答:
Since for $x\ge 0$:
$$\sin x \ge x - \frac{x^3}{6}$$
we have
$$\begin{align}
\|f_n(x) - f(x)\| &= \left\|\frac{\sin x}{n\sin(x/n)} - \frac{\sin x}{x}\right\| \\
&= \color{blue}{\left\|\sin x\right\|}\left\| \frac{1}{n\color{red}{\sin(x/n)}} - \frac{1}{x}\right\| \\
&\le \color{blue}1\cdot \left\|\frac{1}{n\color{red}{\left(\frac{x}{n} - \frac{x^3}{6n^3}\right)}} - \frac{1}{x}\right\| \\
&= \left\|\frac{1}{x - \frac{x^3}{6n^2}} - \frac{1}{x}\right\| \\[1mm]
&\xrightarrow{n\to\infty}0
\end{align}$$

Title: How do you prove that $\frac{\sin x}{n \sin (\frac{x}{n})}$ converges uniformly?

Body:
For $n \in \mathbb N$, define
$$f_n(x) = \frac{\sin x}{n \sin (\frac{x}{n})}.$$
As $n \to \infty$, I have shown that $\{f_n\}$ converges (pointwise) to $\frac{\sin x}{x}$. I am trying to evaluate $\lim\limits_{n \to \infty} \int_0^\pi f_n(x) \, dx$, and my attempt is to show the uniform convergence of the integrand on $[0,\pi]$, so that I can interchange the operations of limit and integral to obtain $\lim\limits_{n \to \infty} \int_0^\pi f_n(x) \, dx = \int_0^\pi \frac{\sin x}{x} \, dx$.

By plotting $f_n(x)$ with a graphing calculator and letting $n \to \infty$, I am quite convinced that $\{f_n\}$ indeed converges uniformly to $\frac{\sin x}{x}$. Nevertheless, I can't seem to prove this proposition, whether by the definition of uniform convergence or by Cauchy criterion. Perhaps my difficulty stems from the fact that the bounds on $f_n$ depends on the interaction between $\sin x$, $\sin(\frac{x}{n})$, and $n$. That is, finding a bound on one or two expressions can't help in establishing a bound for the whole function. So far I have tried applying the inequalities $0 < \sin x \leq 1$ and $\sin x \leq x$ for $0 < x < \pi$, but without any luck.

Any idea on how to prove this statement? Or is there a theorem on uniform convergence that is applicable for this case?

(Note: This question is related to the Dirichlet kernel in Fourier analysis, which can be expressed as $D_N(x) = \frac{1}{2\pi} + \frac{1}{\pi} \sum_{n=1}^N \cos nx = \frac{1}{2\pi} \frac{\sin(N+\frac{1}{2})x}{\sin (\frac{x}{2})}$. However, I don't think this is relevant at this point.)


问:
What is the meaning of the function $f:B\to A$ for sets $A$ and $B$?
Consider $A=\{1,2,3,4,5\} \text{and}$ $B=\{1,2,3,4,5,6,7,8,9,10\}$. then what is the meaning of the function $f:B\to A$ ?

Does that mean it is a function which its domain should contain All the elements of set $B$ and its range should contain All the elements of set $A$?

Can $f$ be the function with following values?

$$f(2)=1, f(4)=2,f(5)=3,f(9)=4,f(10)=5$$
Or we should use all elements of both $A$ and $B$?

答: You only need to have all elements of $B$ in the domain. If all elements of $A$ are in the range, then it'll be called a surjective function.

ContentLicense: CC BY-SA 4.0


答: The general convention I am most familiar with is that a function $f:A\to B$ is usually understood to be defined for all elements of $A$ but does not necessarily map to all elements of $B$. That is, for such an $f$, $f(a)$ will be defined for all $a\in A$, but there may exist some (or many) $b\in B$ for which there exist no $a\in A$ with $f(a)=b$.

Title: What is the meaning of the function $f:B\to A$ for sets $A$ and $B$?

Body:
Consider $A=\{1,2,3,4,5\} \text{and}$ $B=\{1,2,3,4,5,6,7,8,9,10\}$. then what is the meaning of the function $f:B\to A$ ?

Does that mean it is a function which its domain should contain All the elements of set $B$ and its range should contain All the elements of set $A$?

Can $f$ be the function with following values?

$$f(2)=1, f(4)=2,f(5)=3,f(9)=4,f(10)=5$$
Or we should use all elements of both $A$ and $B$?


问:
How to compute the closure of a set of binary strings in term of the "AND" and "OR" operators?
Given a set of $n$ binary strings of length $m$, how many binary strings at least should we add to the set to make sure that the binary "AND" and "OR" operators are closed on the set? The output of the problem is the number of binary strings needed to add.

For example, given the set {001, 010, 110}, we should at least add {000, 011, 111} to the set to make both "AND" and "OR" operators closed. So the output is 3.

Is there any good algorithm or idea related to the problem? Thanks a lot.

答:
You can look for cases where one bit implies another.  In your example bit 2 (the MSB) implies bit 1-there are no examples in your given set with bit 2 on and bit 1 off.  So you can never get rid of this and 100 and 101 will not be required.  This is $n^2m$ in complexity

Added in response to comment:  You are looking for cases where bit $i$ is set in all your input words whenever bit $j$ is set.  So you could make an array $m \times m$ and initialize it to all $1$.  Go through all $n$ input words.  For each word, check if bit i=0 and bit j=1.  If so, set the $(i,j)$ element to $0$.  At the end, if any entries in the array are still $1$, having bit $j$ on means for sure bit $i$ is on in your input words.  As AND and OR cannot destroy this, none of your output words will have $i$ off and $j$ on.  As you check $m(m-1)/2 \approx m^2$ bits in $n$ input words, this is of order $m^2n$

ContentLicense: CC BY-SA 3.0


问:
Exercise III.5.10 in Grillet's Abstract Algebra

Let $R$ be commutative, with characteristic either $0$ or greater than $m$. Show that a root $r$ of $A \in R[X]$ has multiplicity $m$ if and only if $A^{(k)}(r)=0$ for all $k<m$ and $A^{m}(r) \neq 0$. Show that the hypothesis about the characteristic of $R$ cannot be omitted from this result.

By definition of multiplicity $A(x)=(x-r)^{m}B(x)$, where $B(r) \neq 0$. It's easy to see, that $A^{k}(r)=0$ for all $k<m$ and $A^{m}(r)=m!B(r)$. But why $m!B(r) \neq 0$? Shouldn't $R$ necessarily be a domain?

答:
I think you need the characteristic to be prime, not necessarily that $R$ be a domain.

If the characteristic is prime, then note that:

Proposition. Let $R$ be a commutative ring, $r\in R$, $a,b\in\mathbb{Z}$. If $ar=br=0$, then $\gcd(a,b)r = 0$. In particular, if $R$ has prime characteristic $p\gt 0$, and $ar=0$ for some $0\lt a\lt p$, then $r=0$.

Proof. Write $\gcd(a,b) = ax+by$ with $x,y\in\mathbb{Z}$. Then $\gcd(a,b)r = (ax+by)r = x(ar)+y(br) = 0+0=0$.

If $0\lt a\lt p$, then $\gcd(a,p)=1$. Since $ar=pr=0$, then $r = \gcd(a,p)r = 0$. $\Box$

If the characteristic is prime, then we know $\gcd(m!,p)=1$, so the result would follow as well (in that direction, in any case).

If the characteristic $d\gt 0$ is not prime, then there exists $m$, $1\lt m\lt d$, such that $\gcd(m!,d)=k\gt 1$. Write $d=kn$, and let $A(x) = x^m(x+n)$. Then $x=0$ is a root of multiplicity $m$, and $A^{(m)}(0) = m!(n) = 0$, because $d=kn|m!(n)$.

In characteristic $0$ you will have problems if there are elements $a\in R$, $a\neq 0$,  such that $na=0$ for some $n\gt 0$. (Note that characteristic $0$ just means that the only positive integer such that $na=0$ for all $a$ is $n=0$). If such an $a$ and $n$ exist, then you can take $A(x) = x^n(x+a)$ to get the same problem.


问:
How Can I Evaluate $\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$
Question:- Evaluate $$\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$$

Where $a,b,c \in Z$

In order to find Definite Integrals of type

$$\int \frac{ln(x) ln(1+x) ln(1-x)}{(ax+b)(cx+d)}$$

Let $$I_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$$

Differentiate both sides w.r.t $a$

$$I^{'}_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x) dx$$

Now Differentiate w.r.t $b$

$$I^{''}_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x)ln(1+x) dx$$

Again differentiating w.r.t $c$

$$I^{'''}_{a,b,c}=-\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x)ln(1+x)ln(1-x) dx$$

Letting different integral values of $a$, $b$ and $c$.we can evaluate numerous integral of above type. But I find that It is very hard to find $I_{a,b,c}$.

答:
It is not surprising that you feel the problem difficult. For
$$J= \int x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$$ a CAS gives
$$2a(1+a)x^{-a} J=$$ $$2 (a+1) F_1(a;-c,-b;a+1;x,-x)+$$ $$a x
   (F_1(a+1;1-c,-b;a+2;x,-x)-F_1(a+1;-c,1-b;a+2;x,-x))$$
where appear three Appell hypergeometric functions of two variables.

For the definite integral $I_{abc}$ replace $x$ by $\frac a 2$.


问:
Projection of $\mathbf{X}$ in the direction of $\mathbf{a}$: $\mathbf{a}^T \mathbf{X} = \sum_{j = 1}^d a_j X_j$
I am currently studying principal component analysis in statistics. PCA uses the "projection of $\mathbf{X}$ in the direction of $\mathbf{a}$":

$$\mathbf{a}^T \mathbf{X} = \sum_{j = 1}^d a_j X_j$$

I know that $\dfrac{\mathbf{a} \cdot \mathbf{b}}{\vert \vert\mathbf{b} \vert \vert} = \vert \vert \mathbf{a} \vert \vert \cos(\theta)$ is the vector $\mathbf{a}$ in the direction of the vector $\mathbf{b}$, but I'm struggling to see how $\mathbf{a}^T \mathbf{X} = \sum_{j = 1}^d a_j X_j$ is $\mathbf{X}$ in the direction of $\mathbf{a}$. Analogously to the $\dfrac{\mathbf{a} \cdot \mathbf{b}}{\vert \vert\mathbf{b} \vert \vert} = \vert \vert \mathbf{a} \vert \vert \cos(\theta)$ case, how is $\mathbf{a}^T \mathbf{X} = \sum_{j = 1}^d a_j X_j$ $\mathbf{X}$ in the direction of $\mathbf{a}$?

答:
When we talk about PCA, $a$ here is a unit vector.

If $x$ is a vector, then  $$\frac{x \cdot a}{\|a\|}=\frac{x\cdot a}{1}=x\cdot a =a^Tx$$


问:
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$
$\lim_{n\to\infty} \frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}}\,.$

I used the ratio criterion for the calculation and I got to this, can I say now that it is zero or is it still an undefined expression?

$\frac{10+(10/\sqrt{(n+1)})}{\sqrt{(n+2)}+1}$

10/inf < 1 ----> lim an=0

答:
I would like to completely eliminate the doubt regarding why " n! is 'stronger', then any ns power", with the aid of Sterling's approximation, and solve the question even without using the ratio test.....

$\frac{10^n}{\sqrt{(n+1)!} + \sqrt{n!}} \lt \frac{10^n}{2\sqrt{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{n!}}$

This inequality true because the denominator of right term is smaller than left term...

By the Sterling's approximation, we have

$n!\sim \frac{\sqrt{2\pi n}n^n}{e^n}$

Hence, at the limit to infinity, we can write
$\frac{1}{2}\sqrt{\frac{100^n}{n!}}=\frac{1}{2}\sqrt{\frac{100^n}{\sqrt{2\pi n}~n^n}}=\frac{1}{2}\sqrt{\frac{1}{\sqrt{2\pi n}}~.({\frac{100}{n})}^n}=0$,  as the limit$\frac{100}{n}$ goes to 0..

Here the important fact is then 100, can be replaced by any real finite number..., hence proving why n! is stronger


问:
How can we make four distinct lists of books so that the total number of pages in the books on each list are equal?
I have $15$ books each with between $300$ to $400$ pages. How can we make four distinct lists of books  so that the total number of pages in the books on each list are equal ?

Can anyone please give me a hint to proceed ?

答: A list of four books has at least $4\cdot300=1200$ pages. A list of three books has at most $3\cdot400=1200$ pages. So there can only be one solution in that case.


答:
Just expand $(1+x)^{b-1}$ and $(1-x)^{c-1}$ using the Binomial theorem, and we get that our integral is equal to

$$\sum_{j=0}^{b-1}\sum_{k=0}^{c-1}{b-1\choose j}{c-1 \choose k}\frac{(-1)^k}{a+j+k}\left(\frac{a}{2}\right)^{a+j+k}$$

If you feel like compactifying this representation using properties of binomial cooeficients, then feel free.

Title: How Can I Evaluate $\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$

Body:
Question:- Evaluate $$\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$$

Where $a,b,c \in Z$

In order to find Definite Integrals of type

$$\int \frac{ln(x) ln(1+x) ln(1-x)}{(ax+b)(cx+d)}$$

Let $$I_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1} dx$$

Differentiate both sides w.r.t $a$

$$I^{'}_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x) dx$$

Now Differentiate w.r.t $b$

$$I^{''}_{a,b,c}=\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x)ln(1+x) dx$$

Again differentiating w.r.t $c$

$$I^{'''}_{a,b,c}=-\int_{0}^{\frac{a}{2}}x^{a-1} (1+x)^{b-1}(1-x)^{c-1}ln(x)ln(1+x)ln(1-x) dx$$

Letting different integral values of $a$, $b$ and $c$.we can evaluate numerous integral of above type. But I find that It is very hard to find $I_{a,b,c}$.


问:
Why can $(m,m)$ and $(m,n)\oplus(n,m)$ representations of $\mathfrak{so}(1,3)$ be represented over a real vector space?
I understand that they are similar to their conjugate representations, but this does not seem sufficient to me: all representations of $\mathfrak{su}(2)$ are similar to their conjugate representations but even-dimensional ones do not admit a natural real structure.

Just to clarify, some background: I'm talking about the real linear representations of the real Lie algebra $\mathfrak{so}(1,3)$, given by:

$$\rho_{m,n}(J_i)=J_i^{(m)}\otimes1_n+1_m\otimes \overline{J_i^{(n)}}\,,\\\rho_{m,n}(K_i)=-i\left[J_i^{(m)}\otimes1_n-1_m\otimes \overline{J_i^{(n)}}\right]\tag{1}$$

where $J_i,K_i$ is the usual basis (generators of rotations and generators of boosts), and $J_i^{(m)}$ is the $m$-th representation of $\mathfrak{su}(2)$.

(1) have a unique real linear extension to the whole $\mathfrak{so}(1,3)$ which is the real span of $J_i,K_i$. They map to a real vector space of complex matrices.

If I understand correctly, when $n=m$, or when taking the direct sum of the $(m,n)$ representation with the $(n,m)$ representation, there is a change of basis in this vector space such that the matrices become purely real.

I can see that there is a change of basis mapping these representations into their complex conjugate (namely, the interchange of the two factors of the tensor product), but still it looks like this does not mean they are real in this basis.

So, is there such a basis, or am I mistaken? If yes, what is it?

答:
$\DeclareMathOperator{\g}{\mathfrak g}$

A: Complex, Real and Quaternionic Representations

Let $\g$ be a semisimple Lie algebra over $\mathbb R$, and $V$ a (finite-dimensional, complex) representation of it, i.e. a finite dimensional $\mathbb C$-vector space with a homomorphism of real Lie algebras $\rho: \g \rightarrow \mathrm{End}_{\mathbb C} V$. There are several ways to define its conjugate representation $\overline V$, one of which would be to choose a basis of $V$, express the above endomorphisms as matrices, and then just complex-conjugate their entries.

As you correctly say, the existence of an equivalence of representations $V \simeq \bar V$ is a necessary, but not sufficient criterion for $V$ to have a real structure, i.e. for the existence of a real vector space $V_1$ together with a $\mathfrak g$-action such that $V$ identifies with the scalar extension ("complexification") $\mathbb C \otimes_{\mathbb R} V_1$ (or equivalently, the existence of a basis of $V$ such that the matrices of all $\rho(x), x \in \g$ have all real entries).

Namely, let's first assume that $V$ is irreducible. The existence of such an equivalence $V \simeq \bar V$ can also be expressed by saying there is a map $\alpha: V \rightarrow V$ which is antilinear (i.e. $\alpha(cv) = \overline{c} \alpha(v)$ for all $c \in \mathbb C, v \in V$), bijective, and commutes with all $\rho(x), x \in \g$. If that is the case, then $\alpha \circ \alpha$ is a (complex-linear!) automorphism of $V$, i.e. by Schur's Lemma, it's just multiplication by some scalar $b_\alpha \in \mathbb C^*$. Expressing $\alpha \circ \alpha \circ \alpha$ in two different ways we see that that scalar $b_\alpha$ must actually be real; since further we can scale $\alpha$ with any non-zero real number, which scales $\alpha \circ \alpha$ with a square, i.e. any positive real number, we basically have two possible cases: $b_\alpha = \color{red}\pm 1$ or in other words,

$$\alpha \circ \alpha = id_V \text{ or } -id_V.$$

In the $+$ case, we do have a real structure. In fact, since $\alpha\circ \alpha = id$, the map $\alpha$ viewed as endomorphism of $V$ as real vector space has two eigenvalues, $\pm 1$, and if $V \simeq V_1 \oplus V_{-1}$ is the corresponding decomposition into eigenspaces, then $V_1$ gives a real subspace as wanted (I chose the terminology $V_1$ above for that!).

But in the $-$ case, there is no such real structure. (Rather, one can then give $V$ the structure of a vector space over $\mathbb H$, the Hamilton quaternions with standard basis $\{1,i,j,k\}$, and the map $\alpha$ corresponds to multiplication "from the other side" with, say, $j$.)

One can further show that these two cases are mutually exclusive, i.e. there cannot exist two different isomorphisms $\alpha: V \rightarrow \overline{V}$ such that for one of them $\alpha \circ \alpha$ is a positive and for the other one it is a negative scalar.

So in total there are three possible, mutually exclusive cases:

$V \not \simeq \overline V$. We call such $V$ "complex".

$\exists \alpha: V \simeq \overline V$ and $\alpha \circ \alpha = id$. We call such $V$ "real".

$\exists \alpha: V \simeq \overline V$ and $\alpha \circ \alpha = -id$. We call such $V$ "quaternionic".

One can characterize these cases equivalently via the [1: non-existence of any /2:  existence of a symmetric / 3: existence of an alternating] $\g$-invariant bilinear form on $V$, or also via the "commutant" of $V$, i.e. $\mathrm{End}_{U_{\mathbb R}(\g)}(V)$ being 1: $\mathbb C$ / 2: $\mathbb R$ / 3: $\mathbb H$. Confer Bourbaki, Lie Groups and Algebras, chapter 9, Appendix II.

I point out that I have seen some sources which use  "pseudoreal" for what I called "quaternionic", while other sources use "pseudoreal" for both cases 2 and 3. To make things worse, as soon as we remove our assumption that $V$ is irreducible, and/or look at representations of groups, there exists different non-interchangeable nomenclature conventions for these things. Actually, even the above seems to be not generally agreed upon as soon as our $\g$ is not compact. This is highly unfortunate. See e.g. answers and comments to https://mathoverflow.net/q/47492/27465 and https://mathoverflow.net/q/323969/27465.

B: Application to your question

Well, that means that if you are convinced that you have a map (flipping the tensors) which gives an isomorphism from your representation to its conjugate, then all you have to do is to see that if you translate that map into an $\alpha$ as above, and compose it with itself, you get (a positive real number times) the identity map, and not its negative. Seems highly likely from your candidate. I leave the details to you and use the rest of this answer to write down more theory for reference.

C: The standard example $\mathfrak{su}_2$; and the general compact case

Back to the three cases in part A. The basic example (for which almost all nomenclatures agree, and which takes us far!) for the distinction between cases 2 and 3 is the real Lie algebra $\g = \mathfrak{su}_2$, the compact real form of $\mathfrak{sl}_2$. As one learns, up to equivalence, the irreducible representations of this are just indexed by positive integers (mathematicians) or half-integers (physicists). In fact, up to iso there is

one with $\mathrm{dim}(V) = 1$ (the trivial one),

one with $\dim V = 2$ which can be realized by letting $g \in \mathfrak{su}_2 = \{\pmatrix{ai &b+ci\\-b+ci&-ai}: a,b,c \in \mathbb R\}$ act on $V:=\mathbb C^2$ by matrix multiplication,

one with $\dim V = 3$ which can be realized by letting $g \in \mathfrak{su}_2 \simeq \mathfrak{so}_3 = \{\pmatrix{0 &x &y\\-x&0&y\\-y&-z&0}: x,y,z \in \mathbb R\}$ act on $V:=\mathbb C^3$ by matrix multiplication,

and then one each for $\dim V = 4,5,6, ...$ which are not that easily written down with matrices, but instead via general $\mathfrak{sl}_2$-representation theory with ladder operators etc. tied together.

The way I wrote the first three examples, it's obvious that for $\dim V =1,3$ we are in the case that $V$ is real (for $\dim V=3$, I kind of cheated by assuming the isomorphism $\mathfrak{su}_2 \simeq \mathfrak{so}_3$ known, and the matrices in $\mathfrak{so}_3$ already have all real entries). The case $\dim V=2$ however is the standard example for a quaternionic representation. Indeed, check that $\alpha: \pmatrix{c_1\\c_2} \mapsto \pmatrix{-\overline{c_2}\\ \overline{c_1}}$ defines a bijective, antilinear, $\g$-equivariant map with $\alpha \circ \alpha = -id_V$.

Cool fact 1: It can be shown that the parity distinction continues, i.e. of the irreducible $\mathfrak{su}_2$-representations $V$, the ones with odd $\dim V$ are real, while the ones with even $\dim V$ are quaternionic. (Depending on whether you're in math or physics jargon, this will translate into either parity of your classifying parameter, or your classifying parameter being an integer or half-integer. But because that parameter is usually $\dim V-1$ (or $\dfrac{\dim V -1}{2}$ (so that the trival representation has parameter $0$), the parity distinction will be "the other way around": real for even (or integer), quaternionic for odd (or half-integer).

Cool fact 2: It can be shown that actually, this principle generalises to all compact (semi)simple $\mathfrak g$, except that now also the case of "truly complex" reps can occur. Namely, this is the content of Bourbaki's Lie Groups and Algebras, chapter 9, §7 no.2 proposition 1, cf. the answer to What property of the root system means a Lie algebra has complex structure? :

For $\mathfrak g$ the compact (!) real form of a semisimple Lie algebra, the irreducible representation $V$ of highest weight $\lambda$ (for $\lambda$ dominant w.r.t. a chosen set of simple roots $\Delta$) is

complex if and only if $$-w_0(\lambda) \neq \lambda$$ where $w_0$ is the longest element of the Weyl group (w.r.t. that set of simple roots $\Delta$);

real if $-w_0(\lambda) = \lambda$, and a certain invariant is even;

quaternionic if $-w_0(\lambda) = \lambda$, and a certain invariant is odd.

Of course, that "certain invariant" generalises the integer (or half-integer) parameter from the $\mathfrak{su}_2$-case, has several definitions which are equivalent (although maybe not obviously so), and in general needs a little computation. One way to define it is as twice the sum of coefficients of $\lambda$ written in the chosen root basis  $\Delta$,

$$2 \cdot \sum c_\alpha \text{ where } \lambda = \sum_{\alpha \in \Delta} c_\alpha \alpha$$

(The $2$ is there to make the invariant an integer, matching the math convention in the $\mathfrak{su}_2$ case (where the possible highest weights $\lambda$ for $\Delta= \{\alpha\}$ are $0, \frac12 \alpha, \alpha, \frac32 \alpha, ...$).)

D: But we are not in the compact case!

And this is bad, because as seen in the MathOverflow posts linked in A, different people use different nomenclatures now. Also, people mistakenly use the criteria from part C for this case, where they are in general no longer valid. There are actually two questions on this site where the accepted answer seems to fall into this trap, which has made me write lengthy answers outlining what I think is a correct approach: See If fundamental and antifundamental representations of a Lie algebra are inequivalent, can we deduce that all conjugate representations are? and Conjugate Representations of Lie Algebra of Lorentz Group.

In nuce, the issue is that if $\mathfrak g$ is not compact, then complex conjugation does not act on the weight lattice via $-w_0$, the element that plays the crucial role in part C.

Indeed, Jacques Tits undertook a vast generalisation of the idea of part C, not only for arbitrary forms, but also for arbitrary ground fields (not just $\mathbb R$), in his article Représentations linéaires irréductibles d'un groupe réductif sur un corps quelconque. I do not claim to understand half of it, but the gist is that we have to take into account how the full Galois group operates on our weights, and then on the set of those dominant weights $\lambda$ which are "sufficiently stable" under Galois (this generalises the $-w_0(\lambda) =\lambda$ criterion) we have a map going to the Brauer group of the ground field $k$ (this generalises the parameter which decides between $\mathbb R$ and $\mathbb H$, the only elements of the Brauer group of $\mathbb R$).

Now looking into the end of section 5 and the beginning of section 6 in that article, it seems too good to be true, but I would currently bet he shows there that for any quasi-split form (and such is every $\mathfrak{so}(q-2,q)$), if the highest weight of an irrep is stable under Galois (here: similar to its conjugate) at all, then the parameter gives out the trivial class in the Brauer group (here: $\mathbb R$, not $\mathbb H$, i.e. case 3 from part A cannot occur at all). I hope the computation in part B verifies this. It could then be amended to the calculations in the answer to $SO(p,q)$ Fundamental Weights?, which deal with more (but significantly different) irreps of $\mathfrak{so}_{p,q}$.


问:
Determine whether the given points are interior point of the given set
Let $B = \{(x,y) \in \Bbb{R}^2 \mid -1 \le x \lt 2, 0 \lt y \le 2 \}
\cup \{(x,y) \in \Bbb{R}^2 \mid 5 \lt x \le 7, y = 1 \}$. Determine whether a point $(0,1)$
is an interior point of $B$.

I got a little bit confuse here, since a point $(0,1)$ exactly lies on vertical line $y=1$, in which it does mean that $(0,1)$ be an interior point of not.

Here's my attempt:

Fix $0 < r = \frac{1}{2}$. Note that
\begin{equation*}
B((0,1), \frac{1}{2}) = \{(x,y) \in \Bbb{R} \mid x^2 + (y-1)^2 < \frac{1}{4} \}.
\end{equation*}
Let $(x,y) \in B((0,1), \frac{1}{4})$. Then,
\begin{align*}
x^2 + (y-1)^2 < \frac{1}{4} \\
x^2 < \frac{1}{4} \wedge (y-1)^2 < \frac{1}{4} \\
-\frac{1}{2} < x < \frac{1}{2} \wedge -\frac{1}{2} < y-1 < \frac{1}{2} \\
-\frac{1}{2} < x < \frac{1}{2} \wedge \frac{1}{2} < y < \frac{3}{2}.
\end{align*}

Hence, $x \in (-\frac{1}{2}, \frac{1}{2}) \subseteq [-1,2)$ and $y \in (\frac{1}{2}, \frac{3}{2})
\subseteq (0,2]$. Thus, forall $(x,y) \in B((0,1), \frac{1}{2})$, we have $(x,y) \in B$.
In another words, $B((0,1), \frac{1}{2}) \subseteq B$. Therefore, there exists $r>0$
such that $B((0,1), \frac{1}{2}) \subseteq B$. Hence, $(0,1)$ is an interior point of $B$.

Am I true? If not, any explain how to show it?
Thanks in advanced.

答: Your proof is fine, but verbose. You could  also note that $U=(-1,2) \times (0,2)$ is open (open rectangle) and a subset of $B$ so that any point of $U$ is an interior point of $B$, including $(0,1)$.


问:
Exact definition of SAT problem in S4
I kind of know what SAT problem is and I need the exact formal description of this problem for S4 logic.
I am trying to prove this problem is PSPACE-Complete!
Any help would be very much appreciated!</p>

答:
Welcome to MSE!

In general, if we are given a logic $\mathcal{L}$ (which includes a notion of semantics), we can ask the $\mathcal{L}\mathsf{SAT}$ problem1:

If you give me a $\mathcal{L}$-formula $\phi$, is there a model $\mathfrak{M}$ which thinks that $\phi$ is true?

A solution for $\mathcal{L}\mathsf{SAT}$ takes in a formula $\phi$ as input and either outputs a model $\mathfrak{M} \models \phi$ or outputs a ($\mathcal{L}$-)proof that $\phi$ is not satisfiable.

So in your case, a solution to the $\mathsf{S4~SAT}$ problem takes in a sentence
in $\mathsf{S4}$ and either outputs

a (reflexive and transitive) model $\mathfrak{M}$ with a world $w \in \mathfrak{M}$ so that $(\mathfrak{M},w) \models \phi$

an $\mathsf{S4}$ proof of $\lnot \phi$

The classic reference for proving $\mathsf{PSPACE}$ hardness of this problem is
Ladner's paper "The computational complexity of provability in systems of modal propositional logic".

1: There is definitely a law school joke to be made here...

I hope this helps ^_^

Title: Exact definition of SAT problem in S4


问:
Inner product Proof,
So $V$ is an inner product space (finite dimensional) with inner product defined.

If $v$ and $w$ are vectors in $V$, how would one go about proving this?

$\langle \phi_\beta (x), \phi_\beta (y) \rangle ' = \langle [x]_\beta, [y]_\beta \rangle ' = \langle x, y \rangle$

The $\langle \rangle '$ is the standard inner product on $F^n$.
$\beta$ is an orthonormal basis for V.

Attempt:
It seemed to me that the first half of the equality would be obvious, because $\phi_\beta (x) =[x]_\beta$ and $\phi_\beta (y) =[y]_\beta$, or at least that is how I know  $\phi_\beta$ to be defined. I don't know if that's rigorous though.

答:
Probably this is what you are looking for. If $ \beta = \{e_1,e_2,...,e_n\} $ then for all $ v \in V $, you have $ v = \sum_i \langle v,e_i\rangle e_i$ and $ [v]_\beta = (\langle v,e_1\rangle,\langle v,e_2\rangle,.....,\langle v,e_n\rangle) \in \mathbb{F}^n $ Hence immediately using $ \langle e_i,e_j\rangle = \delta_{ij} $ you obtain
$$ \langle x,y\rangle = \sum_{i=1}^n \langle x,e_i\rangle\langle y,e_i\rangle = \langle [x]_\beta,[y]_\beta\rangle' $$</p>

Title: Inner product Proof,


问:
Confusion with the concept "finite almost everywhere"
My issue is this,

say $f: \Omega \mapsto \mathbb{R}$ is a measurable function, for which $\int f d\mu$ exists. What does it then mean to say that $f$ is finite almost everywhere?

Some material I looked at, suggests that it is same as implying that the set $\{ x\in \Omega: |f(x)| = \infty\}$ is contained in a null set, this however is meaningless isn't it ? since $f$ is never going to take the value $\pm \infty$, so no such $x$ can be in $\Omega$.

Then I thought that we're working with the extended real line, $\mathbb{\bar R}$, but now, something like $f(x) = \infty$ is integrable, but never finite.

Sorry if this was too long, where am I wrong?

答:
If $f$ takes values in $\mathbb  R$ the $f$ is finite everywhere, hence also almost everywhere whether or not $\int f d\mu$ exists.
Suppose $f$ takes values in $[-\infty, \infty]$. Then we say that $\int f d\mu $ exists if either $f^{+}$ or $f^{-}$ is integrable. This does not guarantee that $f$ is finite almost eveywhere. For example we can have $f(x)=\infty$ for all $x$.

If $\int |f| d\mu$ is finite then $f$ is finite almost everywhere: $\int |f| d\mu \geq \int_{\{f=\infty\}} |f| d\mu$ so we must have $\mu \{x: f(x)=\infty\} =0$. Similarly, $\mu \{x: f(x)=-\infty\} =0$ and $\mu \{x: f(x) \notin \mathbb R\}=0$.


问:
Finding the number of "good functions"
Consider $A=\{1,2,3,4,5,6,7\}$ and $B=\{1,2,3,4,5,6,7,8,9,10,11,12,13\}$ we call the function $f: B\to A$ "good function" if we have: for every element of $A$ like $x$ be at least one element of $B$ like $y$ so that $f(y)=x$ and $f(1)\le f(2)\le \ldots f(13)$. what is the number of good functions?

$$1)924\quad\quad\quad\quad\quad\quad 2)1296\quad\quad\quad\quad\quad\quad3)3888\quad\quad\quad\quad\quad\quad4)7560\quad\quad\quad\quad\quad\quad5)8440$$

To solve this problem I translated it as this problem:

Considered the digits of $f(1),f(2),\ldots,f(13)$ make a $13$ digits number so that this number contain all the digits $1,2,\ldots,7$ and its first digit is $1$ and last digit is $7$ also the digits are increasing. how many thirteen digits numbers have this condition?

So we know the first and last digits therefore eleven digits left. I am not sure how I should solve this combinatoric problem, but I think I should consider the number $1234567$ and add $6$ more digits between it so that the digits be increasing. I don't know how to count the number of the cases.

答:
There's an easier way to do this. When you pick a good function, you're partitioning the numbers from $1$ to $13$ into $7$ nonempty blocks. You can think of these blocks as being separated by walls, so for example $1|2\;3\;4|5|6\;7|8\;9\;10|11|12\;13$ represents a good function, where the numbers in the first block map to $1$, the numbers in the second block map to $2$, etc.

In other words, you need to choose the number of ways you can place the $6$ walls that divide the elements of $B$ into $7$ blocks. The walls need to go between elements of $B$ (if a wall goes to the left of $1$, or to the right of $13$, then the leftmost or rightmost block will be empty), and no two walls can go between the same elements of $B$, otherwise the block enclosed by those walls will be empty. Hence, you must choose $6$ places to put the walls among the $12$ places possible, so the answer is ${12\choose 6}=924$.

Title: Finding the number of "good functions"


问:
Problem in Understanding Baby Rudin : Chapter 2 : Basic Topology, page number : 28, 2.10(b)
I'm reading the "Baby Rudin" book and one particular example (which is very trivial, I guess but) I couldn't understand properly. I'm stating the problem and the answer here.

I want the visual and logical understanding of this problem.

[Chapter 2 : Basic Topology, page number : 28, 2.10(b)] :

Let $A$ be the set of real numbers such that $0 < x \leq 1$. Fro every
$x \in A$, let $E_{x}$ be the set of real numbers $y$ such that $0 <y <x$. Then

(i) $E_{x} \subset E_{z}$ if and only if $ 0 <x \leq z \leq 1$;

(ii) $\bigcup_{x \in A} E_{x} = E_{1}$;

(iii) $\bigcap_{x \in A} E_{x}$ is empty;

(i) and (ii) are clear. To prove (iii), we note that for every $y>0$,
$y \notin E_{x}$ if $x<y$. Hence $y \notin \bigcap_{x \in A} E_{x}$.

My (wrong?) undertsanding :

(i) From the concept of equality it's obvious. Am I right? What's the proper explanation of this?

(ii) How it's $\bigcup_{x \in A} E_{x} = E_{1}$? Didn't understand.

(iii) For this one with the given hint, I tried to think as points in the real line $x$, but I can't convince myself enough with imagination and logic.

P.S : There are already some answers on this same topic (but not on all 3 points) , but I need the answers in more details in all 3 points, like visually with some sketches in real line, I hope this question will not get duplicated.

答:
Let me try to offer some explanation, unfortunately without sketches, but I hope it will help nonetheless.

(i) The set $E_x$ are all real numbers $y$ such that $0<y<x$. This is the same as the open interval $(0,x),$ so it's really easy to draw: If, say, $x = \frac{1}{2}$ then it's just the line from $0$ to $\frac{1}{2}$ (without the endpoints).

Now the statement of (i) may be more clear: When is a line from 0 to $x$ part of the line from 0 to $z$? Well, the line to $z$ must be at least as long, i.e. $x\leq z$ must be true. But of course $x$ and $z$ must be between 0 and 1 (by the definition of the sets $E_x$ and $E_z$), so finally we get:

$$ E_x \subseteq E_z \iff 0<x\leq z\leq1$$

(ii) Let now $x = 1$, so $E_x = E_1$ is the open interval $(0,1)$. For any other $z \in A$, the open interval $(0,z) = E_z \subseteq E_1$ by (i). Now if we take the union $\bigcup_{x \in A}E_x$ there is not much happening: since $E_1$ is in this union, and all other $E_z \subseteq E_1$, the union is just going to end up to be $E_1$ itself. You simply don't add any points to $E_1$ that are not already in $E_1$.

(iii) To see that the intersection is empty, it is maybe easiest to 'try' to put an element into the intersection and see that it is impossible. What I mean by that is the following:

Let's say $\bigcap_{x \in A} E_x$ is not empty, so it contains some number $y$. We have that $0<y$, just because that's the way we defined our set $A$ (it doesn't include the point 0). Then let us define $x = \frac{y}{2}$. But now, the point $y$ is not in the interval $(0,x) = E_x$ because it is simply too big. (You can think of an explicit example, like $\frac{1}{2}$ is not in the interval $(0, \frac{1}{4})$.)

But $y \in \bigcap_{x \in A} E_x$ means that $y$ must be in all of the $E_x$, which it is not since we explicitly found one that does not contain $y$. So contrary to our assumption, $y \notin \bigcap_{x \in A} E_x$. Since we chose $y$ arbitrarily, this shows that no point can be in $\bigcap_{x \in A} E_x$ which therefore must be empty.

I hope this sheds some light into the matter!


问:
If $U$ is dense in $H$, $(e_i)$ is an orthonormal basis of $U$ and $⟨u,e_i⟩_U=λ_i⟨u,e_i⟩_H$, is $(\sqrt λ_ie_i)$ an orthonormal basis of $H$?
Let $U,H$ be complex Hilbert spaces, $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $U$, $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ be nondecreasing and $$f_n:=\sqrt{\lambda_n}e_n\;\;\;\text{for }n\in\mathbb N.$$

Assume that $U$ is compactly and densely embedded into $H$ and $$\langle u,e_i\rangle_U=\lambda_i\langle u,e_i\rangle_H\tag1\;\;\;\text{for all }u\in U\text{ and }i\in\mathbb N.$$ Are we able to conclude that $(f_n)_{n\in\mathbb N}$ is an orthonormal basis of $H$?

We would need to show that if $x\in\{f_i:i\in\mathbb N\}^{\perp_H}$, then $x=0$. Since $U$ is densely embedded into $H$, there is a $(u_n)_{n\in\mathbb N}$ with $$\left\|u_n-x\right\|_H\xrightarrow{n\to\infty}0\tag2.$$ So, $$0=\langle x,f_i\rangle_H=\lim_{n\to\infty}\langle u_n,f_i\rangle_H=\frac1{\sqrt\lambda_i}\lim_{n\to\infty}\langle u_n,e_i\rangle_U\tag3.$$ But I'm not able to proceed from here ...

答: Let $E:=span\{ e_i:i\in \mathbb N\}$. Then by assumption $E$ is dense in $U$ with respect to $\|\cdot\|_U$. Since $U$ is continuously embedded into $H$, it follows $E$ is dense in $U$ with respect to $\|\cdot\|_H$. By assumption $U$ is dense in $H$ with respect to $\|\cdot\|_H$. This implies $E$ is dense in $U$ with respect to $\|\cdot\|_H$. And $E^{\perp H}=\{0\}$.


问:
First Homology Group and Abelianization
On the Wolfram Mathworld article for Commutator Subgroup, it states that the first homology group is the abelianization, $$H_{1}(G) = G \big/ [G,G]$$
which totally blows my mind because I've only seen the commutator subgroup in the context of Lie algebra representation theory, and I've only seen the first homology group in the context of simplicial homology. This leads me to the following question,

Can you give an explanation (rigorous, or intuitive) for why the first homology group is the abelianization? Furthermore, if you happen to know of how simplicial homology may be related to Lie algebra representation theory (or know of a reference regarding it), please tell!

答:
$H_1(G)$ is the first homology group of the corresponding Eilenberg-Maclane space $K(G,1)=X$ (which is by  definition the space X such that $\pi _1(X)=G$ and all other homotopy groups of X are trivial). Denote $X=K(G,1)$ Then your question reduces to the following

$H_1(X)=\pi_ 1(X)/[\pi _1(X),\pi _1(X)].$

This is a standard algebraic topology fact which you can find in ant standard Algebraic topology textbook (e.r. Algebraic topology by Hatcher Theoem 2A.1)

Title: First Homology Group and Abelianization


问:
A very basic question about geometric construction

I am reading Dummit and Foote's section on the geometric construction,
the author claims that the two triangles gives you $ab$ and $\frac{a}{b}$, but I really don't see how this gives you $ab$ and $a/b$.

Can someone give me some explanations?
I am really bad at this kind of geometry on the plane.

答:
Given a line segment of length $a$ and one of length $b$, then you can construct $ab$ by drawing an (almost) arbitrary angle, marking off a distance of $1$ and $a$ on one ray, and $b$ on the other.  Draw a line segment forming a triangle with legs $1$ and $b$.  Then draw a line parallel to this line segment through the endpoint of $a$, intersecting the other ray and forming a similar triangle.  Then by similarity of triangles, the length cut off from the other ray is $ab$.

The same principle applies to the second diagram, just in reverse.  In both cases, you are drawing a triangle with specified side lengths, then constructing a parallel line to make a similar triangle whose corresponding side has the desired product or ratio.


问:
Why is a triangle a rigid shape?
Triangles are used in structures because they are the most rigid shape. And they are many descriptions of this on network and, if I am not mistaken, this all comes down to the law of cosines.But why this holds? Is it maybe because if we connect three random points on a planes their internal angles will add to half revolution? Is there a more fundamental reason for the rigidness of triangles?

Engineer here, please don't mind the naive question

答:
By definition, a shape is rigid when the distances between its points is constant. In a triangle there are three pairs of points and if you fix the side lengths the shape is perforce rigid.

In a quadrilateral, there are six distances, but is is enough to fix five of them because they define two triangles, each rigid. Any point that you add takes two extra fixed distances.

If you consider the average number of distances per vertex,

$$\frac{2n-3}{n}$$ this is minimized by $n=2$ (a rigid bar). But a bar is a 1D stucture. Hence the minimum for a 2D structure is achieved by the triangle.

In 3D, that ratio is

$$\frac{3n-6}{n}$$ and the minimum (3D) is achieved by the terahedron. The first coefficient in the numerator is the number of dimensions $d$, and the second is a... triangular number, $T_d$.

Title: Why is a triangle a rigid shape?


问:
Financial math- Calculating Perpetuity problem

A investor is hesitating between two projects. The first will yield steady
returns of $X$ every $6$ months for the first $10$ years and $X$ every year
after. The second will return $500$ per month for $5$ years, then will
yield $500$ per year in perpetuity. If the effective annual rate is $10$%,
for what $X$ would the investor be indifferent between the two projects?

I am having trouble starting off the problem.

So far I understand that the cash flows for the first project is $X$ for every $6$ months (semi-annual), for the first $10$ years that means $2(10) = 20$ payments for compounding semi-annually, and $X$ every year going on to infinity. This has "semi-annual" interest so $\frac{0.1}{2} = 0.05$

The second project has a cash flow of $500$ per month for $5$ years so it will be $5(12)=60$ months for compoundings and then there is the $500$ per year in perpetuity (goes to infinity). This project has "monthly" interest rate of $\frac{0.1}{12} = 0.833333$

As for timeline(s), I'm guessing you need two? The first project would have a line going from $0$ to $10$ years with 20 intervals of X, and then infinity years afterward. While the 2nd project would have a line from $0$ to $5$ with subintervals of 12 months for sixty months of 1000 each, then after 5 years is 1000 per year to infinity.

答:
If a payment is made and compounded m times the effective interest rate is $r=(1+i_m)^m-1$. We can solve the equation for $i_m=\sqrt[m]{r+1}-1$. That means in the case of semianully payments $i_2=\sqrt[2]{1.1}-1\approx 0.0488088$. And in the case of monthly payments $i_{12}=\sqrt[12]{1.1}-1\approx 0.00797414$

The present value of $n$ payments in the amount of $X$ and a interest rate of $i_m$ (in one period) is

$$PV_n=X\cdot \frac{(1+i_m)^n-1}{i_m\cdot (1+i_m)^n}$$

The present value of a perpetuity in the amount of $X$ and a interest rate of $i_m$ (in one period) is

$$PV_{\infty}=X\cdot \frac{1}{i_m}$$

This preliminary considerations lead me to the following equation:

$$X\cdot \frac{1.0488088^{20}-1}{0.0488088\cdot 1.0488088^{20}}+\frac{X}{0.1\cdot 1.0488088^{20}}$$ $$=500\cdot \frac{1.007974^{60}-1}{0.007974\cdot 1.007974^{60}}+\frac{500}{0.1\cdot 1.007974^{60}}$$

I get $X\approx 1634.22$


问:
Is a bounded sublinear operator continuous?
We know that, if $T$ is a linear operator between normed spaces $X$ and $Y$, then $T$ is continuous $\iff$ $T$ is bounded.

I was wondering if the result holds when $T$ is a sublinear operator. It seems that the proof used for linear operators still works when you use a sublinear operator, yet in every book I read boundedness doesn't seem to be used as synonymous for continuity, and this sounds strange. So I thought maybe I'm not seeing something.

Any help will be appreciated!

答:
The conditions on sublinearity imply
$$
\left| \ \|T(x_1)\| - \|T(x_2)\|\ \right| \le \|T(x_1-x_2)\|.
$$
To see this, use $(f,g):=(x_1-x_2,x_2)$ and $(f,g):=(x_2-x_1,x_1)$
in those conditions.
So if $T$ is bounded then it is continuous at zero. In addition, $x_n\to x$ implies $\|Tx_n\|\to \|Tx\|$. But not much more.

With the following observation it is quite easy to construct a sublinear function:
If there is a linear and continuous $L$ such that $\|Tx\|=\|Lx\|$ for all $x$ then $T$ is sublinear and bounded.

Hence the following mapping $T:\mathbb R\to \mathbb R$ is bounded, sublinear, discontinuous:
$$
T(x) = \begin{cases} x & \text{ if } x\ne 1\\ -1 & \text{ if } x=+1.\end{cases}$$

Another class of sublinear functions into $\mathbb R$ is given by the Minkowski gauge functional of convex sets.


问:
Bounding the difference between consecutive terms in the sequence $a_n=(1+1/n)^n$
I'm pretty sure the following estimate holds for $C>1$, but I can't prove it:

$$a_n:=\left(1+\frac{1}{n}\right)^n-\left(1+\frac{1}{n-1}\right)^{n-1}\leq\frac{C}{n^2}$$

For context, I was looking for some such bound in order to prove that $na_n\to0$.

I tried using the fact that

$$\left(1+\frac{1}{n}\right)^n\leq\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}+\cdots+\color{red}{\frac{1}{n!}}:=S_n$$

because, by some algebra with the binomial theorem,

$$\left(1+\frac{1}{n}\right)^n=\color{red}{1}+\color{red}{1}+\color{red}{\frac{1}{2}}\color{blue}{\left(1-\frac{1}{n}\right)}+\cdots+\color{red}{\frac{1}{n!}}\color{blue}{\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)}$$

where the blue factors are all less than one.

But I also need a good lower bound on the terms $(1+n^{-1})^{n}$, and I can't see how to pick the right lower bound to get, say,

$$a_n\leq S_n-\text{lower bound applied to the term $\left(1+\frac{1}{n-1}\right)^{n-1}$}\leq\frac{C}{n^2}$$

答:
Suppose that $p\geq 1$. Then
$$
\left( {1 + \frac{1}{p}} \right)^p  = \exp \left( {p\log \left( {1 + \frac{1}{p}} \right)} \right) \ge \exp \left( {1 - \frac{1}{{2p}}} \right) \ge e\left( {1 - \frac{1}{{2p}}} \right)
$$
and
\begin{align*}
&\left( {1 + \frac{1}{p}} \right)^p  \le \exp \left( {p\log \left( {1 + \frac{1}{p}} \right)} \right)\le \exp \left( {1 - \left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right)} \right)  \\ & \le e\left( {1 - \left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right) + \frac{1}{2}\left( {\frac{1}{{2p}} - \frac{1}{{3p^2 }}} \right)^2 } \right) \le e\left( {1 - \frac{1}{{2p}} + \frac{{11}}{{24p^2 }}} \right).
\end{align*}
These inequalities follow from the fact that for all $x>0$,
$$
x - \frac{{x^2 }}{2} < \log (1 + x) < x - \frac{{x^2 }}{2} + \frac{{x^3 }}{3},\quad e^{ - x}  < 1 - x + \frac{{x^2 }}{2}.
$$
In the last step, I re-arranged the terms and observed that when multiplied by $p^2$, the sum of the terms beyond $-\frac{1}{2p}$ tend monotonically from below to $\frac{11}{24}$. Thus, for all $n\geq 2$,
\begin{align*}
\left( {1 + \frac{1}{n}} \right)^n  - \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1} & \le e - \frac{e}{{2n}} + \frac{{11}}{{24n^2 }} - e + \frac{e}{{2(n - 1)}}
\\ &
 = \frac{e}{2}\frac{1}{{n(n - 1)}} + \frac{{11}}{{24n^2 }} \le \frac{{24e + 11}}{{24}}\frac{1}{{n^2 }} < \frac{4}{{n^2 }} .
\end{align*}
Addendum: By the mean value theorem
\begin{align*}
&\left( {1 + \frac{1}{n}} \right)^n  - \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1}  = \left( {1 + \frac{1}{\xi }} \right)^\xi  \left( {\log \left( {1 + \frac{1}{\xi }} \right) - \frac{1}{{\xi  + 1}}} \right)
\\ &
 \le e\left( {\log \left( {1 + \frac{1}{\xi }} \right) - \frac{1}{{\xi  + 1}}} \right) \le \frac{e}{{2\xi ^2 }} \le \frac{e}{{2(n - 1)^2 }},
\end{align*}
where $n-1<\xi<n$ is a suitable number.


问:
Does $\mathbb{R}^2$ Contain Uncountably Many Disjoint Copies of the Warsaw Circle?
The Warsaw Circle is defined as the closed topologist's sine curve, with an additional arc attached at its free end point and one of the end points of the critical line:

Since we don't have an uncountable collection of disjoint open sets in the plane - such as the bounded component of its complement - we'll have to have an uncountable sequence $W_r$ such that if $r < s$ then $W_r$ is contained in the bounded component of $W_s^c$.  Trying to draw it, it's actually kind of hard for me to tell for sure if it's true or not; it LOOKS true, but jeeze I think showing it with analytic, coordinate representations of each one is gonna be a bear.

Anyone really feeling it?

To be honest, I don't know the answer for the closed topologist's sine curve itself, either!  Probably both are possible or both are impossible; have never seen it proved for either space.

答:
Yes, there exists such a collection.  Any arc-like or circle-like planar continuum (all arc-like continua are planar) admits a mutually disjoint, uncountable collection of copies of itself in the plane.  The proof is a bit long but elementary: https://link.springer.com/article/10.1007/BF01696773

Logan Hoehn has given an example of a tree-like continuum which is not arc-like and also admits uncountably many mutually disjoint copies in the plane.  It's harder but still readable: https://uts.nipissingu.ca/loganh/CopiesOfX.pdf

A classification of continua which admit such embeddings is not known, and there isn't any generally accepted "main conjecture" on the problem.  There's a broad feeling that if current tools are sufficient for such a classification then it will center around one of the various notions of 'span.'


问:
Number of possible outcomes when rolling indistinguishable dice?
So I know the answer is

$$N(n) = \frac{( n + 5 )!}{ n! \cdot 5! }$$

where $n$ is the number of dice.

but I'm not able to prove it either mathematically or intuitively,
although I'm able to verify for various cases.

I tried to prove it by first assuming the dice are different, and then eliminating duplicates but that method doesn't look very promising.

How do I prove the above result.

答:
Each outcome can be represented as a tuple of $6$ nonnegative numbers indicating how many times each side appeared (the numbers must sum to $n$). For instance, $(1,3,2,0,1,0)$ represents a roll of seven dice where "1" appears once, "2" appears three times, "3" appears two times, "4" appears zero times, "5" appears once, and "6" appears zero times.

You can count this using the stars-and-bars method: there are $\binom{n+5}{5}$ ways to arrange $5$ bars and $n$ stars in a row.


问:
Subsets - is it possible or not?
This question is taken from a coding practice problem.

Given $n$ numbers, you can perform the following operation any number of times : Choose any subset of the numbers (possibly empty), none of which are $0$. Decrement the numbers in the subset by $1$, and increment the numbers not in the subset by $K$.

Is it possible to perform operations such that exactly $n - 1$ numbers become $0$?

答:
Let $(a_{rk})_{1\le r\le n}$ be the set of numbers in the $k$-th step of the game, and let $S_k\subseteq[n]$ be the set of numbers which are chosen in the $k$-th step. Then by definition, $a_{r(k+1)}=a_{rk}-1$ if $r\in S_k$ and $a_{r(k+1)}=a_{rk}+K$ otherwise.

Now define $b_{rk}=(a_{rk}+k)\bmod (K+1)$. Then if $r\in S_k$,

$$b_{r(k+1)}=(a_{rk}-1+k+1)\bmod (K+1)=(a_{rk}+k)\bmod (K+1)=b_{rk}$$

and if $r\notin S_k$,

$$b_{r(k+1)}=(a_{rk}+K+k+1)\bmod (K+1)=(a_{rk}+k)\bmod (K+1)=b_{rk}$$

so the $b_{rk}=b_r$ are invariant under any number of moves of the game. Since in the final state of the game, $b_r=k\bmod (K+1)$ is the same for $n-1$ of the positions, any solvable game must have, if you evaluate $b_{rk}$ in the initial state, that at least $n-1$ of them are the same.

This gives a necessary, but maybe not sufficient, condition for the game to be solvable.

A minimal unsolvable game: $K=2$, $n=3$, $\langle a_{r1}\rangle=\langle1,2,3\rangle$.

Edit: Actually, this is a sufficient condition as well. Suppose we know, as stated above, that all except one of them (let's call this one $R$) has equal $b_r$. Let $T=[n]\setminus\{R\}$ (which is to say, the set of all indexes except for $R$). These are the indexes we will try to force to $0$ (and $R$ is the unlucky one left out). Our algorithm is to select on step $k$ the set $S_k=\{r\in T\mid \forall s\in T\ a_{sk}\le a_{rk}\}$, that is, the set of all maximal $a_{rk}$ (except $a_{Rk}$, which we don't care about). If in step $1$ the maximum element in $T$ has value $m$, we will finish in step $m+1$.

The reason is because in each step, the maximum elements in $T$ go down by $1$, and I claim that the maximum value of all elements in $T$ also goes down by $1$. Clearly this depends on the non-maximal elements not shooting past the current maximal elements when they go up by $K$, but this follows from the assumption that $b_r$ is the same for all $r\in T$: this means that for any $r,s\in T$, $a_{rk}\equiv a_{sk}\pmod{K+1}$, so if $a_{sk}<a_{rk}$, then $a_{rk}-a_{sk}\ge K+1$, so the non-maximal elements will go up by $K$ and the maximal elements will go down by $1$, but since the non-maximal elements are at least $K+1$ less than the maximal ones, they will not exceed the new maximum, so as a result the maximum over all elements in $T$ goes down by $1$.

This essentially completes the proof. After each step, the maximum goes down by $1$ until it is $0$ and we are done.

As an example of the operation of the algorithm, let's use the OP's original example, with $K=2$ and $\langle a_{r1}\rangle=\langle2,3,5\rangle$: First we identify the odd one out. $\langle b_r\rangle=\langle0,1,0\rangle$, so the game is solvable (since all but one is $0$), and $R=2$. Thus we should be done on step 6.

$$\begin{array}[lll]{}
\langle a_{r1}\rangle=\langle2,3,5\rangle & \quad & S_1=\{3\} \\
\langle a_{r2}\rangle=\langle4,5,4\rangle & \quad & S_2=\{1,3\} \\
\langle a_{r3}\rangle=\langle3,7,3\rangle & \quad & S_3=\{1,3\} \\
\langle a_{r4}\rangle=\langle2,9,2\rangle & \quad & S_4=\{1,3\} \\
\langle a_{r5}\rangle=\langle1,11,1\rangle & \quad & S_5=\{1,3\} \\
\langle a_{r6}\rangle=\langle0,13,0\rangle
\end{array}$$

(Note that if the initial condition contains negative numbers, this algorithm should be preceded by a "waiting" period where $S_k=\emptyset$ is repeatedly selected until all numbers are nonnegative.)

Title: Subsets - is it possible or not?


答:
Let $n_1,\cdots,n_6$ be respectively the numbers of dice showing up $1,\dots,6$. We are looking for the  number of non-negative solutions to the equation : $$n_1+\cdots+n_6=n.$$
Using stars and bars it is:
$$
\binom {n+6-1}{6-1}.
$$

Title: Number of possible outcomes when rolling indistinguishable dice?

Body:
So I know the answer is

$$N(n) = \frac{( n + 5 )!}{ n! \cdot 5! }$$

where $n$ is the number of dice.

but I'm not able to prove it either mathematically or intuitively,
although I'm able to verify for various cases.

I tried to prove it by first assuming the dice are different, and then eliminating duplicates but that method doesn't look very promising.

How do I prove the above result.


问:
In a von Neumann regular commutative ring with unity every finitely generated ideal is principal

Let $R$ be a commutative, von Neumann regular ring with unity. How to show that every finitely generated ideal in $R$ is principal?

I can see, in view of mathematical induction, it suffices to show that any ideal generated by two elements of $R$ must be principal.

Let $I=(a,b)$ be an ideal of $R.$ Since $I$ is commutative with unity, $I=\{xa+yb:x,y\in R\}.$ Also since $R$ is regular there exist $r,s\in R$ such that $a=ara=ra^2$ and $b=bsb=sb^2.$

However I cannot figureout which element would generate $I.$ Please help.

答:
Since $ara=a$, the element $e=ar$ is idempotent (i.e., $e^2=e$) and generates the same ideal as $a$.

Similarly, $f=bs$ is idempotent and generates the same ideal as $b$.

These idempotent elements are easier to deal with than $a$ and $b$, and in particular
$$e+f-ef = ar + bs -arbs$$
is another idempotent element that generates the ideal $\langle e,f\rangle=\langle a,b \rangle$,
since $e(e+f-ef)=e$ and $f(e+f-ef)=f$.


问:
Orthonormal basis for the set of polynomials at most degree $n$

Let $\langle \cdot, \cdot\rangle \colon P_n \times P_n \to \mathbb{R}, \langle p, q\rangle = a_nb_n + \dots+a_1b_1+a_0b_0$ where $P_n$ is the set of all polynomials at most degree $n$ and $p(x) = a_nx^n + \dots +a_1x+a_0,q(x) = b_nx^n+\dots+b_1x+b_0$. Show that for every $j \in \mathbb{N}$ the basis $(p_0, \dots,p_n)$  defined by $p_j : \mathbb{R} \to \mathbb{R}, x\mapsto x^j$ is a orthonormal basis in the inner product space defined by $\langle \cdot, \cdot\rangle \colon P_n \times P_n$.

It seems that I want to show that $(x^0, x^1,\dots,x^n)$ is an orthonomal basis in $\langle \cdot, \cdot\rangle \colon P_n \times P_n$. I would need to show that $$\langle p_j, p_i\rangle = 0 \text{ for all $j\ne i$} \\ \|p_j\| = 1 \text{ for all $j\in \{1, \dots, n\}$}$$

The first condition seems trivial since the polynomials $(x^0, x^1,\dots,x^n)$ doesn't have any coefficients and $\langle \cdot, \cdot\rangle$ was defined to be the sum of the coefficients of the two input vectors. The second condition confuses me a bit if $\|p_j\| = \sqrt{(x^j)^2} = x^j$ then how can I be sure that this evaluates to $1$?

答: $x^{i}=\sum\limits_{k=1}^{n} a_kx^{k}$ where $a_k=0$ if $ k \neq i$ and $a_i=1$. Once you realize this it follows immediately that $ \langle p_i, p_j \rangle =0$ for $i \neq j$ and  $\|p_i\|=1$  for all $i$. [In the sum defining the inner product of $p_i$ with itself  one term is $1$ and all other terms are $0$].


问:
Pages 6 and 27 are on the same (double) sheet of a newspaper, how many pages are there in the news paper altogether
I was never really good at maths, trying to get back into it.

I have this question:

 Pages 6 and 27 are on the same (double) sheet of a newspaper, what are the page numbers on opposite sides of the sheet and how many pages are there in the newspaper altogether

This seems like a trick question... I am not too sure how to take the logic, but I suppose pages 5 and 28 are the numbers on opposite sides of the page?

How do I get the total pages of the newspaper altogether? I am assuming we add the missing pages, 1 thru 4 (4 pages in total) to 27 to get 31?

Not sure how to solve this, and I have no answer key.

答:
Assume the paper is formed by taking a pile of (printed) sheets of paper and folding it in half.

Look at the outside cover - this pairs up the first page and the last page on the outside. If there are $n$ pages altogether, the sum is $n+1$.

The other side of the same sheet has pages $2$ and $n-1$ - sum $n+1$.

In fact it is easy to see that the two pages on the same side of any of the original pieces of  paper have numbers which add to $n+1$.

Here $n+1=33$ so $n=32$ - and the paper consists of $8$ folded sheets.


问:
Are zero knowledge proofs probabilistic?
Does the interaction between prover and certifier yield a 100% certain proof? If I understood correctly this procedure is based on iterative checks by the certifier which keeps trying until it's satisfied  with the reliability of the certificates. So it can be satisfied to arbitrarily high degree but not fully satisfied. Did I get it right?</p>

答:
Yes, they are probabilistic in nature.

As quoted from wikipedia:

Zero-knowledge proofs are not proofs in the mathematical sense of the term because there is some small probability, the soundness error, that a cheating prover will be able to convince the verifier of a false statement. In other words, zero-knowledge proofs are probabilistic "proofs" rather than deterministic proofs. However, there are techniques to decrease the soundness error to negligibly small values.

With more iterations, the probability of cheating decreases but it is not really a proof. It is not $100\%$.


问:
Finding a function as a solution to a system of integral equations
In my research, I reached a point where I need to find a nonnegative function, that satisfies the following:

For any $\phi\in[0,\pi]$:

$[\int^\pi_\phi f(\theta)d\theta][ \int^{\pi-2\phi}_{-\phi} f(\theta)d\theta]+[\int^0_{\pi-\phi} f(\theta)d\theta ][\int^{2\phi}_{\pi+\phi} f(\theta)d\theta]-[\int^\phi_0 f(\theta)d\theta]^2-[\int^{\pi-\phi}_\pi f(\theta)d\theta]^2 = \pi^2\cos^2\frac{\phi}{2}$

I would be very glad if somebody can help me, and thank you in advance for your help!

答:
I think that condition can be satisfied only when $\sin^2\frac{\phi}{2}=\frac{1}{2}$.  However, I suggest a possible, slight modification of the constraint, so that a solution can be found, see later.

Let us start by writing $f(\theta)$ as a Fourier series involving only cosines (which are even functions):

\begin{equation}
f(\theta)=\sum_{n=0}^{+\infty}f_n\cos(n\theta)
\end{equation}

Let us impose the normalization:

\begin{equation}
\int_{0}^{\pi}d\theta \,\,f(\theta)=\sum_{n=0}^{+\infty}f_n\int_{0}^{\pi}\cos(n\theta)=\sum_{n=0}^{+\infty}f_n\frac{\sin{(\pi n)}}{n} \stackrel{!}{=}\frac{\pi}{2}
\end{equation}

this expression vanishes except when $n=0$, in that case
\begin{equation}
\frac{\sin{(\pi n)}}{n}=\pi
\end{equation}
or, in other words, $f_0=\frac{1}{2}$. Hence,

\begin{equation}
f(\theta)=\frac{1}{2}+\sum_{n=1}^{+\infty}f_n\cos(n\theta)
\end{equation}

Now we notice that, for $n\neq 0$,
\begin{equation}
\int_{0}^{\pi}d\theta\,\,\cos\left(n(\theta-\phi)\right)=\frac{\sin(n\phi )}{n}[1-(-1)^n],
\end{equation}

\begin{equation}
\int_{0}^{\pi}d\theta\,\,\cos(n\theta)=0
\end{equation}

Hence the first terms become

\begin{equation}
\int_{0}^{\pi}d\theta\,\,f(\theta)+\int_{0}^{\pi}d\theta\,\,f(\theta-\phi)=\frac{\pi}{2}+\frac{\pi}{2}+\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]
\end{equation}

Now let us consider the other term.

\begin{equation}
\begin{split}
\int_0^{\pi}\int_0^{\pi}d\theta dt\,\,\left(\frac{1}{2}+\sum_{n= 1}^{+\infty}f_n\cos(n\theta)\right)\left(\frac{1}{2}+\sum_{m=1}^{+\infty}f_m\cos(m(t-\phi))\right) \\ \\
=\int_0^{\pi}\int_0^{\pi}d\theta dt\,\,\bigg\{\frac{1}{4}+\frac{1}{2}\sum_{n=1}^{+\infty}f_n\cos(n(t-\phi))+\frac{1}{2}\sum_{n=1}^{+\infty}f_n\cos(n\theta)\\ \\
+\sum_{n,m\neq0}f_nf_m\cos(n\theta)\cos(m(t-\phi))\bigg\}
\end{split}
\end{equation}

Let us integrate first over $\theta$. After the integration the previous integral becomes

\begin{equation}
\int_0^{\pi}dt\,\,\left[\frac{\pi}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\cos(n(t-\phi)) \right]=\frac{\pi^2}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]
\end{equation}

Your condition then is

\begin{equation}
\begin{split}
\pi+\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]-\frac{2}{\pi}\bigg\{\frac{\pi^2}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]\bigg\}\\
\\
=\pi\sin^2\frac{\phi}{2}=\frac{\pi}{2}(1-\sin\phi)
\end{split}
\end{equation}

Now you see that the terms involving the sine cancel out, leaving

\begin{equation}
\sin^2\frac{\phi}{2}=\frac{1}{2} \hspace{3mm} \text{or} \hspace{3mm} \sin\phi=0
\end{equation}

However, if you could modify your condition as

\begin{equation}
\begin{split}
\int_0^{\pi}d\theta\,\,f(\theta-\phi)+\int_0^{\pi}d\theta\,\, f(\theta)-\frac{2}{\pi}\xi\int_0^{\pi}\int_0^{\pi}d\theta dt\,\, f(\theta)f(t-\phi)\\
=\pi\sin^2\frac{\phi}{2}+\frac{\pi}{2}(1-\xi)
\end{split}
\end{equation}

(Notice that you initial model has $\xi=1$), then you have

\begin{equation}
\begin{split}
\pi+\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]-\frac{2}{\pi}\xi \bigg\{\frac{\pi^2}{4}+\frac{\pi}{2}\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]\bigg\}\\
\\
=\pi\sin^2\frac{\phi}{2}+\frac{\pi}{2}(1-\xi)=\frac{\pi}{2}(1-\sin\phi)+\frac{\pi}{2}(1-\xi)
\end{split}
\end{equation}

which implies

\begin{equation}
(1-\xi)\sum_{n=1}^{+\infty}f_n\frac{\sin(n\phi )}{n}[1-(-1)^n]=-\frac{\pi}{2}\sin\phi
\end{equation}

This condition is satisfied by setting $f_n=0$ for all $n>1$ and by choosing $f_1$ such that

\begin{equation}
2(1-\xi)f_1\sin\phi=-\frac{\pi}{2}\sin\phi
\end{equation}

from which

\begin{equation}
f_1=\frac{\pi}{4}\frac{1}{1-\xi}
\end{equation}

I hope the modified model could work for you.


问:
Notation for a function definition
I came across the following notation in a research paper

Suppose we have a function $f(x) \in [1,l] \rightarrow \mathbb{R}$

It is the first time, I am looking such a notation. And the paper is using the same notation for all functions used in it.

Is it same as
$f(x) : [1,l] \rightarrow \mathbb{R}$

If yes, is it an abuse of notation?

If no, what does the notation mean?

答:
In the paper you're using, I understand it means that $f$ belongs to the set of all functions defined from the set $[1,l]$ to the set $\mathbb{R}$, to be said,
$$[1,l]\to\mathbb{R} := \{f \text{ function }: f \text{ is defined from $[1,l]$ to $\mathbb{R}$}\}$$
so it works the same way as defining $f$ the usual way, to be said:
$$f\in[1,l]\to\mathbb{R} \equiv f:[1,l]\to\mathbb{R}$$

Title: Notation for a function definition


问:
How do you find the area under a curve only for the area above the x-axis?
Let's say I have function $f(x) = -x^2 + 1$ and I want to find the area between that curve and the $x$-axis

I use definite integrals obviously.

Let's say I use the $-3$ and $3$ as my definite values for my integral, meaning I would be looking at both areas under and over the $x$-axis

Besides changing my definite values, is there any way to have my integral only measure the area above the $x$-axis?

答:
You have to find the area of the following set:

$$S=\{(x,y) \in \mathbb R^2: x \in [-1,1], 0 \le y \le 1-x^2\}.$$

The area of $S$ is given by

$$ \int_{-1}^1(1-x^2) dx.$$


问:
Homotopy equivalence between pointed CW-complexes
Let $f : (X, x_0) \to (Y, y_0)$ a morphism in the category of pointed, connected CW-complexes. Let $A= [1/2, 1]^{+} \wedge X \cup Y \subset C_f$, where $C_f$ is the mapping cone and $X, Y$ are pointed, connected CW-complexes.

I was trying to prove that $A$ is homotopy equivalent to $Y$: I am pretty sure it is the inclusion of $Y$ in $A$ that determines the equivalence, but I have trouble finding an explicit expression of it.

Any hints?

答: $A$ is a copy of the reduced mapping cylinder of $f$ of which $Y$ is a strong deformation retract.


问:
Some problems on locally finite sets
This is a series of problems on locally finite sets with a few I am stuck on. Can the community provide some help on how to proceed?

Definition. Let $(X,\tau)$ be topological space. A set $\mathcal S$ of subsets of $X$ is said to be locally finite in $(X,\tau)$ if each point $x \in X$ has a neighborhood $N_x$ such that $N_x \cap S= \varnothing$ for all but a finite number of $S \in \mathcal S$.

Prove the following statements:

$(\rm i)$ If the set $\mathcal S$ of subsets of $X$ is finite, then $\mathcal S$ is locally finite.

This is trivial.

$(\rm ii)$ If the set $\mathcal S$ is such that every point of $X$ lies in at most one $S \in \mathcal S$, then $\mathcal S$ is locally finite.

I am not sure if this statement is true. Let us assume that $\mathcal S$ is infinite, otherwise, from part $(\rm i)$, it is locally finite. Now $X$ must also contain an infinite number of elements. If $T$ is the indiscrete topology, where the only open sets are the whole set and the empty set, then even if every point of $X$ lies in one $S$, $\mathcal S$ is not locally finite because the whole space is the only open neighborhood of each point, and its intersection with any $S$ is non-empty because they are all part of the whole space, so I think $(\rm ii)$ is not a true statement.

$(\rm iii)$ Let $X$ be an infinite set and $\tau$ the finite-closed topology on $X$. If $\mathcal S$ is the set of all open sets in $(X,\tau)$, then $\mathcal S$ is not locally finite.

Every open set in the cofinite topology has a non empty intersection with every other open set so $\mathcal S$ cannot be locally finite.

$(\rm iv)$ Let $\mathcal S$ be a locally finite. Define $\mathcal T$ to be the set of all closed sets $T=\operatorname{cl}(S)$ for $S \in \mathcal S$. Then $\mathcal T$ is locally finite.

So, this is saying that the closure of each $S \in \mathcal S$ is still locally finite. Is this a proof by contradiction or is a direct proof possible? If a open neighborhood of a point has an empty intersection with all but finite number of $S$, is there a reason to believe this would change if we took the closure of all $S$?

$(\rm v)$ If $\mathcal S$ is an infinite set of subsets of a infinite set $X$, and $(X,\tau)$ is a compact space, then $\mathcal S$ is not locally finite.

Suppose $\mathcal S$ is locally finite. Then for each $x \in X$, there exist a $U_i$ such that $U_i$ has an empty intersection with all but finite number of $S \in \mathcal S$. The set $U_i$ covers $X$, so there exist a finite subcover. How can I show a contradiction from here?

$(\rm vi)$ Let $\mathcal S$ be an uncountable set of subsets of $X$. If $\mathcal S$ is a cover of the space $(X,\tau)$ and $(X,\tau)$ is either a Lindelöf space or a second countable space, then $\mathcal S$ is not locally finite.

Let us assume $X$ is second countable first and also assume $\mathcal S$ is locally finite. Then for each $x \in X$, there exist a $U_i$ such that $U_i$ has an empty intersection with all but finite number of $S \in \mathcal S$. How should I proceed to show a contradiction?

答:
Yes, (ii) is false. There are even counterexamples in $\Bbb R^2$. For $n\in\Bbb N$ let

$$S_n=\left\{\langle x,y\rangle\in\Bbb R^2:x>0\text{ and }y=nx\right\}\,,$$

and let $\mathscr{S}=\{S_n:n\in\Bbb N\}$; then $\mathscr{S}$ is pairwise disjoint, but every open nbhd of the origin meets every $S_n$, so $\mathscr{S}$ is not locally finite at the origin.

For (iv) just note that if $U$ is open, and $U\cap S=\varnothing$, then $X\setminus U$ is a closed set containing $S$, so $X\setminus U\supseteq\operatorname{cl}(S)$, and therefore $U\cap\operatorname{cl}(S)=\varnothing$.

You’ve done most of the work for (v). If $\mathscr{S}$ is locally finite, each $x\in X$ has an open nbhd $U_x$ that meets only finitely many members of $\mathscr{S}$; let $\mathscr{U}=\{U_x:x\in X\}$. $X$ is compact, so there is a finite $F\subseteq X$ such that $\mathscr{U}_0=\{U_x:x\in F\}$ covers $X$. For each $x\in F$ let $\mathscr{S}_x=\{S\in\mathscr{S}:U_x\cap S\ne\varnothing\}$, and let $\mathscr{S}_F=\bigcup_{x\in F}\mathscr{S}_x$. Each $\mathscr{S}_x$ is finite, so $\mathscr{S}_F$ is finite. But every member of $S$ meets at least one member of $\mathscr{U}_0$, so $\mathscr{S}=\mathscr{S}_F$, and $\mathscr{S}$ is therefore finite.

Finally, (vi) is really the same argument. First, though, note that every second countable space is automatically Lindelöf, so it suffices to prove (vi) for Lindelöf spaces. Just imitate the preceding argument: this time you get a countable $F\subseteq X$ such that $\mathscr{U}_0$ covers $X$. Define $\mathscr{S}_x$ for $x\in F$ as before, and likewise $\mathscr{S}_F$. Then $\mathscr{S}_F$ is the union of countably many finite sets, so it is countable, and $\mathscr{S}=\mathscr{S}_F$ just as before.

Title: Some problems on locally finite sets


问:
$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}$ and $\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}{2n\choose k}\frac{1}{4^{n}}$ is?
$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}=\lim_{n\rightarrow\infty}(1+\frac{1}{4^n})^{2n}$

using $(1+x)^n=1+nx+\frac{n(n-1)x^2}{2!}....$

$\lim_{n\rightarrow\infty}=1+\frac{2n}{4^n}+\frac{2n(2n-1)}{4^{2n}2!}....$

$\lim_{n\rightarrow\infty}=\frac{2n}{4^n}=\frac{\infty}{\infty}=\frac{2}{4^n\ln4}$

all terms vanish we are left with $1$ only.

Am I correct that limit is $1$?

答:
If you want to find the limit of $ \sum\limits_{k=0}^{2n} \binom {2n} {k} \frac  1{4^{k}}$ then you can write the sum  as $(1+\frac 1 {4})^{2n}=(\frac 5  4 )^{2n}$ and the limit is $\infty$

However, if you want to find the limit of $ \sum\limits_{k=0}^{2n} \binom {2n} {k} \frac  1{4^{n}}$ then the answer is obviously $1$ since the value is $1$ for every $n$.


问:
Probability that a divisor of $10^{99}$ is a multiple of $10^{96}$
What is the probability that a divisor of $10^{99}$ is a multiple of $10^{96}$? How to solve this type of question. I know probability but  I'm weak in number theory. </p>

答: HINT: $10^n=2^n5^n$, so the divisors of $10^n$ are the numbers of the form $2^k5^m$, where $0\le k,m\le n$. Since there are $n+1$ choices for each of $k$ and $m$, $10^n$ has $(n+1)^2$ divisors. And $10^\ell\mid 2^k5^m$ if and only if $\ell\le\min\{k,m\}$.</p>


问:
Solving $e^{-x} + x e^{-x} + \frac12x^2e^{-x} + \frac16x^3 e^{-x}=0.99$
Equation:

$$e^{-x} + x e^{-x} + \frac12 x^2 e^{-x} + \frac16x^3e^{-x}=0.99$$

I tried simplifying it and got it down to

$$e^{-x}\left(6+6x+3x^2+x^3\right)=5.94$$

although I am not sure how I would solve the $(6+6x+3x^2+x^3)$ portion.

Thanks to Wolfram Alpha I know the roots (solution for $x$), so I am really looking for guidance on how you would approach this equation and how far you can simplify it before having to use a calculator.

答:
Beside numerical method, having noticed that
$$f(x)=e^{-x} \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)-\frac{99}{100}$$ is small when $x=1$ (which is normal since the quantity inside parentheses is smaller than $e^x$), we can build the infinite series
$$f(x)=\left(\frac{8}{3 e}-\frac{99}{100}\right) +\frac 1{6e}\sum_{n=1}^\infty (-1)^{n+1}\frac{n^3-9 n^2+23 n-16 }{n!} (x-1)^n$$ Truncated to some order $p$, it is then the Taylor series to $O\left((x-1)^{p+1}\right)$.

Now we can use series reversion to get
$$x=1-t-t^2-\frac{11 }{6}t^3-4 t^4-\frac{383 }{40}t^5-\frac{364 }{15}t^6-\frac{21481
   }{336}t^7+O\left(t^8\right)$$ where $t=\frac{297 e-800}{50} $.

Using this truncated series would lead to the estimate
$x\sim 0.823274$ while the "exact" solution, obtained using Newton method, is $x=0.823249$.

For sure, adding more terms will give better and better approximations.


问:
Can a piecewise $C^1$ mapping can define an absolutely continuous function
Let $f,g\in C^1([0,1],\mathbb{R}^n)$, and then we define $h$ as a peicewise $C^1$ mapping as follows
$$h(t)=\begin{cases}
f(t) & t\in[0,\frac{1}{2}[\\
g(t) &t\in]\frac{1}{2},1]
\end{cases}
$$
I know that if I define the following function : $x(t)=\int_0^t f(s)ds$ for all $t\in [0,\frac{1}{2}[$ then $x$ is an absolutely continuous function because $f \in L^1([0,1],\mathbb{R}^n)$.

So my question is that can I say that
$$y(t)= \int_0^t h(s)ds,$$
is an absolutely continuous function on $[0,1]$  ?

答: $\int_0^{1} |h(s)|ds=\int_0^{1/2} |f(s)|ds+\int_{1/2}^{1} |g(s)|ds <\infty$. The indefinite integral of any integrable function is absoluetly continuous. So the answer is YES,


问:
Variance of expression
I consider a random variable

$$A_t = \frac{X_t + C}{e^{gt}}$$

we know that $X_t \sim N(0,\frac{e^{2gt}}{gt})$

I want to know mean and variance of $A_t$

mean is easy, since

$$ \mathbb{E}(A_t) = \frac{\mathbb{E}(X_t) + \mathbb{E}(C)}{\mathbb{E}(e^{gt})} $$

which yields in

$$ \mathbb{E}(A_t) = Ce^{-gt}$$, since $ \mathbb{E}(X_t) = 0$ and the rest are constants.

For the variance I tried

$$\mathbb{V}(A_t) = \mathbb{E}(A_t^2)-\mathbb{E}(A_t)^2 $$

Now: $\mathbb{E}(A_t)^2 = C^2 e^{-2gt}$

and

$$\mathbb{E}(A_t^2) = \mathbb{E}(\frac{X_t + C}{e^{gt}})^2 = \mathbb{E}(\frac{X_t^2 +2X_tC+C^2}{e^{2gt}})$$

since $\mathbb{E}(X_t) = 0$ this should be

$$\mathbb{E}(A_t^2) = \frac{C^2}{e^{2gt}} \mathbb{E}(X_t^2) = \frac{C^2}{e^{2gt}} \cdot (\frac{e^{2gt}}{gt} +0)$$

and so

$$\mathbb{V}(A_t) = \frac{C^2}{gt} - C^2 e^{-2gt}$$

but this is wrong due to my solution....

Can you tell me where my mistake is?

答:
The mean looks wrong to me (EDIT: after you edit it's correct)

Using linearity of expectation you get

$$\mathbb{E}[A_t]=C\cdot e^{-gt}$$

As its variance is concerned, simply remember that

$$\mathbb{V}[aX+b]=a^2\mathbb{V}[X]$$

You get

$$\mathbb{V}[A_t]=\frac{1}{gt}$$

Title: Variance of expression


答:
For the first sum we have
$$\begin {align}
S_n&=\frac1{4^n}\sum_{k=1}^n \binom{2n}k\\
&=\frac1{4^n}\times\frac12\left[{2^{2n}+\binom{2n}n}-2\right]\\
&=\frac12\left[1+\binom{2n}n\frac1{4^n}-\frac2{4^n}\right],
\end {align}$$
which implies:
$$
\lim_{n\to\infty}S_n=\frac12.
$$
Similarly the second sum is 1.

The proofs of $$\lim_{n\to\infty}\binom{2n}n\frac1{4^n}=0\quad\text{and}\quad\lim_{n\to\infty}\frac2{4^n}=0 $$
are left as exercises.

Title: $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}$ and $\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}{2n\choose k}\frac{1}{4^{n}}$ is?

Body:
$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}=\lim_{n\rightarrow\infty}(1+\frac{1}{4^n})^{2n}$

using $(1+x)^n=1+nx+\frac{n(n-1)x^2}{2!}....$

$\lim_{n\rightarrow\infty}=1+\frac{2n}{4^n}+\frac{2n(2n-1)}{4^{2n}2!}....$

$\lim_{n\rightarrow\infty}=\frac{2n}{4^n}=\frac{\infty}{\infty}=\frac{2}{4^n\ln4}$

all terms vanish we are left with $1$ only.

Am I correct that limit is $1$?


Title: Probability that a divisor of $10^{99}$ is a multiple of $10^{96}$

Body: What is the probability that a divisor of $10^{99}$ is a multiple of $10^{96}$? How to solve this type of question. I know probability but  I'm weak in number theory. </p>

Body:
As $10^{99}=2^{99}\cdot 5^{99},$

using divisor function 1,2,  the number of divisors is $(1+99)(1+99)$

which are the product of  $$\{1, 2^1,\cdots,2^{98},2^{99}\}\text{ and }\{1, 5^1,\cdots,5^{98},5^{99}\}$$

To be multiple of $10^{96}$ we need to take $$\{ 2^{96},2^{97},2^{98},2^{99}\}\text{ and }\{ 5^{96},5^{97},5^{98},5^{99}\}$$ i.e., there are $4\cdot4=16$ of them

So, the required probability is $$\frac{4^2}{100^2}=\frac1{625}$$


问:
Radius of convergence for series of $\frac{1}{x^2+\cos^2x}$
Given
$$
f(x) = \frac{1}{x^2+\cos^2(x)}
$$
We can use the binomial expansion to rewrite $f(x)$ as
$$
f(x) = \sum_{k=0}^\infty {-1\choose{k}}\frac{\cos^{2k}x}{x^{2k+2}}
$$
Or, in other terms
$$
f(x) = \sum_{k=0}^\infty \frac{(-1)^k \cos^{2k}x}{x^{2k+2}}
$$
If I remember correctly the formula for R.O.C is:
$$
(x+y)^n \Longrightarrow \, -1<\frac{x}{y} < 1
$$
How would I go about solving this for:
$$
-x^2 < \cos^2x < x^2
$$
Thanks.

答:
The term radius of convergence is reserved for power series, such as $\sum_n a_nx^n$.  Your series does not fit this format.  However, it is a geometric series in $\cos^2 x/x^2$.  It will converge if and only if
$$ \left\lvert \frac {\cos^2 x}{x^2} \right \rvert < 1.$$
At a minimum this will be the case whenever $|x| > 1$ since (for real values of $x$) $|\cos x| < 1$. But in fact, still concentrating on real $x$, for $0 < x < 1$, the ratio $\cos^2x/x^2$ is decreasing and greater than $1$ for small $x$.  Therefore the inequality is true for all larger $x$ once $x > \theta$ where $\theta$ is the solution to $\cos \theta = \theta$, about $0.7$.  Because each term is an even function, the sign of $x$ is unimportant and convergence of your sum arises for all $|x|> \theta$, which is the solution to the inequality you have written.


问:
Probability of two people meeting up at a specific time period
I have recently read the following problem:

Bob and Jane agree to meet at a known location between 1pm and 2pm. The problem is that neither of them know what time the other will arrive.

We may assume that both of them are equally likely to arrive at any time between 1pm and 2pm.

Assume Bob is prepared to wait 15 minutes for Jane to arrive before leaving. And Jane is prepared to wait 15 minutes for Bob to arrive before she leaves.

What is the probability that they meet?

I tried my hand at the problem but I don't know what did I do wrong:

Let $X$ and $Y$ be the arriving time of Bob and Jane $t$ minutes after 1pm respectively, then $X,Y\sim U[0,60]$. Let $A$ be the event that they meet, then by the law of total probability, we have

$$P(A)=\int_0^{60}P(A|X=x)f(x)\,dx$$

Now $f(x)=F'(x)=(x/60)'=1/60$, hence

$$P(A)=\frac{1}{60}\int_0^{60}P(A|X=x)\,dx$$.

$$P(A)=\frac{1}{60}\left( \int_0^{45}P(Y\in [x,x+15])\,dx + \int_{45}^{60}P(Y\in [x,60])\,dx\right)$$

$$P(A)=\frac{1}{60}\left( \int_0^{45}\frac{x+15-x}{60-0}\,dx + \int_{45}^{60}\frac{60-x}{60-0}\,dx\right)=\frac{7}{32}.$$

But the answer is $7/16$. I would like to ask what's wrong with my argument, thank you so much in advance. I would also like to ask is there any other ways to solve this? Thank you.

答:
A simple drawing will get you the right solution, without integrals

The probability they meet is the white area dvided by the total area of $60^2$. To calculate the interested area is easier to calculate the area of the two purple triangle...thus simply

$$1-\left(\frac{45}{60}\right)^2=\frac{7}{16}$$


答:
Draw a square in the Cartesian coordinate plane with vertices at $(0,0), (1,0), (1,1), (0,1)$.  The horizontal value represents the random time when Bob arrives, in hours after 1 PM.  The vertical value represents the random time when Jane arrives, also in hours after 1 PM.  So the point $(0.5, 0.25)$ represents Bob arriving at 1:30 PM and Jane at 1:15 PM.

Now, what is the set of points in this square such that Bob and Jane meet?  Clearly, they meet if and only if $|X - Y| \le 0.25$, where $X$ and $Y$ are the arrival times of Bob and Jane, respectively.  What does the region defined by this inequality look like?  What is its area?  What is its area as a fraction of the total area of the square?

Now, if you perform this geometric exercise, it might give you insight as to where your computation went wrong.  If you were to perform it correctly, the integral would look like this, where I am using units of hours rather than minutes:

$$\int_{x=0}^{1/4} \Pr[0 \le Y \le x + 1/4] \, dx + \int_{x=1/4}^{3/4} \Pr[x - 1/4 \le Y \le x + 1/4] \, dx \\+ \int_{x=3/4}^1 \Pr[x - 1/4 \le Y \le 1] \, dx.$$
But of course, it is easier to exploit symmetry and compute the complementary probability via the single integral
$$1 - 2 \int_{x=0}^{3/4} \Pr[x + 1/4 < Y \le 1] \, dx.$$

In essence, your calculation only considers the case where Bob arrives before Jane, thus the condition in your integrand $Y \in [x, x + 15]$.  Jane could arrive before Bob and still meet, so $Y \in [x - 15, x + 15]$ is appropriate whenever $x \in [15, 45]$.  So this is why your probability evaluates to $7/32$:  you are missing exactly half of the total probability, the missing component being the outcomes in which Jane arrives before Bob.

Title: Probability of two people meeting up at a specific time period

Body:
I have recently read the following problem:

Bob and Jane agree to meet at a known location between 1pm and 2pm. The problem is that neither of them know what time the other will arrive.

We may assume that both of them are equally likely to arrive at any time between 1pm and 2pm.

Assume Bob is prepared to wait 15 minutes for Jane to arrive before leaving. And Jane is prepared to wait 15 minutes for Bob to arrive before she leaves.

What is the probability that they meet?

I tried my hand at the problem but I don't know what did I do wrong:

Let $X$ and $Y$ be the arriving time of Bob and Jane $t$ minutes after 1pm respectively, then $X,Y\sim U[0,60]$. Let $A$ be the event that they meet, then by the law of total probability, we have

$$P(A)=\int_0^{60}P(A|X=x)f(x)\,dx$$

Now $f(x)=F'(x)=(x/60)'=1/60$, hence

$$P(A)=\frac{1}{60}\int_0^{60}P(A|X=x)\,dx$$.

$$P(A)=\frac{1}{60}\left( \int_0^{45}P(Y\in [x,x+15])\,dx + \int_{45}^{60}P(Y\in [x,60])\,dx\right)$$

$$P(A)=\frac{1}{60}\left( \int_0^{45}\frac{x+15-x}{60-0}\,dx + \int_{45}^{60}\frac{60-x}{60-0}\,dx\right)=\frac{7}{32}.$$

But the answer is $7/16$. I would like to ask what's wrong with my argument, thank you so much in advance. I would also like to ask is there any other ways to solve this? Thank you.


问:
Is the graph of a Brownian motion over an interval measurable?
Let $n \in \mathbb{N}_1 := \{1, 2, \dots\}$ and let $B:\Omega \times [0, \infty) \rightarrow \mathbb{R}^n$ be a standard, $n$-dimensional Brownian motion over the probability space $(\Omega, \mathcal{F}, P)$. Let $t \in (0, \infty)$ and consider the following set:

$$
S := \{(\omega, x) \in \Omega\times\mathbb{R}^n \mid: \exists s \in [0,t], x = B(\omega, s)\}
$$

Is $S$ $\mathcal{F} \otimes \mathcal{B}_n$-measurable? ($\mathcal{B}_n$ denotes the Borel field over $\mathbb{R}^n$ generated by the Euclidean topology.)

MOTIVATION

I'm trying to understand why equation (2) in saz's answer makes sense; specifically, why the integrand $1_{B([0,1])}(x,y)$ is measurable.

答:
Here is an alternative approach to show that $1_{B[0,1]}(x)$ is measurable as a function from $\Omega\times \mathbb{R}^n$ to $\mathbb{R}.$

Define $\mathcal{A}_\varepsilon:\mathscr{C}_0^n\times \mathbb{R}^n\rightarrow \{0,1\}\subseteq\mathbb{R}$ by letting $\mathcal{A}_\varepsilon(f,x) = 1_{N(f[0,1],\varepsilon)}(x)$. Here $N(\cdot,\varepsilon)$ is the $\varepsilon$-neighborhood so that $N(A,\varepsilon) = \{x\in \mathbb{R}^n: d(x,A)<\varepsilon\}$.

One can easily verify that $\{(f,x): \mathcal{A}_\varepsilon(f,x) = 1\}$ is open in the product topology of $\mathscr{C}_{0}^n\times \mathbb{R}^n$ and hence $\mathcal{A}_\varepsilon$ is measurable. Taking the pointwise limit $\lim_{\varepsilon\downarrow 0}\mathcal{A}_\varepsilon$ we obtain, the necessarily measurable function, $(f,x)\mapsto 1_{f[0,1]}(x)$, since $f[0,1]$ is closed.

Given an $n$-dimensional Brownian motion $B(t)$ let $\Psi:\Omega\rightarrow \mathscr{C}_0^n$ be defined by $\Psi(\omega) = B(\cdot,\omega)$. Then $(\omega,x)\mapsto 1_{\Psi(\omega)[0,1]}(x)$ is measurable since $\Psi$ is.

Here is an alternative approach to proving directly that $\mathcal{L}_2(B([0,1]))$ is $\mathcal{F}$-measurable

Let $\mathcal{L}_2$ denote the area function on $\mathscr{B}(\mathbb{R}^2)$ and $\mathcal{C}_0$ denote the space of $\mathbb{R}^2$-valued continuous functions on $\mathbb{R}$, given the uniform topology. We claim that the function
\begin{equation*}
  \mathcal{C}_0\ni f \mapsto \mathcal{L}_2(f[0,1])\in [0,\infty)
 \end{equation*}
is measurable. Indeed suppose that
\begin{equation*}
  f_0\in\{f\in \mathcal{C}_0:\mathcal{L}_2(f[0,1])<a\}
 \end{equation*}
using regularity of Lebesuge measure, pick an open set $U$ such that $f_0[0,1]\subseteq U$ while $\mathcal{L}_2(U)<a$. Now the Euclidean distance from $f_0[0,1]$ to $\partial U$ is positive, say $>\delta$ and hence if $\sup_{t\in [0,1]}|f(t)-f_0(t)|<\delta$ then $f[0,1]\subseteq U$ implying that the $\delta$-ball around $f_0$ is contained in $\{f\in \mathcal{C}_0:\mathcal{L}_2(f[0,1])<a\}$ and hence the function $f\mapsto \mathcal{L}_2(f[0,1])$ is measurable.

Let $(B(t))$ denote a planar Brownian motion and let $\Psi:\Omega\rightarrow \mathcal{C}_0$ be the function $\Psi(\omega) = B(\cdot,\omega)$. Then $\Psi$ is measurable and hence $\mathcal{L}_2\circ \Psi = \mathcal{L}_2(B([0,1],\cdot))$ is measurable.


答:
$10^{99}=2^{99} \times 5^{99}$

There are $100$ ways to choose the exponent for $2$ and $100$ ways to choose the exponent for $5$, so that $10^{99}$ has $10000$ distinct factors.

To be divisible  by $10^{96}$ both exponents must be at least $96$, which leaves the four cases $96, 97, 98, 99$ for each exponent. This amounts to $4 \times 4= 16$ cases.

Assuming the distribution is such that any factor is equally likely to be chosen, this gives $\frac {16}{10000}$.


问:
Prove an inequality with ODE's
I need to prove that $\int_0^1 (u(x))^2 dx \leq \frac{1}{3} \int_0^1 (f(x))^2dx$ knowing that u solves the ODE $-u''(x) + u'(x) = f(x)$ for $x \in (0;1)$, and $u\in C_0^2((0;1))$ (u is $C^2$ and $u(0)=u(1)=0$).
I tried two methods but I can't conclude... First I showed that $\int_0^1 u^2(x)dx \leq \frac{1}{2}\int_0^1 u'(x)^2dx$ and then I used the PDE so $u' = u'' + f$ and $u''= f - u'$, put it in the integral : $\frac{1}{2}\int_0^1f(x)^2 + u''(x)(2f(x) + u''(x))dx \leq \int_0^1 f(x)^2  + \frac{1}{2}\int_0^1u'(x)^2dx$ but then I start to run around in circles again.
I also to tried to do something more direct and using integration by parts $\int_0^1 u^2(x)dx = [ (\int_0^xu(t)dt)u(x))]_0^1 - \int_0^1 u'(x)\int_0^x u(t)dtdx$ and then to develop with the ODE by same problem. Someone have an idea? Thank you very much!

答:
Hint: First show that

$$\int_0^1 u'(x)^2 \, dx = \int_0^1 u(x) f(x) \, dx$$

and apply the inequality you have proven and Young's inequality.

Title: Prove an inequality with ODE's


问:
Find a closed simply-connected $4$-manifold with closed non-exact $2$-form
Let $M$ be a closed simply-connected smooth $4$-manifold with a closed non-exact 2-form on it. $H^2_{dR}(M) \ne 0$.

By de Rham theorem, $H_{dR}^2(M) \cong H^2(M;\mathbb R) \cong \text{Hom}(H_2(M;\Bbb Z),\Bbb R)$.

Let $M = S^2 \times S^2$ with coordinates $((\theta_1,\varphi_1), (\theta_2,\varphi_2))$, how can we find such closed but non-exact form on it?

Edit:

How can we find such $2$-form explicity in coordinates $((\theta_1,\varphi_1), (\theta_2,\varphi_2))$?

答:
Let $\omega$ be a closed non exact 2-form on $S^2$. Let $\pi:S^2\times S^2\to S^2$
be the usual projection on the first factor.
Define $\hat \omega$ as the pullback of $\omega$ through $\pi$.
$\hat\omega$ is closed, since $d(\pi^*\omega)=\pi^*(d\omega)=0$.
It is not exact, since its restriction to the embedded submanifold $S^2\times \{(0,0,1)\}\simeq S^2$ is not exact.

To find an explicit example, take $\omega=\iota^*(xdy\wedge dz-ydx\wedge dz+z dx\wedge dy)$,
where $\iota$ is the inclusion of $S^2$ in $\mathbb{R}^3$


问:
Multivariable Taylor series expansion for Lucas-Kanade optical flow derivation
I am learning about Lucas-Kanade optical flow which is an algorithm used in computer vision to track movement of objects between frames of images.

So it starts with a point in an 2D image with an intensity $I$, given by the function $I(x,y,t)$, where $t$ is the time. In the next time step, the point will have an intensity of $I(x+\Delta x,\ y+\Delta y,\ t+\Delta t)$

Now a Taylor expansion is done on the new intensity and truncated to the linear terms
$$I(x+\Delta x,\ y+\Delta y,\ t+\Delta t) \approx I(x,\ y,\ t)+I_x\Delta x+I_y\Delta y+I_t\Delta t $$
where $I_x, I_y$ and $I_t$ are the partial derivatives of $I$ with respect to $x, y, t$ respectively.

I don't understand how this Taylor expansion was made. I know for a multivariable function, for e.g. 2 variables, the Taylor expansion centered around $(a, b)$ up to the linear terms is:
$$f(x,y) \approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$$
Applying that to the intensity equation, if I center around $x+\Delta x$, I get
$$I(x+\Delta x,\ y+\Delta y,\ t+\Delta t) \approx I(x+\Delta x,\ y+\Delta y,\ t+\Delta t)+I_x|_{x+\Delta x}\cdot \Delta x+I_y|_{y+\Delta y}\cdot \Delta y+I_t|_{t+\Delta t}\cdot\Delta t $$

If I center around $x$ instead, then the partial derivatives of $I$ would be w.r.t $x+\Delta x$, $t+\Delta t$, $t+\Delta t$. I'm not sure how to get the expression desired

答:
You should rewrite your formula
$$
f(x,y) \approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)
$$
for the case where $f$ is a function of three variables:
$$
\tag{1} f(x,y,t) \approx f(a,b,c) + f_x(a,b,c)(x-a) + f_y(a,b,c) (y - b) + f_t(a,b,c)(t - c).
$$
If you understand that equation, you're in good shape. This is basically identical to the equation you want to derive. The only difference is that things are named differently. Just rename your variables so that $x$ is replaced with $x + \Delta x$, $y$ is replaced with $y + \Delta y$, $t$ is replaced with $t + \Delta t$, $a$ is replaced with $x$, $b$ is replaced with $y$, $c$ is replaced with $t$, and $f$ is replaced with $I$. Then equation (1) becomes
$$
I(x + \Delta x, y + \Delta y, t + \Delta t) \approx I(x,y,t) + I_x(x,y,t) \Delta x + I_y(x,y,t) \Delta y + I_t(x,y,t) \Delta t.
$$


问:
Farthest distance map continuous
Let $(X,d)$ be a compact metric space, and let $A$ be a subset of $X$.

Define $ sd(x) = sup_{a \in A} d(x,a)$.  (Note the supremum, not infimum. The equivalent function with infimum is well known to be continuous)

Is $sd$ a continuous function?

答:
Fix $x,y \in X$ and let $a\in A$ be arbitrary.

Then $d(x,a) \le d(x,y) + d(y,a) \le d(x,y) + \text{sd}(y)$, so the RHS is a fixed upperbound for all distances $d(x,a)$ so that $$\text{sd}(x) \le d(x,y) +\text{sd}(y)$$ and

$$\text{sd}(x) - \text{sd}(y) \le d(x,y)$$

And by interchanging the rôles of $y$ and $x$ we get

$$|\text{sd}(x) - \text{sd}(y)|   \le d(x,y)$$

Which shows that sd is in fact a contraction so uniformly continuous.

Title: Farthest distance map continuous


问:
At least how many different numbers are there from $a_1,a_2,\cdots,a_{1394}$?

Consider positive integers $a_1,a_2,\ldots,a_{1394}$ so that neither
of two numbers of
$\dfrac{a_1}{a_2},\dfrac{a_2}{a_3},\cdots,\dfrac{a_{1393}}{a_{1394}}$
be equal to each other.
at least how many different numbers are there
from $a_1,a_2,\cdots,a_{1394}$?

$$1)38\quad\quad\quad\quad\quad\quad
 2)45\quad\quad\quad\quad\quad\quad3)49\quad\quad\quad\quad\quad\quad4)53\quad\quad\quad\quad\quad\quad5)55$$

To solve this problem in order to have unique $\dfrac{a_m}{a_{m+1}}$ I considered $a_m$ be prime numbers. so I started form $2$ and wrote few terms of the sequence: $2,3,2,5,3,5,2$ from here I should use new number ($7$). $2,3,2,5,3,5,2,7,3,7,5,7,2$ and here should continue with $11$. I don't know if writing numbers help or not.

答:
You are on the right track.

What you need is to realize that essentially you are looking for the smallest $n$ such that $2{n\choose 2}\geq 1394$.

See, the first sequence $2,3,2$ has length of pairs $2=2{2\choose 2}$.

The second sequence $2,3,2,5,3,5,2$ has length of pairs $6=2{3\choose 2}$

The third sequence $2,3,2,5,3,5,2,7,3,7,5,7,2$ has length of pairs $12=2{4\choose 2}$

The result is pretty obvious because you use every pair of numbers exactly twice, once the bigger one first and once the smaller one first.

Thus $2{38\choose 2}=1406$ is the smallest $n$.


问:
compute line integral using Greens theorem
The question is:
$$\int_\gamma \frac{(x^2+y^2-2)\ dx+(4y-x^2-y^2-2)\ dy}{x^2+y^2-2x-2y+2}$$
$$\gamma:y=2\sin\frac{\pi x}{2} \quad \text{from}\  (2,0)\ \text{to}\ (0,0)$$
Here how i have tried to solve it:
I thought that due to the singularity at $(1,1)$ it would be best to use Greens theorem with one circle around the singularity call it $\sigma$ and the line segment between $(0,0)$ to $(2,0)$ and call it $\varphi$ and then $\int_\gamma=-\int_\sigma-\int_\varphi$ but the problem is that the first integral (circle around the point$(1,1)$) is very complicated with $x=1+Rcost$ and $y=1+Rsint$. Do any of you guys have any suggestion or am i on the right track at all?

答:
Hint:

$\frac{\partial}{\partial y} \bigg(\frac{x^2+y^2-2} {(x-1)^2 + (y-1)^2}\bigg) - \frac{\partial}{\partial x} \bigg(\frac{4y - x^2 - y^2 - 2} {(x-1)^2 + (y-1)^2}\bigg) = 0, (x,y) \ne (1,1)$

The vector field is conservative in a domain that does not contain point $(1,1)$.

Point $(1,1)$ is below path $\lambda: y = 2 \sin \frac{\pi x}{2}$. So if we choose a path above $\lambda$, we will have a domain combined with $\lambda$, that does not contain point $(1,1)$. That will allow us to take advantage of path independence.

We can choose circular path $(x-1)^2 + (y-1)^2 = 2$ in anti-clockwise direction from $(2,0)$ to $(0,0)$. We can parametrize it as

$x = 1 + \sqrt 2 \cos t, y = 1 + \sqrt2 \sin t, -\frac{\pi}{4} \leq t \leq \frac{5 \pi}{4}$.

$dx = - \sqrt2 \sin t \ dt, dy = \sqrt2 \cos t \ dt$

The denominator of the vector field reduces to $2$.

Can you take it from here?


问:
Can we show $\sum_{n\in\mathbb N}e^{-\lambda_n}<\infty$ if $\lambda_n\to\infty$?
If $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ in nondecreasing with $\lambda_n\xrightarrow{n\to\infty}\infty$, how can we show that $\sum_{n\in\mathbb N}e^{-\lambda_n}<\infty$?

It clearly holds $1>e^{-\lambda_n}\xrightarrow{n\to\infty}0$. And $\sum_{n\in\mathbb N}e^{-\lambda_n}\le\sum_{n\in\mathbb N}\frac1{\lambda_n^2}$. Maybe it's easier to show that the right-hand side is finite.

答: No.  Let $\lambda_n= \ln n$.  Then your sum is the harmonic series.


问:
Prove more generally that if $X$ is Hausdorﬀ and I is a set, then the space of constant maps $I \to X$ is closed in $X^I$.
We know of that if $X$ is a Hausdorff Space, then the diagonal $\Delta=\{(x,x): x \in X\}$  is closed.
Can we generalize it for if $X$ is Hausdorff and then the space of constant maps from $I$ to $X$ is closed in $X^{I}$
if so, how can we start?

答: If $I$ is any index set, WLOG $|I| \ge 2$, and $f:I \to X$ is nonconstant we can find  $i\neq j$ in $I$ such that $f(i) \neq f(j)$. As $X$ is Hausdorff we find disjoint open subsets $U$ and $V$ around $f(i)$ resp. $f(j)$. Then use the open neighbourhood $p_i^{-1}[U]\cap p_j^{-1}[V]$ of $f$ which misses the set $C$ of constant functions entirely. So no point outside of $C$ is an adherent point of $C$ so the set is closed. For the reverse we only need the case where $I=\{0,1\}$ e.g. and we use that $X^I \simeq X^2$ and the case you already know applies.