问:
Is the function representation $T_{\infty} : U(1) → GL(L^2(S^1))$ unitary?
Let $S^1$ be a $U(1)$-set, and let $L^2(S^1)$ be the Hilbert space of square-integrable complex-valued functions $f : S^1 → \Bbb C$. It can be shown that the map

$$
T_{\infty} : U(1) → \text{GL}(L^2(S^1))
$$
$$ \hspace{4cm} e^{i\theta} ↦ \left(f ↦ (e^{i\phi} ↦ f\left(\left(e^{i\theta}\right)^{-1}e^{i\phi}\right))\right)$$

is a representation of U$(1)$. I'm trying to determine whether it is unitary. By that I mean, that for all $e^{i\theta} ∈ U(1)$, we have

$$T_{\infty}(e^{i\theta})^* = T_{\infty}(e^{i\theta})^{-1} = T_{\infty}(e^{-i\theta}).$$

To this end, let $f ∈ L^2(S^1)$ and $e^{i\phi} ∈ S^1 $ be arbitrarily given. If I interpret the $^*$ (i.e., the Hermitian adjoint) correctly, that would mean that we have, for all $f ∈ L^2(S^1), e^{i\phi} ∈ S^1$

$$ (T_{\infty}(e^{i\theta})^*f)(e^{i\phi}) = (T_{\infty}(e^{-i\theta})f)(e^{i\phi})= f\left(\left(e^{-i\theta}\right)^{-1}e^{i\phi}\right) = f\left(e^{i(\theta+\phi)}\right).$$

In evaluating the LHS, however, I get confused, as I find several definitions of the adjoint. In some texts, for $f : V → W$, we have $f^* : W → V$, while in others (and on Wikipedia), $f^* : W^* → V^*$.. I would say that in this case, the former actually makes more sense, as we still want to plug in functions from $L^2(S^1))$ not from its dual. Secondly, the texts say for example that "there is at most one function $f^*$ that preserves the inner product" (in the sense that $\langle f(v), w\rangle = \langle v, f^*(w)\rangle$), this function $f^*$ being the adjoint. But they don't actually give a function rule for $f^*$, so how can I evaluate $(T_{\infty}(e^{i\theta})^*f)(e^{i\phi})$..?

(Also, is $T_{\infty}$ generally called that function representation? Or are there other canonical names for it?)

答:
$$\langle F(.+a),H\rangle=\langle F,H(.-a)\rangle$$ so the adjoint of $F\to F(.+a)$ is $H\to H(.-a)$ (and hence those operators are unitary since the adjoint is the inverse).

This works for any $L^2(G)$ where we are using the Haar measure on the (locally compact abelian) group $G$. If $G$ is not abelian it works too with the right Haar measure and $F\to F(a .),H\to H(a^{-1}.)$.


问:
Proof for angle between two curves formula.
Could someone please provide the proof for the formula attached in the image below? The formula is to find the angle between $f(x)$ and $g(x)$ upon intersection.

$\tan a = g'(x)$

$\tan b = f'(x)$

Let $c =a-b$

then $\tan c = \tan (a-b) = \frac{\tan a - \tan b}{1 + (\tan a)(\tan b)}$

Any help is much appreciated!

答: After a bit of intuition I've derived: (please let me know if this is incorrect)

ContentLicense: CC BY-SA 4.0


问:
Find a mistake in finding limit, please.
I wanted to find the limit: $\lim_{n \to +\infty}(\frac{1^p + 2^p + ... + n^p}{n^p} - \frac{n}{p + 1})$, where $p \in \mathbb{N}$.

I saw that I could use Stolz–Cesàro theorem, where $x_n = (p + 1)(1^p + 2^p + ... + n^p) - n^{p + 1}$, and $y_n = n^p(p + 1)$ tends to +$\infty$.

So, here we are:

$\lim _{ x\rightarrow \infty  }{ \frac { { x }_{ n }-{ x }_{ n - 1 } }{ { y }_{ n }-{ y }_{ n - 1 } } } = \lim _{ x\rightarrow \infty  }{ \frac { ( p+1 ) { n  }^{ p }-{n}^{ p+1 }+{ (n-1) }^{ p+1 } }{ ( p+1 ) ( { n }^{ p }-{ (n-1) }^{ p } )  } } = \lim _{ x\rightarrow \infty  }{ \frac { ( p+1 )-n+(n-1)(\frac{n - 1}{n})^p  }{ ( p+1 ) ( 1-{ (\frac{n-1}{n}) }^{ p } )  } } = \lim _{ x\rightarrow \infty  }{ \frac { p  }{ 0 } } $ (because $(\frac{n - 1}{n})^p$ tends to 1).

But as I understood here, limit of this sequence is $\frac{1}{2}$.

Where did I go wrong in my reasoning and why is my answer not correct?

答:
The final identity
$$
\lim_{n\to \infty  }{ \frac { ( p+1 )-n+(n-1)(\frac{n - 1}{n})^p  }{ ( p+1 ) ( 1-{ (\frac{n-1}{n}) }^{ p } )  } } = \lim _{n\to \infty  }{ \frac { p  }{ 0 } }
$$
is wrong. It is correct that $(\frac{n - 1}{n})^p$ tends to one, but that does not imply that $1 - n + (n-1)(\frac{n - 1}{n})^p$ tends to zero and and can be ignored in the numerator. Also the denominator tends to $p+1$, not to zero.

For a correct solution using the Stolz–Cesàro theorem  see for example haqnatural's answer here.


问:
Learning math under time pressure
I'm taking my first and only university math course, discrete mathematics.

We have about a week and a half to learn a full chapter (around 50 pages) from the book and do the related assignment.

As I'm working full-time, I have about 2 hours a day to achieve the weekly mission.

I would like to know the best learning strategy for the current situation.

答:
Discrete mathematics is a wide field of knowledge. You could start by narrowing as much as possible what the topics of interest are. Ask the teachers, tell them your situation. Teachers are also human...

Then you could decide what's more important: keep working full time, or take a few holidays to study. In my opinion, one learns better without (time) pressure, having time from morning to evening to really enter the topics, deeply. $4 \times 2$ hours is not the same as a straight $8$ hours.

You say "we", meaning probably that you are a group of people facing the same situation (minus the full time job, maybe). What do the others? Could you study with them? Even if you keep working, in the 2 hours evening you could benefit from their studying experience.

Title: Learning math under time pressure


问:
Flipping a Coin 10 Times and Getting a Sequence of Heads
I toss a fair coin $10$ times resulting in a sequence of heads and tails. Let $X$ be the number of times that the sequence $HH$ appears. $HH$ appears thrice here.

Solution:

The example goes, say we flip a coin $10$ times and you get all heads then we have $HH$ appears $10$ times. (I would write it out, but my LateX Skills are not up to par.)

And we see $P(HH) = \frac{1}{4}$ because we have $.5$ for heads and tails for the two coins.

$X = HH$ appears.

This is where I don't understand what happens: $\sum_{n=0}^9P(X)\cdot X = \frac{9}{4}$.

When I thought of it, do you factor out $P(X)$ because it is always $\frac{1}{4}$, but I am wrong. Any help would be awesome!

答:
Let $X_j$ be the indicator random variable of HH appearing at position $j$ for $j=1,2,\dots, 9$. Notice that these aren't quite independent from each other so I don't know if the binomial solution is entirely accurate, although it gives the correct expected value. But we can anyway use linearity of the expectation. As $X=\sum_{j=1}^9 X_j$, in another words $X$, which is the total number of appearances of HH, is the sum of the ones that did occur, we get that

$$\mathbb{E}[X] = \mathbb{E} \left[\sum_{j=1}^9 X_j \right] =\sum_{j=1}^9  \mathbb{E}[X_j] = \sum_{j=1}^9 \frac{1}{4} = \frac{9}{4}.$$


问:
Measurable set in product space.
I  am stuck on the following question.  Any helps would be highly appreciated!
Let $(X, \mathcal{M}, \mu)$ be a $\sigma$-finite measure space and let $(\mathbb{R},\mathcal{L}, m)$ denote he set of real numbers with the Lebesgue measure. Let $f:X \to \mathbb{R}$ be such that $f \in L^1(X)$. Let $\Gamma= \{((x,f(x))): x\in X\} \subset X \times \mathbb{R}$. Is $\Gamma$ measurable with respect to $ \mathcal{M} \times  \mathcal{L}$?  How do  we compute $(\mu  \times m)(\Gamma)$?

答:
HINT:

Can reduce to the case $X$ is bounded and $f$ is also bounded, $0<f <1$.

Consider a sequence of simple functions $f_n$ converging pointwise increasingly to $f$. If
$$U(f) = \{ (x,t)\ | \ 0< t< f(x)\}$$
then $U(f) = \cup_n U(f_n)$, and $U(f_n)$ is a finite union of rectangles.  Therefore, $U(f)$ is measurable. Moreover, its measure is $\int_X f$.  Now do the same for the function $1-f$.  Notice that

$$U(f) \cup U(1-f) = (X\times (0,1)) \backslash\Gamma(f)$$
We conclude that $\Gamma(f)$ is measurable, of measure $0$.

Title: Measurable set in product space.


问:
For every natural number $n$, prove that $n^n > 1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n-1)$.

For every natural number $n$, prove that $n^n > 1 \cdot 3 \cdot 5 \cdots (2n-1)$.

I tried solving this by assuming that the above equation is true for $n$ and tried proving it is also true for $n+1$, but did not reach anywhere.

答:
Alternatively, try AM-GM:
$$n = \frac{1+3+5+\dots+(2n-1)}n \geqslant \sqrt[n]{1\cdot3\cdot5\cdots(2n-1)}$$

Equality is possible iff $n=1$.


问:
What quadratic equation can be obtained from following question?
Sorry if I have to copy past the question but I can't understand where I'm wrong in this specific problem:

Two drivers are testing the same model car at speeds that differ by 20 km/hr.
The one driving at the slower rate drives 70 kilometers down a speedway and
returns by the same route. The one driving at the faster rate drives 76
kilometers down the speedway and returns by the same route. Both drivers
leave at the same time, and the faster car returns $\frac{1}{2}$ hour earlier than the
slower car. At what rates were the cars driven?

I get two equations from this problem which is: $(v-20)(t)=70$ and $(v)(t-\frac{1}{2})=76$

after doing substitution I get to the equation: $v^2-8v-3040=0$ which gives faster car 59.28 and slower car 39.28 but the answers in textbook is: 56 and 76. So am I wrong or this is just an error in the book?

答:
As player3236 has noted, the distances should be doubled, as the total distance is the distance to there and back.

Using speed $\times$ time = distance like before, we should have:

$$(v-20)(t) = 140 \Rightarrow t = \frac{140}{v-20} \tag{1}$$
$$(v)(t - 1/2) = 152 \Rightarrow t = \frac{152}{v}+\frac{1}{2} \tag{2}$$

Equating $(1)$ and $(2)$, we find that:

$$\frac{140}{v-20} (v)(v-20) = \frac{152}{v} (v)(v-20) + \frac{1}{2}(v)(v-20)$$
$$\Rightarrow 140v = 152(v-20) + \frac{1}{2}v(v-20)$$
$$\Rightarrow 280v = 304(v-20) + v(v-20)$$
$$\Rightarrow 280v = 304v - 6080 + v^2 - 20v$$
$$\Rightarrow v^2 + 4v - 6080 = 0$$

Use the quadratic formula to find that $v = 76, -80$. Now discard the negative solution, so the faster car has a speed of $76$ km/h, and the slower car is $20$ km/h slower, which is $56$ km/h.


问:
The failure of Arzela-Ascoli over non-archimedean fields
Let $K$ be a complete, algebraically closed non-archimedean field.
The Arzela-Ascoli theorem fails over such a field $K$.

I found an example on a book, but not quite get it.
Here is an example :

$\color{blue}{Let\,\,\, \{x_n\} \,\,be \,\,a\,\, sequence \,\,of\,\, points\,\, in\,\, \mathbb{P}^1(K)\,\, with\,\, no\,\, accumulation\,\, points.\,\, Thus\,\, the\,\, family\,\, of\,\, constant\,\, functions\,\, \{x_n: n\geq 0\}\,\, is\,\, equicontinuous\,\, but \,\,not\,\, normal\,\, on\,\, any\,\, open\,\, subset\,\, U\subseteq \mathbb{P}^1(K).}$

It would be good if someone could explain this to me.

By any chance, is it possible to write such sequence $\{x_n\}\subset\mathbb{P}^1(K)$ concretely?

So the normality and equicontinuity of family of functions cannot be used interchangeably over non-archimedean field, right?

答:
We need that $K$ has an infinite residue field, it doesn't work with $K=\Bbb{Q}_p$.

(For $p\ne 2$) with $K= \Bbb{Q}_p(\zeta_{2^\infty})$ (well its completion in fact) take $$x_n=[\zeta_{2^n}:1]$$
$$d(x_n,x_m)=\inf_j |x_n(j)-x_m(j)|_p=1$$


问:
Does the cartesian product have a neutral element?
Let $A$ be any set. Is there a set $E$ such that $A \times E = E \times A = A$?

I thought of the empty set, but Wikipedia says otherwise. This operation changes dimension, so an isomorphism might be needed for such element to exist.

答:
It's unfortunate (but not your fault at all) that one needs to place a lot of explanations and specifications in place to be able to give a definite answer to what seems like a reasonable question. Unfortunate because in fact there are a lot of assumptions behind the notation.

The first assumption is what is meant by equality? Do we mean set equality, or do we mean set equivalence (up to a bijection)?

The second question is what is meant by $A\times B$; it is the set of ordered pairs, sure, but what is an ordered pair? There are many different definitions. One of the most standard ones is the Kuratowski definition of ordered pair:
$$(a,b) = \bigl\{ \{a\},\{a,b\}\bigr\}.$$
The precise definition of ordered pair will matter if we are asking about equality of sets rather than simple equivalence up to a bijection.

The third question is whether you meant to ask if there is a set $E$ such that for all sets $A$, $E\times A = A\times E = A$; or if you meant to ask if for every set $A$ there exists a set $E$ such that $E\times A = A\times E = A$. Formally, they are different questions (though an affirmative answer for the former implies an affirmative answer for the latter).

And the final question is what Set Theory you are working in.

In its weakest version, if equality of sets means "up to a bijection", then the answer is yes: any singleton set $E$ will satisfy the stronger question: if $E$ is a singleton, then for every set $A$, there is a bijection between $A\times E$ and $E\times A$, and both are bijectable with $A$. The projection map onto $A$ is a bijection in this case, and the precise definition of ordered pair does not matter.

What if by equality we mean true set equality, and we are asking for an identity element for the cartesian product? The answer is no, regardless of your precise definition of ordered pair, as long as your Set Theory has some of the basic axioms (Specification/Separation and Power Sets will do). If $E$ is empty, then for any nonempty $A$ we have $E\times A\neq A$. If $E$ is not empty, then let $A$ be a set that contains an element that is not an element of $E$ (such sets exist by the Axiom of Specification: if $X=\{e\in E\mid e\notin e\}$, then it is an easy exercise to show that $X$ is not an element of $E$; and then the element of the power set of $E$ that contains only $X$ is a set that contains an element that is not an element of $E$). Then $E\times A\neq A\times E$, because if $x\in A$ is an element not in $E$, and $e\in E$ is any element, then $(x,e)\in A\times E$, but $(x,e)\notin E\times A$, because $x\notin E$.

What if by equality we mean true set equality, but the set $E$ is allowed to depend on $A$? In Set Theory with the Axiom of Foundation and the Kuratowski definition of ordered pair, the only set $A$ for which there is a set $E$ with $A\times E = A$ is the empty set. Indeed, if $A$ is empty, then for any set $E$ we have $A\times E = \emptyset = A$. Now, assume the Axiom of Foundation/Regularity:

For every nonempty set $C$, there exists $x\in C$ such that $x\cap C = \emptyset$.

Assume that $A$ is nonempty, and there exists a set $E$ such that $A\times E = A$. For every $a\in A$ there exists $a'\in A$ and $e\in E$ such that $(a',e)=a$. That means that $a = \bigl\{ \{a'\},\{a',e\}\bigr\}$. Hence, for every $a\in A$ there exists $a'\in A$ such that $\{a'\}\in a$. Let $C$ be the set
$$C = A\cup\bigl\{ x\in\mathcal{P}(A)\mid x\text{ is a singleton}\bigr\}.$$
By the Axiom of Foundation, there exists $c\in C$ such that $c\cap C=\emptyset$. If $c\in A$, then by the argument above there exists $a'\in A$ such that $\{a'\}\in c$; but $a'\in A$, so $\{a'\}\in C$, hence $\{a'\}\in c\cap C$, contradicting the choice of $c$. Therefore, we conclude that $c\in\mathcal{P}(A)$ and $c$ is a singleton, $c=\{a\}$ for some $a\in A$. But then $a\in c\cap C$, again contradicting the choice of $c$.

This contradiction arises from assuming that there exists a set $E$ such that $A\times E = A$ when $A$ is nonempty, so we conclude that if $A$ is nonempty, then no such $E$ exists.

However, if you do not have the Axiom of Foundation, then it's possible that there exist some nonempty sets $A$ for which there exist sets $E$ with $A\times E = A$. For example, if you take ZF with Aczel's anti-foundation axiom, then there exists a set $A$ such that $A=\{A\}$. Then $A\times A = A$, since
$$(A,A) = \bigl\{ \{A\},\{A,A\}\bigr\} = \bigl\{\{A\},\{A\}\bigr\} = \bigl\{\{A\}\bigr\} = \{A\} = A.$$
So $A\times A = \{(A,A)\} = \{A\} = A$.

But even in this situation there always exist sets that do not satisfy the desired property (that is, sets $A$ for which there does not exist any $E$ with $A\times E = E\times A = A$); simply take a well-founded non-empty set and proceed as above (a "well-founded set" just means a set that does satisfy the statement in the Axiom of Foundation).

ContentLicense: CC BY-SA 3.0


问:
A Boundary crossing result for discrete brownian bridge
Let $S_n$ be a random walk with gaussian increments with $S_0=0$, i.e. $S_n-S_{n-1}\sim N(0,1), n\geq 1$. Fix $a>0,b\in \mathbb{R}$ and $c<a+bn$. Define the new process
$$
B_{n,c}(k):=S_k-\frac{k}{n}\left (S_n-c\right)
$$
to be a discrete bridge ending at $c$. I want to find an expression for the probability
$$
\mathbb{P}\left(\max_{0\leq k\leq n}B_{n,c}(k)-bk\geq a\right),
$$
which is only the probability that the bridge crosses the linear boundary by time $n$.

This expression is known in the continuous case (for the brownian bridge), see for example Scheike. The probability is given by
$$
\mbox{exp}(-2a(a+bt-c)/t).
$$
I have no reasons to believe that such an equivalent form would not hold in the discrete case.
The proof given in the article above doesn't "discretize" well because of the time inversion and time scaling properties used in the continuous case that dont translate into the discrete case.

I'm looking for some insight on how to compute this probability. I would like to get an exact expression, but I could settle for an upper bound that goes in $1/n$.

If it can be of any help, I have shown that the probability (found in Scheike as well)
$$
\mathbb{P}(\max_{0\leq s\leq t}W_s-bs\geq a)=1-\Phi(a/\sqrt{t}+b\sqrt{t})+exp(-2ab)\Phi(b\sqrt{t}-a/\sqrt{t})
$$
holds for the discrete brownian motion $S_n$.

答:
I haven't managed a full answer for this, so this is just a sketch with a few ideas in it.

For notation I'm going to let $W(t)$ be a standard Brownian motion.
I can then consider $W(k)$ for integer k.  Similarly I'm going to consider a Brownian bridge $B_{n,c}(t)$ sampled at integer times.

Firstly I definitely think your estimate for the non bridge case is a strict upper bound and the error doesn't go away as $n\to\infty$.

To see this recall that with probability one for all positive $b$ there exists a time $T$ such that  $W(t) T$. It is east to see that there is a positive probability that there is a non integer $t$ with $W(t)> a+bt$ but $W(k)  a + bk$ if and only if $B_{n,0}(k) > a + \left(b-\frac cn\right)k$ so we can focus on the $0$ to $0$ bridge. For small $n$ you can work this out exactly by writing down the correlation matrix of the $B_{n,0}(k)$'s and integrating over the multivariate Gaussian. Not very nice and there's probably no closed form, but it's the only way to get an exact answer.

For large $n$ I can construct my bridge $\tilde B_n(t)$ By letting $T$ be the largest time such that $W(T) = \frac{W(n)}2$ then setting $\tilde B_n(t) = W(t)$ for $t<T$ and $\tilde B_n(t) = W(n) - W(t)$ for $t\geq T$.

If $W(n)$ is positive then $\tilde B_n(t) > a+bt$ only if $W(t) > a+bt$ if $W(t)$ is negative the opposite is true, but if there exists $t$ such that $\tilde B_n(t) > a+bt$ but $W(t) < a+bt$ I can define $\tilde T<n$ as the largest time such that $W(T)=0$ and set  $\tilde W(t)$ = $W(t)$ for $t<\tilde T$ and $\tilde W(t) = -W(t)$ for $t>\tilde T$.

You should be able to convince yourself that $\tilde W(t)\geq\tilde B_n(t)$
for every $t$ so if $\tilde B_n(t)$ crosses the boundary so does $\tilde W(t)$. Furthermore as $W(t)$ does not cross the boundary  $\tilde B_n(t)$ can only cross the boundary for $t>T$. So if I create a bridge by reflecting $\tilde W(t)$ it does not cross the boundary.

So there exists a measure preserving map ($W\mapsto\tilde W$) that takes the set of paths where the bridge crosses the boundary but the path itself does not onto to the set  of paths that do cross the boundary but the reflection bridge does not. So the probability that the path crosses the boundary is greater than or equal to the probability that the bridge crosses the boundary.

Notice that the argument does not break down if I only count boundary crossings that occur at integer times. I think that the error for the discrete process is bounded above by the error for the continuous process.

Another avenue you might want to consider is to get rid of $b$ by setting $\tilde c = c - nb$ Then $B_{n,c}(t) > a+bt$ if and only if $B_{n,\tilde c}(t) > a$.  Now I can create a bridge from $0$ to $\tilde c$ the same way, by conditioning my bridge to hit $\tilde c$ before time $n$ and then reflecting as before to create a bridge. So now I'm interested in the probability that
there exists a pair $k < t<n$ with $n\in\mathbb N$ and $t\in\mathbb R$ such that
$W(k) > a$ and $W(t) = $\tilde c$. (or an alternative case where the bridge crosses but the Brownian motion doesn't I think these probabilities will drop off pretty rapidly. )

This might be a way of getting a handle on your discretisation error. Although it's essentially the same problem (What's the probability that a Brownian motion goes above a certain boundary at real times, but not integer times) it should be easier in this case because you can use the reflection principle and other tricks in this situation.

Sorry it's not an answer, but hopefully there's an idea or two in there that you can use.

Body:
I haven't managed a full answer for this, so this is just a sketch with a few ideas in it.

For notation I'm going to let $W(t)$ be a standard Brownian motion.
I can then consider $W(k)$ for integer k.  Similarly I'm going to consider a Brownian bridge $B_{n,c}(t)$ sampled at integer times.

Firstly I definitely think your estimate for the non bridge case is a strict upper bound and the error doesn't go away as $n\to\infty$.

To see this recall that with probability one for all positive $b$ there exists a time $T$ such that  $W(t) < a +bt$ for every $t>T$. It is east to see that there is a positive probability that there is a non integer $t$ with $W(t)> a+bt$ but $W(k) < a+bk$ for every integer $k$. the difficult part of the question is not the bridge, it's the discretisation error.

Notice that the , your constant $c$ is superfluous, $B_{n,c}(k) > a + bk$ if and only if $B_{n,0}(k) > a + \left(b-\frac cn\right)k$ so we can focus on the $0$ to $0$ bridge. For small $n$ you can work this out exactly by writing down the correlation matrix of the $B_{n,0}(k)$'s and integrating over the multivariate Gaussian. Not very nice and there's probably no closed form, but it's the only way to get an exact answer.

For large $n$ I can construct my bridge $\tilde B_n(t)$ By letting $T$ be the largest time such that $W(T) = \frac{W(n)}2$ then setting $\tilde B_n(t) = W(t)$ for $t<T$ and $\tilde B_n(t) = W(n) - W(t)$ for $t\geq T$.

If $W(n)$ is positive then $\tilde B_n(t) > a+bt$ only if $W(t) > a+bt$ if $W(t)$ is negative the opposite is true, but if there exists $t$ such that $\tilde B_n(t) > a+bt$ but $W(t) < a+bt$ I can define $\tilde T<n$ as the largest time such that $W(T)=0$ and set  $\tilde W(t)$ = $W(t)$ for $t<\tilde T$ and $\tilde W(t) = -W(t)$ for $t>\tilde T$.

You should be able to convince yourself that $\tilde W(t)\geq\tilde B_n(t)$
for every $t$ so if $\tilde B_n(t)$ crosses the boundary so does $\tilde W(t)$. Furthermore as $W(t)$ does not cross the boundary  $\tilde B_n(t)$ can only cross the boundary for $t>T$. So if I create a bridge by reflecting $\tilde W(t)$ it does not cross the boundary.

So there exists a measure preserving map ($W\mapsto\tilde W$) that takes the set of paths where the bridge crosses the boundary but the path itself does not onto to the set  of paths that do cross the boundary but the reflection bridge does not. So the probability that the path crosses the boundary is greater than or equal to the probability that the bridge crosses the boundary.

Notice that the argument does not break down if I only count boundary crossings that occur at integer times. I think that the error for the discrete process is bounded above by the error for the continuous process.

Another avenue you might want to consider is to get rid of $b$ by setting $\tilde c = c - nb$ Then $B_{n,c}(t) > a+bt$ if and only if $B_{n,\tilde c}(t) > a$.  Now I can create a bridge from $0$ to $\tilde c$ the same way, by conditioning my bridge to hit $\tilde c$ before time $n$ and then reflecting as before to create a bridge. So now I'm interested in the probability that
there exists a pair $k < t<n$ with $n\in\mathbb N$ and $t\in\mathbb R$ such that
$W(k) > a$ and $W(t) = $\tilde c$. (or an alternative case where the bridge crosses but the Brownian motion doesn't I think these probabilities will drop off pretty rapidly. )

This might be a way of getting a handle on your discretisation error. Although it's essentially the same problem (What's the probability that a Brownian motion goes above a certain boundary at real times, but not integer times) it should be easier in this case because you can use the reflection principle and other tricks in this situation.

Sorry it's not an answer, but hopefully there's an idea or two in there that you can use.


问:
Inequality relying on integer partitions and dominance ordering
Let $\lambda$, $\mu$ be two partitions of a natural number $n$, such that $\lambda$ dominates $\mu$ in the usual dominance order on partitions.

I would like to prove that if $q\geq 2$ is a natural number, then
$$\sum_{j\in \mu} q^j \leq  \sum_{i\in \lambda} q^i $$

I'm sure this must be well-known if it is true, but I cannot find it in the literature. Even a reference for this (or indeed a counterexample!) would be most welcome.

答: This is an immediate consequence of Karamata's inequality since $x \mapsto q^x$ is convex for $q \geq 1$. (Credit where credit is due: I essentially learned this from Fedor Petrov.)


问:
Show that $\phi:\mathbb{Q}(\sqrt{3}) \rightarrow M_{2}(\mathbb{Q})$ maps identity to identity.
Here is the function I am speaking about:

Recall that the field $\mathbb{Q}(\sqrt{3})$ has basis $\{1 ,\sqrt 3\}$ as a vector space over $\mathbb{Q}.$ Let $\mathbb{Z}[\sqrt{3}] \subset \mathbb{Q}(\sqrt{3})$ be the subring $\{ a_{1} + a_{2}\sqrt{3}| a_{1},a_{2} \in \mathbb{Z}\}.$

Show that the function $$\phi:\mathbb{Q}(\sqrt{3}) \rightarrow M_{2}(\mathbb{Q}), \phi (a_{1} + a_{2}\sqrt{3}) =  \begin{pmatrix}
a_{1} & 3a_{2} \\
a_{2} & a_{1}
\end{pmatrix} $$ is an injective ring homomorphism.

My question is:

I have no problem in showing the following:

1- Showing that it is injective.

2- Showing that it preserves addition.

3- Showing that it preserves multiplication.

But I have a problem in

4- Showing that it maps the identity element of $\mathbb{Q}(\sqrt{3})$ to the identity element of $M_{2}(\mathbb{Q}).$

I know the identity element of $M_{2}(\mathbb{Q})$ which is $$\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} $$ but it is not clear for me what is the identity element of $\mathbb{Q}(\sqrt{3}),$ could anyone clarify this for me please?

答: The ring $\Bbb{Q}(\sqrt{3})$ contains $\Bbb{Q}$ as a subring. Hence the identity (additive or multiplicative) of $\Bbb{Q}(\sqrt{3})$ is that of $\Bbb{Q}$.


问:
How to write a proof directly from the definition of derivative as a limit without using any of the differentiation rules?
I have a practice problem I have worked through but don't have an answer for it. How would you answer this?

Let a ∈ R. Let f be a function which is differentiable at a. Assume that f is always
positive. We define a new function g by g(x) = 1/
f(x)
. Prove that g is differentiable
at a and that
g
0
(a) = −f
0
(a)
(f(a))2
.
Write a proof directly from the definition of derivative as a limit without using any
of the differentiation rules (such as quotient rule or chain rule).

答:
You have
$$g(x+h)-g(x)={1\over f(x+h)}-{1\over f(x)}={f(x)-f(x+h)\over f(x+h) f(x)}\ ,$$
and therefore
$${g(x+h)-g(x)\over h}=-{1\over f(x+h)f(x)}\cdot{f(x+h)-f(x)\over h}\ .$$
Now take the limit $h\to0$, and use that $f$ is continuous at $x$.


问:
$x^3-6x^2+2x-2=0$ solve for x
I solved it by approximation but its very long way to solve, so i was wondering if there is a quicker way to solve this kind of questions.</p>

答:
You could just use General cubic formula

Or for another way:

$$x^3-6x^2+2x-2=0$$
$$\implies (x-2)^3 - 10(x-2) -14 = 0$$

Let $t = x - 2$, then:

$$t^3 - 10t -14 = 0$$

And then apply Cardano's formula

Either way, the result will just be:

$$x = 2+\sqrt[3]{\dfrac{63+\sqrt{969}}{9}}+\sqrt[3]{\dfrac{63-\sqrt{969}}{9}}$$

Title: $x^3-6x^2+2x-2=0$ solve for x


问:
For which positive integers will the following set be an ideal?

Find all the positive integers $m$ for which the zero divisors
together with $0$ form an ideal in the ring $\mathbb{Z}/(m)$.

We need the set $\{ x \in \mathbb{Z}/(m) \mid x \text{ is a zero divisor}\} \cup \{ 0 \}$ to be an ideal in the ring $\left( \mathbb{Z}/(m), +, \cdot \right)=\left( \mathbb{Z}_m, +, \cdot \right)$.

I've been struggling with this problem, but I don't even know how to start the solution. I've tried some small examples, but that's it. Can anyone help me?

答: Hint: The best way to start on a problem like this is to try to find examples. If you try the first few numbers, you should get that $\{1,2,3,4,5,7,8,9,11,\dots\}$ work, but $\{6,10,12,\dots\}$ don't. See if you can use the mini-proofs you come up with for these small cases to generalize to larger classes of numbers.


问:
Proving a relation between $\sum\frac{1}{(2n-1)^2}$ and $\sum \frac{1}{n^2}$
I ran into this question:

Prove that:

$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$

Thank you very much in advance.

Body: Hint: $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}+\sum_{n=1}^{\infty}\frac{1}{(2n)^2}=\sum_{n=1}^{\infty}\frac{1}{n^2}$$


问:
Munkres' Topology - Theorem 15.2
Theorem 15.2 of Topology by Munkres states:

The collection
$$\mathcal{S} = \{\pi_1^{-1}(U)|U \text{ open in X}\} \cup \{\pi_2^{-1}(V)|V \text{ open in Y}\}$$
is a subbasis for the product topology on $X \times Y$.

He defined $\pi_1:X \times Y \rightarrow X$ by $\pi_1(x,y)=x$ and similarly $\pi_2: X \times Y \rightarrow Y$ by $\pi_2(x,y)=y.$

It seems to be there is no restriction on the set from $X \times Y$ that is being mapped. So for an example, consider the set $X = \{1,2\}$ and $Y=\{3,4\}$ with topologies $\mathcal{T}_X = \{\phi,\{1\}, \{1,2\}\}$ and $\mathcal{T}_Y = \{\phi, \{3\}, \{3,4\}\}$.

Consider $\{2\} \times \{4\} = \{(2,4)\}$. The mapping $\pi_1(2,4)=2$ and $\pi_2(2,4)=4$. For one thing, these mappings do not give sets, rather an element of $X$ or $Y$. Shouldn't they be sets? Since only sets like $\{2\}$ or $\{4\}$ can be open but not an element like $2$ and $4$.

Moreover, $\{2\}$ and $\{4\}$ are not open in their respective topologies. So their pre-image would just be the empty set, and an empty set cannot be a subbasis which means I am applying this theorem wrong. I would like to know where am I going wrong? When he defines the mapping $\pi_1:X \times Y \rightarrow X$ does he mean the set $X \times Y$ or any subset of $X \times Y$?

Also, could you please provide an example where the subbasis generated by this theorem is not a basis? It is appreciated if the example is for finite sets.

答:
Note first that your post (in my version of Munkres this is correct) contains a typo. $\mathcal{S}$ is a subbasis of $X \times Y$ and not of $X$.

In your case:
$$\{\pi_1^{-1}(U): U \in \tau_X\}= \{\pi_1^{-1}(\emptyset),\pi_1^{-1}(\{1\}), \pi_1^{-1}(\{1,2\})\}= \{\emptyset, \{1\}\times Y, X \times Y\}$$
Similarly

$$\{\pi_2^{-1}(U): U \in \tau_X\}= \{\pi_2^{-1}(\emptyset),\pi_2^{-1}(\{1\}), \pi_2^{-1}(\{1,2\})\}= \{\emptyset, X\times \{3\}, X \times Y\}$$

Thus, $$\mathcal{S}= \{\emptyset, X \times Y, \{1\}\times Y, X\times \{3\}\}$$
$$=\{\emptyset, \{(1,3),(1,4),(2,3),(2,4)\}, \{(1,3),(1,4)\}, \{(1,3),(2,3)\}\}$$
is a subbasis of $X \times Y$.

It is not a basis, because a basis must contain $\{(1,3)\}= \{1\}\times \{3\}$.

Title: Munkres' Topology - Theorem 15.2


问:
How can we view monoids as lax functors?
Let $*_2$ be the terminal $2-$category, $V$ be a monoidal category with unit object $I$ and $\mathbf BV$ be its delooping, i.e the $2-$category with $ob(\mathbf BV) = \{*\}$ and $\text{Hom}_{\mathbf BV}(*,*) = V$ and composition given by

$$\text{Hom}_{\mathbf BV}(*,*) \times \text{Hom}_{\mathbf BV}(*,*) \rightarrow \text{Hom}_{\mathbf BV}(*,*)$$

$$ = $$

$$\otimes:V \times V \rightarrow V$$

The unit of this composition is given by the unit object in $V$.

I am told that a lax $2-$functor $F:*_2 \rightarrow \mathbf BV$ is the same as a monoid in $V$. The problem is that I just can't get it to work out on paper.

Let $F$ be such a lax functor, denote the identity $1-$morphism in $*_2$ by $id_1$ and the identity $2-$morphism by $id_2$.

Now, a monoid in $V$ consists of an object $M$, a morphism $\mu: M \otimes M \rightarrow M$ and a morphism $\iota:I \rightarrow M$

I understand that we get $M = Fid_1$ but how do we get $\mu$?

The only thing I can think of is that we have an "associator"

$$\mu: M = F(id_1) = F(id_1 \circ id_1) \rightarrow F(id_1) \circ F(id_1)= M \circ M = M \otimes M$$

$$\mu: M \rightarrow M \otimes M $$

but it goes the wrong way?

The "unitor" $\iota : F(id_1) \rightarrow id_{F*}$

($\{ * \} = ob(*_2)$)

gives us a map $\iota: M \rightarrow I$

...that also goes the wrong way!

It would seem that lax functors $*_2 \rightarrow \mathbf B V$ are really equivalent to $co$monoids in $V$!

What is going on here?

Source for definitions etc: https://ncatlab.org/nlab/show/2-functor

答: I asked this question on overflow and got the answer that the definitions on the nLab are wrong. I'm right but have the definitions wrong basically.</p>


问:
Find a mistake in finding limit, please.
I wanted to find the limit: $\lim_{n \to +\infty}(\frac{1^p + 2^p + ... + n^p}{n^p} - \frac{n}{p + 1})$, where $p \in \mathbb{N}$.

I saw that I could use Stolz–Cesàro theorem, where $x_n = (p + 1)(1^p + 2^p + ... + n^p) - n^{p + 1}$, and $y_n = n^p(p + 1)$ tends to +$\infty$.

So, here we are:

$\lim _{ x\rightarrow \infty  }{ \frac { { x }_{ n }-{ x }_{ n - 1 } }{ { y }_{ n }-{ y }_{ n - 1 } } } = \lim _{ x\rightarrow \infty  }{ \frac { ( p+1 ) { n  }^{ p }-{n}^{ p+1 }+{ (n-1) }^{ p+1 } }{ ( p+1 ) ( { n }^{ p }-{ (n-1) }^{ p } )  } } = \lim _{ x\rightarrow \infty  }{ \frac { ( p+1 )-n+(n-1)(\frac{n - 1}{n})^p  }{ ( p+1 ) ( 1-{ (\frac{n-1}{n}) }^{ p } )  } } = \lim _{ x\rightarrow \infty  }{ \frac { p  }{ 0 } } $ (because $(\frac{n - 1}{n})^p$ tends to 1).

But as I understood here, limit of this sequence is $\frac{1}{2}$.

Where did I go wrong in my reasoning and why is my answer not correct?

Body:
$$\lim _{ n\rightarrow \infty  }{ \frac { ( p+1 ) { n  }^{ p }-{n}^{ p+1 }+{ (n-1) }^{ p+1 } }{ ( p+1 ) ( { n }^{ p }-{ (n-1) }^{ p } )  } } $$
set $n=\frac{1}{x}$ the limit becomes
$$\lim _{ x\rightarrow 0  }\frac{(p+1) \left(\frac{1}{x}\right)^p-\left(\frac{1}{x}\right)^{p+1}+\left(\frac{1}{x}-1\right)^{p+1}}{(p+1) \left(\left(\frac{1}{x}\right)^p-\left(\frac{1}{x}-1\right)^p\right)}$$
dramatic simplification
$$\lim _{ x\rightarrow 0  }\frac{\frac{p}{1-(1-x)^p}+1-\frac{1}{x}}{p+1}$$
MacLaurin expansion
$$\frac{\frac{p}{1-(1-x)^p}+1-\frac{1}{x}}{p+1}\sim \frac{1}{2}+\frac{1}{12} (p-1) x+O\left(x^2\right)$$
as $x\to 0$ we have $$\frac{\frac{p}{1-(1-x)^p}+1-\frac{1}{x}}{p+1}\to \frac{1}{2}$$


问:
Exercise 5.24, Isaacs's Character Theory

Let $G$ be a doubly transitive permutation group on $\Omega$ and let $\alpha,\beta\in \Omega$ with $\alpha\neq \beta$. Let $\phi\in Irr(G_\alpha)$ and assume that $\phi_{G_{\alpha\beta}}\in Irr(G_{\alpha\beta}).$ Show that $[\phi^G,\phi^G]\leq 2$.(Isaacs exercise 5.24)

The hint says to use the Mackey theorem.

Here is what I have gotten so far:

Since $G$ is doubly transitively on $\Omega$, we can decompose $G$ as $G={G_\alpha}\dot\cup{G_\alpha}g{G_\alpha}$ for any $g\notin G_\alpha$.

By the Mackey theorem, we can write
$$ (\phi^G)|_{G_\alpha}=\phi+((\phi_g)|_{G_\alpha^g\cap G_\alpha})^{G_\alpha} $$
where $G_\alpha^g=g^{-1}G_\alpha g$ and $ \phi_g: G_\alpha^g\ni x\mapsto\phi(gxg^{-1}) $.

Since $\phi\in Irr(G_\alpha)$, we have by the Frobenius reciprocity:
\begin{align}
[\phi^G,\phi^G]_G&=[\phi,(\phi^G)|_{G_\alpha}]_{G_\alpha}\\
&=1+[\phi,((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}.
\end{align}
It suffices to show that $[\phi,((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}\le 1$.

Note that $\phi_{G_{\alpha\beta}}\in Irr(G_{\alpha\beta})$, we have
\begin{align}
1=[\phi_{G_{\alpha\beta}} ,\phi_{G_{\alpha\beta}}]_{G_{\alpha\beta}}&=[\phi,(\phi_{G_{\alpha\beta}})^{G_{\alpha}}]_{G_\alpha}.
\end{align}
We know that
\begin{align}
[\phi,((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}&\le [(\phi_{G_{\alpha\beta}})^{G_{\alpha}},((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha}]_{G_\alpha}\\
&=[\phi_{G_{\alpha\beta}},(((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha})|_{G_{\alpha\beta}}]_{G_\alpha\beta}.
\end{align}
Now it is sufficient to show that $(((\phi_g)_{{G_\alpha}^g\cap {G_\alpha}})^{G_\alpha})|_{G_{\alpha\beta}}$ has at most one irreducible constituent $\phi_{G_{\alpha\beta}}$ in $G_{\alpha\beta}$. Then I got stuck... Any hint? Thank you!

答:
It would simplify notation if you wrote $G_{\alpha\beta}$ rather than $G_\alpha^g \cap G_\alpha$, where $\beta = g^\alpha$.

Then we want to prove that $[\phi,(\phi_g)_{G_{\alpha\beta}}^{G_\alpha}] \le 1$.

By Frobenius Reciprocity again, we have
$[\phi,(\phi_g)_{G_{\alpha\beta}}^{G_\alpha}] = [\phi_{G_{\alpha\beta}},(\phi_g)_{G_{\alpha\beta}}]$.

But $\phi_{G_{\alpha\beta}}$ and $(\phi_g)_{G_{\alpha\beta}}$ are irreducible by assumption, so the result follows.


Title: Proving a relation between $\sum\frac{1}{(2n-1)^2}$ and $\sum \frac{1}{n^2}$

Body:
I ran into this question:

Prove that:

$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}=\frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}$$

Thank you very much in advance.


答:
As an addition to @The 2nd's excellent answer, you can ask a computer to find the solution for you. It is very fast. Here is an exemple in the Python language with the sympy module. It gives you a list of the three complex solutions</p>
```<code>>>> from sympy.solvers import solve
>>> from sympy import Symbol
>>> x = Symbol('x')
>>> solve(x**3-6*x**2+2*x-2, x)
[2 + (-1/2 - sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3) + 10/(3*(-1/2 -
sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3)), 2 + 10/(3*(-1/2 + sqrt(3)*I/2)*
(sqrt(969)/9 + 7)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3),
10/(3*(sqrt(969)/9 + 7)**(1/3)) + 2 + (sqrt(969)/9 + 7)**(1/3)]
</code>```

Title: $x^3-6x^2+2x-2=0$ solve for x

Body: I solved it by approximation but its very long way to solve, so i was wondering if there is a quicker way to solve this kind of questions.</p>

Body:
As an addition to @The 2nd's excellent answer, you can ask a computer to find the solution for you. It is very fast. Here is an exemple in the Python language with the sympy module. It gives you a list of the three complex solutions

>>> from sympy.solvers import solve
>>> from sympy import Symbol
>>> x = Symbol('x')
>>> solve(x**3-6*x**2+2*x-2, x)
[2 + (-1/2 - sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3) + 10/(3*(-1/2 -
sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3)), 2 + 10/(3*(-1/2 + sqrt(3)*I/2)*
(sqrt(969)/9 + 7)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(969)/9 + 7)**(1/3),
10/(3*(sqrt(969)/9 + 7)**(1/3)) + 2 + (sqrt(969)/9 + 7)**(1/3)]


问:
How do you show that the class of diffeomorphisms from a compact manifold is stable?
This is pretty much a copy of this question. However, it is unanswered, which makes my question not a duplicate according to this meta answer. I need to show that the class of diffeomorphisms from $N$ to $M$ is stable, where $N$ is compact. I.e., I need to show that for a smooth family of functions $\{F_s : N \longrightarrow M\}_{s\in S}$ with $F_0$ a diffeomorphism, there is a neighborhood $T \ni 0$ such that $F_s$ is a diffeomorphism whenever $s \in T.$ This is part of Problem 6-16 from Lee's Introduction to Smooth Manifolds.

So far, I can show that $F_s$ is injective and a local diffeomorphism. If I could show it is surjective, then a proposition from Lee would give me diffeomorphism. But I can't think of how to do this. As an alternative, since $N$ is compact, I thought of getting a finite collection of domains on which $F_s$ restricts to a diffeomorphism and using a partition of unity to put them together, but the functions of a partition of unity are only guaranteed to be smooth, not diffeomorphisms, so the resulting composition would not necessarily be a diffeomorphism. I also considered using the fact that $dF_0$ is an isomorphism, but I got stuck here. I think that trying to show surjectivity sounds most promising, unless someone knows how to make the partition of unity work. Any help is appreciated.

答:
So far, you have a connected neighborhood $U\in S$ such that $F_s$ is an injective local diffeomorphism for $s\in U$. You can show that these $F_s$ are also surjective for entirely topological reasons. Throughout, fix $s\in U$

$F_s$ is an open map, since it is a local diffeomorphism, and a closed map since $N$ is compact. Thus, if $C\subseteq N$ is a connected component, then $F(C)$ is as well. Additionally, $F_s$ must map connected components of $N$ to connected components of $M$ in the same way as $F_0$, since $U$ is connected, and thus we can find a path from $F_0(x)$ to $F_s(x)$ for any $x\in M$. Since $F_0$ is surjective, $F_s$ is as well.


答:
Let $m$ be a positive integer with $m\ge 2$, and let $R$ be the ring $\mathbb{Z}/(m)$.

Let $S$ be the set of nonzero zerodivisors of $R$, and let $I= S\cup \{0\}$.

If $m$ is prime, then $R$ has no nonzero zerodivisors (why?), hence $I=\{0\}$ which is an ideal of $R$.

Next assume $m$ is composite, and suppose $I$ is an ideal of $R$.

If there are distinct primes $p,q$ such that $p,q$ are both divisors of $m$, then $p,q$ are zerodivisors (why?), hence $p,q\in I$, but then since $(p,q)=(1)$ (why?), we would have $1\in I$, contradiction (why?).

Hence $m$ must have the form $p^k$ where $p$ is prime and $k\ge 2$ (why?).

Thus for the case where $m$ is composite, we have a necessary condition on $m$.

To see that this condition is sufficient, suppose $m=p^k$ where $p$ is prime and $k\ge 2$. Now show that the zerodivisors of $R$ (including $0$) are precisely the elements of the ideal $(p)$. I'll leave that verification to you.

Title: For which positive integers will the following set be an ideal?

Body:
Find all the positive integers $m$ for which the zero divisors
together with $0$ form an ideal in the ring $\mathbb{Z}/(m)$.

We need the set $\{ x \in \mathbb{Z}/(m) \mid x \text{ is a zero divisor}\} \cup \{ 0 \}$ to be an ideal in the ring $\left( \mathbb{Z}/(m), +, \cdot \right)=\left( \mathbb{Z}_m, +, \cdot \right)$.

I've been struggling with this problem, but I don't even know how to start the solution. I've tried some small examples, but that's it. Can anyone help me?


问:
Limits of integration to find $\int_0^tB_s^2\text{d}s$
I am currently attempting to find the variance of $\int_0^tB_s^2\text{d}s$. In particular, I am confused about the integral of $\mathbb{E}\left(\int_0^tB_s^2\text{d}s\right)^2$. I know that repeated applications of Fubini's theorem leads us to $\mathbb{E}\left(\int_0^tB_s^2\text{d}s\right)^2=\int_0^t\int_0^t\mathbb{E}(B_r^2B_s^2)^2\text{d}r\text{d}s$. I also know that $\mathbb{E}(B_r^2B_s^2)^2=2s^2+rs$. What I don't understand is why we cannot integrate this directly, i.e.
$$\int_0^t\int_0^t\mathbb{E}(B_r^2B_s^2)^2\text{d}r\text{d}s=\int_0^t\int_0^t2s^2+rs\text{d}r\text{d}s=\int_0^t\frac{2t^3}3+\frac{rt^2}2\text{d}r=\frac{11t^4}{12}.$$
I have read from these links here, here and here that I should split the integral when $r<s$ and $s<r$, and use the symmetry of the integrand to obtain
$$\int_0^t\int_0^t\mathbb{E}(B_r^2B_s^2)^2\text{d}r\text{d}s=2\int_0^t\int_0^s2s^2+rs\text{d}r\text{d}s=\frac{7t^4}{12}.$$
Perhaps I have overlooked something from my elementary multivariable calculus, but I really can't think of any reason why. Help is appreciated, thanks!

答:
It is not true that
$$\mathbb{E}(B_r^2 B_s^2) = 2s^2 + rs$$
You can see this because the expression on the left of the equality sign is symmetric in $r$ and $s$ while the expression on the right is not.

Rather, this equality holds for $s \leq r$. I.e.

$$\mathbb{E}(B_r^2 B_s^2) = \begin{cases}2s^ 2 + rs & s \leq r \\ 2r^ 2 + rs & r \leq s\end{cases}$$

Now, it should be obvious why you have to split the limits of integration.


问:
R.v. $X$, $E[X^2]<\infty$. How to prove that $\mathrm{Var}[X\land a] \le \mathrm{Var}[X]$ for any real number $a$?
As the title says, It looks simple, however not very intuitive. I finished it with the step function converging to it. However, it seems too complicated and I wonder would there be some elementary proof.</p>

答:
Since $X=X\wedge a+(X-a)^+$, then
\begin{align*}
 \mathsf{var}[X]&=\mathsf{var}[X\wedge a]+\mathsf{var}[(X-a)^+]+2\cdot\mathsf{cov}(X\wedge a,(X-a)^+)\\
 &\ge \mathsf{var}[X\wedge a]+2\cdot\mathsf{cov}(X\wedge a,(X-a)^+).
\end{align*}
Meanwhile
\begin{align*}
 &\mathsf{cov}(X\wedge a,(X-a)^+)=\mathsf{cov}(X\wedge a-a,(X-a)^+)\\
 &\qquad =-\mathsf{cov}((X-a)^-,(X-a)^+)\\
 &\qquad=-\mathsf{E}[(X-a)^-(X-a)^+]+\mathsf{E}[(X-a)^-]\mathsf{E}[(X-a)^+]\\
 &\qquad=\mathsf{E}[(X-a)^-]\mathsf{E}[(X-a)^+]\ge 0,
\end{align*}
therefore
$$ \mathsf{var}[X]\ge \mathsf{var}[X\wedge a].
$$


问:
Let $a,b>0$ be real numbers satisfying $a^2+b^2=2$. Prove that $a+b \leq 2$.
I've tried messing around with some AM-GM stuff but I can't think of anything else for this. Can I get a proof for Real numbers? (I had a proof for rationals but can't work for reals).</p>

答:
Let $a^2+b^2\le 2$, then
by AM-RMS inequality $$\frac{a+b}{2} \le \sqrt{\frac{a^2+b^2}{2}} = 1 \implies (a+b) \le 2.$$


问:
Translating sentences into quantified logic?
Consider two functions $f : A → B$ and $g : A → B$. Translate each of the following statements
into formal notation using quantificational logic

(a) Every output generated by $f$ is also generated by $g$
$$\forall x \forall y \forall z (f(x)=y=g(x)=z⇒y=z) $$

(b) $f$ is surjective

$$\forall x\in A \forall y \in B(f(x)=f(y) ⇒ x=y)$$

These were my attemps  at translation. If I am off please let me know why thank you.

答:
(a) $\forall x \in A \; \exists y \in A \; f(x) = g(y)$

(b) $\forall y \in B \; \exists x \in A \; f(x) = y$


答:
Use C-S,
$$
2(a^2 + b^2) \geqslant (a + b)^2\Rightarrow4\geqslant (a + b)^2 \Rightarrow2\geqslant a + b
$$
This is true for all real numbers.

Title: Let $a,b>0$ be real numbers satisfying $a^2+b^2=2$. Prove that $a+b \leq 2$.

Body: I've tried messing around with some AM-GM stuff but I can't think of anything else for this. Can I get a proof for Real numbers? (I had a proof for rationals but can't work for reals).</p>


问:
prove that $5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$

prove that $$5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$$

.A little use of calculator shows that  $\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}=5.44$.Thus the inequality is indeed true.

Generalising this result with $$f(x)=x-\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}<0$$ does not help as we see that $$8>\sqrt{8}+\sqrt[3]{8}+\sqrt[4]{8} \tag !$$

Repeated efforts of Bernoullis inequality have failed

Body:
By hand, you can compute :
$$2^2=4$$
$$1.7^3 =4.913$$
$$1.3^4=2.8561$$

So $$\sqrt{5} + \sqrt[3]{5} + \sqrt[4]{5} > \sqrt{4} + \sqrt[3]{4.913} + \sqrt[4]{2.8561}= 2 + 1.7 + 1.3 = 5$$


问:
How to transpose $\left( \frac{\Sigma^{-1} 1 }{1^\top \Sigma^{-1} 1} \right)$? Matrix algebra simplification based on the transpose operator
How does the following matrix algebra reduce dimensionally

$$\left( \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1}}\right)^\top$$
given that $\boldsymbol{\Sigma}$ is the covariance matrix and $\boldsymbol{1}$ is a vector of ones?

I think the solution is a vector, like how its un-transposed self was a vector, but I'm more interested in how the transpose operator changes the dimensions of each individual vector and matrix in the parentheses, derivation-wise. which explains the question title being "how to", instead of "what is".

答:
Assuming your column of ones is a column vector, then the top is a column vector so the transpose is a row vector. So it's not a scalar. Write it as $$\left( \boldsymbol{\Sigma}^{-1} \boldsymbol{1} (\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1})^{-1}\right)^\top$$and it's much easier to see what's going on.

Now you can take out the denominator because it's a scalar. Use $(AB)^T=B^TA^T$ also.

Note that the denominator is a scalar so the transpose doesn't affect it.

Also, a covariance matrix is always symmetric, so you can drop the T on it after you finish.


答:
There are essentially two ways:

Through the properties of matrix multiplication and $(AB)^\top=B^\top A^\top$ and $(A^{-1})^\top=(A^\top)^{-1}$: $$\left(\frac{\Sigma^{-1}1}{1^\top\Sigma^{-1} 1}\right)^\top=(\Sigma ^{-1} 1( 1^\top\Sigma^{-1} 1)^{-1})^\top=(1^\top(\Sigma^\top)^{-1}1)^{-1} 1^\top(\Sigma^\top)^{-1}=\frac{1^\top(\Sigma^\top)^{-1}}{1^\top(\Sigma^\top)^{-1}1}$$

Using the fact that transponse is linear with respect to the action of the scalars over matrices plus the same identities $$\left(\frac{\Sigma^{-1}1}{1^\top\Sigma^{-1} 1}\right)^\top=\frac{\left(\Sigma^{-1}1\right)^\top}{1^\top\Sigma^{-1} 1}=\frac{1^\top(\Sigma^\top)^{-1}}{1^\top\Sigma^{-1} 1}$$

Now, here you should be in a case where $\Sigma^\top=\Sigma$ because it's a covariance matrix, but since the discussion is general we've just re-discovered the fact that $v^\top Aw=w^\top A^\top v$ for all $A\in\Bbb R^{n\times m}$, $v\in \Bbb R^n$ and $w\in \Bbb R^m$.

Title: How to transpose $\left( \frac{\Sigma^{-1} 1 }{1^\top \Sigma^{-1} 1} \right)$? Matrix algebra simplification based on the transpose operator

Body:
How does the following matrix algebra reduce dimensionally

$$\left( \frac{\boldsymbol{\Sigma}^{-1} \boldsymbol{1} }{\boldsymbol{1}^\top \boldsymbol{\Sigma}^{-1} \boldsymbol{1}}\right)^\top$$
given that $\boldsymbol{\Sigma}$ is the covariance matrix and $\boldsymbol{1}$ is a vector of ones?

I think the solution is a vector, like how its un-transposed self was a vector, but I'm more interested in how the transpose operator changes the dimensions of each individual vector and matrix in the parentheses, derivation-wise. which explains the question title being "how to", instead of "what is".


问:
Show that $\forall n \in \mathbb N^+:G^{(n)}\trianglelefteq G^{(n-1)}$
Given a group $G$,the derived subgroup of $G$ denoted by $[G,G]$ is defined as $[G,G]=\langle [a,b]:a,b \in G\rangle$ ,where $[a,b]$ is called the commutator of $a$ and $b$.

One can define: $G^{(0)}=G$ and for $n \in \mathbb N^+:G^{(n)}=[G^{(n-1)},G^{(n-1)}]$

Then the series $$...G^{(3)}\trianglelefteq G^{(2)}\trianglelefteq G^{(1)}\trianglelefteq G^{(0)}=G$$

Is called the derived series.

I'm trying to show that $\forall n \in \mathbb N^+:G^{(n)}\trianglelefteq G^{(n-1)}$ using induction on $n$.

Base case: By the definition $G^{(1)}=[G,G]$,take $a \in G$ and $b \in [G,G]$ then $[a,b]=aba^{-1}b^{-1} \in [G,G]$,since $[G,G]$ is a subgroup it implies that $[a,b]b=aba^{-1} \in [G,G]$ which shows that $G^{(1)}\trianglelefteq G^{(0)}=G$.

Now assume $G^{(n)}\trianglelefteq G^{(n-1)}$ does hold and consider $G^{(n+1)}$.

This is where do I have a problem and cannot finish the proof.

答: Just take $H=G^{(n-1)}$ and use your proof in the case $n=1$ to show $[H,H]=[G^{(n-1)}, G^{(n-1)}]=G^{(n)}$ is a normal subgroup of $H$. I think you are a bit confused about the notation but after you think about it the answer you are looking for is that, simply, the commutator subgroup of a group is normal.


问:
questions about a term and a notation from a linear algebra exercise
I have the following exercise:

here

when they write $V^T_1$, do they mean $(V^T)_1$ - which is the first column of $V^T$, or $(V_1)^T$, which is a row?

when they write "outer product expansion" - what do they mean? they did not explain it in the book, and I could not find anything about the expansion in google - only regular, vector outer product.

答:
They write $\mathbf{v}_1^T$, not $V_1^T$. The former is the transpose of the first column of $V$.

The outer product of two $n$-dimensional column vectors $\mathbf{u}, \mathbf{v}$ is $\mathbf{u}\mathbf{v}^T$. Note that this is an $n \times n$ matrix. (Contrast it with the inner product, $\mathbf{u}^T\mathbf{v}$, which is a scalar.) They don't mean anything more than the explicit expansion they've already written. You can ignore the words "outer product" for the purposes of actually doing the exercise.


问:
What is the probability that the second roll shows an even sum, given that the first roll shows a sum greater than 7?
As the title states: 'If two fair dice are rolled twice, what is the probability that the second roll shows an even sum, given that the first roll shows a sum greater than 7?'

My textbook's answer is 0.208
∧
My answer is 0.6, and my working out:

P(B|A) = P(A ∧ B)/P(A)

P(A ∧ B) = P(A)*P(B)

P(A) = 15/36 {(2,6), (3,5), (3,6), (4,4), (4,5), (4,6),(5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)}

P(B) = 1/2

Therefore, 15/36 * 1/2 = 15/72

p(B|A) = P(A ∧ B)/P(A) = (15/72) / (15/36) => 15/72 * 36/15 => 1/2

I am not sure how the official answer is 0.208.

Please help me with this one.

答:
'If two fair dice are rolled twice, what is the probability that the second roll shows an even sum, given that the first roll shows a sum greater than 7?'

As per independence, the probability that the second roll (second roll of the two dice) shows an even sum is always 0.5. This probability is not affected by the result of the previous two dice's roll.

The answer 0.208 of your textbook is just $\frac{15}{72}$, that can be, for example, the answer to the following similar question:

'If two fair dice are rolled twice, what is the probability that the second roll shows an even sum AND that the first roll showed a sum greater than 7?'

So probably the solution is referred to another exercise... but there is a certain confusion also in your reasoning.


问:
How to model the arc tangent like function in a high school way, given the points on its graph?
I'm trying to come up with a regression function for a student of mine that best fits the the attached graph, but using a method for high school students, who're only familiar with trigonometry and calculus of one variable.

To be precise, they're given $n=30$ pair of points $(x_i,y_i)$ (which I'm not write here, but they do have them!) and they're also given the attached graph that the function generates. They're asked to model the function.

If it was not high school, then I'd have modeled it like:

$$y= f(x):= a \hspace{1mm} arctan (bx + c) + d,  $$
and I'd have tried the minimize the sum of squared errors $\sum_{i=1}^{n} (y_i - f(x_i))^2$, and would've minimized it w.r.t. $a,b,c,d$, form the normal equations and either analytically or numerically solve them.

But for high school, I can't do all that - they've barely seen regression and they don't know partial derivatives, so I'm not completely sure how to proceed here. Any help appreciated!

答:
For a simpler approach in the particular case of the figure joint to the question, see the answer from Claude Leibovici.

The answer below can be applied in a more general case :
$$Y=A\:\arctan(B\:X+C)+D$$
is equivalent to :
$$X=\frac{1}{B}\tan\left(\frac{1}{A}Y-\frac{D}{A}\right)-\frac{C}{B}$$
With change of notations :$\quad\begin{cases}
a=\frac{1}{B}\\
b=\frac{1}{A}\\
c=-\frac{D}{A}\\
d=-\frac{C}{B}\\
y=X\\
x=Y
\end{cases}\quad$ the equation is equivalent to :
$$y=a\tan(bx+c)+d$$

The usual way is to use a nonlinear regression method of fitting. They are several softwares. The calculus is iterative starting from "guessed" values of the parameters. Sometimes the numerical calculus fails due to bad initial guessed values and/or not convergent iteration.

The principle of a non conventional method (no iteration, no initial guess) is explained in this paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales , partially translated in : https://scikit-guess.readthedocs.io/en/latest/appendices/references.html. An application to the case of tangent function is shown below.

Thanks to this method one can compute adjusted values of the parameters $a,b,c,d$ and then $A,B,C,D$.

If a well defined criteria for fitting is specified ( least mean absolute deviation or least relative deviation or others) the above method is not sufficient. An iterative process is required with guessed values to start. The above calculus can be used to get a set of good initial values.

CAUTION :

In the above method numerical integrations are carried out for the computation of $S_k$ and $T_k$. Such a simplified process cannot be used if the function tends to infinity around a point on the range of the data $(x_k\, ,\,y_k)$. For example if the points are distributed as on the next figure the above method will not be convenient.

FOR INFORMATION :

An integral equation to which $\quad y(x)=a\tan(bx+c)+d\quad$ is solution is :
$$y(x)=\frac{b}{a}\int (y(x))^2dx-2\frac{bd}{a}\int y(x)\,dx+\left(\frac{b\,d^2}{a}+ab\right)x+\text{constant}$$
This allows the linearisation of the regression and the construction of the first 4x4 matrix.


问:
When you have variables in the entries of a matrix, what does it mean if a certain value of the variable gives an undefined result?
I'm given the matrix: \begin{bmatrix}-a-7&-2a-6 & 0\\a+3&a+3 &3a-3\\ -2a-14 & -4a-12 & a-1\end{bmatrix} and I wanted to determine which values of $a$ give the matrix only $1$ free variables, so I got the matrix into this echelon form: \begin{bmatrix}-a-7&-2a-6&0\\0&\left(\frac{(a+3)(1-a)}{a+7}\right)&3a-3 \\ 0 & 0 & a-1\end{bmatrix}

so clearly $a=1$ gives the matrix two free variables, so $3$ is not a solution,

and $a=-3$ gives the matrix a single free variable as desired, so $1$ is a solution.

Where I'm unclear is with $a=-7$ since that would make the top left entry $=0$, but it would also make the middle entry (second diagonal entry) be $(\frac{-32}{0})$ which is undefined since you  cannot divide by zero. But when I let $a=-7$ in the original matrix, I get $0$ free variables.

Am I missing something?

答:
Before we divide by a number, we check if it is zero and we can handle them separately. When $a=-7$.

The original matrix becomes

$$\begin{bmatrix} 0 & 8 & 0 \\ -4 & -4 & -24 \\ 0 & 16 & -8\end{bmatrix}$$ which is clearly a ful rank matrix, hence your computation for $a=-7$ is right.

Note that when you mean $1$ free variable, if you mean the number of free parameter in a linear system which is the nullity. Then when $a=1$, the rank of your matrix is $1$ and hence the nullity is $2$.


问:
Integration of the product of two polynomials
Consider this integration:

$$\int (x-\mu_x)^2  \cdot (mx+q) dx$$

I can square the first element, do the products and then integrate but I was wondering, is there a smarter way to perform this integration?

答:
There is.

$$mx+q=m(x-\mu_x)+(m\mu_x+q)$$ and an antiderivative is

$$m\frac{(x-\mu_x)^4}{4}+(m\mu_x+q)\frac{(x-\mu_x)^3}{3}.$$


问:
Understanding the proof the iterative improvement method for the result of linear equation solving using LU decomposition in numerical recipes
The question is from section 2.5 Iterative Improvement of a Solution to Linear Equations in Numerical Recipes book.

When we solve $\mathbf{Ax = b}$ using LU decomposition numerically, the result is not exact due to the round off error accumulation. So we use a trick called interative improvement of the solution to restore the full machine precision. Suppose that a vector $\mathbf{x}$ is the exact solution of the linear set $$\mathbf{Ax = b}$$
You don't however know $\mathbf{x}$. You only know slightly wrong solution $\mathbf{x} + \delta\mathbf{x}$, where $\delta\mathbf{x}$ is the unknown error, when multiplying by $\mathbf{A}$, it becomes $$\mathbf{A\cdot(x + \delta x) = b + \delta b}\tag{1}$$
Then we have $$\mathbf{A\cdot\delta x = \delta b}\tag{2}$$ $\delta b$ is easy to get by equation (1), thus we can iterative improve the result using (1) and (2).

But we neglected the fact that the LU decomposition of $\mathbf{A}$ is itself exact. A different analytical approach starts with some matrix $\mathbf{B_0}$ that is assumed to be an approximate inverse of the matrix $\mathbf{A}$, so that $\mathbf{B_0\cdot A}$ is approximately the identity matrix $\mathbf{1}$. Define the residual matrix $\mathbf{R}$ of $\mathbf{B_0}$ as $$\mathbf{R\equiv 1 - B_0\cdot A}$$ which is supposed to be "small", therefore $$\mathbf{B_0\cdot A = 1 - R}$$ Next consider the following formal manipulation $$\mathbf{A^{-1} = A^{-1}\cdot (B_0^{-1}\cdot B_0) = (A^{-1}\cdot B_0^{-1})\cdot B_0 = {(B_0\cdot A)}^{-1}\cdot B_0 \\ = {(1 - R)}^{-1}\cdot B_0 = (1 + R + R^2 + R^3 + \cdots)\cdot B_0}$$ We can define the nth partial sum of the last expression by $$\mathbf{B_n \equiv (1 + R + \cdots + R^n)\cdot B_0}\tag{3}$$ so that $\mathbf{B_\infty \to A^{-1}}$, if the limit exits.

It now is straight to verify that equation (3) satisfies some interesting recurrence relations. As regards solving $\mathbf{Ax = b}$, where $\mathbf{x}$ and $\mathbf{b}$ are vectors, define $$\mathbf{x_n \equiv B_n\cdot b}\tag{4}$$ It can be show that $$\mathbf{x_{n+1} = x_n + B_0 \cdot (b - A\cdot x_n)}\tag{5}$$
Here is what I cannot understand, how to deduce equation (5) from above equations.

And the book implies that $$\mathbf{B_{2n+1} = 2B_n - B_n\cdot A\cdot B_n} \qquad n = 0, 1, 3, 7 ...$$ can be drived from equation (3), and I don't know how. Can some one make these two deduction more clear, Thank you!

答:
For (5), we can proceed as follows.
You can check using (3) that
$$
(I-R^{n+1})B_0=(I-R)B_n=B_0AB_n.
$$
Then we have
$$
\begin{split}
x_{n+1}-x_n&=(B_{n+1}-B_n)b=R^{n+1}B_0b
\\
&=[I-(I-R^{n+1})]B_0b=B_0b-B_0AB_nb\\
&=B_0(b-Ax_n).
\end{split}
$$

For the last identity, we have
$$
\begin{split}
B_nAB_n&=(I+R+\cdots+R^n)B_0A(I+R+\cdots+R^n)B_0
\\&=(I+R+\cdots+R^n)(I-R)(I+R+\cdots+R^n)B_0
\\&=(I+R+\cdots+R^n)(I-R^{n+1})B_0
\\&=(I+R+\cdots+R^n-R^{n+1}-\cdots-R^{2n+1})B_0
\\&=[2(I+R+\cdots+R^n)-(I+R+\cdots+R^{2n+1})]B_0
\\&=2B_n-B_{2n+1}.
\end{split}
$$
Note that the identity holds for any non-negative integer.


问:
Convert the triple integral from rectangular coordinates to both cylindrical and spherical coordinates
The given integral is:
$$\int_{-7}^7\int_{-\sqrt{49-x^2}}^\sqrt{49-x^2}\int_{x^2+y^2}^{49}xdxdydx$$
Converting to cylindrical coordinates:
$$\int_{0}^{2\pi}\int_{0}^7\int_{r^2}^{49}r^2\cos(\theta)dxdrd\theta$$
It's the spherical coordinates I'm having trouble converting too because I'm don't know how to sketch the given limits. I know that one of the equations is $x^2+y^2+z^2 = 49$ but because the limit of $\theta$ is $2\pi$ I am lost and the example my professor did only went to $\frac{\pi}{2}$. I can't find $\rho$ because I don't know where the two shapes intersect. I calculated the triple integral for cylindrical coordinates and got zero so I'm very confused.

答:
If the integral is

$\displaystyle \int_{-7}^7\int_{-\sqrt{49-x^2}}^\sqrt{49-x^2}\int_{x^2+y^2}^{49} x \, dz \, dy \, dx$

We are basically integrating over the region bound between paraboloid $z = x^2 + y^2$ and plane $z = 49$.

It is a bit complicated in spherical coordinates and you have to split it into two regions.

$x = \rho \cos \theta \sin \phi$

$y = \rho \sin \theta \sin \phi$

$z = \rho \cos \phi$

In spherical coordinates, $\phi$ is the angle made with $z$ axis so at $z = 49$, $\phi$ varies from $0$ to $\cot \phi = \frac{49}{7} = 7 \implies \cot^{-1}(7)$ with no change in value of $z$.

Also as $z$ is constant, $\rho$ as a function of $\phi$ will be $0 \leq \rho \leq 49 \sec \phi$ (from equation of $z$).

Now for $\cot^{-1}(7) \leq \phi \leq \frac{\pi}{2}$, we are traversing till the paraboloid boundary

So, $z = x^2 + y^2 \implies \rho \cos \phi = \rho^2 \sin^2 \phi \implies \rho = \cot \phi \csc \phi$

So, $I = \iiint \rho \cos \theta \sin \phi.\rho^2 \sin \phi \, dr d\phi d \theta$

with two integrations in two parts with limits of integration -

i) $0 \leq \rho \leq 49 \sec \phi, 0 \leq \phi \leq \cot^{-1}(7), 0 \leq \theta \leq 2\pi$.

ii) $0 \leq \rho \leq \cot \phi \csc \phi, \cot^{-1}(7) \leq \phi \leq \frac{\pi}{2}, 0 \leq \theta \leq 2\pi$.


问:
Graph theory problem.
In a country called Sevenland there are 15 cities. And every city has at least 7 highways connected to it. Prove that from any city we can go to any city using the highways</p>

答:
This is a special case of showing that a graph has a Hamilton cycle if $\delta(G)\geq \lfloor \frac{n}{2} \rfloor$, i.e. minimum degree (here 7) is greater or equal than half the number of vertices rounded down (here also 7). This is also known as Dirac's theorem, for a proof see e.g. here: https://proofwiki.org/wiki/Dirac's_Theorem
(This proof is typically covered in any graph theory introduction book.)

Title: Graph theory problem.


问:
Plot any irrational number on number line.
I have a basic question:

Can we plot any irrational number on a number line? I can plot all integers and rational numbers on it but how can I plot any irrational number, such as $\sqrt2$,$\sqrt3$ etc.?

答:
Hints:

Draw a line of length $\,1\,$ , from one of its end points draw another line perpendicular to the first one (how? Can you use a compass...?), and now draw the diagonal between the other end of the original line and the farthest end of the perpendicular one. This perpendicular has length $\,\sqrt 2\,$ (why?) , so with a compass measure it and then place the compass on the origin *the "zero: of your line) and "plot" $\,\sqrt 2\,$ ...you can do something similar with "any" non-rationa square root.


答:
Just another graphic approach.

If you plot $x^2+y^2=2$ you'll get the circumference of center the origin and radius $\sqrt2$.

If you calculate the tangent of this curve at $(1,1)$, you'll get $x+y=2$.

So the whole circle $x^2+y^2=2$ lies on the semiplane $x+y\le2$ (in fact, the circle on "touch" the line). So $x^2+y^2=2$ implies $x+y\le 2$.

MOREOVER.

With the same technique, plotting the tamnget line to the circle at $(-1,-1)$, you'll get $x+y=-2$. So $x^2+y^2=2$ implies $|x+y|\le 2$ and the equality olds if and only if $x=y$.

ANALITICALLY:

$x^2+y^2=2$ $\iff$ $2x^2+2y^2\pm 2xy =4$ $\iff$ $(x^2+y^2+2xy)+(x^2+y^2-2xy)=4$ $\iff$ $(x+y)^2+(x-y)^2=4$.

But $(x+y)^2\le (x+y)^2+(x-y)^2=4$, so $(x+y)^2\le 4$ whence $-2\le x+y\le 2$ and you're (another time), done.


问:
Overdetermined linear system: how can the nullspace help me to find one working solution?
I have an overdetermined linear system and found a general solution to it with four of the twelve variables being free.

I have this assignment to use that solution and nullspace to find a working solution to the system, a solution that has only positive integers. I know that each null space vector corresponds to a free variable, right - but how to use this to find a working solution that I do not know.

I figured out a working solution that has just positive integers by solving the reduced row echelon.

Here are some examples from my Matlab code and outputs:

```<code>%The system
AB =

     1    -1     0     0     0     0     0     0     0     0     0     0     0
     0     1     1     0     0     0     0     0     0     0    -1     0   150
     0     0    -1     1     0     0     0     0     0     0     0     0    20
     0     0     0     1     1     0     0     0     0     0     0    -1   410
     0     0     0     0     1    -1     0     0     0     0     0     0   180
     0     0     0     0     0     1     1     0    -1     0     0     0   210
     0     0     0     0     0     0    -1     1     0     0     0     0    80
     1     0     0     0     0     0     0     1     0    -1     0     0   230
     0     0     0     0     0     0     0     0     1    -1     1    -1     0

%General solution by linsolve(A,b) - has 4 free variables

x =
  -60.0000
  -60.0000
  210.0000
  230.0000
  180.0000
         0
  210.0000
  290.0000
         0
         0
   -0.0000
         0

%My reduced row echelon
     1     0     0     0     0     0     0     1     0    -1     0     0   230
     0     1     0     0     0     0     0     1     0    -1     0     0   230
     0     0     1     0     0     0     0    -1     0     1    -1     0   -80
     0     0     0     1     0     0     0    -1     0     1    -1     0   -60
     0     0     0     0     1     0     0     1     0    -1     1    -1   470
     0     0     0     0     0     1     0     1     0    -1     1    -1   290
     0     0     0     0     0     0     1    -1     0     0     0     0   -80
     0     0     0     0     0     0     0     0     1    -1     1    -1     0
     0     0     0     0     0     0     0     0     0     0     0     0     0

%By hands I found that this was a working solution and all positive integers
x1 =

   210
   210
    10
    30
   440
   260
    20
   100
    70
    80
    70
    60

%the null space vectors
nullspace =

    0.4145    0.0926    0.3329    0.0217
    0.4145    0.0926    0.3329    0.0217
   -0.4035   -0.1754    0.2937    0.1090
   -0.4035   -0.1754    0.2937    0.1090
    0.1665    0.1341   -0.1503    0.4726
    0.1665    0.1341   -0.1503    0.4726
   -0.3065    0.4078    0.0292   -0.1748
   -0.3065    0.4078    0.0292   -0.1748
   -0.1400    0.5420   -0.1211    0.2978
    0.1080    0.5005    0.3621   -0.1531
    0.0110   -0.0827    0.6266    0.1306
   -0.2370   -0.0412    0.1435    0.5816
</code>```

答:
Let $K$ be the $12 \times 4$ representing the kernel above.

By right multiplying it by the invertible matrix

$$M=\begin{pmatrix}
    1 &  0 & 0   &  0\\
    4 &  2 & 1   &  1\\
    4 &  1 & 2   &  1\\
    0 &  1 & 1   &  1\end{pmatrix},$$

one gets the following matrix:

$$K'=KM=\begin{pmatrix}
2.1165&0.5398&0.7801&0.4472\\
2.1165&0.5398&0.7801&0.4472\\
0.0697&0.0519&0.5210&0.2273\\
0.0697&0.0519&0.5210&0.2273\\
0.1017&0.5905&0.3061&0.4564\\
0.1017&0.5905&0.3061&0.4564\\
1.4415&0.6700&0.2914&0.2622\\
1.4415&0.6700&0.2914&0.2622\\
1.5436&1.2607&0.5976&0.7187\\
3.5584&1.2100&1.0716&0.7095\\
2.1866&0.5918&1.3011&0.6745\\
0.1722&0.6427&0.8274&0.6839
    \end{pmatrix}$$

whose columns $K'_1,K'_2,K'_3,K'_4$ constitute as well a basis of the kernel (by considering its columns), with the advantage to have all its coordinates positive.

Therefore, you can get an infinite number of solutions with positive entries by adding to your particular solution any combination $a_1K'_1+a_2K'_2+a_4K'_3+a_4K'_4$ with sufficiently large positive coefficients $a_k$.

Remark: I have obtained $M$ by multiplication of $K$ by thousands of random matrices, followed by a "tuning' of the results.


问:
What does $g(x) = \frac{1}{2} (f(x) + f(-x))$ mean?
Suppose any function $f(x)$ defines $g(x)=\frac{1}{2} (f(x)+f(-x))$. What does this mean essentially? The question is to show $g(x)$ is even, so I am doing
$$g(-x) = g(x) \Rightarrow g(-x)= \frac{1}{2} (f(-x)+f(-(-x))).$$
However, I don't understand why the $x$ in the $f(x)$ function inside the $g(x)$ also has to be $-x$ when I choose to do $g(-x)$, I am doing it just because and it works but I don't know why. What does it mean for a function to be defined in terms of another function?

答:
Take $f(x) = x + 1$, for example. Then
$$g(x) = \frac{1}{2}\big(f(x)+f(-x)\big)$$
means that $g(x) = \frac{1}{2}\big((x+1) + (-x+1)\big) = 1$ (even function).

If a function is defined in terms of another function, say, $g(x) = F(f(x))$ for some rule (function) $F$, then we can find any value of $g$ by plugging the corresponding values of $f$ into the rule $F$. In the above example, $f(x) = x + 1$ and $F_x = \frac{1}{2}(f_x + f_{-x})$.

The following chain can describe the process.
$$x \overset f\to y \overset F\to z$$
We start by taking some value $x$, which obeys the rules. The first rule $f$ orders $x$ to go to $y$, then the rule $F$ sends $y$ to $z$, which is the final value obtained by two rules $F(f(x))) = g(x)$.


答:
The function is NOT $f(x)$, the same way $f(a)$ is not a function. The function is simply $f$. A function
$$
f: A \rightarrow B
$$
has a domain $A$ and a co-domain $B$. When we write $f(a)$, for some $a \in A$, we are talking about a value in $B$ which is associated to $a \in A$.

But how can I tell someone exactly what the function is? How to determine $f(a)$ from $a$? One way to do that is to give a formula. For example:
$$
\begin{align*}
  f: \mathbb{R} &\rightarrow \mathbb{R}
  \\
     x &\mapsto 2x.
\end{align*}
$$
The function described above has NOTHING to do with the variable $x$. It could as well be written as
$$
\begin{align*}
  f: \mathbb{R} &\rightarrow \mathbb{R}
  \\
     a &\mapsto 2a.
\end{align*}
$$
Both are the same, with $x$ or with $a$. What both formula say is that to calculate $f$, for a given $k \in A$, all you have to do is multiply it by $2$. That is, the function $f$ takes its input, doubles its value and present the result as its output.

If we are to write a formula for the function given its input, we usually name the input $x$ and express how to calculate the output based on $x$. If you want to make it less confuse, you can, for example, rewrite $g$'s definition to read
$$
  g(y) = \frac{1}{2}(f(y) + f(-y)).
$$
What it says is that

$g$ takes some value, puts this value and also minus its value into $f$, adds the results and divide by $2$.

Those words would be easier if we give a name to "some value". We called it $y$, but we could have called it $x$... and this $x$ has nothing to do with any other $x$ you might be using elsewhere. We just happened to have labeled this "variable" $x$ so we can express $f$'s formula.

So, to calculate $g(-x)$, what you do is substitute $y$ above by $-x$:
$$
  g(-x) = \frac{1}{2}(f(-x) + f(-(-x))).
$$

Title: What does $g(x) = \frac{1}{2} (f(x) + f(-x))$ mean?

Body:
Suppose any function $f(x)$ defines $g(x)=\frac{1}{2} (f(x)+f(-x))$. What does this mean essentially? The question is to show $g(x)$ is even, so I am doing
$$g(-x) = g(x) \Rightarrow g(-x)= \frac{1}{2} (f(-x)+f(-(-x))).$$
However, I don't understand why the $x$ in the $f(x)$ function inside the $g(x)$ also has to be $-x$ when I choose to do $g(-x)$, I am doing it just because and it works but I don't know why. What does it mean for a function to be defined in terms of another function?


问:
Find derivative of $f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x...}}}$
Find $f(x)$ for $f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x...}}}$

I found that the right expression is an infinite series of $\sum^{\infty}_{n=2}{\frac{1}{n!}}$. I know the series converge but how can I evalute $\sum^{\infty}_{n=2}{\frac{1}{n!}}$?

Any hint or idea will be appreciated. Thank you!

答:
$$f(x)=\tan^2x \sqrt{\tan x\sqrt[3]{\tan x\sqrt[4]{\tan x...}}}=\tan^2 x\cdot \left(\tan x\right)^{\sum _{n=2}^{\infty } \frac{1}{n!}}$$

We have $$\sum _{n=2}^{\infty } \frac{1}{n!}=e-1-1=e-2$$

$$f(x)=\tan^2 x\cdot \left(\tan x\right)^{e-2}=\left(\tan x\right)^e$$

Derivative is $$f'(x)=e \,\frac{\tan ^{e-1} x}{ \cos ^2 x}$$


答:
... as in how do you physically carry out the plotting? Same as plotting any other number.

Pick up a pencil and use a ruler to locate the point at about the right distance from the origin. Use the pencil to apply a short straight mark on the paper, perpendicular to the number line and intersecting it at the point you found with the ruler.

It your pencil is of a somewhat orthodox design, the pencil mark will be at least 0.1 milimeters wide, corresponding to an uncountable infinity of different numbers on the line. If one of those points is the one you set out to plot, you're good.

Title: Plot any irrational number on number line.

Body:
I have a basic question:

Can we plot any irrational number on a number line? I can plot all integers and rational numbers on it but how can I plot any irrational number, such as $\sqrt2$,$\sqrt3$ etc.?


答:
We can convert this problem to a graphs and restate it as follows:

Let $G$ be a simple graph with $|G| = 15$. If $d(v_i) \ge 7,\ \forall i \in \{1,2,...,15\}$, then $G$ is connected.

And to prove such an argument, suppose for a contradiction that $G$ is disconnected. Then $c(G) \ge 2$ where $c(G)$ is the number of components of $G$. But then by Pigeonhole Principle, one of these components has at most $7$ vertices. But we know that $G$ is simple and $d(v_i) \ge 7,\ \forall i \in \{1,2,...,15\}$. But in a component with at most $7$ vertices, we cannot have a vertex with degree $7$ since $G$ is simple; and this is a contradiction.

Title: Graph theory problem.

Body: In a country called Sevenland there are 15 cities. And every city has at least 7 highways connected to it. Prove that from any city we can go to any city using the highways</p>


问:
Given joint pdf $f(x,y,z) = \frac{2}{3}(x+y+z)$, how to find $E(X+Y|Z)$?
As the title says, how do you find $E(X+Y|Z)$ given $f(x,y,z) = \frac{2}{3}(x+y+z)$?

Not sure how to go about approaching this problem. The furthest I got was:

$$
E(X+Y|Z) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x+y)f(x+y|z)dxdy
$$

And now I'm stuck because I'm not sure how to get $f(x+y|z)$.

答:
I think in general you can do as follows:

$\mathbf{E}[X+Y|Z] = \mathbf{E}[X|Z] + \mathbf{E}[Y|Z]$

$\mathbf{E}[X|Z]$ can be found from $f_{X,Z}(x,z)$ and $f_Z(z)$. Same for $\mathbf{E}[Y|Z]$.

Here you are missing the bound of the domain.


问:
Check whether a complex function is conformal or not.
Let
$$f_1(z) = \exp(\sin(z)),\quad f_2(z) = \exp(\log(z)),\quad f_3(z) = \exp(\tan(z))$$ be complex functions defined on an open unit disc centered at origin.

Now $f_1$ and $f_3$ is conformal as it is holomorphic and it's derivative is non zero in the given domain.

But is $f_2$ a conformal map? Derivative is non zero but $\log(z)$ is not defined at $z=0$.

Also I read, since a branch of $\log(z)$ is holomorphic and derivative is non zero, $\log(z)$ is conformal.

So got confused to conclude whether $f_2$ is conformal or not.

Please provide a clarity about these two.

Is the above argument for telling $f_1$ and $f_3$ conformal correct ?

What and how can we conclude whether $f_2$ is a conformal map or not.

Thanks in advance.

答:
A holomorphic function $f$ with nonvanishing derivative is locally conformal. This means that each point $z_0$ in the domain of $f$ has a small neighborhood which is bijectively and conformally mapped onto a neighborhood of $f(z_0)$. This is the case for $f_1$ and $f_3$ in your example, after you have verified that $\sin$ and $\tan$ have no critical point in the unit disc $D$. (Of course $\exp$ has no critical points.)

The "function" $f_2$ is not  defined as a holomorphic function in $D$, since "$\>\log\>$" has no ${\mathbb C}$-valued holomorphic representant there.

Whether $f_1$ and $f_3$ are actually "conformal maps" of $D$ to certain other regions requires further study. It is commonly accepted that a conformal map is not only locally conformal, but a bijective map between two regions. This means that you have to check whether $\sin(D)$, resp. $\tan(D)$, contains two points $w_i=\sin (z_i)$, $z_1\ne z_2$, differing by an integer multiple of $2\pi i$. This check is not so easy, and is maybe not required in your source problem.


答:
When you have any function say $g(x)$,

you get $g(-x)$ by substituting in (-x) whereever x is in g(x). This is just by definition. Similarly if you were trying to get $g(y^2+2)$, it's the same idea. Replace every occurence of x in $g(x)$ with $(y^2+2)$

So if $g(x) = -3x^3-4x^2+6x+2$

To get $g(-x)$, you replace every appearance of x with (-x)

So $g(-x)$ in this case is $-3(-x)^3-4(-x)^2+6(-x)+2$

and

$g(y^2+2) = -3(y^2+2)^3-4(y^2+2)^2+6(y^2+2)+2$

In your case $g(x) = 0.5(f(x)+f(-x))$

What is $g(-x)$ in this case?

Replace the occurence of x with (-x) everywhere... including the spots inside other functions like f.

$g(-x) = 0.5(f((-x)) + f(-(-x)))$

I can replace $f((-x))$ by $f(-x)$ because $(-x) = -x$. I can simplify $f(-(-x))$ by $f(x)$ because $-(-x) = x$

$g(-x) = 0.5(f(-x) + f(x))$

$g(-x) = 0.5(f(x) + f(-x))$

So since we can see that $g(-x) = g(x)$, by definition this $g(x)$ is even.

As André Caldas describes the function g takes a 'something' and does certain operations on that something... The best way to illustrate this is using a dummy variable like x and writing out $g(x)$. g(something else) replaces occurence of x with (something else)


问:
Is $D_8$ a normal subgroup of $S_4$?
I am trying to show that $D_8$ is not a normal subgroup of $S_4$, I have shown that $D_8$ is  Subgroup of $S_4$.Then,  let $r=\langle(1234)\rangle$ and $s =\langle(14)(23)\rangle$.Then,  we see that $D_8$ is a subset of $S_4$ and since it a group so $D_8$ is a subgroup of $S_4$.

Proving that $D_8$ is not a normal subgroup. In $D_8$ there are two elements of order $4$.So,  we take $\sigma (1234) \sigma^{-1}$ which should be in $H$.Now conjugacy preserves the cycle structure so we choose $\sigma$ in such a way so that conjugate element does not belong to $H$.Choosing such a cycle is possible as there are $6$ 4 cycles in $S_4$ and there are only two of them in $D_8$.Since I have shown it for one element, that the conjugate is not in $D_8$.Do I need to show it for the other elements?

答:
A subgroup $H$ of a group $G$ is normal if and only if for all $g\in G$ and $h\in H$, $ghg^{-1}\in H$.

Since you showed that there exists $g\in S_4$ such that $g(1234)g^{-1}\notin D_8$, it is sufficient to conclude that $D_8$ is not a normal subgroup of $S_4$.

Title: Is $D_8$ a normal subgroup of $S_4$?


问:
Understanding proof related to Hungarian algorithm
I am looking at the Hungarian algorithm's matrix interpretation(and more specifically the trick that is needed for steps 3 and 4 on the Wikipedia page on the https://en.wikipedia.org/wiki/Hungarian_algorithm) and trying to figure out the following proof linked to this.

I just have one question about this proof since it is not explained in the text: why is it exactly that the matrix $\boldsymbol{E}$ has $b$ independent zeroes(and $\boldsymbol{D}$ has $a$)? I can't understand this so if someone can explain I would be grateful.

答:
There are $a$ rows, and this is indicated by the matrices $C$ and $D$.

There are $b$ columns, and this is indicated by the matrices $C$ and $E$.

They are not multiplied, as the matrix notation used is augmented.

So, $D$ has $a$ zeroes, and $E$ has $b$ zeroes.

A far simpler proof of what the author is trying to prove is that there can be at most $l=\min(m,n)$ zeroes by the pigeon-hole principle (if there are more than $l$, at least one row/column has two zeroes in it), and then supply an example of a matrix with $l$ zeroes, e.g. a diagonal line of zeroes from a corner.


问:
Sum of squares of characters
Given a finite Abelian group $G$, I am interested in the sum $$ \sum_{\chi} \chi^2(g) $$ for some element $g \in G$ and the sum runs over the characters of $G$.

I tried using the identity $\chi^2(g) = \chi(g^2)$ and changing the above into a usual character sum. Is this correct or am I missing something?

答:
In general, the Second Orthogonality Relation tells you for a general finite group $G$ and $g,h \in G$
$$\sum_{\chi \in Irr(G)}\chi(g)\overline{\chi(h)}= 0 \text{  if  }g \nsim h \text{  and  } = |C_G(g)| \text{  if  } g \sim h. $$
(Here $\sim$ denotes conjugation) Hence if $G$ is abelian and $\chi$ linear, we have
$$\sum_{\chi \in Irr(G)} \chi^2(g)=\sum_{\chi \in Irr(G)}\chi(g)\overline{\chi(g^{-1})}=0 \text{  if  } g \neq g^{-1} \text{and  } =|G| \text{  if } g^2=1.$$
If you want to do some calculations: see for example here (character table of $C_6$), and try an element of order $2$ and of order $3$).

There is even a generalization to groups $G$ not necessarily being abelian. If $g \in G$ and $g \sim g^{-1}$, then $g$ is called a real element. The trivial element (identity) is obviously real, also all elements of order $2$ of the center $Z(G)$. One has
$$\sum_{\chi \in Irr(G)} \chi^2(g)=0 \text{  if  } g \text{  is not real  } \text{  and  } =|C_G(g)| \text{ if } g \text { is real  }.$$
If $|G|$ is odd, then $G$ does not contain any non-trivial real elements. See also here.

Title: Sum of squares of characters


问:
If $AB = I$ then $BA = I$

If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.

I do not understand anything more than the following.

Elementary row operations.

Linear dependence.

Row reduced forms and their relations with the original matrix.

If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?

P.S.: Please avoid using the transpose and/or inverse of a matrix.

答:
If $\rm\,B\,$ is a linear map on a finite dimensional vector space $\rm\, V\,$ over field $\rm\,K,\,$ then easily by finite dimensionality (cf. Note below) there is a polynomial $\rm\,0\ne p(x)\in K[x]\;$ with $\rm\,p(B) = 0.\,$ W.l.o.g.$\rm\,p(0) \ne 0\,$ by canceling any factors of $\rm\,B\;$ from $\rm\;p(B)\;$ by left-multiplying by $\rm A,\,$ using $\rm\, AB = 1.$

Notice $\rm\ AB=1  \, \Rightarrow\,   (BA-1)\, B^n =\, 0\;$  for $\,\rm\;n>0\;$

so by linearity $\rm\, 0 \,=\, (BA-1)\ p(B)\, =\, (BA-1)\ p(0) \;\Rightarrow\;  BA=1 \quad\quad$
 QED

This is essentially a special case of computing inverses by the Euclidean algorithm - see my Apr 13 1999 sci.math post on Google or mathforum.

Note $ $ The existence of $\rm\;p(x)\;$ follows simply from the fact that $\rm\,V\;$ finite-dimensional implies the same for the vector space $\rm\,L(V)\,$ of linear maps on $\rm\,V\,$ (indeed if $\rm\,V\;$ has dimension $\rm n$ then a linear map is uniquely determined by its matrix of $\,\rm n^2\,$ coefficients). So $\rm\, 1,\, B,\, B^2,\, B^3,\,\cdots\;$ are $\rm\,K$-linearly dependent in $\rm\, L(V)$ which yields the sought nonzero polynomial annihilating $\rm\,B.$

Title: If $AB = I$ then $BA = I$


问:
Showing that a union of the subsets of two $\sigma$-algebras is a $\sigma$-algebra.
I got back an assignment for a first course in analysis and I have made a very basic error, and I'm having a lot of trouble pinpointing exactly what piece of information I'm missing.

You have two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\mu)$. Assume that $X$ and $Y$ are disjoint. Let $Z=X\cup Y$

Show that $\mathcal{C}=\{A\cup B:A\in \mathcal{A} \,and\,B\in \mathcal{B}\} $ is $\sigma$-algebra on $Z$.

Showing that $\emptyset \in \mathcal{C}$ and that $\mathcal{C}$ is closed under countable unions was easy, but it's the

If $C\in \mathcal{C}$, then $C^c \in \mathcal{C}$

which I've gotten wrong.

My thinking is like this: We have $\mathcal{A} \subset X$ and $\mathcal{B} \subset Y$, where $A^c \in \mathcal{A}$ and $B^c \in \mathcal{B}$. Since $X$ and $Y$ are disjoint, the intersection between two elements that are respectively in $X$ and $Y$ is just the empty set.

$C=(A\cup B)$, so then $C^c=(A \cup B)^c=A^c \cap B^c=\emptyset$

I guess I'm making a very basic error because the feedback I got was just "Wrong!", so I'd really appreciate if someone pointed it out for me.

答:
For $C \subseteq Z$ you have $$C^c = Z \backslash C$$ i.e. the complement of the set is taken with respect to the set $Z$. Hence $$A^c = Z \backslash A$$ since you consider $A$ as a subset of $Z$ (and not $X$!). Therefore in general $A^c \cap B^c \not= \emptyset$.

Example: $X=[0,\frac{1}{2})$, $Y= \left[\frac{1}{2},1 \right]$, $A=\left(\frac{1}{4},\frac{1}{2}\right)$, $B=\left[\frac{1}{2},\frac{3}{4}\right]$, then $$A^c \cap B^c = \left[0,\frac{1}{4}\right] \cup \bigg(\frac{3}{4},1 \bigg]$$ where the complement is taken with respect to to $Z=[0,1]$.


问:
Integration of $\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz = - \frac{\pi^4}{90}$
Using wolfram alpha,

$$-\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz = \frac{\pi^4}{90}$$

How do you solve this integral, and why is it equal to $\zeta(4)$? I can see that

$$-\frac{\ln(1-z)}{z} = 1 + \frac{z}{2} + \frac{z^2}{3} + \ldots$$

Body:
$$I=\int_0^1 \int_0^1 \int_0^1 \frac{\ln(1-xyz)}{xyz} dx dy dz $$
$$I=-\int_{0}^{1} \int_0^1 \int_0^1 \int_0^1 \frac{1}{1-xyzt} dx dy dz dt$$
$$I=-\int_{0}^{1} \int_0^1 \int_0^1 \int_0^1 \sum_{n=0}^{\infty} x^ny^nz^nt^n dx dy dz dt$$
$$\implies I=-\sum_{n=0}^{\infty}\left(\int_{0}^{1} u^n du \right)^4=-\sum_{n=0}^{\infty}\frac{1}{(n+1)^4}=-\zeta(4)=-\frac{\pi^4}{90}.$$


答:
If we view the elements $1,2,3,4$ as vertices of a square, and consider the dihedral group of that square, this gives us one copy of $D_8$. We can obtain other copies by reordering.

There are $6$ ways to order $4$ vertices (assuming we view cyclic shifts such as $1,2,3,4$ and $2,3,4,1$ as the same ordering), resulting in $3$ distinct copies of $D_8$ (because "opposite" pairs of orderings such as $1,2,3,4$ and $1,4,3,2$ result in the same group; just flip the square upside-down to go from one ordering to the other).

These three distinct copies of $D_8$ are conjugate to each other, e.g. because they are Sylow $2$-subgroups (they have the correct order), and all Sylow subgroups of a given order are conjugate. In particular they are not normal.

Title: Is $D_8$ a normal subgroup of $S_4$?

Body:
I am trying to show that $D_8$ is not a normal subgroup of $S_4$, I have shown that $D_8$ is  Subgroup of $S_4$.Then,  let $r=\langle(1234)\rangle$ and $s =\langle(14)(23)\rangle$.Then,  we see that $D_8$ is a subset of $S_4$ and since it a group so $D_8$ is a subgroup of $S_4$.

Proving that $D_8$ is not a normal subgroup. In $D_8$ there are two elements of order $4$.So,  we take $\sigma (1234) \sigma^{-1}$ which should be in $H$.Now conjugacy preserves the cycle structure so we choose $\sigma$ in such a way so that conjugate element does not belong to $H$.Choosing such a cycle is possible as there are $6$ 4 cycles in $S_4$ and there are only two of them in $D_8$.Since I have shown it for one element, that the conjugate is not in $D_8$.Do I need to show it for the other elements?


问:
Why is $e$ irrational?
I'm not looking for a proof, but rather an explanation, because I know there's something wrong with my thinking.

So, I know that
$$ e = \lim_{n\to\infty} (1+\dfrac{1}{n})^n $$
And also
$$ e = \sum_{k=0}^{\infty} \dfrac{1}{k!} $$

And I'm confused as to why $e$ can be irrational, since both of those definitions are a rational number(?).

I know that addition under rationals is closed, so I'm confused as both of these can be rearranged to something with rationals (e.g. sum of rationals, or $(\dfrac{n+1}{n})^n$), so I guess my question is, is why these rationals converge to an irrational?

Thanks.

答: That the rationals are closed under addition means that the sum of any two rationals is a rational again. By induction you can prove that this works for any finite number of rationals. But the rationals are not closed under infinite sums.

Title: Why is $e$ irrational?


答: The error lies in the assertion that “both of those definitions are a rational number”. Not true. Both of them express $e$ as a limit of a sequence of rational numbers. Your error lies in the hidden (and false) assumption that the limit of a convergent sequence of rational numbers must be rational too.

Title: Why is $e$ irrational?

Body:
I'm not looking for a proof, but rather an explanation, because I know there's something wrong with my thinking.

So, I know that
$$ e = \lim_{n\to\infty} (1+\dfrac{1}{n})^n $$
And also
$$ e = \sum_{k=0}^{\infty} \dfrac{1}{k!} $$

And I'm confused as to why $e$ can be irrational, since both of those definitions are a rational number(?).

I know that addition under rationals is closed, so I'm confused as both of these can be rearranged to something with rationals (e.g. sum of rationals, or $(\dfrac{n+1}{n})^n$), so I guess my question is, is why these rationals converge to an irrational?

Thanks.


答:
Just because every partial (finite) sum is a rational number, that does not mean the limit is.

Take the well known constant as an example: $\pi = 3+0.1+0.04+...$

Every partial sum is rational, but the limit is a well known irrational number.

Similarly, for every finite natural number $n$, $1/n > 0$ but the limit $\lim_{n\to\infty} 1/n \not > 0$.

Title: Why is $e$ irrational?


问:
Show that $\Gamma(\Omega)\leq \Gamma\Big(\operatorname{W}\Big(x^{x}\Big)\Big)<2$ on $(0,1]$
Working with Geogebra I have found :

Let $0<x\leq 1$ :
$$\Gamma(\Omega)\leq f(x)=\Gamma\Big(\operatorname{W}\Big(x^{x}\Big)\Big)<2$$
With the Gamma function and the Lambert's function .

For the RHS the minimum of the difference is  $2-\Gamma\Big(\operatorname{W}\Big(e^{-e^{-1}}\Big)\Big)\simeq0.005167\cdots$

Using the derivative of $f(x)$ it's not hard to show that $e^{-1}$ is a maximum on $(0,1]$ .

The problem is :

$$f(e^{-1})<2\quad (1)$$

My question :

How to show $(1)$ ?

Thanks in advance !

答:
For the time being, consider the function
$$f(y)=y\,\Gamma(y)\qquad \text{for} \qquad 0 \leq y \leq 1$$ and, as I did in this question of mine, let me approximate $f(y)$ by
$$g(y)=1+y(1-y) \sum_{k=0}^p d_k\, y^k$$

Using $\color{red}{p=3}$, matching the function and first derivative values at $x=0$, $x=\frac 12$ and $x=1$ , we have
$$d_0= -\gamma$$ $$ d_1=-17+6 \gamma +4 \sqrt{\pi } (\gamma +2\log (2))$$
$$d_2=4 \left(5-3 \gamma +4 \sqrt{\pi }-3 \sqrt{\pi } (\gamma +2\log (2))\right)$$
$$d_3=-4+8 \gamma -16 \sqrt{\pi }+8 \sqrt{\pi } (\gamma +2\log (2))$$

This gives
$$\int \big[f(y)-g(y)\big]^2\,dy =1.28\times 10^{-8}$$
$$\int \Big[\frac{f(y)-g(y)}y\Big]^2\,dy =3.24\times 10^{-7}$$

Now, assuming that we know the exact value of $W\left(e^{-1/e}\right)$, we have
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq1.994799$$ which seems to be quite good for a first approximation when compared to the exact $1.994832$.

If we do not know  the exact value of $W\left(e^{-1/e}\right)$,  we can have a reasonable approximation of it using the $[5,4]$ Padé approximant of Lambert function $W(x)$ built around $x=0$. This is
$$W(x) \sim x \frac {1+\frac{7430297 }{1597966}x+\frac{1018440443 }{156600668}x^2+\frac{1260595681
   }{469802004}x^3+\frac{974868241 }{9396040080}x^4 } {1+\frac{9028263 }{1597966}x+\frac{1668309215 }{156600668}x^2+\frac{3536864687
   }{469802004}x^3+\frac{14189787721 }{9396040080}x^4 }$$ which gives

$$W\left(e^{-1/e}\right)\sim 0.444019 \qquad \text{while} \qquad \text{exact}=0.444016$$

Using this approximation leads to
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq1.994784$$

Using $\color{red}{p=6}$, matching the function,  first and second derivative values at $x=0$, $x=\frac 12$ and $x=1$, the above norms become $2.72\times 10^{-12}$ and $5.88\times 10^{-11}$ which is a very significant improvement.

Using  the exact value of $W\left(e^{-1/e}\right)$, this gives
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq 1.99483224 \qquad \text{while} \qquad \text{exact}=1.99483209$$

Using $\color{red}{p=9}$, matching the function, first, second and third derivative values at $x=0$, $x=\frac 12$ and $x=1$, the above norms become $6.19\times 10^{-16}$ and $1.23\times 10^{-14}$ which is another significant improvement.

Using  the exact value of $W\left(e^{-1/e}\right)$, this gives
$$\Gamma\Big(W\left(e^{-1/e}\right)\Big)\simeq 1.99483209100 \qquad \text{while} \qquad \text{exact}=1.99483209170$$

Remark

As in the linked question, using $p=9$ does not correspond to any overfit of function $f(y)$. A simple linear regression gives

$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error}  \\
 d_0 & -0.577215 & 1.1 \times 10^{-8}  \\
 d_1 & +0.411823 & 3.7 \times 10^{-7}\\
 d_2 & -0.495316 & 5.0 \times 10^{-6} \\
 d_3 & +0.482941 & 3.5 \times 10^{-5}  \\
 d_4 & -0.477680 & 1.4 \times 10^{-4}  \\
 d_5 & +0.429199 & 3.6 \times 10^{-4}\\
 d_6 & -0.325588 & 5.6 \times 10^{-4}\\
 d_7 & +0.184385 & 5.3 \times 10^{-4}  \\
 d_8 & -0.066496 & 2.8 \times 10^{-4} \\
 d_9 & +0.011163 & 6.1 \times 10^{-5} \\
\end{array}$$ which show the high statistical significance of the parameters.


问:
Embedding of Klein bottle in $\mathbb{R}^4$ and Klein bottle as quotient
Disclaimer: This question is somewhat of a duplicate. However, I do not feel it was answered adequately in previous questions.

We wish to show that the mapping $$ H(x,y) = ((r \cos{y} + a) \cos{x}, (r \cos{y} + a)\sin{x}, r \sin{y} \cos{\frac{x}{2}},r \sin{y} \sin{\frac{x}{2}}) $$ induces an embedding of the Klein bottle into $\mathbb{R}^4$. Here the Klein bottle is defined as the quotient space $T^2 /G$ where $G$ is the group of diffeomorphisms on $T^2$ (torus of revolution) generated by the identity map and the antipodal map $A(p) = -p$.

Is this a valid definition of the Klein bottle? I've seen someone say this is not an accurate description.

Is it enough to 'restrict' $G$ to the equivalence classes $$ [(x,y)]:= \{ \pm (x+2\pi z_1, y+2 \pi z_2) \} $$ and show that this mapping is continuous, an immersion, and injective, if we assume compactness of the Klein bottle?

答:
We have known that the $2$-torus $T^2\simeq S^1\times S^1\simeq\mathbb{R}^2/\mathbb{Z}^2$. To see that $T^2$ is diffeomorphic to the torus of revolution in $\mathbb{R}^3$ obtained as the inverse image of zero of the function $f:\mathbb{R}^3\to\mathbb{R}$
$$
 f(x,y,z)=z^2+(\sqrt{x^2+y^2}-a)^2-r^2,
 $$
we construct parametrization of $S:=f^{-1}(0)$ as follows. Define $\Psi:[0,1]\times[0,1]\to\mathbb{R}^3$ by:
$$
 (x,y)\mapsto\left(
 (a+r\cos 2\pi y)\cos2\pi x,(a+r\cos 2\pi y)\sin2\pi x, r\sin 2\pi y
 \right).
 $$
Then the antipodal mapping $A:S\to S$ can be expressed as $(x, y)\mapsto(x+1/2,-y)$. We slightly twist $G:\mathbb{R}^2\to\mathbb{R}^4$ by multiplying $4\pi$ for $x, y$ such that it can be well-defined on $T^2$ and we still denote it by $G$:
$$
 \begin{aligned}
  G(x,y)=(&(r\cos 4\pi y+a)\cos4\pi  x,(r\cos4\pi  y+a)\sin4\pi  x,\\
 &r\sin4\pi  y\cos(2\pi x),r\sin4\pi  y\sin(2\pi x)).
 \end{aligned}
    $$
Then for any $p\in S$, we have
$$
 G(p)=G(\Psi(x,y))=G(\Psi(x+1/2,-y))=G(-p).
 $$
Thus $G$ induces an mapping $\tilde{G}:K\to\mathbb{R}^4$, where $K=S/\{A,\operatorname{Id}_{\mathbb{R}^3}\}$ is the Klein bottle, by defining $\tilde{G}([p])=G(p)$ for each $p\in S$. It is easily testified that $\tilde{G}$ is an injective immersion. By compactness of $K$, we conclude that $\tilde{G}$ is an embedding of $K$ in $\mathbb{R}^4$.


问:
Is $\mathrm{d} \pi (x,y)$ or $\pi(x,y) \mathrm{d}x \mathrm{d}y$ correct in Optimal Transport formulation?
The Kantorovich formulation of optimal transport sometimes looks like this

$$\mathrm{inf} \int_{E \times E} C(x, y) \ d \pi(x, y)$$

and sometimes looks like this

$$\mathrm{inf} \int_{E \times E} C(x, y) \  \pi(x, y) \hspace{1mm} dx \hspace{1mm} dy $$

both being subject to the usual constraints
$$\int_E \pi(x, y) dy = \mu(x) \qquad \int_E \pi(x, y) dx = \nu(y)$$

What is the difference, and which is correct?

答: You get from the first formula to the second if you assume that the measure $\pi$ has a Lebesgue density on $E\times E$, called by abuse of notation again $\pi$. Then $d\pi(x,y)=\pi(x,y)dxdy$. It is however, not a good idea to impose that π should have a Lebesgue density, because in the most important examples, related to Monge's problem the optimal $\pi$ will actually be singular. In short, the first formula is the better.