答:
The preimage of $U$ is a set in $\mathbb{R}$ and since $f$ is the identity function in each coordinate, it is given by all the points in $\mathbb{R}$ such that they are smaller than $\frac{1}{n}$ for every $n$. Hence you easily see that the preimage is the singleton {0}, which is not open since if it were open, it would contain some interval $(- \delta , \delta)$ for $\delta > 0$. But this would mean that $f( (- \delta , \delta) )\subset U$, getting for each coordinate $(- \delta , \delta) \subset (- 1/n , 1/n)$ for every $n$, reaching a contradiction (since if it holds for every $n$, then it holds also in the limit that here is zero).

Note that the difference with the product topology is that with the product topology we have only finitely many intervals of the type $(- 1/n , 1/n)$, hence we will always a be able to find a small $\delta > 0$ such that $(- \delta , \delta) \subset (- 1/n , 1/n)$ since now $n$ can take only finitely many values, and you cannot take the limit as $n$ going to infinity.

Title: Why is the function $f(t)=(t,t,t,t,t,...) $continuous in the product topology but not the box topology?

Body:
The function $f:\mathbb R \to \mathbb R^\infty$ given by $f(t)=(t,t,t,t,t\dots)$ is said to be not continuous in the box topology but continuous in the product topology.

The example given is $U=(-1,1)\times(-1/2,1/2)\times (-1/3,1/3)\times \dots $. Why does this set show the box topology is not continuous by $f$? We can just take the preimage of $U$ and it would just be $U$ correct? I can see that $(-1/n,1/n)$ converges to $0$ which is closed in $\mathbb R$, but for any specific $\mathbb R_i$ in $\mathbb R^\infty$, we actually do have an open interval even if it is extremely small.


问:
How does the set $U_p\cap S^2$ Is Defined by the Vanishing of the First Coordinate $f$ for the unit Sphere $S^2$ (Smooth Manifold)
Hi am reading An introduction to smooth manifold by Loring and have 1 doubt in example 9.6. It is written that

In this chart, the set $U_p\cap S^2$ is defined by the vanishing of the first coordinate $f$.

My doubts are as follows:

How do we know that the in this chart the set $U_p\cap S^2$ is defined by the vanishing of the first coordinate $f$? As you can see in the screenshot we assumed $\frac{\partial f}{\partial x}(p)\neq0$, which means that f is not a constant value (otherwise derivative of constant is $0$). So why are we saying it is defined by the vanishing of the first coordinate?

For reference i am attaching the screenshot where i have highlighted the part where this is mentioned.

Example 9.6 (The $2$-sphere in $\mathbb{R}^3$). The unit $2$-sphere
$$S^2=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2 =1\}$$
is the level set $g^{-1}(1)$ of level $1$ of the function $g(x,y,z)=x^2+y^2+z^2$. We will use the inverse function theorem to find adapted charts of $\mathbb{R}^3$ that cover $S^2$. As the proof will show, the process is easier for a zero set, mainly because a regular submanifold is defined locally as the zero set of coordinate functions. To express $S^2$ as a zero set, we rewrite its defining equation as
$$f(x,y,z)=x^2+y^2+z^2-1=0.$$
Then $S^2=f^{-1}(0)$.

Since
$$\frac{\partial f}{\partial x}=2x,\quad\frac{\partial f}{\partial y}=2y,\quad\frac{\partial f}{\partial z}=2z,$$
the only critical point of $f$ is $(0,0,0)$, which does not lie on the sphere $S^2$. Thus, all points on the sphere are regular points of $f$ and $0$ is a regular value of $f$.

Let $p$ be a point of $S^2$ at which $(\partial f/\partial x)(p)=2x(p)\ne0$. Then the Jacobian matrix of the map $(f,y,z):\mathbb{R}^3\longrightarrow\mathbb{R}^3$ is
$$\left[\begin{array}{ccc}
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\\[2mm]
\frac{\partial y}{\partial x} & \frac{\partial y}{\partial y} & \frac{\partial y}{\partial z}\\[2mm]
\frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} & \frac{\partial z}{\partial z}
\end{array}\right]=\left[\begin{array}{ccc}
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right],$$
and the Jacobian determinant $\partial f/\partial x(p)$ isnonzero. By Corollary 6.27 of the inverse function theorem (Theorem 6.26), there is a neighborhood $U_p$ of $p$ in $\mathbb{R}^3$ such that $(U_p,f,y,z)$ is a chart in the atlas of $\mathbb{R}^3$. In this chart, the set $U_p\cap S^2$ is defined by the vanishing of the first coordinate $f$. Thus, $(U_p,f,y,z)$ is an adapted chart relative to $S^2$, and $(U_p\cap S^2,y,z)$ is a chart for $S^2$.

答:
Tu considers the smooth map $F : \mathbb R^3 \to \mathbb R^3, F(x,y,z) = (f(x,y,z),y,z)$ and shows that on some $U_p$ it restricts to a chart $F \mid_{U_p} : U_p \to F(U_p)$ on $\mathbb R^3$. In this special case this means nothing else than that $F \mid_{U_p}$ is diffeomorphism between open subsets of $\mathbb R^3$. The three coordinate functions of $F$ are $F_1= f$, $F_2 = p_2$, $F_3 = p_3$, where $p_i : \mathbb R^3 \to \mathbb R,p_i(x_1,x_2,x_3) = x_i$, is the projection onto the $i$-th coordinate.

But now by definition of $f$ we have have $S^2 = f^{-1}(0)$, i.e. $S^2 =  \{(x,y,z) \in \mathbb R^3 \mid f(x,y,z) = 0 \}$. Therefore
$$U_p \cap S^2 =  \{(x,y,z) \in U_p \mid f(x,y,z) = 0 \} = \{(x,y,z) \in U_p \mid F_1(x,y,z) = 0\} = \{(x,y,z) \in U_p \mid F(x,y,z) \in \{ 0 \} \times \mathbb R^2\} = (F \mid_{U_p})^{-1}(\{ 0 \} \times \mathbb R^2) .$$

This means that $U_p \cap S^2$ is the set of points $\xi \in U_p$ where the first coordinate of $F(\xi)$ vanishes, i.e. where $f(\xi)$ vanishes. This is what Tu expresses in the form

In this chart, the set $U_p \cap S^2$ is defined by the vanishing of the first coordinate $f$.

Note that this shows that (cf. Definition 9.1) $S^2 \subset \mathbb R^3$ is a regular submanifold of dimension $2$ because for each $p \in S^2$ we found the chart $(U_p,F \mid_{U_p})$ in the maximal atlas of $\mathbb R^3$ such that $U_p \cap S^2$ is defined by the vanishing of $n-2 = 1$ of the coordinate functions of $F \mid_{U_p}$.

ContentLicense: CC BY-SA 4.0


问:
How to simplify $\arcsin(2\sin (x))$?
I know that $\arcsin(\sin(x))$ would be $x$ (in the right domain). But, I'm unsure how to handle the scalar multiplication. Any thoughts on how to proceed?

Title: How to simplify $\arcsin(2\sin (x))$?

Body:
You can write
$$\arcsin(a\sin (x))=\sum_{n=0}^\infty \frac{ (2 n)!\,\, a^{2 n+1} }{4^n\,(2 n+1)\, (n!)^2}\,\sin ^{2 n+1}(x)$$


问:
Limit of fourier series
I want to compute the limit $f(x)$ of

$$
f(n,x)=\sum_{k=1}^n\frac{1}{k^2+1}\sin(kx)
$$

and didn't succeed. I tried with the means used to compute the >>easy<< limits of trigonometric series like those listed in (e.g.) Bronstein, however, the above series is not in Bronstein. I plotted the graph with sagemath (see my post on ask.sagemath.org: https://ask.sagemath.org/question/55596/limit-of-fourier-series/). At least this suggests that f(x) exists and is not a polynomial.

Is it possible to compute the above limit and if so, how?

答:
The limit can be expressed via hypergeometric functions as:
$$\lim_{n \to \infty} f(n,x) = \\\left(\frac{1}{8}+\frac{i}{8}\right) e^{-i x} \left(\, _2F_1\left(1,1-i;2-i;e^{-i x}\right)+i \, _2F_1\left(1,1+i;2+i;e^{-i x}\right)-e^{2 i x} \left(\, _2F_1\left(1,1-i;2-i;e^{i x}\right)+i \, _2F_1\left(1,1+i;2+i;e^{i x}\right)\right)\right).$$

Title: Limit of fourier series


问:
Show that the midpoints of two segments coincide
Let $ABC$ be a triangle. $A_1,B_1$ and $C_1$ are the midpoints of $BC,AC$ and $AB$, respectively. Show that the midpoints of $CC_1$ and $A_1B_1$ coincide.

Let $S_1$ and $S_2$ be the midpoints of $A_1B_1$ and $CC_1$. We have $$\vec{OS_1}=\dfrac12\left(\vec{OA_1}+\vec{OB_1}\right)=\dfrac12\left[\dfrac12\left(\vec{OB}+\vec{OC}\right)+\dfrac12\left(\vec{OA}+\vec{OC}\right)\right]\\=\dfrac14\left(\vec{OA}+\vec{OB}+2\vec{OC}\right)$$
and $$\vec{OS_2}=\dfrac12\left(\vec{OC}+\vec{OC_1}\right)=\dfrac12\left[\vec{OC}+\dfrac12\left(\vec{OA}+\vec{OB}\right)\right].$$
Are these equal?

答:
Another way you could have done is using position vectors -

If positions vectors of $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ resp., then

$A_1 = \frac{\vec{b}+\vec{c}}{2}, B_1 = \frac{\vec{a}+\vec{c}}{2}, C_1 = \frac{\vec{a}+\vec{b}}{2}$

Midpoint of $A_1B_1 = \frac{1}{2} (\frac{\vec b+ \vec c}{2} + \frac{\vec a+ \vec c}{2}) = \frac{\vec a+ \vec b+ 2 \vec c}{4}$

Midpoint of $CC_1 = \frac{1}{2}(\vec c +\frac{\vec a+ \vec b}{2}) = \frac{\vec a+ \vec b+ 2 \vec c}{4}$

which shows the midpoints of both $A_1B_1$ and $CC_1$ segments coincide.

Edit: on your comment if there is a way without vector - if you are allowed to directly apply midline theorem, then here is the way but if you need to first prove $A_1B_1$ is parallel to $AB$ then see that proof here on wiki

Now given $A_1B_1 \parallel AB, \triangle ACC_1 \sim \triangle B_1CM$ (let $M$ be the point where $A_1B_1$ and $CC_1$ intersect).

So, $\frac{B_1C}{AC} = \frac{B_1M}{AC_1} = \frac{CM}{CC_1}$

As, $B_1C = \frac{AC}{2}, B_1M = \frac{AC_1}{2}, CM = \frac{CC_1}{2}$. So $M$ is the midpoint of $CC_1$.

Similarly you can show, $MA_1 = \frac{C_1B}{2}$ (or just use the fact that $A_1B_1 = \frac{1}{2}AB)$.

As $AC_1 = C_1B, B_1M = M A_1$. So the intersection of line segments $CC_1$ and $A_1B_1$ is the midpoint of both segments.


答:
Yes, they are equal.

Note that since $C_1$ is the midpoint of $AB,$ the vector $\vec{OC_1}$ must be half the sum of the vectors $\vec{OA}$ and $\vec{OB},$ in other words;
$$\vec{OC_1}=\frac{\vec{CA}+\vec{CB}}{2}$$
$$\implies 2\vec{OC_1}=\vec{OA}+\vec{OB}$$
$$\implies 2\vec{OC_1}+2\vec{OC}=\vec{OA}+\vec{OB}+2\vec{OC}$$
$$\implies \frac{2\vec{OC_1}+2\vec{OC}}{4}=\frac{\vec{OA}+\vec{OB}+2\vec{OC}}{4}$$
$$\implies \frac{\vec{OC_1}+\vec{OC}}{2}=\frac{\vec{OA}+\vec{OB}+2\vec{OC}}{4}$$
$$\implies \frac{1}{2}\left(\vec{OC_1}+\vec{OC}\right)=\frac{1}{4}\left(\vec{OA}+\vec{OB}+2\vec{OC}\right).$$

Thus, they are equal, as required.

Title: Show that the midpoints of two segments coincide

Body:
Let $ABC$ be a triangle. $A_1,B_1$ and $C_1$ are the midpoints of $BC,AC$ and $AB$, respectively. Show that the midpoints of $CC_1$ and $A_1B_1$ coincide.

Let $S_1$ and $S_2$ be the midpoints of $A_1B_1$ and $CC_1$. We have $$\vec{OS_1}=\dfrac12\left(\vec{OA_1}+\vec{OB_1}\right)=\dfrac12\left[\dfrac12\left(\vec{OB}+\vec{OC}\right)+\dfrac12\left(\vec{OA}+\vec{OC}\right)\right]\\=\dfrac14\left(\vec{OA}+\vec{OB}+2\vec{OC}\right)$$
and $$\vec{OS_2}=\dfrac12\left(\vec{OC}+\vec{OC_1}\right)=\dfrac12\left[\vec{OC}+\dfrac12\left(\vec{OA}+\vec{OB}\right)\right].$$
Are these equal?


问:
If $l$ is an integer such that $k \le l \le m$, show that there exists subspace $X$ of $V$ such that $U \subseteq X \subseteq W$ and $\dim(X)=l$.
Let $U$ and $W$ be subspaces of a vector space $V$ such that $U \subseteq W, \dim(U)=k,$ dan $\dim(W)=m$
with $k \lt m$.

If $l$ is an integer such that $k \le l \le m$, show that there exists subspace $X$ of $V$ such that
$U \subseteq X \subseteq W$ and $\dim(X)=l$.

Any idea how to starting with? What theorem or properties or other in dimension of vector space could I use? Thanks in advanced.

答:
Here are the basic ingredients:

Every linearly independent set in a vector space can be extended to a basis (hint: $U$ is a subspace of $W$).

Every subset of a linearly independent set is linearly independent, and is a basis for its span.

Let me know if you need more help!


问:
Given $H,K$ finite index subgroups of $G$, prove that the order of $G/(H\cap K)$ obeys these bounds
I am given that $H,K$ are finite index subgroups of a group $G$ (which may not be finite), and that $|G:H|=m,\ |G:K|=n$. I am asked to show that $$\text{lcm}(m,n)\leq |G:H\cap K|\leq mn$$

I'm a bit at a loss as to how to proceed, because all the relevant tools I'm aware of (Like LaGrange's theorem) require $G$ to be finite, which we don't have here. So I've tried a reductio in which I assume that $G$ can be decomposed into $p>mn$ disjoint sets $g_i(H\cap K)$, but this doesn't seem to go anywhere; I can't find any contradiction.

答:
Consider any $g∈a(H∩K)$, then $g=ah_1=ak_1$, so $g∈aH∩aK$. Every left coset of $H∩K $ is an intersection of a left coset of $H$ and a left coset of $K$. As the number of left cosets of $H$ and $K$ are finite, number of distinct left cosets of $H∩K$ must be finite.

As there are $[G:H][G:K]$ possible intersection of left cosets of $H$ and $K$, we get:

$$[G:H∩K]≤[G:H][G:K]$$

Also $H∩K≤H,H∩K≤K$, so by tower law of subgroups:
$$[G:H∩K]=[G:H][H:H∩K]$$
$$[G:H∩K]=[G:K][K:H∩K]$$
so $[G:H],[G:K]$ both divide $[G:H∩K]$, hence

$$lcm([G:H],[G:K])≤[G:H∩K]≤[G:H][G:K]$$

Equality occurs if $[G:H]$ and $[G:K]$ are coprime.

Notation: $[G:H]$ is the index of subgroup $H$ in group $G$

ContentLicense: CC BY-SA 3.0


问:
Is this function an exponential decay function
Consider a function $f(t) = e^{-t + \log t}$. I am not sure this type of function can be seen as an exponential decay function as it does not have the regular form $ce^{-at}$ with $a > 0$. But it obviously obeys
\begin{equation}
\underset{t \rightarrow \infty}{\lim} \frac{e^{-t + \log t}}{e^{-0.5t}} = 0,
\end{equation}
proving that $f(t)$ does not grow faster (or maybe equivalently decay faster) than $e^{-0.5t}$.
If this is not an exponential decay function, does it have any particular name?

答:
$$f(t) = e^{-t + \log t}$$
Suppopsing that your symbol log means the natural logarithm (not the logarithm base 10) then :

$$f(t)=t\,e^{-t}$$
This is a function frequently encountered. Of course all functions have not a specific name.

Note for information :  A particularity of this function is to be related to the inverse of the Lambert W function.
$$f(t)=-W^{-1}(-t)$$


问:
Proof of dimension of Polyhedron

Before proving the theorem the book introduces a new definition called implicit equality.

What does exactly the below equality mean?
$$
A^{=}x=b^{=}
$$
What is inside this set?
$$
\{x \in \mathbb{R}^n \colon A^{=}x = b^{=}\}
$$
I encountered this notation in Michele Conforti, Integer Programming Book, published by Springer

答:
The book first defines within the n-dimensional space of consideration an included subspace, which is defined by all equalities out of $Ax ≤ b$ (which as subsytem gets denoted by $A^= x = b^=$). This is the space where the polytope, being then defined therein by the inequalities out of $Ax ≤ b$ (which as subsytem gets denoted by $A^< x ≤ b^<$), truely lives.

Consider this example: try to describe the triangle with vertices $(0,0,0)$, $(1,0,0)$, and $(0,1,0)$. I.e. a 2D polygon being described within a 3D space. It is defined by the system $x≥0$, $y≥0$, $x+y≤1$, and $z=0$. Here the first 3 relations are the defining inequalities within the subspace defined by the fourth relation.

--- rk

Title: Proof of dimension of Polyhedron


问:
Logarithmic integral (Green function)
I am using a Green function to find the solution of an ODE and I have stuck in finding the solution of the following integral. Could someone please help me with this?

The integral is

$$\int_{0}^{r} d r^{'} \int_{0}^{2 \pi}  \ln\Biggl(\frac{C}{\sqrt{(x-r^{'} \cos{\theta^{'}})^2+(y-r^{'} \sin{\theta^{'}} )^2}} \Biggr)\, d \theta^{'}$$

and $C$ is a positive constant.

答:
We can rewrite  $\ln\Biggl(\frac{C}{\sqrt{(x-r^{'} \cos{\theta^{'}})^2+(y-r^{'} \sin{\theta^{'}} )^2}} \Biggr)=\ln{C}-\frac{1}{2}\ln\left((x-r^{'} \cos{\theta^{'}})^2+(y-r^{'} \sin{\theta^{'}} )^2 \right)=$$\ln{C}-\frac{1}{2}\ln\left(x^2+y^2+r'^2-2xr'cos\theta-2yr'\sin\theta\right)$

If we introduce vector $\vec{a}=(x,y)$, then  $xr'cos\theta+yr'\sin\theta=(\vec{a},\vec{r'})$ at some choice of the direction of $\vec{r'}$.

Because we integrate over $\vec{r'}$ this is up to us to choose the system of coordinates - the final result will not depend on this choice. We can direct the axis $X$ along the vector $\vec{a}$. At this choice the integral of logarithm becomes (accurate to constant factors) $\int_{0}^{r} d r^{'} \int_{0}^{2 \pi}  \ln\Biggl(\frac{1}{\sqrt{(x-r^{'} \cos{\theta^{'}})^2+(y-r^{'} \sin{\theta^{'}} )^2}} \Biggr)\, d \theta^{'}\sim{I}(a)=\int_0^{2\pi}d\phi\int_0^{r}dr\ln(a^2+r'^2-2ar'\cos\phi)$, where $a=\sqrt{x^2+y^2}$

Now we have to consider two cases:

$r<a$

$I(a)=\int_0^{2\pi}d\phi\int_0^{r}dr'\ln\left((a-r'e^{i\phi})(a-r'e^{-i\phi})\right)=$$\int_0^{2\pi}d\phi\int_0^{r}dr'\left(2\ln{a}+\ln(1-\frac{r'}{a}e^{i\phi})+\ln(1-\frac{r'}{a}e^{-i\phi})\right)$

But $\ln(1-\frac{r'}{a}e^{i\phi})=-\frac{r'}{a}e^{i\phi}-\frac{r'^2}{a^2}e^{2i\phi}-...$, and all these terms wanish after integration over $\phi$.

$$I(a)=4\pi{r}\ln{a}$$

$r>a$

In this case we have to split the integrale into to parts: $I(a)=\int_0^a+\int_a^{r}$

$\int_0^a=4\pi{a}\ln{a}$ (the same calculation as in the case 1.)

$\int_a^{r}=\int_0^{2\pi}d\phi\int_a^{r}dr'\ln(a^2+r'^2-2ar'\cos\phi)=\int_0^{2\pi}d\phi\int_a^{r}dr'\ln\left((r'-ae^{i\phi})(r'-ae^{-i\phi})\right)=$$\int_0^{2\pi}d\phi\int_a^{r}dr'\left(2\ln{r'}+\ln(1-\frac{a}{r'}e^{i\phi})+\ln(1-\frac{a}{r'}e^{-i\phi})\right)$

Two last terms vanish, and we get $2\pi\int_a^{r}dr'{2\ln{r'}}=4\pi(r\ln{r}-r-a\ln{a}+a)$

$$I(a)=4\pi(r\ln{r}-r+a)$$

Title: Logarithmic integral (Green function)


问:
If $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$
I'm struggling to proof that if $H$ and $K$ are subgroups of  finite index of a group $G$ such that $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$. I don't know why I can't answer it, because this question seems easy. I'm stuck maybe because I've studied so far just Lagrange's theorem and some of its consequences. But I think we don't need much more, because this is the material covered so far by the Hungerford's book.

I need help.

Thanks.

答:
If $[G:H]$ and $[G:K]$ are coprime, $[G:H∩K]=[G:H][G:K]$ (see here for a proof)

Observe that a coset of $H\cap K$ is just an intersection of a left coset of $H$ and a left coset of $K$.

Thus, every intersection of cosets of $H$ and $K$ are non-empty and make a coset of $H∩K$. In particular,
$$xK∩H≠\emptyset $$ $∀x∈G$. So if $g\in G$, then there exists a $h\in H$ such that $gK=hK$

So for every there is an element of $H$ in every left coset of $K$. Hence
$$G=\bigcup_{g\in G} gK=\bigcup_{h\in H} hK=HK$$

Notation: $[G:H]$ is the index of subgroup $H$ in group $G$


问:
Prove that $m_{*}(A)=\lim_{n\rightarrow \infty}m_{*}(A\cap (-n,n))$
How does one go about proving $m_{*}(A)=\lim_{n\rightarrow \infty}m_{*}(A\cap (-n,n))$ where $m_{*}$ is the outer measure? Here $n\in \mathbb{N}$.

I was considering the notation

$A=\bigcup_{n=1}^{\infty}A_n$ where $A_n=A\cap(-n,n)$ along with the property that

$m_{*}\Big(\bigcup_{n=1}^{\infty}A_{n} \Big) \le 2\sum_{n=1}^{\infty}n$

since $m_{*}(A\cap (-n,n))\le m_{*}((-n,n)) = 2n$. Here I get stuck and I am not sure what to do.

答:
So, I assume we are considering an outer measure $m_*$ on $\mathbb{R}$. Set $S_n := (-n,n)$. Then, $S_n \uparrow \mathbb{R}$ as sets. By the definition of monotonicity, it suffices to prove that
$$ {m_*} (A) \le \lim_{n \mathop \to \infty} {m_*} (A \cap S_n)$$
Assume that $m_*({A \cap S_n})$ is finite for all $n \in N$, otherwise the statement is trivial by the monotonicity of $m_*$.

Clearly, $\mathbb{R} = \bigcup_{n} S_{n+1}\setminus S_n$ and hence $A = \bigcup_{n} A \cap (S_{n+1} \setminus S_n)$. Assuming that $S_n$ is $m_*$-measurable, this yields \begin{align} m_*(A) &\leq \sum_{n}m_*(A \cap (S_{n+1} \setminus S_n)) \\
& =\sum_{n} m_*(A \cap S_{n+1}) - m_*(A \cap S_{n+1} \cap S_{n})\\
&=\sum_{n} m_{*}(A \cap S_{n+1}) - m_*(A \cap S_n)\\
&= \lim_{n \to \infty}m_*(A \cap S_n) - m_*(A \cap \emptyset)\\
&= \lim_{n\to \infty} m_*(A \cap S_n) \end{align}


问:
Isometry group of $N$ using $f:M\to N$ and the isometry group of $M$
I am curious about isometry group of new Riemannian manifolds out of old Riemannian manifolds.  for example what we can say about isometry group of $N$ if we know the isometry group of $M$ and $f:M\to N$ a smooth Riemannian quotient map or smooth Riemannian submersion or a Riemannian covering map? Can one compute the isometry group of $N$ exactly? (for simplicity consider all manifolds are closed.)

Any (Book as) reference?

答:
Even in the case of Riemannian covering maps, there is no simple description. One can get some information, for instance, if $X\to Y$ is the universal covering with the deck-transformation group $G$: Then the isometry group $Isom(Y)$ of $Y$ is isomorphic to the quotient of the normalizer of $G$ in $Isom(X)$ by $G$:
$$
N_{Isom(X)}(G)/G\cong Isom(Y).
$$
But computing the normalizer is not easy either. Sometimes, one can glean information by looking at symmetries of a suitably chosen fundamental domain of $G$ in $X$. But frequently it happens that $Y$ has more symmetries than that.


答: $$f(t) = e^{-t + \log t}=t\,e^{-t}$$ $f(t)$ goes through a maximum at $x=1$ and for $x>1$ it decreases slower than $e^{-t}$.

Title: Is this function an exponential decay function

Body:
Consider a function $f(t) = e^{-t + \log t}$. I am not sure this type of function can be seen as an exponential decay function as it does not have the regular form $ce^{-at}$ with $a > 0$. But it obviously obeys
\begin{equation}
\underset{t \rightarrow \infty}{\lim} \frac{e^{-t + \log t}}{e^{-0.5t}} = 0,
\end{equation}
proving that $f(t)$ does not grow faster (or maybe equivalently decay faster) than $e^{-0.5t}$.
If this is not an exponential decay function, does it have any particular name?


问:
What type of input does trigonometric functions take in
I see in my Book that 45 deg is equivalent of π/4 . Ι also do the conversion if I simply convert degrees into radians like this

45* π/180 = π/4 radians

and back again

π/4 * 180/π = 45 deg

.
So I think that I grasp the Idea finally that π/4 is another way of saying 45 degrees. Now if I use my calculator and I say sin(45) witch I assume that 45 is degrees I get the number 0.7071067812. If I do the calculation sqrt(2)/2 I will get the same number hence sin(45)=sqrt(2)/2 . But when I input sin(π/4) I get the number 0.0137073546. So I say to my self that probably that function does not get that type of input. But then again I see in my book that sin(π/4) = sqrt(2)/2 . So this forces me to ask the question, "What type of input do trigonometric functions take in?" Is the calculator not working properly ?

Excuse me if this looks simple to you but I am so bad in math and the calculator is the only validator I have at the moment and this confuses me allot.

答:
When you use $\sin(45)$ in your calculator, you mean $\sin(45^o)$ (notice the degree symbol).

But when you input $\dfrac{\pi}{4}$, you are still telling your calculator that it's in degrees: $\sin\left(\dfrac{\pi}{4}^o\right)$.

To resolve this, you have to change the mode of your calculator from degrees to radians, and reperform the calculation.


问:
How can I find a smooth map with arbitrary degree?
Suppose $k \in \mathbb{Z}$. I would like to find a smooth map $f : \mathbb{S}^n \to \mathbb{S}^n$ such that $\deg(f)=k$.

I tried with taking $k$ sets in $\mathbb{S}^n$ diffeomorphic to $\mathbb{D}^n$ and taking $f$ to be ugual to:
$$g : \mathbb{D}^n \to \mathbb{S}^n \mbox{ with } g(x)=(2\sqrt{1-\|x\|^2}x, 2 \|x\|^2-1)$$

in these sets diffeomorphic to $\mathbb{D}^n$ but I can only prove that $f$ is continuos.

I thought that using approximation with smooth functions could help but I do not know how.

Here the degree of a map is defined as in Milnor's Topology from the differentiable viewpoint.

答:
Here’s a recursive construction. Suppose you know $g: S^{n-1} \rightarrow S^{n-1}$ continuous of degree $k$. Consider $f:S^n \rightarrow S^n$ given by $f((\epsilon,0) \in \{\pm 1\} \times D^n) =(\epsilon,0)$, and $f((u,x) \in (-1,1) \times \sqrt{1-u^2}S^{n-1})=\left(u,\sqrt{1-u^2}g\left(\frac{x}{\sqrt{1-u^2}}\right)\right)$.

(Basically you apply $g$ to each “parallel” of $S^n$).

If $\ell$ is a smooth function on $S^n$, let $l_u(x \in S^{n-1}=l(u,sqrt{1-u^2}x)$. Then, for some (unimportant) weight function $w$: $$\int_{S^n}{\ell \circ f}=\int_{-1}^1{w(u)\int_{S^{n-1}}{(\ell\circ f)_u}\,du}.$$

Now, if $x \in S^{n-1}$ and $-1<u<1$, then $(\ell \circ f)_u(x)=\ell(f(u,\sqrt{1-u^2}x))=\ell(u,\sqrt{1-u^2}g(x))=\ell_u\circ g(x)$. So $$\int_{S^{n-1}}{(\ell \circ f)_u}=\int_{S^{n-1}}{\ell_u \circ g}=k\int_{S^{n-1}}{\ell_u}.$$

Finally, $$\int_{S^n}{\ell \circ f}=\int_{-1}^1{w(u)k\int_{S^{n-1}}{\ell_u}\,du}=k\int_{S^n}{\ell},$$ so $f$ has degree $k$. Of course, even if $g$ is smooth, $f$ might not be smooth – something may happen at the poles (the $(\epsilon,0)$, $\epsilon=\pm 1$). However, even if $g$ is not smooth, any smooth small enough perturbation of $f$ (and they exist) will still be of degree $k$.

To conclude, it’s enough to construct on $S^1$ a function $f$ of degree $k$, but such is $z \longmapsto z^k$.


问:
How to systematically find the domain of this funtion (with respect to $a \in \mathbb{R}$?
This problem is killing me, I don't know how to systematically find the domain of:

$$ f(x) = \frac{2e^{\frac{1}{2x}}-a}{\sqrt{a^2-1+e^{\frac{1}{x}}}}$$

I am stuck specifically on how to determine for which $x \in \mathbb{R}$:
$$a^2-1+e^{\frac{1}{x}} \geq 0.$$

答:
It is actually needed that $$a^2 -1 +e^{1/x} \gt 0$$ to prevent dividing by zero. This is $$e^{1/x} \gt 1-a^2 $$ Now, if $1-a^2 \le 0 \iff |a| \ge 1$ then this inequality is always true and the domain is the set of all reals. Otherwise, take the natural log and use that it is an increasing function: $$\frac 1x \gt \ln (1-a^2) $$

If $\ln(1-a^2)=0 \iff a=0$, the domain is $(0,\infty)$ as in that case you need $\frac 1x \gt 0$.

Otherwise, $\ln(1-a^2)\lt 0 \iff 0\lt |a| \lt 1$ and looking at the graph of $\frac 1x$ you can deduce that either $$x\lt \frac{1}{\ln(1-a^2)} $$ or $$x\gt 0$$


问:
Open, closed, compact
I have the following question.  I know the answers but am struggling with how to write up the work formally.  Any help would be appreciated.

Consider $B=\{1-\frac{1}{n} :n=1,2,\cdots\}$.  Is $B$ open? Closed? Compact?  Justify.

I know that it is not open or closed, thus not compact.

答:
HINT: It may be helpful to write out a few members of $B$:

$$B=\left\{0,\frac12,\frac23,\frac34,\dots\right\}\;.$$

To help you explain why $B$ is not open: is there any $\epsilon>0$ such that $(-\epsilon,\epsilon)\subseteq B$?

To help you explain why $B$ is not closed: can you find a limit point of $B$ that is not in $B$?

Title: Open, closed, compact


问:
compute $\int_\gamma F\cdot dr$ using Stokes

Evaluate
$$ \int_\gamma F\cdot dr$$ where $ F=(ye^x,e^x+x^3,z^5)$ and $ \gamma\ $ is the intersection between $x^2+y^2=1 $ and $z=2xy$ oriented in such a way that the orthogonal projection on the $xy$ plane is oriented counterclockwise.

Here is how I have tried to solve this problem:
I used Stokes and first calculated the curl of F which is $(0,0,3x^2)$ and then  doted this with unit normal but it become zero, however the answer should not become zero.

What am I doing wrong ?
Any suggestion would be great , Thanks

答:
Parameterize the surface as $ \ r(\rho, t) = (\rho \cos t, \rho \sin t, \rho^2 \sin2t)$

$0 \leq \rho \leq 1, 0 \leq t \leq 2\pi$.

$r_{\rho} \times r_{t} = (2 \rho^2 \sin t, 2 \rho^2 \cos t, \rho)$

$\nabla \times \vec{F} = (0, 0, 3 \rho^2 \cos^2t)$

So the integral should be,

$\displaystyle \int_0^{2\pi} \int_0^1 (0,0,3 \rho^2 \cos^2t) \cdot (2 \rho^2 \sin t, 2 \rho^2 \cos t, \rho) \ d\rho \ dt$

$ = \displaystyle \int_0^{2\pi} \int_0^1 3 \rho^3 \cos^2t \ d\rho \ dt = \frac{3 \pi}{4}$


问:
How to determine the maximum area of the triangle given by
I am trying to find the maximum area of a triangle under some given constraints. I have generated a function to do this. If $x$ and $y$ are Real Numbers such that;
$$x^2+y^2=1,$$
then the area of the triangle is given by the function;
$$f(x,y)=x(1+y).$$

So, when  is the value given by this function maximised (subject to $x^2+y^2=1$)?

答:
We know that maximizing $f(x,y)$ is equivalent to maximizing $f(x,y)^2.$ We have;
\begin{align*}
f(x,y)^2
&=\left(x(1+y)\right)^2\\
&=x^2(1+y)^2\\
&=(1-y^2)(1+y)^2\\
&=(1-y^2)(y^2+2y+1)\\
&=-y^4-2y^3+2y+1.
\end{align*}

Let this function be $g(y)=-y^4-2y^3+2y+1.$ We know that the derivative of this function is $-4y^3-6x^2+2,$ which is nonnegative for $y\leq \frac{1}{2},$ and negative for $y>\frac{1}{2}.$

Thus $g(y)$ incresases whenever $y\leq \frac{1}{2},$ and decreases when $y>\frac{1}{2},$ and acheives a maximum value $f(y)=\frac{27}{16}$ when $y=\frac{1}{2}.$

Thus the maximum value of $f(x,y)^2$ is $\frac{27}{16}$ when $y=\frac{1}{2}$ and $x=\sqrt{1-\dfrac{1}{4}}=\frac{\sqrt{3}}{2}.$

So, the maximum value of the original function $f(x,y)$ is; $$\sqrt{\frac{27}{16}}=\frac{3\sqrt{3}}{4}.$$

It is acheived when;
$$\left(x,y\right)=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right).$$


问:
Build isomorphism from subset of non-injective homomorphism?
Given two vector-spaces $V, W$ with $\dim(V)=5$ and $\dim(W)=2$.

If there exists a surjective homomorphism $f:V \to W$, does there exist a subspace of the domain: $V_1 \subset V$ with $\dim(V_1)= 2$ isomorphic to $W$?

Since $V$ is larger than $W$ and $f$ is surjective, it makes sense that we can choose a subset of the domain to reach $W$.

But is this subset necesarilly a subspace?

答:
Choose a basis of $W$, say $\{w_1, w_2\}$ take an element $v_1$ in the inverse image of $w_1$, and take an element $v_2$. (Why aren't these inverse images empty?) Define $V_1$ to be the subspace generated by $v_1$ and $v_2$, this is the subspace you're looking for and restricting $f$ to $V_1$ is an isomorphism.

To finish this proof show that $f$ is injective and surjective.


问:
Convergence of Cauchy sequences in $(\ell^1, \|\cdot\|_\infty)$

Do Cauchy sequences in $(\ell^1, \|\cdot\|_\infty)$ converge in $\ell^1$?

I feel that the answer is No, but I am not able to find a counterexample. If $x^{(n)}$ is a sequence in $\ell^1$, then we have $$\sum_{i=1}^\infty |x^{(n)}_i| < \infty$$ for every $n\in\mathbb N$. Also if $x^{(n)}$ is Cauchy w.r.t the sup-norm, then
$$\forall\epsilon>0\exists N\in\mathbb N\forall m,n>N (\|x^{(m)} - x^{(n)}\|_\infty < \epsilon)$$

That is the definition, and that is all I can see. Any ideas?

答:
Let us take $x^{(n)} \in \ell^1$ such that
$$
x^{(n)}(k) =
\begin{cases}
1/k & \text{if } k \le n,\\
0 & \text{otherwise}.
\end{cases}
$$
For $m \ge n+1$,
$$\|x^{(m)} - x^{(n)}\|_\infty  = \frac{1}{n + 1} \to 0$$
but $x^{(n)}$ does not converges in $\ell^1$ because
$$\sum_{k \in \mathbb N} \frac{1}{k} = +\infty.$$


问:
Conditional regression points
Suppose I have a model $$y=\beta_0+\beta_1x+\beta_3\sqrt{x}e_1,$$ where the "beta" coefficients are known and $e_1$ is Normal(0,1). Now I simulate $n=100$ points from that model and have the scatter-plot. The question is: how could I get (and plot) the points conditioned to a specific value of $x$ in R? That is, given the simulated points $(x_i,y_i)$, how could I extract those $y_i|x=0.5$, for example? Thank you very much in advance!

答:
Since $E[e] = E[e|X] = 0 $, your model is the estimator of
$$
E[Y|x] = \beta_0 + \beta_1 x + \beta_3\sqrt{x}E[e] = \beta_0 + \beta_1 x,
$$
which is given by
$$
\hat{y}= \hat{\beta}_0 + \hat{\beta}_1x.
$$
Hence,  you can extract the predicted value  for certain $x_0$ by plugging in this $x_0$ into the model, i.e.,
$$
\widehat{E[Y|X=x_0]} = \hat{\beta}_0 + \hat{\beta}_1x_0.
$$

In the case you meant to simulate $y_i$ for a given $x=x_0$ using R programming language, then note that the conditional distribution of $Y$ given $X=x_0$ is $N(\beta_0 + \beta_1x, \beta_3^2x)$, hence just simulate value from the aforementioned distribution by $\text{rnorm}(1, \text{mean} = \beta_0 + \beta_1x_0, \text{sd} = \beta_3 \sqrt{x_0})$.

Title: Conditional regression points


问:
How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$
let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that

$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

I find this inequality are same as Crux 3059 Problem.

答:
In addition: Under the same constraints, does the inequality
$$ab+ac+ad+bc+bd+cd\le a+b+c+d+kabcd$$
always hold, for $k<2?$


问:
Solving the system $\sqrt{x/y}+\sqrt{y/z}+\sqrt{z/x}=\sqrt{y/x}+\sqrt{z/y}+\sqrt{x/z}=3$, $\sqrt{xyz}=1$
I am having problems solving this particular system of equations (SOE). The only thing I see that could be useful is the first and second equations are reciprocals of each other. I just don't know how I can use that information to help solve the SOE.

$$
\left\{
\begin{array}{c}
\displaystyle\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}=3 \\
\displaystyle\sqrt{\frac{y\:}{x\:}}+\sqrt{\frac{z\:}{y\:}}+\sqrt{\frac{x\:}{z\:}}=3 \\
\displaystyle\sqrt{xyz}=1
\end{array}
\right.
$$

Is there a way to solve this algebraically other than solving by inspection?
Thank you for any help/assistance you can provide!

答: Note that for $x,y,z >0$ ,we have $$\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{y}{z}}+\sqrt{\dfrac{z}{x}} \geq 3$$ By the AM-GM inequality, and it is given that it is equal to it's minimum value, and that happens only when all the terms are equal, that is, $$\sqrt{\dfrac{x}{y}}=\sqrt{\dfrac{y}{z}}=\sqrt{\dfrac{z}{x}}$$ We can see that now $x=y=z=1$ is the only solution possible in $\mathbb{R}$ which is satisfied by other equations as well.


问:
Can a prime number be bigger than the sum of adding the previous twin primes (other than 13)?
A simple heuristic of the first million primes shows that no prime number can be bigger than the sum of adding the previous twin primes.

Massive update:
@mathlove made a comment that leaves me completely embarrassed. $13 > 7 + 5$ I don’t know how I missed it and I deeply apologize to everyone!

I ask anyone qualified to suggest any edits for the question.

$7 < 5 + 3$

$11 < 7 + 5$

$17 < 11 + 13$

$23 < 17 + 19$

At larger numbers:

$4886639 < 4886489 + 4886491$

$5389451 < 5388869 + 5388871$

$3155597 < 3154757 + 3154759$

I assume that if it could be proved, it would prove the twin prime conjecture of whether twin primes exist forever.

So I am not exactly seeking for a proof, but rather for possible explanations or references for why it is assumed true (or not)?

Also as the list grows, there seems to be a range for how small or big can a prime be in comparison to the sum of adding the previous twin primes.

As the list grows, a prime is usually never bigger or smaller than slightly above $50\%$ of the sum of the previous twin primes. Any references for such a range will be appreciated too.

*Update: When mentioning "the previous twin primes", I am implying to:  $(107, 109), 113, 127, 131, (137, 139)$.

$131 < 107 +109$

答:
Futher to my comments ...

For any $3$ consecutive primes (regardless of if they contain twin primes or not) $p_{n},p_{n+1},p_{n+2}$, it's true that
$$\frac{p_{n+1}+p_{n}}{2}<p_{n+2}<p_{n+1}+p_{n}$$
In fact, $\forall p_{n+k}\geq p_{n+2}$ we have $p_{n+k} > p_{n+1}$ and $p_{n+k} > p_{n}$, thus
$$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}$$
For a fixed $k$, we can show that
$$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+1}}=1 \tag{1}$$
This means that for large enough $n$'s we have
$$p_{n+k}<2p_{n} \text{ and } p_{n+k}<2p_{n+1}$$
Applying AM-GM yields
$$p_{n+1}+p_{n}\geq 2\sqrt{p_{n+1}\cdot p_{n}}=\sqrt{2p_{n+1}\cdot 2p_{n}}>
\sqrt{p_{n+k}^2}=p_{n+k}$$
for large enough $n$'s, or
$$\frac{p_{n+1}+p_{n}}{2}<p_{n+k}<p_{n+1}+p_{n} \tag{2}$$
The range of the fixed $k$ depends on when $(1)$ gets less than $2$.

A few words about $(1)$, it is true because of this limit
$$\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$
then
$$\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n}}=
\lim\limits_{n\to\infty}\frac{p_{n+k}}{p_{n+k-1}}\cdot
\lim\limits_{n\to\infty}\frac{p_{n+k-1}}{p_{n+k-2}}\cdot ...\cdot
\lim\limits_{n\to\infty}\frac{p_{n+1}}{p_{n}}=1$$
again, for a fixed $k$!


问:
From the equations prove $\{a,b\}=\{c,d\}$

If $a+b=c+d$ and $a^2+b^2=c^2+d^2$ (where $a,b,c,d$ are real numbers)
prove $\{a,b\}=\{c,d\}$

From $a+b=c+d$ we have:

$$(a+b)^2=(c+d)^2=a^2+b^2+2ab=c^2+d^2+2cd$$

Since $a^2+b^2=c^2+d^2$ then $ab=cd$. So I choose these two equations:

$${ \begin{cases}{ab=cd} \\ {a+b=c+d}\end{cases} }$$
Although solving the system looks easy, I don't know how to do it.

Body:
Now we have

$$(a-b)^2 = a^2 + b^2 - 2ab = c^2 + d^2 - 2cd = (c-d)^2$$

so $|a-b| = |c-d|$.

Hence:

$$\max\{a,b\} + \min \{a,b\} = a + b = c+ d = \max\{c,d\} + \min \{c,d\}$$

$$\max\{a,b\} - \min \{a,b\} = |a-b| = |c-d| = \max\{c,d\} - \min \{c,d\}$$

Adding up:

$$2\max\{a,b\} = 2\max\{c,d\}$$

Subtracting:

$$2 \min\{a,b\} = 2\min\{c,d\}$$

so $\{a,b\} = \{c,d\}$.


问:
Let $T,S:V \to V$ be a linear operators such that $T\circ S=T$, $S\circ T=S$ then which of these is true?
a) $kerT=kerS$

b) $T=S=1$

c) $T^2=I$

d) $T=S=0$

e) $Im(S) = Im(T)$.

My Solution:

I was able to prove (a) like this:

let $v_1 \in kerT$, so $T(v_1)=0$, but $S(T(v_1))=S(v_1)=0$, so $v_1 \in kerS$. so $kerT \subset ker S$. (and same for other direction).

However, I was not able to disprove the rest of the statements, seems like I'm still weak at finding counter examples for linear transformations.

I would really appreciate any help in how to disprove the statements b,c,d,e with counter examples, or intuition or anything really.

Thanks in advance to everyone.

答:
Let $S = \begin{pmatrix}0&0\\1&1\end{pmatrix}$ and $T= \begin{pmatrix}1&1\\0&0\end{pmatrix}$, this should help you for $e)$.

The way I found this was by trying to find an example of matrices $S$ and $T$ that satisfy the equations other than $I$ and $0$. Especially by thinking about the image of basis vectors.


答:
A general result on vector spaces and linear maps is the following.

Suppose $f\colon V\to W$ is a linear map and that $\{v_1,\dots,v_k\}$ is a set in $V$. If $\{f(v_1),\dots,f(v_k)\}$ is linearly independent, then also $\{v_1,\dots,v_k\}$ is linearly independent.

Just take scalars so that $a_1v_1+\dots+a_kv_k=0$; then also $a_1f(v_1)+\dots+a_kf(v_k)=f(0)=0$ and therefore $a_1=\dots=a_k=0$.

Whenever you have a surjective linear map $f\colon V\to W$, you can choose $\{v_1,\dots,v_k\}$ so that $\{f(v_1),\dots,f(v_k)\}$ is a basis of $W$.

Now you can finish.

Title: Build isomorphism from subset of non-injective homomorphism?

Body:
Given two vector-spaces $V, W$ with $\dim(V)=5$ and $\dim(W)=2$.

If there exists a surjective homomorphism $f:V \to W$, does there exist a subspace of the domain: $V_1 \subset V$ with $\dim(V_1)= 2$ isomorphic to $W$?

Since $V$ is larger than $W$ and $f$ is surjective, it makes sense that we can choose a subset of the domain to reach $W$.

But is this subset necesarilly a subspace?


问:
Results that came out of nowhere.
Most big results in mathematics are built on years and years of groundwork by the author and other mathematicians, such as Wiles' proof of FLT or the classification of finite simple groups.  Nevertheless, there is this big romantic idea in pop culture where a young genius comes up with a brilliant proof of a major conjecture, based on some creative new idea which flies in the face of the mathematical community.

What are some examples where this has actually happened?  That is, which results stem from independent work by a mathematician who came along "out of nowhere" and solved a huge problem by surprise through nonstandard techniques?

I read that the recent proposed proof of the ABC conjecture comes from years of autonomous theory by Prof Mochizuki, most of which ventures far outside of the current literature; this would seem to qualify as one example provided the proof turns out to be true.

答: Smale's proof of sphere eversion and the proof of the Poincaré conjecture for dimensions 5 and higher were unexpected and completely out of the blue.</p>

Title: Results that came out of nowhere.


答: Consider the polynomials$$P(x)=(x-a)(x-b)\quad\text{and}\quad Q(x)=(x-c)(x-d).$$Note that\begin{align}P(x)&=x^2-(a+b)x+ab\\&=x^2-(c+d)x+cd\text{ (since $a+b=c+d$ and $ab=cd$)}\\&=Q(x).\end{align}Since they are equal, the have the same sets of roots. But the roots of $P(x)$ are $a$ and $b$ and the roots of $Q(x)$ are $c$ and $d$.

Title: From the equations prove $\{a,b\}=\{c,d\}$

Body:
If $a+b=c+d$ and $a^2+b^2=c^2+d^2$ (where $a,b,c,d$ are real numbers)
prove $\{a,b\}=\{c,d\}$

From $a+b=c+d$ we have:

$$(a+b)^2=(c+d)^2=a^2+b^2+2ab=c^2+d^2+2cd$$

Since $a^2+b^2=c^2+d^2$ then $ab=cd$. So I choose these two equations:

$${ \begin{cases}{ab=cd} \\ {a+b=c+d}\end{cases} }$$
Although solving the system looks easy, I don't know how to do it.


答:
One you got the system;
\begin{cases}{ab=cd} \\ {a+b=c+d},\end{cases}
then rest is easy.

Let the common sum be;
$$a+b=c+d=s,$$
and the common product be;
$$ab=cd=p.$$

We know that if we consider the polynomials;
$$(x-a)(x-b) \text{ and } (x-c)(x-d),$$
then both of them are equal to;
$$x^2-sx+p.$$

Thus, we know that $(x-a)(x-b)=(x-c)(x-d),$ implying that $\{a,b\}=\{c,d\};$


问:
Solving a Third order Differential Equation
How would I solve:

$$
v_{yyy} + vv_y + \frac{1}{3}yv_y + \frac{2}{3}v = 0
$$

where $v$ is a function of $y$.

Is there a way to reduce it to a solveable second order equation?

This equation is found when looking for a solution along one of the Lie Symmetries of the ODE:
$$
u_t - uu_x - u_{xxx}
$$

From what I have read, there is a transformation discovered by Miura, for the equation.  We are to set:
$$
v = \frac{-1}{6}w^2 + \frac{dw}{dy}
$$

I now need to find $v_{yyy}, v_y$ in terms of $\frac{dw}{dy}$ and sub that into the initial equation.

I have:
\begin{align}
v &= \frac{-1}{6}w^2 + w_y\\
v_y &= \frac{-1}{3}w w_y + w_{yy}\\
v_{yy} &= \frac{-1}{3}(w_y)^2 - \frac{1}{3}ww_{yy}+w_{yyy}\\
v_{yyy}&=-w_yw_{yy}-\frac{1}{3}w_yw_{yy}-\frac{1}{3}ww_{yyy}+w_{yyyy}\\
\end{align}

When I substitute this into the third order equation I get:

$$
w_{yyyy}-\frac{1}{3}ww_{yyy}-\frac{1}{3}w(w_y)^2-\frac{1}{6}w^2w_{yy}+\frac{1}{18}w^3w_y+\frac{1}{3}yw_{yy}-\frac{1}{9}yww_y+\frac{2}{3}w_y-\frac{1}{9}w^2=0
$$

According to what I have read, this can now be factored and then integrated.
The example I am following and using to apply to my equation is on page 195 of Olver.

In following the examplefrom Olver and applying the technique to my equation I get:
$$
(D_y - \frac{1}{3}w)(w_{yyy} - \frac{1}{6}w^2w_y + \frac{1}{3}yw_y + \frac{1}{3}w) = 0
$$

Now I want to integrate and solve:
$$
w_{yyy} - \frac{1}{6}w^2w_y + \frac{1}{3}yw_y + \frac{1}{3}w = 0
$$

When I integrate I get:
$$
w_{yy} - \frac{1}{18}w^3 + \frac{1}{3}yw + k = 0
$$

How do I go about solving this second order NonLinear Differential Equation?

The example in Olver, which differs slightly from my equation, has a positive $w^3$ term and states that the equation is a from of the Second Painleve Transcendent.

答:
The answer to your question depends on what you meant by solving the equation. To clarify, I will write each part separately.  (Let me know if any parts are not clear.)

First, let me first label your equation:
$$ w_{yy}=\frac{w^3}{18} -\frac{1}{3}yw - k. \qquad (1)$$

How can we recognize Equation (1) as the second Painlevé equation?

Equation (1) can be transformed to the standard form for the second Painlevé equation by taking new variables:
$$  w(y)=\mu p(t),\quad y=\nu\,t, $$
where $\mu$ and $\nu$ are constant parameters, which leads to
$$ \frac{\mu}{\nu^2} p_{tt}=\frac{\mu^3}{18} p^3-\frac{\mu\nu}{3} t p -k .$$
Dividing by $\mu/\nu^2$, we arrive at
$$ p_{tt}=\frac{\mu^2\nu^2}{18} p^3 -\frac{\nu^3}{3} t p - \frac{k\nu^2}{\mu}.$$

Compare this with the standard form of the second Painlevé equation:
$$ {\rm PII:} \quad p_{tt}= 2 p^3 + t p +\alpha.$$
We get this form if we take
$$ \nu^3=-3, \ \mu^2\nu^2=36, \alpha=-k\nu^2/\mu ,$$
which is achieved by
$$\nu=-3^{1/3}, \mu=-6/3^{1/3}, \alpha=k /2.$$

How can we solve the second Painlevé equation in general?

This is a difficult question because general solutions of PII are higher transcendental functions that are known not be expressible in terms of a finite number of earlier known functions, like polynomials of $t$, or rational functions of $t$, exponentials, trigonometric, or other classical functions like hypergeometric functions of $t$.

So the best  we can do is to give a new name to a solution of PII and write that down as an answer to your question. To identify a solution, we need to provide an initial or a boundary value. For example, a famous higher transcendental solution called the Hastings-McLeod solution is defined as a solution of PII for $\alpha=0$ that vanishes (faster than $\exp{-(2t^{3/2}/3)}$) as $t\to +\infty$.

Are there special solutions of the second Painlevé equation that can be expressed explicitly in terms of familiar functions?

Yes, but only if $\alpha$ takes on special values. The starting point for finding these solutions is by observation. Consider $\alpha=0$, it is easy to see that the identically zero function $p(t)\equiv 0$ is a solution.

To get more solutions, we use a well known transformation between solutions of different copies of PII.  Suppose $p_\alpha(t)$ is a solution of PII with parameter $\alpha$, while $p_{\alpha\pm 1}(t)$ is a solution of PII with parameter $\alpha\pm 1$. Then the transformation (called a Bäcklund transformation) is given by
$$p_{\alpha\pm 1}=-p_\alpha-\frac{2\alpha\pm 1}{2p _\alpha^2\pm 2p_\alpha’+t},$$
Where prime refers to a $t$-derivative.
If we apply this transformation to $p_0(t)=0$, we get
$$p_{\pm 1}=\mp\,\frac{1}{t}.$$
Repeated application gives an infinite sequence of rational functions that solve PII with integer values of $\alpha$. (Note that these are not general solutions because no free parameters (or integration constants) arise in these solutions.)

Similarly, there is an infinite sequence of solutions of PII with half-integer values of $\alpha$ given by Airy functions. Take a look at
Chapter 32 of DLMF for more detail.

I hope these give a glimpse of some of the very rich properties of the equation you asked about.


答:
You can set $\sqrt{x/y}=a$, $\sqrt{y/z}=b$, $\sqrt{z/x}=c$, so the first two equations become
$$
a+b+c=3,\qquad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3
$$
and also $abc=1$. Hence the second equation yields $ab+bc+ca=3$.

Therefore the three numbers are the roots of the polynomial $t^3-3t^2+3t-1$. Now you can finish, can't you?

Title: Solving the system $\sqrt{x/y}+\sqrt{y/z}+\sqrt{z/x}=\sqrt{y/x}+\sqrt{z/y}+\sqrt{x/z}=3$, $\sqrt{xyz}=1$

Body:
I am having problems solving this particular system of equations (SOE). The only thing I see that could be useful is the first and second equations are reciprocals of each other. I just don't know how I can use that information to help solve the SOE.

$$
\left\{
\begin{array}{c}
\displaystyle\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}+\sqrt{\frac{z}{x}}=3 \\
\displaystyle\sqrt{\frac{y\:}{x\:}}+\sqrt{\frac{z\:}{y\:}}+\sqrt{\frac{x\:}{z\:}}=3 \\
\displaystyle\sqrt{xyz}=1
\end{array}
\right.
$$

Is there a way to solve this algebraically other than solving by inspection?
Thank you for any help/assistance you can provide!


问:
The ratio of two areas of triangles
On the picture below is a square. The question is to find $s/r$.

I tried to draw a line from the middle of one side of the square to the middle of the opposite one, and it appears that the area of $s$ is $1/4$ the area of our square. So I stopped to calculate the ratio of triangle's area $r$ to the area of whole square.

Any hints?

答:
This is the simplest solution I could come up with. Let's first observe that if we find the following ratios,

$$\frac{AH}{HI} \;and \;\frac{BH}{HG}$$

we will be done (the reason is the well-known area distribution according to the bases).

Let one side of the square be $24k$ units (24 because it is divisible by many numbers). So, $GF$ and $FE$ are $12k$.

First, from many different methods involving similar triangles it can be noted that (for example: connect $AE$ and $DG$ and name their intersection $O$, find $AO/OE$ and use Ceva's theorem in $\bigtriangleup ACE$ ):
$$\frac{AB}{BC}=\frac{2}{1}$$

Therefore, $BC=8k$ and $AB=16k$.

From point $B$, let's draw a line that is parallel to the upper (and lower) sides of the square, and name this point $B'$. From the following similarities, the ratio of $BH$ to $HG$ can be deduced:
$$\bigtriangleup ABB' \sim \bigtriangleup ACD \;\; and \; \bigtriangleup AHG \sim \bigtriangleup B'HB \implies \frac{BH}{HG}=\frac{1}{3}$$

Let's now use Menelaus' Theorem in the following way:
$$\frac{ED}{ED+DC}\cdot\frac{CB}{BA}\cdot\frac{AI}{ID}=1 \implies \frac{12}{12+12}\cdot\frac{8}{16}\cdot\frac{AI}{ID}=1\implies \frac{AI}{ID}=\frac{4}{1}$$

From point $H$, let's draw a line that is parallel to the upper (and lower) sides of the square, and name this point $H'$. Observe:

$$\bigtriangleup BAG \sim \bigtriangleup BH'H \;\; and \; \bigtriangleup AH'H \sim \bigtriangleup ACD \implies \frac{AH}{HD}=\frac{1}{2}$$

This shows us that point $H$ is actually on the midsegment of this square.

Combining the ratios obtained for $AI/ID$ and $AH/HD$:

$$\frac{AH}{HI}=\frac{3}{5}$$

Finally, $$\frac{5}{3}\cdot\frac{3}{1}=5 \; \; \blacksquare $$

These types of area ratio problems are fun, see CTMC (q. 25).

Title: The ratio of two areas of triangles


答:
$B$ is not closed, because it has an accumulation point which is not in $B$ (can you see which one)?

$B$ is not open, because $0 \in B$ but there is no interval $(-\epsilon, \epsilon) \subseteq B$.

It is not compact because it has an open cover by intervals of the form $(0, 1-1/n)$, but this cover has no finite subcover.

Title: Open, closed, compact

Body:
I have the following question.  I know the answers but am struggling with how to write up the work formally.  Any help would be appreciated.

Consider $B=\{1-\frac{1}{n} :n=1,2,\cdots\}$.  Is $B$ open? Closed? Compact?  Justify.

I know that it is not open or closed, thus not compact.


问:
Find orthogonal vector that is in plane
The equation of the plane is $\pi: -x + 2y - z = 2$. I need to find a support vector that is orthogonal to the plane and is also in the plane itself.

To me, that sounds contradictory, because if the vector is orthogonal, it would be parallel to the normal vector $(-1, 2, -1)$, but then it would not be in the plane and if the vector would be in the plane, then how can it be orthogonal to the plane?

Does anyone have a clue?

答:
Your intuition is sort of right: you can't have a direction vector that is simultaneously orthogonal to a plane, but at the same time be a direction between vectors in the plane. It's not quite what they're asking though.

Note that the plane doesn't pass through the origin. When they say they want a vector in the plane, they're looking for a point in the plane, or if you like, a vector from the origin whose tip lies within the plane. Basically, we're looking for the unique point in the plane that is closest to the origin, so that the line between the origin and this point is perpendicular to the plane.

Normal vectors need to be parallel to $(-1, 2, -1)$, as you pointed out. So, the vector (or point) we're looking for takes the form $(-k, 2k, -k)$ for some $k \in \Bbb{R}$, but at the same time satisfies the equation $-x + 2y - z = 2$. Let's use this information to solve for $k$:
$$-(-k) + 2(2k) - (-k) = 2 \iff 6k = 2 \iff k = \frac{1}{3}.$$
So, our vector comes to be
$$\left(-\frac{1}{3}, \frac{2}{3}, -\frac{1}{3}\right).$$
This is the unique vector that is both orthogonal to the plane (i.e. the vector from $(0, 0, 0)$ to that point is orthogonal to the plane), and lies in the plane (i.e. the endpoint lies in the plane).

Title: Find orthogonal vector that is in plane


答:
The problem is that in vector geometry vectors play both the role of directions and of points. Thinking of vectors as directed entities, points are represented by vectors rooted in 0.

What you are supposed to find is a vector, which is parallel to the normal vector (if rooted at a point on the plane), but if rooted in 0 has defines a point on the plane. In 2 dimensions you can depict it like this:

But how do we compute such a vector $\vec p$? Well we know two things:

$\vec p$ is a multiple of the normal vector $\vec n$ ie $\vec p = t \vec n$ for some $t \in \Bbb R$, because this is what parallel means.

$\vec p$ defines a point on the plane, ie satisfies the equation $\vec p \cdot \vec n = \alpha$ defining your plane.

In your case this means $$(p_1,p_2,p_3) = (-t, 2t, -t)$$
and $$-p_1 + 2p_2 -p_3 = 2$$
Thus the only thing to do is to solve for $t$.

Edit I am sorry for this image not being cropped, but somehow the StackExchange App only allows me to upload photos directly from the camera. I will clean this up.

Title: Find orthogonal vector that is in plane

Body:
The equation of the plane is $\pi: -x + 2y - z = 2$. I need to find a support vector that is orthogonal to the plane and is also in the plane itself.

To me, that sounds contradictory, because if the vector is orthogonal, it would be parallel to the normal vector $(-1, 2, -1)$, but then it would not be in the plane and if the vector would be in the plane, then how can it be orthogonal to the plane?

Does anyone have a clue?


问:
Determining whether a piecewise function is odd or even
A piecewise function is defined by $f(x)=-1$ in the interval $ -\pi\lt x \lt 0$ and $f(x)=1$ in the interval $ 0\lt x \lt \pi$. I've been taught that you can check whether the function is even or odd by replacing $f(x)=f(-x)$ if the function remains unchanged then it is even and if $f(-x)=-f(x)$ i.e. negative sign can be taken common then the function is odd. But in this case I am confused because by replacing $f(x)=f(-x)$ the function remains unchanged hence it should be even but I've been told this function is odd

答: It doesn't remain unchanged. If $x\in(-\pi,0)$, then $f(-x)=1$, since $-x\in(0,\pi)$. And, if $x\in(0,\pi)$, then $f(-x)=-1$, since $-x\in(-\pi,0)$. So, $f$ is indeed odd.


问:
Determinant of $\mathbb{F}_2$ square matrices
Let $A_1,A_2,A_3$ and $B$ be in $\mathbb F_2^{n\times n}$. Is $$\mathsf{Det}(A_{1}A_2A_{3}+B)=\mathsf{Det}(A_{1}A_3A_{2}+B)=\mathsf{Det}(A_{3}A_2A_{1}+B)=\mathsf{Det}(A_{2}A_3A_{1}+B)=\mathsf{Det}(A_{2}A_1A_{3}+B)=\mathsf{Det}(A_{3}A_1A_{2}+B)?$$

答:
Here is a quick counterexample. Let $E_{ij}$ be the elementary matrix with exactly one $1$ entry at position $(i,j)$, else zero entries. Then $E_{ij}E_{jk}=E_{ik}$. So in dimension $n=2$
$$
\begin{aligned}
E_{12}E_{21} + E_{22} &= E_{11} + E_{22} = I &\text{ has determinant }1\ ,\\
E_{21}E_{12} + E_{22} &= E_{22} + E_{22} = 0 &\text{ has determinant }0\ .
\end{aligned}
$$
(In higher dimensions add diagonally an identity block.)


问:
Are these two definitions of base for a topology equivalent?
I have some experience with metric spaces, but I never took a dedicated Topology class. I recently encountered the term "base for a topology", and looking on Wikipedia, there seem to be two definitions. The very first line of the article says:

[Definition 1]: In mathematics, a base or basis for the topology $\tau$ of a topological space $(X, \tau)$ is a family $B$ of open subsets of $X$ such that every open set is equal to a union of some sub-family of $B.$

However, the first thing in the "Definition and basic properties" section is:

[Definition 2] A base is a collection $B$ of open subsets of $X$ satisfying the following properties:

The base elements cover $X$.

Let $B_1$, $B_2$ be base elements and let $I$ be their intersection. Then for each $x$ in $I$, there is a base element $B_3$ containing $x$ such that $B_3$ is subset of $I$.

I believe I have proved these definitions to be equivalent. But I want to make sure I haven't made any hidden assumptions, because I know that unlike metric spaces, General Topology can be very unintuitive for the outsider.

Def 1 $\implies$ Def 2:

Suppose $B$ is a base according to definition 1. Take any two base elements $B_1, B_2$. Their intersection $I$ is an open set, so it is a union of base elements. If $x$ is in $I$, then it is in the union of those base elements, so it is in at least one of them. Therefore $x$ is in a base element which is contained in the intersection.

Def 2 $\implies$ Def 1:

Suppose $B = \{B_i|i \in I\}$ is a base according to definition $2$. Let $U$ be an open set of $X$. The set
$$S = \bigcup_{i \in I}U \cap B_i.$$
is a union of oepn sets, and we claim that $S = B$. It is obvious that $S \subseteq B$. To see the reverse, let $x$ be any element of $B$. Since the base covers $X$, there is a base element $b$ such that $x \in b$. Therefore $x \in U \cap b$, so $x \in S$.

答:
I will give a counterexample to Definition 2 implying Definition 1.

Consider any basis $\mathcal{B}$ for $\tau$, and take any open $U$ construct the alternate basis $\mathcal{B}_U$ consisting of every set of the form $B\cup U$ for $B$ any set in $B\in\mathcal{B}$.

The collection $\mathcal{B}_U$ is still a basis satisfying Definition 2, but not Definition 1.


问:
How to derive equation of chord of circle given its midpoint?
There is an equation which instantly gives you the chord having mid point $(x_1,y_1)$ in a circle discussed here:
For a circle,

$$ C(x,y)=x^2 +y^2 + 2gx + 2fy + c = 0$$

The chord having mid point $(a,b)$ is given as:

$$ ax+by+g(x+a)+f(y+b) + c =  C(a,b)$$

and I found a similar result here. However is there a general elegant way to prove this equation for all conics?

答:
Consider the equation of a general conic $$ax^2+2hxy+by^2+2gx+2fy+c=0$$

Now consider a point $(m,n)$ which is to be taken as the midpoint of the chord of the conic.

Now consider the equation $$\begin{aligned}amx+h(my+nx)+bny+g(x+m)+f(y+n)+c \\ =am^2+2hmn+bn^2+2gm+2fn+c\end{aligned}$$

Above is a straight line which passes through $(m,n)$ and of course it would cut the conic at maximum of $2$ points. Now if we solve this equation with the conic, and consider that as a quadratic in $x$, we will get the sum of roots as $2m$ and similarly if we consider it a quadratic in $y$, sum of roots would be $2n$.

Edit: Note that if the roots of the quadratic are real and distinct, then a chord exists, if the roots are equal, then the point $(m,n)$ lies on the conic and hence that line would be a tangent. If the roots are imaginary, no chord exists.


问:
A matrix equation $Ax=0$ has infinite solutions, does $A^Tx = 0$ have infinite solutions?
I'm wondering whether a system with a transpose of a matrix has the same type of solution that the original matrix system has. If an equation $Ax=0$ equation has a unique solution, would a system with $A$ transpose instead of $A$ also have a unique solution? And what about with no solution, and infinite solutions?

答:
If you are dealing with square matrices, then the answer is yes. You are essentially asking if the matrix is invertible or not, and $A$ is invertible iff $\det(A)\ne0$ and $\det(A)=\det\!\left(A^T\right)$.

However,
$$
\begin{bmatrix}
3&1&2\\
2&0&1
\end{bmatrix}
\begin{bmatrix}1\\1\\-2
\end{bmatrix}
=\begin{bmatrix}0\\0
\end{bmatrix}
$$
and any real multiple of this solution is a solution. Yet
$$
\begin{bmatrix}
3&2\\
1&0\\
2&1
\end{bmatrix}
x
=\begin{bmatrix}0\\0\\0
\end{bmatrix}
$$
requires $x=\begin{bmatrix}0\\0\end{bmatrix}$.


问:
find the matrix exponential
Let
\begin{equation*}
A =
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}\end{equation*}

\begin{equation*}
B =
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix}\end{equation*}

\begin{equation*}
C =
\begin{pmatrix}
4 & 1 & 0 & -4 & 0 & 0\\
0 & 4 & -1 & 0 & -4 & 2\\
0 & 0 & 6 & 0 & 0 & -6\\
2 & 0 & 0 & -2 & 1 & 0\\
0 & 2 & -1 & 0 & -2 & 2\\
0 & 0 & 3 & 0 & 0 & -3\\
\end{pmatrix}\end{equation*}

Suppose $\exp(z)=e^z$, for $z\in \mathbb{C}$. Find $\exp(C)$.

Hint:

\begin{equation*}
D =
\begin{pmatrix}
2 & 1 \\
1 & 1 \end{pmatrix}\end{equation*}

is invertible.

I realize that this is already asked. But I'm hoping to get a way of solving this problem in the  Banach algebra prospective.

Attempt:

We know that $\mathbb{C}^n$ is a Hilbert space. and $B(\mathbb{C}^n) \simeq \mathbb{M_n}$. So we can consider each matrix as a function in $B(\mathbb{C}^n)$

And we can write $Ae_j= \sum_{k=1}^n a_k e_k$, where $(e_k)$ is the orthonormal basis for $\mathbb{C}^n$.

Also I know that the spectrum $\sigma(A)= \{\text{eigenvalues of} A \}$

Since the $\mathcal{M_n}$ is over $\mathcal{C}$, it is a matrix ring.

By the hint, I know that

\begin{equation*}
D^{-1}=
\begin{pmatrix}
1 & -1 \\
-1 & 2 \end{pmatrix}\end{equation*}

And we have $D\begin{pmatrix}
x & 0 \\
0 & y \end{pmatrix}D^{-1}=
\begin{pmatrix}
-x+2y & 2x-2y \\
-x+y & 2x-y \end{pmatrix}$.

From here, I wonder if we could perform some sort of induction from $B(\mathbb{C}^2) \simeq \mathbb{M_2}$ to $B(\mathbb{C}^6) \simeq\mathbb{M_6}$.

Any suggestions will be much appreciated.

答:
Rye, since we discussed this a little bit in your previous question: Note that for the case $f=\exp$ there is a standard way of defining $f(a)$ for non-normal elements of a unital Banach algebra:

$$f(a)=\sum_{k=0}^\infty\frac{a^k}{k!}$$
You can find more details about this in the first pages of Murphy's book. This can also be done for any entire function: if $g(z)=\sum_{k=0}^\infty a_kz^k$, then set $g(x)=\sum_{k=0}^\infty a_kx^k$ for $x\in A$. The series converges in the Banach algebra since it converges absolutely and we are in a Banach space. Anyway, back to your question:

We can use block-form: set
Note that $$C=\begin{pmatrix}-A+2B& 2A-2B\\ -A+B&2A-B\end{pmatrix}$$
If $P=\begin{pmatrix}B&0\\0&A\end{pmatrix}$ and $$E=\begin{pmatrix}2I_3&I_3\\I_3&I_3\end{pmatrix}$$
then $E$ is invertible, $E^{-1}=\begin{pmatrix}I_3&-I_3\\-I_3&2I_3\end{pmatrix}$ and $C=EPE^{-1}$. From this we have that $C^n=EP^nE^{-1}$ and from the definition of $\exp(\cdot)$ we will have that $\exp(C)=E\cdot\exp(P)\cdot E^{-1}$.

It is easily verified that $\exp(P)=\begin{pmatrix}\exp(B)&0\\0&\exp(A)\end{pmatrix}$, so the whole problem has been reduced to computing the exponentials of $A$ and $B$ (which are in $M_3(\mathbb{C})$, life is much easier now).

But $A^3=0$, so $\exp(A)=I_3+A+\frac{1}{2}A^2$. On the other hand, for $n\geq1$
$$B^n=\begin{pmatrix}2^n&n\cdot 2^{n-1}&0\\0&2^n&0\\0&0&3^n\end{pmatrix} $$
can be easily verified (use induction). Observe that
$$\sum_{n=0}^\infty\frac{n2^{n-1}}{n!}=\sum_{n=1}^\infty\frac{2^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty} \frac{2^n}{n!}=e^2 $$
so
$$\exp(B)=\begin{pmatrix}e^2&e^2&0\\0&e^2&0\\0&0&e^3\end{pmatrix} $$

Now that we know $\exp(A)$, $\exp(B)$ and $E$, we plug everything back to
$$\exp(C)=E\exp(P)E^{-1}$$
and the result follows.

Title: find the matrix exponential


问:
3- regular connected graph $G$ has independent subgraph $H$ such that $G-H$ is still connected.

There are $1998$ cities in Dussia, each being connected (in both directions) by flights to three other cities. Any city can be reached by any other city by a sequence of flights. The DSS (Dussia secret service) plans to close off $200$ cities, no two joined by a single flight. Show that this can be done so that any open city can be reached from any other open city by a sequence of flights only passing through open cities.

I don't know how to find such $H$, probably it should be constructed with some algorithm or perhaps take some extrem versions of $H$ (say with minimal or maximal number of neighbour set $N(H)$). Any hint what to do?

While writing this problem, this problem pop out as similar question: Every 3-connected graph has a cycle C such that G-V(C) is connected

Ohh, I can't read, it is 3-connected in that link not 3 regular.

答:
Definitions

Let a connected graph $G$ have $a,b$ and $c$ vertices of degrees $1,2$ and $3$, respectively. Define its score to be $S(G)=c-3a-2b$ and define $N(G)$ to be the set of vertices of degree $3$ which are not cutpoints.

The problem

Suppose that we have chosen an independent set of size $k\le199$ whose removal has left a connected graph with $a,b$ and $c$ vertices of degrees $1,2$ and $3$, respectively. Then  $$2a+b=3k,a+b+c=1998-k.$$ $$\Rightarrow c-3a-2b=1998-k-4a-3b=1998-10k+2a\ge8.$$

It is therefore sufficient to prove the following result.

Theorem

A simple connected graph $G$, with no vertices of degree greater than $3$ and a positive score has a vertex in $N(G)$.

Proof

Consider a counterexample with $a+b+c$ minimal. Note that if a process of contracting edges produces a graph $H$ whose new vertices have degree less than three, then a vertex in $N(H)$ is necessarily in $N(G)$.

A vertex of degree $1$ contributes $-3$ to the score and contracting its incident edge cannot reduce the score. Therefore the minimal counterexample has no vertices of degree $1$.

If two adjacent vertices of degree $2$ are not both adjacent to the same vertex then we can contract the edge joining them and increase the score. If, on the other hand, they form a triangle with a third point (necessarily of degree $3$), then these three points contribute $-3$ to the score. Contracting this triangle to a single point cannot therefore reduce the score.

Therefore any vertex $x$ of degree $2$ lies between two vertices, $u$ and $v$ say, of degree $3$. If $u$ and $v$ are not adjacent then contracting the edge $xv$ increases the score. Therefore $u$ and $v$ are, by assumption, adjacent cutpoints. Let $l\in N(u)/\{x,v\}$ and $m\in N(v)/\{x,u\}$. Then we can contract all the edges involving $u,v,x$ to a single edge $lm$. This reduces the number of vertices of degree $2$ and $3$ by $1$ and $2$, respectively, and does not change the score.

The minimal counterexample is therefore a connected 3-regular graph. However, all connected graphs have a point which is not a cutpoint and we are finished.


问:
Sheaf cohomology of the structure sheaf of the grasmannian G(2,7)
How can I calculate $H^i(Gr(2,7),\mathcal{O}_{Gr(2,7)})$ over the base field $\mathbb{C}$?

Or in general $H^i(Gr(k,n),\mathcal{O}_{Gr(k,n)})$?

The first idea came to my mind is using Plücker embedding $i:Gr(2,7)\hookrightarrow\mathbb{P}^{20}$ and the long exact sequence associated to the ideal sheaf exact sequence $$0\rightarrow\mathcal{I}_{Gr(2,7)}\rightarrow\mathcal{O}_{\mathbb{P}^{20}}\rightarrow{i_*}\mathcal{O}_{Gr(2,7)}\rightarrow{0}\,.$$

But I don't know how to calculate $H^i(Gr(2,7),\mathcal{I}_{Gr(2,7)})$ either.

答:
Question: "How can I calculate $H^i(Gr(2,7),\mathcal{O}_{Gr(2,7)})$ over the base field $\mathbb{C}$?"

Answer: Let $K$ be a field of characteristic zero and let $V:=K^n$ be an $n$-dimensional vector space over $K$ and $W \subseteq V$ a $k$-dimensional subspace. You may let $P \subseteq SL(V)$ be the (parabolic) subgroup of elements fixing $W$. There is a "canonical" isomorphism $SL(V)/P \cong G(k,V)$ where $G(k,V)$ is the grassmannian variety of $k$-dimensional subspaces of $V$. If you look in Fulton-Harris book "Representation theory: A first course" in Section 15.4 you will find a relation between global sections of invertible sheaves $\mathcal{L}$ on $G(k,V)$ and $SL(V)$-modules: For any $\mathcal{L} \in Pic(G(k,V))$ it follows $H^i(G(k,V), \mathcal{L})$ is an irreducible $SL(V)$-module. The book also gives explicit formulas for the dimension $h^i(G(k,V), \mathcal{L})$. If $\mathcal{L}:=\mathcal{O}$ is the structur sheaf, the book gives explicit formulas for $h^i(G(k,V), \mathcal{O})$.

Example 1.If $K$ is a field of characteristic zero and of $dim_K(V)=n$, let $G(k,V)$ be the grassmannian of $k$-dimensional subspaces in $V$. There is the Plucker embedding

$$ \rho: G(k,V) \rightarrow \mathbb{P}:=\mathbb{P}(\wedge^k V^*)$$

sending a subspace $W \subseteq V$ to the line $\wedge^k W \subseteq \wedge^k V$. The space of homogeneous polynomials of degree $m$ on $\mathbb{P}$ is the $m$'th symmetric power $Sym^m(\wedge^k V^*)$. Let $I(G)_m$ be the polynomials of degree $m$ vanishing on $G(k,V)$. We get an exact sequence of $SL(V)$-modules

$$0 \rightarrow I(G)_m \rightarrow Sym^m(\wedge^k V^*) \rightarrow W_m \rightarrow 0.$$

In Exercise 15.43 they calculate the $SL(V)$-module $I(G)_2$: $SL(V)$ is a semi simple linear algebraic group and any finite dimensional $SL(V)$-module decompose as a direct sum of irreducible modules. Such a decomposition is given in the exercise. In the book they give an elementary introduction to the representation theory of $SL(V), GL(V),..$ and give explicit and elementary examples from geometry.


问:
Open, closed, compact
I have the following question.  I know the answers but am struggling with how to write up the work formally.  Any help would be appreciated.

Consider $B=\{1-\frac{1}{n} :n=1,2,\cdots\}$.  Is $B$ open? Closed? Compact?  Justify.

I know that it is not open or closed, thus not compact.

答: $B\subset \mathbb{R}$ with the usual topology. $B$ is not open since (for example) $0\in B$ is not an interior point since every neighborhood of $0$ intersects $B^c$. $B$ is not closed since $1$ is a limit point not contained in $B$ (a sequence in $B$ converging to 1 is $\{1-1/n\}$). In a metric space, compact sets are closed. $B$ is not closed, hence not compact. (Edit: I took out the bit of B being a metric space itself. That was not relevant.)</p>

Title: Open, closed, compact


问:
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral? Consider the sides indistinguishable.

I thought about starting off by fixing 1 to one of the sides and then choosing the number of its opposite side but I can't go much further than that

答:
Another approach to doing this calculation is to consider the placements of two specific numbers. With two numbers positioned, you have the orientation given uniquely in almost all cases (the exception is when they're opposite each other, in which case a third number is necessary). Note, however, that we must factor in the distance between the two numbers.

For simplicity, we'll use the defining numbers as $1$ and $2$.

Suppose that a specific side is labelled $1$. Now, how many unique placements are there for the $2$? Well, there are three ways that the $2$ can be adjacent to the $1$... but they're equivalent, so we treat all three as identical, and thus only one position.

The $2$ could be in a position that is adjacent to one of those three sides. There are six such faces. You might be tempted to say that they're all identical... but they're not. A close examination reveals that half of them are clockwise from a $1$-adjacent side, while the other half are anti-clockwise.

The same applies for the next two "layers", giving a total of 6 possible placements of the $2$ that aren't opposite each other.

For these positions, we can then place the remaining 18 numbers in any arrangement we choose, so that's a total of $6\times 18!$ - but we haven't considered the case that the $2$ is opposite the $1$.

In that situation, we choose the placement of the $3$, with the $2$ locked into place. This gives us $6\times 17!$ arrangements.

And so, the total number of unique arrangements is $6\times 18!+6\times 17!$, which can be simplified to $6\times 19\times 17!$ - this is the same as the answers given by Mathew and Karl


问:
Prove $a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$

If $a,b,c,d$ are real numbers and $a+b+c+d=0$. prove:

$$a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$$

Using Euler's identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$.  in this problem substitute $x=a+b,y=c,z=d:$

$$(a+b)^3+c^3+d^3=3cd(a+b)$$
$$a^3+b^3+c^3+d^3=3(a+b)(cd-ab)=3(acd-a^2b+bcd-ab^2)$$
Although I am sure my approach is right, but I should have $3(abc+bcd+cda+dab)$ at the RHS. I don't know how to get it.

Body: $$(a+b)^3=a^3+b^3+3ab(a+b)=a^3+b^3-3ab(c+d)$$


问:
complex analysis- Liouville's theorem
1).Let $f: \mathbb{C}\to \mathbb{C}$ be a holomorphic function. Prove that if $f(0) = f(1) =0 $ and $f:\lvert f'(z) \rvert\leq 1 $ for all $z\in \mathbb{C}$, then $f(z)=0$ for all $z\in \mathbb{C}$.

2).True or false: If $f:\mathbb{C}\to\mathbb{C}$ is holomorphic, then f must be either a constant function or must be surjective.

I know that Liouville's theorem states that Every bounded entire function is constant.
if $f'(z) \leq 1$ then $F(z) = cz+D.$ Where D is a constant.
by the condition $f(0) = f(1) =0 $. substituting 0 and 1 for z. I found that c and D is equal to 0. So would it be true for $f(z)=0$ for all $z\in \mathbb{C}$.

Please correct me if im heading in the wrong direction
*Not sure of Part 2*.By Liouville's theorem i think its true for a constant but cannot identify a surjective function.

答:
For 1. $f'(x)$ is holomorphic, and if it is bounded it is constant.

For 2. Think of the exponentialfunction.

Title: complex analysis- Liouville's theorem


问:
Complex Limit with indeterminate form
I have this limit but have no idea how to show it. Obviously, we can plug in i for z but then we get an indeterminate form. Our professor has stated we can't use l'hospital so I was wondering if anyone had any insight. I gave also tried to factor the numerator but that does not work either.
$$
\lim_{z\to i} \frac{iz^3-1}{z-i}
$$

答:
We have : $$\frac{iz^3-1}{z-i} = i \frac{z^3+i}{z-i}$$

Since $z=i$ is a solution to $z^3+i=0$, all you have to do here is to use long division.

$$\frac{z^3+i}{z-i} = z^2+iz-1$$

$$\therefore \lim_{z \to i}{\frac{iz^3-1}{z-i}} = i \lim_{z \to i}{(z^2+iz-1)} = i (i^2 + i^2 -1)$$

$$\lim_{z \to i}{\frac{iz^3-1}{z-i}} = -3i$$

I hope this will help

Title: Complex Limit with indeterminate form


Title: Prove $a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$

Body:
If $a,b,c,d$ are real numbers and $a+b+c+d=0$. prove:

$$a^3+b^3+c^3+d^3=3(abc+bcd+cda+dab)$$

Using Euler's identity: if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$.  in this problem substitute $x=a+b,y=c,z=d:$

$$(a+b)^3+c^3+d^3=3cd(a+b)$$
$$a^3+b^3+c^3+d^3=3(a+b)(cd-ab)=3(acd-a^2b+bcd-ab^2)$$
Although I am sure my approach is right, but I should have $3(abc+bcd+cda+dab)$ at the RHS. I don't know how to get it.

Body: $$3(acd-a^2b+bcd-ab^2) = 3(cd(a + b) - ab(a+b)) = 3(cd(a + b) + ab(c+d))  = 3(abc + abd + acd + bcd)$$


问:
Transform an AR(2) to MA($\infty$)
So I have the following AR(2) process:

$z_t = \delta + \phi_2z_{t-2} + \epsilon_t$

where $\epsilon_t$ is white noise $(0, \sigma^2)$

How can I transform this process to an MA($\infty$)?

My thought process is to recursively substitute as follows:

$z_t = \delta + \phi_2(\delta + \phi_2z_{t-4} + \epsilon_{t-2}) + \epsilon_t$

$z_t = \delta +\phi_2\delta + \phi_2^2z_{t-4} + \phi_2\epsilon_{t-2} + \epsilon_t$

Then substitute in for $z_{t-4}$ as follows:

$z_t = \delta +\phi_2\delta + \phi_2^2(\delta + \phi_2z_{t-6} + \epsilon_{t-4}) + \phi_2\epsilon_{t-2} + \epsilon_t$

$z_t = \delta + \phi_2\delta + \phi_2^2\delta + \phi_2^3z_{t-6} + \phi_2^2\epsilon_{t-4} + \phi_2\epsilon_{t-2} + \epsilon_t$

And then keep doing the same thing over and over again. Hopefully you see the pattern of what I am doing.

So this is kind of where I am stuck. If this is correct, how would I write the MA($\infty$) process in sigma notation?

答:
Your $\delta + \phi_2\delta + \phi_2^2\delta + \cdots$ is just a geometric series so let's call this $\mu$ and we have $\mu= \frac\delta{1-\phi_2}$, at least when $|\phi_2|<1$

Meanwhile  your $\cdots +\phi_2^3z_{t-6} + \phi_2^2\epsilon_{t-4} + \phi_2\epsilon_{t-2} + \epsilon_{t}$  can be written as $\sum\limits_{i=0}^\infty \psi_i \epsilon_{t-i}$ where  $\psi_{2n}=\phi_2^n$ and $\psi_{2n+1}=0$, noting $\psi_{0}=1$.  You will have $\sum\limits_{i=0}^\infty |\psi_i| = \frac\delta{1-|\phi_2|} < \infty$ when $|\phi_2|<1$.

So, providing that $|\phi_2|<1$, you can write $$z_t=\mu + \sum\limits_{i=0}^\infty \psi_i \epsilon_{t-i} = \frac\delta{1-\phi_2} + \sum\limits_{n=0}^\infty \phi_2^n \epsilon_{t-2n} $$  which is the form of an $\mathrm{MA}(\infty)$ process

Title: Transform an AR(2) to MA($\infty$)


问:
What is the remainder when $4^{96}$ is divided by 6
What is the remainder when $4^{96}$ is divided by 6

My approach :
When 4 is divided by 6 gives 4 remainder, when $4^2$ is divided by 6 gives remainder 4 ,... is it for the complete series.. that means when $4^{96}$ is divided by 6 gives 4 as remainder or is there any short cut or other answer of this question.

Please suggest if there is any sort of short cut to such problems.

Thanks.

答:
By induction, we can prove $4^N\equiv4\pmod6$. The base case is $N=1$, which is trivial. Then suppose for $N=k\in\mathbb{N}$, the statement is true. Then for $N=k+1$, we have
$$4^N=4^{k+1}=4^k \cdot4\equiv4\cdot4=16\equiv4\pmod6$$
So, we have proved the induction case. So sub $N=96$ into above statement then we are finished with $4^{96}\equiv4\pmod6$.


答:
Your answer is in the right direction. Here is the argument I came up with. (I avoided using modular arithmetic.)

Let's prove that $4^n$ has a remainder of $4$ when divided by $6$.
The base case for this argument is $4^1=4$ is obvious.
Then assume $4^n$ has a remainder of $4$ when divided by $6$, this means $4^n = 4 + 6*m$ for some integer $m$. This means that $4^{n+1}=4*(4+6*m)=16+6*(4*m)$. Because $6*4*m$ is divisible by $6$ the remainder of $16+6*4*m$ when dividing by $6$ is equal to that of $16$, which is $4$. This proves that $4^{n+1}$ has a remainder of $4$ when divided by $6$.
This concludes the proof.

The case for $n=96$ is a special case. That follows from the general case.

Body:
What is the remainder when $4^{96}$ is divided by 6

My approach :
When 4 is divided by 6 gives 4 remainder, when $4^2$ is divided by 6 gives remainder 4 ,... is it for the complete series.. that means when $4^{96}$ is divided by 6 gives 4 as remainder or is there any short cut or other answer of this question.

Please suggest if there is any sort of short cut to such problems.

Thanks.


Title: What is the remainder when $4^{96}$ is divided by 6

Body:
$$4^{96}=(6-2)^{96} \equiv 2^{96} \pmod{6}$$

Now, if we find the remainder of $2^{95}=(3-1)^{95}$ by $3$,it is simply $(-1)^{95}=-1$ or $2\pmod{3}$ , so we have $$2^{95}=3k+2$$ Now multiply the above equation by $2$ on both sides to get $$2^{96}=6k+4$$


Title: Complex Limit with indeterminate form

Body:
I have this limit but have no idea how to show it. Obviously, we can plug in i for z but then we get an indeterminate form. Our professor has stated we can't use l'hospital so I was wondering if anyone had any insight. I gave also tried to factor the numerator but that does not work either.
$$
\lim_{z\to i} \frac{iz^3-1}{z-i}
$$

Body:
Hint:

The fraction is

$$\frac{(-iz)^3-1}{i(-iz-1)}=\frac{(-iz)^2+(-iz)+1}i.$$


答:
If $f'$ is bounded, then Liouville says $f'$ is constant, and the fact that $f(0)=f(1)$ tells you WHICH constant it is.

For you second question, consider that $\exp$ is not surjective since there is one number that is not in its image.

What I wrote above should tell you the answers to these questions.  But if you write them verbatim and hand them in as homework solutions, you won't get much if any credit, if the person grading them has any sense.  Take them as "hints".

Title: complex analysis- Liouville's theorem

Body:
1).Let $f: \mathbb{C}\to \mathbb{C}$ be a holomorphic function. Prove that if $f(0) = f(1) =0 $ and $f:\lvert f'(z) \rvert\leq 1 $ for all $z\in \mathbb{C}$, then $f(z)=0$ for all $z\in \mathbb{C}$.

2).True or false: If $f:\mathbb{C}\to\mathbb{C}$ is holomorphic, then f must be either a constant function or must be surjective.

I know that Liouville's theorem states that Every bounded entire function is constant.
if $f'(z) \leq 1$ then $F(z) = cz+D.$ Where D is a constant.
by the condition $f(0) = f(1) =0 $. substituting 0 and 1 for z. I found that c and D is equal to 0. So would it be true for $f(z)=0$ for all $z\in \mathbb{C}$.

Please correct me if im heading in the wrong direction
*Not sure of Part 2*.By Liouville's theorem i think its true for a constant but cannot identify a surjective function.


答:
No need for induction. We have
$$4^{N}=1^{N}=1\mod 3,$$
for all $N\in\mathbb N$, as well as
$$4^{N}=0\mod 2.$$
From the former it follows that $4^N$ is either $1$ or $4\mod6$, and from the latter it follows that it's $0$, $2$, or $4\mod 6$. So it must be equal to $4\mod 6$.


答: Now the factorise-the-cubic approach has been discussed, I may as well mention a use of L'Hôpital's rule,$$\lim_{z\to i}\frac{iz^3-1}{z-i}=\lim_{z\to i}3iz^2=-3i.$$The definition of $\left.\frac{d}{dz}(iz^3-1)\right|_{z=i}$ is an equivalent approach.

Title: Complex Limit with indeterminate form


问:
Circle of radius 1 is inscribed in an equilateral triangle PQR. The point of contacts of C with PQ, QR, RP are D,E,F....

Circle of radius 1 is inscribed in an equilateral triangle PQR. The point of contacts of C with PQ, QR, RP are D,E,F respectively. If PQ is $\sqrt 3 x + y-6=0$ and D is $(\frac{\sqrt 3}{2} , \frac 32)$ and origin and centre of the C are on the same side, find equation of C

Now according to the answer, the center of the given circle is $(\sqrt 3,1)$

Let the center be $(h,k)$

Then
$$-\sqrt 3 \times \frac{k-\frac 32}{h-\frac{\sqrt 3}{2}}=-1$$
But the answer doesn’t seem to satisfying $(h,k)$ in this equation.

I feel like this question will end up embarrassing me for being too simple, but I really don’t know why this formula is not working

答:
I think the writers of this question confused the order while naming the line segments and that's the main problem. Clearly, $D$ is not on $PQ$ as the question states:

```<code>The point of contacts of C with PQ, QR, RP are D,E,F respectively.
</code>```

See the diagram and it will be more clear.

Body:
I think the writers of this question confused the order while naming the line segments and that's the main problem. Clearly, $D$ is not on $PQ$ as the question states:

The point of contacts of C with PQ, QR, RP are D,E,F respectively.

See the diagram and it will be more clear.


问:
A cubic equation with only one root
I have this equation $f(x) = \frac{1}{4}x^4-2x^3+6x^2-13x+4$

I am asked to calculate by hand the one real root out of it. When derivated $x^3-6x^2+12x-13=0$

This is where I am stuck at. I know the root is $\sqrt[3]{5}+2$.

I have been trying to calculate the answer in many different ways and still not getting the correct answer. Using common divisor etc. What am I doing wrong?

答:
$x^3-6x^2+12x-13= (x - 2)^3 -5=0$

Therefore, the only real root is  $\sqrt[3]{5}+2$

Basically, you had to complete the "cube" which is a pretty standard technique.

Title: A cubic equation with only one root


问:
Considerable Change of Geodesics on Some Surfaces
In $\mathbb{R}^2$, beginning with two points $A$ and $B$ and moving $A$ in a continuous manner, the segment $AB$ changes slightly. Look at the figure below:

However it's not the case on some other surfaces, e.g. the geodesic between $A$
and $B$ compared to the one between $A'$ and $B$ on $S^2$, as is shown in this figure:

What's behind this fact? How can the surfaces of the former kind be distinguished from the latter?

Thanks.

答: Look up conjugate points on Wikipedia. If you had taken any non-antipodal pair of points on the sphere, you would still have been OK. Basically, it's positive curvature together with compactness that are causing your problem.</p>


问:
Prove that $\{Y:Y \subseteq X\}$ is a set.
I am a beginner and self-learning Real Analysis. Problem 3.4.6 from Tao's Analysis-I poses the following question. I'd like to someone to verify my proof, and help me write a technically correct proof.

Let $X$ be a set. Then the set $\{Y : Y \text{ is a subset of } X\}$ is a set.

(Hint given with the question: start with the set $\{0, 1\}^X$ and apply the replacement axiom, replacing each function $f$ with the object $f^{-1}(\{1\})$.)

My Attempt.

Using the Power set Axiom, if $X$ and $Y$ are sets, then $Y^X$ consists of all functions for
$X \to Y$

I assume for simplicity that set $X$ has two elements $\{x_1,x_2\}$.

By definition of the power set axiom, the set $Y^X$ consists of 4 functions as follows :

$f_1 \text{ maps } x_1\to 0 \text{ and } x_2 \to 0$

$f_2 \text{ maps } x_1\to 0 \text{ and } x_2 \to 1$

$f_3 \text{ maps } x_1\to 1 \text{ and } x_2 \to 0$

$f_4 \text{ maps } x_1\to 1 \text{ and } x_2 \to 1$

I want to apply the replacement axiom, replacing each function $f_i$ with the object $f_i^{-1}(\{1\})$.

$\{ f_1^{-1}(\{1\}),f_2^{-1}(\{1\}),f_3^{-1}(\{1\}),f_4^{-1}(\{1\})\}$

$\implies$ $\{ \emptyset,\{x_2\},\{x_1\},\{x_1,x_2\} \} $  which consists of all subsets of $X$.

*** My Question***

I have simply replaced the functions$(f_1,f_2,f_3,f_4)$ with the object $f^{-1}(\{1\})$. In order to use the replacement axiom, I must show that for each $x$, the statement $P(x,y)$ is true for atmost one $y$. What should be the proposition $P(x,y)$?

$\{x_1,x_2\}$ consists of all the elements of $X$, but from the definition of a function we know that bijective functions are invertible, but in the last function both  $\{x_1,x_2\}$ map to $1$, in that case, I don't think I can write
$f_4^{-1}(\{1\})$=$\{x_1,x_2\}$.

答:
To prove: $\{Y : Y\subseteq X\} = \{ f^{-1} [\{1\}] : f \in \{0,1\}^X\}$ is true .

Proof :

Definition of Inverse Images : If U is a subset of Y, we  define the set
$f^{-1}(U):= \{x \in X :f(x) \in U\}.$ ,

where we call $f^{-1}(U)$ the inverse image of $U$.

Mapping our statement, to the definition of Inverse images,we understand,

If $U = \{1\}^X$, is a subset to the larger set $f \in \{0,1\}^X$ thus, $U\subseteq f$.

The statement $f^{-1}[\{1\}]$ corresponds to all the subsets of $X$ for which $U$ is true.

Hence, the statement $\{ f^{-1} [\{1\}] : f \in \{0,1\}^X\}$ is well-defined as per the definition of inverse images.

To prove the equivalence, we first prove $\{Y : Y\subseteq X\}\subseteq  \{ f^{-1} [\{1\}] : f \in \{0,1\}^X\}$

Let $\{x\} \in Y \implies \{x\} \subseteq X \implies \{1\}^{\{x\}} \subseteq \{0,1\}^X \implies U \subseteq Y \implies f^{-1}(U) \subseteq f^{-1}(Y)$.

Thus proving, $\{Y : Y\subseteq X\} \subseteq \{ f^{-1} [\{1\}] : f \in \{0,1\}^X\}$

To prove the other subset equivalent.

Let $\{x\} \subseteq f^{-1}(U) \implies \{x\} \in Y$ as stated $\{Y \subseteq X\} \implies \{x\} \in \{Y : Y \subset X \}$

Thus proving,  $\{ f^{-1} [\{1\}] : f \in \{0,1\}^X\}  \subseteq \{Y : Y\subseteq X\}$

Therefore, proving the equality  $\{Y : Y\subseteq X\} = \{ f^{-1} [\{1\}] : f \in \{0,1\}^X\}$


问:
Searching for a counter example for this statement.
Let $A\in \mathbb{C}^{n\times n}$ be similar to $A^2$, then $A=A^2$.

This was in a multiple choice question, I'm trying to construct a counter example, but it's getting complicated for me since I'm trying to use upper triangular or diagonal matrices, I'm trying to fill in the diagonal with different numbers so I can get different eigen values which makes my life easier to decide similarity between $A$ and $A^2$. but It's not really working for me.

I would appreciate any help, and it would mean alot to explain to me your steps of constructing the counter example.

Thanks in advance.

答:
Maybe take, $A=\begin{pmatrix}1 &0\\1&1\end{pmatrix}$, $A^2=\begin{pmatrix}1 &0\\2&1\end{pmatrix}$. The matrices are similar but not equal.

I am not sure if you can use diagonal matrices as their squares will not give you similarity.


问:
Weak converge sequence problem in Banach space

Let $X$ be a Banach space  $x \in X $.$(x_n)_n \subseteq X$ $$x_n\rightharpoonup x$$ $\exists(y_n)_n \subseteq X \ni y_n$ is a linear combinations of $x_n$ then  $$y_n \rightarrow x$$

答:
Let $Y=Spann\{(x_n)\}$ we will show $ x \in Cl(Y)$. We know $f(t)=inf_{y \in Cl(Y)}||t-y||$ functional element of $ X'$ we known this from Hann-Bannach theorem. Also we know $||f||=1$ and $\forall y \in Y f(y)=0$. $x_n \rightharpoonup x$ then $\forall g \in X'$ $g(x_n)\rightarrow g(x)$ so $f(x_n) \rightarrow f(x) $ and this says
$\lim_{n \to \infty} |f(x_n)-f(x)|=0$ then $f(x)=0$ and this means $x\in Cl(Y)$. Finally we get $\exists(y_m) \subseteq Y $ and $y_m \rightarrow x$


问:
How to derive the formula for Zeta function at even intergers
Recently, I have been able to derive the power series for functions which contain the Bernoulli numbers, namely the $\dfrac{x}{e^x-1}$, $\cot(x)$, $\coth(x)$, $\tan(x)$, $\tanh(x)$. This leads me to consider the formula of Zeta function at even integers, which is:

$$(-1)^{n+1}\dfrac{B_{2n}2^{2n-1}\pi^{2n}}{(2n!)}$$

I find this formula very interesting because many parts of it are similar to the kernels of those trigonometric and hyperbolic functions that I mentioned above.

Looking around, I can only find one elementary "proof" of this formula in wikiproof

It is no surprise to me that the function $\pi x\cot(\pi x)$ is related to the  formula by the equation:

$$\pi x\cot(\pi x)=1-2\sum_{n=1}^{+\infty}\zeta(2n)x^{2n}$$

But the proof of this formula looks kind of tricky. You add $-2$ and $-\dfrac{1}{2}$ to derive this formula.

There is another proof on wikiproof which is much longer and use Fourier analysis to derive. This is way over my head and I don't think I can understand it at this moment.

I wonder if someone can list or provide resources for a simple and elementary derivation of this recurrent formula.

I have found some papers that I cannot access, one is:

Another Proof of the Famous Formula for the Zeta Function at Positive Even Integers

Euler's formula for the zeta function at the positive even integers

If you have free access to these articles, may you mind share them with me?

答:
$$f(z)=\frac{\pi^2}{\sin^2(\pi z)}-\sum_n \frac1{(z+n)^2}$$ is a $1$-periodic thus bounded entire function, therefore it is constant: $f(z)=f(i\infty)=0$.

Next expand $\sum_n \frac1{(z+n)^2}$ in Laurent series to make $\zeta(2k)$ appear and relate $\frac{\pi^2}{\sin^2(\pi z)}$ to $\frac{z}{e^z-1}$ to make the Bernouilli numbers appear.


问:
Which definitions of builder notation exist for multiset theory?
Interesting cases would be

$A=[1,1], B=[2]$

$[(a,b) \mid a \in A \wedge b \in B] = [(1,2),(1,2)]$ ?

or

$C=[1,2,3]$

$[x \mid c \in C \wedge x = c \mod 2] = [1,0,1]$ ?

The only kind of informal definition for multiset-builder notation I could find was a part in the preliminaries section of A Potpourri of Reason Maintenance Methods.

Are there other (more rigorous) studies on that topic?

答:
I'd like to describe a notation I used myself a bit. It's really straight-forward and is quite useful for certain cases.

$A = \{1,2,3\}, B=[1,1,2]$

$[x_{[y]} \mid x \in A \wedge y=2x] = [1,1,2,2,2,2,3,3,3,3,3,3]$

$[x_{[2n]} \mid x \in^n B] = [1,1,1,1,2,2]$

So, in general, $[x_{[y]} \mid P(x,y)]$ is the multiset of all elements $x$ with multiplicity $y$ that satisfy the predicate $P(x,y)$.


问:
Calculate distribution of a random variable using characteristic function
Let $X$ be a random variable such that its characteristic function is

$$\varphi_X(t)=e^{-|t|} \qquad \forall t \in \mathbb R$$

Calculate its distribution function.

My attempt at a solution:

The integral $$\int_{- \infty} ^{\infty} |\varphi_X(t)| \,dt$$ it's finite. So $X$ is absolutely continuous and, fixed $x \in \mathbb R$ , it's distribution function calculated in $x$ is equal to

\begin{align}
f_X(x)  &= \frac{1} {2\pi} \int_{- \infty} ^{\infty} e^{-itx} \varphi_X(t)\,dt \\
       &= \frac{1} {\pi} \int_{0} ^{\infty} e^{-itx} e^{-t} \,dt
 \\
       &= \frac{1} {\pi} \int_{0} ^{\infty} \cos(tx) e^{-t} \,dt
        - i \frac{1} {\pi} \int_{0} ^{\infty} \sin(tx) e^{-t} \,dt
\\
       &= \frac{1} {\pi} \frac {1}  {1+x^2}
        - i \frac{1} {\pi} \frac {x}  {1+x^2} \\
\end{align}

But the solution of the book is:

$$ f_X(x) = \frac{1} {\pi} \frac {1}  {1+x^2} $$

So my doubt is: why the imaginary part disappears?

答: There is a mistake in line 2 of your proof. The integral from $-\infty$ to $+\infty$  is twice the integral form $0$ to $\infty$ for even functions. For odd functions the integral is actually $0$. So  integral of $\sin (tx) \phi_X (t)$ is $0$.


问:
Do antilinear maps have adjoints?
Let $T: H \to H$ be a continuous antilinear map on a Hilbert space $H$, that is $T(\alpha \xi + \eta)= \overline{\alpha} T \xi + T \eta$ for $\xi, \eta \in H$ and $\alpha \in \mathbb{C}$. Does there exist an antilinear map $T^*: H \to H$ such that
$$\langle T^* \xi, \eta\rangle = \langle \xi, T\eta\rangle$$
for all $\xi, \eta \in H$ (or a variation on this).

I could try to mimique the "usual" proof for bounded linear maps, but maybe there is a smart way to avoid this kind of work by reducing to this case?

答:
Well, $$\langle T^*\xi,\eta\rangle=\langle\xi, T\eta\rangle$$
cannot stand if $T,T^*$ are antilinear: the LHS is anti-linear on $\xi$ while RHS is linear on $\xi$. But if written as
$$\langle\eta,T^*\xi\rangle=\langle\xi,T\eta\rangle$$
then yeah, check this out:

Forget about scalar multiplication and inner product on $H$ for a bit and define new ones as $\lambda* x:=\bar{\lambda}\cdot x$ where in the RHS  the scalar multiplication is the old one. Also define $\langle x,y\rangle_2:=\langle y,x\rangle_H$ where the RHS inner product is the old one and note that $H$ endowed with the new scalar multiplication and $\langle-,-\rangle_2$ is a new Hilbert space. Also, $T$ is a bounded linear operator when seen acting from $H$ with the old structure towards $H$ with the new structure, thus it has an adjoint operator $T^*$ which acts from $H$ with the new structure to $H$ with the old structure, so
$$\langle Tx,y\rangle_2=\langle x,T^*y\rangle_H$$
i.e.
$$\langle y,Tx\rangle_H=\langle x,T^*y\rangle_H $$

which is what you want. Since $T^*$ is a linear operator for the new scalar multiplication, $T^*$ is antilinear for the original one.

Title: Do antilinear maps have adjoints?


问:
$f(x;\theta)=\dfrac{1}{\theta}e^{\frac{-x}{\theta}}$ , $x>0$ Type $1$ error ,Rejection region is $\overline{X} >6.296.$

Let $X_1,X_2,X_3 $ denote a random sample of size $3$ from population
$X$ with probability density function.

$f(x;\theta)=\dfrac{1}{\theta}e^{\frac{-x}{\theta}}$ , $x>0$

The null hypothesis $H_0:\theta=3$ is to be rejected in favor of the
alternative $H_\alpha:\theta\ne3$ if and only if $\overline{X}
 >6.296.$ What is the significance level of the test?

I tried to do it as follows:

$1.\ \ \ X\sim exp(\alpha) , Y=\sum_{i}X_i \sim G(\alpha,n) $

Using gamma PDF $f_y(y)=\dfrac{a^{\lambda}e^{-ax}x^{n-1}}{\Gamma{n}}$

$\overline{X}\sim(na,n)$

$P(\overline{X}>6.296)=\int_{6.296}^{\infty}\dfrac{e^{-x}x^{2}}{2}dx$

$2. \text{If} \ \  \overline{X}\sim(na,n) ,2\overline{X}a \sim \chi_{n}^2 $

$P(Reject H_0|\theta=3)=P( \overline{X} >6.296)=P( 2\overline{X}\cdot\frac{1}{3} >6.296)$

Using both methods I get $\alpha=0.01$ but the answer given is $0.05$ I don't know what am I doing wrong please correct me and is there any other quick method to solve this problem?

Please try to use the same notations as me because I have an exam coming I don't want to get confused.

答:
For convenience, let $c = 6.296$.  Then $$\begin{align}\Pr[\bar X > c] &= \int_{x=c}^\infty \frac{1}{2} x^2 e^{-x} \, dx \\
&= \left[ -\frac{1}{2} x^2 e^{-x} \right]_{x=c}^\infty + \int_{x=c}^\infty x e^{-x} \, dx \\
&= \frac{1}{2} c^2 e^{-c} + \left[ -x e^{-x} \right]_{x=c}^\infty + \int_{x=c}^\infty e^{-x} \, dx \\
&= \frac{1}{2} c^2 e^{-c} + c e^{-c} + e^{-c} \\
&\approx 0.0499925.
 \end{align}$$

I do not know how you got $0.01$.  It is worth noting (this is not a coincidence) that if $X \sim \operatorname{Gamma}(n, \theta)$, then $$\Pr[X > \lambda \theta] = \Pr[Y < n],$$ where $Y \sim \operatorname{Poisson}(\lambda)$.  Note that the gamma parametrization is by shape and scale.


问:
Brownian motion and second derivatives.
Let $\{B_t\}$ be Brownian motion with $B_0=x\in\mathbb{R}$. Let $f\in C_c^2(\mathbb{R})$. According to page 11 of this document, we have that

$$ \lim_{t\to 0}\frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\frac{1}{2}f''(x).$$
If you follow the reasoning, there is a point where the author obtains through 2 change of variables that
$$ \frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\int_\mathbb{R}\frac{1}{\sqrt{2\pi }}\frac{f(x+\sqrt{t}\cdot y)-f(x)}{t}e^{-y^2}dy.$$
He then uses Taylor's formula to get that
$$f(x+\sqrt{t}y)-f(x)=f'(x)\sqrt{t}\cdot y+\frac{1}{2}f''\big(x+\xi\sqrt{t}\cdot y\big)ty^2.$$
for some $\xi\in[0,1]$ and concludes by dominated convergence that
$$\lim_{t\to 0 }\int_\mathbb{R}\frac{1}{\sqrt{2\pi }}\frac{f(x+\sqrt{t}\cdot y)-f(x)}{t}e^{-y^2}dy =\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\lim_{t\to 0}f''\big(x+\xi\sqrt t y\big)y^2e^{-y^2}dy.$$
I don't get the last step. First of all, why was he able to use dominated convergence in the end and not from the start since the integrand is the same? Second, how did he get rid of the term $f'(x)\sqrt{t} y$ in the Taylor expression?

Either something is weird about this whole argument or I'm missing a main point.

答:
Let's do this slowly.  We start with
$$
\frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\mathbb{E}^x\left[\frac{f(B_t)-f(x)}{t}\right]
=\int_{\mathbb{R}}\frac{f(x+\sqrt{t}\cdot y)-f(x)}{t}\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}
$$
and we expand $f(x+\sqrt{t}\cdot y)$ to get
$$
\dots=\int_{\mathbb{R}}\frac{f'(x)\sqrt{t}\cdot y}{t}\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}
+\int_{\mathbb{R}}\frac{\frac12f''(x+\xi\sqrt{t}\cdot y)ty^2}{t}\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}.
$$
The first integrand is an odd function of $y$ so its integral over $\mathbb{R}$ is 0 and we are left with the second integral
$$
\frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}=\int_{\mathbb{R}}\frac12f''(x+\xi\sqrt{t}\cdot y)\frac{y^2e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}.
$$

Now comes the DCT part.  Since $f\in C^2_c(\mathbb{R})$, we have $\sup\lvert f''\rvert=M<\infty$ and so the integrand on the RHS is dominated by the integrable function $\frac12 M e^{-y^2}$.  Hence we have
$$
\lim_{t\to 0}\int_{\mathbb{R}}\frac12f''(x+\xi\sqrt{t}\cdot y)\frac{y^2e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}
=\int_{\mathbb{R}}\lim_{t\to 0}\frac12f''(x+\xi\sqrt{t}\cdot y)\frac{y^2e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}
$$
i.e.,
$$
\lim_{t\o 0}\frac{\mathbb{E}^x[f(B_t)]-f(x)}{t}
=\int_{\mathbb{R}}\lim_{t\to 0}\frac12f''(x+\xi\sqrt{t}\cdot y)y^2\frac{e^{-y^2}\,\mathrm{d}y}{\sqrt{2\pi}}=\frac12f''(x).
$$

Title: Brownian motion and second derivatives.


问:
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral
In how many ways can we enumerate from 1 to 20 the sides of a icosahedral? Consider the sides indistinguishable.

I thought about starting off by fixing 1 to one of the sides and then choosing the number of its opposite side but I can't go much further than that

答:
Label one arbitrary face with number $1$.

Choose three of the remaining $19$ numbers; that can be done in $\binom{19}3$ ways.

Apply the three chosen numbers to the three faces bordering face number $1$; that can be done in $2$ distinguishable ways, clockwise or counterclockwise.

Apply the $16$ remaining numbers to the other $16$ faces; this can be done in $16!$ different ways.

The final answer is $$\binom{19}3\cdot2\cdot16!=\boxed{\frac{19!}3}$$ which agrees with the other answers.