问:
A unique question in combinatorics and non-commutative variables.
Let $x,y$ be two variables that satisfy:

$xy=yx+1$ (they are not commutative).

Find $(xy)^2 ,(xy)^3, (yx)^2, (yx)^3$ as a linear  combination in terms of $y^jx^j$.
Then find a formula for $(xy)^n$ and $(yx)^n$.

After some calculations we get:

$(xy)^2=y^2x^2+3yx+1$.

$(xy)^3=y^3x^3+6y^2x^2+7yx+1$.

$(xy)^4=y^4x^4+10y^3x^3+25y^2x^2+15yx+1$.

$(yx)^2=y^2x^2+yx$.

$(yx)^3=y^3x^3+3y^2x^2+y$

Now, given $xy=yx+1$ we can use the binomial formula:

$(xy)^n=\sum_{k=0}^{n} {n \choose k} (yx)^k$.

Denote:

$(yx)^{n}=\sum a_{n,k} y^kx^k$.

Then,

$(yx+1)^n=(yx)(yx)^n=\sum {n\choose k} yxy^kx^k$.

After computing we get,

$(yx)^{n+1}= \sum a_{n,k} (ky^kx^k+y^{k+1}x^{k+1})$.

By comparing the coefficient of $[y^kx^k]$ in $(yx)^{n+1}$:

$a_{n,k}=ka_{n-1,k}+a_{n-1,k-1}$

答:
The following is valid for $n\geq 0$:
\begin{align*}
\color{blue}{(xy)^n=\sum_{k=0}^n{n+1\brace k+1}y^kx^k}\tag{1}
\end{align*}
where ${n\brace k}$ are the Stirling numbers of the second kind. We show (1) by deriving the recurrence relation
\begin{align*}
&{n+1\brace k}=k{n\brace k}+{n\brace k-1}\qquad\qquad n, k\geq 1\\
&{0\brace 0}=1,\qquad\quad {n\brace 0}={0\brace n}=0\qquad \ n\geq 1\tag{2}
\end{align*}
for the Stirling numbers of the second kind. We do this in two steps and start with the relation $xy=yx+1$. We get for $k\geq 1$:
\begin{align*}
\color{blue}{xy^k}&=(xy)y^{k-1}\\
&=(yx+1)y^{k-1}=yxy^{k-1}+y^{k-1}\\
&=y(yx+1)y^{k-2}=y^2xy^{k-2}+2y^{k-1}\\
&=y^3xy^{k-3}+3y^{k-1}\\
&=\cdots\\
&\,\,\color{blue}{=y^kx+ky^{k-1}}\tag{3}
\end{align*}
Now we consider
\begin{align*}
(xy)^n=\sum_{k=0}^na_{n,k}y^kx^k\qquad\qquad n\geq 0\tag{4}
\end{align*}
and derive a recurrence relation for $a_{n,k}, n,k\geq 0$ with the help of (3).

We obtain for $n\geq 0$:
\begin{align*}
\color{blue}{\sum_{k=0}^{n+1}a_{n+1,k}y^kx^k}&=(xy)^{n+1}=xy(xy)^n\\
&=xy\sum_{k=0}^na_{n,k}y^kx^k\tag{5}\\
&=\sum_{k=0}^na_{n,k}\left(xy^{k+1}\right)x^k\\
&=\sum_{k=0}^na_{n,k}\left((k+1)y^{k+1}x^k+y^{k+1}x^{k+1}\right)\tag{6}\\
&=\sum_{k=0}^na_{n,k}(k+1)y^kx^k+\sum_{k=1}^{n+1}a_{n,k-1}y^kx^k\tag{7}\\
&\,\,\color{blue}{=\sum_{k=0}^{n+1}\left((k+1)a_{n,k}+a_{n,k-1}\right)y^kx^y}\tag{8}
\end{align*}

Comment:

In (5) we use the representation (4).

In (6) we apply (3).

In (7) we split the sum and shift the index of the right-hand sum to also have terms $y^kx^k$.

In (8) we collect the sums by setting $a_{n,-1}=a_{n,n+1}=0, n\geq 0$.

We obtain from (8) the following recurrence relation for $a_{n,k}, n,k\geq 0$
\begin{align*}
a_{n+1,k}&=(k+1)a_{n,k}+a_{n,k-1}\\
\\
{n+2\brace k+1}&=(k+1){n+1\brace k+1}+{n+1\brace k}
\end{align*}
and since the boundary conditions in (2) can also be shown, the claim (1) follows.

Notes:

We can easily derive from (1) for $n\geq 1$ the identity
\begin{align*}
\color{blue}{(yx)^n}&=y(xy)^{n-1}x\\
&=y\left(\sum_{k=0}^{n-1}{n\brace k+1}y^kx^k\right)x\\
&=\sum_{k=0}^{n-1}{n\brace k+1}y^{k+1}x^{k+1}\\
&\,\,\color{blue}{=\sum_{k=1}^n{n\brace k}y^kx^k}
\end{align*}

A well-known instantiation of (1) is connected with the product rule of differentiation: $D(fg)=(Df)g+f(Dg)$. Considering the differential operator $(Df)(x)=\frac{d}{dx}f(x)$ and the multiplication operator $X$ with $(Xf)(x)=xf(x)$, we have
\begin{align*}
(xD)^n=\sum_{k=1}^n{n\brace k}x^kD^x
\end{align*}
See also chapter 4 in the referenced paper below.

A nice paper providing many informations and insights around this theme is Combinatorial models of creation-annihilation by P. Blasiak and P. Flajolet.


问:
Showing a function defined on a finite field is a bijection
Let $p$ be a prime and $d \in \mathbb{N}$.  If $(p-1,d)=1$ show that $f:F_p^x\to F_p^x$ given be $f(x)=x^d$.

When I first started this proof I considered two elements $a,b\in F_p^x$ where $a\neq b$.  Then $f(a)=a^d$ and $f(b)=b^d.$ However, this being in a finite field, could it be the case that $a^d=b^d$?  It's been a while since I have thought deeply about this topic, and now its bothering me that my answer wasn't satisfactory then or now.

答: $(d,p-1)=1\implies$ that for any generator $g$ of the multiplicative group of the field, $f(g)=g^d$ also generates.  Thus $f$ is surjective.  Thus bijective, by finiteness.

ContentLicense: CC BY-SA 4.0


问:
A circle of radius 1 is inscribed inside a regular octagon (a polygon with eight sides of length b). Calculate the octagon’s perimeter and its area.

A circle of radius 1 is inscribed inside a regular octagon (a polygon with eight sides of length b). Calculate the octagon’s perimeter and its area.

Hint: Split the octagon into eight isosceles triangles.

What I did is divide the entire circle into 16 parts like the following:

Side A has length 1.

Side C has a length of 1 + x (x being the difference between the circle and the vertex of the octagon).

Side B has length b/2.

Angle AC is $\frac{360}{12}=22.5$.

Angle AB is 90 degrees.

Angle BC is 67.5 degrees.

I know that

$\tan{(22.5)}=\frac{b}{2}\times\frac{1}{1}$

The perimeter then is 16 times $\frac{b}{2}:

$16\times2\tan{(22.5)}=P_b$

$32\tan{(22.5)}=P_b$

However, the answer in the textbook is:

What am I doing wrong?

答: Note that $\frac{\frac{b}{2}}{1} = \tan(22.5^{\circ})$. Then, $b = 2\tan(22.5^{\circ})$, so $\boxed{P = 16\tan(22.5^{\circ}).}$ Also, since $b = 2\tan(22.5^{\circ})$, the area of one of the isosceles triangles is $\frac{b}{2} = \tan(22.5^{\circ})$. Thus, $\boxed{A = 8\tan(22.5^{\circ}).}$


问:
When dealing with not simply connected sets how can I prove that a function has/doesn't have a primitive?
Contex:
Recently I asked a question about if $\frac{1}{z} - \frac{1}{z-1}$ is holomorphic on $0<|z|<1$ and it got me thinking say I wanted to prove that its primitive exists/doesn't exist. I know techniques through which to prove the function has a primitive, for example, if it was on an open disc and is holomorphic then the primitive exists or for example if the set is simply connected I could use Caychy's Integral Theorem and then Morera's theorem. However, I do not know about any techniques where we are dealing with sets that are not necessarily simply connected.

My question: What are techniques to check if a function has a primitive on a set that is not simply connected? (for example what can I do to prove $\frac{1}{z} - \frac{1}{z-1}$ does/nt have a primitive on $0<|z|<1$)

答:
For holomorphic functions, the existence of a primitive is equivalent to path-independence of all contour integrals. If a primitive $F$ of $f$ exists, then $\int_\gamma f(z)\mathrm dz=F(b)-F(a)$, where $a$ and $b$ are the start and end points of $\gamma$ (you can show this using the definition of the contour integral and the fundamental theorem of calculus). And for the reverse, if the contour integral is independent of the path, only depending on the start and end points, and the domain is connected, then $F(z):=\int_{z_0}^z f(w)\mathrm dw$ is a primitive, where the integral goes along any arbitrary path connecting a fixed $z_0$ to $z$. You you can show this directly via the definition of differentiability, which will give you $F'(z)=f(z)$. For disconnected domains, you can do the same construction for each connected component of the domain separately and define the primitive piecewise.

Also, path independence of the integral is equivalent to the property that integrals along closed contours all vanish. So a holomorphic function $f:D\to\mathbb C$ has an antiderivative if and only if all integrals along closed contours vanish. To show that an antiderivative doesn't exist, you can simply try to find such a contour (and such a contour is guaranteed to exist if there is no antiderivative).

To show that an antiderivative does exist is slightly more involved on the theoretical side (unless you can guess/already know a suitable primitive).  You need to show that the integral along all closed paths vanishes, and all paths are a lot. The homotopy version of Cauchy's theorem helps reduce that number by guaranteeing that integrals along homotopic paths are equal (homotopic paths are those which you can continuously deform into each other without tearing or glueing). That's fancy vocabulary for the statement that it's sufficient to test a single closed curve per hole in the domain that goes around the hole once. If those all yield vanishing integrals, then a primitive exists.


问:
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano

Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$

Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.

Also, can we prove that there is only one real solution without using the discriminant?

答:
As an alternative to the Cardano’s method, substituting $x=3+2\sqrt{14}\cosh t$ into $x^3-9 x^2-15x-6=0$ to get
$$4\cosh^3t-3\cosh t = \cosh 3t = \frac{15}{4\sqrt{14}}$$
which leads to $t=\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}$ and, then, the solution
$$x=3+2\sqrt{14}\cosh\left(\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}\right)$$
which is numerically equal to $3+2\sqrt[3]{7}+\sqrt[3]{49}$.


问:
"Fundamental Theorem" $\int_a^bF'(x)dx=F(b)-F(a)$ when $F$ is not everywhere differentiable
When $F$ is not everywhere differentiable but almost everywhere differentiable, can we still write:

$\int_a^bF'(x)dx=F(b)-F(a)?$

In my opinion there is no problem since $F'(x)$ does not exist for only a measure zero set, which cannot affect the result of integration. To formalize this, do we need to use Lebesgue integration?

答:
Consider
$$
F(x)=\left\{
\begin{array}{ccc}
0 & \text{if} & x\in [-1,0], \\
1 & \text{if} & x\in (0,1], \\
\end{array}
\right.
$$
Then $F$ is differentiable everywhere but $x=0$, and $F'(x)=0$, $F'$ is a measurable function, which is Lebesgue integrable and
$$
\int_{-1}^1 F'(x)\,dx=0\ne F(1)-F(-1)=1.
$$

However, if $F$ is absolutely continuous, then $F$ is differentiable almost everywhere and $\int_a^b F'\,dx=F(b)-F(a)$.


问:
Burgers vortex: finding the angular velocity
I'm trying to solve exercise 5.19 from Elementary Fluid Dynamics by D. J. Acheson. The exercise is as follows:

Seek an exact, steady solution to the Navier-Stokes equations of the form

$$
\mathbf{u} = -\frac{1}{2}\alpha r \mathbf{e}_r + u_\theta (r) \mathbf{e}_{\theta} + \alpha z \mathbf{e}_z
$$

where $\alpha$ is a positive constant. Note that $\boldsymbol{\omega} = \omega \mathbf{e}_z$, where

$$
\omega = \frac{1}{r}\frac{d}{dr}(ru_{\theta})
$$

Verify that $\nabla \cdot \mathbf{u} = 0$, and show that the equations of motion imply

$$
-\frac{1}{2}\alpha r \omega = \nu \frac{d \omega}{d r}
$$

Deduce that

$$
u_{\theta} = \frac{\Gamma}{2\pi r}\left(1 - e^{-\alpha r^2 / 4\nu}\right)
$$

where $\Gamma$ is an arbitrary constant.

So, here's what I've done so far. It's easy enough to verify that $\nabla \cdot \mathbf{u} = 0$. The divergence in cylindrical coordinates is:

$$
\nabla \cdot \mathbf{u} = \frac{1}{r}\frac{\partial}{\partial r}(ru_r) + \frac{1}{r}\frac{\partial u_{\theta}}{\partial \theta} + \frac{\partial u_z}{\partial z}
$$

Inserting the velocity components I get:

$$
\nabla \cdot \mathbf{u} = \frac{1}{r}(-\alpha r) + \alpha = 0
$$

To show that $-\frac{1}{2}\alpha r \omega = \nu \frac{d \omega}{d r}$ I insert the velocity components into the momentum equation of the $\theta$-component:

$$
\frac{\partial u_{\theta}}{\partial t} + u_r \frac{\partial u_{\theta}}{\partial r} + \frac{u_{\theta}}{r} \frac{\partial u_{\theta}}{\partial \theta} + u_z\frac{\partial u_{\theta}}{\partial z} + \frac{u_r u_{\theta}}{r} = -\frac{1}{\rho r}\frac{\partial P}{\partial \theta} + \nu \left[ \frac{1}{r}\frac{\partial}{\partial r}\left( r \frac{\partial u_{\theta}}{\partial r} \right) + \frac{1}{r^2} \frac{\partial^2 u_{\theta}}{\partial \theta^2} - \frac{u_{\theta}}{r^2} + \frac{2}{r^2}\frac{\partial u_r}{\partial \theta} \right]
$$

The pressure term on the right side equals zero since the pressure doesn't vary in the $\theta$-component. Inserting for the velocity components yields:

$$
-\frac{\alpha r}{2} \left[ \frac{\partial u_{\theta}(r)}{\partial r} + \frac{1}{r} u_{\theta}(r) \right] = \nu \left[ \frac{1}{r} \frac{\partial u_{\theta}(r)}{\partial r} + \frac{\partial^2 u_{\theta}(r)}{\partial r^2} - \frac{1}{r^2}u_{\theta}(r) \right]
$$

Since

$$
\omega = \frac{1}{r}\frac{d}{dr}(ru_{\theta}) = \frac{u_{\theta}}{r} + \frac{\partial u_{\theta} (r)}{\partial r}
$$
$$
\frac{d \omega}{d r} = -\frac{1}{r^2}u_{\theta}(r) + \frac{1}{r}\frac{\partial u_{\theta}(r)}{\partial r} + \frac{\partial^2 u_{\theta}(r)}{\partial r^2}
$$

we have that

$$
-\frac{1}{2}\alpha r \omega = \nu \frac{d \omega}{d r}
$$

What I need help with figuring out is how to get the following expression for the $\theta$-component velocity:

$$
u_{\theta} = \frac{\Gamma}{2\pi r}\left(1 - e^{-\alpha r^2 / 4\nu}\right)
$$

答:
The vorticity is obtained by solving  $\,\,\displaystyle \nu \frac{d \omega}{dr} = -\frac{\alpha }{2} r \omega$. This is a first-order linear differential equation that can be solved using seperation of variables

$$\frac{d \omega}{\omega} = -\frac{\alpha }{2\nu} r \, dr,$$

to obtain $\omega = C_1e ^{- \alpha r^2/4 \nu}$ where $C_1$ is a constant of integration.

Next solve for $u_\theta$ using $\displaystyle \frac{1}{r}\frac{d}{dr}(ru_{\theta}) = \omega$.

We have

$$\frac{d}{dr}(ru_{\theta}) = r\omega = C_1 r e ^{- \alpha r^2/4 \nu}$$

Integrating both sides we get (introducing another integration constant $C_2$)

$$r u_\theta = C_2 - \frac{2\nu C_1}{\alpha}e^{- \alpha r^2/4 \nu},$$

which can be rewritten in terms of arbitrary constants $\Gamma$ and $B$ as

$$u_\theta = \frac{\Gamma}{2\pi r}\left(1 - B e^{- \alpha r^2/4 \nu} \right)$$

In order that $u_\theta$ remain finite as $r \to 0$ we require $B= 1$.


问:
Check if linear functional on $\ell^1$ space is bounded
I'm a bit stuck with this question and I would appreciate any help.

Let's assume we have $\ell^1$ space of sequences over field $\mathbb{K} \in \{\mathbb{R},\mathbb{C}\}$ with norm inherited from $\ell^2$ space. We define (linear) functional $T: l^1 \rightarrow \mathbb{K}$ by
$$T(x) = \sum_{n=1}^{\infty}\ \frac{x_n}{\sqrt{n}}.$$

I would like to check if this operator is bounded or unbounded. Now, I'm not sure if I approach this question correctly. I can show that this operator is unbounded on the space $\ell^2$ by considering sequence $x = (x_n) = \frac{1}{\sqrt{n}ln(n+1)}, n \in \mathbb{N}$, since then $x \in \ell^2$ and after normalizing this vector we will get that $T(x) = \infty$.

Nevertheless, if I understand the question correctly, we need to find such $(x_n) \subset \ell^1$ that $\|x_n\|_2 = 1$ and $T(x_n) \rightarrow \infty$ (so my previous example doesn't work since $x \notin \ell^1$).

Is it possible to find such a sequence? My initial intuition was that this function should be unbounded since I couldn't find a good bound for this operator (again, with respect to $\ell^2$ norm, since with respect to $\ell^1$ norm I would just bound it by $\|x\|_1$, and Hölder's inequality didn't work since $(\frac{1}{\sqrt{n}}) \notin \ell^2$) but perhaps there is a better bound for this operator?

Thanks!

答:
You are on the right track, you noticed that for the sequence $x = \left(\frac1{\sqrt{n}\ln (n+1)}\right)_n \in \ell^2$, $Tx$ is not even well-defined.

Now we can approximate $x$ by vectors $(v_n)_n$ in $\ell^1$, namely set
$$v_n := \left(\frac1{\sqrt{1}\ln 2}, \frac1{\sqrt{2}\ln 3}, \ldots, \frac1{\sqrt{n}\ln (n+1)}, 0, 0, \ldots\right).$$

We have
$$\|v_n\|_2 = \sqrt{\sum_{k=1}^n\left(\frac1{\sqrt{k}\ln (k+1)}\right)^2} \le \underbrace{\sqrt{\sum_{k=1}^\infty \left(\frac1{\sqrt{k}\ln (k+1)}\right)^2}}_{M:=} <+\infty$$
but
$$|Tv_n| = \sum_{k=1}^n \frac{1}{k\ln (k+1)} \xrightarrow{n\to\infty} \sum_{k=1}^\infty \frac{1}{k\ln (k+1)} = +\infty$$
so $T$ maps a bounded sequence $(v_n)_n$ in $\ell^1$ to an unbounded sequence in $\mathbb{K}$. Hence $T$ is unbounded.


问:
How do you show without a graph that the first set $T$ is not compact and that the second set $C$ is compact?
So I'm a little stuck here.

Initially, the question was to determine if either of these sets were compact, and I had concluded that the first was and that the second wasn't.

I then went to check Desmos to verify my answers and found quite the opposite to be true, so I'm curious about how to prove this analytically?

I tried again but I'm entirely lost. Here were the sets:

Let $T$ be a subset of $R^2$ such that, $T = \{ (x,y) \in R^2 : x^3 + y^3 + xy \leq 25\}$

Let $C$ be a subset of $R^2$ such that, $C = \{ (x,y) \in R^2 : x^4 + y^{18} \leq 25 \}$

Thanks in advance, any advice or direction would be awesome. I'm aware of the Heine–Borel theorem but can't figure out how to prove we can use it for the set $C$

答: Hint: A subset of $\mathbb{R}^2$ is compact iff it is both closed and bounded. Since each is defined by a single inclusive inequality, each is closed. Now it only remains to decide which is bounded, I leave that to you.


答: Look at  Cantor function : it is continuous over $[0,1]$, diffeentiable almost everywhere, but one has $$\int_0^1 F'(x) dx = 0 \neq 1 = F(1)-F(0)$$

Title: "Fundamental Theorem" $\int_a^bF'(x)dx=F(b)-F(a)$ when $F$ is not everywhere differentiable

Body:
When $F$ is not everywhere differentiable but almost everywhere differentiable, can we still write:

$\int_a^bF'(x)dx=F(b)-F(a)?$

In my opinion there is no problem since $F'(x)$ does not exist for only a measure zero set, which cannot affect the result of integration. To formalize this, do we need to use Lebesgue integration?


问:
How should I calculate the intersection of subspaces?
How should I calculate the intersection of subspaces? In Z5.

span{(1, 4, 4), (2, 3, 4)} ∩ span{(1, 1, 4), (2, 4, 0)}

And how many vectors it includes?

Body:
Suppose that $(x,y,z)$ is an element of the interseaction. So you can express it in two ways:

$$\alpha_1(1,4,4)+\beta_1(2,3,4)=(x,y,z)=\alpha_2(1,1,4)+\beta_2(2,4,0)$$

Thus:
$$\alpha_1(1,4,4)+\beta_1(2,3,4)-\alpha_2(1,1,4)+\beta_2(2,4,0)=0$$

Solve this system of the equations (over $\mathbb{Z}_5)$, and you will get that:
$$\alpha_1=0$$$$3\alpha_2=\beta_2$$$$\beta_1=\alpha_2$$
So vectors in the intersection are of the form:

$$(x,y,z)=\beta_1(2,3,4)$$
Check all the values of $\beta_1$ from $0$ to $4$:

$\beta_1=0:(x,y,z)=(0,0,0)$

$\beta_1=1:(x,y,z)=(2,3,4)$

$\beta_1=2:(x,y,z)=(4,1,3)$

$\beta_1=3:(x,y,z)=(1,4,2)$

$\beta_1=4:(x,y,z)=(3,2,1)$

So the intersection is:

$$\{(0,0,0),(2,3,4),(4,2,3),(1,4,2),(3,2,1)\}$$


答:
$T$ is not bounded. You can calculate and see that $(-n, 0) \in T$ for any $n \in \mathbb N$. So $T$ compact is impossible.

$C$ is bounded. Indeed, for a crude bound, $C$ is contained in the disk $D(0; 5) = \{(x, y) \in \mathbb R^2 : x^2 + y^2 \leq 25\}$. To see $C$ is closed, note that the function $f_C(x, y) = x^4 + y^{18}$, $\mathbb{R}^2 \to \mathbb{R}$ is continuous and $C$ is the preimage of the closed set $[0, 25]$ in $\mathbb{R}$. Hence, $C$ is closed also. So by Heine-Borel, compactness follows.

Title: How do you show without a graph that the first set $T$ is not compact and that the second set $C$ is compact?

Body:
So I'm a little stuck here.

Initially, the question was to determine if either of these sets were compact, and I had concluded that the first was and that the second wasn't.

I then went to check Desmos to verify my answers and found quite the opposite to be true, so I'm curious about how to prove this analytically?

I tried again but I'm entirely lost. Here were the sets:

Let $T$ be a subset of $R^2$ such that, $T = \{ (x,y) \in R^2 : x^3 + y^3 + xy \leq 25\}$

Let $C$ be a subset of $R^2$ such that, $C = \{ (x,y) \in R^2 : x^4 + y^{18} \leq 25 \}$

Thanks in advance, any advice or direction would be awesome. I'm aware of the Heine–Borel theorem but can't figure out how to prove we can use it for the set $C$


问:
Partial derivatives question?
I have to find $dw$ if $w=f(u,v,z)$ where $u=x^2+y^2,v=x^2-y^2,z=2xy$.

Now,I know that  $dw=  ( ∂w/ ∂z)*dz  + (∂w/ ∂u)*du +  (∂w/ ∂v)*dv$
The problem is that for example,if I want to find  $∂w/ ∂z$ I don't know how to relate $w$ to $z$?

答:
(I withdraw my earlier approach.)

We can "solve" the $u$ and $v$ equations for  $x$ and $y$ as

$$2x^2 \ = \ u + v \  \ \ , \ 2y^2  \ = \ u  -  v  \ ,$$

which will allow us to write $ \ z \ $ as

$$ \ z \ = \ 2xy \ = \ (4x^2y^2)^{1/2} \ =  \ [ \ (u+v)(u-v) \ ]^{1/2} \ = \ [ \ (u^2 -v^2) \ ]^{1/2}, $$

from which we can now extract

$$dz \ = \ \frac{\partial z}{\partial u} \cdot du \  + \  \frac{\partial z}{\partial v} \cdot dv \ , $$

to replace in your chain-rule equation for $dw$ .

ContentLicense: CC BY-SA 3.0

Title: Partial derivatives question?


答:
Both sets are closed and therefore each of them is compact if and only if it is bounded.

The set $T$ is unbounded because, for each $y\in\Bbb R$, the equation (in the variable $x$) $x^3+y^2+xy=0$ has some solution $x$ (it's a cubic equation) and therefore $T$ has points whose second component is arbitrarily large. Therefore, $T$ is unbounded.

But if $(x,y)\in\Bbb R^2$ is such that $|x|,|y|\geqslant3$, then $x^4+y^{18}>25$ and therefore $(x,y)$ doesn't belong to $S$. Therefore, $C$ is bounded.


问:
Flag Stacking Problem
I am confused on part b). I am pretty sure part a) is correct. If someone could explain/show where I am going wrong, it would be much appreciated.

Part a) : Find a recurrence relation for the number of ways to arrange three types of flags on a flagpole n feet high: red flags (1 foot high), gold flags (1 foot high), and green flags (2 feet high).

My solution to this was $A_n = 2A_{n-1} + A_{n-2}$ with $A_0 = 1, A_1 = 2$

Part b) Repeat part (a) with the condition of no red above gold above green (in a row).

My Solution:
 $A_0 = 1, A_1 = 2, A_2 = 5$

My thinking is it would be the same as part a), but when last flag in sequence is red, subtract 1 for the scenario when it matches the forbidden subsequence.

 So, my solution would be $A_n = (A_{n-1} - 1) + A_{n-1} + A_{n-2}$ with $A_0 = 1, A_1 = 2$

答:
Personally in (a) I would say $A_n=2A_{n-1}+A_{n-2}$ starting with $A_0=1$ and $A_n=0$ for $n<0$.  This gives the same result as yours: $1,2,5,12,29,70,169,\ldots$.

Then for (b) I would say $A_n=2A_{n-1}+A_{n-2}-A_{n-4}$ starting with $A_0=1$ and $A_n=0$ for $n<0$.  This gives: $1,2,5,12,28,66,155,\ldots$

Title: Flag Stacking Problem


问:
Prove a inequality with positive numbers
Let $a,b,c>0$ positive real numbers. Moreover, let $C>0.$ I have the following inequality,
\begin{equation}
a^2 + b^2 \leq Cab + c.
\end{equation}
My goal is to absorb the numbers $a,b$ which are on the right hand side with those on the left hand side, in order to prove the following inequality,
\begin{equation}
a^2 + b^2 < \tilde{C}c,
\end{equation}
where $\tilde{C}>0$ is a constant.

My try. I use the Cauchy Schwartz with $\epsilon >0,$ and then I have
\begin{equation}
a^2 + b^2 \leq \epsilon Ca^2 + \frac{1}{4\epsilon}b^2 + c.
\end{equation}
Now, I can choose sufficiently small $\epsilon>0$ in order to absorb $\alpha.$ Can I somehow absorb the $b?$

Body:
You have$$a^2 + b^2 \leq Cab + c = \frac{C}{2}\left(a^2+b^2 - (a-b)^2\right)+c \leq \frac{C}{2}(a^2+b^2 )+c $$

so $$\left(1- \frac{C}{2} \right)(a^2+b^2) \leq c$$

and finally, if $C< 2$, $$a^2+b^2 \leq \frac{2c}{2-C}$$


问:
What group is $U(21)/\langle4\rangle$ isomorphic to?
Given $G = U(21)$ and $H=\langle4\rangle$, what is the group $G/H$ isomorphic to?

I've calculated $G/H$ to be $\{\langle4\rangle, 2\langle4\rangle, 5\langle4\rangle, 10\langle4\rangle\}$, but I can't figure out what group would be isomorphic to it.

My guess is to look for a group with elements $\{1, 2, 5, 10\}$ or some similar elements, but I'm struggling to come up with anything.

Is there some formula or theorem I've missed?

答: You're off to a good start; you've found that $U(21)/\langle 4\rangle$ has four elements, and you've found representatives for the cosets in $U(21)$. This allows you to make a multiplication table. There aren't many groups of $4$ elements, so you will quickly see which group it is.


问:
Detecting closed walks that visit a given edge
Is it possible to indicate whether an edge is part of a closed walk in an undirected graph only by looking at its adjacency matrix and using only matrix powers and linear algebra?

My approach:

Given adjacency matrix $A$ for graph $G$, we denote the edge between nodes $v_i$ and $v_j$ as $\varepsilon_{i, j}$. For $\varepsilon_{i, j}$ not to be in a closed loop, we need to have
$$
A_{i,j}=1
$$
and
$$
A^2_{i,j}+A^3_{i,j}+\dots+A^n_{i,j}=0
$$
where $n$ is the size of graph. However, that includes paths that contain $\varepsilon_{i, j}$. To that end, we can use Roberts approach below:

Let $B$ be obtained from the adjacency matrix $A$ by setting  the
$(i,j)$ and $(j,i)$ entries to $0$.

Then I need to compute a new matrix $B$ for each element $(i,j)$ in A, compute $\sum_{k=1}^{n-1} B^k$. However, that is way too expensive. Is there a method that is computationally less expensive, at least to be able to compute it without n^3 loops?

答:
We need to know if there's a $v_i, v_j$-walk in the graph $G - \varepsilon_{i,j}$ whose adjacency matrix is $B$.

The method that adds up $B + B^2 + \dots + B^{n-1}$ works because the $(i,j)$ entry of $B^k$ counts the number of walks of length $k$ from $v_i$ to $v_j$. If any such walk exists, there is one of length $n-1$ or less.

Without using that last fact, we could keep going forever, and take the infinite sum $I + B + B^2 + B^3 + \dots$ (throwing in the identity to make things simpler). The infinite sum is just equal to $(I-B)^{-1}$, if it converges; unfortunately, it's not going to converge.

However, since we don't care about the value of the $(i,j)$ entry and only whether it's nonzero, we can replace $B$ by a small multiple of $B$. In particular, if we replace $B$ by $\frac1n B$, then $I + \frac1n B + (\frac1n B )^2 + (\frac1n B )^3 + \dots$ is guaranteed to converge, and it will converge to $(I - \frac1n B)^{-1}$.

Therefore it's enough to check if $(I - \frac1n B)^{-1}$ has a nonzero $(i,j)$ entry.

Some further simplifications: instead of $(I - \frac1n B)$, we could invert $nI - B$. Instead of computing the $(i,j)$ entry of $(nI - B)^{-1}$, we can compute the $(i,j)$ entry of adjugate matrix $\operatorname{adj}(nI - B) = \det(nI - B)(nI - B)^{-1}$. This is given by the $(i,j)$ cofactor of $nI - B$, which is the determinant of an $(n-1)\times(n-1)$ submatrix.


问:
Showing a set of the form $\bigcup_{x \in E} [x - 1 , x + 1]$ is measurable
Let $E \subseteq \mathbb{R}$ be a subset of $\mathbb{R}$, and consider $F = \bigcup_{x \in E} [x - 1, x + 1]$. I know that under the assumption that $E$ is measurable, the set $F$ is measurable, but I can't think of an example where $F$ is not measurable. Is $F$ always measurable, independent of whether $E$ is measurable? If not, what would be a counterexample?

Thanks.

EDIT: Measurability is meant with respect to Lebesgue measure

答:
$F$ is always measurable, but it is a bit tricky to argue that this is tha case. The point in here is that if $d(x,y)<2$ then $[x - 1, x + 1] \cup [y - 1, y + 1]$ is a single closed interval.

From here, you can deduce that $F$ is a disjoint union of intervals of diameter at least 2. This can probably be faster proven by looking at the connected components. Show that any such union has only countably many terms.


问:
Proof using the converse.
For all Y⊆N, if Y is finite then Y complement (A^c) is infinite
Prove this question using the converse method and decide whether this statement is true or false.

My method-
Was to show that the intersection of A and A complement is the empty set, but I don’t think that shows if they’re finite or not.

答:
I would do this using proof by contradiction, which I assume is meant by proof using the converse.

Assume there exists some finite $Y \subseteq \mathbb{N}$ so that $Y^c$ is also finite.

We know that the union of two finite sets is another finite set, so $Y \cup Y^c$ is finite, we also know by the definition of the complement that $Y \cup Y^c = \mathbb{N}$ which has to be infinite, since $\mathbb{N}$ is infinite. This is a contradiction!

Because assuming there exists a finite set $Y \subseteq \mathbb{N}$ so that $Y^c$ is also finite leads to contradictions which must mean that that assumption is false, meaning that there either exist no finite sets that are subsets of $\mathbb{N}$ - obviously wrong since $\{1\}$ is a finite subset of $\mathbb{N}$ - or for all finite subsets of $\mathbb{N}$ the property holds that if the subset is finite, the complement of the subset is infinite.

As a remark, depending on who is asking you to prove this and in what context, you might not be allowed to state that the union of two finite sets is finite or that the union of a set and its complement in $\mathbb{N}$ are equal to $\mathbb{N}$ or you might even be required to prove that $\mathbb{N}$ is itself infinite at all, or possibly they might want you to prove that it really is a contradiction for a set to be both finite and infinite.

Title: Proof using the converse.


问:
Is the absolute value on matrices well-defined with respect to PSD cone?
Consider for two matrices $A, B$ we say $A \geq B $ if $ A - B$ is positive semidefinite.
Also, there exists a definition of absolute value for a matrix $A$ that is $|A| = \sqrt{A^\dagger A}$.

I was studying Hermitian matrices(matrices like $A$ with complex entries such that $A^\dagger = A$) and faced with a problem, is the following statement true for two arbitrary Hermitian matrices?
$$ A \geq B \geq -A \implies |A| \geq |B|$$

Note that if $A$ is a Hermitian matrix then $|A|$ is PSD and the eigenvalues of $|A|$ are equal to the absolute value of the eigenvalues of $A$.

答:
No, because the weaker result
$$
A\succeq\left\{\matrix{X\\ Y}\right. \succeq0
\ \text{ and }\ XY=YX=0
\ \Longrightarrow\ A\succeq X+Y
$$
(with $B=X-Y$) doesn't hold. For a counterexample, consider
$$
A=\pmatrix{2&\sqrt{2}\\ \sqrt{2}&2},\ X=\pmatrix{1\\ &0},\ Y=\pmatrix{0\\ &1}
$$
(and $B=X-Y=\operatorname{diag}(1,-1)$).


问:
Neat way to prove that $x^2\gt (1+x)(\ln(1+x))^{2}, ~\forall x\gt 0$
I am supposed to prove the following inequality. One standard method that can be put to use is considering a function $f(x)=x^2-(1+x)(\ln(1+x))^2$ and take its derivative to comment on the behaviour of its increase and arrive at concluding the inequality using the fact that $f(0)=0$.

$$x^2\gt (1+x)(\ln(1+x))^2,~~ \forall x\gt 0$$

I was wondering if there was a neater way to come up with the proof, especially as this is a contest-practice problem. Any hints are appreciated. Thanks.

Body:
Note

\begin{align}
\frac x{\sqrt{1+x}}-\ln(1+x)
&=\int_0^x \frac{t+2}{2(1+t)^{3/2}}dt-\int_0^x \frac1{1+t}dt\\
&=\int_0^x \frac{t+2-2\sqrt{1+t}}{2(1+t)^{3/2}}dt
=\int_0^x \frac{(\sqrt{1+t}-1)^2}{2(1+t)^{3/2}}dt >0
\end{align}
Thus, $\frac x{\sqrt{1+x}}>\ln(1+x)$ and
$$x^2\gt (1+x)\ln^2(1+x)$$


问:
Elliptic definite integral $\int_0^1 x \left(\left(7+x^2\right) E\left(x\right) - 4 \left(1-x^2\right) K(x)\right)^2 \mathrm{d}x$
Can this integral

$$\int_0^1 x \left(\left(7+x^2\right) E\left(x\right) - 4 \left(1-x^2\right) K(x)\right)^2 \mathrm{d}x$$

be found in a closed form?
As usual
$K(x) = \int_0^{\pi /2} \mathrm{d}\theta/\sqrt{1-x^2 \sin^2\theta}$ and $E(x) = \int_0^{\pi /2} \sqrt{1-x^2 \sin^2\theta} \,\,\mathrm{d}\theta$.

EDIT: ANSWER FOUND BY THE AUTHOR

$\frac{945 \zeta (3)}{1024}+\frac{10149}{512}$

答:
Heureka!: I was unsure whether it is possible to compute this, after series of calculations, I found the result myself. This integral is related to an expeted product of two sides of a random triangle formed by three independently uniformly chosen points from a unit disk. And since

$$\mathbb{E}\left[a b\right] = \frac{6766 + 315 \zeta (3)}{864 \pi ^2},$$

it is concluded that

$$\int_0^1 x \left((7+x^2)E(x)-4(1-x^2)K(x)\right)^2 \,\mathrm{d}x = \frac{945 \zeta (3)}{1024}+\frac{10149}{512} \approx 20.931585716294742...$$

EDIT:
Briefly, I will comment why this is related to the expectation given above. We first calculate the expectation $\mathbb{E}[a]$. Pick a random point $X$ in a unit disk, then we pick another point $Y$ and average out the distance $a$ fixing the first point, this expectation we call $g$. This procedure is done for all points $X'$. Due to symmetry, this must be a function of $t = |OX|$ only, so $g = g(t)$. We shift the coordinate system such that $X=O$, the disk moves wlog. to the right by $t$ with polar equation $r = t \cos\phi + \sqrt{1-t^2 \sin^2\phi}$, for the distance we have simply $|XY|=|OY|=r$ polar radius, and thus

$$g(t) = \mathbb{E}\left[|X Y|, X\text{ fixed}\right] = \frac{1}{\pi}\int_0^\pi \int_0^{t \cos\phi + \sqrt{1-t^2 \sin^2\phi}} r^2 \,\mathrm{d}r\mathrm{d}\phi = \ldots = \frac{4}{9\pi}\left[(7+t^2)E(t)-4(1-t^2)K(t)\right]$$

For the expectation $\mathbb{E}[a]$ we have, due to circular symmetry,
$$\mathbb{E}[a]= \mathbb{E}[|XY|]=\mathbb{E}[\mathbb{E}[|XY|, X\text{ fixed}]] = \mathbb{E}[g(t)] = 2\int_0^1 g(t) t \, \mathrm{d}t = \ldots =\frac{128}{45\pi}.$$

Similarly, if we pick three points $X,Y,Z$, then
$$\mathbb{E}[|XY||XZ|] = \mathbb{E}[\mathbb{E}[|XY||XZ|,X\text{ fixed}]] = 2\int_0^1 \mathbb{E}[|XY|]\mathbb{E}[|XZ|] \, t\,\mathrm{d}t = 2\int_0^1 t g(t)^2 \,\mathrm{d}t,$$

which is my original integral upto a scalar multiple.


问:
Finite group of order $mn$ with $\gcd(m,n) = 1$ .
Let $G$ be a finite group of order $mn$ with $\gcd(m,n) = 1$ and $H$ be a normal subgroup of order $m$. Prove that it is the only subgroup of order $m$.

This is somewhat different from this question: Finite group of order $mn$ with $m,n$ coprime and I don't think we are allowed to use homomorphisms(it is in the next section)

I let $K$ be a subgroup of order $m$. If $K = H$, I am done.

Otherwise, I supposed $K \neq H$. Based on previous results, I know that $HK = \{hk | h\in H, k \in H\}$ would be another subgroup of $G$ if $H$ is normal. We would have that $\exists k \in K$ such that $k \notin H$ or $\exists h \in H$ such that $h \notin K$. In either case, $|HK| > m$ which may cause problems with some theorems of number theory and Lagrange's theorem but I am unsure of the specifics.

I don't know if I can say more about $|HK|$ or whether I am even on the right track. Thoughts?

答:
Let $G$ be a group and $1_G$ denotes the identity of this group.

Fact 1. Is $H$ is a normal subgroup of $G$ and $[G:H]=n$, then we have $$x^m\in H$$ for every $x\in G$.

Fact 2. If $K$ is a subgroup such that $|K|=n$, then we have $$x^n=1_G$$ for each $x\in K$.

Now let us assume that $K$ is any subgroup of order $m$ in the situation described in the question. From $\gcd(m,n)=1$ we get existence of integers $u$ and $v$ such that $$mu+nv=1$$ and thus, for any $x\in K$, we get $$x=x^{mu+nv}=(x^m)^u\cdot (x^n)^v = (x^n)^v \in H$$
using $x^m=1$ and $x^n\in H$.

So we see that $K\subseteq H$ and together with $|K|=|H|$ this implies $K=H$.

Fact 2 is consequence of Lagrange's theorem applied to the group $K$. (We are basically saying that if $x\in K$, then order of $x$ divides $|K|$.|

The proof of Fact 1 can be found, for example, here: Let $H$ be a normal subgroup of index $n$ in a group $G$. Show that for all $g \in G, g^n \in H$ and Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in H$ for all $g \in G$.


答:
Hint:

$$\begin{align}
\langle 4\rangle^2 &=\langle 4\rangle\\
(2\langle 4\rangle)^2&=4\langle 4\rangle\\
&=\langle 4\rangle\\
(5\langle 4\rangle)^2&=4\langle 4\rangle\\
&=\langle 4\rangle\\
(10\langle 4\rangle)^2&=16\langle 4\rangle\\
&=\langle 4\rangle.
\end{align}$$

Title: What group is $U(21)/\langle4\rangle$ isomorphic to?

Body:
Given $G = U(21)$ and $H=\langle4\rangle$, what is the group $G/H$ isomorphic to?

I've calculated $G/H$ to be $\{\langle4\rangle, 2\langle4\rangle, 5\langle4\rangle, 10\langle4\rangle\}$, but I can't figure out what group would be isomorphic to it.

My guess is to look for a group with elements $\{1, 2, 5, 10\}$ or some similar elements, but I'm struggling to come up with anything.

Is there some formula or theorem I've missed?


问:
Given the sets $A, B, C, S$
It is easy to solve for x
in $A,B,C$, but I don't understand $S=\{x∈R\mid x∈A↔x∈B−C\}$

Find the set S in terms of the interval. Give as an answer the sum of the finite extremes of each of the intervals that make it up.

$A=\{x∈ R / \dfrac{x}{3} ∈[−1;4]\}$

$B=\{x∈ R / (x+3) ∈[4;7]\}$

$C=\{x∈ R / \dfrac{1−2x}{2} ∈[−1;2]\}$

$S=\{x∈R / x∈A↔x∈B−C\}$

source: Lumbreras Editors

答:
Simplifying the expressions leads to,

$A=[-3,12]$

$B=[1,4]$

$C=[-1.5,1.5]$

Now, $B\setminus C=(1.5,4]$(everything in $B$ that is not in $C$).

We can evaluate the condition $x\in A\iff x\in B\setminus C$ by thinking along the lines: for what values of $x$ is it true that if it belongs to $A$ then it must belong to $B\setminus C$ and vice-versa?

You can try to attempt before looking at the solution below.

 We can see that if $x\in A\cap(B\setminus C)$ then the above condition holds.
 Hence $S=A\cap(B\setminus C)=(1.5,4].$

Title: Given the sets $A, B, C, S$


问:
Existence and uniqueness of homoclinic orbit

Consider the following 2D autonomous system of ODEs:
$$
\left\{
\begin{array}{ll}
\dot{x} = x^2 + 2y - x \\
\dot{y} = 3xy/2 - 3x^2 - y + 2x
\end{array}
\right.
$$

How can we prove the existence and uniqueness of a homoclinic orbit (i.e. a solution $X$ of the system
for which $\lim_{t \to \infty} X(t) = \lim_{t \to -\infty} X(t) = x_0,$ where $x_0$ is an equilibrium point of the system) for the system?

It is not hard to determine the equilibrium points: $(0,0), (-3, -6)$ and $(2/3, 1/9)$. The latter two are asymptotically stable equilibrium points, so there cannot be a homoclinic orbit for those points.

However, $(0,0)$ is a saddle equilibrium, so perhaps there is a homoclinic orbit here.
We can also see that the function $H(x,y) = - y^2 + x^2(1-x)$ is constant on the solutions of the system.
The set $H(x,y) = 0$ seems to be comprised of three solutions of the system, one of which is indeed a homoclinic orbit, so this proves existence.

However, how do we prove that this is unique? I don't really know how to approach this part.

答:
Let me explain the idea from my previous answer with few more additional details.

As it was mentioned in the question, the system has a smooth first integral $H(\mathbf{x})$ — a (locally non-constant) function that is constant along the system's trajectories. The key observation is that if such system has a trajectory $\gamma(t)$ homoclinic to a saddle $p$, then $H(p) \equiv H(\gamma(t))$. To prove that we can pick a sequence of moments $t_i \rightarrow +\infty$. By continuity, since $\gamma(t_i) \rightarrow p$ and $H(\mathbf{x})$ is smooth, then $H(\gamma(t)) \rightarrow H(p)$. But since $H(\mathbf{x})$ is constant at any point of $\gamma(t)$, we have that $H(\gamma(t)) \equiv H(p)$.

This gives us the following method to find all homoclinic trajectories for a 2D system of differential equations. First, find all saddle equilibria. Second, find the level sets of $H(\mathbf{x})$ that contain these saddles. What we have proven before means that a homoclinic trajectory must lie in the same level set as the saddle, to which it is homoclinic. Just take a look at level sets after that and you can find homoclinic (and even heteroclinic) trajectories this way.


问:
Homeomorphism between open unit ball and $\mathbb R^n$
Let $B=\{x\in\mathbb R^n : ||x||<1\}$ the open unit ball with the subapce topology of $\mathbb R^n$. I want to show that $B^n\cong\mathbb R^n$ with the map $F(x)=\tan(\frac{\pi ||x||}{2})\frac{x}{||x||}$ for $x\not=0$ and $F(0)=0$

Well $\frac{x}{||x||}$ is more or less the direction of the vector $x$, the norm is a number out of $[0,1)$ and the scaled tangent is a Homeomorphism from $[0,1)$ to the positive real numbers.

I want to show that $F(x)$ is contionous, bijective and there is an inverse map.

Injectivity: Let $F(x)=F(y)\leftrightarrow ||F(x)||=||F(y)||\leftrightarrow \tan(\frac{\pi ||x||}{2})||\frac{x}{||x||}||=\tan(\frac{\pi ||y||}{2})||\frac{y}{||y||}||$

I know that the tangent is injective but I do not see how $x=y$ can be followed.

Surjectivity: Let $y\in\mathbb R^n$, now I need to show $\exists x\in B^n: F(x)=y$, i.e $\tan(\frac{\pi ||x||}{2})\frac{x}{||x||}=y$ I do not see how surjectivity can be followed form this.

Inverse map: I know that the tangent has an inverse map, namely the arctangent, but how can this be used here?

Continuity: Well I know that the tangent is only contin. on $\mathbb R\setminus\{(n+\frac{1}{2})\pi: n\in \mathbb Z\}$

答:
Injectivity

$F(x)=F(y)$, with $x\ne0$ and $y\ne0$ means

$$
\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}=
\tan\biggl(\frac{\pi \|y\|}{2}\biggr)\frac{y}{\|y\|}
$$

So we can assume $x\ne0$ and $y\ne0$. Then, taking norms,
$$
\tan\biggl(\frac{\pi \|x\|}{2}\biggr)=\tan\biggl(\frac{\pi \|y\|}{2}\biggr)
$$
because $x/\|x\|$ has norm $1$ and the same for $y/\|y\|$; moreover $\tan t>0$ if $t>0$. By the injectivity of the tangent you get
$$
\frac{\pi \|x\|}{2}=\frac{\pi \|y\|}{2}
$$
so $\|x\|=\|y\|$ and, finally $x=y$.

It's clear that $F(x)=0$ if and only if $x=0$, so the proof is complete.

Surjectivity

$0=F(0)$, so we only need to show that, for $z\ne0$, we can find $x$ with $F(x)=z$. We should have
$$
z=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)\frac{x}{\|x\|}
$$
so
$$
\|z\|=\tan\biggl(\frac{\pi \|x\|}{2}\biggr)
$$
hence
$$
\frac{\pi\|x\|}{2}=\arctan\|z\|
$$
that is,
$$
\|x\|=\frac{2\arctan\|z\|}{\pi}.
$$
Thus the candidate is
$$
x=\frac{2\arctan\|z\|}{\pi\|z\|}z
$$
Verify it's the right one.

Continuity

The continuity of $F$ is obvious in the points different from $0$, because it's obtained by continuous functions. Since your domain consists of vectors with norm $<1$, there's no problem with the tangent function, because you consider only arguments in $(0,\pi/2)$. The continuity of the inverse is obvious as well
outside $0$

Are $F$ and its inverse continuous at $0$?

A well known fact is that $\lim_{x\to0}F(x)=0$ if and only if $\lim_{x\to0}\|F(x)\|=0$; now
$$
\|F(x)\|=\biggl\|\tan\biggl(\frac{\pi\|x\|}{2}\biggr)\frac{x}{\|x\|}\biggr\|
=\tan\biggl(\frac{\pi\|x\|}{2}\biggr)
$$
that obviously satisfies the requested property. The same for the continuity at $0$ of $F^{-1}$.


问:
Topological space generated by a segment of a curve inside of $\mathrm{SL}(n,\Bbb R)$
I am interested in to see what is the topological space generated by a segment of a curve inside of $\mathrm{SL}(n,\Bbb R)$? i.e. Suppose $A\neq B\in \mathrm{SL}(n,\Bbb R)$  and $\gamma$ an arbitrary path in $\mathrm{SL}(n,\Bbb R)$ joining these two points such that $\gamma$ does not lay entirely in $\mathrm{SO}(n)$. Then what is the space generated by segment $\gamma$? i.e.
$$\langle\gamma \rangle=\{C_1^{n_1}C_2^{n_2}\dots C_k^{n_k}\  |\ C_i\in\gamma, n_i\in\Bbb Z, k\in \Bbb N\}?$$

How to prove that this space is connected?

I think this space should be all of $\mathrm{SL}(n,\Bbb R)$. But I don't know how. By the above definition, actually it is a group because $C^s\in \langle\gamma \rangle$ and $C^{-s}\in \langle\gamma \rangle$ and $I=C^0\in \langle\gamma \rangle$.

答:
The space you are looking for is a submonoid of SL$(n,{\mathbb{R}})$.
It contains at least two elements.

Note that if $A = I$ (the identity matrix) then $\langle \gamma \rangle = \langle B \rangle = \{ B^{n}, \ n \in {\mathbb{Z}} \}$ which is distinct from SL$(n, {\mathbb{R}})$. So your conjecture is wrong in general.

I take a path in SL$(n, {\mathbb{R}})$ from $A$ to $B$ to mean a continuous map $\gamma:[0,1] \to {\rm{SL}}_{n}(\mathbb{R})$ such that $\gamma(0) = A$ and $\gamma(1) = B$. Since SL$(n, {\mathbb{R}})$ is connected, there is always such a path.

Now fix such a map. Then $\langle \gamma \rangle = \{ \gamma(t)\cdot \gamma(t') \ | \ t, t' \in [0,1] \}$. Thus $\langle \gamma \rangle$ is the image by the
map
$$ [0,1] \times [0,1] \to {\rm{SL}}(n,{\mathbb{R}}), \ (t,t') \mapsto \gamma(t) \cdot \gamma(t').$$
This map is continuous and $[0,1] \times [0,1]$ is connected, so $\langle \gamma \rangle$ is also connected.


问:
Memorylessness of geometric distribution- computating mistake
I would like to show that

$X-n|X\ge n\sim G(p)$, where $X$ has a geometric distribution $G(p)$.

Hence

$P(X-n=k|X\ge n)=\frac{P(X=n+k)}{P(X\ge n)}=\frac{p(1-p)^{n+k-1}}{\sum_{k=n}^{\infty}p(1-p)^{k-1}}=\frac{p(1-p)^{n+k-1}}{(1-p)^{n-1}}=p(1-p)^k$.

Where is the mistake?

答: The problem is there are two definitions of a geometric distribution. You've calculated as if the infimum of $X$'s support is $1$, but it needs to be $0$. (The case $k=0$ is an easy sanity check to verify this; clearly $P(X=n|X\ge n)\ne0$.) Your calculation strategy is fine but, because we start at $0$, the $-1$s in exponents shouldn't be there. Then $p(1-p)^k$ is, in fact, the value of $P(G(p)=k)$.


问:
Problem with understanding notation in exact differential equations
I am studying this book on differential equations and the author just introduced a new notation I was having trouble trying to understand. I'm not sure if I'm not getting the notation or the concepts themselves.
From what I understand, the exact differential equations are those that have the form

$$M(x,y)+N(x,y)\frac{dy}{dx}=0$$
In which

$$M(x,y)=\frac{\partial F(x,y)}{\partial x} , N(x,y)=\frac{\partial F(x,y)}{\partial y}$$

for some function $F$, that we don't necessarily know. And the partial derivatives of $M$ and $N$ have to be equal. But then they say a differential equation can be written as

$$M(x,y)dx+N(x,y)dy=0$$

What do "$dx$" and "$dy$"  mean? And where did the
$$\frac{dy}{dx}$$
go to?

And then they go on to use this notation on every exercise. When you use this notation is it because you already know the equation you're writing down is an exact differential equation?

Then in another part they say:

"the differential
$$dF=F_xdx+F_ydy$$

of F(x,y) is exactly
$$Mdx+Ndy$$

So what does "$dF$" mean?

答:
For your purposes $Mdx+Ndy=0$ is just another notation for the pair $(M,N)$ of functions. In more geometric terms is denotes the direction field that is orthogonal to $(M(x,y),N(x,y))$ at each point $(x,y)$. Solutions are curves that are tangent to the direction field in every point. You can find such solutions by solving the scalar equation $M+Ny'=0$ or by solving a system
$$
\dot x(t)=-N(x(t),y(t)),\\
\dot y(t)=M(x(t),y(t)).
$$
You can also add some non-zero factor to this system.

Of course the specific notation is chosen for its connection to differential or Pfaffian forms, but these do not offer any further insights at the level of an introductory ODE course.

See also

Exact differential equations, what are we trying to find exactly?,

Exact Differential Equation Geometry.


问:
good books on Abstract Algebra and Cryptography for self-study
I want to self-study some abstract algebra and cryptography during the summer, so what are some of books that are suitable for self-study? I have very limited background in algebra and none in cryptography, and like to learn by examples. Also, please let me know if there are any books on these subjects with solutions to most of the exercises, so I can check on myself. (I saw yunone's post about good algebra books, but I would like to focus more on those with good exercises and detailed solutions.)

All suggestions are welcome! Thank you!!

答: Stinson's book is very readable and clear, and develops the algebra needed by itself.</p>


问:
how to calculate the real probability
Say I have a 50% probability of winning a game if I play against person 1, and a 50% probability of winning a game against person 2. I will play with both people, one after another. Before the matches, what was the probability that I would win at least 1 match? It has to be something between 90-99% I think. Any suggestions? Thank you!</p>

答:
Note: as Austin Mohr's helpful follow-up comment explains, the answer relies on the fact that the likelihood of a win and a loss are the same: we have $50\%$ chance of a win $= 50\%$ chance of a loss in each game, so every outcome is equally likely.

W: win... L: lose

Four possible outcomes: (Game 1 vs. person 1) followed by (Game 2 vs. person 2)

W W <--

W L <--

L W <--

L L

In $3$ of $4$ outcomes, you win one or both (at least one) of the two games. In only one outcome, will you lose both (and hence not win at least once).

Therefore: The probability of winning at least one game: $P(\text{win at least one game}) = \dfrac 34 = 75\%$

Title: how to calculate the real probability


问:
Partial derivative of a two argument function
What I have to do is to get a value of partial derivative at a $(0 , 0)$ point in a following function $\\ f(x,y) = \sqrt[3]{xy} \\ \\$

What I struggle with is that I get different results while using different methods. When I use definition I get $\frac{\partial f}{\partial x}(0, 0) =  \lim_{x \to 0} \frac{f(x, 0) - f(0, 0)}{x} = 0\\ \\$, however I can clearly see that $\frac{\partial f}{\partial x} = \frac{y^{\frac{1}{3}}}{3x^\frac{2}{3}}$ but then x can't be equal to $0$ so the derivative shouldn't have any value at a $(0, 0)$ point. Can someone point out my mistake? Which method is right?

Body: Since $f(x,0)=f(0,0)=0$ for every $x\in\Bbb R$,$$\frac{\partial f}{\partial x}(0,0)=\lim_{x\to0}\frac{f(x,0)-f(0,0)}x=0.$$


答:
The limit is the definition. It always gives the right answer, because it is the right answer.

The other approach gives the correct partial derivative for most of the plane, but the fact that the resulting expression is undefined does not let you conclude that the partial derivative doesn't exist.

Title: Partial derivative of a two argument function

Body:
What I have to do is to get a value of partial derivative at a $(0 , 0)$ point in a following function $\\ f(x,y) = \sqrt[3]{xy} \\ \\$

What I struggle with is that I get different results while using different methods. When I use definition I get $\frac{\partial f}{\partial x}(0, 0) =  \lim_{x \to 0} \frac{f(x, 0) - f(0, 0)}{x} = 0\\ \\$, however I can clearly see that $\frac{\partial f}{\partial x} = \frac{y^{\frac{1}{3}}}{3x^\frac{2}{3}}$ but then x can't be equal to $0$ so the derivative shouldn't have any value at a $(0, 0)$ point. Can someone point out my mistake? Which method is right?


问:
Specific unit conversion
This could as easily be a physics or chemistry question, so I would like to avoid this conversation in the comments.

I would like to know the general method of converting between units. For example converting Joules to $\frac{g\cdot cm^2}{s^2}$

I assume it's possible because the SI unit of energy is $\frac{kg\cdot cm^2}{s^2}$ so is it as simple as going from J to $\frac{kg\cdot cm^2}{s^2}$ and then times 1000 to get $\frac{g\cdot cm^2}{s^2}$?

If so then how does one know how many J in 1 $\frac{kg\cdot cm^2}{s^2}$? Is it something you have to learn like 1kg is 1000g?

答:
The idea is to look at some equations which involve the quantities you are interested in. So the definition of Joule is the work done by a force of $1N$ moving an object for $1m$. Then you need to know what a Newton is. From Wikipedia "One newton is the force needed to accelerate one kilogram of mass at the rate of one metre per second squared in the direction of the applied force." Therefore
$$1J=1N\cdot 1m=1kg\cdot1\frac{m}{s^2}\cdot 1m=1kg\frac{m^2}{s^2}$$

Title: Specific unit conversion


问:
Show that each connected graph that is not a block has at least $2$ blocks, each containing exactly one cut-vertex.
Show that each connected graph that is not a block has at least $2$ blocks, each containing exactly one cut-vertex.

I know:

A block of graph $G$ is the maximal connected sub-graph of $G$ that has no cut-vertices.

So "connected graph that is not a block" means that graph has cut-vertices.

What can I say now?

答:
Given a graph $G$, we can associate to it a tree $T_G$, sometimes called its block-cut tree, in the following way: the vertex set of $T_G$ consists of one vertex for each block of $G$, as well as one vertex for each cut-vertex in $G$, and for each cut-vertex $v$ of $G$, we connect the vertex in $T_G$ corresponding to $v$ to the vertices corresponding to blocks containing $v$.

One can show that, as the naming suggests, $T_G$ is indeed a tree. A block that contains exactly one cut-vertex corresponds to a vertex of degree one in $T_G$; thus, your problem amounts to showing that $T_G$ has at least two leaves. This is clearly true under the assumption that $G$ itself was not $2$-connected (so that $T_G$ has at least two vertices).


问:
Number of zeros of a real polynomial
Let $p(x)=x^{10}+a_1 x^9 +...+a_10$ be a polynomial with real coefficients. If $p(0)=-1,p(1)=1,p(2)=-1$ let $n$ be the number of real zeros of $p(x)$ then -

4 $\leq$ n

n=3

n=2

n=1

Option 2 and 4 are wrong as even degree real polynomial cannot have odd number of real roots (because complex roots are always in conjugates). (Correct me if I'm wrong)

I'm confused in option 1 and 4.
By Descartes Rule of signs $n$ can be 2. I cannot discard the option 1 as I don't know the polynomial and so I don't know if it is increasing/decreasing on the left side of 0 or on the right side of 2.

答:
There must be at least $1$ zero in each of $x\in(0,1)$ and $x\in(1, 2)$ by the Intermediate Value Theorem.

Because the coefficient of $x^{10}$ is positive, $p(x)\to \infty$ as $n\to\pm\infty$. Then, by the Intermediate Value Theorem, there is a zero in $x\in(-\infty, 0)$. Similarly, there must be a zero in $x\in(2,\infty)$.

Thus, there must be at least $4$ zeros, so the correct answer is $\boxed{(1)\ 4\leq n.}$

Title: Number of zeros of a real polynomial


问:
Algebraic Geometry text explaining and motivating the transition from the classical to modern point of view
I have a precise reference question: can you recommend me a book (or more probably a chapter of a book, or a paper) that explains and motivates the transition from the classical (more geometrical) point of view on the algebraic geometry to Grothendieck's point of view? In my course we started with some results about functors and representability, in order to introduce the affine spaces as functors on algebras, state the Hilbert's Nullstellensatz and see some basic results of classical algebraic geometry. Then, we defined the geometric points, and began to study the spectrum of rings as a topological spaces. I would like a reference that explains well the sense of this last "generalization", and that gives some motivation, because the notes of my course are very dry.

I don't know if this question is off topic or not; it could be subjective, but I don't think that there are so much books to choose from, so it shouldn't be too subjective. Thank you in advance

答: If you haven’t seen it before, Ravi Vakil’s book is full of fantastic explanations, motivations and examples http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf</p>


问:
Is $y=|x^3|$ a smooth function?
Is this a smooth function? $y=|x^3|$

The graph of this function has no sharp cuts or corners, so I think it is a smooth function but someone told me that it's not.

答: Look at the origin: what is the behaviour at $0$? Is it differentiable? How many times? ($0$ is the only potentially "problematic" point).</p>

Title: Is $y=|x^3|$ a smooth function?


问:
How can I show that the set of decimal representations of the natural numbers divisible by4 is regular?
How can I show that the set of decimal representations of the natural numbers divisible by four is regular?
I'm stack on this problem. I tried some solutions but I know It is incorrect. Please help me.
If you use more details in your explanation, would be better.

Thank you"

I added a larger image of what I tried so far the image

Let $L$ be the language of natural numbers divisible by $4$. We want a DFA $M(L)=\{Z,\Sigma,\delta,z_A,E\}$ with $Z=\{z_A,z_S,z_R,z_0,z_1,z_2,z_3\}$ and $\Sigma=\{0,1,2,3,4,5,6,7,8,9\}$ that recognizes $L$. $E=\{z_S,z_0\}$.

$$\begin{array}{c|cc}
\delta&z_A&z_S&z_R&z_0&z_1&z_2&z_3\\\hline
r_0=0&z_S&z_R&z_R&z_0&z_2&z_0&z_2\\
r_0=8,4&z_0&z_R&z_R&z_0&z_2&z_0&z_2\\
r_1=9,5,1&z_1&z_R&z_R&z_1&z_3&z_1&z_3\\
r_2=6,2&z_2&z_R&z_R&z_2&z_0&z_2&z_1\\
r_3=7,3&z_3&z_R&z_R&z_3&z_1&z_3&z_1
\end{array}$$

$\delta:Z\times\Sigma:\langle z_i,r_k\rangle\mapsto z_j$ with $i\in\{0,1,2,3\}$ and $2i+k\equiv j\pmod 4$, $k,j\in\{0,1,2,3\}$, except that all words beginning with $0$ are thrown away ($z_R$) by the machine except the word $0$ itself, which is accepted by its own state $z_S$.

We want to prove that each word $w$ over the alphabet $\Sigma$ that does not begin with $0$ takes the automaton $M$ to $z_i$ when $w\equiv i\pmod 4$. We prove this by induction on the length $n$ of $w$.

When $n=1$ we have $$\begin{align*}&w=8,4\equiv 0\pmod4\implies z_0\\&w=1,5,9\equiv1\pmod4\implies z_1\\&w=2,6\equiv2\pmod4\implies z_2\\&w=3,7\equiv3\pmod4\implies z_3\end{align*}$$

Our induction hypothesis is that the result holds for words of length $n$.

答:
A natural number whose decimal representation is $d_nd_{n-1}\ldots d_1d_0$ is divisible by $4$ if and only if the numbers whose decimal representation is $d_1d_0$ is divisible by $4$, since

$$d_nd_{n-1}\ldots d_200=100\cdot d_n\ldots d_2=4(25\cdot d_n\ldots d_2)\,.$$

It’s not hard to show that $d_1d_0$ is divisible by $4$ if and only if

$d_1$ is even and $d_0\in\{0,4,8\}$, or

$d_1$ is odd and $d_0\in\{2,6\}$.

You can do it with a DFA $M$ that has an initial state $z_e$ ($e$ for even) and just two other states, $z_{ea}$ ($ea$ for even acceptor) and $z_o$ ($o$ for odd). See if you can complete the transition table: it will have just three types of input. If you get stuck, leave a question for me.


问:
If $a+b$ is even, then $a-b$ is even
Prove that for all $a,b \in\mathbb Z$, if $a+b$ is even then $a-b$ is even.

I started off by assuming that a and $b$ are both odd integers, $2k+1$ and $2m+1$ such that their sum is $2(k+m+1)$ which is an even integer. I then concluded that $a-b$ is $2(k-m)$ which is also even, hence the statement is true.
My question is, is this enough to prove that the statement is true or do I also have to proof it when both $a$ and $b$ are even, and when one of $a$ and $b$ is even and the other is odd. Thanks.

答: So far you've only proved that the sum of two odd integers is even.  Try dropping that assumption and see what you can do, i.e. suppose $a + b = 2n$, for some integers $a,b,n$.  Can you get from that equation to an equation involving $a - b$?

Title: If $a+b$ is even, then $a-b$ is even


问:
For which a and b does latin square exist?
Latin square $L$ is an $n\times n$ array filled with $n$ different symbols, each occurring exactly once in each row and exactly once in each column.

Les us $L = l_{ij} = (a i + b j) \mod n$

For which $a,b$ is $L$ really a latin square?

It is easy to say, that for $a=b=1$ is true, but how to describe all possible values of $a,b$?

答:
As indicated in Berci's comment, it is necessary that $\gcd(a,n)=\gcd(b,n)=1$. This condition is also sufficient.

To prove this equivalence, it is enough to show that the number of distinct symbols in each row is $\frac{n}{\gcd(b,n)}$ and the number of distinct symbols in each column is $\frac{n}{\gcd(a,n)}$.


答:
I think this is definitely a good start, but it is always a good idea to cover all possible cases, which means mentioning even and even, but also even and odd. I hope this helps!

Another way to proof it, is without using cases:
If $a+b$ is even, than there exists a whole number $n$ such that $a+b=2n$. If you substract $2b$ from both sides, you'll see that the result is $a-b=2n-2b=2(n-b)$. Which is even and that is exactly what you were trying to prove.

Title: If $a+b$ is even, then $a-b$ is even

Body:
Prove that for all $a,b \in\mathbb Z$, if $a+b$ is even then $a-b$ is even.

I started off by assuming that a and $b$ are both odd integers, $2k+1$ and $2m+1$ such that their sum is $2(k+m+1)$ which is an even integer. I then concluded that $a-b$ is $2(k-m)$ which is also even, hence the statement is true.
My question is, is this enough to prove that the statement is true or do I also have to proof it when both $a$ and $b$ are even, and when one of $a$ and $b$ is even and the other is odd. Thanks.


答:
You can check that
$$
f'(x)=3x^2\operatorname{sign}(x)\\
f''(x)=6x\operatorname{sign}(x)\\
f'''(x)=6\operatorname{sign}(x)
$$
Since $f'''$ is discontinuous at zero then $f\notin C^2(\mathbb{R})$. Hence $f$ is not smooth. </p>

Title: Is $y=|x^3|$ a smooth function?

Body:
Is this a smooth function? $y=|x^3|$

The graph of this function has no sharp cuts or corners, so I think it is a smooth function but someone told me that it's not.


问:
Is the union of a collection of uniformly continuous sets uniformly continuous?
Suppose $f$ is uniformly continuous on $X_1,X_2,...,X_n$.

Let $X=\cup_{i=1}^n X_i.$

Is $f$ necessarily uniformly continuous on $X$? Is $f$ even necessarily continuous on $X$?

Let $f$ be the constant function given by $f(x)=c$ $ $ where $c \in \mathbb{R}$

Let $X_i =[a_i,b_i)$

Then there is a $\delta>0$ such that for each $x_0 \in X_i$ and $|x-x_o|,$ $$|f(x)-f(x_o)|<\varepsilon$$

$$\Rightarrow |c-c|<\varepsilon$$

$$\Rightarrow0<\varepsilon$$

Now, consider $\cup_{n=1}^\infty X_i=[a_1,b_1)\cup[a_2,b_2)\cup...\cup[a_n,b_n)$

Can we say that there exists $\delta>0$ such that for each $x_0 \in X$ and $|x-x_o|,$ $$|f(x)-f(x_o)|<\varepsilon?$$

答:
If the sets $X_i$ are compact, you will be in good shape.  0XLR gave a counterexample in $\mathbb R$ where one of the sets was not closed.  Here is a counterexample where the sets are not bounded.

Let $$
X_1 = \{1,2,3,\dots\},
\\
X_2 = \left\{1+\frac{1}{1}, 2+\frac{1}{2}, 3+\frac{1}{3}, \dots\right\}
$$
Set $f(x) = 0$ on $X_1$ and $f(x) = 1$ on $X_2$.  Then $f$ is uniformly continuous on $X_1$ and on $X_2$, but not on $X_1 \cup X_2$.


问:
Finding the volume of the cap of a sphere

Let's say I have a sphere of radius 2 and we drill a hole with the
shape of a sphere of radius 1. How shall we find the volume of the cap
of the sphere?

What I did was finding the spherical coordinates for the surface:

$\Phi(\theta, \phi)=2(\cos\theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$ with $\theta \in [0, 2\pi]$

To find the boundaries for $\phi$, I used the Pythagorean theorem:

To find $h$: $(\frac{h}{2})^2=2^2+1^2$, so $h=\sqrt{20}$

So I end up with this triangle:

Then $\sin \phi = \frac{1}{2}$ so, $\phi = \frac{\pi}{6}$.

Setting up the integral:

$\int _{0} ^{\pi/6} \int _{0} ^{2\pi} \int _{0} ^{2} \rho^2 \sin \phi d\rho d\theta d\phi=-\frac{8}{3}(\sqrt{3}-2)\pi$

Which is wrong... I've done a lot of steps, so I can't really tell where I made a mistake

答:
We need to find what is the height of the spherical cap. By the Pythagorean theorem, half of the height of the cylinder is
$$2^2=1^2+\left(\frac{h}{2}\right)^2$$
so $\tfrac{h}{2}=\sqrt{3}$. Then, the height of the spherical cap is
$$H=2-\sqrt{3}.$$
Plugging this value into the formula (see https://en.wikipedia.org/wiki/Spherical_cap), we get
$$V = \frac{\pi H^2}{3}(3R-H)=\frac{\pi(2-\sqrt{3})^2}{3}(4+\sqrt3)=\frac{\pi}{3}(16-9\sqrt{3})$$


问:
Locally quadratic convergence
Show that the iteration method $$x_{n+1}=\frac{1}{2}x_n+\frac{a}{2x_n}, \ \ a>0$$ $n=1, 2, \ldots $ converges locally quadratic to $\sqrt{a}$.

Could you explain to me how we could show that?

$$$$

EDIT:

We have that $|x_{n+1}-\sqrt a|=\frac1{2x_n}|x_n-\sqrt a|^2$.

To get local convergence we want that the starting point is near the root. Then this should hold for all approximations. So it must hold that $\left |x_n-\sqrt{a}\right |<\epsilon$, with $\epsilon>0$. Then we get: $$\left |x_n-\sqrt{a}\right |<\epsilon  \Rightarrow \frac{1}{2(\sqrt{a}+\epsilon)}<\frac{1}{2x_n}<\frac{1}{2(\sqrt{a}-\epsilon)}$$ So we get $$|x_{n+1}-\sqrt a|  = \frac1{2x_n}|x_n-\sqrt a|^2 < \frac{1}{2(\sqrt{a}-\epsilon)}|x_n-\sqrt a|^2$$ If we choose for example $\epsilon=\frac{\sqrt{a}}{2}$ we get $$|x_{n+1}-\sqrt a|   < \frac{1}{\sqrt{a}}|x_n-\sqrt a|^2 \text{ with } \frac{1}{\sqrt{a}}>0$$
So the method converges locally and quadratic to $\sqrt{a}$.

Is that correct?

答:
One needs to assume $a\ne 0$, otherwise the convergence is linear ($\frac{x_{n+1}}{x_n}=\frac12$). From the iteration procedure
$$
x_{n+1}-\sqrt a = \frac1{2x_n}(x_n^2 + a - 2\sqrt a x_n) = \frac1{2x_n}(x_n-\sqrt a)^2
$$
This is quadratic convergence if $(x_k)$ converges to $\sqrt a$.

Now, if the iteration is started at $x_1$ close to $\sqrt a$, i.e.,
$x_1 \in (\sqrt a-\delta,\sqrt a+\delta)$, then
$$
\frac1{2x_1}|x_1-\sqrt a| \le \frac\delta{2(\sqrt a-\delta)}
$$
This factor is equal to $\frac12$  if $\delta = \frac12\sqrt a$. Per induction one can show that $x_k \in  (\sqrt a-\delta,\sqrt a+\delta)$ for all $k$ and $x_k \to \sqrt a$. This is local convergence.

This iteration is nothing else than Newtons method applied to the equation $x^2=a$.

Title: Locally quadratic convergence


答:
person 1 person 2

win      win

person 1 person 2

win      lose

person 1 person 2

lose     win

person 1 person 2

lose     lose

I see the answer being 75% for winning at least one match.

Title: how to calculate the real probability

Body: Say I have a 50% probability of winning a game if I play against person 1, and a 50% probability of winning a game against person 2. I will play with both people, one after another. Before the matches, what was the probability that I would win at least 1 match? It has to be something between 90-99% I think. Any suggestions? Thank you!</p>


问:
Line integral of differential form $\omega = 3dx_1+(2x_1+x_2^2)dx_2$ with $\gamma(t)=(|\cos(2t)|,|1-2\sin(t)|+1)$

$\omega = 3dx_1+(2x_1+x_2^2)dx_2$

$\gamma(t)=(|\cos(2t)|,|1-2\sin(t)|+1), t \in [0,\pi/3]$

Calculate $\int\limits_{\gamma}\omega$

So far I showed that $\omega$ is not exact with partial derivatives. And I know that $\int\limits_{\gamma}\omega=\int\limits_{0}^{\pi/3}(3\gamma_1'+(2|\cos(2t)|+(|1-2\sin(t)|+1)^2)\gamma_2')dt$

In class this week we went over line integrals and that kind of stuff, so I feel like I should use $\Gamma(\gamma(t))$ somehow. But we didn't see any examples of that in class, so I'm very confused about everything. Where should I start? Any help is appreciated!

答:
This is just a straight-forward exercise in applying a definition.

Note that $\cos(2t) \ge 0$ on $[0, \pi/4]$, and $\le 0$ on $[\pi/4, \pi/3]$, while $1 - 2\sin(t) \ge 0$ on $[0,\pi/6]$ and $\le 0$ on $[\pi/6,\pi/3]$.
So $$\gamma(t) = \begin{cases}(\cos 2t, 2 - 2\sin t),&t\in\left[0,\frac\pi 6\right)\\(\cos 2t, 2\sin t),&t\in\left[\frac\pi 6, \frac \pi 4\right)\\(-\cos 2t, 2\sin t),&t\in \left[\frac\pi 4, \frac \pi 3\right]\end{cases}$$

Break your integral into three parts:

$$\int_\gamma \omega = \int_0^{\pi/6}\omega(\gamma(t)) + \int_{\pi/6}^{\pi/4}\omega(\gamma(t))+ \int_{\pi/4}^{\pi/3}\omega(\gamma(t))$$

In the first part $\gamma'(t) = (-2\sin 2t, - 2\cos t)$, so the integral is
$$\int_0^{\pi/6}\omega(\gamma(t)) = \int_0^{\pi/6} 3(-2\sin 2t)\,dt + \int_0^{\pi/6}[2\cos 2t + (2-2\sin t)^2](-2\cos t)\,dt$$
Using a trig identity to write $\cos 2t$ in terms of $\sin t$, these become fairly simple to calculate.

I'll leave the other two parts to you.


问:
If $f:\mathbb R\to \mathbb N$ is continuous, then it's constant.
I have to prove that if $f:\mathbb R\to \mathbb N$ is continuous, then it's constant. Th result looks intuitive, but I can't prove it. I try by contradiction : suppose there is $x,y\in\mathbb R$ s.t. $f(x)\neq f(y)$. So in somehow $|f(x)-f(y)|\geq 1$, and thus, using the $\varepsilon -\delta $ definition with $\varepsilon <1$ I tried to get a contradiction, but since there is no reason to have $|x-y|\leq \delta $, I couldn't find a way to write it properly. Can someone give hint ?

答:
Method 1

Using your method : Suppose $x<y$ and $f(x)\neq f(y)$. Set $z=\inf\{t>x\mid f(t)\neq f(y)\}$. Using continuity of $f$ at $z$ should allow you to get a contradiction.

Method 2

$f^{-1}(\{f(0)\})$ is open and closed in $\mathbb R$.


问:
Suppose $H$ is the only subgroup of order $o(H)$ in the finite group $G$. Prove that $H$ is a normal subgroup of $G$.

Suppose $H$ is the only subgroup of order $o(H)$ in the finite group $G$. Prove that $H$ is a normal subgroup of $G$.

I've been trying this problem for quite a while but to no avail. What I can't understand is, how do you relate the subgroup being normal to its order?

This question is from I.N. Herstein's book Topics in Algebra, page 53, Problem no. 9. This is NOT a homework problem!! I'm studying this book on my own.

答: For every $g\in G$, the map $h\mapsto ghg^{-1}$ is a bijection from $H$ to $gHg^{-1}$. If $H$ is the only subgroup of order $o(H)$, then $\forall g\in G, gHg^{-1}=H$, i.e. $H\unlhd G$.


问:
Evaluating real integral using complex analysis.
I'm trying to compute the following integral:
$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx$$

I'll not write down everything I've done, but choosing the branch cut on the positive real axes we have that:

$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{\pm \sqrt{i},\pm\sqrt{-i}\}$$

So we have to compute four residues.
My thought was changing the branch cut by putting it on the negative imaginary axes. We can do it by choosing $arg(z) \in (-\frac{\pi}{2},\frac{3\pi}{2}]$. So we have that:

$$(1+i)\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=2\pi i \sum_{z_i}Res(f,z_i) \qquad z_i\in\{e^{i\frac{\pi}{4}},e^{i\frac{3\pi}{4}}\}$$

By doing this, we now need to compute only two residues. But I'm really finding difficulties in computing those residues: in fact I can't obtain the result I'm expecting. Can you please show me the computation and tell me if my argument was clear and correct?

Thanks in advance.

答:
Under $x^4\to x$,
$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\frac14\int_0^\infty\frac{1}{x^{5/8}(1+x)}dx. $$
Let
$$ f(z)=\frac{1}{z^{5/8}(1+z)}. $$
Let $C_r, C_R$ be circles at $0$ cut from $r$ to $R$, respectively, and $C_1, C_2$ be the top and bottom parts of the segment from $r$ to $R$. Then, for big $R>0$ and small $r>0$,
$$ \int_{C_R}f(z)dz+\int_{C_r^-}f(z)dz+\int_0^{R}f(x)dx-\int_0^{R}f(xe^{2\pi i})dx=2\pi i\text{Res}(f,z=-1). $$
Clearly
$$ \bigg|\int_{C_R}f(z)dz\bigg|\le\frac{1}{R^{5/8}(R-1)}2\pi R=\frac{2\pi R^{3/8}}{R-1}, \bigg|\int_{C_r^-}f(z)dz\bigg|\le\frac{1}{r^{5/8}(1-r)}2\pi r=\frac{2\pi r^{3/8}}{1-r} $$
and
$$ \int_0^{R}f(xe^{2\pi i})dx=e^{-5\pi i/4}\int_0^\infty f(x)dx, \text{Re}(f,z=-1)=e^{-5\pi i/8}. $$
So letting $R\to\infty, r\to 0^+$, one has
$$ (1+e^{-5\pi i/4})\int_0^\infty f(x)dx=2\pi i e^{-5\pi i/8} $$
or
$$ \int_0^\infty f(x)dx=\frac{2\pi i e^{-5\pi i/8}}{1+e^{-5\pi i/4}}=\frac{\pi}{\cos(\pi/8)}. $$
Thus
$$\int_0^{\infty}\frac{\sqrt{x}}{1+x^4}dx=\frac14\int_0^\infty\frac{1}{x^{5/8}(1+x)}dx=\frac{\pi}{4\cos(\pi/8)}. $$


问:
Closed form for $\int_0^1\log\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\mathrm dx$
Please help me to find a closed form for the following integral:
$$\int_0^1\log\left(\log\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)\right)\,{\mathrm d}x.$$

I was told it could be calculated in a closed form.

答:
$$\boxed{\displaystyle\int_0^1\log\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)\mathrm dx=-\gamma-2\ln\frac{2\Gamma(3/4)}{\Gamma(1/4)}}\tag{$\heartsuit$}$$

Derivation:

After the change of variables $x=\frac{1}{\cosh u}$ the integral becomes
$$\int_0^{\infty}\ln u \frac{\sinh u}{\cosh^2 u}du,$$
as was noticed above by Eric. We would like to integrate by parts to kill the logarithm but we get two divergent pieces. To go around this, let us consider another integral,
$$I(s)=\int_0^{\infty}u^s \frac{\sinh u}{\cosh^2 u}du,$$
with $s>0$. The integral we actually want to compute is equal to $I'(0)$, which will be later obtained in the limit.

Indeed, integrating once by parts one finds that
\begin{align}
I(s)&=s\int_0^{\infty}\frac{u^{s-1}du}{\cosh u}=s\cdot 2^{1-2 s}\Gamma(s)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]=\\ &=2^{1-2 s}\Gamma(s+1)\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right],
\end{align}
where $\zeta(s,a)=\sum_{n=0}^{\infty}(n+a)^{-s}$ denotes Hurwitz zeta function (in the way we have used its integral representaion (5) from here).

Now to get ($\heartsuit$), it suffices to use
\begin{align}
&\frac{\partial}{\partial s}\left[2^{1-2 s}\Gamma(s+1)\right]_{s=0}=-2\gamma-4\ln 2,\\
&\zeta\left(0,\frac14\right)-\zeta\left(0,\frac34\right)=\frac12,
\\
&\frac{\partial}{\partial s}\left[\zeta\left(s,\frac14\right)-\zeta\left(s,\frac34\right)\right]_{s=0}=-\ln\frac{\Gamma(\frac34)}{\Gamma(\frac14)}.
\end{align}
[See formulas (10) and (16) on the same page].


问:
Trivializing the vector bundle of $S^n$
It's know that in general we do not have that $TS^n\cong S^n \times \mathbb{R}^n$. But I saw the interesting fact that is that we will have $TS^n \times \mathbb{R}\cong S^n \times \mathbb{R}^{n+1}$, so we get kinda of a trivialization . Now I have been trying to see why this is true but I got nowhere, does anyone have any suggestions ? Thanks in advance.

答: Consider the standard embedding $\iota:S^n\to\mathbb{R}^{n+1}$, and in particular the tangent/normal splitting $T\mathbb{R}^{n+1}|_{S^n}\cong TS^n\oplus NS^n$. Can you use this to show your statement is true?

Title: Trivializing the vector bundle of $S^n$


问:
Showing that $y_2$ is a constant multiple of $y_1$ if Wronskian $W[y_1,y_2](t^*)=0$ and $y_1(t^*)=0$.
Specifically, I want to show that $$y_2(t) = [y_2 (t^*)/y_1'(t^*)]y_1(t)$$

Here's how far I got.

$W[y_1,y_2](t^*)=y_1'(t^*)y_2(t^*)= 0$. Now, we have two choices.

If $y_1'(t^*)=0$ then $y_1(t)=0$, because of a theorem.

If, on the other hand, $y_2(t^*)=0$, we can rewrite $y_2$ using the Wronskian. Since the Wronskian is either always zero or never, it is always zero since $W(t^*)=0$;

$$y_1(t)y'_2(t)-y_1'(t)y_2(t)=0$$ and
$$y_2(t)=[y_2'(t)/y_1'(t)]\ y(t).$$

Now, my question. Why is the answer $[y_2 (t^*)/y_1'(t^*)]y_1(t)$? That is, with $t^*$ filled in.

Context: this is question 2.1.19 from Braun's Differential Equations and their Applications.

答:
If $y'_1(t^*)=0$ as well, then $y_1\equiv 0$, because the IVP
$$
y''+p(x)y'+q(x)y=0, \quad y(t^*)=y'(t^*)=0
$$
possesses a unique solution, the zero one, and the conclusion holds.

If $y'_1(t^*)=a\ne0$ then
$$
0=w(y_1,y_2)(t^*)=y_1(t^*)y_2'(t^*)-y_1'(t^*)y_2(t^*)=-y_1'(t^*)y_2(t^*)
$$
and hence $y_2(t^*)=0$. But $y_2$ is the unique solution of the IVP
$$
y''+p(x)y'+q(x)y=0, \quad y(t^*)=0,\,\,y'(t^*)=y'_1(t^*)\ne 0, \tag{1}
$$
Whereas $y_2$ is the unique solution of the IVP
$$
y''+p(x)y'+q(x)y=0, \quad y(t^*)=0,\,\,y'(t^*)=y_2'(t^*). \tag{2}
$$
Observe now the if $y$ is a solution of $(1)$, then
$\big( y'_2(t^*)/y_1'(t^*)\big)y(t)$ is a solution of $(2)$.
Hence, uniqueness of solutions of both IVPs implies that
$$
y_2(t)=\big( y'_2(t^*)/y_1'(t^*)\big)y_1(t)
$$


问:
Weak convergence implies almost everywhere?
Suppose that $L^2(R^n;dx) \ni f_n\geq 0$ and $\int f_ng\to 0$ for all continuous function $g$ with compact  support. Then, does $f_n$ convergence to $0$ almost everywhere?

Intuitively, it convergence to $0$ by taking function $g$ s.t., $g=1$ around $x$ for almost every $x$. However I cannot prove.

答:
Another example,  taking place in $L^2[0, 1]$,  which is perhaps easier to understand from a geometric standpoint, is to choose the first 10 functions
in the sequence to be the charachteristic functions of  intervals of length 1/10, sliding from left to right, starting
with  $[0.0,0.1]$  and ending with $[0.9,1.0]$.
Then do the same for the next 100 functions with intervals from
$[0.00,0.01]$  to $[0.99,1.00]$, and so on.

This sequence does not converge pointwise anywhere, but still it converges weakly!


问:
Differentiation of $\sin x$ using Pythagoras theorem:
So, I just tried using Pythagoras theorem to prove:
$\frac{d}{dx}\sin x= \cos x$ but it got me nowhere.
Here is my work.
$$\sin x= p/h.$$; where p and h are lengths of perpendicular and hypotenuse in a right angled triangle.

Differentiating w.r.t $x$, we get
$$\frac{d}{dx}\sin x= \frac{1}{h^2} \left( h\frac{dp}{dx}-p\frac{dh}{dx}\right)\tag{i}$$

We know $h^2=p^2+b^2$. (Where b is the length of the base of the right angled triangle.) Differentiating w.r.t $x$,
$$h\frac{dh}{dx}=p\frac{dp}{dx}$$
(If we put $b$ constant). Substituting in $(i)$,

\begin{align}
\frac{d}{dx} \sin x&= \frac{1}{h^2} \left(h\frac{dp}{dx}-\frac{p^2}{h}\frac{ dp}{dx}\right)\\
       & = \frac{1}{ph^2} \left(b^2\frac{dp}{dx}\right) \tag{ii}
\end{align}
Now, I lost track here because I didn't know how is that equal to $\cos x$. Can anyone help me continue from here?

(Note: I know how to derive this from first principle. I just wanted to try using Pythagoras theorem and trigonometric identities. I kept the base length constant because it appeared to me that if I changed the x, base length won't be effected. But as I can see in the answer, the hypotenuse remains constant. However, can we keep the base constant and end up getting the same result? Maybe not visualising a circle of radius h but just a right angled triangle whose base remains constant w.r.t the reference angle?

Ps: The called duplicate one is out of my understanding. I request a simpler version.

答:
So, the way to think about this is to recognize three important facts:

On a unit circle, the hypotenuse will always be 1

On a unit circle, the distance around the circle is the same as the angle in radians

The change angle x, as the distance on the unit circle approaches zero, closer and closer approximates a LINEAR change

So, we will use $a$ for the adjacent, $p$ for the opposite, and $x$ for the angle in radians.

The definition of sine is $\frac{\text{opposite}}{\text{hypotenuse}}$ and the definition of cosine is $\frac{\text{adjacent}}{\text{hypotenuse}}$.  Since the hypotenuse is $1$, this means that $\sin(x) = p$ and $\cos(x) = a$.  The derivative of $\sin(x)$ with respect to $x$, therefore, is $\frac{dp}{dx}$.  If we can show that $\frac{dp}{dx} = a$, then we win.

Okay, the pythagorean theorem (plus #1 above) gives us:
$$ a^2 + p^2 = 1 \\ p^2 = 1 - a^2 $$

Differentiating yields:
$$2p\,dp = -2a\,da \\ dp = -\frac{a}{p}da \\ da = -\frac{p}{a}dp$$

Now, because of #2 and #3, we can use the distance formula to say:
$$dx = \sqrt{dp^2 + da^2} \\
dx = \sqrt{dp^2 + (-1)^2\frac{p^2}{a^2} dp^2} \\
dx = \sqrt{dp^2 + \frac{1 - a^2}{a^2} dp^2} \\
dx = \sqrt{dp^2 + \frac{dp^2}{a^2} - dp^2} \\
dx = \sqrt{\frac{dp^2}{a^2}} \\
dx = \frac{dp}{a}$$

So, this means that we can transform $\frac{dp}{dx}$ like this:
$$ \frac{dp}{dx} = \frac{dp}{\frac{dp}{a}} = \frac{dp}{1}\frac{a}{dp} = a$$

QED


问:
Why is $\sum_{i=0}^{n - 1}2^i = 2 ^ n - 1$?
Probably another simple question and possibly not asked too well. I'd like to know why this is true (I'm looking for a simple, possibly combinatorial explanation with a real-world example):

$$\sum_{i=0}^{n - 1}2^i = 2 ^ n - 1$$

To add more detail, I don't get why $2^{16}$ is the same as $2^{15}+2^{14}+2^{13}...+2^{0} + 1$.

Body:
Generally,
$$
\sum_{k=0}^{n-1} x^k = \frac{x^{n}-1}{x-1} \tag{*}\label{*}
$$
Taking $x=2$ gives the sum your ask about.

Proof of \ref{*}:

Let $s_n(x) = \sum_{k=0}^{n-1} x^k.$ Then
$$
xs_n(x)
= x\sum_{k=0}^{n-1} x^k
= \sum_{k=0}^{n-1} x^{k+1}
= \sum_{k=1}^{n} x^{k}
= \left( \sum_{k=0}^{n-1} x^{k} \right) - x^0 + x^n
= s_n(x) - 1+x^n.
$$
Thus, $xs_n(x)-s_n(x) = x^n-1,$ i.e.
$$
s_n(x) = \frac{x^n-1}{x-1}.
$$


问:
Convergence of $\left(\ln\left(1+ \frac{1} {x}\right)\right) ^n $
Consider the following sequence of functions defined by $$ f_n(x)= \left[\ln\left(1+ \frac{1} {x}\right)\right] ^n $$ with two different domains, first for $x\in (0,1]$ and then $x\in [\frac{1} {2} ,1]$
What can I say about its convergence?

I've found it similar to the geometric sequence, but I'm not very sure.

I find the problem a bit puzzling.
I don't see a difference between the two intervals (in this type of problems the sequence behaves differently depending on the compactness/openness of the interval).
Please check it and feel free to give any comment.

答:
My attempt

$$ - 1 \lt \ln\left(1+ \frac{1} {x}\right) \le 1 $$
for
$$ 1+ \frac{1} {x} \in \left( \frac{1} {e}, e \right] $$ which happens when $$ x\ge  \frac{1} {e-1},
x\lt \frac{e} {1-e}$$

Restricting to the values in the domains of the problem, we have $\frac{1} {e-1} \gt \frac{1} {2}$, so I can say that the sequence converges pointwise but not uniformly for $ x \in [\frac{1} {e-1}, 1]$. For the other values $x \in (0, \frac{1} {e-1}) $ the sequence does not converge.


问:
Closed graph theorem and continuity
Let $V,W$ be banach spaces and $T:V\to W$ be linear map.

The closed graph theorem says $T$ is continuous iff the graph of $T$ is closed .

Graph of $T$ is closed is same as saying, if ( $x_n\to x$ and $Tx_n\to y$ ), Then $Tx=y$.

It's very similar to continuity and difference is here we say if $Tx_n\to y$ (in other words, we can assume the convergence for granted) , but in continuity we have to show $Tx_n$ converges (and to $Tx$ ).

I want to look at the counter example where one doesn't implies other. I know in hausdroff space continuity implies closed graph. But I couldn't find an example in which closed graph doesn't implies continuity.  Please help.

Also if my understanding of closed graph theorem is wrong,then please correct me.

答:
In order to find a counter-example, one has to give up completeness.

Let $V=W=c_{00}$ be the space of real-valued sequences with at most finitely many non-zero entries. Supplied with $\sup$-norm. This is a normed space but not complete.

Define $T$ by
$$
Tx = (x_1, 2x_2, \dots, nx_n, \dots),
$$
which is a linear mapping from $c_{00}$ to $c_{00}$. It is not continuous, since it is unbounded. However its graph is closed: $x_n \to x$ and $Tx_n \to y$ imply
$x_{n,k}\to x_k$ for all $k$, as well as $kx_{n,k} \to y_k$, hence $kx_k=y_k$ for all $k$, and $Tx=y$.

(We could have chosen $W=l^\infty$ as well. Then the example still works.)

Title: Closed graph theorem and continuity


问:
Prove that $\lim_{n \rightarrow \infty } \int_{a}^{b}f_{n}=0$
Suppose
$f_{n}:[a,b]\rightarrow \mathbb{R}$ are continuous,  $f_{n}\rightarrow 0$ pointwise and $\exists C: \int_{a}^{b}f^{2}_{n} <C$ $\forall n$. I need to prove that $lim_{n \rightarrow \infty } \int_{a}^{b}f_{n}=0$.

My attemp. Using Cauchy-Schwarz inequality: $$\int_{a}^{b}|f_{n}| \leq \int_{a}^{b}1 \int_{a}^{b} f_n^{2}=(b-a)^{1/2}\int_{a}^{b}f_{n}^{2}.$$ So, it's enough to prove that $\lim_{n \rightarrow \infty}\int_{a}^{b}f_{n}^{2}=0$.

Let's denote $g_{n}:=f_{n}^{2}$, $M_{n}:=\max_{x\in [a,b]}(g_{n}(x))$. I want to show that $\{M_{n}\}$ is bounded (to use  Lebesgue's dominated convergence theorem). Suppose $\forall k \exists N: n>N \Rightarrow |M_{n}|>k.$  $g_{n}(x_{n})=M_{n}$, $g_{n}(x)$ is continious $\Rightarrow (\exists U_{n}: x \in U_{n} \Rightarrow g_{n}(x)>k$). So, $k\mu(U_{n})<\int_{U}g_{n}<C$. There should be some contradiction...

Thank you for any help.

答:
Following @gerw 's hint of Egorov's theorem:

Let $\epsilon>0$. Then there exists a measurable set $A\subset [a,b]$ such that $m(A)<\epsilon$ and $f_n$ converges to $f\equiv 0$ uniformly on $B:=[a,b]\backslash{A}$.

On $B$, by definition of uniform convergence, there exists $N$ such that $|f_n(x)-f(x)|=|f_n(x)|<\epsilon$ for all $x \in B$ and all $n\geq N$. Hence,
$$
\left|\int_B f_n \right|\leq \int_B |f_n|\leq \epsilon\, m(B)\leq \epsilon (b-a).
$$
On $A$, we have
\begin{align}
\left|\int_A f_n \right|\leq \int_A |f_n|\leq \sqrt{m(A)}\sqrt{\int_A f_n^2}\leq \sqrt{\epsilon}\sqrt{C}.
\end{align}


答: If $2^n$ people enter an elimination tournament of $2$-person games to determine an overall winner, the left-hand side of your equation counts the people eliminated in each round, while the right-hand side counts the people who lose. This might be the "combinatorial proof by subset counting" @PrasunBiswas referred to.

Title: Why is $\sum_{i=0}^{n - 1}2^i = 2 ^ n - 1$?

Body:
Probably another simple question and possibly not asked too well. I'd like to know why this is true (I'm looking for a simple, possibly combinatorial explanation with a real-world example):

$$\sum_{i=0}^{n - 1}2^i = 2 ^ n - 1$$

To add more detail, I don't get why $2^{16}$ is the same as $2^{15}+2^{14}+2^{13}...+2^{0} + 1$.


问:
Is it even possible to rigorously define sine and cosine using the "traditional" definition?
Most rigorous definitions of the sine and cosine function use power series or some other analytic definition. In geometry class, we learn about the geometric, right-triangle definition of sine and cosine. It is hard to formalize the geometric definition in a rigorous, non-circular (pardon the pun) way. But has anyone done it? Is it even possible to do so? I would be interested in a text or paper that defines the trigonometric functions in a geometric way, but rigorously.

答: Yes. The integral expressing arc length on a unit circle from the point $(1,0)$ up to $(x,\sqrt{1-x^2})$ is arc-sine, etc.


问:
Change the order of integration in Spherical coordinate and Cylindrical Coordiate
Let D be the region bounded below by the plane z = 0, above by the sphere $x^2 +y^2 +z^2 =4$, and on the sides by the cylinder $x^2 + y^2 = 1$. Set up the triple integrals in cylindrical coordinates that enter preformatted text heregive the volume of D using the following orders of integration.

a. $dzdrdu$ b. $drdzdu$ c. $dudzdr$

Let D be the region in Exercise 33. Set up the triple integrals in spherical coordinates that give the volume of D using the follow- ing orders of integration.

a. $d\rho d\phi d\theta $ b. $d\phi d\rho d\theta$

And I'm also wondering if it is correct that we can switch the order of integration directly if the bounds are all constant and independent on each other

答:
There is a mistake in your cylindrical coordinates working and your spherical coordinate is wrong. Bounds of $r$ is from $0$ to $1$ and not from $0$ to $2$. None of the discs that you take will have radius more than $1$. Please also note the upper bound of $z = \sqrt{4-r^2} = 2 \,$ when $r = 0$ and $z = \sqrt{3} \,$ when $r = 1$.

So, $V = \displaystyle \int_{0}^{2\pi} \int_{0}^1 \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$

Now in spherical coordinates, for simplicity, your integral will be two parts - one which is spherical cone between $0 \leq \phi \leq \frac{\pi}{6}$ and then rest of the part of the cylinder which is between $\frac{\pi}{6} \leq \phi \leq \frac{\pi}{2}$.

$V = \displaystyle \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{0}^{2} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \, \, + \int_{0}^{2\pi} \int_{\pi/6}^{\pi/2} \int_{0}^{csc (\phi)} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

You should check your sketch to understand why the second integral goes from $0$ to $\csc \phi$ (if you draw a line from the center (origin in this case) to any point on the cylinder, it is hypotenuse of a right angled triangle with angle $\phi$ to the $z$ axis and distance along $xy$ axis being $1$. So $\rho$ is $1/\sin \phi$.)

Now for the change of order of integral in cylindrical coordinates, please note $(c)$ is pretty straightforward. Now for (b), you need to use your sketch to think how to go about it - you may have to split it into two parts ($0 \leq r \leq 1, 1 \leq r \leq \sqrt{4-z^2}$). Think similarly for the other integral in spherical coordinates - how to represent $\phi$ in in terms of $\rho$.


问:
Complement of halting set is not r.e.
suppose we don't know that Halting problem is not recursive.

I want to prove that complement of halting set is not r.e. then we can find halting problem is not recursive.

Can you direct prove that complement of halting set is not r.e.??

答:
HINT: If a set and its complement are both recursively enumerable, the set is ... ?

Added: This hint was for the original version of the question, which assumed that the halting problem was undecidable and did not ask for a direct proof.

Title: Complement of halting set is not r.e.


问:
$max - \frac{1}{x} - \frac{1}{y}$ s.t. $2x + y ≤ 10, x ≥ 0, y ≥ 0$
$\max    - \frac{1}{x} - \frac{1}{y}$

s.t. $2x + y ≤ 10,\quad  x ≥ 0,\quad y ≥ 0$

I set up the lagrangian and take FOC.

$ \frac{\frac{1}{x^2}}{\frac{1}{y^2}}$=$\frac{2}{1}$

$y=\sqrt2$ x
Substitute in λ partial derivative
and we get $x* = 2.92$ and $y* = 4.14$?

I get $y^* =4.14$ and $x^*=2.92$. Am I correct? Why don't I need to add lagrangrian for the other two constraints as below?

$L= - \frac{1}{x} - \frac{1}{y}$ + $λ_1(10-2x-y)+λ_2x+λ_3y$?

答:
You do not need to add the positivity constraints because here those constraints are unbinding i.e. the candidates for your optimum values are positive $\Rightarrow$ your constraint is unbinding $\Rightarrow$ $\lambda_{2}$ & $\lambda_{3}$ are zero. Therefore, they are not added.

In general case when we use Lagrange Multiplier method and derive Kuhn Tucker Conditions then we test our candidates for optimum values on 2 conditions related to the positivity constraint $x\leq0$ and $\lambda_{2}x=0$. For further information on this topic, I would suggest you to read Chapter 18 of Simon & Blume "Mathematics for Economists" & Chapter on Optimization from "Intermediate Micro" by Thomas Nechyba.


问:
Upper function/bound for $y'(x)=x^3+y^3(x)$
The function is $y'(x)=x^3+y^3(x)$ for $x>0$ and $y(0)=1$

I was asked to find an upper & lower function so that $y_1'(x)<y'(x)<y_2'(x)$

I've found a lower bound for $y'(x)$ by seeing that $y_1'(x)=y^3(x)<x^3+y^3(x)=y'(x)$ with $y_1(0)=1$

With that, I got $y_1(x)=\frac{1}{\sqrt{1-2x}}$ as solution and now I can deduce that $0<x<\frac{1}{2}$

But not I'm having problems constructing the upper function for that problem.

My initial guess was $y_2(x)=1+y^3(x)$ but that proofed hard to solve. Can someone give me a hint for a better estimate?

答:
Let $I = [0, a)$ denote the maximal interval on which a solution of the initial value problem exists.

You already showed that $a \le 1/2$. It follows that $2x < 1 \le y(x)$ for all $x \in I$, and therefore
$$
 y'(x) < \left( \frac{y(x)}{2} \right)^3 + y(x)^3 = \frac 98 y(x)^3 \, .
$$

That leads to
$$
 y(x) < y_2(x) = \frac{2}{\sqrt{4-9x}} \, .
$$
In particular, $a \ge \frac 4 9 \approx 0.444$.

For a better estimate you can use your lower bound
$$
y(x) > \frac{1}{\sqrt{1-2x}}
$$
and verify that
$$
 \frac{1}{\sqrt{1-2x}} > 5x
$$
for $0 \le x <1/2$. It follows that $x < \frac 15 y(x)$, and therefore
$$
 y'(x) < \left( \frac{y(x)}{5} \right)^3 + y(x)^3 = \frac {126}{125} y(x)^3 \, .
$$
That leads to the upper bound
$$
y(x) < \frac{{{5}^{3/2}}}{\sqrt{125-252x}}
$$
and $a \ge \frac{125}{252} \approx 0.496$.


答:
It is clear that the sequence $(f_n(x))_{n\in\Bbb{N}}$ converges if and only if
$$-1<\ln\big(1+\tfrac1x\big)\leq1.$$
Of course given that $x>0$ we have $\ln(1+\tfrac1x)>0$, and $\ln(1+\tfrac1x)\leq1$ if and only if $x\geq\tfrac{1}{e-1}$.

Title: Convergence of $\left(\ln\left(1+ \frac{1} {x}\right)\right) ^n $

Body:
Consider the following sequence of functions defined by $$ f_n(x)= \left[\ln\left(1+ \frac{1} {x}\right)\right] ^n $$ with two different domains, first for $x\in (0,1]$ and then $x\in [\frac{1} {2} ,1]$
What can I say about its convergence?

I've found it similar to the geometric sequence, but I'm not very sure.

I find the problem a bit puzzling.
I don't see a difference between the two intervals (in this type of problems the sequence behaves differently depending on the compactness/openness of the interval).
Please check it and feel free to give any comment.


问:
Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$
Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$

a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$

$a_n =  \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$

$b_n =  \frac{1}{2\pi}\int_0^r r\sin(nr)\,dr = \frac{(-1)^n + 1}{2n}$

$f(x) = \frac{\pi}{8} - \sum_n [\frac{((-1)^n - 1)\cos(nx)}{2\pi n^2} + \frac{((-1)^n + 1)\sin(nx)}{2n}]$

The prof asked us to use this Fourier series to prove that $\pi^2/8 = 1+1/3^2+1/5^2+1/7^2+\cdots$. How do I do this?

答:
This would be one way of going about it but using a different fourier series:
$$f(x)\ = 1+\sum_{n=1}^{\infty}\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}\cos\left(\frac{\pi nx}{2}\right)$$
Notice how $$\frac{4\left(\left(-1\right)^{n}-1\right)}{n^{2}\pi^{2}}$$ is 0 when n is even, so we can just  use odd numbers instead with $(2n+1)$.
So we can make a new formula with only odds and we can do this by letting $n =(2n+1)$, and because we don't want the even results we can resolve the numerator to be equal to $2$.
$$f(x)\ =\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}\cos\left(n\pi x\ +\frac{\pi x}{2}\right)$$
Then setting $x = 0$ gives:
$$0=\ 1\ -\ \frac{8}{\pi^{2}}\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$
After rearranging:
$$\frac{\pi^{2}}{8}=\sum_{n=0}^{\infty}\frac{1}{\left(2n+1\right)^{2}}$$


问:
Why for some $r_{i} \in R$ and not for some $r_{i} \in R[X]$?
Here is the question I am reading the answer of @Xam to it, but I am wondering why

Prove that if $R$ is an integral domain and has ACCP, then $R[X]$ has ACCP

1-I am wondering in his answer in the second paragraph, specifically when he said "As $P_{n+i+1}\mid P_{n+i}$ it follows that $P_{n+i}=r_iP_{n+i+1}$ for some $r_i\in R$." why he said for some $r_{i} \in R$ and not for some $r_{i} \in R[X],$are not we speaking about divisibility of 2 polynomials? could anyone explain that to me please?

2-Also, I did not get the relation between the two leading coefficients in the paragraph following it. why they should be related? the two polynomials could have the same degree but the leading coefficients no one of them is a multiple of the other. could anyone explains this also to me?

3-My last question, why we are adding $n$ to $k,$ why we need to do that? can not $k$ be inside $n$?

答:
He had concluded before that $\deg P_n = \deg P_n+i$ for all $i\in \mathbb{N}$. Now as $P_{n+i+1}∣P_{n+i}$ it follows that $P_{n+i}=r_iP_{n+i+1}$ for some $r_i\in R[X]$. But then $$\deg P_{n+i} = \deg P_{n+i+1}=\deg r_i + \deg P_{n+i+1}$$ (here we used that $R$ is an integral domain). This leads to the conclusion that $r_i$ is constant and hence $r_i\in R$.

As for the relation of the leading coefficients: If $P_{n+i}=r_iP_{n+i+1}$ and the corresponding leading coefficients are $a_{n+i}$ (for $P_{n+i}$) and $a_{n+i+1}$ (for $P_{n+i+1}$), then, by the definition of polynomial multiplication, we must have $a_{n+i}= r_ia_{n+i+1}$.

At $n$, the degree becomes stationary allowing for the argument above. Then, maybe at a later time, the chain on $R$ becomes stationary. The rest of the argument needs that both are stationary from some index on. The chain of ideals generated by leading coefficients could become stationary earlier while stile having a varying degree (think of multiplication with monic polynomials).


问:
Finding the $\sigma$-algebra $\mathbf{X}$ = $\sigma$(S) generated by the collection of subsets S in these two questions.
a) $X = \mathbb{R}$ , $S = \{(-\infty, n): n \in \mathbb{Z} \} $     (i.e. all intervals in the form $(-\infty, n)$).

b) $X = \mathbb{R}^2$ , $S = \{\{(x, y)\} : (x, y) \in \mathbb{R}^2\} $     (i.e all singleton sets in $\mathbb{R}^2$)

for part (a) I feel like the answer is the Borel Algebra on $\mathbb{Z}$ ? but I don't know how to show it.
and for part (b) I amn't sure where to start.

答:
We use the term "countable" to means finite or countably infinite.

a) Let  $S = \{(-\infty, n): n \in \mathbb{Z} \} $ and let
$S_f = \{[m, n): m,n \in \mathbb{Z} \textrm{ and } m\leqslant n \} $

Note that $\emptyset \in S_f$, since $\emptyset =[n,n)$.

Now let $\Sigma= \{\bigcup_{i\in \mathbb{N}} A_i: \textrm{for all }  i \in \mathbb{N}, A_i \in S \cup S_f \}$.

It is easy to see that $S_f \subseteq \sigma(S)$, and so,   $S \cup S_f \subseteq \sigma(S)$. So, countable unions of sets in  $S \cup S_f$ are also in $\sigma(S)$. It means $\Sigma \subseteq \sigma(S)$.

On the other hand, it is easy to prove that $\Sigma$ is a $\sigma$-algebra and, since $S\subseteq \Sigma$, we have that $\sigma(S) \subseteq \Sigma$.

So we have

$$\sigma(S)= \Sigma=\left \{\bigcup_{i\in \mathbb{N}} A_i: \textrm{for all }  i \in \mathbb{N}, A_i \in S \cup S_f \right \}$$

Remark: Since the set in $S$ are countable unions of sets in $S_f$, we can simplify the expression of $\sigma(S)$:

$$\sigma(S)= \Sigma=\left \{\bigcup_{i\in \mathbb{N}} A_i: \textrm{for all }  i \in \mathbb{N}, A_i \in S_f \right \}$$

b) Let  $S = \{\{(x, y)\} : (x, y) \in \mathbb{R}^2\} $. Let $\Sigma=\{A \subseteq \mathbb{R}^2 : A \textrm{ is countable or }  \mathbb{R}^2 \setminus A \textrm{ is countable} \}$.

It is easy to see that $\Sigma \subseteq \sigma(S)$.

On the other hand, it is easy to prove that $\Sigma$ is a a $\sigma$-algebra and all singletons are in  $\Sigma$. So  $\sigma(S) \subseteq \Sigma$. So we conclude that
$$\sigma(S) = \Sigma=\{A \subseteq \mathbb{R}^2 : A \textrm{ is countable or }  \mathbb{R}^2 \setminus A \textrm{ is countable} \}$$


问:
For a complex measure m, why is m(B = E - A) = m(E) - m(A)?
So, this should be a little obvious, but not to me at the moment.

Why is $m(B = E - A) = m(E) - m(A)$?

Of course, all given sets are measurable.

Thanks.

答:
This is false. For example, let $m$ be counting measure on (the power set of) $\{1,2\}$, let $B=E=\{1\}$ and let $A=\{2\}$.

You can add extra hypotheses which make the conclusion true. For example, if $A\subseteq E$ and $m(A)<\infty$ then $m(E-A)=m(E)-m(A)$. Indeed, since $A\subseteq E$ the (measurable) sets $E-A$ and $A$ are disjoint and have union $E$, so $m(E-A)+m(A)=m(E)$, and since $m(A)<\infty$ we can subtract it to get $m(E-A)=m(E)-m(A)$.


问:
To construct a set with a limit point.
I learned how to construct a Cantor Set, and I am asked to do the following.

"Construct a bounded set with exactly 3 limit points."

Since the Cantor set contains infinitely many points, I don't think something like it will not work.
But this is the only thing that I have learned from the book that tells me anything about constructing a set that has a limit point.

I am also considering the interval $[0,1]$ and constructing a set so that the limit points are $\{0,1/2,1\}$.

If possible, I would like to see more than one simple examples because I am new to analysis and I have no teacher. It's very tough.

答:
"Construct a bounded set with exactly 3 limit points."

consider this set:
$$\{0.1,\space 0.2,\space0.3,\space\space0.11,\space0.22,\space0.33,\space\space0.111,\space0.222,\space0.333,\space\space0.1111,\space0.2222,\space0.3333,...\}$$

Title: To construct a set with a limit point.