问:
Determine the inverse function of $f(x)=3^{x-1}-2$
Determine the inverse function of $$f(x)=3^{x-1}-2.$$ I'm confused when you solve for the inverse you solve for $x$ instead of $y$

so would it be $x=3^{y-1}-2$?

答: it's right, but I think you need simplify it. $y=log_3(x+2)+1$</p>


问:
Vector containing matrix
I have a list/vector of $n$ elements where each element is a matrix of size $a\times b$.

What is the conventional name of this and how do I define or represent it mathematically?

答:
I am assuming your matrices are over the reals. We write $\mathbb{R}$ to denote the ("set of") real numbers. Then $M_{a\times b}(\mathbb{R})$ is the set (actually, vector space) of $a\times b$ matrices with entries in $\mathbb{R}$.

So you are asking about this set:

$(M_{a\times b}(\mathbb{R}))^n$
(where $"-^n"$ is the cartesian product n times; so this is the cartesian product n times of the set of $a\times b$ matrices over the reals; "over the reals" meaning that the entries of the matrices are in the reals)

But actually, as far as we are concerned in linear algebra, this set is really just:

$M_{(an)\times b}(\mathbb{R})$
(the set of $(an)\times b$ matrices over the reals)

The reason is that $(M_{a\times b}(\mathbb{R}))^n$ is a vector space of dimension nab over the reals, and so is $M_{(an)\times b}(\mathbb{R})$, which implies that they are isomorphic as vector spaces. It is because of the coordinate map, which, for our purposes in this context, takes matrices, say $(an)\times b$ matries (which are really the same as vectors of n entries, each entry being an $a\times b$ matrix), to vectors with abn entries like this:

$\begin{pmatrix} a_1\\ a_2\\ \vdots \\ a_{abn-1}\\ a_{abn}\end{pmatrix}$.

These vectors are really the same as $(an)\times b$ matrices (as far as linear algebra is concerned).

ContentLicense: CC BY-SA 4.0

Title: Vector containing matrix


问:
Iterated and Double limits exisitence
The existence of the double limit of a multi variable function at a point implies that the iterated limits will also be the same in value and exist, given that we know that that the individual single limits of the function, in x and y exist.
$$\lim_{(x,y) \to (\alpha, \beta)} f(x,y) = L\implies \lim_{x \to \alpha} (\lim_{y \to \beta} f(x,y)) = \lim_{y \to \beta} (\lim_{x \to \alpha} f(x,y)) = L$$
Given that we know that :
$$\lim_{x \to \alpha} f(x,y) \mbox{  and  } \lim_{x \to \beta} f(x,y) \mbox{  exist  } $$
My question is why is this so? Why is the existence of single limits necessary for the above mentioned implication.

As an example take the following function:
$$ f(x,y) = xsin(1/y) $$
$$\lim_{(x,y) \to (0,0)} f(x,y) = 0 $$
$$\therefore \mbox{    as expected :} $$
$$ lim_{y \to 0} (\lim_{x \to 0} f(x,y)) = 0  $$
$$ \mbox{but} $$
$$ lim_{x \to 0} (\lim_{y \to 0} f(x,y)) = \mbox{Does not exist} $$

答:
One of the iterated paths doesn't exist because the line $y = 0$ is not in the domain of $f(x,y)$. So, $\lim_{{y}\to{0}}f(x,y)$ doesn't exist. But, we can still travel through every other path within the domain of $f(x,y)$ and the limit remains $0$.

You're not meant to follow paths outside the domain to check if a limit exists.

Title: Iterated and Double limits exisitence


问:
Shortcut solution for $f''(t)+c^2f(t)=0$?
I am trying to prove that the general solution to $f''(t)+c^2f(t)=0$ is $f(t) = acos( ct) + bsin (ct)$

Could you please tell me if it is possible to solve it this way?
$$
\frac{d}{dt}(\frac{df}{dt})=-c^2f(t)
$$

$$
df* \frac{d}{dt}(\frac{df}{dt})=-c^2f(t)*df
$$
$$
d(\frac{df}{dt})^2=-c^2 f(t)df
$$
If not for the constant C resulting from integration, this would lead to the desired solution, but I think I am not doing it right somehow.

Thanks in advance for any suggestions.

答:
Another way to do this is via the method of Frobenius. We seek solutions of the form $y=\sum_{k\geq 0} a_k x^k$. Differentiating and collecting like terms, we have

$$\sum_{k\geq 2} k(k-1)a_k x^{k-2}+c^2 \sum_{k\geq 0} a_k x^k = \sum_{k\geq 0}\left[(k+2)(k+1)a_{k+2}-c^2 a_k\right]x^k=0.$$

Matching term-by-term, we have $(k+2)(k+1)a_{k+2}=-c^2 a_k$ for all $k\geq 0$. This is solved as $$a_{2k}= \frac{(-c)^k}{(2k)!}a_0,\quad a_{2k+1}= \frac{(-c)^k}{(2k+1)!}a_1.$$

Hence
\begin{align}
y(x)&=a_0\left(1-\frac{(cx)^2}{2!}+\frac{(cx)^4}{4!}+\cdots\right)+a_1 \left(cx-\frac{(cx)^3}{3!}+\frac{(cx)^5}{5!}-\cdots\right)\\&=a_0 \cos(cx)+a_1 \sin(cx)
\end{align}
as expected.


问:
Singular Value Inequality
I am trying to show that for $A \in \mathbb{R}^{m\times n}$, where $m > n$, $\sigma_{min}(A) \leq \sigma_{min}([A | x])$ for some $x \in \mathbb{R}^m$. My approach is using the two norm, but am not sure if that is correct. If this is correct, I believe it involving showing that $||A||_2 \leq ||[A | x]||_2$. How can this be shown?

Would anyone be able to point me in the right direction? Thanks.

答:
It's the other way round, $\sigma_{min}([A|x])\le\sigma_{min}(A)$.

Let $B:=[A|x]$. Since $B^TB$ is positive semi-definite, for any vector $v\in\mathbb{R}^{n+1}$, $$v^TB^TBv\ge\sigma^2_{min}(B)|v|^2.$$
In particular pick $v=(w,0)$ where $w$ is a unit singular vector satisfying $Aw=\sigma_{min}(A)w'$, then $$\sigma^2_{min}(B)\le v^TB^TBv=(w,0)\begin{pmatrix}A^TA&A^Tx\\x^TA&x^Tx\end{pmatrix}\begin{pmatrix}w\\0\end{pmatrix}=w^TA^TAw=\sigma_{min}^2(A)$$

Title: Singular Value Inequality


问:
n x n Matrix Puzzle
You are given an n×n square matrix to be ﬁlled with numerals so that no two adjacent cells have the same numeral. Two cells are called adjacent if they touch each other horizontally, vertically or diagonally. So a cell in one of the four corners has three cells adjacent to it, and a cell in the ﬁrst or last row or column which is not in the corner has five cells adjacent to it. Any other cell has eight cells adjacent to it.

What is the minimum number of different numerals needed to ﬁll a 5×5 square matrix?

I tried, and I came up with the answer 5 but the answer that has been provided is 4. Can anyone please tell me how one can get to 4? What will be the arrangement?

Title: n x n Matrix Puzzle

Body:
Consider
$$\begin{bmatrix}
0 &3  &0  &3  &0 \\
2 &1  &2  &1  &2 \\
0 &3  &0  &3  &0 \\
2 &1  &2  &1  &2 \\
0 &3  &0  &3  &0
\end{bmatrix}$$
which uses only $0,1,2,$ and $3$.


问:
$c \in C \setminus R$ third unity root, $z \in C$. If $|z - 1| \le 1$ and $|z - c| \le 1$, then $|z| \le 1$
Let $c \in C \setminus R$ a third root of the unity. If $a \in C$ and $|z - 1| \le 1$ and $|z - c| \le 1$, then prove $|z| \le 1$. This problem was taken from a Romanian mathematical magazine. I have solved $c ^ 3 = 1$, with $c \in C \setminus R$ and got $c = -\frac{1}{2} \pm \frac{\sqrt3}{2}i$. I tried dealing with the second inequality involving $z$ anc $c$, but I couldn't get anywhere. Any help?

答:
$\newcommand\abs[1]{\lvert#1\rvert}$Fix $a_i \in \mathbb C$.  You can see the following fact by drawing a picture:  if $\abs{a_1} = \abs{a_2}$, then the circles of radius $\abs{a_i}$ centred at $a_i$ intersect at $0$ and $a_1 + a_2$, and the intersection of the corresponding discs is contained in the circle of radius $\abs{a_1 + a_2}$ centred at $0$.

Here's an algebraic proof, which shows that we don't even need $\abs{a_1} = \abs{a_2}$ for the second part:  If $\abs{z - a_i}^2 = \abs z^2 - 2\Re(\overline{a_i}z) + \abs{a_i}^2 \le \abs{a_i}^2$, then $\abs z^2 \le 2\Re(\overline{a_i}z)$, so $2\abs z^2 \le 2\Re((\overline{a_1 + a_2})z) \le 2\abs{a_1 + a_2}\,\abs z$.  That is, $\abs z \le \abs{a_1 + a_2}$ (either because $\abs z \ne 0$ and we divided, or because $\abs z = 0$ and it's obvious).

Now apply this with $a_1 = 1$ and $a_2 = c$.


问:
What is the error hiding in this trivial integral with spherical polar coordinates and divergence theorem?
I am working in spherical polar coordinates $(r,\theta,\phi)$ where $\theta$ and $\phi$ are the polar and azimuthal angles respectively. I define a vector field $\mathbf{V}=\phi \hat{\boldsymbol{\phi}}$, which has divergence
$$ \boldsymbol{\nabla}\cdot\mathbf{V}=\frac{1}{r^2}\frac{\partial}{\partial r}(0)+ \frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}(0) + \frac{1}{r\sin\theta}\frac{\partial}{\partial \phi}(\phi) =  \frac{1}{r\sin\theta} .$$

Also define $u=e^{-r^2}$, with gradient
$$ \boldsymbol{\nabla}u=\Bigg(\hat{\mathbf{r}}\frac{\partial}{\partial r} + \frac{1}{r}\hat{\boldsymbol{\theta}}\frac{\partial}{\partial \theta} + \frac{1}{r\sin\theta} \hat{\boldsymbol{\phi}} \frac{\partial}{\partial \phi}\Bigg) e^{-r^2} = -2r e^{-r^2} \hat{\mathbf{r}} . $$

Then I can write
$$ \boldsymbol{\nabla}\cdot(u\mathbf{V}) = u\boldsymbol{\nabla}\cdot\mathbf{V} + \boldsymbol{\nabla}u\cdot\mathbf{V} = \frac{e^{-r^2}}{r\sin\theta}  $$

I want to integrate the above expression over all $\mathfrak{R}^3$. I apply divergence theorem to the left hand side, evaluated at the surface of a sphere at infinity. So I get

$$ LHS= \int\boldsymbol{\nabla}\cdot(u\mathbf{V}) dV= \int u \mathbf{V}\cdot d\mathbf{S} = \int u \mathbf{V}\cdot \hat{\mathbf{r}} dS = 0,$$

as expected. But for the RHS I get
$$ RHS=\int \frac{e^{-r^2}}{r\sin\theta} dV = 2\pi \int_{0}^{\infty} dr\, r e^{-r^2} \int_{0}^{\pi} d\theta = \pi^2 .$$

Obviously this cannot be right! I have pored over this but I cannot figure out what is happening. Thanks in advance.

答:
What the volume integral is really calculating is the following limit. Take the spherical "slice" which goes from $\phi = 0$ to $\phi = \beta$. The volume integral calculation is altered by

$$\beta\int_0^\infty re^{-r^2}dr\int_0^\pi d\theta = \frac{\pi\beta}{2}$$

The surface integral now has three components: The spherical surface itself is $0$ as before, what's new is the addition of the semicircular planes at $\phi = 0$ and $\phi = \beta$. The contribution from the first plane is $0$, but the contribution from the second plane is

$$\int_{\phi=\beta} u\mathbf{V}\cdot d\mathbf{S} = \int_0^\infty \int_0^\pi e^{-r^2}\beta\hat{\phi}\cdot\hat{\phi}(rdrd\theta) = \beta\int_0^\infty re^{-r^2}dr\int_0^\pi d\theta = \frac{\pi\beta}{2}$$

Ordinarily as you take the limit $\beta\to2\pi$, the two planes would close up and since the function takes on the same values but with faces oriented opposite ways, they would cancel. But the vector field has a jump discontinuity at the $xz$ plane, $x>0$, so this does not happen. Instead,

$$\lim_{\beta\to2\pi}\int_{S_{\beta}} u\mathbf{V}\cdot d\mathbf{S} = \pi^2 \neq 0$$

Since the vector field had a discontinuity, it's divergence was hiding a secret Dirac delta on the "prime meridian" of the sphere. The divergence theorem is valid if the delta is considered properly.


问:
Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

Solve, $[3x + 1] = 2x - \frac{1}{2}$ , and find the sum of all roots.

What I Tried: I have :-
$$\rightarrow [3x + 1] = \frac{4x - 1}{2}$$
Now, the LHS is an integer, so $2|(4x-1)$ .
I can see that if $x$ is an integer, then $(4x - 1)$ is odd, so there are no integer solutions. I am not sure about fractions, as $x = \frac{1}{2}$ still makes it odd, but $x = \frac{1}{4}$ makes it even, although $2$ does not divided $0$.

This is where I am stuck, how should I proceed next?

Can anyone help me?

答:
I'm leaving this here, but it's wrong. I added a correct answer below.

\begin{align}
   \lfloor 3x + 1 \rfloor &= 2x - \frac{1}{2} \\
   \lfloor 3x \rfloor + 1 &= 2x - \frac{1}{2} \\
   \lfloor 3x \rfloor &= 2x - \frac 32
\end{align}

So, for some integer, $n$,

\begin{array}{ccccc}
                       &&\lfloor 3x \rfloor = n \\
\hline
      n           &\le  &3x  &< &n+1 \\
      \frac n3    &\le  &x   &< &\frac n3 + \frac 13 \\
\hline
      \frac {2n}3            &\le  &2x                   &<   &\frac {2n}3 + \frac 23 \\
      \frac {2n}3 - \frac 32 &\le  &2x - \frac 32        &<   &\frac {2n}3 - \frac 56 \\
      \frac {2n}3 - \frac 32 &\le  &n                    &<   &\frac {2n}3 - \frac 56 \\
      -\frac 32              &\le  &\frac n3             &<   &-\frac 56 \\
      -\frac 92              &\le  & n                   &<   &-\frac 52 \\
      -4                     &\le  & n                   &\le &-3 \\
\hline
      -4                     &\le  & \lfloor 3x \rfloor  &\le &-3 \\
      -4                     &\le  & 2x - \frac 12       &\le &-3
\end{array}

Hence $x \in \left\{ -\dfrac 74, -\dfrac 54 \right\}$

The sum of the roots is $-3$.

This is a correct answer.

Let $x = n + \epsilon$ where $n \in \mathbb R$ and $\epsilon \in [0,1)$.

\begin{align}
   \lfloor 3x \rfloor &= 2x - \frac 32 \\
   3n + \lfloor 3\epsilon \rfloor &= 2n + 2\epsilon - \frac 32 \\
   6n + 2\lfloor 3\epsilon \rfloor &= 4n + 4\epsilon - 3 \\
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
\end{align}

Since the LHS is an integer, then so too must $4\epsilon$ be an integer.
It follows that $4\epsilon \in \{0,1,2,3\}$

If $\epsilon = 0$.

\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 3 &= 0 \\
   &\text{No solution.}
\end{align}

If $\epsilon =  \dfrac 14$.

\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 3 &= 1 \\
   n &= -1 \\
   x &= -\dfrac 34
\end{align}

If $\epsilon =  \dfrac 12$.

\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 5 &= 2 \\
   &\text{No solution.}
\end{align}

If $\epsilon =  \dfrac 34$.

\begin{align}
   2n + 2\lfloor 3\epsilon \rfloor + 3 &= 4\epsilon \\
   2n + 7 &= 3 \\
   n &= -2 \\
   x &= -\dfrac 54
\end{align}


答:
In case of inverse you find $x$ in terms of y.

First you note that the inverse exists(monotonic function) then you do the following.

$y=3^{x-1}-2\Rightarrow y+2=3^{x-1}\Rightarrow \log_3(y+2)=x-1\Rightarrow x=\log_3(y+2)+1$

And you are done.(As a customary one needs to replace x with y as we generally denote a function by y with independent variable x)

ContentLicense: CC BY-SA 3.0

Title: Determine the inverse function of $f(x)=3^{x-1}-2$

Body:
Determine the inverse function of $$f(x)=3^{x-1}-2.$$ I'm confused when you solve for the inverse you solve for $x$ instead of $y$

so would it be $x=3^{y-1}-2$?


问:
An intriguing differential equation
I recently came across the following differential equation:

$$(y')^4 = 1 - y^2$$

I am unable to see any way forward, having only found the trivial $y=1$ as a solution. How can one solve this equation?

Any help is much appreciated.

答:
Switch variables to make
$$[x']^4=\frac 1{1-y^2}$$ which will give the "nice"
$$x=c_1\pm y \, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{2};y^2\right)$$ where appears the gaussian hypergeometric function. This would be almoast impossible to inverse.

We can make decent approximations for the case $c_1=0$ and the positive branch since
$$y \, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{2};y^2\right)=\frac{1}{\Gamma \left(\frac{1}{4}\right)}\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{4}\right)}{(2 n+1) \Gamma (n+1)}y^{2n+1}$$  Truncating to some order and using series reversion
$$y=\sum_{n=0}^\infty a_n \,x^{2n+1}$$ the $a_n$'s forming the sequence
$$\left\{1,-\frac{1}{12},-\frac{1}{96},-\frac{23}{8064},-\frac{1195}{1161216},-\frac{21
   985}{51093504},-\frac{242057}{1226244096},-\frac{9943859}{103004504064},-\frac{1
   63085399}{3296144130048},-\frac{59258145865}{2254562584952832}\right\}$$  This is probably much faster than a series solution.

Title: An intriguing differential equation


问:
Are all commutative, associative binary operations isomorphic to addition?
Addition and multiplication are the two classic commutative, associative binary operations on the reals. They satisfy a striking property: they are equivalent up to unary operations. By taking a logarithm of the arguments to addition and exponentiating the result, we get multiplication:

$xy = \exp(\log x + \log y)$

In fact, we can create a new commutative, associative operation from every invertible unary function $f$ in the same way:

$(x,y) \mapsto f^{-1} (f(x) + f(y))$

I actually haven't been able to come up with a commutative, associative operation that doesn't follow this form. Are all commutative, associative operations on the reals isomorphic to addition in a similar way?

EDIT - The answers and comments below give several good counterexamples to the problem as stated. Requiring $f$ to be invertible is too strong a condition. What if we allow this slightly more general functional form

$(x,y) \mapsto g (f(x) + f(y))$

where $f$ maps reals to some field and $g$ maps that field to the reals? This seems to capture the counterexamples. For example:

$(x, y) \mapsto 0$ matches with arbitrary $f$ and $g(z) = 0$

$(x, y) \mapsto x + y - \lfloor{x+y}\rfloor$ matches with $f(x) = x$ and $g(z) = z - \lfloor{z}\rfloor$

I think $(x, y) \mapsto max(x,y)$ matches with $f(x) = (x + a)^n$ and $g(z) = z^{1/n} - a$, where $a, n$ are constants we take to infinity. That choice is inspired by the relationship between the max function and the infinity norm. I get that including limits in $f$ and $g$ is pretty suspect, but even with that hack, it's interesting to me that it's possible.

Modifying the question slightly, can every commutative, associative binary operation be cast in the form $(x,y) \mapsto g (f(x) + f(y))$?

答: "Isomorphic" refers to entire mathematical structure, not to an operation in that system. One could ask whether, given any set $S$ and an operation $*$, whether $(S,\star)$ is isomorphic to  $(\mathbb R, +)$, but the answer would be a rather trivial; for instance, $(\mathbb C,+)$ is not isomorphic to $(\mathbb R, +)$. If you want to restrict it to $(\mathbb R, \star)$ being isomorphic to  $(\mathbb R, +)$,  that doesn't really change anything essential; $\mathbb C$ has the same cardinality as $\mathbb R$, so there exists bijections between them. Take any such bijection $\phi: \mathbb R \rightarrow \mathbb C$, and we can define $a\star b$ for real numbers $a, b$ as $a\star b = \phi^{-1}(\phi(a)+ \phi(b))$ , and $(\mathbb R, \star)$ will not be isomorphic to $(\mathbb R, +)$.


问:
Is diffirentiability in finite dimensional space is equivalent to the existence of partial derivatives
Let $f$ be a continuous mapping, defined on $\mathbb{R}^2$. I want to prove that $f$ is differentiable at a point $(a,b)\in \mathbb{R}^2$ : Is it sufficient to prove that $f$ has first order derivative at $a$ and $b$ i.e $f_x(a,b)$ and $f_y(a,b)$ exists (since we have a finite dimensional space)?

Is there a reference book I could look at to find such result ?

答:
You’ll have hard time to prove what you want... as it is wrong.

A counterexample is $g : \mathbb R^2 \to \mathbb R$

$$g(x,y)=\begin{cases}
\frac{xy}{\sqrt{x^2+y^2}} & (x,y) \neq (0,0)\\
0 & (x,y) =(0,0)
\end{cases}$$ which has partial derivatives at $(0,0)$ at that point but is not differentiable there.

More here.


答:
Define $f(x) = \min(x_1,x_2)$ for $x\ge 0$ and zero otherwise.
$f$ is not differentiable at zero but the partials exist at $x=0$.

Title: Is diffirentiability in finite dimensional space is equivalent to the existence of partial derivatives

Body:
Let $f$ be a continuous mapping, defined on $\mathbb{R}^2$. I want to prove that $f$ is differentiable at a point $(a,b)\in \mathbb{R}^2$ : Is it sufficient to prove that $f$ has first order derivative at $a$ and $b$ i.e $f_x(a,b)$ and $f_y(a,b)$ exists (since we have a finite dimensional space)?

Is there a reference book I could look at to find such result ?


问:
Proving $||A||_2 \leq || [A|x]||_2$
I would like to show $||A||_2 \leq || [A|x]_2||$ where $x$ is any vector of appropriate size. My thinking is using column spaces and the fact that the column space can only increase, not decrease, but am not confident on this idea.

答:
Let $A$ be $m \times n$, and let $v$ be any non-$0$ vector in $\mathbb R^n$.  Then, with hopefully obvious ‘padding’ notation, $[A\vert x](v; 0) = A v$, so
$$
\frac{\lVert A v\rVert_2}{\lVert v\rVert_2}
= \frac{\lVert[A\vert x](v; 0)\rVert_2}{\lVert(v; 0)\rVert_2}
\le \lVert[A\vert x]\rVert_2.
$$
Taking suprema gives the result.

Title: Proving $||A||_2 \leq || [A|x]||_2$


问:
How to reconcile two definitions of the gradient (partial derivative vs Frechet derivative)?
There are two definitions of the gradient of a function $f$ at $x$

Stacked partial derivatives of $f$
$$ \nabla f(x) = \left(\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n}\right)(x)$$

The Frechet derivative

$$
    \lim_{h \to 0} \frac{f(x+h) - f(x) - v^Th}
    {\|h\|} = 0
$$
The unique vector $v$ is the gradient of $f$ at $x$.

How can I reconcile between these two definitions? It is not obvious to me that this vector $v$ must be the stacked partial derivatives of $f$.

Can someone help?

答:
In the second definition if you take $h=(h_1,0,..,0)$ and let $h_1 \to 0$ you see immediately that $v_1=\frac {\partial f} {\partial x_1}$. Similarly, $v_i=\frac {\partial f} {\partial x_i}$ for all $i$.

However, the existence of partial derivatives does not guarantee the existence of Frechet derivative. For counter-examples see Is diffirentiability in finite dimensional space is equivalent to the existence of partial derivatives


问:
Are logarithms and nth root of numbers same?
We write $2^3 = 8$ and express the same as $\log (8)$ to the base $2$ is $3$. Isn't this the same as finding nth root of a number? What is the significance of logarithms actually and what if we don't have the concept of Logarithms?

答: The b-th root is the answer to Which number should we raise to the power $b$ to get $c$ ? ($a = \sqrt[b]{c}$), while the logarithm is the answer to To which power should we raise the number $a$ to get $c$ ? ($b=\log_a c$).


问:
Energy functional in Poisson's equation: what physical interpretation?
Let's consider this boundary-value problem:

$$\begin{cases} -\Delta V = \rho & \rm{in}\ \Omega
\\
V=0 & \rm{on}\ \partial \Omega \end{cases}.$$

We know that this problem has a variational formulation: its solutions are critical points of the energy functional

$$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx,$$

(cfr. Evans Partial differential equations 2nd ed., §2.2.5).

Now let us look at one (of the many) possible physical interpretations of this problem, the electrostatic one. If units are chosen so that $\varepsilon_0=1$, this equation describes the electric potential $V$ in a region $\Omega$ filled with a charge distribution $\rho$ and whose boundary is grounded.

It would be nice if the energy functional introduced above coincided with the total energy of this physical system. Unfortunately, it seems to me not to be so. In fact, as I read in Feynman's Lectures on physics, vol.II, §8-5, the total energy of this system consists of two additive parts: one is

$$\frac{1}{2}\int_{\Omega} \lvert \nabla V \rvert^2\, dx$$

and is due to the electric field; the other is

$$\frac{1}{2}\int_{\Omega} V\rho\, dx$$

and is due to the charge distribution.

This induces me to think that the
  energy functional should be

$$U[u]=\int_{\Omega} \left(\frac{1}{2}\lvert \nabla u \rvert^2 + \frac{1}{2}u\rho \right)\, dx,$$
  instead of the correct
  $$I[u]=\int_{\Omega} \left( \frac{1}{2}\lvert \nabla u \rvert^2 - u\rho \right)\, dx.$$
  Why am I wrong?

答:
I think this has to do with what you treat as the dynamic variables. In your first formulation, $\rho$ was given and fixed, and you wanted to calculate $V$ by varying only $V$, not $\rho$. In the second formulation, $V$ is just a quantity derived from $\rho$, and this is the energy suitable for obtaining $\rho$ by varying $\rho$, with $V$ just a shorthand for a certain linear transformation of $\rho$. A third possibility is to consider $V$ as given and find the motion of a charge distribution in this potential -- here, again, there would not be a factor of $\frac{1}{2}$. This is the case, for instance, if we calculate the motion of the electron in the potential of a hydrogen nucleus, or of the Earth in the gravitational potential of the Sun.

Mathematically speaking, we need a factor of $\frac{1}{2}$ in front of quadratic terms (where $V\rho$ is quadratic in $\rho$ if $V$ is considered as a derived quantity derived from the dynamical variable $\rho$) but not in front of linear terms. Physically speaking, the factor of $\frac{1}{2}$ avoids double-counting when the energy is regarded as the interaction energy of a charge distribution with itself.

Now you may ask: But surely there is some well-defined energy and we can't just pick and choose the factors according to expedience? That's true, but consider again the hydrogen atom: If we consider the potential of the nucleus as given, then we're not counting the potential energy of the nucleus in the field of the electron in the integral over $\rho V$. But the entire energy is there and has to be accounted for in the integral, hence the factor $1$. On the other hand, in treating a helium atom, where both electrons are treated as dynamic and their interaction energy enters into the Hamiltonian, we do include a factor of $\frac{1}{2}$ to avoid double-counting the energy in the double integral over $\rho_1\rho_2/r_{12}$.


答:
Yes, you are right till here. Recall that you have to solve for $y$ and finally replace $y $ with $f^{-1}(x)$ and $x$ with $y$.
$$x = 3^{y-1} - 2$$
Just add $2$ and take log base 3 on both sides.$$\begin{align} &x + 2 = 3^{y-1} \\  \Rightarrow & \log_3(x + 2) = y - 1 \\ \Rightarrow & y = \log_3(x + 2) + 1  \end{align}$$And you have $f^{-1}(x) = \log_3(x + 2) + 1$. Does it make sense now?</p>


问:
Why is the Sorgenfrey Line not second countable?
A topology is second countable if there exist a countable basis. For the Sorgenfrey line, each basis are of the form [a,b) where a and b are both real numbers. This is uncountable, but if I restrict them to only rationals, they it would be countable right?</p>

答: If $\mathcal{B} $ is a basis for the Sorgenfrey line, then for any $b \in \mathbb R$, since $b \in [b,\infty)$ and the latter set is open, we can find $V_b \in \mathcal B$ such that $b \in V_b \subseteq [b,\infty)$. Notice that $\inf V_b =b$. Thus, for $b_1 \neq b_2$ we get that $V_{b_1}\neq V_{b_2} $. In other words, the map $b \mapsto V_b$ is injective and $\mathcal{B}$ has cardinality at least $\mathfrak{c}=|\Bbb R|$.


问:
Question - Differentiability of functions of several variables
I'm currently studying some definitions of the differentiability of functions of several variables. But I have trouble understanding.

May I ask why the limit below is equal to zero?

[![enter image description here][1]][1]

Is it because we consider $y = y_0$?

答: That's a definition. You don't prove a definition. That's how we define a differentiable function from $\Bbb R^2$ into $\Bbb R$. We could alse define a differentiable function from $\Bbb R$ into $\Bbb R$ as a function $f$ such that, for each $x_0\in\Bbb R$, there is some $a\in\Bbb R$ such that$$\lim_{x\to x_0}\frac{f(x)-f(x_0)-a(x-x_0)}{x-x_0}=0.$$


答:
$2^3 = 8$ and $\sqrt[3] 8 = 2$ and $\log_2 8 = 3$ all express the same relationship between $2,3$ and $8$.  But the each refer to different aspect of the relation ship.

$\sqrt[3] 8 = \color{red} x$ means to solve $\color{red}x^3 = 8$.  We know what the final result is, and we know the power, but we don't know the base.  That is what $\sqrt[3] 8$ is supposed to express; the base.

$\log_2 8 = \color{green} y$ means to solve $2^{\color{green} y} = 8$.  We know what the final result is and we know the base, but we don't know the power.  That is what $\log_2 8$ is supposed to express; the power.

And $2^3 = \color{blue} z$ means to solve $2^3 = \color{blue}z$.  We know the base and we know the power, but we don't know what the final result.  That is what $2^3$ is supposed to express; the final result.

So $2^3 = 8, \sqrt[3]8 = 2; \log_2 8 = 3$.  One relationship with three different components, and an expression to describe each possible component.

Title: Are logarithms and nth root of numbers same?

Body: We write $2^3 = 8$ and express the same as $\log (8)$ to the base $2$ is $3$. Isn't this the same as finding nth root of a number? What is the significance of logarithms actually and what if we don't have the concept of Logarithms?


答:
One historical significance of logarithms is that they change multiplication into addition, and that was used in practical computation (and inherent in the slide rule as a calculating device) before electronic calculators.

Here are some more mathematically advanced aspects of the logarithm, which give a clue as to why the function emerges so often and across a range of mathematical applications:

Logarithms solve the problem of integrating $\frac 1x=x^{-1}$. Other integer powers of $x$ can be integrated by taking the inverse of obvious derivatives.

The natural logarithm is the inverse function of the exponential function ($e^x$), which is a fixed point of the operation of differentiation.

The discrete logarithm (and associated primitive root of a finite field) are significant in coding theory. This replicates the multiplication to addition aspect.


问:
Is my reasoning about Borel Lebesgue theorem correct?
I was thinking alone about the Borel Lebesgue theorem in R: "Any open cover of a compact set contains a finite open sub cover"

My first thought was: "Why doesn't it work for a closed cover?" And my insight was this:
intervals within a closed cover doesn't need to "overlap" in both endpoints, so you might end up in a situation where you cannot afford removing any of the infinite number of sets in the cover. For instance:

$I = [a,b]$

$U_k = [a, b-\frac{1}{k}]$

Then the cover $\cup^{k=1}_{\infty} U_k$ is a cover because the series $X_n = b - \frac{1}{n}$ converges to $b \in I$, because I is closed. So removing any of the $U_k$ will make some element out.

The thing with open intervals is that they need to overlap so you can cover any specific interval with more than one open interval.

Is that insight correct?

答:
$U_k$'s do not cover $[a,b]$ since $b$ is not in any of them.

$[a,b]$ is the union of all singleton sets $\{\{x\}: a \leq x \leq b\}$ and this gives  a closed cover with no finite subcover.


问:
Linear Map extension from subspace to vector space
$V$ is a finite dimensional vector space. How to prove that any linear map on subspace of $V$ can be extended to linear map on $V$.

I attempted it by taking the basis of the subspace $W$ with $\text{Dim}(m)$ and extending it to basis of $V$ with $\text{Dim}(n)$. Since mapping of all vectors of basis of $W$ exists and taking the remaining $m-n$ vectors as zero - we will get a linear map to an element that is in $V$. But is it sufficient to prove that all vectors in $V$ will also be having a linear map?

答:
Suppose $L:W\to W$ is the linear map that you want to extend to whole of $V$. First consider a basis $\{w_1, \ldots, w_m\}$ of $W$ and extend this to a basis $\{w_1, \ldots, w_m, v_{m+1}, \ldots, v_n\}$ of $V$. Now define $\widetilde{L}:V\to V $ as follows: we define the action of $\widetilde{L}$ on basis vectors by,

$$\widetilde{L}(w_1)= L(w_1),$$
$$\widetilde{L}(w_2)= L(w_2),$$
$$\vdots$$
$$\widetilde{L}(v_{m+1})=0,$$
$$\vdots$$
$$\widetilde{L}(v_n)= 0,$$
and on any arbitrary vectors $v\in V$, we define the action of $\widetilde{L}$ on $v=\alpha_1w_1+ \ldots + \alpha_mw_m + \alpha_{m+1}v_{m+1} + \ldots + \alpha_n v_n$, by
$$\widetilde{L}(v):= \alpha_1\widetilde{L}(w_1) + \ldots + \alpha_m\widetilde{L}(w_m) + \alpha_{m+1}\widetilde{L}(v_{m+1})+\ldots+ \alpha_n\widetilde{L}(v_n).$$

Its clear that $\widetilde{L}|_W = L$ and it is easy to check that $\widetilde{L}$ is a linear map (defined on whole of $V$).


问:
Show that the case $A$ is true
Let $A,B,C$ three logical cases. Assuming that $B$ is false and $C$ is true. Assuming that we have the following implications:

$C⇒A$

$C⇒B$

$C⇒A$ or $B$

Then show that the case $A$ is true.

答:
First, it does not make sense for $C$ to be true and $B$ false, since you have said that the property
$$C\implies B$$
is true.

Then, since $C\implies A$, we got that
$$C \text{ true} \implies A \text{ true},$$
and you said $C$ is true so we conclude $A$ is also true.

We can also deduce the same from the property $C\implies A \vee B$, since we are assuming $B$ is false, so in order for $A \vee B$ to be true, then $A$ must be true.

Title: Show that the case $A$ is true


问:
Determine the inverse function of $f(x)=3^{x-1}-2$
Determine the inverse function of $$f(x)=3^{x-1}-2.$$ I'm confused when you solve for the inverse you solve for $x$ instead of $y$

so would it be $x=3^{y-1}-2$?

答:
To find the inverse function $f^{-1}$ in the form $y = f^{-1}(x)$, there are two things you can do:

1) Switch $x$ and $y$, and then solve for $y$.

2) Solve for $x$, and then switch $x$ and $y$.

You seem to be trying to switch $x$ and $y$ and then solve for $x$.  You need to solve for $y$ if you've already switched the variables.  See the methods (1) and (2) above.


问:
Forcibly making a node a source in a directed graph
I have a directed acyclic graph (DAG) and I want to perform the operation of forcibly making a given node a source node, by reversing all its incoming edges, and the incoming edges of its parents, grandparents, etc. It seems that equivalently one can simply reverse the subgraph induced by its ancestors + the node itself

Does this operation have a name?

Is the result a DAG again?

What if I want to make a whole set of nodes sources? Do I need to apply the operation sequentially in reverse topological order?

Example:

The target node is (3). Going in topological order, we first reverse the incoming arrows of (1), then the incoming arrows of (2) and finally the incoming arrows of (3).

```<code>(1)→(2) | (1)→(2) | (1)←(2) | (1)←(2)
   ↘ ↓  |    ↘ ↓  |   ↘  ↓  |    ↖ ↑
    (3) |     (3) |     (3) |     (3)
</code>```

The result is the same as simply reversing the arrows of the original graph.

Pseudocode:

```<code>invert_on_node(Graph G, node n):
   A = anscestors(G, n)
   H = G.subgraph(A+n)

   G.delete_edges(H.edges)

   for node in topological_sort(H):
       reverse_incoming_edges(H, node)

   G.add_edges(H.edges)
<span class="math-container">```
```

答:
Regarding question 2, it turns out the proof is trivial:

Claim: Let $G=(V, E)$ be a DAG. Then, for all $v\in V$, the graph $G' = (V, E_v^R \dot\cup (E\setminus E_v))$ obtained by reversing all edges contained in the subgraph $G[A(v)] = (V_v, E_v)$ induced by the set of ancestors $A(v) = \{u\in G\mid u\le v\}$ is also a DAG.

Proof: Let $B=V\setminus A(v)$ and $p$ be an arbitrary path in $G'$. We observe that once $p$ enters $B$, it can never go back into $A(v)$, since all edges with only one node in $A(v)$ must point outwards into $B$, otherwise $A(v)$  would not include all ancestors. But that means we can split the path into 3 parts: one part contained in $A(v)$, a single edge from $A(v)$ into $B$ and a part contained in $B$. Since both $A(v)^R$ and $B$ are DAGs, this implies $p$ must be circle free.


问:
Solve the ODE $y'^{2}+y^2=1$
I can figure out solutions $y=\sin x$ or $\cos x$ and trivial solution $y=1$. But how to get all solutions?

答:
Hint

If you're looking for the solutions around $x=0$ and supposes that $\vert y(0) \vert \lt 1$ then the ODE is equivalent to

$$\frac{y^\prime(x)}{\sqrt{1 - y^2(x)}} = \pm 1$$ which can be integrated as

$$\arcsin\left(y(x) \right) = \pm x + C$$ and leads to
$$y(x) = \sin\left(\pm x +C\right).$$

While if $\vert y(0) \vert \gt 1$ there is no solution and the solution is constant when $\vert y(0) \vert =1$.

Title: Solve the ODE $y'^{2}+y^2=1$


答:
To find the inverse start by writing $y=3^{x-1}-2$. Now, can you rearrange it so you get $x=\ldots$ ?

For example $y+2 = 3^{x-1}$. Then how do you "unpick" a power? You take the logarithm:

\begin{array}{ccc}
y+2 &=& 3^{x-1} \\ \log_3(y+2) &=& \log_3(3^{x-1}) \\ \log_3(y+2) &=& x-1
\end{array}

Finally, $x=\log_3(y+2)+1$. Your inverse function is then $\operatorname{f}^{-1}(x) = \log_3(x+2)+1$.


问:
epsilon delta limit proof verification of $\lim_{x \to 0} \frac{x-1}{x^2-1}=1$
First at all,

Let $\lim x→0 \frac{x-1}{x^2-1}=1$  then there exists $\delta>0$ such that if $0<|x|<\delta$ then $|f(x)-1|< \varepsilon $

Proof:

$\bigg|\frac{x-1}{x^2-1}-1\bigg|=\bigg|\frac{x}{x}\big(\frac{1-1/x}{x-1/x} \big)-1\bigg|=\bigg|\frac{1}{x-1/x}-\frac{1/x}{x-1/x}-1 \bigg|=\bigg|\frac{1}{x-1/x}-x^2+1-1 \bigg|=\bigg|\frac{1}{x-1/x}-x^2 \bigg|$

I use the triangular inequality,

$\bigg|\frac{1}{x-1/x}-x^2 \bigg|\leq\bigg|\frac{1}{x-1/x}\bigg|+\bigg|-x^2 \bigg|=\bigg|\frac{1}{x-1/x}\bigg|+\bigg|x^2 \bigg|=\bigg|\frac{1}{x-1/x}\bigg|+\bigg|x \bigg|^2$

Let say
$\frac{1}{x-1/x}<1$. If that's the case then:

$1<x-\frac{1}{x} \Longrightarrow 1+\frac{1}{x}<x \Longrightarrow \frac{x+1}{x}<x \Longrightarrow x+1<x^2 \Longrightarrow 0<x^2-x-1$

Using the quadratic formula, I know that inequality is true for $x:(-1.6 ; 0.6)$ approximately, which is good knowing that I'm interested in those $x$ near $0$.
Then,

$\bigg|\frac{1}{x-1/x}\bigg|+\bigg|x \bigg|^2\leq1+|x|^2<\varepsilon$

I take $\delta=\sqrt{\varepsilon -1}$ and QED, which is weird because $\varepsilon$ should be bigger than $1$, and isn't that a bit big?

I'm sure there's a simpler way to do this. But I'd like to know if it's ok, or where did I go wrong...

答: In your first line you replaced $\frac {1/x} {x-\frac 1 x}$ by $x^{2}-1$ which is not correct. Let $0 <\epsilon <1$ Then $|f(x)-1|=|\frac 1  {x+1} -1|=\frac {|x|} {|x+1|} \leq \frac {|x|} {1-|x|}<\epsilon$ if $|x| <\frac {\epsilon} {1+\epsilon}$. Take $\delta =\frac {\epsilon} {1+\epsilon}$.


问:
A question about constructible real numbers
This question was asked in my Field theory quiz and I was unable to solve it. So, I am asking for help here.

Let c and d be constructible real numbers. Show that c+d and c-d are also constructible.

I am clear with the definition of constructible real numbers and related theorems but still I was unable to solve the question. I have studied related results from Algebra by THomas HUngerford.

Can you please give solution proving any 1 of them to be constructible. That would really help me.

Thank you.

答:
Let $l$ be a line containing two points $A$ and $B$ that are $c$ apart. Draw a circle with radius $d$ centred at $B$; it intersects $l$ at $C$ (closer to $A$) and $D$ (farther from $A$). Then $\overline{AC}=c-d$ and $\overline{AD}=c+d$.

This presumes $c\ge d\ge0$; other cases can be handled by basic arithmetic properties and interpreting distances along $l$ as signed, with $A$ at the origin.


问:
Function on a star domain has a primitive
Let $U \subset \Bbb C$ be a star domain and $f:U \to \Bbb C$ continuous, s.t.
$$\int_{\partial \bigtriangleup}f(z)dz=0$$
for all triangles $\bigtriangleup$ which are in $U$. It is to be shown that $f$ has a primitive.

Let $z_0 \in U$ be a center of the star domain, so for $z \in U$ the segment between $z_0$ and $z$ is in $U$. We denote the segment between $z_0$ and $z$ with $[z_0,z]$. Let
$$F(z):=\int_{[z_0,z]}f(w)dw$$
Now $U$ is open and so it exists $r>0$ s.t. $B(z,r) \subset U$. If we take $\vert u \vert <r$, then the segment $[z,z+u]$ lies in $U$. Because the integral over a triangle is $0$ it follows for the triangle $[z_0,z,z+u]$ that
$$F(z+u)-F(z)=\int_{[z,z+u]}f(w)dw.$$
But now how can I conclude the proof? I should show that
$$\lim_{u \to 0}\frac{F(z+u)-F(z)}{u}=f(z)$$
But I got stuck...

答:
With the parametrization $[0, 1] \ni t \mapsto z +tu$ of the segment $[z, z+u]$ we get
$$
F(z+u)-F(z)= u  \int_0^1 f(z +t u) \, dt
$$
and therefore
$$
\frac{F(z+u)-F(z)}{u}  - f(z) = \int_0^1 \bigl(f(z +t u) -f(z) \bigr) \, dt \, .
$$
$f$ is continuous at $z$, so given $\epsilon > 0$ there is a $\delta > 0$ such that $|f(w) - f(z)| < \epsilon $ if $|w-z| < \delta$. It follows that if $|u| < \delta$ then
$$
\left|\frac{F(z+u)-F(z)}{u}  - f(z) \right| \le \int_0^1 \underbrace{|f(z +t u) -f(z) |}_{< \epsilon} \, dt < \epsilon
$$
and that proves the desired limit.


问:
For $x \in \Bbb R$, for all $y (|y|<x \implies y<5)$
Given that $x$ is a free variable beloging to the set of real numbers, find the set of values for $x$ for which the following statement is true:

$\forall y (|y|<x \implies y<5)$

My attempt:

If $x < 0$ then the hypothesis $|y|<x$ is False, so the statement is True.

If $x \geq 5$ then if $|y|$ is between $5$ and $x$, the hypothesis is True but the conclusion is False, so the statement is False.

If $0 \leq x < 5$ then the hypothesis is True for all $|y| < x$ and the conclusion $y < 5$ is also True.

So my answer would be the set $x < 5$. However the answer in the back of the book says the set $(-\infty,0] \cup [5,\infty)$. Where am I wrong?

答:
Consider the contrapositive which is logically equivalent and easier to work with:

If $\forall y( y \geq 5 \implies (|y| \geq x))\Longleftrightarrow \forall y $$( y \geq 5,\implies  (y \geq x))$ since $y \geq 0$.

If $x > 5$, then choose $y = 5$ which contradicts the statement.

Hence we must have all $x \leq 5$, specifically $(-\infty,5]$ as BrianO commented.

You missed $x = 5$, here is why it works for your original statement:

Note that $y \leq |y|$.  If $x = 5$, then the statement becomes $\forall y (y\leq|y|<5 \implies y < 5)$, which is seen to be true immediately.


答:
Let $I$ an interval in $ \mathbb R$ and let $y:I \to \mathbb R$ be a solution of the ODE.

Furthermore we assume that $y'(x) \ne 0$ for all $x \in I$. This gives $|y(x)|<1$ for all $x \in I.$ From $y'(x)= \pm \sqrt{1-y(x)^2}$, we see that $y'$ is differentiable on $I$ and

$$2y'(x)y''(x)+2y'(x)y(x)=0$$

for all $x \in I.$ Hence

$$ y''+y=0.$$

The last ODE has the general solution $y(x)=a \cos x+ b \sin x.$ It is now easy to see that $a^2+b^2=1.$

Conclusion:

Title: Solve the ODE $y'^{2}+y^2=1$

Body: I can figure out solutions $y=\sin x$ or $\cos x$ and trivial solution $y=1$. But how to get all solutions?


问:
Odd and even functions- a direct sum?
Question:
Let V be the vector space of all functions $\Bbb R\to \Bbb R$.
Show that $V=U \oplus W$
for $U=${$f | f(x)=f(-x) \forall x$}$, $W={$f | f(x)=-f(-x) \forall x$}

What I did:

I did prove that $U \cap W$={$0$}. But proving that any function from R to R can be displayed as a sum of odds and evens wasn't a success. I tried saying that for $v \in V, w \in W: v=v-w+w$ and proving that $v-w \in U$ but that didn't work (That trick worked with some linear transformations we saw, but this isn't a linear transformation).

Title: Odd and even functions- a direct sum?

Body: Hint: $$f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}$$


问:
Finding a maximal cylinder to be placed in a sphere
Take both volumes, viz.: pi r^2 h to be < 4/3 pi a^3, to distinguish between radii of the 2 figures, let us call the radius of the sphere a

pi is divided out

r^2 h < 4/3 a^3

It is time to take the derivate in order to find the maximal r, but the 4/3 a^3 in case of differentiating on r, is a constant, and is lost in the derivative. With thus all useful info is lost. But I see no way to retain the constant in the derivative. Or is there a cleverer way? thank you

答:
What you're missing is you haven't actually made any link between r,h and a.

I presume a is constant here, and you want the r and h of a cylinder that (a) fits inside the sphere, and (b) has maximal volume. Yes?

Intuitively, the top and bottom of the cylinder will be jammed against the internal walls of the sphere. If you draw a side-on diagram, you'll see a right-angle triangle with sides $r$ and $h/2$, and hypotenuse $a$. So, you get $a^2=r^2 + h^2/4$.

Now, you can either

rearrange that formula, substitute it into the equation for the cylinder's volume to get a formula involving either $r$ only or $h$ only, then differentiate that to find the $r$ or $h$ that maximises the volume

differentiate both $V=\pi r^2 h$ and $a^2=r^2+h^2/4$ implicitly, to get two equations for $dV/dh$ and $dr/dh$ (or $dV/dr$ and $dh/dr$). Then set $dV/dh$ (or $dV/dr$) to zero, eliminate the other derivative and one of $h$ and $r$ (using $a^2=r^2+h^2/4$), and you'll have the other of $h$ and $r$.


答:
No, no proof of $A$ being true is possible, because as others have pointed out, the premises are contradictory.

When you have contradictory premises, no implications can be meaningful.

For example, suppose that you have two premises:

[1] $P$

[2] $\neg P.$

Suppose that $Q$ is any other assertion.

Then, from [1], you have $(P \vee Q).$

Then, you have, (from the previous result + [2]),

that $Q$ must be true.

You can similarly prove that $\neg Q$ must also be true.

Clearly the situation becomes nonsensical.

Title: Show that the case $A$ is true

Body:
Let $A,B,C$ three logical cases. Assuming that $B$ is false and $C$ is true. Assuming that we have the following implications:

$C⇒A$

$C⇒B$

$C⇒A$ or $B$

Then show that the case $A$ is true.


问:
Conditional Probability (Subset)
"Suppose $A\subseteq B$ and $Pr(B)\gt0$, show that $Pr(A|B)\geq Pr(A)$." (There are no other antecedents given.)

$Pr(A|B)=\frac{Pr(A)}{Pr(B)}$. Which implies, $Pr(A|B)Pr(B)=Pr(A)$. I don't think the inequality is true. I think the inequality should have been the other way around. Am I missing something?

答:
$Pr(A|B)=\frac {\Pr(A \cap B)} {Pr(B)}=\frac {Pr(A)} {Pr(B)} \geq Pr(A)$.

The first equality is the definition.  The second equality follows from the fact  that $A \subseteq B$ which implies $A \cap B=A$. The ineqility at the end follows from the fact that $P(B)\leq 1$.

Title: Conditional Probability (Subset)


问:
How do I find the value of k based on the fact that x+k is a tangent to a parametric equation?
I have been given the question of:

A curve has the parametric equations x=2$t^2$ and y=4t. Find the value(s) of k if y=x+k is a tangent to the curve.

Being the first question I've gotten of this nature I have no idea on how to go about it.

答: So first find cartesian equation of parametric equation: y=4t so y^2=16t^2 then y^2/x=8 so x=y^2/8. also y=x+k so x=y-k. now y^2/8=y-k so y^2-8y+8k=0. it is clear we want one solution for y or else the line will intersect at more than one point and not be tangent. so discriminant =0, (-8)^2-32k=0 so k=2. Final answer y=x+2 only one solution


问:
How to find the perimeter of a shaded region when one side isn't given?
The problem is as follows:

Use the figure from below to find the orange shaded regions.

$\begin{array}{ll}
1.&16(4+\pi)\,\textrm{in}\\
2.&8(12+\pi)\,\textrm{in}\\
3.&4(16+\pi)\,\textrm{in}\\
4.&12(8+\pi)\,\textrm{in}\\
\end{array}$

I'm confused exactly how to approach this problem?. The thing is how to get that side for the little square which is on the top left?.

Because the side has some displacement or off set from the top makes it difficult to get the radius.

I've attempted to do the following:

Assuming $a$ is the size of one side on the little square:

Where $P$ is perimeter:

$P=(8+4\sqrt{2}-a)\sqrt{2}\cdot \pi+4(8+4\sqrt{2}-a)+2(4)(a)$

But this expression is not possible to simplify. I mean the term $a$ doesn't cancel.

What can it be done here?. Can someone help me please?. Is there any trick here or what?.

答:
Note that the diagonal of the inner square is the diameter of the circle. With $r$ being the radius of the circle and $i$ the side length of the inner square, you get according to Pythagoras

$$2i^2 =(2r)^2$$

Hence, you get the following system of equations for $a$ and $r$:

\begin{eqnarray*}
\sqrt 2 r + a & = & 8+4 \sqrt 2\\
2 r - a & = & 8+4 \sqrt 2\\
\end{eqnarray*}

Solving this gives $r=8$ and $a=8-4\sqrt 2$. So, the perimeter $P$ of the shaded region is $P = 16(\pi + 4)$.


答:
It is $$\int \frac{dy}{\sqrt{1-y^2}}=\pm \int dx$$
$$\implies \sin^{-1} x=\pm x+C \implies y(x)=\pm \sin (x\pm C)$$
So the general solution can also be written as $y=\pm [\sin x \cos C\pm \cos x \sin C]$ If $C=0$ Solution is $y=\pm \sin x$, if $C=\pi/2$ the solution is $y=\pm \cos x$.$y(x)=1$ is singular (essential) solution of this ODE which is free of constant.

Title: Solve the ODE $y'^{2}+y^2=1$


问:
Show that $\frac{ \sin(z) }{z(1 + z^2)} = \sum_{n = 0}^\infty (-1)^n z^{2n} \sum_{k = 0}^n \frac{1}{(2k + 1)!}$ for $|z| < 1$
Show that $$\frac{ \sin(z) }{z(1 + z^2)} = \sum_{n = 0}^\infty (-1)^n z^{2n} \sum_{k = 0}^n \frac{1}{(2k + 1)!}\text{ for }|z| < 1.$$

I'm doing the exercise 2.7.17 from the book Mathematical Methods in the Physical Sciences, Third Edition. I'm trying to multiply the series for $\sin(z)/z$ to the one for $(1 + z^2)^{-1}$, but i'm struggling with the change of indices. How do I obtain the upper limit $n$ for the sum on $k$?

答: Let $x:=-z^2$. We need to rearrange $\prod_{i\ge0}\frac{x^i}{(2i+1)!}\prod_{j\ge0}x^j$, in which the $x^n$ coefficient is $\sum_{j=0}^n\frac{1}{(2(n-j)+1)!}$, to $\prod_{n\ge0}x^n\sum_{k=0}^n\frac{1}{(2k+1)!}$. Just identify $k:=n-j$.


答:
$$y'^2=1-y^2.$$

If $y^2=1$, we get $y'=0$ and we have two constant solutions $y=\pm1$.

Otherwise,

$$\frac{y'^2}{1-y^2}=1$$ or $$\frac{y'}{\sqrt{1-y^2}}=\pm1^*.$$

Then by integration,

$$\arcsin(y)-\arcsin(y_0)=\pm x$$ and

$$y=\sin(\arcsin(y_0)\pm x).$$

$^*$Note that we cannot switch from $+1$ to $-1$ at some arbitrary $x$, as this would introduce a discontinuity in the derivative. Hence the sign remains constant for all $x$.

Title: Solve the ODE $y'^{2}+y^2=1$


Body:
$$
y=3^{x-1}-2
$$

$$
y+2=3^{x-1}/\cdot\log_3
$$

$$
\log_3(y+2)=\log_3 3^{(x-1)}
$$
$$
\log_3(y+2)=x-1
$$
$$
x=\log_3 (y+2)+1
$$

Now we take the substituon $x=y^{-1}$, and we have
$$
y^{-1}=\log_3 (x+2)+1
$$


问:
$ \int_A \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} = 0 \implies \int_A \frac{1}{\sqrt{2 \pi}} e^{-\frac{(x-h)^{2}}{2}} = 0 $
I'm trying to show the following: For all $A \in \mathcal{B}(\mathbb{R})$ and $h \in \mathbb{R}$,
$$
\int_A \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} = 0 \implies \int_A \frac{1}{\sqrt{2 \pi}} e^{-\frac{(x-h)^{2}}{2}} = 0.
$$
This seems so easy intuitively, but I don't know how to solve it. Intuitively, as $\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}}$ is a strictly positive continuous function, and the Riemann integral is subsumed into the Lebesgue integral, I may hypothesize that $A$ is "disconnected" at every element of $A$, whatever that is supposed to mean here. The picture in my mind is that that "disconnectedness" carries on to the integral on the right, giving zero. I know this is (definitely) not a proof, not even close, but can anyone provide a nice solution to the above?

For the context of where this question came from, I had this question while doing the exercise:

For all $h \in \mathbb{R},$ let $N(h, 1)$ be the probability measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ given by
$$
N(h, 1)(A)=\int_{A} \frac{1}{\sqrt{2 \pi}} e^{-\frac{(x-h)^{2}}{2}} d x
$$
Show that the measures $N(h, 1)$ and $N(0,1)$ are equivalent, and compute the corresponding Radon-Nikodym derivatives.

Proof: For all $x \in \mathbb{R}$,
$$
\frac{1}{\sqrt{2 \pi}} e^{-\frac{(x-h)^{2}}{2}}=\exp \left(x h-h^{2} / 2\right) \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}}
$$
so $N(h, 1)$ is absolutely continuous w.r.t. $N(0,1)$.... (The rest of the proof is irrelevant)

答: Since the integrand is strictly positive, the hypothesis implies that $A$ has measure $0$. The integral of any measurable function over  a set of measure $0$ is $0$. (This is a complete and logically correct proof.)


问:
Determining the orthogonal projection a subset of $L^2(\mathbb{R})$
Define $K \subset L^2(\mathbb{R})$ (with $L^2(\mathbb{R})$ being the set of all square-integrable functions) as
\begin{align} K = \{ f \in L^2(\mathbb{R}) \ | \ f(-x) = 2f(x) \ \text{for almost all} \ x \geq 0 \}.\end{align}

I have to prove and find an expression for $K^{\perp}$ and an explicit formula for the orthogonal projection $p_K(f)$ for all $f \in L^2(\mathbb{R})$. However, I have troubles finding an explicit orthonormal basis for $K$ (with this I mean a set of orthonormal functions such that $K$ is equal to the  closure of the linear span of that set) as that's the only way I know to find a formula for $p_K(f)$ (and thus also find $K^\perp$). I haven't yet learned anything about Fourier series, so I don't think I will need that.

答:
Suppose $g\in L^2$ is orthogonal to $K$. Then
$$
       \int_{-\infty}^{\infty}g(x)\overline{f(x)}dx =0,\;\; f\in K,
$$
which may be written as
$$
       0=\int_{-\infty}^{0}g(x)\overline{f(x)}dx+\int_{0}^{\infty}g(x)\overline{f(x)}dx \\
        =\int_{0}^{\infty}g(-x)\overline{f(-x)}+g(x)\overline{f(x)}dx\\
        =\int_{0}^{\infty}g(-x)2\overline{f(x)}+g(x)\overline{f(x)}dx \\
        =\int_{0}^{\infty}(2g(-x)+g(x))\overline{f(x)}dx.
$$
This holds for arbitrary $f\in L^2[0,\infty)$, which is equivalent to $g(x)=-\frac{1}{2}g(-\frac{x}{2})$ for a.e. $x \in (-\infty,0]$.


问:
Number of minors of a full column rank matrix
Let a matrix $A \in \mathbb{R}^{m \times n}$ be given, with $m \geq n$ and $\mathrm{rank}(A) = n$. We know that there exists at least one non-vanishing minor of order $n$.

Now, let $A$ be appended with a matrix $B \in \mathbb{R}^{p \times n}$, getting $\left[ \begin{array}{c} A\\ B \end{array}\right] \in \mathbb{R}^{(m+p) \times n}$, with $\mathrm{rank}(B) < n \leq p$. Does the number of non-vanishing minors of order $n$ possibly increase?

答:
Yes, it's possible. To see this, let's pick an extreme example: $A$ is the identity matrix appended vertically to a zero matrix. Hence there is exatly one nonzero minor of order $n$.

You then append vertically a matrix $B$ that is made of one column vector of ones and all the other column vectors are null. $B$ has rank $1<n$ as required, as long as $n\ge2$.

Then, by taking the rows $2\dots n$ together with any of the $p$ last rows, you get a nonzero minor (the submatrix is the identity, apart from a permutation of the rows).

More generally, with any $A$ and $B$ satisfying the conditions in the question, any nonzero row $u$ of $B$ will add at least one more nonzero minor: you can always remove one row of the known nonzero minor and replace it with $u$, to get a nonzero minor. But not always an arbitrary one: $u$ is a linear combination of the rows of the minor (since these row vectors form a base of $\Bbb R^n$), i.e. $u=\sum \lambda_i r_i$ with at least one nonzero $\lambda_i$. Then for any $\lambda_i\ne0$, if you exchange $u$ and $r_i$ you will get a nonzero minor.


答:
Failed attempt:

As the $(y,y')$ trajectory is a circle, we can write the solution as $$y(x)=\sin(u(x)),y'(x)=\cos(u(x)).$$ This requires

$$y'(x)=\cos(u(x))=u'(x)\cos(u(x))$$ and $u'(x)=1$ or $\cos(u(x))=0$.

The first case yields $u(x)=x+c$, or $y(x)=\sin(x+c)$, and the second is $y(x)=\pm1$.

Unfortunately, this method does not yield the most general solution because it does not cover the case

$$y(x)=\cos(u(x)),y'(x)=\sin(u(x))$$ that results in $u'(x)=-1$.

Title: Solve the ODE $y'^{2}+y^2=1$


问:
Vector calculus for ellipse in polar coordinates
I'm having trouble with this question, can somebody please help me with it! I'll thanks/like your comment if help me =)

I know that for a ellipse the parametric is $x=a\sin t$ , $b= b \cos t$, $t\in (0, 2\pi)$ (?)

For part a) I drew up the graph but not sure if it's right. The circle has radius 1 and for the ellipse I'm able to find $x= 2 $.
$$
(x-1)^2 =9  \\
\sqrt{(x-1)^2}=\sqrt 9\\
x-1=3\\
x=4$$

For $y=2\sqrt 2$
$$
y^2=8\\
\sqrt y=\sqrt 8\\
y=\sqrt 4 \sqrt 2\\
y= 2\sqrt 2 \text{ or } 2.8 $$

So the region should be the circle? since the ellipse covers the whole circle?
Thank you very much for helping!
Cheers.

答:
I am noting you a useful Theorem which I found it for you. So by using it you can defeat the second part of your question by yourself.

Theorem: The proper conic with focus-directrix distance $p$ and eccentricity $e$ has equation $$r=\frac{ep}{1-e\cos \theta}$$ in polar coordinates, if the focus is at the pole and the polar axis is perpendicular to the directrix $L$ in the direction pointing away from $l$. (Fig below)

Additional points here are $e=\frac{c}a<1,p=\frac{a^2}{c}-c$ where your ellipse is $$x^2/a^2+y^2/b^2=1$$ Here we have $a=3,b=\sqrt{8}$ so $$c=\sqrt{a^2-b^2}=1,~~e=\frac{1}3,~~p=\frac{9}{1}-1=8$$ Now put these to the formula in the theorem. This solves the second part.


问:
Can we combine this two Lyapunov functions (which imply local stability by separate) to conclude global stability?
Let $x(t)\in\mathbb{R}^n$ constrained to a dynamical system
$$
\dot{x}(t) = f(x(t))
$$
for some vector field $f:\mathbb{R}^n\to\mathbb{R}^n$. Moreover, the dynamical system has a unique equilibrium point at the origin.

My goal is to conclude that for this system (with a particular $f$ I have) the origin is globally asymptotically stable. So far, I have found two Lyapunov function candidates $V_1(x),V_2(x)$ which both are positive definite and radially unbounded. Moreover, I managed to show that they comply

$\dot{V}_1(x)<0$ $\ \forall x : \|x\|<r_1$

$\dot{V}_2(x)<-c$ $\ \forall x : \|x\|>r_2$ and some constant $c>0$

$0<r_1<r_2$

If it was the case that $r_1>r_2>0$ instead of item 3), we would be done: global stability follows since item 2) ensures that $x$ reaches $\|x\|\leq r_2$ in finite time from any initial condition, and then one can use 1) to conclude $x$ reaches the origin asymptotically from here.(as pointed out in the comments, this argument requires nuance, nevertheless, the important part is the following). However, from item 3) this is not the case. Thus, I have this disk $r_1<\|x\|<r_2$ in which my Lyapunov stability test is inconclusive. My hope is to show that trajectories $x(t)$ cannot stay in the disk $r_1<\|x\|<r_2$ forever but ultimately reach $\|x\|<r_1$.

My question is: from the information given here, do you think is possible to conclude global asymptotic stability either by constructing a new Lyapunov function from $V_1,V_2$ or by checking some condition on $f$ to conclude trajectories cannot stay in $r_1<\|x\|<r_2$?

Indeed, there are many counterexamples in which this is not possible. But my question here is if there is a way to check if it is possible: what properties do you suggest me to check for $f$. I'm looking for suggestions, ideas or relevant bibliography that you think can help me.

EDIT: In case this problem is easier, or adds useful information, I can also show the following instead of the original items:

$\dot{V}_1(x)<0$ $\ \forall x \in M_1 = \{x : V_1(x)<r_1'\}$

$\dot{V}_2(x)<-c$ $\ \forall x \in M_2=\{ V_2(x)>r_2'\}$ and some constant $c>0$

$D:=\mathbb{R}^n\setminus(M_1\cup M_2)\neq \emptyset$

We are left to show that there are no invariants in $D$.

EDIT 2: After some discussion in the comments, I think the main essence of my question, is more on what happens when the Lyapunov arguments are inconclusive on this non empty set $D$ (bounded, but not containing the origin). A concrete diagram for the state space of what $r_1,r_2,M_1,M_2,D$ looks in the case I am interested is the following:

答:
Bounds on $V$

We start considering the case when we have bounds on the levels of $V$, and we will use these results to study the original question later.

Let us say that there exist $C_1$ such that $\dot{V}_1(x)<0$ for all $x$ such that $V_1(x)>C_1$ and $C_2$ such that $\dot{V}_2(x)<0$ for all $x$ such that $V_2(x)<C_2$. I.e., the $C_1$ constant defines the level set for $V_1$ such that $x$ is converging for large initial conditions, and $C_2$ defines the level set for $V_2$ such that $x$ is converging for small initial conditions. When the system is converging? The intuition is that we need a safe switch from $V_1$ to $V_2$.

One way...

Choose an arbitrary small constant $\delta>0$ and define $\Omega = \{x: V_1(x)\le C_1+\delta\}$. From the properties of $V_1$ we know that $x$ reaches $\Omega$ in finite time, and since $V_1$ is radially unbounded, the set $\Omega$ is bounded and contains the origin. Then to prove the convergence we need that $V_2(x)<C_2$ for all $x\in\Omega$. Then for any $x_0\in\Omega$ we have $V_2(x_0)<C_2$ and due to the properties of $V_2$ we conclude that $V_2\to 0$. Note that the trajectory $x$ can leave the set $\Omega$ as it is not invariant.

We need the constant $\delta$ because otherwise, it is possible that $x\to\Omega$ asymptotically and we never switch to $V_2$.

...or another

Another way to see the same idea is to define the set $\mathcal{M} = \{x:V_2(x)<C_2\}$. From the properties of $V_2$ we know that the set $\mathcal{M}$ is invariant ($V_2$ is monotonic decaying for all $x\in\mathcal{M}$ and thus $V_2$ will always remain below $C_2$), contains the origin, and the origin is attractive for all initial conditions in $\mathcal{M}$ since $V_2\to 0$.

Then to prove the convergence we need to show that the set $\Omega = \{x: V_1(x)\le C_1+\delta\}$ lies inside $\mathcal{M}$, $\Omega\subset\mathcal{M}$. Then any initial condition outside $\mathcal{M}$ converges to $\Omega$ (and thus to $\mathcal{M}$) due to the $V_1$ analysis, and for all trajectories starting in $\mathcal{M}$ the origin is attractive due to the $V_2$ analysis.

NB

We do not say that $C_1>C_2$ or $C_2>C_1$. What is important is the relation between the sets $\Omega$ and $\mathcal{M}$.

UPD: Bounds on $x$

Ok, let us see the original question. Suppose there exists $r_1$ such that $\dot{V}_1<0$ for all $\|x\|>r_1$ and $r_2$ such that $\dot{V}_2<0$ for all $\|x\|<r_2$, and $r_2<r_1$. Let us replace $r_1$ with $R_1 = r_1+\delta$, then we know for sure that $x$ enters the ball $\|x\|<R_1$ in finite time. Choose $C_1$ as the largest value of $V_1$ on the boundary $\|x\|=R_1$. Then the set $\Omega=\{x:V_1(x)\le C_1\}$ is attractive and invariant, and the ball $\|x\|<R_1$ lies inside $\Omega$. Note also that the ball $\|x\|\le R_1$ is not necessarily invariant.

Next, choose $C_2$ as the smallest value of $V_2$ on the boundary $\|x\|=r_2$. Then the set $\mathcal{M} = \{x:V_2(x)<C_2\}$ lies inside the ball $\|x\|\le r_2$, this set is invariant and the origin is attractive for all trajectories starting in $\mathcal{M}$.

So what we know at the moment is that the set $\Omega$ is invariant and attractive for all trajectories starting outside of $\Omega$ and that the set $\mathcal{M}\subset \Omega$ is invariant and all trajectories starting in $\mathcal{M}$ converge to the origin.

What is left is the set $\mathcal{D} = \Omega-\mathcal{M}$. As far as I see, it is sufficient to show that no trajectory staring in $\mathcal{D}$ can stay in $\mathcal{D}$, i.e., there are no invariants in $\mathcal{D}$.


问:
Geometric Distribution vs Exponential Distribution Curve & relationship
The curve of Geometric Distribution is shaped according to parameter $\alpha$ using equation as
$P_{r}=\frac{ (1-\alpha) \alpha^{CW}} { 1- \alpha^{CW} }. \alpha^{-r}$. Where CW is a parameter (contention window size like 16). So probability for occurrence of r=0 to 15 is is represented as shown in figure for different values of $\alpha$. Here 'number starting from 0 are put at higher side so r='0' has highest probability and r='15' has lowest probability.

Question1. So my question is that exponential distribution also gives curve that matches shape of one of the curves of geometric distribution so how can we find value of $\alpha$ that give same curve as given by exponential distribution.
I have also read in Wikipedia that "The exponential distribution is the continuous analogue of the geometric distribution."  https://en.wikipedia.org/wiki/Geometric_distribution

Question 2. So how can I prove any relationship between them like when exponential distribution equals geometric distribution?.  It is some times also referred as "truncated geometric probability  distribution"

答:
The geometric and exponential distributions are not the same, since they aren't even defined on the same domain. The geometric distribution lives on a discrete domain, the exponential distribution on a continuous domain. Consequently, you can't make the graphs of their density functions match perfectly. You can at most get them to look similar.

However, there is a reason why they look so similar: both are memoryless distributions, so they're both characterized by their median (at least approximately in the case of the geometric distribution, since its median is not unique in general). The best match between a geometric and exponential function should be when their medians match. The median of the exponential distribution with density function $f:[0,\infty)\to\mathbb R,~f(x)=\frac1\lambda\exp(-\lambda x)$ is $\frac{\ln2}{\lambda}$. Now calculate the median of your geometric distribution and choose $\alpha$ such that the medians match.


问:
The two equations $F (x, y, u, v) = 0$ and $G (x, y, u, v) = 0$ determine $x$ and $y$ implicitly as functions of $u$ and $v$
The two equations  $F (x, y, u, v) = 0$ and  $G (x, y, u, v) = 0$  determine $x$  and $y$  implicitly as functions of u and v, say  $ x = X (u, v) $ and $ y = Y (u, v) $.   Show that
$$ {\partial X \over \partial u} = {{\dfrac{\partial (F, G)}{\partial (y, u)}} \over \dfrac{\partial (F, G)}{\partial (x, y )}} $$
at the points where the Jacobian $ {\partial (F, G) \over \partial (x, y)} \neq 0 $ and finds similar formulas for the partial derivatives $ {\partial X \over \partial v} , {\partial Y \over \partial u}, {\partial Y \over \partial v} $.

I need help with this exercise, I don't know how to take the function, should I define a composition or another function? Like
$h(u,v)=F(x,y,u,v)=F(X(u,v),Y(u,v),u,v)$?

答:
Let $h(u,v):=F(X(u,v),Y(u,v),u,v)$ and $k(u,v):=G(X(u,v),Y(u,v),u,v)$. We know that $h,k$ are identically $0$, so their derivatives are also $0$

Then by the chain rule

$$
\begin{align}
0&=&h_u &=& F_x X_u +F_y Y_u + F_u+0\\
%0&=&h_v &=& F_x X_v +F_y Y_v + 0+ F_v\\
\text{and}\\
0&=&k_u &=& G_x X_u +G_y Y_u + G_u+0\\
%0&=&k_v &=& G_x X_v +G_y Y_v + 0+ G_v\\
\end{align}
$$
which we can write as
$$
\begin{pmatrix}
F_x & F_y\\
G_x & G_y
\end{pmatrix}
\begin{pmatrix}
X_u\\
Y_u
\end{pmatrix}
=-
\begin{pmatrix}
F_u\\
G_u
\end{pmatrix}.
$$
Inverting the $2\times 2$ matrix we have
$$
\begin{pmatrix}
X_u\\
Y_u
\end{pmatrix}
=
\frac{1}{\dfrac{\partial (F,G)}{\partial (x,y)}}
\begin{pmatrix}
-G_y & F_y\\
G_x & -F_x
\end{pmatrix}
\begin{pmatrix}
F_u\\
G_u
\end{pmatrix},
$$
and we can then see that
$$
X_u
=
{{\dfrac{\partial (F, G)}{\partial (y, u)}} \over \dfrac{\partial (F, G)}{\partial (x, y )}},
$$
with a similar expression for $Y_u$.

Differentiating $h,k$ with respect to $v$ will give the formulae for $X_v, Y_v$.


问:
A question from my textbook (AS LEVEL STATS, PERMUTATIONS)
How many ways 4 men and 2 women can stand in a line if no two men may stand next to each other

The answer to the question is 0.

答: Suppose the person on the end of the line is a woman. Then, the person standing next to her must be a man, who in turn must to be standing next to a woman. Continuing this line of reasoning suggests that there is no way for this to be achieved if a woman is on the end of a row. The same can be done for when a man is standing on the end of a row.</p>


问:
Applying conditions inside and outside a probability.
For random variables $X_n$ for $n\in \mathbb N$ and a random variable $X$, consider the statements

(A) $P(X_n=X\text{ for all }n\in \mathbb N)=1$;

(B) $P(X_n=X)=1$ for all $n\in \mathbb N$.

I thought A and B were equivalent statements, but instead A implies B. Why is this the case?

答: They are indeed equivalent! This follows from the fact that a countable union of events of probability $0$ has probability $0$. (Consider the events $(X_n \neq X)$).


问:
Find the minimum of $x_n$ where $n\in \mathbb{N}^{*}$.
I denote with $x_n$ the sum of the digits of $2771^n$. I have to find the minimum of $x_n$ where $n\in \mathbb{N}^{*}$.

I computed few terms of the sequence and I think that $x_1$ is the minimum. I notice that $x_n$ should be congruent with $1$ or $8$ modulo $9$. I tried to find an argument why it can't be $8$ or $10$ but I didn't succeed.

答:
Consider $2771^n$ for $n>0$ and let $O$ be the sum of the digits located in odd position. Let $E$ be the sum of the remaining digits. (For $n=1$ we have $O=8,E=9$.) Then

$$O+E\equiv 2771^n \equiv(-1)^n \pmod 9$$
$$O-E\equiv 2771^n \equiv(-1)^n \pmod {11}$$

For a total digit sum of less than 17 we require $O+E=9+(-1)^n$ but then
$$2E\equiv 9\pmod {11} \implies E\ge 10\implies O+E> 10.$$

Therefore the digit sum cannot be less than $17$.


问:
A question based on algebraically closed field and extending an isomorphism to monomorphism over different fields
The following question was asked in mind term exam (held in november) and I was unable to solve it. Now I was trying to solve it again and failed this time too.

So, I am in need of help.

F is an algebraic closure of K if and only if F is algebraic over K and for every
algebraic field extension E of another field $K_1$ and isomorphism of fields
$\sigma : K_1 \to K , \sigma$ extends to a monomorphism $E  \to F$.

If given is that F is an algebraic closure of K then it is clear that F is algebraic over K but I am not clear on how should I prove that $\sigma$ extends to a monomorphism $E\to F$. Actually I am bad at proving results when any type of morphism is to be extended.

For the  converse, I am not able to get any idea.

It is my humble request to you to guide me in this.

答:
The part about extending $\sigma$ is a bit lengthy and usually involves two steps:

Proving the case where $E=K(\alpha)$ is a simple extension field

Concluding the general case by Zorn's lemma.

This should be covered in pretty much any algebra/field theory book.

For the other direction. Take an algebraic closure $C$ of $F$. By applying the hypothesis to $\sigma=\operatorname{id}_K:K\to K$ we get a $K$-homomorphism $f:C\to F$. The image $f(C)$ will then contain $K$ and is still algebraically closed. Since $F/F(C)/K$ is algebraic it follows that $F=F(C)$, hence $F$ is algebraically closed.

(Also note that a homomorphism of fields is automatically a monomorphism)


问:
"Old-Fashioned" O-level Maths Book
I'm looking for recommendations for an "old-fashioned" style of maths textbook for O-level/GCSE, i.e., one which is concise (not full of pictures) with plenty of exercises. Ideally something which is available in PDF format online. I don't necessarily mean an old book, more a book written in a style which is typical of the early 20th century or earlier. (For instance, Gwynne's Latin is a modern book for learning Latin which I would consider to fall under this category.)

I am quite fond of this one which I own a physical copy of, but am struggling to find anywhere online in PDF format.

Also I would appreciate if it is just one book rather than many volumes. A lot of maths textbooks made for school come in many volumes with titles like "School Maths 1A, 1B, 2A, 2B", etc.

To clarify, by "O-level/GCSE", I mean a book with a table of contents which should at least tackle all of the following:

答:
I would recommend something like this. It's not specifically for O Level's or GCSE's but I've found it to be an excellent textbook.

Alternatively, if you are just looking for the facts and how to use the facts (so without any, or with very few proofs) then I can recommend the CGP GCSE Maths study guides. They are extremely cheap and extremely good, complete with plenty of examples and exercises.

You can also buy a CGP GCSE Maths questions book that is meant as a companion to the main study guide; it contains hundreds of questions, and is once more also very cheap. The links are below:

https://www.cgpbooks.co.uk/secondary-books/gcse/maths/mxhr47-new-2021-gcse-maths-edexcel-revision-guide

https://www.cgpbooks.co.uk/secondary-books/gcse/maths/mxhq42-gcse-maths-edexcel-exam-practice-workbook-h

They also have an option of an all-in one maths GCSE book, complete with practice as well:

https://www.cgpbooks.co.uk/secondary-books/gcse/maths/mxhs44-new-2021-gcse-maths-edexcel-complete-revisi

Personally, I would go for the book that is linked directly above, as well as the practice exam workbook as a bonus; thay are what I used (mainly the all in one book) for my GCSE Maths and they served me very well.

Title: "Old-Fashioned" O-level Maths Book


答:
This doesn't need mathematics. Just look at the picture below

$M-M-M-M$

If no two men are to stand next to each other, the $3$ gaps between them must be filled, but there are only $2$ women available, ergo it is impossible.

Title: A question from my textbook (AS LEVEL STATS, PERMUTATIONS)

Body:
How many ways 4 men and 2 women can stand in a line if no two men may stand next to each other

The answer to the question is 0.


问:
Computing expected value of a function along a Brownian Path
I have been reading a paper by Desmond J. Higham  titled "An Algorithmic Introduction to
Numerical Simulation of Stochastic Differential Equations." Here is the link:

https://epubs.siam.org/doi/pdf/10.1137/S0036144500378302

Suppose $W(t) , t \geq 0$ is a Brownian motion. The author wants to evaluate the function $u\big(W(t)\big)  = \exp\big(t + \frac{1}{2}W(t)\big)$ along $1000$ discretized Brownian paths. He writes some MATLAB code which I 'm clear with. He then wishes to compare the mean of the discretised paths with the expected value of $u\big(W(t)\big).$

On page $528,$ fourth line from the last, he mentions that the expected value of $u\big(W(t)\big)$ is $\exp (9t/8).$ I'm not clear as to how he got this. Any help highly appreciated.

答:
Write
$$E[\exp\big(t + \frac{1}{2}W(t)\big)]=e^{t} \cdot E[\exp\big( \frac{1}{2}W(t)\big)]$$
at this point remember the fact that $W(t)$ is a centered Gaussian r.v. with variance equal to $t$, and use the moment generating function to obtain
$$=e^t\cdot e^{1/2 \cdot (1/2)^2 \cdot t}=e^{9t/8 }$$


答:
By Ito you get
$$
X_t=F(t,W_t)=\exp(t+\frac12W_t)
\\
\implies
dX_t=(F_t+\frac12F_{ww})dt+F_wdW_t
=(X_t+\frac12\frac1{2^2}X_t)dt+\frac12X_tdW_t
$$
and the expectation of $X$ follows the equation
$$\newcommand{\E}{\mathbb{E}}
d\E(X_t)=\frac98\E(X_t)dt\implies \E(X_t)=e^{\frac98t}.
$$

Title: Computing expected value of a function along a Brownian Path

Body:
I have been reading a paper by Desmond J. Higham  titled "An Algorithmic Introduction to
Numerical Simulation of Stochastic Differential Equations." Here is the link:

https://epubs.siam.org/doi/pdf/10.1137/S0036144500378302

Suppose $W(t) , t \geq 0$ is a Brownian motion. The author wants to evaluate the function $u\big(W(t)\big)  = \exp\big(t + \frac{1}{2}W(t)\big)$ along $1000$ discretized Brownian paths. He writes some MATLAB code which I 'm clear with. He then wishes to compare the mean of the discretised paths with the expected value of $u\big(W(t)\big).$

On page $528,$ fourth line from the last, he mentions that the expected value of $u\big(W(t)\big)$ is $\exp (9t/8).$ I'm not clear as to how he got this. Any help highly appreciated.


问:
the minimum of the sum of two convex functions
Context: Here is a question jump into my mind, when I met a problem in 1-dim case, I'm wondering if there is a generalization to higher dimensional $\mathbb R^n$ space:

Is the following statement true or false? If it's false, please give counter-example, if it's true, please prove that:

Suppose $f, g$ are two convex function $\mathbb R^n \rightarrow \mathbb R$ with unique minimum in $\mathbb R^n$. The minimum of function $f$ is denoted as $\vec{x^{(1)}}$; the minimum of function $g$ is denoted as $\vec{x^{(2)}}$. Suppose $\vec{x^*}$ is the minimum of $f + g$, then we have
\begin{align}
\min \{\vec{x^{(1)}_j}, \vec{x^{(2)}_j} \} \leq \vec{x^*_j} \leq \max \{\vec{x^{(1)}_j}, \vec{x^{(2)}_j} \} ~~~~~~~~~~\forall 1 \leq j \leq n.
\end{align}

For $1$ dim case, the above statement is of course true; but what if in the high dimensions case $n \geq 2$?

答:
Let $f(x,y)=2(x+1)^2+((x+y)+1)^2$ and
$g(x,y)=(x-1)^2+((x+y)-1)^2$. Since the second partial derivatives of
$f$ and $g$ are positive and both $H_f=\begin{bmatrix} 6 & 2 \\ 2 & 2
\end{bmatrix}$ and $H_g=\begin{bmatrix} 4 & 2 \\ 2 & 2
\end{bmatrix}$ are positive definite, $f$ and $g$ are strictly
convex. The minimum for $f$ is at $(-1,0)$ and for $g$ at
$(1,0)$. Since $0=\partial(f+g)/\partial y$ implies $x+y=0$ and
$\partial(f+g)/\partial x=6x+2+4(x+y)$, $f+g$ achieves its minimum
at $(-1/3,1/3)$. Thus the statement is false.


问:
Is the countable complement topology connected if the set is infinite
There are two cases, I need to work on. If X is countably infinite and uncountably infinite. The second case is easy to show its connected by contradiction. If the topology is connected, then there are two sets A and B both closed whose union is X. But that is impossible because X is uncountable and all closed sets are countable. Hence the set is connected for uncountable case.

Now I need to work on the countably infinite case and I'm stuck on this one. I'm not sure whether this is also connected or not.

答: If $X$ is countable with countable complement topology then every point is open, since its complement is countable. Therefore $X$ is discrete and so disconnected as every discrete space with at least $2$ points is.


问:
Two triangles of equal area

Claim. Given a convex quadrilateral $ABCD$ that isn't parallelogram. The Newton line intersects sides $AD$ and $BC$ of the quadrilateral at the points $H$ and $G$ , respectively . Then $Area(\triangle HGD)=Area(\triangle HBG)$ and $Area(\triangle HGC)=Area(\triangle HAG)$.

GeoGebra applet that demonstrates this claim can be found here.

Proof. Observe that diagonal $HG$ of the quadrilateral $HBGD$ lie on the Newton line of the quadrilateral $HBGD$. If we apply Anne's theorem on point $H$ and quadrilateral $HBGD$ we can write $Area(\triangle HGD)+0=Area(\triangle HBG)+0$ , hence $Area(\triangle HGD)=Area(\triangle HBG)$. Similarlly we can show that $Area(\triangle HGC)=Area(\triangle HAG)$.

Question. Is this proof acceptable? Can we use Anne's theorem in the case when the point lies on the side of the quadrilateral? Can you provide an alternative proof?

答:
Drop from $B$ and $D$ the altitudes $BB'$ and $DD'$ to common base $HG$: triangles $BB'I$ and $DD'I$ are congruent by $ASA$. Hence $BB'\cong DD'$ and $Area(HGD)=Area(HGB)$.

An analogous proof can be repeated for triangles $HGC$ and $HGA$.

Title: Two triangles of equal area


问:
why integrating only alternating forms?
Hello I was reviewing some concepts of differential forms. I cannot recall why only multilinear alternating forms can be integrated on manyfolds and not general multilinear forms... Why is the hypothesis of being alternating so important? In an intuitive way I'd like to understand it... Is it for having a covariant concept? Or for making some riemann sums convergent?</p>

答:
Let $\alpha$ be a covariant $k$ -tensor on a finite-dimensional $n$ vector space $V .$ The following are equivalent:

$\alpha$ is alternating.

$\alpha\left(v_{1}, \ldots, v_{k}\right)=0$ whenever the $k$ -tuple $\left(v_{1}, \ldots, v_{k}\right)$ is linearly dependent.

$\alpha$ gives the value zero whenever two of its arguments are equal:
$$
\alpha\left(v_{1}, \ldots, w, \ldots, w, \ldots, v_{k}\right)=0
$$

The determinant can be defined as an $n$ alternating form such that $\det(I_n)=1$. This hints that alternating forms behave well with volume. As you want lower dimensional subspace to have zero volume in space.

For a $C^{\infty}$ function $f(x)$ on $\mathbb{R}^n$ with compact support, its Riemann integral
\begin{equation}\label{eq:Riemann_integral}
    \int_{\mathbb{R}^{n}} f(x) d x_{1} \cdots d x_{n}=\lim _{\left|\sigma_{j}\right| \rightarrow 0} \sum_{j} f\left(x_{j}\right)\left|\sigma_{j}\right|
\end{equation}
is defined. Here the $\sigma_{j}$ are $n$ -dimensional small cubes which altogether cover supp $f$, $x_{j}$ is a point on $\sigma_{j},$ and $\left|\sigma_{j}\right|$ is the volume of $\sigma_{j}$. we have that translating $\sigma_i$ to the origin gives $|\sigma_i|=|\det \sigma_i|$.
However, if we change the viewpoint slightly and integration of $n$-forms (putting in the symbols $\wedge$ also), and set
$$
\omega=f(x) d x_{1} \wedge \cdots \wedge d x_{n}
$$
Then this is nothing but an $n$ -form on $\mathbb{R}^{n}$. We may consider that the integral $\int_{\mathbf{R}^{n}} \omega$ of $\omega$ on $\mathbb{R}^{n}$ is defined by the riemann integral However, two problems arise. One is that if we change the order of $d x_{1}$ and $d x_{2},$ for example, in the formula of $\omega,$ the sign changes, while in the formula for the riemann integral the integral value does not change. This is where the covariant concept comes to play.

In another set of coordinates we have
$$\omega=f(x(y)) \operatorname{det}\left(\frac{\partial x_{i}}{\partial y_{j}}\right) d y_{1} \wedge \cdots \wedge d y_{n}$$
By the formula of variable change, we have
$$
\int_{\mathbf{R}^{n}} f(x) d x_{1} \cdots d x_{n}=\int_{\mathbf{R}^{n}} f(x(y))\left|\operatorname{det}\left(\frac{\partial x_{i}}{\partial y_{j}}\right)\right| d y_{1} \cdots d y_{n}
$$
This means that if we take coordinates $y_1, \dots y_n$ that have the same orientation the integral will have the same value. The change in order of $dx_1$ and $dx_2$ was given by $(x_1,x_2,x_3 \dots,x_n) \mapsto (x_2,x_1,x_3\dots,x_n)$ an orientation reversing diffeomorphism thus we solved the problem of sign ambiguity of the riemann integral once we choose an orientation.

Bodo has a nice answer and this answer tries to focus on the isomorphism to the density bundle he wrote about and connect it to the concepts you talked about. An added bonus of this discussion is that you may understand why alternating forms of lower degrees are also interesting as you can integrate them on cycles the content of the celebrated de rham theorem De Rham's theorem.

Title: why integrating only alternating forms?


问:
Let $(G,*)$ be an abelian group.
Prove that, for all $a,b \in G$, then $(ab)^n = a^nb^n$ for three consecutive integers $n$.
Let $(G,*)$ be an abelian group.
Prove that, for all $a,b \in G$, then $(ab)^n = a^nb^n$ for three consecutive integers $n$.

How to bring $ab=ba$ to $(ab)^n=a^nb^n$? Any ideas?
Is it need to prove firstly $(ab)^n=a^nb^n$ for all integers $n$? Then the result follows for $n+1$ and $n+2$?

答:
The converse statement is much more interesting: let $n$, $n+1$ and $n+2$ be three consecutive integers $k$ with $(ab)^k=a^kb^k$.  From $a^{n+1} b^{n+1}  = (ab)^{n+1} = (ab)(ab)^n = aba^n b^n$ it follows $a^n b = b a^n$. A similar argument with $n$ replaced by $n+1$ yields
$a^{n+1} b = b a^{n+1}$. Now $a b = a b a^n a^{-n} = a a^n b a^{-n} = a^{n+1} b a^{-n} = b a^{n+1} a^{-n} = b a$. Hence $G$ must be abelian.


问:
The diagrams of the following unary operators of temporal logic
I am trying to learn about the following temporal operators: $X,F,G$
where $X$ is for neXt (sometimes denoted by $N$), $F$ is Future
and $G$ stands for Globally.

I believe I understand what each one means, but I am having a problem
understanding the diagrams at Wikipedia.

For $X$:

I am not sure about which color denotes True (they use white and red),
but $Xp$ should be true at the next state, hwo did they say this in the diagram ?

Also (and this is a question for all $3$ diagrams): $Xp$ does not depends on if $p$ is currently true or not, so why do they sketch the truth values of $p$ above $Xp$ ?

For $F$ -

again, I don't understand how the diagram tells us what $Fp$ means.

For $G$:

This one is the most confusing, I would of expected $Gp$ to be in the same color (the one that denotes True), why some parts are white while the others are red ?

I would appriciate any help in understanding those diagrams!

答:
Your $Xp$ is their $Np$, for "Next, $p$."  The red indicates truth along that fixed $y$-coordinate. So for instance, the semantics for $Np$ state "$Np$ is true just in case $p$ is true at the next time interval." In your first diagram, $Np$ is true at 1-2, so at the next time interval (2-3), $p$ must be true. If, for example, $p$ were true at intervals 3-5 and 7-8, we'd have $Np$ being true at times 2-4 and 6-7.

$Fp$ is true just in case $p$ is true at some future time. So $p$ since $p$ is true at intervals 2-3 and 4-5, $Fp$ will be true up to 5. If $p$ is not true at another other intervals, then $Fp$ will not be true past 5.

$Gp$ is true just in case $p$ will be true at all future times. In this case, since the intervals only go up to 6, if $p$ is true from 4-6, then $Gp$ is true from 4-6. But at intervals 1-4, $p$ isn't always true at all future times, so $Gp$ isn't true.


问:
How will the plot of CDF and PDF look like for $A\ =\ \sqrt{X_1^2\ +\ X_2^2}$ where $X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.
I know Chi-Square distribution. Here the expression is similar to Chi-square distribution with the degree of freedom being 2, except for the square root.

Given:
$X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.

$A\ =\ \sqrt{X_1^2\ +\ X_2^2}$

How to find the CDF $F_A(x)$ and PDF $f_A(x)$?

The PDF of $\chi^2$ distribution for $\nu$ degrees of freedom is given
by -

$f(\nu^2)\ =\ \frac{1}{2^{\nu / 2}.\Gamma(\nu /
 2)}.e^{-\chi^2/2}.(\chi^2)^{\nu/2-1}\ \ ;\ \chi^2\ \geq\ 0$

Can I substitute 2 in place of $\nu$ to get the expression for PDF $f_A(x)$?

答:
You forgot to mention independence. Without independence of $X_1$ and $X_2$ you cannot find hhe distribution of $A$.

Let $Y=X_1^{2}+X_2^{2}$. Then $f_{\sqrt Y}(z) =2zf_Y(z^{2})$ [since $P(\sqrt Y \leq z)=P(Y \leq z^{2})$ whose deirvative is $2zf_Y(z^{2})$]. Since you know the density of $Y$ you can write down the denisty of $\sqrt Y$. By definition, the disrtibution of $A$ is a  chi-distribution with two degrees of freedom.


Body:
Let $(G,*)$ be an abelian group.
Prove that, for all $a,b \in G$, then $(ab)^n = a^nb^n$ for three consecutive integers $n$.

How to bring $ab=ba$ to $(ab)^n=a^nb^n$? Any ideas?
Is it need to prove firstly $(ab)^n=a^nb^n$ for all integers $n$? Then the result follows for $n+1$ and $n+2$?

Body:
No, it is not necessary to use induction over all $n$.

Without induction:

$ab=ab$,

$(ab)^{2} = (ab)(ab)=a(ba)b=a(ab)b=(aa)(bb)=a^{2}b^{2}$,

$(ab)^{3} = ((ab)(ab))(ab)=(a(ba)b)(ab)=(a(ab)b)(ab)=((aa)(bb))(ab)=(aa)((bb)(ab))=(aa)((b(ba)b))=(aa)((b(ab)b))=(aa)((ba)(bb))=((aa)(ba))(bb)=((aa)(ab))(bb)=((a^{2})(ab))(bb)=((a^{2}a)b))(bb)=(a^{3}b)(bb)=(a^{3}b)(b^{2})=a^{3}(bb^{2})=a^{3}b^{3}.$

Then $(ab)^{n} = a^{n}b^{n}$ for three consecutive integers, namely $n = 1,2$, $3$.


答: Take a look at Chi distribution — it's distribution of $\sqrt{X}$ for $X \sim \chi^2$.

Title: How will the plot of CDF and PDF look like for $A\ =\ \sqrt{X_1^2\ +\ X_2^2}$ where $X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.

Body:
I know Chi-Square distribution. Here the expression is similar to Chi-square distribution with the degree of freedom being 2, except for the square root.

Given:
$X_1\ \sim\ N(0,1)\ and\ X_2\ \sim\ N(0,1)$.

$A\ =\ \sqrt{X_1^2\ +\ X_2^2}$

How to find the CDF $F_A(x)$ and PDF $f_A(x)$?

The PDF of $\chi^2$ distribution for $\nu$ degrees of freedom is given
by -

$f(\nu^2)\ =\ \frac{1}{2^{\nu / 2}.\Gamma(\nu /
 2)}.e^{-\chi^2/2}.(\chi^2)^{\nu/2-1}\ \ ;\ \chi^2\ \geq\ 0$

Can I substitute 2 in place of $\nu$ to get the expression for PDF $f_A(x)$?


问:
How to prove that the stereographic projection of the $S^2$ is conformal
I read on Wikipedia that the stereographic projection of the $S^2$ sphere is conformal. However, there was no proof provided.
How can one show it?

答:
We can stereographically project from any sphere $S$ to any plane $\Pi$ from a pole, so long as $\Pi$ is parallel to (but distinct from) the pole's tangent space. Moving $\Pi$ (along the direction of $S$'s surface normal at $p$) simply has the effect of a homothety on the stereographic projection, which is conformal, and reflecting the plane $\Pi$ across the pole's tangent space has the effect of a central inversion on the projection, which is also conformal, so without loss of generality $\Pi$ is tangent to $S$ at the poles antipode, or equivalently $\Pi$ is the tangent space of the pole's antipode.

Also, in $n$ dimensions, any two intersecting lines are contained in a $3$-dimensional vector subspace, so if we can show stereographic projection is conformal for $n=3$ then it is true for all $n>3$ too.

Stereographic projection between a sphere minus a pole and the pole's antipode's tangent plane is actually the restriction of an involution (self-inverse function) on all of $\mathbb{R}^3$. Namely, it is inversion across the larger sphere centered at the pole and in contact with pole's antipode (geometry exercise). Spherical inversion is the 3D version of circular inversion in the plane. The fact that lines invert into circles through the origin carries over into 3D.

Given two lines $L_1,L_2$ intersecting at a point $P$ (tangent to two curves, say), they invert to two circles $C_1,C_2$ at $Q$. The tangent lines $T_1,T_2$ of the two circles at $Q$ are the same angle apart as the circles' tangent lines $U_1,U_2$ at the circle center $O$, but the lines $U_1,U_2$ are parallel to the lines $L_1,L_2$ by symmetry. Thus, we can say $\angle T_1T_2=\angle U_1U_2=\angle L_1L_2$. This means inversion preserves the angles between tangent lines.

To see why circular inversion inverts circles into lines into circles through the origin, notice in the picture below similar triangles $\triangle AOB\sim\triangle B'OA'$ (geometry exercise):

There's also a boring way to show inversion is conformal by showing its differential is an orthogonal transformation between tangent spaces. That is, show its Jacobian is an orthogonal matrix. The formula for inversion across a sphere centered at $\mathbf{a}$ of radius $r$ is

$$ f(\mathbf{x})=\frac{\mathbf{x}-\mathbf{a}}{\|\mathbf{x}-\mathbf{a}\|^2}r+\mathbf{a}. $$

Without loss of generality, $r=1$ and $\mathbf{a}=\mathbf{0}$.


问:
Compute $F_V(t)$

Given $f_{X,Y}(t)=\left\{\begin{array}{rcl} 1/2& x\in(-1,0],-1-x\leq y\leq1+x\\1/2& x\in(0,1),x-1\leq y\leq1-x\end{array}\right.$

Compute $F_V(t)$ V=X-Y

I've tried using $F_V(t)=\int_{-\infty}^{t}\int_{-\infty}^{\infty}f(x+y,y)dydx$ and got $F_V(t)=\left\{\begin{array}{rcl} \frac{t^2+2t+2}{2}& t\leq0\\
\frac{-t^2+2t+2}{2}& t\geq0\end{array}\right.$

But in the answers the result is $V\sim U(-1,1)$

答:
Observe that the region in which $f_{X,Y}$ is non-zero is a square $\huge\diamond$ with vertices $(\pm1, \pm1)$ and $(\pm1, \mp1)$. i.e. $(X, Y) \sim U\left({\huge \diamond}\right)$ since the value of this given pdf is $1/2$ on the square.

Rewrite the inequalities

\begin{cases}
x \in (-1,0], & -1-x \le y \le 1 + x &\implies x - y \ge -1, x + y \ge -1 \\
x \in (0,1), & x-1 \le y \le 1 - x &\implies x - y \le 1, x + y \le 1
\end{cases}

Introduce the auxiliary variable $W = X+Y$, and do something similar for variables in small letters.  It's clear that the pdf $f_{V,W}$ is $1/4$ on $[-1,1] \times [-1,1]$, from which it's clear that $V \sim U([-1,1])$.

Title: Compute $F_V(t)$


问:
Existence of a path between two sides of a square, which is contained in the intersection of two open subsets of it
Denote by $I$ the interval $[0,1]$. Let $A,B$ be two open subsets of the square $I\times I=I^2$ such that $A\cup B=I^2$, $\{0\}\times I\subseteq A$ and $\{1\}\times I\subseteq B$. In more geometric terms, the left edge of the square is contained in $A$ and the right edge is contained in $B$.

My question is: does there exist a continuous path $\gamma:I\to A\cap B$ such that $\gamma(0)\in I\times\{0\}$ and $\gamma(1)\in I\times\{1\}$? In other words, is there a curve connecting the two horizontal sides of the square, whose trace is contained in $A\cap B$ ?

答:
Yes. Indeed, shrinking sets $A$ and $B$, if necessary, we can assure that $A\cup B$ covers $I^2$, $A\subseteq [0,1)\times I$, and
$B\subseteq (0,1]\times I$. Pick a number $0<\delta<1/2$ such that $3\delta$ is smaller than Lebesgue’s number of the cover of the square bu $A$ and $B$.
Triangulate the square as shown in the picture, assuring that each side of each small triangle is at most $\delta$.

Mark triangle vertices belonging to horizontal sides of the square by $0$’s and belonging to vertical sides of the square by $1$’s (the corners of the square mark arbitrarily by $0$’s or $1$’s). Mark an inner triangle vertex $v$ by $0$ if all triangles incident to $v$ are covered by the set $A\cap B$, and by $1$, otherwise. Now, starting from the left bottom corner of the square, begin to walk traversing small triangles, entering and leaving them through edges marked by distinct numbers. It is easy to check that we never traverse any triangle twice. Also it is easy to check that we can stop because we came to a blind alley, only at a corner of the square. It follows that our walk connects either the bottom and the top sides (1) of the square of its left and right sides (2). Consider these cases.

(1) This case is shown at the picture. It is easy to show that the walk has a subwalk $S$ which begins in a triangle adjacent to the bottom side of the square, ends in a triangle adjacent to the top side of the square, and has all intermediate traversed triangles adjacent neither to the bottom nor to the top sides of the square. Each time when $S$ traverses a triangle side, this side has a vertex marked $0$, which follows that both triangles incident to the side are contained in $A\cap B$. Thus $S$ provides a path from the bottom to the top side of the square, contained in $A\cap B$.

(2) It is easy to show that the walk has a subwalk $S$ which begins in a triangle adjacent to the left side of the square, ends in a traingle adjacent to the right side of the square, and has all intermediate traversed triangles adjacent neither to the left nor to the right sides of the square. Let $s_1,\dots, s_n$ be the sequence of triangle sides traversed by $S$. For each $i=1,\dots, n$, a side $s_i$ has a vertex $v_i$ marked $1$. It means that there exists a point $p_i\in I^2\setminus (A\cap B)$ contained in a triangle incident to $v_i$. Let $d$ be the usual distance on $I^2\subseteq\Bbb R^2$. Then $d(v_i,p_i)\le\delta$. Since $d(v_i,v_{i+1})\le\delta$ for each $i=1,\dots, n-1$, by the triangle inequality we have that $d(p_i,p_{i+1})\le 3\delta$ for each $i=1,\dots, n-1$. Pick a point $p_0$ contained in the intersection of the beginning triangle of $S$ and the left side of the square and a point $p_{n+1}$, contained in the intersection of the ending triangle of $S$ and the right side of the square. Then $d(p_0, p_1)\le 2\delta$, $d(p_n, p_{n+1})\le 2\delta$, $p_0\in A$, $p_{n+1}\in B$. So there exists an index $0\le i\le n$ such that $p_i\in A$ (and so $p_i\not\in B$) and $p_{i+1}\in B$ (and so $p_{i+1}\not\in A$). But since $d(p_i, p_{i+1})\le 3\delta$ is smaller than Lebesgue’s number of the cover of the square by $A$ and $B$, a set $\{p_{i}, p_{i+1}\}$ must be contained either in $A$ or in $B$, a contradiction.

PS. The picture and the idea of the proof were taken from Problem 8 of [Sh], on the draw impossibility in the hex game.

References

[Sh] Shashkin Yu. A., Fixed points, Moscow, Nauka, 1989, in Russian.


问:
System of two simple modular equations
$$x \equiv -7 \mod 13$$
$$x \equiv 39 \mod 15$$

I need to find the smallest x for which these equations can be solved. I've been always doing this using Chinese Reminder Theorem, but it seems that it doesn't work here, I'm not sure why though. Let's see:

$N = 13*15 = 195$

$N_{1} = \frac{195}{13} = 13$

$N_{2} = \frac{195}{15} = 15$

$GCD(13, 13) = 1*13 + 0*13$

$GCD(15,15) = 1*15 + 0*15$

$$x = 1 * (-7) * 13 + 1 * 39 * 15 = 480$$

The answer however is supposed to be $x = 84$...What's wrong?

答:
Below is a correct form of that version of the CRT = Chinese Remainder Theorem

$ \begin{eqnarray}x\equiv \color{#0a0}{-7}\!\!\pmod{\color{#c00}{13}}\\ x\equiv\ \color{#c00}{39}\!\!\pmod{\color{#0a0}{15}}\end{eqnarray} \!\iff  x \equiv 15\left[\color{#0b0}{\dfrac{-7}{15}}\ {\rm mod}\ 13\right] +  13\left[\color{#c00}{\dfrac{39}{13 }}\ {\rm mod}\ 15\right]  \pmod{13\cdot 15}$

Computing the $\rm\color{#0b0}{frac}\color{#c00}{tions\!\!:}$ $ $ mod $13\!:\, \color{#0b0}{\dfrac{-7}{15}\equiv \dfrac{6}2\equiv 3},\, $ so $\ x \equiv 15\cdot \color{#0b0}3\! +\! 13\cdot \color{#c00}3\equiv 84\pmod{\!195}$

Title: System of two simple modular equations


问:
How to prove $ \lim_{x\rightarrow 0^+} \frac{1}{\sqrt{x}} \cos(1/x) $ does not exist
How do I prove that this limit does not exist?

$$
\lim_{x\rightarrow 0^+} \frac{1}{\sqrt{x}} \cos(1/x)
$$

When plugging in $0^+$ in $x$ we get $-\infty \times \cos(\infty)$ which is undefined. however I am trying to prove that it doesn't exist and am not capable? Do I use the squeeze rule or toast rule? I have tried a few different ways and haven't managed anything

答:
As for any $\delta$, $0<x<\delta\implies \dfrac 1x$ is unbounded,

$$\max_{0<x<\delta}\left|\frac1{\sqrt x}\cos\dfrac1x\right|=\max_{0<x<\delta}\frac1{\sqrt x},$$ which is obviously unbounded.


问:
Is the integral of computable functions computable?
For each $n$, let $f_n$ be a computable function which takes an infinite binary sequence $\omega\in 2^\mathbb{N}$ as its input. $2^\mathbb{N}$ has the standard measure (for a finite string $x=x_1x_2\ldots x_n$, the set $\Omega_x$ of all $\omega$ starting with prefix $x$ has measure $2^{-n}$).

Assuming $(n,\omega)\mapsto f_n(\omega)$ is computable, is the function
$$ n\mapsto \int f_n(\omega) d\omega $$
computable?

I'm pretty sure that this is true, but I can't find a proof (by myself or by looking online). I know that the Cantor space $2^\mathbb{N}$ is compact and that computable functions are continuous in this space, but I don't fully see how this can be used to show that the integral is always computable.

Any help would be appreciated!

答:
Answering my own question:

First of all, I need to specify the range of $f_n$, which are reals. This means that we can think of $f_n$ as a Turing machine which given oracle $\omega$ and a rational $\epsilon>0$, outputs a rational $r$ such that $|r-f_n(\omega)|<\epsilon$. Denote this $r$ with $f_n^{(\epsilon)}(\omega)$.
Given that $\int f_n(\omega)d\omega$ will be a real, we need to show that for a given $\epsilon$, we can compute in finite time a rational $r$ such that $|r-\int f_n(\omega)d\omega|<\epsilon$.

Now some notation: For any $\omega\in 2^{\mathbb{N}}$, let $\omega| m$ be the string containing the first $m$ digits of $\omega$. Also, for a finite string $x$ of length $m$, let $\Omega_x = \{ \omega\in 2^{\mathbb{N}}: \omega | m=x\}$. Finally, for a finite string $x$, let $x0_{\infty}$ be the infinite sequence which starts with $x$ and simply adds an infinite amount of zeros at the end.

For any $\omega$, the computation of $f_n^{(\epsilon)}(\omega)$ uses only a finite part of $\omega$, say the first $m$ digits. This means that $f_n^{(\epsilon)}$ is constant on $\Omega_{\omega | m}$. As this holds for any $\omega$, these sets $\Omega_{\omega | m}$ ($m$ can dependent on $\omega$) form an open covering of the whole space $2^{\mathbb{N}}$. Since $2^{\mathbb{N}}$ is compact, there are finitely many $\Omega_{\omega_1 | m_1}, \Omega_{\omega_2 | m_2},\ldots,\Omega_{\omega_k | m_k}$ which cover $2^{\mathbb{N}}$. This means that for any $\omega$, the computation of $f_n^{(\epsilon)}$ uses at most $M=\max\{m_1,m_2,\ldots,m_k \}$ digits.

The final ingredient is that we can effectively check which digits of $\omega$ were used for the computation of $f_n^{(\epsilon)}(\omega)$.

Now for the final algorithm: The inputs are $n$ and a rational $\epsilon>0$.
For any $m\in \mathbb{N}$, consider the sum
$$ \sum_{\text{string }x \text{ of length }m} 2^{-m}  f_n^{(\epsilon)}(x0_{\infty}).$$
At each step, check if the computation required more than $m$ digits, and if it did, move on to $m+1$. If for all $x$, this did not happen, than $m=M$ and the sum above is equal to  $\int f_n^{(\epsilon)}(\omega)d\omega$, which differs at most $\epsilon$ from $\int f_n(\omega)d\omega$