问:
Prove that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)=\sigma(S_2)$.
QUESTION: Given the following collection of subsets of $\mathbb{R}^n$,

$$S_1=\{F\subset\mathbb{R}^n; F\; \text{is closed}\;\}$$ and
$$S_2=\{(a_1, b_1)\times\cdots \times (a_n, b_n)\in \mathbb{R}^n; \; \text{where}\; (a_i, b_i)\subset \mathbb{R}, \; i=1, \cdots, n. \}.$$ Prove that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)=\sigma(S_2)$.

MY ATTEMPTY:

First let's prove that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)$.

We have $\mathcal{B}(\mathbb{R}^n)=\sigma(\mathcal{O})$ where $\mathcal{O}$ is a collections of open subsets in $\mathbb{R}^n$. Now, remembering that a subset in $\mathbb{R}^n$ is open iff its complement is closed. Considering $S_1=\mathcal{O}^c$, since the Borel $\sigma$-algebra in $\mathbb{R}^n$ is also a $\sigma$-algebra then
$$\mathcal{B}(\mathbb{R}^n)=\sigma(\mathcal{O})=\sigma(\mathcal{O}^c)=\sigma(S_1).$$

Now let's prove that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$.
Remembering that open rectangles $(a_i, b_i)\times\cdots\times (a_j, b_j)\in \mathbb{R}^n$ provides a generator bases of the topology of $\mathbb{R}^n$, such that any open subset can be represented by a countable union of rectangles, hence, writting
$$E_i=\displaystyle\bigcup_{i=1}^{\infty}\left[(a_i, b_i)\times \cdots \times (a_j, b_j)\right]_i$$
and
$$\displaystyle\prod_{k=1}^{n}(a_k, b_k)=\left[(a_i, b_i)\times \cdots \times (a_j, b_j)\right].$$
Thereby,
$$E_i=\displaystyle\bigcup_{i=1}^{\infty}\prod_{k=1}^{n}(a_k, b_k).$$
On the one hand ones has
$$S_2\subset S_1^c\implies \sigma(S_2)\subset\sigma (\mathcal{O})=\mathcal{B}(\mathbb{R}^n).$$
On the other hand, if $E_i\in S_1^c=\mathcal{O}$ then exists open rectangles $\prod_{k=1}^{n}(a_k, b_k), \; k=1, 2, \cdots, n$ such that $$E_i=\displaystyle\bigcup_{i=1}^{\infty}\prod_{k=1}^{n}(a_k, b_k).$$
Thus, $E_i\in \sigma(S_2)$, this is, $S_1^c=\mathcal{O}\subset\sigma(S_2)$, therefore
$$\mathcal{B}(\mathbb{R}^n)=\sigma(S_1^c)\subset\sigma(\sigma(S_2))=\sigma(S_2).$$

MY DOUBT: Would someone scan for mistakes in my proof? I feel that it is not completely right. For example, I'm not sure about the first step if I can conclude that equality. And didn't persuade myself with the proof I've provided in the second equality.

答:
Your proof that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_1)$ is fine.

The proof that  $\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$ is OK, but it can be done in a simpler way.

Proof that $\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$ :

Since $S_2 \subset \mathcal{O}$, it follows immediately that $$\sigma(S_2) \subseteq \sigma(\mathcal{O}) = \mathcal{B}(\mathbb{R}^n)$$

Since, any open set can be written as a countable union of rectangles, we have that
$ \mathcal{O} \subseteq  \sigma(S_2)$. So, we have
$$\mathcal{B}(\mathbb{R}^n)= \sigma( \mathcal{O}) \subseteq  \sigma(S_2)$$
So, we have $\mathcal{B}(\mathbb{R}^n)=\sigma(S_2)$.


问:
A problem about using characteristic equation to solve recursive sequence
I'm working on a recursive sequence. However, I found that the roots of the characteristic equation are complex numbers. Does it mean there isn't a general term formula for this sequence, thanks a lot.</p>

答:
The solution goes through as normal, just using exponentials of complex quantities. I.e., if you've got:

$\begin{align*}
  \exp(a + b n + (c + d n) \mathrm{i})
    &= \exp(a + b n) \cdot (\cos (c + d n) + \mathrm{i} \sin(c + d n))
\end{align*}$

If the original coefficients are all real, the zeros will show up in complex conjugate pairs, and the sinus terms end up cancelling. Thus you end up with solutions of the form $\alpha \beta^n$ times a periodic term.

ContentLicense: CC BY-SA 4.0


问:
A nontrivial everywhere continuous function with uncountably many roots?
This is my first post on SE, forgive any blunders.

I am looking for an example of a function $f:\mathbb{R} \to \mathbb{R}$ which is continuous everywhere but has uncountably many roots ($x$ such that $f(x) = 0$). I am not looking for trivial examples such as $f = 0$ for all $x$. This is not a homework problem. I'd prefer a nudge in the right direction rather than an explicit example.

Thanks!

Edit: Thanks all! I've constructed my example with your help.

答: I just thought I'd mention that a Brownian motion has this property (almost surely).</p>

ContentLicense: CC BY-SA 3.0


问:
Is this set countable? (intervals in $\mathbb{R}$)
A set consists of intervals in $\mathbb{R}$. They have no intersection, their starting point and end point can be any numbers(not limited to rational number), and the union all intervals should cover all real numbers. For example, the first interval is $[0,1)$ then the next interval can be $[1,6)$, but not $[0.5,2)$. How many such intervals? Is it countable infinity?

答:
The same rational number cannot be in two of these intervals. That gives us an injection from the set of intervals to $\mathbb Q$. Thus there are at most the cardinality of the rationals.  But there are only countably many rationals.

Note:  the axiom of choice may rear its head here.


问:
Help with derivation (just a single step)
I'm trying to understand the following step in a derivation:

Step where I need help

Could anyone explain the process?

Thank you so much!!

答:
Using the power series $(1-x)^{-1} = 1 + x + x^2 + \dots$, we have

$$(a-b\epsilon)^{-1} = \frac {1/a}{1-b\epsilon/a} = \frac1a\left(1 + \frac {b\epsilon}a + \left(\frac {b\epsilon}a\right)^2 + \dots\right) = a^{-1}(1+o(\epsilon))$$

Therefore

\begin{align}
\Delta_\epsilon &\approx -[\nabla^2R(\hat \theta) + \epsilon\nabla^2L(2,\hat\theta)]^{-1}[\nabla R(\hat\theta) + \epsilon\nabla L(z,\hat\theta)]\\
&=-(\nabla^2R(\hat \theta))^{-1}(1 + o(\epsilon))[0+\epsilon\nabla L(z,\hat\theta)]\\
&\approx-\nabla^2R(\hat \theta)^{-1}\nabla L(z,\hat\theta)\epsilon
\end{align}


问:
Prove that any 2-independent set of $\{0,1\}^n$ has size at least $n+1$
Definition: Set S is called 2-independent subset of $\{0,1\}^n$ if for every two different positions $i_1, i_2 \in \{1,2, \dots , n\}$ and for every two numbers $a_1, a_2 \in \{0,1\}$ the probability
$$ Pr_{x \sim U(S)}[x_{i_1} = a_1, \; x_{i_2} = a_2] = \frac{1}{4}$$
The probability is calculated over $x$ taken from uniform distribution over the set $S$.

It is easy to see that this definition is equivalent to: for every two positions every combination of bits (00, 01, 10, 11) occur in $n/4$ vectors from $S$ in this positions.

The problem is to prove lower bound on size of $S$.

Question: Prove that $|S| \ge n + 1$.

I've found a hint, that didn't help a lot:

Hint: Find a set of linearly independent vectors in $R^{|S|}$ of cardinality at least $n + 1$. Specifically, for $i \in \{1, 2, . . . , n\}$, consider the vector $\langle \chi_i (x)\rangle_{x\in S}$ where $\chi_i(x) = (−1)^{x_i}$ and vector $\langle 1,1, \dots,1 \rangle$.

答:
A hint on how to use the hint:

Consider the $A$ of $n + 1$ vectors in $\Bbb R^{|S|}$, which are the vectors $\langle \chi_i(x)\rangle_{x \in S}$ for $i = 1, \dots, n$, together with the vector $\langle 1, \dots, 1\rangle$.

Try to show that

each vector in $A$ has length $\sqrt n$;

any two different vectors in $A$ are orthogonal to each other.

This proves that the set $A$ is linearly independent over $\Bbb R$ and hence $|S| = \operatorname{dim}(\Bbb R^{|S|}) \geq |A| = n + 1$.


答: It seems that step (12) $\epsilon\nabla^2 L(z,\hat{\theta})$ can be dropped. Then the equation becomes $\nabla_{\epsilon} \approx -[\nabla^2 R(\hat{\theta})]^{-1} \cdot [\nabla R(\hat{\theta}) + \epsilon \nabla^2 L(z,\hat{\theta})]$. '$\hat{\theta}$ minimizes R' means that the gradient of R, $\nabla R(\hat{\theta})$, shoule be $0$. As a result $\nabla_{\epsilon} \approx -[\nabla^2 R(\hat{\theta})]^{-1} \cdot [\epsilon \nabla^2 L(z,\hat{\theta})]$

Title: Help with derivation (just a single step)

Body:
I'm trying to understand the following step in a derivation:

Step where I need help

Could anyone explain the process?

Thank you so much!!


问:
In definite integration, do I have to watch out for the behaviour of a function between the limits?

Calculate -
$$\int^{1}_{-1}\frac{e^\frac{-1}{x}dx}{x^2(1+e^\frac{-2}{x})}$$

This is my approach-

Substitute $t=e^\frac{-1}{x}$

$$\therefore \int^{\frac{1}{e}}_{e}\frac{dt}{1+t^2}$$

which gives us $\frac{\pi}{2}- 2\arctan{e}$, but the answer given is $\pi- 2\arctan{e}$. I'm guessing that my wrong answer has something to do with the exponential term.

答:
The answer to the question in the header is yes, and this is a really interesting example.

When you substitute a variable, the substitution applies to the whole interval of integration. If you say $t=e^{\frac{-1}{x}}$, you mean that all the values that $x$ takes on are related to the values that $t$ takes by that relation. Really you're applying the function $e^{\frac{-1}{x}}$ to the whole interval of the $x$'s, i.e. $[-1,1]$. Normally, when we do substitution the function you use maps an interval to an interval, so it suffices to just look at the endpoints. Not so with this function! What is the image of $[-1,1]$ under $e^{-\frac1{x}}$? It's $[0,\frac1e]\cup [e,\infty]$, which you can see in a number of ways, e.g. the image of $[-1,1]$ under $1/x$ is $[-\infty,-1]\cup [1,\infty]$, then map it through $e^{-x}$ to get the right interval. Thus the correct substitution ought to be \begin{eqnarray}
&&\int_{-1}^1\frac{e^{-\frac1x}}{x^2(1+e^{-\frac2x})}dx = \int_{[0,\frac1e]\cup [e,\infty]} \frac1{1+t^2} dt = \int_0^{\frac1e} \frac1{1+t^2} + \int_e^\infty \frac1{1+t^2} dt\\ &=& \arctan{\frac1e} - \arctan0 + \arctan{\infty} - \arctan{e} = \arctan\frac1e + \frac\pi2 - \arctan(e) \\&=& \pi-2\arctan(e)
\end{eqnarray}


问:
If invertible matrices $A$, $B$ satisfy $AB=cBA$, then $c^n=1$
Suppose $A$ and $B$ are invertible $n\times n$ matrix such that $AB=cBA$ where $c\in\mathbb{C}$. Show that $c^n=1$.

I have tried this question from different angles. $n=1$ is obvious. But even for $n=2$, I am not able to show this, even after multiplying, squaring the equations etc.

Body: $$AB=cBA\implies \det(AB)=c^n\det(BA)\implies c^n=1.$$


问:
Solving simultaneous equations using a determinant
Upon revisiting the lecture notes from the first year undergraduate mathematics course I took, I found the following problem

Factorise the determinant $$\text{det}\begin{pmatrix}z & 1 & 2 \\ 1 & z & 3 \\ 1 & 1 & z+1 \end{pmatrix}$$ and hence solve the simultaneous equations $$ zx + y = 2, \ \ \  x + zy = 3, \ \ \ x + y = z+1.$$

The determinant is $(z-1)(z^2+2z-4)$, but how does this help to solve the simultaneous equations?

One way to solve the equations is by substituting $y = z + 1 - x$ into the first two equations, yielding $$z(x+1) = x + 1 $$ and $$x + z(z + 1 - x) = 3. $$ Then if $x \neq -1$ we get $z = 1$, implying that $x + 2 - x = 3$, a contradiction. Hence, $x = -1$.

Substituting this into the three simultaneous equations we get $zy = 4$ and $y = z + 2$. Hence $z \neq 0$, and so $\frac{4}{z} = z + 2$, yielding $z^2 + 2z - 4 = 0$ (one of the factors of the determinant!). Solving this quadratic equation for $z$ and using the fact that $y = z+2$ we obtain $z = -1 \pm \sqrt{5}$ and $y = 1 \pm \sqrt{5}$.

I'm not seeing the connection to the determinant in the statement of the problem. How does factoring the determinant lead to a solution of the simultaneous equations?

答:
Let $y=z+1-x$. Then the two remaining equations are
\begin{align*}
f(x,z) & =xz - x + z - 1=0,\cr
g(x,z) & =- xz + x + z^2 + z - 3=0.
\end{align*}
Now the resultant of $f$ and $g$ with respect to $x$ is given by
$$
{\rm res}(f,g,x)=z^3 + z^2 - 6z + 4=(z-1)(z^2+2z-4),
$$
which is exactly the determinant in question. Of course, the resultant gives the common roots of the two polynomials, i.e., it resolves the equations.

The result is not surprising, since we know that the resultant of two polynomials can be computed by the determinant of the Sylvester matrix, see here.


问:
Which Statement is True and Which One is False?
Which of the following statements is true or false and why?

Image Provided

答: The first one is wrong, since the global minimum of $x^2 - 2x -3$ is -4. The second one is right, just pick $y = -5$ (whatever number smaller than -4).


问:
possible dimensions of a nilpotent endomorphism of a K-Vectorspace
I have the following exercise in linear algebra that I think I know what they're looking for, however have not really got an idea which theorems to use / how to approach this.

Let $V$ be a 5-dimensional $K$-Vectorspace, and $N$ a nilpotent endomorphism of $V$. For $r \in \mathbb{N}$, we set $V^{(r)} =$ Ker$(N^{\circ r})$ and $d_r =$ dim$V^{(r)}$. What are the possible sequences of integers that one may get in this way$?$ For each possible sequence, give an example of a nilpotent endomorphism $N$ realising it.

From my script I know that Matrix $A$ is nilpotent if $A^k = 0$ for some integer $k$.

So if I understood this correctly, we are looking for possible sequences for dim(Ker$(N^{\circ r})$).

My guess is that based off of the definition of a nilpotent matrix, the sequence will always end with $0$. However I am not sure if there is a theorem to calculate the first elements of the sequence.

How can I approach calculating these possible sequences?

Body:
If you know the theory of the Jordan Form this is reasonably straightforward. (But I suspect that you are expected to do the problem bare-handed as it were.)

As $N$ is nilpotent we have that $N^k=0$ for some $k$, and as $\dim V=5$ we see (by Cayley-Hamilton) that $k\leqslant 5$, and $m_A(X)=X, X^2, X^3, X^4, \text{ or } X^5$.

The eigenvalues of $N$ are all $0$, and by the general theory $N$ is similar to one of

$$
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix}.
$$
The ranks of these are visibly $0,1,2,2,3,3,4$.

As for $N^2$ we have the following (in the same order)
$$
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix}
$$
whose ranks are $0,0,0,1,1,2,3$.

Now we get for $N^3$
$$
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix}
$$
whose ranks are $0,0,0,0,0,1,2$.

For $N^4$ we have
$$
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix},
\begin{pmatrix}
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0\\
\end{pmatrix}
$$
whose ranks are $0,0,0,0,0,0,1$.

For all $k\geqslant 5$ we have $N^5=0$ whose rank is $0$.

By the Rank-Nullity Theorem you can now write down the possible sequences $(\dim\ker N^k)_{k=0}^{\infty}$


问:
Orthogonal projections and being trace class
On some Hilbert space $\mathcal{H}$, I have two orthogonal projections $P,Q:\mathcal{H}\rightarrow \mathcal{H}$. In fact, $Q=UPU^*$ for some unitary $U$ (Note that $Q$ is indeed another orthogonal projection).

I have shown that the following operator
\begin{align}
A:=P-PQP=P-PUPU^*P\,,
\end{align}
is trace-class. However, in this particular setting, I am very interested in whether or not this operator is trace-class,
\begin{align}
B:=P-Q = P-UPU^*\,.
\end{align}
Notice how $A = PBP$. It might not be true at all that $B$ is trace-class, but I was wondering if there are any results out there about trace-class products of operators that I can exploit to investigate whether or not $B$ is trace-class. Intuitively, I think the trace-class property of $A$ must have something to do with $B$, because in my case the projection $P$ is the projection of $L^2(\mathbb{R}^2)$ onto $L^2(\mathbb{R}_+\times \mathbb{R})$, that is, it restricts functions to the upper half-plane (I defined this projection $P$ by means of multiplication operators). I have showed that an orthogonal projection is only trace-class if the dimension of its range is finite, and this really doesn't seem to be the case for $P$, so my guts are telling me that it must have something to do with $B$.

Any links to material are appreciated, or arguments for either cases. Thanks.

答:
Consider $\mathcal{H}= \mathcal{K}\oplus \mathcal{K}$, where $\mathcal{K}$ is an arbitrary Hilbert space and let $P$ be the orthogonal projection onto the first component and $Q$ on the second component. Let $U: \mathcal{H}\to \mathcal{H}$ be the unitary that switches components. Then
$$Q = UPU^* = UPU$$
Note further that $PQ = QP = 0$. Thus $B= P-Q$ is also a projection. However, if $\mathcal{K}$ has infinite-dimension, then $B$ does not have finite-dimensional range and thus $B$ is not trace-class by the observation you have made.

Thus, in general, $B$ not not be trace-class.


问:
What is the relationship between $H(A)$ (entropy of $A$) and $H(B)$ when $B$ is defined in terms of $A$?
I'm trying to solve the following problem:

Given a discrete random variable $A$ with alphabet $\chi \subset \{ 1, 2, \dots\}$, what is the relationship between $H(A)$, the entropy of $A$, and $H(B)$, the entropy of $B$, when $B$ is defined as follows:

(a) $B = \log_2(A).$

(b) $B = A^2.$

As I understand, the problem is solved by looking at the range of the variables $A$ and $B$. Let $\chi$ be the range of $A$, then, by how $B$ is defined in both cases, we have that: $|\chi|=|range(B)|$. That is, we have a one-to-one correspondence and for that reason $H(A)=H(B)$.

Is my reasoning correct? Any suggestion is welcome. Thanks in advance.

答:
It's true that, for discrete variables, a one-to-one function does not change the entropy. This should be obvious conceptually (entropy as a measure of information). A small proof follows.

In general (from the chain rule) we have $H(A,B)= H(A)+ H(B|A)  = H(B) + H(A|B)$.

If $B=f(A)$ ($B$ is a deterministic funcion of $A$), then $H(B|A)=0$. If, besides, the function is one-to-one, so that $A=f^{-1}(B)$, then we also have  $H(A|B)=0$.

Hence $H(A)=H(B)$


问:
Express the integral as beta function
How can this definite integral be represented in the form of $\beta$ function ?

$ \int_0^1 \frac{x dx}{\sqrt{1-x^5}} $

The answer is $\frac{1}{5}\beta (\frac{2}{5}, \frac{1}{2})$ but I am not sure how to reach here

Title: Express the integral as beta function

Body: Perform the substitution $u= x^5$, $x = u^{1/5}$, $dx = (1/5)u^{-4/5}$.


问:
Simple random walk reaching one value before terminating
Setup: I have a Simple Random Walk (SRW) that starts in 1 and will terminate whenever it reaches 0. The probability of increasing and decreasing by $1$ is $\frac 1 2$ each.

Question: I'm trying to figure out the probability for the event that the SRW reaches a value $K\in\mathbb N$ before it terminates i.e. reaches 0. In my paper it says that the probability is given by $(1-\frac 1 K).$

Background: I came across this problem while reading a paper about Brownian Motion that is in some way reduced to a Simple Random Walk (I belive the details are unnecessary for this question, if you disagree please ask for further information).

At first, I thought that this question is quite helpful, however it does not include that termination aspect with which I am struggling.

答:
I think I have just got some sort of answer thanks to this question.

So, let's say $S_n=1 + \sum_{k=0}^n X_k$ where $(X_k)k$ are i.i.d. random variables with $\mathbb P (X_k = 1) = \frac 1 2 = 1 - \mathbb P (X_k = -1).$

Then, set $S_T = \inf\{ n\in\mathbb N :S_n \in \{0,K\} \}$ and calculate
$$
\mathbb E[S_T] = 0 \ \mathbb P (S_T = 0) + K\ \mathbb P (S_T = K) = K (1 - \mathbb P (S_T = 0)) = K(1-\mathbb P(T_0 < T_K)),
$$
where
$$
T_0 = \inf\{n: S_n = 0\},\ T_K = \inf\{ n:S_n = K \}.
$$
With the Optional Stopping Theorem, you get $\mathbb E[S_T]= \mathbb E[S_0] = 1$, and thus
$$
K(1-\mathbb P(T_0 <T_K)) = 1,
$$
which then leads to $\mathbb P(T_0 <T_K) = 1-\frac 1 K.$


问:
Closed formula to count number of binary numbers of length $x$ having at least $y$ $1$ bits
I'm interested in solving a sub problem of the algorithm related question from SO How many binary numbers having given constraints .... The sub problem being, having $x \geq y$

determine how many binary numbers of length $x$ have at least $y$ $1$ bits

The immediate algorithm is to go over all $2^x$ binary numbers and count the ones having at least $y$ $1$ bits. Time complexity of such algorithm is $O(2^x)$ or even $O(x2^x)$ since for each number, one has to count the $1$ inside using a loop through all bits (actually $x/2$ on average, but this is not the point).

My target is ideally to find a $O(1)$ algorithm, i.e. a closed formula $f$ such that $\textrm{numbers} = f(x,y)$.

I started with some observations. For $x = 5$, with x being either $0$ or $1$:

```<code>y = 1         y = 2    y = 3
1xxxx (2^4)   11xxx    111xx
01xxx (2^3)   101xx    (...)
001xx (2^2)   011xx    Σ = 1x2^2+3x2^1+6x2^0
0001x (2^1)   1001x
00001 (2^0)   0101x
              0011x
Σ = 2^5 - 1   10001
              01001
              00101
              00011

    Σ = 1x2^3+2x2^2+3x2^1+4x2^0
</code>```

The number of terms is $C_x^y$ and from there it is possible to improve the initial algorithm into parsing only $C_x^y$ elements instead of $2^x$, and as a developer that could make me happy, but, and this is the question,

is there a closed formula that gives directly the number of binary numbers of length $x$ having at least $y$ $1$ bits?

答:
The required value should be the following:$$\sum_{i=0}^{x-y}\binom{x}{y+i}$$

This is because you need to select $y, (y+1), ..., x$ positions from the total $x$ positions to place $1$'s, and the rest zeroes.

This expression is just a sum of the last $y$ binomial coefficients. As far as I know, there is no general expression for the sum of $k$ binomial coefficients to give you an $O(1)$ answer.

You may find this link helpful for finding a efficient algorithm or a good bound on the sum.


问:
Show that $K=\{T\in\mathcal{L}(V,W): T\ \text{not injective}\}$ is not a subspace of $\mathcal{L}(V,W)$
Suppose $V$ and $W$ are finite-dimensional with $2\le\dim(V)\le\dim(W)$. Show that $K=\{T\in\mathcal{L}(V,W): T\ \text{not injective}\}$ is not a subspace of $\mathcal{L}(V,W)$. My attempt is to pass by fundamental theorem of linear maps. I would like to know if my solution holds, please.

For me, to show that $K$ is not a subspace of $\mathcal{L}(V,W)$ is equivalent to show that $\forall S\in K$ $S$ is not linear. Is it correct?

Suppose by absurd that $S\in K$ is linear. So, by fundamental theorem of linear map we have that $\dim(V)=\dim(\text{range} \ S)+\dim(\text{null}\ S)$

But, $\dim(\text{range} \ S)+\dim(\text{null}\ S)\ge\dim(V)+\dim(\text{null} \ S)\ge\dim(V)+1\neq \dim(V)$ as $\dim(\text{null} \ S)\ \neq 0$ because $S$ is not injective. Thus, each element $S$ of $K$ is not a linear application. We conclude that $K$ is not a subspace of $\mathcal{L}(V,W)$.

答:
As Bargabbiati wrote in his comment, your proof is not valid.

Hint: Let $P \in \mathcal{L}(V,W)$ be a projection (this means $P^2=P$) with $0 \ne P \ne I.$

Then $P \in K.$ Now let $Q:= I-P$.

Show that $Q \in K$, but $P+Q \notin K.$


问:
Conditional probability of triple joint density
The joint density function of X,Y and Z is
$$f(x,y,z) = \begin{cases}6(z-y)& \text{for } 0<x<1,
\; 0<y<z<1,\\
0 &\text{otherwise} \end{cases}$$

So finding the probability $P(X + Y + Z  \le 1)$ gives us
$$P(X + Y + Z  \le 1) = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 6(z-y) \ \mathrm dz \ \mathrm dy \ \mathrm dx$$

For a conditional density $P(X \le Z | X + Y + Z  \le 1)$ this is equal to $$\frac{P(X \le Z , X + Y + Z  \le 1)}{P(X + Y + Z  \le 1)}$$

I am unsure what bounds of integration to use for the top part of this.

答:
Since  $y < z$, your denominator could be $$P(X + Y + Z  \le 1) = \int_{x=0}^1 \int_{z=0}^{1-x} \int_{y=0}^{\min(z,1-x-z)}  6(z-y) \ \mathrm dy \ \mathrm dz \ \mathrm dx $$ or $$P(X + Y + Z  \le 1) = \int_{z=0}^{1} \int_{y=0}^{\min(z,1-z)}  \int_{x=0}^{1-y-z}  6(z-y) \ \mathrm dx \ \mathrm dy \ \mathrm dz $$

Your numerator could then be
$$P(X\le Z, X + Y + Z  \le 1) = \int_{z=0}^{1} \int_{y=0}^{\min(z,1-z)}  \int_{x=0}^{\min(z,1-y-z)}  6(z-y) \ \mathrm dx \ \mathrm dy \ \mathrm dz $$


问:
One-liner probability problem
Two people flip a coin $n$ times each. What is the probability that they both get the same number of heads?

If I let $k$ be the number of heads, I can get $$\frac{{n\choose k}^2}{2^{2n}}$$ which is correct so far: but I kind of don't understand why this is correct. Why doesn't $\frac{1}{2}$ show up anywhere? I would've expected $$(\frac{1}{2})^k (\frac{1}{2})^{n-k}$$ to appear somewhere. What I initially proposed was $\frac{\left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{n-k} {n\choose k}}{\left(\frac{1}{2}\right)^{2n}}$, why is this wrong?

Anyway, i could continue with the first equation to produce $$\frac{{2n\choose n}}{2^{2n}}$$ as the final probability. But I think there should be a logical argument that leads straight to the final answer in one line, given the equiprobability events. What could that be?

Edit: perhaps, the equiprobability explains my flawed second argument? I don't need to consider $\frac{1}{2}$ because both events are equally likely, so choosing is sufficient?

答:
Because of the equiprobability we have:
$$
\left(\frac12\right)^k\left(\frac12\right)^{n-k}=\left(\frac12\right)^n
$$
so that $k$ magically disappears.

And you are right: one can directly obtain the final formula. Imagine $2n$ cells which you need to fill with heads and tails, the first half belonging to the first player and the second half - to the second one. We need to count all sequences with the equal number of heads in both halves. To do this choose arbitrary $n$ cells for heads and then invert heads and tails in the second half, which results in the equal numbers of heads in both halves. Since this procedure is obviously bijective, it gives us immediately the final expression.

Title: One-liner probability problem


答:
i) Bounds for $P(X + Y + Z  \le 1)$

We also have $0 \le y \le z, 0 \le z \le 1, 0 \le x \le 1$.

Intersection of $y = z$ and $x+y+z = 1$ is given by $\displaystyle y = \frac{1-x}{2}$

So the bounds should be,

$y \leq z \leq (1-x-y), 0 \leq y \leq \frac{1-x}{2}, 0 \leq x \leq 1$. The integral will be,

$\displaystyle \int_0^{1} \int_0^{(1-x)/2} \int_y^{1-x-y}  6(z-y) \ dz \ dy \ dx$

ii) Now for second integral we have additional condition of $x \le z$. We will have to split the integral into two - for $y \ge x$ and for $x \ge y$.

Intersection of $x = z, y = z, x + y + z = 1$ is $x = y = z = \displaystyle \frac{1}{3}$

For $y \ge x$,

Upper bound of $y$ is intersection of $x+y+z = 1$ and $y = z$. So the integral is,

$I_1 = \displaystyle \int_0^{1/3} \int_{x}^{(1-x)/2} \int_y^{1-x-y}  6(z-y) \ dz \ dy \ dx$

For $x \ge y$,

Upper bound of $x$ is intersection of $x+y+z = 1$ and $x = z$. So the integral is,

$I_2 = \displaystyle \int_0^{1/3} \int_{y}^{(1-y)/2} \int_x^{1-x-y}  6(z-y) \ dz \ dx \ dy$

Title: Conditional probability of triple joint density

Body:
The joint density function of X,Y and Z is
$$f(x,y,z) = \begin{cases}6(z-y)& \text{for } 0<x<1,
\; 0<y<z<1,\\
0 &\text{otherwise} \end{cases}$$

So finding the probability $P(X + Y + Z  \le 1)$ gives us
$$P(X + Y + Z  \le 1) = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 6(z-y) \ \mathrm dz \ \mathrm dy \ \mathrm dx$$

For a conditional density $P(X \le Z | X + Y + Z  \le 1)$ this is equal to $$\frac{P(X \le Z , X + Y + Z  \le 1)}{P(X + Y + Z  \le 1)}$$

I am unsure what bounds of integration to use for the top part of this.


问:
Evaluate $ f(a) = \int^a_{-a}|x|dx$ of $a=2$
I want to evaluate this integral $$ f(a) = \int^a_{-a}|x|dx$$between $a$ and $-a$ and I know that  $\int^a_{-a}|x|dx$ is divided into two: $$\int^a_{-a}|x|dx \rightarrow a>0 = \int^a_{0}xdx  $$
$$\int^a_{-a}|x|dx \rightarrow a<0 = \int^0_{-a}-xdx  $$

after I evaluate it I get zero, so I dont know if I`m right.

What I did is :

$$a<0 \rightarrow \int^0_{-2}-xdx = \frac{-x^2}{2}|^{0}_{-2} = -2 $$
$$a>0 \rightarrow \int^2_{0}xdx = \frac{x^2}{2}|^{2}_{0} = 2 $$
Thanks!

Body:
You've got the sign of the first integral wrong:
$$\int^0_{-2}-xdx = \frac{-x^2}{2}|^{0}_{-2} = 0 - \frac{-(-2)^2}{2}=2$$


答:
It can be done using elementary calculations (without using optional stpping theorem) as well. Here is a sketch:
Write $X_n$ as your MC. You want to compute $\mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_0=1)$. Write $l_i=\mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_0=i)$. Now do a one step conditioning

$$
l_1=\mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_0=1)\\ = \mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_1=2,X_0=1)\mathbb{P}(X_1=2|X_0=1) \\
+\mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_1=0,X_0=1)\mathbb{P}(X_1=0|X_0=1).
$$
Therefore you get
$$
\mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_0=1) =\frac{1}{2} \mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_1=2,X_0=1) \\
=\frac{1}{2} \mathbb{P}(\text{MC visits }K \text{ before visiting 0}|X_0=2) =l_2
$$

Now using the similar argument you get for $i\geq 2$,
$$
l_{i}= 0.5l_{i-1}+ 0.5l_{i+1},
$$
simplify and get
$$
l_{i}-l_{i-1}= l_{i+1}-l_{i},
$$
Then use telescopic sum etc.

Title: Simple random walk reaching one value before terminating

Body:
Setup: I have a Simple Random Walk (SRW) that starts in 1 and will terminate whenever it reaches 0. The probability of increasing and decreasing by $1$ is $\frac 1 2$ each.

Question: I'm trying to figure out the probability for the event that the SRW reaches a value $K\in\mathbb N$ before it terminates i.e. reaches 0. In my paper it says that the probability is given by $(1-\frac 1 K).$

Background: I came across this problem while reading a paper about Brownian Motion that is in some way reduced to a Simple Random Walk (I belive the details are unnecessary for this question, if you disagree please ask for further information).

At first, I thought that this question is quite helpful, however it does not include that termination aspect with which I am struggling.


问:
Prove the inequalites of complex vectors
Let $x(t), y(t)\in \mathbb{C}_{N}$, where
$$   \mathbb{C}_{N}=\{x:\mathbb{Z}\to \mathbb{C}|x(j)=x(j+N)~\text{for any}~j\in \mathbb{Z}\} $$
Define the functions
$$ R_{xy}(j)=\sum_{k=0}^{N-1}x(k)\overline{y(k-j)} $$
and
$$ R_{xx}(j)=\sum_{k=0}^{N-1}x(k)\overline{x(k-j)} $$
$$ R_{yy}(j)=\sum_{k=0}^{N-1}y(k)\overline{y(k-j)} $$

Show that
\begin{equation}
\sum_{j=0}^{N-1}|R_{xy}(j)|^2\geq R_{xx}(0)R_{yy}(0)-\left(\sum_{j=1}^{N-1}|R_{xx}(j)|^{2}  \right)^{\frac{1}{2}}\left(\sum_{j=1}^{N-1}|R_{yy}(j)|^{2}  \right)^{\frac{1}{2}}
\end{equation}
and
\begin{equation}
\sum_{j=0}^{N-1}|R_{xy}(j)|^2\leq R_{xx}(0)R_{yy}(0)+\left(\sum_{j=1}^{N-1}|R_{xx}(j)|^{2}  \right)^{\frac{1}{2}}\left(\sum_{j=1}^{N-1}|R_{yy}(j)|^{2}  \right)^{\frac{1}{2}}.
\end{equation}
When does the equality holds?

I have trying the Cauchy-Schwarz inequality, but I stuck here. Can someone help me with above inequalites? Thank you very much!

答:
Since the functions $R$ are convolutions, we might use the Fourier transform. Let's define it as
$$\hat{x}(k) := \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}x(j)e^{-2\pi ijk/N}.$$
You may see that
$$\begin{align*}
(i)\quad &x(j) = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\hat{x}(k)e^{2\pi i jk/N}; \\
(ii)\quad &\lVert x\rVert_2 = \lVert \hat{x}\rVert_2; \\
(iii)\quad &\hat{R}_{xy}(k) = \sqrt{N}\hat{x}\bar{\hat{y}}(k).\end{align*}$$
By homogeneity we can assume that $\lVert \hat{x}\rVert_2 = \lVert \hat{y}\rVert_2 = 1$, and using these properties both inequalities get into
$$N\lVert \hat{x}\hat{y} \rVert_2^2 \geq 1-(N\lVert \hat{x}\rVert_4^4-1)^{1/2}(N\lVert \hat{y}\rVert_4^4-1)^{1/2}$$
and
$$N\lVert \hat{x}\hat{y} \rVert_2^2 \leq 1+(N\lVert \hat{x}\rVert_4^4-1)^{1/2}(N\lVert \hat{y}\rVert_4^4-1)^{1/2}.$$
Now we can think in terms of multiplication instead of convolution. We can write these equations more briefly, changing notation also, as
$$\lvert N\lVert fg \rVert_2^2-1\rvert \le (N\lVert f\rVert_4^4-1)^{1/2}(N\lVert g\rVert_4^4-1)^{1/2}. \qquad(*)$$
We don't lose anything if we think of $f$ and $g$ as positive functions.

Now we exploit some probabilistic intuition. Since $\lVert f\rVert_2 = 1$, we have that the mean value of $f$ is $1/\sqrt{N}$, so let's rewrite the left hand side of (*) as

$$N\Big(\sum f^2g^2 \Big) - 1 = N\sum\Big(f^2-\frac{1}{N}\Big)\Big(g^2-\frac{1}{N}\Big).$$
We use Cauchy-Schwarz to see that
$$\begin{align*}
\lvert\sum\Big(f^2-\frac{1}{N}\Big)\Big(g^2-\frac{1}{N}\Big)\rvert &\le \Big[\sum\Big(f^2-\frac{1}{N}\Big)^2\Big]^{1/2}\Big[\sum\Big(g^2-\frac{1}{N}\Big)^2\Big]^{1/2} \\
&= [\lVert f\rVert_4^4-1/N]^{1/2}[\lVert g\rVert_4^4-1/N]^{1/2},
 \end{align*}$$
which implies (*). I hope there is no mistake.


问:
Help to resolve an integral
I am reading an article (effect of carbon taxes and subsidies on optimal forest rotation age and supply of carbon services) that has the following integral:

$$ PV(c)= \int_0^T pcαQ'(t)e^{-rt}\, dt$$

Later on, the article states that "integrating by parts gives":

$$ PV(c)= pcα \left(Q(T)e^{-rT} + r∫_0^T Q(t)e^{-rt} dt\right) $$

I have been trying to resolve this integral but cannot arrive at the primitive function, that is, at the second equation. I admit my math is rusty. Any help will be most welcome. Thanks in advance.

Title: Help to resolve an integral

Body:
$$\int_{0}^T f^{\prime}(t)g(t)dt = f(t)g(t)  \Bigg| ^{T}_{0}  -\int_0^T f(t)g^{\prime}(t)dt$$

with $f(t)=Q(t)$ and $g(t)=e^{-rt}$. So $g^{\prime}(t)=-re^{rt}$.


问:
Why $\nabla\ln x=\nabla x/x$ include the $\nabla x$ in the numerator?
Reading a book I came across this identity:

$\nabla  \ln(x) = \frac{\nabla x}{x}$

I understand that the derivative of $\ln(x)$ is $x$, but I don't understand why the gradient (which is a vector of partial derivatives) of $\ln(x)$ is equal to $\nabla x$ divided by $x$. Is it because of the chain rule?

答: Yes:$$\nabla f(x)=\sum_i\frac{\partial f}{\partial y^i}e^i=f^\prime(x)\sum_i\frac{\partial x}{\partial y^i}e^i=f^\prime(x)\nabla x. $$


问:
Syndrome Decoding - Coset Leader
Currently, I am studying linear coding and came across the syndrome decoding. I have some difficulties trying to solve this problem:

Let $C$ be  the  linear $[10,5]$-code  over $\mathbb{F}_3$ with generator matrix
$$G=
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 2 & 2 & 1 & 1\\
0 & 1 & 0 & 0 & 0 & 2 & 0 & 1 & 2 & 1\\
0 & 0 & 1 & 0 & 0 & 2 & 1 & 0 & 1 & 2\\
0 & 0 & 0 & 1 & 0 & 2 & 1 & 2 & 0 & 1\\
0 & 0 & 0 & 0 & 1 & 2 & 2 & 1 & 1 & 0
\end{bmatrix}$$
By syndrome decoding, decode
$u=
\begin{pmatrix}
  1 & 0 & 2 & 2 & 0 & 1 & 2 & 1 & 0 & 0\\
\end{pmatrix}.$

My approach: I know the parity-check matrix to be, in this case, $H=[-G^T\,I]$. That is,
$$H=\begin{bmatrix}
0 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0\\
1 & 0 & 2 & 2 & 1 & 0 & 1 & 0 & 0 & 0\\
1 & 2 & 0 & 1 & 2 & 0 & 0 & 1 & 0 & 0\\
2 & 1 & 2 & 0 & 2 & 0 & 0 & 0 & 1 & 0\\
2 & 2 & 1 & 2 & 0 & 0 & 0 & 0 & 0 & 1
\end{bmatrix}.$$

The syndrome of $u$ is
$\text{syn}(u)=Hu^T=
\begin{pmatrix}
2 & 2 & 1 & 0 & 2
\end{pmatrix}^T.$

Now, I am not sure on how to proceed here. How do I find a coset leader, $\alpha$, corresponding to $\text{syn}(u)$ so that I may decode $u$ as $u-\alpha$? Many thanks in advance!

答:
We have $3^5=243$ codewords, hence the standard array would have $243$ columns and $3^{10}/3^5=243$ rows (that is, $243$ cosets, each having $243$ elements).

It would be too cumbersome to do it by brute force.

By inspection, we can see that your computed syndrome $s$ equals the sum of columns $4$ and $6$ (one-based indexing). Hence the tuple

$$e_1 = (0 0 0 1 0 1 0 0 )$$

verifies $H e_1^t= s$ and is one of the elements of the coset. You could find all the other elements by suming to it the $242$ non-zero codewords. The coset leader (asuming a channel with small probability of tribit change) would be the element with smallest weight. I'd bet that $e_1$ is the one.

Title: Syndrome Decoding - Coset Leader


问:
Meaning of $\inf$ in proof of fundamental theorem of algebra
$f = a_0 + a_1x+..+a_nx^n \in \Bbb C[x] $, $\deg(f)\geq1$
$$\mu = \inf(|f(z)|,z\in \Bbb C[x])$$

I am reading proof of fundamental theorem of algebra and can't understand this part.$|f(z)|$ is a number yes?How can we define $inf$ of number.Sorry if it is silly question.

答:
For any given $z\in \Bbb C$, yes, $|f(z)|$ is a non-negative real number. However, that number is not what were $\inf$-ing. We are $\inf$-ing the set of all possible values $|f(z)|$ can take, for all possible $z\in \Bbb C$. That's what
$$
(|f(z)|,z\in\Bbb C)
$$
refers to: all possible values, not one arbitrary value. It is more conventional to use curly brackets for this, rather than round. But it's clear either way.

And I strongly suspect that's a typo in your $\inf$. We have $f\in \Bbb C[x]$, but when evaluating $f(z)$, the $z$ comes from $\Bbb C$.


答:
Let $A_k$ be the event "reaching $k$ before reaching $0$". We will show $P(A_k) =\frac{1}{k}$.

Clearly, $P(A_1)=1$.

Next, $P(A_{k+1} ) = P(A_{k+1} | A_k) P(A_k)$.
The conditional probability on the RHS is equal to starting from $k$ and getting to $k+1$ before getting to $0$.

If we reverse the picture and look from $k+1$ to the left,  the conditional probability is the same as starting from $(k+1)-k=1$ and getting to $(k+1)-(k+1)=0$ before getting to $(k+1)-0=k+1$. That is, $1-P(A_{k+1})$. In other words,

$$P(A_{k+1}) = (1-P(A_{k+1})) P(A_k)$$

We solve by induction, assuming $P(A_k)=\frac 1k$. Then

$P(A_{k+1}) = (1-P(A_{k+1}))/k$, or

$P(A_{k+1})(1+\frac1k)=\frac 1k \Rightarrow P(A_{k+1})= \frac{1}{k+1}$.


答:
You can build your closed formula by adding "number of terms with exactly x '1' bits" and "number of terms with exactly x-1 '1' bits" and so on.

The number of terms with length x and exactly k '1' bits (and hence x-k '0' bits) is
$$
 \binom {x} {k}=\frac{x!}{(x-k)!k!}
$$
Now you just have to add the terms up you are interested in.

Title: Closed formula to count number of binary numbers of length $x$ having at least $y$ $1$ bits

Body:
I'm interested in solving a sub problem of the algorithm related question from SO How many binary numbers having given constraints .... The sub problem being, having $x \geq y$

determine how many binary numbers of length $x$ have at least $y$ $1$ bits

The immediate algorithm is to go over all $2^x$ binary numbers and count the ones having at least $y$ $1$ bits. Time complexity of such algorithm is $O(2^x)$ or even $O(x2^x)$ since for each number, one has to count the $1$ inside using a loop through all bits (actually $x/2$ on average, but this is not the point).

My target is ideally to find a $O(1)$ algorithm, i.e. a closed formula $f$ such that $\textrm{numbers} = f(x,y)$.

I started with some observations. For $x = 5$, with x being either $0$ or $1$:

```<code>y = 1         y = 2    y = 3
1xxxx (2^4)   11xxx    111xx
01xxx (2^3)   101xx    (...)
001xx (2^2)   011xx    Σ = 1x2^2+3x2^1+6x2^0
0001x (2^1)   1001x
00001 (2^0)   0101x
              0011x
Σ = 2^5 - 1   10001
              01001
              00101
              00011

    Σ = 1x2^3+2x2^2+3x2^1+4x2^0
</code>```

The number of terms is $C_x^y$ and from there it is possible to improve the initial algorithm into parsing only $C_x^y$ elements instead of $2^x$, and as a developer that could make me happy, but, and this is the question,

is there a closed formula that gives directly the number of binary numbers of length $x$ having at least $y$ $1$ bits?


问:
Long Division of polynomials
How can we prove that $x^n$  isn't divisible by $g(x)=x^4+x+1$ without remainders?

I understand why for all $n<4$ it's working, but how with $n \ge  4$? How we can prove it formally?

Note: $x^n,\:g\left(x\right)\in \mathbb{Z}_2\left[x\right]$

答: $x^n$ splits completely in $\mathbb{Z}_2$, so if $x^4+x+1$ were a factor, it would split completely as well. But it has no roots in $\mathbb{Z}_2$.

Title: Long Division of polynomials


问:
Calculate area of plane in 3D space using X, Y, Z
I have a polygon that's positioned on the earth. For each point, I have:

The latitude in degrees (x)
The longitude in degress (y)
The altitude above sea level in meters.

An example of one of my points is  { x: 144.99533074458003, y: -32.39730248504104, z: 263.24070871460185 }

Now, in my application, a user draws a polygon in space, and I need to determine the area of that polygon. The user could draw a polygon around a door, or a roof maybe. Sometimes even a sidewalk.

Once the user is done, I have a list of all the points they selected.

An example could be this:

```<code>DOOR = [
  { x: 144.99533074458003, y: -32.39730248504104, z: 263.24070871460185 }, // First point, bottom left
  { x: 144.99533197184215, y: -32.39730172327194, z: 267.1472276731874 }, // Top left
  { x: 144.99534689346532, y: -32.39726245542652, z: 267.04165748822226 }, // Top right
  { x: 144.99534752390096, y: -32.39726407569114, z: 263.24884951333394 }, // Bottom right
  { x: 144.99533074458003, y: -32.39730248504104, z: 263.24070871460185 }, // Last point, same as first to close the loop.
];
</code>```

The issue now, is that I need to figure out the approximate area between all these points. I know that the polygon won't be perfectly flat, but I can even it out if needed to give an approximation.

My current strategy is as follows:

Find the distance between each of the points using the Haversine Formula.

Determine the height difference between the two points by subtracting z2 from z1.

Find the actual length of the line using Pythagorus.

Get the angle of the line relative to the horizontal plane using (Math.atan2(end.y - start.y, end.x - start.x) * 180) / Math.PI (Please excuse the Javascript, that's what I'm working in).

I then have a list of all the lengths and the angles of each line.

I then convert each angle to Radians.

Then I generate x and y values like this:

```<code>  endX = start.x + length * Math.cos(angle)
  endY = start.y + length * Math.sin(angle)
</code>```

This give me a list of the X.Y values of the points.

I then calculate the area using this formula

This strategy works when my plane is nearly horizontal, but as soon as I try use it on a vertical surface, I get incorrect results.

I guess there's a few questions here, and any help will be greatly appreciated.

Edit: This would need to work for any number of coordinates greater than 3, not only 4 sided polygons.

Thanks!

答: An easier approach would be calculating the projected area using the same formula you provided but ignoring the z-coordinate (let's call it $A_p$ for demonstration). Then you can calculate the area of the original polygon by multiplying $A_p$ by $\sec \theta$, where $\theta$ is the angle between the polygon and the horizontal.


问:
What could differential operator be?
Suppose $M$ is a differential manifold, and $C^\infty(M)$ denotes all the smooth functions on $M$. A linear operator $D:C^\infty(M) \rightarrow C^\infty(M)$ satisfies the Leibniz law: $D(f \cdot g)=D(f) \cdot g + f \cdot D(g)$. Must $D$ be the derivative alongside a given smooth vector field?

Bonus: What is the situation when we consider complex manifolds and holomorphic functions/vector fields?

答:
For smooth manifolds, the answer is yes: indeed, it is trivial to see that $D$ is a rough vector field. To prove that it is a smooth vector field, let $D^i(x)\partial_i$ be a description of $D$ in local coordinates $(x_\alpha,U_\alpha)$. It is known that $D$ is smooth (on $U_\alpha$) iff $D^i$ is smooth for all $i$. Now let $\varphi$ be a bump function such that, outside the chart $U_\alpha$, $\varphi\equiv 0$ and $\varphi=1$ on a neighbourhood of $x_0\in U_\alpha$ fixed. Then, for every $j$, $D(x_{\alpha,j}\cdot \varphi)=D^j$.
On the other hand, $D(x_{\alpha,j}\cdot \varphi)\in C^\infty(M)$ and so $D^j$ is smooth.
Thus $D$ is a vector field locally smooth everywhere, which means it is smooth.

This proof breaks down for complex manifolds, since we do not have partitions of unity and or bump functions.

Title: What could differential operator be?


答:
Since you haven't really showed any attempt, I'll just give a hint:

Suppose there were a polynomial $h(x)$ such that $h(x)\cdot g(x)=x^n$. Now think of the lowest and highest degree terms of the product $h(x)\cdot g(x)$.

Title: Long Division of polynomials

Body:
How can we prove that $x^n$  isn't divisible by $g(x)=x^4+x+1$ without remainders?

I understand why for all $n<4$ it's working, but how with $n \ge  4$? How we can prove it formally?

Note: $x^n,\:g\left(x\right)\in \mathbb{Z}_2\left[x\right]$


问:
Proving an example doesn't violate the Dominated Convergence Theorem
I just worked through the following example from Shreve's Stochastic Calculus for Finance.

For each positive integer $n$, define $f_n$ to be the normal density
with mean $0$ and variance $n$, i.e.,

$$f_n(x)=\frac{1}{\sqrt{2n\pi}}e^{-\frac{x^2}{2n}}\text{.}$$

He defines $f(x) = \lim_{n \to \infty} f_n(x)$. Then he poses the problem

(iii) Note that

$$ \lim_{n \to \infty} \int _{-\infty} ^{\infty} f_n(x)\, dx \neq \int_{-\infty}^{\infty}f(x)\,dx\text{.}$$

Explain why this does not violate the Monotone Convergence Theorem,
Theorem 1.4.5.

It's easy to see that these functions aren't monotonically decreasing, so the assumptions for the Monotone Convergence Theorem aren't satisfied. I now want to understand why this example does not violate the Dominated Convergence Theorem. I believe that it suffices to show that

$$\int_{-\infty}^{\infty} \sup_{n \in \mathbb{N}} f_n(x) \, dx \nless \infty \text{.}$$

I have two questions about this.

Can you confirm, yes or no, that the above (essentially) proves the $f_n$ do not satisfy the assumption of the Dominated Convergence Theorem?

Do you have an explanation for why the above display is true? A proof is preferable, the details aren't important.

答: Hint: Let $x>1$. $\sup_n f_n(x)\geq f_{[x^{2}]}(x)$ where $[,]$ is the greatest integer function. $f_{[x^{2}]}(x) \geq \frac  1 {(2\pi)^{1/2} |x|}e^{-x^{2}/2[x^{2}]}$. Note that  $x^{2}/[x^{2}] \to 1$ as $x \to \infty$. Conclude that $\int_0^{\infty} \sup_n f_n(x)dx=\infty$.


问:
Is there any mathematical operation from which the operands can be traced back from the result without knowing one of them?
hope everyone will be doing good.

I don't have any Mathematical background, so this question may seem awkward.

Can I perform some operation on two or more integers, by the result of which can be determined the operands, without knowing any of them.

For example, if I perform some mathemical operation on 5, 6, and 7, and I get 33(for example) as a result, how can I determine what were the original numbers before this operation was performed.

答:
If by operation, you mean "any rule taking two numbers and producing a third", then yes.

Essentially, what you are asking is if there exists an injective (i.e., one to one) function from $\mathbb N\times\mathbb N$ to $\mathbb N$, and indeed, such a function exists. For example, one such function would be $$(a,b)\mapsto 2^a\cdot 3^b$$

For example, if I take the number $48$, I can calculate what the numbers $a$ and $b$ were that matched this number. Because $48=3\cdot16=2^4\cdot3^1$, I know that $a$ was $4$ and $b$ was $1$.

However, I cannot get $33$ as a result of this operation, but there are plenty of operations where all integers can be their result. For example, if the operation is $$(a, b)\mapsto 2^{a-1}\cdot (2b - 1),$$ then I know that the result $33$ can only be the result of $a=1, b=17$.


问:
What is a proof of Euler's formula $e^{ix}=\cos(x)+i \sin(x)$ that doesn't use differentiation or De Moivre's theorem, if it exists at all?
The Taylor series expansions of $e^x$, $\sin(x)$ and $\cos(x)$ are the standard way to prove it, but does there exist an elementary way that doesn't use limits, calculus or De Moivre's theorem?

答:
The relationship between $e^x$ and $\sin(x)$ and $\cos(x)$

You know know the the $\text{exp()}$ function is a special one, and has many interesting properties among them is that expression $ e^{x} = \lim_{n \to \infty} (1+\frac{x}{n})^n $

If you expand this with binomial theorem, and put $n \rightarrow \infty$, you would get the series for $\text{exp()}$

$$ e^{x} = \lim_{n \to \infty} (1+\frac{x}{n})^n $$
$$ e^{x} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040} + \dots \dots $$
$$ e^{x}= \sum_{k=0}^{\infty}\frac{x^k}{k!}$$
From this series, replace $x$ with $xi$, where $i = \sqrt{-1}$ and split the result into real and imaginary section
$$ e^{x} = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{x^7}{5040} + \dots \dots $$
$$ e^{xi} = 1+xi+\frac{(xi)^2}{2}+\frac{(xi)^3}{6}+\frac{(xi)^4}{24}+\frac{(xi)^5}{120}+\frac{(xi)^6}{720}+\frac{(xi)^7}{5040} + \dots \dots $$
$$ e^{xi} = \{ 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+ \dots \} + \{ x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040} + \dots \} \cdot i$$
$$\sin(x) = x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040} + \dots $$
$$\cos(x) = 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+ \dots $$

Therefore $e^{xi} = \cos(x)+i\sin(x)$


问:
normally distributed mean 110cm variance $1050cm^2$. 2 plants chosen random. probability that the height is at least 1.5 times the height of the other
Height of a certain plants are normally distributed with mean $110\text{ cm}$ and variance $1050\text{ cm}^2$.

2 plants are chosen at random. What is the probability that the height
of one plant is at least 1.5 times the height of the other one?

This question Is the first I have seen doing past paper questions in statistics and probability its worth 7 marks I don't know how to solve it I know the answer is 0.346 but don't know how to get to it and what to I have no teacher as I am self studying so any help is much appreciated thank you

答:
If you select at random 2 plants it is self evident that one must be higher than the other, thus you have to calculate

$$\mathbb{P}[X>1.5Y]+\mathbb{P}[Y>1.5X]=2\mathbb{P}[X>1.5Y]=2\mathbb{P}[X-1.5Y>0]=$$

$$=2\cdot\left[1-\Phi\left(\frac{55}{\sqrt{3412.5}}\right)\right]\approx2\times 0.173=0.346$$

This because

$$(X-1.5Y)\sim N(110-1.5\times110;1050+1.5^2\times1050)=N(-55;3412.5)$$

To understand the reasoning think at the following question:

Selecting at random 2 plants what is the probability that one of the 2
is higher than the other?

the answer is obviously 1, but you can calculate it

$$\mathbb{P}[X>Y]+\mathbb{P}[X<Y]=0.5+0.5=1$$


问:
Optimal probability measure
Let $A$ be a finite set and let $\Bbb P$ be a probability measure on $A^{\Bbb N_0}$. Further, let $x_i:A^{\Bbb N_0}\to A$ be projection maps, so that $(x_i)_{i=0}^\infty$ can be treated as a stochastic process. Suppose that for some integer number $m\geq 1$ the measure $\Bbb P$ satisfies the following condition:
$$
  (x_i,\dots,x_{i+m})_*\Bbb P = (x_0,\dots,x_m)_*\Bbb P\quad \forall i\in \Bbb N,\tag{1}
$$
that is any sequence of the length $m+1$ has equal probability of appearance anywhere along the trajectory of the process. For example, $\Bbb P$ is shift-invariant (stationary) if and only if it satisfies $(1)$ for any choice of $m$. In my case, the number $m$ is fixed though, so $\Bbb P$ does not have to be a stationary measure.

Let us denote $\Bbb P^m := (x_0,\dots,x_m)_*\Bbb P$ to be a measure on $A^{m+1}$, and let $F:A^{m+1}\to\Bbb R$ be some arbitrary function. Suppose, I need to maximize
$$
  \Bbb P^m[F]:=\sum_{s\in A^{m+1}}F(s)\cdot\Bbb P^m(\{s\}) \tag{2}.
$$
where the optimization is done over all possible measures $\Bbb P$ that satisfy $(1)$. Have such problem been already studied, perhaps the solution is already known?

答:
Consider the following counterexample.

Let $A=\{0,1\}$ and $m=2$.
Consider the optimization problem with
$$F(110) = F(100) = F(001) = F(011) = 1,\qquad F=0\; \text{otherwise}.
$$

This problem is solved by
$$\mathbb{P}^m(110) = \mathbb{P}^m(100) = \mathbb{P}^m(001) = \mathbb{P}^m(011) = 1/4,
\qquad \mathbb{P}^m=0\; \text{otherwise}.$$
and the optimal value is $1$.

The stochastic process can be build by choosing with equal probability $1/4$ one of the
realizations

$(00110011\dots), (01100110\dots), (11001100\dots), (10011001\dots)$

Note that $\{(110),(100),(001),(011)\}$ is not a shift-invariant class and the average
of the objective function is less than 1 for all the possible shift-invariant classes.

The problem in the proof (I guess) is assigning equal probability $\frac{1}{m+1}$ to all the elements
inside a shift-invariant class (it does not need to be the case).

Title: Optimal probability measure


问:
How to see that existence non-trivial parallel $p$-form implies $b_p\geq 1$ using De Rham cohomology?
Suppose $(M,g)$ be a (for simplicity consider closed) Riemannian manifold. Because every parallel $p$-form $\omega$ is harmonic so the $p$-th Betti number should be positive i.e. $b_p\geq 1$. How to see this using De Rham cohomology (of course without using properties of Hodge theory like $\delta$, 1-1 correspondence between De Rham and Harmonic forms, etc.)? It is well-known that parallel forms are closed. What is the next step?

Such an easy argument based on harmonic forms I think there must be similar easy argument based on De Rham cohomology or it needs somewhat difficult argument?

答:
You should read a reference on Hodge theory.

The conclusion is not hard, that the map $\mathcal H^k \to H^k_{dR}$ is injective (that is, that harmonic forms are closed and exact harmonic forms are zero). The hard part is arguing that this map is surjective.

Lemma: A $p$-form is harmonic, $\Delta \alpha = 0$, if and only if $d \alpha = 0$ and $d^* \alpha = 0$.

Proof: Recall that $\Delta = \pm (d^* d + d d^*)$, where the sign depends on dimension of manifold and degree $p$ of the form. $$\langle \pm \Delta \alpha, \alpha \rangle = \langle d^* d \alpha, \alpha \rangle + \langle d d^* \alpha, \alpha \rangle = \langle d^* \alpha, d^*\alpha \rangle + \langle d\alpha, d\alpha \rangle = |d^*\alpha|^2 + |d\alpha|^2,$$ here using that $d^*$ is the $L^2$-adjoint of $d$; the inner product is $\langle \alpha, \beta \rangle = \int \alpha \wedge *\beta$.

The RHS can only be zero if $d^* \alpha$ and $d\alpha$ are both zero. QED.

Corollary: if $\alpha$ is an exact harmonic form, then $\alpha = 0$.

Proof: we know that $d^*\alpha = 0$. If $\alpha = d\beta$, it follows that $d^* d\beta = 0$. But then $$0 = \langle d^* d\beta, \beta \rangle = \langle d\beta, d\beta \rangle = |d\beta|^2,$$ so that $0 = d\beta = \alpha,$ as desired. The same argument shows that coexact harmonic forms ($\alpha = d^* \beta$) are also zero.


问:
Article suggestion for Russian reading
I apologize if this question does not really fit here.

I appreciate if someone can suggest math research articles in Russian (to practice reading, learning some math related terminologies etc.)

I don't want to restrict any subject or level of difficulty, but I prefer an original work which will be interesting to read. Many thanks.

答:
Try Успехи математических наук
(which is translated in English as Russian Mathematical Surveys).

http://www.mathnet.ru/php/journal.phtml?jrnid=rm&option_lang=rus

and in particular their archive, where you get free access to articles three years old or older:

http://www.mathnet.ru/php/archive.phtml?jrnid=rm&wshow=contents&option_lang=rus

It publishes research on various math areas, and as to picking work interesting to read, you are on your own, browse the archive, titles and abstracts.

I have occasionally read math papers in Russian in the past (learned Russian in school in Bulgaria years ago). In particular some interesting ones to me in the area of general topology, a very old Memoir (from the 1920's) by Alexandroff and Urysohn, a more recent 1970's paper by Mischenko about Linerly Lindelof spaces (finally compact, forgot the entire term, in Russian). In general, at least for older papers, but also new ones, some of the terminology is different, e.g. bicompact (instead of compact), finally compact (instead of Lindelof), and I don't know much about other areas. I am pretty sure you could find more journals like that, but this was the first one for me to think about (I may look for more later).

Well, actually here is another well-known one,
Математические заметки (Mathematical Notes):

http://www.mathnet.ru/php/journal.phtml?jrnid=mzm&option_lang=russian

and its archive

http://www.mathnet.ru/php/archive.phtml?jrnid=mzm&wshow=contents&option_lang=rus

And (on the same general website) they have a list of other journals.

http://www.mathnet.ru/ej.phtml?option_lang=eng

or (in Russian)
http://www.mathnet.ru/ej.phtml?option_lang=russian

Good luck!

Title: Article suggestion for Russian reading


问:
How to prove $A+I_n$ reversibility
Let $n$-order square matrix $A$ satisfy $A^m=A$,where $m$ is a positive even number.prove that $A+I_n$ reversibility.I want to use
$$(I_n+A)(I-A+A^2-\cdots+(-1)^{m-1}A^{m-1})=I_n+(-1)^mA^m.$$
But it felt like something was wrong.

答:
This is only true if your coefficient field has characteristic $\neq 2$, which I will assume from now on.

Suppose that $A + I_n$ is not invertible. This means that it has non-trivial kernel, and hence there exists a nonzero vector $v$ such that $(A+I_n)v = 0$, or $Av = -v$.

Applying $A$ repeatedly, we see that $A^mv = (-1)^mv = v$ since $m$ is even.

However, we also have $A^m v = Av = -v$. We now have $2v = 0$, which leads to $v = 0$, a contradiction.

It is easy to see that the condition of characteristic $\neq 2$ is necessary for the question, as shown by the example $A = I_n$ over a field of characteristic $2$.

Title: How to prove $A+I_n$ reversibility


问:
Evaluate $ f(a) = \int^a_{-a}|x|dx$ of $a=2$
I want to evaluate this integral $$ f(a) = \int^a_{-a}|x|dx$$between $a$ and $-a$ and I know that  $\int^a_{-a}|x|dx$ is divided into two: $$\int^a_{-a}|x|dx \rightarrow a>0 = \int^a_{0}xdx  $$
$$\int^a_{-a}|x|dx \rightarrow a<0 = \int^0_{-a}-xdx  $$

after I evaluate it I get zero, so I dont know if I`m right.

What I did is :

$$a<0 \rightarrow \int^0_{-2}-xdx = \frac{-x^2}{2}|^{0}_{-2} = -2 $$
$$a>0 \rightarrow \int^2_{0}xdx = \frac{x^2}{2}|^{2}_{0} = 2 $$
Thanks!

答:
When you have an even function $g$, that is $g(-x)=g(x)$ for all $x$, you can always do
$$
f(a)=\int_{-a}^a g(x)\,dx=
\int_{-a}^0 g(x)\,dx+
\int_{0}^a g(x)\,dx
$$
and do the substitution $x=-t$ in the first integral, getting
$$
f(a)=-\int_{a}^0 g(-t)\,dt+\int_{0}^a g(x)\,dx=
\int_{0}^a g(t)\,dt+\int_{0}^a g(x)\,dx=
2\int_{0}^a g(x)\,dx
$$</p>


问:
How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?
How can the following inequation be proven?

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

答:
$a^2+b^2+c^2-ab-bc-ca\\
=(a, b, c)\begin{pmatrix}1&-1/2&-1/2\\-1/2& 1&-1/2\\-1/2&-1/2&1\end{pmatrix}\left(\begin{array}{c}a\\b\\c\end{array}\right)$

$=(a, b, c)A\begin{pmatrix}a\\b\\c\end{pmatrix}$

It is sufficient to prove $A$ is a positive semi-definite.

It follows from that the minor determinant of $A$ is one of $1,\dfrac{3}{4},0$


问:
A proof of J.L. Lions' Lemma
I started a study on the Navier Stokes equations and my supervisor gave me a first result to work on which is the J.L. Lions Lemma which basically states the following:

If $f\in H^{-1}(\Omega)$ and $\nabla f\in H^{-1}(\Omega)$. Then $f\in L^2(\Omega)$

My supervisor told me that since $\Omega$ is a Lipschitz domain and from the proof of the Global approximation by functions smooth up to the boundary theorem and its Evans's Partial Differential Equations:

making the convolution of $f\in H^{-1}(\Omega)\, \nabla f\in H^{-1}(\Omega)$, we are in $L^2$. We can then use the Necas inequality.

My ideas:

I think that the step 1 and 2 of proof is for build the open covers $\{V_i\}_{i=0}^{q}$ where $V_i=\Omega\cap B(x_0,\frac{r_i}{2})$ and we can choose $\Omega = \cup_{i=0}^q V_i$.

We can also take a classic regularizing sequence $(\rho_n)$ for $\epsilon>0$, let $\eta_\epsilon = \rho_{[1/\epsilon]+1}$ or again $f_\epsilon(x) = f(x^\epsilon) = f(x+\lambda\epsilon e_n)$ and I can take $$g_\epsilon = \eta_\epsilon * f_\epsilon.$$
I think it must be proven that $g_\epsilon \in L^2(V_i)$ and with that we can later use the Necas' inequality

Then, we seek to provide hypotheses to ensure the existence of the unit partition $\theta_0,\,\theta_2\,...\,\theta_q$ associated with the open cover $\{V_i\}$ and we take $h = \theta_i g_\epsilon$ and then prove that (by Necas' inequality) $$\|h\|_{L^2(\Omega)} = \|h\|_{L^2(V_i)}\leq C(\|g_\epsilon\|_{H^{-1}(\Omega)}+\|\nabla g_\epsilon\|_{H^{-1}(\Omega)}),$$
and passing to the limit when $\epsilon \rightarrow 0 $ we would get $$\|f\|_{L^2(\Omega)}\leq C(\|f\|_{H^{-1}(\Omega)}+\|\nabla f\|_{H^{-1}(\Omega)})<\infty.$$ We can conclude that $f\in L^2(\Omega)$

Any indications or suggestions are welcome

答:
The approximation argument is a lot simpler than you might think; for negative order spaces the zero extension $H^{-1}(\Omega) \hookrightarrow H^{-1}(\Bbb R^n)$ is continuous, so you you can simply extend by zero and mollify.

The key ingredient is the estimate
$$ \lVert g \rVert_{L^2(\Omega)} \leq C \left( \lVert g \rVert_{H^{-1}(\Omega)} + \lVert \nabla g \rVert_{H^{-1}(\Omega)}\right) $$
you mention in the comments, which is valid for $g \in L^2(\Omega).$ This is where one needs $\Omega$ to be a Lipschitz domain, as the proof I involves using Fourier analysis in the full space and the continuity of the restriction $H^{-1}(\Bbb R^n) \to H^{-1}(\Omega),$ and the latter is where one needs some regularity of the boundary. From here the strategy as you noted is to regularise $f,$ apply the above, and conclude by passing to the limit. For this we need the following.

Lemma: If $g \in H^{-1}(\Omega),$ extending by zero let $g_{\varepsilon} = g \ast \rho_{\varepsilon}$ where $\rho_{\varepsilon}$ is a standard mollifier. Then $g_{\varepsilon} \in C^{\infty}(\overline\Omega)$ and $g_{\varepsilon} \to g$ in $H^{-1}(\Omega).$

Proof: Since $g \in H^{-1}(\Omega),$ there exists $g_0,g_1,\dots,g_n \in L^2(\Omega)$ such that $g = g_0 + \sum_{i=1}^n D^ig_i.$ Then extending each $g_i$ by zero to $\Bbb R^n$ we have
$$ g_{\varepsilon} = g_0 \ast \rho_{\varepsilon} + \sum_{i=1}^n D^i(g_i \ast \rho_{\varepsilon}) $$
which is smooth on $\overline\Omega.$ Now we can estimate
$$ \lVert g-g_{\varepsilon} \rVert_{H^{-1}(\Omega)} \leq \sum_{i=0}^n \lVert g_i \ast \rho_{\varepsilon} - g_i \rVert_{L^2(\Omega)} \to 0 $$
as $\varepsilon \to 0,$ which proves the lemma.

Now the rest is straightforward; since $f_{\varepsilon} = f \ast \rho_{\varepsilon} \in C^{\infty}(\overline\Omega) \subset L^2(\Omega)$ we can apply Necas' inequality to get
\begin{align*}
   \liminf_{\varepsilon \to \infty}\lVert f_{\varepsilon} \rVert_{L^2(\Omega)} &\leq \lim_{\varepsilon \to \infty}C \left( \lVert f_{\varepsilon} \rVert_{H^{-1}(\Omega)} + \lVert \nabla f_{\varepsilon} \rVert_{H^{-1}(\Omega)}\right) \\
& = C \left( \lVert f \rVert_{H^{-1}(\Omega)} + \lVert \nabla f \rVert_{H^{-1}(\Omega)}\right) < \infty,
\end{align*}
so by Fatou's lemma we conclude that $f \in L^2(\Omega).$

Title: A proof of J.L. Lions' Lemma


问:
A question about limit
Can anyone help this questions?

Find the limit of
$$1+\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\sqrt[5]{5+....+\sqrt[n]{n}}}}}$$

I can only solve that this formula is less than 3 but can not find the exact answer for this.

答:
It can be shown that $1.9<a_0<2$:
We use following inequalities:

$\sqrt[k]{k+1}>1; \sqrt[k]{k}<\sqrt[k]{k+2}<2$

Seris of $a_n$ is increasing and from first inequality we have:

$a_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n]n}}<\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n-1]{n-1+2}}}<\cdot\cdot\cdot<\sqrt{2+\sqrt[3]5}$

Therefore the series has a limit like $a_n $ such that :

$a_0\leq\sqrt{2+\sqrt[3]5}<2$

Now using first inequality we find:

$a_n>\sqrt{2+\sqrt[3]{3+\cdot\cdot\cdot+\sqrt[n-1]n}}>\cdot\cdot\cdot>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$

Therefore:

$a_0>\sqrt{2+\sqrt[3]3+\sqrt[4]5}$

It can easily be seen rhat:

$\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}>1,9$

Therefore:

$1.9<a_0<2$

Title: A question about limit


问:
Empty function finite support
If $I=\emptyset$ is the empty set and $G$ a group with trivial element 1. Then $G^I=\lbrace f:I\to G \rbrace$ is the trivial group with the empty function as an element. Am I right that $G^{(I)}=\lbrace f:I\to G \mid f \, \textrm{finite support} \rbrace$ is just $G^I$ itself?

I want to describe the condition "$f \, \textrm{finite support}$" via quantors. How to do that? One idea was $\exists J\subset I: J \, \textrm{finite} : f|_{I\setminus J} = 1$. But this is not compatible with $I=\emptyset$, since $f|_{I\setminus J} = 1$ is not satisfied (if I am not wrong).

答:
How about:

A function $f:I\to G$ has finite support if there exists finite $J\subseteq I$ such that for every $x\not\in J: f(x)=1$.

Title: Empty function finite support


问:
Show that the following process based on exponential waiting time is Markov wrt natural filtration
For a parameter $\lambda>0$ consider an exponential random variable: $T\sim \operatorname{Exp}(\lambda)$ and define a process $X_{t}=(0)1_{T> t}+(t-T)1_{T\leq t}$. Show that $X_{t}$ is $\mathcal{F}_{t}^{X}$ - Markov process with transition function:

$P_{t}(0,A)=\exp(-\lambda t)1_{A}(0)+\int_{0}^{t}1_{A}(t-s)\lambda\exp(-\lambda s)ds$

and for $x>0$:

$P_{t}(x,A)=1_{A}(x+t)$

Basically all I need to show is that for $t,h\geq 0, \; A \in \mathcal{B}((0,\infty))$ and $n\in \mathbb N$ with $0=t_{0}<...<t_{n}=t$ and $A_{i}\in \mathcal{B}((0,\infty))$ for all $i = 0,...,n$ that:

$E[1_{\{X_{t+h}\in A\}}\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}}]=E[P_{h}(X_{t},A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}}]\; (*)$

Now in order to evaluate $(*)$ I will consider the two cases $\{T\leq t\}$ and $\{T> t\}$, thus considering

$E[P_{h}(X_{t},A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]=E[1_{A}(X_{t}+h)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]=E[1_{A}(t+h-T)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]=E[1_{A}(X_{t+h})\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T\leq t]$ and so we are done in the case that $T\leq t$. But now in the case that $T> t$:

$E[P_{h}(X_{t},A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T> t]=E[P_{h}(0,A)\prod\limits_{i=0}^{n}1_{\{X_{t_{i}}\in A_{i}\}},T> t]=\prod\limits_{i=0}^{n}\delta_{0}(A_{i})P_{h}(0,A)\mathbb P(T> t)$

So I supppose this problem would reduce itself to showing that $P_{h}(0,A)\mathbb P(T> t)=1_{A}(X_{t+h})$, so now clearly:

$P_{h}(0,A)\mathbb P(T> t)=\exp(-\lambda t)P_{h}(0,A)=\exp(-\lambda t)(\exp(-\lambda h)1_{A}(0)+\int_{0}^{h}1_{A}(h-s)\lambda\exp(-\lambda s)ds)$

This is where I am unsure of how to proceed, to show that the above is equal to $1_{A}(X_{t+h})$, any ideas?

答:
I think there is a typo in your formula. I feel like we should have $P_t(0,A)=e^{\lambda t}1_A(0)+\int_0^t1_A(t-s)e^{\lambda s}\textrm{d}s$ (or, indeed, the calculation below should show it!).

Well, $P_h(0,A) \mathbb{P}(T>t)$ is definitely equal to neither $0$ nor $1$, so that's not gonna be equal to an indicator function. However, we have
\begin{align}
&\mathbb{P}[X_{t+h}\in A, X_i\in A_i\; \forall i, T>t]=\mathbb{P}[X_{t+h}\in A, 0\in A_i \; \forall i, T>t]\\
=&\mathbb{E}[1_A(X_{t+h}) 1_{T>t} \prod_i 1_{A_i}(0)]=\prod_{i} 1_{A_i}(0) \mathbb{E}[1_A(X_{t+h}) 1_{T>t}]
\end{align}
And, furthermore,
\begin{align}
\mathbb{E}[1_A(X_{t+h}) 1_{T>t}] = & \mathbb{E}[ 1_A(t+h-T)1_{T\in (t,t+h]}+1_A(0)1_{T\geq t+h}] \\
=& \int_0^{\infty} e^{-\lambda s} \left(1_A(t+h-s)1_{[t,t+h]}(s)+1_A(0)1_{(t+h,\infty)}(s)\right)\textrm{d}s\\
=& \int_{t}^{t+h} e^{-\lambda s} 1_A(t+h-s)\textrm{d}s+\int_{t+h}^{\infty} e^{-\lambda s}1_A(0)\textrm{d}s \\
=& e^{-\lambda t}\int_0^h e^{-\lambda s}1_A(h-s)\textrm{d}s+1_A(0) e^{-\lambda(t+h)}\\
=& P_h(0,A)e^{-\lambda t}\\
=& P_h(0,A)\mathbb{P}(T>t),
\end{align}
which should finish the proof for you.


问:
Property of inverse function
If a function $f$ from $\mathbb{R}$ to $\mathbb{R}$ is one-to-one and bounded is it true that $f^{-1}$ is also one-to-one and bounded? I believe the answer is no but I'm not sure.</p>

答: Inverse function itself may not exist. For example the function $f(x) = \frac{1}{1+e^x}$ is defined from $\mathbb{R}$ to $\mathbb{R}$ which is one to one and bounded. But its inverse $\log\left(\frac{1}{y} -1\right)$ is not defined for all real numbers. </p>

Title: Property of inverse function


问:
Why do limits have a strict inequality?
Why is the limit of a function $f(x)$ defined as $$f(x)-L<\varepsilon$$$$x-a<\delta$$but not like $$f(x)-L \leq \varepsilon$$ $$x-a \leq \delta$$What problems would arise if it was defined like this as compared to the standard definition?

P.S.- Some may point out these as silly questions, but I want to understand every bit that I read, not just know. Thanks.

答:
Assume that $A \subset \mathbb R$ and $f: A \to \mathbb R$ and $a$ is a limit point of $A$. Let $L \in \mathbb R$.

You can show that $\lim_{x \to a} f(x) = L$ if and only if for all $\epsilon > 0$ there exists $\delta > 0$ such that if $x \in A$ and $0 < | x - a | \leq \delta$ then $| f(x) - L | \leq \epsilon$.

So, the alternate definition you're suggesting is equivalent to the usual definition.

Title: Why do limits have a strict inequality?


问:
Long Division of polynomials
How can we prove that $x^n$  isn't divisible by $g(x)=x^4+x+1$ without remainders?

I understand why for all $n<4$ it's working, but how with $n \ge  4$? How we can prove it formally?

Note: $x^n,\:g\left(x\right)\in \mathbb{Z}_2\left[x\right]$

答:
Write $x^n = q_n(x) (x^4+x+1) + r_n(x)$.
Then $$x^{n+4} = x^4 q_n(x) (x^4+x+1) + x ^4 r_n(x) = x^4 q_n(x) (x^4+x+1) + (x+1) r_n(x)$$ and so $r_{n+4} \equiv (x+1) r_n \bmod (x^4+x+1)$.

The result follows by induction because $r_n = x^n \ne 0 $ for $n=0,1,2,3$ and $x+1$ is a unit mod $(x^4+x+1)$.

Title: Long Division of polynomials


问:
General approach to finding number of significant figures in mixed operations
There are ways to doing operations with significant figures, but not really for mixed operations. I have looked all over the Internet, Stack Exchange, and the textbook I use. I'm surprised, but in any case, I would like a general approach for dealing with the precision, uncertainty, and accuracy of a calculation.

Each operation has its own set of rules. See below:

Addition/subtraction - Use the significant figures of the number with the fewest decimal places (digits) for the calculation.
$$1.145 + 1.5 = 2.6$$

Multiplication/division - Use the significant figures of the number with the fewest significant figures for the calculation. $$1.145 × 1.50 = 1.72$$

However, problems arise in these cases:

Trigonometry
$$sin(\fracπ{3.1}) = 0.018$$
Here, I used the significant figures of the number with the fewest significant figures for the calculation. In other words, I took 3.1 (2 sig figs) to calculate 0.018.

Logarithms
$$\log_{10}(0.00002734) = \log_{10}(2.734) + \log_{10}(10^{-5}) = 0.46379 + 0.00001 = 0.4630$$
Here, the significant figures of addition turned out to be how you would do significant figures of multiplication/division, but in general, you calculate by calculating the operation as if it were multiplication/division.

But I’m just generally confused. What if you have a situation as I saw above with the logarithms? Or something like this:

$$\frac{[(0.23 × 5) + (8.521)^5]}3$$

There is not much consensus.

Any clarification would be golden.

答:
Good question. Don't mix significant figures with the error propagation concept and the output of transcendental functions like $\log x, \sin x, \cos x$ etc. These are three different things.

For example $\sin \left(\pi/4\right) = 0.70710678118...$(never ending). There is no experimental or measurement "error" here. The concept of significant figures is meaningless here. Similarly, the output of a logarithm for pure numbers is also exact (or rather it never ends). There is no error. No need to apply significant figure concept there. The easy trick is that you use the maximum numbers in all steps which your calculator is showing and round off only at the end.

Significant figures is just a common sense approach of retaining answers to a realistic experimentally possible value. For example, if we weigh a few crystals of salt on an analytical balance, which can measure up to 4 decimal places of grams, we should record the weight as 0.1074 g. It would be a bad practice to record it as 0.11 g or 0.10740 gram. Both are wrong because we did not retain the four significant figures.

So the Rules 1 and 2 are just qualitative common sense type statements.

In real cases, we must estimate the errors as standard deviations of our measurements by repeatedly making the same measurements.

Scenario I: In reality, if you wish prepare a solution of 1 M NaCl in one liter. You would know the standard deviation of the mass you weighed, and you would also know the standard deviation of your 1 L volumetric flask, then the uncertainty in the molarity of 1 M NaCl would be

$$ (1 M \pm ?)=\frac{58.4427\pm0.0034}{(1.000\pm0.050)(58.4427)}$$

Scenario II: Suppose you measure the hydrogen ion concentration of a solution by titration 10 times. The hydrogen ion concentration is $0.058 \pm0.002$ M. Now I wish to calculate the pH of this solution, which is given by

$$-\log [H^+]=-\log[0.058 \pm0.002]$$.

What is the pH with its experimental error?

How do we determine the error in both scenarios?

As you stated problems start when experimental measurements become more complex involving more complex functions. You need to know partial derivative formula for error propagation.

Suppose your function is $$y = f(X, Z, \ldots \, )$$, where $X$, $Z$ are measured numbers with their associated standard deviations $s_x$, $s_z$ and so on. The the standard deviation of the final calculation should be

$$s_y = \sqrt{
\left( \frac{\partial y}{\partial X} \right)^2 s_x^2 +
\left( \frac{\partial y}{\partial Z} \right)^2 s_z^2 + \cdots +}$$

This formula becomes even more involved when the measurement errors depend on each other. See the NIST page

https://www.itl.nist.gov/div898/handbook/mpc/section5/mpc55.htm

OwnerDisplayName: Alex

OwnerDisplayName: M. Farooq


问:
Characteristics for nonlinear waves governed by $u_t + c(u)u_x = 0$
I have some confuse about the characteristic of nonlinear wave propagation. I read the PDF from A. Salih at IIST about the Inviscid Burgers’ Equation, where the original PDF can be find at the (https://www.iist.ac.in/sites/default/files/people/IN08026/Burgers_equation_inviscid.pdf)

So, I am not fully understand some statement about the characteristic line in this PDF.

Consider the 1D nonlinear advection equation
$$ u_t + c(u)u_x = 0$$
where the wave speed is not constant but a nonlinear term $c(u)$.

As above PDF state, we defined the characteristic curve as
$$ \frac{dx}{dt} = c(u).$$
Then Let $x = x(t)$, we have
$$\frac{d}{dt}u(x(t),t) = \frac{\partial u}{\partial t} + \frac{\partial u}{\partial x}\frac{dx}{dt} = u_t +c(u)u_x = 0$$
Therefore the $u$ is constant long the characteristic curve, and the characteristic curve is straight line since
$$ \frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{dc(u)}{dt} = c'(u)\frac{du}{dt} = 0$$

I didn't understand 3 places,

(1) why we can just assume the $x$ is dependence of $t$.

(2) why we say the $u$ is constant long the characteristic curve. For sure, the $\frac{d}{dt}u(x(t),t) = 0$ shows that the solution $u$ does not change along time, but I don't understand what logic shows that the $u$ is constant along $ \frac{dx}{dt} = c(u)$.

(3) why the characteristic curve is straight line because
$ \frac{d^2x}{dt^2} =  c'(u)\frac{du}{dt} = 0$. How do I know the derivative of $c(u)$ is equal to zero?

Could someone can help me?

答:
There is some confusion with the derivation's steps and notation, which OP presents in the wrong logical order. Below, the proper way is shown (see also this related post).

First things first, we write the initial-value problem for the quasi-linear conservation law
$$
u_t + c(u) u_x =0 , \qquad u(x,0) = \phi(x).
$$
The dependent variables are position $x$ and time $t$, and the unknown is $u(x,t)$.
The method of characteristics consists in seeking a parametrisation $s \mapsto \big(x(s),t(s),u(s)\big)$
of these quantities, in such a way that the PDE transforms into ordinary differential equations which we might be able to solve. Note in passing that the dependence of $u$ w.r.t. $s$ can be expressed as $u = u(x(s), t(s))$, and similar notation can be used for the partial derivatives. Using the chain rule, the evolution of $u$ is governed by
\begin{aligned}
\frac{d}{ds}u(s) &= x'(s) u_x(s) + t'(s) u_t(s) \\
&= \left[x'(s) - c(u(s))t'(s)\right] u_x(s)
\end{aligned}
where we have used the PDE.
As shown in Wikipedia, we may write the system
$$
\frac{dt}{ds} = 1, \quad \frac{dx}{ds} = c(u), \quad \frac{du}{ds} = 0
$$
with initial condition $t(0) = 0$, $x(0) = x_0$ and $u(0) = \phi(x_0)$, which resolution can be tackled by hand. Here, we find
$$
t=s, \quad x = x_0 + c(\phi(x_0))\, s,\quad u = \phi(x_0).
$$

Now let us go back to OP's questions. Given that $t=s$, this can be rewritten as
$$
x = x_0 + c(\phi(x_0))\, t,\quad u = \phi(x_0),
$$
as a consequence of the choice $x' = c(u)$ with $u'=0$. Now, we can express the unknown as $u = u(x(t), t)$, and similar notation can be used for the partial derivatives.
We note that $u = \phi(x_0)$ is constant along the characteristic curve $t \mapsto x(t)$ starting at $(x_0, 0)$, and that those curves are straight lines in the $x$-$t$ plane. In fact, computation of the derivative of $u$ along those lines gives
$$
\frac{d}{dt}u(t) = x'(t) u_x(t) + u_t(t) = 0
$$
according to the definition of the characteristic curves and the PDE itself. Since the slope $x' = c(u)$ of these lines is a function of $u$ with $u$ constant, we can conclude that $x'$ is constant too.


问:
What is $\mathbb{Z}_5[x]/((x-1)^3)$ isomorphic to?
What is $\mathbb{Z}_5[x]/((x-1)^3)$ isomorphic? ($\mathbb{Z}_5=\mathbb{Z}/5\mathbb{Z}$.)

Since $(x-1)^3$ is not irreducible in $\mathbb{Z}_5[x]$, it is not a prime ideal and the quotient ring is not a field. Thus $\mathbb{Z}_5[x]/((x-1)^3)\ncong\mathbb{Z}_5$.

In this case, $(x-1)+(x-1)^2\neq(1)$ so I can't apply Chinese Remainder theorem here.

Body:
$$
\mathbb{Z}_5[x]/((x-1)^3)
=
\mathbb{Z}_5[x-1]/((x-1)^3)
\cong
\mathbb{Z}_5[t]/(t^3)
\cong
\mathbb{Z}_5[u], \ u^3=0
$$


问:
If $s=a+b+c,p=ab+bc+ac, r=abc, s^2-2p+r=4$ and $a,b,c>0,$ prove that $2sp+15r^2\geq 33r$
If $s=a+b+c$,$p=ab+bc+ac$, $r=abc$, $s^{2}-2p+r=4$ and $a,b,c>0,$

prove that $$2sp+15r^{2}\geq 33r.$$
I tried to use some inequalities from here, but nothing works. Suggestions ?

答:
If $a,b,c \gt 0$ and $s=a+b+c$,$p=ab+bc+ac$, $r=abc$

This looks like the coefficient of some polynomial in $x$, where $ x \in \{ a,b,c \} $
$$x^3-sx^2+px-r = 0$$
From $s^2-2p+r=4$ , say $ r =4+ 2p-s^2$
$$2sp+15r^2 \geq 33r$$
$$2sp+15(4+2p-s^2)^2 - 33(4+2p-s^2) \geq 0$$
$$ 15s^4-60ps^2-87s^2+2ps+60p^2+174p+108 \geq 0$$
Noticing the number of positive constant, you can substitute back the value of $s$ and $p$, Or use that fact that $x^3-sx^2+px-4-2p+s^2= 0$, where $ x \in \{ a,b,c  \} $ to reduce the problem by removing powers of $s$ or $p$ and finish it up
$$ 15s^4+2ps+60p^2+174p+108 \geq 60ps^2+87s^2$$


问:
possible dimensions of a nilpotent endomorphism of a K-Vectorspace
I have the following exercise in linear algebra that I think I know what they're looking for, however have not really got an idea which theorems to use / how to approach this.

Let $V$ be a 5-dimensional $K$-Vectorspace, and $N$ a nilpotent endomorphism of $V$. For $r \in \mathbb{N}$, we set $V^{(r)} =$ Ker$(N^{\circ r})$ and $d_r =$ dim$V^{(r)}$. What are the possible sequences of integers that one may get in this way$?$ For each possible sequence, give an example of a nilpotent endomorphism $N$ realising it.

From my script I know that Matrix $A$ is nilpotent if $A^k = 0$ for some integer $k$.

So if I understood this correctly, we are looking for possible sequences for dim(Ker$(N^{\circ r})$).

My guess is that based off of the definition of a nilpotent matrix, the sequence will always end with $0$. However I am not sure if there is a theorem to calculate the first elements of the sequence.

How can I approach calculating these possible sequences?

答:
In general, there are a few important this to know about the sequence $(\dim\ker(N^k))_{k\in\Bbb N}$

It starts at $0$ at $k=0$ (since $N^0=I$ by definition of exponentiation);

It ultimately stabilises at the value $n=\dim(V)$, the size of $N$ (since $N^k=0$ ultimately);

It is weakly increasing (since $N^kv=0$ implies $N^{k+1}v=0$, so $\ker(N^k)\subseteq\ker(N^{k+1})$);

In is concave, meaning the sequence of non-negative differences $(\dim\ker(N^{k+1})-\dim\ker(N^k))_{k\in\Bbb N}$ is weakly decreasing; in particular when the difference becomes$~0$ (the original sequence has two equal consecutive terms) it remains$~0$ (the original sequence becomes stationary).

The final point is slightly more difficult to show: first, applying $N$ to vectors maps the subspace $\ker(N^{k+1})$ to $\ker(N^k)$ for every $k\in\Bbb N$, second, this induces a map of quotient spaces $\ker(N^{k+2})/\ker(N^{k+1})\to\ker(N^{k+1})/\ker(N^k)$ that is well defined, and third, that induced (linear) map is injective, which gives the desired inequality.

Thus the sequence of differences is a weakly decreasing sequence of non-negative integers, ultimately becoming $0$, and whose sum is$~n$. Such sequences are called a partition of $n$, and they are very well studied. There are $7$ partitions of $5$, namely (with the trailing zeros omitted) $(5),(4,1),(3,2),(3,1,1),(2,2,1),(2,1,1,1),(1,1,1,1,1)$. The corresponding (original) sequences of dimensions can be found by taking partial sums left to right, starting from the partial sum$~0$ of the empty subsequence, giving respectively

$(0,5,5\ldots)$

$(0,4,5,5,\ldots)$

$(0,3,5,5,\ldots)$

$(0,3,4,5,5,\ldots)$

$(0,2,4,5,5,\ldots)$

$(0,2,3,4,5,5,\ldots)$

$(0,1,2,3,4,5,5,\ldots)$

It is not hard to see that these sequences can all be realised for appropriate nilpotent matrices, and that this is the case for all$~n$.


问:
How to prove that for any real number $a$, $A+aI_n$ is invertible
Let an $n$-order real square matrix $A$ satisfy $A^{2m}+I_n=0$, where $m$ is a positive integer. Prove that for every real number $a$, the matrix $A+aI_n$ is invertible. I don't know how to start.

答:
If by absurdum we had $A+\alpha I$ not invertible, then there would exist a $v \ne \mathbf{0}$ such that :

$(A+\alpha I)v=\mathbf{0}$

so that $v$ is an eigenvector with eigenvalue $-\alpha$.

Now:

$\mathbf{0}=(A^{2m}+I)v=((-\alpha)^{2m}+1 )v \ne \mathbf{0}$

, which is a contradiction.


答:
It's best to consider this problem geometrically, it's far more simple.

$$\int_{-a}^{a} |x|dx = a^2$$

because it's just the area under the curve $f(x) = |x|$ which is "two triangles with base and height $a$".

Thus it's just $$2*\frac{1}{2}\text{(base)}\text{(height)}=2*\frac{1}{2}*a*a$$

Since $a=2$, the answer is 4.

Algebraically speaking, the reason why your answer did not work is because

$$|x| = \begin{cases} & -x \quad (x<0) \\
&x \quad \quad (x\ge0)\end{cases}$$

So the $-2$ you got should have been a $+2$.

Title: Evaluate $ f(a) = \int^a_{-a}|x|dx$ of $a=2$

Body:
I want to evaluate this integral $$ f(a) = \int^a_{-a}|x|dx$$between $a$ and $-a$ and I know that  $\int^a_{-a}|x|dx$ is divided into two: $$\int^a_{-a}|x|dx \rightarrow a>0 = \int^a_{0}xdx  $$
$$\int^a_{-a}|x|dx \rightarrow a<0 = \int^0_{-a}-xdx  $$

after I evaluate it I get zero, so I dont know if I`m right.

What I did is :

$$a<0 \rightarrow \int^0_{-2}-xdx = \frac{-x^2}{2}|^{0}_{-2} = -2 $$
$$a>0 \rightarrow \int^2_{0}xdx = \frac{x^2}{2}|^{2}_{0} = 2 $$
Thanks!


答:
Like EH2019 did, I would like to reinterpret my attempt in the beginning. We will do the following:

$\exists J\subset I : J \, \text{is finite} : \forall x \, (x\in I\setminus J \implies f(x)=1)$.

Since $\forall x\in \emptyset$ is always false, the implication is always true.

Title: Empty function finite support

Body:
If $I=\emptyset$ is the empty set and $G$ a group with trivial element 1. Then $G^I=\lbrace f:I\to G \rbrace$ is the trivial group with the empty function as an element. Am I right that $G^{(I)}=\lbrace f:I\to G \mid f \, \textrm{finite support} \rbrace$ is just $G^I$ itself?

I want to describe the condition "$f \, \textrm{finite support}$" via quantors. How to do that? One idea was $\exists J\subset I: J \, \textrm{finite} : f|_{I\setminus J} = 1$. But this is not compatible with $I=\emptyset$, since $f|_{I\setminus J} = 1$ is not satisfied (if I am not wrong).


问:
Why do we use G, H, and 'K' for groups?
In many cases of mathematics, we tend to set several variables in ascending alphabetical order.

$x,y,z \dots$ for normal variables

$A,B,C\dots$ for sets, coefficients, matrices

$V,W\dots$ for vector spaces

${I,J}\dots$ for indices, ideals

$\mathcal{C,D}\dots$ for categories

But when we need 3 groups, we tend to use $G, H$ and suddenly, $K$, Which is unexpected.

Here are some examples that groups are denoted by $K$.

Prove that quotient group K/H is normal subgroup of quotient group G/H

https://en.wikipedia.org/wiki/Group_homomorphism#The_category_of_groups

https://proofwiki.org/wiki/Pullback_of_Quotient_Group_Isomorphism_is_Subgroup

Why do we denote groups by $K$ regardless of the alphabetical order?

答:
These are choices that mathematical writers have to make. There's not enough letters for all the stuff we need them to represent, and they often clash. So nice clean alphabetic sequences reserved for special purposes sometimes get in the way of our notational needs.

The letters $I$ and $J$ are also very commonly used for intervals in the real line. In particular $I$ is often reserved for one very special interval, namely $I = [0,1] \subset \mathbb R$. It is even somewhat common to be writing about groups and intervals at the same time, for example when writing about the fundamental group $G = \pi_1(X)$ of a topological space $X$ in which the elements of the group are represented by certain continuous paths $f : I \to X$ where $I = [0,1]$.

So, due to its many other busy duties, to avoid clash of variables the letter $I$ is rarely used for groups.

And $J$ usually refuses to be used too, in protest of the insult to his buddy $I$.

On the other hand $K$ is happy to be used for a group, particularly as a normal subgroup of a group $G$, because in that case $K$ is the Kernel of the quotient homomorphism $G \mapsto G / K$. And we sometimes jump wayyyyy further in the alphabet just to use $Q = G / K$ for the Quotient group.

Title: Why do we use G, H, and 'K' for groups?


问:
Vector space $U/V$
Let $U$ a vector space of dimension $n$ and $V\subsetneq U$ a subspace of dimension $m<n$. We define $$U/V:=\{u+V\mid u\in U\},$$
where $$u+V:=\{u+v\mid v\in V\}.$$

Define law $\bar +$ and $\bar \cdot $ on $U/V$ that make $(U/V,\bar +,\bar \cdot )$ is a vector space, and prove that $U/V$ has dimension $m-n$.

I defined $$(u+V)\bar +(w+V)=(u+w)+V\quad \text{and}\quad \alpha \bar \cdot (u+V)=\alpha u+V,$$
and I proved that $(U/V,\bar +,\bar \cdot )$ is a vector space. But I don't see how to prove that it has dimension $n-m$.

Attempt

Let $V=Span\{v_1,...,v_m\}$ and we prolonge it in a basis of $U$, i.e. $U=Span\{v_1,...,v_n\}$. Now, what is going to be a basis of $U/V$. I guess it will be $\{v_{m+1},...,v_{n}\}$, but how can I prove it ?

答:
Hint

It should rather be $\{v_{m+1}+V,\cdots,v_n+V\}$. If $u+V\in U/V$, then $$u+V=\alpha _1v_1+...+\alpha _nv_n+V=\alpha _{m+1}\bar \cdot (v_{m+1}+V)\bar+\cdots \bar +\alpha _n\bar\cdot (v_n+V).$$
What remains to be proved is that $$\alpha _1\bar \cdot (v_{m+1}+V)\bar+\cdots \bar+\alpha _m\bar\cdot (v_n+V)=V\implies \alpha _i=0,$$
for all $i$.

Title: Vector space $U/V$


问:
If three dice are rolled, what is the probability that all three are the same number?
The dice are fair.

You have a $1\over6$ chance of getting the first number. A $1\over6$ chance of the second and so on.  Is it just $({1\over6})^3$ (1/216) or is that not accounting for the second and third roll properly?

答: total no of outcomes will be 6x6x6=216.....favourable otcomes are (1,1,1), (2,2,2), ... ,(6,6,6)...i.e. 6 favouracle outcomes.....so probability will be 6/216=1/36</p>


问:
How to proceed with this complex equation?
I don't have experience with solving these types of equations.

After looking around; I learned that I should try substituting a + bi in for a complex number in these types of equations.

$$w^2 + 3w^* + 2 = 0$$

I found that the real part is $a^2 - b^2 + 3a + 2$

I also found the imaginary part to be $2ab - 3b$

I'm not sure how to proceed.

Note, w* is the complex conjugate.

答:
So, equating the imaginary parts,

$$b(2a-3)=0$$

Equating the real parts, $$0=a^2-b^2+3a+2$$
Can you take it from here?


问:
Why is the solution of dy/dx = y/x, y = Cx and not y = C|x|?
Since the derivative of |x| is |x|/x, does it follow that y = |x| can be differentiated to give dy/dx = |x|/x = y/x meaning that the solution of this ODE is y = C|x| rather than the standard y = Cx which one would get by the separation of variables?

This meant that I got stuck with the following question as I do not know whether it is A or D. Thank you in advance!
MC question

答: Exactly, when you exclude $x=0$, function $y=c|x|$ falls into $y=cx$ categories.


问:
Euclidean Models, Finite Model Property and Filtrations (Modal Logic)
I have partially solved Exercise $1$ from this problem set by Yde Venema, and I need some help in completing it.

Blackburn and Venema's Modal Logic 2.3.8: Call a frame or model euclidean if it satisfies $\forall x\forall y \forall z ((Rxy ∧ Rxz) → Ryz)$, and let $E$ be the
class of euclidean models. Fix a formula $ξ$, and let $Σ$ be the smallest subformula closed set
of formulas containing $ξ$ that satisfies, for all formulas ψ$:$ if $\diamond ψ ∈ Σ$, then $\square\diamond ψ ∈ Σ$. Note that in general, $Σ$ will be infinite.

(a) Prove that $E \Vdash \diamond ψ → \square\diamond ψ$.

(b) Prove that every euclidean model can be filtrated through $Σ$ to a euclidean model.

(c) Show that every euclidean model satisfies the following modal reduction principles, $\diamond\diamond\diamond\leftrightarrow \diamond\diamond, \diamond\diamond\square\leftrightarrow \diamond\square, \diamond\square\diamond\leftrightarrow \diamond\diamond$ and $\diamond\square\square\leftrightarrow \diamond\square$.

(d) Prove that the basic modal similarity type has the finite model property with respect
to the class of euclidean models.

I was able to prove the first two parts (I need help with c and d), and here's what I did:

(a) Pick an arbitrary $\mathfrak M \in E$, and $w\in W$ where $\mathfrak M = (W,R,V)$. Suppose $\mathfrak M,w\Vdash \diamond\beta$. We want $\mathfrak M,w\Vdash\square \diamond\beta$. Suppose instead that $\mathfrak M,w\Vdash\lnot\square \diamond\beta$, i.e. $\mathfrak M,w\Vdash\diamond \square\lnot\beta$. $\mathfrak M,w\Vdash \diamond\beta$ iff there is $v\in W$ such that $wRv$ and $\mathfrak M,v\Vdash \beta$. $\mathfrak M,w\Vdash\diamond \square\lnot\beta$ iff there exists $z\in W$ such that $\mathfrak M,z\Vdash \square\lnot\beta$. Now since $\mathfrak M$ is Euclidean, we must have $wRz \land wRv \to vRz\land zRv$. Now, clearly, $\mathfrak M,z\Vdash \square\lnot\beta$ and $\mathfrak M,v\Vdash \beta$ cannot hold together. This is a contradiction, and the proof is complete.

(b) This follows from the definition of $R^f$ in a filtrated model. We know that $wRv\implies [w]R^f[v]$. Suppose $\mathfrak M$ is Euclidean. Then, $\forall x\forall y \forall z ((Rxy ∧ Rxz) → Ryz)$. Pick arbitrary $x,y,z$ such that $xRy \land xRz$. Then, $[x]R^f[y] \land [x]R^f[z]$ hold. Also, since $\mathfrak M$ is Euclidean, we have $yRz$ and hence $[y]R^f[z]$. Since $x,y,z$ were arbitrary to begin with, the filtrated model is also Euclidean, and we are done.

I have trouble proving (c) and (d). The approach to (c) is straightforward though, to show that $\phi_1\leftrightarrow\phi_2$ holds, I should assume $\phi_1$ at an arbitrary world, prove $\phi_2$ and vice versa. However, I am unable to complete the arguments in this case. Could someone help? Also hints for (d) would be good!

Thanks a lot.

答:
For the first part of (c).
If $\diamond \diamond \diamond \phi$ holds at $w$ then we have this picture for $R$
$$w \to x \to y \to z$$
where $\phi$ holds at $z$, and $\diamond \phi$ holds at $y$, and $\diamond \diamond \phi$ at $x$

From the Euclidean property we have
$$R(w,x) \wedge R(w,x) \to R(x,x),$$
$$R(x,y) \wedge R(x,x) \to R(y,x),$$
$$R(y,x) \wedge R(y,z) \to R(x,z).$$
As $\phi$ holds at $z$ and $R(x,z)$ we get $\diamond \phi$ at $x$ and so $\diamond \diamond \phi$ at $w$.

For the converse: Suppose $\diamond \diamond \phi$ holds at $w$
with $\phi$ at $y$ in this picture
$$w \to x \to y$$
using the above argument that $R(x,x)$ we get
$$w \to x \to x \to y$$
and thus $\diamond \diamond \diamond \phi$ at $w$.

For the next part of (c): $\diamond \diamond \Box \leftrightarrow \diamond \Box$,
suppose $\diamond \diamond \Box \phi$ holds at $x$. Then there is $y$ with
$R(x,y)$ and $z$ with $R(y,z)$ where $\Box \phi$ holds at $z$. Now $\Box \phi$ must hold at $y$, because if $R(y,w)$ then because $R(y,z)$ we have $R(z,w)$ so that $\phi$ holds at $w$. Hence $\diamond \Box \phi$ holds at $x$.

Conversely, suppose that $\diamond \Box \phi $ holds at $x$ with $R(x,y)$ and $\Box \phi$ holds at $y$. As observed earlier $R(y,y)$ must hold, and hence $\diamond \Box \phi$ holds at $y$. Thus $\diamond \diamond \Box \phi$ holds at $x$.

I think the rest of part (c) will be rather similar


问:
Why there's a sin and cos here? Shouldn't the obvious guess be $Ae^{-x}$?
image shows the solution for the differential equation y double prime minus 4 y prime plus 5 y equal to e powered to the minus x

I solved this $y'' - 4y' + 5y = e^{-x}$ equation with the guess of:

$Ae^{-x}$, such that A ended up being $A = 1/2$;

The solution given by math calculator websites was the one of the image —

$$y(x)=(C_1 \sin(x)+C_2 \cos(x))e^{2x}+\frac{e^{-x}}{10},$$

but I can't understand how could I ever get to that solution on my own.
Can someone explain the logic to me?

答:
Let us pretend that we know nothing about the linear ODE with constant coefficients nor complex exponentials.

We will try by factoring $y$ and get some simplification.

$$y=zh,\\y'=z'h+zh',\\y''=z''h+2z'h'+zh''.$$

We plug this in the ODE:

$$z''h+2z'(h'-2h)+z(h''-4h'+5h)=e^{-x}$$ and let the coefficient of $z'$ vanish.

From $h'-2h=0$, which is separable,

$$\frac{h'}h=(\log(h))'=2$$ and $h=e^{2x}$ is a solution.

Plugged in the ODE gain,

$$z''e^{2x}+ze^{2x}=e^{-x}$$ or

$$z''+z=e^{-3x}.$$

Now if we try a solution of the form $e^{-3x}$ by educated guess, we get that

$$z''+z=10\,e^{-3x}$$ and it suffices to correct with a suitable factor.

So, using linearity of the derivative, we can move the RHS as

$$z''-\frac9{10}e^{-3x}+z-\frac1{10}e^{-3x}=0$$ or

$$\left(z-\frac1{10}e^{-3x}\right)''+\left(z-\frac1{10}e^{-3x}\right)=0$$

which is of the form $$w''+w=0.$$ We make it integrable by multiplying by $2w'$:

$$2w'w''+2ww'=(w'^2+w^2)'=0$$ and for some (positive) constant $a^2$,

$$w'^2=a^2-w^2$$ then

$$\frac{w'}{\sqrt{a^2-w^2}}=\pm1.$$

After integration,

$$\arccos\frac wa=\pm t+c$$

or

$$w=a\cos(\pm t+c)=c_c\cos t+c_s\sin t.$$

Finally,

$$y=ze^{2x}=\left(w+\frac{e^{-3x}}{10}\right)e^{2x}=\left(c_c\cos t+c_s\sin t+\frac{e^{-3x}}{10}\right)e^{2x}
\\=\left(c_c\cos t+c_s\sin t\right)e^{2x}+\frac{e^{-x}}{10}$$ as claimed.