答:
$$y - b = -\frac{a-x_0}{b-y_0}(x-a)$$

(Tangents are perpendicular to radii, and then just use the standard equation of a line of known gradient through a point)

Title: What is the equation of a tangent to a circle at a point?

Body:
Suppose I have the following circle
$$
(x - x_0)^2 + (y - y_0)^2 = r^2
$$
where the center $(x_0, y_0)$ and the radius $r$ are known. Is there a general formula for the equation of a tangent at a point $(a, b)$ on the circle, with $a, b$ known?

Idea

Since the point $(a, b)$ is on the circle, we must have
$$
(x - a)^2 + (y - b)^2 = r^2
$$
and since the tangent will pass through $(a, b)$ we must have
$$
\frac{y - b}{x - a} = m
$$

How do you incorporate the fact that the radius is perpendicular to the tangent?

Following the comments, I think then one notices that the radius is perpendicular to the tangent. The line through the center $(x_0, y_0)$ and the point $(a, b)$ has slope

$$
m' = \frac{b - y_0}{a - x_0}
$$

Since this line must be perpendicular to the tangent,  we must have that

$$
m'm = -1
$$

Therefore we can already find the slope as
$$
m = -\frac{1}{m'} = - \frac{a - x_0}{b - y_0}
$$

Now we just need to find the intercept. Plugging $(a, b)$ into $y = mx + q$ with the $m$ that we've just found we find
$$
b = ma + q
$$
and so
$$
b + a\left(\frac{a - x_0}{b - y_0}\right)
$$
The final equation is
$$
y = -\frac{a - x_0}{b - y_0} x + b + a\left(\frac{a - x_0}{b - y_0}\right)
$$


问:
Writing Probability distribution in terms of a trace over a density matrix
I have been given and expression for a probability distribution precisely,

$P(x,y,z)= \sum_\lambda P(x|y,\lambda)P(y|\lambda,z)P(z)P(\lambda)$

and I have been asked to show that the above expression can be written in the forma of the following (with correct changes)

$P(x,y,z)= Tr(\rho_{AB}(E_A^{x|y} \otimes E_B^{y|z}))P(z)$

where $\rho_{AB}, E_A^{x|y}, E_B^{y|z}$ are the density matrix and POVM’s on system AB.

I have no clue of even how the trace comes into picture. I have no clue as to how the first expression can be simplified to get a trace, leave alone getting the whole expression correctly.

Cross-posted on quantumcomputing.SE

答:
It's not exactly clear from the question exactly what you want but here's maybe something to get you started.

If you define some density matrices $\rho_A^\lambda$ such that $\mathrm{Tr}[\rho_A^{\lambda} E_A^{x|y}] = p(x|y,\lambda)$ and similarly define $\rho_B^{\lambda}$ such that $\mathrm{Tr}[\rho_B^{\lambda} E_B^{y|z}] = p(y|z,\lambda)$. Then you can define the joint state
$$
\rho_{AB} = \sum_\lambda p(\lambda) \rho_A^\lambda \otimes \rho_B^\lambda.
$$

Then you have
\begin{align*}
\mathrm{Tr}[\rho_{AB} (E_A^{x|y}\otimes E_B^{y|z})]p(z) &= \mathrm{Tr}[(\sum_\lambda p(\lambda) \rho_A^\lambda \otimes \rho_B^\lambda) (E_A^{x|y}\otimes E_B^{y|z})]p(z) \\
&= \sum_\lambda p(\lambda)\mathrm{Tr}[(\rho_A^\lambda \otimes \rho_B^\lambda) (E_A^{x|y}\otimes E_B^{y|z})]p(z) \\
&= \sum_\lambda p(\lambda)\mathrm{Tr}[\rho_A^\lambda E_A^{x|y}] \mathrm{Tr}[   \rho_B^\lambda E_B^{y|z}]p(z) \\
&= \sum_\lambda p(\lambda) p(x|y,\lambda) p(y|z,\lambda) p(z)\\
&= p(x,y,z)
\end{align*}

ContentLicense: CC BY-SA 4.0


问:
identity theorem to prove/disprove the existence of a holomorphic function
Does there exist a holomorphic function $f$ the unit disk $\{z :|z|<1 \}$ such that
$f(\frac{1}{n}) \leq \frac{1}{2^n} $ for all $ n \in \mathbb{N}$?

I know by identity theorem we can prove that $f$ not exists with $f(\frac{1}{n}) = \frac{1}{2^n} $ but not sure if this inequality can change the conclusion.

答: Conjecture:  The function $f(z)=0$ suggested by @Desperado is the unique holomorphic function that satisfies $$|f(1/n)| \le \frac{1}{2^n}.$$


问:
What is the equation of a tangent to a circle at a point?
Suppose I have the following circle
$$
(x - x_0)^2 + (y - y_0)^2 = r^2
$$
where the center $(x_0, y_0)$ and the radius $r$ are known. Is there a general formula for the equation of a tangent at a point $(a, b)$ on the circle, with $a, b$ known?

Idea

Since the point $(a, b)$ is on the circle, we must have
$$
(x - a)^2 + (y - b)^2 = r^2
$$
and since the tangent will pass through $(a, b)$ we must have
$$
\frac{y - b}{x - a} = m
$$

How do you incorporate the fact that the radius is perpendicular to the tangent?

Following the comments, I think then one notices that the radius is perpendicular to the tangent. The line through the center $(x_0, y_0)$ and the point $(a, b)$ has slope

$$
m' = \frac{b - y_0}{a - x_0}
$$

Since this line must be perpendicular to the tangent,  we must have that

$$
m'm = -1
$$

Therefore we can already find the slope as
$$
m = -\frac{1}{m'} = - \frac{a - x_0}{b - y_0}
$$

Now we just need to find the intercept. Plugging $(a, b)$ into $y = mx + q$ with the $m$ that we've just found we find
$$
b = ma + q
$$
and so
$$
b + a\left(\frac{a - x_0}{b - y_0}\right)
$$
The final equation is
$$
y = -\frac{a - x_0}{b - y_0} x + b + a\left(\frac{a - x_0}{b - y_0}\right)
$$

答:
(A bit different take on this from the other answers...)

Yes.

$$(x-a)^2+(y-b)^2+r^2=(x-x_0)^2+(y-y_0)^2$$

This formula comes from Pythagoras' theorem applied to the right-angled triangle with corners $(x,y)$, $(x_0, y_0)$ and $(a,b)$. It is a formula for a line because $x^2$ and $y^2$ will cancel when you expand LHS and RHS.


问:
Differential Equation Problem finding the Value
$(y'''-y"-4y'+4y)=2x^2-4x-1+2x^2e^{2x}+5xe^{2x}+e^{2x}$

I've already get the complementary function $y_{c}$$=$ $C$$_1$$e^x$$+$ $C$$_2$$e^{2x}$$+$$C$$_3$$e^{-2x}$
Also y$_p$$=$$A+Bx+Cx^2+Dxe^{2x}+Ex^2e^{2x}+Fx^3e^{2x}$

What  is the value of A,B,C,D,E,F?

Body:
$y_p=A+B x+C x^2+D e^{2 x} x+E e^{2 x} x^2+F e^{2 x} x^3$

$y'_p=B+2 C x+2 D e^{2 x} x+D e^{2 x}+2 E e^{2 x} x^2+2 E e^{2 x} x+2 F e^{2 x} x^3+3 F e^{2 x} x^2$

$y''_p=2 C+4 D e^{2 x} x+4 D e^{2 x}+4 E e^{2 x} x^2+8 E e^{2 x} x+2 E e^{2 x}+4 F e^{2 x} x^3+12 F e^{2 x} x^2+6 F e^{2 x} x$

$y^{(3)}=8 D e^{2 x} x+12 D e^{2 x}+8 E e^{2 x} x^2+24 E e^{2 x} x+12 E e^{2 x}+8 F e^{2 x} x^3+36 F e^{2 x} x^2+36 F e^{2 x} x+6 F e^{2 x}$

Plugging into $$y^{(3)}-y''-4 y'+4 y$$
Must be $$4 A-4 B-2 C+(4 B-8 C) x+4 C x^2+ e^{2x}\left(4 D+10 E+6 F+(8 E+30 F) x+12 F x^2\right))\equiv$$
$$\equiv 2 x^2-4 x-1+e^{2 x}\left(2 x^2+5 x+1\right) $$
$$
\begin{cases}
4 A-4 B-2 C=-1\\
4 B-8 C=-4\\
4C=2\\
4 D+10 E+6 F=1\\
8 E+30 F=5\\
12F=2\\
\end{cases}
$$
$$\left(A= 0,B= 0,C= \frac{1}{2},D= 0,F= \frac{1}{6}\right)$$
$$y_p=\frac{1}{6} e^{2 x} x^3+\frac{x^2}{2}$$


问:
If $A\in M_{n}(\mathbb{C})$ is a Hermitian matrix  so $I-iA$ is Invertible
I'd love your help with this.

Let $A\in M_{n}(\mathbb{C})$ be a Hermitian matrix $(A^{*}=A)$.

How can I prove that $I-iA$ is Invertible?

Thank you

答:
$I$ has only eigenvalue 1, and $A$ has only real eigenvalues, thus $iA$ will only have imaginary eigenvalues. Say $A$ has eigenvalues $\lambda_k$, where $\lambda_k$ could possibly be zero, with corresponding eigenvector $v_k$. Then:

$$(I-iA)v_k= v_k - i\lambda_k v_k = (1-i\lambda_k)v_k$$

Thus the eigenvalues for $(I-iA)$ will be on the form $1-i\lambda_k$, $\lambda_k \in \mathbb R$. Since all eigenvalues are non-zero, the matrix will be invertible.

Edit: Hermitian matrices have only real eigenvalues since, if $Av = \lambda v$, then
$$v^* A v = \lambda v^*v$$
taking the hermitian conjugate of both sides yields:
$$v^* A^* v = \lambda^* v^* v$$
Since $A = A^*$ we get $\lambda = \lambda^*$ from the two equations above, where $\lambda^*$ is the complex conjugate of $\lambda$. Thus $\lambda$ has to be real.

ContentLicense: CC BY-SA 3.0


问:
A problem: Exists a proposition $r \in Prop(A \cup B)$ such that $\vDash (p \rightarrow r) \land (r \rightarrow q)$
Let $A,B$ two arbitrary iof propositions. Show that if $p \in Prop(A)$ and $q \in Prop(B)$ are two propositions such that $\vDash p \rightarrow q$ then exists a proposition $r \in Prop(A \cup B)$ such that $\vDash (p \rightarrow r) \land (r \rightarrow q)$.

I have a hint wich says that I should use induction in the numbers of letters wich are in p but not are in q.

My attemp:

Base case:
Since we do not have letters on $p$ wich are not in $q$, we have that $p = q$. As $\vDash p \rightarrow q$ we have that for all model $t: A \cup B \rightarrow \{0,1\}$ it follows that $t(p \rightarrow q) = 1$; then, $t(p)\leq t(q)$. If we consider the same model for:
$$\begin{array}{r c l c r}
            t(R) & = & t[(p \rightarrow r) \land (r \rightarrow q)] & \hspace{1cm} & \left(\text{$R$ definition}\right)\\
                 & = & \min\{t[(\neg p \lor r), t(\neg r \lor q)\} & \hspace{1cm} & \left(\text{$\land$ definition $\rightarrow$abreviation}\right)\\
                 & = & \min\{\max\{(1 - t(p), t(r), 1 - t(r), t(p)\}\} & \hspace{1cm} & \left(\text{$\lor$ definition and $\neg$ definition}\right)\\
\end{array}$$

And then I do not want to do. Can any one help me with the following step.

答:
As stated, your question is trivial: is enough to take $r=p$; you certainly have both $p\rightarrow r$ (tautology) and $r\rightarrow q$ (from the hypothesis), and since $p$ is a formula with variables in $A$, $r$ is clearly a formula with variables in $A\cup B$.

What you probably wanted to prove is Craig's interpolation theorem (of which Wikipedia presents the standard proof), stating that you can find a formula $r$ satisfying $\vDash(p\rightarrow r)\wedge(r\rightarrow q)$ and that lies in $Prop(A\cap B)$: notice it is an intersection instead of an union.

The hint you received leads to what is probably the most standard proof of the theorem. If we denote the set of propositional variables appearing in the formula $p$ by $var(p)$, what you are doing is induction on the cardinality of $var(p)\setminus var(q)$.

Base case: I don't think you can actually prove, from the fact that $|var(p)\setminus var(q)|=0$, that $p=q$, but what you can prove is that by making $r=p$ you are done (since, whenever $var(p)\setminus var(q)=\emptyset$, $var(p)\subseteq var(q)$).

Inductive step: Assume, as induction hypothesis, that the result holds for any formulas $s$ and $t$ such that $|var(s)\setminus var(t)|\leq n$ and $\vDash s\rightarrow t$, and suppose $|var(p)\setminus var(q)|=n+1$. Then, take a propositional variable $x$ in $p$, but not in $q$, and define
$$p'=p[\top/x]\vee p[\bot/x],$$
where $p[\top/x]$ is the formula obtained from $p$ by replacing every occurrence of $x$ by the top element (if you do not want to add a symbol to your language, just make $\top=x\vee \neg x$), and $p[\bot/x]$ is obtained from $p$ by replacing every occurrence of $x$ by $\bot$ (equal to $x\wedge\neg x$, if you want to). Now, we may prove a few properties of $p'$.

$p\vDash p'$: To see this, take any valuation $\nu$ such that $\nu(p)=1$; if $\nu(x)=1$, $\nu(p[\top/x])=1$ (since $\nu(\top)=1$) and so $\nu(p')=1$; if $\nu(x)=0$, $\nu(p[\bot/x])=1$ and again $\nu(p')=1$.

$|var(p')\setminus var(q)|=n$: Since $p'$ has the same variables as $p$, except for $x$.

$p'\vDash q$: Take a valuation $\nu$ such that $\nu(p[\top/x])=1$; if you define the valuation such that $\nu^{*}(y)=\nu(y)$, for every propositional variable $y\neq x$, and $\nu^{*}(x)=1$, one sees that $\nu^{*}(p)=\nu(p[\top/x])=1$, and since $p\vDash q$, $\nu^{*}(q)=1$. Given that $x$ is not a variable in $q$, $\nu^{*}(q)=\nu(q)$, meaning $p[\top/x]\vDash q$. Analogously you prove that $p[\bot/x]\vDash q$, and so $p'\vDash q$.

Now, from induction hypothesis, there exists $t$ such that $p'\vDash t$, $t\vDash q$ and $var(t)\subseteq var(p')\cap var(q)$; but since $p\vDash p'$ and $var(p')\subseteq var(p)$, you get that $p\vDash t$, $t\vDash q$ and $var(t)\subseteq var(p)\cap var(q)$, what ends the proof!


问:
Meaning of G(k) being dense in G for an algebraic group G/k
Let $G$ be an algebraic group over a field $k$ of characteristic 0. I have read that: if $G$ is connected, then $G(k)$ is dense in $G$ for the Zariski topology. I do not understand what kind of density this might be since $G(k)$ is abstractly defined as the set of morphisms $\text{Spec}(k) \to G$. If $G$ is an affine variety in $\mathbb{A}_k^n$ for instance, one might think of $G(k)$ as a subset of $k^n$. But how does this connect to the abstract definition of $G$ as a scheme or as a functor from $k$-algebras to groups? I do not see a direct relation or any mean to relate the group $G(k)$ (a set of points) to the abstract scheme (or functor) $G$, let alone saying that $G(k)$ is dense in $G$ for the Zariski topology. How is this topology even defined on $G(k)$ and how is $G(k)$ even a subset of $G$? Could someone please resolve this issue for me? Thank you in advance.

答:
It's good to learn the functor-of-points approach inside out, but for affine algebraic groups we can use embeddings to get our head around things.

First, $G$ had better be affine or this is utterly false (for instance an elliptic curve might have only one rational point). Once you know $G$ is affine, you can pick an embedding in affine space and drop the functor-of-points abstraction.

Then $G(k)$ just means the points whose coordinates are all in $k$, and the claim is that this subset of $G(\overline{k})$ is dense in it, for the usual zariski topology.


问:
Prove that $f(z) $ is an odd function when $z \in \mathbb{R}$
Given an entire function which is real on the real axis and imaginary on the imaginary axis, prove that it is an odd function i,e $f(-z)= -f(z)$

My attempt :  Here two cases will be possible

$1.$ when $z= x$

$2.$ when $z = ix $

I have no any issue when $z=ix$  and solution is alreaday given here

but i have some issue when $z=x$

$$f(x) = \overline{ f(\bar x) }$$

$$\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\bar a_nz^n \implies a_n=\bar a_n$$

\begin{align}
f(x)=\sum_{n=0}^{\infty}a_nx^n&=\sum_{n=0}^{\infty}a_{2n}x^{2n}+\sum_{n=1}^{\infty}a_{2n+1}x^{2n+1}
\end{align}

Here  $f(x)$  is both  even and odd

so here  we are getting  contradiction that  $f(z)$  is odd that mean $f(x)$ is not even function

答:
On the solution here

The guy first assumed z is only imaginary ($z = iy$). Then he used that to show that all even $a$'s should be zero. That is $a_{2n}=0$.
This means
$$f(z) = a_1z^1 + a_3z^3 + a_5z^5 + ...$$
for all $z = x +iy $

Then he used that to show that the function is odd for any $z$ ($z = x+iy$)

That is:
$$
f(-z) = a_1(-z)^1 + a_3(-z)^3 + a_5(-z)^5 + ... = - (a_1z^1 + a_3z^3 + a_5z^5 + ...) = -f(z)
$$


问:
Good introductory book on mathematical morphology
What is a good introductory book on mathematical morphology? It should outline the basic theory and a focus on applications would be nice. If possible, it should be approachable for a determined undergrad.

Thank you.

EDIT: The field is relatively new I guess, so if there indeed doesn't exist any introductions yet, please do let me know.

答: A good book is "Mathematical Morphology" edited by Laurent Najman and Hugues Talbot, Wiley, 2010. It's translated from a French version (2008, 2010). The introductory chapter is a good overview, and the other chapters are more specialized accounts.</p>


问:
Simplify Fraction with variables
Simplify -270y^2/189x^3

Not too sure how to go about simplifying this. I googled it, and google showed me the steps but no explanation.

Step 1: Cancel terms that are in both numerator and denominator

```<code>     -270y^2/189 * x^3
</code>```

Reduced: -10y^2/7 * x^3

Step 2: combine multiplied terms into a single fraction

```<code>     -10y^2/7 * x^3

     -10y^2 * x^3 / 7
</code>```

Solution: -10y^2 * x^3 / 7

Why did 7x^3 get separated into 7 and x^3?

Why did the equation get multiplied or looks multiplied by x^3?

Why do we combine x^3 with 10y^2 in the numerator? I.E. 10y^2*x^3 / 7?

Can someone go over simplifying this in a more detailed manner?

TIA

答:
It is often useful to use parentheses in order to avoid ambiguity. We have
$$
(-270y^2) / (189x^3) = \frac{-270y^2}{189x^3} = \frac{-10y^2}{7x^3}
$$
since
$$
270 = 2 \times 3^3 \times 5 \qquad \text{and} \qquad 189 = 3^3 \times 7.
$$

Title: Simplify Fraction with variables


问:
Let $k \geq 3$; prove $2^k$ can be written as $(2m+1)^2+7(2n+1)^2$
Prove:

If $k \geq 3$, then $2^k$ can be written as $(2m+1)^2+7(2n+1)^2$, where $k, m, n \in \mathbb{N}$.

Body: Hint: $2=(\frac{1+\sqrt{7}i}{2})(\frac{1-\sqrt{7i}}{2})$.


问:
Why are my probabilities from binomial distribution not a sum of 1?
I am holding a best of 3 series against two basketball teams: $Team\,A$ and $Team\,B$. Either team must win or lose the series; a tie is impossible. I must calculate the probability of either team winning the series.

During their exhibition games (outside of the 3-game series), $Team\,A$ beat $Team\,B$ in $3$ out of $5$ games.

Through experimental probability, $Team\,A$ has a $3/5$ chance of winning a game against $Team\, B$.

Therefore, $Team\,B$ must have a $2/5$ chance of winning a game against $Team\, A$.

Using these values, I determined the binomial probability of $Team\, A$ winning 2 out of 3 series games ($P_{2_A}$) as follows, assuming that losing is the only other outcome:

$P_{2_A}={3\choose2} \times \left(\frac{3}{5}\right)^2 \times \left(1- \frac{3}{5}\right)^{3-2}$

$P_{2_A} = 3\times \frac{9}{25}\times \frac{2}{5}$

$P_{2_A} = \frac{54}{125}$

I also did the same math for $Team\, B$ ($P_{2_B}$), using the assumptions as prior:

$P_{2_B}={3\choose2} \times \left(\frac{2}{5}\right)^2 \times \left(1- \frac{2}{5}\right)^{3-2}$

$P_{2_B} = 3\times \frac{4}{25}\times \frac{3}{5}$

$P_{2_B} = \frac{36}{125}$

When adding $P_{2_A}$ and $P_{2_B}$ together, the probabilties do not add to $1$. I do not believe this should happen as there are only two possible endgames: $Team\, A$ wins or $Team\, B$ wins.

$\frac{54}{125} + \frac{36}{125} = \frac{90}{125} \, or \, \frac{18}{25}$

Is it incorrect to assume that all binomial probabilties must add to 1? Could binomial distribution be the incorrect approach for this type of probability scenario?

答: You missed the possibility that one team would win all three games.  You assumed that the series would end two games to one.  The easiest way to compute these possibilities is to assume they play all three games, but you can also condition on whether one team wins the first two games.  Yes, the probabilities will sum to $1$ if you get all of them.


问:
Pattern for all the binary chains divisible by 5
For instance, $x = 101$ is divisible by $5$ because it is the integer 5. Same thing for $x=1111$ is also divisible by 5 as it is the integer 15. However, $x=1100$ is not divisible by $5$ as it is the integer 12.

Is there a pattern the recognise the binary chains divisible by 5?

答:
$5 = 2^2 +1$ so whatever rule we can use in base $10$ to tell if a number is divisible by $10^2 + 1 = 101$ should be similar.

But .... consider this.

$2^4 \equiv 1 \pmod 5$

So $2^{4k+1}\equiv 2^1\equiv 2 \pmod 5$.

$2^{4k+2}\equiv 2^2\equiv 4 \equiv -1 \pmod 5$

$2^{4k+3}\equiv 2^3\equiv 8\equiv - 2\pmod 5$

$2^{4k}\equiv 1^k\equiv 1\pmod 5$.

So consider the digits in position.  Subtract the digits in the $4k+2$ positions for the one in the $4k$ positions. Subtract the $4^k+3$ from the $4^k+1$ positions.  Multiply result from the odd positions by $2$ and add to the even; and mod by $5$.

Example:  I'm typing at random: $101110101000111010111$

Group it into four sections:

$\color{blue}1\color{orange}0\color{purple}1\color{red}1\color{blue}1\color{orange}0\color{purple}1\color{red}0\color{blue}1\color{orange}0\color{purple}0\color{red}0\color{blue}1\color{orange}1\color{purple}1\color{red}0\color{blue}1\color{orange}0\color{purple}1\color{red}1\color{blue}1$

The $\color{blue}{4k}$  positions (these are all $\equiv 1\pmod 5$) I count/add: $\color{blue}6$.

In the $\color{purple}{4k+2}$ positions (these are all $\equiv -1\pmod 5$) I count/add $\color{purple}4$.  Subtracting I get $6-4\equiv 2\pmod 5$.

In the $\color{red}{4k+1}$ positions (these are all $\equiv 2\pmod 5$) I count: $\color{red}2$.

And int the $\color{orange}{4k+3}$ positions ($\equiv -2\pmod 5$) I get: $\color{orange}1$.  Subtracting I get $2-1=1$ which is equiv $1\times 2\equiv 2\pmod 5$.

So multiply tho odd positions by $2$ to get $2$ and add to the even positions of $2$ to get $2+2 = 4$ and I conclude:

$101110101000111010111_2\equiv 4\pmod 5$.

I google for "binary decimal converter" and find this cute little page: https://www.rapidtables.com/convert/number/binary-to-decimal.html

And I see $101110101000111010111_2 = 1528279$ which is $\equiv 4\pmod 5$.

....

This is essentially the "casting out every other digit" rule to determine if a number in base $10$ is divisible by $11$.

.....

To show how quick this can be:

I convert $35765$ to $1000101110110101$ and do

$1000|1011|1011|0101$ and calculate $(3-1) + 2(2-3)=2+(-2)=0$ so it is divisible by $5$.


问:
Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2.
So far, I have denoted
$$\frac{MA}{MB}=a,\frac{BN}{NC}=b,\frac{CP}{PD}=c,\frac{DQ}{QA}=d,\frac{AR}{RC}=e,\frac{BS}{SD}=f$$
and then I have computed the volumes of the "small" tetrahedra
$$V_1=V_{AMRQ}=\frac{ae}{(a+1)(e+1)(d+1)};$$
$$V_2=V_{BMSN}=\frac{bf}{(b+1)(f+1)(a+1)};$$
$$V_3=V_{CRNP}=\frac{c}{(c+1)(e+1)(b+1)};$$
$$V_4=V_{DPQS}=\frac{d}{(d+1)(c+1)(f+1)}.$$
Also, by Menelaus theorem for tetrahedra, one can deduce that
$$abcd=1;~~~ af=ce; ~~~b=fde.$$
It remains to show that $$V_1+V_2+V_3+V_4\geq \frac{1}{2}.$$

The problem has a well known analogue in 2-D: for a triangle $ABC$ of area 1 and $AM, BN, CP$ concurrent cevians, the area of the triangle $MNP$ is less than 1/4.

Later edit/update:

I have managed to rewrite $V_k$ as follows:
$$V_1={\frac {a}{ \left( a+1 \right)  \left( d+1 \right)  \left( cd+1
 \right) }};$$
$$V_2={\frac {b}{ \left( a+1 \right)  \left( b+1 \right)  \left( ad+1
 \right) }};$$
$$V_3= {\frac {c}{ \left( c+1 \right)  \left( b+1 \right)  \left( ab+1
 \right) }};$$
$$V_4= {\frac {d}{ \left( c+1 \right)  \left( d+1 \right)  \left( bc+1
 \right) }},$$
Plus, we still have $abcd=1$.

答:
Well this not a rigorous proof for this problem, but I decided to express some of my ideas that may lead to a geometric proof.

First I managed to find set of points of concurrency $S_i$ inside an arbitrarily triangle $T$ which make same area of triangles shaped by their three concurrent segments as you mentioned in your 2-D case. the sets of such points can also be defined by the level sets of a function $f$ that maps each point inside of the triangle $T$ to the area of its mentioned triangle. by use of continuity we can deduce that every level set is a jordan closed loop $J_i,i\in [0,1]$ and even it seems that they are convex loops which doesn't intersect eachother and their value by function $f$ (which resembles the mentioned triangles area) become bigger and bigger as the loops become smaller and smaller until they retract to centroid of triangle $T$.

I think this process can be easily generalized to the case of the the 3-d tetrahedron and the level sets here would become border of convex bodies which retracts to the centroid of tetrahedron when it gains its highest value by the mapping function sending each point inside of tetrahedron to the volume of its octahedron you mentioned.

While searching for a proof of this claim ,I was curious about a transformation that convert an arbitrarily tetrahedron to a regular one as it seems good idea too for the 2-D case to convert a given triangle to the most symmetrical triangle which is equilateral triangle. I've just find a kind of such transformation here. I still don't know this transformation would be useful from the perspective of my approach for the sake of the statement of inequality of areas in this problem.
for example if the jordan loops $J_i$ in $T$ becomes circles while transformed, then the proof would become easier.

My second idea is about to prove this statement:

The volume of mentioned octahedron $MNPQRS$ will be maximum if and only if the centroids
of $ABCD$ and $MNPQRS$ coincide.


问:
The sum of subspaces of $V$ is the smallest subspace of $V$ containing each of the subspaces
Let $V$ be a vector space with subspaces $U_1,\ldots,U_k$. Then $U_1+\ldots+U_k$ is the smallest subspace of $V$ containing each of $U_1,\ldots,U_k$.

Here's my approach (using Einstein summation) where I have a doubt:

Let $v\in U_1+\ldots+U_k$. Then $v=u_1+\ldots+u_k$ which is a linear combination of elements from $U_1\cup\ldots\cup U_k$, which means $v\in\text{span}(U_1\cup\ldots\cup U_k)$.

Let $v\in\text{span}(U_1\cup\ldots\cup U_k)$. Then $v$ is a linear combination of vectors $u_{ij}$ from subspaces ($j$ subscript denotes which subspace that vector is from and $i$ runs over vectors from a particular subspace):
$$v=b^ja^iu_{ij}=b^jw_j$$ where $a^iu_{ij}=w_j$ lies in $U_j$ due to closure under linear combinations.

My doubt is that any linear combination can contain arbitrarily many terms. Is the above approach valid in the general sense?

答: Let $W$ be any subspace of $V$ containing each of  $U_1,U_2,\cdots, U_k$. Then for any $u_i \in U_i~~~i=1, \cdots, k$, $u_1+u_2+\cdots+u_k \in W~~~(\because \textit{subspace})$ implying $U_1+U_2+\cdots+U_k \subseteq W$, hence $U_1+U_2+\cdots+U_k$ is the smallest.


问:
Find the range of: $y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$
Find the range of:
$$y=\sqrt{\sin(\log_e\frac{x^2+e}{x^2+1})}+\sqrt{\cos(\log_e\frac{x^2+e}{x^2+1})}$$

What I tried:

Let:$$\log_e\frac{x^2+e}{x^2+1}=X,$$
then $$y=\sqrt {\sin X}+\sqrt{\cos X}$$
$$y_{max}at X=\pi/4$$

The rest is too complicated. I am stuck up.

Can someone please give a analytical solution.

Even Wolfram alpha doesn't help.

答:
let $f(x)=\frac{x^2+e}{x^2+1} \to 1<f(x)\le e \to 0<X\le 1$

$y^2=sinX+cosX+\sqrt{2sin2X}=\sqrt{2}\left(sin(X+\dfrac{\pi}{4})+\sqrt{sin2X}\right)$

when $0<X<\dfrac{\pi}{4}$,$sin2X$ and $sin(X+\dfrac{\pi}{4})$ are mono increasing,$\to y^2>1 \to y>1$

I left another part for you to discuss and you should be able to find out the range now.


问:
$\lim_{y \to 0}\frac{1}{\ln(y)}\int_{y}^1\frac{1-\cos(x)}{x^3}dx=-1/2$
We want to find an equivalent to $\int_{y}^1\frac{1-\cos(x)}{x^3}dx$ when $y \to0.$ It seems that $\lim_{y \to 0}\frac{1}{\ln(y)}\int_{y}^1\frac{1-\cos(x)}{x^3}dx=-1/2.$

After Trying an integration by parts, change of variables, dominated convergence theorem.... Didn't find anything.

Is there a way to solve it, maybe a change of variables which I missed, dominated convergence theorem...

Body:
We know that $\cos(x)=1-\frac{x^{2}}{2!}+O\left(x^{4}\right)$, so

\begin{align*}
\lim_{y\to0}\frac{1}{\ln(y)}\int_y^{1}\frac{1-\cos(x)}{x^3}dx&=\lim_{y\to0}\frac{1}{\ln(y)}\int_y^{1}\frac{x^2/2}{x^3}dx+O\left(\lim_{y\to 0}\frac{1}{\ln(y)}\int_{y}^{1}xdx\right)\\
&=\lim_{y\to0}\frac{1}{\ln(y)}\int_y^{1}\frac{x^2/2}{x^3}dx\\
&=\lim_{y\to0}\frac{1}{2\ln(y)}\int_y^{1}\frac{1}{x}dx\\
&=\lim_{y\to0}\frac{1}{2\ln(y)}(1-\ln(y))\\
&=\frac{-1}{2}
\end{align*}

And so we have

$$\lim_{y\to0}\frac{1}{\ln(y)}\int_y^{1}\frac{1-\cos(x)}{x^3}dx=-\frac{1}{2}$$.

Not $-1$.


问:
Prove that $|{xy-x_0y_0}| < \epsilon$
Supposing that
$|x-x_0| < min(1, \frac{\epsilon}{2(|y_0|+1)})$ and $|y-y_0| < \frac{\epsilon}{2(|x_0|+1)}$
it follows that  $|{xy-x_0y_0}| < \epsilon$

(A friend told me it was helpful to prove this in hopes of proving that $\lim_{x \to c}(f*g)=L*M$ where $\lim_{x \to c}f(x) = L$ and $\lim_{x \to c}g(x) = M$)

I think it's supposed to end something like $|{xy-x_0y_0}| <|x-x_0||y-y_0|<(2(|x_0|+1))*\frac{\epsilon}{2(|x_0|+1)}<\epsilon$ and you're finished, but I get stuck in the steps before.

Title: Prove that $|{xy-x_0y_0}| < \epsilon$

Body:
just to complete the proof:

$$|x-x_0|<min\left(1, \frac{\epsilon}{2(|y_0|+1)}\right)\leq 1$$ then $|x-x_0| <1$.

$\implies |x|-|x_0| \leq |x-x_0|<1$. Therefore $|x| < 1+|x_0|$.

Also $|y_0|<|y_0|+1$, from the hint we have:

$$|xy - x_0y_0| \leq |x||y-y_0| + |y_0||x-x_0| < (1+|x_0|)\frac{\epsilon}{2(|x_0|+1)} + (|y_0| + 1)\frac{\epsilon}{2(|y_0|+1)} = \epsilon$$


答:
Numerator goes to $\infty$ and denominator to $-\infty$

L'Hopital rule
$$\lim_{y\to 0}\frac{\int_y^1 \frac{1-\cos x}{x^3} \, dx}{\ln y}=\lim_{y\to 0}\frac{\frac{\cos y}{y^3}-\frac{1}{y^3}}{\frac{1}{y}}=\lim_{y\to 0}\frac{\cos y-1}{y^2}=-\frac12$$

Title: $\lim_{y \to 0}\frac{1}{\ln(y)}\int_{y}^1\frac{1-\cos(x)}{x^3}dx=-1/2$

Body:
We want to find an equivalent to $\int_{y}^1\frac{1-\cos(x)}{x^3}dx$ when $y \to0.$ It seems that $\lim_{y \to 0}\frac{1}{\ln(y)}\int_{y}^1\frac{1-\cos(x)}{x^3}dx=-1/2.$

After Trying an integration by parts, change of variables, dominated convergence theorem.... Didn't find anything.

Is there a way to solve it, maybe a change of variables which I missed, dominated convergence theorem...


问:
A question about infimum and supremum.
Let $f,g : \mathbb{R}\rightarrow [0,1]$ two functions. My question is whether or not we have that

$$|\inf_{x} f(x) - \inf_{x} g(x)|\leq \sup_{x} |f(x) - g(x)|$$.

If it is not true in general, what are the cases that it is true?.

Thanks in advance.

答:
Yes the inequality always holds.

Let $F:=\inf_x f(x)$, $G:=\inf_x g(x)$.
We can
assume w.l.o.g. that $F<G$.

For any $0<\varepsilon<G-F$ there exists a $y$ such that
$$f(y)<F+\varepsilon < G \leq g(y)$$
so
$$\sup_x |f(x)-g(x)| \geq \underbrace{g(y)}_{\geq G} - f(y) \geq G-F - \varepsilon .$$
Letting $\varepsilon\to 0$ shows the claim.

Title: A question about infimum and supremum.


问:
What is the equation of a tangent to a circle at a point?
Suppose I have the following circle
$$
(x - x_0)^2 + (y - y_0)^2 = r^2
$$
where the center $(x_0, y_0)$ and the radius $r$ are known. Is there a general formula for the equation of a tangent at a point $(a, b)$ on the circle, with $a, b$ known?

Idea

Since the point $(a, b)$ is on the circle, we must have
$$
(x - a)^2 + (y - b)^2 = r^2
$$
and since the tangent will pass through $(a, b)$ we must have
$$
\frac{y - b}{x - a} = m
$$

How do you incorporate the fact that the radius is perpendicular to the tangent?

Following the comments, I think then one notices that the radius is perpendicular to the tangent. The line through the center $(x_0, y_0)$ and the point $(a, b)$ has slope

$$
m' = \frac{b - y_0}{a - x_0}
$$

Since this line must be perpendicular to the tangent,  we must have that

$$
m'm = -1
$$

Therefore we can already find the slope as
$$
m = -\frac{1}{m'} = - \frac{a - x_0}{b - y_0}
$$

Now we just need to find the intercept. Plugging $(a, b)$ into $y = mx + q$ with the $m$ that we've just found we find
$$
b = ma + q
$$
and so
$$
b + a\left(\frac{a - x_0}{b - y_0}\right)
$$
The final equation is
$$
y = -\frac{a - x_0}{b - y_0} x + b + a\left(\frac{a - x_0}{b - y_0}\right)
$$

答:
Label $(x_0, y_0)$ as $O$, $(a,b)$ as $T$ and let $P$ be an arbitrary point on the tangent line at $(x,y)$.

Then the dot product $OT.TP = 0$ because the vectors need to be perpendicular, which calculates out as

$(a-x_0)(x-a) + (b-y_0)(y-b) = 0$

$(y-b) = \dfrac{(a-x_0)}{(b-y_0)}(a-x)$


问:
Is the ideal $I = \{f\mid f (0) = 0\}$ in the ring $C [0, 1]$ of all continuous real valued functions on $[0, 1]$ a maximal ideal?
Is the ideal $I = \{f \mid f (0) = 0\}$ in the ring $C [0, 1]$ of all continuous real valued functions on the interval $[0, 1]$ a maximal ideal?</p>

答:
As suggested by Sigur, you can show that $\varphi : \left\{ \begin{array}{ccc} C([0,1]) & \to & \mathbb{R} \\ f & \mapsto & f(0) \end{array} \right.$ induces an isomorphism between $C([0,1])/I$ and $\mathbb{R}$. Therefore, $C([0,1])/I$ is a field and $I$ is a maximal ideal.

Otherwise, you can prove it directly from the definition: Let $J$ be an ideal of $C([0,1])$ such that $I \subsetneq J$. In particular, there exists $g \in J \backslash I$ (ie. $g(0) \neq 0$). For every $f \in C([0,1])$:

$$f= \underset{ \in I \subset J}{\underbrace{\left( f- \frac{f(0)}{g(0)}g \right)}}+ \underset{\in J}{\underbrace{\frac{f(0)}{g(0)}g}} \ \in J$$

Therefore, $J=C([0,1])$ and $I$ is a maximal ideal.


问:
Upper bound on trace formula
I am trying to prove that this quantity: $f(W) = trace((C^T\circ(W^T - W'^T))(W'-W)(XX^T))$ is smooth. $\circ$ is the Hadamard product. Smoothness is satisfied if  $ f(W)\le H\|W - W' \|^2$. $H$ is a constant. $\| .\|$ is any norm.

I am very close to do it, however, I am struggling in dealing with Hadamard product inside a trace.

答:
One approach is to use vectorization. If $\|\cdot\|$ denotes the Frobenius norm, then we have $\|W\| = \|\operatorname{vec}(W)\|$, and we can apply the following facts:

$\operatorname{vec}(A \circ B) = \operatorname{diag}(\operatorname{vec}(A)) B$,

$\operatorname{tr}(A^T B) = \operatorname{vec}(A)^T\operatorname{vec}(B)$,

$\operatorname{vec}(AB) = (B^T \otimes I)\operatorname{vec}(A)$.

With that, we can rewrite
$$
\operatorname{tr}((C^T\circ(W^T - W'^T))(W'-W)(XX^T)) =\\
\operatorname{vec}(W' - W)^T
\operatorname{diag}(\operatorname{vec}(C))((XX^T) \otimes I)
\operatorname{vec}(W' - W).
$$
With the inequality $x^TAx \leq \sigma_{\max}(A) \|x\|^2$, we can therefore conclude that
$$
\operatorname{tr}((C^T\circ(W^T - W'^T))(W'-W)(XX^T)) \leq
\\
\sigma_{\max}(\operatorname{diag}(\operatorname{vec}(C))((XX^T) \otimes I)) \cdot \|W - W'\|^2.
$$

Another less computation-intensive approach: for $W \neq W'$, note that $\frac{f(W)}{\|W - W'\|^2} = f(Y)$, where $Y = W' + \frac{W - W'}{\|W - W'\|}$. Note that every matrix of the from $A = \frac{W - W'}{\|W - W'\|}$ satisfies $\|A\| = 1$.

With that in mind, let $S$ denote the set
$$
S = \{Y: Y = W' + A \text{ and } \|A\| = 1\}.
$$
Note that $S$ is compact (closed and bounded). Thus, the continuous function $f$ must attain a maximum over $S$. In other words, there exists a $k > 0$ such that for all $W \neq W'$, we have
$$
\frac{f(W)}{\|W - W'\|^2} \leq k \implies f(W) \leq k \cdot \|W - W'\|^2.
$$
Trivially, this second inequality holds for $W = W'$ as well.

Regarding the explicit computation of an upper bound: note that
$$
\sigma_{\max}(\operatorname{diag}(\operatorname{vec}(C))((XX^T) \otimes I)) \leq \\
\sigma_{\max}(\operatorname{diag}(\operatorname{vec}(C)))
\cdot \sigma_{\max}(((XX^T) \otimes I)) = \\
\left[\max_{i,j} |C_{ij}| \right]\cdot \sigma_{\max}(X)^2.
$$
Most computational software will have a reasonably efficient method for computing $\sigma_{\max}(X)$.

Title: Upper bound on trace formula


问:
Making $100$ ordered lists with the numbers $1 - n$ $5$ times such that no two lists are the same at more than one index.
What is the minimum $n$ such that it is possible to make at least $100$ ordered lists with the numbers $1-n$ (can repeat numbers) of length $5$ with the property that any two ordered lists are the same at one index?

Ex:
If we considered $n =10$, then it would be impossible to have the lists $\{1, 2, 6, 8, 3 \}$ and $\{1, 8, 6, 6, 4 \}$ because they both share a $1$ at the first index and a $6$ at the third index.

However, we could have the lists $\{1, 2, 3, 4, 5 \}$ and $\{6, 7, 8, 9, 5\}$ because they only share a $5$ at the last index and one common element is allowed.

I honestly have no idea how to begin with this problem. I was thinking of derangements at first but the casework got very complex. Does anyone have any ideas?

答:
The minimum $n$ is either $10$ or $11$, and the answer is unknown by the greater math community.

Obviously, you need $n\ge 10$; if you look at the first two entries of each list, you will have a sequence of $100$ ordered pairs. These pairs must be pairwise distinct, otherwise the two corresponding lists would agree in their first two places. Since there are at most $n^2$ such ordered pairs, you need $n^2\ge 100$.

This would be possible with $n=10$ if and only if there existed $3$ mutually orthogonal Latin squares of order $10$. Whether these exist is an open problem as of 2005.

You can succeed with $n=11$ using all $121$ lists of the form
$$
(x,y,x+y,x+2y, x+3y)
$$
where $x,y$ range over $\{0,1,\dots,10\}$, and all arithmetic is taken modulo $11$.

In terms of coding theory, you are looking for the smallest number $n$ such that $A_n(5, 4)\ge 100$, where $A_q(\ell, d)$ is the size of the largest code with length $\ell$, distance $d$, and an alphabet of size $q$.


问:
Solve for x Given Two Square Roots: Algebra Problem
I am trying to solve for all real numbers for $x$ given $5=\sqrt{9-x^2}+\sqrt{16-x^2}$.

The answer is that $x=\pm \frac{12}{5}$. I am looking for a clean way to do it. I am stumped, and I feel like I am missing something about conjugates here.

Body:
There is no clean way that I know of. We have

$5=\sqrt{9-x^2}+\sqrt{16-x^2} \implies $

$5-\sqrt{9-x^2}=\sqrt{16-x^2} \implies $

$(5-\sqrt{9-x^2})^2=16-x^2 \implies $

$25+9-x^2-10\sqrt{9-x^2} = 16-x^2 \implies $

$34-x^2-10\sqrt{9-x^2}  = 16-x^2 \implies $

$18 =10\sqrt{9-x^2} \implies $

$\frac{9}{5} = \sqrt{9-x^2} \implies $

$\frac{81}{25} = 9-x^2 \implies $

$ \frac{-144}{25} = -x^2  \implies $

$ x = \pm \frac{12}{5} $


问:
period of a function $\lvert\sin x\rvert+\lvert\cos x\rvert$
I have read that $$y=\lvert\sin x\rvert+ \lvert\cos x\rvert $$ is periodic with fundamental period $\frac{\pi}{2}$.

But Wolfram says it is periodic with period $\pi$.

Please tell what is correct.

答:
You should try this and figure it out. It is not that hard, and it is a bad habbit to give up on something that you can do. Wolfram alpha is not always correct, it's written by humans.

It's true that this function is $\pi$ periodic, just that its prime period is $\frac{\pi}{2}$.


问:
Eigenvalues of $8\times8$ non-diagonal matrix $A$ satisfying $A^2 - 6A + 9I_8 = 0_8$
In an exam today I had the following problem:

Let $A \in M_{8}(\mathbb{R})$ ($8\times8$ matrix with real entries) satisfying $$A^2 - 6A + 9I_8 = 0_8$$ Find the eigenvalues of $A$.

My approach

If $\lambda$ is an eigenvalue and $v$ is an eigenvector with eigenvalue $\lambda$, dotting both sides of the equation with $v$ gives
$$\lambda^2 + 6\lambda + 9 = 0 \implies (\lambda - 3)^2 = 0 \implies \lambda = 3.$$

On the other hand the given relation reads $(A - 3I_8)^2 = 0_8$ so
$$\det(A - 3I_8)^2 = 0 \implies \det(A - 3I_8) = 0.$$

Thus $3$ is always an eigenvalue of A, so we get $\text{Spec} A = {3}.$

My question

Why is it relevant that $A$ is nondiagonal? The only implication I could find is that considering the minimal polynomial $p(x)$ of $A$, we have
$$p(x) \mid (x - 3)^2 \Rightarrow p(x) = x-3 \text{  or  } p(x) = (x-3)^2.$$

But since $A$ is nondiagonal we can't have $p(x) = x - 3$ so $p(x) = (x-3)^2$. But I don't see why this would influence the eigenvalues of $A$.

So is the fact that $A$ is diagonal relevant?

答:
The matrix is similar to one or more little blocks on the diagonal, each
$$
\left(
\begin{array}{cc}
3&1 \\
0&3
\end{array}
\right)
$$
Whatever is left over is diagonal $3$


问:
identity theorem to prove/disprove the existence of a holomorphic function
Does there exist a holomorphic function $f$ the unit disk $\{z :|z|<1 \}$ such that
$f(\frac{1}{n}) \leq \frac{1}{2^n} $ for all $ n \in \mathbb{N}$?

I know by identity theorem we can prove that $f$ not exists with $f(\frac{1}{n}) = \frac{1}{2^n} $ but not sure if this inequality can change the conclusion.

答:
I guess the conjecture of @mjw is right, here a tentative proof. Let me now if there are mistakes.

The zero-function satisfies the conditions, and looking for a contradiction,
let $f$ be a non-zero holomorphic function satisfying the condition too. The function $f$must have a zero of order $m\in\mathbb N_{>0}$ in $z=0$, hence
$g(z):=f(z)/z^m$ is also holomorphic and $g(0)\neq 0$. On the other hand, it holds
$$|g(0)|= \lim_{n\to \infty}|g(1/n)|=\lim_{n\to \infty} |n^m f(1/n)|\leq \lim_{n\to \infty}\frac{n^m}{2^n}=0$$
a contradiction.


问:
Range of: $\sqrt{\sin(\cos x)}+\sqrt{\cos(\sin x)}$
Range of:
$$\sqrt{\sin(\cos x)}+\sqrt{\cos(\sin x)}$$

Any help will be appreciated.

答:
$$u=\cos(x); v=\sin(x)$$

$$\sqrt{\sin(u)}+\sqrt{\cos(v)}$$

$\sin(u)$ must be greater than or equal to 0, and $\cos(v)$ must be greater than or equal to 0:

$$u\in{[2\pi{k},\pi+2\pi{k}],k\in\mathbb{Z}}$$

$$v\in{[-\frac{\pi}{2}+2\pi{k},\frac{\pi}{2}+2\pi{k}]},k\in\mathbb{Z}$$

Since $v=\sin{x}$, $v\in{[-1,1]}$, $v$ will always be in the necessary interval. But $u$ will only be in the necessary interval when $\cos{x}\ge0$. That is, when $x\in[-\frac{\pi}{2}+2\pi{k},\frac{\pi}{2}+2\pi{k}]\bigcap[2\pi{k},\pi+2\pi{k}]$.

So the domain is $[2\pi{k},\frac{\pi}{2}+2\pi{k}],k\in\mathbb{Z}$. Now find the set of function values over that domain. Note that the function is not strictly increasing or decreasing over the domain, so the minimum and maximum are not the endpoints of the domain.


问:
Efficient way for calculating the sum of all products of m elements in vector.
Let's introduce a bit of notation:

Say we had a vector $x = (x_1, x_2, x_3,\dots,x_n)^{T}$, then

$$\Sigma^{1}x=\sum_{i}^{n} x_{i}$$
$$\Sigma^{2}x=\sum_{i}^{n}\sum_{j}^{n} x_{i}x_{j}$$
$$\Sigma^{3}x=\sum_{i}^{n}\sum_{j}^{n}\sum_{k}^{n} x_{i}x_{j}x_{k}$$
$$\vdots$$

Is there an easy (or more efficient) number-theory-like way of doing $\Sigma^m x$?
Or Is there a fast algorithm for calculating it?

I've realised for example, that when $n = 2$, $\Sigma^m x$ follows pascal's triangle:

$$\Sigma^{1}x = x_1 + x_2$$
$$\Sigma^{2}x = x_1^2 + 2x_1x_2 + x_2^2$$
$$\Sigma^{3}x = x_1^3 + 3x_1^2x_2 + 3x_1x_2^2 + x_2^3$$
$$\Sigma^{4}x = x_1^4 + 4x_1^3x_2 + 6x_1^2x_2^2 + 4x_1x_2^3 + x_2^4$$

答:
So to answer my own question, as an extension of WA Don's answer, it's easy to see that

$$\Sigma^mx = (\sum x)^m$$

Since the opposite is true for:

$$(a+b)^m = \Sigma^{m} (a, b)^T$$

and

$$(a+b+c)^m = \Sigma^{m} (a, b, c)^T$$

and so on.


问:
Define the open balls of this metric
The post office metric space, $P$ has the distance function defined as follows:

$$
d_P (\mathbf{x},\mathbf{y}) := \begin{cases}
0 & \mathbf{x} = \mathbf{y}\\
\Vert \mathbf{x-y}\Vert_2 & \mathbf{x}, \mathbf{y}, \mathbf{0} \mathbf{ \ are \ aligned } \\
\Vert \mathbf{x}\Vert_2+\Vert \mathbf{y}\Vert_2 & \mathbf{x}\neq \mathbf{y}
\end{cases}
$$

Where $\Vert \mathbf{x}\Vert_2 = \sqrt{x_1^2+x_2^2}$ is the Euclidean distance from $\mathbf{x}=(x_1,x_2) \in \mathbb{R}^2$ to the origin.

I am interested in drawing the balls of this metric centered at point $\mathbf{p}$, having a radius $r$:

$$B_r(\mathbf{p}) \triangleq  \{\mathbf{x} \in P\vert\ d(\mathbf{x},\mathbf{p})<r  \}$$

Does anyone have any tips on how to do this?

答:
The definition for the $\textit{post office metric}$ I know is slightly different than yours, namely:

$$d_P(\textbf{x}, \textbf{y}) =
\begin{cases}
0 & \textbf{x} = \textbf{y} \\
\|\textbf{x}\|_2 + \|\textbf{y}\|_2 & \textbf{x} \neq \textbf{y}
\end{cases}
$$

The way I've been told to think about this metric is, that a letter has to pass through the post office (the origin) in all cases, even if the house i want to send my letter to lies in a straight line with my own house. (In fact, when I learned about this metric it was called the 'French Railway Metric', since every train has to go through Paris, but that's the same concept...).

But nonetheless we can figure out how open balls in your metric look like! First the easy case, if $\textbf{p} = 0$ then the open ball around $\textbf{p}$ with radius $r$ is the same as the open ball in the euclidean metric, because any point is aligned with just 0.

Let's now look at $\textbf{p} \neq 0$. We have two cases, either $\|\textbf{p}\| \leq r$, or $\|\textbf{p}\| > r$. In the first case, we never reach the origin, so the open ball is just the straight line from $p$ in direction of the origin of length $r$, and also in the opposite direction (away from the origin, also of length $r$) not including the endpoints. So geometrically this open ball is just a straight line. If $\|\textbf{p}\|<r$, the open ball is the union of the line segment from $p$ to the origin, and the open disc centered at the origin of radius $r-\|\textbf{p}\|$ (with respect to the usual euclidean metric).

Title: Define the open balls of this metric


问:
How Many Uniquely Enumerated $4\times4$ Sudoku Grids Exist?
I'm looking to find how many uniquely enumerated $4\times4$ Sudoku grids exist. I am aware that there are other questions with solutions to this question, however I am asking again as there seem to be conflicting answers online. Some say $384$ whilst some say $288$.

I make my solution to be 384, as I see that there are $4!$ ways of enumerating the first (top left) $2\times2$ block, and then there are $2!$ ways of enumerating the remaining cells of each of the first column, second column, first row and second row. Once the first block and the adjacent blocks (blocks two and three, if you will) are fully enumerated, the fourth block (bottom right) can only be enumerated in one way due to the restrictions on number placement caused by the enumeration of blocks 2 (top right) and 3 (bottom left).

From this, I therefore make my total number of possible enumerations to be:
$$4!\times2!\times2!\times2!\times2!=384.$$

It is clear that quite a few others have calculated it in a slightly different way and made their solution to be 288. This doesn't make sense to me as it seems that the total should be divisible by 8, as the any given enumerated grid should have 8 symmetries (rotations and flips) which are all equally valid sudoku enumerations.

Is my calculation incorrect or are some of the answers online incorrect?

Edit: I made a typo initially and wrote 388 instead of 288. I was concerned that 8 did not divide 388, but it does in fact divide 288 - which makes the issue of divisibility by 8 go away.

答:
It's easy to write a program to find all valid grids by brute force, and verify that there are indeed only $288$ grids. There is no slick counting argument to arrive at this number that I know of.

I do now know who claims that there are $384$ grids, but perhaps there are people who simply claim that $384$ is an easily calculable upper bound.

Here's my quick and dirty code.

```<code>def is_valid(grid):
    for col in range(4):
        if len(set([grid[i][col] for i in range(4)])) < 4:
            return False
    for q in range(4):
        (i,j) = (q % 2, q // 2)
        box = grid[2*i][2*j:2*j+2] + grid[2*i+1][2*j:2*j+2]
        if len(set(box)) < 4:
            return False
    return True

from itertools import permutations, product

count = 0
for r1,r2,r3,r4 in product(permutations(range(4)), repeat = 4):
    grid = [r1,r2,r3,r4]
    if is_valid(grid):
        count += 1
print(count)
</code>```

Body:
It's easy to write a program to find all valid grids by brute force, and verify that there are indeed only $288$ grids. There is no slick counting argument to arrive at this number that I know of.

I do now know who claims that there are $384$ grids, but perhaps there are people who simply claim that $384$ is an easily calculable upper bound.

Here's my quick and dirty code.

def is_valid(grid):
    for col in range(4):
        if len(set([grid[i][col] for i in range(4)])) < 4:
            return False
    for q in range(4):
        (i,j) = (q % 2, q // 2)
        box = grid[2*i][2*j:2*j+2] + grid[2*i+1][2*j:2*j+2]
        if len(set(box)) < 4:
            return False
    return True

from itertools import permutations, product

count = 0
for r1,r2,r3,r4 in product(permutations(range(4)), repeat = 4):
    grid = [r1,r2,r3,r4]
    if is_valid(grid):
        count += 1
print(count)


问:
Modified version of generelized Hechler forcing
I'm looking at a modified version of the generalized Hechler forcing:

Let $\kappa$ be inaccessible, and fix some $s:\kappa \to \kappa$ strictly monotone taking value in the regular cardinals.* Conditions in the forcing are of the form $(f,g)$ where for some $\alpha<\kappa$, $f$ is a function from $\alpha$ to $\kappa$, $g$ is a function from $\kappa \smallsetminus \alpha$ to $\kappa$, and for all $\xi < \kappa$, either $s(\xi) > f(\xi)$ or $s(\xi) > g(\xi)$ (whichever of them makes sense). A condition $(f,g)$ is stronger then $(f',g')$ if $f' \subset f$ and for all $\xi \in \mathrm{dom}{g'}$ either $f(\xi) \ge g'(\xi)$ or $g(\xi) \ge g'(\xi)$ (again, whichever of them makes sense).

If we remove the restriction given by $s$, the forcing adds a function dominating all functions from $V$ in $^\kappa\kappa$ (mod bounded).

Say we force with the restriction given by $s$. Then the forcing adds a function dominating all functions from $V$ in $\prod_{i<\kappa}s(i)$. Does it also add a dominating function for $^\kappa\kappa$? What about the other way around - when we force without restricting using $s$, do we also add a dominating function for $\prod_{i<\kappa}s(i)$?

* It might be similar if we take $\kappa$ to be any regular cardinal, and ask that $s$ takes values in limit ordinals. But I defined it this way because I'm not sure, and because the broader setting  in which this question came up has $\kappa$ being supercompact.

答:
The restricted version never adds a dominating function in $\kappa^\kappa$ (mod bounded) (Edit: if $\kappa$ is weakly compact) and it is consistent (relative to a Mahlo cardinal) that the unrestricted forcing does not add a dominating function in $\prod_{i<\kappa}s(i)$ (mod bounded).

Lets first check the first part. Let us call the restricted by $s$ forcing $\mathbb P_s$. The crucial points are that

Any $(f, g), (f^\prime, g^\prime)\in\mathbb P_s$ with $f=f^\prime$ are compatible.

For any $\alpha<\kappa$ there are ${<}\kappa$-many $f$ with $\mathrm{dom}(f)\subseteq\alpha$ such that $(f, g)\in\mathbb P_s$ for some $g$.

Note that the second point is not true for the unrestricted forcing. Now let $\dot h$ be a name for a function $h:\kappa\rightarrow\kappa$. For $\alpha<\kappa$ let $\mathbb P_s^\alpha$ be the set of $(f, g)\in\mathbb P_s$ with $\mathrm{dom}(f)\subseteq\alpha$. The above implies that for any $i,\alpha<\kappa$ there are ${<}\kappa$-many $\beta$ such that some $(f, g)\in\mathbb P_s^\alpha$ decides $\dot h(\check i)$ to be $\beta$. Let us define
$$F_i:\mathbb P_s\rightarrow\kappa,\ F_i((f, g))=\mathrm{min}\{\alpha\mid \exists (f^\prime, g^\prime)\in\mathbb P_s^\alpha\text{ deciding }\dot h(\check i)\wedge (f^\prime, g^\prime)\leq (f, g)\}$$
and
$$H:\kappa\rightarrow\kappa,\ H(i)=\mathrm{sup}\{\beta\mid \exists (f, g)\in\mathbb P_s^{\mathrm{sup}F_i[\mathbb P_s^i]}\text{ deciding }\dot h(\check i)\text{ as }\beta\}+1$$

We must check that $H$ is well-defined. As j3M points out, it seems non-trivial that $F_i[\mathbb P_s^i]$ is always bounded in $\kappa$ (see the comments). We make use of the weak compactness of $\kappa$ to prove this:

So suppose $\mathrm{sup}F_j[\mathbb P_s^j]=\kappa$. There must then be a single first component $f$ so that $F_j((f, g))$ gets arbitrarily large by varying $g$. For convenience, say $f=\emptyset$. Let
$$X_\alpha=\{g\in\prod_{i<\kappa}s(i)\mid F_j(f, g)\geq\alpha\}$$
Observe that membership of $g$ in$ X_\alpha$ depends only on $g\upharpoonright\alpha$, so (abusing notation) we will say $g\upharpoonright\alpha\in X_\alpha$.
Consider the tree $T$ whose $\alpha$-th level $T_\alpha$ is exactly
$$T_\alpha=X_\alpha\cap\prod_{i<\alpha}s(i)$$
and is ordered by end-extension. Then $T$ is indeed a tree, all levels have size ${<}\kappa$ and by our choice of $f$, $T$ has height $\kappa$. By weak compactness, there is a cofinal branch through $T$. We may think of that branch as a function $g\in\bigcap_{\alpha<\kappa} X_\alpha$. But now $F_j((f, g))\geq\kappa$, meaning $(f, g)$ has no extension deciding the value of $\dot h(\check j)$. This is clearly a contradiction.

Finally $H$ is not dominated (mod bounded) by $\dot h^G$ for any generic $G$ (as the generic will often "pick minimal extensions deciding some value of $\dot h$").

For the second part, suppose that $\kappa$ is a Mahlo cardinal and $s(i)$ is simply the $i$-th regular cardinal. The set $S=\{i<\kappa\mid i=s(i)\}$ is stationary in $\kappa$ and note that any $f\in\prod_{i<\kappa}s(i)$ is regressive on $S$. Assume that after forcing with the unrestricted generalised Hechler forcing $\mathbb P$ at $\kappa$, there is $f\in\prod_{i<\kappa}s(i)$ dominating all such functions in $V$ (mod bounded). Then $f^{-1}(\{\xi\})$ is bounded and thus non-stationary for any $\xi<\kappa$. By Fodor's Lemma this means that $S$ is no longer stationary. However, $\mathbb P$ is ${<}\kappa$-closed and so preserves the stationarity of $S$, contradiction.


问:
Show that a certain map $V \to \mathbb R$ is an element of the dual Space $V^*$
Given a vector space $V = \operatorname{Mat}_{2\times2}(\mathbb{R})$.

Define $\varphi:V\to\mathbb{R}$ by
$$\varphi\left(\begin{pmatrix}a & b \\c & d\end{pmatrix}\right) = 2a+b$$

Prove that $\varphi\in V^*$

Thank you for your help.

答:
I will try to answer my own question, please edit if wrong.

Lets have $\begin{pmatrix}a_1 & b_1 \\c_1 & d_1\end{pmatrix},\begin{pmatrix}a_2 & b_2 \\c_2 & d_2\end{pmatrix}\in V$ and $k,l \in \mathbb{R}$.

$$\varphi\left(k\begin{pmatrix}a_1 & b_1 \\c_1 & d_1\end{pmatrix}+l\begin{pmatrix}a_2 & b_2 \\c_2 & d_2\end{pmatrix}\right)=2(a_1k+a_2l)+(b_1k+b_2l)$$
$$=k(2a_1+b_1)+l(2a_2+b_2)$$
$$=k\varphi\left(\begin{pmatrix}a_1 & b_1 \\c_1 & d_1\end{pmatrix}\right)+l\left(\begin{pmatrix}a_2 & b_2 \\c_2 & d_2\end{pmatrix}\right)$$

So $\varphi$ is linear map, hence $\varphi \in V^*$


问:
Let $g$ be a continuous function on $[0,1]$ with certain conditions. Show that $g(x) = x$
Let $g$ be a continuous function on $[0,1]$ s.t.
$$\int_{0}^{1} g^2(x) dx = \frac{1}{3}$$
$$\int_{0}^{1} xg(x)dx = \frac{1}{3}$$

Show that $g(x) = x \forall x \in [0,1]$.

My work so far.
I wrote it as
$$\int_{0}^{1} (g^2(x) - xg(x))dx = 0$$
and
$$\int_{0}^{1} (xg(x) - g^2(x)) dx = 0$$

So, if one of these is negative, then the other must be positive. I know that if integral of a non-negative function is $0$, then the function is $0$. However, I am not quite sure how to show that the function is non-negative. In fact I'm not even convinced it has to be. I believe this could work if the function is $g(x) = -x$. I must be forgetting an important property here since I'm not seeing it. Can anyone point me in the right direction?

Body:
Notice that $(g(x)-x)^2\geq 0$ for all $x$, so:\begin{align}0&\leq \int_{0}^{1} (g(x)-x)^2 dx \\
&= \int_{0}^{1} g^2(x) dx-2\int_{0}^{1} xg(x) dx +\int_{0}^{1} x^2 dx \\&
= \frac{1}{3} -2\frac{1}{3} +\frac{1}{3}\\ &=0\end{align}

so $$\int_{0}^{1} (g(x)-x)^2 dx =0$$

so $g(x)=x$, for all $x$, since $x$ is continuous.


问:
Multiplying a Probability distribution function by a constant
This is a pretty basic question, I understand, but in my head there is something that doesn't make sense.

Let's say I have a uniform probability density function X that goes from [0,6]. The pdf would thus be 1/(6-0), which is just 1/6.

Now, let's say I multiply this pdf X by 2, to get Y= 2X. Would the pdf then be 2*(1/6)? If so, wouldn't this break the condition that the integral of a pdf has to be equal to 1? What would I do? And for whatever answer I get, would that work for all pdf's (binomial, Poisson, uniform, normal, etc).

答:
You can explain it in this way:
$$X \sim Uni(0,6) \Rightarrow F_X(t)=P(X\leq t) = \frac{t}{6} \\
Y = 2X \Rightarrow F_Y(s) = P(Y\leq s) = P(2X \leq s ) = P (X\leq \frac {s}{2})
=_{s\in(0,12)} \frac{s}{12}$$

Hence it is similar to what you thought, but opposite, we got that the density function of $Y$ is $\frac{1}{12}$ which is half, and the explanation of this is in the computation above

The same explanation goes for every other distribution, BUT that doesn't mean you will get the property of anti-homogeneous we got here.

For example, if $X ~ \exp(1)$ then $f_X(x)=e^{-x}$ and for $Y=2X$ we get
$$ F_Y(t) = P(Y\leq t) = P(X \leq \frac{t}{2}) = 1-e^{-\frac{t}{2}}$$
Hence, in this case $f_Y(y) = e^{-\frac{y}{2}}$ which means $Y\sim\exp(\frac{1}{2})$ and NOT $(\frac{1}{2}\exp{1})$ as you could (maybe) have expected


问:
Sufficient condition for $\mathrm{ord}(a)$ divides exponent of finite abelian group?
In Niederreiter and Winterhof, Applied Number Theory, Proposition 1.3.24 (p. 18) is:

If $G$ is a finite abelian group of exponent $E$, then
$\mathrm{ord}(a)$ divides $E$ for all $a \in G$.

They're using the multiplicative notation for the group operation.  The proof begins like this:

We consider a fixed element $a \in G$. Let $p$ be any prime number.
Then we can write $E=p^ef$ with integers $e\geq 0$ and $f\geq 1$
satisfying $\gcd(p,f)=1$. It suffices to show that if $p^r$ divides
$\mathrm{ord}(a)$ for some integer $r\geq 0$, then we must have $r\leq e$.

I don't understand why proving the last statement would be sufficient to prove the theorem.

I understand that the order of $a$ is the least $m$ s.t. $a^m = 1_G$, and that the exponent is the maximum of $\mathrm{ord}(a)$ for all $a \in G$.  We're trying to show that $\mathrm{ord}(a)$ divides $E=p^ef$ (where $f$ of course is a product of primes).  If $p^r$ divides $\mathrm{ord}(a)$ and $r\leq e$, then I see that $p^r$ divides the $p^e$ part of $E$, but why can I assume that there are not other components in the product $\mathrm{ord}(a)$ that will not divide into $f$?

(I'm also confused about the logical form of the sufficient condition.  Why is it "IF $p^r$ divides $\mathrm{ord}(a)$, THEN $r\leq e$"?)

答:
That last statement is an instance of this fact, which is a consequence of the fundamental theorem of arithmetic:

Let $x,y \in \mathbb Z$. Then $x$ divides $y$ iff $v_p(x) \le v_p(y)$ for all primes $p$.

Here, $v_p(x)$ is the exponent of $p$ in the factorization of $x$.

The proof given in that book seems too complicated. The proposition follows directly from $\{ k \in \mathbb Z : a^k =1 \} = \operatorname{ord}(a) \mathbb Z$, which is a simple consequence of Euclidean division. This holds for elements of finite order in every group, not just finite abelian ones.


答: We know that $$\int_0^1g^2(x)dx=\dfrac{1}{3};\quad \int_0^1g(x)\cdot x dx=\dfrac{1}{3}\Longrightarrow\int_0^12g(x)\cdot x dx=\dfrac{2}{3};\quad \int_0^1x^2dx=\dfrac{1}{3}.$$ Then $$\int_0^1\underbrace{g^2(x)-2xg(x)+x^2}_{(g(x)-x)^2}dx=0\Longrightarrow\int_0^1(g(x)-x)^2dx=0.$$ Since $(g(x)-x)^2$ is a positive function whose integral is equal to $0$, then we conclude that $$g(x)-x=0\quad a.e. x\in[0,1]\Longrightarrow g(x)=x\quad a.e. x\in[0,1].$$ Finally, thanks to $g$ being continous, we get that $$g(x)=x\quad \forall x\in[0,1].$$

Title: Let $g$ be a continuous function on $[0,1]$ with certain conditions. Show that $g(x) = x$

Body:
Let $g$ be a continuous function on $[0,1]$ s.t.
$$\int_{0}^{1} g^2(x) dx = \frac{1}{3}$$
$$\int_{0}^{1} xg(x)dx = \frac{1}{3}$$

Show that $g(x) = x \forall x \in [0,1]$.

My work so far.
I wrote it as
$$\int_{0}^{1} (g^2(x) - xg(x))dx = 0$$
and
$$\int_{0}^{1} (xg(x) - g^2(x)) dx = 0$$

So, if one of these is negative, then the other must be positive. I know that if integral of a non-negative function is $0$, then the function is $0$. However, I am not quite sure how to show that the function is non-negative. In fact I'm not even convinced it has to be. I believe this could work if the function is $g(x) = -x$. I must be forgetting an important property here since I'm not seeing it. Can anyone point me in the right direction?


问:
Calculate the number of pages required form expanding report design into multip pages
I am programmer, I am writing a report designer and the problem is
report contains five section
1 - Report Header visible in the first page only
2 - Page Header visible in all pages
3 - Details Section visible in all pages
4 - Page Footer visible in all pages
5 - Report Footer visible in last page only

givens:
1 - Page size is known
2 - Each section size is known
3 - the number of data rows to be expanded in the details section is known
4 - each row height in the details section is known

required:
How many number of pages required to output a report given the previous parameters,
I need an equation, or mathematical model to govern the rules of this thing

答:
In real life one usually faces additional problems such as detail rows of different height, intermediate headings requiring layout decisions, or orphan/widow avoidance such that simulating the whole process in a first run is not one of the worst ideas.

But apart from this, we can say the follwing:
Let $p$ be the page size, $h_0$ exceptional first page header size, $h$ generic header size, $f$ generig footer size, $f_0$ exceptional last page footer size, $r$ detail row height, $n$ detail row count. (I assume here that the generic header does not appear on the first page and the generic footer does not appear on the last page; in other words, if these assumptions are not correct, then $h_0$ and $f_0$ are to be understood as including the generic header/footer).

The page fits on one page if $$nr+h_0+f_0\le p.$$
Otherwise, we can fit $m_0:=\left\lfloor\frac{p-h_0-f}{r}\right\rfloor$ rows on the first, $m_1:=\left\lfloor\frac{p-h-f_0}{r}\right\rfloor$ on the last and $m:=\left\lfloor\frac{p-h-f}{r}\right\rfloor$ on any intermediate page.
Thus $k$ pages suffice if
$$ m_0+m_1+(k-2)m\ge n.$$
Therefore, we find the needed number $k$ of pages as
$$k=\begin{cases}1&\text{if }nr+h_0+f_0\le p,\\\left\lceil\tfrac{n-m_0-m_1}{m}\right\rceil+2&\text{otherwise}.\end{cases} $$


问:
Prove that $f(x)$ has either a biggest or lowest value.
Let $f(x)$ be a continuous function in $\mathbb{R}$ and $\lim\limits_{x\rightarrow +\infty}{f(x)}=\lim\limits_{x\rightarrow -\infty}{f(x)}=L$. Prove that $f(x)$ has either a biggest or a lowest value in $\mathbb{R}$.

From the task, we can get:

$\forall \epsilon>0 \exists a>0, \forall x>a: |f(x)-L|<\epsilon$

$\forall \epsilon>0 \exists b>0, \forall x<-b: |f(x)-L|<\epsilon$

But then, what's the next step? The problem reminds me of Weierstrass' theorem:

If $f(x):[a, b]\rightarrow\mathbb{R}$ is a continous function, then it
is limited and reaches its biggest and lowest values.

However, I don't think that interpreting $\mathbb{R}$ as a collection of infinitely numerous intervals will make this theorem usable.
Could someone tell me the method of solving this problem, please?

答:
If $f$ is constant, we're done. Otherwise and without loss of generality (even if it means changing $f$ into $-f$ and $L$ into $-L$), we can suppose that it exists $x_0 \in \mathbb R$ with $f(x_0) \gt L$.

By definition of the limits, one can find $M \gt 0$ such that

$$f(x) \le \frac{f(x_0) + L}{2} \lt f(x_0)$$ for $\vert x \vert \ge M$. $f$ attains its maximum on the compact interval $[-M,M]$ (that contains $x_0$), which is also a global maximum due to above inequality.

We're done.


答:
Using fundamental transformation  theorem you get

$$f_Y(y)=f_X\left(g^{-1}(y)\right)|\frac{d}{dy} g^{-1}(y)|  $$

$$f_Y(y)=\frac{1}{6}\cdot\frac{1}{2}\cdot\mathbb{1}_{(0;12)}(y)$$

Being

$$y=2x\rightarrow x=\frac{y}{2}\rightarrow x'=1/2$$

In other word, $Y\sim U(0;12)$

Important observation: in your case you are not multiplying the distribution by 2 but THE RANDOM VARIABLE. It is not the same.

Title: Multiplying a Probability distribution function by a constant

Body:
This is a pretty basic question, I understand, but in my head there is something that doesn't make sense.

Let's say I have a uniform probability density function X that goes from [0,6]. The pdf would thus be 1/(6-0), which is just 1/6.

Now, let's say I multiply this pdf X by 2, to get Y= 2X. Would the pdf then be 2*(1/6)? If so, wouldn't this break the condition that the integral of a pdf has to be equal to 1? What would I do? And for whatever answer I get, would that work for all pdf's (binomial, Poisson, uniform, normal, etc).


问:
Proof by induction: L'Hospital rule doesn't work, what next?
Show through induction:
$\lim_{x\to\infty} \frac{\ln^k x}x=0$

Well the base case of 1 is rather trivial, but when substituting k for k+1 for the inductive hypothesis step we get,

$\lim_{x\to\infty} \frac{\ln^{k+1} x}x=0$

Applying L'Hospital's rule gets is not helpful as we will get

$\lim_{x\to\infty} \frac{(k+1)\ln^{k} x}x$

Applying l 'hospital's again gets:

$\lim_{x\to\infty} \frac{-(k+1)(\ln^{k-1} x) ({\ln (x)} -k)}{x^2} $

Applying l 'hospital's rule runs into an issue of always having an infinity/infinity issue.

Sometimes when l'hospital's rule gets stuck, there is an algebraic manipulation that can be done to overcome the problem. I am not seeing anything. I am wonder what would be another course of action to prove the k+1 case of induction.

答:
I think you might be mistaken in your application of L'Hospital's rule. Assume that case $k$ is true. That is, $$\lim_{x\rightarrow \infty} \frac{\ln^k x}{x}=0.$$ The case $k+1$ is
$$\lim_{x\rightarrow \infty} \frac{\ln^{k+1} x}{x}=0,$$ which is of the form $\infty/\infty$.

Apply L'Hospital's rule,
$$\lim_{x\rightarrow \infty} \frac{(k+1)\ln^{k} x (1/x)}{1}=(k+1)\lim_{x\rightarrow \infty} \frac{\ln^{k} x}{x}=0,$$ from case $k$.


问:
Prove that $f(z) $ is an odd function when $z \in \mathbb{R}$
Given an entire function which is real on the real axis and imaginary on the imaginary axis, prove that it is an odd function i,e $f(-z)= -f(z)$

My attempt :  Here two cases will be possible

$1.$ when $z= x$

$2.$ when $z = ix $

I have no any issue when $z=ix$  and solution is alreaday given here

but i have some issue when $z=x$

$$f(x) = \overline{ f(\bar x) }$$

$$\sum_{n=0}^{\infty}a_nz^n=\sum_{n=0}^{\infty}\bar a_nz^n \implies a_n=\bar a_n$$

\begin{align}
f(x)=\sum_{n=0}^{\infty}a_nx^n&=\sum_{n=0}^{\infty}a_{2n}x^{2n}+\sum_{n=1}^{\infty}a_{2n+1}x^{2n+1}
\end{align}

Here  $f(x)$  is both  even and odd

so here  we are getting  contradiction that  $f(z)$  is odd that mean $f(x)$ is not even function

答:
From this answer we have  show that

$a_{2n}=0$ for all $n$.

now  $f(x+iy)=
\sum_{n=0}^{\infty}a_n(x+iy)^n= a_0+a_1(x+iy)^1 + a_2(x+iy)^2+.......
$
give  odd  function


问:
obtain parity check matrix in different ways
We have the code words 111000, 100110, 110101 which result in the following generator matrix.

$G=\begin{bmatrix}1&1&1&0&0&0\\1&0&0&1&1&0 \\ 1&1&0&1&0&1\end{bmatrix}$

According to my book, the rows of the control matrix are solution vectors of the system of equations whose rows are formed by the base of the given code, i.e.

\begin{eqnarray}
y_1 + y_2 + y_3 = 0\\
y_1 + y_4 + y_5      = 0 \\
y_1 + y_2 + y_4 + y_6 = 0 \\
\end{eqnarray}

If we choose $y_1$, $y_2$ und $y_4$ as free variables, we get the rows of the following matrix as the basis of the solution space:

$H=\begin{bmatrix}1&0&1&0&1&1\\0&1&1&0&0&1 \\ 0&0&0&1&1&1\end{bmatrix}$
where columns 1, 2 and 4 are the identity matrix.

Why are the colums 1, 2 and 4 the identity matrix?

How did he use the generator matrix to obtain the parity check matrix?

Why did he choose $y_1$, $y_2$ und $y_4$ and what does this mean?

I thought the way to get the parity check matrix is the following:

If the generator matrix is in the form $G = [I_k | P]$, then its parity check matrix is $H = [P^T | I_{n-k}]$

So we swap columns 1&3, 2&5 and 3&6

$G'=\begin{bmatrix}1&0&0&0&1&1\\0&1&0&1&0&1 \\ 0&0&1&1&1&1\end{bmatrix}$

$H'=\begin{bmatrix}0&1&1&1&0&0\\1&0&1&0&1&0 \\ 1&1&1&0&0&1\end{bmatrix}$

Then we permutate it back (3&6, 2&5 and 1&3) and obtain:

$H= \begin{bmatrix}0&0&0&1&1&1\\0&1&1&0&0&1 \\ 1&0&1&0&1&1\end{bmatrix}$

Does it matter in which order the identity matrix appears?

答:
Having the generator matrix in systematic form, that is $G=[I_k|P]$, has of course the advantage as stated, which is the easy construction of the parity check matrix $H=[P^T|I_{n-k}]$.

Consider any word $x$ as a row vector, then the syndrome $s=Hx^T=0$ if and only if $x$ is a codeword of the given linear code. Since $H$ is a $(n-k)\times n$ matrix, $x^T$ is a $n \times 1$ vector. It is easy to see, that swapping rows does not have any effect, but swapping columns changes the code, since bit positions are permuted.

Consider the valid codeword $100011$ (first row of $G'$). Using $H$ to check: $H\cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\1\end{pmatrix}=\begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}$, hence the syndrome is not equal to the zero vector. However, performing the same for $H'$ yields to $H'\cdot \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 1 \\ 1\end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}$ as intended.


问:
Collatz Conjecture confusion about the following patterns:
According to the rules of the Collatz Conjecture if $n$ is odd then execute $3n + 1$ and then when $n$ is even execute $n/2$ and then we are supposed to repeat, however for the purpose of this experiment I have just initiated the process of staring with an odd number until it becomes odd again, examples:

$1 → 4 → 2 → 1$ with the end result being $1 → 1$ (starting with $1$, ending with $1$)

$13 → 40 → 20 → 10 → 5$ with the end result being $13 → 5$ (starting with $13$, ending with $5$)

Let us denote it as $X → Y$

I have then decided to sort the data by $X$ ending in a specific digit, and doing so has revealed many patterns.

In order to better present this question , I will focus on a consecutive list where $X$ ends in $1$, such as $1 , 11, 21 , 31 ...$ (all the way to $1001$).

I will present the list as an Image to include different colors, to better visualize the patterns (with their repetitions being evenly distributed). My question in regards to these patterns is broken into 3 parts:

A) As you will notice each color represents increments of $+30$ in $Y$, and its repetition becomes evenly distributed through out the list, but some how there is never an overlap of colors (more than 1 color in a row). Could there ever be one (a row that has more than 1 color)?

B) Do the listings that aren't colored yet, represent an upcoming pattern (additional colors)?

C) Will there be a finite or infinite of patters possibilities (colors) ?

Note: The best way that came to my mind in order to present this question was by presenting the image and using the colors to ask for logic. If anyone thinks there could be a better way, feel free to edit.

Note: To better understand the patterns, please notice that Yellow repeats every $2$ rows, Red every $4$ rows. ...

答:
Answer to Question A: All your patterns will continue indefinitely and there won't be any overlapping. As @GottfriedHelms pointed out, you are simply grouping the results by how many divisions by $2$ is required to yield an odd number.

Answer to Question B: The listings that aren't colored yet do indeed represent upcoming patterns.

Answer to Question C: There will be an infinite list of colors since each division by $2^k$ where $k\in\mathbb{N}$ will need a different color, that is, one color for $k=1$ (yellow in your case), another color for $k=2$ (red in your case), and so on.


问:
Collinearity in bicentric pentagon
Can you provide a proof for the following claim:

Claim. The circumcenter, the incenter, and the incenter of the pentagon formed by diagonals in a bicentric pentagon are collinear.

GeoGebra applet that demonstrates this claim can be found here.

My idea is to show that $|OJ|=|OI|+|IJ|$ . By Fuss formula we know that $$r(R-|OI|)=(R+|OI|)\sqrt{(R-r+|OI|)(R-r-|OI|)}+(R+|OI|)\sqrt{2R(R-r-|OI|)}$$

where $R$ and $r$ are circumradius and inradius of bicentric pentagon ,respectively. But how to express lengths $|OJ|$ and $|IJ|$ in terms of $R$ and $r$ ?

答:
Consider the effect of a  continuous movement of $A_1,A_2$ to $B_1,B_2$, keeping the points on the circumcircle and maintaining tangency with the incircle. Let $A_1A_2$ be tangential at $P_1$ and let $B_1B_2$ be tangential at $P_2$.

Let line $P_1P_2$ cut $A_1B_1$ at $Q_1$ and cut $A_2B_2$ at $Q_2$. Then $A_1B_1$ and $A_2B_2$ are equally inclined to line $Q_1Q_2$ and so there is a circle $\mathscr{C}$ tangential at $Q_1$ and $Q_2$ to the lines $A_1B_1$ and $A_2B_2$.

Let the line $Q_1Q_2$ make successive angles $p,q,q,p$ with the four tangents at $Q_1,P_1,P_2,Q_2$. Then, applying the sine rule, the ratio of the powers of each of $A_1,A_2,B_1,B_2$ with respect to the two circles is $$\frac {\sin q}{\sin p}\text{, a constant}.$$
The circumcircle, through these four points, is therefore coaxal with $\mathscr{C}$ and the incircle.

Similarly, we can determine a further coaxal circle, $\mathscr{C'}$, for the movement of $A_2,A_3$ to $B_2,B_3$. The coaxal circles $\mathscr{C}$ and $\mathscr{C'}$ have a common tangent, $A_2B_2$ and so are either equal or on opposite side of the radical axis of the system but the latter possibility is impossible by continuity.

We can now apply the same idea to the movement of $A_1,A_3$ to $B_1,B_3$ with respect to the circumcircle and $\mathscr{C}$. Then $A_1A_3$ and $B_1B_3$ are both tangential to a  further coaxal circle $\mathscr{D}$.

We can now consider successive moves of each $A_i$ to $A_{i+1}$, considering the indices modulo $5$.Then each edge $A_iA_{i+2}$ is tangential to $\mathscr{D}$ and so the incentre of the pentagon is the coaxal circle $\mathscr{D}$.

Title: Collinearity in bicentric pentagon


问:
Locally Compact Topological Space
Is it true that in a locally compact (not necessarily Hausdorff) space, every point has a neighbourhood bases consisting of precompact subsets? (precompact means closure is compact).

In addition, if the space is also Hausdorff, then we have a local base consisting of compact neighbourhoods.

The first assertion is easy to see in a Hausdorff space as compact subsets of a Hausdorff space is closed. However, I cannot see this without the Hausdorff condition.

Locally compact means every point has a neighbourhood (a set containing an open set containing that point) which is compact.

答:
Let $X$ be an infinite set, $p \in X$ fixed, and as a topology on $X$ we use the included point topology on $X$: $O$ is open iff $O = \emptyset$ or $p \in O$.
$X$ is not compact, as the cover by all sets $\{x,p\}, x \neq p$, shows: we cannot spare a single element.

Also, the closed sets are $X$ and all sets missing $p$. This means that the closure of every non-empty open set equals $X$, as every non-empty open set contains $p$ and the only closed set that contains $p$ is $X$. So there are no precompact neighbourhoods at all. But $X$ is locally compact according to your definition, as every point $x$ has a finite (hence compact) open neighbourhood $\{x,p\}$.

Title: Locally Compact Topological Space


问:
$\sum_{n=2}^{\infty} \frac{n^2 -n -1}{n!}$
I would like hint as to how we might solve this $\sum_{n=2}^{\infty} \frac{n^2 -n -1}{n!}$. Convergence is easy to show via the ratio test.  I have tried computing the sequence of partial sums, separating into a sum of "simpler" series. For example, I separate this into $$\sum_{n=2}^m\frac{n^2 -n}{n!}-\sum_{n=2}^m\frac{1}{n!}$$

or $$\sum_{n=2}^m\frac{1}{(n-2)!}-\sum_{n=2}^m\frac{1}{n!}$$

I thought about removing some terms from the right sum, so I could use the fact that $\sum_{n=0}^{\infty}\frac{1}{n!}=e$. Can I do so? Will this method lead anywhere? I don't really know what else to do here.

答: Your difference of two sums is$$\sum_{k=0}^{m-2}\frac{1}{k!}-\sum_{k=2}^m\frac{1}{k!}=2-\frac{1}{(m-1)!}-\frac{1}{m!}.$$I'm sure you can find the $m\to\infty$ limit.


问:
Is a circle a manifold?
I am very much a beginner when it comes to manifolds. I understand that the torus in $\mathbb{R}^3$ is given by
$$
T^2 = \left\{(x,y,z)\in\mathbb{R}^3\, : \, \left(x^2 + y^2 + z^2 + R^2 - r^2\right)^2 - 4R^2(x^2+
 y^2) = 0\right\}
$$
And its dimension is $2$ because we have $1$ constraint so $3 - 1 = 2$.

Then my question is this: what's the dimension of the circle manifold? It comes out as zero to me.

$$
C = \left\{(x, y)\in\mathbb{R}^2\,:\, y = y_0 + \sqrt{r^2 - (x-x_0)^2}\text{ and } y = y_0 - \sqrt{r^2 - (x-x_0)^2} \right\}
$$
So now we have $2$ constraints and so $2-2 = 0$.

答:
The dimension of $C$ is one. If you see it as$$\left\{(x,y)\in\Bbb R^2\,\middle|\,(x-x_0)^2+(y-y_0)^2=r^2\right\},$$then there is only one constraint, and $2-1=1$.

Besides, your description of $C$ should be$$C = \left\{(x, y)\in\mathbb{R}^2\,\middle|\, y = y_0 + \sqrt{r^2 - (x-x_0)^2}\color{red}{\text{ or }} y = y_0 - \sqrt{r^2 - (x-x_0)^2} \right\}.$$

Title: Is a circle a manifold?


问:
Integration question verifying piecewise
I have the following question: from direct integration show

$\displaystyle \int \limits_{-L}^{L} \cos({m πx\over L})\cos({nπx\over L}) \ dx = \begin{cases}0 & m \neq n \\ L & m = n \\  \end{cases} $

I use a trigonometric identity and evaluate individually to produce:

$\displaystyle {L\sin(π(m-n)\over πm-πn} + {L\sin(π(m+n)\over πm+πn}$

I am pretty sure I am correct but I can quite clearly see that whenever m=n the part on the left will not exist let alone equal L. Any hints would be appreciated.

Cheers

答: It's safest to consider the case $m=n$ separately when solving the integral. But also note that $\lim_{x\rightarrow 0}\frac{\sin x}{x}=1$.</p>


问:
Dividing fractions in real life scenario / application
First of all sorry if this question sounds too stupid or offends anyone.

One apple divide by two you get half an apple.
$\large{\frac{1}{2} = 0.5}$

I couldn't get my head around with dividing fractions...

Half an apple divide by half an apple you get one apple?
$\large{\frac{0.5}{0.5} = 1}$

How do I make any sense out of it? Any real life scenarios / applications that I can relate or explain dividing fractions?

答:
Easy example:

(How much you have or need / The size of section desired) = # of sections can be filled

(You have 8/3 apples / 1/3 apple per person) = 8 people can eat


问:
Completing a proof on integrability: $f$ is Riemann integrable under the mesh definition iff $\sup L_\Gamma = \inf U_\Gamma$
$
\newcommand{\II}{\mathcal{I}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\ve}{\varepsilon}
\newcommand{\d}{\delta}
\newcommand{\g}{\Gamma}
\newcommand{\norm}[1]{\left \Vert  #1   \right \Vert}
\newcommand{\volume}{\mathrm{volume}}
\newcommand{\gs}{{\g_\ast}}
$Definitions:

Considering we're working in $\R^n$, and there are a lot of competing/equivalent definitions for Riemann integrability (as I learned the hard way in trying to find some help here and elsewhere), I want to get all of these out of the way first, so we're on the same page. These are the definitions I'm working with, in their essence taken from Measure & Integral: An Introduction to Real Analysis by Richard Wheeden and Antoni Zygmund.

Riemann Integrable: Let $f : \mathcal I \subseteq \R^n \to \R$ be bounded. We say $f$ is Riemann integrable on $\II$ if
$$
\lim_{\norm \g \to 0} R_\g
$$
exists. In particular, we call this value $L$, and say $L = \int_\II f(x) \, \mathrm{d}x$.

Formalization: Formally, the above is equivalent to

$$(\forall \ve >0)(\exists \d > 0)(\forall \g : \norm \g < \d)(|R_\g - L| < \ve)$$

Boxes: $\II$ denotes a box. You can write it a couple of different ways: for given $a_i,b_i$,
$$
\II := \Big\{ (x_1,\cdots,x_n) \in \R^n \; \Big| \; x_i \in [a_i,b_i] \Big\} = \prod_{i=1}^n [a_i,b_i]
$$

Volume of Boxes: The volume of a box is given in the usual, expected way:
$$
\volume(\II) = \prod_{i=1}^n (b_i - a_i)
$$

Partitions: Let $\g$ denotes a partition of $\II$ into finitely many nonoverlapping boxes $\II_k$. (That is, $\mathrm{int}(\II_i) \cap \mathrm{int}(\II_j) = \varnothing$ for all $i \ne j$, and $\bigcup \II_k = \II$.)

Norm/Mesh of Partitions: The mesh or norm of a partition is defined to be
$$
\norm \g := \max_i \mathrm{diam}(\II_i)
$$

Diameter of Set: The diameter of a subset of a metric space is the supremum of possible distances between points. In $\R^n$ under the $2$-norm,

$$\mathrm{diam}(\II) := \sup_{x,y \in \II} \norm{x - y}_2$$

Tags: We take representative points (or tags) $\xi_i \in \II_i$ for each $\II$.

Riemann Sum: We define the Riemann sum to be (implicitly dependent on the partitioning and tags)
$$
R_\g := \sum_i f(\xi_i) \cdot \volume(\II_i)
$$

Upper & Lower Sums: We define the upper and lower sums as so:
$$
U_\g := \sum_i \sup_{x \in \II_i} f(x) \cdot \volume(\II_i) \qquad L_\g := \sum_i \inf_{x \in \II_i} f(x) \cdot \volume(\II_i)
$$
Trivially, $L_\g \le R_\g \le U_\g$ for any choice of tags $\xi_i$.

My Goal: I would like to show that
$$
\text{$f$ is Riemann integrable} \iff \inf_\g U_\g = \sup_\g L_\g
$$
More specifically, I would like to show the backwards direction; I've already proven the forwards I believe. I'd also in particular like as direct a proof as possible (rather than circumnavigating through the other definitions of integrability).

My Attempt So Far:

So far, I feel like I've hit on the right idea, with a minor snag in the way.

So, let $\ve > 0$, and let
$$
\sup_\g L_\g = \inf_\g U_\g =: L
$$
From the definition of supremum/infimum, we may find two partitions $\g_1,\g_2$ such that
$$
0 \le \sup_\g L_\g - L_{\g_1} < \frac \ve 2 \qquad 0 \le U_{\g_2} - \inf_\g U_\g < \frac \ve 2
$$
Form their common refinement $\gs$ of $\g_1$ and $\g_2$. (This would just be their union in the $\R^1$ case where we can equivalently use points as our partition, but not the case in general.) Then we easily see that
$$
0 \le \sup_\g L_\g - L_{\gs} < \frac \ve 2 \qquad 0 \le U_{\gs} - \inf_\g U_\g < \frac \ve 2
$$
Recalling the assignment of $L$, we have
$$
0 \le L - L_{\gs} < \frac \ve 2 \qquad 0 \le U_{\gs} - L < \frac \ve 2
$$

Where I Get Lost:

This is where I get lost. My end goal is to ultimately show $|R_\g - L| < \ve$ (independent of the choice of $\xi_i$ namely). At this point, I feel like I'm meant to use the inequality $L_\g \le R_\g \le U_\g$, but I'm not certain that it would be valid. If it were viable, then I would intend to have
$$
0 \le L - R_{\gs} < \frac \ve 2 \qquad 0 \le R_{\gs} - L < \frac \ve 2
$$
but would it not be possible for $R_\gs$ to fall on one side of $L$ or the other, and thus outside of the bound of $0$? (Though on the other hand, would that even matter? Is the bound of $0$ important for this?)  From a situation like this, I would then like to claim the conclusion.

Another worrying part is that I have no particular $\d$ brought up. Maybe I could just arbitrarily set a partition $\g_0$ such that $\norm{\g_1},\norm{\g_2} < \norm{\g_0} = \d$, but that doesn't really feel right.

Can anyone give me an idea as to how to complete this proof, even if just a nudge in the right direction? Or am I completely off-base? Thanks for any insights you can give!

答:
Let $M = \sup_{x \in \mathcal{I}}|f(x)|$ and define the "width" of $\mathcal{I}$ to be W = $\sup \{b_i-a_i : 1 \leqslant i \leqslant n\}$.  The objective is to show that there exists $\delta >0$ such for any partition $\Gamma$ with $\|\Gamma\| < \delta$, we have

$$ \tag{*}\sup_{\Gamma}L_\Gamma - \varepsilon < L _\Gamma \leqslant U_{\Gamma} < \inf_{\Gamma}U_\Gamma + \varepsilon,$$

Focusing first on the right-hand inequality, you have shown for any $\epsilon > 0$ there is a partition $\Gamma_2$ such that

$$U_{\Gamma_2} < \inf_{\Gamma}U_\Gamma + \frac{\varepsilon}{2}$$

We will use the following

Lemma.  Let $\Gamma = (P_1,\ldots, P_n)$ be a partition of
$\mathcal{I}$ where $P_i$ is a partition of the interval $[a_i,b_i]$
and $\|P \| < \delta $. Suppose  $\Gamma'$ is a refinement of $\Gamma$
obtained by adding a single point $c$ in the subinterval
$(x_{i,j-1},x_{i,j})$ to the "edge" partition $P_i$. Then we have

$$U_\Gamma < U_{\Gamma'} + 2M W^{n-1}\delta$$

It follows easily by induction that if  $q$ such points are added to edge partitions, we have $$U_\Gamma < U_{\Gamma'} + 2Mq W^{n-1}\delta$$

Now take $\delta = \frac{\varepsilon}{4MN\delta W^{n-1}}$ where $N$ is the total number of interior points of the edge partitions of $\Gamma_2$. Let $\Gamma$ be any partition with $\|\Gamma\| < \delta$ and form the common refinment $\Gamma' = \Gamma \cup \Gamma_2$.  Since $\Gamma'$ is a refinement of $\Gamma_2$ we have immediately that $U_{\Gamma'} \leqslant U_{\Gamma_2}$.  Also, in forming $\Gamma'$ at most $N$ points are added to edge partitions of $\Gamma$ and from the Lemma it follows that

$$U_\Gamma < U_{\Gamma'} + 2MNW^{n-1}  \delta = U_{\Gamma'} + 2MNW^{n-1} \cdot \frac{\varepsilon}{4MN\delta W^{n-1}}= U_{\Gamma'} + \frac{\varepsilon}{2} \\< U_{\Gamma_2} + \frac{\varepsilon}{2} < \inf_\Gamma U_\Gamma + \epsilon$$

A similar Lemma and argument can be used to prove that the left-hand inequality in (*) holds.

Proof of Lemma.

Consider the subrectangle $R_{jpq} =S_p \times [x_{i,j-1}, x_{i,j}] \times T_q$ of $\Gamma$, where $S_p \subset \mathbb{R}^{i-1}$ and $T_q \subset \mathbb{R}^{n-i}$.  If we add the single point $c \in (a_{i,j-1}, a_{i,j})$ to $P_i$ in forming the refinement $\Gamma'$, then the contribution to $U_\Gamma - U_{\Gamma'}$ from this subrectangle is

$$\sup_{x \in [x_{i,j-1}, c]}f(x)\cdot vol(S_p)vol(T_q)(c- x_{i,j-1})+  \sup_{x \in [c, x_{i,j}]}f(x)\cdot vol(S_p)vol(T_q)(x_{i,j}-c) \\-  \sup_{x \in [x_{i,j-1}, x_{i,j}]}f(x)\cdot vol(S_p)vol(T_q)(x_{i,j}-x_{i,j-1}),$$

and, leaving the details to you, we can show by summing over indexes $p$ and $q$,

$$U_\Gamma - U_{\Gamma'} < 2MW^{n-1}\delta$$


问:
Let $\xi_{n}$ be a symmetric one-dimensional simple random walk, what is $P( a \leq \frac{\xi_{n}}{\sqrt{n}} \leq b | \xi_{0}=0)$ as $n \to \infty$?
I was curious about how to do the following question:

Let $\xi_{n}$ be a symmetric one-dimensional simple random walk, what is $P( a \leq \frac{\xi_{n}}{\sqrt{n}}  \leq b | \xi_{0}=0)$ as $n \to \infty$?

Here $\xi_{n} := X_{1} + \cdots X_{n}$ where the $X_{i}$ are i.i.d Bernouli random variables taking the values $\pm 1$.

So I feel this should be by the central limit theorem

$\frac{1}{\sqrt{2\pi}}\int^{b}_{a} e^{-x^{2}/2}dx$

am I correct here?

If so, what happens when I add the condition that $\xi_{2n}=0$, that is what is the probability

$\lim_{n \to \infty} P( a \leq \frac{\xi_{n}}{\sqrt{n}}  \leq b | \xi_{0}=0 = \xi_{2n}\}?$

答:
You can make reference to the fact that a simple random walk started from $0$, conditioned to return to $0$ at time $2n$ converges to the appropriate Brownian bridge. However, in this situation the combinatorics are simple enough that you can also just explicitly compute the limiting distribution:

Notice that a simple random walk of length $n$ corresponds to the uniform measure on paths started from $0$ that move either $+1$ or $-1$ with each step. Thus, conditioning on $\{\xi_{2n}=\xi_{0}=0\}$ we are merely picking out those paths that find themselves at $0$ on the $(2n)^{th}$ step. That is,

$$\mathbb{P}(\hspace{5pt}\cdot\hspace{5pt}|\xi_{2n}=\xi_{0}=0)$$

is just the uniform measure on paths that move $\pm{1}$ with each step and end at $0$ after $2n$ steps. These paths must have $n$ "up" steps and $n$ "down" steps in total. There are ${2n}\choose{n}$ such paths. By counting we have:

$$ \mathbb{P}(\xi_{n}=k \hspace{3pt}|\xi_{2n}=\xi_{0}=0)=\frac{|\{\text{$n$- step paths with $k$ more $1$'s than $-1$'s}\}|\cdot|\{\text{$n$- step paths with $k$ more $-1$'s than $1$'s}\}|}{{2n}\choose{n}}=\frac{|\{\text{$n$- step paths with $k$ more $1$'s than $-1$'s}\}|^{2}}{{2n}\choose{n}}=\frac{{{n}\choose{(\frac{n+k}{2})}}^{2}}{{2n}\choose{n}}$$

Thus,

$$ \mathbb{P}(a<\xi_{n}/n^{1/2}<b \hspace{3pt}|\xi_{2n}=\xi_{0}=0)=\mathbb{P}(an^{1/2}<\xi_{n}<bn^{1/2} \hspace{3pt}|\xi_{2n}=\xi_{0}=0)={{2n}\choose{n}}^{-1}\sum_{k=\lceil{an^{1/2}}\rceil}^{\lfloor{bn^{1/2}}\rfloor}{{n}\choose{(\frac{n+k}{2})}}^{2}$$

To compute the behavior of this sum as $n\rightarrow{\infty}$ we need to understand the asymptotics of ${2n}\choose{n}$ and ${n}\choose{\frac{n+k}{2}}$ in the regime where $k=O(n^{1/2})$. Stirling's formula with remainder (https://en.wikipedia.org/wiki/Stirling%27s_approximation#Speed_of_convergence_and_error_estimates) tells us that:

$$ n!=\sqrt{2\pi{n}}n^{n}e^{-n}(1+O(n^{-1}))$$

Applying this to our binomial coefficients then tells us that:

$$ {{2n}\choose{n}} = \frac{2^{2n}}{\sqrt{\pi{n}}}(1+O(n^{-1}))$$

$$ {{n}\choose{\frac{n+k}{2}}}=\frac{\sqrt{2\pi{n}}}{\sqrt{\pi(n+k)}\sqrt{\pi(n-k)}}\frac{n^{n}}{(\frac{n+k}{2})^{(\frac{n+k}{2})}(\frac{n-k}{2})^{(\frac{n-k}{2})}}(1+O(n^{-1}))=\sqrt{\frac{2}{\pi{n}}}(1+O(n^{-1}))\frac{2^{n}n^{n}}{(n+k)^{(\frac{n+k}{2})}(n-k)^{(\frac{n-k}{2})}}(1+O(n^{-1}))$$

$$ {{n}\choose{\frac{n+k}{2}}}^{2}=\frac{2}{\pi{n}}\frac{2^{2n}n^{2n}}{(n+k)^{n+k}(n-k)^{n-k}}(1+O(n^{-1}))$$

To understand the asymptotics of this last term we write:

$$ \frac{n^{2n}}{(n+k)^{n+k}(n-k)^{n-k}}=\big(1+\frac{k}{n}\big)^{-(n+k)}\big(1-\frac{k}{n}\big)^{-(n-k)}=e^{-(n+k)\log{(1+\frac{k}{n})}-(n-k)\log{(1-\frac{k}{n})}}=e^{-n\log{(1-\frac{k^{2}}{n^{2}})}-k(\log(1+\frac{k}{n}) - \log(1-\frac{k}{n}))}=e^{-n(-\frac{k^{2}}{n^{2}}+O(n^{-2}))-k(\frac{2k}{n}+O(n^{-1}))}=e^{-\frac{k^{2}}{n}}(1+O(n^{-1/2}))$$

The fourth equality in the calculations above was attained by Taylor expanding ($\log{(1+x)}=x+O(x^{2})$) Thus we have:

$$ {{n}\choose{\frac{n+k}{2}}}^{2}= \frac{2^{2n+1}}{\pi{n}}e^{-\frac{k^{2}}{n}}(1+O(n^{-1/2})) $$

Plugging all this into our original sum we get:

$$ \mathbb{P}(an^{1/2}<\xi_{n}<bn^{1/2} \hspace{3pt}|\xi_{2n}=\xi_{0}=0)=(1+O(n^{-1/2}))\sum_{k=\lceil{an^{1/2}}\rceil}^{\lfloor{bn^{1/2}}\rfloor}\frac{2}{\sqrt{\pi{n}}}e^{-\frac{k^{2}}{n}}$$

Recognizing this last sum as a Riemann sum for the integral $\int_{a}^{b}\frac{2}{\sqrt{\pi}}e^{-x^{2}}dx$ we get that:

$$ \lim_{n\rightarrow{\infty}}\mathbb{P}(an^{1/2}<\xi_{n}<bn^{1/2} \hspace{3pt}|\xi_{2n}=\xi_{0}=0)=\int_{a}^{b}\frac{2}{\sqrt{\pi}}e^{-x^{2}}dx$$


问:
Hypothesis Testing concerning difference in means. Misprinting Error?
I was solving this question in my book and I realized the answer at the back and my answer doesn't match. I wanted to verify whether my process is correct or not?

The question is as follows:

Samples are taken from two different types of honey and the viscosity is measured.

Honey A:

Mean: 114.44

S.D : 0.62

Sample Size: 4

Honey B:

Mean: 114.93

S.D: 0.94

Sample Size: 6

Assuming normal distribution, test at 5% significance level whether there is a difference in the viscosity of the two types of honey?

Here's what I did:

I took my null hypothesis as μB - μA = 0 and alternative hypothesis as μB - μA ≠ 0

Then I did my calculations which were as following:

Test Statistic = (B -A ) - (μB - μA) / sqrt {(variance B / sample size B) + (variance A / sample size A)}

This gave me test statistic as = 0.49/0.49332 that is equal to 0.993

However the test statistic in the book solution is given as 0.91. What am I doing wrong? Help!

答:
The book is using a pooled standard error, one of two possible methods. The central limit theorem says that $\bar x_1-\bar x_2$ is approximately normal with mean $\mu_1-\mu_2$ and standard deviation $\sqrt {\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$, but when the population sizes are small and it can be reasonably assumed the populations have the same variance, then a pooled standard error is used with $s_p=\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}}=\sqrt{\frac{.62^3*3+.94^2*5}{8}}$ giving a standard error of $\bar x_1-\bar x_2$ of $s_p\sqrt{\frac 1 {n_1}+\frac 1 {n_2}}$. This is to combine more data into the calculation of the standard error of each population mean, which are assumed to be equal for these purposes.

So the t-distribution with $n_1+n_2-2=8$ degrees of freedom has test statistic $$t^*=\frac{(\bar x_1-\bar x_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac 1 4+\frac 1 6}}=0.9096458$$


问:
Kerodon exercise 1.1.2.8
I'm doing the exercise 1.1.2.8. from Lurie's Kerodon: https://kerodon.net/tag/000T

The link is straight to the exercise does not contain anything else but the exercise itself, so I see no reason to retype it here. However, if someone thinks this should be done, I will do it.

Let $\Sigma = \{ (\sigma_0,\dots, \sigma_n) \in S^n_{n-1} \mid S_{\partial^{n-1}_j}(\sigma_k) = S_{\partial^{n-1}_{k-1}}(\sigma_j) \text{ for } 0 \leq j < k \leq n \}$. We have an injection $\Phi\colon \mathsf{Hom}_{\mathsf{Set}_{\Delta}}(\partial \Delta^n, S) \to \Sigma$ given by $f \mapsto (f\circ \Delta^{\partial^n_j})_{0 \leq j \leq n}$. It is asked to prove that $\mathrm{im}(\Phi) = \Sigma$. The implication $\mathrm{im}(\Phi) \subseteq \Sigma$ is straightforward. The implication $\Sigma \subseteq \mathrm{im}(\Phi)$ is doesn't seem hard as well, but I can't complete a proof due to my lack of experience in combinatorial matters, probably.

We clearly have $\partial \Delta^n = \bigcup_{0 \leq j \leq n} \Delta^{[n]\setminus\{j\}}$. There are unique order-preserving bijections $a_j\colon [n]\setminus\{j\} \to [n-1]$. Let $(\sigma_0,\dots, \sigma_n) \in \Sigma$. Via the Yoneda lemma, we can idenity $\sigma_i$ with a map $\tilde{\sigma}_i\colon \Delta^{n-1}\to S$. Explicitly, $(\tilde{\sigma}_i)_m(u) = S_u(\sigma_i)$ for any $[m] \in \Delta$ and any morphism $u\colon [m]\to [n-1]$. Thus, we can take the union $\bigcup \tilde{\sigma}_i$ to obtain a map $f\colon \partial \Delta^n\to X$ with sends $u\colon [m]\to [n]$ such that $\mathrm{im}(f) \subseteq [n]\setminus\{j\}$ to $(\tilde{\sigma}_i)(a_j\circ u)$. My idea (of which I'm almost sure) is that the condition $S_{\partial^{n-1}_j}(\sigma_k) = S_{\partial^{n-1}_{k-1}}(\sigma_j)$ for $j < k$ should guarantee that the map $f$ is well-defined, that is, that it agrees on $\Delta^{[n]/\{j\}}\cap \Delta^{[n]\setminus\{k\}}$. Of course, $u$ factors through $[n]\setminus\{j,k\}$ by assumption.

What can be done next? I've spent some time playing with simplicial operators, but it led me nowhere.

答:
Your solution looks to me like it works, but here is a comment on how you
might simplify the combinatorics.  Note that we can rephrase the exercise as
the statement that we have an equalizer
$$
    \hom_{\mathrm{Set}_{\Delta}}(\partial\Delta^{n}, S_{\bullet})
    \to
    \prod_{0 \leq j \leq n}
    S_{n-1}
    \rightrightarrows
    \prod_{0 \leq j < k \leq n}
    S_{n-2}
  $$
where one of the parallel maps acts as $\vec{\sigma} \mapsto
  d_{j}(\sigma_{k})$ and the other as $\vec{\sigma} \mapsto
  d_{k-1}(\sigma_{j})$.  This is a rephrasing because the subset of $\prod_{j}
  S_{n-1}$ described in the exercise is the standard construction in $\mathrm{Set}$ of
the equalizer of these two maps, and the exercise is asking you to show that
$f \mapsto \{f \cdot \delta^{j}\}$ gives a bijection between
$\hom_{\mathrm{Set}_{\Delta}}(\partial\Delta^{n}, S_{\bullet})$ and this subset.

Because $\hom_{\mathrm{Set}_{\Delta}}(-, S_{\bullet})$ preserves
limits, it suffices by Yoneda to show that the preimage diagram
$$
    \coprod_{0 \leq j < k \leq n}
    \Delta^{[n-2]}
    \rightrightarrows
    \coprod_{0 \leq j \leq n}
    \Delta^{[n-1]}
    \to
    \partial\Delta^{n}
  $$
is a coequalizer in $\mathrm{Set}_{\Delta}$.  Colimits are created pointwise in
$\mathrm{Set}_{\Delta}$, so this in turn amounts to showing that
$$
    \coprod_{0 \leq j < k \leq n}
    \Delta^{[n-2]}_{m}
    \rightrightarrows
    \coprod_{0 \leq j \leq n}
    \Delta^{[n-1]}_{m}
    \to
    (\partial\Delta^{n})_{m}
  $$
is a coequalizer for all $m$.

The proof of this last statement is the meat of this exercise, and it's based
on the observation in your answer.  It's perhaps simplest to show that
$(\partial\Delta^{n})_{m}$ is in bijection with the standard construction of
the coequalizer as the quotient $Q$ of $\coprod_{j}\Delta^{[n-1]}_{m}$ by the
equivalence relation generated by
$$
    (j, \delta^{k-1} \cdot \alpha)
    \sim
    (k, \delta^{j} \cdot \alpha)
  $$
for all $j < k$ and $\alpha : [m] \to [n-2]$.  An element of
$(\partial\Delta^{n})_{m}$ is a non-surjective map $\alpha : [m] \to [n]$.
Since it's non-surjective there's some $j$ that's not in its image, and so
there's some $\alpha_{j} : [m] \to [n-1]$ such that $\alpha = \delta^{j}
  \cdot \alpha_{j}$.  This means that the map $Q \to (\partial\Delta^{n})_{m}$
is surjective.  To show that it's injective, consider some pairs $(j,
  \alpha)$ and $(k, \beta)$ in $\coprod_{j} \Delta^{[n-1]}_{m}$ such that
$\delta^{j} \cdot \alpha = \delta^{k} \cdot \beta$.  If $j = k$ then $\alpha
  = \beta$ by the injectivity of $\delta^{j}$, and so $(j, \alpha) = (k,\beta)$.  If $j \not= k$ then we can assume without loss
of generality that $j < k$.  Since neither $j$ nor $k$ lies in the image of
$\delta^{j} \cdot \alpha = \delta^{k} \cdot \beta$, there's some $\gamma :
  [m] \to [n-2]$ such that
$$
    \delta^{j} \cdot \alpha
    = \delta^{k} \cdot \beta
    = \delta^{k} \cdot \delta^{j} \cdot \gamma
  $$
By the injectivity of $\delta^{j}$ and $\delta^{k}$ and the fact that
$\delta^{k} \cdot \delta^{j} = \delta^{j} \cdot \delta^{k - 1}$ we therefore
have
$$
    (j, \alpha)
    =
    (j, \delta^{k-1} \cdot \gamma)
    \sim
    (k, \delta^{j} \cdot \gamma)
    =
    (k, \beta)
  $$
This means that $Q \to (\partial\Delta^{n})_{m}$ is injective, hence
bijective, giving $(\delta\Delta^{n})_{m}$ the universal property of the
coequalizer.

Title: Kerodon exercise 1.1.2.8