 a-2 c=d\\
2 a+3 b-c=d\\
c=d\\
\end{cases}
$$
solutions are
$a= \frac{3 d}{4},b= \frac{d}{6},c= d$

So the equation of the plane is
$$\frac{3 }{4}d x+\frac{d}{6}y+dz=d$$
simplify and rearrange
$$9 x+2 y+12 z=12$$

Title: Setting up linear system


问:
Proving x and y is divisible by p (prime).
If p is a prime number and x and y are integers, how do I prove "if xy and x+y are both divisible by p, then x and y is divisible by p"?

I started like this..

1) p divides xy, so p divides x or p divides y, since p is a prime number.

and then I'm already stuck :/

Help me to get to the next step please.

答: If $p$ divides $x$ but not $y$, can $p$ divide $x+y$?</p>


问:
Convergence of (-1)^n * sqrt(n)/n series
What I am having problems with is the following series.

$$\sum_{n=1}^\infty (-1)^n\frac{\sqrt{n}}{n+907}$$

So far I have tried Abel's test and while It is apparent that the series $$(-1)^n \frac{1}{n+907}$$ converges, I couldn't prove that $\sqrt{n}$ is bounded above.

The next thing I tried was Leibniz test for alternating series and while I could prove that
$$\lim_{n \to\infty}\frac{1}{n+907}=0$$ I couldn't prove that $\frac{1}{n+907}$ is monotonically decreasing.

Up to now I have run out of ideas about what to try. Note that I cannot test for absolute convergence, I must do it by another method.

答:
If you don't want to prove that monotonicity (or in cases where it is difficult to prove it), a small expansion can be used :
$$(-1)^n\frac{\sqrt{n}}{n+907} =\frac{(-1)^n}{\sqrt{n}}\left(\frac{1}{1+\frac{907}{n}} \right)=\frac{(-1)^n}{\sqrt{n}}\times\left(1-\frac{907}{n} + o \left(\frac{1}{n} \right) \right)$$ so $$ (-1)^n\frac{\sqrt{n}}{n+907} =\frac{(-1)^n}{\sqrt{n}}-\frac{907(-1)^n}{n\sqrt{n}} + o \left(\frac{1}{n\sqrt{n}} \right) $$

and all the terms here are the terms of convergent series, so the initial series converges.


问:
Decompose into simple fractions $\frac{f'}{f}$
Let $f(x) = (x-a_1)(x-a_2)...(x-a_n)$.

Find a decomposition into simple fractions of $\frac{f'}{f}$.

Where $f'$ is a derivative of our polynomial.

As I understand, we have to find a pretty-format of $f'$ and then task will be easy. But I have no idea how to find this.

Note that some $a_i$ can be equal to each other.

Body:
Following Martin's idea,

Let for $ x\ne a_1,a_2,...,a_n $,
$$g(x)=\ln(|f(x)|)=$$
$$\sum_{k=1}^n\ln(|x-a_k|)$$

thus

$$g'(x)=\frac{f'(x)}{f(x)}$$
$$=\sum_{k=1}^n\frac{1}{x-a_k}$$


问:
How to calculate Z-transform from a series
If we have the sequence $\{1,1,\dots\}$, how would I go about calculating the Z-transform? Such that we find $Z\{1,1 \dots\}(z)$.

Body: $1+1/z+1/z^2+\cdots= 1/(1-1/z)=z/(z-1)$


问:
Given $T_1 = 0$ and for $i > 1$: $T_i = 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j$ Prove via induction that $T_i = \sum\limits_{j=1}^{i-1} \frac{1}{j}$

Given $T_1 = 0$ and for $i \in \mathbb{N}, i > 1$:
\begin{align*}
  T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j \\
\end{align*}
Manually computing terms $2,3,4$:
\begin{align*}
  T_2 &= 1 \\
  T_3 &= 1 + \frac{1}{2} \\
  T_4 &= 1 + \frac{1}{3} \left(1 + 1 + \frac{1}{2} \right) = 1 + \frac{1}{2} + \frac{1}{3} \\
\end{align*}

Prove by induction that this is:

\begin{align*}
  T_i &= \sum\limits_{j=1}^{i-1} \frac{1}{j} \\
\end{align*}

The given equation is clearly satisfied for terms $1,2,3,4$. We need to show the inductive step. Assuming it's true for all $j < i$, show that it is also true for $i$.

\begin{align*}
  T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} \sum\limits_{k=1}^{j-1} \frac{1}{k} \\
  T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-2} \frac{i-1-j}{j} \\
\end{align*}

Body:
Let's say it is true for $i$, and we want to prove for $i+1$ ($i>1$):

I think you can develop the left hand:
$$T_{i+1} = 1 + \frac{1}{i}\sum_{j=1}^{i}T_j = 1 + \frac{1}{i}\sum_{j=1}^{i-1}T_j + \frac{T_i}{i}$$

$$iT_{i+1} = i + \sum_{j=1}^{i-1}T_j + T_i$$
$$\frac{i}{i-1}T_{i+1} = \frac{1}{i-1} + 1 + \frac{1}{i-1}\sum_{j=1}^{i-1}T_j + \frac{T_i}{i-1}$$
$$\frac{i}{i-1}T_{i+1} = \frac{1}{i-1} + T_i + \frac{T_i}{i-1}$$
$$iT_{i+1} = 1 + (i-1)T_i + T_i$$
$$T_{i+1} = T_i + \frac{1}{i}$$

And that is the right hand:
$$T_{i+1} = \sum_{j=1}^{i}\frac{1}{j} = \sum_{j=1}^{i-1}\frac{1}{j} + \frac{1}{i} = T_i + \frac{1}{i}$$


问:
If $P(b\mid a) >P(b)$ and $P(a)>0$ then $P(\neg a \cap \neg b)>P(\neg a)\cdot P(\neg b)$
I tried to break the conditional probability down and also substitute $P(a) = 1 - P(\neg a)$, however, I wouldn't get what I want. I am struggling with this question and I don't know how should I approach it.

答:
I would start by noticing that

\begin{align*}
\mathbb{P}(B|A) = \frac{\mathbb{P}(B\cap A)}{\mathbb{P}(A)} > \mathbb{P}(B) \Longleftrightarrow \mathbb{P}(B\cap A) > \mathbb{P}(A)\mathbb{P}(B)
\end{align*}

Then we may approach the second part as follows:

\begin{align*}
\mathbb{P}(A^{c}\cap B^{c}) - \mathbb{P}(A^{c})\mathbb{P}(B^{c}) & = 1 - \mathbb{P}(A\cup B) - (1 - \mathbb{P}(A))(1 - \mathbb{P}(B))\\\\
& = -\mathbb{P}(A\cup B) + \mathbb{P}(A) + \mathbb{P}(B)- \mathbb{P}(A)\mathbb{P}(B)\\\\
& = \mathbb{P}(A\cap B) - \mathbb{P}(A)\mathbb{P}(B) > 0
\end{align*}

and we are done.

Hopefully this helps!


Title: Remainder when $20^{15} + 16^{18}$ is divided by 17

Body:
What is the reminder, when $20^{15} + 16^{18}$  is divided by 17.
I'm asking the similar question because I have little confusions in MOD.
If you use mod then please elaborate that for beginner.
Thanks in advance.</p>

Body:
Hints:

$$20=3\pmod{17}\;,\;\;16=-1\pmod{17}\implies$$

$$20^{15}+16^{18}=3^{15}+(-1)^{18}=(**)$$

But for any integer $\,a\;,\;\;(a,17)=1\,$ , we have that $\,a^{16}=1\,$ , so

$$(**)=3^{16}\cdot 3^{-1}+1=1\cdot 6+1=7\ldots$$

and so the claim is false: the remainder is $\,7\,$ .


问:
How to find tangent of a an equation
I am trying to solve this question by using the slope of a curve at a point formula. How do you find an equation for each tangent to the curve y = 1 / (x-1) that has a slope of -1? Thanks.</p>

答: Solve $\frac{df(x)}{dx} = -1$ for $x$.

Title: How to find tangent of a an equation

Body: Solve $\frac{df(x)}{dx} = -1$ for $x$.


问:
Maximum souvenirs possible with given conditions.
(I was not sure where to post this question whether here or in the CS one, but I felt it was rather math'y)

This was a problem from a local programming contest held at my school (I couldn't solve it and I am asking for a solution with detailed explanations please.)

the problem states:

You and a few friends went traveling to Russia, and you wanted some souvenirs. Luckily, you met the king of making souvenirs. He makes his unique souvenir "Lambda" using one of the following:

-2 eyes and a body

-2 eyes, a mouth and a body

-an eye, a mouth and a body

He isn't very good at math, so he asked you to tell him what is the maximum number of Lambdas he can make if he has $e$ eyes, $m$ mouths and $b$ bodies.

Constraints:

$0 \le e,m,b \le 10^8$

Examples:

$e = 1, m =2, b = 3$ Then $MAX = 1$

$e = 0, m =11, b = 2$ Then $MAX = 0$

$e = 14, m =21, b = 23$ Then $MAX = 14$

$e = 90, m =24, b = 89$ Then $MAX = 57$

(Thanks in Advance :D)

答:
Note that the second configuration is dominated by the other two, so ignore it.  Let $x_1$ and $x_2$ be the numbers of Lambdas of each of the remaining two types.  The problem is to maximize $x_1+x_2$ subject to
\begin{align}
2x_1 + 1x_2 &\le e &&(y_e \ge 0) \\
0x_1 + 1x_2 &\le m &&(y_m \ge 0) \\
1x_1 + 1x_2 &\le b &&(y_b \ge 0) \\
x_1, x_2 &\ge 0
\end{align}
The dual problem is to minimize $e y_e + m y_m + b y_b$ subject to
\begin{align}
2y_e + 0y_m +1y_b &\ge 1 &&(x_1 \ge 0) \\
1y_e + 1y_m +1y_b &\ge 1 &&(x_2 \ge 0) \\
y_e, y_m, y_b &\ge 0
\end{align}

Here are optimal solutions for your examples:
\begin{matrix}
e & m & b & x_1 & x_2 & y_e & y_m & y_b & \max =x_1 + x_2 = e y_e + m y_m + b y_b \\
\hline
1 &2 &3 &0 &1 &1 &0 &0   &1 \\
0 &11 &2 &0 &0 &1 &0 &0  &0 \\
14 &21 &23 &0 &14 &1 &0 &0 &14 \\
90 &24 &89 &33 &24 &1/2 &1/2 &0 & 57 \\
\end{matrix}
Because we have only two decision variable $x_1$ and $x_2$, it is reasonable to solve this linear programming problem by enumerating all extreme points.  By considering two constraints at a time, we find the following possibilities:
\begin{matrix}
x_1 & x_2 \\
\hline
(e-m)/2 & m \\
e-b & 2b-e \\
0 & e \\
e/2 & 0 \\
b-m & m \\
0 & m \\
0 & b \\
b & 0 \\
0 & 0
\end{matrix}
To maximize $x_1+x_2$ for given $e,m,b$, it suffices to compute this list of candidates, check feasibility for each one, and keep one that has the largest sum.


问:
A Symmetric Numeric System of Multiplicative Inverses?
I was just thinking about how strange it is that multiplicative inversion bifurcates the rational numbers into such asymmetric segments.

Additive inversion bifurcates the number line cleanly into a left half and a right half, but multiplicative inversion bifurcates it into a finite length segment ($-1$ to $1$) and an infinite length segment (everything else).  I understand that it is a paradox of infinity that all multiplicative inverses of elements in an infinite length interval can fit into a finite length interval, but would it have been possible for civilization to have chosen a numeric paradigm where multiplicative inversion was more symmetric, i.e., via some combination of...

A different total ordering of numbers

A different number base

As I understand, some (all?) modular arithmetic systems have the property that their multiplicative inverses are spread across the entire finite range of numbers; perhaps we could start to arrive at an infinite numeric paradigm similar to our own by thinking of the $n$ in a mod-$n$ arithmetic system as a variable and taking its limit as it approaches infinity?

...or is this asymmetry a logically necessary fact of mathematics?

答: Let $A=(-1,0)\cup (1,\infty)$. Then $\{\,\frac1x\mid x\in A\,\}=\{-x\mid x\in A\,\}$ and all we left out are $\{-1,0,1\}$.


问:
Is it possible turn the Dirichlet ring into a Banach algebra?
The set of all arithmetic functions $f:\mathbb{Z}^{+}\to\mathbb{C}$, under pointwise addition and Dirichlet convolution, is a commutative ring, not all functions are Dirichlet invertible.

So my question is:
To what subset of the arithmetic functions and under what norms can one attach a Banach algebra structure to the Dirichlet ring? Is it even possible?

答:
If you restrict to the convergent Dirichlet series (those with polynomially bounded coefficients) then $$\|a\|=4\sup_n |a_n| e^{-n}$$ should be a ring norm, as for some $n$

$$\|a\ast b\|=4 |\sum_{d| n}a_d b_{n/d}|e^{-n} \le 4  \sum_{d|n} \frac1{4} \|a\| e^{n/d}\frac1{4} \|b\| e^d e^{-n}\le \frac1{4} \|a\|\|b\|(1 + 2 e^{-n}\sum_{d=2}^{\sqrt n} e^{n/d}e^{d})$$ $$\le \frac1{4} \|a\|\|b\| (1 + 2 \sum_{d=2}^{\sqrt n} e^{d-n/2})\le \|a\|\|b\| $$

The completion is clearly the Banach algebra of formal Dirichlet series with $O(e^n)$ coefficients. It is an integral domain.

Note that the inverse of a convergent Dirichlet series $k^{-s}+\sum_{n=k+1}^\infty c_n n^{-s}$ exists as a convergent generalized  Dirichlet series $k^s \sum_{l=0}^\infty (-\sum_{n=k+1}^\infty c_n (n/k)^{-s})^l=\sum_{u\in \Bbb{Q}_{> 0}} d_u u^{-s}$, the obtained ring is a field. Does anyone want to check how $\|a\ast b\|\le \|a\|\|b\|$ fails when extending the norm to those generalized Dirichlet series ?