问:
Strong characterization of $\mathbb C$ with respect to $\mathbb R$

$\mathbb R^2$ is not a field, but $2$-tuple arithmetic rules like $(a,b)(c,d)=(ac-bd,ad+bc)$ coupled with $\mathbb R^2$ make it a field, but are there other rules than $(a,b)(c,d)=(ac-db,ad+bc)$ combined with $\mathbb R^2$ that make it a field?

Is $\mathbb C$ the only field that contains $\mathbb R$ as a subfield or are there other fields that contain $\mathbb R$ as subfields?

Is $\mathbb C$ the only that can be made by coupling $\mathbb R^2$ and rules like $(a,b)(c,d)$ $=$ $(ac-db,ad+bc)$? Are there other rules that can be coupled with $\mathbb R^2$ to make it a field?

PS: Just trying to pin down some properties that make $\mathbb C$ special/unique from other fields that can/may be constructed using some specific algebraic structure of 2 tuples in $\mathbb R$ e.g. $(\sqrt 2,\pi)$. At this point this is just a wide net as I lack the terminology to make this question specific. But any hints on how to make this question specific to listing properties that make $\mathbb C$ unique or gives strong characterisation of relationship between  $\mathbb C$ and $\mathbb R$ are welcome.

答:
This is not really an answer to your question, but you seem to be interested in the extent to which the complex structure on $\mathbb{C}$ is 'special' or 'unique'. I think the Gelfand-Mazur theorem might be of interest to you. It comes up in the study of spectral theory of operators on Banach spaces:

If $A$ is a commutative complex Banach algebra in which every nonzero element is invertible, then $A$ is isometric and isomorphic to $\mathbb{C}$.

In other words, if I give you some Banach algebra $A$ over $\mathbb{C}$ which satisfies the multiplicative and additive inequalities $\|xy\| \leq \|x\|\|y\|$ and $\|x+y\| \leq \|x\|+\|y\|$, so long as the space is i) complete and ii) a ''field'', that space must be the complex numbers.

The proof is also very simple if you assume the (non-trivial) result that the spectrum of any element $x \in A$ non-empty. To say that $\lambda \in \mathrm{sp}(x)$ is to say that $x-\lambda e$ is not invertible. The only non-invertible element in $A$ is 0, so $x-\lambda e = 0$. The isomorphism $\phi(x) = \lambda$ is also an isometry.

This theorem becomes useful in defining the Gelfand transform (see Rudin's functional analysis, chapter 11).


问:
how many ways can we arrange As, Bs, and Cs
In how many ways can we arrange

$3$ As, $2$ Bs, and $5$ Cs with no restriction?

$$P(3,2,5) = \frac{10!}{3!2!5!}$$

$2$ As, $4$ Bs, and $3$ Cs in such a way that the $2$ As are one next to the other?

I am considering the AA as one symbol:

$$P(1,4,3) = \frac{8!}{1!4!3!}$$

$5$ As, $2$ Bs, and $6$ Cs in a way that between the $2$ Bs there are exactly $3$ letters.

The position of the first B determines the position of the second one

$$P(5,1,6) = \frac{12!}{5!1!6!}$$

$3$ As, $4$ Bs, and $5$ Cs in a way that not all $5$ Cs are one next to the other.

All the arrangements:  $$P(3,4,5) = \frac{12!}{3!4!5!} \tag{1}$$

5C's together:  $$P(3,4,1) = \frac{8!}{3!4!1!} \tag{2}$$

solution $= (1)-(2)$

$10$ As, $6$ Bs, and $5$ Cs in such a way that there are no Bs one next to the other.

Total          $$P(10,6,5)= \frac{21!}{10!6!5!}  \tag{1}$$

$2$ Bs together  $$P(10,4,1,5) = \frac{20!}{10!4!1!5!}  \tag{2}$$
$4!$ corresponds to the remaining Bs and $1!$ to the group BB.

$3$ Bs together $$P(10,3,1,5)= \frac{19!}{10!3!1!5!}  \tag{3}$$
$3!$ corresponds to the remaining Bs and $1!$ to the group BBB.

$4$ Bs together $$P(10,2,1,5)= \frac{18!}{10!2!1!5!}  \tag{4}$$
$2!$ corresponds to the remaining Bs and $1!$ to the group BBBB.

$5$ Bs together $$P(10,1,1,5)= \frac{17!}{10!1!1!5!} \tag{5}$$
$1!$ corresponds to the remaining Bs and the other $1!$ to the group BBBBB.

$6$ Bs together $$P(10,1,5)= \frac{16!}{10!1!5!} \tag{6}$$
$1!$ corresponds to the group BBBBBBB.

Solution $$(1) - [(6)+[(5)-(6)]+[(4)-(5)-(6)]+[(3)-(4)-(5)-(6)]+[(2)-(3)-(4)-(5)-(6)]$$
$$ = (1) + (3) \times (6)+ (2) \times (5) + (4) - (2)$$

Can somebody check, especially the last question?

答:
Your solutions to the first, second, and fourth questions are correct.

In how many ways can be arranged $5$ As, $2$ Bs, and $6$ Cs in such a way that between the two Bs there are exactly three letters?

There are a total of $5 + 2 + 6 = 13$ positions.  Since there must be exactly three letters between the two Bs, the first B must be in one of the first $10$ positions.  Once the position of the first B is selected, the position of the second B is also determined.  That leaves $11$ spaces in which to place the $5$ As and $6$ Cs.  Choosing which $5$ of these $11$ spaces will be filled with As completely determines the arrangement since the Cs must fill the remaining six positions.  Hence, there are
$$\binom{10}{1}\binom{11}{5}$$
such arrangements.

Your count is too large since you have not restricted the position of the first B to the first ten positions.

In how many ways can we arrange $10$ As, $6$ Bs, and $5$ Cs so that no two of the Bs are next to each other?

We could use the Inclusion-Exclusion Principle to solve the last problem, but we would have to account for all the ways we could have one or more pairs of consecutive Bs.  By only considering how many Bs were consecutive, you overlooked cases in which there are consecutive Bs in separate blocks such as AABCABBACCABBBAAACAAC.  Also, when you apply the Inclusion-Exclusion Principle, you first subtract cases with a pair of consecutive Bs, then add cases with two pairs of consecutive Bs, then subtract cases with three pairs of consecutive Bs, and so forth until you exhaust the restrictions.

An easier method is to first arrange the $10$ As and $5$ Cs in a row, then insert the Bs in the $16$ spaces this creates, $14$ between successive letters and $2$ at the ends of the row so that no two of the Bs are consecutive.

There are
$$\binom{15}{10}$$
distinguishable ways to arrange $10$ As and $5$ Cs in a row since selecting which $10$ positions will be filled with the As also determines which five spaces will be filled with the Cs.  In each such arrangement, there are $16$ spaces in which to place the Bs, where $L$ represents a letter and $\square$ represents a space in which a B could be inserted.

$$\square L \square L \square L \square L \square L \square L \square L \square L \square L \square L \square L \square L \square L \square L \square L \square$$

To ensure that no two of the six Bs are consecutive, we select six of these sixteen spaces in which to insert a $B$, which can be done in
$$\binom{16}{6}$$
ways.

Hence, there are
$$\binom{15}{5}\binom{16}{6}$$
arrangements of $10$ As, $6$ Bs, and $5$ Cs in which no two of the Bs are consecutive.

ContentLicense: CC BY-SA 4.0


问:
Consistency of non-trivial elementary embedding $j: Ord \to Ord$
What is the consistency strength of the existence of an elementary embedding $j: Ord\to Ord$ from the class of ordinal to itself? For example, if there is an elementary embedding $j: V\to M$, then $j\upharpoonright Ord$ is such an embedding. So the consistency strength of such an elementary embedding is below a measurable. Similarly, if $j: L\to L$ is an elementary embedding, then $j\upharpoonright Ord$ again satisfies the requirement.

答:
$Ord=(Ord;\in)$, as a (class-sized) substructure of $V$, is actually surprisingly simple: while arbitrary transitive sets, and in particular levels of the $V$- or $L$- (or related) hierarchies, can have quite intricate structure, there's very little expressive power in $Ord$ or its initial segments. For an initial taste of this, one can show (Theorem $6.21$ in Rosenstein's lovely book) that for every ordinal $\alpha$ there is some $\beta<\omega^\omega\cdot 2$ such that $(\alpha;\in)\equiv(\beta;\in)$, and that $Ord$ itself is elementarily equivalent to $(\omega^\omega\cdot 2;\in)$. Meanwhile, quick back-of-the-napkin arguments establish that the least ordinal $\alpha$ such that $L_\alpha\equiv L_\beta$ for some $\beta>\alpha$ is truly gigantic (while still being countable of course).

As it turns out, $Ord$ has so little structure that there are definitions of operations on ordinals which $\mathsf{ZFC}$, or indeed much less, outright prove yield nontrivial elementary embeddings $Ord\rightarrow Ord$. This uses the same techniques as in the proof of the theorem mentioned above: via EF-games, we show that as long as $\alpha$ is "sufficiently closed" then the map $$\beta\mapsto \begin{cases}
\beta & \mbox{ if $\beta<\alpha$,}\\
\alpha+\beta & \mbox{ otherwise.}
\end{cases}$$ is in fact elementary. I don't recall at the moment what the least such $\alpha$ is, though (a natural guess is $\epsilon_0$ but I think it's actually much smaller than that - $\omega^{\omega^\omega}$ seems more plausible to me). And variations on the same idea can build elementary embeddings with richer behavior (e.g. the one above is eventually the identity, but we can fix that if we want to).

Of course $Ord$ is often more than just a linear order. What if we consider $Ord$ with the ordinal arithmetic operations of addition, multiplication, and exponentiation? What about additionally throwing on higher Veblen functions, or the map $\omega\mapsto\omega_\alpha$, or etc.?

Now things get progressively more complicated of course (already once we throw on addition and multiplication, the theory we get is no longer decidable), but I believe for any "reasonable" additional structure on $Ord$ we still get nontrivial elementary embeddings provably in $\mathsf{ZFC}$.

At this point a general theorem to that effect would be great. I vaguely recall that one exists, either by Mostowski or Harvey Friedman, but I can't remember the details at the moment. Of course we have to use a notion of "reasonable" which is sufficiently strict to outlaw operations/relations which let us code $L$ into the ordinals, so there's some finickiness here. Perhaps for any specific set of $\mathcal{L}_{\infty,\infty}$-formulas $(\varphi_i)_{i\in I}$ each defining an operation on ordinals, there is a nontrivial elementary embedding of $(Ord; (\varphi_i)_{i\in I})$ into itself? The point is that with only set-many formulas, we're really living in some fixed $\mathcal{L}_{\kappa,\kappa}$, and at ordinals "vastly bigger than" $\kappa$ things should start to blur together.

EDIT: I still don't know what theorem I was thinking of in the previous paragraph, but in answer to a question of mine Gabe Goldberg observed that there's a very strong embedding principle which appears consistent with $\mathsf{ZFC}$, although at the extreme edge of plausibility: specifically, that there is a nontrivial second-order elementary embedding of $Ord$ into itself, that is, a nontrivial map which preserves the truth of second-order formulas. This is absolutely not what I was thinking of originally, but it's cool.


问:
How to solve this fredholm equation $u(x)=x+\int_0^2e^{x+t}u(t)dt$?
$u(x)=x+\int_0^2e^{x+t}u(t)dt$

I tried adm and couldnt find a geometric series. ($e^x(1+e^2),e^{2x}(1+e^2)(e^2-1),e^{3x}(1+e^2)(e^2-1)(e^2-1)....etc $
and also tried noise term phenomenon and direct computation method which gave me $u(x)=x+e^x\frac{(e^2+1)}{3/2+e^4/2)}$ and it didnt work.

答: The equation shows $u(x)=x+ce^{x}$ for some constant $c$. (Pull out $e^{x}$ from the integral to see why this is true).  Put this back into the equation to get the value of $c$.  You get $c=\frac {2(e^{2}+1)} {3-e^{4}}$ so the solution is $u(x)=x+\frac {2(e^{2}+1)} {3-e^{4}} e^{x}$.


问:
The boundary of a closed Mobius band
The Mobius band is defined by the quotient: $r:I\times I\rightarrow M$, with the equivalence relation $(0,x)\sim (1,1-x)$ for all $x\in I$.

The boundary of Mobius band $M$ is defined as the set of points that have an open neighbourhood which is homeomorphic to the closed half space.

I know its boundary is $r(B)$ where $B=\{(x,y)\in I\times I: y\in\{0,1\}, x\in I\}$. We need to find an open neighbourhood for each point in $r(B)$ s.t. homeomorphic to the closed half space $H=\{(x,y)\in\mathbb{R}^2:y\geq 0\}$.

It is easy to prove $r((0,1)\times (0,1))$ is not in the boundary $\partial M$, and $r((0,1)\times\{0,1\})\subset\partial M$.

But how to prove $r(\{0,1\}\times (0,1))$ is not in the boundary, and $r(0,0)$ $r(1,0)$ are in the boundary?

答:
You gave a description of the Möbius band as
$$
M = (I\times I)/{\sim}_1 \quad\text{where}\quad (0,y)\sim_1(1,1-y) \text{ for $t\in I$.}
$$
To allow for less messy operations on points of $M$, you can also describe this as
$$
M' = (\mathbb R\times I)/{\sim}_2 \quad\text{where}\quad (x,y)\sim_2(x+1,1-y) \text{ for $x\in\mathbb R$ and $y\in I$.}
$$
Check that the inclusion $I\times I\hookrightarrow \mathbb R\times I$ induces a homeomorphism $M\to M'$.

The benefit of this description is that it doesn't treat the points $(x,y)$ with $x\in\{0,1\}$ any different from those with $x\in (0,1)$. In particular, we get an automorphism $M'\to M'$ given by $[(x,y)]_2\mapsto[(x+\tfrac 1 2, y)]_2$.

Since you have shown that $r_1((0,1)\times (0,1))$ does not contain boundary points of $M$, the above automorphism yields that $r_2(\mathbb R\times (0,1))$ does not contain any boundary points of $M'$ and hence $r_1(\{0,1\}\times(0,1))$ doesn't either.

You can think of the automorphism $M'\to M'$ in terms of $M$ as gluing $I\times I$ along $\{0,1\}\times I$ as you did, then cutting it open along $\{\tfrac 1 2\}\times I$ to obtain a new description as a quotient of $I\times I$, where points with $x\in\{0,1\}$ in the old quotient are now points with $x=\frac 1 2$ in the new one.

Title: The boundary of a closed Mobius band


问:
A tricky probability question.
I have been asked the following question, and unfortunately I have no idea how to proceed. Here is the question:

Suppose we have 99 empty papers and we wrote numbers from 1 to 99(using each number) on one side of the papers randomly. Then we mixed all the papers randomly and started to write numbers from 1 to 99 on the other sides (empty sides) of each paper randomly. What is the probability of having the same number on both sides of at least one paper?

Thanks!

答:
Building off what's been said in the comments and answers, here's a solution using inclusion/exclusion. Hope this helps.

Let's let $X = \text{# of papers with matching numbers}$. Then $P\left(X\geq 1\right)$ is the probability that at least 1 paper has matching numbers.

Now, let $E_i = \left\{\text{$i$th paper has matching numbers}\right\}$. Then we have:

\begin{align*}
 P\left(X \geq 1\right) &= P\left(\bigcup_{i=1}^{99} E_i \right) \\[5pt]
 &= P\left(E_1 \cup E_2 \cup E_3 \cup \dots \cup E_{99}\right) \\[5pt]
 &= \sum_{i} P\left(E_i\right) - \sum_{i<j}P\left(E_i \cap E_j\right) + \dots + (-1)^{99+1} P\left(E_1 \cap E_2 \cap \dots \cap E_{99}\right) \\[5pt]
% &= N\left(\frac{1}{N}\right) - \binom{N}{2}\left(\frac{(N-2)!}{N!}\right) + \binom{N}{3}\left(\frac{(N-3)!}{N!}\right) - \dots +(-1)^{N+1}\left(\frac{1}{N!}\right)
\end{align*}

which comes from inclusion/exclusion.

Now, by symmetry we have that $P\left(E_1\right) = P\left(E_2\right) = \dots = P\left(E_{99}\right)$, and the same is true for the intersections in the inclusion/exclusion summations.

So the various probabilities we need in inclusion/exclusion are:

\begin{align*}
 P\left(E_i\right) &= \frac{1}{99} \\[5pt]
 P\left(E_i \cap E_j \right) &= \frac{1}{99}\times \frac{1}{98} \\[5pt]
 P\left(E_i \cap E_j \cap E_k \right) &= \frac{1}{99}\times \frac{1}{98} \times \frac{1}{97}  \\
 \vdots & \\
 P\left(E_1 \cap E_2 \cap \dots \cap E_{99}\right) &= \frac{1}{99!}
\end{align*}

And there are $99$ different $E_i$'s, and $\binom{99}{k}$ intersections of size $k$. So, plugging into the inclusion/exclusion equation above we get:

\begin{align*}
 P\left(X \geq 1\right) &= P\left(\bigcup_{i=1}^{99} E_i \right) \\[5pt]
 &= 99\left(\frac{1}{99}\right) - \binom{99}{2}\left(\frac{1}{99\cdot 98}\right) + \binom{99}{3}\left(\frac{1}{99\cdot 98 \cdot 97}\right) - \dots +(-1)^{99+1}\left(\frac{1}{99!}\right) \\[5pt]
 &= 1 - \frac{1}{2!} + \frac{1}{3!} - \frac{1}{4!} + \dots + \frac{1}{99!} \\[5pt]
 &\approx 1 - e^{-1}
\end{align*}

which comes from the Taylor series for $e^{x}$.

Note that the Taylor series is an infinite series and we are cutting it off at $\frac{1}{99!}$. However, since we have an alternating series we can also say that our error is less than the magnitude of the next term, i.e. error less than $\frac{1}{100!}$, so it's quite a good approximation.

ContentLicense: CC BY-SA 3.0

Title: A tricky probability question.


问:
Argument for Grandi's Series equaling zero? Or flawed logic?
$1-1+1-1+1-1+.....=S$

$1-S=S=1/2$    (this is the reasoning I've learned before)

but if I try continuing this reasoning, I get

$$1-1+S=S$$
$$1-1+1-S=S=1/2$$
$$1-1+1-1+S=S$$
$$1-1+1-1+1-S=S=1/2$$
$$1-1+1-1+1-1+S=S$$
$$1-1+1-1+1-1+1-S=S=1/2$$
$$\cdots$$
$$S-S=S$$
$$S+S=S$$

Therefore, $S$ must equal zero? But that seems contradictory to what I've been taught? Is my logic flawed? Am I missing something here?

答: Grandi's series is, of course, divergent, and what you have done only confirms it. Different ways of manipulating the series yield different results; the usually given sum $\frac12$ is the result of a certain summation method needed for some physical interpretation to make sense.


Title: How to solve this fredholm equation $u(x)=x+\int_0^2e^{x+t}u(t)dt$?

Body:
$u(x)=x+\int_0^2e^{x+t}u(t)dt$

I tried adm and couldnt find a geometric series. ($e^x(1+e^2),e^{2x}(1+e^2)(e^2-1),e^{3x}(1+e^2)(e^2-1)(e^2-1)....etc $
and also tried noise term phenomenon and direct computation method which gave me $u(x)=x+e^x\frac{(e^2+1)}{3/2+e^4/2)}$ and it didnt work.

Body:
$$u(x)=x+e^x\int_{0}^{2} e^t u(t) dt~~~(1)$$
Let $\int_{0}^{2} e^t u(t) dt=A$, then
$$u(x)=x+Ae^x.$$
For consistency, $$\int_{0}^{2} e^t (t+Ae^t) dt =A$$
$$\int_{0}^{2}te^t dt+A\int_{0}^{2} e^{2t} dt=A$$
$$[te^t-e^t+\frac{A}{2} e^{2t}]_{0}^{2}=A \implies A=\frac{2(1+e^2)}{3-e^4}.$$
So the solution of (1) is
$$u(x)=x+\frac{2(1+e^2)}{3-e^4}e^x$$


问:
Stronger version of Taylor's Theorem
Show that Taylor's Theorem may be strengthened as follows:

Let $f$ be a continuous real-valued function on the closed interval in R of extremities $a$ and $b$. That is, (n + 1) times differentiable on the open interval with these same extremities and suppose that $\lim_{x \to a} f'(x), \lim_{x \to a} f''(x), \dots, \lim_{x \to a} f^{(n)}(x)$ exist and that $f',f'',\dots,f^{(n)}$ are bounded.

Then,
$$f(b) = f(a) + (\lim_{x \to a} f'(x)) \frac{(b - a)}{1!} + \dots | (\lim_{x \to a} f^{(n)}(x)) \frac{(b - a)^n}{n!} + f^{(n + 1)}(c) \frac{(b - a)^{n + 1}}{(n + 1)!}$$
for some c between a and b.

I'm not too familiar with Taylor's Theorem in an Analysis sense. We briefly talked about it in class, but we sort of moved on quickly. This was said to be a "cool problem" to complete, so I would like to see it. Can anyone help me with this one? Thank you!

答:
Theorem: Let $a,b\in\mathbb{R}$ with $a<b$. Let $n\geq 0$ be an integer. Let $f:[a,b]\rightarrow\mathbb{R}$ be a function
that satisfies the following conditions:

(1) $f$ is continuous on $[a,b]$,

(2) $f^{(k)}$ exists on $(a,b)$ for $k=1,2,\ldots,n+1$.

(3) $\lim_{x\rightarrow a+}f^{(k)}(x)$ exists for $k=1,2,\ldots,n$.

(4) $\lim_{x\rightarrow b-}f^{(k)}(x)(b-x)^{k}=0$ for $k=1,2,\ldots,n.$

(Remark: Note that condition (4) is implied by the condition that
$f^{(k)}$ is bounded on $(a,b)$. However, the condition that $f^{(k)}$
is bounded on $(a,b)$ is too strong and we can replace it by a much
weaker condition.)

Denote $f^{(k)}(a+)=\lim_{x\rightarrow a+}f^{(k)}(x)$. Then there
exists $\xi\in(a,b)$ such that
$$
f(b)=f(a)+\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}+f^{(n+1)}(\xi)\cdot\frac{(b-a)^{n+1}}{(n+1)!}.
$$

Proof: Define $\theta:[a,b]\rightarrow\mathbb{R}$ by
$$
\theta(t)=\begin{cases}
f(b)-f(t)-\sum_{k=1}^{n}f^{(k)}(t)\cdot\frac{(b-t)^{k}}{k!}, & \mbox{if }t\in(a,b)\\
f(b)-f(a)-\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}, & \mbox{if }t=a\\
0, & \mbox{if }t=b
\end{cases}.
$$
Clearly $\theta$ is continuous on $(a,b)$. By condition (3), $\theta(a)=\lim_{t\rightarrow a+}\theta(t).$
By condition (4), $\theta(b)=\lim_{t\rightarrow b-}\theta(t)$. In
short, $\theta$ is continuous on $[a,b]$. It is also clear that
$\theta$ is differentiable on $(a,b)$. By direct calculation (observe
the telescope cancellation), we have
$$
\theta'(t)=-f^{(n+1)}(t)\cdot\frac{(b-t)^{n}}{n!}.
$$
Define $g:[a,b]\rightarrow\mathbb{R}$ by $g(t)=(b-t)^{n+1}$. Clearly
$g$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $g'(t)\neq0$
for all $t\in(a,b)$. By Cauchy's mean-value theorem, there exists
$\xi\in(a,b)$ such that
\begin{eqnarray*}
\frac{\theta(b)-\theta(a)}{g(b)-g(a)} & = & \frac{\theta'(\xi)}{g'(\xi)}\\
 & = & -f^{(n+1)}(\xi)\cdot\frac{(b-\xi)^{n}}{n!}\cdot\frac{1}{-(n+1)(b-\xi)^{n}}\\
 & = & f^{(n+1)}(\xi)\cdot\frac{1}{(n+1)!}.
\end{eqnarray*}
On the other hand,
\begin{eqnarray*}
\frac{\theta(b)-\theta(a)}{g(b)-g(a)} & = & -\frac{f(b)-f(a)-\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}}{-(b-a)^{n+1}}.
\end{eqnarray*}
Rearranging terms, we obtain,
$$
f(b)=f(a)+\sum_{k=1}^{n}f^{(k)}(a+)\cdot\frac{(b-a)^{k}}{k!}+f^{(n+1)}(\xi)\cdot\frac{(b-a)^{n+1}}{(n+1)!}.
$$

Title: Stronger version of Taylor's Theorem


问:
Rolling $20$ dice at least three six
If we roll $20$ dice, what is the probability of at least three six?

I solved by using $P(A)=1-P(A^C)$ property.
Considering the complement of problem, there is a $5/6$ probability of not rolling a six for any given die.
So the result is $1-{\left(\frac{5}{6}\right)}^{20}$ Am I thinking wrong? Any help will be appreciated.

答:
$$1-\binom{20}{0}\Big(\frac{1}{6}\Big)^0\Big(\frac{5}{6}\Big)^{20}-\binom{20}{1}\Big(\frac{1}{6}\Big)^1\Big(\frac{5}{6}\Big)^{19}-\binom{20}{2}\Big(\frac{1}{6}\Big)^2\Big(\frac{5}{6}\Big)^{18}$$

That is

$$1-\text{Probability of zero six's}-\text{Probability of 1 six}-\text{Probability of 2 six's}$$

Title: Rolling $20$ dice at least three six


问:
Exercise in Beauville, about the blowup of $P^2$ at seven points
I am trying to solve: (this is an exercise in Beauville's book, Complex Algebraic Surfaces, page 52.)

Consider $7$ points of $P^2$ in general position. Let $P_7$ denote the blowup of $P^2$ at these points and $H$ the strict transform of a cubic containing them. Show that $|H|$ defines a two-to-one map $P_7 \to P^2$ branched along a smooth quartic.

I showed that $|H|$ is base-point-free, so there is a well-defined map $P_7\to P^2$, but I can't show the rest. How can we show that this map is branched along a quartic?

答:
Expanding on my comment because I figured out what errors I was making before: let $L$ denote the pullback of the class of a general line in $P^2$ to $P_7$. The degree of a finite map over $\mathbb C$ is the cardinality of a general fiber. Since a general point is the intersection of two general lines, and H is by definition the class of the pullback of $\mathcal O_{P^2}(1)$, we can compute this as the intersection number
$$
H^2 = (3L - E_1 - \cdots -E_7)^2 = 9L^2 + E_1^2 + \cdots + E_1^2 = 9-1-\cdots -1=2,
$$

where $L.E_i$ and all cross terms $E_i.E_j$ vanish because $L$ is general, and because blowing up general points produces disjoint exceptional divisors.

As for the branching, this is essentially a topological question. The topological Euler characteristic $\chi$ of $P^2$ is $3$, and blowing up a point (since it replaces a point with a $2$-sphere) increases $\chi$ by $1$, so $\chi(P_7) = 10$. Now when we have a branched double cover $X\to Y$, we can think of $X$ as being constructed by cutting away the branch locus $C$ from $Y$ to get an affine variety $U$, taking two copies of $U$, and gluing back together along a single copy of the branch locus, so $\chi(Y) = 2\chi(U) + \chi(C)$. Filling in what we know we get
$$
10 = 2(3 - (2-2g(C))) + (2-2g(C)) = 2(1 + 2g(C)) + 2 - 2g = 4 + 2g(C)
$$

hence $g(C) = 3$. Now assuming that $C \subset P^2$ is smooth, it must be a quartic by the degree-genus formula.

To see that $C$ is smooth requires a local calculation that I will leave to you. You want to show that the surface defined by $w^2 = F(x,y,z)$ in $P^2 \times A^1$ is smooth if and only if the plane curve $F=0$ is smooth. Since we know $P_7$ is smooth, we conclude that $C$ is smooth.


问:
Factorization of $a^{n}-b^{n}$.
Let $X$ be a Banach space and let $A, B$ be two bounded linear operators on $X$ such that $AB \neq BA$. Can we write $A^{n}-B^{n}$ as follows and what is the expression of $K$:
$$A^{n}-B^{n}=C(A-B)+K(AB-BA).$$
Where $C=A^{n-1}+BA^{n-2}+\dots+B^{n-1}$ and $K \in \mathbb{R}\backslash\{0\}$.

答:
For $n=3$, $$A^3-B^3-(A^2+BA+B^2)(A-B)=A^2B+BAB-BA^2-B^2A$$

And you can't reasonably expect this to be a scalar multiple of $AB-BA$: there must be counterexamples with $\dim X=3$. I would guess you can make do with $A=\begin{pmatrix}1&0&0\\ 0&2&0\\0&0&3\end{pmatrix}$, $B=\begin{pmatrix}0&1&0\\ 0&0&1\\ 0&0&0\end{pmatrix}$.

Title: Factorization of $a^{n}-b^{n}$.


问:
Construction of a right uniformly continuous function which is not left uniformly continuous.
Consider the locally compact topological group
$$
G=
  \begin{pmatrix}
    a & b \\
    0 & 1
  \end{pmatrix}
$$
Where $a>0$ and $b\in \mathbb{R}$. In the Book Measure Theory by Cohn (2nd edition, page 284, problem 12), it is asked to construct a function $f$ from $G$ to $\mathbb{R}$ that is right uniformly continuous but not left uniformly continuous. According to his definitions, a function $f$ will be left (right) uniformly continuous if for a given $\epsilon$, we can to find an open neighbourhood $U$ of $id$ such that
$$| f(y)-f(x) |<\epsilon \,\,\text{whenever}\,\,y\in xU\,\, (y\in Ux).$$

As a hint, it has been told to find a suitable function $\phi :\mathbb{R}\rightarrow \mathbb{R}$ and define

$$f: \begin{pmatrix}
    a & b \\
    0 & 1
  \end{pmatrix} \mapsto \phi(b).$$

I can’t find any such function $\phi$ with which the expressions of $x^{-1} y$ and $y x^{-1}$ are working. I can’t even find a function $\phi$ (not even the identity function) for which $f$ can be shown to be right uniformly continuous.

A little help will be very much appreciated. Thank you.

答:
Consider an open $\epsilon$ neighbourhood of the identity $I_2$
$$B(I_2, \epsilon)=\Big\{\begin{pmatrix} a & b \\
    0 & 1 \end{pmatrix}:|a-1|< \epsilon,\, |b|< \epsilon\Big\}.$$
Also consider the function $f:G\rightarrow \mathbb{R}$ sending $\begin{pmatrix} a & b \\
    0 & 1 \end{pmatrix}$ to $\frac{b}{1+ |b|}.$

If we set $x_n = \begin{pmatrix} e^n & 0 \\
    0 & 1 \end{pmatrix}$ and $y_n = \begin{pmatrix} e^n & n \\
    0 & 1 \end{pmatrix}$, then $x_n ^{-1} y_n$ tends to $I_2$ but $|f(x_n) - f(y_n)|=\frac{n}{n+1}$ tends to $1$ as $n$ tends to $\infty$. So $f$ is not left uniformly continuous.

Now consider $x=\begin{pmatrix}x_1 & x_2 \\
0 & 1 \end{pmatrix}$ and $y=\begin{pmatrix}y_1 & y_2 \\
0 & 1 \end{pmatrix}$ such that $yx^{-1}\in B(I_2, \epsilon),\,\,1>\epsilon.$ Then it is easy to see the following:

$1-\epsilon <\frac{y_1}{x_1}$, $|x_2|<\frac{\epsilon}{1-\epsilon}$, $|y_2|<\epsilon$ and $\frac{|x_2 -y_2|}{1+|x_2|}<\epsilon$.

If $x_2 y_2 \geq 0$, then $x_2 |y_2| -|x_2| y_2=0$.

If $x_2 y_2 <0$, then $|\,x_2 |y_2| -|x_2| y_2\,| <2\frac{\epsilon ^2}{1-\epsilon}$.

$|f(x)-f(y)|\leq \frac{|x_2 -y_2|}{1+|x_2|} + |\,x_2 |y_2| -|x_2| y_2\,| $.

From these claims, it is easy to deduce that $f$ is right uniformly continuous.


问:
Find the direction cosines of the a vector in the plane of $\vec b=<2, 1, 3>$ and $\vec c=<-1, 2, 1>$ and perpendicular to $\vec c$
Let the vector $\vec a =\langle x, y, z\rangle$

Since it is perpendicular to $\vec c$
$$-x+2y+z=0$$

And $$\begin{vmatrix} x &y&z \\ 2&1&3 \\ -1&2&1 \end{vmatrix}$$

$$\implies -5x-7y+5z=0$$ since a, b, c are coplanar

So $y=0$

Then $x=y$

So the answer should be $\left\langle\frac{1}{\sqrt 2}, 0, \frac{1}{\sqrt 2}\right\rangle$ , but that is not correct.

What is wrong with my solution?

The answer is $\left \langle\frac{15}{\sqrt {369}} , 0, \frac{12}{\sqrt {369}} \right \rangle $

答:
Let $\vec r=p \vec b+ q \vec c$, take dot product of this with $\vec c$, and use $\vec r. \vec c=0$, then
$$\vec r. \vec c\vec =p \vec b. \vec c+q\vec c. \vec c \implies p=-q \frac{c^2}{\vec b. \vec c}=-q \frac{6}{3}=-2q.$$
So the required vector $\vec r$ is:
$$\vec r=q[-2\vec b+ \vec c]=q(-5\vec i- 5\vec k).$$
$q$ is arbitrary real number. Next $\hat r= \frac{\vec i+\vec k}{\sqrt{2}}$

It also means that $$\vec r= \alpha~ \vec c \times(\vec b \times \vec c),$$
namely, the vector triple product.


问:
Recursive function with code $n$, whose output is the number $n$ itself

There is a recursive function with code $n$, whose output is the
  number $n$ itself. That is $\phi_n$ is the constant function with value
  $n$.

I've been toying with the idea of using s-m-n theorem applied to a particular function and then applying the recursion theorem to get the constant function/value.
I cannot go any further. How should I prove the above statement? There's another question with four answers but they don't answer my question.

答:
This is not entirely just s-m-n theorem. You need to use the Recursion theorem, which is a fairly easy consequence of the s-m-n theorem.

Define the partial computable function

$\Psi(x,y) = x$

Since it is partial computable, there exists an $e$ such that

$\Psi(x,y) = \Phi_e(x,y)$

By applying the s-m-n theorem (just like I described in one of your earlier questions, i.e. letting $f(x) = s(e,x)$, where $s$ is the function coming from the s-m-n theorem), there exists a computable function $f$ such that

$\Psi(x,y) = \Phi_e(x,y) = \Phi_{f(x)}(y)$

By the recursion theorem, there exists a $n$ such that

$\Phi_{f(n)} = \Phi_n$

Hence on all input $y$,

$\Phi_n(y) = \Phi_{f(n)}(y) = \Phi_e(n,y) = \Psi(n,y)  = n$

So $\Phi_n$ is a program with code $n$ which is also constant function taking input $n$.


问:
Discrete Mathematics-logic
We let $(A,\geq)$ be an ordered set. We define a relation $\preceq$ on AxA by:
$$(a_1,a_2)\preceq(b_1,b_2) <=>[(a_1\neq b_1) ∧(a_1\leq b_1)]∨ [(a_1=b_1)∧(a_2\leq b_2)]$$
First I have to show that $(a_1,a_2)\preceq(b_1,b_2)$ involve that we have $a_1 \leq b_1$. I think it makes sense but I when I try to prove it formally by the distributive probaties, I don't get a statement, which finally proves that. Can someone help me to proove that?

答:
A "formal proof" must be: assume $(a_1,a_2)\preceq(b_1,b_2)$.

This implies (bi-conditional from left to right):

$\text { (1)   } [(a_1\neq b_1) ∧ (a_1\leq b_1)] ∨ [(a_1=b_1) ∧ (a_2\leq b_2)]$.

Proof by Cases with $(a_1=b_1) \lor \lnot (a_1=b_1)$:

(i) if $(a_1=b_1)$, then by Addition we have $(a_1=b_1) \lor (a_1 \lt b_1)$ which is abbreviated as $(a_1 \le b_1)$.

(ii) If $\lnot (a_1=b_1)$, we have $\lnot [(a_1=b_1)∧(a_2\leq b_2)]$. Thus, applying Disjunctive Syllogism to (1) we have:

$[(a_1\neq b_1) ∧ (a_1\leq b_1)]$,

from which, by Simplification: $(a_1\leq b_1)$.

Thus, in both cases (i) and (ii) we have proved that $(a_1\leq b_1)$ holds.

Title: Discrete Mathematics-logic


问:
Solution/ argument verification of a question related to product topology
This question is part of my assignment in topology and I am not sure whether I am thinking along the right lines for the solution.

Let {${X_\alpha : \alpha \in \Lambda} $} be an indexed family of sets , let $\beta \in \Lambda$ , and let $A_\beta \subseteq X_\beta$ . Prove that $\pi_{\beta}^{-1}(A_\beta)= \prod_{\alpha\in \Lambda} B_\alpha$ , where $B_\beta =A_\beta $ and $B_\alpha = X_\alpha$ for all $\alpha\neq \beta $.

I think that there is not much to prove as inverse image of $A_\beta$ under projection is given by $X_\alpha  \times ... A_\beta \times ...X_\alpha $.

So, i think it was trivial.

Am i right? In case I am wrong kindly outline correct method!

答:
You are right, I see this as trivial, but if you want a formal demonstration:

It is clear that $\prod_{\alpha\in\Lambda}B_\alpha$ as you defined is contained in $\pi_{\beta}^{-1}(A_{\beta})$, now suppose that $\pi_{\beta}((x_{\alpha})_{\alpha\in\Lambda})\in A_{\beta}$, then $x_{\beta}\in A_{\beta}$ so $(x_{\alpha})_{\alpha\in\Lambda}\in\prod_{\alpha\in\Lambda}B_{\alpha}$, and we have the doble inclusion.


问:
Prove that the tensor product of unitary matrices is also unitary
Show that if $A$ and $B$ are unitary matrices, then $C = A \otimes B$ is unitary.

答: Hint: $(A\otimes B)^{\dagger} = A^\dagger \otimes B^\dagger$, where $A^\dagger$ denotes the conjugate transpose of $A$.


问:
Does the domain of a function affect the meaning of limit
I recently was reading in stewarts calculus book the definition of limits and continuity.from what I understand for a limit to exsist for a function $f(x)$ at a point a means that
$\lim\limits_{x \to a^{+}} f(x)$ = $\lim\limits_{x \to a^{-}} f(x)$.
However if I had the function say $y=x^2$ and I restricted its domain so that it was $-2\leq x \leq 2$. If I was to say that the $\lim\limits_{x \to 2} x^2 = 4$ is this incorrect as the function is not defined for x greater than $2$ however since the functions domain does not consider values of x greater than $2$ then is it true that only the one sided limit is needed.my question is if the domain of a function has been restricted than can we consider the limit at its end point defined and to be the true limit not one sided as $x$ approaches the end point from within the domain?.

答: The rule “$\lim_{x\to a}f(x)$ exists if and only $\lim_{x\to a^+}f(x)$ and $\lim_{x\to a^-}f(x)$ both exist and they are equal” holds when (and only when) there are points in the domain of $f$ arbitrarily close to $a$ and smaller than $a$ and there are points in the domain of $f$ arbitrarily close to $a$ and greater than $a$. So, it doesn't apply to $\lim_{x\to 2}f(x)$ when the domain of $f$ is $[-2,2]$. In this case, there is no $\lim_{x\to 2^+}f(x)$ and then$$\lim_{x\to 2}f(x)=\lim_{x\to 2^-}f(x).$$


问:
Evaluating $\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$
How to evaluate the following integral?

$$\int_0^\infty \frac{e^{-kx}\sin x}x\,\mathrm dx$$

Body: One more option:$$\begin{align}\int_0^\infty\int_0^\infty e^{-(k+y)x}\sin x\mathrm{d}x\mathrm{d}y&=\Im\int_0^\infty\int_0^\infty e^{-(k+y-i)x}\mathrm{d}x\mathrm{d}y\\&=\int_0^\infty\tfrac{1}{(k+y)^2+1}\mathrm{d}y\\&=[\arctan(k+y)]_0^\infty\\&=\tfrac{\pi}{2}-\arctan k.\end{align}$$


答:
You could take this definition of a limit:

Take a function $f:X\subset\mathbb R\to \mathbb R$ then for $a\in X'$ (the set of accumulation points of X) $\lim_{x\to a}f(x)=L$ if for every sequence $\lbrace x_n\rbrace\subset X\setminus \lbrace a\rbrace$ $\lim_{n\to\infty} f(x_n)=L$.

If you take a point $a$ in the interior of the domain then the definition above implies that $\lim_{x\to a^{+}} f(x)=\lim_{x\to a^-}f(x)$.

Title: Does the domain of a function affect the meaning of limit

Body:
I recently was reading in stewarts calculus book the definition of limits and continuity.from what I understand for a limit to exsist for a function $f(x)$ at a point a means that
$\lim\limits_{x \to a^{+}} f(x)$ = $\lim\limits_{x \to a^{-}} f(x)$.
However if I had the function say $y=x^2$ and I restricted its domain so that it was $-2\leq x \leq 2$. If I was to say that the $\lim\limits_{x \to 2} x^2 = 4$ is this incorrect as the function is not defined for x greater than $2$ however since the functions domain does not consider values of x greater than $2$ then is it true that only the one sided limit is needed.my question is if the domain of a function has been restricted than can we consider the limit at its end point defined and to be the true limit not one sided as $x$ approaches the end point from within the domain?.


问:
Interesting intermediate value problem and fixed point theorem
A thread of length $\ell$ is tied inside a rectangle between sides $M1$ and $M2$, parallel to the remaining sides (as in the figure). The thread is released from both sides, maintaining its length, and wraps arbitrarily inside the rectangle. Show that there is a point on the wound thread that maintains the same distance to the sides $M1$ and $M2$ as before.

I have already solved it, but I think it is interesting because it is a graphic way to see the fixed point theorem, and I'm looking for more problems like this. Do you know of some book where I can find similar problems?

Thanks a lot.

答:
I can't find any similar problems, but can offer an attempt to prove the one you have....

With the obvious choice of $X, Y$ the initial position of the thread can be viewed as the line $[0,  1] \times \{1/2\}$ in the unit square $[0,  1] \times [0, 1]$ and can apply the Euclidean metric to this space..

The second position can be seen as a path joining its two endpoints, and therefore has a representation $p(t) = (x(t), y(t))$ a continuous function mapping the unit interval into the unit square.

In particular we can parameterise the path by arc length, where we know from the initial position that the total arc length  $= 1$.

Then if we consider just the $x$ component, it is also a continuous function

$x: [0, 1] \to [0, 1]$ where the arc length $t$ in the $x(t)$ represents the distance from $M1$ in the initial position.

So, $x$ is a continuous function mapping the compact convex set $[0, 1]$ to (not necessarily onto) itself and so by the Brouwer Fixed Point Theorem

https://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem#One-dimensional_case

it has a point with $x(t) = t$.

I.e. there is at least one point in the second position at the same distance from $M1$ as in in the initial position.


问:
Finding a closed form for a recursively defined sequence.
So I have this definition: $T(n,k) = \sum_{i=0}^{k} T(n-1, k-i) $. Is there any way to get its closed form? I have tried looking for a pattern but I get really complicated sums.

Edit: Initial condition is  $T(2,k)= k+1.$ For any positive integer $k$.

答:
Notice that $T(n,k)=\sum_{i=0}^k T(n-1,k-i)$ is the same as $T(n,k)=\sum_{i=0}^k T(n-1,i)$

Using the recursive definition:

$$T(3,k)=T(2,0)+T(2,1)+\dots+T(2,k)=1+2+\dots+(k+1)=\frac{(k+1)(k+2)}2\\ T(4,k)=T(3,0)+T(3,1)+\dots+T(3,k)=\sum_{i=0}^k\frac{(i+1)(i+2)}2=\frac{(k+1)(k+2)(k+3)}6\\\vdots$$

By induction on $n$, you can show that $$T(n,k)=\dfrac{(k+1)(k+2)\cdots (k+(n-1))}{(n-1)!}=\dbinom{k+n-1}{n-1}$$

Proof sketch:

Show that the base case holds ($n=2$ by the given initial condition)

Assume the inductive hypothesis, ie, $T(n,k)=\dbinom{k+n-1}{n-1}$

Now, for the inductive step, note that,

$$\begin{align}T(n+1,k)\overset{(\#)}{=}\sum_{i=0}^k T(n,i)=\sum_{i=0}^k\binom{n-1+i}{n-1}&\overset{(\ast)}{=}\binom{(n-1)+(k+1)}n \\&=\binom{n+k}n\\&=\binom{k+(n+1)-1}{(n+1)-1}\end{align}$$

where the step $(\#)$ is using the given recurrence relation and the step $(\ast)$ follows by the hockey-stick identity.


问:
Evaluate $\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x\cos \left( \sin \left( x \right) -1 \right)}$
Evaluate the limit
$$
\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin  x}-1}{\sinh x(\cos \left( \sin \left( x \right)  \right)-1)}
$$

My Attempt: I tried to use L'Hôpital's rule to evalute it, however I found that the $1$st and $2$nd derivative of the numerator is $0$ at $x=0$, and the $3$rd derivative is very complicated. And the method of Taylor's Series is too complicated here. So, my question is, is there any easier way to evaluate this limit?

The desired answer is $-1$.

答:
Let's handle the denominator first. We can write it as $$x^3\cdot\frac{\sinh x} {x} \cdot \frac{\sin^2x}{x^2}\cdot \frac {\cos(\sin x) - 1}{\sin^2x}$$ and hence it can be replaced by $-x^3/2$.

To deal with numerator let us write it as $$(\cosh x) ^{\sin x} - 1=\frac{\exp(\sin x\log \cosh x) - 1}{\sin x\log \cosh x} \cdot\frac{\sin x} {x} \cdot\frac {\log\cosh x} {\cosh x - 1}\cdot\frac{\cosh x - 1}{x^2}\cdot x^3$$ and hence it can be replaced by $x^3/2$. It follows that the desired limit is $-1$.

In the above process we have used the following standard limits $$\lim_{x\to 0}\frac{\sin x} {x} =\lim_{x\to 0}\frac {\sinh x} {x} =1\\ \lim_{t\to 0}\frac{\exp(t)-1}{t}=\lim_{t\to 1}\frac {\log t} {t-1}=1\\ \lim_{x\to 0}\frac {1-\cos x} {x^2}=\lim_{x\to 0}\frac {\cosh x - 1}{x^2}=\frac{1}{2}$$

Often a little algebra combined with well known/standard limits helps to simplify the expression under limit to a great extent and gives the limit value in direct manner. L'Hospital Rule and Taylor series should be used  when they are damn simple to use.


问:
To show $f(x)=e^{x^2} +2x(xe^{x^2}-e^{-x^2})$ is positive in the interval $x\in[0,1]$
$f(x)=e^{x^2} +2x(xe^{x^2}-e^{-x^2})$ and $x\in[0,1]$

I want to know how $f(x)$ is positive in the interval.

$e^{x^2}$ is +ve but $(xe^{x^2}-e^{-x^2})$ is not always +ve.

So how can I show that $e^{x^2}$ is always greater than $2x(xe^{x^2}-e^{-x^2})$

Thanks.

答:
It is enough to show that $\frac {f(x)} {e^{x^{2}}} > 0$. This reduces to $1+2x^{2}-2xe^{-2x^{2}} >0$. Re-write this as $e^{2x^{2}} >\frac {2x} {1+2x^{2}}$. To prove this use the following:

a) $e^{2x^{2}}>1+2x^{2}$ (from the series expansion)

b) $(2x-1)^{2} \geq 0$.


问:
Find equation of this plane
Find the equation of a plane that goes through the intersection of the planes $x_1+5x_2+x_3-5=0$ and $x_1-3x_3-2x_3=0$ and contains the point $(1,1,0)$ .

I calculated the cross product of the given plane to obtain the intersection and it gave me $-7x_1+3x_2-8x_3$, but I'm stuck there.

答: You need two vectors for the plane you want, the one you got is$(-7,3,-8)$, so you need to find another one, choose a point P which lies both on $x_1+5x_2+x_3-5=0$ and $x_1-3x_2-2x_3=0$ (I assume you had a type mistake), and the vector from P to $(1,1,0)$ will be the second vector, do cross product then you will have the normal vector of the plane you want, $(1,1,0)$ must be contained so you can get the constant from $(1,1,0)$.

Title: Find equation of this plane


问:
Open ball around a set is open
Let $(E,d)$ be a metric space, $A\subseteq E$ and $r>0$. Can we show that $$C:=\left\{x\in E:d(x,A)<r\right\}$$ is open? Or do we need to assume that $A$ is closed/open?

Intuitively, the claim is rather trivial if we think about $A$ as a sufficiently nice set.

By definition of the infimum, we know that $$\forall x\in E:\forall\varepsilon>0:\exists a\in A:0\le d(x,a)-d(x,A)<\varepsilon\tag1.$$

If $x\in C$, we need to show that there is a $\varepsilon>0$ such that $B_\varepsilon(x)\subseteq C$. How can we do that or what do we need to assume to show that?

答: If $d(x,A) <r$ and $d(x,y) <r-d(x,A)$  then $d(y,A) \leq d(x,A)+d(x,y) <r$.  So $B(x,r-d(x,A)) \subseteq C$.

Title: Open ball around a set is open


问:
A calculation of the norm of an ideal
Let $L$ be a number field of degree $n$ over $\mathbb{Q}$ and $\mathfrak{a}$ a non-zero ideal of the ring of integers $\mathcal{O}_L$. Suppose that $X=\{x_1,...,x_n\}$ is a $\mathbb{Z}$-basis of $\mathcal{O}_L$, and $Y=\{y_1,...y_n\}$ is a $\mathbb{Z}$-basis of $\mathfrak{a}$.

How can one show that the norm of $\mathfrak{a}$ is equal to the absolute value of the determinant of the change of basis matrix $M=(m_{ij})$ from $Y$ to $X$?

I can show this in the special case of a simultaneous basis, but I'm having trouble with the general case. The problem is trying to show that $$\vert\frac{\oplus_{i=1}^{n}\mathbb{Z}x_{i}}{\oplus_{i=1}^n\mathbb{Z}y_i}\vert=det(M)$$

where $y_i=\sum_{j}m_{ji}x_j$.

Many thanks for your answers.

答:
So you're trying to show this?

Theorem. Let $G$ be a free abelian group of rank $r$, $H\le G$. Then $G/H$ is finite if and only if $G$ and $H$ are of the same rank. In this case, given two $\mathbb{Z}$-bases $X$ and $Y$ for $G$ and $H$ respectively, we can write $Y=SX$ for some invertible matrix $S$, and $|G/H|=|\text{det}(S)|$.

Let $G$ and $H$ have rank $r$ and $s$ respectively. Then we can choose $\mathbb{Z}$-bases $u_1,\dots,u_r$ of $G$ and $v_1,\dots,v_s$ of $H$ with $v_i=\alpha_iu_i$. Clearly $G/H$ is the direct product of finite cyclic groups of orders $\alpha_1,\dots,\alpha_s$, and $r-s$ infinite cyclic groups. Hence $G/H$ is finite if and only if $r=s$, and in that case $|G/H|=\alpha_1\cdots\alpha_r$.

Now
$$
\begin{align*}
u_i&=\sum b_{ij}x_j, & v_i&=\sum c_{ij}u_j, & y_i&=\sum d_{ij}v_j
\end{align*}
$$
and the matrices $B=(b_{ij})$ and $D=(d_{ij})$ are unimodular. Now $C=(c_{ij})$ is the diagonal matrix with the $\alpha_i$ down the diagonal. So $S=BCD$, and taking determinants the result follows.

Title: A calculation of the norm of an ideal


问:
The morphism $\text{Hom}_A(E,G)\otimes_BF\rightarrow\text{Hom}_A(E,G\otimes_BF)$ is injective when $F$ is free
Let $A,B$ be two rings. Suppose $E$ is a left $A$-module, $G$ an $(A,B)$-bimodule and $F$ a left $B$-module. Let
$$\nu_F:\text{Hom}_A(E,G)\otimes_BF\rightarrow\text{Hom}_A(E,G\otimes_BF)$$
be the unique $\mathbb{Z}$-linear mapping such that $\nu_F(u\otimes y)(x)=u(x)\otimes y$. I want to show that $\nu_F$ is injective when $F$ is free.

Attempt:

Fix a basis $(b_\mu)_{\mu\in M}$ of $F$. Then $u\mapsto\sum_{\mu\in M}u_\mu\otimes b_\mu$ is an isomorphism of $\text{Hom}_A(E,G)^{(M)}$ onto $\text{Hom}_A(E,G)\otimes_BF$. Suppose $\nu_F(X)=0$. Then there exists a unique $u$ such that $\sum_{\mu\in M}u_\mu\otimes b_\mu=X$. Thus $\nu_F(X)=\sum_{\mu\in M}\nu_F(u_\mu\otimes b_\mu)$. By assumption
$$\sum_{\mu\in M}u_\mu(x)\otimes b_\mu=0$$
for all $x\in E$. Assume there exists a $\lambda\in M$ such that $u_\lambda\ne0$. This implies that there exists $x\ne0\in E$ such that $u_\lambda (x)\ne 0$. Obviously
$$0=\sum_{\mu\in M}u_\mu(x)\otimes b_\mu=\sum_{\mu\ne\lambda}u_\mu(x)\otimes b_\mu+u_\lambda(x)\otimes b_\lambda.$$
How can I derive a contradiction from all of this?

答:
You have a basis $(b_\mu)_{\mu\in M}$ of $F$ over $B$. It follows that for all $A$-$B$-bimodules $Z$ we have an isomorphism of $A$-modules $Z\otimes_BF\cong Z^{(M)}$ via $\sum_\mu z_\mu\otimes b_\mu \leftrightarrow (z_\mu)_\mu$.

You already applied this to $Z=\mathrm{Hom}_A(E,G)$, but you can equally apply it to $G$.

So, if $\sum_\mu u_\mu\otimes b_\mu$ lies in the kernel of $v_F$, then for each $e\in E$ we have $\sum_\mu u_\mu(e)\otimes b_\mu=0$ in $G\otimes_BF$. Using the identification with $G^{(M)}$ we see that $u_\mu(e)=0$ for all $\mu$. Since this happens for all $e\in E$, we get $u_\mu=0$ for all $\mu$, and hence $\sum_\mu u_\mu\otimes b_\mu=0$ in $\mathrm{Hom}_A(E,G)\otimes_BF$.

This proves that $v_F$ is injective.


问:
Subgroups of index 2 of $\mathbb{Z}_2 \times S_3$
I know that the group $G=\mathbb{Z}_2 \times S_3$ has $3$ subgroups of index $2$ (i.e. of $6$ elements). It is easy to see that $2$ of them are $\{0\} \times S_3$ and $\mathbb{Z}_2 \times A_3$, but I am struggling to find the third one, that probably isn't a direct product of a subgroup of $\mathbb{Z}_2$ and a subgroup of $S_3$ like the first $2$. What can I do?

答:
Hint: We know that $\Bbb Z/2\times S_3\cong D_6$, and we know all subgroups of dihedral groups. Then we just pick out the ones of order $6$.

See for example How many subgroups of order $n$ does $D_n$ have?


问:
Checking if the mean obtained is within the interval $\mu_{ \bar X} ± 2\sigma_{ \bar X }$
I have the following question:

A soft-drink machine is regulated so that the amount of drink
dispensed averages $240$ milliliters with a standard deviation of $15$
milliliters. Periodically, the machine is checked by taking a sample
of $40$ drinks and computing the average content. If the mean of the
$40$ drinks is a value within the interval $\mu_{ \bar X}  ± 2\sigma_{
\bar X }$, the machine is thought to be operating satisfactorily;
otherwise, adjustments are made. The com- pany official found the mean
of $40$ drinks to be $\bar x = 236$ milliliters and concluded that the
machine needed no adjustment. Was this a reasonable decision?

The reason I am posting this is that I solved it in a different way than the solution provided and I want someone to verify my solution, I did the following:

I applied the central limit theorem since the central limit theorem as I think the central limit theorem tells how many standard deviations away from the mean, I got a value of $-1.68$ which is between $-2$ and $2$ so the decision is correct.

In the solution given they find the sample standard deviation, multiply it by $2$ and add/subtract it from the population mean to see if $236$ falls in this range, the decision was correct too but I am not very sure about my reasoning of the central limit theorem so I want someone to correct me if I am wrong and the right decision answer was just by luck.

答:
I don't see how it is possible to calculate a sample standard deviation from the information that is provided in the question, so either:

there is additional information not stated in the question

the provided solution is incorrect

you have misunderstood the provided solution.

The way I would answer the question is as follows.  For a sample of $n = 40$ randomly selected drinks, the observed mean volume is $\bar x = 236$ milliliters, which is $4$ milliliters less than the hypothesized mean $\mu = 240$ milliliters.  The standard error of the sample mean is $$\sigma_{\bar x} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{40}} \approx 2.37171,$$ hence the permissible margin of error is $2\sigma_{\bar x} \approx 4.74342$ milliliters.  Since $4$ milliliters does not exceed this margin of error, the machine is not adjusted:  $$|\bar x - \mu| \le 2\sigma_{\bar x}.$$
My suspicion is that you have mistaken $\sigma_{\bar x}$ as the sample standard deviation.  This would be incorrect because you are given that $\sigma = 15$ milliliters, consequently the sampling distribution of the sample mean for $n = 40$ is approximately normally distributed with mean $\mu$ and standard deviation $\sigma_{\bar x}$.  The latter is not a sample standard deviation because it is not a statistic estimated from the sample:  it is a value inherited from the known standard deviation of individual drink volumes.

That said, your approach is not materially different because in both cases, an asymptotically normal distribution is assumed for the sampling distribution.  You are calculating a $z$-score for the sampling distribution and comparing it against the critical value $2$.  In other words, my solution is working on the scale of  milliliters; your solution is working on the scale of $z$-scores, which is measured in standard deviations and is unitless.


问:
Calculate the sum of all the six-digit numbers you can build using $1,2,3,4,5,6$
Consider all the six-digit numbers you can form using the numbers $1,2,3,4,5,6 $(Each digit is included exactly once in each number). Calculate the sum of these numbers.

b) Calculate the sum of these numbers where each digit can be included several times in each number.

I haven't found a good way to calculate this. Especially b, I know there are $720$ numbers and $120$ of them have $1$ in unit digit, and I think $120$ has $6$ in unit digit? (I'm not sure if it's correct tho)

答:
For problem a), each digit is going to appear exactly 120 times as the first digit, second digit and so on. So the answer is

$120*111111*(1+2+3+4+5+6)$.

For problem b), it is the same but instead of $120 = 5!$, we take $7776 = 6^5$.


问:
How to determine the Green’s function of $y''=f(x)$ with the boundary conditions: $y(0)=0$, $y'(1)=0$
I want to determine the Green’s functions of the following BVP:

Let $y''=f(x)$ with the boundary conditions: $y(0)=0$, $y'(1)=0$

So the homogeneous solution for this is: $y_h(x)=C_1+C_2x$,  has the fundamental system $ \{u_1(x),u_2(x)\}=\{1,x\}$

Now I'm stuck here, I want to determine the new FS $\{v_1(x),v_2(x)\}$:

the first boundary conditions gives us:

$y(0)=0 \Rightarrow C_1+C_2.0=0\Rightarrow C_1=0 \Rightarrow v_1(x)=x$

the second boundary conditions gives us:

$y'(1)=0 \Rightarrow C_2=0  \Rightarrow v_2(x)=1$

Hence $\{v_1(x),v_2(x)\}=\{x,1\}$ and furthermore satisfy the boudary conditions $v_1(0)=0,v'_2(1)=0 $ and we can easly show the Wronskian determinant $W(x)=-1$ is that correct? I think it should be different from  $ \{u_1(x),u_2(x)\}=\{1,x\}$.

And I don't know how to continue this to find the green's function.

Thanks in advance for any help!

答:
Taking it up on my comment, the Green's function associated to the problem $L(y) = y'' = f$, $y(0)=y'(0)=0$ is given by the solution to $L(G)=G''=\delta(x-\xi)$, $G(0) = G'(1) =0$, with $0<\xi<1$. One can apply integration by parts to show that $y(\xi) = \int^1_0 f(x) G(x;\xi)\,\mathrm{d}x$.

The task at hand is then to integrate $G''=\delta(x-\xi)$. Integrate once with respect to $x$ to have:

$$\int^{0}_{G'} \mathrm{d}G' = \int^1_x \delta(x-\xi)\,\mathrm{d}x \implies G' = H(x-\xi) - 1,$$

where $H$ is the Heaviside (or unit step) function. Note that the BC $G'(1)=0$ has been imposed. Integrate once more to find

$$\int^G_0 \,\mathrm{d}G = \int^x_0 \left[ H(x-\xi) -1 \right]\,\mathrm{d}x \implies G(x;\xi) = (x-\xi)H(x-\xi) - x,$$

where the BC $G(0)=0$ has been imposed.

I have checked that indeed this is the Green's function to the problem for the particular case $f(x) = x$, with the help of Mathematica.

I hope you find this useful.


问:
Find the direction cosines of the a vector in the plane of $\vec b=<2, 1, 3>$ and $\vec c=<-1, 2, 1>$ and perpendicular to $\vec c$
Let the vector $\vec a =\langle x, y, z\rangle$

Since it is perpendicular to $\vec c$
$$-x+2y+z=0$$

And $$\begin{vmatrix} x &y&z \\ 2&1&3 \\ -1&2&1 \end{vmatrix}$$

$$\implies -5x-7y+5z=0$$ since a, b, c are coplanar

So $y=0$

Then $x=y$

So the answer should be $\left\langle\frac{1}{\sqrt 2}, 0, \frac{1}{\sqrt 2}\right\rangle$ , but that is not correct.

What is wrong with my solution?

The answer is $\left \langle\frac{15}{\sqrt {369}} , 0, \frac{12}{\sqrt {369}} \right \rangle $

答:
Let the unknow vector be $\vec a =(x,y,z)$ since $\vec a, \vec b, \vec c$ are coplaner their box product must vanish $$\implies \vec a. (\vec b \times \vec c)=0 \implies -5x-5y+5z=0 ~~~~(1)$$ Next the perpendicularity of $\vec a$ and $\vec c$ gives $vec a. \vec c=0 \implies -x+2y+z=0~~~~(2)$
From (1)(2) we get $y=0, x=z$, hence $\vec a= (x,0,x)$ $$ \implies \vec a= \frac{(1,0,1)}{\sqrt{2}}$$


问:
Why is boundary information so significant? -- Stokes's theorem
Why is it that there are so many instances in analysis, both real and complex, in which the values of a function on the interior of some domain are completely determined by the values which it takes on the boundary?

I know that this has something to do with the general version of Stokes's theorem, but I'm not advanced enough to understand this yet -- does anyone have a (semi) intuitive explanation for this kind of phenomenon?

答:
Well it might be easier to start with the version of stokes theorem you probably know best, the fundamental theorem of calculus: $\int_a^b df = f(b) - f(a)$ (when applicable).

A sketch of (a) proof is that $\int_a^b df = \lim_{N \to \infty} \Sigma_1^N (x_i - x_{i-1})df(y_i)$, where $y_i \in [x_{i-1}, x_i]$. By the mean value theorem, we can choose $y_i \in [x_{i-1}, x_i]$ such that $f(x_i) -f(x_{i-1}) = (x_i - x_{i-1})df(y_i)$, and so in the sum you cancel a lot. But you can't continue this process by virtue past the edges, the buck stops there.

Somehow, in some of these things, you can think of the problem has being pushed off to the sides, because a little tiny "cell" and their interactions in the interior have certain nice properties that lets you do so, but the edge lacks this.

Another example is the residue theorem; nothing exciting happens except at the poles.

This might be a bit tangental, is in de Rham cohomology, contractible sets, the building blocks of manifolds are boring, but how they patch together is not. Interesting cohomology is a emergent phenomenon.


答:
For part b), there are $6^6$ possible numbers in total, as there are $6$ choices for the first digit, $6$ for the second and so on. Thus there are $\frac{6^6}{6}  = 6^5$ possible numbers with the first digit being $1$, and this is the same for $2,3$ and the other digits.

Apply the same logic to the second and other digits, except now the place value is $10^4$ instead of $10^5$. Doing this for the other digits gives:

$$6^5 \cdot (1 + 2 + 3 +4 +5+6) \cdot (10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 10^0)$$

Title: Calculate the sum of all the six-digit numbers you can build using $1,2,3,4,5,6$

Body:
Consider all the six-digit numbers you can form using the numbers $1,2,3,4,5,6 $(Each digit is included exactly once in each number). Calculate the sum of these numbers.

b) Calculate the sum of these numbers where each digit can be included several times in each number.

I haven't found a good way to calculate this. Especially b, I know there are $720$ numbers and $120$ of them have $1$ in unit digit, and I think $120$ has $6$ in unit digit? (I'm not sure if it's correct tho)


问:
Argument for Grandi's Series equaling zero? Or flawed logic?
$1-1+1-1+1-1+.....=S$

$1-S=S=1/2$    (this is the reasoning I've learned before)

but if I try continuing this reasoning, I get

$$1-1+S=S$$
$$1-1+1-S=S=1/2$$
$$1-1+1-1+S=S$$
$$1-1+1-1+1-S=S=1/2$$
$$1-1+1-1+1-1+S=S$$
$$1-1+1-1+1-1+1-S=S=1/2$$
$$\cdots$$
$$S-S=S$$
$$S+S=S$$

Therefore, $S$ must equal zero? But that seems contradictory to what I've been taught? Is my logic flawed? Am I missing something here?

答: The Grandi series is divergent. The reasoning that you claim to have learnt before is not quite correct as you cannot assign a particular value to a divergent series. A divergent series cannot be assigned a particular finite value this way.</p>


答: $d(x,A)<r$ iff there exists $a\in A$ such that $d(x,a)<r$ so $B(A,r)=\bigcup_{a\in A}B(a,r)$, since arbitrary union of open sets is an open set we have that $B(A,r)$ is open.

Title: Open ball around a set is open

Body:
Let $(E,d)$ be a metric space, $A\subseteq E$ and $r>0$. Can we show that $$C:=\left\{x\in E:d(x,A)<r\right\}$$ is open? Or do we need to assume that $A$ is closed/open?

Intuitively, the claim is rather trivial if we think about $A$ as a sufficiently nice set.

By definition of the infimum, we know that $$\forall x\in E:\forall\varepsilon>0:\exists a\in A:0\le d(x,a)-d(x,A)<\varepsilon\tag1.$$

If $x\in C$, we need to show that there is a $\varepsilon>0$ such that $B_\varepsilon(x)\subseteq C$. How can we do that or what do we need to assume to show that?


答:
The third subgroup is a "diagonal". It consists of all elements of the form
$$
\{(\operatorname{sgn}\sigma, \sigma)\mid \sigma\in\Bbb S_3\}
$$
Each of these subgroups appear as the kernel of a surjective homomorphism onto $\Bbb Z_2$ (index 2 subgroups always do). This probably would've helped you find the final subgroup.

Title: Subgroups of index 2 of $\mathbb{Z}_2 \times S_3$

Body: I know that the group $G=\mathbb{Z}_2 \times S_3$ has $3$ subgroups of index $2$ (i.e. of $6$ elements). It is easy to see that $2$ of them are $\{0\} \times S_3$ and $\mathbb{Z}_2 \times A_3$, but I am struggling to find the third one, that probably isn't a direct product of a subgroup of $\mathbb{Z}_2$ and a subgroup of $S_3$ like the first $2$. What can I do?


问:
convergence in distribution $(u_n,\varphi_n) \to (u,\varphi)$
Define a sequence of distribution $u_n$.

Let $u_n\to u$ in $D'(X)$ and assume we have a seuqence $\varphi_n\in C^\infty_c(X)$ such that $\varphi_n\to \varphi $ in $C_c^\infty(X)$.

Can we show $$(u_n,\varphi_n) \to (u,\varphi)$$

I know we can show $(u_n,\phi) \to (u,\phi)$ for any $\phi\in C_c^\infty(X)$,and $(u_n,\varphi_j) \to (u_n,\varphi)$ for each $n$.How to combine them togethor?

$$\lim_k\lim_n (u_n,\varphi_k) = (u,\varphi)$$

But not exactly two same variable?

答:
I assume that $X$ is an open subset of $\mathbb{R}^n$. For any compact subset $K$ of $X$, let $C_K^{\infty}(X)$ denote the Frechet space of all $f \in C_c^{\infty}(X)$ such that $\text{supp}(f) \subset K$.

A non-trivial Theorem about convergence in the strict inductive limit topology of $C_c^{\infty}(X)$ implies that there must be $n_0 \geq1$ and a compact subset $K \subset X$ so that each $\varphi_n$ with $n \geq n_0$ and $\varphi$ itself belong to $C_{K}^{\infty}(X)$ and that $\varphi_n \rightarrow \varphi$ in this space. The restriction map $C_{c}^{\infty}(X)^{\ast} \rightarrow C_K^{\infty}(X)^{\ast}$ is continuous for the weak-star topologies and hence the sequence of restricted distributions $u_n|_{C_K^{\infty}}$ converges to the restricted distribution $u|_{C_K^{\infty}}$ in the weak-star topology on $C_K^{\infty}(X)^{\ast}$.

We have thus reduced our problem to proving that in every Frechet space $V$, for every convergent sequence of vectors $\varphi_n \rightarrow \varphi$ and weak-star convergent sequence of continuous linear functionals $\ell_n \rightarrow \ell$, we have $\ell_n(\varphi_n) \rightarrow \ell(\varphi)$ in $\mathbb{C}$, as $n \rightarrow \infty$.

By a further easy reduction, it suffices to prove this in the case $\varphi=0$ and $\ell = 0$.

This in turn follows from the uniform boundedness-principle in Frechet spaces, as explained in this  answer. This Theorem implies that the family $\{\ell_n \}_{n \in \mathbb{N}}$ is automatically equi-continuous, meaning that, given any $\varepsilon >0$, there is $U \subset X$ open, $0\in U$, so that for all $(n,v) \in \mathbb{N} \times U$ we have $|\ell_n(v)| < \varepsilon$. So given $\varepsilon$, first choose such such $U$ and then take $n$ sufficiently large so that $\varphi_n \in U$.


问:
Integrate: $\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)}dx$
How to evaluate the following
$$\int_0^{\infty} \frac{\sin (ax)}{e^{\pi x} \sinh(\pi x)} dx $$
Given hints says to construct a rectangle $0\to R\to R+i\to i \to 0$ and consider $\displaystyle f(z):=\frac{e^{iaz}}{e^{2\pi z}-1} $ and evaluate around it but that does not help.

ADDED:: I need to evaluate it with method of contour.

Particularly using given hint on book, when I integrate $i \to 0$, the integral does not converge. Also, real part of $\displaystyle f(z):=\frac{e^{iaz}}{e^{2\pi z}-1} $ does not converge around $0$.

Also I considered $\displaystyle f(z):=\frac{e^{iaz}-e^{-iaz}}{e^{2\pi x}-1} $, and integrate from $-R \to R \to R+i \to -R+ i \to R$ and I get $\displaystyle \int_{-\infty}^\infty f(z) dz$ but the function is not symmetric about $0$.

If I consider $\displaystyle f(z):=\frac{e^{iaz}}{e^{2\pi x}-1} $ around the path $-R - i \to R - i \to R + i \to -R +i \to -R -i $, I again end up with $\displaystyle \int_{-\infty}^{\infty}f(z) dz$ and I cannot extract $\int_0^{\infty}$ due to  function is not symmetric at $0$ because of $e^{2\pi z}$ at the denominator.

Please help!!

答:
For the contour you describe in your text, you have to indent about the poles at $z=0$ and $z=1$.  In that case, the contour integral

$$\oint_C dz \frac{e^{i a z}}{e^{2 \pi z}-1}$$

is split into $6$ segments:

$$\int_{\epsilon}^R dx \frac{e^{i a x}}{e^{2 \pi x}-1} + i \int_{\epsilon}^{1-\epsilon} dy \frac{e^{i a R} e^{-a y}}{e^{2 \pi R} e^{i 2 \pi y}-1} + \int_R^{\epsilon} dx \frac{e^{-a} e^{i a x}}{e^{2 \pi x}-1} \\+ i \int_{1-\epsilon}^{\epsilon} dy \frac{ e^{-a y}}{e^{i 2 \pi y}-1} + i \epsilon \int_{\pi/2}^0 d\phi \:e^{i \phi} \frac{e^{i a \epsilon e^{i \phi}}}{e^{2 \pi \epsilon e^{i \phi}}-1}+ i \epsilon \int_{2\pi}^{3 \pi/2} d\phi\: e^{-a} e^{i \phi} \frac{e^{i a \epsilon e^{i \phi}}}{e^{2 \pi \epsilon e^{i \phi}}-1}$$

The first integral is on the real axis, away from the indent at the origin.  The second integral is along the right vertical segment. The third is on the horizontal upper segment.  The fourth is on the left vertical segment.  The fifth is around the lower indent (about the origin), and the sixth is around the upper indent, about $z=i$.

We will be interested in the limits as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.  The first and third integrals combine to form, in this limit,

$$(1-e^{-a}) \int_0^{\infty} dx \frac{e^{i a x}}{e^{2 \pi x}-1}$$

The fifth and sixth integrals combine to form, as $\epsilon \rightarrow 0$:

$$\frac{i \epsilon}{2 \pi \epsilon} \left ( -\frac{\pi}{2}\right) + e^{-a} \frac{i \epsilon}{2 \pi \epsilon} \left ( -\frac{\pi}{2}\right) = -\frac{i}{4} (1+e^{-a}) $$

The second integral vanishes as $R \rightarrow \infty$.  The fourth integral, however, does not, and must be evaluated, at least partially.  We rewrite it, as $\epsilon \rightarrow 0$:

$$-\frac{1}{2} \int_0^1 dy \frac{e^{-a y} e^{- i \pi y}}{\sin{\pi y}} = -\frac{1}{2} PV\int_0^1 dy \: e^{-a y} \cot{\pi y} + \frac{i}{2} \frac{1-e^{-a}}{a}$$

where

$$PV\int_0^1 dy \: e^{-a y} \cot{\pi y} = \lim_{\epsilon \to 0} \int_{\epsilon}^{1-\epsilon} dy \: e^{-a y} \cot{\pi y}$$

is the Cauchy principal value of that integral.  By Cauchy's theorem, the contour integral is zero because there are no poles within the contour.  Thus,

$$(1-e^{-a}) \int_0^{\infty} dx \frac{e^{i a x}}{e^{2 \pi x}-1} -\frac{i}{4} (1+e^{-a}) -\frac{1}{2} PV \int_0^1 dy \: e^{-a y} \cot{\pi y} + \frac{i}{2} \frac{1-e^{-a}}{a}=0$$

Now take the imaginary part of the above equation - note that the nasty Cauchy PV integral drops out - and get

$$(1-e^{-a}) \int_0^{\infty} dx \frac{\sin{ a x}}{e^{2 \pi x}-1} = \frac{1}{4} (1+e^{-a})-\frac{1}{2} \frac{1-e^{-a}}{a}$$

or, after a little algebra and simplifying things, we get:

$$\int_0^{\infty} dx \frac{\sin{ a x}}{e^{2 \pi x}-1} = \frac14 \coth{\left (\frac{a}{2}\right )} - \frac{1}{2 a}$$


答:
Consider an arbitrary metric space $(X, d)$. For arbitrary nonempty subset $\varnothing \neq M \subseteq X$ and for arbitrary $x, y \in X$ the relation $|d(x, M)-d(y, M)| \leqslant d(x, y)$ is valid. On these grounds it is easy to establish the continuity of the following map:
$$\begin{align}
f \colon X &\to [0, \infty)\\
f(x) &= d(x, M).
\end{align}$$
For given $\rho>0$ it is obvious that $\{x \in X \mid d(x, M)<\rho\}=f^{-1}[[0, \rho)]$ and your claim follows immediately since obviously $[0, \rho)$ is open in the relative topology of $[0, \infty)$ (which is none other than the order topology on this positive semiaxis).

Title: Open ball around a set is open


答:
For part b) again, the mean number is $\frac{111 \ 111+666 \ 666}{2}$. Since this is not an integer, and there are an even number of integers, the pairs of numbers will cancel out ($111\ 111$ and $666 \ 666$; $111\ 112$ and $666 \ 665$ etc.) and sum to twice the mean.

Therefore, the sum of all these numbers is $6^6 \cdot \frac{111 \ 111+666 \ 666}{2}$, which gives the same number as before.


问:
Fourier Transform - Dirac Delta
How can I compute this Fourier transform?

$$\int \frac{d^3q}{(2\pi)^3}(\vec q \cdot \vec a)(\vec q \cdot \vec b)\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }\,$$

My idea was to write it as

$$-( \vec a \cdot \vec \nabla)(\vec b \cdot \vec \nabla)\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }$$

and use the fact that $\int \frac{d^3q}{(2\pi)^3}\frac{1}{q^2}\,e^{i \vec q \cdot \vec r }=\frac{1}{4\pi r}$, together with $\Delta \frac{1}{r} = -4\,\pi \,\delta(\vec r)$ but is is not exactly the same.

答:
So the other answers are correct only for $x≠0$. In the sense of distributions, however, there is indeed a Dirac delta appearing in the result.

The first steps in your approach are correct. both functions $\frac{1}{q^2}$ and $\frac{(q·a)\,(q\cdot b)}{q^2}$ are indeed locally integrable functions and bounded outside a compact set, so are tempered distributions: one can take their Fourier transform in the sense of distributions (remark however that these functions are not in $L^1$ or in $L^2$, so one cannot take the Fourier transform in the usual sense).

In the sense of distributions (writing with abuse of notation the Fourier transform as an integral)
$$
\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q = -(a\cdot\nabla)(b\cdot\nabla)\, \frac{1}{4π|x|}.
$$
Taking a first derivative of $1/|x|$, one obtains $-x/|x|^3$ (even in the sense of distributions) which is still a locally integrable function. However, taking two derivatives leads to a non locally integrable function, and so one has to be careful as you suspect. A formula that generalizes the formula of the Laplacian $-\Delta \frac{1}{4\pi|x|} = \delta_0$ is the Formula of the Hessian (where $\nabla^2 = \nabla\nabla$)
$$
\nabla^2\left(\frac{1}{4\pi|x|}\right) = \frac{1}{4π} \,\mathrm{pv.}\,\frac{3x\otimes x - |x|^2 \,\mathrm{Id}}{|x|^5} - \frac{1}{3}\, \delta_0 \,\mathrm{Id}
$$
where pv. denotes the principal value, see e.g. p.55 here. (In particular, if you sum the coordinates $(j,j)$ of this matrix, you obtain the formula for the Laplacian). Now remark that $(a\cdot\nabla)(b\cdot\nabla) = (a\otimes b):\nabla^2$ (if you prefer coordinates, $\sum_{ij} a_i\partial_i\, b_j\partial_j = \sum_{ij} a_ib_j\,\partial_i\partial_j$) and so
$$
\begin{align}
\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q &= -(a\otimes b):\nabla^2\left(\frac{1}{4\pi|x|}\right).
\\
&=  \frac{1}{4π} \,\mathrm{pv.}\,\frac{|x|^2 (a·b) - 3\,(a\cdot x)(b\cdot x)}{|x|^5} + \frac{a\cdot b}{3}\, \delta_0
\end{align}
$$

The physicist usually do not write the principal value, and so in shorter notations (with $r=|x|$)
$$
\boxed{\frac{1}{(2\pi)^3}\int_{\mathbb R^3} \frac{(q·a)\,(q\cdot b)}{q^2} \,e^{i\,q\cdot x} \,\mathrm d q =  \frac{1}{4πr^3} \left(a·b - 3\,(a\cdot \tfrac{x}{r})(b\cdot \tfrac{x}{r})\right) + \frac{a\cdot b}{3}\, \delta_0}
$$

Title: Fourier Transform - Dirac Delta


问:
Probability of I passengers sitting randomly but in their actual assigned seats
There was a question that came up in my practice:

If there are N passengers on a full flight who sit in random seats,
what is the probability that i passengers sit in their assigned seats?

The answer is supposed to be $ \sum_{j=0}^{n-1} (-1)^j {n\choose i+j} \frac{(n-(i+j)!)}{n!}$

But I don't understand where the expression came from. I'd love if someone could explain it step-by-step.

答:
This is a modified version of the Derangement problem. Recall that the derangement formula tells us that the number of ways $n$ objects can be arranged in $n$ places so that none of the objects land up in their respective places is $$D(n)=!n=n!\sum_{i=0}^{n} \frac{(-1)^i}{i!}$$

Now we use this in our problem. The number of ways that exactly $i$ people sit at their respective places and the rest $n-i$ people sit at incorrect places will hence turn out to be $${n \choose i}D(n-i)={n \choose i}(n-i)!\sum_{j=0}^{n-i} \frac{(-1)^j}{j!}$$ (choosing any $i$ objects from $n$ and deranging the rest)

Now the number of ways that at least $i$ people sit at their correct places will simply be summing this up from $i$ till $n-1$. You can finish the rest.


问:
Primitive Root Sum Requiring Element for Rearrangement
I have a question that I have solved the first part for, but the solution to the second part seems elusive.

Let p be an odd prime and let n be a natural number. Define $S_n(p)=$ Sum between $k=1$ and $p-1$ of $k^n$. Show that $S_n(p)$ is congruent to:

$-1$ mod $p,$ if $n$ is congruent to $0$ mod $p-1$

$0$ mod $p,$ if n is not congruent to $0$ mod $p-1$

I have solved part 1 by rewriting the sum $S_n(p)$ in terms of a primitive root $g$ mod $p$. So the sum becomes $g^n + g^2n + ... + g^{p-1}n$. From this, we know that as $n$ is congruent to $0$ mod $p-1,$ the sum is congruent to $1 + 1 + ... + 1, p-1$ times so it is congruent to $p-1$ and hence $-1$ mod $p$.

With part 2, I am fairly confident I need to find something which is non-zero to multiply the sum by so that I can then rearrange the sum so that it's congruent to 0 mod p, but I can't figure out what it could be. Any ideas?

Many thanks in advance for any help!

答: Hint: $k \mapsto gk$ is a bijection of the classes mod $p$. Conclude that $g^n S_n = S_n$.


问:
Boolean algebras without atoms
Why is the theory of Boolean algebras without atoms $\omega$-categoric?</p>

答:
It isn't true generally that all atomless Boolean algebras of the same cardinality are isomorphic, so we don't expect $\kappa$-categoricity, and the diversity of such Boolean algebras give rise to all the various distinct forcing notions.

But in the countable case, it turns out that there is just one atomless Boolean algebra up to isomorphism. There are evidently a variety of proofs.

Here is a 1972 article by Abian that gives a brief topological proof, as well as a detailed proof for the case of atomless Boolean rings.

This book by Givant seems to have an explanation of the proof using the back-and-forth technique, which I believe is probably how you would prefer to understand it. (Here is a link to [the Google Books version][3], where you can see a complete and fully detailed proof with exercises afterward.)

[3]: https://books.google.com/books?id=ORILyf8sF2sC&pg=PA134&lpg=PA134&dq=countable+atomless+Boolean+algebra&source=bl&ots=YL2jQrXiS9&sig=PTWu4vLDZO2TqSgHj_ksKjZ6oFU&hl=en&ei=fYHRTYCqKILe0QGK86yODg&sa=X&oi=book_result&ct=result#v=onepage&q=countable atomless Boolean algebra&f=false

Title: Boolean algebras without atoms


答:
You have to be a bit carefull with that comparison. In complex analysis, you really have (for holomorphic functions) that the values on the boundary completely determine the values within. My interpretation of why this works there is that being differentiable is a much stronger property for complex functions than for real ones. For complex functions, it implies being representable (locally) by a power series, and by one which converges in a way that allows component wise integration and differentiation. So what this comes down to is that once you know the results of evaluating the power series on all points on the boundary, you can determine the individual coefficients and the function is thus uniquely determined.

Stoke's theorem, on the other hand, speaks about the relationship of an integral over a boundary vs. the integral of some kind of derivative over the whole volume. You can view this as the (pretty vast) generalization of one-dimensional anti-derivatives. If $F$ is the anti-derivative of $f$, i.e. $\frac{dF}{dx} = f$, then $ \int_a^b f(x) \,dx = F(b) - F(a)$. You the view the left-hand side as an integral over the "volume" [a,b] and the right-hand-side as an integral over the boundary "{a,b}".

Title: Why is boundary information so significant? -- Stokes's theorem

Body:
Why is it that there are so many instances in analysis, both real and complex, in which the values of a function on the interior of some domain are completely determined by the values which it takes on the boundary?

I know that this has something to do with the general version of Stokes's theorem, but I'm not advanced enough to understand this yet -- does anyone have a (semi) intuitive explanation for this kind of phenomenon?


问:
In a finite semigroup, does each ideal (esp. the minimal ideal) have an idempotent?
Given a finite semigroup $S$, let $I$ be a $\mathcal{J}$-minimal ideal, where $\mathcal{J}$ is equivalence relation brought about by the Green relation $\leq_{\mathcal{J}}$, where given $a,b \in S$, $a \mathcal{J} b$ if and only if $a \leq_{\mathcal{J}} b$ and $b \leq_{\mathcal{J}} a$.

I would like to ask, does $I$ have an idempotent ? ... and since $I \subseteq I'$ for any other ideal $I' \subseteq S$, does each ideal have an idempotent ?

答:
Every finite semigroup contains an idempotent, and ideals are subsemigroups.

To see the first statement note that it's enough to find $a\in S$ with $a^n=a$ for some $n\in\Bbb N$: if $n=2$ we are done, otherwise $a'=a^{n-1}$ is idempotent.

Now consider any $a\in S$ and look at the sequence $a,a^2,a^4,\ldots$. Since $S$ is finite there must be a repetition, so for some $m,l$ we have $a^{2^m}=a^{2^l}$. Assume wlog $m>l$ and consider $a'=a^{2^{l}}$ and $n=2^{m-l}$, so that $a'^n=a'$ and we are done.

The "using a nuke for a mosquito" approach to the first statement is to note that every finite semigroup is a compact topological semigroup in the discrete topology all of whose ideals are closed, and then use Ellis theorem that a compact semitopological semigroup has an idempotent.


问:
Taylor expansion in linear stability analysis of diffusion-driven instability
We are shown an (apparently trivial) Taylor expansion for a system of reaction-diffusion equations (re Turing stability), where $D$ is a constant:
$$
\frac{\partial\vec u}{\partial t} = D\nabla^2\vec u\,+\,\bar F (\vec u)
$$
For a small perturbation $\vec w$ about a steady-state $\vec u_s$ (s.t. $\vec F(\vec u_s)=0$), we have:
$$
\vec u = \vec u_s+\vec w
$$
And the taylor series linear expansion is given as:
$$
\frac{\partial\vec w}{\partial t} \approx D\nabla^2\vec w\,+\,\vec F (\vec u_s) + \mathbf{J}\vec w
$$
Where $\mathbf{J}$ is the jacobian of $\vec F$ evaluated at $\vec u_s$, and of course the second term above is zero.

I'm a little confused as to how this final equation is reached - I tried doing out the taylor expansion myself but got $\vec u_s$ in the first term instead of $\vec w$.

答: Solved the issue myself. Taking the TS expansion of $\mathbf{F}(\mathbf{u_s}+\mathbf{w})$, we achieve the second two terms in the final expression above. Fine. What I didn't understand was why the final expression ommited the term $D\nabla^2\mathbf{u_s}$. In the Turing stability case, we have a spatially uniform steady state, therefore this term is zero and we get the result stated.


问:
Why are there so many Probability law and when do we use them?
I'm not using maths professionally but I'm curious about Data Sciences and Probability in general.

There are so many Probability laws but I don't understand the motivation.

I guess they will be used in Statistics but are they chosen by hand or are there more rigorous process ?

Why do we need them ?

答: By probability laws, I think you meant about probability distributions. Apart from the theoretical perspective, the reason we study these distributions in great detail is because we can model many of our real-world scenarios with the help of these distributions. For example, the normal distribution can model the different heights in a population, a poison distribution can model the number of emails you'll receive in the next hour, etc. Hope this answers your question</p>


答:
Let's first present Stokes' theorem symbolically:

$$\int_{\partial\Omega}\omega=\int_\Omega d\omega$$

Intuitively, $\Omega$ is some 'domain', with boundary $\partial\Omega$, and $\omega$ is some 'function' with derivative $d\omega$.  So the theorem is roughly stating "the integral of a function on a boundary is the integral of the derivative on the enclosed domain."

For more concreteness, and to see intuitively why this theorem is true, that is, why the boundary values are so important, let's consider two analogous examples from vector calculus.

Divergence Theorem

$$\iint_{\partial V}{\mathbf {F}}\cdot d\mathbf n=\iiint_V(\nabla\cdot\mathbf F)dV$$

Notice this takes the form of Stokes' theorem, where our domain $\Omega$ is now some volume $V$ in $\mathbb{R}^3$, our function $\omega$ is now a vector field $\mathbf F$, and our derivative $d$ is the divergence of a vector field.  In terms of fluid flow, what does divergence measure?  Imagine a small cube placed at a point $(x,y,x)$ in $V$.  The divergence of $\mathbf F$ at this point is a scalar number measuring the tendency for fluid to move into or out of the cube.  The right hand side of the Divergence theorem integrates this value throughout the volume $V$.

Now, if we fill the region $V$ with small cubes, it makes sense that boundaries of cubes touching each other on the interior will cancel out (fluid flow out of one is fluid flow into another), so all we need to calculate is the fluid that moves out across $\partial V$.  This is exactly what the integral on the left hand side computes.

Kelvin-Stokes Theorem

$$\oint_{\partial S}{\mathbf {F}}\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot d\mathbf n$$

Notice this takes the form of Stokes' theorem, where our domain $\Omega$ is now some surface $S$ in $\mathbb{R}^3$ and our derivative $d$ is the curl of a vector field.  In terms of fluid flow, what does curl measure?  Imagine a small pinwheel placed at a point $(x,y,x)$ in $S$.  The curl of $\mathbf F$ at this point is a vector measuring the tendency for the pinwheel to rotate.  The vector has direction according to the right hand rule.  The right hand side of the Kelvin-Stokes theorem integrates these vectors throughout the surface $S$.

Now, if we fill the surface $S$ with small pinwheels, it makes sense that boundaries of pinwheels touching each other on the interior will cancel out (a flow causing a pinwheel to rotate clockwise will contribute to a counterclockwise rotation of an adjacent pinwheel), so all we need to calculate is the fluid flow that moves around $\partial S$.  This is exactly what the integral on the left hand side computes.


问:
Minimum of $\tan x+\tan y+\tan z$ using Lagrange multipliers for acute angled triangle $\Delta xyz$
Consider an acute angled triangle $\Delta xyz$. Suppose we want to minimise the function $f(x,y,z)=\tan x+\tan y+\tan z$. So as per the method of Lagrange multipliers here is what I did.

$$\begin{aligned}&\text{minimise}: f(x,y,z)=\tan x +\tan y+\tan z \\ &\text{given: }x+y+z=\pi \end{aligned}$$

$$\begin{aligned}\sec^2 x&=\lambda\\ \sec^2 y&=\lambda \\ \sec^2 z&=\lambda\end{aligned}$$

Now we have the constraint $x+y+z=\pi$ and $0\lt x,y,z\lt \pi/2$. Also do have $4$ unknowns and $4$ equations to solve for $x,y,z,\lambda$. But I am not clear on how to use these equations, managing the $\pm$ signs when we take square roots. Besides if it was not given that $\Delta xyz$ is acute angled, then the minimum would have been in the case of a degenerate triangle when any one of the angles is $0$ and others add up to $\pi$, in which case, $f(x,\pi-x,0)=0$, but I do not understand the jump from $0$ to $3\sqrt{3}$ either for the minimum. Can it not attain values in $(0,3\sqrt{3})$? Why? . Any hints are appreciated. Thanks.

答:
To handle this problem with the Lagrange Multipliers method, we need first of all, convert the inequalities into equations so with the help of slack variables $(s_x,s_y,s_z,e_x,e_y,e_z)$ the formulation could be

$$
\min_{x,y,z} \tan x + \tan y + \tan z\ \ \  \text{s. t.}\cases{x+y+z=\pi\\
x = s_x^2\\
y = s_y^2\\
z = s_z^2\\
\frac{\pi}{2}-x = e_x^2\\
\frac{\pi}{2}-y = e_y^2\\
\frac{\pi}{2}-x = e_z^2\\
}
$$

and the corresponding lagrangian

$$
L = \tan x + \tan y + \tan z + \lambda(x+y+z-\pi)+\sum_{k=\{x,y,z\}} \eta_k(k-s_k^2)+ \sum_{k=\{x,y,z\}} \mu_k\left(\frac{\pi}{2}-k-e_k^2\right)
$$

so the set of stationary points for $L$ could contain the solution. Now the minimization problem has the variables $\{x,y,z,\lambda,\eta_k,\mu_k,s_k,e_k\},\ \ k=\{x,y,z\}$. Also, to avoid algebraic problems to obtain the solutions for equations involving trigonometric functions, we can use

$$
\tan x\approx x + \frac{x^3}{3}
$$


问:
Evaluating $\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$.
My attempt.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$
$$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
The first sum can be evaluated easily if one uses the central binomial coefficient generating function
, the closed form is $2\zeta \left(2\right)$.

For the remaining sum consider the $\arcsin$ series expansion.
$$\sum _{k=0}^{\infty }\frac{x^{2k+1}}{4^k\left(2k+1\right)}\binom{2k}{k}=\arcsin \left(x\right)$$
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{\arcsin \left(\sqrt{x}\right)}{\sqrt{x}}-1$$
$$-\sum _{k=1}^{\infty }\frac{1}{4^k\left(2k+1\right)}\binom{2k}{k}\int _0^1x^{k-1}\ln \left(1-x\right)\:dx=-\int _0^1\frac{\arcsin \left(\sqrt{x}\right)\ln \left(1-x\right)}{x\sqrt{x}}\:dx$$
$$+\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=-2\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx-\zeta \left(2\right)$$

But I got stuck with:
$$\int _0^1\frac{\arcsin \left(x\right)\ln \left(1-x^2\right)}{x^2}\:dx$$
Anything I try yields more complicated stuff, is there a way to calculate the main sum or the second one (or the integral) elegantly\in a simple manner?

答:
It seemed something was missing, so with the right tools the proof isn't difficult.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}$$

Consider:
$$\sum _{k=1}^{\infty }\frac{x^k}{4^k}H_k\binom{2k}{k}=\frac{2}{\sqrt{1-x}}\ln \left(\frac{1+\sqrt{1-x}}{2\sqrt{1-x}}\right)$$
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k}\binom{2k}{k}\int _0^1x^{2k}\:dx=2\int _0^1\frac{\ln \left(1+\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(\sqrt{1-x^2}\right)}{\sqrt{1-x^2}}\:dx$$
$$-2\ln \left(2\right)\int _0^1\frac{1}{\sqrt{1-x^2}}\:dx$$
$$=2\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx-2\int _0^1\frac{\ln \left(x\right)}{\sqrt{1-x^2}}\:dx-\pi \ln \left(2\right)$$

$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=\frac{\pi }{2}\ln \left(2\right)-\int _0^1\frac{\arcsin \left(x\right)}{1+x}\:dx$$
$$=\frac{\pi }{2}\ln \left(2\right)-\int _0^{\frac{\pi }{2}}\frac{x\cos \left(x\right)}{1+\sin \left(x\right)}\:dx=\int _0^{\frac{\pi }{2}}\ln \left(1+\sin \left(x\right)\right)\:dx$$
$$=4\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(1+t^2\right)}{1+t^2}\:dt$$
This means that:
$$\int _0^1\frac{\ln \left(1+x\right)}{\sqrt{1-x^2}}\:dx=-\frac{\pi }{2}\ln \left(2\right)+2G$$

Thus:
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=-\pi \ln \left(2\right)+4G$$

Bonus.
$$\sum _{k=1}^{\infty }\frac{H_k}{4^k\left(2k+1\right)}\binom{2k}{k}=\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k}\binom{2k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}$$
And so we find that:
$$\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}=2\zeta \left(2\right)+2\pi \ln \left(2\right)-8G$$
And in the body of the question we had:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{H_k}{k\:4^k\left(2k+1\right)}\binom{2k}{k}-\frac{1}{2}\zeta \left(2\right)$$
Hence:
$$\int _0^1\frac{\ln \left(1-x^2\right)\arcsin \left(x\right)}{x^2}\:dx=-\frac{3}{2}\zeta \left(2\right)-\pi \ln \left(2\right)+4G$$


问:
Trying to understand Theorem 2.27 in a recent paper on the Chebyshev function
In February 2013, Sadegh Nazardonyavi and Semyon Yakubovich posted on arxiv: Sharper estimates for Chebyshev's functions $\vartheta$ and $\psi$.

I have a question about Theorem 2.27 on page 22.

My question regards the argument for this:

$$\vartheta(x) < 1.000027651, \;\;(x > 0)$$

I can follow the beginning of the proof:

Let:

$$8\cdot10^{11} \le x < 10^{16} $$

From Theorem 2.2.5 on p21 (which is taken from N. Costa Pereira.  Estimates for the Chebyshev function $\psi(x) - \theta(x)$.  Math. Comp.,44(169):211-221,1985.):

$$\psi(x) - \vartheta(x) > \sqrt{x} + \frac{6}{7}\sqrt[3]{x}, \;\;\;(2,036,329 \le x \le 10^{16})$$

Then:

$$\vartheta(x) < \psi(x) - \sqrt(x) - \frac{6}{7}\sqrt[3]{x}$$

I am unclear on the next step:

$$ \psi(x) - \sqrt(x) - \frac{6}{7}\sqrt[3]{x} < \left\{1.0000284888 - \frac{1}{\sqrt{x}} - \frac{6}{7}\frac{1}{\sqrt[3]{x^2}}\right\}x $$

I can see that:

$$- \sqrt(x) - \frac{6}{7}\sqrt[3]{x} < \left\{- \frac{1}{\sqrt{x}} - \frac{6}{7}\frac{1}{\sqrt[3]{x^2}}\right\}x $$

I am unclear how it is established that:

$$\psi(x) < 1.0000284888x$$

If anyone can help me to understand this, that will be very helpful.

Thanks,

-Larry

答: It seems that $2.84888\times 10^{-5}$ is the $\varepsilon$ you compute using Theorem 2.12 for $x\ge e^{27.407\cdots}=8\times 10^{11}$. For some reason this appears as a fragment on page 36.</p>


问:
Proving set $W$ is neither open nor closed
This question is from page 86 Wayne Patty's Topology and I was unable to solve 1 part of th efollowing question.

Question: Let $X_n =\mathbb{R}$ and let $T_n$ denote the usual topology on $X_n$ and $T$ be the product topology on $X=\prod_{n\in \mathbb{N}}$ .Define for each $n \in \mathbb{N}$, $W_n$ the open interval $(0,1)$ and let $W=\prod_{n \in \mathbb{N}} W_n$.

Show that $W$ is neither open nor closed in $(0,1)$.

well $W_i$ are open sets in $T_i$ and so why is product not open in T?

It means that this argument doesn't extends in product topology.

But Why(kindly give a complete argument or add a source where I can look)? and how can I disprove the question ?

I am adding another relevant problem which I have in mind: Let $X_n =\mathbb{R}$ and let $T_n$ denote the usual topology on $X_n$ and T be the product topology on $X=\prod_{n\in \mathbb{N}}$ .Define for each $n \in \mathbb{N}$, $W_n$ the closed interval [0,1] and let $W=\prod_{n \in \mathbb{N}} W_n$.

Is $W$ closed in $(X,T)$?

Please forgive me if I am asking many questions which are not of advanced level but i did a course on topology last year and that was totally marred by a non-serious professor and I read the book again in free time. I wasn't aware of this website earlier.

答:
If $\lbrace(X_\alpha,T_\alpha):\alpha\in\Lambda\rbrace$ is a family of topological spaces, then the product topology of $\prod_{\alpha\in\Lambda}X_{\alpha}$ is the topology genereted by the sets of the form $\prod_{\alpha\in\Lambda}B_{\alpha}$, such that $B_\alpha\in T_{\alpha}$ and $B_\alpha=X_\alpha$ except for a finite subset of $\Lambda$. Since $W_n\neq R$ for all $n\in\mathbb N$ it is clear that $W$ doesn't contain any of the generators of the product topology, hence it cannot be an open set.

Now for your other question indeed $W$ is a closed set, if we define the sets $W'_i=\prod_{n\in\mathbb N}B_n$ such that $B_n=\mathbb R$ for $n\neq i$ and $B_i=[0,1]$ it is clear that this set is closed (fast-check that the compliment is open by the above definition), and $W=\bigcap_{i\in\mathbb N}W_i'$, since arbitrary intersection of a closed set is closed we have that $W$ is closed.


问:
Inner product of dual basis

Assume $V$ is an inner product space over $\mathbb{R}$ with inner product $\left<\cdot,\cdot\right>$. Let $u_1,\cdots,u_n$ be a basis of $V$ and $v_1,\cdots,v_n$ be the dual basis, i.e., $\left<u_i,v_j\right> = \delta_{ij}$. Prove that if $\left<u_i,u_j\right>\leq 0$ for all $1\leq i < j \leq n$, then $\left<v_i,v_j\right>\geq 0$ for all $1\leq i < j \leq n$.

If we let $B=(b_{ij})$ be the matrix from basis $\{v_i\}$ to $\{u_i\}$, that is, $(u_1,\cdots,u_n) = (v_1,\cdots,v_n)B$, then $u_j = \sum_{i=1}^n b_{ij}v_i$, hence $\left< u_j, u_k \right> \leq 0$ and the duality imply $\left< u_j, u_k \right> = \left< \sum_{i=1}^n b_{ij}v_i, u_k \right> = \left<  b_{kj}v_k, u_k \right> = b_{kj} \leq 0$ for $k\neq j$.

Now $(v_1,\cdots,v_n) = (u_1,\cdots,u_n) B^{-1}$, if we let $B^{-1} = (c_{ij})$ then $v_j = \sum_{k=1}^n c_{kj}u_k$, so $\left< v_i, v_j \right> = \left< v_i, \sum_{k=1}^n c_{kj}u_k \right> = \left< v_i, c_{ij}u_i \right> = c_{ij}$ for $i\neq j$.

Since $BB^{-1} = I_n$, we have $\sum_{k=1}^n b_{ik}c_{kj} = \delta_{ij}$. But it seems difficult to determine whether $c_{ij}\geq 0$ from this condition. How should I do next?

答:
Not in the same way as you suggested but here is a simple computation based answer. Some trivial calculations are needed. I hope you will be able to figure it out.

Consider $n=2$ and without loss of generality let $u_1,u_2$ be unit vectors.

Let $a=(u_1,u_2)$. Then $0\geq a>-1$.

Let $v_i=a_{i1}u_1+a_{i2}u_2$ for $i=1,2$.

Using $(u_i,v_j)=\delta_{ij}$ we get,

$$a_{11}=a_{22}=\frac{1}{1-a^2},  ~~a_{12}=a_{21}=-\frac{a}{1-a^2} \tag{1} $$

Now $$ (v_1,v_2)=(a_{11}a_{21}+a_{12}a_{22})+(a_{11}a_{22}+a_{12}a_{21})a \tag{2} $$

Using (1) and (2), we get

$(v_1,v_2)=\frac{a}{a^2-1}\geq 0$ since $0\geq a>-1$.

Let us consider the general case now.

By induction hypothesis, $(v_i,v_j)\geq 0$ for $i,j\leq n-1$. It's now sufficient to show that $(v_n,v_i)\geq 0$ for $i<n$. But it's also trivial since $u_n,u_i$ generate a two dimensional subspace of $V$ and $v_n,v_i$ restricted to that subspace form a dual basis of $u_n,u_i$. Hence $(v_n,v_i)\geq 0$ by $n=2$ case.

Title: Inner product of dual basis


问:
Graph of a morphism of affine schemes
Let $f:X\to Y$ be a morphism of schemes (here smooth, separated, of finite type over a field). Then the graph $\Gamma_f$ is defined as the image of the morphism $(id,f):X\to X\times Y$.

Claim: The structure sheaf of the graph is a coherent $\mathcal O_{X\times Y}$-module.

I tried to convince myself of this by considering the affine case first and got embarassingly confused:

So if $X=Spec\ (R)$ and $Y=Spec\ (S)$, then $X\times Y=Spec\ R\otimes S$. Let $f':S\to R$ be the corresponding ring homomorphism, then the graph should be just $\{s\otimes f'(s)\}$. And I guess I did something wrong here, because this is certainly no $S\otimes R$-module. The module generated by the set $\{s\otimes f'(s)\}$ on the other hand is the whole ring, because $f'(1)=1$.

It is easy to see that the graph is a closed subscheme, so it should be possible to write it as a quotient of $R\otimes S$.

答: The (pushforward of the structure sheaf of the) graph is $R$ regarded as an $S$-module via the homomorphism $f$ and regarded as an $R$-module in the obvious way. </p>

Title: Graph of a morphism of affine schemes


问:
Reduction of a type of hyperelliptic integrals to elliptic integrals.
In [1] (refered to as "the handbook"), it is said that

... the more general integral (Eq 575.16)

$$ \int R(\tau)\sqrt{(\tau-r_1)(\tau-r_2)(\tau-r_3)(\tau-r_4)(\tau-r_5)(\tau-r_6)}\,d\tau $$
  can be reduced to elliptic integrals if the six branch-points
  $r_1,...,r_6$ form three pairs of points of an involution.

First of all, I think there is a typo; it should be $R(\tau)/\sqrt{\cdots}$.
Then, the handbook cites [2] for reference.  The relavant context seems to be from the bottom of page 289 until the top of page 291.  However, my poor German knowledge does not find there a satisfying discussion about roots of polynomials of degree six.

So, where can I find a proof of this statement?  How does this reduction looks like?

For reference, the integral that I'm looking at looks like
$$ \int \frac{R(\tau)}{\sqrt{a_0\tau^6+a_1\tau^5+a_2\tau^4+a_3\tau^3-a_2\tau^2+a_1\tau^1-a_0}}\,d\tau.$$
Similar to Eqs. 579.00 and 581.00 of the handbook, but not exactly the same.  In particular, the polynomial in the square root has three pais of roots, each pair of the form $rr'=-1$.

[1] Byrd, P.F.; Friedman, M.D., Handbook of elliptic integrals for engineers and scientists. 2nd ed., revised, Die Grundlehren der mathematischen Wissenschaften, 67. Berlin-Heidelberg-New York: Springer-Verlag. XVI, 358 p. with 22 fig. (1971). ZBL0213.16602.

[2] Königsberger, L., On the reduction of hyperelliptic integrals to elliptic integrals., Borchardt J. LXXXV, 273-294 (1878). ZBL10.0323.01.

答:
There's no typo in 575.16, for $R(t)\sqrt{P(t)}=R(t)P(t)/\sqrt{P(t)}=S(t)/\sqrt{P(t)}$. Regarding the main question, suppose the $r_i$'s form three pairs of points of an involution (i.e. three points, and the images of those three points under a self-inverse Möbius transformation). Then there is a linear fractional transformation mapping those $r_i$'s onto the points
$$z,\frac1z,ze^{2\pi i/3},\frac1ze^{2\pi i/3},ze^{-2\pi i/3},\frac1ze^{-2\pi i/3}$$
where $z$ is some complex number. When this transformation is applied to the integral, it takes the form $\int R(t)/\sqrt{t^6+at^3+1}\,dt$. The polynomial under root being a palindromic sextic, it can be reduced to elliptic integrals by substituting $u=t+\frac1t$ (B&F 579.01, 581.02):
$$2\int_0^x\frac{R(t)}{\sqrt{P(t)}}\,dt=\int_{1+1/x^2}^{\infty}\frac{R_1(t)+(t+2)R_2(t)}{\sqrt{(t+2)Q(t)}}\,dt+\int_{1+1/x^2}^{\infty}\frac{R_1(t)+(t-2)R_2(t)}{\sqrt{(t-2)Q(t)}}\,dt$$
$$2\int_1^x\frac{R(t)}{\sqrt{P(t)}}\,dt=\int_2^{1+1/x^2}\frac{R_1(t)+(t-2)R_2(t)}{\sqrt{(t-2)Q(t)}}\,dt-\int_2^{1+1/x^2}\frac{R_1(t)+(t+2)R_2(t)}{\sqrt{(t+2)Q(t)}}\,dt$$
where $P(t)=t^3Q(t+1/t)$ and

in the first case, $x\le1$ and $R((t-\sqrt{t^2-4})/2)=R_1(t)+R_2(t)\sqrt{t^2-4}$

in the second case, $x\ge1$ and $R((t+\sqrt{t^2-4})/2)=R_1(t)-R_2(t)\sqrt{t^2-4}$

Since the polynomials under root are now quartics ($Q$ is cubic, the "folding" of palindromic $P$ onto itself), elliptic integrals can be used. I reduced the sextic further than what B&F requires for convenience's sake of the desired proof.

For your specific integral:
$$\int\frac{R(t)}{\sqrt{a_0t^6+a_1t^5+a_2t^4+a_3t^3-a_2t^2+a_1t-a_0}}\,dt$$
the reduction to elliptic integrals requires a complex substitution $u=-it$:
$$=\int\frac{iR(iu)}{\sqrt{-a_0u^6+(a_1i)u^5+a_2u^4-(a_3i)u^3+a_2u^2+(a_1i)u-a_0}}\,du$$
You will have to consider branches carefully, but the reduction is still possible. (Elliptic integrals accept arbitrary complex numbers for their arguments.)


问:
Complete the square for $f(x) = 2x^2 + 4x - 6$
I'm studying for a math test. This is the question:

$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
$$2(x^2 + 2x + 1^2 - 1^2 - 3)$$
$$2((x + 1)^2 - 4)$$

But I get stuck here. Can someone complete it and explain?

I'm sorry i forgot a important piece off the question
it says: Complete the square and derived to get the minimum value off $f(x)$

答:
$f(x) = 2x^2 + 4x - 6$. complete the square.

This is how much I get out of the question:
$$2x^2 + 4x - 6$$
$$2(x^2 + 2x - 3)$$
For ease of understanding at this stage ignore the 2 and consider

$$(x^2 + 2x - 3)$$
Next step
$$(x+1)^2=x^2+2x+1$$
Here we have divided the coefficient of the x term by $\frac{1}{2}$ because when we times out we get the term twice as you have just seen there.

Next note we now have 1 instead of -3 thus to balance this we -4. That is,
$$(x+1)^2+1-4 = (x+1)^2-3$$

Now a good point about completing the square is we can find the minimun point and value easily. Note that with the right hand side RHS we have a squared term which contains the $x$ since all squared numbers are greater than or equal to zero the smallest we can make this term is $0$ and this is done at $x=-1$.

Now we have this point we can substitute $x=-1$ into our earlier equation:

$$2(x^2 + 2x - 3)$$

to get

$$2((-1)^2 + 2(-1) - 3) = -8$$


问:
On the principal value of $\ln(1+z)$
A complex number $a$ is said to be a logarithm of $c$ if $e^a=c$. It follows that the complex number $w=\sum_{n=1}^{\infty}(-1)^{n-1}z^n/n$ satisfies $e^w=1+z$. I want to prove that this number $w$ is the principal value of the logarithm. By principal value of $\ln(x)$ we mean the number $\ln|x|+i\arg(x)$ right? There are also other solutions, specifically those obtained by adding $2k\pi i$ for $k$ integer. Now I want to prove, if I'm not wrong, that $k$ is zero? How can I do it formally? For example, if $z=0$ then $w=0=\ln(1+0)$ implies I think $k=0$. But for all the other values of $z$? Thank you! I'm new to this subject.

答: The Principal logarithm, usually denoted by $Log \,z$, is an  analytic function on $\mathbb C \setminus (-\infty,0]$. In particular it is analytic in the open disk of radius $1$ around $1$. So $Log (1+z)$ has power series expansion around $0$. The successive deriavtives of $Log (1+z)$ are $\frac {(-1)^{n-1} (n-1)!} {(1+z)^{n}}$ so the power series becomes exactly the one you have.

Title: On the principal value of $\ln(1+z)$