问:
Proving manifold dimension theory using homology
The text I am using stated that any manifold with different dimension cannot be homeomorphic, but did not prove it (only proved the case when $n = 2$) until now, and it is given as an exercise in the homology chapter, but I am not sure how to approach this problem. Can anyone provide a solution or hint?

答:
Let $h : M \to N$ be a homeomorphism and $p \in M$. Then $h$ restricts to a homeomorphism $h'  : M \setminus \{p\} \to N \setminus \{h(p)\}$. Thus long exact sequences of $(M,M \setminus \{p\})$ and $(N, N \setminus \{h(p)\})$ are connected by the map $h$ and its restrictuion $h'$. But $h_* : H_k(M) \to H_k(N)$ and $h'_* : H_k(M \setminus \{p\}) \to H_k(N \setminus \{h(p)\})$ are isomorphisms, therefore the five lemma gives us an isomorphism $H_k(M,M \setminus \{p\}) \to H_k(N, N \setminus \{h(p)\})$.

By excision we see that $H_k(M,M \setminus \{p\}) \approx H_k(U,U \setminus \{p\})$, where $U$ is an open neigbhorhood of $p$ which is homeomorphic to an open ball in $\mathbb R^n$,  $n = \dim M$. Thus $H_k(M,M \setminus \{p\}) \approx \mathbb Z$ for $k = 0,n$ and $H_k(M,M \setminus \{p\}) = 0$ for all other $k$.

This shows that we must have $\dim M = \dim N$.


问:
Information about series sum function
I used an online tool that solves symbolic series sums and gives a breakdown of the steps. One of the steps that it gave was:

$$
\sum_{k=0}^\infty \frac{(1-p)^k}{k+t+1}
$$
is a known series. It is $\sum_{k=0}^\infty z_0^k(k+r_0)^{-n_0} = \Phi(z_0,n_0,r_0)$ for $z_0=1-p, n_0=1, r_0=t+1$.

I am unfamiliar with this $\Phi(z_0,n_0,r_0)$ - I am looking for information about it, in particular any lower and upper bounds that would be useful in deriving simpler bounds for my problem.

答:
$$\sum_{k=0}^\infty \frac{(1-p)^k}{k+t+1}=\Phi (1-p,1,t+1)$$ which is  the Hurwitz-Lerch transcendent function.

Have a look here for more information.

ContentLicense: CC BY-SA 4.0

Title: Information about series sum function


问:
Solving task with probability
I am trying to solve this task but I can't find solution at the moment.Task is:

Only two international airlines fly daily into an airport. UN Air has 70 flights a day and IS Air has 65 flights
a day. Passengers flying with UN Air have an 18% probability of losing their luggage and passengers flying
with IS Air have a 23% probability of losing their luggage. You overhear someone in the airport complain
about her luggage being lost. Find the probability that she travelled with IS Air.

So can anybody help me how to solve this?

答:
This is a conditional probability problem, your question can be rephrased to "Find the probability that she traveled with IS Air given that she lost her luggage"

We should use one form Bayes' formula since someone losing their luggage can occur in two different ways (either traveling in the first plane or the other)

$P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B|A)\cdot P(A)+P(B|C)\cdot P(C)}$

Where $P(A|B)$ means the probability that event $A$ happens given that event $B$ has already happened

Here we have the following:

Event A: She traveled with IS Air, here we have $P(A)=\frac{65}{135}$ since we have $65+70=135$ flights a day and $65$ of them are with IS Air

Event B: She lost her luggage, probability someone loses their luggage is the following

Event C: She flew with UN Air which would be $\frac{70}{135}$

Can you continue? I advise you to do so alone as you would get used to applying the formula however, if we continue we get the following:

$\frac{0.23\cdot\frac{65}{135}}{0.18 \cdot \frac{70}{135}+0.23 \cdot\frac{65}{135}}=\frac{299}{551}=0.543$

Title: Solving task with probability


问:
Hochschild cohomology and extension (counterexample?)
Let $k$ be a field, $R$ a unital, associative $k$-algebra and $M$ an $R$-$R$-bimodule; it is a classical fact that $\mathrm{HH}^2(R, M)$ classifies Hochschild ($k$-algebra) extensions of $R$ by $M$. Anyway it seems strange to me not to require some compatibility on $M$ such as $cm = mc$ for every $c \in k \subset R$: for example, if we put $k = R = \mathbb{C}$ and $M = \mathbb{H}$ (quaternions) with left and right multiplications bimodule structure, how is it possible to give a $\mathbb{C}$-algebra structure on $E = \mathbb{H} \oplus \mathbb{C}$ with product
$$
  (x, y) (x', y') = (xy' + yx', yy')
$$
(this would correspond to $0 \in \mathrm{HH}^2$)? For sure $z \cdot (x, y) = (zx, zy)$ or the same on the right does not work, since
$$
  \left\{
    \begin{aligned}
      (ij, i) (k, 1) &= (k-j, i) \\
      (j, 1) (ik, i) &= (-k-j, i)
    \end{aligned}
  \right.
  \implies i \cdot [(j, 1) (k, 1)] = {?} \,.
$$

ADDENDUM: there is not such $\mathbb{C}$-algebra structure: it would be a ring homomorphism $\mathbb{C} \to Z(E)$, where the center $Z(E)$ is the subring $\mathbb{C} \oplus \mathbb{R}$ (if my calculations are correct); now since necessarily $1 \mapsto (0, 1)$, we would have
$$
  i \mapsto (z,x)
  \quad\text{s.t.}\quad
  (0, -1) = (z, x)^2 = (2zx, x^2)
$$
and this is impossible. So, where is the problem?

答: I notice now that my example is misleading, since before being a $R$-$R$-bimodule, $M$ must be a $k$-module. Now it’s natural to ask whether the module structure and the induced $k$-$k$-bimodule must be compatible in order to have an algebra structure on the extension, but this answers to my question. I leave it online just for anybody who may ask the same question.


问:
Derivative of numbers
I am not talking here about deriving a constant in the usual sense (which gives $0$ as everyone knows).

I have once heard of a derivative operator with relaxed conditions that was defined over the real numbers whose purpose at first was to study number theory.

Usually, two of the essential properties that we demand on a differential operator is its linearity and its way of acting over products, i.e. if I call my differential operator $D$, :

$$
D(f+g) = Df + Dg, \quad D(fg) = (Df)g + fDg.
$$

Let's suppose we delete the linearity condition and keep only the product rule. Define $D(1) = 0$, $D(p) = 1$ if $p$ is prime and suppose $D$ satisfies the product relation over products. It is quite unclear in my memory at which point this operator could do (i.e. on which things could we apply it, was it restricted to algebraic numbers, or could it go over any real, complex?) Between the two lines I'm detailing what I know about this operator.

For instance, $D(4) = 2D(2) + 2(D(2)) = 2(1) + 2(1) = 4$. Hence $D(4) = 4$ is equal to its own derivative. Right now my definition only make sense with integers, since we can factor an integer $n$ in its prime decomposition and then apply the product rule to find the answer. We can also find the derivative of a fraction :
$$
0 = D(1) = D(q/q) = q D(1/q)  + 1/q D(q)  \quad \Rightarrow \quad D(1/q) = -\frac{D(q)}{q^2}.
$$
hence we can deduce the usual formula $D(p/q) = \frac{qD(p) - pD(q)}{q^2}$ by a similar argument. By an inductive argument, we can also show things as $D(a^n) = na^{n-1}$, and define this operator on algebraic numbers (at least for some that I know), for instance if $a$ is positive,
$$
D(a) = a^{1/2} D(a^{1/2}) + a^{1/2} D(a^{1/2}) = 2a^{1/2} D(a^{1/2}), \quad \Rightarrow D(a^{1/2}) = \frac{D(a)}{2a^{1/2}}
$$
and note here the similarity for the formula for deriving the function $1/f(x)$ in the real differentiable functions system.

If anyone has heard of such an operator over numbers which I have tried to describe as much as possible, can anyone tell me if there are any known results related to number theory that uses this tool to lead to some demonstrations that are useful? Some facts about primes, disivibility, writing a number as a sum of things, I don't know, just tell me what you know. I'd love to hear about it.

答:
Several references at the OEIS.

Added: I especially like this one.

ContentLicense: CC BY-SA 3.0

Title: Derivative of numbers


问:
What conditions do this polynomial satisfy?
Be p(x) a monic polynomial with real and positive coefficients, degree n >= 2 and with n real roots. What of this alternatives are right?

$(a) p(2) < 2(2^{n−1} + 1)\\ (b) p(1) < 3 \\(c) p(1) ≥ 2^{n} \\(d) p(3) < 3(2^{n−1} − 2)$

I have been seeking for consequences of this conditions, and how does them impact in the polynomial, but i am finding anything. Particularly i think the conditions n real roots and monic polynomial is the key to answer this question, but i don't know what to do with this properties...

Would appreciate any help.

答:
Let $p(x) = x^n + a_{n-1}x_{n-1} + \dots + a_1x + a_0$ with $a_i > 0,\ a_i\in\mathbb{R}$. Since all the coefficients are positive, for any $x>0$, we have $p(x)>0$. Hence, all the $n$ real roots are negative. Let $-\alpha_i$ be the $i$-th root of $p(x)$. Note that here $\alpha_i$ will be strictly positive. Then, we can rewrite $p(x)$ as
$$ p(x) = (x + \alpha_1)\dots(x + \alpha_n).$$

Now,

(1). $p(1) = \prod_{i=1}^n (1+\alpha_i)$ tells us that $p(1) > 1$.

(2). $p(2) = \prod_{i=1}^n (2+\alpha_i)$ tells us that $p(2) > 2^n$.

(3). $p(3) = \prod_{i=1}^n (3+\alpha_i)$ tells us that $p(3) > 3^n$.

One can choose suitable $\alpha_i$'s so that $p(1) > 3$. Hence, (b) is false. For example, $\alpha_1 = 100$.

Next, consider option (d).

$p(3) < 3(2^{n-1} - 2) < 3(2^{n-1}-2) + 6 < 3\cdot2^{n-1} < 3\cdot3^{n-1} < 3^n$ contradicts (3) above. Hence, (d) is false.

For option (c), set $\alpha_i = 0.5$ for all $1\leq i\leq n$. Then, $p(1) < 1.5^n < 2^n$. Hence, (c) is also false.

But even option (a) seems to be a bad bound because I can simply choose my $\alpha_i$'s to be equal to $4$ which will make the $p(2) = 6^n > 2(2^{n-1} + 2)$ for $n\geq 2$.

Choosing a set of $\alpha_i$'s means there exists a polynomial with real coefficients and $n$-real roots such that the above conditions do not hold. This is a general way to check. Maybe you need to check the question. Might have messed up some inequality. Hope this helps.


问:
Euler's summability method for series convergence
I was going through the book on Functional Analysis by Erwin Kreyzig, and I came across this as one of the exercises. As an application of functionals to summability of sequences, the following theorem has been established in the book:

Theorem: If $A = \left[ \alpha_{nk} \right]_{n, k \in \mathbb{N}}$ is an infinite real matrix and $\left( x_n \right)_{n \in \mathbb{N}}$ is a real sequence, then the summability method which takes $\left( x_n \right)_{n \in \mathbb{N}}$ to the real sequence $\left( y_n \right)_{n \in \mathbb{N}}$, given by
$$y_n = \sum\limits_{k = 0}^{\infty} \alpha_{nk} x_k,$$
is regular if and only if the following holds:

(a) $\lim\limits_{n \to \infty} \alpha_{nk} = 0$, for all $k \in \mathbb{N}$.

(b) $\lim\limits_{n \to \infty} \sum\limits_{k = 1}^{\infty} \alpha_{nk} = 1$.

(c) There is some $\gamma > 0$ such that for all $n \in \mathbb{N}$, we have $\sum\limits_{k = 1}^{\infty} \left| \alpha_{nk} \right| \leq \gamma$.

The exercise which I was talking about, transforms a given series $\sum\limits_{j = 0}^{\infty} \left( -1 \right)^j a_j$ to the series $\sum\limits_{n = 0}^{\infty} \dfrac{\Delta^n a_0}{2^{n + 1}}$, where $\Delta^0 a_j = a_j$ and $\Delta^n a_j = \Delta^{n - 1} a_j - \Delta^{n - 1} a_{j + 1}$ for all $j \in \mathbb{N} \cup \left\lbrace 0 \right\rbrace$.

I want to prove that this summability method is regular. So, I thought of using the above theorem. So, if we consider

$$x_n = \sum\limits_{j = 0}^{n} \left( -1 \right)^j a_j,$$

and

$$y_n = \sum\limits_{k = 0}^{n} \dfrac{\Delta^k a_0}{2^{k + 1}},$$

we can do some algebraic maipulations to obtain,

$$y_n = \sum\limits_{k = 0}^{n} \dfrac{1}{2^{k + 1}} \sum\limits_{j = 0}^{k} \binom{k}{j} \left( -1 \right)^j a_j.$$

Now, this is almost of the form required in the theorem. However, I am unable to manipulate the binomial coefficient outside the inner sum. Any hints to do so?

Edit: A summability method is called regular if it does not change the limits of convergent sequences.

答:
You need $y_n$ in terms of $x_k$ (not $a_k$!) with $0\leqslant k\leqslant n$. We're going to show that $$y_n=\frac{1}{2^{n+1}}\sum_{k=0}^n\binom{n+1}{k+1}x_k;$$ having this, the conditions (a), (b), (c) of the theorem are very easy to check.

Recall that $x_n=\sum\limits_{k=0}^n(-1)^k a_k$. Introducing $x_{-1}:=0$, we have $(-1)^ja_j=x_j-x_{j-1}$, and $$\Delta^k a_0=\sum_{j=0}^k\binom{k}{j}(x_j-x_{j-1})=\sum_{j=0}^k\left[\binom{k}{j}-\binom{k}{j+1}\right]x_j,$$ so that, doing the interchange $\sum\limits_{k=0}^n\sum\limits_{j=0}^k=\sum\limits_{j=0}^n\sum\limits_{k=j}^n$, we get $$y_n=\sum_{j=0}^n\alpha_{nj}x_j,\quad\alpha_{nj}:=\sum_{k=j}^n\frac{1}{2^{k+1}}\left[\binom{k}{j}-\binom{k}{j+1}\right].$$

Finally, $\alpha_{nj}=\frac{1}{2^{n+1}}\binom{n+1}{j+1}$ may be shown using induction on $n$. It clearly holds at $n=j$ (both formulae give $\alpha_{jj}=2^{-j-1}$), and the induction step $(n-1)\to n$ for $n>j$ goes like this: $$\alpha_{nj}=\underbrace{\frac{1}{2^n}\binom{n}{j+1}}_{=\alpha_{(n-1)j}}+\frac{1}{2^{n+1}}\left[\binom{n}{j}-\binom{n}{j+1}\right]=\frac{1}{2^{n+1}}\left[\binom{n}{j}+\binom{n}{j+1}\right]=\frac{1}{2^{n+1}}\binom{n+1}{j+1}.$$


问:
How to prove this identity: $4\cdot(\frac12)!=\pi$
Progress: I recently found this identity on an YouTube Video: $$4\cdot\left(\frac{1}{2}\right)!=\pi$$

But it didn't provide any kind of rigorous proof or simply a 'proof' in that regard. It only showed some Calculus identities to give some insight. Also importantly this is the only identity or equation I know that exactly equals to '$\pi$'. The part which is very interesting is $(\frac{1}{2})!$, the factorial of a fraction not an integer.

So if anyone can give a rigorous proof and even elementary proof or share more heuristics and insights, it would be useful. I'll really appreciate your efforts, thanks.

答:
Note that gamma function is defined as$$\Gamma{(n)}=\int x^{n-1}e^{-x}dx$$
for $n>0$, $n\in\mathbb{R}$.

Gamma function is generalization of factorial function.

$$\Gamma (n)=(n-1)!.$$

Gamma function has property
$$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$
and
$$\Gamma(n+1)=n\Gamma(n).$$

So,

\begin{align}
4\left(\dfrac{1}{2}\right)!&=4\left(\dfrac{3}{2}-1\right)!\\
&=4\Gamma\left(\dfrac{3}{2}\right)\\
&=4\Gamma\left(1+\dfrac{1}{2}\right)\\
&=4\cdot \dfrac{1}{2} \Gamma\left(\dfrac{1}{2}\right)\\
&=2\sqrt{\pi}.
\end{align}

I guess you want to prove
$$4\left(\left(\dfrac{1}{2}\right)!\right)^2=\pi.$$

\begin{align}
 4\left(\left(\dfrac{1}{2}\right)!\right)^2&=4\left(\left(\dfrac{3}{2}-1\right)!\right)^2\\
 &=4\left(\Gamma\left(\dfrac{3}{2}\right)\right)^2\\
 &=4\left(\Gamma\left(1+\dfrac{1}{2}\right)\right)^2\\ &=4\cdot
 \left(\dfrac{1}{2} \Gamma\left(\dfrac{1}{2}\right)\right)^2\\ &=4\cdot
 \left(\dfrac{1}{2}\sqrt{\pi}\right)^2\\ &=4\cdot \dfrac{1}{4}\pi\\
 &=\pi. \end{align}


答:
Not a answer, but important to know:
As Qiaochu Yuan mentioned, the Gamma function gives your equality. But in the usual definition of the factorial it's undefined. That means the Gamma function is only one way to extend the factorial, even if it seems to be the most natural and useful way.</p>

Title: How to prove this identity: $4\cdot(\frac12)!=\pi$

Body:
Progress: I recently found this identity on an YouTube Video: $$4\cdot\left(\frac{1}{2}\right)!=\pi$$

But it didn't provide any kind of rigorous proof or simply a 'proof' in that regard. It only showed some Calculus identities to give some insight. Also importantly this is the only identity or equation I know that exactly equals to '$\pi$'. The part which is very interesting is $(\frac{1}{2})!$, the factorial of a fraction not an integer.

So if anyone can give a rigorous proof and even elementary proof or share more heuristics and insights, it would be useful. I'll really appreciate your efforts, thanks.


问:
Evaluating $2\log_{4} (\cos 67^\circ) + \log_{2}(\cos 23^\circ) + \log_{1/2}(\cos 44^\circ)$
How to calculate the following?

$$2\log_{4} (\cos 67^\circ) + \log_{2}(\cos 23^\circ) + \log_{1/2}(\cos 44^\circ)$$

答:
Hint:

$$\log_4(\cos67^\circ)=\dfrac{\log\cos67^\circ}{\log4}=\dfrac{\log\cos67^\circ}{2\log2}$$

Use $\log a+\log b=\log(ab)$

finally, $$2\cos67^\circ\cos23^\circ=2\sin(90-67)^\circ\cos23^\circ=2\sin(2\cdot23^\circ)$$


问:
Oriented manifolds and the intermediate value theorem
Working through a proof that any connected orientable manifold has two orientations (From Tu's manifolds), I'm having a difficulty with a small nuance in the proof. Consider a manifold $ M $ with two orientations $ \mu $ and $ \nu $, and some arbitrary point $ p \in M $. $ \mu_p $ and $ \nu_p $ are the same or opposite orientations of $ T_p M $. The strategy of the proof is to show that the function $ f:M \rightarrow \{1,-1\} $ (where $ f(p) = 1 $ if $ \mu_p = \nu_p $ and $ f(p) = -1 $ if $ \mu_p = - \nu_p $) is locally constant, and therefore globally constant since $ M $ is connected.

Okay, all well and good. Now, the orientations are continuous, so for any point there is a connected open neighborhood of $ p $, $ U $, where $ \mu = [(X_1,...,X_n)] $ and $ \nu = [(Y_1,...,Y_n)] $ are equivalence classes of continuous frames on $ U $. Of course, there is a transition matrix between these frames, $ A:U \rightarrow GL(n,\mathbb{R}) $ with continuous entries, meaning the nonsingular $ \text{det} A:U \rightarrow \mathbb{R}^\times $ is continuous.

Here's the kicker, though: Tu says that because of the Intermediate Value Theorem, $ (\text{det} A)(p) $ is always positive or always negative on $ U $, being unable to change signs without smoothly crossing over zero---thus showing that $ f $ is constant on U. However, I have no idea what the intermediate value theorem means when not talking about functions from some closed interval of $ \mathbb{R} $ to $ \mathbb{R} $. Naturally, I'd say $ \text{det} A $ being continuous means that $ \text{det} A \circ \phi^{-1}:\phi(U) \rightarrow \mathbb{R} $ is continuous for a chart $ (U,\phi) $. But, how is the intermediate value theorem extended to a function $ \mathbb{R}^n \rightarrow \mathbb{R} $? My intuition is that, like in one dimension, the connectedness of $ U $ is key to an analogous theorem, but can a generalization of the IVT be made without introducing notions of curves along $ U $ and other differential geometry machinery into the mix? Can I apply the 1-D IVT to all possible curves between two points in $ U $ or something? I get the intuitive sense of what he means by it, but how can I put it on more mathematical footing?

答:
Usually the IVT is understood as a theorem of elementary calculus which deals with functions $f : [a,b] \to \mathbb R$. However, there is a generalized version.

Note that if $f : X \to Y$ is a continuous map and $X$ is connected, then $f(X)$ is connected. If $Y = \mathbb R$, then the connected subsets of it are precisely the intervals. Therefore

Let $f : X \to \mathbb R$ be a continuous map and $X$ be connected. If $y_1 = f(x_1) < y_2 = f(x_2) $, then for each $y \in (y_1,y_2)$ there exists $x \in X$ such that $f(x) = y$.


问:
Some remarks/questions from Primary Decomposition Theorem to get verified
In course of self-studying the Canonical Form in Linear Algebra I'm trying to put some remarks from the concept I acquired from Primary Decomposition Theorem which reads as follows:

Let $T:V_F\to V_F~(\dim V<\infty)$ be a linear transformation and let $m(t)$ be the minimal polynomial (existence of such polynomial is guaranteed due to Cayley-Hamilton Theorem) of $T$ over $F$ which is expressed in it's totally factored form (uniqueness follows from the observation that $F[x]$ is a UFD) as $m(t)=\prod_{i=1}^k [f_i(t)]^{n_i}$ for some $k\in\mathbb Z^+$ where $f_i$'s are monic irreducible polynomial over $F.$ Then $V=\oplus_{i=1}^k \ker (f_i(T))^{n_i}$ ($(f_i(T))^{n_i}\in L(V,V)$ since $L(V,V)$ is a linear algebra).

Here're my remarks/questions which I want to get verified/answered:

If $T$ is non-decomposable then for every collection of $T$-invariant subspaces $\{W_i\}_{i=1}^m$ of $V$ satisfying $V=\oplus_{i=1}^m W_i$ one has $W_i=(0)$ for some $i$. (A linear operator $T:V\to V$ is called decomposable if $\exists$ a collection of non-zero $T$-invariant subspaces $\{W_i\}_{i=1}^m$ (with $m>1$) of $V$ such that $V=\oplus_{i=1}^m W_i$)

However if $T$ is decomposable then $\exists$ a collection of non-zero $T$-invariant subspaces $\{W_i\}_{i=1}^m$ of $V$ such that $V=\oplus_{i=1}^m W_i.$

Q. Does for decomposable $T,$ the collection $\{\ker (f_i(T))^{n_i}\}_{i=1}^k$ serve the purpose, i.e. whether $\ker (f_i(T))^{n_i}\neq (0)$ for all $i$?

Of course, even if $T$ is decomposable, for every collection $\{W_i\}_1^n$ of $T$-invariant subspaces of $V$ satisfying $V=\oplus_1^n W_i$ the condition the $W_i\ne(0)~\forall ~i$ doesn't hold since $V=V\oplus(0)$ where both $V,(0)$ are $T$-invariants.

I can see that (for the sake of simplicity I use $W_i=\ker (f_i(T))^{n_i}$) $(f_i(T|_{W_i}))^{n_i}=\underline 0:W_i\to W_i~\forall~i=1(1)k$ whence $\ker (f_i(T|_{W_i}))^{n_i}=W_i.$ Consequently, $\ker ((f_i.T)^{n_i}|_{W_i}))=\ker (f_i(T|_{W_i}))^{n_i}=\ker (f_i(T))^{n_i}.$ So $(f_i(T))^{n_i}$ can't send any element outside $W_i$ to $0.$

答:
$\newcommand{\ker}{\operatorname{ker}}$
0) Claiming the existence of the minimal polynomial from Cayley-Hamilton is overkill.  The powers $T^k$ of $T$ lie in the finite-dimensional vector space $L(V,V)$ so they cannot all be linearly independent.  A nontrivial linear dependence relation between them gives a nonzero polynomial satisfied by $T$.

1) Writing $[f_i(t)]^{n_i}$ is unnecessarily notationally heavy.  Better would be $f_i^{n_i}(t)$ or even just $f_i^{n_i}$.

2) I'm not sure whether you understand the definition of a decomposable endomorphism.  It just means that you can write $V_F$ nontrivially as a direct sum of invariant subspaces, i.e., $V_F = W_1 \oplus W_2$ in which $W_1$ and $W_2$ are both proper and nonzero.  It would be logically equivalent to say that both are proper; it would also be logically equivalent to say that both are nonzero.  Let's also assume that $V_F$ has positive dimension to exclude trivialities.

From this definition it is immediate that -- since $V_F$ is finite-dimensional -- $V$ can be written as a finite direct sum of indecomposable invariant subspaces.

You seem to be asking about the converse: namely, if $k = 1$ must $V$ be indecomposable?  The answer is no.  In fact $V$ is indecomposable iff $V$ is primary -- i.e., $k = 1$ -- and $V$ is cyclic, i.e., spanned by the powers of $T$ acting on a single vector $v$ . To see this in general requires another big theorem, the Cyclic Decomposition Theorem, which indeed says (in particular) that any $V$ is a direct sum of cyclic, primary invariant subspaces and that the number of factors in this decomposition is unique.  However, you can get an easy example just by taking $T = 0$: then the minimal polynomial is $T$, which is a prime power, so $k = 1$.  But as long as $\operatorname{dim} V > 1$ there are plenty of decompositions into direct sums of invariant subspaces, since every subspace is invariant in this case.

3) Your third assertion is true and is really a tautology: if $W_i$ is the kernel of a certain linear operator, then indeed that operator does not send any vector outside of $W_i$ to $0$.


问:
I am really stuck with this problem, let b(x) be a system of neighoruhoods of point x of the infinite set X with cofinite
let b(x) be a system of neighoruhoods of point x of the infinite set X on which there is some topology, that is stronger than cofinite topology. Prove that ⋃U∈b(x)(X\U) = X{x}</p>

答:
So $\mathcal{B}(x)$ is a base of neighbourhoods for $x$.

We have to show that $$\bigcup_{U \in \mathcal{B}(x)} (X\setminus U) = X\setminus\{x\}$$

As $x \in U$ for all $U \in \mathcal{B}(x)$, it’s clear that $X\setminus U \subseteq X\setminus \{x\}$, and so the union on the left is also a subset of the right hand side.

If $y \in X\setminus \{x\}$, we use the assumption that $X$ has a finer topology then the cofinite one to observe that $X\setminus \{y\}$ is open in $X$ (it’s cofinite) and a neighbourhood of $x$ (as $y \neq x$). So for some $U \in \mathcal{B}(x)$ we have $x \in U \subseteq X\setminus \{y\}$. It follows that $y \in X\setminus U$ for that $U$ and so $y$ is a member of the left hand union.

This shows two inclusions hence equality.


问:
Alternative to complex numbers - Study numbers $j^2=1$, $j \neq 1$
On p. 24 "Clifford Algebra and Spinors" (2.2 Double Ring of  $\:{}^2\mathbb{R}$ of $\mathbb{R}$) by Petri Lounesto the author mentions Study numbers as an equivalent alternative to complex numbers.

The Study numbers are pair of two real numbers (a,b) $\in \mathcal{\mathbb{R}}^2$ such as
$$a+jb, \quad j^2=1 ,\quad j \neq 1$$

He introduces Study conjugate as $(a+jb)^-=a-jb$, then they can be written in hyperbolic polar form e t c.

It seems that the use of such numbers can be as powerful as the use of complex numbers then.

My questions are

Why Study numbers are not widely known and used?

Can someone point me to the reference to the Study numbers? I can not find it either in Wikipedia, neither in literature.

PS The mathematician is Eduard Study. He approached the dual quaternions with the dual numbers.

答:
I haven't heard of these being called the Study numbers; the seemingly more common name is split-complex numbers. Equivalently, it is the Clifford algebra $\operatorname{Cl}_{1,0}(\mathbb R)$ of a one-dimensional real vector space with a positive-definite quadratic form.

As a real algebra, these are isomorphic to a direct product $\mathbb R \times \mathbb R$, thus it is hardly an interesting research subject (anything about it follows from properties of $\mathbb R$ and direct products of rings).


问:
Supremum and Infimum of two variable function
A Two variable function
$$f(x,y)=(x-2)^2 +(y-2)^2 $$
The Domain of x is $[0,4]$ and Domain of $y$ is $[0,4]$.

How I can determine the supremum  $$ g(x) = \sup_y f(x, y)$$ and infimum $$h(y) = \inf_xf(x, y)$$ of the function.

For $g(x)$ I guess I find derivate of $f(x,y)$ with respect to $x$ which turns out to be $2(x-2)$ and for $h(y)$ i find derivate of $f(x,y)$ with respect to which turns out to be $2(y-2)$

答:
Notice that this is a separable problem, the $x$ and $y$ can be optimized independently.

$$g(x) =\sup_y f(x,y) = (x-2)^2 + \sup_y (y-2)^2 =(x-2)^2+4$$

$$h(y) =\inf_x f(x,y) = \inf_x(x-2)^2 + (y-2)^2 =(y-2)^2$$


问:
A non-nilpotent transformation $A$ of a vector space $V$ such that for every $v\in V$ there exist a $n\in\mathbb{N}$ for which $A^n(v)=0$.
Give an example of a vector space $V$ (dim$V$ may or may not be infinite) and a transformation $A$ of $V$ such that $A$ is NOT nilpotent and for every vector $v\in V$ there exist a natural number $n\in\mathbb{N}$ for which $A^n(v)=0$.

I really don't have any clue on how to approach this problem.

答:
For an infinite-dimensional vector space, the classical example is the left shift operator on $\{(a_1,a_2,\ldots):a_i\in\mathbb R,0=a_k=a_{k+1}=\cdots\text{ for some } k\}$ (the vector space of all sequences of real numbers with finitely many nonzero entries).

For a finite-dimensional vector space, no counterexamples exist. Given the assumption on $A$, when $n$ is sufficiently large, $A^n$ is zero on a basis of $V$. Therefore $A^n=0$ on $V$ and $A$ is nilpotent.


问:
How to make a plot of function in MATLAB
I have a function ${\dot{\varphi } = \gamma - F(\varphi )}$ (where $F(\varphi)$ - is 2${\pi }$-periodic function) and graph of function $F(\varphi)$.

So I need to create a program that outputs 6 plots of φ(t) for different meaning of $\gamma (\gamma= 0.1, 0.5, 0.95, 1.05, 2, 5)$, and t ∈ [0, 100]

I've got a following F(φ) function:
$$F(\phi)=\begin{cases}
 & {-\frac{\phi}{a}-\frac{\pi}{a}},  &\text{if } {\phi \in [-\pi, -\pi + a]},\\
 & -1, &\text{if } {\phi \in [-\pi + a, - a]}, \\
 & -\frac{\phi}{a},  &\text{if }  {\phi \in [- a,  a]}, \\
 & 1, &\text{if } {\phi \in [a, \pi- a]},\\
 & {-\frac{\phi}{a}+\frac{\pi}{a}},&\text{if }  {\phi \in [\pi-a, \pi]}.
\end{cases}$$

```<code>                 ^
                 |
                 |1   ______
                 |   /|     \
                 |  / |      \
                 | /  |       \
__-π_______-a____|/___|________\π____>
   \        |   /|0    a
    \       |  / |
     \      | /  |
      \     |/   |
       ¯¯¯¯¯¯    |-1
</code>```

I've tried to do it in MATLAB but I faced with some problems. I don't know what to put in ode45 function (on which section should each part of the graph be displayed and what is the initial value to take). Because the evolution of  $\phi(t)$ must be continuous.
It's a code for $\gamma= 0.1$

```<code>hold on;
df1dt=@(t,f1) 0.1 - f1 - 3.14;
df2dt= @(t,f2)- 1;
df3dt=@(t,f3) 0.1 + f3;
df4dt= @(t,f4)+1;
df5dt=@(t,f5) 0.1- f5 + 3.14;
[T1,Y1] = ode45(df1dt, ...);
[T2,Y2] = ode45(df2dt, ...);
[T3,Y3] = ode45(df3dt, ...);
[T4,Y4] = ode45(df4dt, ...);
[T5,Y5] = ode45(df5dt, ...);
plot(T1,Y1);
plot(T2,Y2);
plot(T3,Y3);
plot(T4,Y4);
plot(T5,Y5);
hold off;
title('\gamma  = 0.1')
</code>```

答:
As a first step, you have to implement the periodicity of $F$ by reducing the argument to a basic period.
Your function has odd symmetry, so use $F(-x)=-F(x)$ to reduce the number of cases.

```<code>function y = F(x,a)
    n = round(x/(2*pi));
    x = x-n*2*pi; % now in [-pi,pi]
    s = 1;
    if x<0  x=-x; s=-1; end%if % now in [0,pi]
    y = s*min([1,x/a,(pi-x)/a]);
end%function
</code>```

Then you can solve your ODE in just one call,

```<code>[t,phi] = ode45(@(t,x)gamma-F(x,a), [t0,tf], phi0, options);
plot(t,phi)
</code>```

The full script that contains the main program and the functions as subroutines in one file can look like this:

```<code>function mathse3954790_ballbeam_heun()

  g = 10;
  tau = 0.2;
  J = 1.2;
  m = 5;
  x1_ini = 0.5;
  t0 = 0;
  tf = 10;
  h = 0.05;

  x0 = [x1_ini, 0, 0, 0 ];
  t = t0:h:tf;
  x = odeheun(@(t,x)ballbeam(t,x),t,x0); % possibly @ballbeam
  plot(t,x(:,1));

  function dotx = ballbeam(t,x)
      dx2 = x(1)*x(4)^2 - g*sin(x(3));
      dx4 = (-2*m*x(1)*x(2)*x(4)-m*g*x(1)*cos(x(3))+tau)/(m*x(1)^2+J);
      dotx = [ x(2), dx2, x(4), dx4 ];
  end

  function x = odeheun(f,t,x0)
      x = [x0];
      for i=1:(length(t)-1)
          h = t(i+1)-t(i);
          k1 = h*f(t(i),x(i,:));
          k2 = h*f(t(i+1),x(i,:)+k1);
          x(i+1,:) = x(i,:)+0.5*(k1+k2);
      end%for
  end%function

end
</code>```

There are some tricks that allow to leave out the enclosing function, but it is not very intuitive as the default behavior on loading the script is to call the first function in the script without arguments.

Body:
As a first step, you have to implement the periodicity of $F$ by reducing the argument to a basic period.
Your function has odd symmetry, so use $F(-x)=-F(x)$ to reduce the number of cases.

function y = F(x,a)
    n = round(x/(2*pi));
    x = x-n*2*pi; % now in [-pi,pi]
    s = 1;
    if x<0  x=-x; s=-1; end%if % now in [0,pi]
    y = s*min([1,x/a,(pi-x)/a]);
end%function

Then you can solve your ODE in just one call,

[t,phi] = ode45(@(t,x)gamma-F(x,a), [t0,tf], phi0, options);
plot(t,phi)

The full script that contains the main program and the functions as subroutines in one file can look like this:

function mathse3954790_ballbeam_heun()

  g = 10;
  tau = 0.2;
  J = 1.2;
  m = 5;
  x1_ini = 0.5;
  t0 = 0;
  tf = 10;
  h = 0.05;

  x0 = [x1_ini, 0, 0, 0 ];
  t = t0:h:tf;
  x = odeheun(@(t,x)ballbeam(t,x),t,x0); % possibly @ballbeam
  plot(t,x(:,1));

  function dotx = ballbeam(t,x)
      dx2 = x(1)*x(4)^2 - g*sin(x(3));
      dx4 = (-2*m*x(1)*x(2)*x(4)-m*g*x(1)*cos(x(3))+tau)/(m*x(1)^2+J);
      dotx = [ x(2), dx2, x(4), dx4 ];
  end

  function x = odeheun(f,t,x0)
      x = [x0];
      for i=1:(length(t)-1)
          h = t(i+1)-t(i);
          k1 = h*f(t(i),x(i,:));
          k2 = h*f(t(i+1),x(i,:)+k1);
          x(i+1,:) = x(i,:)+0.5*(k1+k2);
      end%for
  end%function

end

There are some tricks that allow to leave out the enclosing function, but it is not very intuitive as the default behavior on loading the script is to call the first function in the script without arguments.


问:
Find the variance of $U[a,b]$ from $U[0,1]$
I know the variance of uniform distribution $U[a,b]$ is $(b-a)^2/12$, but I'm having trouble to prove it from $U[0,1]$.

i.e. let $X = a+(b-a)U[0,1], Var(X) = \mathbb{E}(X^2)+\mathbb{E}(X)^2 = \mathbb{E}(a^2+(b-a)^2U^2+2a(b-a)U)-(a+b)^2/4 = a^2+a(b-a)+(b-a)^2\mathbb{E}(U^2)-(a+b)^2/4 = \frac{-3(a^2+b^2)+4b-4a+10ab}{12}$

where $\mathbb{E}(U[0,1]^2)=\frac{1}{3(b-a)}$

what did I do wrong?

Body:
You calculated it wrong :(
Please try again.

For reference,

$Var (X)$

$= \mathbb{E}(a^2 + (b-a)^2U^2 - 2a(b-a)U) - \mathbb{E}(a + (b-a)U)^2$

$= a^2 + \frac{a^2+b^2-2ab}{3} + a(b-a) - \frac{a^2+b^2+2ab}{4}$

$= a^2 + \frac{4a^2+4b^2-8ab}{12} + ab-a^2-\frac{3a^2+3b^2+6ab}{12}$

$= (12ab + 4a^2+4b^2-8ab - 3a^2-3b^2-6ab)/12$

$=( a^2+b^2+(12-8-6)ab)/12$

$=( a^2+b^2-2ab)/12$

$= (b-a)^2/12.$

Q.E.D.


问:
Filter and ultrafilter properties in Boolean algebra
In a Boolean algebra $(\mathbb{B}, \vee, \wedge, 0, 1, \neg)$, a ultrafilter $U \subseteq \mathbb{B}$ satisfies

If $a \wedge b \in U \iff a \in U$ and $b \in U$.

If $a \vee b \in U \iff a \in U$ or $b \in U$.

The first one is actually true for all filters ($\Rightarrow$ upward closure, $\Leftarrow U$ downward directed). For the second one, $\Leftarrow$ can be proven by upward closure, but how about the $\Rightarrow$ direction which seems to be true only for ultrafilter ?

NOTE: The only definition of ultrafilter I have at this stage is: an ultrafilter $U$ is a filter satisfying $\forall x \not\in U, \neg x \in U$ (i.e. $\forall x \in \mathbb{B}, x \in U \leftrightarrow \neg x \not\in U$).

答: To add to Brian’s answer: yes that property implies it is an ultrafilter as any filter contains $1$ (being non-empty and upper-closed) and so for any $a \in \Bbb B$, $a \lor \lnot a = 1 \in U$, so the right to left implication of 2 then implies $a \in \mathcal{U}$ or $\lnot a \in U$. These cannot hold at the same time as $0 = a \land \lnot a \notin U$. This implies that $U$ is an ultrafilter in your definition.


问:
Halfspace Theorem in Differential Geometry
I'm sutdying the Classical Halfspace Theorem, it says that :"There is no non-planar, complete, minimal
surface properly immersed in a halfspace of $\mathbb{R}^3$". But in this theorem, I don't know the definition of "complete". By the way, I would like to understand what does Theorem mean via some simple examples.

答:
For a Riemannian manifold without boundary $(M,g)$, completeness means one of the following equivalent conditions:

all (unitary parametrized) geodesics are defined on $\mathbb{R}$

the metric space $(M,d_g)$ is complete (all Cauchy sequences converge)

the compact subsets are the closed and bounded subsets

This is called the Hopf-Rinow theorem. Here, complete is to be understood as "every surface endowed with the induced Riemannian metric that is complete".

The half space theorem says that if a surface in $\mathbb{R}^3$ is non planar (thus, "curved"), complete (thus, geodesics do not end somewhere: there is no "missing point" or "boundary"), properly immersed (thus, the surface "extends to infinity"), and if it is minimal, then it goes "in every direction": there is no half-space in which you can enclose the surface. It has to go out of it at some time. Equivalently, it intersects every affine hyperplane.


答:
By "from" they mean if $X\sim U(a,\,b)$ then $X=a+(b-a)Y$ with $Y\sim U(0.\,1)$, so $\operatorname{Var}X=(b-a)^2\operatorname{Var}Y=(b-a)^2/12$.

Edit, since @BenjaminWang has suggested proving $\operatorname{Var}Y=\tfrac{1}{12}$: since $\Bbb EY^n=\int_0^1y^ndy=\tfrac{1}{n+1}$ for $n>-1$, $\operatorname{Var}Y=\tfrac13-(\tfrac12)^2=\tfrac{4-3}{12}=\tfrac{1}{12}$.

Title: Find the variance of $U[a,b]$ from $U[0,1]$

Body:
I know the variance of uniform distribution $U[a,b]$ is $(b-a)^2/12$, but I'm having trouble to prove it from $U[0,1]$.

i.e. let $X = a+(b-a)U[0,1], Var(X) = \mathbb{E}(X^2)+\mathbb{E}(X)^2 = \mathbb{E}(a^2+(b-a)^2U^2+2a(b-a)U)-(a+b)^2/4 = a^2+a(b-a)+(b-a)^2\mathbb{E}(U^2)-(a+b)^2/4 = \frac{-3(a^2+b^2)+4b-4a+10ab}{12}$

where $\mathbb{E}(U[0,1]^2)=\frac{1}{3(b-a)}$

what did I do wrong?


问:
Show that the characteristic function of a finite signed measure on a normed vector space is uniformly continuous
Let $E$ be a normed $\mathbb R$-vector space, $\mu$ be a finite signed measure on $(E,\mathcal B(E))$ and $$\hat\mu:E'\to\mathbb C\;,\;\;\;\varphi\mapsto\int\mu({\rm d}x)e^{{\rm i}\varphi}$$ denote the characteristic function of $\mu$.

Replying to a previous formulation of this question, Kavi Rama Murthy
 has shown that if $E$ is complete and separable and $\mu$ is nonnegative, then $\hat\mu$ is uniformly continuous.

It is easy to see that his proof still works in the general case as long as we are assuming that $\mu$ is tight$^1$, i.e. $$\forall\varepsilon>0:\exists K\subseteq E\text{ compact}:|\mu|(K^c)<\varepsilon\tag1.$$

Taking a closer look at the proof, I've observed the following: Let $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the duality pairing between $E$ and $E'$ and $$p_x(\varphi):=|\langle x,\varphi\rangle|\;\;\;\text{for }\varphi\in E'$$ for $x\in E$. By definition, the weak* topology $\sigma(E',E)$ on $E'$ is the topology generated by the seminorm family $(p_x)_{x\in E}$.

Now, if $K\subseteq E$ is compact, $$p_K(\varphi):=\sup_{x\in K}p_x(\varphi)\;\;\;\text{for }\varphi\in E'$$ should be a seminorm on $E'$ as well. And if I'm not missing something, the topology generated by $(p_K:K\subseteq E\text{ is compact})$ is precisely the topology $\sigma_c(E',E)$ of compact convergence on $E'$.

What Kavi Rama Murthy
 has shown is that, since $\mu$ is tight, for all $\varepsilon>0$, there is a compact $K\subseteq E$ and a $\delta>0$ with $$|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)|<\varepsilon\;\;\;\text{for all }\varphi_1,\varphi_2\in E'\text{ with }p_K(\varphi_1-\varphi_2)<\delta\tag2.$$

Question: Are we able to conclude that $\hat\mu$ is $\sigma_c(E',E)$-continuous?

EDIT:

In order to conclude that $\hat\mu$ is (uniformly) $\sigma_c(E',E)$-continuous, we need to that $(2)$ holds for $K$ replaced by an arbitrary compact $\tilde K\subseteq E$. Given $\varepsilon>0$, we can show $(2)$ by choosing the compact subset $K\subseteq E$ such that $$|\mu|(K^c)<\varepsilon\tag3.$$

We may then write \begin{equation}\begin{split}\left|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)\right|&\le\underbrace{\int_{K\cap\tilde K}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|}_{<\:\varepsilon}\\&\;\;\;\;\;\;\;\;\;\;\;\;+\int_{K\cap\tilde K^c}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|\\&\;\;\;\;\;\;\;\;\;\;\;\;+\underbrace{\int_{K\cap\tilde K}\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}\left|\mu\right|}_{<\:2\varepsilon}\end{split}\tag4\end{equation} for all $\varphi_1,\varphi_2\in E'$ with $p_{\tilde K}(\varphi_1-\varphi_2)<\delta$, where $$\delta:=\frac\varepsilon{\left\|\mu\right\|},$$ but I have no idea how we can control the second integral.

EDIT 2

A "proof" of this claim can be (found in Linde's Probability in Banach Spaces), but I have no idea why this proof is correct, since he is concluding the continuity immediately from $(2)$ (for a single $K$):

Maybe we need to assume that $\mu$ is even Radon, i.e. that for all $B\in\mathcal (E)$, there is a compact $C\subseteq E$ with $C\subseteq B$ and $|\mu|(B\setminus C)<\varepsilon$. The author is actually imposing this assumption, but he obviously doesn't make use of it in his proof (he would need to consider an arbitrary compact $\tilde K\subseteq E$, as I did above).

$^1$ On a complete separable metric space, every finite signed measure is tight.

答:
Hopefully, I didn't make any stupid mistake, but I think I figured out why the argument in the excerpt is correct.

First of all, let's establish a common understanding of the definitions:

Definition 1

If $(E,\tau)$ is a topological space, then $$\mathcal N_\tau(x):=\{N:N\text{ is a }\tau\text{-neighborhood of }x\}\;\;\;\text{for }x\in E.$$

If $(E_i,\tau_i)$ is a topological vector space, then $f:E_1\to E_2$ is called uniformly $(\tau_1,\tau_2)$-continuous if $$\forall N\in\mathcal N_{\tau_2}(0):\exists M\in N_{\tau_1}(0):\forall x,y\in E_1:x-y\in M\Rightarrow f(x)-f(y)\in N.$$

If $(E_i,\tau_i)$ is a topological vector space, then $\mathcal F\subseteq E_2^{E_1}$ is called uniformly $(\tau_1,\tau_2)$-equicontinuous if $$\forall N\in\mathcal N_{\tau_2}(0):\exists M\in N_{\tau_1}(0):\forall f\in\mathcal F:\forall x,y\in E_1:x-y\in M\Rightarrow f(x)-f(y)\in N.$$

Definition 2: If $(E,\mathcal E)$ is a measurable space, then $$\mathcal M(E,\mathcal E):=\{\mu:\mu\text{ is a finite signed measure on }(E,\mathcal E)\}.$$ If $\mu\in\mathcal M(E,\mathcal E)$, then $|\mu|$ denotes the total variation of $\mu$. Thte total variation norm $\left\|\;\cdot\;\right\|$ on $\mathcal M(E,E)$ is defined by $$\left\|\mu\right\|:=|\mu|(E)\;\;\;\text{for }\mu\in\mathcal M(E,\mathcal E).$$ If $E$ is a Hausdorff space, then $\mathcal F\subseteq\mathcal M(E):=\mathcal M(E,\mathcal B(E))$ is called tight if $$\forall\varepsilon>0:\exists K\subseteq E\text{ compact}:\sup_{\mu\in\mathcal F}|\mu|(K^c)<\varepsilon.$$

Now, it is important to remember the following fact:

Lemma 1: If $(X,\tau)$ is a topological vector space and $p$ is a seminorm on $X$, then

$p$ is $\tau$-continuous;

$p$ is $\tau$-continuous at $0$;

$U_p:=\{x\in X:p(x)<1\}$ is a $\tau$-neighborhood of $0$

are equivalent.

We are ready to establish the following result:

Theorem 1: If $\mathcal F\subseteq\mathcal M(E)$ be $\left\|\;\cdot\;\right\|$-bounded and tight, then $\{\hat\mu:\mu\in\mathcal F\}$ is uniformly $\sigma_c(E',C)$-equicontinuous.

ProofI: Let $\varepsilon>0$. Since $\mathcal F$ is $\left\|\;\cdot\;\right\|$-bounded, $$c:=\sup_{\mu\in\mathcal F}\left\|\mu\right\|<\infty.$$ And since $\mathcal F$ is tight, there is a compact $K\subseteq E$ with $$\sup_{\mu\in\mathcal F}|\mu|(K^c)<\frac\varepsilon3.\tag5$$ Assume $c\ne0$. Then $$\delta:=\frac\varepsilon{3c}$$ is well-defined. Let $$N:=\{\varphi\in E':p_K(\varphi)<\delta\}.$$ Now, $$\int_K\underbrace{\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|}_{\le\:|\varphi-1-\varphi_2|}{\rm d}|\mu|\le\left\|\mu\right\|p_K(\varphi_1-\varphi_2)<\frac\varepsilon3\tag6$$ and hence \begin{equation}\begin{split}|\hat\mu(\varphi_1)-\hat\mu(\varphi_2)|&\le\int\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}|\mu|\\&=\underbrace{\int_K\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|{\rm d}|\mu|}_{<\:\frac13\varepsilon}+\underbrace{\int_{K^c}\underbrace{\left|e^{{\rm i}\varphi_1}-e^{{\rm i}\varphi_2}\right|}_{\le\:2}{\rm d}|\mu|}_{<\:\frac23\varepsilon}<\varepsilon\end{split}\tag7\end{equation} for all $\mu\in\mathcal F$ and $\varphi_1,\varphi_2\in E'$ with $p_K(\varphi_1-\varphi_2)<\delta$; i.e. $$\forall\mu\in\mathcal F:\forall\varphi_1,\varphi_2\in E':\varphi_1-\varphi_2\in N\Rightarrow\hat\mu(\varphi_1)-\hat\mu(\varphi_2)\in B_\varepsilon(0)\tag8.$$

By definition of $\sigma_c(E',E)$, the seminorm $p_K$ is $\sigma_c(E',E)$-continuous. Thus, by Lemma 1, $$N=\delta U_{p_K}\in\mathcal N_{\sigma_c(E',\:E)}(0)\tag9$$ and hence we should have shown the claim.

Remark: I would highly appreciate any confirmation of my proof or any hint to a mistake in the comment section below


问:
Proving integrability in integration by parts in Rudin's text
Integration by parts, as stated in W. Rudin's Principles of Mathematical Analysis, Theorem 6.22, goes as follows:

Suppose F and G are differentiable functions in $[a,b]$, $F'=f\in \mathcal{R}$, and $G'= g\in \mathcal{R}$. Then $\int_a^bF(x)g(x)dx = F(b)G(b) - F(a)G(a) - \int_a^bf(x)G(x)dx$.

The proof is to put $H(x)=F(x)G(x)$ and apply the fundamental theorem of calculus to H and its derivative. The fundamental theorem of calculus is stated as:

If $f \in \mathcal{R}$ on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ s.t. $F'=f$, then $\int_a^bf(x)dx=F(b)-F(a)$.

We also have, from an earlier theorem (6.13), that:

If $f \in \mathcal{R}$ and $g \in \mathcal{R}$ on $[a,b]$, then $fg \in \mathcal{R}$.

Rudin notes that "$H' \in \mathcal{R}$, by Theorem 6.13" (above). Is that theorem really enough to prove this? It doesn't seem to be. After all, if $H(x)=F(x)G(x)$ then $H'(x)=F(x)g(x)+f(x)G(x)$, and we only have $f(x)g(x) \in \mathcal{R}$, while $F(x)g(x) \neq f(x)g(x)$. Where do we get integrability of $F(x)g(x)$ and $f(x)G(x)$?

I'd be grateful for any pointers. Thanks!

答: As $F,G$ are given to be differentiable so they are continuous. And all continuous functions in a closed inteval is integrable so $F,G$ are integrable. Integrability of $Fg,Gf$ follows from this.</p>


问:
Find a cyclic sum
Given that

$$ \frac{a}{bc-a^2}+\frac{b}{ca-b^2}+\frac{c}{ab-c^2}=0,\quad a,b,c\in \mathbb{R}, $$

find the value of

$$ \frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}. $$

I solved it via direct computation which is clearly not the best way to do. Is there any simple transformation I could perform or geometical meaning of this equation?

答:
From first equation (get rid of the denumerators) we get $$abc(a^2+b^2+c^2+ab+bc+ca)= a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(c+a)$$ which can be easly transforemd to $$abc(a^2+b^2+c^2+2ab+2bc+2ca)= (a^2b^2+b^2c^2+c^2a^2)(a+b+c)$$ so $$abc(a+b+c)^2= (a^2b^2+b^2c^2+c^2a^2)(a+b+c)$$

Now if $a+b+c\neq 0$ then we have $$\color{red}{abc(a+b+c)= a^2b^2+b^2c^2+c^2a^2}$$

But since \begin{align}a^2b^2 +b^2c^2&\geq 2b^2ac \\
b^2c^2 +c^2a^2&\geq 2c^2ab \\
c^2a^2 +a^2b^2&\geq 2a^2bc \\
\end{align}
we have always $$abc(a+b+c)\leq  a^2b^2+b^2c^2+c^2a^2$$

with equality iff $a=b=c$. So, since we have equality sign in red equation, we deduce that $a=b=c$, but this can not hold since then fractions are not defined.

So $\boxed{a+b+c =0}$ and now I suppose it is not difficult to finish....

 Let $E$ be the expression. Then $$E = b\Big(\frac{1}{(ca-b^2)^2}-\frac{1}{(bc-a^2)^2}\Big)+      c\Big(\frac{1}{(ab-c^2)^2} -\frac{1}{(bc-a^2)^2}\Big) $$ Value in each big bracket is $0$ so $E=0$.

Title: Find a cyclic sum


问:
Why the sum of the squares of the roots of the $n$th Hermite polynomial is equal to $n(n-1)/2$?

How to prove that the sum of the squares of the roots of the $n$th Hermite polynomial is $\frac{n(n-1)}{2}$?

I tried with Vieta formulas, but it's hard. I appreciate a proof or reference to it.
An idea is to use the definition of sum of Hermite polynomials, but do not know.

答:
I would suggest to use Statisticians Hermite polynomials instead.  Of course it does not make the answer different, just imho easier not having to deal with all the 2s.  Also monic polynomials have a simpler Vieta formula.

The answer given is correct - with one minor typo
$$\operatorname{He}_n(x) = n!\sum_{m=0}^n \frac{(-1)^m}{2^m m!(n-2m)!}x^{n-2m}$$
Giving the coefficient in front of the $x^{n-2}$ term
$-\frac{n(n-1)}{2}$ (the 2 is just due to the statisticians Hermite, it is the 1 in the numerator that is the typo).

The fact that there is no $n-1$ coefficient means that the sum of the roots are zero.  Implying that they always come in symmetric pairs, for every root $x_k$ there is a value that is $-x_k$.  For odd Hermite polynomials, this implies zero is always a root.

This also means that $e_2$ is explicitly the negative sum of all squares itself, no need for the subtraction step in the earlier response.
$$e_2(\{x_k\}) = x_0x_1 + x_1x_2 + ... x_{n-1}x_n$$

Since every $x_k$ has a negative pair, you can pair them up as you wish and these become negative squares of $x_k$.  If $n$ is even you have accounted for all of them, if odd, the unpaired root is zero, so its entries do not contribute to the sum.

However, the sum $\sum_k x_k^2$ actually counts each value twice - since each root has a symmetric (negative) partner, the value will appear twice in this sum.  So, the sum of squares is $-2$ times the coefficient of the $n-2$ polynomial expansion of $He_n(x)$.

Now all you have to do is read off the coefficient in front of $x^{n-2}$ as in the original response.

$$\sum_k x_k^2=-2e_2(\{x_k\})$$
$$ \sum_k x_k^2= n(n-1)$$

Therefore the sum of the squares is not $n \choose 2$, but twice this amount.  Look at the first few for confirmation.  $n=2, x_k\in\{-1,1\}\implies\sum_k x_k^2=2$, $n=3,x_k\in\{-\sqrt{3},0,\sqrt{3}\}\implies\sum_k x_k^2=6$.


问:
Associativity of multiplication in a field
I have the following definition of a field:

A field is a set $F$ together with two commutative binary operations $(+
 ,\times)$ and distinct elements $0$ and $1$ such that $F$ is a group under $+$ with identity $0$, $F\setminus\{0\}$ is a group under $\times$ with identity $1$ and the distributive law

$x \times (y + z) = x \times y + x \times z$

holds

My question is, since  $F\setminus\{0\}$ is a group under $\times$, associativity of $\times$ holds for $F\setminus\{0\}$, but does that mean that associativity of $\times$ does also holds for $F$ itself? Or do you have to prove that associativity for $\times$ in $F$ also holds in the case where $0$ is one of the elements you are multiplying by?

I saw an alternative definition of a field on wikipedia where $\times$ is defined to be associative. So is this property actually already part of the above definition?

答:
Multiplicative associativity also holds if you include $0$, since $0$ is absorbing, i.e., for all $x\in F$,

$x\cdot 0 = 0$.

Absorption can be proved as follows:

$0=0+0$ and so by multiplying with $x$, $x0 = x0+x0$. Adding $-x0$ on both sides gives $0=x0$.


问:
calculating $\lim_{x\to1} \space\space \space \dfrac{a}{1-x^{a}}-\dfrac{b}{1-x^{b}}$
If $\space a,b \in \mathbb{N}$ ,then calculate the limit $$\lim_{x\to1} \space\space \space \dfrac{a}{1-x^{a}}-\dfrac{b}{1-x^{b}}$$ and hence find $$ \lim_{x\to1} \space\space \space \dfrac{40}{1-x^{40}}-\dfrac{30}{1-x^{30}}$$
My approach: simplified
$\lim_{x\to1} \space\space \space \dfrac{40}{1-x^{40}}-\dfrac{30}{1-x^{30}}$

to $\lim_{x\to1} \space\space \space \dfrac{5}{1-x^{5}}-\dfrac{15}{1-x^{15}}$

Unable to solve further. and also not able to generalize the result.

Any help would be appreciated .

THANKS!

答: Using L'Hôpital's rule twice, we seek$$\begin{align}\lim_{x\to1}\frac{a-b+bx^a-ax^b}{1-x^a-x^b+x^{a+b}}&=\lim_{x\to1}\frac{abx^{a-1}-abx^{b-1}}{-ax^{a-1}-bx^{b-1}+(a+b)x^{a+b-1}}\\&=\lim_{x\to1}\frac{ab(a-1)x^{a-2}-ab(b-1)x^{b-2}}{-a(a-1)x^{a-2}-b(b-1)x^{b-2}+(a+b)(a+b-1)x^{a+b-2}}\\&=\frac{ab(a-b)}{-a(a-1)-b(b-1)+(a+b)(a+b-1)}\\&=\frac{a-b}{2}.\end{align}$$While the limit you seek is $5$, the limit you thought was equivalent is in fact $-5$.


问:
An application of taylor formula
Let $\phi\in \mathcal{S}(\mathbb{R}^n)$, i.e. $\phi$ is a Schwartz function, is true that:
$$|2\phi(x)-\phi(x+y)-\phi(x-y)|\leq |D^2\phi(x)||y|^2,\quad\forall y\in\mathbb{R}^n???$$
I have no idea on how to proceed, any help is appreciated.

答:
No, this inequality is not true. For a counterexample, let $n=1$ be the dimension and consider the function $\phi(x) = (x^2+x^4)\rho(x)$, where $\rho(x)$ is some smooth compactly-supported bump function with $\rho \equiv 1$ on $(-1,1)$. Then clearly $\phi\in\cal S(\Bbb R)$ since it is smooth and compactly-supported.

Now, in a neighborhood of $0$ we have $\phi''(x) = 2+12x^2$, meaning $\phi''(0)=2$. Since $\phi(0)=0$ as well, your inequality at $x=0$ becomes
$$|-\phi(y)-\phi(-y)|\leq 2|y|^2.$$
On the other hand, for $y$ small we can compute the left-hand side to be $2(y^2+y^4)$. Since this exceeds $2y^2$ for $y\neq 0$, we conclude your inequality is false.

Title: An application of taylor formula


问:
tensor rotation
why does tensor rotation require multiplication by the rotation matrix twice, once from the right and once from the left by the inverse?

if $T$ is the tensor I wish to rotate and $R$ is the rotation matrix, why isn't $T'=RT$ but is $T=RTR^{-1}$?

I have seen and understood the construction of tensor transformation, but I am intuitively uncomfortable with it.

thank you

答:
Consider the following example

Let's say we have a certain vector v that we would like to first transform somehow and then rotate.(Lets say that the transformation we want to apply to v is a simple scaling, which is simpler to describe wrt to orthodox v i.e. before we rotate it) This leads us to having two options for the operations:

First transform our vector v with Tv and then apply the rotation i. e RTv

First apply the rotation and then transform the vector i.e. TRv

Doing this would yield 2 different results, in order to avoid this if we go down the 2nd pathway we would have to adjust our transformation matrix T by "rotating" it i.e. in order to get the same result from the first pathway RTv = b we would have to adjust our transformation T to T' s.t. T'Rv = b which leads to the simple description of how we should adjust T' i.e.

RT = T'R

multiply on the right with R^(-1) would yield the following relationship between the transformed T tensor (T') and the original one T

RTR^(-1) = T'

Title: tensor rotation


问:
Condition For No Existence Of Real Root
$2x^{4}+5ax^{3}-2bx^{2}+1=0$ has no real root in $(5,2014)$

Find the conditions for $a$ and $b$

I am suspicious of even the existence of its solution and at a loss.

答:
This answer assumes $a,b>0$. (If it were otherwise, the presence of the minus sign in the $-2bx^2$ term seems unnecessary). If this is not a valid assumption, my result at least treats one of the four possibilities; note that if $a>0,b<0$ the equation has no positive roots at all, so really only two cases remain.

Anyway let $p(x)=2x^4+5ax^3-2bx^2+1$ and note that since $p(0)=1$ we may define $q(x)=p(x)/x^2$ and $q(x)$ will have the same zeroes as $p(x).$
$$q(x)=5ax-2b +2x^2+x^{-2},\\ q'(x)=5a+4x-2x^{-3}.$$
Since $a>0$ we see that $q'(x)>0$ for $x \in [5,2014].$ [in fact $q'>0$ on $((1/2)^{1/4},\infty)].$ This means that $q$ is increasing on $[5,2014]$ so that $q$ maps this interval bijectively (and continuously) to the interval $[q(5),q(2014)].$ Now the condition that there be no zero in the open interval $(5,2014)$ becomes equivalent to saying that either $q(5) \ge 0$ or $q(2014) \le 0.$ Luckily both expressions for $q(5)$ and for $q(2014)$ involve the same subtracted term $-2b$; in fact
$$q(5)=25a-2b+1251/25,\ \ q(2014)=10070a-2b+k,$$
where $k$ is an ungainly fraction. We can then express the conditions on $a,b$ for no zero in $(5,2014)$ by saying that $b$ satisfies, in terms of $a$, one of the inequalities
$$b \le (25/2)a+1251/50,$$
which corresponds to $q(5) \ge 0$, or else
$$b \ge 5035a+k_2,$$
where $k_2 \approx 4.056\cdot 10^6$, which corresponds to $q(2014) \le 0.$

Title: Condition For No Existence Of Real Root


答:
You can prove that $(F, \times)$ is associative using the field axioms you cite. Associativity states that for any $a,b,c \in F$, $(ab)c=a(bc)$. If none of $a$, $b$, or $c$ are zero then this equation is in $F^\times$, where it is true. Suppose then that at least one of $a$, $b$, or $c$ are zero. We'll need the following lemma.

Lemma. Let $a \in F$. Then $0a=0$.

Proof. $0a + 0a = (0+0)a =0a$ by distributivity. Then as $(F,+,0)$ is a group, $0a=0$ (subtract $0a$ from both sides).

Using this lemma, we can see that if any of $a,b,c$ were zero then $(ab)c=0$ and $a(bc)=0$. To be rigorous, we can do the 3 cases.

$a=0$: $a(bc)=0(bc)=0$ and $(0b)c=0c-0$.

$b=0$: $a(0c)=a0=0$ and $(a0)c=0c=0$.

$c=0$: $a(b0)=a0=0$ and $(ab)0=0$.

Title: Associativity of multiplication in a field

Body:
I have the following definition of a field:

A field is a set $F$ together with two commutative binary operations $(+
 ,\times)$ and distinct elements $0$ and $1$ such that $F$ is a group under $+$ with identity $0$, $F\setminus\{0\}$ is a group under $\times$ with identity $1$ and the distributive law

$x \times (y + z) = x \times y + x \times z$

holds

My question is, since  $F\setminus\{0\}$ is a group under $\times$, associativity of $\times$ holds for $F\setminus\{0\}$, but does that mean that associativity of $\times$ does also holds for $F$ itself? Or do you have to prove that associativity for $\times$ in $F$ also holds in the case where $0$ is one of the elements you are multiplying by?

I saw an alternative definition of a field on wikipedia where $\times$ is defined to be associative. So is this property actually already part of the above definition?


问:
Existence of $\phi\in C_{0}^{\infty}(\Omega)$
I have the following problem that I am struggling to prove :

If $K \subset \Omega$ and $K$ is a compact set, then there exists $\varphi \in C_{0}^{\infty}(\Omega)$ such that $0 \leq \varphi(x) \leq 1$ for all $x \in \Omega$ and $\varphi(y)=1$ for all $y \in K$

I am completely clueless in proving it. I have another lemma that states that If $\varphi \in C_{0}^{m}\left(\mathbb{R}^{n}\right), m \geq 0, f \in L_{l o c}^{1}\left(\mathbb{R}^{n}\right),$ then $\varphi * f \in C^{m}\left(\mathbb{R}^{n}\right)$ and
$$
D^{\alpha}(\varphi * f)=\left(D^{\alpha} \varphi\right) * f
$$
I don't know how to use this in the proof as well.

答:
If $K=\emptyset$, then take $\varphi=0$ and we are done. Now Assume that $K\neq\emptyset$ and define :
$$
d:=
\begin{cases}
1&\text{if $\Omega=\mathbb{R^{n}}$} \\
\displaystyle\frac{\text{dist}(K,\Omega^{c})}{3}&\text{if otherwise}
\end{cases}
$$

Let $f$ be the characteristic function of the set $C=K+B(0, d)$ and define $\varphi:=J_{d} * f,$ where $J_{d}$ is a mollifier. We have that $\varphi \in C^{\infty}\left(\mathbb{R}^{n}\right)$ by the lemma you provided and by :
$$
\varphi(x)=\int_{B(x, d) \cap C} J_{d}(x-y) d y, \quad x \in \mathbb{R}^{n}
$$
since $B(x, d) \cap C=B(x, d)$ for $x \in K$ and $B(x, d) \cap C=\emptyset$ when :
$$
x \notin K+B(0,2 d) \subset \Omega
$$
$$
$$

[Note] : We say that $J_{d}$ is a mollifier, if for $d\in(0,+\infty)$ we  have :
$$J_{\varepsilon} \in C_{0}^{\infty}\left(\mathbb{R}^{n}\right), \quad J_{\varepsilon} \geq 0, \quad J_{\varepsilon}(x)=0 \text { for }|x| \geq \varepsilon, \quad \int_{\mathbb{R}^{n}} J_{\varepsilon}=1$$
I assume you are aware of this and of the method involvded in constructing mollifiers. Also note that your definition did not specify what the range of  $\alpha$ is but your lemma would be valid if $$|\alpha|\leq m$$


答:
A better definition (imho) is this:

A ring is a tuple $(R,+,0,\cdot,1)$ with a set $R$ and binary operations $+,\cdot$ such that $(R,+,0)$ is a commutative group and $(R,\cdot,1)$ is a monoid (so a group without the existence of an inverse as an axiom). In addition, multiplication distributes over addition (from the left and from the right).

A ring is called commutative if its multiplicative monoid is commutative.

A field is a commutative ring where $(R\backslash\{0\},\cdot,1)$ is also a group.

So multiplication by $0$ is still associative by definition, since $(R,\cdot,1)$ is still a monoid, though not a group. But we could also deduce associativity, if we were so inclined, since multiplication by $0$ always yields $0$, so $(0a)b=0b=0=0(ab)$.


问:
Doubt in proof of integrating by part
In the proof of integration by parts given in the photo below I see f(X)•g(x) get differntiated and I see the dx below them move to the RHS, so I left hand side we should have d(f(X)•g(X)) right? I am not able to understand how did dx come in the LHS

答: This proof keeps repeating the same typo. When they write $\int fgdx$, it should be $fg$. Or if we work with definite integrals, $\int_a^b(fg^\prime+gf^\prime)dx=[fg]_a^b=f(b)g(b)-f(a)g(a)$. However, the later special case $g=\int vdx$ has been handled correctly.

Title: Doubt in proof of integrating by part


问:
Is this true lemma in the proof of momentum map $J$ is a Poisson map?
Suppose $(M, \omega)$ be a symplectic manifold equipped with $G$-action $\Phi : G \times M \to M$. Assume $\Phi^*_g \omega = \omega$.

Let $J$ be an equivariant momentum map of $M$, $J : M \to \mathrm{Lie}(G)^*$. In here, equivariance of $J$ means
\begin{align}
 \mathrm{Ad}_g^* J = J \circ \Phi_{g}
\end{align}

Then, does it always hold,
\begin{align}
J_{\mathrm{d}_{J(x)}f}(x) = f \circ J(x)
\end{align} for any $x \in M$ and $f : C^\infty(\mathrm{Lie}(G)^*, \mathbb{R})$? In here, I use the identification,
\begin{align}
 \mathrm{d}_{J(x)}f : T\mathrm{Lie}(g)^* \cong \mathrm{Lie}(g)^*  \overset{linear}{\to} \mathbb{R}
\end{align}

If $f$ is linear functional, i.e. $f \in \mathrm{Lie}(G)$, then I can show this. But, does it holds for general case?

If it's not true, how can I approach to prove
$$
\{ f \circ J, g \circ J \}_M = \{ f, g \}_{\mathrm{Lie}(G)^*} \circ J
$$ where
$$
\{ f, g\}_M = \omega(X_f, X_g)\\
\{ f, g\}_{\mathrm{Lie}(G)^*} = \mu \mapsto \mu([\mathrm{d}_\mu f , \mathrm{d}_\mu g])
$$?

The reason why I tried to prove the above lemma is,
I get the result that It's enough to prove
$$
\{f \circ J, g \circ J\}_M = \{J_{\mathrm{d}_{J(x)}f}(x), J_{\mathrm{d}_{J(x)}g}(x)  \}_M
$$

答:
Your claim that for all $f\in C^{\infty}(\mathfrak{g}^{*})$ and $x\in M$,
$$
J_{d_{J(x)}f}(x)=f(J(x))
$$
is not true. For instance, if $f\equiv 1$ then the LHS is zero, while the RHS is equal to $1$.

Note that $J:(M,\{\cdot,\cdot\}_{M})\rightarrow (\mathfrak{g}^{*},\{\cdot,\cdot\}_{\mathfrak{g}^{*}})$ matches the Poisson brackets exactly when for all $f\in C^{\infty}(\mathfrak{g}^{*})$, the Hamiltonian vector fields $X_{f}$ and $X_{J^{*}f}$ are $J$-related. In fact, it suffices to check this condition for functions $f\in C^{\infty}(\mathfrak{g}^{*})$ whose differentials span the cotangent spaces of $\mathfrak{g}^{*}$. Such a collection is given by $\{h_{v}:v\in\mathfrak{g}\}$, where
$h_{v}\in C^{\infty}(\mathfrak{g}^{*})$ denotes the linear function on $\mathfrak{g}^{*}$ corresponding with $v\in\mathfrak{g}$.

For all $v\in\mathfrak{g}$, we have $J_{v}=J^{*}(h_{v})$. Taking Hamiltonian vector fields, we get
$$
v^{\sharp}=X_{J^{*}(h_{v})},
$$
where $v^{\sharp}$ denotes the fundamental vector field of the action corresponding to $v$.

Show that $v^{\sharp}$ and $X_{h_{v}}$ are $J$-related. This is where equivariance of $J$ comes into play.

We now know that $X_{J^{*}(h_{v})}$ and $X_{h_{v}}$ are $J$-related, so the conclusion follows.


问:
Showing that some point is in a compact set
$(X, d)$ is a metric space. $W$ is a compact subset of $X$.

Some $x \in X$ satisfies that
$$
\forall \epsilon > 0: \exists y \in W: d(y, x) < \epsilon
$$

Then, can we conclude that $x \in W$?

If the proposition is modified to $\exists y \in W:y\neq x \land d(y,x) < \epsilon$, the point $x$ is a limit point of $W$, hence in $W$.

答: For each $n\in\Bbb N$, there is some $w_n\in W$ such that $d(w_n,x)<\frac1n$. Therefore, $x=\lim_{n\to\infty}w_n$, and so $x\in\overline W=W$, since $W$ is compact and therefore closed.


问:
Showing the inequality for $L^2( \mu)$ function
This is the problem from analysis. (In Lebesgue theory).

Question : Suppose $\int_{\mathbb R^n} |f|^2 d \mu < \infty$. Show that $\mu (\{x \in \mathbb R^n  | |f(x)|>a\})\leq \frac{1}{a^2} \int_{\mathbb R^n} |f|^2 d\mu$.

I tried to solve it using Bessel's inequality...but it is not quite clear... How can I solve it?

答:
Chebyshev's Inequality

This is known as Chebyshev's Inequality.

Let $E_a=\{x:|f(x)|\ge a\}$, then for $p\gt0$
$$
a^p|E_a|\le\int_{E_a}|f(x)|^p\,\mathrm{d}x\le\int_{\mathbb{R}^n}|f(x)|^p\,\mathrm{d}x\tag1
$$
Thus,
$$
|E_a|\le\frac1{a^p}\int_{\mathbb{R}^n}|f(x)|^p\,\mathrm{d}x\tag2
$$
For your inequality, take $p=2$.

Slightly Better Result

If $f\in L^p$,
$$
\lim_{a\to\infty}\int_{E_a}|f(x)|^p\,\mathrm{d}x=0\tag3
$$
Thus, using the left hand inequality from $(1)$, in addition to $(2)$, we get that
$$
\lim_{a\to\infty}a^p|E_a|=0\tag4
$$


问:
How to read and execute $\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}}$

How to read and execute this sum?
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}}$$

I am having trouble to understand where is my error.

The question does not say, but I am assuming that $\ell$ starts at $1$, $m$ at $2$, and so $n$ at $3$.

This is essential a product of pg:

$$\sum_{1 \leq \ell<m<n} \frac{1}{5^{\ell}3^{m}2^{n}} = \sum_{\ell=1}\frac{1}{5^{\ell}}\sum_{m=2}\frac{1}{3^{m}}\sum_{n=3}\frac{1}{2^{n}} = \frac{1/5}{1-1/5}\frac{1/9}{1-1/3}\frac{1/8}{1-1/2}$$
But this does not agree with the answer :/

答:
Assuming $n$ is not fixed:

Rewrite as triple sum
$$\sum_{1 \leq \ell <m<n} \frac{1}{5^{\ell}3^{m}2^{n}} = \sum_{\ell=1}^\infty\sum_{m=\ell+1}^\infty\sum_{n=m+1}^\infty\frac{1}{5^{\ell}3^{m}2^{n}} =\sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \sum_{n=m+1}^\infty \frac{1}{2^n} $$
Then, this should be obvious in terms of how to read and execute.

Indeed, since each of them are Geometric Progressions, we have
$$\begin{align}\sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \sum_{n=m+1}^\infty \frac{1}{2^n} &= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{3^m} \left(\frac{1}{2^m}\right) \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell} \sum_{m=\ell+1}^\infty \frac{1}{6^m} \\&= \sum_{\ell=1}^\infty \frac{1}{5^\ell}\left(\frac{1}{5\cdot 6^\ell}\right) \\& = \frac{1}{5}\sum_{\ell=1}^\infty \frac{1}{30^\ell} \\&= \frac{1}{5}\cdot\frac{1}{29} \\&= \frac{1}{145}\end{align} $$

Assuming $n$ is fixed:

The sum will be a finite double sum and you can exclude $1/2^n$ from the sum(treating that as a constant). Then, again use the GP formula to calculate each $\ell$-sum and $m$-sum.


问:
Show that $f : X → \Bbb{R}$ is measurable if and only if $\{x ∈ X : f(x) > r\}$ is measurable for every $r ∈ \Bbb{Q}$
$f  : X → \Bbb{R}$ is a function on a measurable space $(X, χ)$.  I want to show that $f : X → \Bbb{R}$ is measurable if and only if $\{x ∈ X : f(x) > r\}$ is measurable for every $r ∈ \Bbb{Q}$.

I know, given alpha, I need to choose $r_n ∈ \Bbb{Q}$ with $r_n$ bring a decreasing sequence in $\Bbb{Q}$ but I'm unsure where to go from here. Any help at all would be appreciated!

答:
Given $r \in \mathbb{R}$, choose a sequence $(q_n)$ of rationals such that $q_n \searrow r$. Then
$$\{f > r\} = \bigcup_n \{f > q_n\}$$ so $\{f  > r\}$ is measurable.


问:
Maximum flow (Ford-Fulkersons)
Hi I tried to use the algoritm, but im not sure if my answer in correct. I will show the solution and my answer. I got that the maximum flow 32, but the flow on the edges are diffrent in some cases.Is my answer correct also?

答: Yes, your answer is also correct. If you modify your answer by sending an extra flow of $1$ around the loops $\ a\rightarrow b\rightarrow g \rightarrow h \rightarrow a\ $ and $\  b\rightarrow g \rightarrow d \rightarrow b\ $, thus increasing the flow on the forward edges $\ (a,b)\ $ and $\ (g,d)\ $ by $1$, decreasing it on the backward edges $\ (h,g)$,$\ (a,h)\ $, and $\ (b,d)\ $ by $1$, and on the backward edge $\ (g,b)\ $ by $2$, you'll end up with the answer given in the book. This doesn't change the net flow leaving any node, which remains at $32$ for node $a$, $-32$ for node $z$, and $0$ for all other nodes.

Title: Maximum flow (Ford-Fulkersons)


问:
Finding the values of $\cos \frac{n\pi}{2}$ and $\sin \frac{n\pi}{2}$.
i know that the values of $\cos n\pi=(-1)^{n}$ and $\sin n\pi=0$. Now i want to know that what is the general expressions of $\cos \frac{n\pi}{2}$ and $\sin \frac{n\pi}{2}$.</p>


问:
Conics meeting 8 general lines
I am trying to show that the number of plane conics in $\mathbb{P}^3$ meeting $8$ general lines is $92$, using what I know about intersection theory.

I started considering the tautological bundle $S$ of $G=G(2,3)$, and I took $P=P(\operatorname{Sym}^2S^{\vee})$. $P$ is a projective bundle of rank $5$ over $G$ whose fiber over the plane $H$ consists of the conics lying on $H$.

I also computed the Chow group of $P$, i.e. $$\mathbb{Z}[h,t]/(h^4,t^6+4ht^5+10h^2t^4+20h^3t^3)$$

What I cannot do is to find the cycle of the conics meeting a given line $L$ (I cannot even prove that it is a closed subscheme!). After that, I just need to elevate it to the eighth power and I should get $92$. Could anyone help?

答:
Let $D$ be the divisor corresponding to conics in $P$ which meet a fixed line $L$, and I will show that $[D]=2h+t$. This is discussed in detail in section 9.7 of the book 3264 and all that, and actually I write nothing new.

There are two ways to compute this: a smart one and a stupid one.

The smart way:
Let $U\subset P$ be the open set of pairs $(H,\xi)$ such that $H$ does not contain $L$. Note the complement of $U$ has codimension $2$, so it does not change the divisor classes. Consider the natural morphism
$$\alpha\colon U \to L$$
which sends $(H,\xi)$ to $p=H\cap L$. Note that $\mathcal O_U(-1)$ is the tautological bundle, i.e. the fibre at $(H,\xi)$ is precisely the $1$-dimensional linear space of quadratic form $\mathcal Q \in \xi$. Therefore, the evaluation
$$\mathcal Q \mapsto \mathcal Q(p)$$
defines a morphism of line bundles
$$ \mathcal O_U(-1) \to \alpha^* \mathcal O_L(2)$$
and the desired divisor $D$ is precisely the zero locus of this morphism. This is equivalent to the zero locus of some section of the line bundle $\mathcal O_U(1) \otimes\alpha^* \mathcal O_L(2)$, which is exactly $t+2h$.

The stupid way:
Since you have already computed the Chow ring, we can write
$$[D]=a h + b t$$
for some (undetermined) integers. To solve $a$ and $b$, we test how it intersects with some special curves in $P$. Precisely, we can look at the following two curves:

Fixed a general $H$ and let $\{C_\lambda \subset H\}$ be a general pencil of conics. This defines curve $\Gamma\subset P$.

Fixed a general quadric surface $Q\subset \mathbb P^3$, and let $\{ H_\lambda\subset \mathbb P^3\}$ be a general pencil of (hyper)planes. Then the intersections $\{ H_\lambda \cap Q \}$ are conics, hence defines a curve $\Phi \subset P$.

We compute the intersection numbers between the divisors classes $h$, $t$, $[D]$ and the curves $[\Gamma]$, $[\Phi]$.

$[\Phi]\cdot h$. On $\mathbb P^{3*}$, the image $\Phi$ meets $\mathcal O_{\mathbb P^{3*}}(1)$ at precisely one point, and hence easy to see $[\Phi]\cdot h=1$.

$[\Phi]\cdot t$. This is the degree of the restriction of $\mathcal O_P(1)$ to the curve $\Phi$. We compute with its dual $\mathcal O_P(-1)$, as it has more clear geometric meaning: the fibre of $\mathcal O_P(-1)$ at $(H,\xi)$ is precisely the $1$-dimensional linear space $\xi$. Consider the $F\in H^0(\mathcal O_{\mathbb P^3}(2))$ which defines $Q$, then its restrictions to $H_\lambda$ give an nowhere-zero section of  $\mathcal O_P(-1)|_\Phi$. Therefore, the restriction is a trivial bundle, and hence $[\Phi]\cdot t=0$

$[\Phi] \cdot [D]$. This is equivalent to compute $[Q]\cdot[L]$, which is clearly to be $2$.

I will omit the computation with $[\Gamma]$ which is in the same sprit. After all this, you can get the intersetion numbers as follows:

h
t
[D]

$[\Gamma]$
0
1
1

$[\Phi]$
1
0
2

Therefore,
$$1= [\Gamma] \cdot [D] =[\Gamma] \cdot (a h+ b t)=b$$
and
$$2=[\Phi] \cdot [D] = [\Phi] \cdot (ah+b t) =a.$$

Title: Conics meeting 8 general lines


问:
Constructing a triangle given an angle, the side opposed to this angle, and the median to the given side.
I came across the following problem in my Euclidean Geometry text:

Construct a triangle having given an angle, the side opposed to this angle, and the median to the given side.

I have tried solving this problem using method of loci, with Locus 1 being the circumcircle of the sought for triangle and Locus 2 being the circle of radius of the given  median. However I don't know how to start this construction.

答:
I found this method experimentally. I am still looking for some theorem to prove following statement:

For a certain measure of BC which is the base of triangle ABC and radius of a circle center O at mid point of BC, with radius equal to measure of median $m_a$ we have:

$\angle BAC - \angle BPO= constant$

In figure AC||PO is the condition of above relation.

We use this fact to construct a triangle it's base, median and angle $\alpha $ opposite to base is given:

Draw a circle center on midpoint O of base BC. Draw PO and PB such that:

$\angle BPO=\frac{\alpha}2$

Draw a line from C parallel with PO, it intersect the circle at A. Connect A to B , The measure of  $\angle BAC$  will be:

$\angle BAC- \angle BPO=d$

Now to get $\angle BAC=\alpha$, construct $\angle BPC = \frac {\alpha}2 + d$ and draw a line from C parallel with PO to get vertex A.

In figure givens are $\alpha=40^o$, $BC=110 mm$ and $m_a=115 mm$


问:
If $f$ is integrable, show that $f(x)$ is finite almost everywhere
If $f$ is integrable, show that $f(x)$ is finite almost everywhere.

I know that $f$ is integrable/finite if and only if $\vert f \vert$ is integrable/finite, so I only have to show it for $f \geq 0$. I think I should use the set $E = \{x \mid f(x)= \infty\}$ and show that its measure is zero, but I'm not sure why this shows that integrable implies finite a.e. or how to show that its measure is zero.

答:
It suffices to show that $|f| < \infty$ almost everywhere since this implies $f < \infty$ almost everywhere.

Let $A = \{x: |f(x)| = +\infty\}$ and suppose $\mu(A) = M > 0$. For each $n \in \mathbb{N}$, $n \chi_{A}$ is a simple function and we have $0 \leq n \chi_{A} \leq |f|$, and so by definition of the Lebesgue integral for a non-negative function

$$
\int |f| \geq \sup_{n} \int n \chi_{A} = \sup_{n} nM = +\infty
$$
since $M > 0$. This contradicts the fact that $\int |f| < \infty$ so we conclude that $\mu(A) = 0$.


问:
Pushforward of pullback of a sheaf
Are there any reasonable hypotheses on a map $f: X \to Y$ and a sheaf $E$ on $Y$ so that $f_* f^* E \cong E$?</p>

答:
This question may have general interest to students in algebraic geometry.

If $f:X:=Spec(B)\rightarrow S:=Spec(A)$ is a map of affine schemes and if $E$ is an $A$-module with $\mathcal{E}$ the corresponding $\mathcal{O}_X$-module,
it follows $f_*f^*(\mathcal{E})$ is the sheafification of $B\otimes_A E$ as left $A$-module. There are canonical $A$-linear map $i:E\rightarrow B\otimes_A E$ and $\rho:B\otimes_A E \rightarrow E$ defined by $i(e):=1\otimes e$ and $\rho(b\otimes e):=be$. The maps $i,\rho$ are seldom isomorphisms.

Question. "Are there any reasonable hypotheses on a map $f:X\rightarrow Y$ and a sheaf $\mathcal{E}$ on $Y$ so that $f_*f^*\mathcal{E} \cong \mathcal{E}$?"

Answer: In the affine situation it follows the sheaf $f_*f^*(\mathcal{E})$ is seldom isomorphic to $\mathcal{E}$. In the trivial case when $X=S$ it is true.

Title: Pushforward of pullback of a sheaf


问:
A limit on binomial coefficients
Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.

What I can do is just use Stolz formula. But I could not proceed.

答:
The sum does not diverge, but converges. I suspect it converges to $1/2$ based on the value at $n=5000$ [link], but I've only managed to bound it by $3/4$.

From Wikipedia,
$$\binom{n}{k} \leq \left(\frac{n \cdot e}{k}\right)^k$$

Then,
$$\begin{align} x_n=\frac{1}{n^2}\sum_{k=0}^n\ln \binom{n}{k}&\leq \frac{1}{n^2}\sum_{k=0}^n\ln \left(\frac{n \cdot e}{k}\right)^k\\
&=-\frac{1}{n^2}\sum_{k=0}^n k\ln \left(\frac{k}{n \cdot e}\right)\\
&=-\frac{1}{n}\sum_{k=0}^n \left(\frac{k}{n}\right)\ln \left(\frac{k}{n \cdot e}\right)\end{align}$$

Looking at this as a Riemann Sum,

$$\begin{align}
\lim_{n \to \infty} x_n &\leq \lim_{n \to \infty}-\frac{1}{n}\sum_{k=0}^n \left(\frac{k}{n}\right)\ln \left(\frac{k}{n \cdot e}\right) \\
&= \int_0^1 -x\ln\left(\frac{x}{e}\right)\mathrm{d}x
\\
&=\left.\frac{x^2}{4}-\frac{x^2}{2}\ln x+\frac{x^2}{2}\ln e\right|_0^1
\\
&=\frac{3}{4}
\end{align}$$

You can also bound it from below by $\frac{1}{4}$ by taking $\binom{n}{k} \geq  \left(\frac{n}{k}\right)^k$.

Title: A limit on binomial coefficients


问:
Poisson Distribution to estimate the probability of consecutive cards in terms of the same "face value" on a Deck
We were given this problem but I'm not sure I understood the term "face value". Does it mean cards with "faces" like Jack, Queen, King? or Does it pertain to the value a card holds like 4 of Spades has the same value with 4 of hearts? Maybe someone else had a similar problem before, please tell me what "face value" means in context of these types of problems.

Here's the original problem:

Suppose you have a well-shuffled deck of 52 cards.  On the average, how many pairs of consecutive cards on the deck have the same face value? Estimate the probability that such a pair exists using Poisson distribution.

I'm also confused with what a well-shuffled deck means. Initially, I am thinking that it suggests that all cards have the same chances of being drawn.

答:
Let $X_i$ be an indicator random variable that is equal to $1$ if the card following it has the same face value, and $0$ otherwise

Then $\Bbb P(X_i) = \frac3{51}$

Now the expectation of an indicator r.v. is just  the probability of the event that it indicates, so $\Bbb E(Xi) = \frac 3{51}$

and by linearity of expectation, $\Bbb E(X) =\Bbb E(X_1) + \Bbb E(X_2) +... \Bbb E(X_{51}) = 51\cdot \frac3{51} = 3$


问:
Prove that intersection with each sub vector space is contained with the intersection of the sum of those vector spaces.
I'm trying to find a way to prove the following:

Let there be $U, W_1, W_2$ sub vector spaces of linear space V

Prove that: $(U\cap W_1 ) + (U\cap W_2 ) \subseteq U \cap (W_1+W_2 )$

Logically, I can think see it, as the sum of $W_1 +W_2 $ is larger, but I can't think of the steps to take to prove it.

Help will be appreciated.

答:
To show that $(U \cap W_1) + (U \cap W_2) \subseteq W \cap (W_1 + W_2)$, we will take an arbitrary element of $(U \cap W_1) + (U \cap W_2)$ and show that it is in the set $U \cap(W_1+W_2)$.

Let $x \in (U \cap W_1) + (U \cap W_2)$. By definition, there exist vectors $ x_1 \in (U \cap W_1)$ and $x_2 \in (U \cap W_2)$ such that $x=x_1+x_2$. Observe the following:

Because $x_1 \in U \cap W_1$, then $x_1 \in U$ and $x_1 \in W_1$.

Because $x_2 \in U \cap W_2$, then $x_2 \in U$ and $x_2 \in W_2$.

$U$, $W_1$ and $W_2$ are vector spaces and thus are closed under vector addition. So,

$x_1 \in U$, $x_2 \in U$, therefore $x=x_1+x_2 \in U$

$x_1 \in W_1$, $\textbf{0}_\textbf{V} \in W_2$, therefore $x_1=x_1+\textbf{0}_\textbf{V} \in (W_1+W_2)$

$x_2 \in W_2$, $\textbf{0}_\textbf{V} \in W_1$, therefore $x_2=x_2+\textbf{0}_\textbf{V}\in (W_1 + W_2)$

$x_1 \in (W_1 + W_2)$, $x_2 \in (W_1 + W_2)$, therefore $x = x_1 + x_2 \in (W_1 + W_2)$

$x \in (W_1 + W_2)$, $x \in U$, therefore $x \in U \cap (W_1+W_2)$

This concludes the proof.


问:
What does the notation (p,q) = 1 mean for a rational number p/q?
I'm beginning to work through Murty and Rath's Transcendentals and came across this notation which I've never seen before.

The theorem states:

Given a real algebraic number $\alpha$ of degree $n>1$, there is a positive constant $c=c(\alpha)$ such that for all rational numbers $p/q$ with $(p,q)=1, q>0$, we have $\mid \alpha - \frac{p}{q} \mid  >  \frac{c(\alpha )}{q^n}$.

So I was wondering what the notation $(p,q)=1$ meant.

Thanks.

答: Basically, it means $p$ and $q$ are coprime (do not have any common factor other than $1$) and, therefore, the fraction $\frac{p}{q}$ is not reducible anymore.


答: $(p,q)$ a common notation for $\gcd(p,q)$ and so $(p,q)=1$ means that the fraction is irreducible.

Title: What does the notation (p,q) = 1 mean for a rational number p/q?

Body:
I'm beginning to work through Murty and Rath's Transcendentals and came across this notation which I've never seen before.

The theorem states:

Given a real algebraic number $\alpha$ of degree $n>1$, there is a positive constant $c=c(\alpha)$ such that for all rational numbers $p/q$ with $(p,q)=1, q>0$, we have $\mid \alpha - \frac{p}{q} \mid  >  \frac{c(\alpha )}{q^n}$.

So I was wondering what the notation $(p,q)=1$ meant.

Thanks.


答: This is probably the notation for the greatest common divisor. Many authors, like Apostol, prefer to use the notation $(a,b)$ rather than $\gcd(a,b)$.The notation $(p,q)=1$ means that $p$ and $q$ are relatively prime.


问:
why not restrict codomain?
I have some confusion regarding restriction of function

In wikipedia it is written that

Restriction of function

let $f : E \rightarrow F$ be  a  function from a set $E$ to a set  $F$. If a set $A$ is  a subset  of $E$ , then  the restriction  of  $f$ to $ A$ is the function $f|_{A}: A \to F$

Here it's only restrict  domain part   , why  not restrict  codomain ?

My thinking :

Suppose  $B \subset  F$, then the restriction  of  $f$ to $ A$ is the function $f|_{A}: A \to B$

Is this statement is true/false ?

答: For instance $A=\lbrace 0,1\rbrace\subseteq E=\lbrace 0,1,2\rbrace$ and $f:E\rightarrow F=\lbrace 0,1\rbrace$ such that $f(0)=0, f(1)=f(2)=1$. In this example, the restriction on codomain is $F$ itself.

Title: why not restrict codomain?


问:
When Max is 8m from a lamp post which is 6m high his shadow is 2m long. When he is 3m from the lamp post, what is the length of his shadow?
When Max is 8 m from a lamp post which is 6 m high his shadow is 2 m long. When Max is 3 m from the lamp post, what is the length of his shadow?</p>

答:
Let the top of the lamp post be at $A = (0,6)$, and Max be at $B = (8,0)$, with his head being at height $h$ or $H = (8, h)$.

The gradient of line $AH$ is $\frac{h-6}{8-0} = \frac{h-6}{8}$. When Max moves to the point $B' = (3,0)$, with his head at $H' = (3,h)$, the gradient of line $AH'$ is $\frac{h-6}{3}$.

Now $\text{gradient} = \frac{\text{change in } y}{\text{change in } x}$. Since Max's height does not change, the change in $y$ is the same for both lines. Therefore, for both lines, gradient $\times$ change in $x$ are equal, and the change in $x$ is the length of Max's shadow ($s$):

$$\frac{h-6}{8} \cdot 2 = \frac{h-6}{3} \cdot s\Rightarrow \frac{1}{8} \cdot 2=\frac{1}{3}s \Rightarrow s = \boxed{\frac{3}{4}}.$$

and here is a diagram (look at triangles $H'B'E$ and $HBD$):

Note that we don't need to know Max's height.


答:
That definition doesn't always work. For example, if we have $f:\Bbb R\to\Bbb R$ defined as $f(x)=x^2$, we couldn't restrict the codomain to $(0,+\infty)$, since we'd miss $f(0)=0$. Remember a function $f:A\to B$ takes elements $a\in A$ and give us elements $f(a)=b\in B$ in return.

The restriction of a function on a subset $A$ of the domain is different, since the function is already defined in a bigger set (the domain), the only thing to worry about is to only take the elements of the domain that are also in $A$, whose images were already in the codomain.

In general, you can define a function $f$ on a set $A$ as you please, taking each element of $A$ and deciding where to map it, but then you need to make sure your codomain B contains all images, or otherwise the notation $f:A\to B$ wouldn't make much sense. For example, if I want to define $f$ on $[0,1]$ as the constant map giving $1$, I can define $f:[0,1]\to\{1\}$, or $f:[0,1]\to\Bbb [0,1]$, or $f:[0,1]\to\Bbb C$ (note that they are different maps; the first one is onto for instance), but I cannot define $f:[0,1]\to\{0\}$, since I need $1$ to be in the codomain.

Title: why not restrict codomain?

Body:
I have some confusion regarding restriction of function

In wikipedia it is written that

Restriction of function

let $f : E \rightarrow F$ be  a  function from a set $E$ to a set  $F$. If a set $A$ is  a subset  of $E$ , then  the restriction  of  $f$ to $ A$ is the function $f|_{A}: A \to F$

Here it's only restrict  domain part   , why  not restrict  codomain ?

My thinking :

Suppose  $B \subset  F$, then the restriction  of  $f$ to $ A$ is the function $f|_{A}: A \to B$

Is this statement is true/false ?


问:
Prove that connected clopen set of a topological space is a component of connectedness of this space
Prove that connected clopen set of a topological space (X,) is a component of connectedness of this space.
Clopen set is closed and open, Connected space is a space that we cannot  represent as unition of 2 non-empty non-intersected open sets</p>

答:
Let $G$ be a connected subspace of $X$ which is both open and closed. Since set of  components forms partition of $X$, there is a component $C$ which contains $G$. Observe that $G\cap C$ and $(X-G) \cap C$ are open sets in $C$ and disjoint and $C$ is union of both of them. Since $C$ is connected we get a contradiction  and we conclude that $G=C$.

Edit : With respect to comment made by OP

Generally components are taken to be subspaces of topological space. A nonempty subset $Y$ of $X$ is subspace of $X$ if the topology on $Y$ consists sets of the form $Y \cap H$ where $H$ is open in $X$. So above since $G \subset C$ we have $G \cap C = G$ which is open and since $X-G$ is also open in $X$ we must have $C \cap (X-G)$ is open in $C$


答:
This is actually a good example where I think it is simpler working with the definition of the integral. Assume $\mu(E)>0$, and notice that for all $n$
$$ \phi_n(x)=n\cdot \mathbf{1}_E(x) $$
is a simple function satisfying that $\phi_n\leq f$. Now consider what the definition of $\int f d\mu$ for non-negative functions, and you can conclude that $\int fd\mu>M$ for all $M>0$.

Just to give some an alternative approach.

Body:
If $f$ is integrable, show that $f(x)$ is finite almost everywhere.

I know that $f$ is integrable/finite if and only if $\vert f \vert$ is integrable/finite, so I only have to show it for $f \geq 0$. I think I should use the set $E = \{x \mid f(x)= \infty\}$ and show that its measure is zero, but I'm not sure why this shows that integrable implies finite a.e. or how to show that its measure is zero.


问:
A $1-1$ function is called injective. What is an $n-1$ function called?
A $1-1$ function is called injective. What is an $n-1$ function called ?

I'm thinking about homomorphisms. So perhaps homojective ?

Onto is surjective. $1-1$ and onto is bijective.

What about n-1 and onto ? Projective ? Polyjective ?

I think $n-m$ and onto should be hyperjective as in hypergroups.

答:
IMHO, an n to 1 function should be called "an n to 1 function."
Simple, decriptive, to the point.  Adding a new term in this case just muddies the waters.</p>

ContentLicense: CC BY-SA 2.5


问:
How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
Could you provide a proof of Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?</p>

答:
Well this question actually boils down to "How is the complex exponential defined?"

Here is my view of this problem:

Let

$$f(x+iy)= e^{x}(\cos(y)+i\sin(y) ) \,.$$

Then $f$ has continuous partial derivatives $f_x$ and $f_y$, and verifies the Cauchy-Riemann equations, thus it is analytic.

Moreover, for any $z_1,z_2 \in {\mathbb C}$ we have

$$f(z_1+z_2)=f(z_1) f(z_2) \,.$$

Last but not least $f(x)=e^x$ for all $x \in {\mathbb R}$.

In particular we showed that $e^x$ can be extended to an analytic complex function, and the theory tells us that such extension is unique.

Now, since $f(z)$ is the unique analytic extension of $e^x$ to the complex plane, and it also satisfies the exponential relation $f(z_1+z_2)=f(z_1) f(z_2)$, we call this function $e^z$.


问:
Every ball in different urn
We've got $n$ urns, I have to show that for some $c>0$ there are $c \sqrt{n}$ balls, that after putting them inside urns, probability that all the urns have at most 1 ball is equal or lower than $\frac{1}{e}$.

I know that there are $\frac{n!}{(n-c\sqrt{n})!}$ different cases of putting $c\sqrt{n}$ balls following mentioned assumption (first ball can be thrown to $n$ different urns, second to $n-1$, $c \sqrt{n}$ ball to $n-c\sqrt{n}+1$ urns) and there are $n^{c\sqrt{n}}$ all cases of putting these balls inside $n$ urns.

Probability is the division of these two components, but how it can be compared to $e^{-1}$? I've got a hint to use inequality $e^{-x} \ge 1-x$, but after determining probabilities of these two events and dividing first by second, I don't know what to do next.

Edit
So, using Stirling formula I've got $\frac{\frac{n!}{(n-c\sqrt{n})!}}{n^{c\sqrt{n}}}$ which is about $\sqrt{2\pi n} (\frac{n}{e})^n \cdot (\sqrt{2\pi (n-c\sqrt{n})} (\frac{n-c\sqrt{n}}{e})^{c\sqrt{n}})^{-1} \cdot (n^{c\sqrt{n}})^{-1}$

I simplified it into $\sqrt{\frac{n}{n-c\sqrt{n}}} \cdot (\frac{n}{n-c\sqrt{n}})^{n-c\sqrt{n}} \cdot e^{-c\sqrt{n}}$

But now I'm stuck again and I'm not certain if that Stirling approximation is greater or lower than previous fraction.

答:
The number of possibilities of putting $c\sqrt n$ balls into $n$ urns s.t. each has his own urn is actually $n\choose c\sqrt n$.

The number of different ways of putting $c\sqrt n$ balls into $n$ urns is given by the probability of dividing $c\sqrt n$ objects to $n$ cells, which is known to be given by $n+c\sqrt n-1\choose n-1$.

From these two quantities we obtain:
$$
{n\choose c\sqrt n} / {n+c\sqrt n-1\choose n-1}=\prod_{i=1}^{c\sqrt n - 1}\frac{n-c\sqrt n+i} {n+i}=\prod_{i=1}^{c\sqrt n - 1}1 -\frac{c\sqrt n} {n+i}
\leq\prod_{i=1}^{c\sqrt n - 1}1 -\frac{c\sqrt n} {2n}\leq(1-c/2\sqrt n)^{c\sqrt n}\\
\leq e^{\frac{-c} {2\sqrt n}\cdot c\sqrt n}=e^{-c^2/2}
$$

Title: Every ball in different urn


问:
Can any open subset of $\overline{\mathbb{R}}$ be written as a countable union of disjoint open intervals.
The context of this question is that I am trying to prove that $\{[-\infty,a]\}_{a\in\mathbb{R}}$ generates the Borel set of the extended reals. And I used this property of the real line to prove a similar argument but for the Reals.

答:
Yes, that is true. Let $A$ be an open subset of $\overline{\Bbb R}$. Then:

If $A\cap\Bbb R$ is bounded, then it is an open subset of $\Bbb R$ and, since $A$ is open, $A=A\cap\Bbb R$. And you know that then $A$ is a disjoint union of open intervals.

Otherwise, $A\cap\Bbb R$ is unbounded. Write it as a disjoint union of open intervals. If $A\cap\Bbb R$ has no lower bound, then one of those intervals will be of the form $(-\infty,a)$; replace it with $[-\infty,a)$. Do a similar thing if $A\cap\Bbb R$ has no upper bound. And then you will have your disjoint union of open intervals.


问:
Is there an uncomputable number between any two real numbers?
I know close to nil about uncomputable numbers, so perhaps it doesn't even make sense to ask this question. All the information I can find about them is unaccessible with my level of education, but I have heard that "most numbers are uncomputable."</p>

答: Well since computable numbers are numbers that can be calculated by a finite computer program (in an infinite amount of time), now a finite computer program can be expressed as a finite string of charachters such as a,b,c...1,2....,+,-,*... etc, the space is also a character, so "hello, how are you?" would be a string of 19 characters, let's say there are $n$ characters in total, so the amount of finite strings of $m$ characters would be $n^m$, a finite number lets define the set of strings of $m$ characters as $Str(m)$, it is clear that any finite computer program would belong to $\bigcup_{m\geq 1}Str(m)$, but this is the countable union of finite sets, so the set of computable numbers is countable, since any interval $(a,b)$ is uncountable there would exist an uncomputable number between $a$ and $b$. In fact since the Lebesgue measure of a countable set is $0$ you can say that almost every number is uncomputable.


问:
Find analytic function such that $u = \phi \left( \frac{x^2+y^2}{x} \right) $
Find analytic function $w= u+iv$  such that $u = \phi \left( \frac{x^2+y^2}{x} \right) $

My idea

$\Delta u = 0$

$u_{xx} + u_{yy} = 0$

but it gets complicated...

Body:
From equation $u_{xx} + u_{yy} = 0$ we get ode
$$\phi''(t)=-\frac{2\phi'(t)}{t},\quad t=\frac{x^2+y^2}{x}.$$

Then $$\phi(t)=\frac{c_1}{t}+c_2,$$
$$u=\frac{x}{x^2+y^2}.$$


问:
Upper bound for $|3x - 2|$ given $|x+3| < 2$?
I am given a condition $|x+3| < 2$ to find the upper bound of $|3x - 2|$. I am not quite sure if my process is correct.
\begin{align*}
|x+3|<2 & \Longleftrightarrow -2 < x + 3 < 2\\\\
& \Longleftrightarrow -6 < 3x + 9 < 6\\\\
& \Longleftrightarrow -17 < 3x - 2 < -5
\end{align*}

Is it safe to assume that the upper bound of $|3x - 2|$ is $-5$?

Body:
Continuing your argument:

$-17 < 3x - 2 < -5 \implies 3x-2<0 \implies |3x-2|=-(3x-2)$

$-17 < 3x - 2 < -5 \implies 17 > -(3x - 2) > 5 \implies 5 < |3x - 2| < 17 $


问:
What does X^k mean in mathematical logic
I am reading this book https://www2.karlin.mff.cuni.cz/~krajicek/mendelson.pdf and it writes

If X is a set and k is a positive integer, we denote by $X^k$
the set of all ordered k-tuples $(b_1, ... , b_k)$ of elements $b_1, ... , b_k$ of $X$. In particular, $X^1$ is $X$.

I've looked online and I can't find any other source explain this.

Can someone explains how this works? For example, is $X^2$, the set of ordered pairs?

Also, since I can't find anything else about this, I am wondering if this is standard or whether this source is just making up some formatting rules?

答:
$X^n$ denotes the $n$-th Cartesian power of the set $X$. Your guess is correct, $X^2$ is the set of ordered pairs $(x_1,x_2)$ with $x_1, x_2 \in X$. Similarly, $X^3$ is the set of ordered triples $(x_1,x_2,x_3)$ with $x_1, x_2, x_3 \in X$. Unless you choose to define pairs, triples and $k$-tuples in a very uncommon way, ordered pairs are 2-tuples and ordered triples are 3-tuples. Some examples that you might already be familiar with are $\mathbb{R}^2$,$\mathbb{R}^3$ and $\mathbb{R}^n$.

This notation, Cartesian powers and a $k$-tuple as a mathematical object are very universal in mathematics. Here is the Wikipedia entry about them.


问:
Prove that in a graph a group of even nodes there are two of degree at least $2$
We have just started learning graphs, and I understand the concept clearly, but when it comes to proving something I just don't know how to start!

Prove that in in a group of an even number of people there are at least two of them that the number of the common people that they know is even.

So I translated it to a Graphs problem:

Given a graph $G$ (not directed; knowing someone is mutual), prove that in every group of an even number of vertices there are at least two vertices that the number of their common neighbors is even

I understand that this is not of the hard questions, but I just can not think the methods of proving such things in Graphs in my head, yet. I would be happy for a direction!

答: You can find a nice solution here.</p>

Body: You can find a nice solution here.


问:
Find the area bounded by the curve $y=\ln x$, $y=\frac xe$ and the x y axis
$\ln x$ intersects the y axis at $-\infty$

So
$$A=\int_{-\infty}^{e} \ln x \,dx-\int_{-\infty}^{e} \frac xe dx$$

Which isn’t really a proper definite integral since I am having trouble applying the limits

What is the correct way to do it?

答:
Notice that due to $\ln x$ being one of the function, an immediate constraint is $ x>0$ , so we only have to consider half of the $x-y$ plane.

Sketching the graph:

Legend:
Red = ln(x), blue = $ \frac{x}{e}$

Observe both curves intersect at x=e.

Hence, we can by inspection write down area under curve as:

$$ |\int_0^1 \ln x dx| + \int_0^1 x dx + \int_1^e ( x - \ln x ) dx$$


答:
Considering the graphs of $y = \frac x e$ and $y = \ln x$, the bounds should be $(0,e)$, and $\frac x e > \ln x$. Hence you are looking for the integral:

$$\int_0^e \left(\frac xe - \ln x\right) \ dx$$

Title: Find the area bounded by the curve $y=\ln x$, $y=\frac xe$ and the x y axis

Body:
$\ln x$ intersects the y axis at $-\infty$

So
$$A=\int_{-\infty}^{e} \ln x \,dx-\int_{-\infty}^{e} \frac xe dx$$

Which isn’t really a proper definite integral since I am having trouble applying the limits

What is the correct way to do it?


答:
Your lower bounds should be $0$ since the $y$-axis is stated as a bound. Then the integral is
$$\int_0^e(x/e-\ln x)\,dx=[x^2/2e-x\ln x+x]_0^e=\frac e2$$


问:
Spivaks chapter 7 theorem 9
I am a student trying to get a better understanding of calculus and I've been working my way through Spivak's Calculus book. I am stuck on his proof on chapter 7, theorem 9, pg. 123, that all polynomials of odd degree have a root. What I cannot grasp is that he assumes that $|x|>1$, $2n|a_{n-1}|,...,2n|a_0|$. What I don't understand is why the input value x has to be greater than all of these terms. I know that he later shows the inequality to be less than $\frac{1}{2}$, but why?

答: At that point, Spivek has already stated that$$\left|\frac{a_{n-1}}x+\frac{a_{n-1}}{x^2}+\cdots+\frac{a_0}{x^n}\right|\leqslant\frac{|a_{n-1}|}{|x|}+\frac{|a_{n-1}|}{|x^2|}+\cdots+\frac{|a_0|}{|x^n|}.$$The RHS of this is$$\frac{|a_{n-1}|}{|x|}+\frac{|a_{n-1}|}{|x|^2}+\cdots+\frac{|a_0|}{|x|^n}.$$So, if $|x|\geqslant1$,$$\left|\frac{a_{n-1}}x+\frac{a_{n-1}}{x^2}+\cdots+\frac{a_0}{x^n}\right|\leqslant\frac{|a_{n-1}|}{|x|}+\frac{|a_{n-1}|}{|x|}+\cdots+\frac{|a_0|}{|x|}.$$And now, if $|x|\geqslant2n|a_k|$, for each $k\in\{0,1,\ldots,n-1\}$, it follows from the previous inequality that$$\left|\frac{a_{n-1}}x+\frac{a_{n-1}}{x^2}+\cdots+\frac{a_0}{x^n}\right|\leqslant\frac12.$$That's his goal.

Title: Spivaks chapter 7 theorem 9


问:
A regular polygon expressed as inequality
A square can be described as

$$
\max(x - x_0, y - y_0) < r
$$

or

$$
|x - x_0| + |y - y_0| < r
$$

I am looking for a similar way to describe the set of all points inside a regular $n$-gon, so that $f(x,y) < r$ is true if and only if $(x, y)$ is inside the polygon. $r$ is expected to be the distance between the origin and one of the polygon vertices. I have a parameterization of one of the line segments:

The angle between two edges:
$$
\alpha = \frac{\pi}{2n}\left(n - 2\right) = \pi\left(\frac{1}{2} - \frac{1}{n}\right)
$$

Angle between ray through a vertex, and the line segment:
$$
\theta = \pi - \alpha = \pi\left(1 - \left(\frac{1}{2} - \frac{1}{n}\right)\right) = \pi\left(1 - \frac{1}{2} + \frac{1}{n}\right) =  \pi\left(\frac{1}{2} + \frac{1}{n}\right)
$$

The length of an edge:
$$
s = \sqrt{ 2r^2 -2r^2\cos\left(\frac{2\pi}{n}\right)} = \sqrt{2}r \sqrt{1 - \cos\left(\frac{2\pi}{n}\right)}
$$

Point on the edge:
$$E(t) = (r, 0) + \frac{t}{s}\begin{bmatrix}
\cos\left(\theta\right) \\
\sin\left(\theta\right)
\end{bmatrix}, t\in[0, 1]
$$

Would the next step be to solve for $t$, such that the ray from $O$ to $P$ intersects $E(t)$. If $t$ is outside $[0, 1]$, rotate the coordinate system by $\frac{2\pi}{n}$ and try the next edge. Take the distance as the shortest result.

答:
One way of encoding mathematically a regular polygon of "radius" $r$, centered at $(x_0,y_0)$, is as the set of points $$\sum_{i=1}^na_i^2\begin{pmatrix}x_0+r\cos\frac{2\pi i}{n}\\y_0+r\sin\frac{2\pi i}{n}\end{pmatrix}$$ subject to the constraint of $\sum_{i=1}^na_i^2\le1$, ($a_i\in\mathbb{R}$).

More generally, any convex polygon with vertices $z_i=(x_i,y_i)\in\mathbb{R}^2$ is the set of points $\frac{\sum_ia_i^2z_i}{\sum_ia_i^2}$, ($a_i\in\mathbb{R}$). This can be seen as the convex hull of the set of vertices.


问:
Third central moment Bernoulli variable
I'm looking for a proof of the third central moment of a Bernoulli variable $X$ with probability $p$. I know it must be $p(1-p)(1-2p)$, but I'm looking for a way to show this. Any ideas? Thanks!</p>

答: This is just an application of the definition of the third central moment. What's the mean of a Bernoulli variable? Subtract that from the two possible values, take the third powers, multiply by the probabilities, add, factor out $p(1-p)$, and you're done.</p>

Title: Third central moment Bernoulli variable