问:
Function for n-th term in sequence that doubles in length on reaching modulus
Does a $f(x)$ exist that could return the n-th term for the sequence below?

```<code>1, 2, 3, 4, 5, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 1, 1, 1, 1, …
</code>```

For example, I am able to solve it in an iterative manner for $f(15) = 5$ using the following pseudo-code:

```<code>let i = 0;
let j = 0;
let k = 1;
let n = 14;

while n > 0 {
    j = (j + 1) % k;

    if (j + 1) % k == 0 {
        i = (i + 1) % 5;

        if (i + 1) % 5 == 0 {
            k = k * 2;
        }
    }

    n = n - 1;
}

// i + 1 = 5
</code>```

However, this is highly inefficient and I wondered if there is a non-iterative way to find the n-th term of this sequence?

It feels as if there are many ways to solve this, e.g. perhaps even using trigonometry, but I'm somewhat out of my depth here. Pointers would very much be appreciated.

Thank you in advance for your time.

答:
Let $$a=\Big\lceil\log_2\left(\Big\lceil\frac{x}{5}\Big\rceil+1\right)\Big\rceil$$

$$f(x)=\Bigg\lceil\frac{x-5*2^{a-1}+5}{2^{a-1}}\Bigg\rceil$$

$2^{a-1}$ is the number of repetitions of each number in the block you are in.  Starting with $x=1 \rightarrow a=1 \rightarrow 2^{a-1}=1$


问:
Ring homomorphism $f: \Bbb{Z}_p[X] \to \Bbb{Z}_p^{n\times n}$. Prove it is never bijective and find the amount of elements in Im$(f)$.
Let $f: \mathbb{Z}_p[X] \to \mathbb{Z}_p^{n \times n}$
with $n \geq 2$ be a ring homomorphism with $f(1) = I_n, f(X) = M$.

a) Prove it is never injective and never surjective.

b) Find the minimal polynomial and the amount of elements $ \operatorname{Im}(f)$ with $p=3, n=6$ and with $M$ given:
$f(X)= M  = \left( \begin{matrix} 2 & 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} \right).$

a) I first proved that $f$ is fully determined by $f(X)$.
$f(p(X)) = f(\sum_{i=1}^n a_iX^i) = \sum_{i=1}^n f(a_i)f(X^i) =  \sum_{i=1}^n a_if(X)^i$, so we see that indeed $f$ is determined by the value of $f(X)$. (These operations are allowed since $f$ is a ring homomorphism).

Then I tried proving that it is never injective.
Since $M \in \mathbb{Z}_p^{n \times n}$, I know that there exists a $q \in \mathbb{N}$ such that $M^q = 0$ with $q$ a divisor of $p$.
Consider $p(X) = X^i$, then $f(p(X)) = p(f(X)) = p(0) = 0$, which means that $p(X) \in \text{ker}(f)$ and since the kernel is not equal to $\{0\}$ we know it is not injective.

I am not sure how to prove that it is never surjective.

b) I think that the minimal polynomial is equal to $\phi_M(X) = (X-2)^2(X-1)$ since there are 2 Jordan blocks for eigenvalue 2 and 1 for eigenvalue 1.

I don't really know how to calculate the amount of elements that $ \operatorname{Im}(f)$ has.

答:
A ring homomorphism $\mathbb{Z}_p[X]\to\mathbb{Z}_p^{n\times n}$ can be surjective, precisely when $n=1$. In this case, send a polynomial to its constant term.

For $n>1$ such a homomorphism cannot be surjective, because the domain is commutative and the codomain isn't.

So this generalizes to the case $A[X]\to A^{n\times n}$, where $A$ is any commutative ring.

Proving that such a homomorphism cannot be injective is the same as proving that for every $M\in A^{n\times n}$, there is a nonzero polynomial $a_0+a_1X+\dots+a^kX^k$ such that
$$
a_0I+a_1M+\dots+a_kM^k=0
$$
This is easier when $A$ is a field, because in this case the set $\{I,M,\dots,M^{n^2}\}$ has more elements than the dimension of $A^{n\times n}$ as $A$-vector space; a nontrivial linear combination will provide the required polynomial.

For general commutative rings one can observe that Hamilton-Cayley holds, but this requires proving it. If you know a proof for matrices over a field, then you can consider the field of fractions $F$ of $\mathbb{Z}[X_{ij}:1\le i,j\le n]$ (polynomials over $n^2$ indeterminates). Let $\xi$ be the matrix having $X_{ij}$ as its $(i,j)$ entry. Then the characteristic polynomial $p(X)$ of $\xi$ (where $X$ is another indeterminate) is a polynomial in $\mathbb{Z}[X_{ij},X]$ and we know that $p(\xi)=0$. Given a matrix $M=(m_{ij})\in A^{n\times n}$, consider the homomorphism $\mathbb{Z}[X_{ij}]\to A^{n\times n}$ defined by $X_{ij}\mapsto m_{ij}$ and finish up.

ContentLicense: CC BY-SA 4.0


问:
How does changing the domain of a function change it’s bijectivity?
I have this function

$f(x)=\frac{x+5}{x-8}$

Why is it different when I’m working from $\mathbb R\setminus8$ to $\mathbb R$ and from $\mathbb R\setminus8$  to $\mathbb R\setminus1$  when trying to prove bijectivity?

Thank you.

答:
$F(x)=(x+5)/(x-8)$ is not surjective onto $\mathbb R$,

because $1\in\mathbb R$ and
$(x+5)/(x-8)=1$ leads to a contradiction.


问:
Find the flaw in my proof that $z^2 =1$ has more than $2$ distinct solutions.
Let $z \in \mathbb{C}$ be any number that satisfies the equation $z^2=1$. Certainly, $z=\pm1$ are two possible solutions to this equation. I claim that $z^k$ is also a solution to this equation for any $k \in \mathbb{R}$, resulting in (probably) at least one other distinct solution (for example, the solution $z=(-1)^{\pi}$).

Proof: Choose any $k \in \mathbb{R}$. Then $(z^k)^2=z^{2k}=(z^2)^k=1^k=1$, as desired.

So, can someone tell me what's wrong with my proof? I know that this has something to do with roots of unity and that this equation should only have 2 distinct solutions, so something must be wrong with my proof. Please bear in mind that I understand very little about complex numbers besides the definition that $i^2 = -1$. Thanks!

答:
$$1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i^2=-1$$

The rule $\sqrt{x} \sqrt{y}=\sqrt{xy}$ is generally valid only if both $x,y \in +ve \text {  real numbers}$

Look at one more:

$$e^{i 2\pi}=1$$

$$(e^{2 \pi i})^i=1^i$$

$$e^{-2 \pi}=1$$

The error here is that the rule of multiplying exponents as when going to the third line does not apply unmodified with complex exponents, even if when putting both sides to the power i only the principal value is chosen.

ContentLicense: CC BY-SA 3.0


问:
Why does $\int_{-1}^{1}\frac{dx}{\sqrt{-x^2+1}} = \pi$?
Can you please show me the process that determines $\displaystyle\int_{-1}^{1}\frac{dx}{\sqrt{-x^2+1}} = \pi$ ? Unfortunately I only know that the derivative of $\sin^{-1}(x)$ is equal to the integrand of that, but I don't actually know what to do with it. Can you help me?

答:
Because you know that $$\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{-x^2+1}},$$ you are already most of the way there.  This tells you that $$\int_{-1}^{1}\frac{dx}{\sqrt{-x^2+1}} = \sin^{-1}(1) - \sin^{-1}(-1) = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi.$$

This is due to the Fundamental Theorem of Calculus, which says that for a continuous function $f$ on $[a,b]$, with antiderivative $F$, then $$\int_{a}^{b}f(x)\,dx = F(b) - F(a).$$


问:
sobolev spaces - product of two functions
I am working in a exercise, to my solution works I need the following affirmation is true:

Let $\varphi : R \rightarrow R$ a convex and smooth function. Let $u \in H^{1}(U)$ a bounded function and $v \in H^{1}_{0}(U)$ a non negative function. Then $\varphi^{'}(u)v \in H^{1}_{0}(U)$.

I am trying to use the definiton , but it's not working. Someone can give me a hint?
Thanks in advance

答: You need some more assumptions. Let us consider the simple case $\varphi(x) = x^2$, then $\varphi'(x) = x$. You want to have $u \, v \in H^1(\Omega)$ for $u, v \in H^1(\Omega)$. Using the product rule one can show that the weak derivative of the product is $\nabla (u \, v) = u \, \nabla v + v \, \nabla u$. However, this function may not belong to $L^2(\Omega)$. You may want to assume $u, v \in H^1(\Omega) \cap L^\infty(\Omega)$.</p>


问:
Making vector subject of equation
I have the following question

Solve the vector equation $\bf{a \times r}\ +\ \lambda\bf{r = c}$ for
r where $\lambda \ne 0$.

So far my approach has been

 Dot both sides with c, which with some rearranging leads to $\bf{r}.(\bf{c \times a}\ +\ \lambda \bf{c}) = |{\bf{c}}|^2$. Leading me to believe that r is a point on a plane. I then converted this equation to Cartesian form and then to Parametric for so r is the subject.

However my answer looks very different to the provided answer and no working is provided.

The solution given was

 $$\bf{r} = N^{-1} \left(\bf{(a.c)a} + \lambda^2\bf{c} - \lambda \bf{a} \times \bf{c}\right)\ \text{where}\ N = \lambda(k^2 + \lambda^2)$$

Is my answer equivalent to this and how was this solution reached?

答:
First off, I assume that in the provided solution, $\mathbf{k}$ should be $\mathbf{a}$, since the former doesn't appear in the problem statement, and after swapping it for $\mathbf{a}$ the answer coincides with mine.

The provided answer contains the term $\mathbf{a}\times\mathbf{c}$, which gives us a hint that we need to apply $\mathbf{a}\times$ to the equation. Indeed doing so gives us
$$(\mathbf{a}\cdot\mathbf{r})\mathbf{a} - (\mathbf{a}\cdot\mathbf{a})\mathbf{r} + \lambda\mathbf{a}\times\mathbf{r} = \mathbf{a}\times \mathbf{c}\ .$$
Now observe that we can use the original equation to substitute $\mathbf{a}\times\mathbf{r}$ with $\mathbf{c} -\lambda \mathbf{r}$, so that $(\mathbf{a}\cdot \mathbf{r})\mathbf{a} - |\mathbf{a}|^2 \mathbf{r} + \lambda\mathbf{c} - \lambda^2 \mathbf{r} = \mathbf{a}\times \mathbf{c}$, or
$$
(\mathbf{a}\cdot\mathbf{r})\mathbf{a}-(|\mathbf{a}|^2+\lambda^2)\mathbf{r} = \mathbf{a}\times\mathbf{c} - \lambda\mathbf{c}\ .
$$
At this point we still have a pesky $\mathbf{a}\cdot\mathbf{r}$ factor remaining, but since it's the coefficient of $\mathbf{a}$, it gets annihilated if we take the vector product with $\mathbf{a}$ again,
$$
-(|\mathbf{a}|^2 + \lambda^2)\mathbf{a}\times\mathbf{r} = (\mathbf{a}\cdot \mathbf{c})\mathbf{a} - (\mathbf{a}\cdot\mathbf{a})\mathbf{c} - \lambda\mathbf{a}\times \mathbf{c}\ .
$$
This might alarm you, because we seem to be reintroducing $\mathbf{a}\times \mathbf{r}$, but we already know how to transform it into something simpler. Making the same substitution for $\mathbf{a}\times\mathbf{r}$ as before, we get
$$
-(|\mathbf{a}|^2 + \lambda^2)(\mathbf{c} - \lambda\mathbf{r}) = (\mathbf{a}\cdot \mathbf{c})\mathbf{a} - |\mathbf{a}|^2\mathbf{c} - \lambda\mathbf{a}\times \mathbf{c}\ .
$$
Finally the $-|\mathbf{a}|^2\mathbf{c}$ terms will cancel, and rearranging the equation gives us the provided answer.

There are two main ideas behind the solution. The first is to use the properties of the cross product and vector triple product to get rid of factors that involve products of the unknown $\mathbf{r}$ with other vectors, so that we end up with an equation containing only a linear factor of $\mathbf{r}$ with a constant coefficient. The second is to use the original equation to substitute $\mathbf{a}\times\mathbf{r}$ repeatedly by an expression where $\mathbf{r}$ is only multiplied by a constant scalar.

Your attempt sounds correct in principle, but perhaps a little too "coordinate-y". In essence, reversing dot products is not a particularly easy thing to do. However, at least the expression $\mathbf{r}\cdot(\mathbf{c}\times\mathbf{a} + \lambda\mathbf{c}) = |\mathbf{c}|^2$ is consistent with the solution. To see this, take the dot product of the provided solution with $\mathbf{c}\times\mathbf{a} + \lambda\mathbf{c}$. In the denominator, all the triple products of $\mathbf{a}$, $\mathbf{a}$, $\mathbf{c}$ or of $\mathbf{a}$, $\mathbf{c}$, $\mathbf{c}$ vanish, so what remains is
$$
\lambda|\mathbf{a}\cdot\mathbf{c}|^2 + \lambda^3|\mathbf{c}|^2 + \lambda|\mathbf{a}\times\mathbf{c}|^2\ .
$$
Here the term $|\mathbf{a}\times\mathbf{c}|^2$ can be rewritten as a scalar triple product $(\mathbf{a}\times\mathbf{c})\cdot(\mathbf{a}\times\mathbf{c})$. We can cyclically permute it into $\mathbf{a}\cdot(\mathbf{c}\times(\mathbf{a}\times\mathbf{c}))$, and expand the result using the vector triple product relation, i.e. $\mathbf{a}\cdot[(\mathbf{c}\cdot\mathbf{c})\mathbf{a} - (\mathbf{c}\cdot\mathbf{a})\mathbf{c}] = |\mathbf{a}|^2|\mathbf{c}|^2 - |\mathbf{a}\cdot\mathbf{c}|^2$. Notice that second term here leads to a cancellation, so we end up with
$$
\lambda|\mathbf{a}|^2|\mathbf{c}|^2 + \lambda^3|\mathbf{c}|^2\ .
$$
After dividing by $N = \lambda(|\mathbf{a}|^2 + \lambda^2)$, you get $|\mathbf{c}|^2$, which is the right hand side of the equation you obtained.

Title: Making vector subject of equation


Title: Why does $\int_{-1}^{1}\frac{dx}{\sqrt{-x^2+1}} = \pi$?

Body: Can you please show me the process that determines $\displaystyle\int_{-1}^{1}\frac{dx}{\sqrt{-x^2+1}} = \pi$ ? Unfortunately I only know that the derivative of $\sin^{-1}(x)$ is equal to the integrand of that, but I don't actually know what to do with it. Can you help me?

Body:
Denote $x = \sin(t)$.

Because $-1 \le x \le 1$ then $-\frac{\pi}{2} \le t \le \frac{\pi}{2}$. We have:
\begin{align}
I &=\int_{-1}^{1}\frac{1}{\sqrt{-x^2+1}}dx \\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{\sqrt{-(\sin(t))^2+1}} d\sin(t)\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos(t)}{\cos(t)} dt\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1 dt\\
I&=\pi. \\
\end{align}


答:
You are right in saying that indeed, ${\int \frac{1}{\sqrt{1-x^2}}dx = \arcsin(x)}$. Now, you need to use the Fundamental Theorem of Calculus, which says that
$$
\int_{a}^{b}f(x)dx = F(b) - F(a)
$$
where $F$ is a function satisfying
$$
F'(x) = f(x)
$$
Can you take it from here?


问:
Optimality conditions in linear programming problems with lower- and upper-bounded variables
Background:

According to a paper by Andersen and Andersen (https://www.davi.ws/doc/Andersen.pdf), any linear programming problem can be formulated as:

min $c^Tx$,

s. t. $Ax = b$,

$l \le x \le u$,

$ ...(1)$

where $x, c, l, u \in \Bbb R^n$, $b \in \Bbb R^m$ and $A \in \Bbb R^{m \times n} $, and some of the simple bounds $l_j$ and $u_j$ may be $-\infty$ and  $+\infty$ respectively. With $L = \{j: l_j \gt -\infty \}$ and $U = \{j: u_j \gt +\infty \}$, then the optimality conditions to 1) can be stated as follows:

$Ax^* = b$,

$A^Ty^* + z^* = c$,

$(x^*_j - l_j) z^*_j = 0$ $\forall j \in L $,

$(u_j - x^*_j) z^*_j = 0$ $\forall j \in U $,

$z^*_j \ge 0$ $\forall j \in \{j \in L: x^*_j = l_j$ $\land$ $l_j \le u_j  \} $,

$z^*_j \le 0$ $\forall j \in \{j \in U: x^*_j = u_j$ $\land$ $l_j \le u_j  \} $,

$z^*_j = 0$ $\forall j \notin L\cup U $,

$l \le x^* \le u$,

$...(2)$

where $y \in \Bbb R^m$ and $z \in \Bbb R^n$. $y$ and $z$ are the Lagrange multipliers corresponding to the linear constraints and linear inequalities in (1), respectively.

My problem:

I am having difficulty seeing whether the optimality conditions (2) came from. I have attempted to derive them as follows:

Introduce a new variable $x'= x - l$ and a new (slack) variable $x'' = u - x'$. Then the primal problem can be cast as:

min $c^T(x' + l)$

s. t. $A(x' + l) = b$

$x' + x'' = u - l$

$x', x'' \ge 0$.

That is:

min $c^T x' + c^T l$

s. t. $\bigl( \begin{smallmatrix} A & 0 \\ 1 & 1 \end{smallmatrix} \bigr)$  $\bigl( \begin{smallmatrix} x' \\ x'' \end{smallmatrix} \bigr)$ $=$  $\bigl( \begin{smallmatrix} b - A l \\ u -l \end{smallmatrix} \bigr)$,

$ x', x'' \ge 0$.

Then, with dual variables $y'$ and $y''$, complementary slackness gives, for optimality:

$x' (c - (y'A + y'')) = 0$, and $x''(-y'') = 0$

$...(3)$

Also, for dual feasibility:

$c \ge y'A + y''$, and $0 \ge y''$

$...(4)$

Is my working correct so far, and if it is, how can I derive (2) from (3) and (4)? Or is there a simpler way to prove (2)?

I would be very grateful for any help.

答:
The easiest way is probably directly via the Lagrangian, but your approach is (almost) correct if all variables have upper and lower bounds, and you have almost solved it. The only error is how you defined $x''$ (it should be $u-x$, not $u-x'$).

You just have to add a slack variable $s$ to $c \geq A^Ty + y''$. Then $z$ from the paper is $y'' + s$ (with $y'' < 0$, so $z$ is not restricted in sign). Your (elementwise) complementarity condition $x'(c-A^Ty'+y'')=0$ means that if $x_j'>0$, i.e., $x_j^*>l$, then $s=0$ so $z_j^* \leq 0$. Similarly, the condition $x''(-y'') = 0$ means that if $x_j'' > 0$, i.e., $x_j^*<u$, then $y''=0$, which means $z_j^* \geq 0$.


问:
Derivative of symmetric inverse of a matrix
I have an expression involving the inverse of a symmetric matrix, which I want to differentiate with respect to one of the elements of that matrix.  I've been working through the problem with help from the matrix cookbook (https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf) but am unsure if I've correctly worked through - I'd be glad of advice as to whether the below is correct, that I have correctly adjusted for the fact that the matrix is symmetric.

(This is for some algebra I am working out in order to implement the move step of an MCMC algorithm)

I have an expression $\mathbf{A}\mathbf{B}^{-1}$, where $\mathbf{A}$ and $\mathbf{B}$ are both square matrices, and $\mathbf{B}$ is symmetric.  I want to evaluate the expression:

$\frac{d\mathbf{A}\mathbf{B}^{-1}}{dB_{gh}}$

Where $B_{gh}$ is the elementof matrix $\mathbf{B}$ in the $g$th row, $h$th colummn.

So far I have:

$\frac{d\mathbf{A}\mathbf{B}^{-1}}{dB_{gh}} = \{\frac{d\mathbf{A}}{dB_{gh}}\mathbf{B}^{-1}\}+\{\mathbf{A}\frac{d\mathbf{B}^{-1}}{dB_{gh}}\} $

With $\frac{d\mathbf{A}}{dB_{gh}}=\mathbf{0}$ as $\mathbf{A}$ doesn't contain any elements of $\mathbf{B}$.

Then, that

$\frac{d\mathbf{B}^{-1}}{dB_{gh}}= -\mathbf{B}^{-1}\frac{d\mathbf{B}}{dB_{gh}}\mathbf{B}^{-1}$

Then that as $\mathbf{B}$ is symmetric we have:

$\frac{d\mathbf{B}}{dB_{gh}}=Tr\{\{\frac{d\mathbf{B}}{d\mathbf{B}}\}^T \frac{d\mathbf{B}}{dB_{gh}}\}$

As $\mathbf{B}$ is symmetric, $\frac{d\mathbf{B}}{dB_{gh}}$ evaluates to $\mathbf{S}^{gh}$, where:

$\mathbf{S}^{gh}=\mathbf{J}^{gh}+\mathbf{J}^{hg}-\mathbf{J}^{gh}\mathbf{J}^{gh}$

Where $\mathbf{J}^{gh}$ is a matrix with a $1$ in the $g$th row and $h$th column, with $0$ elsewhere.

However, from this question (How to take the derivative of a matrix with respect to itself?), am I correct in understanding that $\frac{d\mathbf{B}}{d\mathbf{B}}$ evaluates to $\mathbf{B}$?

Which would leave the solution to be the following?

$\frac{d\mathbf{A}\mathbf{B}^{-1}}{dB_{gh}}=-\mathbf{B}^{-1}\{\mathbf{B}\mathbf{S^{gh}}\}\mathbf{B}^{-1}$

答:
$\def\v{{\rm vec}}\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\E{{\cal E}}$Let $F=AB^{-1}\,$ denote the matrix-valued function and calculate its differential.
$$\eqalign{
dF &= A\,dB^{-1} \\
 &= AB^{-1}\,dB\,B^{-1} \\
 &= F\,dB\,B^{-1} \\
}$$
At this point you have several choices.

If you are comfortable with tensors, then the fourth order identity tensor
$$\E = \p{B}{B} \quad\implies\quad
\E_{ijk\ell} = \delta_{ik}\delta_{j\ell}$$
can be used rearrange the differential and extract the tensor-valued gradient
$$\eqalign{
dF &= F\E B^{-T}:dB \\
\p{F}{B} &= F\E B^{-T}
\;\;\in\,{\mathbb R}^{n\times n\times n\times n} \\
}$$
You can also use a Kronecker product to flatten the matrices into vectors and extract a matrix-valued gradient
$$\eqalign{
\v(dF) &= (B^{-T}\otimes F)\;\v(dB) \\
\p{\,\v(F)}{\,\v(B)} &= B^{-T}\otimes F
\;\;\in\,{\mathbb R}^{n^2\times n^2} \\
}$$
$\big[$NB: These first two approaches product identical components, but in different shapes.$\big]$

You could also define a set of indexed (single-entry) matrices by
$$\eqalign{
J_{ik} &= \p{B}{B_{ik}} \;=\; e_ie_k^T \\
}$$
where {$e_k$} are the standard basis vectors. Then substitute this directly into the differential expression to obtain
$$\eqalign{
\p{F}{B_{ik}} &= FJ_{ik}B^{-1}
\;\;\in\,{\mathbb R}^{n\times n} \\
}$$
As for  the symmetry constraint, please read this post. Afterward, if you still feel compelled to pursue the idea, then one common (but very misleading) interpretation is
$$\eqalign{
S_{ik} &= \p{B}{B_{ik}}
\;=\; e_ie_k^T + e_ke_i^T - I\odot e_ie_k^T \\
\p{F}{B_{ik}} &= FS_{ik}B^{-1}
\;\;\in\;{\mathbb R}^{n\times n} \\
}$$
where $\odot$ is the elementwise/Hadamard product and $I$ is the identity matrix.


答: Surjectivity is the property that every element of the destination set of the function has at least one preimage by the function. Obviously, it depends on the destination set.</p>

Title: How does changing the domain of a function change it’s bijectivity?

Body:
I have this function

$f(x)=\frac{x+5}{x-8}$

Why is it different when I’m working from $\mathbb R\setminus8$ to $\mathbb R$ and from $\mathbb R\setminus8$  to $\mathbb R\setminus1$  when trying to prove bijectivity?

Thank you.


问:
How to estimate distance from root to nearest immediate basin boundary for Newton's method in one complex variable?
Context: I want to check that the atom domain size estimate is smaller than the inradius of the Newton immediate basin, for centers of hyperbolic components in the Mandelbrot set, and thus justify using Newton's method to find the center given an initial guess inside an atom domain.

Initial approach: I am using Newton's method for finding roots of polynomial (and sometimes rational) functions in one complex variable: $$z_{n+1} = z_{n} - \frac{f(z_n)}{f'(z_n)}$$  Once I have found each root, I wish to estimate each distance to the nearest (fractal) boundary of the immediate basin of attraction/convergence, like the radii of the white circles shown on the image (background coloured by root, shaded by number of iterations to converge), so that I can check that the atom domain size estimate is smaller:

(Image originally from https://commons.wikimedia.org/wiki/File:Newtroot_1_0_m3i_m5m2i_3_1.png modified to add white circles, under CC BY-SA license)

How do I estimate the circle radius given the root?  I can evaluate the function and all derivatives as necessary.

One approach I tried is based loosely on formulas from Wikipedia: for a root $\alpha$, first find the distance $r_0$ from $\alpha$ to the nearest root of $f'(z) = 0$ (this is because of condition 1 from the page; this step may be tricky for the high degree polynomials I will be using in practice...).  Then use condition 3c to get $$r_C = \frac{1}{C}\left|\frac{f'(\alpha)}{f''(\alpha)}\right|$$ Then the desired $$r = \frac{1}{2} \min(r_0, r_C)$$ where the $\frac{1}{2}$ is empirically chosen.  I also used $C = \frac{1}{2}$ for want of better ideas.  Here is the result for one cubic polynomial:

It has the problem of the middle radius of three roots in a row getting very small, and for my actual functions finding the nearest root of the derivative is problematic: using Newton's method gives a root, but not necessarily the nearest, at least I think that is why my code calculates a too-large radius for the 3rd root from the left in this picture:

Can this approach be cleaned up and justified formally?

Alternatively, I am considering tracing a path along the steepest ascent gradient of a smoothed iteration count, would this work?  I suppose I would stop tracing when the Newton method step size starts to go down again (as it approaches the boundary).

I derived a smooth iteration count (number of iterations required to get from $z_0$ to within $\epsilon$ of a root) as $$\mu_n = n + \log_2 \frac{\log\epsilon - \log|z_{n+1}-z_n|}{\log|z_{n+2}-z_{n+1}| - \log|z_{n+1}-z_n|}$$ which I think is valid for $z_n$ in the immediate basin region of quadratic convergence and $\epsilon \to 0$, and for the spatial derivative to find the steepest slope I got as far as $$(\mu_n)_x = \frac{1}{\log 2}\left( \frac{-\frac{|z_{n+1}-z_n|_x}{|z_{n+1}-z_n|}}{\log\epsilon - \log|z_{n+1}-z_n|} - \frac{\frac{|z_{n+2}-z_{n+1}|_x}{|z_{n+2}-z_{n+1}|}-\frac{|z_{n+1}-z_n|_x}{|z_{n+1}-z_n|}}{\log|z_{n+2}-z_{n+1}| - \log|z_{n+1}-z_n|}\right)$$ and similarly for $\cdot_y$, which may be computable numerically using dual numbers with two dual dimensions.

But there is also a problem of where to start tracing from: from the root itself wouldn't work as $z_{n+2}=z_{n+1}=z_n$ there, so I suppose to pick an $r>\epsilon$ (not too small, not too big...) and try probe points on the boundary of the circle radius $r$...

Another idea: perhaps I don't need to compute the Newton basin inradius at all: if I could show that the Newton step of $B(\alpha, R_a(\alpha))$ (the ball centered on the root with atom domain size estimate as radius) is contained entirely within that self-same ball, then would it be guaranteed to be entirely inside the immediate basin of the root $\alpha$?  Could the Kantorovich theorem help here?

For example some period 6 components in the Mandelbrot set: radius as described above with second derivatives in red, atom domain radius in green, Mandelbrot set boundary in blue, Newton basins in black:

The radii computed are not totally ordered (sometimes red is outside green, sometimes the opposite), what I really want to show is that the green circles are inside the immediate Newton basins.

答:
Not really an answer, just contributing some relevant pictures from a modification of the code for the first Newton-fractal image.

The last image looks quite optimal. Doing directly what you proposed, determine the atom period and apply Newton to try to find the next hyperbolic center for this period gives the Newton-Fractal

X:-1.7578 Y:0.0138 R:5.0e-03

with the full set having the coloration

As usual, color indicates the roots and shading the number of steps needed to reach a neighborhood of the root. The color is white if the root is already reached in the first step, black if no convergence was achieved.


问:
Choice function and well ordering
Let A ≠ ∅ and f : P(A) \ {∅} → A is a choice function for P(A) \ {∅}.

Prove that the following two conditions are equivalent:

∀B∀C(B ≠ ∅ & B ⊆ A & C ≠ ∅ & C ⊆ A ⇒ f(B∪C) = f({f(B), f(C)}));

exists such a well order ≤ in A, that for each non empty B in A it holds that f(B) = min ≤ B

My attempt:

From f : P(A) \ {∅} → A ⇒ Dom(f)=P(A)\ {∅}.

Hence Let F={f|f : P(A) \ {∅} → A (∀a∈Dom(f))[f(a)∈a]}.
<F,⊆F > is partial ordered set in which every chain has an upper bound.

Indeed, let F'⊆F is a chain. Then for every f∈F',f⊆∪F'.

Since F' is a chain of functions, then ∪F' also is a function and Dom(∪F')=∪(Dom(f)|f∈F'}⊆P(A)\ {∅} and for every  a∈Dom(∪F'),∪F'(a)∈a.
Then (∪F')∈F  and as such it is an upperbound for the chain F'.

And now ∪(B,C)∈F'⊂f. So f((B∪C))=f({f(B),f(C)}).

But I do not know how to go further.
Any help is welcome.

答:
I cannot make any sense of your argument. If you’re trying to prove that the first statement implies the second, you cannot look at a whole family of choice functions: you’re given one specific choice function, and you have to use it to construct a well-ordering of $A$. And if you’re trying to prove the other direction, you must simply define $f(S)=\min_\le S$ for each non-empty $S\subseteq A$ and show that this $f$ has the desired property. That direction is quite trivial, and I’ll leave it to you.

For the other direction, assume that such a choice function $f$ exists. Define a relation $\le$ on $A$ by setting $a\le b$ iff $f(\{a,b\})=a$. We first prove that $\le$ is a partial order on $A$. Reflexivity and antisymmetry are easy.

For each $a\in A$ it is certainly true that $f(\{a,a\})=f(\{a\})=a$, so $a\le a$, and $\le$ is reflexive. And if $a,b\in A$, $a\le b$, and $b\le a$, then $a=f(\{a,b\})=b$, so $\le$ is antisymmetric.

Proving transitivity takes a bit more work.

If $a,b,c\in A$, $a\le b$, and $b\le c$, then $f(\{a,b\})=a$ and $f(\{b,c\})=b$. Thus,

$$\begin{align*}
f(\{a,b,c\})&=f\big(\{a,b\}\cup\{b,c\}\big)\\
&=f\big(\big\{f(\{a,b\}),f(\{b,c\})\big\}\big)\\
&=f(\{a,b\})\\
&=a\,.
\end{align*}$$

On the other hand,

$$\begin{align*}
f(\{a,b,c\})&=f\big(\{a,c\}\cup\{b,c\}\big)\\
&=f\big(\big\{f(\{a,c\}),f(\{b,c\})\big\}\big)\\
&=f\big(\big\{f(\{a,c\}),b\big\}\big)\,,
\end{align*}$$

so $f\big(\big\{f(\{a,c\}),b\big\}\big)=a$, and therefore $f(\{a,c\})=a$ and hence $a\le c$.

Clearly $\le$ is linear: for any $a,b\in A$, $f(\{a,b\})\in\{a,b\}$, so either $a\le b$, or $b\le a$. It only remains to show that $\le$ is a well-order. Let $\varnothing\ne S\subseteq A$, and for convenience let $s_0=f(S)$. Suppose that $s_0\ne\min_\le S$; then there is an $s\in S$ such that $s<s_0$, i.e., $f(\{s,s_0\})=s$. This is impossible if $S=\{s,s_0\}$ (why?), so $S\setminus\{s,s_0\}\ne\varnothing$. Let $R=S\setminus\{s,s_0\}$; then

$$\begin{align*}
f(\{s,f(R)\})&=f\big(\big\{f(\{s,s_0\}),f(R)\big\}\big)\\
&=f(\{s,s_0\}\cup R)\\
&=f(S)\\
&=s_0\,,
\end{align*}$$

which is impossible, since $s\ne s_0\ne f(R)$. Thus, $s_0\le s$ for each $s\in S$, and hence $s_0=\min_\le S$. Every non-empty subset of $A$ therefore has a $\le$-minimum element, and $\le$ well-orders $A$.

Title: Choice function and well ordering


问:
$\int \frac{2^{\sin \left(\sqrt{x}\right)} \cos \left(\sqrt{x}\right)}{\sqrt{x}} \, dx$
I have been asked to integrate:

$$\int \frac{2^{\sin \left(\sqrt{x}\right)} \cos \left(\sqrt{x}\right)}{\sqrt{x}} \, dx$$

In such a small integration you dont have to write it down but to see where I am struggling I have provided a step by step approach:

$$u=\sin \left(\sqrt{x}\right)$$

$$2 \text{du}=\frac{\cos \left(\sqrt{x}\right)}{\sqrt{x}}dx$$

$$2 \int 2^u \, du$$

Know this is where I get stuck cause I do not see that the answer from this should be:

$$\frac{2^{u+1}}{\log (2)}$$ Is there systematic approach to solving this and if not how do you reason?

Please notice it s not the substitution I am struggling with.

答:
$$I=\int\frac{2^{\sin\sqrt{x}}\cos\sqrt{x}}{\sqrt{x}}dx$$
$u=\sin\sqrt{x}\Rightarrow \frac{du}{dx}=\frac1{2\sqrt{x}}\cos\sqrt{x}$ and so: $dx=\frac{2\sqrt{x}}{\cos\sqrt{x}}du$ which converts our integral into:
$$I=2\int2^{u}du$$
now to integrate this notice that:
$$2^{u}=e^{\ln(2^{u})}=e^{u\ln(2)}$$
now if we make the substitution: $v=u\ln(2)\Rightarrow du=\frac{dv}{\ln(2)}$ and so:
$$I=\frac{2}{\ln 2}\int e^vdv$$
now this should be easy, just remember the constant of integration and then back-substitute :)


问:
How do you deal with units when normalizing a Gaussian?
I have a Gaussian with unitless output, but which is a function of time. The formula for this function is

$$
f(t)=\exp{(-\frac{t^2}{2c^2})}
$$
Where both $t$ and $c$ are in units of time. I want to force the condition that the FWHM is equal to a specific time, $\tau$, so I enforce the condition $c=\frac{\tau}{\sqrt{2\ln{2}}}$ and now I proceed to normalize the function by adding some (presumably unit less) parameter A
$$\int_{-\infty}^\infty \! A\exp{(-\frac{t^2}{2c^2})}  \, \mathrm{d}t=Ac\sqrt{2\pi}=1$$
$$A=\frac{1}{c\sqrt{2\pi}}$$

However now I run into a problem, which is that since c has units of time, A has units of 1/time, and the output of the original function is no longer unitless. One way to rectify this is to instead integrate not with respect to time but with respect to some unitless parameter (such as $t/c$). Is it possible to normalize the original function (with unitless output) without changing the units of the output?

Thanks

答:
If $f(t)$ is unitless, and $t$ has units of time, then $dt$ also has units of time, as does
$$\int_{-\infty}^{\infty} f(t)\,dt.$$
Probabilities are themselves unitless, so if you have a probability density over a variable with some unit $U$, the probability density has to have units of $U^{-1}$, since it is probability per $U$.

So $f(t) = e^{-t^2/c^2}$ fails to be a probability density not just because its integral is not $1$, but also because it has the wrong units for a probability density over time. When you multiply $f(t)$ by $A = \frac 1{c\sqrt{2\pi}}$, you are correcting both problems and turning $f(t)$ into an actual probability density.


问:
Convergence on a Piecewise $f_n$ convergence
$f_n:[0,1]\rightarrow \mathbb{R}$ and $f_n(x)=\begin{cases}
n^2x & \text{ if } 0\leq x< \frac{1}{n} \\
 \frac{1}{(n+1)x-1}& \text{ if }  \frac{1}{n}\leq x\leq 1
\end{cases}$

Study convergence and the uniformly convergence.

for $x=0$ we have $f_n(0)=0$ for $0< x\leq 1$
it exists a $n_o \in\mathbb{N}$ such that $\frac{1}{n_0}<x \Rightarrow $  for every $n>n_0$  $\frac{1}{n}<x\Rightarrow $ $f_n(x)=\frac{1}{(n+1)x-1}\rightarrow 0$ as $n\rightarrow \infty $

$\Rightarrow  f_n\rightarrow0$

I want to show that $f_n $ doesn't convergence uniformly, I calculated $\int_{0}^{1}f_ndx=1/2\neq 0 $ and that mean that indeed it doesn't convergence uniformly. But it $\int_{0}^{1}f_ndx=0$ does that mean that
$f_n $  convergence uniformly? Are there any other ways to show that it is uniformly converges or not. how can I use the $sup$ on a piecewise $f_n $?

答:
I think that it is much simpler to say that$$\sup_{x\in[0,1]}|f_n|=n.\tag1$$So, you don't have $\lim_{n\to\infty}|f_n-0|=0$, which means that the convergence is not uniform.

Note that $f_n\left(\frac1n\right)=n$, and this is enough to prove that $\sup_{x\in[0,1]}|f_n|\geqslant n$, which, in turn, is enough to prove that the convergence is not uniform. You actually have $(1)$ because $f$ is increasing on $\left[0,\frac1n\right]$ and decreasing on $\left[\frac1n,1\right]$.

On the other hand, even if $(f_n)_{n\in\Bbb N}$ is a sequence of continuous functions from $[0,1]$ into $\Bbb R_+$ which converges pointwise to the null function and that $\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm dx=0$, you cannot deduce that the convergence is uniform. Take, for instance$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}nx&\text{ if }x\leqslant\frac1{2n}\\1-nx&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ otherwise.}\end{cases}\end{array}$$The convergence is not uniform because$$(\forall n\in\Bbb N):\sup f_n=f_n\left(\frac1{2n}\right)=\frac12.$$


答:
For all $n$ you have
$$f_n(1/n) = n$$
Hence
$$\sup_{x \in [0,1]} |f_n(x)| \ge n$$
This means that $f_n$ does not converge uniformly.

Title: Convergence on a Piecewise $f_n$ convergence

Body:
$f_n:[0,1]\rightarrow \mathbb{R}$ and $f_n(x)=\begin{cases}
n^2x & \text{ if } 0\leq x< \frac{1}{n} \\
 \frac{1}{(n+1)x-1}& \text{ if }  \frac{1}{n}\leq x\leq 1
\end{cases}$

Study convergence and the uniformly convergence.

for $x=0$ we have $f_n(0)=0$ for $0< x\leq 1$
it exists a $n_o \in\mathbb{N}$ such that $\frac{1}{n_0}<x \Rightarrow $  for every $n>n_0$  $\frac{1}{n}<x\Rightarrow $ $f_n(x)=\frac{1}{(n+1)x-1}\rightarrow 0$ as $n\rightarrow \infty $

$\Rightarrow  f_n\rightarrow0$

I want to show that $f_n $ doesn't convergence uniformly, I calculated $\int_{0}^{1}f_ndx=1/2\neq 0 $ and that mean that indeed it doesn't convergence uniformly. But it $\int_{0}^{1}f_ndx=0$ does that mean that
$f_n $  convergence uniformly? Are there any other ways to show that it is uniformly converges or not. how can I use the $sup$ on a piecewise $f_n $?


问:
Conditional constraint LP
My question is fairly basic: how can I express a conditional constraint in a LP?

Example:
I have a minimization of cost of production at a company. My company either can produce up to 100 bikes with a minimum of 25 for that month or doesn't produce anything at all that month.

How could I express this as a constraint in a LP?

答:
You have to extend your LP to a mixed-integer model. What you have in the example is known as a semicontinuous variable and can be written as:

$$25z\leq p\leq 100z$$
where $p$ is the production that month and $z\in\{0,1\}$ is a binary variable.

Title: Conditional constraint LP


问:
What are morphisms of functors
I am not been able to understand, what is a morphism between two functors. I have gone through the formal definition involving a commutative diagram. Can someone explain that to me in a bit more details or in pictures?</p>

答: Rather than give you an unproductive definition, I would suggest you to take a look at the paragraph I.4 of Saunders Mac Lane's Categories for the Working Mathematician, where the author writes about natural transformations (morphisms of functors) and give (so to say) an intuitive idea of them as ways of naturally translating a diagram into a category $C$ (given by a functor with codomain $C$) into another. He also spends some time in explaining some examples which might help you in understanding deeper and better the concept (in particular, I'd suggest you to try and solve exercise number 6 at the end of that paragraph: I found it quite useful to comprehend the idea behind natural transformations!). I hope this might help.

Title: What are morphisms of functors


问:
Inequality in Sobolev spaces that allows to prove weak differentiability
Let $f \in L^2([0,1])$.

Assume that for every $\varphi \in W^{1,2}([0,1])$ we know that
$$
\int_{0}^1 \varphi'(x) f(x) dx \leq \|\varphi \|_{W^{1,2}}.
$$
Does this imply that $f$ belongs to $W^{1,2}([0,1])$?

My guess is yes because we can define $f'$ as
$$
 \int_0^1 \varphi(x) f'(x) dx  = - \int_0^1 \varphi'(x) f(x) dx,
$$
but I am not able to formalize this idea.
Any hint is welcome, thank you

答:
The answer is no. Consider the following function $f: [0,1] \to \Bbb R$:
$$f(x) = \begin{cases}
0 & x < 1/2 \\ 1 &x \ge 1/2
\end{cases}
$$
You can easily check that $f \in L^2$.

Moreover, for all $\varphi$
$$\int_0^1 \varphi'(x) f(x) \ \mathrm dx \le \int_0^1 | \varphi'(x) | \cdot 1 \ \mathrm dx \le ||\varphi||_{W^{1,2}}$$
However $f$ is a step function: it has no weak derivative.


问:
Bipartite graph with even degrees
Given a bipartite graph with an equal number of vertices in its parts. Each vertex in the first part is connected with an even number of vertices in the second part. Prove that we can remove some (maybe none, but NOT all) vertices from the first part so that every vertex in the second part will have an even degree?

I tried to build an eulerian subgraph of such graph using some traverses like in Kuhns algorithm but got nothing good.

答:
This just uses Pigeonhole Principle, but the underlying idea/approach is essentially the linear algebra solution, which is why I didn't write it up initially. Since Aryan added a PP solution, I thought I should add this.

Let the vertices of the bipartite graph be $ A \cup B$.

Let the pigeons be the $ 2^n - 1$ possible subsets of $A$.

Let the holes be the $2^n$ possible parities for the vertex degree of B.

Since each vertex in $A$ has even degree, the sum of vertex degrees in $B$ must also be even, which means that there are actually at most $2^{n-1}$ possible parities for the vertex degree of B.

Hence, by PP, there exist 2 distinct subsets of $A$, (which we call $A_1, A_2$,) that give the same parities for the vertex degree of $B$.
Then, $A_1 \Delta A_2$ is a non-empty subset of $A$ whose vertex degree of $B$ is all even.

Title: Bipartite graph with even degrees


问:
Pure State Definition Confusion
Let me start with some objects. Consider the $\mathrm{C}^*$-algebra $A$ defined by:
$$A=M_1(\mathbb{C})\oplus M_2(\mathbb{C})\subset B(\mathbb{C}^3).$$
Let $x=\mathbb{C}^3$ be given by $(e_1+e_2)/\sqrt{2}\,$  (the one dimensional factor acts on $e_1$).

We have a vector state given by $\rho_x(f)=\langle x,f(x)\rangle$, and

$$\rho_x\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1+\frac{1}{2}c_{11}.$$

If I read something about quantum mechanics for such a state, as it is is given by a unit vector, it is called a pure state.

However this contradicts the answer to this question... and when I pick up Murphy he says that a pure state is such that if $\rho_0\leq \rho_x$ is a positive linear functional, then $\rho_0=t\rho_x$ for $t\in[0,1]$.

However for $\rho_x$ we have the linear functional:

$$\rho_0\left((c_1)+\left(\begin{array}{cc}c_{11} & c_{12} \\ c_{21} & c_{22}\end{array}\right)\right)=\frac12 c_1$$

is a bounded linear functional such that $\rho_x-\rho_{0}$ is a positive linear functional --- half the state given by the vector $e_2$, not $t\rho_x$... and so Murphy would say $\rho_x$ is not pure.

Can you help me with my confusion?

答:
If $\pi :A\to B(H)$ is any non-degenerate representation of the C*-algebra $A$,  and if $x$ is a unit vector in $H$, then the state
$$
  \varphi (a) = \langle x,\pi(a)x\rangle
  $$
is pure iff the cyclic space
$$
  [\pi (A)x] = \overline{\text{span}}\{\pi (a) x: a\in A\}
  $$
is irreducible for the action of $A$ there.

This is because the representation of $A$ on  $[\pi (A)x]$ is equivalent to the GNS representation associated to $\varphi $, and
it is well known that a GNS representation is irreducible iff the state is pure.

In the example given,  one can show that $[Ax]=\mathbb C^3$, which is not irreducible,  so $\rho _x$ is not pure.

On the other hand,
as noted by @MaoWao,
if $A$ is an irreducible algebra of operators on $H$, such as $B(H)$ itself,    $[\pi (A)x] = H$, for every unit vector
$x$, whence every vector state is pure.

Title: Pure State Definition Confusion


问:
Why is it 10*9*8 and not 10 choose 3?
A college has 10 (non-overlapping) time slots for its courses, and blithely assigns
courses to time slots randomly and independently. A student randomly chooses 3 of
the courses to enroll in (for the PTP, to avoid getting fined). What is the probability
that there is a conflict in the student’s schedule?

Solution: The probability of no conflict is $\frac{10·9·8}{10^3} = 0.72$. So the probability of there being
at least one scheduling conflict is 0.28.

Shouldn't the numerator be $\binom{10}{3} = \frac{10*9*8}{3!}$ because we don't care about the order of the time slots. For example, if 1,2,3 are time slots that don't conflict, then isn't that the same as 2,1,3? And so we must divide by 3! because order doesn't matter?

答:
This is a similar problem to this question asked:
Probability question in Harvard Stat110 Homework1 on course time slots overlapping

The reason is that the order matters is because they are not the same classes, they are all different from each other. Taking classes 1, then 2, then 3 would be a different schedule than 3, 2, then 1. That is why you don't divide by the 3!. I hope this helps.

Title: Why is it 10*9*8 and not 10 choose 3?


问:
Left inverse and nullspace
I am doing some independent studying, and came across one note from a professor, which states that if $A$ is an $m \times n$ matrix with zero nullspace, $m \geq n$. I simply have no idea on how to prove it. Hopefully someone can help me out. Thank you. </p>

答:
Finding the nullspace of $A$ is equivalent to finding all solutions to the matrix equation $A\mathbf{x}=\mathbf{0}$.

Solving the matrix equation $A\mathbf{x}=\mathbf{0}$ is equivalent to solving a homogeneous system of linear equations. You have as many equations as there are rows in $A$, and as many unknowns as there are columns in $A$.

So if $A$ is an $m\times n$ matrix, then you have a system with $m$ equations in $n$ unknowns.

Rather than proving the statement, let's prove the contrapositive. Remember that the contrapositive of "If $P$, then $Q$" is "If (not $Q$), then (not $P$)". In this case, rather than proving:

If $A$ has zero nullspace, then $m\geq n$.

we will look at

If $m\lt n$, then the nullspace of $A$ is not just the zero vector.

So, suppose you have a homogeneous system of linear equation with $m$ equations in $n$ unknowns, and there are more unknowns than equations. Do you know anything about solutions to such systems?

Title: Left inverse and nullspace


问:
Continuity of the real and imaginary parts of a continuus complex-valued function
If a complex-valued function is continuous, are the component real and imaginary parts $u(x,y)$ and $u(x,y)$ necessarily continuous? If so, why?</p>

答: A complex valued function $f = u + iv$ is continuous if and only if $u$ and $v$ are continuous, since $|u(z)-u(z')| \le |f(z)-f(z')| \le |u(z)-u(z')| + |v(z)-v(z')|$.</p>


问:
Problem understanding partial derivatives
Question: How is the differentiation of $xy=constant$ equal to $x\text{d}y+y\text{d}x$?

My Approach: I first tried using partial differentiation, which I know very little of. Basically, it's the differentiation of the function with respect to one variable at a time, while keeping the other constant, right?

So using that, shouldn't I get the answer as $x+y$?

All help will be appreciated greatly.

答:
In general, the differential of the function $f(x,y)$ is given by
$$
df=f_x(x,y)dx+f_y(x,y)dy
$$
where $f_x$ and $f_y$ denote the partial derivatives.

In your example, if you set $f(x,y)=xy$, you have
$$
df=ydx+xdy
$$

Notes.

The expression $xy=C$ is not a function, but an equation. You do not differentiate an equation per se, although you can differentiate both sides of an equation.


问:
Geometrically distinguishing between relative positioning of 3 points on a line
Scenario: we have a line in a 2D plane, and we place 3 points on it each having a different color. For example:

we see in this case that the blue point is the first point, the green is the second and the red the last one.

Here's a second scenario where the red point is behind the green point, i.e. green comes last:

I am trying to find out if by a simple geometric criterion one can figure out if the red point is past the green point (as in the 1st image) or if the red point is behind the green one and past the blue (as in the 2nd image).

The drawn vectors are my attempt to solve the mentioned problem: with the vectors defined as: $\vec{x}$ connecting blue to green, $\vec{y}$ connecting green to red, and $\vec{z}$ connecting blue to red, we can form the shown triangles, which relates them as $\vec{x} + \vec{y} = \vec{z}.$ And the triangle inequality tells us the norm of $z$ is smaller or equal to the sum of the two other norms. But I am getting stuck in how to use this triangular relation in order to distinguish between the two cases.

Question:

Is there a simple geometric criterion, possibly based on this vectorial formulation, that for a given set of 3 points, allows us to distinguish between the 1st image and the 2nd? For simplicity we assume the blue point to be always the first point, but the relative positioning of the next two points can change.

答:
There is no way to distinguish order strictly by the vector representations. You have to know something more about the curve to determine the order. For example, in your second example, you would get the exact same three vectors from a curve that headed more directly towards $x$, then turned tightly to head back towards $y$. But in that case the order of the points on the curve are "origin, $x$, $y$", instead of the "origin, $y$, $x$" in your example.

How else you handle this is going to depend on what you are trying to do, and whether there is anything more that you know about the curve. If you just have a collection of points, and are trying to guess the order in which they are traversed, then an "always go to the closest" rule (with some tie-breaker clause) is the best you can hope for. Either that or "minimize the sum of the angles over the whole path".


问:
If $\operatorname{trcl}(x) \in L_\alpha$ , then $x \in L_\alpha$
$L_\alpha$ here denotes the $\alpha$ stage of the constructible universe, and $\operatorname{trcl}(x)$ is the $\in$-transitive closure of $x$. Is this true for every ordinal $\alpha$? I tried proving by induction, and clearly the cases $\alpha = 0$ and limit ordinal are direct, but I'm not sure how to proceed when $\operatorname{trcl}(x) \in L_{\alpha+1} = Def(L_\alpha)$. One can also argue that since $L_\alpha$ is transitive, $x \subset\operatorname{trcl}(x) \subset L_\alpha$, but I can't conclude anything from that.

答:
As discussed in the comments, this claim isn't true in general. The transitive closure of an unbounded set of natural numbers is $\omega\in L_{\omega+1}$. At the same time, the set of natural numbers coding the theory of $L_\omega$ is not an element of $L_{\omega+1}$ by Tarski undefinability, and yet its transitive closure is. Or for that matter, the argument can be carried out for any real, including those not in $L$.

On the other hand, assuming $V=L$, an ordinal $\alpha$ satisfying "$trcl(x)\in L_\alpha\rightarrow x\in L_\alpha$" if and only if $\alpha$ is a cardinal.

To see this, first recall if $\kappa$ is a cardinal, then $L_\kappa=H_\kappa$. And $trcl(x)∈L_\kappa$ just implies $x\in H_\kappa$. Conversely, if $\alpha$ is not a cardinal, then take some $X\subseteq|\alpha|$ which comes in $L$ at a stage later than $L_\alpha$. The transitive closure of such an $X$ is just $|\alpha|$, which is in $L_\alpha$. But by the choice of $X$, we have that $X\notin L_\alpha$.


问:
Is a sum of Brownian motions a Gaussian process?
I was wondering whether the following sum of a Brownian motion $\{W_t, t\geq 0\}$ is also a Gaussian Process:
$$
X_t:= W_t +W_{t/2}
$$
My attempt:

Choose $t_1,\ldots,t_n$ and $v_1,\ldots,v_n$ randomly, if we can show $v_1X_{t_1}+\ldots+v_nX_{t_n}$ is normally distributed, we are done and know that $X_t$ is a Gaussian Process.

To show this, we want to rewrite $v_1X_{t_1}+\ldots+v_nX_{t_n}$ as some sum of independent increments of the Brownian Motion $W_t$, since we know that the sum of independent normals is once again normal.

Note that the set of all points for which we evaluate $W_t$ will be: $\{t_{1/2},t_1,t_{3/2},\ldots,t_{\lfloor{n/2}\rfloor},t_{n/2},t_{\lfloor{n/2}\rfloor+1},\ldots,t_{n-1},t_n\}$. Thus you can just rewrite the sum $v_1X_{t_1}+\ldots+v_nX_{t_n}$ as the sum of independent normals and therefore it is distributed normally and $X_t$ is a Gaussian process.

Thoughts

Is my approach correct? I feel as if I am missing something and my result is not correct. Is there a better way of approaching questions with regard to whether some transformation of Brownian Motions is a Gaussian Process?

答:
If you know that Brownian motion $(W_t)_{t \geq 0}$ is a Gaussian process, i.e. that $\sum_{j=1}^m c_j W_{s_j}$ is Gaussian for any $c_j \in \mathbb{R}$ and $s_1,\ldots,s_m>0$, then there is an easier way to prove the assertion:

Take $0<t_1 < \ldots<t_n$ and $d_1,\ldots,d_n \in \mathbb{R}$, then

$$\sum_{j=1}^n d_j X_{t_j} = \sum_{j=1}^n d_j W_{t_j} + \sum_{j=1}^n d_j W_{t_j/2}. \tag{1}$$

By renaming/-labeling, we can write the sum on the right-hand side in the form $\sum_{j=1}^{2n} c_j W_{s_j}$ for suitable coefficients $c_j$ and times $s_j>0$. Since $(W_t)_{t \geq 0}$ is a Gaussian process, it thus follows that the right-hand side (hence, also the left-hand side) of $(1)$ is Gaussian.


答:
Consider a function in a single variable only, say for example $f(x)=xsinx$. How would you go about finding $f'(x)?$ Certainly, you would have the privilege of using chain rule to differentiate $f(x)$. Try to think of using the same chain rule in case of $f(x,y)=xy$.

```<code>Note: xy can't be a function in a single variable, if you assume both x and y as variables.
</code>```

Assume $y$ to be some function of $x$. Just as you differentiated $xsinx$, try to differentiate $xy$ now(with respect to $x$). You can work it out and see that-
$$x\frac{dy}{dx}+y=0$$ and then, you can reach your coveted equation. If you wanna look up what $dy$ and $dx$ really are, then go for differentials and read it up. It's fairly a good read. As of partial differential, get to know what a differential actually is, and apply your known concepts in 1 variable differentiation to understand it.

```<code>PS: The equation which you got is a "differential equation". This can be solved via integration in order to come to your f(x,y)=xy
</code>```

Coming to the last question, apply the previous concepts again and differentiate with respect to x to get it done! More can be learnt after you go on to deal with multi-variable calculus.

Title: Problem understanding partial derivatives

Body:
Question: How is the differentiation of $xy=constant$ equal to $x\text{d}y+y\text{d}x$?

My Approach: I first tried using partial differentiation, which I know very little of. Basically, it's the differentiation of the function with respect to one variable at a time, while keeping the other constant, right?

So using that, shouldn't I get the answer as $x+y$?

All help will be appreciated greatly.

Body:
Consider a function in a single variable only, say for example $f(x)=xsinx$. How would you go about finding $f'(x)?$ Certainly, you would have the privilege of using chain rule to differentiate $f(x)$. Try to think of using the same chain rule in case of $f(x,y)=xy$.

Note: xy can't be a function in a single variable, if you assume both x and y as variables.

Assume $y$ to be some function of $x$. Just as you differentiated $xsinx$, try to differentiate $xy$ now(with respect to $x$). You can work it out and see that-
$$x\frac{dy}{dx}+y=0$$ and then, you can reach your coveted equation. If you wanna look up what $dy$ and $dx$ really are, then go for differentials and read it up. It's fairly a good read. As of partial differential, get to know what a differential actually is, and apply your known concepts in 1 variable differentiation to understand it.

PS: The equation which you got is a "differential equation". This can be solved via integration in order to come to your f(x,y)=xy

Coming to the last question, apply the previous concepts again and differentiate with respect to x to get it done! More can be learnt after you go on to deal with multi-variable calculus.


问:
Number of possible Scrabble draws
I would like to compute the number of Scrabble draws, when starting a game. Translated to mathematics, this asks for the number of sub-multisets of a multiset.

Lets say we have a multiset $M$ over a set $N=\{1,...,n\}$. This means there are numbers $m_1,..m_n$, which indicate the multiplicity of the numbers $1$ to $n$. For $k \leq m := m_1+...+m_n$ we now want to draw a $k-$multiset from $M$, in other words numbers $k_1,...,k_n$ with the property $k_1+...+k_n=k$ and $0 \leq k_i \leq m_i$.

My Ansatz was drawing a number of $1$'s, this is $k_1$. In order to do so, we have $\min(m_1, 1+k)$ many possibilities.

Then we draw $k_2$, the number of $2$'s. For this, we have $\min(m_2, 1+k - k_1)$ many possibilities.

...

Finally, for $k_n$, there are $\min(m_n, 1+k- (k_1 + ... +k_{n-1}))$ many possibilities.

The problem is the obviously the self referentiality, i.e the number of possibilities for $k_i$ depending on $k_j$ for $j < i$. I'm sure there is a very clean way to do it via binomial coefficients or so but I'm not sure how to do it that way. Any help is appreciated. :-)

EDIT: bump

答:
I don't know any tricks to make this easier.

I got the number of tiles of each letter and the blanks (2) from here and then counted the number of ways with the R program below.  For example, you can draw 0 "A"s and 7 tiles from the remainder, or 1 "A" and 6 tiles from the remainder, or 2 "A"s and 5 tiles from the remainder,... The total can be counted recursively. I got a total of 3,199,724 different ways.

This is related.

```<code>x=sort(c(9, 2, 2, 4, 12, 2, 3, 2, 9, 1, 1, 4, 2, 6, 8, 2, 1, 6, 4, 6, 4, 2, 2, 1, 2, 1,2))

f=function(x,k) {
  if (sum(x)<k) cnt=0 else if (sum(x)==k | (length(x)==1)) cnt=1 else {
    cnt=0
    for (i in 0:min(x[1],k)) cnt=cnt+f(x[-1],k-i)
  }
  return(cnt)
}
f(x,7)
</code>```

Title: Number of possible Scrabble draws

Body:
I don't know any tricks to make this easier.

I got the number of tiles of each letter and the blanks (2) from here and then counted the number of ways with the R program below.  For example, you can draw 0 "A"s and 7 tiles from the remainder, or 1 "A" and 6 tiles from the remainder, or 2 "A"s and 5 tiles from the remainder,... The total can be counted recursively. I got a total of 3,199,724 different ways.

This is related.

x=sort(c(9, 2, 2, 4, 12, 2, 3, 2, 9, 1, 1, 4, 2, 6, 8, 2, 1, 6, 4, 6, 4, 2, 2, 1, 2, 1,2))

f=function(x,k) {
  if (sum(x)<k) cnt=0 else if (sum(x)==k | (length(x)==1)) cnt=1 else {
    cnt=0
    for (i in 0:min(x[1],k)) cnt=cnt+f(x[-1],k-i)
  }
  return(cnt)
}
f(x,7)


问:
Is $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain?

Is  $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain ?

I have some  confusion  that is   what is difference between  euclidean  domain and euclidean Norms ?

My attempt : I thinks  yes

i know  that  $d( a+b \sqrt 8) = |a^2 - 8b^2 | $ as i can show it is  euclidean domain  by same pattern $\mathbb{Z}[{ \sqrt 2 } ]$ is  euclidean  domain

答:
Learn what the property "integrally closed" means and why every UFD is integrally closed. In particular, every Euclidean domain is integrally closed (Euclidean domains are UFDs).

The ring $\mathbf Z[\sqrt{8}] = \mathbf Z[2\sqrt{2}]$ is not integrally closed since $\sqrt{2}$ is in its fraction field and is integral over $\mathbf Z[\sqrt{8}]$ (being a root of $x^2-2$) without being in $\mathbf Z[\sqrt{8}]$.
Therefore $\mathbf Z[\sqrt{8}]$ is not Euclidean or even a UFD, without having to give a specific counterexample to the UFD property: we showed this ring fails to have a property that every UFD (in particular, every Euclidean domain) satisfies: being integrally closed.

By similar reasoning, if $d$ is a squarefree integer and $m \geq 2$, then $\mathbf Z[m\sqrt{d}]$ is not integrally closed and thus can't be Euclidean (or a UFD).


问:
Prime and Primary Ideals in Completion of a ring
Let $(R,\mathfrak m)$ be a local noetherian ring and $\widehat{R}$  its $\mathfrak m$-adic completion.

If $\mathfrak q\in \operatorname{Spec}(\widehat{R})$ then can we find $\mathfrak p\in  \operatorname{Spec}(R)$ such that $\mathfrak q=\mathfrak p\widehat{R}$ ?

Is a primary ideal of  $\widehat{R}$   an extension of a primary ideal of  $R$ ?

答:
This is false: take $R$ a local noetherian integral domain such that $\hat R$ is not an integral domain and $\mathfrak q$ a minimal ideal of $\hat R$. Since $R\to\hat R$ is (faithfully) flat $\mathfrak q$ is lying over $(0)$ and $\mathfrak q\neq(0)$.

Obviously, from 1., this is also false. However one can say something positive: there is a one-to-one correspondence between the set of $\mathfrak m$-primary ideals in $R$ and the set of $\hat{\mathfrak m}$-primary ideals in $\hat R$ given by $\mathfrak a\to\mathfrak a\hat R$.


问:
How to calculate the Rotation Matrix from other known values?
Dear Math Stackexchange community, how can I compute $R$ rotation matrix in this equation ($P=Rp+t$)? (Please see the figure. More detailed explanation is under the figure.)

The equation is for the transformation of $p=p_1,p_2,p_3$ points. $p$ is in Oxyz (local) reference frame.     $p$ consists of 3 points of $p_1,p_2,p_3$. And $p_1,p_2,p_3$ each has 3 coordinates of $x,y,z$. Therefore p is 3x3 matrix. $p=\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}$

$t$ is the distance between Oxyz(local) and OXYZ(global) reference frame.  $t=\begin{bmatrix} t_x \\ t_y \\ t_z \end{bmatrix}$. To be able to do matrix operations, we can take $t$ as $t=\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

$P$ is the coordinates of p_1, p_2, and p_3 in OXYZ global reference frame. $P$ is a 3x3 matrix.
$P=\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix}$

$R$ is the (orthogonal) rotation matrix.
$R=\begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}$

Therefore, all in all:
$P=Rp+t$

$\begin{bmatrix} P_{1x} & P_{2x} & P_{3x} \\ P_{1y} & P_{2y} & P_{3y} \\ P_{1z} & P_{2z} & P_{3z} \end{bmatrix} = \begin{bmatrix} r_{11}&r_{12}&r_{13} \\ r_{21}&r_{22}&r_{23} \\ r_{31}&r_{32}&r_{33} \end{bmatrix}.\begin{bmatrix} p_{1x} & p_{2x} & p_{3x} \\ p_{1y} & p_{2y} & p_{3y} \\ p_{1z} & p_{2z} & p_{3z} \end{bmatrix}+\begin{bmatrix} t_x&t_x&t_x \\ t_y&t_y&t_y \\ t_z&t_z&t_z \end{bmatrix}$

We can derive the following form of the equation of  $P=Rp+t$

$(P-t)=R.p$

$(P-t).p^{-1}=R$

$p$ is an arbitrary matrix, and not always be taken of its inverse. I can't take the inverse of $p$, if last raw of $p$ is zero.

Let's look at an example to make it clear:
$P=Rp+t$

$p_1=(1,1,0)$ , $p_2=(2,4,0)$ , $p_3=(5,10,0)$ is the coordinates of three points. So in matrix form:
$p=\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}$

$t=\begin{bmatrix} 0 \\ 0 \\ 50 \end{bmatrix}$

$\alpha=10^\circ $ (rotation around the x axis)

$\beta=15^\circ $ (rotation around the y axis)

$\gamma=20^\circ $ (rotation around the z axis)

$R_x=\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(\alpha) & -sin(\alpha) \\ 0 & sin(\alpha) & cos(\alpha) \end{bmatrix}$

$R_y=\begin{bmatrix} cos(\beta) & 0 & sin(\beta) \\ 0 & 1 & 0 \\ -sin(\beta) & 0 & cos(\beta) \end{bmatrix}$

$R_x=\begin{bmatrix} cos(\gamma) & -sin(\gamma) & 0 \\ sin(\gamma) & cos(\gamma) & 0 \\ 0 & 0 & 1 \end{bmatrix}$

$R=R_z.R_y.R_x$

Therefore, for $\alpha=10^\circ $, $\beta=15^\circ $, $\gamma=20^\circ $, the rotation matrix is :

$R=\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$

$P=R.p+t$

$P=$$\begin{bmatrix} 0.9077 & -0.2946 & 0.2989 \\ 0.3304 & 0.9408 & -0.0760 \\ -0.2588 & 0.1677 & 0.9513 \end{bmatrix}$$.\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$  $+\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

$P=\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}$

$p_1=(0.6131,1.2712,49.90)$,$p_2=(0.6370,4.4239,50.1533)$,$p_3=(1.5925,11.0597,50.3832)$.

Now let's take opposite example with the same values to find $R$, with the values of $P,p,t$ as known.

$P=R.p+t$

$\begin{bmatrix} 0.6131 & 0.6370 & 1.5925 \\ 1.2712 & 4.4239 & 11.0597 \\ 49.9089 & 50.1533 & 50.3832 \end{bmatrix}=R.$ $\begin{bmatrix} 1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0 \end{bmatrix}+$ $\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 50 & 50 & 50 \end{bmatrix}$

Could you find $R$ here in the above equation?

To repeat the formulation:

$P=Rp+t$

$(P-t)=Rp$

$(P-t).p^{-1}=R.p.p^{-1}$

$(P-t).p^{-1}=R$

For the inverse of $p$ (where $p$ is $\begin{bmatrix}  1 & 2 & 5 \\ 1 & 4 & 10 \\ 0 & 0 & 0  \end{bmatrix}$ ), I can't take the inverse of $p$.

Here I am stuck. It is interesting for me, because for the condition that $R,p,t$ is known, $P$ can be calculated(9 unknown elements in 3x3 matrix of $P$). However, for the condition that $P,p,t$ is known, I am having problem to compute $R$ (9 unknown elements in 3x3 matrix of $R$). I should be able to compute the equivalent $R$ matrix for that inverse problem.

Do you have any idea about how to find $R$ rotation matrix? Any suggestion would be  appreciated.

Thanks in advance.

答:
For  clarity reasons, let's use capital letters to denote Matrices, and lowercase for vectors so let's talk
about
$$
{\bf q}_{\,k}  = {\bf R}\,{\bf p}_{\,k}  + {\bf t}
$$
and
$$
{\bf Q} = {\bf R}\,{\bf P} + {\bf T}\quad \left| {\;{\bf T} = \left( {\matrix{
   {t_{\,x} } & {t_{\,x} } & {t_{\,x} }  \cr
   {t_{\,y} } & {t_{\,y} } & {t_{\,y} }  \cr
   {t_{\,z} } & {t_{\,z} } & {t_{\,z} }  \cr
 } } \right)} \right. = \left( {\matrix{
   {{\bf t}\;,} & {{\bf t}\;,} & {{\bf t}\;}  \cr  } } \right)
$$

A rotation matrix is Orthogonal, which means that it preserves  the dot product  of two vectors and the modulus of their cross product.

Consequently the two triangles $P,Q$ must be congruent, including reflection since we are dealing with a 2D shape embedded in 3D.
If they are congruent then we shall have
$$
\left\{ \matrix{
  \left( {{\bf q}_{\,2}  - {\bf q}_{\,1} } \right) \cdot \left( {{\bf q}_{\,2}  - {\bf q}_{\,1} } \right)
 = \left( {{\bf p}_{\,2}  - {\bf p}_{\,1} } \right) \cdot \left( {{\bf p}_{\,2}  - {\bf p}_{\,1} } \right) \hfill \cr
  \left( {{\bf q}_{\,3}  - {\bf q}_{\,1} } \right) \cdot \left( {{\bf q}_{\,3}  - {\bf q}_{\,1} } \right)
 = \left( {{\bf p}_{\,3}  - {\bf p}_{\,1} } \right) \cdot \left( {{\bf p}_{\,3}  - {\bf p}_{\,1} } \right) \hfill \cr
  \left( {{\bf q}_{\,2}  - {\bf q}_{\,1} } \right) \cdot \left( {{\bf q}_{\,3}  - {\bf q}_{\,1} } \right)
 = \left( {{\bf p}_{\,2}  - {\bf p}_{\,1} } \right) \cdot \left( {{\bf p}_{\,3}  - {\bf p}_{\,1} } \right) \hfill \cr
  \left| {\left( {{\bf q}_{\,2}  - {\bf q}_{\,1} } \right) \times \left( {{\bf q}_{\,3}  - {\bf q}_{\,1} } \right)} \right|
 = \left| {\left( {{\bf p}_{\,2}  - {\bf p}_{\,1} } \right) \times \left( {{\bf p}_{\,3}  - {\bf p}_{\,1} } \right)} \right|
 \hfill \cr}  \right.
$$
where the labels shall be permuted if there is not a  point to point correspondence between $\bf p$ and $\bf q$.

A triangle lies on a plane. If that plane passes through the origin then the position vectors of the vertices are co-planar,
i.e. dependent and so the matrix ($\bf P$ and/or $\bf Q$) has determinant null and is not invertible.

Also, a triangle which is far from the origin, when acted by a rotation matrix will experience a rotation about one of its points plus a translation
of that point.

Then the most practical approach is to construct the matrices
$$
\eqalign{
  & {\bf \Delta P} = \left( {\matrix{
   {{\bf p}_{\,2}  - {\bf p}_{\,1} \;,} & {{\bf p}_{\,3}  - {\bf p}_{\,1} \;,}
 & {\left( {{\bf p}_{\,2}  - {\bf p}_{\,1} } \right) \times \left( {{\bf p}_{\,3}  - {\bf p}_{\,1} } \right)}  \cr
 } } \right)  \cr
  & {\bf \Delta Q} = \left( {\matrix{
   {{\bf q}_{\,2}  - {\bf q}_{\,1} \;,} & {{\bf q}_{\,3}  - {\bf q}_{\,1} \;,}
 & {\left( {{\bf q}_{\,2}  - {\bf q}_{\,1} } \right) \times \left( {{\bf q}_{\,3}  - {\bf q}_{\,1} } \right)}  \cr
 } } \right) \cr}
$$
which, for the congruence identities established above, will give
$$
\overline {{\bf \Delta P}} \,{\bf \Delta P} = \overline {{\bf \Delta Q}} \,{\bf \Delta Q}
$$
and
$$
\left| {\,{\bf \Delta Q}\,} \right| = \left| {\,{\bf \Delta P}\,} \right|
$$

Now we have
$$
{\bf \Delta Q} = {\bf R}\,{\bf \Delta P}
$$
and we can retrive $\bf R$ because $\bf \Delta P$ is invertible (if the triangle is not degenerated).

To determine $\bf t$ we have then
$$
\eqalign{
  & \left( {{\bf q}_{\,2}  - {\bf q}_{\,1} } \right)
 = {\bf R}\left( {{\bf p}_{\,2}  - {\bf p}_{\,1} } \right)\quad  \Rightarrow   \cr
  &  \Rightarrow \quad {\bf q}_{\,2}  = {\bf R}\,{\bf p}_{\,2}  + {\bf q}_{\,1}  - {\bf R}\,{\bf p}_{\,1}
 \quad  \Rightarrow   \cr
  &  \Rightarrow \quad {\bf t} = {\bf q}_{\,1}  - {\bf R}\,{\bf p}_{\,1}
  = {\bf q}_{\,2}  - {\bf R}\,{\bf p}_{\,2}  = {\bf q}_{\,3}  - {\bf R}\,{\bf p}_{\,3}  \cr}
$$

In conclusion and in plain words words

translate both triangles to have one vertex at the origin;

add to each triangle the cross-product of the edges from the origin, to build two tetrahedrons;

rotate one tetrahedron to coincide with the other;

translate back the triangles.

Example

Taking the data of your example
$$
\eqalign{
  & {\bf P} = \left( {\matrix{
   1 & 2 & 5  \cr
   1 & 4 & {10}  \cr
   0 & 0 & 0  \cr
 } } \right)\quad  \Rightarrow \quad {\bf \Delta P} = \left( {\matrix{
   1 & 4 & 0  \cr
   3 & 9 & 0  \cr
   0 & 0 & { - 3}  \cr
 } } \right)\quad \left| {\,{\bf \Delta P}\,} \right| = 9  \cr
  & {\bf Q} = \left( {\matrix{
   {0.6131} & {0.6370} & {1.5925}  \cr
   {1.2712} & {4.4239} & {11.0597}  \cr
   {49.9089} & {50.15330} & {50.3832}  \cr
 } } \right)\quad  \Rightarrow   \cr
  &  \Rightarrow \quad {\bf \Delta Q} = \left( {\matrix{
   {0.0239} & {0.9794} & { - 0.8970}  \cr
   {3.1527} & {9.7885} & {0.2280}  \cr
   {0.2444} & {0.4743} & { - 2.8538}  \cr
 } } \right)\quad \left| {\,{\bf \Delta Q}\,} \right| = 9.0008 \cr}
$$

Then
$$
{\bf R} = {\bf \Delta Q\Delta P}^{\, - \,{\bf 1}}  = \left( {\matrix{
   {0.9077} & { - 0.2946} & {0.2990}  \cr
   {0.3304} & {0.9408} & { - 0.0760}  \cr
   { - 0.2589} & {0.1678} & {0.9513}  \cr
 } } \right)\quad \left| {\,{\bf R}\,} \right| = 1.0001
$$
and
$$
{\bf t} = {\bf q}_{\,1}  - {\bf R}\,{\bf p}_{\,1}
  = \left( {\matrix{   0  \cr    0  \cr    {50}  \cr
 } } \right)
$$
and these check with
$$
{\bf Q} = {\bf R}\,{\bf P} + {\bf T}\quad \quad
 {\bf R} = {\bf R}_{\,{\bf z}} (\gamma ){\bf R}_{\,{\bf y}} (\beta ){\bf R}_{\,{\bf x}} (\alpha )
$$


问:
Restricted power series and tensor product
Let $R$ be a complete adic ring (not necessarily Noetherian) with an ideal of definition $I$. Let $R\left<\zeta\right> := \underset{\longleftarrow}{\lim}_n R/I^n \left[\zeta\right]$ be the ring of restricted power series in $R$. Is it true that $R\left<\zeta\right>\otimes_R R/I \simeq R/I\left[\zeta\right]$?

Edit: I think I see why it is true by the explicit description of $R\left<\zeta\right>$, but for some reason I can't prove it using only formal properties of tensor, limits etc.

答:
Just to get this off the unanswered list.

Assume that $I$ is finitely generated one can show using the Mittag—Leffler condition (e.g. see this post by Matt Emerton) that the natural map

$$(\varprojlim R[z]/I^n)\otimes_R (R/I^m)\to \varprojlim(R[z]/I^n\otimes_R R/I^m)$$

is an isomorphism. But, evidently

$$R[z]/I^n\otimes_R R/I^m$$

forms the constant projective system $R[z]/I^m=(R/I^m)[z]$ and so we see the desired conclusion.

What happens if $I$ is not finitely generated? If $R$ is of finite ideal type, which means that $R$ possesses an ideal of definition which is finitely generated, then one can in practice replace $I$ by this different ideal of definition and thus one can assume WLOG that $I$ is finitely generated. If $R$ is not even of finite ideal type the one gets into a sticky question of what precisely $R\langle \zeta\rangle$ means. Namely, there are two candidates for this adic ring

$$\left\{\sum_n a_n \zeta^n\in R[[\zeta]]: \lim a_n=0\text{ in }R\right\}$$

and

$$\varprojlim R[\zeta]/I^n=\text{the }I\text{-adic completion of }R[\zeta]$$

If $R$ is of finite ideal type, then these are isomorphic. If $R$ is not of finite ideal type, I’m not even sure if these are isomorphic or if, even, the former is complete (e.g. see [1, Example 7.2.10])—but I have not checked in detail.

Note that, in practice, in the context of rigid geometry/formal geometry that appears naturally in nature are all of finite ideal type, so I wouldn’t worry too much.


问:
Homological dimension of a curve
I'm studying Huybrechts' "Fourier-Mukai transforms in algebraic geometry". In the $3^{rd}$ chapter, when he shows that any object in $D^b(C)$, with $C$ a smooth projective curve, is $\oplus \mathcal{E}_i[i]$ (up to isomorphism), he says: "The homological dimension of a curve is one". What does it mean? I know the definition of homological dimension for a category (the maximum $p$ such that there exists two object $X$ e $Y$ with $Ext^p(X,Y) \neq 0$), but not for a curve.

答:
For $X$ a smooth projective variety, the cohomological dimension of $X$ is the cohomological dimension of the Abelian category $\operatorname{Coh}(X)$. We might hope that this agrees with the ordinary notion of dimension, and in this case it does.

Indeed, if $X$ is a smooth projective variety and $\mathcal{F},\mathcal{G}$ are coherent sheaves, then the cohomological dimension of $X$ is $n=\dim X$. If $i>n$ then by Serre Duality,
$$
\operatorname{Ext}^i(\mathcal{F},\mathcal{G})\cong \operatorname{Ext}^{n-i}(\mathcal{G},\mathcal{F}\otimes\omega_X)^*=0
$$
because $n-i<0$. So, the cohomological dimension is $\le n$. On the other hand,
$$0\ne H^0(X,\mathcal{O}_X)=\operatorname{Ext}^0(\mathcal{O}_X,\mathcal{O}_X)\cong \operatorname{Ext}^n(\mathcal{O}_X,\omega_X)^*$$
so that the cohomological dimension is $\ge n$. Hence, $\operatorname{cd}\operatorname{Coh}(X)=\dim X.$

Title: Homological dimension of a curve


问:
derivative function of area function $1-(1-x)^s$
I'm looking for a function $f(z)$ with $\int \limits^x_0 f(z)dz=1-(1-x)^s$ where $0<s=\log_t2<1$ ($t>2$).
The derivative of the integral leads to $f(x)=s(1-x)^{s-1}$, but if one integrates this function, the original area function couldn't be calculated.
What's wrong here with my considerations?

答:
Integrate within limits $0$ to $x$;

$ \int_0^x f(z)dz=\int_0^x s(1-t)^{s-1}dt$

$=\left(-(1-t)^s\right )_0^x$

$=-(1-x)^s+1$


问:
How to construct a chamfered dodecahedron with equilateral or coplanar faces?
According to wikipedia, "The chamfered dodecahedron is a convex polyhedron with 80 vertices, 120 edges, and 42 faces: 30 hexagons and 12 pentagons. It is constructed as a chamfer (geometry) (edge-truncation) of a regular dodecahedron. The pentagons are reduced in size and new hexagonal faces are added in place of all the original edges." And according to this question, the steps to create a chamfered polyhedron are as follows:

shrink the given polyhedron's faces by a factor $r, 0<r<1$

translate the shrunken faces outward [I'll call this distance $s$], so that their edges are coplanar with the
original polyhedron's edges

construct the required hexagons from the original polyhedron's vertices and the vertices of the shrunk faces.

I haven't found any other explicit details regarding the relationship between $r$ and the shape of the hexagons, and so my questions are these (note that the value of $s$ will be directly dependent on the shape of the original face, and so $s$ will need to be represented with some parameter(s) that represent the face's shape somehow):

What is the general relationship between $r$ and $s$ in order to get coplanar vertices? Does this relationship change depending on how many times the polyhedron is chamfered?

What is the general relationship between $r$ and $s$ in order to get equilateral vertices? Does this relationship change depending on how many times the polyhedron is chamfered?

For equilateral and coplanar vertices, what value does $r$ need to be so that the diameter of the resulting hexagon equals the original side length? (ie. given pentagons that share edge $AB$, the resulting hexagon between the two would have a diameter equal to the length of $AB$ - shown in the images below) I assume this relationship could be obtained by setting up the answers to the previous questions as a system of equations

Note that this is different than my question here as it takes a different approach in constructing the chamfered faces - scaling along the relative Z axis vs. translating.

答:
I'll consider the case of faces that are simultaneously equilateral and coplanar, leveraging the work done by David McCooey in generating the coordinates of an equilateral chamfered dodecahedron (text link via dmccooey.com) with unit-length edges.

McCooey gives the coordinates in terms of these values:
$$\begin{align}
c_0 &:= \tfrac14\left(1 + \sqrt{5}\right) \\
c_1 &:= \tfrac1{10}\sqrt{10\left(5 + \sqrt{5}\right)} \\
c_2 &:= \tfrac1{10}\left( 5 +              \sqrt{10\left(5 -\sqrt{5}\right)}\right) \\
c_3 &:= \tfrac1{20}\left( 5 + 5\sqrt{5} + 2\sqrt{10\left(5 +  \sqrt{5}\right)}\right) \\
c_4 &:= \tfrac1{10}\left( 5 +             2\sqrt{ 5\left(5 + 2\sqrt{5}\right)}\right) \\
c_5 &:= \tfrac1{20}\left(15 + 5\sqrt{5} + 2\sqrt{10\left(5 +  \sqrt{5}\right)}\right) \\
c_6 &:= \tfrac1{20}\left( 5 + 5\sqrt{5} + 4\sqrt{ 5\left(5 + 2\sqrt{5}\right)}\right) \\
c_7 &:= \tfrac1{20}\left(15 + 5\sqrt{5} + 4\sqrt{ 5\left(5 + 2\sqrt{5}\right)}\right) \\
\end{align} \tag{1}$$
Of the eighty vertices, we take two sets of five (where I'm assigning "$P$" and "$Q$" names to McCooey's "$V$"s):

$$\begin{align}
P_0 := V_{0} = (0, c_2, c_7)\quad
P_1 &:= V_{72} = (\phantom{-}c_3, c_3, c_3)\quad
P_2 := V_{8}  = (\phantom{-}c_2, c_7,  0)\\
P_3 &:= V_{10} = ( -c_2, c_7,  0)\quad
P_4 := V_{76} = ( -c_3, c_3, c_3) \\[4pt]
Q_0 := V_{60} = ( 0, c_4, c_5) \quad
Q_1 &:= V_{52} = (\phantom{-}c_0, c_6, c_3) \quad
Q_2 := V_{28} = (\phantom{-}\tfrac12, c_7, c_1) \\
Q_3 &:= V_{32} = (-\tfrac12, c_7, c_1) \quad
Q_4 := V_{56} = ( -c_0, c_6, c_3)
\end{align} \tag{2}$$
The $P$ vertices neighbor what would be a face of the dodecahedron in OP's images; such vertices lie on a sphere of approximate radius $2.87463$. The $Q$ vertices neighbor the corresponding pentagonal face of the chamfered dodecahedron; such vertices lie on a sphere of approximate radius $2.86094$. Define $P_\star := \frac15\sum P_i$ and $Q_\star:=\frac15\sum Q_i$, the centers of these faces.

OP's chamfering process scales a dodecahedral face by a factor of $r$ (taking $P_i$ to $P_\star+r(P_i-P_\star)$, then translates it by $z$. It's not clear if OP's "$z$" is a scalar or a vector, so I'll introduce a scalar $s$ and say that the translation is by the vector $(s-1)P_\star$. Then we have

$$Q_i = sP_\star + r (P_i - P_\star) \tag{3}$$

The scaling values are simply
$$\begin{align}
r &= \frac{|Q_0Q_1|}{|P_0P_1|} = \frac{1}{\sqrt{(c_2-c_3)^2 + c_3^2 +(c_3 -c_7)^2}} \approx 0.487457 \tag{4}\\[8pt]
s &= \frac{|Q_\star|}{|P_\star|} = \sqrt{
\frac{
(2c_1+2c_3+c_5)^2 + (c_4+2c_6+2c_7)^2
}{(c_2+2c_3+2c_7)^2 + (2 c_3 + c_7)^2}} \approx 1.19577 \tag{5}
\end{align}$$

Of course, the above numerics are specific to the first iteration of the subdivision process OP is seeking to define in a previous question. What to do after that isn't entirely clear, although it should be "obvious" that every iteration changes the geometry so that particular scale factors don't simply carry-over.

The first iteration is "easy" because all faces are equivalent, as are edges. At the second iteration, we have two types of faces (regular pentagons and non-regular hexagons) and two types of edges (bordering two hexagons, and bordering a hexagon and a pentagon). From there, the types of hexagons and the types of edges will only proliferate, adding complexity.

Consideration of strategies for later iterations would seem to go beyond the scope of this question, so I'll end my discussion here.

Appendix. Here's a bit more of a derivational approach that lends some insight into later chamfering iterations. In what follows, I'll be using $\lambda$ as the scale factor for edges, and $p$ (and also $q$) as the distance by which a scaled face is translated above its forebear.

The equilaterality condition.

Consider an edge $\overline{AB}$ of length $2m$, and let $C$ be on a neighboring face such that $C$ serves as the apex of isosceles $\triangle ABC$ with height $c$. Let $A_\lambda$ be the dilation of $A$ in $C$ by factor $\lambda$; likewise define $B_\lambda$. (Necessarily, $|A_\lambda B_\lambda| = \lambda|AB|=2\lambda m$.) Then, let $A_p$, $B_p$, $C$ be the translates of $A_\lambda$, $B_\lambda$, $C$ by distance $p$ perpendicular to the plane of $|\triangle ABC|$.

Equilaterality of the chamfer requires that $|A_pB_p|=|AA_p|=|BB_p|$. The second equality follows from $\triangle ABC$ being isosceles, so we need only consider the first. A couple of appeals to Pythagoras yields
$$4\lambda^2m^2 = p^2 + (1-\lambda)^2|AC|^2 = p^2 + (1-\lambda)^2 (c^2+m^2) \tag{A.1}$$
Repeating the argument for isosceles $\triangle ABD$ of height $d$ on the other neighboring face of $\overline{AB}$, with $q$ the corresponding translation distance, gives for the same scale factor $\lambda$
$$4\lambda^2m^2 = q^2 + (1-\lambda)^2 (d^2+m^2) \tag{A.2}$$

The coplanarity condition.

Viewing along $\overline{AB}$ (with midpoint $M$), the edge visually collapses to a point and the neighboring faces to segments $\overline{CM}$ and $\overline{DM}$ of lengths $c$ and $d$. Extending $\overline{CC_p}$ to meet the plane of $\square ABB_pC_p$ at $C'$ (and likewise obtaining $D'$), the dihedral angles between the chamfer plane(s) and the faces of the polyhedron become $\gamma := \angle CMC'$ and $\delta := \angle DMD'$.

Writing $\theta$ as the polyhedron's dihedral angle along $\overline{AB}$, we have a planar hexagon $AA_pB_pBB_qA_qA$ when $\gamma+\delta+\theta=180^\circ$. Consequently,
$$\begin{align}
\tan\theta = -\tan(\gamma+\delta) & = -\frac{\tan\gamma+\tan\delta}{1-\tan\gamma\tan\delta} \tag{A.3}\\[4pt]
&= -\frac{\dfrac{p}{(1-\lambda)c}+\dfrac{q}{(1-\lambda)d}}{1-\dfrac{pq}{(1-\lambda)^2cd}} \tag{A.4}
\end{align}$$
Thus,
$$(pq)-(1-\lambda)^2 c d) \tan\theta = (1-\lambda) (cq+dp) \tag{A.5}$$

Equilaterality and coplanarity.

Equations $(A.1)$, $(A.2)$, $(A.5)$ comprise a system of three equations in five unknowns: $\lambda$, $c$, $d$, $p$, $q$. In the specific case of chamfering the dodecahedron, we have $c=d=m\tan54^\circ$ and $p=q$, leaving just the unknowns $\lambda$ and $s$. Taking $m=1$, we can solve to obtain
$$\begin{align}
\lambda &= -5 - 2 \sqrt{5} + \sqrt{2 (25 + 11 \sqrt{5})} \approx 0.487457 \\
p &\approx 0.435995
\end{align} \tag{A.6}$$
Scale factor $\lambda$ here matches $r$ in $(4)$. Distance $p$ is consistent with $s$ in $(5)$ in that $(p+i)/i=s$, for the dodecahedral inradius $i= \frac{m}{10}\sqrt{250+110\sqrt{5}}\approx 2.22703$.

More generally, we don't have enough information to solve for unique values of our five parameters. But our equations only think locally along one edge of the polyhedron being chamfered. Every edge gives rise to a corresponding system; edges sharing a common face share the translation parameter (the analogue of $p$ and $q$); and, for universal equilaterality, the systems share the same scale factor $\lambda$. So, globally, we "should" obtain enough equations to determine unique parameters (or else find that no parameters can do everything we ask of them).

Those considerations are beyond the scope of this discussion, but I will note that we (and by "we", I mean "Mathematica") can eliminate $c$ and $d$ from the single-edge system to get a relation between the common scale factor $\lambda$ and the translation distances $p$ and $q$ for neighboring faces.

$$(1 + \lambda) (3\lambda-1) m^2 \sin^2\theta = p^2 + q^2 \mp 2 p q \cos\theta \tag{A.7}$$
(where I suspect "$\pm$" is always "$-$" (it is for the regular dodecahedron) but haven't verified this). Interestingly, this is exactly a Law of Cosines statement identifying $p$, $q$, and $m\sin\theta\sqrt{(1+\lambda)(3\lambda-1)}$ as sides of a triangle, with $p$ and $q$ surrounding angle $\theta$ (or its supplement). This suggests there might be a clean geometric argument behind all of this. (It also shows that we must have $\lambda \geq 1/3$.) Substituting back into the system, we find

$$
c = \frac{|p\cos\theta\mp q|}{(1-\lambda)\sin\theta} \qquad
d = \frac{|p\mp q\cos\theta|}{(1-\lambda)\sin\theta} \tag{A.8}
$$

That's enough typing for now.


问:
How to compute the probability using Naive Bayes assumption?
I am struggling a bit with this question (and it's on a practice test -- not an actual test).

```<code>Consider that  Y = wealth, and X1 = Gender, X2 = Hours worked.

Let us consider the following data:

P(Y = rich, X1 = Female, X2 < 40.5) = 0.0245895
P(Y = poor, X1 = Female, X2 < 40.5) = 0.253122
P(Y = rich, X1 = Female, X2 >= 40.5) = 0.0116293
P(Y = poor, X1 = Female, X2 >= 40.5) = 0.0421768
P(Y = rich, X1 = male, X2 < 40.5) = 0.0971295
P(Y = poor, X1 = male, X2 < 40.5) = 0.331313
P(Y = rich, X1 = male, X2 >= 40.5) = 0.105933
P(Y = poor, X1 = male, X2 >= 40.5) = 0.134106
Compute P(Y = rich | X1 = Female, X2 < 40.5) using Naiive Bayes assumption

**P(X1 = Female, X2 < 40.5| Y = rich)**

**P(Y = rich)/ P(X1 = Female, X2 < 40.5)**
</code>```

Firstly, I don't even understand what those last 2 equations mean -- how do they relate to the question?

Here is how I went about it (which is obviously not correct since it's not one of the answers):

P(Y = rich | X1 = female, X2 < 40.5)

= (P(X1 = female | Y = rich) * P (X2 < 40.5 | Y = rich) * P(Y = rich)) / P(X1 = female, X2 < 40.5)

= ((0.0245895 + 0.0116293) * (0.0245895 + 0.0971295) * (0.0245895 + 0.0116293 + 0.0971295 + 0.105933)) / (0.0245895 + 0.253122) = 0.003798

Possible answers:

0.07, 0.09, 0.91, 0.38

答:
The last two equations are the probabilities you multiply together to get the answer. They can be got like this:

$P(A|BC)=\displaystyle \frac {P(ABC)}{P(BC)}=\frac {P(BC|A)P(A)}{P(BC)}$, where you use definition of conditional probability twice. Let’s use the first equality and not the one you’re given, the second one.
Here,
$P(ABC)= 0.0245895
P(A^cBC)= 0.253122
P(ABC^c)= 0.0116293
P(A^cBC^c)= 0.0421768
P(AB^cC)= 0.0971295
P(A^cB^cC) = 0.331313
P(AB^cC^c)= 0.105933
P(A^cB^cC^c)= 0.134106 $

You can see $P(ABC)$ from the first line.
$P(BC)=P(ABC)+P(A^cBC)$.
The answer is $\displaystyle \frac {.025}{.025+.25}=.09$


问:
How to proof the convolution product with the Fourier integral?
One should prove the following:

$\mathcal{F}_k^{-1}[\tilde f (k)\tilde g (k)](x)=\frac{1}{\sqrt{2\pi}}(f * g)(x)$

I have absolutely no idea how to solve this exercise. I researched for articles and websites regarding this topic, but couldn't find anything similar :( I would be very thankful for any help!

答:
$$\begin{align}
\int (f*g)(x) \, e^{-ix\xi} \, dx
&= \int \left( \int f(x-y)\,g(y)\,dy \right) \, e^{-ix\xi} \, dx
\\
&= \iint f(x-y)\,g(y)\,e^{-ix\xi}\,dy\,dx
\\
&= \int \left( \int f(x-y) \, e^{-ix\xi} \, dx \right) g(y) \, dy
\\
&\stackrel{z:=x-y}{=} \int \left( \int f(z) \, e^{-i(y+z)\xi} \, dz \right) g(y) \, dy
\\
&= \int \left( \int f(z) \, e^{-iz\xi} \, dz \right) g(y) e^{-iy\xi} \, dy
\\
&= \left( \int f(z) \, e^{-iz\xi} \, dz \right) \left( \int  g(y) e^{-iy\xi} \, dy \right)\end{align}
$$


问:
Inequality of the probability of a sequence of events
Let $(A_n)_{n\ge 1}$ be a sequence of events in a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and
$$ A=\{ \omega\in \Omega: \omega\in A_n \text{ for infinitely many }n \}. $$
Show that
$$ \mathbb{P}(A)\le\sum_{k=n}^{\infty} \mathbb{P}(A_k) $$
for all $n$.

I tried to expresss $A$ as some combinations of $A_k$ but I cannot construct for arbitrary $n$. I also tried induction but had no clues.

Any hints are appreciated.

答:
Hint

$$A=\bigcap_{n\in\mathbb N}\bigcup_{m\geq n}A_m.$$


问:
Cardinality of the set of injective functions from $A$ to $B$
Given two infinite sets $A$ and $B$ with $|A|\leq|B|$, determine the cardinality of the set of injective functions from $A$ to $B$.
I think this set has the same cardinality as $B^A$, but I am not sure how to prove it.
Thanks

答:
The cardinal you are looking for is obviously $\leq B^A$.

To prove it´s $\geq B^A$, notice that as the cardinal of $B$ is the same as the cardinal of $A\times B$, we can find instead the cardinal of the set of injective functions from $A$ to $A\times B$. This is $\geq B^A$ because to each function $f:A\to B$ you can assign the injective function $f':A\to A\times B; a\mapsto(a,f(a))$, and obviously if $f$ and $g$ are different, $f'$ and $g'$ are different.


问:
What is the mathematical equivalent of an axle?
I am talking about axles in machines, on which we can put gears, belts and other devices that help us convert the motion.

So this structure has to have (at least) the following properties:

distance - works like a line, i can 'put' more stuff on it and be able to 'differentiate'(not in a mathematical sense) between them, essentially a $d(x_1, x_2)$ function has to be defined on it

has finite or infinite length

if i were to put a gear on it then the direction and the angular velocity of the rotation of the gear would be equivalent

anticommutativity of its rotations(2 exist in 3D)

the 'speed' of two parallel axes can be compared by a gear placed on each axle, so that they touch

the touching point doesn't move at all, but the gears of a touching point have opposite rotation and same angular velocity

touching gears need not be of same sizes

Of course if you liked to, you could generalize this to any curve.

Initially i thought of the cross-product and normals to a bi-vector. But those don't fulfil all requirements.

Is there a specific object that describes the motions and properties of a physical axle? (even only as a crude model)

答: Screw theory seems like what you want, or is at least highly relevant.  I do not know how to apply it to your problem specifically without more details.</p>


问:
Is there a rational parameter for the cylindrical curve?
Is there a rational parameter for the cylindrical curve?
\begin{align*}
\left\{
\begin{split}
x^2-y^2=2\\
y^2-z^2=3
\end{split}
\right.
\end{align*}
It seems to have something to do with the topic of number theory.

concordant forms

\begin{align*}
\left\{
\begin{split}
\left(\frac{a}{k}\right)^2-\left(\frac{b}{k}\right)^2=2\\
\left(\frac{b}{k}\right)^2-\left(\frac{c}{k}\right)^2=3
\end{split}
\right.
\quad\Rightarrow\quad
\left\{
\begin{split}
b^2+2k^2=a^2\\
b^2-3k^2=c^2
\end{split}
\right.
\end{align*}

答:
An idea which may be of use for you

I'm assuming that you are asking if these equations have a rational solution. If not please ignore my answer and I'll delete this post.

The answer to the rational point question is yes! You can quickly check that there is a solution $$x=\frac{9}{4},y=\frac{7}{4},z=\frac{1}{4}.$$

If this is all you require then that's great but going much beyond this does require a lot of relatively advanced number theory so I will await your response before proceeding much further. If you do know some of the theory of elliptic curves, then the fact that I have given you a rational point means that it is now easy to convert the problem into finding solutions of an elliptic curve which can be done online.


问:
"Large" metric spaces
I'm studying Functional Analysis from Joseph Muscat's book. The author says the non separable spaces are very large. This makes sense too as finite metric spaces and countable spaces are trivially separable and so the only metric spaces which can be non-separable must be uncountable. This is what he says to justify to say that non-separable spaces are too large.

Can someone explain to me what he means by computing distances "precisely", "approximately" and "to any accuracy"? Given two points, we know distance between those points, so, where does the impreciseness occur here?

答:
What Muscat means is that if a metric space $(X,d)$ has finitely many points, then you can specify the metric in a form of a finite list. In fact, if there are $n$ points $x_1, \ldots, x_n$, then this list has $\frac{n(n-1)}{2}$ entries $d_{ij} = d(x_i,x_j)$ with $1 \le i < j \le n$.

Listing is no longer possible if there are infinitely many points.

However, if there are "only" countably many points $x_1, x_2, x_3, \ldots$, then one still has the chance that there is an algorithm determing the exact value of all $d(x_i,x_j)$. This requires a function $f : \mathbb N^2 \to \mathbb R$ which is effectively computable.

If there are uncountably many points, then no algorithm will help to calculate the precise value of $d(x,x')$ for all $x,x' \in X$ - simply because there are uncountably many of these values. Muscat says that in a separable metric space one can at least determine distances $d(x,x')$ approximately up to any desired precision. What does this mean? $X$ contains a countable dense subset $D \subset X$ and there may an algorithm computing $d(p,p')$ for all $p, p' \in D$. But each $x \in X$ can be approximated arbitrarily close by elements of $D$; therefore given $x, x' \in X$ and $\epsilon > 0$, there exist $p, p' \in D$ such that $d(x,p) < \epsilon/2, d(x',p') < \epsilon/2$. Then $\lvert d(x,x') - d(p,p') \rvert < \epsilon$ and if we know the precise value of $d(p,p')$ we have found an approximation for $d(x,x')$ with a deviation smaller $\epsilon$ from the precise value.

At first glance this seems to be an absolutely reasonable approach, but in my opinion it is questionable. The problem is to explicitly find $p, p'$ as above. Sure, we know that such points exist, but in practice it only works if

we can explicitly specify a sequence $(p_n)$ in $D$ such that $p_n \to x$

we can explicitly specify an error estimate for each $n$, i.e an upper bound for $d(x,p_n)$.

Let us consider $X = \mathbb R$ with $d(x,x') = \lvert x - x' \rvert$. The rationals $\mathbb Q$ form a countable dense subset. For many irrational numbers $x$ we can successfully carry out the above steps 1. and 2.; as examples take roots of rational numbers, $e$, $\pi$. But this is a circular dependency: It works for irrational numbers $x$ which can be specified by some algorithm; if we do not have such an algorithm, we do not even have a name for $x$ and are floating in a sort of nirvana.

My conclusion: The approximation approach only works for the subset $X' \subset X$ of points which can be specified algorithmically. All other points are not really tangible, beyond their mere existence we do not know anything concrete about them and will not be able to find approximations for their distance to other points. BUT: This is no drama because we only want to have information about the distance of points which we can explicitly describe.

Title: "Large" metric spaces


问:
Inf of a particular subset of vectors
Let $N\geq2$ an even natural number and $A_N=\{k=(k_1,k_2) \in \mathbb{N} \times \mathbb{N}:1<k_2<k_1\leq N\}$. Let $U_N=\{(k,l): k \in A_N, l \in A_N, \angle(k,l)>0\}$, where $\angle(k,l)$ is the angle between two vectors $k$ and $l$ in the plane. I have to study:
\begin{equation} \alpha_N=\inf_{(k.l)\in U_N}\angle(k,l) \ \ \ \text{and} \ \ \ \lim_{N \to \infty} N\alpha_N. \end{equation}
I tried in this way: $k \cdot l = |k||l|\cos(\angle(k,l))$, so, since $|k|=\sqrt(k_1^2+k_2^2)$ and $|l|=\sqrt(l_1^2+l_2^2)$,
\begin{equation} \cos(\angle(k,l))=\frac{k_1l_1+k_2l_2}{\sqrt(k_1^2+k_2^2)\sqrt(l_1^2+l_2^2)} \end{equation}
Now using that $k,l \in A_N$
\begin{equation} \cos(\angle(k,l))>\frac{2k_2l_2}{\sqrt(2k_1^2)\sqrt(2l_1^2)}>\frac{k_2l_2}{k_1l_1}>\frac{1}{N^2} \end{equation}
Could it be a good way to find $\alpha_N$?

答:
Here is a representation of $A_6$. In the picture it is considered the triangle with vertices $(6,5), (5,4), (0,0)$.

Recall the formula for the area of a triangle:
$$A = \frac{1}{2} |k| |l| \sin ( \alpha )$$
Note that for all $(k,l) \in U_N$ the area formed by the two vectors is the area of a triangle whose vertices have integer coordinates. This means that, by Pick's Theorem, the area of such a triangle is at least
$$A \ge \frac{1}{2}$$
Hence
$$\sin ( \angle (k,l)) = \frac{2A}{|k||l|} \ge \frac{1}{|k||l|}$$
If we want to minimize $\angle (k,l)$, we have to minimize the sine of $\angle (k,l)$: that's because $\sin: [0, \pi/2] \to \Bbb R$ is increasing.

This means that we need to minimize
$$\frac{1}{|k||l|}$$ or, equivalently, we need to maximize its inverse $|k||l|$.
It happens that the maximum is achieved for:
$$l= (N, N-1) \\ k= (N-1, N-2)$$
This is intuitive, but it is not trivial. I have to think about how to prove this.

After we have established this, we have
$$\alpha_N= \arcsin \left( \frac{1}{\sqrt{(N^2 + (N-1)^2)((N-1)^2 + (N-2)^2)}} \right)$$

Asymptotically, since $\sin ( \theta ) \approx \theta$ we have
$$\alpha_N \approx \frac{1}{\sqrt{(2N^2)(2N^2)}} = \frac{1}{2N^2}$$
This means that
$$\lim_{N \to \infty} N \alpha_N = 0$$

Title: Inf of a particular subset of vectors


问:
Prove or disprove that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is a principal ideal in $\mathbb{Z}[X]$

Prove or disprove that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is a principal ideal in $\mathbb{Z}[X]$

Assume that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is principal. Then exist generator $f(X)\in \mathbb Z[X]$ such that: $$\Big((X+1)^2,(2X+1)(X^2-2)\Big)=f(X)\mathbb Z[X]$$
However I don't have idea how to (dis)prove that $f$ exist.

Body:
$\begin{align}{\bf Hint}\ \  I =&\ ((x\!+\!1)^2,\ f(x))\\[.2em]
= &\ ((x\!+\!1)^2,\ f(x)\bmod (x+1)^2)\ \ \ {\rm as\ in\ Euclidean\ Algorithm}\\[.2em]
= &\  ((x\!+\!1)^2,\ \color{#c00}{f(-1)} + \color{#0a0}{f'(-1)} (x\!+\!1))\ \ \ {\rm by\ Taylor\ expansion}
\end{align}$

Your $\,f(x)\,$ has $\,\color{#c00}{f(-1) = 1}\,$ and $\,\color{#0a0}{f'(-1) = 0}\,$ so $\,\color{#c00}{1}\in I\Rightarrow\, I = (\color{#c00}{1})$

Remark $\ $ i.e. $f\equiv 1\pmod{(x\!+\!1)^2}\,$ by double root test shows $-1$ is a double root of $\,f-1$


问:
Variation of Parameters to solve
Told to solve $y''-2y'+y=e^xcos(x)$ using variation of parameters method. Solution then uses Wronskian, $W$, as $e^x \begin{vmatrix}1 & x\\
1 & 1+x
\end{vmatrix}$ which I understand. However, the next step is that $W\begin{bmatrix}c'_{1}\\c'_{2}\end{bmatrix}=\begin{bmatrix}0\\cos(x)\end{bmatrix}$ which I don't understand at all. Obviously comes out as the right answer for $c_{1}$ and $c_{2}$ as the coefficients of the two solutions of the homogeneous equation but I am not sure how they knew it was $0$ and $cos(x)$. If anyone could explain this step that would be great thanks!

答:
We are given

$$y''-2y'+y= g(x) = e^x \cos(x)$$

Finding the complementary solution

$$y_c(x) = e^x(c_1 + c_2 x)$$

Thus

$$y_1(x) = e^x, ~~~~y_2(x) = x e^x$$

The Wronskian of these two functions is

$$W(y_1, y_2) = \begin{vmatrix} e^x & x e^x \\ e^x & e^x(x+1) \end{vmatrix} = e^{2x}$$

Using VoP, we have

$$y_p(x) = -y_1\int \dfrac{y_2~ g}{W}~dx + y_2 \int \dfrac{y_1~ g}{W}~dx$$

This is

$$y_p(x) =  -e^x \int \dfrac{x e^x( e^x \cos(x))}{e^{2x}}~dx + x e^x \int \dfrac{e^x(e^x \cos(x))}{e^{2x}}~dx$$

Can you proceed? You should get

$$y_p(x) = -e^x \cos(x)$$

The final solution is

$$y(x) = y_c(x) + y_p(x) = e^x(c_1 + c_2 x - \cos(x))$$

Update Maybe the derivation of this process is causing issues. For the DEQ

$$y'' + p(x) y' + q(x) = g(x)$$

Suppose $y_1(x), y_2(x)$ are two linearly independent solutions to  homogeneous and $W(y_1, y_2) \ne 0$.

We seek to determine two functions $u_1(x)$ and $u_2(x)$ such that

$$y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x)$$

To determine two functions $u_1(x)$ and $u_2(x)$ uniquely, we need to impose two independent conditions.

First, we shall require $y_p(x)$ above be a solution and second

$$u_1'(x) y_1 + u_2'(x) y_2 = 0$$

We thus have

$$u_1' y_1 + u_2' y_2 = 0 \\ u_1' y_1' + u_2' y_2' = g$$

This can be compactly written as

$$W\begin{bmatrix} u_1' \\ u_2' \end{bmatrix}=\begin{bmatrix} 0 \\ e^x \cos(x) \end{bmatrix}$$

Lastly, this is solved as I show in the answer.

This can be extended to higher order non-homogenous DEQs.

Title: Variation of Parameters to solve


问:
Is the answer to this summation correct?
Consider the summation below:
$$\sum_{k=1}^\infty(2k+1)x^{2k}$$
The problem asks to find what the mentioned summation is equal to. The solution provided in the book starts like the following:
$$x^3+x^5+x^7+...=\frac{x^3}{1-x^2}$$
and then takes the derivative of both sides and uses that to find the solution.

My question is shouldn't $x^3+x^5+x^7+...=\frac{x^3(1-x^{n+1})}{1-x^2}$?

How does the book assume that $x^3+x^5+x^7+...$ equals $\frac{x^3}{1-x^2}$ when we're not told whether $\lvert x\rvert<1$ or not?

答:
Let's make some order.

As @Crostul and @cosmo5 pointed out in the comments we clearly have to suppose $|x| < 1$ otherways the series diverges.
Then using only some known properties of geometric series:
$$ x^3 + x^5 + \dots + x^{3+2n} = \sum_{n=0}^N x^{3 + 2j} = x^3 \cdot \sum_{n=0}^N (x^2)^j = x^3 \cdot \frac{1-(x^2)^{N+1}}{1-x^2} = \frac{x^{3}(1-x^{2(N+1)})}{1-x^2}$$

Since we have supposed $|x| < 1$ we can state
$$ \sum_{n=0}^\infty x^{3 + 2n} = \frac{x^{3}}{1-x^2} \qquad \qquad (1)$$

Now you can take what suggest your book [see somewhere why you can do the $\star$ passage!]

Differentiating the left part in $(1)$ we get:
$$ \frac{d}{dx}\sum_{n=0}^\infty x^{3 + 2n} \stackrel{\star}{=} \sum_{n=0}^\infty \frac{d}{dx} \left(x^{3 + 2n} \right)= \sum_{n=0}^\infty (3+2n)x^{2n+2} = 3x^2 + \sum_{n=1}^\infty (3+2n)x^{2n+2}$$

Differentiating the right part in $(1)$ we get:
$$ \frac{d}{dx} \left( \frac{x^{3}}{1-x^2} \right)=  \frac{x^2(3 - x^2)}{(1-x^2)^2}$$

Dividing both results for $x^2$ we get:
$$ 3 + \sum_{n=1}^\infty (3+2n)x^{2n} = \frac{(3 - x^2)}{(1-x^2)^2}$$

Split
$$\sum_{n=1}^\infty (3+2n)x^{2n} = \sum_{n=1}^\infty (1+2n)x^{2n} + 2\sum_{n=1}^\infty x^{2n}$$

and you get:
$$\begin{align}\sum_{n=1}^\infty (1+2n)x^{2n} & = \frac{(3 - x^2)}{(1-x^2)^2} -3 - 2\sum_{n=1}^\infty x^{2n} \\[10pt]
& = \frac{(3 - x^2)}{(1-x^2)^2} - 3- \frac{2x^2}{1-x^2} \\[10pt]
& = \frac{3x^2 - x^4}{(1-x^2)^2}
\end{align}$$