答:
Some facts and notations:

Trace and Frobenius product relation $$\left\langle A, B C\right\rangle={\rm tr}(A^TBC) := A : B C$$

Cyclic properties of Trace/Frobenius product
\begin{align}
A : B C
 &= BC : A \\
 &= B^T A   :  C  \\
 &= {\text{etc.}} \cr
\end{align}

Let $f := x^T A x = x:Ax$.

Compute the differential first, and then the gradient can be obtained from it.
\begin{align}
df
&= dx:Ax + x: A dx \\
&= Ax:dx + A^Tx:dx \\
&= (A + A^T)x:dx
\end{align}

Thus, the gradient is
\begin{align}
\frac{\partial }{\partial x} \left( x^T Ax \right)= (A + A^T)x.
\end{align}

When $A$ is symmetric, i.e., $A^T = A$, then the gradient is $\frac{\partial }{\partial x} \left( x^T Ax \right)= 2Ax$.

Title: Derivative of quadratic form for matrices and vectors

Body:
I'm not very familiar with multivariable calculus as it relates to matrices. Could someone explain, in detail, why
$$\frac{\partial}{\partial x} \left[ x^T A x \right] = (A + A^T)x$$
In the case of a symmetric matrix and
$$\frac{\partial}{\partial x} \left[ x^T A x \right] = 2Ax$$
if the matrix is not symmetric. I'm mainly confused about how we even arrive at the first derivative. However, I understand how the first derivative simplifies to the second in the case that A is symmetric.


问:
Continuous random variable question
$ X $ is a non-negative continuous random variable with density function $f$ and distribution function $F$.

Use integration by parts to show that

$ \int_0^{\infty}  ( 1- F(x)) dx =
\int_0^{\infty}xf(x)dx  $

I'm quite puzzled on how to even integrate $F(X)$ to get $f(x)$  :S

答:
Hint: In the usual calculus notation for integration by parts, let $u=1-F(x)$ and $dv=dx$. Then $du=-f(x)\,dx$ and we can take $v=x$.

Added: As pointed out by  беркай. the fact that $\lim_{x\to\infty}x(1-F(x))=0$ requires justification. This takes a few lines. For details, please see this part of the Wikipedia article on cumulative distribution functions.

ContentLicense: CC BY-SA 3.0

Title: Continuous random variable question


问:
Do I Substitute Values into an Equation Before or After Simplification of the Expression and Why?
Considering the equation:
\begin{align} y&=\dfrac{6x+x^2}{x}
&\text{(before simplification)}\\ y&=6+x &\text{(after simplification).} \end{align}

If Before Simplification of the expression, $x = 0$ is substituted,
$y = undefined$

But after Simlfication, $x = 0$ is substituted, $y =
   6$

Why is this so? Or am I getting something wrong?

答:
This is because you did the simplification almost correct. The actual correct simplification would be
$$y = x + 6, \quad x \neq 0$$
because that's what captures all information in the original equation. If you forget the $x \neq 0$, you lose that piece of information, and that leads to your confusion.

ContentLicense: CC BY-SA 4.0


问:
Does a function for which $f'(x)$ is globally small but $f''(x)$ is globally large exist?
Per title. Does a function for which the magnitude of the first derivative is small (say, $f'(x) < 1$), but the magnitude of the second derivative is large (say, $f''(x) > 10$), exist?

It's easy enough to find functions where this is locally true, e.g. $y = \sin(x)$ around the point $x = \pi/2$, but I'm wondering about the general case. If such functions do not exist, what kind of function allows for the largest possible parameter space such that $f'(x)$ is small but $f''(x)$ is large?

答:
The mean value theorem implies

$$f^\prime(x) \gt f^\prime(a) +M(x-a)$$ if

$$f^{\prime \prime}(x) \gt M>0$$ for all $x \in \mathbb R$.

Therefore what is asked in the question is not possible if above hypothesis hold.


问:
Derivative of quadratic form for matrices and vectors
I'm not very familiar with multivariable calculus as it relates to matrices. Could someone explain, in detail, why
$$\frac{\partial}{\partial x} \left[ x^T A x \right] = (A + A^T)x$$
In the case of a symmetric matrix and
$$\frac{\partial}{\partial x} \left[ x^T A x \right] = 2Ax$$
if the matrix is not symmetric. I'm mainly confused about how we even arrive at the first derivative. However, I understand how the first derivative simplifies to the second in the case that A is symmetric.

答:
An alternative approach, though similar: we have

$$\begin{align}f(x+h)&=(x+h)^TA(x+h)\\
&=x^TAx+x^TAh+h^TAx+h^TAh\\
&=x^TAx+x^T(A+A^T)h+h^TAh\\
&=f(x)+\mathrm Df(x)h+o(\vert h\vert),
\end{align}$$

where $\mathrm Df(x):h\mapsto x^T(A+A^T)h$ is linear and $h^TAh\in o(\vert h\vert)$. And thus the linear map $\mathrm Df(x)$ is the derivative of $f$ at $x$. The version you were given is the transpose of its matrix representation. If $A$ is symmetric, $A+A^T=2A$.


问:
Asif , Aniqa and Saki are perplexed with grid and numbers
Asif has filled in a 3×3 grid with the numbers 1,2, . . . ,9.  Saki writes down the three numbers obtained by multiplying the numbers in each horizontal row.  Aniqa writes down the three numbers obtained by multiplying the numbers in each vertical column.  Can Asif  fill in the grid in such a way that  Saki and Aniqa obtain the same lists of three numbers?  What if the Asif writes the numbers 1,2, . . . ,25 in a 5×5 grid?  Or 1,2, . . . ,121 in a 11×11 grid?  Can you find any conditions that guarantee that it is possible or any conditions that guarantee that it is impossible for Asif to write the numbers 1,2, . . . , $n^2$ in a n×n grid so that  Saki and Aniqa obtain the same lists of numbers?

My observation:

```<code>Asif's grid →  1 2 3       Sakis' three numbers= 6 120 504
               4 5 6      Aniqas' three numbers= 28 80 162
               7 8 9
</code>```

I think it is possible for a magic square if it was addressed as sum. A square array of numbers, usually positive integers, is called a magic square if the sums of the numbers in each row, each column, and both main diagonals are the same.

Then the first grid will be:

```<code>Asif's modified grid →  2 7 6  and so Sakis' three numbers= 15 15 15
                        9 5 1        Aniqas' three numbers= 15 15 15
                        4 3 8
</code>```

I knew a term called multiplication magic square where all number are not needed to be in consecutive. So I can't next what to do?

Or it was asking for it:
\begin{array}{|c|c|c|}
\hline A & B & C \\\hline
D & E & F \\\hline
G & H & I\\\hline
\end{array}

then,

\begin{align}
ABC&=ADG \tag{1}\\
DEF&=BEH  \tag{2}\\
GHI&=CFI \tag{3}\\
\end{align}

Please help me the solution.

Thank you so much for your help.

答:
One thing you know immediately is that $7$ and $5$ must be in one of the $A,E,I$ position.

Using the above fact and some quick trials, the following is found to be a valid construction.

\begin{array}{|c|c|c|}
\hline 7 & 1 & 6 \\\hline
2 & 5 & 4 \\\hline
3 & 8 &9\\\hline
\end{array}

Edit: sorry I overlooked the other two cases asked in the question.

For the $11\times11$ case, there are $13$ primes numbers from $61$ to $121$, namely $61,67,71,73,79,83,89,97,101,103,107,109,113$ must be in the $11$ diagonals so clearly impossible. $5\times5$ seems possible, I will add to the answer if managed to find one.

Update: here's one:
\begin{array}{|c|c|c|}
\hline 13 & 11 & 24 & 9 &  14  \\\hline
22 &17 &1 & 20 &8 \\\hline
6  &5 & 19& 2 & 18\\\hline
12& 4 & 3 & 23 &25\\\hline
21& 16& 15& 10& 7\\\hline
\end{array}

In general, if the number of prime numbers $p$ of the following property:

$p > n$ and $p\leq n^2$ and the number of multiples of $p$ between $n+1$ nad $n^2$ is odd

is greather than $n$ then we can immediately say the $n\times n$ configuration is impossible. This number grows quickly comparing to $n$ (at least at the speed of prime numbers that appears between $n^2\over 2$ and $n^2$ which is $O({n^2\over \ln(n)})$ by Prime Number Theorem which is much larger than $n$). Even when $n=8$ we have $37,41,43,47,53,59,61$ whose multiples appear only once(odd) and $17,19$ whose multiples appear $3$ times(odd) therefore impossible. Smaller cases need to be checked manually.


问:
Reference request, Descriptive set theory
I was wondering what a good text would be to learn descriptive set theory out of? Hopefully something more in the spirit of Kunen's text on the introduction to independence proofs.</p>

答:
Moschovakis' Descriptive Set Theory is the standard text for anyone who wants to learn descriptive set theory. However, this book does not contain

The in-depth inner model theory

Kechris' application of descriptive set theory to the other areas, including the theory of countable Borel equivalence relations.

Having read Moschovakis' book, if you want to know more about 1, I suggest reading books about Woodin's core model (Note that I've been told this studying involves a very very very long way.) If you want to know more about 2, Kechris' Classical Descriptive Set Theory is the standard book. A lot of the theorems in this book are used frequently when reading the latest paper concerning descriptive set theory. If you want to know more about the countable Borel equivalence relation, Kechris' The Theory of Countable Borel Equivalence Relations is available on his own website. This is a survey of up-to-date progress in the countable Borel equivalence relations.


问:
Finding trigonometric integral (challenging)
Integrate:

$$\displaystyle\int \dfrac{(1+\cos^2A\times\cos(2A))^2}{\cos(2A)\times(2\cos^4A+\sin^2A)}dA$$

Again my working with trigonometric identities has not take me anywhere useful, that is I haven't been able to break free of fractions or nasty products. But again, I'm a beginner to calculus and I only know how to integrate the 6 primary trigonometric functions. Perhaps there is a good way to bypass the tedious mess?

Despite the most off-putting look, the teacher claims that the integral is not anywhere as nasty as one would expect. This, of course, is not a part of class work but a challenge she threw at us.

Thanks in advance and as usual, all advice welcome.

答:
$ \frac{(1+cos^2A*cos(2A))^2}{cos(2A)*(2cos^4A+sin^2A)]}dA$ which after some simplification of the denominator part $(2cos^4A+sin^2A)$ gives the numerator part which is : $(1+cos^2A*cos(2A))^2$

Therefore after simplification  original problem = $\frac{(1+cos^2Acos(2A))}{cos(2A)}$
=$\frac{sin^2A + cos^2A +cos^2A(1-2sin^2A)}{cos2A} $

=$\frac{sin^2A(1-2cos^2A)+2cos^2A}{cos2A}$

=$\frac{-sin^2A(cos2A) +2cos^2A}{cos2A}$

=$ -sin^2A +\frac{2(1-sin^2A)}{cos2A}$

=$-sin^2A + 2sec2A -\frac{2-2sin^2A)}{cos2A}$

=$-sin^2A + 2sec2A -2sec2A +\frac{1-cos2A}{cos2A}$

= $-sin^2A +sec2A -1$

This you can can easily integrate.


答:
$$y = \frac{6x+x^2}{x} = \frac{x(6+x)}{x}$$
Now, you cannot cancel out the $x$ if $x = 0$ because division by zero is undefined. Your simplification is valid for any $x \neq 0$
$$  y = 6+x ~~~\forall ~x \neq 0 $$
Another thing to note here is that the domain of $y(x)$ is $\mathbb{R} - \{0\} $ which means that $y$ is not defined at $x=0$. Therefore, the graph of $y(x)$ will be a straight line ($y=x+6$) missing the point $(0,6)$. However, if you plot $y=x+6$ , the graph will be a straight line including $(0,6)$. Apart from the point $(0,6)$, both the graphs are identical.

Title: Do I Substitute Values into an Equation Before or After Simplification of the Expression and Why?

Body:
Considering the equation:
\begin{align} y&=\dfrac{6x+x^2}{x}
&\text{(before simplification)}\\ y&=6+x &\text{(after simplification).} \end{align}

If Before Simplification of the expression, $x = 0$ is substituted,
$y = undefined$

But after Simlfication, $x = 0$ is substituted, $y =
   6$

Why is this so? Or am I getting something wrong?


问:
Why does calculating volume using cubic inches gives different answer than with using cubic centimeters?
Length: $8$'

Width: $2$'

Height: $7$cm

$243.84$cm × $60.96$cm × $7$cm = $104051.4048$ cm³

$104051.4048$ cm³ = $104.051$ liters

$96$" × $24$" × $1.181$" = $2721.024$ inches³

$2721.024$ inches³ = $44.59$ liters

$104$L $\ne$ $44$L

What have I done wrong and which is correct? Please show the appropriate calculation and conversions.

I'm using google calculator and converting by typing in google search bar (i.e. $2721$ inches to liters =). The cm and inch conversions were taken the same (i.e. Google search bar: $8$' in cm =).

I am guessing that there is an error in Google's conversions when asked to convert from the search bar.

答:
As said:
Length: $8'$ or $243.84cm$

Width: $2'$ or $60.96cm$

Height: $0.229659'$ or $7cm$

$8'×2'× 0.229659'=3.674544 cubic foot=104.05149874L$

$243.84cm × 60.96cm × 7cm = 104051.4048 cm³=104.051405L$

accounting for rounding it's relieving to say that our measurement system has not screwed us over.


答:
Suppose $x=(x_1,\ldots,x_n)^T$ and $A=(a_{ij})$, by calculating partial derivative w.r.t the $k^{th}$ component, we have
$$ \frac {\partial x^T A x}{\partial x_k} $$
$$ = \frac {\partial (\sum_{ij} x_i a_{ij}x_j)}{\partial x_k}$$
$$= \sum_j a_{kj}x_j +\sum_i x_i a_{ik} $$
$$ =\sum_j (a_{kj} + a_{jk})x_j$$
$$ =[(A+A^T)x]_k$$
Hence, $\frac {\partial x^T A x}{\partial x}=(A+A^T)x$.

The case when $A$ is not symmetric can be understand with tensor language:

If $A$ is not symmetric, $a_{ij}$ is not equal to $a_{ji}$ in general. In the expression $x_i a_{ij} x_j$, we need to differentiate $x_i$, $x_j$ respectively. This gives $a_{kj}x_j$ and $x_i a_{ik}$ in the above. Notice that $j$ is dummy index, which can be substituted by any symbol, we exchange $i$ with $j$ and combine the two items. It yields  $(a_{jk} + a_{kj})x_j$, where both left and right index of $a_{ij}$ is cancel exactly once. So that when $A$ is not symmetric, $a_{jk}+a_{kj}$ is not in general $2a_{jk}$, which is where the difference takes place.


问:
If $X^*$ is reflexive then so is $X.$

Let $X$ be a normed linear space. Then there is a linear isometry $J_X : X \longrightarrow X^{**}$ such that $J_X(x) (\varphi) = \varphi (x), x \in X, \varphi \in X^*.$ Then $X$ is said to be reflexive if $J_X (X) = X^{**}$ i.e. if $J_X$ is onto.

Prove that $X$ is reflexive if so is $X^*.$

Attempt $:$ Let $\psi \in X^{***}$ be such that $\psi (J_X(x)) = 0,$ for all $x \in X.$ Since $X^*$ is reflexive there exists $\varphi \in X^*$ such that $J_{X^*} (\varphi) = \psi.$ So $0 = \psi (J_X(x)) = J_{X^*} (\varphi) (J_X(x)) = J_X(x) (\varphi) = \varphi (x),$ for all $x \in X.$ Hence $\varphi = 0$ in $X^*.$ Therefore $\psi = J_{X^*} (\varphi) = J_{X^*} (0) = 0$ in $X^{***}.$ So any bounded linear functional on $X^{**}$ which vanishes on $J_X (X)$ is identically zero. So if $\overline {J_X (X)} \subsetneq X^{**}$ then by Hahn-Banach theorem there would exist a $y \in X^{**} \setminus {J_X(X)}$ (so that $\text {dist}\ (y, J_X(X)) \gt 0)$ and a $T \in X^{***}$ such that $T(y) = 1$ and $T(x) = 0,$ for every $x \in J_X (X),$ which will lead us to a contradiction. What will happen if $\overline {J_X(X)} = X^{**}$ i.e. if $J_X (X)$ is dense in $X^{**}\ $?

Any help in this regard will be appreciated. Thanks for your time.

答:
The result is not true without completeness of $X$. Take $X=\ell_0$ the space of all sequences which are eventually $0$ with the $\ell^{2}$ norm. The dual of this space is the usual $\ell^{2}$ space which is reflexive. But $X$ itself is not reflexive since any reflexive space is necessarily complete.

Of course, if $X$ is complete then you don't run into a problem with your proof.

Title: If $X^*$ is reflexive then so is $X.$


问:
Point Order on Elliptic Curve
i got stuck doing the exercise 4.5 d) in the Book Elliptic Curves: Number Theory and Cryptography by L. Washington and would be grateful for hints.

Let $ p \equiv 1 \text{ (mod 4)}$ be prime and let $E$ be given by $y^2 = x^3 -kx$, where $k \not\equiv 0 \text{ (mod $p$)}$.
Let $k$ be a square but not a fourth power mod $p$. Show that exactly one of the curves $y^2 = x^3 -x$ and $y^2 = x^3 -kx$ has a point of order $4$ defined over $\mathbf{F_p}$.

I tried to find a point such that $2P = (\pm\sqrt{k},0)$, but that did not work out and solving the division polynomial $\psi_4$ also didnt work.

答:
$k=d^2$ then $y^2=x^3-kx$ is isomorphic to $E_d:y^2=d(x^3-x)$.

$E_1:y^2=x^3-x$.

Since $d$ is not a square, for all $a\in \Bbb{F}_p$, either $(a,\pm \sqrt{a^3-a})\in E_1(\Bbb{F}_p)$ or $(a,\pm \sqrt{d(a^3-a)})\in E_d(\Bbb{F}_p)$

$\#E_1(\Bbb{F}_p)+\#E_d(\Bbb{F}_p)= 2(p-3)+8\equiv 4\bmod 8$

Both $E_1(\Bbb{F}_p)$ and $E_d(\Bbb{F}_p)$ contain the 2-torsion (the points coming from $a=\infty$ or $a^3-a=0$)

Thus, exactly one of $\# E_1(\Bbb{F}_p)$,$\# E_d(\Bbb{F}_p)$ is $\equiv 0\bmod 8$, the other is $\equiv 4\bmod 8$.

Exactly one of $ E_1(\Bbb{F}_p)$,$E_d(\Bbb{F}_p)$ contains some point of order $4$.

Title: Point Order on Elliptic Curve


问:
Line Integral - Independence of the way (Why these conditions)
Given the following differential form
$$ydx - x^2dy$$
It is desired to obtain the value of the line integral along the segment going from (0,0) to (1,1). In addition, it is stated that for any parameterization $x = y = f (t), 0 \leq t \leq 1, f'(t) > 0, f(0) = 0, f(1) = 1$ the value of the integral will be the same.

Well, the first part I already got, which is the value of the integral equal to 1/6.

However, in the second one, which is to verify the statements, I thought about using the independence of the path. For that, I checked if the field is conservative.

$$\frac{dP}{dy} = \frac{dQ}{dx}$$

But, equality is not satisfied. Implying that I cannot use the independence of the path to verify the claims. I'm wrong? What is the correct way to proceed. Furthermore why the need for the conditions mentioned ($x = y = f (t), 0 \leq t \leq 1, f'(t) > 0, f(0) = 0, f(1) = 1$)?

Thanks in advance!

答:
I show the reparametrizazion exercise using some notation on differential forms. Hope to not mess it up too much since I am rusty.

So the exercise suggests to take the generic parametrization of the path $\gamma$, defined through $f$ :

$p(t):[0,1]\rightarrow \mathbf{R}^2, p(t)=(f(t),f(t)), p(0)=(0,0),p(1)=(1,1)$

and than apply the definition of the line integral. Calling $z=ydx-x^2dy$ the 1-form on $\mathbf{R}^2$ this leads to:

$\int_{\gamma}z=\int_0^1 z(p(t))[p_*(\partial_x)]dt$

where $ z(f(t))$ is the one form evaluated at $f(t)$ and $p_*(\partial_x)$ is the pushforward of the tangent field in [0,1]. $z(x)[v]$ is the value of the differential form at $x$ applied to the tangent vector $v$.

Explicitely:

$z(p(t))=f(t)dx-f(t)^2dy$

$p_*(\partial_x)=f'(t)\partial_x+f'(t)\partial_y$

So putting together:

$\int_{\gamma}z=\int_0^1f(t)f'(t)-f(t)^2f'(t)dt$

And now we can change variables : $y=f'(t)$, so that:

$\int_{\gamma}z=\int_0^1 y-y^2 dy=1/6$

You may want to apply the same calculations using a different notation according to the one you are studying right now.


问:
Probability of getting an ace in the 2nd draw without knowing the first draw
The question is: Given a 52-card deck, two cards are drawn without replacement. What is the probability that the second card is an ace?

So I think I already found the answer (correct me if my answer is incorrect), but I still have a question.

Here's my answer: $$\frac{48}{52}\cdot\frac{4}{51}+\frac{4}{52}\cdot\frac{3}{51}=0.07692$$

So in my answer, I'm just adding the cases. The first case is where the first draw isn't an ace and the second case is where the first draw is an ace. Now to my question. How come the expression below seem to overcount, despite seemingly doing the same thing? $$\frac{\binom{48}{1}\binom{4}{1}}{\binom{52}{2}} + \frac{\binom{4}{2}}{\binom{52}{2}} = 0.1493213$$

My guess is that the first term is double counting, but I really can't grasp why.

答:
In your first calculation, $4/51$ should be $3/51$ because there are only theee aces left.

In the second calculation, since the denominator is $52\choose2$, the order doesn't matter.  You are including an ace first as a win, not just an ace second.


问:
Summing of even binomial coefficients with parity
Is there a general formula for
$$
\sum_{2m\le n}(-1)^m\left( n\atop 2m\right)
$$
where $n$ could be either even or odd?

EDIT. Just realized that this is equal to $\Re(1+i)^n$.

答:
Your edit is correct; equivalently, it's

$$\frac{(1 + i)^n + (1 - i)^n}{2}.$$

When $n$ is even we have $(1 \pm i)^2 = \pm 2i$ so this simplifies to

$$\frac{(2i)^{\frac{n}{2}} + (-2i)^{\frac{n}{2}}}{2}$$

which is $0$ if $n \equiv 2 \bmod 4$ and $(-4)^{\frac{n}{4}}$ if $n \equiv 0 \bmod 4$. (This kind of periodicity is to be expected because $1 + i$ is a real multiple of an $8^{th}$ root of unity.) When $n$ is odd we instead get

$$\frac{(1 + i)(2i)^{\frac{n-1}{2}} + (1 - i)(-2i)^{\frac{n-1}{2}}}{2}$$

which is $(-4)^{\frac{n-1}{4}}$ if $n \equiv 1 \bmod 4$ and, if I haven't made a mistake, $(-2)(-4)^{\frac{n-3}{4}}$ if $n \equiv 3 \bmod 4$.

Plausibly a nicer and less error-prone way to organize this computation is to write $1 + i = \sqrt{2} \zeta_8$ where $\zeta_8 = e^{ \frac{\pi i}{4} }$ is a primitive $8^{th}$ root of unity, which gives

$$\text{Re}((1 + i)^n) = 2^{\frac{n}{2}} \cos \frac{n \pi}{4}.$$


问:
Ramification in local fields
Question: I know how to detect ramified or unramified fields in the case of number fields. But I have no feelings about how to do this for local fields. How should I find that $\mathbb{Q}_p(\sqrt[3]{2})=\mathbb{Q}_p[x]/(x^3-2)$ is ramified or unramified? I can answer that $\mathbb{Q}_p(\sqrt[3]{2})=\mathbb{Q}_p[x]/(x^3-2)$ is (totally) ramified, because $x^3-2$ is Eisenstein at $2$ (This is not correct. See the comments). But then how should I realize that $\mathbb{Q}_p(\sqrt[3]{2}, \zeta_3)=\mathbb{Q}_p[x]/(x^3-2, x^2+x+1)$ is ramified or unramified? How should I find that $\mathbb{Q}_p[x]/(x^5-x-2)$ is ramified or unramified? How can I find the uniformizer in the local field $\mathbb{Q}_p[x]/P(x)$, where $P(x)$ is an irreducible monic polynomial, and how can I find the ramification index?

My attempt: This is not actually an attempt. I just want to say what I know in the case of number fields, and I have some questions and solutions (probably unsuccessful solutions) based on it. In the case of number fields I used the discriminant. Also Dedekind-Kronecker theorem was very helpful to find the ramification indices, $e_i$'s, and reside degrees $f_i$'s for almost all primes. Do we have a version of Dedekind-Kronecker theorem for local fields?

答: You said you know about using the discriminant of a $\mathbf Z$-basis of the integers of a number field to detect ramification.  You can do the same thing in a local field: if $K$ is a finite extension of $\mathbf Q_p$ then the discriminants of all $\mathbf Z_p$-bases of $\mathcal O_K$ are equal to multiplication by the square of a unit in $\mathbf Z_p$, and $K/\mathbf Q_p$ is ramified if and only if the discriminant of a $\mathbf Z_p$-basis of $\mathcal O_K$ is divisible by $p$ (equivalently, the discriminant is not a unit in $\mathbf Z_p$).

Title: Ramification in local fields


问:
Expressing a polynomial in terms of basis
Suppose $\alpha$ is a root of the irreducible polynomial $x^3 -2x-2 \in \mathbb{Q}[x]$.

PROBLEM:

Express $(\alpha +1)(\alpha^2+\alpha+1)^{-1}$ in terms of $\{1, \alpha, \alpha^2\}$ for $\mathbb{Q} (\alpha)$ over $\mathbb Q$.

My attempt: I tried to find $(\alpha^2+\alpha+1)^{-1}$, $(a+b\alpha+c\alpha^2)(\alpha^2+\alpha+1)=1$. Implies that $a=1,b=0$ and $c=0$ . From here I dont know what to do or did I do something wrong here?

Any help and hints would be much appreciated!

答:
There is another technique based on the proof that $\mathbb{Q} [x] /\langle x^3-2x-2\rangle$ is a field.

Let us compute the GCD of polynomials $x^2+x+1$ and $x^3-2x-2$. We have $$x^3-2x-2=(x-1)(x^2+x+1)- 2x-1$$ and $$4(x^2+x+1)=(2x+1)^2+3$$ And then going backwards $$3=4(x^2+x+1)-(2x+1)^2=(5+x-2x^2) (x^2+x+1)+(2x+1)(x^3-2x-2)$$ Replacing $x$ by $\alpha$ we get $$\frac{1}{1+\alpha+\alpha^2}=\frac{5+\alpha-2\alpha^2}{3}$$ Next we can multiply by $(\alpha+1)$ on both sides and replace $\alpha^3$ on right side by $2\alpha+2$.

The above is a direct method for inverting any polynomial in $\alpha$ and helps in proving that $\mathbb{Q} [\alpha] $ is a field. The technique however is not any more efficient compared to other methods.


答:
You can avoid doing any work here with a symmetry argument. If you aren't given any information about the first draw then it doesn't matter what order you draw the cards in, or said another way it's the same as drawing the cards simultaneously. So the probability the second card is an ace is the same as the probability the first card is an ace, which is $\frac{4}{52} = \frac{1}{13}$.

(This argument generalizes - you can draw any fixed number of cards, up to and including the entire deck, and if you aren't given any information about any of the draws the probability that the last card you drew is an ace is still the same $\frac{1}{13}$.)

Your first calculation actually simplifies to this:

$$\frac{48}{52} \cdot \frac{4}{51} + \frac{4}{52} \cdot \frac{3}{51} = \frac{4}{52} \cdot \left( \frac{48}{51} + \frac{3}{51} \right) = \frac{4}{52}.$$

Empy's explanation of the problem with your second calculation is correct. If you just expand out the binomial coefficients you'll see that you've introduced an additional factor of $2$ in the first term but not in the second, and this corresponds exactly to overcounting cases where the first card is an ace. The ${52 \choose 2}$ denominator means you are choosing a set of cards and ignoring order which is only appropriate if the cards can be treated as indistinguishable, as in your second term where the cards are just "two aces."

Title: Probability of getting an ace in the 2nd draw without knowing the first draw

Body:
The question is: Given a 52-card deck, two cards are drawn without replacement. What is the probability that the second card is an ace?

So I think I already found the answer (correct me if my answer is incorrect), but I still have a question.

Here's my answer: $$\frac{48}{52}\cdot\frac{4}{51}+\frac{4}{52}\cdot\frac{3}{51}=0.07692$$

So in my answer, I'm just adding the cases. The first case is where the first draw isn't an ace and the second case is where the first draw is an ace. Now to my question. How come the expression below seem to overcount, despite seemingly doing the same thing? $$\frac{\binom{48}{1}\binom{4}{1}}{\binom{52}{2}} + \frac{\binom{4}{2}}{\binom{52}{2}} = 0.1493213$$

My guess is that the first term is double counting, but I really can't grasp why.


问:
Normal Distribution Probability after finding value C
I am having trouble determining the probability of a $P(X > 8)$ for a problem that requires integrating a function to find the value of $c$.  I believe that I need to determine the mean and sigma in order to use the value of $(8-\mu)/\sigma$ to look up the probability using a $z$-score table.  My problem is not knowing how to determine the mean and sigma?  Here is the question:

The lifetime $X$ (in units of $10$ years) of a certain component of a home-heating furnace is a random variable with p.d.f.

$$f(x) =\left\{\begin{array}{cl} cx^2(1-x) & \text{if}\  0 < x < 1;\\ 0 & \text{otherwise}.
\end{array}\right.$$

Determine the value of $c$ (which I've determined to be $12$ by integration).  Then find the probability that the life of such a component is more than eight years.

答:
As you did for obtaining $c=12$, you also need to integrate $f(x)$ for computing the probability. As Sean said, you can't use a $z$-score table because this is not a normal distribution. Also, you are asked to find the probability that the life is greater than $8$ years and the units of time are $10$ years, so you must find $P(X>0.8)$.
$$P(X>0.8)=\int_{0.8}^\infty f(x)\mathrm{d}x=\int_{0.8}^112x^2(1-x)\mathrm{d}x+\int_1^\infty0\mathrm{d}x=\cdots=0.1808.$$


问:
Proving lower sum inequality
For bounded function $f$ and $g$ on $[a,b]$, $h = f + g$

I have proved that $\underline{\int_a^b}(f) \; + \; \underline{\int_a^b}(g) \; \leq \;  \underline{\int_a^b}(h)$ holds.

But is it always true that $\underline{\int_a^b}(f) \; + \; \underline{\int_a^b}(g) \; = \;  \underline{\int_a^b}(h)$?

I try to prove that $\underline{\int_a^b}(f) \; + \; \underline{\int_a^b}(g) \; \geq \;  \underline{\int_a^b}(h)$ also holds so maybe I can reach the conclusion.

It seems like the statement is true but I had trouble proving it.

答: The latter statement is false. Consider $$f(x) = \begin{cases} 0, & x \in \mathbb{Q} \cap [0,1]\\  1, &x \in (\mathbb{R} \backslash \mathbb{Q}) \cap [0,1] \end{cases}$$ and $g(x) = 1-f(x)$. Then the lower sum of both $f$ and $g$ are $0$ because $\mathbb{Q}$ is dense in $\mathbb{R}$, but the lower sum of $f+g=1$ is $1$.

Title: Proving lower sum inequality


问:
non unique imaginary units
A professor told me that another way to define complex numbers is through the real matrices of the form  $\begin{bmatrix}a & b\\-b &a\end{bmatrix}$. With this definition, the imaginary unit becomes the matrix $I=\begin{bmatrix}0 &1\\-1&0\end{bmatrix}$. By definition $i^2=-1$ and It's pretty easy to check that the square of the matrix $I$ is the negative of the identity matrix.

So apparently, any matrix with square equal to the negative of the identity can be taken as $I$. Why is that? When you think about the $2D$ plane, this corresponds to choosing a vector different from (1,0) as $\bar i$, the unit vector in the x-direction.

Then he says that by some general theorem all complex structures in dimension 2 are isomorphic to each other and that one can allow the imaginary unit to change from point to point.

Does anyone understand what this means and why we can do this?

答:
$(0,1)$ and $(0,-1)$ are equally good to be taken as square root of $-1$. The isomorphism that is being talked about is the map $(x,y) \to (x, -y)$.

Incidentally, the definition of $i$ as $\sqrt {-1}$ found in some books is quite meaningless.

Title: non unique imaginary units


问:
Can two non-isomorphic $G$-sets have the same character?
Let $G$ be a finite group, and let $H_1$, $H_2$ be two non-conjugate subgroups. Can we have $\text{Ind}_{H_1}^G(\mathbf{1})\cong \text{Ind}_{H_2}^G(\mathbf{1})$ as representations?

Using characters, this can be rephrased as, "If for all $g\in G$, $g$ fixes an equal number of left $H_1$ cosets as left $H_2$ cosets, must $H_1$ be conjugate to $H_2$ as subgroups?"

It is easy to show that this holds if either $H_i$ is normal, or cyclic, but the general case doesn't seem clear.

If the answer to this is false, can we say anything about such a pair $H_1$, $H_2$? For instance, by considering the identity, we can see that $|H_1|=|H_2|$.

答: Yes. A triple $(G, H_1, H_2)$ with this property is called a  Gassmann-Sunada triple; the Wikipedia article gives an example with $G = SL_3(\mathbb{F}_2)$. They can famously be used to construct examples of pairs of isospectral Riemannian manifolds which are not isometric, and pairs of number fields with the same Dedekind zeta function which are not isomorphic.


问:
Dealing with numbers bigger than Realmax
I have a code in Matlab, and in one step it calculates Gamma function for different values. But when I try to find Gamma(172.5)  I get inf as it is bigger than realmax. My question:  is there a away that I can get and calculate  big numbera like Gamma (172.5) in Matlab without getting inf?

Thanks

答:
Yes, you can if you have the Symbolic toolbox. Then you can work with the variable-precision arithmetic. E.g., gamma(vpa(172.5)) or vpa(gamma(sym(172.5))) or

```<code>syms x real;
y = subs(gamma(x),x,172.5);
disp(y);
disp(vpa(y));
</code>```

Body:
Yes, you can if you have the Symbolic toolbox. Then you can work with the variable-precision arithmetic. E.g., gamma(vpa(172.5)) or vpa(gamma(sym(172.5))) or

syms x real;
y = subs(gamma(x),x,172.5);
disp(y);
disp(vpa(y));


问:
Eigenvectors of a block matrix
I am trying to find the eigenvectors and eigenvalues of the following block matrix $$M = \begin{bmatrix} I & A\\A^T &I\end{bmatrix}$$ where $A \in \mathbb{R}^{n\times n}$. I know I must use $A$'s SVD and so I have $$M = \begin{bmatrix} I & U\Sigma V^T \\ V\Sigma U^T & I \end{bmatrix}$$ I am unsure how to proceed from here.

I was able to find a similar question (Finding left singular vector of a block matrix) but I am a little lost on the answer.

答: The eigenvector problem $$\begin{pmatrix}I&A\\A^T&I\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}=\lambda\begin{pmatrix}u\\v\end{pmatrix}$$ $$\implies \begin{matrix}u+Av=\lambda u\\A^Tu+v=\lambda v\end{matrix}$$ $$\implies \begin{matrix}Av=(\lambda-1)u\\A^Tu=(\lambda-1)v\end{matrix}$$ Hence $$A^TAv=(\lambda-1)^2v, \quad AA^Tu=(\lambda-1)^2u$$ But $A^TA=V\Sigma^2V^T$ and $AA^T=U\Sigma^2U^T$, so $$\Sigma^2V^Tv=(\lambda-1)^2V^Tv,\quad \Sigma^2U^Tu=(\lambda-1)^2U^Tu$$ Since the eigenvectors and eigenvalues of $\Sigma^2$ are $\sigma^2$ and $e_i$, it follows that $v_i=Ve_i$, $u_i=Ue_i$ (i.e. the columns of the matrices $V$ and $U$) and $$\lambda_i=1\pm\sigma_i$$

Title: Eigenvectors of a block matrix


问:
Cosine and sine rule on a triangle
$\triangle ABC$ is isosceles with legs $AC=BC=b$ and angle $\measuredangle ACB=\gamma$. Let $CD$ and $BM$ be altitudes that intersect at $H$. Find $MD$.

I have been struggling with this problem for an hour now. I really don't know how to start. I was thinking about using cosine or sine rule, but it seems useless at the end. For example
$$AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\gamma=2b^2-2b^2\cos\gamma$$

答: Hint: Show that quadrilateral MDBC is inscribed. Then, show that $\angle MAD=\angle AMD$ so $\triangle ADM$ is isosceles with $MD=AD$. Finally, you have to express $AD$ in terms of $b$ and $\gamma$.

Title: Cosine and sine rule on a triangle


问:
Understanding why this linear operator is diagonalizable.
let me show you the context of my question:

Let $T:V \to V$ a linear operator, where $V$ is a finite dimensional vector space and $m_t(x)=(x - \lambda_1)^{m_1} \cdots(x-\lambda_k)^{m_k}$.
In the proof of Primary decomposition Theorem we take $q_j=\displaystyle\prod_{{i=1}}^{k}(x- \lambda_i)^{m_i}$, with $j \neq i$ then since $gdc(q_1,q_2,\dots,q_k)=1$ there exists $f_1,f_2,\dots,f_k$ s.t $f_1q_1+f_2q_2+\dots +f_kq_k = 1$.

So, if we call $g_j= f_jq_j$ we have that the operator $P_j(T) = g_j(T)$ is a projection s.t

$P_jP_i = 0$ if $j \neq i$

$I = P_1+\cdots+P_k$

$Im Pj = \ker (T - \lambda_jId)^{m_j}= W_j$.

Using that we prove that $V = W_1 \oplus\cdots \oplus W_k$.

Ok, I have a simple question, in the condiction above, why $D = \lambda_1P_1 + \dots+ \lambda_kP_K$ is diagonalizable??

I know that for each $v \in V$, by 2. we have $v = P_1(v)+ \cdots + P_k(v)$ and by 1. we have $$D(v) = D(P_1(v)+ \cdots+ P_k(v))= (\lambda_1P_1(v)+ \cdots+ \lambda_kP_k(v))(P_1(v)+ \cdots+ P_k(v)) = \lambda_1P_1(v)+ \cdots+\lambda_kP_k(v).$$

I have no Idea what can I do with that, so can you help me?

答:
First of all, we have $D = \lambda_1P_1 + \dots+ \lambda_kP_k$, and multiply both sides by $I = P_1+\cdots+P_k$ gives
$$
DI = D = \lambda_1P_1^2 + \dots+ \lambda_kP_k^2
$$
as $P_iP_j = 0$ if $i \neq j$. This shows $P_i = P_i^2$ for all $i$. Furthermore, multiply both sides by $P_i$ gives
$$DP_i = \lambda_iP_i^2 = \lambda_iP_i$$
This says that every vectors that is in the image of $P_i$ is in $\ker(D - \lambda_i)$. Since $P_i \neq O$, the kernel is nonempty, meaning that $D$ does have an eigenvalue $\lambda_i$ for each $i$. In fact, $\{\lambda_i\}_{i=1}^k$ are all of the eigenvaules of $D$, as
$$
D - cI = (c-\lambda_1) P_1 + \cdots + (c-\lambda_k) P_k
$$
and if $(D - cI)x = 0$ for some nonzero $x$, then $(c-\lambda_i)E_ix = 0$ for all $i$, and since $P_jx$ must be nonzero for some $j$ (if it werent, then this means $0 \neq x = Ix = \sum_i P_ix = 0$, which is nonsense), we must set $(c - \lambda_j) = 0$, meaning $\{\lambda_i\}_{i=1}^k$ are all of the eigenvaules of $D$.

It then follows from that for every $x$ we have
$$
x = Ix = P_1x+\cdots+P_kx
$$
and since $P_ix$ are eigenvectors for any $i$, every vector can be written as a linear combination of eigenvectors, meaning $D$ is diagonizable.


问:
A problem on square matrices

If $B,C$ are $n$ rowed square matrices and if $A=B+C, BC=CB, C^2=O$, then show that for every $n \in \mathbb N$,
 $$A^{n+1}=B^n(B+(n+1)C)$$

I tried to prove it using mathematical induction. But I could not get $P(1)$ to be true.
$$P(1): A^2 = B(B+2C)$$
I couldn't equate them.

Please offer some assistance.
Thank you. :)

答:
Note that this is just the Binomial Theorem truncated to the first two terms, i.e.

$$ (B\!+\!C)^{n}\! = B^{n}\! + n B^{n-1} C + C^2(\cdots)\, \equiv\, B^n + n B^{n-1} C\ \pmod{C^2}$$

It is true in any ring where $\,\color{#c00}{BC = CB}.\,$ An inductive proof is easy. The base case $\,n= 0\,$ is true, being $\, 1 = 1.\,$ Suppose as inductive hypothesis that it is true for $\, n = k.\,$ The inductive step is

$$\begin{eqnarray}(B+C)^{k+1} &=\,& (B+C)^{k}(B+C)\\ &=\,& (B^k+kB^{k-1}C)(B+C)\quad {\rm by\ the\ induction\ hypothesis} \\ &=\,& B^{k+1}+ B^k C + kB^{k-1}\color{#c00}{CB} + kB^{k-1}C^2 \\  &=\,& B^{k+1}+(k\!+\!1) B^k C\ \ \ {\rm by}\ \ \color{#c00}{CB = BC},\ \  C^2\equiv 0\end{eqnarray}$$

Title: A problem on square matrices


问:
distribution of linearly transformed variable
Let $X$ be a random variable and it takes values from $\ell_{2}$, which is the space of square-summable sequences. Next, let us define random variable $U=a(X - b)$, for some $a \in \mathbb{R}_{+}$ and $b \in \ell_{2}$, also both $a$ and $b$ are not all zeros.

Next, assume that for some random variables $V$ and $Y$, taking values from $\ell_{2}$, defined on, possibly, another probability space we know that $V = a(Y - b)$ and $V\stackrel{d}{=}U$.

Does this mean that

$$
Y\stackrel{d}{=}X
$$
?

Does the result depend on the space which the variables take values from?

答:
Let $X=(X_1,\ldots,X_n,\ldots)$, where $X_i$ are scalar random variables, similarly for $U$, $V$, $Y$ and $a$, $b$.

Then it follows immediately with

$$P(Y_1\le y_1, \ldots, Y_n \le y_n,\ldots)$$
$$=P(\frac 1a V_1 + b_1 \le y_1,\ldots,\frac 1a V_n + b_n \le y_n,\ldots)$$
$$=P(V_1 \le a(y_1-b_1),\ldots, V_n \le a(y_n-b_n),\ldots)$$
$$=P(U_1 \le a(y_1-b_1),\ldots, U_n \le a(y_n-b_n),\ldots)$$
$$=P(\frac 1a U_1 + b_1 \le y_1,\ldots,\frac 1a U_n + b_n \le y_n,\ldots)$$
$$=P(X_1\le y_1, \ldots, X_n \le y_n,\ldots)$$

Hence the distribution of $X$ and $Y$ is the same.


问:
How to find the length of a segment which connects two triangles in a rhomboid?
The problem is as follows:

The alternatives in my book are as follows:

$\begin{array}{ll}
1.&\textrm{1 cm}\\
2.&\textrm{3 cm}\\
3.&\textrm{5 cm}\\
4.&\textrm{6 cm}\\
\end{array}$

I was only able to spot on:

$\triangle APT$ then $AT= 5$

Then:

$CT=4\cdot5=20$

But that's it. What else to do from here?.

What I attempted to find is some way to get:

$\frac{TO}{TC}=\frac{OR}{QC}$

However none of those segments seem to be known. How can this be solved relying only in euclidean geometry postulates?.

If possible please include a drawing in the answer so I can spot if a construction is needed.

答:
Hints:

$\triangle APT$ is similar to both $\triangle CQT$ and $\triangle ORT$

$O$ is the midpoint of $AC$

Try to find the lengths of $AC$ and thus $CO$ and $OT$

Then find the lengths of $CQ$ and thus $OR$


问:
Invertibility of Matrix with Number Theory
I'm studying for an exam on abstract algebra, and I found a activity which asked you to discuss if the following matrix is invertible:
\begin{equation}
\begin{pmatrix}
54401 & 65432 & 45530 & 45678\\
34567 & 12121 & 11111 & 12345\\
12345 & 76543 & 98760 & 65456\\
43211 & 45678 & 88888 & 98765
\end{pmatrix}
\end{equation}

I can detect certain partterns in the matrix. However, this question was included in the section of the Integer's Ring and as a hint the activity proposes to do a reduction modulo certain integer. First of all, I can't see a direct relation between singular matrices and $\mathbb Z/(n)$ (maybe for finding linear combinations(?), but I'm not sure). I would thank any hint or possible approach to the activity.

答: Hint: Consider $A$ and $\det(A)$ modulo $2$. It is easy to see that $\det(A)\equiv 1 \bmod 2$. In particular, $\det(A)\neq 0$.


问:
Marginal probability density function for each $X_j$
Let $X_1, \ldots , X_n$ i.i.d with $X_1\sim \text{Exp}(5\theta)$. For the statistical distribution of $\theta$ the a priori belief is described by $\text{Gamma}(\alpha,1)$, where $a$ a positive parameter.

Find the Marginal probability density function for each $X_j$.

Find the estimator for $\theta$ according to Bayes considering the square loss function.

For the first question we have to calculate the integral $$f_{X_j}(x_j)=\int_0^{\infty}f_{x_1\ldots x_n}(x_1, \ldots , x_n)\, dx_1, \ldots dx_{j-1} dx_{j+1}\ldots dx_n$$ or not?

So first we have to find the function $f_{x_1\ldots x_n}(x_1, \ldots , x_n)$.

Knowing that $X_1\sim \text{Exp}(5\theta)$ does the following hold?
$$f_{X_1\mid \Theta }(x\mid \theta)=\begin{cases}5\theta e^{-5\theta x_1} & \text{ for } x> 0 \\ 0 & \text{ for } x\leq 0\end{cases}$$

答:
The MSE community suggests me to avoid long conversation in comments.

For the fisrt question below is a integral using gamma function,
$$f_X(x)=\int_0^{+\infty}\prod_{i=1}^{n}(5\theta e^{-5\theta x_i}) \frac{1} {\Gamma(\alpha)}\theta^{\alpha-1}e^{-\theta}d\theta=\frac{5^n}{\Gamma{(\alpha)}}(1+5\sum_{i=1}^{n}x_i)\Gamma(n+\alpha)$$
Foe the second, Bayes estimation will be
$$\theta=\mathbb{E}[\theta|X=x]=\frac{\int\theta f_\Theta(\theta)f_{X|\Theta}(x|\theta)d\theta}{\int f_\Theta(\theta)f_{X|\Theta}(x|\theta)d\theta}=(1+5\sum_{i=1}^{n}x_i)(n+1+\alpha)$$

For the UMVUE，a common method is to use Lehmann–Scheffé theorem, to check the distribution family is complete and from the fisrt question $S=\sum_{i=1}^{n}X_i$ is a complete statistic. And find a statistic $\Theta$ with $\theta$ expectation and then calculate $\mathbb{E}(\Theta|X)$


答:
In a similar way, we can represent the real number $-1$ by the matrix: $\begin{bmatrix}-1 &0\\0&-1\end{bmatrix}$ which can also be thought of as rotation by $180^\circ$.  Is it rotation clockwise or anticlockwise?  Well for $180^\circ$ it doesn't matter.

However, $i$ represents rotation by $90^\circ$ and clockwise and anticlockwise make a difference.  Which is right?  It doesn't matter, just make a choice and stick to it.

Compare driving on the road.  Should we drive on the left or the right?  Clearly both work as some countries choose one and some choose the other.  However, it is important for a country to make a choice and be consistent.  Also, when visiting a country with the opposite convention, care is required to adjust.  The UK and France have similar road rules except that left and right are flipped.  For example, when joining a roundabout in the UK you give way (yield) to traffic already on the roundabout, this traffic is approaching from your right.  In France, you also give way to traffic already on the roundabout but it is approaching from your left.

Please note that this is only a brief description of the differences between driving in the UK and France.  If you actually plan to drive in France, please research the rules more thoroughly.  Swapping left and right is just a first approximation.

Title: non unique imaginary units

Body:
A professor told me that another way to define complex numbers is through the real matrices of the form  $\begin{bmatrix}a & b\\-b &a\end{bmatrix}$. With this definition, the imaginary unit becomes the matrix $I=\begin{bmatrix}0 &1\\-1&0\end{bmatrix}$. By definition $i^2=-1$ and It's pretty easy to check that the square of the matrix $I$ is the negative of the identity matrix.

So apparently, any matrix with square equal to the negative of the identity can be taken as $I$. Why is that? When you think about the $2D$ plane, this corresponds to choosing a vector different from (1,0) as $\bar i$, the unit vector in the x-direction.

Then he says that by some general theorem all complex structures in dimension 2 are isomorphic to each other and that one can allow the imaginary unit to change from point to point.

Does anyone understand what this means and why we can do this?


问:
Statistical probability of being sick
You have a machine that has a $0.9$ probability to give the correct diagnosis of a flu disease (similar for both the negative and positive diagnoses).
The probability of having the disease is $0.01$ for the general public. You know that a person sampled from the population was diagnosed as positive for the flu by this medical device.
Given this observation, what is the probability of that person having the disease?

I have no idea how to solve this, I think the population follows a normal distribution, but I can't think of anything. I would like you to explain to me what I should do. Thank you!

答:
Hint:

Suppose you had $1000$ people.

How many would you expect to have the disease?

How many of those would you expect to test positive?

How many would you expect to not have the disease?

How many of those would you expect to test positive?

So how many in total would you expect to test positive?

What proportion of those testing positive would you expect to have the disease?

Title: Statistical probability of being sick


问:
Minimisation problem on cuboid
A problem asks to find the shortest path on a cuboid with volume $1$ connecting two opposite corners. How does one solve this? I think I have to minimise $(a+b)^2+c^2$ with $abc = 1$  but I have no idea how to continue from this.

答:
Using the AM-GM inequality, we have
$$(a+b)^2 \geqslant 4ab = \frac{4}{c},$$
so
$$(a+b)^2+c^2 \geqslant \frac{4}{c} + c^2 = \frac{2}{c} + \frac{2}{c} + c^2\geqslant 3\sqrt[3]{\left(\frac{2}{c}\right)^2 \cdot c^2} = 3\sqrt[3]{4}.$$
Equality hold for $a = b = \frac{1}{\sqrt[6]{2}}, \, c = \sqrt[3]{2}.$

Therefore, the minimum value is $3\sqrt[3]{4}.$

Title: Minimisation problem on cuboid


问:
How to fit a 2D ellipse to given points and a given focus
This question is similar to the question, How to fit a 2D ellipse (in x-y plane) to given points? (See the link below)
(https://stackoverflow.com/questions/47873759/how-to-fit-a-2d-ellipse-to-given-points)

Now, we know how to use least square method to fit a 2D ellipse with given points by the code provided by Casey. (The code is also provided below.) Based on this code, if I want to fit not only the given points but also a given focus at (0,0), how could I do it? Or is there any better way to do it?

I was thinking about if we can derive the focus (as a function of x and y) based on the ellipse equation,$  Ax^2 + Bxy + Cy^2 + Dx + Ey=1 $, where A, B, C, D, E are coefficients, and use it as a constrain to the least square method. Buts sadly, I don't know how to derive the focus either.

CODE

```<code>import numpy as np
import matplotlib.pyplot as plt
alpha = 5
beta = 3
N = 500
DIM = 2

np.random.seed(2)

# Generate random points on the unit circle by sampling uniform angles
theta = np.random.uniform(0, 2*np.pi, (N,1))
eps_noise = 0.2 * np.random.normal(size=[N,1])
circle = np.hstack([np.cos(theta), np.sin(theta)])

# Stretch and rotate circle to an ellipse with random linear tranformation
B = np.random.randint(-3, 3, (DIM, DIM))
noisy_ellipse = circle.dot(B) + eps_noise

# Extract x coords and y coords of the ellipse as column vectors
X = noisy_ellipse[:,0:1]
Y = noisy_ellipse[:,1:]

# Formulate and solve the least squares problem ||Ax - b ||^2
A = np.hstack([X**2, X * Y, Y**2, X, Y])
b = np.ones_like(X)
x = np.linalg.lstsq(A, b)[0].squeeze()

# Print the equation of the ellipse in standard form
print('The ellipse is given by {0:.3}x^2 + {1:.3}xy+{2:.3}y^2+{3:.3}x+{4:.3}y = 1'.format(x[0], x[1],x[2],x[3],x[4]))

# Plot the noisy data
plt.scatter(X, Y, label='Data Points')

# Plot the original ellipse from which the data was generated
phi = np.linspace(0, 2*np.pi, 1000).reshape((1000,1))
c = np.hstack([np.cos(phi), np.sin(phi)])
ground_truth_ellipse = c.dot(B)
plt.plot(ground_truth_ellipse[:,0], ground_truth_ellipse[:,1], 'k--', label='Generating Ellipse')

# Plot the least squares ellipse
x_coord = np.linspace(-5,5,300)
y_coord = np.linspace(-5,5,300)
X_coord, Y_coord = np.meshgrid(x_coord, y_coord)
Z_coord = x[0] * X_coord ** 2 + x[1] * X_coord * Y_coord + x[2] * Y_coord**2 + x[3] * X_coord + x[4] * Y_coord
plt.contour(X_coord, Y_coord, Z_coord, levels=[1], colors=('r'), linewidths=2)

plt.legend()
plt.xlabel('X')
plt.ylabel('Y')
plt.show()
</code>```

答:
For a ellipse with a focus at $(0,0)$, the ellipse's equation in polar coordinate can be written as the following:

Given $a$ = semi-major axis, $e$ = eccentricity, the other focus at the angle $\phi$,

$r(\theta) = \frac{a(1-e^2)}{1-ecos(\theta-\phi)} $.

Now, we can use the data points $(r_i, \theta_i)$ to fit the equation above by
non-linear least square method (ex: Levenberg–Marquardt algorithm).

i.e. Find the minimum for the following equation:

$\sum_{i=1}^n(\frac{a(1-e^2)}{1-ecos(\theta_i-\phi)}-r_i)^2$


问:
How to "invent" a function?
I'm currently making a game. I was given some numbers the leader wants me to stick close to. They are used for leveling; level 1, for example, needs 100 experience points, level 2 needs 250 experience points, etc.

Is there any way, maybe with a computer program, to calculate a function that will do something like this, instead of having to hard-code it?

Thanks in advance!

Simon.

答:
There are plenty of ways; it all depends on what kind of function you want.

One way is to use Lagrange interpolation to get a polynomial that will have the values you want (e.g., $f(1)=100$, $f(2)=250$, etc). This, however, may not give you the "correct" long-term behavior, and the long-term growth is not going to be exponential. You will "hit" all the right values for the points you are given, but you may not have good control later.

(For example, if you use Lagrange polynomial interpolation to find a function that satisfies
$$f(0)=1,\quad f(1)=2,\quad f(2)=4,\quad f(3)=8,\quad\text{and}\quad f(4)=16,$$
then this function will give $f(5)=31$ (rather than the perhaps expected $32$).

A more likely long-term behavior will be given by exponential functions. If the rate of growth is relatively constant, then that's the way to go. Consider the sequence of quotients of successive values:
$$\frac{f(2)}{f(1)},\quad\frac{f(3)}{f(2)},\quad\frac{f(4)}{f(3)},\quad\frac{f(5)}{f(4)},\ldots$$
and if these are all reasonably close to one another, then take $a$ to be their average. You can approximate the values pretty well with
$$f(n) = f(1)a^{n-1}.$$
This is what Ross gives you, with $a=2.5$ and $f(1)=100$, except he's writing it as
$$f(n) = 100(2.5)^{n-1} = \left(\frac{100}{2.5}\right)2.5^n = 40\times 2.5^n.$$

If the rate of growth is (the successive factors) are changing a lot, though, it's going to be a lot harder to come up with a good exponential approximation.

Title: How to "invent" a function?


问:
Integration using summation
How do you integrate $\sqrt{x}$ from an arbitrary constant $a$ to another $b$ by summation ?</p>

答:
Since the function $x\mapsto\sqrt{x}$ is nondecreasing, every Riemann sum associated to the subdivision $s=(x_i)_{0\leqslant i\leqslant n}$ is between $R^+(s)$ and $R^-(s)$ with
$$
R^-(s)=\sum_{i=1}^n(x_i-x_{i-1})\sqrt{x_{i-1}},\qquad R^+(s)=\sum_{i=1}^n(x_i-x_{i-1})\sqrt{x_{i}}.
$$
If the mesh of $s$ is $\delta(s)$,
$$
R^+(s)-R^-(s)=\sum_{i=1}^n(x_i-x_{i-1})(\sqrt{x_{i}}-\sqrt{x_{i-1}})\leqslant\delta(s)\sum_{i=1}^n\sqrt{x_{i}}-\sqrt{x_{i-1}}.
$$
Thus, $R^+(s)-R^-(s)\leqslant\delta(s)(\sqrt{b}-\sqrt{a})\to0$ when $\delta(s)\to0$, that is, the function is Riemann integrable. Furthermore, for every $i$,
$$
(x_i-x_{i-1})\sqrt{x_{i-1}}\leqslant\tfrac23(x_i\sqrt{x_i}-x_{i-1}\sqrt{x_{i-1}})\leqslant(x_i-x_{i-1})\sqrt{x_{i}},\qquad (\ast)
$$
hence
$$
R^-(s)\leqslant\sum_{i=1}^n\tfrac23(x_i\sqrt{x_i}-x_{i-1}\sqrt{x_{i-1}})\leqslant R^+(s),
$$
and the sum in the middle is telescoping hence all this shows that
$$
\int_a^b\sqrt{x}\,\mathrm dx=\tfrac23(b\sqrt{b}-a\sqrt{a}).
$$
Finally, the result hinges on the algebraic inequalities $(\ast)$, which we now prove.

Dividing $(\ast)$ by $\frac13(\sqrt{x_i}-\sqrt{x_{i-1}})\gt0$ and renaming $\sqrt{x_{i-1}}=x$ and $\sqrt{x_i}=y$, one must show that, for all $0\leqslant x\lt y$,
$$
3x(y+x)\leqslant2(y^2+yx+x^2)\leqslant3y(y+x),
$$
or,
$$
-(y-x)(2y+x)\leqslant0\leqslant(y-x)(y+2x).
$$
Since $y-x\gt0$, $2y+x\gt0$ and $y+2x\gt0$, this proves $(\ast)$.

Title: Integration using summation


问:
total variation, signed measure, inequality
Suppose that $\mu$ is a signed measure on a $\sigma$-algerba of the subsets of X. Show that:

$\frac{1}{2}\lvert{\mu}\rvert(X) \leq \sup[{\vert{\mu(E)}\rvert: E \in \sigma-algebra}]$

also 1/2 is the biggest constant that the ineqality holds.

I'd used the defenition of total varaition but I don't know where this constant comes from.The definition of total varaition is about the $\Sigma\lvert{\mu(E)}\rvert$,but how this can be related to $\sup[\vert{\mu(E)}\rvert: E \in \sigma-algebra]$.

any help would be greatly appreciated.

答:
By Hahn Decomposition Theorem we can write $\mu (E) =\mu^{+}(E)-\mu^{-}(E)=\mu (E \cap A)+\mu (E\cap A^{c})$ and $|\mu| (E) =\mu^{+}(E)-\mu^{-}(E)=\mu (E \cap A)-\mu (E\cap A^{c})$ for some measurable set $A$. Hence, $|\mu|(X) \leq |\mu(X\cap A)|+|\mu(X\cap A^{c})|\leq 2 \sup\{|\mu (F)|: F \in \mathcal F\}$.

Take $\mu (E)=\lambda (E \cap (-1,0))-\lambda (E \cap (0,1))$ for subsets of $(-1,1)$ to get an example where equality holds. Here $\lambda$ is the Lebesgue measure.

See https://en.wikipedia.org/wiki/Hahn_decomposition_theorem
for Hahn Decomposition Theorem.


答: $$\therefore MD = AD = b\sin \gamma$$

Title: Cosine and sine rule on a triangle

Body:
$\triangle ABC$ is isosceles with legs $AC=BC=b$ and angle $\measuredangle ACB=\gamma$. Let $CD$ and $BM$ be altitudes that intersect at $H$. Find $MD$.

I have been struggling with this problem for an hour now. I really don't know how to start. I was thinking about using cosine or sine rule, but it seems useless at the end. For example
$$AB^2=AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\gamma=2b^2-2b^2\cos\gamma$$

Body: $$\therefore MD = AD = b\sin \gamma$$


问:
$SL(2)/U(2)$ is isomorphic to $\mathbb{C}^2 \setminus \left\{ 0 \right\}$?
Let  $G= SL(2)$ and $H=U(2)$ the set of $2x2$ matrices of type $\begin{pmatrix} 1 & * \\ 0 & 1 \end{pmatrix}$. By Chevalley Theorem $G/H$ can be immersed in $\mathbb{P}(\mathbb{C}^2 \oplus \mathbb{C})$ and can be described as the set of $g \cdot (\mathbb{C}e_1 \oplus \mathbb{C})$ with $g \in G$. Why is $G/H$ isomorphic to $\mathbb{C}^2 \setminus \left\{ 0 \right\}$? They are both quasiprojective varieties, so I should find an open covering of affine sets $U_i \subset SL(2)$ such that if $\phi : G/H \to \mathbb{C}^2 \setminus \left\{ 0 \right\}$, $\phi|_{U_i}$ is an isomorphism of affine varieties (therefore regular and invertible). Any hints?

答:
Since Sasha probably won't be more specific I'll add some more details to the answer. Although the content is virtually the same the wording might be a bit more psychologically comforting.

There is a natural action of $G = \mathrm{GL}_2(\mathbb{C})$ on $\mathbb{C}^2 \setminus \{0\}$.

Exercise. This action of $G$ on $\mathbb{C}^2 \setminus \{0\}$ is transitive, meaning that, if you pick two points $x,y \in \mathbb{C}^2 \setminus \{0\}$, you can always find a $g \in G$ such that $gx = y$.

Now consider the point $x = (1, 0)$. By the orbit--stabilizer theorem, there should be an isomorphism between $G / G_x$ (the quotient of $G$ by the stabilizer at $x$) and $Gx$ (the orbit of $x$).

Exercise. In case you've only seen the orbit--stabilizer theorem for discrete groups, prove that it also works in the expected way for Lie groups acting on topological spaces.

Exercise. Prove that the stabilizer of the point $x = (1, 0)$ is isomorphic to $H$. Also prove that the orbit $Gx$ is all of $\mathbb{C}^2 \setminus \{0\}$. Conclude the desired result.


问:
Transforming of a nilpotent matrix to a specific form
I wish to build a question in the field of discrete state spaces representation (control theory). The canonical form has a very unique, but not singular, representation. I am focusing on the $A$ matrix of the representation of the form:

$$\begin{bmatrix}
-a_1   & -a_2   & \cdots & -a_{N-1}& -a_N\\
1      & 0      & \cdots & 0       & 0   \\
0      & 1      & \cdots & \cdots  & 0   \\
\vdots & \vdots & \vdots & \ddots  & \vdots   \\
0      & 0      & \cdots & 1       &0   \\
\end{bmatrix}$$

That is a subdiagonal matrix + a matrix of the same size where only the first row has specific values.

Other representations can be generated using a transformation matrix $T$ by $T^{-1}AT$. In my case, $A\in3\times3$.

I want to use a representation with a nilpotentic matrix $B$. Given, $A$ and $B$ which answer these definitions, is there a way to calculate $T$ which transition between them so that  $B=T^{-1}AT$? I do not mind the values of $a_1,\ldots,a_N$ being dictated by $B$. All I care about is the form of the resulting $A$.

For example:

$$B=\begin{bmatrix}
2 & 2 & -2\\
5 & 1 & -3\\
1 & 5 & -3\\
\end{bmatrix},
A=\begin{bmatrix}
-a_1 & -a_2 & -a_3\\
1 & 0 & 0\\
0 & 1 & 0\\
\end{bmatrix}$$

Is it even possible to transform between the two or am I missing a limitation here? If it is possible, is there a close solution for $T$?

答:
It is impossible for any regular matrix $A$. Proof:

Let's suppose that there is a regular matrix $T$ for such $B = T^{-1}AT$ is nilpotent. Therefore there exists $n$ for such $B^n=0$.
$$
0 = B^n=(T^{-1}AT)^n=T^{-1}A^nT\\
T0T^{-1}=A^n\\
0=A^n
$$
However, $A$ is regular, so $A^n\neq0$. This is a contradiction.


问:
Picture of l'Hopital's rule at infinity
I know that when L'Hopital's rule is taken for the $\frac{0}{0}$ case, we get that the functions start looking at lines, like in this figure from Stewart's Calculus:

But what is the visual when the limit is the $\frac{\infty}{\infty}$ case?

The only thing that comes to mind that is easily visualized is a rational function's asymptotes (and that comes from dividing numerator and denominator by the largest exponent in the denominator)

答: L'Hospital is based on the fact that a continous function can be locally approximated by its tangent. If you want to evaluate the limit at infinity, for example $\lim \limits_{x \to \infty} \frac{\sqrt{x}}{\ln{x}}$, the tangents would also lie there so there is no visual for that.

Title: Picture of l'Hopital's rule at infinity


问:
Conjugacy classes when changing group
Let $A \leqslant B \leqslant G$ subgroups, with $[B:A]<\infty$. I would like to understand conjugacy classes when the subgroup with respect to which we consider them changes. Denote $C(a)\in CL(A)$ the conjugacy classes in $A$ and $C(b)\in CL(B)$ the conjugacy classes in $B$ (I hope no confusion will arise). I would like to understand what $A$-classes give rise to a single $B$-class $C(b)$, i.e.
$$\{C(a) \in CL(a) \ : \ \exists x \in B, x^{-1}ax = b\}$$

This implies $x b x^{-1} = a \in A$. This last condition is obviously invariant by left-translation of $x$ by $A$, so that we can afford to be looking at $x \in A/B$. Is the above set in bijection with
$$\{x \in B/A \ : \ x^{-1}bx \in A\}?$$

I can prove it painfully (looking at the natural map $C(a) \mapsto x$ and checking it is injective and surjective. Is there a more elegant way to prove it?

答:
Not exactly clear what you are looking for but perhaps this helps.

We have $A
\leq B$. Fix an element $b \in B$, and put $T_b=\{ x \in B: xbx^{-1} \in A\}$. This is a subset of $B$ and if it is non-empty we can define an equivalence relation on this set by $x \sim  y$ if and only if $x=yc$ for some $c \in C_B(b)$. (Here $C_B(b)$ denotes the centralizer of $b$, that is, all elements of $B$ that commute with $b$. This is a subgroup by the way). Note that $yc \in T_b$ and indeed $\sim$ is an equivalence relation.

This implies that we can write $T_b$ as a disjoint union as follows: $\bigcup_{i=1}^kx_iC_B(b)$ for some positive integer $k$ and the $x_i \in T_b$ are representatives of the different $k$ equivalence classes.
Denote $Cl_B(b)$ as the conjugacy class of $b$ in $B$. Now we can define a map $f: \{x_1, \cdots, x_k\} \rightarrow Cl_B(b) \cap A$ by $f(x_i)=x_ibx_i^{-1}$. This map is well-defined ($x_i \in T_b$ for all $i$) and it is a bijection.
Injectivity: if $x_ibx_i^{-1}=x_jbx_j^{-1}$, then $x_j^{-1}x_i \in C_B(b)$, whence $x_i \sim x_j$, so $i=j$.
Surjectivity: an element of $C_B(b) \cap A$ is of the form $xbx^{-1} \in A$ for some $x \in B$, so $x \in T_b$, whence $x=x_ic$ for some $i$ and $c \in C_B(b)$. So $f(x_i)=x_ibx_i^{-1}=x_icbc^{-1}x_i^{-1}=xbx^{-1}$.
It follows in particular that $Cl_B(b) \cap A=\emptyset$ or $\#(Cl_B(b) \cap A)=k$.

Title: Conjugacy classes when changing group


问:
Does a topology have a canonical topology?
Consider a topological space $(X,\tau)$. I would like to know whether the set $\tau$ of open sets can be given its own topology in a canonical way. In that respect I'm asking for the same kind of thing as this previous question. However I'd also like to specify a property that this canonical topology should have, which the other question doesn't do, so I hope this won't be considered a duplicate.

Specifically, let $(X,\tau)$ and $(Y,\sigma)$ be topological spaces, and consider a continuous function $f\colon X\to Y$. We can define the pullback map as a function $f^*\colon \sigma\to\tau$ that maps open subsets of $Y$ to their preimages under $f$, which are open subsets of $X$.

I would like to know whether $f^*$ can itself be seen as a continuous function. That would mean defining the open subsets of $\sigma$ and $\tau$. I'm wondering whether these can be defined in a canonical way, such that every continuous map between any pair of topological spaces has a continuous pullback.

Of course this would be the case if we just defined the topology on $\tau$ to be the discrete topology for every topological space $(X,\tau)$, but I'm wondering if it's possible to do it in a way where they can be coarser than that.

答:
There are many ways to topologize the set $\tau$. You can give it the trivial topology or the discrete toplogy, but surely that is not what you expect.

Another way to do it is to observe that $\tau$ is partially ordered by set inclusion. With respect to this partial order it is a lattice. Now have a look at  Compact subspaces of the Poset.


问:
in fundamental theorem of calculus, does the following reverse implication hold true?
In FTC, define $G(t)=\int_a^xg(t)\mathrm{d}t$ on $[a,b]$, if $g$ is continuous, is $G$ continuous? I was not able to come up with a way to prove this implication, any help would be appreciated. Also, does this imply $G$ is differentiable?

答: Not only that $G$ is continous, but also differentiable. When you use the FTC, you make the function "nicer" - so an integrable function become continous, and a continous function becomes differentiable.


问:
A question on the complex grassmann
The set of $k$-dimensional complex-linear subspaces of $\mathbb{C^n}$ is denoted by
$G_k(\mathbb{C}^n)$. Show that $G_k(\mathbb{C}^n)$ has a unique smooth manifold structure making
it into a homogeneous $GL(n,\mathbb{C})$-space, with the action of $GL(n,\mathbb{C})$
induced from its usual action on $\mathbb{C}^n$. Show that $G_k(\mathbb{C}^n)$ is compact, and
determine its dimension.

I already solved this for real numbers could anybody help me solve this for complex numbers?

答:
Define the homogeneous coordinate of $x\in G_k(\mathbb{C}^n)$ as

$$x=\left[\begin{array}{}
x_{11} & x_{12} & \cdots & x_{1n}\\
x_{21} & x_{22} & \cdots & x_{2n}\\
\cdots &        & \cdots &       \\
x_{k1} & x_{k2} & \cdots & x_{kn}
\end{array}\right]
$$
where the matrix is of full rank $k$.

Next, we define an equivalence relation. Two element $x,y\in G_k(\mathbb{C}^n)$ are said to be equivalent if there exists $M\in GL(k, \mathbb{C})$, such that $Mx=y$. The quotient space under this equivalence is clearly $G_k(\mathbb{C}^n)$, since each element consists of $k$ row vectors that together span a $k$-dimensional subspace, and two set of vectors represent the same subspace if and only if they linearly represent each other.

Clearly $G_k(\mathbb{C}^n$), given in this manner, is a $GL(n, \mathbb{C})$-space. Let $x, \in G_k(\mathbb{C}^n), N\in GL(n, \mathbb{C})$, define $N(x)=xN$, where the RHS is matrix multiplication. Since $N$ is nonsingular, $xN$ is of full rank, hence in $G_k(\mathbb{C}^n)$.

To show that $G_k(\mathbb{C}^n)$ is compact, let $S=\{m\in M(k\times n, \mathbb{C}): \text{the rows are orthogonal to each other and of length 1}\}$. $S$ is closed and bounded, hence compact. Observe that every matrix $m\in S$ is of full rank. Hence the quotient space of $S$ is $G_k(\mathbb{C}^n)$, which is compact.

Finally, the dimension of $G_k(\mathbb{C}^n)$ is $k(n-k)$, because the system of equations $Mx=y$ has $k^2$ independent equations. There is a trick to observe this:

Suppose $\Sigma_1,\Sigma_2$ are two $k$-dimensional differentiable manifolds, $p\in \Sigma_1,q\in \Sigma_2$, the rows of $x$ is a basis of $T_p\Sigma_1$ and the rows of $y$ is a basis of $T_q\Sigma_2$. Then $v\mapsto Mv$ is a map from $T_p\Sigma_1$ to $T_q\Sigma_2$, which has $k^2$ degrees of freedom.

From what I have achieved so far, it seems there's no big change if we subsitute $\mathbb{C}$ with $\mathbb{R}$.

Title: A question on the complex grassmann


问:
Number of orbits of an action
I've already showed that the map
$$\overbrace{\Bigg\{\begin{pmatrix}a &b\\-b & a\end{pmatrix}:a,b\in\mathbb Z_7\Bigg\}}^{:=G\le \operatorname{GL}_2(\mathbb Z_7)}\times\overbrace{\Bigg\{\begin{pmatrix}x\\y\end{pmatrix}:x,y\in\mathbb Z_7\Bigg\}}^{:=V}\to\Bigg\{\begin{pmatrix}x\\y\end{pmatrix}:x,y\in\mathbb Z_7\Bigg\}\text{ such that}$$
$$\begin{pmatrix}a &b\\-b & a\end{pmatrix}\cdot \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}ax+by\\-bx+ay\end{pmatrix} $$
is an action of $G$ on $V$. The order of $G$ is $49$ (so $7^2$ and I could use the relation $|Fix_G(V)|=|V|\text{ (mod 7)}$), which is also the order of the group $V$.

I think that in order to compute the number of orbits of the action I should look at the stabilizer of a certain element in $V$.

For example amatrix $g\in G$ stabilizes an element $\begin{pmatrix}x\\y\end{pmatrix}\iff$
$$g\cdot\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}x\\y\end{pmatrix}\implies\begin{cases} ax+by=x\text{ (mod 7)}\\-bx+ay=y\text{ (mod 7)}\end{cases}(?)$$
but it's not so easy to solve this system. Could you help me please?

Thank you for your attention.

答:
In these kind of cases, it might be helpful to consider a particular element (especially if we take an element with a $0$ entry, it is even easier) and find its stabilizer. For instance, here take $\begin{pmatrix}1\\0\end{pmatrix} \in V$. Then, we have

$$\ \ \ \ a \equiv 1\ (\text{mod } 7) \implies a = 1\\
-b \equiv 0\ (\text{mod } 7) \implies b = 0$$

So, stabilizer of $\begin{pmatrix}1\\0\end{pmatrix}$ consists of a single element, which is the identity of $G$. Thus, by Orbit Stabilizer Theorem, orbit of $\begin{pmatrix}1\\0\end{pmatrix}$ consists of $49 = |V|$ elements, which means there is only $1$ orbit.

Of course, you might think that we were lucky here and we got a stabilizer with $1$ element so obtained the result immediately. If we try to do it by using some Number Theory, one can multiply the first congruence by $y$ and the second one by $x$ to obtain
$$\ \ \ \ axy+by^2 \equiv xy\ (\text{mod } 7)\\
\ \ \ \ axy-bx^2 \equiv xy\  (\text{mod } 7)$$
which gives us
$$b(x^2+y^2) \equiv 0\ (\text{mod } 7)$$

and we should use Sum of two squares Theorem to obtain $b = 0$ (note that $x^2+y^2 = 7$ has no integer solutions since $7 \equiv 3\ (\text{mod }4)$). Then again, the result you mentioned in comments follows.

Title: Number of orbits of an action


答:
By the FTC , $G$ is differentiable on $[a,b]$ and $G'=g$ on $[a,b]$

Since $g$ is continuous, we have that $G \in C^1[a,b].$

Title: in fundamental theorem of calculus, does the following reverse implication hold true?

Body: In FTC, define $G(t)=\int_a^xg(t)\mathrm{d}t$ on $[a,b]$, if $g$ is continuous, is $G$ continuous? I was not able to come up with a way to prove this implication, any help would be appreciated. Also, does this imply $G$ is differentiable?


问:
Constructing a conformal map from $\mathbb{D}$ to a cut plane
Source: Oxford Exam $A2 \ 1999$

We want to construct a conformal map $F$ from the unit disc $\mathbb{D}=\{z:|z|<1\}$ to $\mathbb{C} \setminus S$ where $S$ is the half-line $\{x+i:x \in (-\infty,0] \}$ with the additional property that $F(0)=0$.

Here are my thoughts, can someone see if they are correct please:

Use the Möbius transformation $f_1:z \mapsto \frac{1-z}{z+1}$ to send $\mathbb{D}$ to the right half plane.

Use the map $f_2:z \mapsto z+ \alpha$ where $\alpha \in \mathbb{C}$ we will determine later. This simply takes the right half plane to the right half plane.

Use the map $f_3:z \mapsto z^2$ to map the right half plane to $\mathbb{C}$ with a cut along the negative reals.

Use the map $f_4:z \mapsto z+i$ to move the cut up and map to $\mathbb{C} \setminus S$.

Now the composition of these four maps is conformal so we certainly have a conformal map between the sets.

We now need $F(0)=f_4 \circ f_3 \circ f_2 \circ f_1(0)=0$.

$f_1(0)=1$, $f_2(1)=1+\alpha$, $f_3(1+\alpha)=\alpha^2+2\alpha+1$, $f_4(\alpha^2+2\alpha+1)=\alpha^2+2\alpha+1+i$

So we need to solve $\alpha^2+2\alpha+1+i=0$ and this is the $\alpha$ required to ensure $F(0)=0$.

We can find that $\alpha=\sqrt{-i}-1$.

Thus we are done.

Is this correct?

答:
The map $f_4\circ f_3$ as you constructed it will conformally map the right half plane to the desired $\mathbb C\setminus S$. So it should be possible to express any vaid solution as a combination of a conformal map from the unit disk to the right half plane, concatenated with that function. So every valid $F$ can be written as $F=f_4\circ f_3\circ f_{12}$ for some conformal map $f_{12}: \mathbb D\to\{z\in\mathbb C:\operatorname{Re}(z)>0\}$.

So what ways are there to conformally map the unit disk to the right half plane? According to Riemann's mapping theorem, that map is unique up to Möbius transformation. Since it is possible to perform $f_{12}$ using a Möbius transformation, this tells us that it is enough to consider only Möbius transformations in this step.

A Möbius transformation on $\mathbb C\cup\{\infty\}$ is uniquely defined given three distinct points and their images. Since we know that the boundary of the unit disk must map to $S$, we can choose two points on the unit disk as the preimages of $0$ and $\infty$. Without loss of generality, we can even rotate the unit disk such that one of these points will be the point $1$. We can furthermore require the point $0$ to map to $\sqrt{-i}=\frac{1-i}{\sqrt2}$, so that $f_4\circ f_3$ will take it from there back to $0$ as required. All together, our Möbius transformation $M$ (which will play the role of $f_{12}$) is uniquely defined by

\begin{align*}
0 &\mapsto \sqrt{-i}=\tfrac{1-i}{\sqrt2}=e^{-i\frac\pi4} \\
1 &\mapsto 0 \\
e^{i\varphi} &\mapsto \infty
\end{align*}

for some real parameter $\varphi$. Plugging this into a computer algebra system (sage in my case), you can obtain

$$ M: z\mapsto\frac
{e^{i\left(\varphi-\frac\pi4\right)}\cdot (z-1)}
{z - e^{i\varphi}} $$

Now plug an arbitrary point from the boundary of $\mathbb D$ into that function. It should end up on the imaginary axis, the boundary of your half plane. So its real part should be zero. Again with help from computer algebra, I got

$$\operatorname{Re}\bigl(M(-1)\bigr) =
\frac{\sin\varphi + \cos\varphi + 1}{\sqrt2\,\left(\cos\varphi+1\right)}$$

Note: The choice of the example point here is pretty much arbitrary, but not completely so. In my first attempt, I plugged $-i$ instead of $-1$ into this, which is a special case for which this approach will not work. The reason is that later on we will see that $-i$ is the point which gets mapped to $\infty$, and which therefore has no well-defined real part. If in doubt, one should probably perform this computation with two different points.

In any case, requiring that real part to be zero, one can obtain the solution $\varphi=-\frac\pi2$. With this, the Möbius transformation is fully defined:

$$ M: z\mapsto\frac
{e^{-i\frac34\pi}\cdot (z-1)}
{z - e^{-i\frac12\pi}} =
\frac{(1+i)(1-z)}{\sqrt2(z+i)} $$

This transformation will map the boundary of the unit circle onto the imaginary axis, since it does so for three points and a circle (including lines as a special case) is already uniquely defined by three points. So this will map your unit disk to the right half plane, and put $M(0)$ in the right place.
Now you can combine this with $f_3$ and $f_4$ to obtain the complete map $F$:

$$F = f_4\circ f_3\circ M:
z\mapsto
\frac{2 i \, z^{2} - \left(2 i + 2\right) \, z}{z^{2} + 2 i \, z - 1}$$


问:
Equivalence of minimization problems for Lipschitz function

Assume that $f: \mathbb{R}^n \rightarrow \mathbb{R} $ is Lipschitz continuous with constant $M$ and that $S \subseteq \mathbb{R}^n$  is closed. Show that for $\lambda > M$ problem of minimizing $f(x) + \lambda d(x, S)$, where $d(x,S)$ is a distance from $x$ to $S$ have the same solutions as problem of minimizing $f(x)$ over $S$.

I know that because $S$ is closed, then there exists $b \in S$ such that $d(x, S) = d(x, b)$,
Because I now have some $b$ I can use Lipschitz condition to get $d(f(x), f(b)) \leq M d(x, b)$. But I don't see how to proceed further or if this actually gives something useful.

Edit: I have tried the following, but I got stuck

Let $x_S$ be the solution to the problem of minimizing $f(x)$ over $S$. Because $x_S \in S$ we have $d(x_S, S) = 0$.

Assume $\hat x \in S$ is a solution to the problem of minimizing $f(x) + \lambda d(x, S)$. Because $\hat x \in S$ we have $d(\hat x, S) = 0$. Because it is a solution to the problem of minimizing $f(x) + \lambda d(x, S)$ we have that $f(\hat x) < f(x_S)$. But both points are in $S$ and $x_S$ minimizes $f(x)$ over $S$, so we have contradiction.

Assume $\hat x \notin S$ is a solution to the problem of minimizing $f(x) + \lambda d(x, S)$. Then because $S$ is closed we have $d(\hat x, S) > 0$ (If $S$ wasn't closed then points from boundary of $S$ wouldn't be in $S$ and their distance to $S$ would be $0$). So we have $$f(\hat x) + \lambda d(\hat x, S) < f(x_S) + \lambda d(x_S, S) = f(x_S)$$
$$f(\hat x) - f(x_S) < - \lambda d(\hat x, S)$$
Both sides are negative, so by taking abolute value on both sides we get
$$|f(\hat x) - f(x_S)| > \lambda d(\hat x, S) > M d(\hat x, S)$$
Basically I wanted to get a contradiction with Lipschitz condition but for that I would need something like $d(\hat x, S) = d(\hat x, x_S)$ which I am not sure I can get. Buth this is what I have tried.

Edit 2:

Now I understand that I can frame minimization of $f(x)$ over $S$ as
$$\text{minimize } f(x)\\ \text{such that }d(x,S) = 0$$

Then $$\text{minimize} f(x) + \lambda d(x, S)$$ looks like a Lagrangian. Basically I think this changes the question to question about strong duality of this problem. Maybe.

答:
Lipschitz property of $f(\cdot)$ implies that

$$\| f(\mathbf{y})-f(\mathbf{x}^*)\|\leq M\| \mathbf{y}-\mathbf{x}^*\| <\lambda\| \mathbf{y}-\mathbf{x}^*\| $$

Now, let $\mathbf{x}^*$ be the optimal solution and assume $\mathbf{x}^*\notin S$. From the optimality of $\mathbf{x}^*$ we know that:
$$ f(\mathbf{x}^*)+\lambda \overbrace{\|\mathbf{y}-\mathbf{x}^*\|}^{d_s(\mathbf{x}^*)}\leq f(\mathbf{x}^s)$$
where $\mathbf{x}^s\in S$ is an optimal solution if we would restrict the domain to $S$ and $\mathbf{y}\in S$. Rearranging the terms and applying a norm on the inequality we get \begin{aligned}
&\lambda\|\mathbf{y}-\mathbf{x}^*\|\leq f(\mathbf{x}^s)-f(\mathbf{x}^*)\leq f(\mathbf{y})-f(\mathbf{x}^*) \Longrightarrow \\ &\lambda\|\mathbf{y}-\mathbf{x}^*\|\leq \| f(\mathbf{y})-f(\mathbf{x}^*)\|\end{aligned}
where we know from the optimality of $\mathbf{x}^s$ over $S$ that $f(\mathbf{x}^s)\leq f(\mathbf{y})$. However this is contradiction to the Lipschitz property. Therefore the optimal solution must be in $S$, and the problems $\min_{\mathbf{x}\in \mathbb{R}^n} \{f(\mathbf{x})+\lambda d_s(\mathbf{x})\}, \; \min_{\mathbf{x}\in s} \{f(\mathbf{x})\}$ are equivalent.


问:
Expected value of occurrences of a variable under a Poisson distribution
I asked a question on this forum recently, to which I got a very thorough answer that I didn't properly understand. Link to the original question here.

The answer stated:

"If we have N buckets and place M entries in them, the number of elements in each buckets, for large $N, m$, approximately follows ("Poissonization") a Poisson distribution $P_k= e^{-\lambda} \frac{\lambda^k}{k!}$ with $λ=M/N$.

If we denote $X_i=1$ if the bucket i is empty, 0 otherwise, we have $E[X_i]≈e^{\frac{−M}{N}}$"

How did they derive $E[X_i]≈e^{\frac{−M}{N}}$? I have tried and failed to deduce it myself or find online resources.

答:
If you take the Poisson approximation then $$E[X_i] = P_0 \approx e^{-M/N} \frac{(M/N)^0}{0!}= e^{-M/N}$$

If you want the binomial route then $$E[X_i] = P(\text{no balls go in that bucket }) = \left(1-\frac1N\right)^M \approx e^{-M/N}$$ for large $M$ and $N$


问:
Suppose $T,S$ are two non-identity elements in $PSL(2,\mathbb{R})$ and $TS=ST$. Then the number of fixed points of $S$ and $T$ are same.
What I could see is $T$ maps the fixed point set of $S$ to itself, so does $S$ for the fixed point set of $T$. But I can not proceed further. I was actually looking at the proof of a stronger result that says the fixed set is equal if and only if they commute. The proof uses the above statement. I seem to stuck to this point.

答:
Im really just spelling out the comment of reuns in greater detail.

The fixed points of $S$ (resp. $T$) on $\mathbb P^1_{\mathbb R}$ correspond to eigenvectors of (lifts of) $S$ (resp. $T$). This is because for $p \in \mathbb R^2 \setminus \{0\}$, one has

$$[Sp] = [p] \iff \text{there is } \lambda \neq 0 \text
{ such that } Sp = \lambda p.$$

Here, we use square brackets to denote the corresponding point in $\mathbb P^1_{\mathbb R}$.

Can you show that the number of fixed points of $S$ (resp. $T$) is the number of eigenvalues of $S$ (resp. $T$)? (Here, you use that $S$ and $T$ are not the identity.)

Afterwards, compare eigenspaces of $S$ and $T$ using $ST = TS$.


问:
separation of variables to find solution to heat equation (cylinder)
I'm trying to model heat flow in a cylinder using the heat equation PDE where heat flow is only radial:

$$ \frac{\partial u}{\partial t} = \frac{1}{r} \frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial r^2}
$$
The I.C for $0<r<1$:
$$
u(r,0) = 0
$$

The B.Cs for $t>0$:
$$
\lim_{r \rightarrow 0} \frac{\partial u}{\partial r} = 0
$$
$$
u(1,t) = 1
$$
Now, we use separation of variables:
Assume that the solution to the PDE is given by:
$$
u(r,t) = U(r)T(t)
$$
Then by rearranging, we have:
$$  \frac{T'(t)}{T(t)} = \frac{1}{r} \frac{U'(r)}{U(r)} + \frac{U''(r)}{U(r)} = -\lambda = -m^2
$$
Here $m>0.$
The first ODE is given by:
$$
T'(t) = -m^{2}T(t)
$$
which has the following solution:
$$
\sum_{n=1}^{\infty} C_{n} \exp(-m^2t)
$$
The second ODE is given by:
$$
U''(r) +\frac{1}{r}U'(r) +m^{2}U(r) = 0
$$
According to the literature, this is a second order ODE that would result in Bessel's functions.
How do you then find the solution to this ODE and then apply the initial and boundary conditions to then find $u(r,t)$?

答:
I was just working on this same problem. @user142288's answer is on the right track, but I think it overlooks some fundamental aspects and based on the comments regarding the initial and boundary conditions, overlooking these steps can lead to some confusion. Of course, there are many ways to solve a problem, this is just what I did.

The final solution, $u(r,t)$ will be a combination of the homogeneous solution (where the boundary conditions are 0) and the particular solution

$$ u(r,t) = u_h(r,t) + u_p(r,t) $$

The homogeneous case will have boundary conditions

$$ u_h(1, t) = 0 $$

$$ \frac{\partial u_h}{\partial r} |_{r=0} = 0 $$

and initial condition
$$ u_h(r, 0) = f(r) $$

where $f(r)$ is an arbitrary profile (I believe it has to be smooth).

We can solve first for the homogeneous solution using separation of variables as you pointed out

$$ u_h(r,t) = U(r) T(t) $$

which when plugged-into the original PDE (and a little rearrangement) produces

$$ \frac{T'(t)}{T(t)} = \frac{\nabla^2 U(r)}{U(r)} = -m^2 $$

where $\nabla^2$ is the Laplace operator. The solutions to these two ODEs are

$$ T(t) = a \cdot \exp(-m^2 t) $$

and in cylindrical coordinates we get

$$ U(r) = c_1 J_0(mr) + c_2 Y_0(mr) $$

where $ J_0 $ and $ Y_0$ are Bessel functions. Applying your boundary condition at $r = 0$, i.e. $ \frac{\partial u}{\partial r} |_{r=0} = 0 $ gives

$$ 0 = c_1 m J_0'(0) + c_2 m Y_0'(0) $$

$J_0'(0) = 0$, while $Y_0'(0) \neq 0$, therefore $c_2 = 0$

The second homogeneous boundary condition, i.e. $u_h(1,t) = 0$ gives the equation

$$ J_0(m) = 0 $$

Because $J_0$ is oscillatory, there are an infinite number of $m$ values that satisfy this equation. Our homogeneous solution then becomes

$$ u_h(r,t) = \sum_{i=0}^\infty C_i J_0(m_i r) \exp(-m_i^2 t) $$

The initial condition $u_h(r,0)$ states

$$ u_h(r,0) = \sum_{i=1}^\infty C_i J_0(m_i r) $$

and orthogonality allows for the constants, $C_i$ to be solved for

$$ C_i = \frac{\int_V f(r) \cdot J_0(m_i r) dV} {\int_V [J_0(m_i r)]^2 dV} $$

where $V = \pi R^2 L$, the volume of a cylinder, and $dV = 2\pi L r dr$.
The full homogeneous solution is then:

$$ u_h(r,t) = \sum_{i=1}^\infty \left[ J_0(m_i r) \exp(-m_i^2 t) \ \frac{\int_V f(r) \cdot J_0(m_i r) dV} {\int_V [J_0(m_i r)]^2 dV} \right] $$

The case of the homogeneous boundary conditions can be thought of physically as a cylinder with some arbitrary temperature profile (or concentration profile) and at time $t=0$ the surface of the cylinder is held at zero temperature. The homogeneous solution solves for the evolution of this temperature profile in time and space. We can see intuitively that at infinite time, the temperature everywhere will equal the surface temperature, i.e. 0. And
because of the exponential decay this is indeed what is observed, and can be stated mathematically as

$$ \lim_{t \rightarrow \infty} u_h(r, t) = 0 $$

The problem originally asked is totally analogous: instead of tracking the profile as it tends toward 0, we are solving for a profile as it tends toward 1 (because the boundary condition at the surface is $u(1,t) = 1$).

The particular solution $u_p(r,t)$ needs to satisfy the original boundary conditions:

$$ u(1, t) = 1 $$

$$ \frac{\partial u}{\partial r} |_{r=0} = 0 $$

And we can easily verify that

$$ u_p(r,t) = 1 $$ is a solution to these boundary conditions.

To make the original problem homogeneous, we had to subtract the particular solution from the full solution

$$ u_h(r,t) = u(r,t) - u_p(r,t) $$

which includes the boundary and initial conditions. Which means for the homogeneous solution, the initial condition is

$$ u_h(r,0) = u(r,0) - u_p(r,0) = 0 - 1 = -1 $$

or

$$ f(r) = -1 $$

The full solution can then be written

$$ u(r,t) = u_p(r,t) + u_h(r,t)$$

$$ u(r,t) = 1 - \sum_{i=1}^\infty \left[ J_0(m_i r) \exp(-m_i^2 t) \ \frac{\int_V J_0(m_i r) dV} {\int_V [J_0(m_i r)]^2 dV} \right] $$

Some sanity checks confirm that
$$ u(r,0) = 0 $$

and

$$ \lim_{t \rightarrow \infty} u(r,t) = 1 $$


答:
The expected value in each bucket is $\frac MN$. Using the Poisson formula, the probability of getting $0$ in a bucket is
$$
e^{-\frac MN}\frac{\left(\frac MN\right)^0}{0!}=e^{-\frac MN}\tag1
$$
A way to visualize this is to note that the probability that placing an entry misses our bucket is $\left(1-\frac1N\right)$. Thus, the probability that all of $M$ entries miss our bucket is
$$
\left(1-\frac1N\right)^M\approx\left(e^{-\frac1N}\right)^M=e^{-\frac MN}\tag2
$$

Title: Expected value of occurrences of a variable under a Poisson distribution

Body:
I asked a question on this forum recently, to which I got a very thorough answer that I didn't properly understand. Link to the original question here.

The answer stated:

"If we have N buckets and place M entries in them, the number of elements in each buckets, for large $N, m$, approximately follows ("Poissonization") a Poisson distribution $P_k= e^{-\lambda} \frac{\lambda^k}{k!}$ with $λ=M/N$.

If we denote $X_i=1$ if the bucket i is empty, 0 otherwise, we have $E[X_i]≈e^{\frac{−M}{N}}$"

How did they derive $E[X_i]≈e^{\frac{−M}{N}}$? I have tried and failed to deduce it myself or find online resources.


问:
Why does a projection is considered as a disc and calculated by this way?
I have been reading a scientific paper that illustrate a projection method as follow

${F^2} = \left\{ {\begin{array}{*{20}{c}}
  {{f_1}:{x_1}^3 + {x_2}^2 + {x_1}{x_2} - \frac{1}{2}} \\
  {{f_2}:{x_1}^2 + {x_2}^2 - 2}
\end{array}} \right.$

Define the projection order ${x_1} \prec {x_2}$ the projection is $\left( {{x_1},{x_2}} \right) \mapsto {x_1}$.

1/ Why does a projection is considered as a disc and is calculated by this way ?

2/ How to visualize or at least understand the meaning of $lc\left( {{f_2},{x_2}} \right)$ and $res\left( {{f_1},{f_2},{x_2}} \right)$ ?

Please help me understand this !
Thank you !

答:
Answer to question 2):

lc(f,x), means "Leading Coefficient" (coefficient of monomial in $f$ having the highest degree with repect to variable $x$ ).

Besides, the computations re linked to Resultants of two polynomials $Res_x(f,g)$ (denoted here  $Res(f,g,x)$), a particular case being the Discriminant of a polynomial:

$$Disc_x(f)=Res_x(f,f')$$

where $f'$ denotes the derivative of $f$ with respect to variable $x$.

All this and the way to compute a resultant using a determinant can be found here.

This said, there is a simpler (classical) way to compute discriminants in this case. Indeed, polynomial $f_1$, considered as a quadratic in variable $x_2$;

$$x_2^2 +x_1x_2 + (x_1^3-\frac12)$$

has discriminant (in the usual sense)

$$\Delta= x_1^2-4(x_1^3-\frac12)=2+x_1^2-4x_1^3$$

Same kind of computation for $f_2$...


问:
How to get the product of significant digits from $\overline{abc}$?
The problem is as follows:

Let $\overline{abc}$ the maximum amount of microfilms stored in a military hospital. If the digits from such amount follows the rule:

$$2a+3b+c=\overset{\circ}{7}$$

$$a-b+c=\overset{\circ}{11}+7$$

Find the product of the significant digits from the number of microfilms in the hospital.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{224}\\
2.&\textrm{27}\\
3.&\textrm{1}\\
4.&\textrm{56}\\
\end{array}$

Can this problem be solved without modular arithmetics?. I mean using a more precalculus approach.

What I've attempted so far is as follows: (By going in the system of equations as follows)

$2a+3b+c=\overset{\circ}{7}$

$a-b+c=\overset{\circ}{11}+7$

I did multiplied by 3 to the second equation:

$3a-3b+3c=\overset{\circ}{11}+21=\overset{\circ}{11}+10$

$2a+3b+c=\overset{\circ}{7}$

My approach was to eliminate $b$ from the second equation but when attempting to add these equations I came stuck. Can someone help me here?. What's the approach on where to go from here?.

答:
I assume that $\overset{\circ}{7}$ means some multiple of seven.

The equations are essentially the classic divisibility tests. For example $2a+3b+c=\overset{\circ}{7}$ says that $\overline{abc}$ is a multiple of 7 because $\overline{abc}=100a+10+c=2a+3b+c+7(14a+b)$. Similarly $a−b+c=\overset{\circ}{11}+7$ says that $\overline{abc}$ has remainder 7 when divided by 11.

The smallest number that satisfies those two divisibility requirements is 7, but that is not a 3-digit number. You can however add $77=7\cdot 11$ to it any number of times without changing its divisibility properties w.r.t. 7 and 11, so the numbers that satisfy these properties are:

7, 84, 161, 238, 317, 392, 469, 546, 623, 700, 777, 854, 931, and then you get to 4-digit numbers.

The product of their digits is (7, 32,) 6, 48, 21, 54, 216, 120, 36, 0, 343, 160, 27.

The only one of those that is listed in the multiple choice options is 27.


问:
Does the set intersection operation has unit
I'm trying to get better understanding of binary operations, and I came across this problem: namely on one online discussions I saw that set intersection as binary operation doesn't have a unit, however I think it has:

The set intersection operation for any set universe $U$ is defined on the $\mathcal P(U) \times \mathcal P(U)$ set. From the definition of unit for binary operation, wouldn't it be true that the $U$ set is the unit of this operation. Any set in $PU$ intersected with $U$ will give back the set, and it works in both directions.

Am I doing something wrong, or the discussion on that place was wrong?

答:
Your universe $U$ is not a set. You cannot intersect it with anything, and so your proof does not work (for this reason, it is also incorrect to write $\mathcal P(U)$, as the operator $\mathcal P$ - returning the set of all subsets of a given set - can only be applied to, well, sets themselves.

Hence, what was claimed in the other discussion is true: there exist no set $A$ such that for all sets $B$, $A \cap B = B$.

Edit: It is possible to consider some sort of set algebra, in which such things do make sense ($A \cap U = A$, and so on). This is however a choice that has to be made consciously; the typical ZF(C) axioms do not define those operations, and the "collection of all sets, $U$", is explicitly not a set on which you can perform those operations (power set, cardinality, intersection ...).


问:
Proof of cauchy schwarz inequality in inner product space
$0 \le \lVert x-cy \rVert ^2= \langle x-cy,s-cy \rangle = \langle x,x\rangle -\bar{c}\langle x,y\rangle-c\langle y,x\rangle +c\bar{c}\langle y,y\rangle$.

If we set $c=\frac{\langle x,y\rangle}{\langle y,y\rangle}$ then the inequality becomes

$0 \le \langle x,x\rangle - \frac {\lvert \langle x,y\rangle \rvert^2}{\langle y,y\rangle}$.

I calculate the above equation ans wonder why $\langle x,y\rangle \langle y,x\rangle$ becomes $\lvert\langle x,y\rangle\rvert^2$? In complex field that's not true, isn't it?

答: Note that for any complex number $\alpha$, we have: $$\alpha \overline{\alpha}=|\alpha|^2$$where $|\alpha|$ is the modulus (or length) of the complex number $\alpha$.</p>


问:
I can't find my error in solving line integral in two different methods

I have this problem:

given the points $A(-1,-2)$ and $B(2,1)$ the path $T$ from $A$ to $B$ is on the circule $(x-0.5)^{2}+(y+0.5)^{2}=4.5$

caculate: $$\int _{T}\frac{ydx-xdy}{x^{2}+y^{2}}$$

after a short calculate I found out that: $P_{y}=Q_{x}$ and with this following simply connected space $D$, making $\vec{F}$ conservative vector field:

Therefore I can choose a different path between the two points.
Now my main error is that I calculate the integrals with two different paths and I've given different results.

for the path $C$ on the circle $x^{2}+y^{2}=5$ I've got that:
 the red path is part of the circule.
$$\left\{\begin{matrix}
x= \sqrt[]{5}\cos(t) & dx = (-1)\sqrt[]{5}\sin(t)dt \\
y= \sqrt[]{5}\sin(t) & dy = \sqrt[]{5}\cos(t)dt
\end{matrix}\right.$$
and by using caculator:$$\int _{T}\frac{ydx-xdy}{x^{2}+y^{2}} = \int _{C}\frac{((\sqrt[]{5}\sin(t)\cdot (-1)\sqrt[]{5}\sin(t))-(\sqrt[]{5}\cos(t)\cdot\sqrt[]{5}\cos(t)))dt}{5}$$$$=-1\cdot\int _{C}1\cdot dt = -1\cdot\int_{\arccos (\frac{-1}{\sqrt[]{5}})}^{\arccos (\frac{2}{\sqrt[]{5}})} 1 \cdot dt = \frac{\pi }{2}$$

but

for the path $E$ wich is the sum of the following paths: $E1$, $E2$ I'm getting diffrent result.

$$E_{1}(t) = (t,-2)\rightarrow\left\{\begin{matrix}
dx= dt   \\
dy= 0
\end{matrix}\right., -1\leq t\leq 2$$
$$\int _{E_{1}}\frac{ydx-xdy}{x^{2}+y^{2}} = \int_{-1}^{2}\frac{-2}{t^{2}+4}dt$$

$$E_{2}(t) = (2,t)\rightarrow\left\{\begin{matrix}
dx= 0   \\
dy= dt
\end{matrix}\right., -2\leq t\leq 1$$
$$\int _{E_{2}}\frac{ydx-xdy}{x^{2}+y^{2}} = \int_{-2}^{1}\frac{-2}{t^{2}+4}dt$$
in this picture the paths $E_{1}$ and $E_{2}$ are drwan in pink:

and I'm getting using caculator that: $$\int _{T}\frac{ydx-xdy}{x^{2}+y^{2}} = \int_{-1}^{2}\frac{-2}{t^{2}+4}dt + \int_{-2}^{1}\frac{-2}{t^{2}+4}dt \approx -2.49$$

so, where is my error?

答:
The principal range for arccos(x) is from 0 to $\pi$, so when you calculate the angle for the (-1, -2) point, you're actually getting the angle for the (-1, +2) point.  So your first approach's integral should be:

$$
-1\cdot\int_{-\arccos (\frac{-1}{\sqrt[]{5}})}^{\arccos (\frac{2}{\sqrt[]{5}})} 1 \cdot dt = -2.498
$$

The change is in the lower limit of integration.  That doesn't account for all the error though.

The last part of your error is in the final calculation of the second approach.  It should be -2.498, says wolframalpha


问:
Continuous and differentiable function $f(x)$ in $(x_1, x_2)$.
Consider the continuous and differentiable function $f(x)$ in $[x_1, x_2]$.

Let $f'(x)$ be its derivative and

$f'(x_2) = 0$

Show that there exist a number $c >0$ and a $x_3 \in (x_1, x_2)$ such that $f'(x_3) = c(f(x_3)-f(x_1))$.

Apologies but I haven't touched calculus for more than 20 years. One of the few things I remember is that when the 1st derivative is zero, we have a local minimum or maximum.

(I am not a student or anything).

答:
Edited to improve the middle step:

We can assume without loss of generality that $f(x_1) = 0$ (otherwise simply consider $f(x) -f(x_1)$).

Then, if $f(x)$ is identically zero, $x_3 \in (x_1,x_2)$ and $c>0$ may be chosen freely.

Otherwise, the function $\frac{1}{2} f(x)^2$ must be non-zero at some point $\xi $ in the interval $[x_1, x_2]$;  but $\xi > x_1$ because $f(x_1)=0$ and $f(\xi)^2$ is positive because $f(x)^2 \geqslant 0$.  Apply the mean value theorem to obtain $x_3 \in (x_1, \xi)$ satisfying,
\begin{align}
\frac{\frac{1}{2}f(\xi)^2 - \frac{1}{2}f(x_1)^2}
{\xi-x_1} = f'(x_3)f(x_3)
\end{align}

The left side is strictly positive, so we have $x_3 \in (x_1,x_2)$ with
\begin{align}
f'(x_3)f(x_3) > 0 \tag{1}\label{e1}.
\end{align}
It follows neither $f(x_3)$ nor $f'(x_3)$ is zero and both have the same sign.  Accordingly we can define $c = f'(x_3)  / f(x_3) > 0 $ to obtain the required conclusion,
$$f'(x_3) = c f(x_3).$$

I note that this version does not require $f'(x_2)$ to be zero, so I wonder if the OP question ha been mis-transcribed.


问:
Does this two variable limit exist?
Does this limit exist? How do I evaluate it or prove it doesn't exist?

\begin{equation}
\lim_{(x,y)\to(0,0)} \frac{e^{xy^2}-1}{(x^2+y^2)^{\frac{3}{2}}}
\end{equation}

I tried some restrictions to see if I could  prove it does not exist, for example:

\begin{align}
&E_1=\{(x,y)\in\mathbb{R}^2: x=0 \land y\neq0\} \\
&E_2=\{(x,y)\in\mathbb{R}^2: x=y\} \\
&E_3=\{(x,y)\in\mathbb{R}^2: x=y^2\}
\end{align}

but the limit of all those restrictions was $0$.

I tried multiplying and dividing by $xy^2$ so the limit became:
\begin{equation}
\lim_{(x,y)\to(0,0)} \frac{e^{xy^2}-1}{xy^2}\cdot\frac{xy^2}{(x^2+y^2)^{\frac{3}{2}}}
\end{equation}

And I could just compute the limit:
\begin{equation}
\lim_{(x,y)\to(0,0)} \frac{xy^2}{(x^2+y^2)^{\frac{3}{2}}}
\end{equation}

All I know about this function is that its absolute value is less or equal than $1$, so comparison test doesn't help me either.

Title: Does this two variable limit exist?

Body:
The last limit do NOT exists:

$$\lim_{\substack{(x,y)\to(0,0)\\x=0}}\frac{xy^2}{(x^2+y^2)^{3/2}}=0$$
BUT

$$\lim_{\substack{(x,y)\to(0,0)\\x=y}}\frac{xy^2}{(x^2+y^2)^{3/2}}= \lim_{y\to0} \frac{y^3}{(2y^2)^{3/2}}=\frac{1}{2\sqrt2}$$


问:
Asymptotic notation proof question
Note: We defined asymptotic notation for functions $f : \mathbb{N} \to \mathbb{N}$, but the same definitions work for real-valued functions $f : \mathbb{N}\to\mathbb{R}$.

Let $f, g : \mathbb{N}\to\mathbb{R}$ be two real-valued functions greater than 1.

(a) True or false: $n! = \mathcal{O}(n^n)$? Prove your answer. (Hint: use Stirling’s formula.)

Stirling’s formula: $n! \sim \left(\frac ne\right)^n \sqrt{2\pi n}$.

答:
There are two ways to take down the problem:

Method 1: Stirling's formula

As suggested by the question, $n!=\mathcal O(\sqrt nn^ne^{-n})$. Because $e^n$ grows substantially faster than $\sqrt n$, we have $\sqrt ne^{-n}=\mathcal O(1)$., which indicates $n!=\mathcal O(n^n)$.

Method 2: Elementary comparison

If we were to expand $n!$ and $n^n$, we have

$$
n!=1\cdot2\cdot3\cdot4\cdots n
$$

$$
n^n=n\cdot n\cdot n\cdot n\cdots n
$$

By comparing each term, we conclude that $n^n>n!$ for sufficiently large $n$.

Title: Asymptotic notation proof question


问:
Picture of l'Hopital's rule at infinity
I know that when L'Hopital's rule is taken for the $\frac{0}{0}$ case, we get that the functions start looking at lines, like in this figure from Stewart's Calculus:

But what is the visual when the limit is the $\frac{\infty}{\infty}$ case?

The only thing that comes to mind that is easily visualized is a rational function's asymptotes (and that comes from dividing numerator and denominator by the largest exponent in the denominator)

答:
I would rather split it into four cases:

limit of type $\frac00$ in finite point

limit of type $\frac00$ in infinity

limit of type $\frac\infty\infty$ in finite point

limit of type $\frac\infty\infty$ in infinity

The case 2 doesn't seem to be covered by those ilustrations either. I thought how it could work for case 4:

Use the same illustrations but move the point $a$ to $0$. Then you can see that the limit is equal to the ratio of the slopes of the functions approximated by lines.

Title: Picture of l'Hopital's rule at infinity


问:
Is this an integrating factor at all and why does it work?
So I required some help for the differential equation: $y' = y\,\sin(x)+\sin(2\,x)$

I figured that it's not that easy to solve because of the integral $\int\frac{\sin(2\,x)}{e^{cos(x)}}\mathrm{d}x$ that is appearing in the premade formula.

However a kind user suggested me to multiply the particular solution $y_p = \sin(2\,x)$ with $e^{\cos(x)}$ and integrate afterward. As a result the integral becomes pretty handy. To be honest I recognized the beauty of this approach but didn't understand it yet - why this is functioning.
Can someone explain to me afterwards?

答:
Multiplying by $e^{\cos x}$,

$$(y'-y\sin x)\,e^{\cos x}=(y\,e^{\cos x})'=\sin 2x\,e^{\cos x}.$$

Then by parts,

$$\int\sin x\cos x\,e^{\cos x}dx=-\cos x\,e^{\cos x}+\int\sin x\,e^{\cos x}dx=-\cos x\,e^{\cos x}+e^{\cos x}+c.$$