问:
The boundary point of every open set of a metric space is a limit point?
Consider a metric space $X$ and its open subset $A$. $\newcommand{\ball}[2]{B_{#1}(#2)} \newcommand{\eps}{\epsilon}$

Proposition: $\forall x \in X: x \in \partial A \Rightarrow x \in L(A)$ where $L(A)$ is the set of limit points of $A$

My proof: Let $x$ be a boundary point of $A$. Then, $x \in \overline{A}\setminus A^\circ$, i.e., $x \in \overline{A}$ and $x \not\in A^\circ = A$ because $A$ is open. For all $\eps > 0$, there is $y \in A$ such that $y \in \ball{\eps}{x}$. From the fact that $x \not\in A$, we have that for all $\eps >0$, there is $y \in A$ such that $y \in \ball{\eps}{x}$ different from $x$.

Please check if my proof is correct.

答:
Your proof is correct, since your last line is exactly the definition of limit point.

Just as a curiosity, note that every limit point in a metric space (first countable would be enough) can be reached by a convergent sequence. This property seems trivial, but may fail in general topological spaces where the topology is not induced by a metric. Then, more sophisticated tools like nets or ultrafilters are required to characterize the closure of a set.


问:
Alternative method for finding the particular solution when using reduction of order
I'm attempting to solve a Cauchy-Euler form ODE by reduction of order. I've found an alternative solution online that is correct but the exact methodology being performed isn't outlined.

Reduce the order of the following ODE and solve,
$2x^2z'''+xz''-3z'=2/x^3, x>0$

Solution:

Place the equation in Cauchy-Euler form by multiplying by $x$,

$2x^3z'''+x^2z''-3xz'=2/x^2$

We reduce the above equation by making the substitution $x=e^t$ and using differential operators to obtain,

$[2D(D-1)(D-2)+D(D-1)-3D]z=2e^{-2t}$
$[2D^3-5D^2]z=2e^{-2t}$

The homogeneous solution is found by using the roots of the auxiliary equation,

$2D^3-5D^2=0 \rightarrow2r^3-5r^2=0\rightarrow r=0,0,5/2$

$z_h=C_1e^{0t}+C_2xe^{0t}+C_3e^{5/2t}$

Now the next step would be to find the particular solution using typical method undetermined coefficients by comparing the standard linear operator $L[z]=Cx^me^{rx}$ to $g(x)=2e^{-2t}$ and finding $y_p=x^s[A_m+...A_1+A_0]e^{rx}$. From there, we'd find successive directives of $y_p$ to substitute into the original Cauchy-Euler equation to determine the coefficients of the particular solution.

However, a correct answer I found online uses the method below instead:

Consider $[2D^3-5D^2]z=2e^{-2t}$ and solve for $z_p$ as $z_p=(2e^{-2t})/[2D^3-5D^2]$. Now substitute $D=-2$ to obtain,

$z_p=(2e^{-2t})/[2(-2)^3-5(-2)^2]=-e^{-2t}/18$

Substitute $t=ln(x)$ to obtain the final particular solution of,

$z_p=-x^2/18$.

I've seen this method of solving for the particular solution used in other questions but there is never any explanation of what exact method and logic is being used. My textbook has no information either. What is this method called? Why can we simply rearrange for $z_p$ using the differential operator equation? Where does the substitution $D=-2$ come from?

答:
$$2x^2z'''+xz''-3z'=\dfrac 2 {x^3}, x>0$$
Note that you can also rewrite the DE as:
$$2x^2z'''\color {blue}{+4xz''-3xz''}-3z'=\dfrac 2 {x^3}$$
$$(2x^2z''-3xz')'=\dfrac 2 {x^3}$$
$$(2x(xz')'-5xz')'=\dfrac 2 {x^3}$$
You can substitute $u=xz'$. And integrate.

I suppose that $a \ne b$ for simplicity:
$$y_p=\dfrac 1 {D-a}e^{bx}$$
$$(D-a)y_p=e^{bx}$$
$$y'_p-ay_p=e^{bx}$$
We seek an integrating factor. And we find that it is $\mu (x)= e^{-ax}$ mulitply both side by the integrating factor then we have:
$$y'_pe^{-ax}-ae^{-ax}y_p=e^{bx}e^{-ax}$$
More simply:
$$(y_pe^{-ax})'=e^{(b-a)x}$$
Now integrate both sides. We dont need the constant of integration:
$$y_pe^{-ax}=\dfrac {e^{(b-a)x}}{b-a}$$
$$y_p=\dfrac {e^{bx}}{b-a}$$
We just put $D=b$ in the first line. This explains the trick.
$$y_p=\dfrac {e^{bx}}{\color {red}{D-a}}=\dfrac {e^{bx}}{\color {red}{b-a}}$$

ContentLicense: CC BY-SA 4.0


问:
Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$
Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!

Body:
Denote $t:=\sqrt{45x^2-30x+1}$. Then we observe that
$$t^2=45x^2-30x+1=-5(7+6x-9x^2)+36.$$
As a result, it follows that $$t=-\frac{t^2-36}{5}\implies t^2+5t-36=0\implies(t+9)(t-4)=0.$$
Since $t\geq 0$, it follows that $t=4$. Therefore, $$45x^2-30x+1=16\implies45x^2-30x-15=0\implies 3x^2-2x-1=0.$$
Hence $(3x+1)(x-1)=0$. Either $x=1$ or $x=-1/3$.


Title: Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$

Body:
Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!

Body:
$$45x^2-30x+1=(-9x^2+6x+7)^2$$
$$-5(-9x^2+6x+7)+36=(-9x^2+6x+7)^2$$
Substitute $u=-9x^2+6x+7$ and solve the quadratic equation.


问:
Sum of the roots of unity, $z_{1}^p+...+z_{n}^p$
Let $z_1,...,z_n$ be the $n$ roots of unity. I am not able to find a value for the sum: $$z_{1}^p+...+z_{n}^p,\ p \in \Bbb N$$
I know that this sum can also be written as $$\sum_{k=0}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}=1+\sum_{k=1}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}$$
I have been trying with some examples and if I am correct this sum should be equal to $0$, so I think that $1+\sum_{k=1}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}$ should converge to $-1$. That is the furthest I have reached, but I don't know which is the formal mathematical procedure here.

答:
This isn't a formal answer  but it's good foundation for intuition.

First of all, if $Z=\{\zeta\}$ is the set of $n$th roots of unity, the sum $S=\sum\zeta$ of these $n$th roots must be $0$ "by symmetry." That is, since the set $Z$ is symmetric (rotating by $1/n$th of a full rotation doesn't change the set), we can expect the sum to also be symmetric, but the only point that doesn't change under rotation is the origin itself. In other words, the sum ought'nt be biased in any particular direction. More algebraically, if we pick a particular root of unity $\xi\ne1$, multiplying-by-$\xi$ is a bijection from $Z$ to itself, so $\xi S$ is the same sum as $S$ but with the summands cyclically reordered. Writing $S=\xi S$, we can divide $(1-\xi)S=0$ to get $S=0$.

Define $f(x)=x^p$. If we consider $Z^p:=\{f(\zeta)\mid \zeta\in Z\}$ (that is, $z_1^p,\cdots,z_n^p$) as a multiset (that is, a set where elements can be members multiple times, i.e. "with multiplicity") then we can say $Z$ is the $m$th roots of unity, where $m=n/g$ with $g=\gcd(n,p)$, but each $m$th root has $g$ copies in $Z^p$. This is a fact from group theory: the function $f$ is a surjective homomorphism of with kernel the $g$th roots of unity, so we can apply the first isomorphism theorem and Lagrange's theorem.

Thus, the sum of the elements of $Z^p$ counted with multiplicity is actually $g\sum \eta$, where $\eta$ ranges over the set $W$ of $m$th roots of unity. As long as $m\ne 1$ (i.e. as long as $p$ is not a multiple of $n$), the sum $\sum\eta=0$, so the sum over elements of $Z^p$ is also $0$. That is, $z_1^p+\cdots+z_n^p=0$ unless $n\mid p$ in which case the sum is $n$.


Body:
From $45x^2-30x+1=(7+6x-9x^2)^2$, you have that $$81x^{4}-108x^{3}-135x^{2}+114x+48 = 0$$

This factors as $$3(x-1)(3x+1)(9x^2-6x-16) = 0$$

So the roots of that polynomial are $x = 1, -\frac{1}{3}, \frac{1\pm\sqrt{17}}{3}$. Out of these, the only ones in the range $\left[\frac{1-2\sqrt{2}}{3} ,\frac{1+2\sqrt{2}}{3} \right]$ are $x = 1$ and $x = -\frac{1}{3}$.


问:
How can I show this field extension equality?
How can I show this field extension equality   $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?</p>

答: Note that $(\sqrt 2+\sqrt 3)^2=5+2\sqrt 6$ and $(\sqrt 2+\sqrt 3)\sqrt 6=3\sqrt 2+2\sqrt 3$, which you can combine linearly with $\sqrt 2+\sqrt 3$ to obtain $\sqrt 2$ and $\sqrt 3$.</p>

ContentLicense: CC BY-SA 3.0


问:
Radius of convergence for matrix exponential
In the context of systems of linear ODE with constant coefficients, my lecture notes on ODE mention that the matrix exponential $e^{tA}$ has an infinite radius of convergence. This shows up in a proof that $e^{tA}$ is a fundamental matrix of $y'(t)=Ay(t)$, where the derivative of a matrix-valued function can be defined component-wise.

The matrix exponential function is defined by the series

$exp(A):=e^A:=\sum \limits_{n=0}^{\infty} \frac{A^n}{n!}$.

Now I know that in the case of the power series in $\mathbb{C}$ we define the radius of convergence of a power series $\sum \limits_{n=0}^{\infty} c_n z^n$ as

$R:=\sup \{r \geq0:(c_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where bounded means that $|c_n r^n| \leq M$ for some $M>0$.

How can we generalize this to matrix power series like the matrix exponential? In general, it does not make sense to take the supremum over a set of matrices. But if we have a power series where $z$ is still a complex number, and only the coefficients can be matrices, then we could use the same definition as before

$R:=\sup \{r \geq0:(C_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where $C_n$ are matrices and bounded then means that $\|C_n r^n\|| \leq M$ for some $C$.

It follows that each component of the power series has the same radius of convergence as the matrix series.

So we can differentiate term-wise in each component to obtain the derivative of the matrix exponential.

Is this a common definition? I could not find any source that gives a general definition.

Thanks for any help and suggestions!

答:
We can actually keep the same definition for matrix power series since it is in terms of real sequences $(c_n r^n)_{n \in \mathbb{N}}$ only, i.e $r \geq 0$ is a non-negative real number.

For complex series we then have:

Theorem: Let the power series $P(z)=\sum_{n=0}^{\infty} c_n z^n$ have radius of convergence $R$. Then

$(i)$ $P(z)$ diverges for all $z$ with $\lvert z \rvert >R$.

$(ii)$ For each $0<r<R$, the series $P(z)$ converges uniformly on $\bar{B}_r(0)=\{z \in \mathbb{C}:\lvert z \rvert \leq r\}$

$(iii)$ $P(z)$ converges absolutely for $z \in B_R(0)=\{z \in \mathbb{C}:\lvert z \rvert <R\}$

A similar statement follows for matrix power series by replacing the absolute value by the matrix norm (see here).


答:
Let $u=9x^2-6x$ and rewrite the equation as

$$\sqrt{5u+1}=7-u$$

Squaring both sides gives $5u+1=49-14u+u^2$, or

$$u^2-19u+48=(u-16)(u-3)=0$$

We see that $u=16$ is not a solution, since $\sqrt{81}\not=-9$. This leaves us with $u=3$, which is a valid solution, since $\sqrt{16}=4$, and from this we have

$$9x^2-6x=3\implies3x^2-2x-1=(x-1)(3x+1)=0$$

which gives $x=1$ and $x=-1/3$ as the complete solution set.

Remark: What makes this work so nicely is that $45:30=9:6$. If the coefficients of $x^2$ and $x$ on the two sides hadn't been in proportion, the solution would have been much more involved.


问:
Differentiating multiple products/quotients
I have a function

$$y(x) = \frac{u’(x)}{u(x)r(x)}$$

I want to differentiate this function to find $y’(x)$  in terms of $u’(x)$, $u(x)$, $r(x)$.

I am trying to decide whether to use the product or quotient rule but I have never seen a situation where there were three different functions of x all involved at the same time.

Any suggestions please on what approach I should take?

Thanks

答:
$(ln(u(x))^`=\frac{u’(x)}{u(x)} $

$y(x) = \frac{u’(x)}{u(x)r(x)}=\frac{d\ln u}{dx} \frac1{r(x)} \Rightarrow \frac{dy}{dx}=\frac{d^2(\ln u)}{dx^2}\frac1{r(x)}+\frac{d\ln u}{dx}\frac{-r^`(x)}{r(x)^2}$


答:
Notice that this is a sum of a geometric sequence:
$$ \sum_{k=0}^{n-1} e^{i\frac{2\pi}{n}p\,k} = \sum_{k=0}^{n-1} \left(e^{i\frac{2\pi}{n}p}\right)^k$$
Then you can use formula
$$ \sum_{k=0}^{n-1} q^k = \frac{q^n-1}{q-1}$$
which can be applied if $q=e^{i\frac{2\pi}{n}p}\neq 1$, that is, if $\frac{p}{n} \notin \mathbb{Z}$.

If $\frac{p}{n}\in\mathbb{Z}$, that is $p=mn$ for some $m\in\mathbb{Z}$, note that $z_k^p=1$.

Title: Sum of the roots of unity, $z_{1}^p+...+z_{n}^p$

Body:
Let $z_1,...,z_n$ be the $n$ roots of unity. I am not able to find a value for the sum: $$z_{1}^p+...+z_{n}^p,\ p \in \Bbb N$$
I know that this sum can also be written as $$\sum_{k=0}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}=1+\sum_{k=1}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}$$
I have been trying with some examples and if I am correct this sum should be equal to $0$, so I think that $1+\sum_{k=1}^{n-1}e^{i(\frac{2\pi}{n}p\ k)}$ should converge to $-1$. That is the furthest I have reached, but I don't know which is the formal mathematical procedure here.


问:
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

This is how i started solving this limit:

$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$

$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$

$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$

$\left(1- \frac {x+1}{4x+3}\right)^x$

$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$

$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$

$e^{\infty} = \infty$

answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!

答:
Being
$$\lim_{x \to \infty} \left(\frac{3 x + 2}{4 x + 3}\right)^{x} = \lim_{x \to \infty} e^{\ln{\left(\left(\frac{3 x + 2}{4 x + 3}\right)^{x} \right)}}= e^{\lim_{x \to \infty} x \ln{\left(\frac{3 x + 2}{4 x + 3} \right)}}=e^{\lim_{x \to \infty} x \color{blue}{\ln{\left(\lim_{x \to \infty} \frac{3 x + 2}{4 x + 3} \right)}}}$$

If you multiply and divide by $x$ the ratio $$\frac{3 x + 2}{4 x + 3}=\frac{\frac{3 x + 2}x}{\frac{4 x + 3}x}=\frac{3+\frac{2}x}{4+\frac{3}x}$$
you have

$$\lim_{x \to \infty} \frac{3+\frac{2}x}{4+\frac{3}x}=\frac 34$$

But if we remember that $\ln \frac 34<0$, the limit for $x\to \infty$ is $-\infty$ (to the exponent of the Napier's number) i.e.

$$\to e^{-\infty}=0$$


问:
Orthogonal projector in Hilbert spaces
Let $H$ be a Hilbert space and $U\in\mathcal{L}(H)$ an unitary operator. Let $P$ be the orthogonal projection whose image is $\ker(U-I)$. Let, for $n\geq 1$, $$S_n=\frac{I+U+\ldots+U^n}{n+1}.$$

I would like to show that, for all $u\in H$, $S_nu\to Pu$ when $n\to\infty$, but I don't really know how to do it. All I have got is that $H=\ker(U-I)\oplus\overline{\text{Im}(U-I)}$, but I'm not sure that it is really useful to answer my question.

Could anybody help me?

Thanks in advance.

答:
I assume that $limS_n(x)$ exists. (Consider $f(x)=-x$ in $\mathbb{R}$.

It is enough to show that $S_n(x)\in Ker(U-I)$.

We have$(U-I)S_n=-{I\over{n+1}}$, we deduce that $lim_n(U-I)S_n(x)=(U-I)lim_n(S_n(x))=0$ and $lim_nS_n()\in Ker(U-I)$.

Title: Orthogonal projector in Hilbert spaces


问:
Bounded Variation and Sobolev Norms.
I have two questions about weak derivatives and bounded variation, as you can see they are very similar, help with either would be much appreciated. I will start with one confusion I have

In the definition of bounded variation functions https://en.wikipedia.org/wiki/Bounded_variation test functions are taken from $C^1_c(\mathbb{R}^d,\mathbb{R}^d)$. I see some authors take functions from $C^\infty_c(\mathbb{R}^d,\mathbb{R}^d)$ e.g Proposition 3.13 of https://arxiv.org/abs/1512.02783 . Maybe one implies the other though some density argument.

$\underline{Questions :}$

$\textbf{Question 1 :}$

Assume I have a $L^1(\mathbb{R}^d)$ function $f$, where we know

$$ \int_{\mathbb{R}^d}f\text{div}(\phi)dx=\int_{\mathbb{R}^d} \langle g , \phi \rangle dx ~~\forall \phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d)$$

and $\sup_{\phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d)}| \int_{\mathbb{R}^d}f\text{div}(\phi)dx|<c$.  can I conclude $f \in W^{1,1}(\mathbb{R}^d)$ and $\nabla f=g$  $\textbf{?} $ Basically asking does begin in $BV(\mathbb{R}^d)$ imply weak derivatives are in $L^1(\mathbb{R}^d)$ ?

$\textbf{Question 2 :}$

Assume I have an sequence of uniformly bounded $L^1(\mathbb{R}^d)$ functions $\|f_k\|_{L^1(\mathbb{R}^d)}<C~ \forall k $.  Furthermore I know that they are of uniformly bounded variation $L^1(\mathbb{R}^d)$ functions $\|f_k\|_{BV(\mathbb{R}^d)}<C~ \forall k $.  What can I say about the $\|\nabla f_k\|_{L^1(\mathbb{R}^d)}$ weak derivatives of $f_k$ $\textbf{?}$

答:
Answer to Question 1.

For the first part of question 1 the answer is yes i.e. $f\in W^{1,1}(\Bbb R^d)$, while for the second part the answer is no.

For the first part: we assume that
$$\DeclareMathOperator{\dvg}{\mathrm{div}}
\sup_{\substack{\phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d) \\ |\phi(x)|\le 1}}\Bigg| \,\int\limits_{\mathbb{R}^d}f(x)\dvg\phi(x)\,\mathrm{d}x\,\Bigg|=TV(f)<+\infty\label{1}\tag{1}
$$
and that there exists a $\Bbb R^d$-valued function $g$ such that
$$
\int\limits_{\mathbb{R}^d}f(x)\dvg\phi(x)\,\mathrm{d}x =
\int\limits_{\mathbb{R}^d} \langle g(x) , \phi(x) \rangle \mathrm{d}x \qquad\forall \phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d).\label{2}\tag{2}
$$
which is thus at least locally integrable (i.e. $g\in L^1_\text{loc}(\Bbb R^d)$). Now consider the positive and negative parts of the components of $g$, defined as
$$
g^+_i(x)=
\begin{cases}
g_i(x) & g_i(x)> 0\\
0 & g_i(x)\le 0
\end{cases}\qquad
g^-_i(x)=
\begin{cases}
0 & g_i(x)\ge 0\\
-g_i(x) & g_i(x)< 0
\end{cases}\qquad\forall x\in\Bbb R^d, i=1, \ldots, d
$$
and their support sets defined as
$$
\begin{align}
G^+_i & \triangleq \{x\in\Bbb R^d : g_i(x) > 0\} \\
G^-_i & \triangleq \{x\in\Bbb R^d : g_i(x) < 0\}
\end{align} \qquad i=1, \ldots, d.
$$
From \eqref{2} we have that, for each $i=1, \ldots, d$
$$
\int\limits_{\mathbb{R}^d} \langle g(x) , \phi(x) \rangle \mathrm{d}x =
 \sum_{i=1}^d \Bigg[\,\int\limits_{\mathbb{R}^d} g^+_i(x)  \phi_i(x)\,  \mathrm{d}x -
\int\limits_{\mathbb{R}^d} g^-_i(x) \phi_i(x) \, \mathrm{d}x \Bigg]\qquad\forall \phi\in C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d).
$$
Now, for each compact set $K_i^\pm\Subset G^\pm_i$ (possibly one or more of these sets, but not all, can be empty) it is possible construct compactly supported functions $\varphi^\pm_i(x)\in C^\infty_c(\mathbb{R}^d,\mathbb{R})$ by using partitions of unity in such a way that
$$
\begin{split}\DeclareMathOperator{\supp}{\mathrm{supp}}
0\le\varphi^\pm_i &(x) \le 1\quad \forall x\in G^\pm_i\\
\varphi^\pm_i &(x) = 1 \quad  \forall x\in K^\pm_i\\
&\supp \varphi^\pm_i \Subset G^\pm_i
\end{split}
$$
By defining $\varphi_i(x)=\varphi_i^+(x)-\varphi^-_i(x)$ and $\varphi(x)= \big(\varphi_1(x), \ldots, \varphi_d(x)\big)$ we have
$$
\int\limits_{\mathbb{R}^d} \langle g(x) , \varphi(x) \rangle \mathrm{d}x =
 \sum_{i=1}^d \Bigg[\,\int\limits_{\mathbb{R}^d} g^+_i(x)  \varphi_i^+(x)\,  \mathrm{d}x +
\int\limits_{\mathbb{R}^d} g^-_i(x) \varphi_i^-(x) \, \mathrm{d}x \Bigg],
$$
and then
$$
\begin{split}
TV(f) & \ge \sup_{\varphi(x)} \sum_{i=1}^d \Bigg[\,\int\limits_{\mathbb{R}^d} g^+_i(x)  \varphi_i^+(x)\,  \mathrm{d}x +
\int\limits_{\mathbb{R}^d} g^-_i(x) \varphi_i^-(x) \, \mathrm{d}x \Bigg]\\
& = \sum_{i=1}^d \int\limits_{\mathbb{R}^d} |g_i(x)| \,  \mathrm{d}x \ge 0
\iff g_i\in L^1(\Bbb R^d)\quad i=1, \ldots,d
\end{split}
$$
where the supremum on the right side is taken on the subset of $C^{\infty}_c(\mathbb{R}^d,\mathbb{R}^d)$ consisting of functions $\varphi$ constructed as shown above: thus $f\in W^{1,1}(\Bbb R^d)$.

For the second part, as David C. Ulrich also noted in his comment on the $d=1$ case, weak derivatives of functions $f\in L^1(\Bbb R^d)$ do not need to be functions: they can be measures or even more general distributions, and precisely $BV$ functions are those integrable function whose weak derivative are possibly singular (i.e. supported on sets of dimensions $<d$) measures.

Answer to Question 2.

The answer to question 2 is no: as a simple counterexample you may consider the following sequence of monotonically decreasing functions
$$
f_n(x) =
\begin{cases}
1 & x\in [-1,1]\\
{n(1+x) +1 } & x\in\big]-\frac{1}{n}-1, -1\big[\\
n(1-x)+1 & x\in \big] 1,1+\frac{1}{n}\big[\\
0 & \text{elsewere}
\end{cases}
$$
The variation of all these function is constant and equal to $+2$ but the limit function does not belong to $W^{1,1}(\Bbb R)$.

A final Note.

A more rigorous way to define the sets $G_i^\pm$, $i=1,\ldots, d$ is the following:$\DeclareMathOperator{\intr}{\operatorname{int}} \DeclareMathOperator{\closr}{\operatorname{clo}}$
$$
\begin{align}
G^+_i & \triangleq \intr\closr\{x\in\Bbb R^d : g_i(x) > 0\} \\
G^-_i & \triangleq \intr\closr\{x\in\Bbb R^d : g_i(x) < 0\}
\end{align} \qquad i=1, \ldots, d.
$$
where $\closr$ and $\intr$ are respectively the closure and the interior operators. In this way, the partitions of unity can be constructed since $G_i^\pm$ is open and non empty (at least not for all $i$) and thus contains non empty compact sets $K_i^\pm$ for all $i=1, \ldots, d$.

Title: Bounded Variation and Sobolev Norms.


答:
$$\frac{d}{dx}\left(\frac{u'\left(x\right)}{u\left(x\right)r\left(x\right)}\right)$$
Using the product rule with $\frac{u'(x)}{u(x)}$ and $\frac{1}{r(x)}$ gives
$$\left(\frac{d}{dx}\left(\frac{u'\left(x\right)}{u\left(x\right)}\right)\right)\cdot\left(\frac{1}{r\left(x\right)}\right)+\left(\frac{d}{dx}\left(\frac{1}{r\left(x\right)}\right)\right)\left(\frac{u'\left(x\right)}{u\left(x\right)}\right)$$
Now you can differentiate $\frac{u'(x)}{u(x)}$ using the quotient rule to get
$$\frac{\left(u''\left(x\right)\right)\left(u\left(x\right)\right)-\left(u'\left(x\right)\right)^{2}}{\left(u\left(x\right)\right)^{2}}\cdot\left(\frac{1}{r\left(x\right)}\right)+\left(\frac{d}{dx}\left(\frac{1}{r\left(x\right)}\right)\right)\left(\frac{u'\left(x\right)}{u\left(x\right)}\right)$$

You could use the quotient rule for $\frac{1}{r(x)}$, but I think it's easier to use the chain rule. The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$, so using the chain rule $\frac{d}{dx}\left(\frac{1}{r\left(x\right)}\right)=-\frac{r'\left(x\right)}{\left(r\left(x\right)\right)^{2}}$.

Substituting this into what we had earlier gives
$$\frac{u''\left(x\right)\left(u\left(x\right)\right)-\left(u'\left(x\right)\right)^{2}}{\left(u\left(x\right)\right)^{2}}\left(\frac{1}{r\left(x\right)}\right)-\frac{r'\left(x\right)}{\left(r\left(x\right)\right)^{2}}\left(\frac{u'\left(x\right)}{u\left(x\right)}\right)$$
$$=\frac{1}{r\left(x\right)u\left(x\right)}\left(\frac{\left(u''\left(x\right)\left(u\left(x\right)\right)-\left(u'\left(x\right)\right)^{2}\right)}{u\left(x\right)}-\frac{r'\left(x\right)u'\left(x\right)}{r\left(x\right)}\right)$$
$$=\frac{1}{r\left(x\right)u\left(x\right)}\left(\frac{u''\left(x\right)u\left(x\right)r\left(x\right)-\left(u'\left(x\right)\right)^{2}r\left(x\right)-r'\left(x\right)u'\left(x\right)u\left(x\right)}{u\left(x\right)r\left(x\right)}\right)$$
$$=\frac{1}{r\left(x\right)u\left(x\right)}\left(u''\left(x\right)+\frac{-\left(u'\left(x\right)\right)^{2}r\left(x\right)-r'\left(x\right)u'\left(x\right)u\left(x\right)}{u\left(x\right)r\left(x\right)}\right)$$

Title: Differentiating multiple products/quotients

Body:
I have a function

$$y(x) = \frac{u’(x)}{u(x)r(x)}$$

I want to differentiate this function to find $y’(x)$  in terms of $u’(x)$, $u(x)$, $r(x)$.

I am trying to decide whether to use the product or quotient rule but I have never seen a situation where there were three different functions of x all involved at the same time.

Any suggestions please on what approach I should take?

Thanks


问:
Existence of specific sections of vector bundles over a manifold
I am trying to do the following exercise from Hirsch, one could say that it's 3 exercises but they are all related so I believe it's best to treat them together:

Let $\xi=(E,M,p)$ be an $n$-plane bundle over a connected $k$-manifold $M$.

a) If $k<n$ then $\xi$ has a non-vanishing section.

Here I belive the idea is just to use the transversality theorem, since we can approximate the zero section map $s$ by mapping the transversal to the zero section $h_k$ and for $h_k$ close enough we will get that $p\circ h_k$ is a diffeomorphism. So we can consider $h_k\circ (p\circ h_k)^{-1}$ , and this will be a section, transversal to the zero section. But then by a dimension argument we have to have that their intersection is empty.

b) If $k=n$ and $x\in M$ then $\xi$ has a section vanishing only at $x$.

Here let's divide this into two cases. First suppose $M$ has a non-vanishing section $s$ , then we can consider a function $f:M\rightarrow \mathbb{R}$ such that $f^{-1}(0)=x$, I believe this is always possible, and just consider the section $fs$. Now from the next question $c)$ and what we just did, we can assume that $\partial M=\emptyset $ and that $M$ is compact. Now if I remove a point from $M$ I get a non-compact manifold with a non-vanishing section, to be proved, and then we can just do an analogous argument to what we just did.

c) If $k=n$, and $\partial M\neq \emptyset $ or $M$ is non-compact, then $\xi$ has a non-vanishing section.

Now let's first assume that $\partial M\neq \emptyset $ and that $M$ is compact . My idea is to try and do something similiar to what was done when we proved that a compact connected manifold with boundary has a non-vanishing vector field. So let's take the double $M'$ of $M$. This will have a section $s$ with a finite number of zeros, by an analogous argument to what we did in $a)$, which we denote by $F$. Here I assume I can create a vector bundle over $M'$ by just taking the fibers of the vector bundle over $M$. I believe this is possible, but would appreciate some input. We will call this $\xi'$. Since $M'$ is connected there is a diffeomorphism $\phi :M'\rightarrow M'$ that takes $F$ into $M-M'$. Then we can consider the map $s\circ \phi^{-1}|M :M\rightarrow \xi'$ that has no zeros. Now I would like to have a vector bundle map that goes from $\xi'\rightarrow \xi $ and covers the identity when we restrict to $M$. But I'm not sure if this is possible since we want this map to not create zeros from the result we get from $s\circ \phi^{-1}$, and I'm not sure how to create a map without using some partitions of unity.

Now for the non-compact case I am kinda lost, have really no ideas on what to do. I thought about using the trivializing charts and then gluing everything together with partitions of unity, but when we glue things together, how can we do it in a way that doesn't create zeros ? I don't think I can have this control. So I am out of ideas and would appreciate some input. Don't really see where I can use the non-compactness hypothesis.

Thanks in advance.

答:
Assume $M$ is a compact $n$-manifold and that $\xi$ is a $n$-plane bundle on $M$.  By transversality, there exists a section $s$ of $\xi$ that has finitely many isolated zeros.  We shall "push" each zero onto the special point $x$.

For each zero of the section $s$, there is an embedded arc connecting it to $x$ since $M$ is connected.  Take a small open neighborhood $U$ of this arc such that $\bar{U}$ does not contain any other zeros of $s$.  If $U$ is chosen small enough, it will be contractible and $\xi|_U$ is trivial; we use this to identify all the fibers of $\xi|_U$.  We claim that there is a smooth modification $\tilde{s}$ of $s$ such that $\tilde{s} \equiv s$ outside of $U$, and the only zero of $\tilde{s}$ inside $U$ is at $x$.

Without loss of generality, by applying appropriate diffeomorphisms, we may assume that $U$ is the unit ball $B^n$ in $\mathbb{R}^n$ and that $x = 0 \in B^n$.  We define $\tilde{s}$ to be identical to $s$ outside of $U$, and inside of $U \cong B^n$, $$\tilde{s}(r, \theta) = \rho(r) \cdot s|_{\partial B^n}(\theta),$$ where $\rho: [0,1] \to [0,1]$ is a smooth function such that $\rho \equiv 1$ in a neighborhood of $1$, $\rho(r) = e^{-1/r}$ in a neighborhood of $0$, and $\rho(r) = 0$ iff $r = 0$.  Then the only zero of $\tilde{s}$ inside $U$ is at $x$, as desired.

Repeating these modifications for each zero of $s$, we obtain a section whose only possible zero is located at $x$.

It turns out that this procedure of pushing (and possibly merging) zeros of sections is very useful.

For example, suppose now that $M$ has nonempty boundary $\partial M$.  Attach a collar to the boundary (which does not change the topology of $M$), construct a section with isolated zeros, and push all of them onto the collar as above.  Then simply detach the collar to get a section with no zeros.

If $M$ is noncompact, take a compact exhaustion $\emptyset = K_0 \subset K_1 \subset K_2 \subset \cdots \subseteq M = \bigcup_i K_i$.  Given zeros in $K_i \setminus K_{i-1}$, we push them all to $K_{i+1} \setminus K_i$. Observe that this process leaves the section defined on $K_{i-1}$ unchanged, so as we keep pushing all the zeros of a section off to infinity, we obtain a well-defined nonvanishing section of $\xi$.

For full disclosure, I'm not sure whether the given construction of $\tilde{s}$ is the most efficient, and checking that it is indeed smooth at the center is a bit of a hassle.

Also, there is a simple argument to show that nonvanishing sections exist on noncompact manifolds if you know a bit about Euler classes.  The only obstruction to a nonvanishing section of a rank $n$ bundle on a $n$-manifold $M$ is the Euler class, which is necessarily zero since $H^n(M) = 0$.


问:
Prove that there exists a four colored intersection in a four colored $100×100$ grid

A $100×100$ grid is colored with four colors. There are exactly 25 blocks of each color in every row and column. Prove that there exists an intersection between two rows and columns such that all four intersecting blocks have different colors.

I am trying to prove this using invariance. But I don't know how to proceed. I also don't know if this is the correct approach so any ideas is appreciated:)

答:
Notice that there are $\binom{4}{2} \cdot 25^2$ pairs of different colors on each row, so there are $100 \cdot \binom{4}{2} \cdot 25^2$ pairs of different colors which are on the same row in total. Now, notice that $100 \cdot \binom{4}{2} \cdot 25^2 > 75 \cdot \binom{100}{2}$. So, by the generalized pigeonhole principle, there are two columns with $>75$ pairs of different colors which are on the same row. Say there are 76 pairs of different colors which are on the same row. Say the names of the colors are from the set $\{0,1,2,3\}$. Now, if the claim is not true, then either $\{0,1\}$, $\{0,2\}$, $\{0,3\}$ or $\{0,1\}$, $\{0,2\}$, $\{1,2\}$ are the possible pairs we can use to cover these $2$ columns (WLOG). The first case is clearly impossible since we have a limit of $25$ for each color, and the second case is impossible since $3$ colors are not enough to cover a total of $76 \cdot 2=152$ blocks. So the claim is true.

Edit: If you cannot understand what I mean by "pairs of different colors that are on the same row", see the comments below from @Mike.


问:
find out the value of $\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$
If $(x-3)^2+(y-5)^2+(z-4)^2=0$,then find out the value of $$\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$$

just give hint to start solution.

答: HINT: Assuming real solutions, there is only one possibility for values of $x, y$ and $z$. Then put in the corresponding values in the 2nd expression. If they are complex then there are infinite values.</p>


问:
If $n,m \in \mathbb{N} - \{0\}$ such that $n,m \geq 2$ and $n > m$ then $n^{\frac{1}{n!}} < m^{\frac{1}{m!}}$.
As the title says, I'm trying to show that if $n,m \in \mathbb{N} - \{0\}$ such that $n,m \geq 2$ and $n > m$ then $n^{\frac{1}{n!}} < m^{\frac{1}{m!}}$.

I've tried using logarithms, but it is notoriously difficult since logarithm is an increasing function, therefore it doesn't flip the inequality between $n$ and $m$ the way I need it to. Here's what I mean

$$n > m \Rightarrow n! > m! \Rightarrow \frac{1}{n!} < \frac{1}{m!}$$
Since $n > m \Rightarrow \ln (n) > \ln (m)$
$$\Rightarrow \frac{1}{m!} \ln (n) > \frac{1}{n!} \ln (n) > \frac{1}{n!} \ln (m)$$

Here's where I get stuck. I naturally thought of using logarithms since they let us treat exponentiation as multiplication, but I can't seem to get anywhere. Is there a better way to approach this problem?

Thank you for your time.

答:
Maybe I’m missing something, but if you take the function

$$X^{1/X!}$$

And take log then you get

$$\log(X)/X!$$

Since $\log(X)/X$ is a decreasing function for $X$ above $e$ (you can check by differentiating), this function decreases as $X$ increases past $e$.  It also decreases as $X$ goes from $2$ to $3$ as you can see by hand.

So $\log(n)/n!$ is less than $\log(m)/m!$

If you then take exponentials and rely on $\exp$ being an increasing function then you are home.


答:
We will prove this for two consecutive integers first and then perform induction over the larger of the two integers. Say we are working with $a+1$ and $a.$ We want to prove that $$(a+1)^{\frac{1}{(a+1)!}}<a^{\frac{1}{a!}}.$$ This is equivalent to $$1<(a^a - 1)a,$$ which is true because $$2\le (a^a -1)a$$ due to the fact that $a\ge 2.$ Now we wish to show that if $n>m$ for fixed $m$ and $n^{\frac{1}{n!}}<m^{\frac{1}{m!}},$ then $$(n+1)^{\frac{1}{(n+1)!}}<m^{\frac{1}{m!}}.$$ We get what we want because
\begin{align*}
n^{\frac{1}{n!}}&<m^{\frac{1}{m!}}\\
n&<m^{\frac{n!}{m!}}\\
n^{n+1}&<m^{\frac{(n+1)!}{m!}}.
\end{align*}
It suffices to prove that $$n+1<n^{n+1},$$ which we already showed except with $a$ instead of $n$ earlier.

Title: If $n,m \in \mathbb{N} - \{0\}$ such that $n,m \geq 2$ and $n > m$ then $n^{\frac{1}{n!}} < m^{\frac{1}{m!}}$.

Body:
As the title says, I'm trying to show that if $n,m \in \mathbb{N} - \{0\}$ such that $n,m \geq 2$ and $n > m$ then $n^{\frac{1}{n!}} < m^{\frac{1}{m!}}$.

I've tried using logarithms, but it is notoriously difficult since logarithm is an increasing function, therefore it doesn't flip the inequality between $n$ and $m$ the way I need it to. Here's what I mean

$$n > m \Rightarrow n! > m! \Rightarrow \frac{1}{n!} < \frac{1}{m!}$$
Since $n > m \Rightarrow \ln (n) > \ln (m)$
$$\Rightarrow \frac{1}{m!} \ln (n) > \frac{1}{n!} \ln (n) > \frac{1}{n!} \ln (m)$$

Here's where I get stuck. I naturally thought of using logarithms since they let us treat exponentiation as multiplication, but I can't seem to get anywhere. Is there a better way to approach this problem?

Thank you for your time.


问:
When is $C^\infty(M)$ a division ring?
Given a smooth manifold $M$ and its tangent bundle $TM$, the set of vector fields on $M$, denoted $\Gamma(TM)$, is a module over the commutative algebra of smooth functions on $M$, denoted $C^\infty(M)$. Now, since every nonzero module over a division ring has a basis by Zorn's lemma, I was wondering if there are any smooth manifolds $M$ with $C^\infty(M)$ as a division ring. If so, what are the conditions on $M$?

答:
If $m \in M$ is any point, $C^{\infty}(M)$ has a maximal ideal given by the kernel of the evaluation map $C^{\infty}(M) \to \mathbb{R}$ at $m$. $C^{\infty}(M)$ is a field iff it has no nontrivial ideals so this ideal must be zero, and (with a little effort) this implies $m = M$; that is, $M$ must consist of a single point.

As for your motivation, vector fields form not only a module but a finitely generated projective module over $C^{\infty}(M)$ (by the smooth Serre-Swan theorem), so a less restrictive version of your question is to ask when every such module is free. This is true, for example, if $M = \mathbb{R}^n$, but otherwise almost never true, e.g. it is never true if the (even) $K$-theory $K^0(M)$ of $M$ is nontrivial, and $K$-theory is nontrivial whenever even rational cohomology $H^{2\bullet}(M, \mathbb{Q})$ is nontrivial due to the existence of the Chern character isomorphism, at least if $M$ is compact.

Finally, the least restrictive version of your question is to ask when vector fields themselves form a free module. Manifolds with this property are called parallelizable; examples include the spheres $S^1, S^3, S^7$ as well as any Lie group and any orientable $3$-manifold, but generally speaking examples are rare. A parallelizable manifold has the property that all of its characteristic classes vanish which is quite a strong condition.

Title: When is $C^\infty(M)$ a division ring?


答:
Let's define $a(n)=n^{\frac{1}{n!}}$. We want to show that this sequence is decreasing since $n>2$. To obtain this result let's consider the fraction $\frac{a(n+1)}{a(n)}$. We have
$$\frac{(n+1)^{\frac{1}{(n+1)!}}}{n^{\frac{1}{n!}}}=\left(\frac{(n+1)^{\frac{1}{n+1}}}{n}\right)^{\frac{1}{n!}}$$
For $n>2$: $\sqrt[n+1]{n+1}<n$ and $\frac{1}{n!}>0$, so $\frac{a(n+1)}{a(n)}<1$, which implies that the sequence is decreasing.


问:
Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$
Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$

The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!

答: Subtract $x$ from both sides, square it, and then multiply both sides by $x^2-1$.  You’ll get a quartic equation in $x$.


问:
A 7-step algorithm for finding the candy boxes
Problem: there are $15$ boxes, of which $2$ contain candy and $13$ contain  grass. There's also a candy-detector machine that can be used to tell whether a particular set of boxes contains candy boxes in it: e.g. you could take $10$ boxes of your choice and it would say 'yes' if there's a candy box among them and 'no' if there isn't one.

The machine can only be used 7 times. Find an algorithm to locate the candy boxes with 100% certainty.

I tried brute-forcing it, but it seems to be complicated. Is there a smart way of solving this?

答:
Preliminary results.

Result-1

Immediate that if you know that there is exactly 1 candy box from a set of exactly $2^n$ boxes, then you will need exactly $n$ tests to guarantee determining which of the $2^n$ boxes has the candy.

Result-2

Suppose that you know that there are exactly 2 candy boxes from a set of exactly 7 boxes.  Can the 2 boxes be determined in no more than 5 tests?

Label the boxes $1,2,3,4,5,6,7.$

$T_1~$ (i.e. test 1) :  $~\{1,2\}.$

If this test fails, then you have 4 tests to determine which of the 2 boxes from $\{3,4,5,6,7\}$ have the candy.  Simply test boxes $3,4,5,6$ one at a time.

Therefore, WLOG, $T_1 : \{1,2\}$ succeeded.

Now: $T_2 : \{3,4,5,6\}.$

If $T_2$ fails, then you have 3 tests left to check boxes $\{1,2,7\}.$  Simply check boxes $1,2$, one at a time.

Therefore, WLOG

$T_1 : \{1,2\}$ succeeded and

$T_2 : \{3,4,5,6\}$ also succeeded.

By Result-1, the situation can be resolved with the 3 remaining tests.

Result-3

Suppose that you know that there are exactly 2 candy boxes from a set of exactly 10 boxes.  Can the 2 boxes be determined in no more than 6 tests?

Label the boxes $1,2,3,4,5,6,7,8,9,10.$

$T_1~$ :  $~\{1,2,3,4\}.$

If this test fails, then you have 5 tests to determine which of the 2 boxes from $\{5,6,7,8,9,10\}$ have the candy.  By Result-2, that is sufficient.

Therefore, WLOG, $T_1 : \{1,2,3,4\}$ succeeded.

Now: $T_2 : \{5,6,7,8\}.$

If $T_2$ succeeds, then Result-1 can be applied against

$\{1,2,3,4\}$ and $\{5,6,7,8\}$ so the 4 remaining tests are sufficient.

Therefore, WLOG

$T_1 : \{1,2,3,4\}$ succeeded and

$T_2 : \{5,6,7,8\}$ failed.

Now: $T_3 : \{9,10\}.$

If $T_3$ succeeds, then Result-1 can be applied against

$\{1,2,3,4\}$ and $\{9,10\}$ so the 3 remaining tests are sufficient.

If $T_3$ fails, then both boxes of candy are in $\{1,2,3,4\}$ so use the 3 remaining tests by testing $1,2,3$ one at a time.

Result-4

Suppose that you have 6 tests to use against

$\{1,2,3,4,5,6,7,8,9,10\}.$  Further suppose that you have
used the first test as follows:

$T_1: \{1,2,3,4\}.$  Further suppose that this test succeeded.

Can the situation be completely resolved by the 5 remaining tests - here it is unknown whether $\{1,2,3,4\}$ contains 1 candy box or 2 candy boxes.

$T_2: \{5,6,7,8\}.$

This test can be assumed to have failed, since otherwise, Result-1 can be applied (re 4 remaining tests) against

$\{1,2,3,4\}, \{5,6,7,8\}.$

$T_3: \{9,10\}.$

Similar to the above, this test can also be assumed to have failed because otherwise, by Result-1, the 3 remaining tests can be applied against

$\{1,2,3,4\}, \{9,10\}.$

Consequently, with 3 tests remaining, it can be assumed that both candy boxes are in $\{1,2,3,4\}.$  Therefore, simply test $1,2,3$, one at a time.

Label the boxes $1,2,3, \cdots, 14,15.$

$T_1 : \{1,2,3,4,5\}.$

By Result-3, WLOG, $T_1$ succeeds, since otherwise, 6 tests are sufficient against $\{6,7,8, \cdots, 15\}.$

$T_2 : \{5,6,7,8,9\}.$

If $T_2$ fails, then the scenario precisely fits Result-4, with $\{1,2,3,4\}$ having been successfully tested, and $\{10,11,12,13,14,15\}$ still untested.  Note how the hypothetical failure of $T_2$ eliminates $\{5\}$ and allows the supposition that $\{1,2,3,4\}$ has been tested successfully.

Therefore, by Result-4, WLOG, $T_2$ succeeds.

$T_3 : \{5\}.$

If $T_3$ fails, with $T_1$ and $T_2$ both assumed to have succeeded, then there must be 1 candy box in $\{1,2,3,4\}$ and 1 candy box in $\{6,7,8,9\}.$  Then, by Result-1, the 4 remaining tests are sufficient against $\{1,2,3,4\}$ and $\{6,7,8,9\}.$

Therefore, it may be assumed that $T_3$ succeeds.

$T_4 : \{10,11,12,13,14,15\}.$

Regardless of whether $T_4$ succeeds or not, Result-1 kicks in.

If $T_4$ succeeds, then 3 tests are sufficient against $\{10,11,12,13,14,15\}.$

If $T_4$ fails, then 3 tests are sufficient against $\{1,2,3,4,6,7,8,9\}.$


答:
Hint: What values does the function $x^2$ acquire(positive/negarive)? What is the solution of the equation $x^2=0$?

Can you find the solution of the equation $x^2+y^2=0$?

Now, what can you say about the equation $(x-3)^2+(y-5)^2+(z-4)^2=0
$? Can you find the values of $x,y,z?$

Title: find out the value of $\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$

Body:
If $(x-3)^2+(y-5)^2+(z-4)^2=0$,then find out the value of $$\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$$

just give hint to start solution.


问:
Why is Entscheidungsproblem undecidable not semi-decidable?
Let's say checking if a proof is valid is decidable. Then surely we can just enumerate the proofs then check if it's valid. This seems to be a semi-decidable procedure. But why Entscheidungsproblem is undecidable then?</p>

答:
It is both.

"Decidable" means "both semi-decidable and co-semi-decidable". "Undecidable" means "not semi-decidable or not co-semi-decidable or neither".

The Entscheidungsproblem is semi-decidable: We can enumerate all valid inferences. But it is not co-semi-decidable: We can not enumerate all invalid inferences. So it is undecidable but semi-decidable.


问:
Proving or disproving facts about Cauchy Sequences
Exercise 2.6.2 in Understanding Analysis by Stephen Abbott
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

I'd like someone to verify my solution, and check if my arguments are technically correct. Do I need to add a short proof for (c)?

Give an example of each of the following, or argue that such a request is impossible.

(a) A Cauchy sequence that is not monotone.

(b) A Cauchy sequence with an unbounded subsequence.

(c) A divergent monotone sequence with a Cauchy subsequence.

(d) An unbounded sequence containing a subsequence that is Cauchy.

Proof.

(a) By CC(Cauchy Criterion),
\begin{align*}
 \text{Cauchy sequences} \iff \text{Convergent sequences}
\end{align*}
A convergent sequence could be monotone or oscillating. Consider the sequence $(a_n)$ given by,
\begin{align*}
 a_n = \frac{1}{n} \sin \left(\frac{n\pi}{4}\right)
\end{align*}

is an example of a Cauchy sequence that is not monotone.

(b) This request is impossible. Cauchy sequences are bounded. All subsequences of a bounded sequence must be bounded.

(c) This request is impossible. By MCT, a monotone and bounded sequence is convergent. The contrapositive of this statement is; if a sequence is divergent, atleast one of the two possibilities must hold (1) the sequence is oscillating (2) the sequence is unbounded. As we are told, the sequence is monotone, it must be the case, that the sequence is unbounded. An unbounded monotone sequence $(a_n)$ cannot nest inside itself, a bounded subsequence $(a_{2n})$ or $(a_{f(n)})$.

(d) This request is impossible. An unbounded sequence cannot contain a bounded subsequence. And Cauchy sequences are always bounded.

答:
Your example for (a) works, though there are certainly simpler ones, e.g., $a_n=\left(-\frac12\right)^n$.

(b) is fine.

In (c) the assertion that an unbounded monotone sequence cannot have a bounded subsequence requires proof. Without loss of generality suppose that the unbounded sequence $\sigma=\langle a_n:n\in\Bbb N\rangle$ is increasing and has a subsequence $\langle a_{n_k}:k\in\Bbb N\rangle$ bounded by $b$. Since $\sigma$ is unbounded, there is an $m\in\Bbb N$ such that $a_m>b$. But then there is a $k\in\Bbb N$ such that $n_k\ge m$ and hence $b\ge a_{n_k}\ge b_m>b$, which is absurd.

As has been pointed out in the comments, (d) is wrong. Any interlacing of an unbounded sequence with a Cauchy sequence will give you a counterexample; the one in the comments is about as simple as they come.


答: Complete the square.  Then solve to find $x = \dfrac54$ or $x=\dfrac53$.

Title: Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$

Body:
Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$

The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!


答:
If $y \in Ker (U-I)$ then $Uy=y$ so $U^{n}y=y$ for all $n$ and $S_n y=\frac {y+\cdots+y} {n+1}=y$ for all $n$ So $S_n y \to y= Py$.
If $y \in Im(U-I)$ Then $y=Ux-x$ for some $x$. Now $S_ny$ simplifies to $\frac {U^{n+1}x-x} {n+1}$ after cancelling some terms. Hence $S_n y \to 0=Py$. If $y \in \overline {Im (U-I)}$ and $\epsilon >0$ then there exists $z \in Im(U-I)$  such that $\|y-z\|<\epsilon$. Use triangle inequality and the fact that $\|U^{i} \|\leq 1$ for all $i$ to finish the proof.

Title: Orthogonal projector in Hilbert spaces

Body:
Let $H$ be a Hilbert space and $U\in\mathcal{L}(H)$ an unitary operator. Let $P$ be the orthogonal projection whose image is $\ker(U-I)$. Let, for $n\geq 1$, $$S_n=\frac{I+U+\ldots+U^n}{n+1}.$$

I would like to show that, for all $u\in H$, $S_nu\to Pu$ when $n\to\infty$, but I don't really know how to do it. All I have got is that $H=\ker(U-I)\oplus\overline{\text{Im}(U-I)}$, but I'm not sure that it is really useful to answer my question.

Could anybody help me?

Thanks in advance.


答:
Hint:

Multiply for $12\sqrt{x^2-1}$ your initial equation,
you have $$12(\sqrt{x^2-1})x+12x=35\sqrt{x^2-1} \iff 12x(\sqrt{x^2-1}+1)=35\sqrt{x^2-1}$$
and square (two times to delete the sqrt) to the left and the right.

$$144x^2(\sqrt{x^2-1}+1)^2=35^2(x^2-1)$$

$$144x^2(x^2-1+1+2\sqrt{x^2-1})=35^2(x^2-1)$$


答:
After moving the $x$ to the right side and squaring, it's essentially the quartic equation $$144 x^4 - 840 x^3 + 937 x^2 + 840 x - 1225 = 0.$$

This can be factorized over $\Bbb Q$: $$(12x^2 - 35x - 49)(4x - 5)(3x - 5) = 0$$ and you easily get all four real roots from here.

EDIT:

I should confess that I didn't do the factorisation by hand. Instead, I used Sage, with the following code (which you can also try on the linked page):

```<code>R.<x> = ZZ[]
(144*x^4 - 840*x^3 + 937*x^2 + 840*x - 1225).factor()
</code>```

and it gave the result in no time.

Body:
After moving the $x$ to the right side and squaring, it's essentially the quartic equation $$144 x^4 - 840 x^3 + 937 x^2 + 840 x - 1225 = 0.$$

This can be factorized over $\Bbb Q$: $$(12x^2 - 35x - 49)(4x - 5)(3x - 5) = 0$$ and you easily get all four real roots from here.

EDIT:

I should confess that I didn't do the factorisation by hand. Instead, I used Sage, with the following code (which you can also try on the linked page):

R.<x> = ZZ[]
(144*x^4 - 840*x^3 + 937*x^2 + 840*x - 1225).factor()

and it gave the result in no time.


问:
Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number?
Is there a way to show that

$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$

is a rational number?

I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.

答:
Anay's idea of cubing both sides looks promising, but there is a little subtlety in the method.  Let's explore.

Start with

$\alpha=\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$

Cube with the Binomial Theorem:

$\alpha^3=\color{blue}{7+5\sqrt2}+\color{brown}{3(\sqrt[3]{7+5\sqrt2})^2\sqrt[3]{20-14\sqrt2}+3\sqrt[3]{7+5\sqrt2}(\sqrt[3]{20-14\sqrt2})^2}+\color{blue}{20-14\sqrt2}$

Combine the terms that are the same color in the above expression:

$\alpha^3=\color{blue}{27-9\sqrt2}+\color{brown}{3\sqrt[3]{7+5\sqrt2}\sqrt[3]{20-14\sqrt2}(\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2})}$

$\alpha^3=27-9\sqrt2+3\sqrt[3]{2\sqrt2}\alpha$

Render $\sqrt[3]{2\sqrt2}=\sqrt2$:

$\alpha^3-3\sqrt2\alpha-27+9\sqrt2=0$

Now suppose that $\alpha$ is rational in this final equation.  Then the left side is rational only if $\sqrt2$ has a zero overall coefficient, meaning $3\alpha-9=0$ or $\alpha=3$ is the only possible rational root.  So, we try this value and find that the equation holds.  Thus $\alpha=3$ checks out.

We have the possibility that even though $\alpha=3$ is a root of the cubic equation, it may not be the original sum of the cube roots with which we started.  To check that, divide the cubic by $\alpha-3$ to get a quadratic equation whose remaining real roots are other possibilities for $\alpha$.   This gives the factorization

$(\alpha-3)(\alpha^2+3\alpha+9-3\sqrt2)=0$

The discriminant of the quadratic factor is $-27+12\sqrt2=-\sqrt{729}+\sqrt{288}<0$.  Thereby $\alpha=3$ is the only real root and must be the actual sum $\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$.


问:
Standard Bivariate/ Multivariate Gaussians and Independence
Suppose you have $Z=(Z_1,Z_2,\ldots,Z_n)^T$, where $Z$ has the standard Gaussian distribution on $\Bbb R^n$. Is it true that the $Z_i$, $i=1,2,\ldots,n$ are all independent, since they have the Standard Gaussian distribution on $\Bbb R^n$?

答:
It's true that $Z_i$ are independent, and it follows from the equality
$$Ee^{i(\vec{t}, \vec{Z})} = Ee^{i \sum_{k=1}^n t_i Z_k} = \prod_{k=1}^n Ee^{i t_k Z_k}$$
This equality may be checked easily since we know characteristic funcitons of $\vec{Z}$ and $Z_k$.

But it's not true that $Z_i$ are independent since "they have $N(0,1)$ distribution".
They are independent for another reason.


问:
$\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$
I'm trying to prove and compute the limit of this function.

  $\lim_{x→0^+} \frac{\sin(6x)}{\sqrt{\sin(2x)}}$

I've tried converting it into different functions like $\cos(\pi/2-2x)$ or multiplying by the inverse function and so on, but it keep getting back to $0/0$.

I'm sure that the limit does in fact exist because using L'Hôpital's rule it is fairly easy to prove it, but I can't use it because the course I'm attending still hasn't defined the rule's properties, but I couldn't find a different way aside from that.

答: Hint: $\dfrac{\sin(6x)}{\sqrt{\sin 2x}}= \dfrac{\sin 6x}{\sin 2x}\cdot \sqrt{\sin 2x}$


问:
Definition of Sobolev Space
I have a definition that says that the space of functions that satisfy$$\|u\|_{H^m}^2=\sum_{k\in\mathbb{Z}}(1+|k|^2)^m|\hat{u}_k|^2<\infty$$is called Sobolev Space and when $m=1$, this is equivalent to saying,
$$\int_0^{2\pi}(|u(x)|^2+|u'(x)|^2)dx<\infty$$

I have a few questions. The Sobolev Space is about the space of functions whose functions have well behaved derivatives in some sense (??). Or at least the norm incorporates some information about the derivative.

What is $m$ in this context?

I do not see any information about derivative in first statement.

That is obvious in second case. So the natural questions is:
How does second become equivalent to first when $m=1$?

P.S. The last statement is perfectly clear and makes perfect sense. Since, it says both derivative and function are bounded in certain sense.

答:
In this context, $m$ can be a real number.

We can find a link between the Fourier coefficients of a function and of its derivative. Hence in the case $m=1$ it's a consequence of Parseval's equality.

Title: Definition of Sobolev Space


问:
Calculate path of vehicle with two wheels parallel to each other
There's a vehicle that starts at $P(0|0)$ oriented in positive y direction. This vehicle has two wheels that are a certain distance $d$ apart, have a diameter of $\frac{10}\pi cm$, are parallel to each other and can move at different speeds ($v_r$ and $v_l$) backwards and forwards. $d$, $v_l$ and $v_r$ are known constants.

I'm wondering if it is possible to calculate the vehicles coordinates at any point in time.

A concrete example would be for $d=8cm; \; v_l=-1\frac{rev}{s}; v_r=3\frac{rev}{s}$

Here's a visualization of the above description:

Note that this is only a model representation and the scale might not be perfectly correct.

Edit: The black dots in the first and in the second figure represent the center of the vehicle and not the point $P(0|0)$. I just marked the point in the first figure to clarify, that the vehicle's center is starting at $(0|0)$.

I've noticed that there always seems to be a point where both wheels are rotating around. If this is true finding this point could help also finding the location of the vehicle but I'm not 100% sure

答:
The following simple situation as I understood the problem.... There is no forward motion but pure rotation. The wheel velocities in proper units should be cms/sec.

$$ v_{right}=3 \text{ cm/sec},  \;v_{left}= 1 \text { cm/sec}\;$$

but should not be given in revs/sec. Disregarding wheel size since you call it as velocity. The wheel size does not matter as long as it is the velocity that is given. The velocity diagram should be drawn so that the center of stationary rotation is located dividing axle distance in the ratio

$$ 3:1= 6:2 $$

Else the axle would break. The wheels rotate around a stationary vertical axis at $O$, there would be no forward motion of the single axle Segway sort of vehicle. The angular velocity

$$\omega= \dfrac{3}{6}=\dfrac{1}{2}\text{ = 0.5 radians/sec}$$

We have here pure circular motion:

Left/Right wheels motion

$$ (x_{left},y_{left})=\{2,6\} (\cos \omega t,\sin \omega t)\text{units} $$

EDIT1:

The above is repeated to include same direction of inner wheel velocity, for generality to include wheel directions parallel and anti-parallel.

When steering a four wheeled automobile and considering the front pair of wheels the inner wheel has lower speed and radius of curvature compared to the outer..considered in design of steering mechanisms (like the Ackermann's).

We consider velocities in same direction as plus, in opposite direction as minus.

$$ L = 8 \text{cm},v_r=3, v_l= \mp 1 \text{ cm/s}$$

$e$ is distance offset from the common origin to locate center of rotation. from triangle similarity

$$\dfrac{v_l}{v_r}=\dfrac{e}{e+L}\to e=\dfrac{L v_l}{v_r-v_l};  \omega= \dfrac{ v_r}{L+e}= \dfrac{v_r- v_l}{L}; $$

The last can be directly written by consideration of relative velocity of axle ends.

Same direction (not in question)
$$ (e,\omega)=(-2,\dfrac14) $$

Opposite sense direction
$$ (e,\omega)=(4,\dfrac12) $$

The centers of rotation in these cases are $(-2,0), (4,0); $

The calculations are shown in the following vector diagram for $-1 $ cm/s opposite sense of direction:

$$ \omega= \dfrac12 ;  \;(x_{left},y_{left})= (2 \cos \omega t,2\sin \omega t +4)$$

Radius of curvature of inner turning wheel:

$$= \dfrac{v_l=-1}{\omega= \dfrac12}= -2\; cm. $$

i.e., 2 cm for the CW rotating circle.

If $v_r= v_s $  the Segway goes straight and when $v_r<v_l $ it turns left as can be shown in these equations and the associated vector/ locus diagram.


答:
From the equation solve for $\sqrt{x^2-1}$ and get
$$\sqrt{x^2-1} = \frac{x}{\frac{35}{12}-x}$$
now raise to the square and get an equation of degree $4$. Only $2$ of its roots will be roots of the initial equation, since they also include the   roots of $x-\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$.

Worth plotting the graph of the function $x + \frac{x}{\sqrt{x^2-1}}$ on $(1, \infty)$. It it convex, with a minimum at $\sqrt{2}$, which is in between the roots $\frac{5}{4}$ and $\frac{5}{3}$ of our equation.


答:
There is a lot of terminology, and a lot of redundant terminology, in computability theory. Briefly, here's the situation:

Decidable, recursive, and computable are all equivalent.

Semidecidable, recognizable, recursively enumerable (r.e.), and computably enumerable (c.e.) are all equivalent.

In general, the "computation"- and "recursion"-centered terminologies are the standard ones - with the former more modern than the latter. So in the modern literature you'll see references to r.e. or c.e. sets, but rarely to semidecidable sets.

A set with c.e. (or etc.) complement is called co-r.e. or co-c.e.; I've not seen "co-semidecidable" before, but I wouldn't be surprised if it occurred.

There is no specific term for non-c.e. (or etc.). You just say "non-c.e."

Title: Why is Entscheidungsproblem undecidable not semi-decidable?

Body: Let's say checking if a proof is valid is decidable. Then surely we can just enumerate the proofs then check if it's valid. This seems to be a semi-decidable procedure. But why Entscheidungsproblem is undecidable then?</p>


问:
Does descending chain imply well ordering without axiom of choice?
I can prove no descending chain implies well ordering. Take any subset $S$, assume it has no least element, then we can construct a function $f: S \to S$ such that $f(x) < x$, then use recursion theorem to prove there is a descending chain.

However the construction of $f$ requires axiom of choice. Is this needed? Or is there proof without AC?

答:
To remove this from the unanswered queue:

Bof's comment is exactly right. In Cohen's original model of $\mathsf{ZF+\neg AC}$, we have a set of reals $X$ (namely the set of Cohen reals added to the model) which is not well-ordered but has no infinite descending sequence.


问:
Limit of square root
Can you help me to calculate this limit as '$a$' varies in $\mathbb{R}$:
$$\lim _{x\rightarrow +\infty} \sqrt{2x^2 + x + 1} - ax$$

答:
By multiplying numerator and denominator by $\sqrt{2x^2+x+1}+ax$ we obtain
\begin{align*}
\lim_{x\rightarrow+\infty}\frac{\sqrt{2x^2+x+1}-ax}{1}&=\lim_{x\rightarrow+\infty}\frac{2x^2+x+1-a^2x^2}{\sqrt{2x^2+x+1}+ax}\\
&=\lim_{x\rightarrow+\infty}\frac{x^2(2-a^2)+x+1}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+a\right)}
\end{align*}

If $a=\sqrt{2}$, then
$$=\lim_{x\rightarrow+\infty}\frac{x\left(1+\frac{1}{x}\right)}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{2}\right)}=\lim_{x\rightarrow+\infty}\frac{1+\frac{1}{x}}{\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+\sqrt{2}}=\frac{1}{\sqrt{2}+\sqrt{2}}=\frac{1}{2\sqrt{2}}$$

If $a=-\sqrt{2}$, then
$$=\lim_{x\rightarrow+\infty}\frac{x\left(1+\frac{1}{x}\right)}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2}\right)}=\lim_{x\rightarrow+\infty}\frac{1+\frac{1}{x}}{\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{2}}=\left[\frac{1}{0^+}\right]=+\infty$$

If $a\neq \pm\sqrt{2}$, then $(2-a^2)\neq0$, so
$$\lim_{x\rightarrow+\infty}\frac{x^2(2-a^2)+x+1}{x\left(\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+a\right)}=\lim_{x\rightarrow+\infty}\frac{x(2-a^2)+1+\frac{1}{x}}{\sqrt{2+\frac{1}{x}+\frac{1}{x^2}}+a}=\left[\frac{+\infty(2-a^2)+1}{\sqrt{2}+a}\right]=\pm\infty$$
You can notice, that

if $a<-\sqrt{2}$, then $(2-a^2)<0$ and $\sqrt{2}+a<0$, so the limit is equal to $+\infty$

if $a\in(-\sqrt{2},\sqrt{2})$, then $(2-a^2)>0$ and $\sqrt{2}+a>0$, so the limit is equal to $+\infty$

if $a>\sqrt{2}$, then $(2-a^2)<0$ and $\sqrt{2}+a>0$, so the limit is equal to $-\infty$

Title: Limit of square root


答:
For $a=\sqrt{2}$, the limit calculation is more complicated.  $\sqrt{2x^2+x+1}=\sqrt{2}x\sqrt{1+\frac{x+1}{2x^2}}\approx\sqrt{2}x(1+\frac{1}{4x})$  Subtract $ax$ and get $\frac{\sqrt{2}}{4}$ as the limit.

For $a\ne \sqrt{2}$, this calculation leads easily to $\infty$ as the limit for $a\lt\sqrt{2}$ and $-\infty$ for $a\gt \sqrt{2}$.

Title: Limit of square root

Body:
Can you help me to calculate this limit as '$a$' varies in $\mathbb{R}$:
$$\lim _{x\rightarrow +\infty} \sqrt{2x^2 + x + 1} - ax$$


问:
The significance of two symbols
Excuse me for this canonical or simple question. But someone, please, can explain me, also with simplest example for a teacher of an high school, the significance of these symbols/operators?

$$\sum_{\text{cyc}}, \qquad \prod_{\text{cyc}}, \qquad \color{red}{?}$$

PS: I require the simplest explanation.

答:
A cyclic summation cycles through all the variables.

For example, if there are three variables, $a, b,$ and $c$,

then $\sum\limits_{cyc} (a^3+b^2c)$ is the sum of $a^3+b^2c$ and that expression with the variables cycled through

(i.e., $a\mapsto b, b\mapsto c, c\mapsto a$, and also $a\mapsto c, b\mapsto a, c\mapsto b$):

$(a^3+b^2c)+(b^3+c^2a)+(c^3+a^2b)$.

Similarly, the cyclic product $\prod\limits_{cyc}(a^3+b^2c)=(a^3+b^2c)(b^3+c^2a)(c^3+a^2b).$

Title: The significance of two symbols


问:
The number of distinct piles of ranked ballots in Ranked-Choice Voting.
This is about Ranked-Choice Voting (RCV) where the ballot has $C$ candidates, and there are $C$ levels of ranking of preference.  Equal ranking of candidates is not allowed and no voter is required to rank any candidates that they don't want to, but any candidate not ranked is tied for lowest preference with all other unranked candidates on the ballot.

We want to know how many operationally distinct ways a ballot can be marked so that all like-marked ballots can go onto a single pile and handled as a block in the Single Transferable Vote operation that occurs in rounds in an RCV election.  Besides the pile of blank ballots in which no candidates were ranked, how do we prove the number of piles is:

$$ \sum\limits_{n=1}^{C-1} \frac{C!}{n!} $$

Then using nothing more advanced than what we get from calculus, what would be a concise proof that:

$$ \sum\limits_{n=1}^{C-1} \frac{C!}{n!} = \big\lfloor (e-1)C! \big\rfloor - 1 $$

for all $C \in \mathbb{Z} \ge 2$ and $e$ is the base of the natural logarithm?

(I'm hoping for something more concise than what I got.  If no one posts, I'll post my not-so-concise proof.)

答:
Consider the ballots that operationally have $n$ unranked candidates; we’re ignoring the blank ballots, so $1\le n\le C-1$. The remaining $C-n$ candidates must, operationally speaking, occupy ranks $1$ through $C-n$. There are $\binom{C}{C-n}$ ways to choose these candidates and $(C-n)!$ ways to assign ranks to them, so there are

$$\binom{C}{C-n}(C-n)!=\frac{C!}{(C-n)!n!}(C-n)!=\frac{C!}{n!}$$

operationally distinct ballots with $n$ unranked candidates. Summing over the possible values of $n$, we get a total of $\sum_{n=1}^{C-1}\frac{C!}{n!}$ possibilities.

If a more elementary calculation is wanted, argue that there are $C$ choices of candidate for rank $1$, then $C-1$ choice for rank $2$, and so on, with $C-(k-1)$ choice for rank $C_k$. Thus, for ranks $1$ through $C-n$ there are

$$C(C-1)\ldots\big(C-(C-n-1)\big)=C(C-1)\ldots(n+1)=\frac{C!}{n!}$$

ways to pick $C-n$ candidates and rank them.

Granted, even this very basic counting doesn’t usually appear in calculus courses, but it really is very basic, and I see no way to avoid it altogether.

The Maclaurin series $e^x=\sum_{n\ge 0}\frac{x^n}{n!}$, however, does generally appear in calculus courses and gives us the approximation

$$e-1=\sum_{n\ge 1}\frac1{n!}$$

and hence

$$\sum_{n\ge 1}\frac{C!}{n!}=(e-1)C!\,.$$

Thus,

$$\begin{align*}
(e-1)C!&=\sum_{n\ge 1}\frac{C!}{n!}\\
&=\sum_{n=1}^{C-1}\frac{C!}{n!}+\sum_{n\ge C}\frac{C!}{n!}\\
&=\sum_{n=1}^{C-1}\frac{C!}{n!}+1+\sum_{n\ge C+1}\frac{C!}{n!}\\
&=\sum_{n=1}^{C-1}\frac{C!}{n!}+1+\sum_{k\ge 1}\frac{C!}{(C+k)!}\,,
\end{align*}$$

and the problem reduces to showing that

$$\sum_{k\ge 1}\frac{C!}{(C+k)!}<1\,.$$

And finally,

$$\frac{C!}{(C+k)!}=\prod_{i=1}^k\frac1{C+i}\le\prod_{i=1}^k\frac1{1+i}=\frac{1!}{(1+k)!}\,,$$

so

$$\sum_{k\ge 1}\frac{C!}{(C+k)!}\le\sum_{k\ge 1}\frac{1!}{(1+k)!}=\sum_{k\ge 2}\frac1{k!}=e-2<1\,.$$


问:
Field and embedded map
Let $S=\{a\in \mathbb{R}^{\mathbb{Q}}| (\exists k\in\mathbb{Z})(\forall n<k)a_n=0\}$ and let $+$ and $\cdot$ be binary operations defined with:
$$(a+b)_n=a_n+b_n,\quad (a\cdot b)_n=\sum_{i+j=n}a_i,b_j,\qquad a,b\in\mathbb{R}^{\mathbb{Q}}, n\in\mathbb{Z}.$$
Constants $\mathbf{0},\mathbf{1},\varepsilon$ we define with
$\mathbf{0}=\langle 0 | n\in\mathbb{Z}\rangle,$ $\mathbf{1}_n=1$ if $n=0,$ otherwise $\mathbf{1}_n=0,$ and $\varepsilon_n=1$ if $n=1,$ otherwise $\varepsilon_n=0,$ $n\in\mathbb{N}.$
How to prove that $\mathbf{S}=(S,+,\cdot,\mathbb{0},\mathbb{1})$ is a field and how to show that there exists embedded map $h:\mathbb{R}\rightarrow \mathbf{S}?$

答:
(If you really meant $n\in \Bbb{Q}$) then you forgot an important condition: that for any $N$ there are only finitely many $n<N$ such that $a_n\ne 0$.

Your ring is then the ring of formal series $$\Bbb{R}((x^\Bbb{Q}))=\{ \sum_{j\ge 0} c_j x^{r_j}, c_j\in \Bbb{R},r_j\in \Bbb{Q}, \lim_{j\to \infty} r_j=\infty\}$$

It is obviously an integral domain. If $r_0=0,c_0=1$ and all the others $r_j > 0$ then the inverse is $$1+ \sum_{k\ge 1}(- \sum_{j\ge 1} c_j x^{r_j})^k$$  which proves that it is a field.

Note that $\overline{\Bbb{Q}}((x^\Bbb{Q}))$ is algebraically closed and complete, it is abstractly isomorphic to $\Bbb{C}$ but with an alternative topology to that of $\Bbb{C}$ and $\Bbb{C}_p$.

Title: Field and embedded map


问:
Exercise about prime ideals
Let $A$ be a ring. Prove that the following conditions are equivalent:

$i)$  All ideals $I \subsetneq A$ are prime.

$ii)$ The set of all ideals of $A$ is totally ordered by
inclusion and all ideals of $A$ are idempotent.

Please give me a hint. I dont see the relation between these two statemens.

答:
First suppose the ideals are linearly ordered and they are all idempotent.

We will show that a proper ideal $C\lhd R$ is prime: Let $A,B$ be two other ideals such that $AB\subseteq C$. By way of contradiction, suppose that neither $A$ nor $B$ are contained in $C$. By the linear order, the three form a chain with $C$ at the bottom, so, without loss of generality, we suppose $C\subseteq B\subseteq A$. But then $C\supseteq AB\supseteq BB=B$. This contradicts the statement that $B$ is not contained in $C$. Thus, $C$ is prime.

The other direction is easy of course! Suppose all proper ideals of $R$ are prime.

Firstly, for any ideal $A$, $A^2$ is prime. But then by primeness $A\subseteq A^2$, so that $A=A^2$.

Secondly, given two ideals $A,B$, the product $AB$ is a prime ideal. But by primeness, either $A\subseteq AB$ or $B \subseteq AB$. In the first case, $A\subseteq AB\subseteq B$ and in the second, $B\subseteq AB\subseteq A$.

Considering the classic commutative theory result (A commutative ring in which all proper ideals are prime is a field."), this exercise shows that the direct noncommutative analogue ("All ideals prime implies division ring???") is not going to hold." Of course, any full square matrix ring $M_n(F)$  over a field ($n>1$) has only one proper ideal, which is prime, and this shows that such a ring does not have to be a division ring.

But with a suitable definition of a prime right ideal the result can be saved! In Lam and Reyes's excellent paper A one-sided prime ideal principle for noncommutative rings such a definition of "prime right ideal" is given, and it's an elementary result shown there that a ring whose right ideals are all prime in this way is a division ring. (Actually the paper is full of much more interesting results, and I just can't resist plugging it here.)

Title: Exercise about prime ideals


问:
Difference between $\forall n\in\mathbb N$ and $\bigcap_{i = 1}^{\infty}$
Really confused about the difference between $\forall n\in\mathbb N$ and $\bigcap_{i=1}^\infty$.

In Understanding Analysis, I quote from Exercise 1.2.13. that

It is tempting to appeal to induction to conclude $(\bigcup_{i = 1}^\infty A_i)^c = \bigcap_{i=1}^\infty A_i^c$.

but induction does not apply here. Induction is used to prove that a particular statement holds for every value of $n\in\mathbb N$, but this does not imply the validity of the infinite case.

Have done some research on that for a while and understood that eventually the fact that I can point out a $n\in\mathbb N$ means that $n$ is finite. Hence, it cannot apply on the infinite case.

Yes, I understand the rationale. But if $\forall n \in\mathbb N$ doesn't work, then what works on proving infinite case?

Just as I feel comfy about the difference. The confusion is again brought up by the book and I quote in the following, in hopes of making it as short as possible:

The nested interval property assumes that each $I_n$ contains $I_{n+1}$. They are a nested sequence of closed intervals defined as such. $I_n = [a_n, b_n] = \{x\in\mathbb R : a_n\leq x \leq b_n\}$.

The proof focuses on finding a single real number x that belongs to all $I_n$ and it argues it is supA.

In the proof, it said $x\in I_n$, for every choice of $n\in\mathbb N$. Hence, $x\in \bigcap_{n=1}^\infty I_n$ and the intersection is not empty.

Let me know if the missed out details are needed. However, my point is just that:

Why in the infinite de morgan's rule $\forall n\in\mathbb N$ doesn't apply to $\infty$

Why in the nested interval property $\forall n\in\mathbb N$ applies to $\infty$

答:
De Morgan's Rule does happen to work for infinite sets. But this cannot be proven by inducting on the finite version of De Morgan's Rule, since induction is a tool for proving that a statement is true for an arbitrarily large value of $n$ (but $n$ is still finite).

As for the intersection of a countably infinite number of sets, this follows from the definition. We say that $x \in \bigcap_{n=1}^\infty I_n$ iff $x \in I_n$ for all $n \in \mathbb N$.


问:
On a pandiagonal Latin square, what is a broken diagonal?
After reading "No pandiagonal latin squares with order divisible by 3?" I didn't understand what a "broken diagonal" is.

Thanks

答:
My illustration for broken diagonals

Please see this thread
https://boinc.progger.info/odlk/forum_thread.php?id=178


答:
$\forall n\in\Bbb N$ never applies to $\infty$, because $\infty$ is not an element of $\Bbb N$. In the nested interval theorem there is no $I_\infty$. What we know is that $x\in I_n$ for each $n\in\Bbb N$, and therefore by definition $n$ is in the intersection of the sets $I_n$. You could call this intersection $I_\infty$ if you wanted to do so, but that would be an arbitrary choice altogether independent of the induction argument involving the sets $I_n$; you could just as well call it George. (Many years ago a friend of mine did in fact publish a paper about a mathematical object that he named George.)

As for De Morgan’s law, one proves it for arbitrary families of sets simply by showing that each side of the proposed identity is a subset of the other. This is done for arbitrary indexed families of sets here and in this answer (and probably other places at MSE as well). The proof does not depend on the theorem for finite families of sets and does not involve any kind of induction.

Title: Difference between $\forall n\in\mathbb N$ and $\bigcap_{i = 1}^{\infty}$

Body:
Really confused about the difference between $\forall n\in\mathbb N$ and $\bigcap_{i=1}^\infty$.

In Understanding Analysis, I quote from Exercise 1.2.13. that

It is tempting to appeal to induction to conclude $(\bigcup_{i = 1}^\infty A_i)^c = \bigcap_{i=1}^\infty A_i^c$.

but induction does not apply here. Induction is used to prove that a particular statement holds for every value of $n\in\mathbb N$, but this does not imply the validity of the infinite case.

Have done some research on that for a while and understood that eventually the fact that I can point out a $n\in\mathbb N$ means that $n$ is finite. Hence, it cannot apply on the infinite case.

Yes, I understand the rationale. But if $\forall n \in\mathbb N$ doesn't work, then what works on proving infinite case?

Just as I feel comfy about the difference. The confusion is again brought up by the book and I quote in the following, in hopes of making it as short as possible:

The nested interval property assumes that each $I_n$ contains $I_{n+1}$. They are a nested sequence of closed intervals defined as such. $I_n = [a_n, b_n] = \{x\in\mathbb R : a_n\leq x \leq b_n\}$.

The proof focuses on finding a single real number x that belongs to all $I_n$ and it argues it is supA.

In the proof, it said $x\in I_n$, for every choice of $n\in\mathbb N$. Hence, $x\in \bigcap_{n=1}^\infty I_n$ and the intersection is not empty.

Let me know if the missed out details are needed. However, my point is just that:

Why in the infinite de morgan's rule $\forall n\in\mathbb N$ doesn't apply to $\infty$

Why in the nested interval property $\forall n\in\mathbb N$ applies to $\infty$


问:
A linear map $T: \mathbb{R^3 \to \mathbb{R^3}}$ has a two dimensional invariant subspace.

Let $T: \mathbb{R^3 \to \mathbb{R^3}}$ be an $\mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $\mathbb{R^3}.$

I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.

I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$

In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.

答: There is a simpler (and cleaner) solution: the adjoint has a real eigenvalue with associated eigenvector $v$. Now, the orthogonal complement of $span(v)$ is a $2$ dimensional invariant subspace of $T$.


问:
How to prove a tiling of a hexagon must form a 3D cubic stack?
How to prove that a tiling of a big hexagon consisting of triangles, using only $2$-triangle tiles (three possible orientations), must resemble a continuous, convex (for each small cube), manifold, $3$D cubic stack (i.e. a 3D staircase) if we color each orientation differently?

The original question comes from Mathologer's video from Youtube, in which he asks the question why there must be an equal number of tiles for each orientation. The answer is simple by simply projecting each faces to the "wall", if we assume all tilings can resemble a convex continuous manifold $3$D cubic stack. However this assumption does not seem obvious to me. There might be some tiling that leads to strange shapes which we cannot do the $3$D projection and we need to prove that's never happening.

答:
This discrete surface "contained" into an hexagonal box with sidelength $N$ (see an example on Fig. 1) is perfectly known in terms of a set of regularly spaced level lines at heights $k+1/2, k=0,1,... N-1$ (see dotted lines on figure 2, in connection with figure 1).

Fig. 1: A realization.

Fig. 2: The associated paths.

In fact, we can think in terms of an algorithm able to build any such set of regularly spaced level lines, and associate to this set the adequate lozenges (or rhombi), providing the final 3D visual interpretation.

Here is the principle of this algorithm for a "side-$N$" hexagon (here $N=8$):

Initialize the hexagon with a "all blue color".

Build $N$ paths connecting the LHS to the RHS ; a path being a broken line with slopes $\pm 1/2$ with two constraints a) they must be non-intersecting b) for each path, the $k$-th line segment on the LHS is connected to the $k$-th line segment on the right (otherwise said, each broken line has the sum of its $\pm 1/2$ slopes equal to $0$).

Now the graphical part: for each path, in the case of a $+1/2$ slope (resp. $-1/2$ slope) associate to it an ascending yellow rhombus (resp. a descending red rhombus).

Almost magically, in fact very naturally, the blue sectors appear as flat "plateaus" or "mesas".

Remarks:

This issue is connected to important combinatorial problems, for example the so-called "Aztech diamond" configuration; see the very interesting slideshow here where our algorithm is described as well.

I am indebted to one of my students who has written on my demand the program that has given these figures... funnily just some days ago!

The site "Art Of Problem Solving" uses this kind of "cube piles" as a (default) visual code for each of its participants. See for example [here] (https://artofproblemsolving.com/community/c3h1170419p5617947).

Working on hexagonal grids necessitates sometimes specifing coding means: have a look at this site.


问:
Show a partition of $\Bbb{R}$ with the following properties (or just show that one exists)
I need to show a partition of $\Bbb{R}$, call it $P$, s.t.:

$|P| = \aleph$

For all $X\in P$ we have $|X| = \aleph $

My aproch was to define an equivilance relation $S$ of $\Bbb{R}$ s.t. for $x,y\in \Bbb{R}$ whe have $xSy$ if and only if $x-y\notin \Bbb{Q}$ or $x=y$. Hence, for $X\in \Bbb{R}/\Bbb{Q}$ we can take $a\in X$ and so $X=\{a+r|r\in \Bbb{R}-\Bbb{Q}\}$ and $|X| = |\Bbb{R}-\Bbb{Q}| = \aleph$. Now the challenging part is showing that $|\Bbb{R}/\Bbb{Q}|=\aleph$, clearly $|\Bbb{R}/\Bbb{Q}|\geq\aleph_{0}$ since for two rational numbers $a,b\in \Bbb{Q}$ we have $a-b\in \Bbb{Q}$ (since $\Bbb{Q}$ is a field) and hence each rational number has its uniqe equivalence class. Also, $\bigcup_{q\in \Bbb{Q}}[q]_{S}\neq\Bbb{R}$ since, for example, $1\notin \bigcup_{q\in \Bbb{Q}}[q]_{S}$.

And here I'm pretty much stuck so any help would be appreciated, I need to show such a partition exists but I decided to try and find one, So if you can help me with that, I would be very thankful.

答:
Do you need to find such partition? Or just to show it exists?

If you only need to show it exists, note that $\aleph=2^{\aleph_0}=2^{\aleph_0+\aleph_0}=2^{\aleph_0}×2^{\aleph_0}=\aleph\times\aleph$, let $F:ℝ^2→ℝ$ be bijection, define $F_i(x)=F(i,x)$, and let $P=\{F_i''ℝ\mid i\inℝ\}$.

If you need to find such partition, you just need to find $F$, and now that you have $F$ the above construction will give you the desired partition


问:
No pandiagonal latin squares with order divisible by 3?
I would like to prove the claim that pandiagonal latin squares, which are of form

```<code>                  0        a          2a         3a         ...  (n-1)a
                  b        b+a        b+2a       b+3a       ...   b+ (n-1)a
                  .        .          .
                  .        .          .
                  .        .          .
                  (n-1)b   (n-1)b +a  (n-1)b +2a (n-1)b +3a  ...  (n-1)(b+a)
</code>```

for some $a,b\in (0,1...n-1)$ cannot exist when the order is divisible by 3.

I think we should be able to show this if we can show that the pandiagonal latin square of order $n$ can only exist iff it is possible to break the cells of a  $n \times n$ grid  into $n$ disjoint superdiagonals. Then we would show that an $n\times n$ array cannot have a superdiagonal if $n$ is a multiple of $2$ or $3$. I am, however, not able to coherently figure out either part of this proof. Could someone perhaps show me the steps of both parts?

A superdiagonal on an $n \times n$ grid is a collection of $n$ cells within the grid that contains exactly one representative from each row and column as well as exactly one representative from each broken left diagonal and broken right diagonal.

EDIT: Jyrki, could you please explain how the claim follows from #1?

答:
It is weakly pandiagonal Latin square of order 9

http://forumimage.ru/uploads/20171204/151240677122376874.png

Weakly pandiagonal Latin square is pandiagonal as magic square.
I found weakly pandiagonal Latin square of order 12.
But I have not yet found the weakly pandiagonal Latin square of order 15.

It is semi-pandiagonal Latin square of order 9 (pandiagonal in one direction)

http://forumimage.ru/uploads/20171217/151350437058089933.png

I don't know the semi-pandiagonal Latin square of order n > 9 yet.

See more here
https://boinc.progger.info/odlk/forum_thread.php?id=178&postid=7105
https://boinc.progger.info/odlk/forum_thread.php?id=178&postid=7107


问:
Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number?
Is there a way to show that

$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$

is a rational number?

I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.

答:
Note $\sqrt[3]{20-14 \sqrt{2}}=\sqrt2 \cdot \sqrt[3]{5 \sqrt{2}-7}$ and assume $\sqrt[3]{5\sqrt{2}\pm7}=p\sqrt2\pm q$. Then

$$\alpha
=(p\sqrt2+q)+\sqrt2(p\sqrt2-q)=(q+2p)+(p-q)\sqrt2
$$

For $a$ to be rational requires $p=q=\frac{\sqrt[3]{5 \sqrt{2}+7}}{\sqrt2+1}
= \sqrt[3]{\frac{ 5 \sqrt{2}+7}{(\sqrt2+1)^3 }}=1$ to be rational, which is indeed.


答: Notice that the second expression with $m = 1$ says that $u$ and its derivative are both square-integrable. When $m$ is a positive integer, $u \in H^m$ means that $u$ and all its derivatives up to order $m$ are square-integrable. However, this can be extended to $m$ real, by using the fact that the Fourier transform of $D^m u$ is the same as $(ik)^m \hat{u}$ and using the unitarity of the Fourier transform.</p>

Title: Definition of Sobolev Space

Body:
I have a definition that says that the space of functions that satisfy$$\|u\|_{H^m}^2=\sum_{k\in\mathbb{Z}}(1+|k|^2)^m|\hat{u}_k|^2<\infty$$is called Sobolev Space and when $m=1$, this is equivalent to saying,
$$\int_0^{2\pi}(|u(x)|^2+|u'(x)|^2)dx<\infty$$

I have a few questions. The Sobolev Space is about the space of functions whose functions have well behaved derivatives in some sense (??). Or at least the norm incorporates some information about the derivative.

What is $m$ in this context?

I do not see any information about derivative in first statement.

That is obvious in second case. So the natural questions is:
How does second become equivalent to first when $m=1$?

P.S. The last statement is perfectly clear and makes perfect sense. Since, it says both derivative and function are bounded in certain sense.


问:
Orthogonal projector in Hilbert spaces
Let $H$ be a Hilbert space and $U\in\mathcal{L}(H)$ an unitary operator. Let $P$ be the orthogonal projection whose image is $\ker(U-I)$. Let, for $n\geq 1$, $$S_n=\frac{I+U+\ldots+U^n}{n+1}.$$

I would like to show that, for all $u\in H$, $S_nu\to Pu$ when $n\to\infty$, but I don't really know how to do it. All I have got is that $H=\ker(U-I)\oplus\overline{\text{Im}(U-I)}$, but I'm not sure that it is really useful to answer my question.

Could anybody help me?

Thanks in advance.

答:
This comment is too long for the comment section. The other answers make sense but if I use the Functional Calculus, I am getting an actual formula for $S.$ So something is not right, but I don't know what it is.

We have an isometric $∗$-isomorphism $C^∗(U)\cong C(\sigma(U))$ that sends $p(U)$ to $p(z)$ for any finite polynomial in $I$ and $U$. Now $\sigma (U)\subseteq \mathbb T$ and since

$p_n(z)=\frac{1}{n+1}\sum_{k=0}^nz^n=\begin{cases} 1\quad z=1\\\frac{1-z^{n+1}}{(1+n)(1-z)}\quad z\neq 1\end{cases}$ we have that $\lim p_n(z)=\begin{cases} 1\quad z=1\\0\quad z\neq 1\end{cases}\quad $

so if $p_n\to S\in C^∗(U)$ then $p_n$ has to converge to a continuous function on $\sigma(U).$ Therefore, if $1\in \sigma(U)$ then $S=I$ and if $1\notin \sigma(U)$ then $S=0.$

Title: Orthogonal projector in Hilbert spaces


问:
Notation for set of equivalence classes
Suppose we define an equivalence relation $\sim$ on a set $S$. The exact details of the relation and checking that it verifies reflexivity, symmetry, and transitivity is less important, so let's assume it satisfies those properties. Then $S/\sim$, read "S mod $\sim$ denotes the set of equivalence classes.

I am trying to make sense of this notation, specifically the use of "mod." In a quotient $R/S$, we say that $r_1, r_2 \in R$ are equivalent if $r_1 - r_2 \in S$. Similarly, in $\mathbb{Z}/2\mathbb{Z}$, $a$ is equivalent to $b$ if $a - b \in 2\mathbb{Z}$. I do not see any parallel in general to $S/\sim$. The only similarity is that they are equivalent "with respect to $\sim$."

Am I missing something? Is there a better way to think of this notation or even what it means to "quotient by $\sim$"?

答:
Good. In the case of a generic equivalence relationship, this is the same. In the case of $\mathbb{Z} /2\mathbb{Z}$
this mod notation is used to symbolize the congruence modulo p.

Title: Notation for set of equivalence classes


答:
You should resolve the wheel rotation rates into common rotation and differential rotation.  Common rotation will make the vehicle move forward or reverse in the direction it is currently pointing.  Differential rotation will make the vehicle rotate around its center.  In your example the common rotation is $+1 \frac {rev}s$ and the differential rotation is $+2 \frac {rev}s$ where positive common rotation is forward, the direction that starts pointing to $+y$ and positive differential rotation causes the vehicle to rotate counterclockwise.

The linear speed of the wheels caused by differential rotation is $10\frac {\text{cm}}{\text{rev/sec}}$, so here is $20 \frac {\text{cm}}{\text{sec}}$.  One rotation of the vehicle requires $8\pi \text{ cm}$ of motion of each wheel, so the vehicle will rotate $\frac {20\ \text{rev}}{8\pi\ \text{sec}}$ or $5\frac {\text{rad}}{\text{sec}}$.  At the same time it will be moving at a constant rate of $10 \frac {\text{cm}}{\text{sec}}$ in the direction the body is pointing.

If we measure time in seconds, distances in cm, and angles in radians from the starting position we have
$$\theta=5t \\
\dot y=10 \cos (5t)\\
\dot x=-10 \sin (5t)\\
y=2 \sin(5t)\\
x=2 \cos(5t)-2$$

Title: Calculate path of vehicle with two wheels parallel to each other

Body:
There's a vehicle that starts at $P(0|0)$ oriented in positive y direction. This vehicle has two wheels that are a certain distance $d$ apart, have a diameter of $\frac{10}\pi cm$, are parallel to each other and can move at different speeds ($v_r$ and $v_l$) backwards and forwards. $d$, $v_l$ and $v_r$ are known constants.

I'm wondering if it is possible to calculate the vehicles coordinates at any point in time.

A concrete example would be for $d=8cm; \; v_l=-1\frac{rev}{s}; v_r=3\frac{rev}{s}$

Here's a visualization of the above description:

Note that this is only a model representation and the scale might not be perfectly correct.

Edit: The black dots in the first and in the second figure represent the center of the vehicle and not the point $P(0|0)$. I just marked the point in the first figure to clarify, that the vehicle's center is starting at $(0|0)$.

I've noticed that there always seems to be a point where both wheels are rotating around. If this is true finding this point could help also finding the location of the vehicle but I'm not 100% sure


问:
Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$
Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$

The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!

答:
One can use trigonometric functions to avoid squaring both sides to get a degree 4 polynomial. Clearly $x>1$. Let $x=\sec t$, $t\in(0,\frac{\pi}{2})$. Then the equation becomes
$$ \sec t+\csc t=\frac{35}{12} $$
or
$$ \frac{\sin t+\cos t}{\sin t\cos t}=\frac{35}{12}. $$
Squaring both sides gives
$$ \frac{1+\sin(2t)}{\sin^2(2t)}=\frac{35^2}{24^2} $$
which is equivalent to
$$ 35^2\sin^2(2t)-24^2\sin(2t)-24^2=0 $$
or
$$ (25\sin(2t)-24)(49\sin(2t)+24)=0. $$
Since $\sin(2t)>0$, one has
$$ \sin(2t)=\frac{24}{25} $$
which gives
$$ \cos(2t)=\pm\frac{7}{25}. $$
So
$$ \cos t=\sqrt{\frac{1+\cos(2t)}{2}}=\frac{3}{5} \text{ or } \frac{4}{5} $$
and hence
$$ x=\frac{5}{3} \text{ or }\frac{5}{4}. $$


问:
Find minimum of $2x^3+x+\frac3{x^2}$ for $x>0$
Using AM-GM inequality we have:
$$2x^3+x+\frac3{x^2}=2x^3+x+\frac3{2x^2}+\frac3{2x^2}\geq 4\sqrt[4]{2x^3x\frac3{2x^2}\frac3{2x^2}}=4\sqrt[4]{\frac92}$$

As it turns out it is not the minimum, can you spot the mistake?

答:
By using AM-GM inequality we get:

$2x^3+x+\dfrac3{x^2}=x^3+x^3+x+\dfrac1{x^2}+\dfrac1{x^2}+\dfrac1{x^2}\geqslant$

$\geqslant x+5\sqrt[5]{x^3\cdot x^3\cdot\dfrac1{x^2}\cdot\dfrac1{x^2}\cdot\dfrac1{x^2}}=x+5\;.$

By using again AM-GM inequality we get:

$2x^3+x+\dfrac3{x^2}=2x^3+\dfrac{16}9x+\dfrac3{x^2}-\dfrac79x=$

$=x^3+x^3+\dfrac89x+\dfrac89x+\dfrac3{4x^2}+\dfrac3{4x^2}+\dfrac3{4x^2}+\dfrac3{4x^2}-\dfrac79x\geqslant$

$\geqslant8\sqrt[8]{x^3\cdot x^3\cdot\dfrac89x\cdot\dfrac89x\cdot\dfrac3{4x^2}\cdot\dfrac3{4x^2}\cdot\dfrac3{4x^2}\cdot\dfrac3{4x^2}}-\dfrac79x=$

$=8\sqrt[8]{\dfrac14}-\dfrac79x=8\sqrt[4]{\dfrac12}-\dfrac79x>6+\dfrac{163}{225}-\dfrac79x\;.$

Consequently,

$2x^3+x+\dfrac3{x^2}\geqslant x+5\geqslant5+\dfrac{97}{100}\quad$ for all $\;x\geqslant\dfrac{97}{100}\;,$

$\begin{align}2x^3+x+\dfrac3{x^2}&>6+\dfrac{163}{225}-\dfrac79x>6+\dfrac{163}{225}-\dfrac79\cdot\dfrac{97}{100}=\\&=6+\dfrac{163}{225}-\dfrac{679}{900}=6+\dfrac{652-679}{900}=\\&=6-\dfrac{27}{900}=6-\dfrac3{100}=5+1-\dfrac3{100}=\\&=5+\dfrac{97}{100}\quad\text{ for all }\;0<x<\dfrac{97}{100}\;.\end{align}$

Hence,

$2x^3+x+\dfrac3{x^2}\geqslant5+\dfrac{97}{100}\quad$ for all $\;x>0\;.$

The value $\;5+\dfrac{97}{100}=5.97\;$ obtained by applying twice the AM-GM inequality, is very close to the minimum of the function $\;f(x)=2x^3+x+\dfrac3{x^2}\;$ on the interval $\;\bigl]0,+\infty\bigr[\;.$


问:
Is Cesaro summation always resistant to attack by finitely growing 0s
Suppose you have a sum $\sum a_n $  which is Cesaro-summable. I.E. the average of the partial sums of

$$ a_1 + a_2 + a_3 + a_4 ... $$

is well defined and converges. Now suppose you throw a bunch of zeros in there but not too many, every term $a_n$ has finitely many zero's before it, so for example you could transform the series to the following (among many other options which are much more zero dense)

$$ a_1 + 0 + a_2 + 0 + a_3 + 0 + a_4  + ... $$

$$ a_1 + 0 + a_2 + 0 + 0 + a_3 + 0 + 0 + 0 + a_4 + ... $$

It seems that the Cesaro sums of these series still converge and that by throwing lots of 0 you can slow down the convergence arbitrarily high but you can never prevent it from converging, as long as your sprinkling of zeros still allows every summand $a_i$ to appear after a FINITE number of preceding terms.

How to prove this? Or: to find a counter example to it. I.E. a sequence of numbers $n_1, n_2, n_3...$ such that if you put $n_1$ $0$'s after $a_1$ and $n_2$ $0$'s after $a_2$ etc...   that the cesaro sum of this sequence will not be defined.

答:
You have a sequence of partial sums $\{S_n\}_{n=1}^{\infty}$.

By adding an arbitrary but finite number of zeros after each term, you have new partial sums $\{\tilde{S}_n\}_{n=1}^{\infty}$. The new partial sum sequence is the same as the old except each term $S_n$ is repeated however many (finite) times you like.  For example $$\{\tilde{S}_n\} = \{S_1, S_2, S_2, S_3, S_3, S_3, S_3, S_4, ...\}$$

Fix $c \in \mathbb{R}$. If $S_n\rightarrow c$ then $\tilde{S}_n\rightarrow c$.

On the other hand, suppose we only know:
$$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n S_i = c$$
Then $\frac{1}{n}\sum_{i=1}^n S_i$ may not converge.  Try
$$a_i = (-1)^{i+1} \quad \forall i \in \{1, 2, 3, ...\}$$
Then $S_i=1$ if $i$ is odd and $S_i=0$ if $i$ is even, and
$$ \lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n S_i = \frac{1}{2}$$
However, we can just repeat terms to make $\frac{1}{n}\sum_{i=1}^n \tilde{S}_i$ converge to something other than $\frac{1}{2}$, or not converge at all. For example
\begin{align}
&\{S_n\} = \{1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...\}\\
&\{\tilde{S}_n\} = \{1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...\}
\end{align}
We can make $\{\tilde{S}_n\}$ any arbitrary binary-valued sequence that starts with 1 and contains an infinite number of 0s and an infinite number of 1s.


答:
Our experience with modular arithmetic includes a few features:

The formalization of $\mathbb Z / 2 \mathbb Z$, and more generally $\mathbb Z / n \mathbb Z$, as a set of equivalence classes.

The usage of the abbreviation "mod", and the notation of the division symbol "$/$", when referring to an individual equivalence class or to a whole set of equivalence classes

Because of the utility of this formalization/usage/notation, it has been generalized to all equivalence relations: an individual equivalence class, or a set of equivalence classes, is often referred to using the abbreviation "mod" and/or the division symbol "$/$".

What you are seeing in action here is mathematicians borrowing, extending, stretttttccccchhhhhhhing old notation and re-using it in new situations. It happens allllllllll the time in mathematics: usage and notation for some special case being stretched to apply to a much broader case. That's all that's happening here.

Title: Notation for set of equivalence classes

Body:
Suppose we define an equivalence relation $\sim$ on a set $S$. The exact details of the relation and checking that it verifies reflexivity, symmetry, and transitivity is less important, so let's assume it satisfies those properties. Then $S/\sim$, read "S mod $\sim$ denotes the set of equivalence classes.

I am trying to make sense of this notation, specifically the use of "mod." In a quotient $R/S$, we say that $r_1, r_2 \in R$ are equivalent if $r_1 - r_2 \in S$. Similarly, in $\mathbb{Z}/2\mathbb{Z}$, $a$ is equivalent to $b$ if $a - b \in 2\mathbb{Z}$. I do not see any parallel in general to $S/\sim$. The only similarity is that they are equivalent "with respect to $\sim$."

Am I missing something? Is there a better way to think of this notation or even what it means to "quotient by $\sim$"?


问:
Find the magnitude of the product of all complex numbers $c$ such $x_{n+1} = (c^2 - 2c)^2 x_n x_{n-1} + 2x_n - x_{n-1}$
Find the magnitude of the product of all complex numbers $c$ such that the recurrence defined by $x_1 = 1$, $x_2 = c^2 - 4c + 7$, and $x_{n+1} = (c^2 - 2c)^2 x_n x_{n-1} + 2x_n - x_{n-1}$ also satisfies $x_{1006} = 2011$.

My try:

Lemma 1: $x_n$ is a polynomial in $c$, where the first coefficient is 1.

Proof: Induction.

Lemma 2:$$P_{n}(0)=6n-5$$

Since $P_{n+1}(c) = (c^2 - 2c)^2 P_n(c) P_{n-1}(c) + 2P_n(c) - P_{n-1}(c),$ we can plug in zero to get the recursion $P_{n+1}(0) = 2P_n(0) - P_{n-1}(0).$ The starting values are $P_1(0) = 1, P_2(0) = 7,$ and it is easy to prove by induction that $P_n (0) = 6n - 5.$

then I can't find this problem ,because if $P_{n}(x)=2n-1$ have no  double roots.I can do it,But I can't solve this equation have double roots case.so How do it? Thanks

答:
Adopting the notation of function sug,  $x_n$ is a monic polynomial $P_n(c)$ with variable $c$ and integer coefficients.

Consider $Q_n(c) = P_n(c)-P_n(2)$. By induction, we can prove that $Q_n(c)$ has $(c-2)^2$ as a factor (for the factorization over $\mathbb{Z}[c]$).

We have $P_n(0)=6n-5$, $P_n(2)=2n-1$ also by induction. Thus,
$Q_n(0)= 4n-4$. For $n=1006$, $Q_n(0)= 4\cdot 1005 = 4020$. Then we have
$$
Q_{1006}(0)=4\cdot 3 \cdot 5\cdot 67.
$$

The numbers $c$ we are looking for are the roots of the equation
$$
Q_{1006}(c)=0.
$$
We factor the polynomial $Q_{1006}(c)$ into a product of monic irreducible factors over $\mathbb{Q}[c]$ (coefficients are integers by Gauss lemma). Write
$$
Q_{1006}(c)=(c-2)^2 q_1(c)^{e_1} \cdots q_r(c)^{e_r},
$$
where $q_1, \ldots, q_r$ are distinct irreducible factors and $e_i\geq 1$.

Evaluating at $c=0$, we obtain that
$$
4 q_1(0)^{e_1} \cdots q_r(0)^{e_r} = 4 \cdot 3\cdot 5\cdot 67.
$$
As $3,5,67$ are primes,  there is a nonempty set $B$ of indices up to $r$, such that
$$
i\in B \Longleftrightarrow |q_i(0)|\in \{3,5,67, 3\cdot 5, 5\cdot 67, 3\cdot 67, 3\cdot 5\cdot 67\},
$$
and for any other $j\in \{1,\ldots,r\}-B$, we have
$$
|q_j(0)|=1.
$$

Comparing the prime factorization, we must have for the index $i\in B$, $e_i=1$.
Therefore the roots from $q_i(c)$, $i\in B$, are simple.

On the other hand, for $j\notin B$, the product of absolute values of the roots from $q_j$ must be $1$. Regardless of counting with or without multiplicity, the product contributes to $1$.

Hence, the product of absolute values of distinct roots of $Q_{1006}(c)=0$ is
$$
2 \cdot 3\cdot 5\cdot 67 = 2010.
$$


问:
Can two events in probability be both non-disjoint AND non-independent? Confused about interpretation of conditional probabilities?
I am currently reviewing conditional probability in my textbook and have stumbled across this:

The formula for conditional probability is this: Where $A$ and $B$ are two events where $P(B) > 0$ (naturally)

$$P(A \mid B) = \frac{P(A\cap B)}{P(B)}$$

so naturally: $P(A\cap B) = P(B)\cdot P(A\mid B)$

and if $A$ and $B$ are independent, $P(A \mid B) = P(A)$

so $P(A\cap B) = P(B) \cdot P(A)$ which matches the definition of independence.

What I wonder is this. Say that $P(B)$ and $P(A \mid B)$ are non-zero probabilities and $P(A \mid B) \neq P(A)$ so $P(A\cap B) \neq 0$ and $A$ and $B$ are non-disjoint AND not independent events. Is this legal? Have I broken a law of probability? The textbook doesn't specify. If it's legal can someone give me an example in real life of two events that are non-disjoint and not independent?

Also can someone answer my second question?

Does the RHS of $P(A\cap B) = P(B)\cdot P(A\mid B)$ mean that event $B$ occurs first and THEN event $A$ occurs in which event $B$ occurring first affected the probability of event $A$ also occurring?

OR

Does the RHS mean that event $B$ and $A$ occur simultaneously? (If it's this second interpretation, can someone explain how events $A$ and $B$ occurring simultaneously affects event $A$'s probability of occurring?)

答:
Any example where $A \subsetneq B$ (with $P(A) > 0$ and $P(B) < 1$) will satisfy your conditions:

$A$ and $B$ are clearly not disjoint

$P(A \cap B) = P(A) \ne P(A) P(B)$

$P(A \mid B)$ should be interpreted as the probability of $A$ occurring if you already know that $B$ occurred. So you can interpret $P(B) \cdot P(A \mid B)$ as first accounting for the probability that $B$ occurred, and then accounting for the probability that $A$ occurred given that you already know $B$ occurred. This is common in sequential computations, i.e. the probability that two draws without replacement from a deck are both aces is $P(\text{first card is ace}) P(\text{second card is ace} \mid \text{first card is ace}) = \frac{4}{52} \cdot \frac{3}{51}$.