问:
Can this relative notion of cardinality survive ZF-Reg.?
The posting here aims at defining a relative notion of well foundedness and thus of ranks, in relation to a base set, then tries to use Scott's trick to define a relative notion of cardinality to that base set as well.

Let's at first work in ZF-Reg. + the following anti-foundation axiom:

$\exists B \forall x \exists \alpha : x \in P^\alpha (B) $

Where:

$P^\emptyset (B)= B \\ P^{\alpha +1} (B)= P(P^\alpha(B)) \\P^\alpha = \bigcup_{\lambda<\alpha} P^\lambda(B) \text { for a limit } \alpha$

Define: $base(B) \iff \forall x \exists \alpha : x \in P^\alpha(B)$

Now we can define the notion of relative rank to a base set $B$, denoted by $\rho^B$ as: $$\rho^B(x)=\alpha \iff \forall \lambda (x \in P^\lambda (B) \Leftrightarrow \alpha \leq \lambda)$$. Now well-founded sets would here correspond to sets having ranks relative to the empty set. Formally: $$\forall x [wf(x) \iff \exists \alpha: \alpha= \rho^\emptyset(x)]$$. The idea here is that the definition of a set as what has a rank, need not be confined to the well founded sets! $B$ can be a set of Quine atoms for example, and so all of its elements would have zero ranks relative to $B$, while the empty set would have rank 1 relative to $B$.

Now one can define a notion of cardinality relative to a base set as:

$|x|^B = \{y: y \sim x \land  \neg  \exists z (z \sim y \land \rho^B(z) < \rho^B(y))\}$

For that notion of cardinality to be total we need the axiom:

$\exists B \forall x \exists y [\exists \alpha (\rho^B(y)=\alpha) \land x \sim y]$

Which is weaker than the above anti-foundation axiom.

Its obvious that the above line is just another use of Scott's trick, but through extending it to a world outside of regularity and choice, and outside of Coret's axiom of every set being equinumerous to some well founded set. However, here we'll have a modified Coret's axiom of every set being equinumerous to some $B$-well founded set, where the later is any set that has a rank relative to base set $B$.

Are there models of $\sf ZF-Reg.$ which violates the above axiom [the weaker one]?

It is indeed well-known that $\sf ZF-Reg.$ doesn't support a definable notion of cardinality that is total over the whole universe of discourse. But here the mentioned notion is a relative one, so is it the case that even this relative notion of cardinality cannot survive in $\sf ZF-Reg.$ as a total function?

答:
Start with a proper class of Quine atoms, $A$, and suppose that the universe is generated by iterated power sets over $A$ (the axiom of choice holds, and there are infinite sets of atoms, the nicest situation you can think of).

Now consider the permutation model obtained by writing $A=\{a_{\alpha,n}\mid\alpha\in\mathrm{Ord},n<\omega\}$, then considering permutations of $A$ moving set-many atoms and satisfying: $\pi(a_{\alpha,n})=a_{\alpha,m}$ for all $\alpha$. In other words, we partition $A$ into countable blocks and permute them independently.

Finally, consider the filter of groups given by finite stabilisers.

Let $A_\alpha=\{a_{\alpha,n}\mid n<\omega\}$, then in the permutation model $A_\alpha$ is Dedekind-finite for any $\alpha$, that is easy.

Now suppose $\alpha$ is an ordinal, and let $B_\alpha=A\setminus A_\alpha$. Then $A_\alpha$ is not equipotent to any set generated by any subset of $B_\alpha$. If it wasn't the case, there was some $X\subseteq B_\alpha$, an ordinal $\gamma$, and an injective $f\colon A_\alpha\to\mathcal P^\gamma(X)$. And it had a finite support $E$.

Since $E\cap A_\alpha$ is finite, pick some $a,b\in A_\alpha\setminus E$ and let $\pi=(a\ b)$. Trivially, $\pi\restriction\mathcal P^\gamma(X)$ is the identity. Now, $\pi A_\alpha=A_\alpha$ and $\pi f=f$, so $f(a)=\pi(f(a))=f(b)$, so $f$ is not injective.


问:
Example of Normed Spaces $X$ and $Y$ and a linear operator $T$, so that $\ker (T)$ is not closed.
So first, I thought about the zero transformation, however I saw it would not work. After that, I thought about the differentiation transformation, being $X$ and $Y$ the set of all infinitely differentiable functions, however $\ker(T)$ would be closed, since it would be the set of all constant functions, wouldn't it?

Can anyone help or just give a hint of something I am not thinking about? Thanks in advance.

答:
Let $V$ be a normed space. Then if $\tau: V \to \mathbb{C}$ is a linear map, the following two statements are equivalent:

$\tau$ is continuous.

$\ker(\tau)$ is closed in $V$.

This fact can be found in virtually every basic functional analysis book.

Hence, your example will be found once we construct a linear map $\tau: V \to \Bbb{C}$ such that $\tau$ is not continuous! To do this, we proceed as follows: let $V$ be your favorite  infinite-dimensional normed vector space with Hamel basis $\{e_n:n \in \mathbb{N}\}$. By rescaling the basis, we may assume that $\|e_n\| = 1$ for all $n \in \Bbb{N}$. Define a linear map $\tau:  V \to \mathbb{C}$ by
$\tau (e_n) = n$ for all $n \in \mathbb{N}$. Then
$$\|\tau\|= \sup_{\|v \| \leq 1} |\tau(v)| \geq n$$ for all $n \in \mathbb{N}$, so that $\|\tau\| = \infty.$ Hence, $\tau$ is not continuous and $\ker(\tau)$ is not closed in $V$.

ContentLicense: CC BY-SA 4.0


问:
Expression of the exponential integral $E_1$ using standard functions with real arguments
In standard numeric packages (for C++) the function $$E_1(z)=\int_1^\infty \frac{e^{-zt}}{t}dt$$
is only implemented for real arguments. For a specific calculation I need to be able to evaluate this function for all complex $z$. My hope is, that this can be acomplished using "standard" special function with real argument only (where with standard I mean accessible to numerics using e.g. the GSL).

Can this be achieved, and if so, how?

答: Julia's SpecialFunctions package includes exponential integrals of arbitrary complex arguments (and arbitrary orders, even complex orders).  It is callable from C/C++ by linking to libjulia and writing a few lines of C/C++ code to get a compiled function pointer, as explained in this thread on stackexchange

ContentLicense: CC BY-SA 3.0


问:
How the inverse of this matrix be found?
How can the inverse of matrix
$A = \left( \begin{smallmatrix} 6&5\\5&4 \end{smallmatrix} \right)$ be $A^{-1} = \left( \begin{smallmatrix} -4&5\\ 5&-6 \end{smallmatrix} \right)$ where $\frac{1}{ad-bc} = \frac{1}{24-25} = \frac{1}{-1}$?

I thought that an inverse to this matrix was not possible, but my textbox and Wolfram Alpha says otherwise. Can someone tell me how this is possible, or if I have misunderstood the formula?

答:
You can see that this matrix is invertible in couple of ways.

For example:

$|A|=6\cdot4-5\cdot5=-1\neq0$

The rows of $A$ are linearly independent .

The columns of $A$ are linearly independent .

$0$ is not an eigenvalue of $A$ (you can calculate the eigenvalues
by finding the characteristic polynomial of $A$)

You have found a matrix $B$ s.t $AB=BA=I$


问:
Minimizing the Schatten 1-norm over symmetric matrices.
Let $X$ be an $n \times n$ Hermitian matrix, it follows that we can write $X=A+iB$ where $A$ is symmetric and $B$ is skew-symmetric. Let $S$ be the set of $n \times n$ symmetric matrices and define the Schatten $1$-norm (also known as trace norm) as $\lVert M \rVert_1=\sum_j \sigma_j(M)$, where $
\sigma_j(M)$ are the singular values of $M$. To measure how close $X$ is to the subspace of symmetric the matrices, we can define the distance measure
$$ D(X) = \min_{Y\in S} \lVert X-Y \rVert_1 .$$
Expanding $X$ into symmetric and skew-symmetric parts, we get that
$$D(X)= \min_{Y\in S} \lVert (A-Y)+iB \rVert_1$$
From this it seems intuitively that the minimum occurs when $Y=A$, in which case $D(X)=\lVert B\rVert_1$ is simply the $1$-norm of the imaginary part of $X$. How would one show that this quantity is minimized for $A=Y$?</p>

答:
If the distance is measured using the Frobenius norm then the problem is almost trivial.
(This post works through the exercise that Josse suggested at the end of his answer)

Use a real symmetric matrix $A=A^T$ and a real skew-symmetric matrix $B=-B^T$
to construct the hermitian matrix
$$\eqalign{
X &= A + iB \quad\implies\quad
X = X^H,\quad X^T = X^* \\
}$$

Use an unconstrained matrix $U\in{\mathbb C}^{n\times n}$ to construct the complex symmetric matrix
$$\eqalign{
Y &= {\rm Sym}(U) \doteq \tfrac 12(U+U^T) \quad\implies\quad
Y = Y^T,\quad Y^H = Y^* \\
}$$
Calculate the gradient of the objective with respect to $U$
$$\eqalign{
\phi &= \big\|X-Y\big\|_F^2 \;=\; (X-Y)^*&:(X-Y)  \\
d\phi &= (Y-X)^*:dY &\quad+\; conj \\
 &= (Y-X)^*:{\rm Sym}(dU) &\quad+\; conj \\
 &= {\rm Sym}(Y-X)^*:dU &\quad+\; conj \\
 &= (Y-\tfrac 12(X+X^*))^*:dU &\quad+\; conj \\
 &= (Y-{\cal Re}(X))^*:dU &\quad+\; conj \\
\frac{\partial\phi}{\partial U} &= (Y-{\cal Re}(X))^* \quad&
\frac{\partial\phi}{\partial U^*} = Y-{\cal Re}(X) \\
}$$
where "+ $conj\,$" denotes the complex conjugate of the first half of the expression.

Anyway, setting either gradient to zero yields
$$\eqalign{
Y &= {\cal Re}(X) = A \\
}$$

In the above, a colon is used to denote the trace product, i.e.
$$\eqalign{
A:B &= {\rm tr}(A^TB) = \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; B:A \\
X^*:X &= {\rm tr}(X^HX) \;\doteq\; \big\|X\big\|_F^2 \\
}$$
It is interesting that the result obtained using the Frobenius norm is apparently the same as a much more difficult derivation involving the Nuclear norm.


问:
Limit of the Recursive Sequence $X_1=1, X_{n+1}= \frac 1{1 + X_n}$
We have that
$$
X_1 = 1, \quad X_{n+1} = \frac 1{1 + X_n}.
$$
If $X_n$ has a limit find algebraically the value of the limit.

The sequence goes;

$$1,\frac{1}{2},\frac{3}{5},\frac{5}{8}, ...$$
The sequence appears to be converging to 0.6 as $n$ tends to infinity. May you have a say on this one.
Thanks

答:
Hint: Your recurrence is of the form $X(n+1) = f(X(n))$, where $f$ is the continuous function $f(x) = \frac 1{1 + x}$. It follows that if the sequence $X(n)$ approaches a limit $L$, then this limit $L$ must satisfy
$$
L = f(L).
$$
That is, $L$ must be a fixed point of the function.


问:
A question about an inequality
How to get the solution set of the inequality $$\left ( \frac{\pi}{2} \right )^{(x-1)^2}\leq \left ( \frac{2}{\pi} \right )^{x^2-5x-5}?$$</p>

答:
Hint : take logarithms on both sides. This can be done since the logarithm is a monotone function.

then use the property: $\log a^b=b\log a$

Title: A question about an inequality


答:
Hint:

If $X(n) \to L$, where $L$ is the limit in question, then obviously the same is true for $X(n+1)$. Thus in the limit

$$X(n+1) = \frac{1}{1+X(n)} \text{ becomes } L = \frac{1}{1+L}$$

Solve for $L$.

Warning:

You will find two possible values for $L$. Be mindful of which value makes sense based on the fact that $X(1)$. (Which value of $L$ does the sequence $X(n)$ converge to?)

Title: Limit of the Recursive Sequence $X_1=1, X_{n+1}= \frac 1{1 + X_n}$

Body:
We have that
$$
X_1 = 1, \quad X_{n+1} = \frac 1{1 + X_n}.
$$
If $X_n$ has a limit find algebraically the value of the limit.

The sequence goes;

$$1,\frac{1}{2},\frac{3}{5},\frac{5}{8}, ...$$
The sequence appears to be converging to 0.6 as $n$ tends to infinity. May you have a say on this one.
Thanks


问:
Under which condition arbitrary Union of ideals is an ideal?
I was wondering some condition for it as Union of two ideals is ideal if either of them is contained in other! Someone may help.</p>

答:
Let $\mathcal{C}:=\{I_k : k \in K\}$ be a set of ideals in a ring $R$. Put $I:= \bigcup_{k \in K} I_k$. The question is: "When is $I$ an ideal of $R$?"

Here is a sufficient condition, which is a good exercise:

$\mathcal{C}$ is a chain, that is for all $k,l \in K$ we have either $I_k \subseteq I_l$ and $I_l \subseteq I_k$.


答:
If you take a collection $\Sigma=\{\mathfrak{a}_i\}_{i\in I}$ of ideals in a unital ring $A$, then $\bigcup_{i\in I} \mathfrak{a}_i$ is still an ideal if $\Sigma$ is a directed set under $\subseteq$. That is, if for any pair of indices $i,j$, there exists an index $k$ so that $\mathfrak{a}_i\subseteq \mathfrak{a}_k$ and $\mathfrak{a}_j\subseteq \mathfrak{a}_k$.

In particular, a nested union of ideals is an ideal.

Title: Under which condition arbitrary Union of ideals is an ideal?

Body: I was wondering some condition for it as Union of two ideals is ideal if either of them is contained in other! Someone may help.</p>


问:
Find x, y coordinates of a curve by knowing only 3 points?
I was wondering if it's possible to find other coordinates of a curve if I know only 3 points? The points that I know are (1,5), (5.5, 25), (8,50) The curve is exponential I think. It's actually the formula for making a tea. The narrator of a book says that the bigger the container, the less ingredients I will need. In this case, ingredients is x and container size is y

What if I want to find x when the container size = 10 (x, 10)? Is there a formula for this?

答:
The general form of an exponential equation is $y = A + Be^{cx}$.

Given three coordinates that the curve must pass through, we plug in each coordinate into the equation.

This gives us a system of three equations with three unknowns, $A, B$ and $C$, which we should be able to solve and thus determine the specific values of $A, B$ and $C$, which in turn tells us the specific exponential function that our curve satisfies.


答:
The sequence converges to $\frac{1}{\varphi},$ where $\varphi$ is the golden ratio. Indeed, notice that $X_n=\frac{F_n}{F_{n+1}},$ with $F_n$ the $n$-th term in the Fibonacci sequence (with $F_{1}=F_2=1.$) Indeed, we can prove this by induction, as $X_{n+1}=\frac{1}{1+X_n}=\frac{1}{(F_{n+1}+F_n)/F_{n+1}}=\frac{F_{n+1}}{F_{n+1}+F_n}=\frac{F_{n+1}}{F_{n+2}}.$

It is a known fact that the ratio of two consecutive Fibonacci terms converges to the golden ratio, and in this case we have the inverse of that ratio.

P.D. your first terms are wrong. You are missing a 2/3 between 1/2 and 3/5.


问:
What does an outer automorphism look like?
I am working on a project in my group theory class to find an outer automorphism of $S_6$, which has already been addressed at length on this site and others. I have a prescription for how to go about finding this guy, but I have a larger conceptual question - what does an outer automorphism really look like? Is there an intuitive way to understand the difference between an inner and an outer automorphism? Inner automorphisms have always seemed easier for me to understand since we have an explicit representation $ghg^{-1}$ for members of the group. I can also understand this representation in terms of the Rubik's cube - rotate an edge, rotate a perpendicular edge, and the rotate the other edge back (is not the same as just rotating the perpendicular edge). What does an "outer automorphism" look like?</p>

答:
Another rather accessible example of an outer automorphism can be found on matrix groups: Look at the inverse transpose map
$$\operatorname{GL}(n,F) \to \operatorname {GL}(n,F), \quad A \mapsto (A^{-1})^\top.$$
It is an automorphism, and except from all cases with $n=1$ and $\operatorname{GL}(2,\mathbb F_2)$, it is is outer.</p>


问:
Classical number theoretic applications of the $p$-adic numbers
I am sure we can all agree that the $p$-adic numbers are highly fascinating objects in their own right - just as the closely related theory of valuations.

Having independently read up on the $p$-adic numbers for a few weeks now, I have so far only seen one application of them to what I would call classical number theory - namely the proof given in Serre's Cours d'arithmétique that a natural number is expressible as the sum of $\leq 3$ squares if and only if it is not of the form $4^a(8b-1)$ for some $a,b \in \mathbb{N}$.

Since I have a tendency to appreciate the value of the higher theories of mathematics in proportion to their applications to elementary number theory, I immediately found myself wondering if there are any other applications.

So my question to the community is: What are the most delightful applications of the $p$-adic numbers and the theory of valuations to elementary number theory?

Many thanks.

P.s.: I am aware that there are already several posts on the forum about the applications of the $p$-adic numbers, but none that refers to elementary number theory specifically.

Edit: I agree that I have been too vague in what I mean by "elementary number theory", so I will try to be a little more specific: By a classical "elementary" number theoretic proposition, I mean a number theoretic proposition that Fermat might have come up with. Thus, the above proposition about the sum of three squares is an elementary number theoretic proposition, as is e.g. Fermat's Last Theorem and the Twin Prime Conjecture, while e.g. the BSD Conjecture or the Class Number Problem are not.

Edit 2: Thank you for all the answers below - they are all excellent! In case anyone should come up with another one, I should like to say that bonus-points are given for results that have so far only been proven using the theory of $p$-adic numbers, or whose proof using $p$-adic numbers is far more conceptual and insightful than the original / more elementary one.

答:
One well known application of $p$-adic numbers is the Hasse-Minkowski theorem for quadratic forms.
This states that if $P$ is an irreducible quadratic form in any number of variables $n$, then the equation $P(x_1,\dots,x_n)=0$ has rational nonzero solutions if and only if it has nonzero solutions with coefficients in $\mathbb{R}$ and $\mathbb{Q}_p$ for every prime $p$.

This result is a very powerful tool in determining whether such an equation has any rational solutions, since if $n\geq 3$, the Chevalley-Warning theorem implies that the equation $P(x_1,\dots,x_n)=0$ has a nonzero solution modulo $p$ for every prime $p$. Combining this with Hensel's lemma, we see that the only primes that need to be checked are those for which $P$ is reducible modulo $p$.


问:
Derivative of $f(X)=X^2$ for $X \in \mathbb{R}^{n \times n}$
I am not looking for a solution, since this is a task from my assignment - rather a sanity check, since I spent way too much time thinking about this and I am thinking that there is possibly something wrong.  The task goes like this:

Prove that the function: $D f(X): \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n} $, $X \rightarrow X^2$ is differentiable and that the Jacobian matrix $Df(X)$ at $X \in \mathbb{R}^{n \times n}$ satisfies:
$$
Df(X)(H) = HX + XH \text{   for } H \in \mathbb{R}^{n \times n}
$$
Hint: By definition, $Df(X): \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}$ is given via $(Df(X)(H))_{i, j} = \sum_{k,l=1}^n \partial x_{k, l} f_{i, j}(X)H_{k, l}$ for $i, j \in \{1, ..., n\}$ and $H \in \mathbb{R}^{n \times n}$.

Now, here goes my doubt with the statement: How is it possible that the codomain of the derivative $Df(X)$ is $\mathbb{R}^{n \times n}$? I would expect the Jacobian to be a 4D tensor - for each output in the matrix $X^2$ we are looking at how the slight perturbations in the input cause it to change, yielding a tensor of size $\mathbb{R}^{n \times n \times n \times n}$.

Where am I wrong?

答: Your analysis is mostly right. But there's one crucial distinction: while $Df(X)$ is of rank $4$, $Df(X)(H)$ is of rank $4-2=2$ because $H$ is of rank $2$.


问:
How do I extract information about the standard deviation from the following variance matrix
Let $X$ = $\begin{pmatrix}X_{1} \\X_{2} \end{pmatrix}$ with

$$ X \sim N\left(\begin{pmatrix}5 \\ 2 \end{pmatrix},
\begin{pmatrix}
36&-30\\
-30&26
\end{pmatrix}
\right)$$

I have been asked to decompose $Var[X]$ into a product of a lower triangular matrix and it's transpose:
$$\mathrm{Var}[X] = LL^{T}$$

for which I obtained:

$$\mathrm{Var}[X] = \begin{pmatrix}
-6&0\\
5&1
\end{pmatrix}\begin{pmatrix}
-6&5\\
0&1
\end{pmatrix}$$

I am now being asked to write down the formula below and replace the *'s with appropriate numbers:

$$\begin{pmatrix}X_{1} \\X_{2} \end{pmatrix} = \begin{pmatrix}* \\* \end{pmatrix} + \begin{pmatrix}
*&*\\
*&*
\end{pmatrix}\begin{pmatrix}Z_{1} \\Z_{2} \end{pmatrix}$$

where $Z_{1}$ and $Z_{2}$ are two independent $N(0,1)$ random variables.

I understand where the formula is derived from as it is simply a vector and matrix representation of the Z-score standardisation formula given by:

$$Z_{1} = \frac{X_{1}-\mu_{1}}{\sigma_{1}}$$

and by rearranging, we get:

$$X_{1} = \mu_{1} + Z_{1}\sigma_{1}$$ and $$X_{2} = \mu_{2} + Z_{2}\sigma_{2}$$

Then, by using vector and matrix notation, we can re-write both random variables as:

$$\begin{pmatrix}X_{1} \\X_{2} \end{pmatrix} = \begin{pmatrix}5 \\2 \end{pmatrix} + \begin{pmatrix}
\sigma_{1,1}&\sigma_{1,2}\\
\sigma_{2,1}&\sigma_{2,2}
\end{pmatrix}\begin{pmatrix}Z_{1} \\Z_{2} \end{pmatrix}$$

I am struggling to find the values of $\sigma_{1,1}$, $\sigma_{1,2}$, $\sigma_{2,1}$, $\sigma_{2,2}$.

Are they the just the values of the conditional variance?

答:
The matrix you are looking for is $L$!

This is one of the problems where you are led to make a guess and check it. Let us guess that $X = \mu + LZ$ where $\mu = [\mu_1, \mu_2]^T$ and check that both the mean vector and the covariance matrix agree with the values stated at the top of your question. This is sufficient because a Gaussian distribution is completely determined by its mean and covariance matrix.

Note that the distribution of the $Z$ vector is

$$
\begin{pmatrix}
Z_1 \\
Z_2
\end{pmatrix} \sim \mathcal{N}\left(\begin{bmatrix}
0 \\
0
\end{bmatrix}, \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}\right)
$$

i.e. $\mathbb{E}(Z) = 0$ and $\mathrm{Var}(Z) = I$.

Start with the mean

$$
\mathbb{E}[X] = \mathbb{E}[\mu + LZ] = \mathbb{E}[\mu] + \mathbb{E}[LZ] = \mu + L \, \mathbb{E}[Z] = \mu + L \cdot 0 = \mu
$$

which agrees with what you have already found.

Next, the covariance matrix

$$
\begin{align}
\mathrm{Var}[X] &= \mathrm{Cov}(X, X) \\
&= \mathrm{Cov}(\mu + LZ, \mu + LZ) \\
&= \mathrm{Cov}(\mu + LZ, \mu) + \mathrm{Cov}(\mu + LZ, LZ) \\
&= \mathrm{Cov}(\mu, \mu) + \mathrm{Cov}(LZ, \mu) + \mathrm{Cov}(\mu, LZ) + \mathrm{Cov}(LZ, LZ) \\
&= 0 + 0 + 0 + \mathbb{E}[(LZ - 0) (LZ - 0)^T] \\
&= \mathbb{E}[LZZ^TL^T] \\
&= L \, \mathbb{E}[ZZ^T] \, L^T \\
&= L \, \mathrm{Var}[Z] \, L^T \\
&= L L^T.
\end{align}
$$

Thus, our guess that $X = \mu + LZ$ was correct and therefore

$$
\begin{pmatrix}X_{1} \\X_{2} \end{pmatrix} = \begin{pmatrix}5 \\2 \end{pmatrix} + \begin{pmatrix}
-6&0\\
5&1
\end{pmatrix}\begin{pmatrix}Z_{1} \\Z_{2} \end{pmatrix}.
$$

Remark 1: Your attempted solution is fairly close (in particular, it found the mean correctly), but it misses an important fact: that the two components of $X$ may depend on each other. The issue is that in your equations $X_1$ depends on $Z_1$ but not on $Z_2$ and similarly for $X_2$. It is easy to avoid such mistakes by writing equations in terms of vectors and matrices rather than components.

Remark 2: Knowledge provided by this exercise is of great practical value. It suggests a very simple algorithm to generate samples from multivariate Gaussian distribution with arbitrary mean and covariance.


答:
Hint:

$$\text{Since}\;\;0\neq a\implies a^{-1}=\frac1a\;,\;\;\text{ we get}:$$

$$\left(\frac\pi2\right)^{(x-1)^2}\le\left(\frac2\pi\right)^{x^2-5x-5}=\left(\frac\pi 2\right)^{-x^2+5x+5}\implies\ldots$$

Title: A question about an inequality

Body: How to get the solution set of the inequality $$\left ( \frac{\pi}{2} \right )^{(x-1)^2}\leq \left ( \frac{2}{\pi} \right )^{x^2-5x-5}?$$</p>


答: The derivative of $f$ at $X$ is not the $n \times n$ matrix $df(X)(H)$. Rather, the derivative is the linear map $df(X):\Bbb R^{n \times n} \to \Bbb R^{n \times n}$. If we interpret this object as a tensor array, then it is indeed 4-dimensional.

Title: Derivative of $f(X)=X^2$ for $X \in \mathbb{R}^{n \times n}$

Body:
I am not looking for a solution, since this is a task from my assignment - rather a sanity check, since I spent way too much time thinking about this and I am thinking that there is possibly something wrong.  The task goes like this:

Prove that the function: $D f(X): \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n} $, $X \rightarrow X^2$ is differentiable and that the Jacobian matrix $Df(X)$ at $X \in \mathbb{R}^{n \times n}$ satisfies:
$$
Df(X)(H) = HX + XH \text{   for } H \in \mathbb{R}^{n \times n}
$$
Hint: By definition, $Df(X): \mathbb{R}^{n \times n} \rightarrow \mathbb{R}^{n \times n}$ is given via $(Df(X)(H))_{i, j} = \sum_{k,l=1}^n \partial x_{k, l} f_{i, j}(X)H_{k, l}$ for $i, j \in \{1, ..., n\}$ and $H \in \mathbb{R}^{n \times n}$.

Now, here goes my doubt with the statement: How is it possible that the codomain of the derivative $Df(X)$ is $\mathbb{R}^{n \times n}$? I would expect the Jacobian to be a 4D tensor - for each output in the matrix $X^2$ we are looking at how the slight perturbations in the input cause it to change, yielding a tensor of size $\mathbb{R}^{n \times n \times n \times n}$.

Where am I wrong?


问:
Common first term and difference in 3 arithmetic series
Let $AP(a,d)$ denote the set of all the term of an infinite arithmetic progression with first term a and common difference d>0. If $AP(1,3)\ \cap AP(2,5)\ \cap AP(3,7)=AP(a,d)$ then find the value of a+d.

My approach

$1 + \left( {{n_1} - 1} \right)3 = X$

$2 + \left( {{n_2} - 1} \right)5 = X$

$3 + \left( {{n_3} - 1} \right)7 = X$

$1 + \left( {{n_1} - 1} \right)3 = 2 + \left( {{n_2} - 1} \right)5 = 3 + \left( {{n_3} - 1} \right)7 \Rightarrow 3{n_1} - 2 = 5{n_2} - 3 = 7{n_3} - 4 = X$

${n_1} = \frac{{X + 2}}{3};{n_2} = \frac{{X + 3}}{5};{n_3} = \frac{{X + 4}}{7}$

We need to find the value where $n_1, n_2,n_3$ are integer and through excel I found the first term as 52.

How do we calculate the first term a=52, d=LCM(3,5,7)=105

答: $n_1,n_2,n_3$ must be integers, which only happens when $$X=3p+1=5q+2=7r+3$$ for $p,q,r  \in \mathbb Z$. This means $$3p=5q+1=7r+2 $$ The easiest way to proceed from here is to consider $7r+2$,it needs to be a multiple of $3$ and one more than a multiple of $5$. Since $7r+2 \equiv r+2 \equiv 0 \pmod 3$, you need $r\equiv 1 \pmod 3$. Then it is quick and easy to check $r=1,4,7$ and $7$ is the smallest $r$ that meets all the conditions. So, $$a=X=7(7)+3 =52$$


问:
Moebius transformation in upper half plane model
I am studying Hyperbolic Geometry on my own. More especially, I am studying the Moebius transformation of the upper half plane model in hyperbolic geometry. But I got stuck studying it. I am looking for a solution of this problem. The problem is as follows:

Find the Moebius transformation  that rotates the hyperbolic plane about $i$ through an angle of $\frac { \pi}{4}$.

Please help me. Thanking in advanced.

答:
The stabilizer of $i$ in the upper half-plane consists of elliptic linear-fractional transformations $g_A(z)=\frac{az+b}{cz+d}$ corresponding to matrices
$$
A=\left[\begin{array}{cc}
a & b\\
c & d
\end{array}\right]= \left[\begin{array}{cc}
\cos(\phi) & \sin(\phi)\\
-\sin(\phi) & \cos(\phi)
\end{array}\right]
$$
(which makes it easy to remember: Euclidean rotations correspond to hyperbolic rotations). The angle of rotation $\phi$ of the matrix $A$ acting on ${\mathbb R}^2$ correspond to the angle of rotation $2\phi$ for the Moebius transformation $g_A$. (This is again easy to remember since for $A=-I$, rotation by $\pi$, $g_A=id_{\mathbb H^2}$, rotation by $2\pi$.)
The rest you can figure out on your own.


问:
Ultrametric completion does not introduce new norms
I could not solve this problem:

Prove that for a non-Archimedian field $K$ with completion $L$, $$\left\{|x|\in\mathbb R \mid x\in K\right\} =\left\{|x|\in\mathbb R \mid x\in L\right\}$$

I considered a Cauchy sequence in $K$ with norms having limit $l$, but I could not construct an element of $K$ with norm $l$ from the sequence.

Will anyone please show how to prove it?

答:
Here is another solution: Let $(K,| \ |)$ be any non-Archimdean normed field.  If $\alpha \in \mathbb{R}^{> 0}$, then the corona $\{x \in K \ | \ |x| = \alpha \}$ is open (possibly empty).  (The analogous thing holds in any ultrametric space: c.f. Exercise 2.12 in these notes.)  This is a counterintuitive fact even for those who have developed some good intuition for ultrametric spaces: a small part of me still feels instinctively that $|K^{\times}|$ should be discrete in order for this to hold, but it is quite straightforward to prove it in the general case.

Now, let $(\hat{K},| \ |)$ be the completion.  No additional calculations are necessary to observe that $K$ is dense in $\hat{K}$ -- that's the whole point of the completion of a metric space, after all.  But by the above paragraph, if the group of nonzero norms of the completion were any larger than the group of nonzero norms of $K$, there would be a nonempty, open corona in $\hat{K}$ without any $K$-valued points: contradiction.


答:
Hint: Denote $a=\frac{\pi}{2}$. then $\frac{2}{\pi}=\frac{1}{a}$

Recall $$\frac{1}{a^{\large{\text{expression}}}}=a^{-\large{\text{expression}}}$$

Title: A question about an inequality


问:
How is this related to the geometric law?
Assume we have $2^n$ possible outputs but only one of them is the right answer.
On average we need $2^n$ attempts to get the right answer.
Why is that and how is it related to the geometric law?

答:
If you have probability $p$ of success and you keep running independent trials and stop when you succeed, then the probability of succeeding after $n$ trials is the same as the probability of failing $n-1$ times in a row and then succeeding; that is, $$(1-p)^{n-1}p,$$ i.e. the number of required trials is geometrically distributed. Then the expected value of number of trials is

\begin{align*}
\sum_{n=1}^\infty n(1-p)^{n-1}p
\end{align*}

To see that this is $1/p$, we can apply an identity. In general we have

\begin{align*}
\sum_{n=1}^\infty nx^{n-1} &= \sum_{n=1}^\infty \frac{d}{dx}x^{n} \\
&= \frac{d}{dx} \sum_{n=0}^\infty x^n\\
&= \frac{d}{dx}1/(1-x) \\
&= \frac{1}{(1-x)^2}
\end{align*}

for $|x| < 1$. Letting $x = 1-p$,

\begin{align*}
\sum_{n=1}^\infty n(1-p)^{n-1}p &= p/(1 - (1-p))^2\\
 &= p/p^2\\
&= 1/p.
\end{align*}


问:
How to solve inequality equation with ceil operator product
I am trying to solve this inequality for $x$:

$$ \left \lceil{\frac{c_1}{x}}\right \rceil \cdot \left \lceil{\frac{c_2}{x}}\right \rceil \leq N $$

With: $c_1, c_2, x, N \in \mathbb{N}$ and $c_1, c_2, N$ being just constants.

I could make my system work with a solution assuming that $ x \leq c_1, c_2 $ but I would like ot have a general solution if someone can help me. The context behind this equation is how to split an image of $[H,W]=[c_1, c_2]$ in square patches of size $[H_{patch}, W_{patch}]=[x,x]$.

What I have tried yet:

Solve the equation for the lower bound of each inequality: I does not work.

$$\frac{c_1c_2}{x^2} \leq N \rightarrow \sqrt{\frac{c_1c_2}{N}} \leq x$$

Solve the equation for the upper bound: I am stuck getting x out of this expression.

$$ (\frac{c_1}{x^2}+1) \cdot (\frac{c_1}{x^2}+1) =\frac{c_1c_2}{x^2} + \frac{c_1}{x} + \frac{c_2}{x} + 1\leq N$$

Forgive my possible mistakes, I am not a mathematician. Thank you for your help in advance.

答:
Hint: first, let's consider the equation:
$$
\left\lceil \frac{c}{x} \right\rceil = k \text{,}
$$
where $c,k,x \in \mathbb{N}$. This is equivalent to the following inequality
$$
k-1 < \frac{c}{x} \leq k \text{.}
$$
If $k = 1$ then $x \geq c$. For $k>1$ we obtain
$$
\frac{c}{k} \leq x < \frac{c}{k-1}
$$
and since we're interested only in natural solutions, we have
$$
x \in
\left[ \tfrac{c}{k},\; \tfrac{c}{k-1} \right) \cap \mathbb{N} =
\left\lbrace
  \left\lceil \tfrac{c}{k} \right\rceil,\;
  \left\lceil \tfrac{c}{k} \right\rceil + 1,\;
  \ldots,\;
  \left\lceil \tfrac{c}{k-1} \right\rceil - 1
\right\rbrace \text{.}
$$

Now, to solve your problem you need to sum up the solutions of
$$
\left\lbrace
\begin{array}{l}
\left\lceil \frac{c_1}{x} \right\rceil = k_1 \\
\left\lceil \frac{c_2}{x} \right\rceil = k_2 \\
\end{array}
\right.
$$
for all pairs $k_1,k_2\in\mathbb{N}$ such that $k_1 \cdot k_2 \leq N$.


问:
Group Extensions and Couplings
In this post, I will be referencing this paper. Here is proposition 1.2 from said paper:

Proposition 1.2 --- Let $G \neq \{1\}$ be a group that decomposes as an extension:
$$1 \longrightarrow K \longrightarrow G \stackrel{\pi}{\longrightarrow} Q \longrightarrow 1$$
and $\Theta : Q \to \text{Out}(K)$ the associated coupling. A sufficient condition for $G$ to be ICC is (i) $FC_{G}(K) = \{1\}$, and (ii) $\Theta$ restricted to $FC(Q)$ is injective.

I am guessing that a coupling is a group homomorphism, but perhaps it satisfies further conditions. So I guess my first question is, what exactly is a coupling? Second, is this author saying that are taking a group $G$ which decomposes as an extension and comes with associated coupling, or do all group extensions have an associated coupling? If all group extensions come with a coupling, how does one define/construct it?

答:
Suppose that we have a group extension

$1 \to K \to G \xrightarrow{\pi} Q \to 1$

Wlog suppose that $K$ is a subgroup of $G$ to simplify notation.

To define a homomorphism $Q \to \mathrm{Out}(K)$, we can do the following: take $q \in Q$. Then take any lift $q'$ of $q$ under $\pi$. Because $K$ is normal in $G$, conjugation by $q'$ gives an automorphism of $K$. We claim that the image of this automorphism in $\mathrm{Out}(K)$ is independent from the choice of lift $q'$ so that this construction is well-defined. Suppose that $q''$ is another lift so that $\pi(q'')=\pi(q')$. Then we have $\pi(q''q'^{-1})=1$ so by exactness we have $a:=q''q'^{-1} \in K$ now if $C_g:K \to K$ denotes conjugation by $g \in G$, we have $C_{q''}(k)=q''kq''^{-1}=aq'kq'^{-1}a^{-1}=C_a(C_{q'}(k))$ for all $k \in K$ showing that the automorphisms $C_{q'}$ and $C_{q''}$ differ only by the inner automorphism $C_a$, thus the class in $\mathrm{Out}(K)$ is well-defined as claimed.

To see that this is actually a homomorphism just use that if $q'$ is a lift of $q \in Q$ and $p'$ is a lift of $p \in Q$, then $q'p'$ is a lift of $qp$.

Title: Group Extensions and Couplings


问:
Probabilistic geometric problem in high dimensions
In the $n$-dimensional Euclidean space, let $H$ be a random hyperplane selected in way similar to the so called "random radial point" (used to propose a solution for the Bertrand paradox): we select a point $\mathbf{z}$ uniformly at random on the unit $(n-1)$-sphere and we select uniformly at random a hyperplane $H=\{\mathbf{x}\in\mathbb{R}^n:\langle\mathbf{x},\mathbf{z}\rangle=b\}$, for a uniformly random $b\in[-1,1]$.

Question: What is the probability
$\Pr_H$
that $H$ separates point $\mathbf{x}:=(1, 0, 0, \ldots, 0)$ from point $\mathbf{y}=(-1, 0, 0, \ldots, 0)$?

I conjecture that $\Pr_H$ approaches $0$ as $n$ increases, but I do not know how to prove it. I am especially interested in how $\Pr_H$ varies as a function of $n$.

答:
Assume instead that $H$ is the hyperplane $x_1=b$, where $b$ is uniformly distributed on $[0,1]$ (as Misha Lavrov has proposed), and that $x=(x_1,\ldots,x_n)$ is uniformly distributed on the unit sphere $S^{n-1}$. We have a success iff $|x_1|>b$. This means that we have to begin with calculating the portion of the area of $S^{n-1}$ corresponding to $|x_1|\in[b,1]$, then integrate over $b\in[0,1]$.

In fact you don't need to compute this area. It is sufficient to have a formula for the density $f_{X_1}(x_1)$ $(-1\leq x_1\leq 1)$ of the first coordinate under the assumed distribution. This formula is around, but I have not found a quotable source.


问:
Are there any properties that distinguish the decimal numeral system?
Since early childhood we are used to work with the decimal numbers.
Living in the world, where everything is written is terms of decimals people feel most convenient, when performing the calcuations in this system, wherears when dealing with binary, hex, and any other kind of data we usually at first transform and process them as decimals.

However, for purely mathematical point of view there seems nothing to be special about the decimal system. $$10 = 2 * 5$$
The product of the first and the third prime number. Other base, like the binary, looks more natural, writing everything in terms of sum of powers of the smallest prime number.

However, writing the numbers with this base is not very convenient, due to the fact, that even moderately small numbers, like $73$ would require $7$ digits to be written.

The decimal system in this sense, is a compromise, that one doesn't have to remember a lot of symbols, and the numbers, appearing in most real-life applications can be written rather compactly. From the point of view of humans, the historical reason, why this system became dominant is most likely due to the number of fingers on both hands (for normal people). And the hands were often used for representing small numbers.

My question is - do there exist some other properties, or remarkable features, which distinguish the decimal system?

答:
If you're a fan of divisibility tests and wanted a good base, you'd want it divisible by lots of small primes. You wouldn't want to waste size by using a prime more than once.  But even the product of the first $3$ primes is $30,$ which is a bit large.  We can barely get school children to learn the $55$ multiplication facts in base $10.$ (Even though a third of them involve $0$ or $1$.)   I'd hate to think of the size of the flashcard deck if we went base $30.$

So let's just pick two of the primes.  Use base $6$, $10$ or $15$.
In each case, think about how you'd test for divisibility of the missing prime.  In base $15,$ divisibility by $2$ is not obvious, and that seems a serious flaw.   In base $6,$ if you want to test for divisibility by $5$, you'd have the "sum of digits" trick that works for $9$ in base $10$, so that might not be a bad choice.  (And unlike base $10$, you get an easy test for divisibility by $7$.)

Oh, but in base $10$, checking for $3$ gets you $9$ for free, so it wins the "2 primes" contest.  Base 10 gives quick divisibility tests for everything up to $12$, except for $7$.   Hard to beat.


问:
Derivative of Schur function
In his answer to https://mathoverflow.net/questions/129854, R. Stanley says that the partial derivative (over the relevant x[i]) of the Schur function of a partition lambda of n equals the sum the Schur functions over the trimmed partions of lambda:

"Let $s_λ$ denote a Schur function and $p1=s1=x1+x2+\dots$. Then, $\frac{\partial\lambda}{\partial p_1}=\sum_\mu s_\mu$, where the $μ$'s are obtained by removing a single box from $λ$."

If I work this out for $\lambda = \{3,1\}$, I find for the derivative
$12 X[3]+ 108 X[2,1]+ 60 X[1,1,1]$ (substituting $X[i,j,k]$ for terms $x[1]^i*x[2]^j*x[3]^k$) while the trimmed partitions $\{2,1\}$ and $\{3\}$ produce
$4 X[3]+ 24 X[2,1] + 12 X[1,1,1]$.

What am I missing? Can anyone provide an example of what is intended?

答:
Stanley uses a cryptic description whose interpretation is easy for the initiated, but not for others. Remember that the ring of symmetric functions is equipped with an inner product for which the Schur functions form an orthonormal basis (you can take that as its definition, although there are also other descriptions). Now the operation written $\frac\partial{\partial p_1}$ is just the adjoint (in the sense of linear mappings) of the operation of multiplication by $p_1=s_1=\sum_i x_i$. Since that multiplication applied to $s_\lambda$ gives the sum of all $s_\mu$ where the diagram of$~\mu$ can be obtained by adding a square to that of$~\lambda$, its adjoint applied to $s_\lambda$ gives the sum of all $s_\mu$ where the diagram of$~\mu$ can be obtained by removing a square from that of$~\lambda$.

A way to view this operation as applied to any symmetric function $f$ is to single out one of the variables in which $f$ is  a symmetric function (since it is symmetric it doesn't matter which one), call that variable $y$ and the others $x_1,x_2,\ldots$; observing that $f$ is certainly symmetric in the $x_1,x_2,\ldots$, write it as a polynomial in $y$ with as coefficients symmetric functions in the $x_1,x_2,\ldots$, and then take the coefficient of the first power $y^1$ of$~y$  as outcome (we've used up one variable, but given the infinite supply this makes no difference; this is the Hilbert hotel in operation). In practice one doesn't need infinitely many variables: one can write $f$ as symmetric polynomial in variables $x_1,\ldots,x_n$ where $n$ is at least $\deg f$, and choose $y=x_n$ as "ultimus inter pares" which avoids renumbering (the degree conveniently drops by $1$, so $x_n$ is expendable).

If one takes the description of $s_\lambda(x_1,\ldots,x_n)$ in terms of semistandard tableaux, one can partition the set of its terms according to the set of squares containing an entry$~n$ in the tableau. The parts of this partition of terms each define a product $s_\mu(x_1,\ldots,x_{n-1})x_n^d$, where $d$ is the number of squares in the fixed set containing an entry$~n$, and the diagram of $\mu$ consists of the remaining squares. It follows that the coefficient of $x_n^d$ is the sum of all such $s_\mu(x_1,\ldots,x_{n-1})$, where the diagram of$~\mu$ is obtained from that of$~\lambda$ by removing a horizontal strip of size$~d$. In particular, our operator maps $s_\lambda$ to the sum of all $s_\mu$ whose diagram is obtained from that of $\lambda$ by removing one corner.

To see that the operator coincides with formal differentiation with respect to $p_1$ when applied to a polynomial in $p_1,p_2,\ldots$, interpreted as power sum symmetric functions, one can either prove that it is a derviation which vanishes on $p_2,p_3,\ldots$ and sends $p_1\mapsto1$, or one can directly apply our description to a product $p_1^mf$ where $f$ is a factor on which the operator vanishes (a monomial in $p_2,p_3,\ldots$), using $p_1(x_1,\ldots,x_n)=p_1(x_1,\ldots,x_{n-1})+x_n$ and expansion of $(p_1(x_1,\ldots,x_{n-1})+x_n)^m$, giving as result of the operation $mp_1^{m-1}f$.

Title: Derivative of Schur function


问:
Point of circles : Show that $SC$ is parallel to $AA'$
Two circles $M_1, M_2$ with centers $A$ and $B$ respectively intersect outside at a point $C$. .

Let $A'$ be a point on $M_1$ and $B'$ be a point on $M_2$ such that $A'$ and $B'$ are on the same side of the line $AB$ such that $AA'$ and $BB'$ are parallel.

Let $S$ be the intersection point of $AB'$ and $BA'$.

Show that $SC$ is parallel to $AA'$ and calculate the distance from $S$ to $C$ as a function of the two radius.

$$$$

Do we have to show that $\angle A'AS=\angle ASC=\angle AB'B$ to get that $SC$ is parallel to $AA'$ ?

答:
The key is to recognise the similar triangles, and that $ACB$ is a straight line.

Notice that since $AA' // BB'$, we have $\triangle AA' S \sim \triangle B'BS$.

This gives $\dfrac {AS}{SB'} = \dfrac {A'S}{SB} = \dfrac {AA'}{BB'} = \dfrac {AC}{CB}$ (radii).

By "ratio of 2 sides, included angle" (also called the SAS similarity criterion), we see that $\triangle BCS \sim \triangle BAA'$. This gives $CS// AA'$ as desired. Now the length of $CS$ follows from similar triangles.


问:
$P(\text{duck})$ given $P(\text{look})$ and $P(\text{swim})$ and $P(\text{quack})$

"If it looks like a duck, swims like a duck, and quacks like a duck,
then it probably is a duck"

We have the 3 probabilities:

  $P(\text{look}) = 80\%$

  $P(\text{swim}) = 70\%$

  $P(\text{quack}) = 90\%$

I would assume then that
$P(\text{duck}) = P(\text{look})·P(\text{swim})·P(\text{quack})$ $= 0.8 · 0.7 · 0.9$ $= 0.504$

This is obviously dead wrong. Common sense expects the resulted probability close to $1$. What would be right operation among the individual probabilities to represent the original saying correctly?

Update:

Giving some context: We have a problem in an industrial environment, where we have to validate if our calculated values are true or not. We can conduct a few independent measurements, each addresses a different aspect of the same setup and each measurment returns a yes or no with a probability. Then we have to summarize them and compare to the theoretical calculations. What we do is duck typing.

Update 2:

While reading up the answers did hit me: the more independent measurement events we do on properties of the duck, should prove or disprove the "duckness" of the object.
Should we have 6 independent measurements (look, swim, quack, fly, eat, walk), each around probability of 0.1, our calculation should conclude that this is certainly not a duck. Should we get confirming measurements (higher than 0.5), I expect higher "duckness" probability.

答:
Assume look, swim and quack are independent, then the required probability is

$P(L\cup S\cup Q) =P(L)+P(S)+P(Q)-P(L\cap S)-P(S\cap Q)-P(L\cap Q)+P(L\cap S\cap Q)=0.994$
$P(S\cap Q) = P(S)\times P(Q)$ and so are others


问:
Distance between highest and lowest point on rotated square
I have a square and I want to know the vertical distance between the highest point and the lowest point. The square is $10 \times 10$ cm.
The amount of degrees the square is rotated can be anything, but I need a formula that always works no matter the amount of degrees it is rotated.

Does anyone know a formula to calculate this?

So in this picture I would need to know the length of the red line:

答: You need to know the angle $\theta$ that the square has been rotated relative to the horizontal. A small amount of trig then shows that the distance between the two lines is $10( \sin \theta +\cos \theta)$. The smallest the distance can be occurs when $\theta=0$, and is just 10, and the largest value occurs when $\theta = 45^\circ$ at which point it is $10\sqrt2$.


答:
Suppose that the square is rotated by an angle of $\theta$. You can draw a square $B$ that just touches the vertex of your original square, as in here ($\theta$ would be angle $STY$):

Thus, the height you are looking is the sum of $XS$ and $SY$, which are equal to the shortest sides of triangle STY. Thus, the height equals $(\cos(\theta)+\sin(\theta))L,$ where $L$ is the side length of the original square (in your case, 10 cm).

Title: Distance between highest and lowest point on rotated square

Body:
I have a square and I want to know the vertical distance between the highest point and the lowest point. The square is $10 \times 10$ cm.
The amount of degrees the square is rotated can be anything, but I need a formula that always works no matter the amount of degrees it is rotated.

Does anyone know a formula to calculate this?

So in this picture I would need to know the length of the red line:


问:
$\sum p\;$ over consecutive primes $\;p \equiv 3 \pmod 4\;$ is never a square?
Prove or disprove that $\;\sum p\;$ over consecutive primes $\;p \equiv 3 \pmod 4$, starting from $\,p=3$, is never a square.

I think it would be useful to examine the modular equation $\;\sum p \equiv x^2 \pmod 4$.

Many thanks for your help.

答:
I conjecture that the sum is a square (rarely but) infinitely often. (Moreover, this has nothing to do with the precise sequence being summed—the conjecture would be the same for any natural sequence growing at the same rate.)

The sum of the first $n$ primes that are congruent to $3$ (mod $4$) is asymptotic to $n^2\log n$. When adding the next such prime, which is about $2n\log n$ in size, we will jump over approximately $\sqrt{\log n}$ squares, which means that the "probability" of landing on a square should be the same as the probability that a randomly chosen number of size about $n^2\log n$ is a square, which is $\frac1{n\sqrt{\log n}}$. Since $\sum \frac1{n\sqrt{\log n}}$ diverges (at a rate of about $2\sqrt{\log n}$, by the integral test), this suggests that there should be infinitely many examples.


问:
Natural parameterization of a curve
I need the natural parameterization or a geodesic curve contained in the surface $z=x^2+y^2$, that goes through the origin, with x(s=0)=0, y(s=0)=0, dx/ds (s=0)=cos(a) and dy/ds(s=0)=sin(a), with "a" constant, expressed as a function of the arc length, i.e., I need $r(s)=r(x(s),y(s))$.

Thank you very much!

Eva

答:
So you have $$ s(r) = \int_0^r\sqrt{1+4t^2}\,dt$$
set $t = \frac 12\sinh u,\ dt = \frac12\cosh u\,du$
$$s(r) = \frac12\int_0^{\sinh^{-1}2r}\cosh^2u\,du\\=\frac14\int_0^{\sinh^{-1}2r}1 + \cosh 2u\,du\\=
\frac{\sinh^{-1}2r}{4} +\left.\frac18\sinh 2u\right|_0^{\sinh^{-1}2r} \\= \frac{\sinh^{-1}2r}{4} + \frac14\sinh(\sinh^{-1}2r)\cosh(\sinh^{-1}2r)\\=\frac{\sinh^{-1}2r}{4} + 2r\sqrt{1+4r^2}\\
=\frac14\ln(2r + \sqrt{1+4r^2})+ 2r\sqrt{1+4r^2}$$
I agree that finding a nice expression for $r(s)$ from inverting the above fomula does not look possible.

But consider this. Suppose you find some other parametrization in terms of some $t$, so you know $(x(t), y(t), z(t))$ and can solve $$s(t) = \int \sqrt{\dot x^2 + \dot y^2 + \dot z^2}\,dt$$ and invert the result to get $t(s)$. But since the curve is on the surface, $z = r^2$, so $$r(s) = \sqrt{z(t(s))}$$
which by presumption are all known functions. That is, to reparametrize so as to have a nicely invertible arclength function *is to invert the function of $r$ above. If you can do one, you can do the other.

I rather doubt that $r(s)$ has an expression in terms of elementary functions, even though $s(r)$ does. I could be wrong,but generally when you find $r$ occurring both inside a transcendental function and outside it, the inverse is not going to elementary. And if that is the case here, then no other parametrization will give you a better result. They can be easy to determine with respect to $r$ or with respect to $s$, but not with respect to both.

But all is not lost. While actually finding a nice expression for the inverse may not be possible, $s'(r) > 0$, so the function is $1-1$ and therefore invertible. You can define $$f(r) =\frac{\sinh^{-1}2r}{4} + 2r\sqrt{1+4r^2}$$ and then $$r = f^{-1}(s)$$.
This is not just playing with notation. For specific $s$, there are numerous techniques for calculating $f^{-1}(s)$ to any desired accuracy, including developing Taylor series for it. And this is no worse than calculating most functions elementary expressions.

"$r = f^{-1}(s)$" may not be the solution you wanted, but it is still a solution.

Title: Natural parameterization of a curve


问:
Help with the gradient in different co-ordinate systems
Let $L(x,y)$ be the linear Taylor series expansion of some function $f(x,y)$. This can be written as $$L(x,y)=f(x_0,y_0)+f_x(x-x_0)+f_y(y-y_0)$$ Or in more compact form as $$L(x,y)=f(x_0,y_0)+\nabla f ^T \vec{p}$$
where
$$
\vec{p}=
\begin{bmatrix}
x-x_0\\
y-y_0 \\
\end{bmatrix}
$$
and $$\nabla f(x,y)=f_x \vec{i}+f_y\vec{j}$$
Now, we find out that $x'= \alpha_1 x$ and $y'= \alpha_2 y$, which is the linear transformation $A$ of the original co-ordinates and we want to find the Taylor series expansion in terms of $x' \text{ and } y'$, i.e. the new co-ordinate system. Writing the composite function $$g(x',y')=f(x(x'),y(x'))$$
$$L(x',y')=g(x'_0,y'_0)+\frac{\partial f}{\partial x}\frac{\partial x}{\partial x'}(x'-x'_0)+\frac{\partial f}{\partial y}\frac{\partial y}{\partial y'}(y'-y'_0)$$
$\implies$.
$$L(x',y')=f(x_0,y_0)+\frac{1}{\alpha_1}\frac{\partial f}{\partial x}(x'-x'_0)+\frac{1}{\alpha_2}\frac{\partial f}{\partial y}(y'-y'_0)$$
$\implies$
$$L(x',y')=f(x_0,y_0)+\vec{p}'^TA^{-T}\nabla f(x_0,y_0)$$
This function is describing exactly the same function but in the new stretched Cartesian co-ordinate system. By inspection the gradient in terms of the new co-ordinate system is $$\nabla g(x_0',y_0') = A^{-T}\nabla f(x_0,y_0)=g_{x'} \vec{i}'+g_{y'} \vec{j}'= \frac{1}{\alpha_1}f_x \vec{i}' + \frac{1}{\alpha_2}f_y \vec{j}'$$
Now because the Cartesian co-ordinate system has been stretched the basis vectors have changed, such that now $$\vec{i}'=\frac{\vec{i}}{\alpha_1}$$ and $$\vec{j}'=\frac{\vec{j}}{\alpha_2}$$
Substitution of these into $\nabla g(x_0',y_0')$ reveals that $$\nabla g(x_0',y_0')= \frac{1}{\alpha_1^2}f_x \vec{i} + \frac{1}{\alpha_2^2}f_y \vec{j}$$ which is different to the direction in the standard Cartesian co-ordinate system. On my travels to understand this I discovered Pavel Grinfeld's video series on Tensor algebra, where he shows that to make the gradient invariant under transformation I should use the definition $$\nabla f(x,y)=\frac{f_x}{\vec{i}\cdot\vec{i}}\vec{i}+\frac{f_x}{\vec{j}\cdot\vec{j}}\vec{j}$$
My background is in EEE, and frequently use steepest decent to solve numeric problems, I had never considered the transformation until I looked into conditioning the problem. I'm having a hard time understanding why the gradient derived from the transformed system is different to the original one in the first place. If anyone can shed any light on using the basis vectors $\frac{\vec{i}}{\vec{i}\cdot\vec{i}}$ and $\frac{\vec{j}}{\vec{j}\cdot\vec{j}}$ I would be very grateful. There seems to be a contradiction in here, or a subtle detail I a missing. I have seen several posts of people just saying the gradient is variant under transformation and to use the direction in the transformed co-ordinate system. My issue is, there shouldn't be any variation in the gradient, as per the geometric definition, it points in the direction of greatest increase and this doesn't change simply because co-ordinates are transformed.

Example
$$f(x,y)=2x+2y$$
$$\nabla f(x,y)=2\vec{i}+2\vec{j}$$
let $x'=2x$ and $y'=2y$
$$g(x',y')=x'+y'$$
$$\nabla g(x',y')=1\vec{i'}+1\vec{j'}$$
$\vec{i}=2 \vec{i}'$ and $\vec{j}=2 \vec{j}'$
and so this gradient in terms of the old basis vectors but in the new co-ordinate system is
$$\nabla g(x',y')=0.5\vec{i}+0.5\vec{j}$$

so $\nabla g(x',y') \neq \nabla f(x,x) $ but $g(x',y') = f(x(x'),y(y')) $

[Gradient in two co-ordinate systems][1]

I don't understand how the following would help, can anyone help me consolidate this?
$$\nabla g(x',y')=\frac{1\vec{i'}}{\vec{i'} \cdot \vec{i'}}+\frac{1\vec{j'}}{\vec{j'} \cdot \vec{j'}}$$
[1]: https://i.stack.imgur.com/PArGk.png

EDIT:

Just to add, I found this thread on the same subject Definition of the gradient for non-Cartesian coordinates. They define the gradient geometrically as the directional derivative in the unit direction of steepest ascent, dotted with the unit direction of steepest ascent. That is: $$\nabla f = d(\vec{v}_{max})\vec{v}_{max}$$ Why am I getting two different gradients in two different co-ordinate systems? Why do these: http://www.stat.cmu.edu/~ryantibs/convexopt-S15/scribes/14-newton-scribed.pdf lecture notes describe the gradient decent method as affine variant? I understand that the sensitivities will be different because different co-ordinate systems are used, but the basis vectors should compensate for this such that the gradient points in the same direction regardless of the system used.

答:
The discrepancy in your argument:

Your mistake was writing $\nabla g = g_{x'} i' + g_{y'} j'$. It is not true that the gradient has components $\left( \frac{\partial f}{\partial u}, \; \frac{\partial f}{\partial v} \right)$ in every coordinate system $(u,v)$. The gradient is only defined this way for the standard $(x,y)$ coordinates.

The differential:

On the other hand, the more well-behaved object is the differential $df$ (which is a $1$-form, not a vector field):
$$ df = \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy $$
We can change to new coordinates $(u,v)$, and replace $dx = x_u du + x_v dv$, and similarly for $y$:
$$ df = (f_x x_u + f_y y_u) \, du + (f_x x_v + f_y y_v) \, dv $$
By the chain rule, $f_u = f_x x_u + f_y y_u$, so actually this is the same as
$ df = f_u \, du + f_v \, dv$.
And so $df$ does look the same in every coordinate system (unlike the gradient).

What is the relation?

Differential $1$-forms "eat" vectors: for a vector $v = (a,b)$, in ordinary $(x,y)$ coordinates,
$$ df(v) = a f_x + b f_y $$
In more abstract differential geometry and Riemannian geometry, the gradient is defined as the vector $\nabla f$ with the property that for any vector $v$:
$$ df(v) = \left< \nabla f, \; v \right> $$
where $\left<-,-\right>$ is the usual dot product. In usual $(x,y)$ coordinates, the dot product of vectors looks like
$$ \left< (a,b), (c,d) \right> = ac + bd $$
The discrepancy that you are seeing is because the dot product does not look like this in every coordinate system. In a generic coordinate system $(u,v)$, the dot product of vectors (written in the $i', j'$ basis) looks like
$$ \left< (a,b), \, (c,d)\right> = (x_u^2+y_u^2) ac + (x_ux_v +y_uy_v)(bc + ad) + (x_v^2+y_v^2)bd $$
So although $df$ always looks the same, $\nabla f$ will not.

Your Example:

In your example, where $u= x' = \alpha_1 x$ and $v = y' = \alpha_2 y$, the dot product of vectors in the $(x',y')$ coordinates (in your notation, using the $i',j'$ basis) looks like
$$ \left< (a,b), \, (c,d) \right> = \frac{ac}{\alpha_1^2} + \frac{bd}{\alpha_2^2} $$
So in these $(x',y')$ coordinates, the gradient $\nabla f = (p,q)$ is defined by the equation $df(v) = \left< \nabla f, \, v \right>$, which translates to
$$ a f_{x'} + b f_{y'} = \frac{ap}{\alpha_1^2} + \frac{bq}{\alpha_2^2} $$
Plug in $v=(1,0)$ and $v=(0,1)$ to quickly see that
$$ \nabla f = (\alpha_1^2 f_{x'}, \; \alpha_2^2 f_{y'}) $$


问:
A question about an inequality
How to get the solution set of the inequality $$\left ( \frac{\pi}{2} \right )^{(x-1)^2}\leq \left ( \frac{2}{\pi} \right )^{x^2-5x-5}?$$</p>

答:
Multiplying either sides by $\left(\frac\pi2\right)^{(x^2-5x-5)}$

$$\left(\frac\pi2\right)^{(x-1)^2+(x^2-5x-5)}\le 1\iff \left(\frac\pi2\right)^{2x^2-7x-4}\le 1=\left(\frac\pi2\right)^0$$

$\implies 2x^2-7x-4\le0$ as $\pi>2\iff \frac\pi2>1,$

Now, if $(x-a)(x-b)\le0$ where $a\le b,$   $a\le x\le b$

Title: A question about an inequality


问:
Do first-order dynamical systems have any periodic solutions?
From Strogatz's Nonlinear Dynamics and Chaos (2nd ed), Section 2.6:

If a fixed point is regarded as an equilibrium solution, the approach
to equilibrium is always monotonic — overshoot and damped oscillations
can never occur in a first-order system. For the same reason, undamped
oscillations are impossible. Hence there are no periodic solutions to
$\dot{x} = f ( x )$.

I was tinkering around with a logistic map tool that I found here and observed this plot for the values: $x_0 = 0.2, r = 2.7$.

Is this an example of an overshoot / transient oscillations? Am I misunderstanding something?  Why do we observe these transient oscillations at the beginning?

答: Discrete dynamical systems and continuous dynamical systems behave wildly differently. In particular, there are vastly more possible behaviors of discrete dynamical systems in low dimensions than there are for continuous dynamical systems. Prof. Strogatz is referring to the continuous case. The logistic map can produce oscillations of every period, and has periods of chaos. By contrast, chaos is not a phenomenon that's possible for continuous systems until you get to 3 dimensions.</p>


问:
Can we write $\sin(\frac{\pi}{14})$ as an finite expression using only basic operations?
First, let us recall some trigonometric values

$\sin(0)=0$

$\sin(\pi/6)=\frac{1}{2}$

$\sin(\pi/3)=\frac{\sqrt{3}}{2}$

$\sin(\pi/2)=1$

$\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$

$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$

Here, we can observe that for some values of $\theta$, $\sin(\theta)$ can be expressed as a sum of finite radical terms. It is also easy to see that, there exists infinite such $\theta$'s. Say, I have given $\sin(\pi/14)$, can we determine whether it can also be expressed as sum of finite radical terms? What about the general case? Are there any criteria that $\theta$ must obey in order to do that?

答:
$8x^3−4x^2−4x+1=0$

Where does this polynomial come from?

Suppose you take a regular polynomial and plot the coordinates.

Consider a n-gon, with one vertex on $(1,0)$ all the vertices one unit from the center, and the center at $(0,0).$

$(1,0)\\
(\cos \frac {2\pi}{n},\sin \frac {2\pi}{n})\\
(\cos \frac {4\pi}{n},\sin \frac {4\pi}{n})\\
\cdots\\
(\cos 2(n-1)\pi,\sin 2(n-1)\pi)$

If we average, or sum, all of these coordinates we get the center of the circle.

$1 + \cos\frac {2\pi}{n} + \cdots + \cos \frac {2(n-1)\pi}{n} = 0$

and similarly

$1 + \sin\frac {2\pi}{n} + \cdots + \sin {2(n-1)\pi}{n} = 0$

We could rotate the polygon to make one vertex associate with the angle $\frac {\pi}{14},$ but that isn't what I did.  Instead, I kept the same orientation and said:

$\sin \frac {\pi}{14} = \cos (\frac {\pi}{2}-\frac {\pi}{14}) = -\cos (\frac {\pi}{2}+\frac {\pi}{14}) = -\cos \frac {4\pi}{7}$

Working with:

$1 + \cos\frac {2\pi}{7} + \cos\frac {4\pi}{7} + \cdots + \cos \frac {12\pi}{7} = 0$

This polygon has symmetry about the x-axis.

$\cos \frac {2\pi}{7} = \cos \frac {12\pi}{7}\\
\cos \frac {4\pi}{7} = \cos \frac {10\pi}{7}\\
\cos \frac {6\pi}{7} = \cos \frac {8\pi}{7}$

$1 + 2\cos\frac {2\pi}{7} + 2\cos\frac {4\pi}{7} + 2\cos\frac {6\pi}{7}=0$

Now we use multiple-angle identities.

$1 + 2\cos\frac {2\pi}{7} + 2(2\cos^2\frac {2\pi}{7}-1) + 2(4\cos^3\frac {2\pi}{7} - 3\cos\frac {2\pi}{7})$

Say $\cos\frac {2\pi}{7} = x$

$8x^3 +4x^2 - 4x - 1 = 0$

This polynomial has 3 roots.  One equals $\cos\frac {2\pi}{7}$, another equals $\cos\frac {4\pi}{7}$ and the third equals $\cos\frac {6\pi}{7}$

But we wanted the negative of one of those roots.  Then replace $x$ with $-x$ in the equation above, remembering that $(-x)^n = x^n$ when $n$ is even and $-x^n$ when $n$ is odd.

$-8x^3 +4x^2 + 4x - 1 = 0\\
8x^3 -4x^2 - 4x + 1 = 0$

$\sin \frac {\pi}{14}$ is a root of $8x^3 -4x^2 - 4x + 1 = 0$


问:
finding the functions satisfying $f(xy)=f(x)f(y)-f(x+y)+1$
Define a function $f\colon\mathbb{R}\to\mathbb{R}$ which is continuous and satisfies

$$f(xy)=f(x)f(y)-f(x+y)+1$$
for all $x, y \in\mathbb Q$. With a supp condition $f(1)=2$. (I didn't notice that.)

How to show that $f(x)=x+1$ for all $x$ that belong to $\mathbb{R}$?i got the ans from Paul that it is true for all rationals x but I still cannot show that for $x, y \in\mathbb R$ is correct.

答:
You have $f(1)=2$ so $f(x)=2f(x)-f(x+1)+1\iff f(x)=f(x+1)-1$ which implies $f(0)=1$ and $f(-1)=0$ so the function has three collinear points $(-1,0),(0,1)$ and $(1,2)$.

The equation of the line passing through these three points is $y=x+1$. It follows

$$\begin{cases}f(x)=x+1\\f(y)=y+1\\f(xy)=xy+1\\f(x+y)=x+y+1\end{cases}\Rightarrow f(xy)=f(x)f(y)-f(x+y)+1$$
All the conditions are verified.


答:
If our probabilities are independent (arguable).

Then suppose we have a object that looks, swims and quacks like a duck.

What is the probability that this object is not a duck?

$(1-0.8)(1-0.7)(1-0.9) = 0.006$

Which means that the probability it is a duck is $1-0.006 = 0.994$

Title: $P(\text{duck})$ given $P(\text{look})$ and $P(\text{swim})$ and $P(\text{quack})$

Body:
"If it looks like a duck, swims like a duck, and quacks like a duck,
then it probably is a duck"

We have the 3 probabilities:

  $P(\text{look}) = 80\%$

  $P(\text{swim}) = 70\%$

  $P(\text{quack}) = 90\%$

I would assume then that
$P(\text{duck}) = P(\text{look})·P(\text{swim})·P(\text{quack})$ $= 0.8 · 0.7 · 0.9$ $= 0.504$

This is obviously dead wrong. Common sense expects the resulted probability close to $1$. What would be right operation among the individual probabilities to represent the original saying correctly?

Update:

Giving some context: We have a problem in an industrial environment, where we have to validate if our calculated values are true or not. We can conduct a few independent measurements, each addresses a different aspect of the same setup and each measurment returns a yes or no with a probability. Then we have to summarize them and compare to the theoretical calculations. What we do is duck typing.

Update 2:

While reading up the answers did hit me: the more independent measurement events we do on properties of the duck, should prove or disprove the "duckness" of the object.
Should we have 6 independent measurements (look, swim, quack, fly, eat, walk), each around probability of 0.1, our calculation should conclude that this is certainly not a duck. Should we get confirming measurements (higher than 0.5), I expect higher "duckness" probability.


问:
Which is true $A$ is subset of $B$ or $B$ is subset of $A$.

Consider the sets dened by the real solutions of the inequalities $$A=\{(x,y):x^2+y^4\le 1\}$$ and $$B=\{(x,y):x^4+y^6\le 1\}$$Then which is true $A$ is subset of $B$ or $B$ is subset of $A$.

This is Question No. $10.$ of this paper. Please help me.Thanks in advance.

答:
The set A looks more like circle, the B looks more like a squre... and the almost-circle lies inside of almost-square. try that on your own...

A may be expressed as $y\in (-(1-x^2)^\frac{1}{4},(1-x^2)^\frac{1}{4})$

B may be expressed as $y\in (-(1-x^4)^\frac{1}{6},(1-x^4)^\frac{1}{6})$

You have to prove that $(1-x^2)^\frac{1}{4} \leq (1-x^4)^\frac{1}{6}$ so that the almost-circle doesn't reach out of the almost-squre.


问:
Maximizing quadratic function over unit Euclidean ball
I am considering the following maximization problem

$$\begin{array}{ll} \text{maximize} & \| A x - b \|_2^2\\ \text{subject to} & \| x \|_2 \leq 1\end{array}$$

For easiness, let's assume $A\in\mathbb R^{n\times p}$ is a tall matrix
($n>p$), whose rank is $p$. Note that this is not least squares, as we are seeking the maximum instead of the minimum. Intuition is that the maximum must be achieved on the boundary, i.e., when $\|x\|_2=1$.

Is this a well studied problem? Any thought?

答:
Thank @Brian Borchers for your comments.

Here is my reasoning following your comments. let's assume $A$ is a tall full-column-rank matrix. This is also the actual use case of my interest. That means $(A^\top A)^{-1}$ exists. In the following, I add superscript $\ast$ to denote optimum.

By complementary slackness,

$\lambda^\ast (\|x^\ast\|^2-1) = 0$ where $\lambda^\ast \geq 0$.

By stationarity,

$(A^\top A - \lambda I) x^\ast = A^\top b$

If $x^\ast$ is in the interior, then $\lambda^\ast=0$ and $x^\ast=(A^\top A)^{-1}A^\top b$. However we know that this is the least squares solution. It cannot be the maximum. Therefore $\lambda^\ast>0$. Then $x^\ast=(A^\top A-\lambda^\ast I)^{-1}A^\top b$. In other words, we need to seek a strictly positive $\lambda^\ast$ such that $\|x^\ast\|=\|(A^\top A-\lambda^\ast I)^{-1}A^\top b\|=1$. But my question is, does this $\lambda^\ast > 0$ always exist?

An update to my answer, I found this solved question, which is totally relevant. Maximization of quadratic form over unit Euclidean sphere not centered at the origin


问:
When is the quotient algebra of a unital C* algebra helpful?
Let $\mathcal A$ be a unital C* algebra.

Which properties does $\mathcal B \subset \mathcal A$ has to have for it to make sense to form the quotient algebra $\mathcal A / \mathcal B$?

In cases where this construction makes sense, does $\mathcal A / \mathcal B$ have any special structure/properties that are helpful?
Put differently, for what kind of standard questions does it help to consider $\mathcal A / \mathcal B$ because it has desired properties?

答: If $\mathcal{B}$ is a closed $*$-ideal of $\mathcal{A}$, then $\mathcal{A}/\mathcal{B}$ is again a unital C*-algebra. This can be found in every introduction to C*-algebras. And of course this basic construction is used everywhere, so that it doesn't make sense to start a list of applications here. A well-known example is the Calkin algebra. In the commutative case, we have $C(X)/I(A) \cong C(A)$ for closed subspaces $A \subseteq X$ with vanishing ideal $I(A)=\{f : f|_A=0\}$.</p>


问:
What's the sign of $\det\left(\sqrt{i^2+j^2}\right)_{1\le i,j\le n}$?
Suppose $A=(a_{ij})$ is a $n×n$ matrix by $a_{ij}=\sqrt{i^2+j^2}$. I have tried to check its sign by matlab. l find that the determinant is positive when n is odd and negative when n is even. How to prove it？

答: Recall that $e^{-|x|}=\int_0^\infty e^{-sx^2}\,d\mu(s)$ for some non-negative probability measure $\mu$ on $[0,+\infty)$ (Bernstein, totally monotone functions, etc.). Thus $e^{-t|x|}=\int_0^\infty e^{-st^2x^2}\,d\mu(s)$. for $t>0$. In particular, since $(e^{-st^2(i^2+j^2)})_{i,j}$ are rank one non-negative matrices, their mixture $(e^{-t\sqrt{i^2+j^2}})_{i,j}$ is a non-negative matrix for all $t>0$. Hence, considering the first two terms in the Taylor expansion as $t\to 0+$, we conclude that $-A$ is non-negative definite on the $n-1$-dimensional subspace $\sum_i x_i=0$, so the signature of $A$ is $1,n-1$. To finish, we just need to exclude the zero eigenvalue, i.e., to show that the columns are linearly independent, which I leave to someone else (i.e., up to this point we have shown that the determinant has the conjectured sign or is $0$).


问:
Looking for a very gentle first book on Measure Theoretic probability
I am looking for a very gentle first book on measure theory. I want to acquaint myself with the basic ideas in Measure theory. Briefly, my background is as follows.

I am self-learning basic probability theory through a couple of books:

Introduction to Probability theory and its applications, Volume I - William Feller. (Currently reading Chapter X. Law of Large Numbers).

Probability and Random Processes - Grimmett and Stirzaker (Currently studying Chapter II. Random variables and distribution functions).

I am also learning Real Analysis (at the level of Baby Rudin) through the book:

Understanding Analysis, Stephen Abbott. (Currently reading chapter II on sequences and series).

One of the books I have in mind, is Sheldon Axler's book on measure theory (it's modern). Do you think I can jump into it anyway, given my current background? Also, any video playlists(lectures) are welcome.

答: Check out A First Look at Rigorous Probability Theory by Rosenthal. The first five chapters, in particular, seem like exactly what you're looking for: a very gentle introduction to the basics of measure theory with an eye toward probability.


问:
Does an integrable function vanish at infinity in a suitable sense?
I'm searching for a generalization of the following observation: If $f\in L^2(\mathbb R)$, then \begin{equation}\begin{split}\int_{[-n,\:n+1]}f(x)\:{\rm d}x&=\left\langle1_{[-n,\:n+1]},1_{\mathbb R\setminus[-n,\:n)}g\right\rangle_{L^2(\mathbb R)}\\&\le\left\|f-1_{[-n,\:n)}f\right\|_{L^2(\mathbb R)}\xrightarrow{n\to\infty}0\end{split}\tag1\end{equation} by the Cauchy-Schwarz inequality and Lebesgue's dominated convergence theorem.

$(1)$ captures the idea that an integrable function should "vanish at infinity" in a suitable sense.

Can we generalize this result to $f\in L^p(\mu)$, where $(E,\mathcal E,\mu)$ is an arbitrary measure space and $p\ge1$? Maybe assuming that $\mu(E)=\infty$ and/or that $E$ is a metric/normed space with $\mathcal E=\mathcal B(E)$?

In the case of a metric/normed space, the intuition tells me that an integrable function should "essentially be supported in a compact set".

What's obviously true is that $$\forall f\in L^p(\mu):\not\exists B\in\mathcal E:\mu(B)=\infty\text{ and }\inf_{x\in B}|f(x)|>0\tag2.$$

答:
Isn't this almost trivial by contradiction? Let $E$ be a normed vector space and denote $B_r := \{x \in E : \|x\|_{E} \le r \}$, let also $f \in L^1(E)$ with respect to the measure $\mu$. We will now prove that
$$ \tag{1} \lim_{n \to \infty} \int_{B_{n+1} \setminus B_n} f \, d\mu = 0 \,.$$
First, by splitting $f$ into positive and negative part, we can suppose without loss of generality that $f \ge 0$. Suppose that $(1)$ does not hold, then
$$ \limsup_{n \to \infty} \int_{B_{n+1} \setminus B_n} f \, d\mu := L > 0 \,,$$
because if $L=0$ then $(1)$ holds. Extract a subsequence $\Lambda \subset \mathbb{N} $ such that
$$  \lim_{ \Lambda \ni n \to \infty} \int_{B_{n+1} \setminus B_n} f \, d\mu = L \,,$$
then by the limit definition with $\varepsilon = \frac{L}{2}>0$ there exists $N \in \Lambda$ such that
$$ \int_{B_{n+1} \setminus B_n} f \, d\mu > \frac{L}{2} \, , $$
for every $n \in S := \{n \in \Lambda : n \ge N  \}$. Then
$$ \int_{E} f \, d\mu \ge \sum_{n \in S} \int_{B_{n+1} \setminus B_n} f \, d\mu = \infty \,,$$
but this contradicts $f \in L^1(E)$. Long story short, you are just saying that the general term of a convergent sequence must tend to zero. For sure there's no such result in arbitrary measure space without further structure, as the notion of "going to infinity" is definitely not well defined in a general measure space.


问:
Cube roots of the complex numbers 1+i?
I cant get any good reference in my books regarding cube of complex numbers. Please help me find cube roots of the Complex number i+1??</p>

答:
The way for finding the nth roots of a complex number, is to express it in the form of $r.cis(\theta)$.  The function $cis()=cos()+i.sin()$.

For example, if one has $x+iy$, one can find $\theta=arctan(y/x)$, with appropriate quadrant adjustment, and $r^2=x^2+y^2$.

The value of the nth roots, are then $r^{1/n} cis(2\pi*m/n+\theta/n)$, for $m = 0, \ldots, n-1$.

In your case 1+i gives $r=\sqrt{2}$, and $\theta=45°$.

The angles of the cube root are at $15°$, $135°$, and $255°$, with a radius of $2^{1/6}$.

Title: Cube roots of the complex numbers 1+i?


问:
Adjoint of the linear transformation over a vector space over complex numbers.
Consider the vector space $V:=P_2(\mathbb{C})$ consisting of polynomials of degree at most $2$ with complex coefficients together with the following inner product $$\langle f, g\rangle=\int_{-1}^1f(t)\overline{g(t)}\, dt$$
Evaluate the adjoint of the linear transformation $T:V\to V$ by $$T(f)=if'+2f.$$
My effort:
I know that $T^*$ is a linear transformation which satisfies
$$\langle Tf, g\rangle=\langle f, T^*g\rangle.$$
Above implies that $$i\int_{-1}^1f'(t)\overline{g(t)}\, dt+2\int_{-1}^1f(t)\overline{g(t)}\, dt=\langle f, T^*g\rangle.$$
Please tell how to proceed next?

答:
It is enough to consider the transformation $D(f)=f'$, since $T=iD+2\operatorname{id}$ yields $T^*=-iD^*+2\operatorname{id}$.

Instead of constructing an orthonormal basis for $P_2(\mathbb C)$ first, we can do the computation in our favorite basis $B=(1,t,t^2)$ using the Gram matrix of the given inner product.

The Gram matrix of $\langle -,-\rangle$ with respect to $B$ is obtained from the integrals $\int_{-1}^1 t^k\,\mathrm dt$ for $k=0,1,2,3,4$ as
$$
G_B = (\langle t^i, t^j\rangle)_{i,j=0,1,2} =
\begin{pmatrix}
2 & 0 & 2/3 \\
0 & 2/3 & 0 \\
2/3 & 0 & 2/5
\end{pmatrix}.
$$
Denoting the coordinate vector of $f\in P_2(\mathbb C)$ with respect to $B$ by $[f]_B$ this describes our inner product as
$$
\langle f,g\rangle = [f]_B^T \ G_B \ \overline{[g]_B}.
$$

The matrix of $D$ with respect to $B$ is given by
$$
[D]_B =
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 2 \\
0 & 0 & 0
\end{pmatrix}.
$$

Now by comparing the descriptions
\begin{align*}
\langle Df,g\rangle &= [f]_B^T\ [D]_B^T \ G_B \ \overline{[g]_B} \qquad\text{and} \\
\langle f,D^*g\rangle &= [f]_B^T\ G_B \ \overline{[D^*]_B}\ \overline{[g]_B},
\end{align*}
one obtains the general formula for the matrix of the adjoint in non-orthonormal bases:
$$
[D^*]_B = \overline{G_B^{-1} [D]_B^T \ G_B} =
\begin{pmatrix}
0 & -5/2 & 0 \\
3 & 0 & 1 \\
0 & 15/2 & 0
\end{pmatrix}.
$$
Hence $D^*$ is given by
$$
D^*(1) = 3t,\quad D^*(t) = -\frac 5 2+ \frac{15}2 t^2,\quad D^*(t^2) = t
$$
which translates to $T^*$ as
$$
T^*(1) = -3it+2,\quad D^*(t) = \frac 5 2i - \frac{15}2 i t^2 + 2t,\quad D^*(t^2) = -it+2t^2.
$$


问:
What is wrong with this solution if $\Omega$ is not convex?
I want to solve the equation $\partial_{xy} \psi = 0$ with $(x,y) \in \Omega$. In class we studied that a solution is $\psi(x,y) = \phi(x)+\varphi(y)$ when $\Omega$ is a convex set.
I don't see the problem if $\Omega$ is not convex.

答:
For example, in $\mathbb{R}^2$ with $\Omega = \left\{(x,y) ~|~ x \neq 0 \text{ and } y \neq 0 \right\}$, which is the plane with the axes removed, the function
$$
\psi(x,y) = \left\{\begin{array}{rcl}
x & \text{if} & x>0, y>0 \\
x+y & \text{if} &x>0, y <0 \\
x-y & \text{if} & x<0, y>0 \\
y  & \text{if} & x<0, y<0
\end{array}\right.
$$
is solution of $\partial_{xy}\psi = 0$, but there is no functions $\varphi$ and $\phi$ such that $\psi(x,y)= \phi(x) + \varphi(y)$.

The problem is that while trying to solve $\partial_{xy}\psi=0$, you integrate, thus, you fix a constant of integration. This is possible only on an interval, and hence, this uses convexity.


问:
question about summation?
Are there any general rules to find

$???\leqslant \sum_{n=t}^{m}f(n)\leqslant ???$

when $m$ and $t$ $\in $ R

答:
There is always :

$$\sum_{n=t}^m f(n)\le (m-t+1)\max_{x\in E}\left| f(x)\right|\; \text{ where }\; E = \{t,t+1,…,m-1,m\}$$

Title: question about summation?


问:
Basis determining a unique topology
When I read Munkres' Topology, I have a feeling that if we have a basis $\mathscr{B}$ on a set $X$, then the basis determines uniquely a topology on $X$; that is, if we have two topologies $\mathscr{T}_1, \mathscr{T}_2$ with the same basis $\mathscr{B}$, then $\mathscr{T}_1=\mathscr{T}_2$. I'm not sure if I'm right because I can't see this in the definition, which is as follows:

If $X$ is set, a basis for a topology on $X$ is a collection $\mathscr{B}$ of subsets of $X$ (called basis elements) such that for each $x\in X$, there is at least one $B\in \mathscr{B}$ such that $x\in B$ and if $x\in B_1\cap B_2$, where $B_1, B_2\in \mathscr{B}$, then there exists $B_3\in \mathscr{B}$ such that $x\in B_3\subset B_1\cap B_2$.

Moreover, the basis $\mathscr{B}$ generates a topology

$\mathscr{T}_\mathscr{B}=\left\{ U\subset X: \text{for each $x\in U$, there exists $B\in \mathscr{B}$ such that $x\in B\subset U$}\right\}$,

which is the smallest topology containing $\mathscr{B}$. Hence, I guess it's likely that those topologies whose bases are $\mathscr{B}$ should be equal to $\mathscr{T}_\mathscr{B}$.

By the way, I have consulted the article
Uniqueness of Topology and Basis and one of comments (left by Henno) seems to justify my hunch and they mentioned any open set $O$ is a union of the elements of $\mathscr{B}$, so $O$ is already in the topology $\mathscr{T}_\mathscr{B}$, but how could they know $O$ can be written this way just by the definition of a basis? I mean, in Munkres' book, he mentioned in lemme 13.1, from my understanding, that $\mathscr{T}_\mathscr{B}=\left\{\cup_\alpha B_\alpha:B_\alpha \in \mathscr{B}\right\}$, as opposite to saying it holds for any topology with basis $\mathscr{B}$. Perhaps I'm misunderstanding at this point.

Any help is really appreciated!!

答: I would start from the definition of topology as the collection of all open sets. Note now that every single open set can be written as the set-theoretic union of every basis element that contains a point $x \in U$, that is, $U = \bigcup_{x\in U} B_x $. Note now that, by the assumptions of a basis of a topology, you can always take two basis elements $B_1, B_2$ with nonempty intersection and find a third basis element in them (call it $B_3$). Nevertheless, the topology generated by the collection without $B_3$ and the one with $B_3$ is exactly the same and this comes from the fact that the set-theoretical union is not changing if we add a set that is already taken into account considering the sets $B_1$ and $ B_2$. This is the meaning for when Munkres writes that a basis for a topology is not like a basis for a vector space. So, from this point of view you can see that since the set-theoretic union of all (fixed) open sets is a unique object, then you can say that a basis determines the topology but not the converse.

Title: Basis determining a unique topology


问:
Prove or disprove that $x_1 = x_2= \cdots = x_n$, if $x_1^{x_2} = x_2^{x_3} = \cdots = x_{n-1}^{x_n} = x_n^{x_1}$, $x_i > 1, \forall 1 \le i \le n$.
The case $n=3$ is from here. It's straightforward to prove it's true:

First we notice that if any two of $x_1, x_2, x_3$ are equal then all must be equal.

Denote $a=x_1^{x_2}=x_2^{x_3}=x_3^{x_1}$ then
$$x_1 = a^{1/x_2}, x_2=a^{1/x_3}, x_3=a^{1/x_1}$$
WLOG assume $x_1 \ge x_2, x_3$. There are two cases:

$x_1 \ge x_2 \ge x_3 \implies \frac{1}{x_2} \ge \frac{1}{x_3} \ge \frac{1}{x_1} \implies x_1 \ge x_3 \ge x_2 \implies x_2=x_3 \implies x_1=x_2=x_3$.

$x_1 \ge x_3 \ge x_2 \implies \frac{1}{x_2} \ge \frac{1}{x_1} \ge \frac{1}{x_3} \implies x_3 \ge x_1 \ge x_2 \implies x_1=x_3 \implies x_1=x_2=x_3$.

If $n$ is even, then $x_{2i-1} = 2, x_{2i}=4$ is a counterexample.

When $n=5$ the above method will need to examine $4!=24$ cases. Basically we map $x_i$ to $x_{(i \mod 5) +1}$ and reverse the order. In many of cases we can deduce $4$ or all $5$ of the $x_i$'s are equal. For example, if $$x_1 \ge x_3 \ge x_2 \ge x_5 \ge x_4 \tag 1$$ then
$$
x_5 \ge x_1 \ge x_3 \ge x_4 \ge x_2 \tag 2
$$
Since the order of $x_2$ and $x_5$ reversed from $(1)$ to $(2)$, they must be equal and so are everything in between them from both $(1)$ and $(2)$ hence all $x_i$'s are equal.

There are other cases that are different.

Example 2: $$x_1 \ge x_3 \ge x_4 \ge x_2 \ge x_5 \tag 3$$ then
$$x_1 \ge x_3 \ge x_5 \ge x_4 \ge x_2 \implies x_1 \ge x_3 \ge x_2=x_4=x_5 \tag 4$$

Example 3: $$x_1 \ge x_3 \ge x_4 \ge x_5 \ge x_2 \tag 5$$ then
$$x_3 \ge x_1 \ge x_5 \ge x_4 \ge x_2 \implies x_1=x_3 \ge x_2=x_4=x_5 \tag 6$$

But they all lead to the conclusion that $x_1=x_2=x_3=x_4=x_5$.

Now my questions:

Question #1: Is it always true if $n>1$ and $n$ is odd?

Question #2: What if we allow $x_i>0$ instead of $x_i>1$?

答:
Consider the question as a sequence. It starts with two values and has the recursive definition $x(n) = x(n-1)\frac{\log x(n-2)}{\log x(n-1)}$. For $x_1^{x_2} = x_2^{x_3} = \cdots = x_{n-1}^{x_n} = x_n^{x_1} $ to hold, this sequence  has to be periodic with some period length. This gives rise to the hypothesis (to be proved) that the sequence will indeed be monotonously rising / falling for odd/even $n$, hence they cannot be periodic. If this can be established, indeed $x_1 = x_2= \cdots = x_n$ is the only solution.

Let $x(1) = z > 1$ and $x(2) = z^q$.  Then $x(3) = \frac{z^q}{q}$. For $x(3) = x(1)$ to hold we require that $z =   q^{\frac{1}{q-1}}$. One can for example choose $q=2$ which gives $z=2$, hence we obtain $x(1)= 2$ and $x(2) =4$ which gives the series $2,4,2,4, \cdots$ which OP already quoted. Infinite other choices of $q$ are possible as well. For $q=3$ we would have $z = \sqrt 3$ and hence $x(1)= \sqrt 3$ and $x(2) =\sqrt{27}$ which gives the series $\sqrt 3, \sqrt{27}, \sqrt 3, \sqrt{27}, \cdots$. All of the series constructed this way are periodic with (even) period length 2 or integer multiples thereof.

Now let's prove that there are no series possible with odd period length. Start with the situation that  the initially (for $x(1)$ and $x(2)$) chosen $z$ and $q$ with  $z \ne  q^{\frac{1}{q-1}}$. Then the first case is that $x(3) > x(1)$ and hence we have   $x(4) = x(2)\frac{\log x(1)}{\log x(3)} < x(2)$. This gives $x(5) = x(3)\frac{\log x(2)}{\log x(4)} > x(3)$. Continuing like this, the odd sequence is rising whereas the even sequence is falling. If initially $x(1) > x(2)$, as the odd sequence values will always be larger than  $x(1)$, and the even sequence values will always be smaller than    $x(2)$, hence no periodicity will be possible.  If initially $x(1) < x(2)$, hence the odd sequence (rising) and the even sequence (falling)  will converge to some $x^*$ in between $x(1)$ and $x(2)$, which means no periodicity will be possible.

In the second case, $x(3) < x(1)$. With the same reasoning as above,   $x(4) = x(2)\frac{\log x(1)}{\log x(3)} > x(2)$ and $x(5) = x(3)\frac{\log x(2)}{\log x(4)} < x(3)$. Continuing like this, the odd sequence is falling whereas the even sequence is rising. This again means that the sequences either diverge from each other or converge to some $x^*$ in between $x(1)$ and $x(2)$. In both cases, no periodicity will be possible.

Note that the condition $x_i >1$ is not guaranteed  with all initial conditions. The lower values can fall rapidly, consider $x(1) = 1.1, x(2) = 2, x(3) = 0.275007, \cdots$

In summary, in the cases where $x(1) = z > 1$ and $x(2) = z^q$ with $z =   q^{\frac{1}{q-1}}$,  the period is $2$ or any integer multiple of that, hence even. In all other cases, there is no periodicity, hence a constant $x(i) =z$ is the only solution.


问:
Homogeneous ideal vs. cone
Let $k$ be a field and $I$ an ideal of $k[X_1, ..., X_n]$. By definition,
$V(I)$ is a cone if $V(I) = V(I') \ne \emptyset$ for some homogeneous ideal
$I'$ (generated by homogeneous polynomials).

If $k$ is infinite and $V(I)$ is a cone, is then $I$ necessarily homogeneous ?
Is the additional condition that $I$ is prime necessary ? Or is there a
counter-example, even if $I$ is prime ?

答:
I assume you are working in a classical setting, not a scheme theoretic one, so that $V(I)$ is a subset of $k^n$. Following @KReiser's question, let me know if this is not right.

Consider the ideal $I=(x^2+y^2, z-x^2)$ in $\mathbb{Q}[x,y,z]$. Clearly $V(I)=\{(0,0,0)\}$; indeed, $x^2+y^2=0$ forces $x=y=0$, and hence forces $z=x^2=0$ as well. Thus $V(I)=V((x,y,z))$. However, $I$ is not a homogeneous ideal; since $I$ contains elements with terms of degree $1$, any homogeneous generators of $I$ would have to include a non-zero polynomial of degree $\leqslant 1$, and no such polynomials lie in $I$.

To see that $I$ is prime, first consider the map $f:\mathbb{Q}[x,y,z]\rightarrow\mathbb{Q}[x,y]$ given by $z\mapsto x^2$. We claim that $\ker f=(z-x^2)$; the $\supseteq$ direction is clear, and for the $\subseteq$ direction suppose $f(p)=0$. Then $p(x,y,x^2)=0$, and so $p$ is divisible by $z-x^2$ in $\mathbb{Q}(x,y)[z]$. By Gauss' lemma, this means that $p$ is divisible by $z-x^2$ in $\mathbb{Q}[x,y][z]$, as desired.

So ${\mathbb{Q}[x,y,z]}\big/{(z-x^2)}\cong\mathbb{Q}[x,y]$, with isomorphism induced by $f$, and $I\supseteq\ker f$. Thus $$\frac{\mathbb{Q}[x,y,z]}{I}\cong\frac{\mathbb{Q}[x,y]}{f(I)}=\frac{\mathbb{Q}[x,y]}{(x^2+y^2)}.$$ Since $x^2+y^2$ is irreducible in the UFD $\mathbb{Q}[x,y]$, the ideal $(x^2+y^2)<\mathbb{Q}[x,y]$ is prime, so the ring above is a domain and hence $I<\mathbb{Q}[x,y,z]$ is prime, as desired.

Title: Homogeneous ideal vs. cone


问:
How to configure a sigma function so that it only counts every third number
I have a sigma function where the start value is k and the operation can only be counted when k is a multiple of 3. I know how to differentiate between even and odd numbers in a sigma function, but how do I differentiate between numbers that are multiples of 3, and numbers that are not?

答: I think your question needs more clarity. I am not completely sure if this is what you meant to ask, but to denote the sum of the values of a function at all natural numbers divisible by 3 greater than or equal to $k$, you can use $\sum_{n=\left \lceil {k \over 3} \right\rceil}^{\infty}f(3n)$. Alternatively, you can use an index set $S=\{ n \in \mathbb{N} | (\exists m \in  \mathbb{N} ,n=3\cdot m) \wedge (k \leq n)\}$ and denote the sum as $\sum_{i \in S}f(i)$.


问:
How do I create the set necessary to prove the following statement: If $X$ is a non-empty set of ordinals, then $\bigcup X = \text{sup}(X)$
In the following paragraphs, I will outline my proof for the claim:

If $X$ is a non-empty set of ordinals, then $\bigcup X = \text{sup}(X)$

I've taken note of the post here ($\bigcup X$ = $\sup(X)$ for a set $X$ of ordinals),but my question is NOT related to "How Do I Prove This".

I have included my proof of this statement for reference...but for the actual question, you can scroll to the bottom.

This proof amounts to demonstrating that $\bigcup X$ is the smallest ordinal $\alpha$ such that $\alpha \geq \xi$ for all $\xi \in X$.

At the end of my proof, I will clarify the specifics of my question.

As an important side note, my book (The Foundations of Mathematics By Kenneth Kunen) uses the following conventions:

When discussing ordinals, Greek letters (especially $\alpha, \beta, \gamma, \delta, \zeta, \eta, \xi, \mu$) "range over the ordinals"; that is, $\forall \alpha \varphi(\alpha)$ abbreviates $$ \forall x [x \text{ is an ordinal} \rightarrow \varphi(x)] $$ Also, $\alpha \lt \beta$ means (by definition) $\alpha \in \beta$, and $\alpha \leq \beta$ means $\alpha \in \beta \lor \alpha = \beta$

Also, the following definition will be used:

Let $R$ be a relation. $y \in X$ is $R$-maximal in $X$ iff $\neg \exists z (z \in X \land yRx))$

This proof is broken up into two parts:

Part $1$: Demonstrate that $\neg \exists \zeta (\zeta \lt \bigcup X \land \forall \alpha (\alpha \in X \rightarrow \alpha \leq \zeta ))$

Part $2$: Demonstrate that $\bigcup X$ satisfies the property that $\forall \alpha ( \alpha \in X \rightarrow \alpha \leq \bigcup X)$

The following lemmas are useful:

$\bigcup X$ is an ordinal

$\alpha \text{ is an ordinal } \land \delta \in \alpha \rightarrow \delta \text{ is an ordinal}$

Part $1$: Proof by Contradiction - show that $\exists \zeta (\zeta \lt \bigcup X \land \forall \alpha (\alpha \in X \rightarrow \alpha \leq \zeta ))$ leads to a contradiction.

By definition, $\bigcup X = \{ \alpha \in A \  | \ \exists \beta (\beta  \in X \land \alpha \in \beta)\}$ where $A$ is the set that is posited to exist by my book's Union Axiom.

Let $\xi$ be an ordinal that satisfies the claim we want to produce a contradiction with.

Therefore, $\xi \in \bigcup X$. By definition of $\bigcup X$, this implies that $\exists \beta (\beta \in X \land \xi \in \beta)$. Call this object $\beta'$.

But $\xi \in \beta' \iff \xi \lt \beta'$ means that there is an ordinal in $X$ (i.e. $\beta'$) that is strictly greater than $\xi$...a contradiction. Therefore,  $\neg \exists \zeta (\zeta \lt \bigcup X \land \forall \alpha (\alpha \in X \rightarrow \alpha \leq \zeta ))$

Part $2$:

Let $\alpha \in X$.

Show that either $\alpha \in \bigcup X$ or $\alpha = \bigcup X$.

There are two scenarios that are possible with respect to $\alpha$:

i) $\alpha$ is not $R$-maximal in $X$ $\iff \exists \beta ( \beta \in X \land \alpha \in \beta)$

ii) $\alpha$ is $R$-maximal in $X$ $\iff \neg \exists \beta (\beta \in X \land \alpha \in \beta)$

Scenario i: Prove that $\alpha \in \bigcup X$

By definition of $\bigcup X$, if $\exists \beta ( \beta \in X \land \alpha \in \beta)$ then $\alpha \in \bigcup X$.

Scenario ii: Prove that $\alpha = \bigcup X$ (i.e. prove that they are subsets of each other)

Let $\gamma \in \alpha$. By definition of $\bigcup X$, $\gamma \in \bigcup X$. Therefore, $\alpha \subseteq \bigcup X$.

Let $\gamma \in \bigcup X$. By definition of $\bigcup X$, $\exists \beta (\beta \in X \land \gamma \in \beta)$. Call this $\beta^*$. (i.e. $\gamma \in \beta^*$)

$\alpha$ is maximal and $\beta^* \in X$. By trichotomy of ordinals, $\beta^* \leq \alpha$.

Breaking this into two cases:

a) $\beta^* \lt \alpha \iff \beta^* \in \alpha$. Therefore, because $\beta^*$ is an ordinal, $\beta^* \subset \alpha$. But that means $\gamma \in \alpha$. Thus, $\bigcup X \subseteq \alpha$.

b) $\beta^*=\alpha$. Well, in such a case, it is immediately clear that $\gamma \in \alpha$. Thus, $\bigcup X \subseteq \alpha$.

I believe what I carried out above is correct and demonstrates that

If $X$ is a non-empty set of ordinals, then $\bigcup X = \text{sup}(X)$

My question is as follows:

Throughout this proof, it dawned on me that what I am effectively doing with my argument is creating a set that looks like this:

$\{\eta\in ON \ | \forall \alpha (\alpha \in X \rightarrow \alpha \leq \eta) \}$ (...where $ON$ is the set of all ordinals...)

and subsequently proving that $\bigcup X$ must be the minimal element of this set.

After doing a little digging, I found that $ON$ is not a set, and actually forms a Proper Class, which means that the notation "$\eta \in ON$" is invalid syntax in formal set theory.

How then should I describe the aforementioned set?

The only thing I can think to do is the following:

$\{\eta\in S(\bigcup X) \ | \forall \alpha (\alpha \in X \rightarrow \alpha \leq \eta) \}$

Where $S(\bigcup X)=\bigcup X \cup \{\bigcup X\}$

Is this the proper convention (or at least acceptable in and of itself?)

答:
Your concern is unfounded: you are not looking at some set of ordinals and showing that $\bigcup X$ is the least element of that set. The axioms ensure the existence of the set $\bigcup X$, and you have presumably proved that $\bigcup X$ is an ordinal. Now you are simply showing $(1)$ that $X$ has no upper bound that is less than the ordinal $\bigcup X$, and $(2)$ that $\bigcup X$ is an upper bound for $X$. It follows immediately from $(1)$ and $(2)$ that $\bigcup X=\sup X$.

It’s true that once you’ve shown that $\bigcup X=\sup X$, you get as an immediate corollary that $\bigcup X=\min Y$ whenever $Y$ is a non-empty set of upper bounds for $X$, but the proof that $\bigcup X=\sup X$ does not make use of any such $Y$.


答: A trivial, yet sometimes useful, inequality is $$(m-t+1)\min_{i\in\{t,t+1,...,m\}}f(i)\leq\sum_{n=t}^{m}\ f(n) \leq(m-t+1)\max_{i\in\{t,t+1,...,m\}}f(i)$$</p>

Title: question about summation?

Body:
Are there any general rules to find

$???\leqslant \sum_{n=t}^{m}f(n)\leqslant ???$

when $m$ and $t$ $\in $ R


问:
Osculating circle given parametric coordinates in polar form
I have the following parametric equation for a curve in polar form:
\begin{align}
r(t) &= r_0-kt\\
\phi(t)&=h\ln(r_0-k_0t)+a
\end{align}
where $(r_0,h,a,k_0,k)\in\mathbb{R}$
what is the radius of the osculating circle $R(t)$?
Thanks in advance.

答:
Use Frenet- Serret relations with vectors $(N,T)$ in normal plane in the usual differential form after converting  from polar to rectangular coordinates.

```<code>rz = 0.75; kz = 0.4; k = 0.25; h = 0.2; a = 1.25;
rho[t_] = (rz - k t); th[t_] = h Log[rz - kz t] + a ;
ParametricPlot[rho[t] {Cos[th[t]], Sin[th[t]]}, {t, -5, 2},
 GridLines -> Automatic, AspectRatio -> 1.25]
</code>```

In the above  Mathematica used function definition f[t_].

Body:
Use Frenet- Serret relations with vectors $(N,T)$ in normal plane in the usual differential form after converting  from polar to rectangular coordinates.

rz = 0.75; kz = 0.4; k = 0.25; h = 0.2; a = 1.25;
rho[t_] = (rz - k t); th[t_] = h Log[rz - kz t] + a ;
ParametricPlot[rho[t] {Cos[th[t]], Sin[th[t]]}, {t, -5, 2},
 GridLines -> Automatic, AspectRatio -> 1.25]

In the above  Mathematica used function definition f[t_].


问:
classifying space induces a equivalence of categories between PBun$_G(M)$ and $\Pi(M,BG)$ for finite groups $G$
Let $G$ be a finite group, $BG$ its classifying space and M a manifold. Then it is mentioned in https://arxiv.org/abs/1705.05171 (Remark 2.3 d) that there is an equivalence of categories
$$
\Pi (M,BG) \cong PBun_G(M)
$$
between the groupoid of maps from $M$ to $BG$ (with homotopy-classes of homotopies as morphisms) and the groupoid of principal $G$-bundles over $M$. How can one prove that? Where in the litterature can I find a proof of that?

Yet I could only find proofs that homotopy classes of maps correspond to isomorphism classes of principal bundles. But in order to prove that it gives an equivalence of categories I also need that the morphisms correspond to each other.

That they correspond to each other is also mentioned in https://www.uni-due.de/~hm0002/stacks.pdf (In the beginning of chapter 1), but no proof is given.

Thank you in advance for your help

答:
To have an equivalence of groupoids we need an equivalence on components and on automorphism groups of each component. The equivalence on components is the classical story. The reference I like best for this is Stephen Mitchell's notes https://www3.nd.edu/~mbehren1/18.906/prin.pdf

To get equivalence on automorphism groups is not difficult. Suppose $H: X \times I \rightarrow BG$ is a homotopy from the map $f: X \rightarrow BG$ to itself. Then this can instead be written as $H: X \times S^1 \rightarrow BG$ with the restriction to $X \times \{*\}$ equal to $f$.

We may pull back the universal bundle to get a vector bundle over $X \times S^1$. Because our group is discrete, by going in the counter-clockwise direction we have a unique path associated to $x \in X$ starting at $(x,\{*\})$ and returning to $(x',\{*\})$ for some $x' \in X$. Think about what happens on the boundary of the Mobius strip.

This assignment $x \rightarrow x'$ then gives an automorphism of the pullback bundle along $f$. Hence, we have a map from the self homotopies of a map to the automorphism group of the pullback bundle. To see that this is an isomorphism, we construct the inverse map.

Given an automorphism of the principal bundle $P \rightarrow X$, pullback along the projection to get a principal bundle over $X \times I$. Now we may glue along the automorphism to get a principal bundle over $X \times S^1$. This is classified by a map $X \times S^1 \rightarrow BG$ which gives rise to a self homotopy of the map classifying $P \rightarrow X$. It is not hard to check these are inverse.

Notably, the first map requires discreteness and the second does not.