问:
Characteristic Property = Universal Property?
Problem

They seem to be the same -almost! But are they really or is it just unlucky accident that they look so similar however describe totally different notions?

Example

I was trying to set the characteristic property of the initial resp. final topology into the framework of initial resp. terminal objects; however, I wasn't able to do so. So my question arose wether they are really the same.

答:
I think there is a subtle difference. In the first place, I want to thank the author John Lee of Introduction to Topological Manifolds, which clarifies the two notions. All the excerpts below are from this book.

The following excerpt gave a definition of a universal property: categorical product.

Then the example (b) in the following excerpt gave an example of a product space in the product topology. It said that the product space is the categorical product in the category Top

The characteristic property of product topology is given below

Here is the point, note the words above the red line in the 2nd excerpt. When we prove the characteristic property of product topology, we use only topological methods. Then, based on this characteristic property, we get that a product space with the product topology satisfies the language of universal property. In other words, it can be interpreted as a universal property in the category theory. In short, we can use characteristic property(a property has nothing to do with category theory), to construct many things, guaranteeing these things satisfying the universal property.


问:
Question on $\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]]$
I have a question on these lecture notes:
http://page.math.tu-berlin.de/~scheutzow/WT3main.pdf Page 46-47 Lemma 4.15

In this proof, we use, amongst other things that:

$\mathbb E_{x}[1_{B_{n-1}}(X_{t_{n-1}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]\lvert \mathcal{F}_{t_{n-2}}]=\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]](*)$

which comes from the Markov Property which states for any and bounded measurable function $f$ we have:

$\mathbb E_{x}[f((X_{t+h})_{t\geq 0})\lvert \mathcal{F}_{h}]=\mathbb E_{X_{h}}[f((X_{t})_{t\geq 0})]$

But my issue in $(*)$ would be that in our case the function $f$ is indeed: $f(X_{t_{n-1}})=1_{B_{n-1}}(X_{t_{n-1}})\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]$ and thus

$\mathbb E_{x}[f(X_{t_{n-1}})\lvert \mathcal{F}_{t_{n-2}}]=\mathbb E_{X_{t_{n-2}}}[f(X_{t_{n-1}-t_{n-2}})]=\mathbb E_{X_{t_{n-2}}}[1_{B_{n-1}}(X_{t_{n-1}-t_{n-2}})\mathbb E_{X_{t_{n-1}-t_{n-2}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]]$

Note the difference in this computation compared to $(*)$. My question is rather why would

$\mathbb E_{X_{t_{n-1}}}[1_{B_{n}}(X_{t_{n}-t_{n-1}})]$ remain the same when evaluated under $\mathcal{F}_{t_{n-2}}$ even though it is a function of $f(X_{t_{n-1}})$. Is this a mistake or am I simply missing something?

答:
The notation is a bit confusing (although, it is typical to these settings). Here is the same proof presented differently. First, for $g\in \text{b}\mathcal{E}$, a transition semigroup $(P_t)_{t\ge 0}$, and $s<t$, the Markov property is
$$
\mathsf{E}_x[g(X_{t})\mid \mathcal{F}_{s}]=\mathsf{E}_{X_s}[g(X_{t-s})]=P_{t-s}g(X_s) \quad(\mathsf{P}_x\text{-a.s.}),
$$
which is a function of $X_s$. Applying this property recursively, for $f_1,\ldots,f_n\in \text{b}\mathcal{E}$, one has

\begin{align}
&\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-1}(X_{t_{n-1}})f_n(X_{t_n})] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-1}(X_{t_{n-1}})\mathsf{E}_x[f_n(X_{t_n})\mid \mathcal{F}_{t_{n-1}}]] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots (f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-1}})] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots f_{n-2}(X_{t_{n-2}})\mathsf{E}_x[(f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-1}})\mid \mathcal{F}_{t_{n-2}}]] \\
&\qquad=\mathsf{E}_x[f_1(X_{t_1})\cdots (f_{n-2}P_{t_{n-1}-t_{n-2}}f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_{n-2}})] \\
&\qquad =\ldots \\
&\qquad=\mathsf{E}_x[(f_1P_{t_2-t_1} f_{2}\ldots P_{t_{n-1}-t_{n-2}}f_{n-1}P_{t_{n}-t_{n-1}}f_{n})(X_{t_1})] \\
&\qquad= P_{t_1}[f_1P_{t_2-t_1}f_2 \ldots P_{t_{n}-t_{n-1}}f_n](x).
\end{align}

For example, in the fourth line we apply the Markov property to
$$
g(x)=(f_{n-1}P_{t_{n}-t_{n-1}}f_n)(x) \in \text{b}\mathcal{E},
$$
evaluated at $X_{t_{n-1}}.$

ContentLicense: CC BY-SA 4.0


问:
For every monoid $M$ with zero is there a group $G$ such that $\mathrm{Grp}(G,G)\cong M$?
All monoids that I will consider here have identities. A monoid $M$ is said to have a zero iff $\exists z\in M \forall x\in M (zx=xz=z)$.

Let $M$ be any monoid with a zero. Must there exist a group $G$ such that $\mathrm{Grp}(G,G)\cong M$ ?

I know that there is no reason to suppose that there must exist such a group. I also expect that there will exist a counterexample, however the fact that I can't find one is irritating me.

Thank you

答:
The group of invertible elements of $\operatorname{Grp}(G,G)$ contains the automorphism group of $G$, and in particular the group of inner automorphisms, which is isomorphic to $G/Z(G)$.

So take the monoid $\{ z \} \cup C_{p}$, where $p > 2$ is a prime, $C_{p}$ is a multiplicative cyclic group of order $p$, and $z$ is the zero.

So $G/Z(G)$ is isomorphic to a subgroup of $C_{p}$, and thus cyclic. Therefore $G = Z(G)$ is abelian. Then $\operatorname{Grp}(G,G)$ is nothing else but the (multiplicative part of the) endomorphism ring of $G$, and the automorphism group of $G$ has odd order $p$.

Now in any abelian group inversion is an automorphism, so inversion must be trivial here, that is, $G$ has exponent $2$. Clearly $\lvert G \rvert = 2$ is impossible, and if $\lvert G \rvert > 2$, then $G$ is a vector space of dimension $\ge 2$ over the field with two elements, and thus swapping two basis elements would be an automorphism of order $2$, a contradiction.

ContentLicense: CC BY-SA 3.0


问:
Only finite group that acts on even spheres is $\mathbb Z_2$
If $G$ is acting freely on $\mathbb S^{2n}$ we can associate to $\rho\in G, \rho\neq0$ it's degree $deg\rho=(-1)^{n+1}$ as any map without fixed points has that degree. This shows that every non-trivial element is homotopic to the antipodal map. Why does that imply that $G\subset \mathbb Z_2$?

答:
Recall that $\{-1, 1\}$ is a group with respect to the standard number multiplication, isomorphic to $\mathbb{Z}_2$.

With that we have the following group homomorphism

$$f:G\to\{-1, 1\}$$
$$f(g)=deg(x\mapsto gx)$$

What you've shown is that $f(g)=-1$ for any $g\neq e$. In particular $\ker f=\{e\}$ and thus $f$ is injective.


问:
$X$ and $Y$ be i.i.d $U(0,1)$ random variables. Then $E(X|X>Y)$

Let $X$ and $Y$ be i.i.d $U(0,1)$ random variables. Then $E(X|X>Y)$
equals?

The way I interpreted it is as follows:

$E(X|X>Y)$ is $E(X)$ realizing $X>Y$

so $E(X)=\int_{0}^{1}\int_{y}^{1}xdxdy=\int_{0}^{1}(\frac{x^2}{2})_{y}^{1}dy=\frac{1}{2}\int_{0}^{1}(1-y^2)dy=\frac{1}{2}(y-\frac{y^3}{3})_{0}^{1}=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$

What can be a better alternative to solve this problem? I tried so hard but couldn't get the answer using the proper definition of expectation, can anyone help me?And can we use law of total expectation here?

答:
This is an answer how to use strictly the proper definition of conditional expectation to justify the interpretation in the question:

The conditional expectation conditioned by the event $\{X>Y\}$ is the conditional expectation conditioned by the $\sigma$-algebra generated by the event: $$\mathcal M:=\sigma(\{X>Y\})=\{\emptyset,\Omega,\{X>Y\},\{X>Y\}^\complement\}.$$ This $E(X|\mathcal M)$ is a $\mathcal M$-measureable random variable satisfying
$$\int_MX\mathrm dP=\int_ME(X|\mathcal M)\mathrm dP
$$ for all $M\in\mathcal M$.
So you have to find out which $\mathcal M$-measureable RV satisfies
$$\int_{\{X>Y\}}X\mathrm dP\color{blue}{=}\int_{\{X>Y\}}E(X|\mathcal M)\mathrm dP,\\
\int_{\{X\leq Y\}}X\mathrm dP\color{blue}{=}\int_{\{X\leq Y\}}E(X|\mathcal M)\mathrm dP,\\
\int_\Omega X\mathrm dP\color{blue}{=}\int_\Omega E(X|\mathcal M)\mathrm dP.
$$
As being $\mathcal M$-measureable it should be constant on the sets $\{X>Y\}$ and $\{X>Y\}^\complement$, i.e. $E(X|\mathcal M)(\omega)=c_1$ for $\omega\in\{X>Y\}$ and $E(X|\mathcal M)(\omega)=c_2$ for $\omega\in\{X\leq Y\}$. So you want to have (calculated for this special case; everything written above holds general)
\begin{align}
\frac{1}{3}=\int_{\{X>Y\}}X\mathrm dP&\color{blue}{=}\int_{\{X>Y\}}c_1\mathrm dP=c_1P(\{X>Y\})=c_1/2,\\
\frac{1}{6}=\int_{\{X\leq Y\}}X\mathrm dP&\color{blue}{=}\int_{\{X\leq Y\}}c_2\mathrm dP=c_2P(\{X\leq Y\})=c_2/2,\\
\frac{1}{2}=E(X)=\int_\Omega X\mathrm dP&\color{blue}{=}\int_\Omega (c_1 \cdot1_{\{X>Y\}}+c_2\cdot1_{\{X\leq Y\}})\mathrm dP\\&=c_1P(\{X>Y\})+c_2P(\{X\leq Y\})=(c_1+c_2)/2.
\end{align}
Usually when you determine this random variable $E(X|\mathcal M)$ you are just interested in the values for $\omega\in\{X>Y\}$, since this is the inital intention of $E(X|X>Y)$. But to write it completely
\begin{align}
E(X|\mathcal M)(\omega)=\begin{cases}c_1=\frac{2}{3}, &\text{for }\omega\in\{X>Y\}\\
c_2=\frac{2}{6}=\frac{1}{3}, &\text{for }\omega\in\{X\leq Y\}.
\end{cases}
\end{align}


问:
Finding the amount of time it takes to complete a job
I was given the following problem:

Of three workmen, the second and third can complete a job in 10 days. The first and third can do it in 12 days, and the first and second can do it in 15 days. In how many days can each of them do the job alone?

This is my work:

Let a = the first workman, b= the second workman, and c = the third workman.
$$\frac{a+b}{2}=15\rightarrow a+b=30\ \rightarrow a=30-b\\
 \frac{b+c}{2}=10\rightarrow b+c=20\rightarrow c=20-b\\
 \frac{a+c}{2}=12\rightarrow a+c=24$$
Substitute using above identities: $30-b+20-b=24$
$$50-2b=24\\
  26=2b\\
  b=13$$
Substituting in the value for b in the first two equations gives: a=17 and c=7.
Solution: The first workman takes 17 days to do the job. The second workman takes 13 days, and the third takes 7.

The obvious problem with this is that it can't be that when working together it can take one of the workmen longer to complete the work than when he works alone. However, this is what happens with workman c. And so obviously this technique does not work. Why not?

*Edit: My thought now is to divide each sum by four. This is because when they work together we need to half BOTH of their times. Otherwise if two people that take the same amount of time work together, they won't go any faster than usual. Does this make sense?

答:
Assume that they each need $a$, $b$ and $c$ days individually.

Let $W$ be the total amount of work. Then
$$
\frac{W}{W/b+W/c}=10,\quad \frac{W}{W/a+W/b}=15,\quad \frac{W}{W/a+W/c}=12
$$

The "formula" you need is that
$$
\frac{\textrm{total amount of work}}{\textrm{"work efficiency"}}=\textrm{time}
$$


问:
An $\omega$-stable theory that is not uncountably categorical
Morley's theorem states that a countable complete theory that's $\omega$-stable and has no Vaughtian pair is uncountably categorical. From the statement, I guess the "no Vaughtian pair" condition is necessary. In searching for examples of theories that are $\omega$-stable (with Vaughtian pair) but not uncountably categorical, I stumbled upon this article. An example is given in Example 3.1.3, namely the theory in the language $\{E\}$ saying that $E$ is an equivalence relation and that for each natural number $n$, there is one $E$-class of size $n$. However, I don't understand the argument in the article, so I might just ask for an explanation here.

Why is this theory $\omega$-stable but not uncountably categorical?

答:
Call this theory "$T$."

Showing that $T$ is not uncountably categorical is straightforward. Keep in mind that a model of $T$ is gotten by adjoining to the prime model (= one equivalence class of size $n$ for each finite $n$) an arbitrary number (possibly $0$) of equivalence classes of arbitrary infinite cardinalities. Now given $\kappa$ infinite, the following are some examples of non-isomorphic models of $T$ with cardinality $\kappa$:

Exactly one class of size $\kappa$, no other infinite classes.

Exactly two classes of size $\kappa$, no other infinite classes.

For $\kappa>\aleph_0$: exactly one class of size $\kappa$, exactly one class of size $\aleph_0$, no other infinite classes.

And so on. Personally I think the last example is especially important; it's good to keep in mind that similar-looking "pieces" of the model (e.g. infinite classes) can behave very differently.

As for $\omega$-stability, let's start by considering the no-parameters case: can we show that there are only countably many $1$-types over $\emptyset$?

(As usual we're working in some "sufficiently saturated" monster model $\mathfrak{M}$ here.)

Well, intuitively the $1$-type over $\emptyset$ of some element $a$ is determined by a simple cardinal invariant $size(a)$:

 Take $size(a)$ to be the size of the equivalence class $a$ lives in.

After proving that in fact $size(a)=size(b)$ implies $tp_\emptyset(a)=tp_\emptyset(b)$, we can now count the number of possible values of this invariant within the monster model $\mathfrak{M}$:

 Either $size(a)$ is finite, or by saturation $size(a)=\vert\mathfrak{M}\vert$. So there are $\aleph_0$-many possible values of $size(a)$.

We now need to think about folding in parameter sets larger than $\emptyset$ and tuple lengths larger than $1$ - that is, counting $n$-types over $A$ for arbitrary finite $n$ and countable $A\subseteq\mathfrak{M}$. EDIT: not really, see Alex Kruckman's comment below; I'm making things harder than they need to be here. This gets a bit tedious, but the basic idea is still the same: we want to show that there are only a few different possible behaviors for a given tuple.

For example, here's how to count the $2$-types over a one-element parameter set $\{c\}$. The type of a pair $(a_0,a_1)$ in $\mathfrak{M}$ over $\{c\}$ is determined by the following data.

First, we have the "$\emptyset$-part" of each individual element of the pair $(a_0,a_1)$:

 What are the cardinalities $\alpha_0$, $\alpha_1$ of the classes of $a_0,a_1$ in $\mathfrak{M}$? As before, there are $\aleph_0$-many possibilities here.

Next, we consider the structure within the tuple $(a_0,a_1)$:

 Are $a_0$ and $a_1$ equal? Are $a_0$ and $a_1$ equivalent? There are finitely many possibilities here.

Finally, we bring the parameter into the question:

 How do $a_0$ and $a_1$ compare with $c$, both with respect to equality and equivalence? Again, there are only finitely many possible behaviors.

Putting all this together we get as hoped for only countably many $2$-types over a one-element parameter set in $\mathfrak{M}$. And it should be reasonably clear how to continue this to get $\omega$-stability.


问:
Finding a continuous function satisfying $f^{(k)}(k)=k$ for any whole number $k$.
I am having some knowledge in ordinary differential equations. I think my knowledge is not low, but average. However, I could not use my knowledge to solve, or even to start solving the following problem;

Find a continuous function, $f(x)$, that satisfies $f^{(k)}(k)=k$ for any whole number $k$, i.e. $f(0)=0, f'(1)=1, f''(2)=2$, and so on.

I am not sure if that is a difficult problem, an average-difficulty problem, a simple problem, or an impossible problem.

If this problem can be solved, where the function is expressed as infinite sum, then is it possible if we have the restriction that $f$ has finitely many terms?

I had not have this question from a book or exam, but it popped up into my mind, and I am keen to get to the solution.

Forgive my poor English, hopefully my problem is clear.

I am not asking for a solution, but just asking for the way to start.

Your help is highly appreciated. THANKS!

答:
Not quite the same function but we can solve $f^{(k)}(a)=k$ by using taylor series:
$$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$
then using the requirements above we get:
$$f(x)=\sum_{n=1}^\infty\frac{(x-a)^n}{(n-1)^!}$$
$$f(x)=(x-a)\sum_{n=1}^\infty\frac{(x-a)^{n-1}}{(n-1)!}=(x-a)e^{x-a}$$

maybe you can use taylor series or some form of approximation to line it up into a useable form e.g.
$$f(0)=0$$
$$f'(1)=1=f'(0)+...$$


问:
A so-called total-variation inequality

Given real polynomial $P\left ( x \right )$ so that $P\left ( 0 \right )= P\left ( 1 \right )= 0, \int\limits_{0}^{1}\left | {P}'\left ( x \right ) \right |{\rm d}x= 1.$ For $x\in\left ( 0, 1 \right )$ prove or disprove that $\left | P\left ( x \right ) \right |< \frac{1}{2}.$

I read on Twitter a fleet that proves the following inequality with strictly decreasing $f\left ( x \right )$ for all $x\in \left [ 0, 1 \right ]$ by Cauchy-Schwarz
$$\int_{0}^{1}f\left ( x \right ){\rm d}x+ \left ( \int_{0}^{1}xf\left ( x \right ) \right )^{2}+ 1> 0$$
I'm trying to find a strictly decreasing function from the Original Problem.. of course there exists no such $P\left ( x \right ), {P}'\left ( x \right ).$ Then replace it for $f\left ( x \right ).$

答:
As $P(0) = 0$, we get $P(x) = \int_0^x P'(t) \, \mathrm{d}t$. Moreover, $0 = P(1) = P(x) + \int_x^1 P'(t) \, \mathrm{d}t$. Taking absolute values and adding these two equations up, we get
$$2|P(x)| \leq \int_0^x |P'(t)| \, \mathrm{d}t + \int_x^1 |P'(t)| \, \mathrm{d}t = 1$$
and therefore $|P(x)| \leq \frac{1}{2}$.

Title: A so-called total-variation inequality


问:
Fundamental group of wedge sum not equal to the free product
Wedge sum's Wikipedia article states that:

"Van Kampen's theorem gives certain conditions (which are usually fulfilled for well-behaved spaces, such as CW complexes) under which the fundamental group of the wedge sum of two spaces X and Y is the free product of the fundamental groups of X and Y."

Is there a counterexample to the general statement, for path-connected spaces?

Namely - two path-connected topological spaces $X, Y$ such that $\pi_1(X \vee Y) \ne \pi_1(X)\ *\ \pi_1(Y)$

答:
Observe that the Hawaiian Earrings ${\mathbb H}$ is homeomorphic to ${\mathbb H}\vee {\mathbb H}$. On the otber hand, it is proven by Sam Corson (see the answer here) that if $G=\pi_1({\mathbb H})\cong G_1\star G_2$, then one of the free factors is finitely generated. Since $G$ is not finitely generated, you get your example.

Edit. Incidentally, it is a nice exercise to find example of a loop in ${\mathbb H}$ which is not homotopic to a (finite) product of loops contained the summands of the wedge sum decomposition ${\mathbb H}\vee {\mathbb H}$.


问:
Calculate this limit without using L'Hôpital rule
the limit is, I can't find where to start

$$\lim_{x\rightarrow 0} \frac{\arcsin\frac{x}{\sqrt{1-x^{2}}} }{\ln (1-x)}$$

your help is appreciated

Body:
$$ \lim_{x \to 0} \frac{ \arcsin (x) }{ x} = 1$$ $$\lim_{x \to 0} \frac{x}{\ln (1-x)} = -1$$

Then:

$$
\begin{split}
\lim_{x\rightarrow 0} \frac{\arcsin\frac{x}{\sqrt{1-x^{2}}} }{\ln (1-x)}
 &= \lim_{x\rightarrow 0} \frac{x}{\ln(1-x)} \cdot \lim_{x \to 0} \frac1x\arcsin\frac{x}{\sqrt{1-x^{2}}}  \\
 &= - \lim_{x \to 0} \frac{\frac{1}{\sqrt{1-x^2}}}{x \frac{1}{\sqrt{1-x^2}}} \cdot\arcsin\frac{x}{\sqrt{1-x^{2}}} \\
 &= - \lim_{x \to 0} \frac{1}{\sqrt{1-x^2}} = -1
\end{split}
$$

Proof of the second limit:

$$ \lim_{x \to 0} \frac{x}{\ln(1-x)} = \lim_{x \to 0} \frac{1}{\frac1x \cdot \ln(1-x)} = \lim_{x \to 0} \frac{1}{\ln( (1-x)^{\frac1x})} \equiv \frac{1}{\ln(e^{-1})} = -1$$


问:
Nolinear system of equations.
$$
\left\{\begin{array}{l}
x^y = y^x \\
 x-y\cdot\log_xy=(x+y)\cdot\log_xy
\end{array}
\right.
$$

Thanks for your time!

答:
With rules for logarithms the second equation translates to
$$
x - y\log_xy = \log_x y^x + y \log_x y
$$
and using the first equation this simplifies to
$$
x= y+2y (1+\log_xy)
$$
and then
$$
\log_xy = \frac{x-y}{2y}
$$
From first equation you can express
$$
y = x^{\frac{y}{x}}
$$
implying
$$
\frac{y}{x} = \frac{x-y}{2y}
$$
which translates to
$$
x^2 -xy - 2y^2 = 0
$$
and factoring this you get
$$
(x+y)(x-2y)=0
$$
that gives $2$ solutions which, as usual with equation with logarithms, need to be checked. Try.</p>

Title: Nolinear system of equations.


答:
You want to set up your equations in terms of work rate. To make life easier, I'll pretend that the job is painting one house.

That is, suppose that worker 1 can paint one house in $a$ days. This means that they have a work rate of $(1/a)$ houses per day. And similarly say worker 2 has work rate $(1/b)$ houses per day.

Now, the work rate of a team is the sum of the individual work rates. Therefore your first equation should be
$$\left(\frac 1a + \frac 1b\right) \text{ [houses / day] } * 15 \text{ [days] }  = 1 \text{ [houses]}$$

Title: Finding the amount of time it takes to complete a job

Body:
I was given the following problem:

Of three workmen, the second and third can complete a job in 10 days. The first and third can do it in 12 days, and the first and second can do it in 15 days. In how many days can each of them do the job alone?

This is my work:

Let a = the first workman, b= the second workman, and c = the third workman.
$$\frac{a+b}{2}=15\rightarrow a+b=30\ \rightarrow a=30-b\\
 \frac{b+c}{2}=10\rightarrow b+c=20\rightarrow c=20-b\\
 \frac{a+c}{2}=12\rightarrow a+c=24$$
Substitute using above identities: $30-b+20-b=24$
$$50-2b=24\\
  26=2b\\
  b=13$$
Substituting in the value for b in the first two equations gives: a=17 and c=7.
Solution: The first workman takes 17 days to do the job. The second workman takes 13 days, and the third takes 7.

The obvious problem with this is that it can't be that when working together it can take one of the workmen longer to complete the work than when he works alone. However, this is what happens with workman c. And so obviously this technique does not work. Why not?

*Edit: My thought now is to divide each sum by four. This is because when they work together we need to half BOTH of their times. Otherwise if two people that take the same amount of time work together, they won't go any faster than usual. Does this make sense?


问:
How to compare log and exponential

$x^{1/3}$

$　 (\log_{2}{x})^3$

How to compare them?

Title: How to compare log and exponential

Body:
Cubing, consider
$$f(x)=x-\frac{\log ^9(x)}{\log ^9(2)}$$
$$f'(x)=1-\frac{9 \log ^8(x)}{x \log ^9(2)}$$
$$f''(x)=\frac{9 (\log (x)-8) \log ^7(x)}{x^2 \log ^9(2)}$$
The first derivative cancels at two  points
$$x_{1,2}=\exp\Big[-8 W(\pm k) \Big]\qquad \qquad k=\frac{\log ^{\frac{9}{8}}(2)}{8 \sqrt[4]{3}}$$ where $W(.)$ is Lambert function (numerically, $x_1=0.622419$ and $x_2=1.71274$).
$$f(x_1) \sim 0.655209 \qquad \qquad  f''(x_1)=+28.7147$$
$$f(x_2) \sim 1.610341 \qquad \qquad  f''(x_2)=-8.09651$$

So, $x_1$ corresponds to a minimum and $x_2$ to a maximum.

Now, by inspection, $f(2)=1$. So, using Newton method, the iterates will be
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 2.00000 \\
 1 & 2.18208 \\
 2 & 2.13183 \\
 3 & 2.12495 \\
 4 & 2.12483
\end{array}
\right)$$
Then $f(x) >0$ for any $x < 2.12483$


问:
Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$
Prove that $(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}, \forall$ n, k nonnegative integers, $k\le n$.

I know that has something to do with Bernoulli's inequality $(1+\alpha)^x\ge 1+\alpha x, \alpha \ge -1, n\ge1$.

If I reconsider Bernoulli's inequality with $\alpha=\frac{1}{n}$ and $x=k$ it follows that

$(1+\frac{1}{n})^k\ge 1+\frac{k}{n}$, but I don't know how to continue.

I also tried to prove it with induction where I consider
$p(n):(1+\frac{1}{n})^k < 1+ \frac{k}{n}+\frac{k^2}{n^2}$ to be true and prove that $ p(n+1):(1+\frac{1}{n+1})^k < 1+ \frac{k}{n+1}+\frac{k^2}{(n+1)^2}$ to be also true, but it didn't work.

Thank you!

答:
Proof without induction/calculus:

It's easy to prove the inequality is true for $k=1,2$. If $k\ge 3$ we have

$$\left(1+\frac{1}{n}\right)^k - 1- \frac{k}{n}-\frac{k^2}{n^2} $$ $$= - \frac{k^2}{n^2} + \binom{k}{2} \frac{1}{n^2} + \sum_{i=3}^k \binom{k}{i} \frac{1}{n^i} = -\frac{k(k+1)}{2n^2}+\sum_{i=3}^k \binom{k}{i} \frac{1}{n^i} \tag1$$

Notice that
$$\frac{\binom{k}{i+1}/n^{i+1}}{\binom{k}{i}/n^i}=\frac{k-i}{n(i+1)} \le \frac{k-3}{4n}, \forall i \ge 3$$

Then $(1)$ is less then
$$-\frac{k(k+1)}{2n^2} + \frac{\binom{k}{3}/n^3}{1-\frac{k-3}{4n}}$$ and it suffices to prove
$$ \frac{\frac{k(k-1)(k-2)}{6n^3}}{1-\frac{k-3}{4n}} < \frac{k(k+1)}{2n^2}$$
Or equivalently
$$\frac{4(k-1)(k-2)}{4n-k+3} < 3(k+1) \iff 4(k-1)(k-2) < 3(k+1)(4n-k+3)\\
\stackrel{k\le n}{\Leftarrow} 4(k-1)(k-2) < 3(k+1)(4k-k+3) = 9(k+1)^2$$ which is trivially true.


问:
Why do we need to consider $ \hat{0} = \hat{m} $ separately in the proof for the union of all residue classes being $ \mathbb{Z} $?
The book Introduction to Analytic Number Theory by Apostol states in Theorem 5.10 (c):

The $ m $ residue classes $ \hat{1}, \hat{2}, \dots, \hat{m} $ are disjoint and their union is the set of all integers.

The proof goes like this:

Hence by part (b) the residue classes
$$ \hat{0}, \hat{1}, \hat{2}, \dots, \widehat{m - 1} $$
are disjoint. But every integer $x$ must be in exactly one of these classes because $ x = qm + r $ where $ 0 \le r < m $, so $ x \equiv r \pmod{m}$ and hence $ x \in \hat{r} $. Since $ \hat{0} = \hat{m} $ this proves (c).

Why is the last sentence necessary for the proof. What if we did not write "Since $ \hat{0} = \hat{m} $ this proves (c)."? Will the proof remain incomplete then and why?

答: The statement is that the classes $[1], \cdots, [m]$ are disjoint, but the author actually proved that $[0], \cdots [m - 1]$ are disjoint. So to connect these two they observed that $[0] = [m]$ and so the two sets of residue classes are the same.


问:
Integers $n$ for which $|2^n + 5^n – 65|$ is a perfect square

Find the sum of all positive integers $n$ for which $|2^n + 5^n– 65|$ is a perfect square.

It is easy to check for $n<3$ only $n=2$ works. Looking at $n≥3$, $2^n + 5^n– 65= a^2$
For odd $n$, LHS has unit digit $2$ or $8$ but RHS has unit digit $0,4$ or $6$. Hence $n$ cannot be odd. For even $n$ we can rewrite our expression as follows ($n=2k$):  $$(a – 2^k)(a+2^k)=5(5^{2k–1} – 13)$$

But I wasn't able to finish off after this. Both LHS and RHS have the same unit digit ($0$). Through trial and error I found that $n=4$ works, but $n=6,8$ don't. Since the questions asks for the sum as is common in Olympiad number theory it's highly unlikely that any other larger values satisfy. I also tried some alternative ways like looking at the equation modulo $4$ but that didn't help.

答:
NOTE: This is not a full answer and is intended to give a boost.

Taking modulo $4$,
$$a^2-(0)^k\equiv 0\ (\text{mod}\ 4)$$
$$a\equiv0\ (\text{mod}\ 4)$$
Taking modulo $3$,
$$a^2\equiv(-1)^n+(-1)^n-2\equiv 0\ (\text{mod}\ 3)$$
$$a\equiv0\ (\text{mod}\ 3)$$
Therefore, $a$ must be divisible by $12$.

Hope this part helps :)

EDIT: @CalvinLin has a wonderful (intended too I guess), so this will be left as intact, or if I should delete this, tell me.


问:
Vector Line Integral Question
I need to compute the line integral for the vector $\vec{F} = \langle x^2,xy\rangle$, for the curve specified: part of circle $x^2+y^2=9$ with $x \le0,y \ge 0$,oriented clockwise.

Once again, I'm stuck at the setup (this happens a lot with me).  I know that I need to parameterize F, but how would I go about doing this exactly?

答: We have $$F(x,y)=x^2\textbf{i}+xy\textbf{j}$$ and $$C: x^2+y^2=1, x\le0, y\ge0$$ and want to evaluate $$\oint_CF\cdot dr=\int_{\pi}^{\pi/2} F(\cos t,\sin t)\cdot(-\sin t,\cos t)dt=\int_{\pi}^{\pi/2}(\cos^2 t,\sin t\cos t)\cdot(-\sin t,\cos t)dt\\\\\\ =\int_{\pi}^{\pi/2}(-\sin t\cos^2 t+\sin t\cos^2 t)dt=0$$

Title: Vector Line Integral Question


问:
How to find probability in binomial distribution with n, p and q?
Consider this question:

Screws are sold in packets of 15. Faulty screws occur randomly. A large number of packets are tested
for faulty screws and the mean number of faulty screws per packet is found to be 1.2.
Damien buys 8 packets of screws at random.
Find the probability that there are exactly 7 packets in which there is at least 1 faulty screw.

With this information given, I found out the Variance = 1.104

The probability = p = 0.08

The probability that a packet contains at most 2 faulty screws = 0.887. (I found this because it was a subpart of the question...)

How do I go about solving this question? Any help is appreciated

答:
Let $p$ be the probability of a screw being faulty. Then (as you have already calculated)
\begin{align}
p &= \frac{1.2}{15} = 0.08\\
1 - p &= 0.92
\end{align}

Probability of finding no faulty screws in a packet is $15 \choose 0$$(0.92)^{15}$. Hence the probability of finding at least faulty screw in a packet is $1-$$ 15 \choose 0$$(0.92)^{15}$.

Coming to your question, I am a bit confused, but this is what I think -

From 8 packets, we can choose 7 packets in $8\choose7$ ways. The probability of these packets having at least one faulty screw is $1-$$ 15 \choose 0$$(0.92)^{15}$ (as before). The remaining packet must have no faulty screws, with probability $15 \choose 0$$(0.92)^{15}$. Hence the total probability is - $8\choose7$$(1-$$ 15 \choose 0$$(0.92)^{15})$ + $15 \choose 0$$(0.92)^{15}$.


问:
Does anyone have an idea how to integrate this functions?

$$\int \frac{1}{x\sqrt{x^2+x+1}}\mathrm{d}x\; ;$$
$$\int\frac{6\sin(2x) + 6 + \ln(x^6)}{x\ln(x) + \sin^2(x)}\mathrm{d}x$$

I have done the first one using Euler's substitution, but I am not sure if it is correct,  I have used the substitution for $a>0$, for the second one I have no idea how to start?

答:
$$I=\int \frac{1}{x\sqrt{x^2+x+1}}\mathrm{d}x$$
Let $x=\dfrac{1}{t} \implies dx=-\dfrac{dt}{t^2}$
$$I=-\int \frac{dt}{\sqrt{t^2+t+1}}=-\int \frac{dt}{\sqrt{\left(t+\dfrac12 \right)^2+\dfrac34}} $$ $$=-\ln\left[t+\frac12+\sqrt{t^2+t+1}\right]+C, ~ \text{where} ~t=1/x$$

$$J=\int\frac{6\sin(2x) + 6 + \ln(x^6)}{x\ln(x) + \sin^2(x)}\mathrm{d}x$$
Let $$x\ln x+\sin^2 x=t \implies  (\ln x+1+\sin 2x)dx=dt$$
$$\implies J=\int \frac{6dt}{t}=6\ln (x\ln x+\sin x)+C$$


问:
How can I calculate infinite series $\frac{1}{n.2^n}$ for $n > 2$?
a few days ago I was confronted with this series, and my attempts to solve it by raising exponential powers failed. I know I should try with exponentials but dont know how!

$$A=\sum_{n=3}^{\infty}\frac{1}{n.2^n}$$
$$e^A =\prod_{n=3}^{\infty}e^\frac{1}{n.2^n}$$

any hint would be appreciated.

Body:
Hints:

$\begin{align}\left(\sum_{n=3}^{\infty} \frac{2^{- n}}{n}\right)=\left(\sum_{n=1}^{\infty} \frac{2^{- n}}{n} + \sum_{n=1}^{2} - \frac{2^{- n}}{n}\right)\end{align}$

Or,

$\begin{align}\left(\sum_{n=1}^{2} - \frac{2^{- n}}{n}\right) + \sum_{n=1}^{\infty} \frac{2^{- n}}{n}=\left(- \frac{5}{8}\right)+ \sum_{n=1}^{\infty} \frac{2^{- n}}{n}\end{align}.$

$\sum_{n=1}^{\infty} \frac{2^{- n}}{n}$ is a known series as $\text{ln}(2)$.


问:
When is $AB=I$ for not necessarily square matrices $A$, $B$?
I have the following matrix equation:

$$
\begin{pmatrix}
0 & 3 & 0 & 0 \\
0 & 0 & 4 & 0 \\
0 & 0 & 0 & 5
\end{pmatrix} \cdot \begin{pmatrix}
0 & 0 & 0 \\
\frac 1 3 & 0 & 0 \\
0 & \frac 1 4 & 0 \\
0 & 0 & \frac 1 5
\end{pmatrix} = \begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
$$

Is there a special name or adjective to describe the relationship between $A$ and $B$?

I also want to know, except orthogonal and unitary matrices, with what else matrices can a identity matrix be produced?

Many thanks!

答:
If $A$ is an $n\times m$ matrix and $B$ an $m\times n$ matrix, we say $A$ is a left inverse of $B$ and $B$ is a right inverse of $A$ if $AB=I_n$. In this situation, it is not necessarily true that $BA=I_m$ as well, as you can verify in your example.

When $n=m$, we have $AB=I_n$ if and only if $BA=I_n$ and call $A$ the inverse of $B$ and $B$ the inverse of $A$, since in this case there is only one possible (both sided) inverse matrix. The inverse is then denoted by $B=A^{-1}$, $A=B^{-1}$.

In the case $n\neq m$, left and right inverses are not unique, that is, $AB=I_n$ and $A'B=I_n$ does not imply $A=A'$ (and the same for right inverses).

Given an $n\times m$ matrix $A$, we can find a $B$ such that $AB=I_n$ if and only if the rank of $A$ is equal to $n$.

Given an $m\times n$ matrix $B$, we can find $A$ such that $AB=I_n$ if and only if the kernel of $A$ is trivial, that is, the nullity of $A$ is equal to $0$.


问:
$\rho(A \circ A) \ge \rho(A)^2.$
Suppose we have a nonnegative matrix $A$. Can we claim the following:
$$\rho(A \circ A) \ge \rho(A)^2.$$ Here $A \circ A$ is the Hadamard product.

Context: For a nonnegative matrix $A$ I know the spectral radius. Can we use that to give a bound for the spectral radius of $A \circ A$.

I computationally verified for a few examples. It seems to be true.

答:
If $ \rho(A)$ denotes the spectral radius, then no. Take $A = \begin{bmatrix} 1 &2 \\ 2 &1 \end{bmatrix}.$ Then we have

$\rho(A)^2=9$ and $\rho(A \circ A)=5.$

Title: $\rho(A \circ A) \ge \rho(A)^2.$


问:
A bounded set is contained in an open ball
Given metric space $ (X, d) $ .

Statement : A set $ E \subset X $ is bounded iff $ \exists $ $ x \in E $ and $ r \in R^{+} $ such that $ E \subset B_{r}(x,d) $.

In the above statement, the set  $ B_{r}(x,d) $ is an open ball.

Is $ B_{r}(x,d) = \{y \in X | d(x, y) < r \}  $ or $ B_{r}(x,d) = \{ y \in E | d(x, y) < r \}  $ , in context of the above statement ?

答:
The correct characterization of a bounded set in a metric space $(X,d)$ is:

$E \subset X$ is bounded iff there exists an $r>0$ and a point $x \in E$ (can be also just $x \in X$) such that $E \subset B_r(x) = \{ y \in X : d(x,y)<r \}$. I.e. $E$ must be contained in some open ball.


问:
Area of a Semi-Regular Decagon
I'm trying to calculate the area of the floor plan of an oddly shaped cabin my in-laws own.

The cabin is a semi-regular decagon with sides 10 and 20 alternating.

I recognize that if I could solve for the distance from the center to the mid-point of the sides (apothem) then this becomes pretty easy by decomposing the problem into 2 sets of 5 equal triangles.

Edit: Thanks for the help here. My previous assertion on the bounding of the area was incorrect so I have removed it.

答:
Assuming the decagon is circumscribable (i.e. there's a circle going through all its vertices)... Just find the area of a triangle with sides $a=20$, $b=10$ and angle between these two sides of $144$ degrees. Also find the third side $c$ of this triangle (e.g. by using cosine theorem). Then your decagon is simply composed of 5 such triangles plus a regular pentagon with side $c$. So finding its area presents no problems.

I can try to make a drawing too.

Title: Area of a Semi-Regular Decagon


问:
Show that $\frac{1}{x^{1/p}(\ln(2/x))^{2/p}}$ is only in $L^p(0,1)$
Problem: Let $\mu$ be the Lebesgue measure.
Define $f:(0, 1) \to \mathbb{R}$,
$$
f(x) = \frac{1}{x^{1/p}(\ln(2/x))^{2/p}}.
$$
Show that $f \in L^p(0, 1)$. Show  for every $q>p$ that $f \not\in L^q(0, 1)$.

The first claim is very easy as it is just a simple calculation. But I can't show the second. I have tried to come up with different kinds of lower bounds for $f$ like $\frac{1}{x\ln(2/x)}$, but I can't prove this for all $q>p$.

答: Let $k=q/p$ and $u=\ln(2/x)$. Then $$\|f\|_q^q=\int_0^1\frac{dx}{\left(x\ln^2\frac2x\right)^{q/p}}=\int_\infty^{\ln2}\frac{-2e^{-u}\,du}{(2e^{-u}u^2)^k}\propto\int_{\ln2}^\infty\frac{e^{u(k-1)}}{u^{2k}}\,du.$$ Whenever $k>1$ the integrand is monotonically increasing as $u\to\infty$.


问:
Solving a matrix equation $AX=XB$ in a CAS
I have the following computational problem. Let $N$ be a positive integer and $A\in \mathbb{C}^{2N\times 2N}$, $X\in \mathbb{C}^{2N\times 4}$ and $B\in  \mathbb{C}^{4\times 4}$. I want to solve the following equation for the matrix $X$
$$
A\cdot X= X\cdot B.
$$
I found the following paper online (http://www.jstor.org/stable/2099652) but I haven't been able to implement it in Maple or Magma and I don't know what I should be looking for.
Does someone know if it is implemented somewhere already?</p>

答:
This is a Sylvester equation, $AX-XB = 0$. This can be solved in Maple by SylvesterSolve and in Matlab by lyap and in Mathematica by LyapunovSolve.

In general, you can get $X$ in a vectorized form by solving
$$((I_4 \otimes A) - (B^T \otimes I_{2n})) \operatorname{vec} X = 0$$
for example by Gaussian elimination. $\otimes$ is the Kronecker product and $\operatorname{vec} X$ is the columns of $X$ put on top of each other in a vector.


问:
In the construction of Brownian Motion: Show that the increment vector is a Gaussian vector
In constructing a Brownian motion in the book with the same name by Mörters and Peres, they come to a point on page 10 where they want to show that $\{B(d)-B(d-2^{-n})\}_{d\in \mathcal{D}_n}$ are independent. Here $\mathcal{D}_n = \{i/2^n, i = 0,\dotsc, 2^n\}$ and $B$ is constructed iteratively so that on step $n$ $B(d)-B(d-2^{-n})$ has normal distribution with mean $0$ and variance $2^{-n}$. They say:

that it suffices to show that they are pairwise independent, as the vector of these increments is Gaussian.

I see how independence follows if $X = (B(d_1)-B(d_1-2^{-n}), B(d_2)-B(d_2-2^{-n}),\dotsc, B(d_{\mathcal{|D_n|}}-B(d_{|\mathcal{D}_n|}-2^{-n})))$ is shown to be a Gaussian vector. But for this to be the case we need to find a standard Gaussian vector $Y$, a matrix $A$ and a constant vector $b$ such that $X = AY+b$. I'm wondering how this can be done.

答:
$A$ is a diagonal matrix with diagonal entries ${2^{-1/2},...,2^{-n/2}}$.

The variance of $X=AY$ is $YY'$, a diagonal matrix with diagonal entries ${2^{-1},...,2^{-n}}$.


答:
Per the Euler substitution $\sqrt{x^2+x+1}=x+t$
$$\int \frac{1}{x\sqrt{x^2+x+1}}{d}x=2\int \frac1{t^2-1}dt=\ln\frac{t-1}{t+1}+C
$$

Title: Does anyone have an idea how to integrate this functions?

Body:
$$\int \frac{1}{x\sqrt{x^2+x+1}}\mathrm{d}x\; ;$$
$$\int\frac{6\sin(2x) + 6 + \ln(x^6)}{x\ln(x) + \sin^2(x)}\mathrm{d}x$$

I have done the first one using Euler's substitution, but I am not sure if it is correct,  I have used the substitution for $a>0$, for the second one I have no idea how to start?


问:
Pullback of hermitian line bundle
let $f:X\to Y$ be a holomorphic map between complex manifolds. Let $(L,h)$ be a hermitian line bundle on $Y$. Namely $L$ is a holomorphic line bundle on $Y$ and $h$ is a $C^{\infty}$-hermitian inner product on $L$.

What is the pullback $f^\ast(L,h)$? I guess that it should be a hermitian line bundle of the form $(f^\ast L, f^\ast h)$. I have no problem in the definition of $f^\ast L$, but what is $f^\ast h$?

In other words assume that $s$ and $t$ are local sections of $f^\ast L$, how is it defined $f^\ast h\,\big(s(x),t(x)\big)$? As far as I know it is not true that any local section of $f^\ast L$ is of the form $r\circ f$ for a local section $r$ of $L$.

答:
Note that the formal definition of the pullback of $\pi : L \to Y$ by $f : X \to Y$ is as follows: the total space is given by $f^*L = \{(x, \ell) \in X\times L \mid f(x) = \pi(\ell)\}$ and the projection $\pi' : f^*L \to X$ is given by projection onto the first factor; i.e. $\pi'(x, \ell) = x$. There is also a map $F : f^*L \to L$ given by projection onto the second factor, i.e. $F(x, \ell) = \ell$. These maps give rise to a commutative square:

$$\require{AMScd}
\begin{CD}
f^*L @>{F}>> L\\
@V{\pi'}VV @VV{\pi}V \\
X @>{f}>> Y
\end{CD}$$

That is, $\pi\circ F = f\circ\pi'$.

Note that for $x_0 \in X$, we have $$(f^*L)_{x_0} = \pi'^{-1}(x_0) = \{(x_0, \ell) \in X\times L \mid f(x_0) = \pi(\ell)\} \cong \pi^{-1}(f(x_0)) = L_{f(x_0)}.$$

The pullback metric $f^*h$ on $(f^*L)_{x_0}$ is just the original metric $h$ on $L_{f(x_0)}$. More precisely, $$(f^*L)_x(s(x), t(x)) = h_{f(x)}(F(s(x)), F(t(x))).$$

Title: Pullback of hermitian line bundle


问:
What's difference between Penrose Arc Decoration and Ammann Line Decoration for tiling kite and dart?
Penrose Arc for edge tiling rule

Ammann line for edge tiling rule

The result seemingly the same. It could be use to join the tile to be penrose tiling. So what is the difference between these two method? I try to look for detail but only could conclude that Ammann decorated line is somehow stronger rule or something like that. Is it true and why it is?

答: They give the same tilings, but the decorations are highlighting different aspects of the global structure of the tilings (radial vs rotational structures, which are essentially orthogonal to one another).</p>


问:
Best way to write the characteristic polynomial
In linear algebra I've seen that there are two major different ways
to write the characteristic polynomial of a mapping $f$: As
$$
p_{f}=\pm\left(T-\lambda_{1}\right)^{m_{1}}\cdot\ldots\cdot\left(T-\lambda_{k}\right)^{m_{k}}
$$
with distinct $\lambda_{i}$ or as
$$
p_{f}=\pm\left(T-\lambda_{1}\right)\cdot\ldots\cdot\left(T-\lambda_{n}\right)
$$
where the $\lambda_{i}$ are not necessarily distinct.

The only advantage I can see the latter expression has over the former
is that the eigenvalues are already given in an order from $1$ to
$n$, so it's easier to user the latter expression when doing something
like induction over the eigenvalues.

But this argument is not very convincing to me...what kind of situations (theorems)
do you know, where it is definitely better to use one expression
over the other ? What are the disadvantages of each of the expressions
?

答:
In bringing the matrix to Jordan canonical form (see http://en.wikipedia.org/wiki/Jordan_canonical_form) it is crucial to have the multiplicities you denoted $m_i$ so as to describe the blocks into which the matrix decomposes (up to conjugation).

On the other hand, computationally speaking the coincidence of eigenvalues is not stable, and from this point of view the significance of the Jordan canonical form is purely theoretical.  From the viewpoint of applications the second form may be more relevant.


问:
What does $\omega+\omega*$ denote?(model theory)
In Kees Doets' Basic Model Theory, there is a lemma(3.12) as follows:

$$k,m\geq2^n-1\implies Sy(k,m,n).$$

$$m\geq2^n-1\implies Sy(\omega+\omega*,m,n).$$

where:

$Sy(A,B,n)$ is a notation for the situation that the Spoiler has a winning strategy in the Ehrenfeucht–Fraïssé game of n rounds, in which $A$ and $B$ are the two structures.

$k,m,\omega$ are order types.

What confuses me is the notation $\omega+\omega*$. $\omega$ denotes the order type of natural numbers, and $\omega*$ denotes the order type of the dual of natural numbers(the reversed order). Then what about $\omega+\omega*$? Does it denote

$\{0，1，2，3，…,…,ω+3,ω+2,ω+1,ω\}$, that is, $\omega$ followed by $\omega*$

or

$\{…,−3,−2,−1, 0，1，2，3，…\}$, that is, $\omega*$ followed by $\omega$?

答:
It's the former. In general, for $A,B$ linear orders, "$A+B$" is the linear order consisting of a copy of $A$ followed by a copy of $B$.

(Incidentally, "$\zeta$" is used to denote the ordertype of the integers, so $\omega^*+\omega=\zeta$. This is your second possibility.)


答:
Once you know that $n\geq 3$ must be even, let $ n = 2k$.

Then, apart from finitely many values of $k$,

$$5^{2k} < 5^{2k} + 2^{2k} - 65 < 5^{2k} + 2 \times 5^k + 1. $$

This bounds the expression strictly between 2 consecutive perfect squares, so it cannot be a perfect square.

So, we just need to check those finitely many values, which is left as an exercise to the reader.

Title: Integers $n$ for which $|2^n + 5^n – 65|$ is a perfect square

Body:
Find the sum of all positive integers $n$ for which $|2^n + 5^n– 65|$ is a perfect square.

It is easy to check for $n<3$ only $n=2$ works. Looking at $n≥3$, $2^n + 5^n– 65= a^2$
For odd $n$, LHS has unit digit $2$ or $8$ but RHS has unit digit $0,4$ or $6$. Hence $n$ cannot be odd. For even $n$ we can rewrite our expression as follows ($n=2k$):  $$(a – 2^k)(a+2^k)=5(5^{2k–1} – 13)$$

But I wasn't able to finish off after this. Both LHS and RHS have the same unit digit ($0$). Through trial and error I found that $n=4$ works, but $n=6,8$ don't. Since the questions asks for the sum as is common in Olympiad number theory it's highly unlikely that any other larger values satisfy. I also tried some alternative ways like looking at the equation modulo $4$ but that didn't help.


问:
What's the motivation of the ideal?
I'm reading a book on Algebra, it introduces the concept of ring after some examples, the concept of ideal.

Definition I.1.8. Let $(A,+,\cdot)$ be a ring and $I$ a non-empty subset of $A$. We say that $I$ is a ideal if:

$x+y\in I,\;\;\;\;\forall \; x,y \in I$

$ax\in I, \;\;\;\;\forall \;x\in I, \;\;\;\;\forall a\in A$

I can't understand the motivation of the ideal: The sum of any elements of $I$ are elements of $I$ and the product of any element of $I$ with any element of $A$ is an element of $I$. I'm confused: What's that useful for?

答:
The motivation is to characterize what subsets of a ring can be the kernel of homomorphisms.

An equivalent way is to define congruences as for the integers: $x \equiv y \bmod I$ iff $x-y \in I$. Then congruences are compatible with the ring operations iff $I$ is an ideal. Being compatible allows one to define a natural ring structure for the set of equivalence classes $R/I$.

Title: What's the motivation of the ideal?


问:
Evaluating limits within limits
Suppose $f(h)$ is a function in the variable $h$ such that the limit $\lim_{h\to 0}\frac{\lim_{h\to 0}f(h)}{h}$ is known to exist. Then does the following relation hold?

$$\lim_{h\to 0}\frac{\lim_{h\to 0}f(h)}{h}=\lim_{h\to 0}\frac{f(h)}{h}$$

Intuitively this seems to make sense to me but I am unable to come up with a rigorous mathematical proof to either prove or disprove this. Can someone tell me if this is a valid equation and please justify it with a proof?

答:
I think it is better to write

$\lim_{h\to 0}\frac{\lim_{t\to 0}f(t)}{h}$ instead of $\lim_{h\to 0}\frac{\lim_{h\to 0}f(h)}{h}.$

Inorder to answer your question look at $f(x)=x.$

Title: Evaluating limits within limits


问:
Calculating expected value of a function
Consider the function :
$f(L) = \sum\limits_{x:L} g(x)$  where L is a list of integers where each element is randomly chosen between $1$ and $20$

$g(x) = 0$ if $x < 10$

$g(x) = 1 $ if $10 <=x <=19$

$g(x) = 2$  if $x =20$

What is the expected value of $f(L)$ for a list of size $5$ ?

As per my understanding I did it like :

$E(f(L)) = \sum\limits_{x:L} E(g(x)))$

$E(g(x)) = 1 * 10/20 + 2 * 1/20$

So it will just be 5 times the above value.

I am not sure if I am solving it correct?

答:
Each element is generated by choosing the maximum of two independent uniformly randomly generated numbers.

The chance of one of the numbers being $20$ is

$$p_2 = 1 - (19/20)^2.$$

The chance of both of the numbers being less than 10 is

$$p_0 = (9/20)^2.$$

Set

$$p_1 = 1 - (p_0 + p_2).$$

Then, the expected value of each number is

$$S = (1 \times p_1) + (2 \times p_2).$$

If there will be $n$ numbers (e.g. $n = 5$), then the expectation will be

$$S \times n.$$

Again, assuming that I am interpreting correctly the Linearity of Expectation article referenced in my comment, then it is irrelevant whether each pair of numbers is drawn with or without replacement.  Here, I am interpreting [drawing without replacement] to signify that if the first drawing is $(19,20)$, which results in a score of $2$, then no subsequent drawing can contain either of the numbers $19$ or $20$.


答:
In groups, we have the notion of a normal subgroup.  If $G$ is a group, then we have the result that every normal subgroup $N$ is the kernel of some homomorphism $\phi: G \rightarrow H$.

If $R$ is a ring and $\psi: R \rightarrow S$ is a ring homomorphism, its kernel $I$ will certainly be a subring of $R$, but it will also have the property that if $r \in R$ is any element of the ring and $i \in I$ is any element of the kernel, then $ri$ will have $\phi(ri) = \phi(r)\phi(i) = \phi(r) \cdot 0 = 0$, so $ri$ is also in the kernel.

In short, ideals are the natural substructures arising from homomorphisms.  You can check that you need the ideal property to get a well-defined quotient for the same reason.

Title: What's the motivation of the ideal?

Body:
I'm reading a book on Algebra, it introduces the concept of ring after some examples, the concept of ideal.

Definition I.1.8. Let $(A,+,\cdot)$ be a ring and $I$ a non-empty subset of $A$. We say that $I$ is a ideal if:

$x+y\in I,\;\;\;\;\forall \; x,y \in I$

$ax\in I, \;\;\;\;\forall \;x\in I, \;\;\;\;\forall a\in A$

I can't understand the motivation of the ideal: The sum of any elements of $I$ are elements of $I$ and the product of any element of $I$ with any element of $A$ is an element of $I$. I'm confused: What's that useful for?


问:
Compute $\sum_{n=1}^\infty (\frac34)^n \frac{7n+32}{n(n+2)}$

Question: Compute $$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{7n+32}{n(n+2)}$$

I first did the partial fraction decomposition into:
$$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{16}n - \sum_{n=1}^\infty \left(\frac34\right)^n \frac9{n+2}$$
Here I am stuck because I wrote out a couple terms and I don't see a way that terms can cancel out. Can I get some help?

答:
There is no need to cancel terms across both series because of the $(3/4)^n$. Evaluating both series individually (using the Maclaurin series of $\log(1-x)$) gives
$$S=\frac{16}{\log4}-\left(-\frac{33}2+\frac{16}{\log4}\right)=\frac{33}2$$


答: In this this type of problems just do telescoping now make the second terms in your partial fraction in powers of n+2 that is just divide and multiply by 16 in second term then do telescoping and possible answers are 33/2</p>

Title: Compute $\sum_{n=1}^\infty (\frac34)^n \frac{7n+32}{n(n+2)}$

Body:
Question: Compute $$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{7n+32}{n(n+2)}$$

I first did the partial fraction decomposition into:
$$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{16}n - \sum_{n=1}^\infty \left(\frac34\right)^n \frac9{n+2}$$
Here I am stuck because I wrote out a couple terms and I don't see a way that terms can cancel out. Can I get some help?


答:
Hint:

Let $\dfrac{7n+32}{n(n+2)}\cdot\left(\dfrac34\right)^n=f(n)-f(n-2)$

where $f(m)=\dfrac a{m+2}\cdot\left(\dfrac34\right)^m$

so that for $r\ge4,$ $$\sum_{n=2}^r\dfrac{7n+32}{n(n+2)}\cdot\left(\dfrac34\right)^n=\sum_{n=2}^r(f(n)-f(n-2))=f(r)+f(r-1)-f(1)-f(0)$$

$$f(n)-f(n-2)=\left(\dfrac34\right)^n\left(\dfrac a{n+2}-\dfrac an\cdot\dfrac{16}9 \right)$$

We need $\dfrac{7n+32}{n(n+2)}=\dfrac a{n+2}-\dfrac an\cdot\dfrac{16}9=-a\cdot\dfrac{7n+32}{9n(n+2)}\implies a=-9$


问:
Interval of Convergence for $\sum_{n=1}^\infty \frac{n4^n}{n^2+1}x^{2n}$
I'm trying to understand how to retrieve the interval of convergence for the following series: $$\sum_{n=1}^\infty \frac{n4^n}{n^2+1}x^{2n}$$

Using the ratio test I'm seeing that the interval is $-\frac{1}{2} \le x \le \frac{1}{2}$. I'm struggling with evaluating the endpoints, as our lecturer stated that the series at both $x=\frac{-1}{2}$ and at $x=\frac{1}{2}$ is the same as $\sum_{n=2}^\infty \frac{n}{n^2+1}$ which I don't quite understand. He ended up taking the integral test to then evaluate the endpoints. I would really appreciate someone explaining how we can evaluate these endpoints, as I'm very confused to it.

Thanks in advance!

答: If $x=\pm\frac12$, $\require{cancel}x^{2n}=\left(\pm\frac12\right)^{2n}=\frac1{4^n}$. Therefore, for each such $x$,\begin{align}\sum_{n=1}^\infty\frac{n4^n}{n^2+1}x^{2n}&=\sum_{n=1}^\infty\left(\frac{n\cancel{4^n}}{n^2+1}\times\frac1{\cancel{4^n}}\right)\\&=\sum_{n=1}^\infty\frac n{n^2+1}.\end{align}And\begin{align}\int_1^\infty\frac x{x^2+1}\,\mathrm dx&=\frac12\lim_{M\to\infty}\left(\log(M^2+1)-\log(2)\right)\\&=\infty,\end{align}and therefore, by the integral test, your series diverges.


问:
Sum of two floors is equal

Let $a, b, c$, and $d$ be real numbers. Suppose that $\lfloor na\rfloor +\lfloor nb\rfloor =\lfloor nc\rfloor +\lfloor nd\rfloor $ for all positive integers $n$. Show that at least one of $a+b$, $a-c$, $a-d$ is an integer.

My Progress:
Since $$1=\lim_{n\to \infty}{\frac{ \lfloor na\rfloor +\lfloor nb\rfloor }{\lfloor nc\rfloor +\lfloor nd\rfloor}}=\frac{a+b}{c+d}$$ Thus $a+b=c+d=x$.

We have $\lfloor na\rfloor+\lfloor nx-na\rfloor=\lfloor nc\rfloor +\lfloor nx-nc\rfloor$. This means that $\{nx\}\geq \{na\} \equiv \{nx\}\geq\{nc\}$.
I think some clever bounding should do it but I am not able to do so...

答:
$n=1$ shows that one of $(a,b)$ is $\geq 1$ iff one of $(c, d)$ is $\geq 1$. Since we can add or subtract $1$ from one of $(a,b)$ and one of $(c,d)$ and the equation will still be satisfied, we may assume without loss of generality that $0 \leq a,b,c,d < 1$.

Write the base-$B$ representations of these numbers as $a = 0.a_{1,B}a_{2,B}a_{3,B}\cdots$ where each $a_{i,B}$ is a single digit,  and similarly for $b, c, d$. Using $n = B^k$ with induction on $k$, $a_{i,B} + b_{i,B} = c_{i,B} + d_{i,B}$ for every base $B$ and $i \geq 1$. [1]

For $B=2$ this shows that when the binary digits of $a$ and $b$ agree or disagree, the corresponding digits in $c$ and $d$ also agree or disagree.

If $a = b$ then this shows that $a=c$ and we are done.

Now assume $a \neq b$. Let $i_1, i_2, i_3, \ldots$ be the indices of binary digits on which $a$ and $b$ disagree (this may either be finitely long or an infinite sequence).

Let $A_k=0.a_{1,2}\cdots a_{k,2}$ be the number given by the binary representation of $a$ truncated to $k$ digits, and similarly for $b,c,d$.

Suppose by contradiction that $a \neq c$ and $a \neq d$. Then there must be some minimum $k$ such that $A_{i_k} \neq C_{i_k}$ and $A_{i_k} \neq D_{i_k}$, since $a$ must agree with $c$ and $d$ on all other binary digits. Since $c$ and $d$ also must disagree on binary digit $i_k$, $C_{i_k} \neq D_{i_k}$. Since $A_{i_k} + B_{i_k} = C_{i_k} + D_{i_k}$, $B_{i_k}$ also has a distinct value.

Let $e$ be the unique member of $\{a,b,c,d\}$ with maximum truncated value $E_{i_k}$. In base $Q=2^{i_k} E_{i_k}$, $E_{i_k}$ has representation $0.1000\cdots$. The remaining three of $A_{i_k},B_{i_k},C_{i_k},D_{i_k}$ have representations $0.0?????\cdots$ for some base-Q digits $?$. Since besides $e$ the other members of $a,b,c,d$ must have values less than $E_{i_k}$, it follows that $e$ has first base-Q digit $1$ while the others have first base-Q digit $0$, contradicting formula [1] for $B=Q$. Therefore either $a=c$ or $a=d$. QED

Title: Sum of two floors is equal


问:
Does this mean that one can construct the cube root of two in three dimensions?
The idea here is to extend to three dimensions what ordinary compass-and-straightedge constructions do in two dimensions.  The first thing is to define the tools and rules for their use.  For instance, in two dimensions, the tools are a compass and straightedge (like a ruler, but with only one edge and no markings), and with them, one may:
Draw a line between any two distinct points.
Draw a circle with one point as the center, and any other point on its circumference.
Draw an arbitrary point on a line or a circle, or off it.
Draw the point at the intersection of two lines (if they intersect).
Draw the point (or two) at the intersection of two circles (if they intersect).
Draw the point (or two) at the intersection of a line and a circle (if they intersect).

In three dimensions, the canvas is not a flat plane, as it is in two dimensions, but all of space.  And we introduce a new tool, which I will call a flatiron, which permits you to draw planes.  The flatiron rules are as follows; in addition to the above, one may:
Draw the unique plane containing any three non-collinear points.
Draw a sphere with one point as the center, and any other point on its surface.
Draw an arbitrary point on a plane or a sphere, or off it.
Draw the line at the intersection of two planes (if they intersect).
Draw the circle (or point) at the intersection of two spheres (if they intersect).
Draw the circle (or point) at the intersection of a plane and a sphere (if they intersect).
Draw the point (or two) at the intersection of a line or circle with a plane or sphere (if they intersect).
As an example of what one might do in a three-dimensional construction, consider the following fairly simple task: Given points P and Q, construct a regular tetrahedron with PQ as edge.  We proceed as follows:
Draw spheres of radius PQ around both P and Q.
Draw the circle C at the intersection of spheres P and Q.
Draw R, an arbitrary point on circle C.
Draw a sphere of radius PR around R.
Draw S, one of the two points of intersection between circle C and sphere R.  PQRS is then a regular tetrahedron.

Given these new abilities, would it be possible construct a cuberoot?

答:
A good way to approach the problem is to define algebraic equations which describe all of your tools and the rules governing the use of each tool. The structure of the algebraic equations will determine what can be constructed.

Tool 1 lets you: Draw a line between any two distinct points. So:

Given $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. Use the line equation.

Tool 2 the Flatiron lets you: Draw the unique plane containing any three non-collinear points. So:

Given $(x_1, y_1, z_1), (x_2, y_2, z_2)$ and $(x_3, y_3, z_3)$. Use the plane equation.

Tool 3 lets you: Draw a sphere with(I assume 'given' is better than with) one point as the center, and (taking) any other (constructed or known)point (to be)on its surface. So:

Given $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. Use the sphere equation with center $(x_1, y_1, z_1)$ and radius as the distance between the points.

Tool 4 lets you: Draw an arbitrary point on a plane or a sphere, or off it. So:

I am stumped as to the meaning of tool 4 to be honest.

Tool 5 lets you: Draw the line at the intersection of two planes (if they intersect). This is a given. When you draw the planes you automatically draw that line.

Tool 6 lets you: Draw the circle (or point) at the intersection of two spheres (if they intersect). Another given with the same reasoning as above.

Tool 7 lets you: Draw the circle (or point) at the intersection of a plane and a sphere (if they intersect). Another given with the same reasoning as above.

The following two families of equations govern what values can be constructed under this ruleset:

$$(x-x_1)^2 + (y-y_1)^2 + (z-z_1)^2 - (x_2-x_1)^2 - (y_2-y_1)^2 - (z_2-z_1)^2 = 0$$

$$(x-x_1)[(y_2-y_1)(z_3-z_1)-(z_2-z_1)(y_3-y_1)] - (y-y_1)[(x_2-x_1)(z_3-z_1)-(z_2-z_1)(x_3-x_1)] + (z-z_1)[(x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1)] = 0$$

You can show that you don't need Tool 1 since anything constructed by Tool 1 can be constructed by Tool 2. It isn't too much work to show that.

Before continuing the analysis I would like to clarify the mysterious tool 4. I don't understand it. Sounds like it lets you construct anything which isn't very interesting. I must be misunderstanding your wording.


问:
Why is $agHg^{-1}a^{-1}=gHg^{-1}$ for $a \in H$ (with $H$ as a subgroup)
I got to read this in Rotman's Introduction to Group Theory:

I don't understand the part that $agHg^{-1}a^{-1}=gHg^{-1}$ for $a \in H$ (with $H$ as a subgroup) ... any help ?

Here, $N_g(x)$ is the normalizer of $x$

答: This is a different scenario. As $\{gHg^{-1}\}$ is an orbit of size 1 with the operation being the conjugation, $agHg^{-1}a^{-1}=H$, or else the orbit wouldn't be of size 1.


问:
Non-uniqueness of a Dirichlet's problem
I found a statement in the book "Random Perturbations of Dynamical Systems" written by M.I.Freidlin and A.D. Wentzell that is not justified. Specifically, in a smooth domain $D \subset \mathbb{R}^n$ let us consider the Dirichlet's problem $$\mathcal{L}u(x) + c(x)u(x) = f(x),~~x\in D;\quad u(x)|_{x\in \partial D} = g(x),$$ in which $\mathcal{L}u(x) = \frac 12\sum_{i,j} a^{ij}(x)\frac{\partial^2 u}{\partial x^i\partial x^j}(x) + \sum_i b^i(x)\frac{\partial u}{\partial x^i}(x) $ with $a(x) = (a^{ij}(x))=\sigma(x)\sigma^T(x)$. Here $c(x)$, $f(x)$, for $x \in \mathbb R^n$, and $g(x)$, for $x\in \partial D$, are bounded continuous functions. If we suppose that $\mathcal{L}$ is uniformly elliptic, then under the additional assumption that $c(x) \leq 0$, this problem admits a solution provided by the Feynman-Kac formula, i.e., $$u(x) = -\mathbb{E}_x \int_0^\tau f(X_t)\exp\left\{\int_0^t c(X_s) ds\right\} + \mathbb{E}_x~ g(X_\tau)\exp\left\{\int_0^t c(X_s) ds\right\},$$ where $\tau = \inf\{t\colon X_t\notin D\}$ is the first exit time of the process defined via $dX_t = b(X_t)dt + \sigma(X_t)dW_t$ from the domain $D$. The authors claimed that "If $c(x) > 0$, then as is well known, Dirichlet's problem can "go out to the spectrum"; the solution of the equation $\mathcal{L}u + c(x)u = 0$ with vanishing boundary values may not be unique in this case." May I know why the last statement holds? Can you provide a baby example showing such non-uniqueness of solutions?

答:
The simplest example is to just take $c$ to be a (constant) eigenvalue of the operator.  For example, if we take $\mathcal{L} = \Delta$, then we can find an infinite sequence of eigenfunction/eigenvalue solution to
$$
\begin{cases}
-\Delta u_k = \lambda_k u_k &\text{in } D \\
u_k = 0 & \text{on } \partial D.
\end{cases}
$$
These can be found in a number of ways.  For example, we can minimize the functional $E(u) =   \int_D |\nabla u|^2$ over $H^1_0(D)$ subject to the constraint that $J(u) = \int_D |u|^2 =1$ in order to find the principal eigenvalue, $\lambda_0>0$.

Once we have these, we immediately see that we cannot have unique solutions to
$$
\begin{cases}
\Delta u + \lambda_k u = f &\text{in } D \\
u = 0 & \text{on } \partial D
\end{cases}
$$
for any $k$.

Title: Non-uniqueness of a Dirichlet's problem


问:
Proving Property of Bernoulli's polynomial $B_n(x+1)-B_n(x)=nx^{n-1}$
I am trying to show that $B_n(x+1)-B_n(x)=nx^{n-1}$, where $B_n(x)$ is the Bernoulli polynomial. In order to avoid circular reasoning, I define the Bernoulli's number $B_n$ as the coefficient in the following generating function:
$$\frac{x}{e^x-1}=\sum_{n=0}^\infty \frac{B_n}{n!}x^n$$
and define the Bernoulli's polynomial as
$$B_n(x)=\sum_{k=0}^n \begin{pmatrix}n\\ k\end{pmatrix}B_{n-k}x^k$$
From the above definition, I am able to derive the translational property:
$$B_n(x+y)=\sum_{r=0}^n \begin{pmatrix}n\\ r\end{pmatrix}B_r(x)y^{n-r}$$
Then by putting $y=1$, we have
\begin{align*}
B_n(x+1)-B_n(x)&=\sum_{r=0}^n\begin{pmatrix}n\\ r\end{pmatrix}B_r(x)-B_n(x) \\
&= \sum_{r=0}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}B_r(x) \\
&= \sum_{r=0}^{n-1}\sum_{k=0}^r\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k}x^k \\
&= \sum_{k=0}^{n-1}\sum_{r=k}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k}x^k \\
&= \sum_{k=0}^{n-1}x^k\sum_{r=k}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k} \\
\end{align*}
It is easy to see that $\displaystyle \sum_{r=k}^{n-1}\begin{pmatrix}n\\ r\end{pmatrix}\begin{pmatrix}r\\ k\end{pmatrix}B_{r-k}=n$ when $k=n-1$, but I have no idea how to show that this is equal to $0$ when $k$ is not $n-1$. Any help would be appreciated.

答:
From your main definition of the Bernoulli numbers, for $ x\in\mathcal{B}\left(0,2\pi\right) $, multiplying both sides by $ \mathrm{e}^{x} $, we have : \begin{aligned}\frac{x\,\mathrm{e}^{x}}{\mathrm{e}^{x}-1}&=\left(\sum_{n=0}^{+\infty}{\frac{B_{n}}{n!}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{\frac{x^{n}}{n!}}\right)\\ \iff\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{-x}{\mathrm{e}^{-x}-1}&=\sum_{n=0}^{+\infty}{\frac{1}{n!}\left(\sum_{k=0}^{n}{\binom{n}{k}B_{k}}\right)x^{n}}\\ \iff\sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}B_{n}}{n!}x^{n}}&=\sum_{n=0}^{+\infty}{\frac{1}{n!}\left(\sum_{k=0}^{n}{\binom{n}{k}B_{k}}\right)x^{n}}\end{aligned}

Thus, for all $ n\in\mathbb{N} $, we have : $$ \sum_{k=0}^{n}{\binom{n}{k}B_{k}}=\left(-1\right)^{n}B_{n} $$

Now : \begin{aligned} \sum_{r=k}^{n-1}{\binom{n}{r}\binom{r}{k}B_{r-k}}&=\binom{n}{k}\sum_{r=k}^{n-1}{\binom{n-k}{r-k}B_{r-k}}\\ &=\binom{n}{k}\sum_{r=0}^{n-k-1}{\binom{n-k}{r}B_{r}}\\ &=\binom{n}{k}\sum_{r=0}^{n-k}{\binom{n-k}{r}B_{r}}-\binom{n}{k}B_{n-k}\\ &=\binom{n}{k}\left(-1\right)^{n-k}B_{n-k}-\binom{n}{k}B_{n-k}\\ \sum_{r=k}^{n-1}{\binom{n}{r}\binom{r}{k}B_{r-k}}&=\left(\left(-1\right)^{n-k}-1\right)\binom{n}{k}B_{n-k}\end{aligned}

Let $ k\in\left[\!\left[0,n-2\right]\!\right] $, If $ k $ is such that $ n-k $ is even than $ \left(-1\right)^{n-k}-1 =0 $, but if $ k $ is such that $ n-k $ is odd than $ B_{n-k}=0 $, because besides $ B_{1} $, all $ B_{n} $ such that $ n $ is odd are equal to $ 0 $. Thus, in all cases $ \left(\left(-1\right)^{n-k}-1\right)B_{n-k}=0 $ for all $ k\in\left[\!\left[0,n-2\right]\!\right] $. Hence the result.


问:
is $(x+1)^4-x^4$ non-prime for all natural positive integers $x$
Looking at difference between two neighbouring positive integers raised to the power 4, I found that all differences for integer neighbours up to $(999,1000)$ are non-prime.

Does this goes for all positive integers?
And can someone please prove?

Body: Hint : you can write $(x+1)^4-x^4=((x+1)^2-x^2)((x+1)^2+x^2)=(x+1-x)(x+1+x)((x+1)^2+x^2)=(2x+1)((x+1)^2+x^2)$


问:
Finding if the group of isometries of a cube is a normal subgroup of $S_{12}$
Let $G$ be the group of isometries of a cube and let X be the set of edges of the cube.
This action defines an injective homomorphism $\phi:G \rightarrow S_{12} $, injective since stabilisers of adjacent edges overlap only at ${e}$.
I am struggling to determine if $Im(\phi)$ is a normal subgroup of $S_{12}$. I have tried considering the action of G on its left cosets and a few other ideas but nothing has yielded useful progress.

Thanks a lot in advance.

答:
The unique non-identity proper normal subgroup of $\mathfrak S_{12}$ is the alternating group $\mathfrak A_{12}$.

$G$ is isomorphic to $\mathfrak S_{4} \times \mathbb Z / 2\mathbb Z$ and therefore has $2 \times 4!= 48$ elements. As $ \vert \mathfrak A_{12} \vert \gt 48$, $\text{Im}(\phi)$ isn't normal in $\mathfrak S_{12}$.


问:
Merge fraction with x in it
I have a problem. My teacher made an online explanation about merging fractions. She wrote the following:
$$\frac{1}{x+1}+3$$
Then she wanted that to merge in one fraction, so she did this:
$$\frac{1}{x+1}+\frac{3(x+1)}{(x+1)}=\frac{1+3x+3}{x+1}=\frac{3x+4}{x+1}$$

Now I understand why she does this, but calculating with $1$ is easy. In my case I have $24$ like this:
$$\frac{24}{x+2}+4$$

How do I write this in one fraction?

答:
$$y=\frac{24}{x+2}+4=\frac{24+4\cdot(x+2)}{x+2}=\frac{24+4x+8}{x+2}=\frac{4x+32}{x+2}$$

Please take in consideration to see some basic LaTeX commands.

Title: Merge fraction with x in it


问:
How to find the minimum of $xy$ from the equation $2x+y+6=xy$ , $x,y>0$
I have tried doing factoring and completing the square and also AM-GM

$(2-y)(x-1)=-8$

$xy=2x+y+6≥\sqrt{12xy} $

答:
AM-GM works perfectly fine with a small trick.
$2x+y+6=xy \Rightarrow y(x-1)=2x+6 \Rightarrow y = \dfrac{2x+6}{x-1} \Rightarrow xy =\dfrac{2x^2+6x}{x-1}$.

As @Yves Daoust points out $\dfrac{2x^2+6x}{x-1} = 2x+8+ \dfrac{8}{x-1} = 2(x-1)+10+\dfrac{8}{x-1} $.

But by AM-GM,  $ 2(x-1)+10+\dfrac{8}{x-1} \geq 10 + 2\left(\sqrt{2(x-1) \cdot \dfrac{8}{x-1} } \right)  = 10+2\sqrt{16}=18$. This is indeed the minimum of $xy$.


问:
Converting "n is an odd negative integer" Into Formal Logic
I am learning how to think about mathematical truth at the level of formal logic and I am tasked with converting the following basic statement into something formal: "n is a negative integer that is odd"

My first attempt at this was the following: Let $n \in \mathbb{Z}$ such that $2n + 1 < 0$

From my point of view this statement is saying that "n is an integer and it's values are restricted by the inequality $2n + 1 < 0$". But how this actually translated to a mathematician was "$n$ is an integer satisfying the inequality $2n + 1<0$" and after some thought, I concluded my initial statement was indeed wrong.

So I went back to the drawing board and came up with this:

$$\exists k \in \mathbb{N} \quad n = -2k + 1 $$

But I feel that this formulation is still missing something, and I just can't put my finger on it. Should something be said about $n$ as well? Another way I thought of writing this was:

$$ \exists k \in \mathbb{N} \quad (n = -2k + 1) \Rightarrow n \in \mathbb{Z}  $$

$$\exists n \in \mathbb{Z} \quad \exists k \in \mathbb{N} \quad n = -2k + 1 $$
Any help thinking through this would be appreciated.

答:
We can write "$n$ is a negative integer" as
$$n\in\mathbb{Z}^-$$
And "$n$ is odd" as
$$\exists k\in\mathbb{Z}.\ n=2k-1$$
Combine both:
$$\exists k\in\mathbb{Z}.\ n=2k-1 \wedge n\in\mathbb{Z}^-$$
Or alternatively,
$$\exists k\in\mathbb{Z}.\ \mathbb{Z}^-\ni n=2k-1$$

Hope this helps. :)


问:
modular arthmetic
To prove that a number p is prime we use this formula $a^{p-1} \pmod p = 1$

In the case of high powers can we derive a formula to solve higher power orders:
$a^{p-1} \pmod p = 1$

$(a^{p-2}) (p-1) \pmod p = 1$

taking log to both sides we get:
$\log\left(\left(a^{p-2}\right)(p-1)\pmod p\right) = \log(1)$

From the property $\log(uv) = \log (u ) + \log(v)$, we get:

$\log(a^{p-2})+\log(p-1 \pmod p) = \log 1$

From another property of log : $\log(a^n) = n \log a $

we have
$(p-2) \log(a)+\log(p-1)=\log 1 $

wheres the part that has gone wrong here?
why is it not mathematically true?

答:
The primality tests in regular use (e.g. Miller-Rabin) are different from this Fermat test (which can be fooled, but rarely; practical primality tests use the Fermat test to weed out composites as a first step).

To compute $a^n$ for large $n$ a practical solution is to use repeated squaring:

$\begin{align*}
   a^{2 k}
     &= (a^k)^2 \\
   a^{2 k + 1}
     &= (a^k)^2 \cdot a
\end{align*}$

Computing $a^n$ is not that expensive.

To use logarithms (you need the exact power!) you'd need to compute $\log a$ and then $10^{n \log a}$ to a lot of significant figures, much more than is practical.

Title: modular arthmetic


答:
This answer follows the comment provided by Mauro ALLEGRANZA in the comments:

The statement "$n$ is an odd negative integer" can be translated to "$n$ is an integer AND $n$ is odd AND $n$ is negative". Converting this to formal logic we can write:

"$n$ is an integer" == $n \in \mathbb{Z}$

"$n$ is negative" == $n < 0$

"$n$ is odd" == $\exists k \in \mathbb{Z} \quad (n = 2k +1) $ (edit : if one takes $k \in \mathbb{N}$, then $n > 0$)

and combining these statements about $n$:

"n is an odd negative integer" == ($n \in \mathbb{Z}) \quad \land \quad (n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$

edit : Since $\exists k \in \mathbb{Z} \thinspace (n = 2k +1)$ already implies that $n \in \mathbb{Z}$, one can omit it and state :

$$ (n<0) \quad \land \quad \exists k \in \mathbb{Z} \thinspace (n = 2k +1)$$

Title: Converting "n is an odd negative integer" Into Formal Logic

Body:
I am learning how to think about mathematical truth at the level of formal logic and I am tasked with converting the following basic statement into something formal: "n is a negative integer that is odd"

My first attempt at this was the following: Let $n \in \mathbb{Z}$ such that $2n + 1 < 0$

From my point of view this statement is saying that "n is an integer and it's values are restricted by the inequality $2n + 1 < 0$". But how this actually translated to a mathematician was "$n$ is an integer satisfying the inequality $2n + 1<0$" and after some thought, I concluded my initial statement was indeed wrong.

So I went back to the drawing board and came up with this:

$$\exists k \in \mathbb{N} \quad n = -2k + 1 $$

But I feel that this formulation is still missing something, and I just can't put my finger on it. Should something be said about $n$ as well? Another way I thought of writing this was:

$$ \exists k \in \mathbb{N} \quad (n = -2k + 1) \Rightarrow n \in \mathbb{Z}  $$

$$\exists n \in \mathbb{Z} \quad \exists k \in \mathbb{N} \quad n = -2k + 1 $$
Any help thinking through this would be appreciated.


问:
Why do we need to consider $ \hat{0} = \hat{m} $ separately in the proof for the union of all residue classes being $ \mathbb{Z} $?
The book Introduction to Analytic Number Theory by Apostol states in Theorem 5.10 (c):

The $ m $ residue classes $ \hat{1}, \hat{2}, \dots, \hat{m} $ are disjoint and their union is the set of all integers.

The proof goes like this:

Hence by part (b) the residue classes
$$ \hat{0}, \hat{1}, \hat{2}, \dots, \widehat{m - 1} $$
are disjoint. But every integer $x$ must be in exactly one of these classes because $ x = qm + r $ where $ 0 \le r < m $, so $ x \equiv r \pmod{m}$ and hence $ x \in \hat{r} $. Since $ \hat{0} = \hat{m} $ this proves (c).

Why is the last sentence necessary for the proof. What if we did not write "Since $ \hat{0} = \hat{m} $ this proves (c)."? Will the proof remain incomplete then and why?

答:
Hence by part (b) the residue classes
$$ \hat{0}, \hat{1}, \hat{2}, \dots, \widehat{m - 1} $$
are disjoint. But every integer $x$ must be in exactly one of these classes because $ x = qm + r $ where $ 0 \le r < m $, so $ x \equiv r \pmod{m}$ and hence $ x \in \hat{r} $. Since $ \hat{0} = \hat{m} $ this proves (c).

The above analysis (including the unspecified part [b]) demonstrates that

$$ \hat{0}, \hat{1}, \hat{2}, \dots, \widehat{m - 1} $$

are disjoint and comprehensive (i.e. that every integer belongs to exactly one of the above equivalence classes).

In order to then demonstrate that the equivalence classes

$$ \hat{1}, \hat{2}, \dots, \widehat{m - 1}, \widehat{m}$$

are disjoint and comprehensive you must then make the observation that $\hat{0} = \widehat{m}.$


问:
Is there a way to check if a group extension is split or not using GAP?
I use the GAP command 'NormalSubgroups' to obtain the list of normal subgroups of a given group. Then I use 'FactorGroup' to construct the quotient of the given group by a normal subgroup.
An example is given below.

G:=AlternatingGroup(4);;
list:=NormalSubgroups(G);;
N:=list[2];;
H:=FactorGroup(G,N);;

Here, G is an extension of H by N. Is there a way to check if this extension is split using GAP commands?

答:
The function ComplementClassesRepresentatives(G,N) finds representatives of the $G$-conjugacy classes of complements to $N$. It returns an empty list if the extension is not split. For example (in your case and a nonsplit case:

```<code>gap> ComplementClassesRepresentatives(G,N);
[ Group([ (2,4,3) ]) ]
gap> G:=SL(2,5);N:=Center(G);
SL(2,5)
<group of 2x2 matrices over GF(5)>
gap> ComplementClassesRepresentatives(G,N);
[  ]
</code>```

The current methods available require that $N$ is solvable (calculated using 1-cohomology, respectively "Fox derivatives"), or that $G/N$ is solvable (using cyclic factors in a composition series) -- these seem to be the cases relevant in applications.

Body:
The function ComplementClassesRepresentatives(G,N) finds representatives of the $G$-conjugacy classes of complements to $N$. It returns an empty list if the extension is not split. For example (in your case and a nonsplit case:

gap> ComplementClassesRepresentatives(G,N);
[ Group([ (2,4,3) ]) ]
gap> G:=SL(2,5);N:=Center(G);
SL(2,5)
<group of 2x2 matrices over GF(5)>
gap> ComplementClassesRepresentatives(G,N);
[  ]

The current methods available require that $N$ is solvable (calculated using 1-cohomology, respectively "Fox derivatives"), or that $G/N$ is solvable (using cyclic factors in a composition series) -- these seem to be the cases relevant in applications.


答:
This can be written as:

$[(x+1)^2-x^2][(x+1)^2+x^2]$

Which can be written as:

$[(x+1)-x][(x+1) +1 ][(x+1)^2+x^2] = (1)(2x+1)[(x+1)^2+x^2]$

which clearly has two factors other than 1 and is $\therefore $ not prime for all x's.

Title: is $(x+1)^4-x^4$ non-prime for all natural positive integers $x$

Body:
Looking at difference between two neighbouring positive integers raised to the power 4, I found that all differences for integer neighbours up to $(999,1000)$ are non-prime.

Does this goes for all positive integers?
And can someone please prove?


问:
How many different lottery tickets are there in which at least three numbers match numbers in the draw?

A lottery ticket consists of six numbers between $1$ and $49$. A holder of a ticket wins the jackpot if the numbers match the six randomly drawn numbers. Any ticket with at least three matching numbers from the draw also qualifies for some prize money.  How many different lottery tickets are there in which at least three numbers match numbers in the draw?

Given answer: $\displaystyle{\binom 66 + \binom65\binom{43}{1} + \binom64\binom{43}{2} + \binom63\binom{43}{3}}$.

I understand the presence of summation in the answer. It's due to "at least" bit in the statement of the problem. Can't seem to make sense of how they obtained the terms in the sum.

Consider the product $\displaystyle{\binom63\binom{43}{3}}$. How do we interpret it? If $6$ in the coefficient $\displaystyle{\binom63}$ is the number of places in a ticket (where three places are fixed), then shouldn't $\displaystyle{\binom63\binom{43}{3}}$ actually be $\displaystyle{\binom63\binom{46}{3}}$ given that only three numbers are taken out of $\{1, 2, \ldots, 49\}$ to fix anywhere within the six spots on a ticket? Suppose, then, $6$ in the coefficient $\displaystyle{\binom63}$ is the number of integers from  $\{1, 2, \ldots, 49\}$ in which case $\displaystyle{\binom{43}{3}}$, kind of, makes sense, but then, shouldn't we choose those six integers from $\{1, 2, \ldots, 49\}$ implying presence of something like  $\displaystyle{\binom{49}{6}}$ in a product term, not to mention what happens to three of six chosen integers?

Anyway, my question is How do we interpret the terms in the sum in the given answer? Thanks.

答: To have exactly three numbers correct, you choose three of the winning numbers which you can do in $6 \choose 3$ ways, then choose three losing numbers, which you can do in $43 \choose 3$ ways.  As each choice of winning numbers can go with any choice of losing numbers you multiply them, getting ${6 \choose 3}{43 \choose 3}$.  The other terms work the same way.  To get exactly $n$ winning numbers you have ${6 \choose n}{43 \choose 6-n}$ ways because there are $6-n$ losing numbers.


问:
Eigenvalues and (Singleton)-Borel Functional Calculus
I am having some confusion that I hope people might be able to help me with. To be concrete I will use two (isomorphic) $\mathrm{C}^*$-algebras.

The first, $A_1$, is the $\mathrm{C}^*$-group algebra of the (amenable) infinite dihedral group $D_\infty$ with order two generators $u$ and $v$ (it is isomorphic to the free product of cyclic groups) $C_2\star C_2$. Consider it represented by the regular representation as $\pi_1(\mathrm{C}^*(D_\infty))\subset B(\ell^2(D_\infty))$. The particular element I am interested in is the sum of two projections:

$$a_1:=\frac12\delta^e+\frac12 \delta^u+\frac12\delta^ e+\frac12 \delta^v=\delta^e+\frac{1}{2}\delta^u+\frac{1}{2}\delta^v.$$

I understand (via the isomorphism with $A_2$ below that the spectrum  $\sigma(a_1)=[0,2]$.

Question 1: is one an eigenvalue of $\pi_1(a_1)$? If yes what is an eigenvector?

The second $\mathrm{C}^*$-algebra $A_2$ is the universal $\mathrm{C}^*$-algebra generated by projections $p$ and $q$. It is isomorphic that it is isomorphic to $A_3$, where $A_3$ is the $\mathrm{C}^*$-algebra:
$$A_3:=\left\{f\in C([0,1],M_2(\mathbb{C})):f(0) \text{ and }f(1)\text{ are diagonal}\right\}.$$

There is an isomorphism that carries $p$ and $q$, respectively, to
$$p(x)=\left(\begin{array}{cc} 1 & 0 \\ 0 & 0\end{array}\right)\quad \text{ and }\quad q(x)=\left(\begin{array}{cc}x & \sqrt{x(1-x)} \\ \sqrt{x(1-x)} & 1-x\end{array}\right).$$

This can be represented $\pi_2(A_3)\subset B(L^2([0,1])\oplus L^2([0,1]))$.

Consider
$$a_2:=x\mapsto p(x)+q(x)=\left(\begin{array}{cc}1+x & \sqrt{x(1-x)} \\ \sqrt{x(1-x)} & 1-x\end{array}\right).$$

The determinant of $a_2(x)-\lambda I_{\pi(A_3)}$ is $\lambda^2-2\lambda-x+1$ which is an issue when $\lambda=1\pm\sqrt{x}$, and so the spectrum $\sigma(a_2)=[0,2]$. As far as I can see it $\pi_2(a_2)$ has no eigenvalues at all.

Question 2: Is this correct --- $\pi_2(a_2)$ has no eigenvalues?

This leads onto a third question (possibly giving an answer to this question).

Question 3: Unless I am mistaken, the spectrum of an element $a$ of a $\mathrm{C}^*$-algebra $A$ is the same in every faithful
representation of $A$. Is it possible for a $\lambda\in \sigma(a)$ to
be an eigenvalue in one representation $\pi(A)$ but not in another
$\pi'(A)$?

Final question. Take a faithful representation of $A_1$, $A_2$ or $A_3$ (which I think are all isomorphic) and consider, as appropriate, the $a_1$ or $a_2$ given above. Let us call it by $a$. Is there any particular relation between an element of the spectrum $\lambda\in\sigma(a)$ being an eigenvalue or not, and the bounded operator given by the Borel functional calculus, $\mathbf{1}_{\{\lambda\}}(a)$? Is this $\mathbf{1}_{\{\lambda\}}$ a projection if $\lambda$ is not an eigenvalue?

In particular,

Question 4: For either $a_1$ or $a_2$ above, if it is true that $1\in\sigma(a)$ is not an eigenvalue, is $\mathbf{1}_{\{1\}}(a)$ equal
to zero?

I fully expect there to be some errors/confusions above and would appreciate any help.

答:
Here are a few answers.  I'll be happy to elaborate more upon request.

Question 2: Is this correct --- $\pi_2(a_2)$ has no eigenvalues?

I think you are correct. For almost all $x$, the matrix $a_2(x)-\lambda $ is invertible, so a potential eigenvector will have to vanish a.e.

Question 3: Unless I am mistaken, the spectrum of an element $a$ of a C$^*$-algebra $A$ is the same in every faithful
representation of $A$. Is it possible for a $\lambda\in \sigma(a)$ to
be an eigenvalue in one representation $\pi(A)$ but not in another
$\pi'(A)$?

Whenever $A$ is a C*-algebra, $a\in A$ is self adjoint, and $\lambda \in \sigma (A)$, there exista a faithful representation $\pi $ of $A$
such that $\lambda $ is an eigenvalue of $\pi (a)$.  To prove it, start by considering a character on $C^*(1, a)$  sending $a$ to $\lambda $, extend this to a representation of $A$ (on a larger Hilbert space) and add on any faithful representation of $A$.

Question 4: For either $a_1$ or $a_2$ above, if it is true that $1\in\sigma(a)$ is not an eigenvalue, is ${1}_{\{1\}}(a)$ equal
to zero?

Whenever $T$ is a bounded self-adjoint operator on a Hilbert space, one has that $1_{\{\lambda \}}(T)$ is the projection onto the eigenspace $\text{Ker}(T−\lambda )$. So, yes, $1_{\{\lambda \}}(T)$ is nonzero iff $\lambda $ is an eigenvalue.


问:
A Polar Coordinates Computation
I'm looking at the following computation:

Here: $u_k = u(x_0 + \rho_k x)/\rho_k$

I'm having trouble understanding the change of variables.

For the first integral if I consider $x' \mapsto x_0 + \rho_k x$ then $dx'/\rho_k^n = dx$ -- there's an extra factor of $\rho_k$ in the denominator of the first integral though.

For the second integral if I set $r' \mapsto \rho_k r$ then the conversion works because there's a hidden factor of $r^{n-1}$ in the $d\mathcal{H}^{n-1}$, is that correct?

$Q^2$ should be evaluated at $0$ in the second integral, instead of $x_0$, is that correct?

答:
For (1), the factor of $\rho_k^n$ in the new differential cancels out with the one in front of the integral.

For (3), $Q_2(x_0)$  is a constant, so it remains the same when doing the change of variables.

For (2), note that you also have $\rho_k$ in the denominator in the definition of $u_k$. Think about a 1D example: $v(x)=\frac{u(2x)}{2}$. You have
$$
|v'(x)|=|u'(2x)|
$$
so you have no extra factor of $2$ when you take the gradient.

Lebesgue integral is shift-invariant. So without loss generality, we can assume that $x_0=0$. Now for any $x\in\mathbb{R}^n$ you have by direct calculation
$$
\nabla u_k(x)=\nabla u(\rho x)\,.
$$
For simplicity, just write $\rho$ for $\rho_k$.
Now you have

\begin{align}
\frac{1}{\rho^nR^n}\int_{B_{\rho R}(0)}|\nabla u(x)|^2dx
&=\frac{1}{\rho^nR^n}\int_{B_R(0)}|\nabla u(\frac{y}{\rho})|^2\rho^ndy
&(y=\rho x,\quad dy=\rho^n dx)\\
&=\frac{1}{\rho^nR^n}\int_{B_R(0)}|\nabla u_k(y)|^2\rho^ndy
\end{align}

If you are confused with notations, it is very useful to calculate a 1D example.

Title: A Polar Coordinates Computation


问:
What's the motivation of the ideal?
I'm reading a book on Algebra, it introduces the concept of ring after some examples, the concept of ideal.

Definition I.1.8. Let $(A,+,\cdot)$ be a ring and $I$ a non-empty subset of $A$. We say that $I$ is a ideal if:

$x+y\in I,\;\;\;\;\forall \; x,y \in I$

$ax\in I, \;\;\;\;\forall \;x\in I, \;\;\;\;\forall a\in A$

I can't understand the motivation of the ideal: The sum of any elements of $I$ are elements of $I$ and the product of any element of $I$ with any element of $A$ is an element of $I$. I'm confused: What's that useful for?

答:
The notion of ideal comes from a generalization of modular arithmetic. It is a refinement (by Dedekind) of Kummer's notion of ideal number, which arose from attempts to prove Fermat's Last Theorem (or some special cases).

When you have a number $n\in {\bf Z}$, then the multiples of $n$ form an ideal in ${\bf Z}$ (a principal ideal denoted by $(n)$ or $n{\bf Z}$). The ring ${\bf Z}_n$ of integers modulo $n$ is then actually the quotient ring ${\bf Z}/n{\bf Z}$. If $p$ is prime, then $p{\bf Z}$ is a prime (even maximal) ideal and ${\bf Z}_p={\bf Z}/p{\bf Z}$ is a domain (even a field).

Similarly, if you have any ring $R$ and an ideal $I\unlhd R$, then $R/I$ makes sense and has a natural ring structure, which is a domain iff $I$ is prime and a field iff $I$ is maximal.

For Dedekind domains, there is a further analogy: if $R$ is a Dedekind domain (for example, the integer subring of a number field), then any nonzero ideal $I\unlhd R$ factors uniquely into a product of prime ideals and nonzero prime ideals are exactly the maximal ideals.

In ${\bf Z}$, this corresponds to the fact that any natural number decomposes uniquely into a product of prime numbers (as ${\bf Z}$ is a principal ideal domain, every ideal is principal).

Title: What's the motivation of the ideal?


问:
Quick Arc Length Question
So I was given the following prompt:

Let $f$ be the function satisfying $f(0)=0$ and $f'(x)=\frac{\ln(x+2)}{x^2+1}$ for $x>-2$. What is the length of the graph of $y=f(x)$ over the closed interval $[0,3]$?

I guess I'm a bit confused over what the length formula would look like in this example and how this formula would be evaluated. I understand that the length formula would look something like the following: $$L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx,$$ but I'm not seeing an easy integration when this formula is plugged into the equation. I ended up getting something like: $$L=\int_0^3\sqrt{1+\left[\frac{\ln(x+2)}{(x^2+1)}\right]^2}\,dx.$$ I'm not really understanding how I'd go about integrating this or what an answer would look like here. Any help/clarification would be appreciated!

答:
Basically you get the integral
$$\int_0^3 \sqrt{1+\left(\frac{\ln(x+2)}{x^2+1}\right)^2}\mathrm{d}x$$
Numerical integration yields $\approx 3.3197.$

Title: Quick Arc Length Question


问:
If $a$ is an odd integer, then prove that $24\mid a(a^2 - 1)$

If $a$ is an odd integer, then prove that $24\mid a(a^2 - 1)$

So, I simplified $a(a^2 -1)$ to $a(a+1)(a-1)$.

Now, I have seen some answers which first say that this is clearly divisible by $6$ and go on to express this as a multiple of $4$.

But such answers ignore that $2$ could be a common multiple. So that they ultimately prove $12\mid a(a^2 -1)$

Now, back to my unfinished proof.
I let $a = 2k+1$ and substituted this value of $a$ to get:
$4(2k+1)(k)(k+1)$

And here I am stuck...

答:
If $a\equiv3\pmod4$ then $4|a+1$ and $2|a-1$, so $8|(a+1)(a-1)$.

If $a\equiv1\pmod4$ then $2|a+1$ and $4|a-1$, so $8|(a+1)(a-1)$.

Either way, $8|a(a+1)(a-1)$.

You should be able to see that at one of $a-1$, $a$, $a+1$ is divisible by $3$. Once you show that then you can conclude $24|a(a+1)(a-1)$.


答:
Now you just need to prove that $4(2k+1)(k)(k+1)$ is divisible by $24$.

Note that one of $k$ or $k+1$ is always even. Hence the expression is divisible by $8$.

Now, if either of $k$ or $k+1$ is divisible by $3$, you are done.

But if not, $k+2$ must be divisible by $3$ and so should be $k-1$, and hence so should $(k+2)+(k-1)=2k+1$. Hence it is divisible by $3$ as well.

Body:
If $a$ is an odd integer, then prove that $24\mid a(a^2 - 1)$

So, I simplified $a(a^2 -1)$ to $a(a+1)(a-1)$.

Now, I have seen some answers which first say that this is clearly divisible by $6$ and go on to express this as a multiple of $4$.

But such answers ignore that $2$ could be a common multiple. So that they ultimately prove $12\mid a(a^2 -1)$

Now, back to my unfinished proof.
I let $a = 2k+1$ and substituted this value of $a$ to get:
$4(2k+1)(k)(k+1)$

And here I am stuck...


问:
Intuition why the volume and surface area of the unit sphere eventually decrease
The volume formula for a unit sphere, $$\frac{\pi^{n/2}}{\Gamma{(1 + n/2)}},$$
and the surface area formula, $$\frac{2\pi^{n/2}}{\Gamma{(n/2)}},$$

both attain maximum values for finite $n$. We can see from the Wikipedia page on Unit sphere that (for integer $n$ at least), those values are 5 and 7, respectively.

Now, I know that in a strict sense, these values are not really comparable, because they represent measures in different dimensions.  But I was wondering if there is nonetheless some good intuitive reason why this happens, both why they both decrease eventually, and why the values where they attain their maximums are those (apparently the actual maxima are fractional, from WolframAlpha, though it didn't give a closed form).

答:
One should consider that the formula for a sphere can be presented as a product of fractions of the form for even and odd dimensions

```<code>  1     3     5     7     9                0     2    4    6    8
 ---------------------------              -------------------------
  2    2pi   2pi   2pi   2pi      and      1    2pi  2pi  2pi  2pi
       --    --    ---   ---                    --  ---  ---   ---
        3     5     7     9                      2    4    6    8
</code>```

One sees that the numerator is stuck on $2\pi$, but the denominator accelerates away.  The intuitive answer is not so much that the sphere becomes smaller, but that the cube goes very large.  The intuition follows exactly, a square, cube, etc that has a specific margin (ie 'dihedral') angle in hyperbolic space.  The body stays around, but the vertices go to infinity etc.  This tells us that it's the cube's fault, not the sphere's, that the ratio runs away.

Volume is measured as the moment of surface.  What this means is that we construct a vector pointing out from the solid, and then integrate the dot product of a ray from a point or plane, over the element of the surface, that is this integrated over the whole surface. thus:

$d \text{ volume} = \text{coordinate}\cdot d\text{ surface}$

When the coordinate is radial, we need to divide the value by $n$.  When it is done from a plane (eg relative to the x-axis), there is no need to divide by $n$.  The sum by coordinate by radial, is exactly the sum of doing it from the $X$ axis, and the $Y$ axis etc.  That's where the division by $n$ comes in.

When one uses coordinates in radial terms, and uses the above as a definition of volume, one gets measures in tegmic units as in my Polygloss.  The ordinary measure-polytope units (square, cube, ...), derive from considering just the x-coordinate.  The radial scheme is equivilant to considering all of the perpendiculars, so the tegmic volume is n times the prismic volume for the same surface.  And since surface is a volume of N-1 dimensions, one can prove by induction, that the prismic unit is $n!$ of the tegmic one.

The tegmic radian, which is $1/n!$ of the spheric one, is slightly les than the solid angle of the simplex of the same dimension.

So after roughly 19 dimensions, one can no longer crowd $n!$ simplexes at a point.

Another approach

One should remember that a sphere will sit in any prism of lesser spheres.  So, for example, a sphere sits in a circle-prism (or cylinder).  This means that the maximum is by steps of 2d, $\pi^{n/2}$.  But this is a maximum.  The sphere in four dimensions, occupies just half of this space (ie $\pi^2/2$).  If we suppose not $\pi$ in the numerator, but $2\pi$, then the divisor corresponds to $n$: the sphere of $2\pi$ is $1/n$ of the cylinder formed by a circle, and an N-2 sphere, which leads to the resluts above.  In general, for the even dimensions, it might be writen as $\pi^{n/2}/(n/2)!$, and with odd dimensions, this relation is used to set eg $(1/2)!$ so that $2 = \pi^{1/2}/(1/2)!$, gives $(1/2)! = \pi^{1/2}/2$.

Body:
One should consider that the formula for a sphere can be presented as a product of fractions of the form for even and odd dimensions

  1     3     5     7     9                0     2    4    6    8
 ---------------------------              -------------------------
  2    2pi   2pi   2pi   2pi      and      1    2pi  2pi  2pi  2pi
       --    --    ---   ---                    --  ---  ---   ---
        3     5     7     9                      2    4    6    8

One sees that the numerator is stuck on $2\pi$, but the denominator accelerates away.  The intuitive answer is not so much that the sphere becomes smaller, but that the cube goes very large.  The intuition follows exactly, a square, cube, etc that has a specific margin (ie 'dihedral') angle in hyperbolic space.  The body stays around, but the vertices go to infinity etc.  This tells us that it's the cube's fault, not the sphere's, that the ratio runs away.

Volume is measured as the moment of surface.  What this means is that we construct a vector pointing out from the solid, and then integrate the dot product of a ray from a point or plane, over the element of the surface, that is this integrated over the whole surface. thus:

$d \text{ volume} = \text{coordinate}\cdot d\text{ surface}$

When the coordinate is radial, we need to divide the value by $n$.  When it is done from a plane (eg relative to the x-axis), there is no need to divide by $n$.  The sum by coordinate by radial, is exactly the sum of doing it from the $X$ axis, and the $Y$ axis etc.  That's where the division by $n$ comes in.

When one uses coordinates in radial terms, and uses the above as a definition of volume, one gets measures in tegmic units as in my Polygloss.  The ordinary measure-polytope units (square, cube, ...), derive from considering just the x-coordinate.  The radial scheme is equivilant to considering all of the perpendiculars, so the tegmic volume is n times the prismic volume for the same surface.  And since surface is a volume of N-1 dimensions, one can prove by induction, that the prismic unit is $n!$ of the tegmic one.

The tegmic radian, which is $1/n!$ of the spheric one, is slightly les than the solid angle of the simplex of the same dimension.

So after roughly 19 dimensions, one can no longer crowd $n!$ simplexes at a point.

Another approach

One should remember that a sphere will sit in any prism of lesser spheres.  So, for example, a sphere sits in a circle-prism (or cylinder).  This means that the maximum is by steps of 2d, $\pi^{n/2}$.  But this is a maximum.  The sphere in four dimensions, occupies just half of this space (ie $\pi^2/2$).  If we suppose not $\pi$ in the numerator, but $2\pi$, then the divisor corresponds to $n$: the sphere of $2\pi$ is $1/n$ of the cylinder formed by a circle, and an N-2 sphere, which leads to the resluts above.  In general, for the even dimensions, it might be writen as $\pi^{n/2}/(n/2)!$, and with odd dimensions, this relation is used to set eg $(1/2)!$ so that $2 = \pi^{1/2}/(1/2)!$, gives $(1/2)! = \pi^{1/2}/2$.


问:
Precison on atomless $\sigma$-algebras and their cardinality
Suppose I have an infinite $\sigma$-algebra $\mathcal{A}$ generated by a fixed set of generators whose cardinality is stricly larger than $\aleph_0$. In other words, the set of generators of $\mathcal{A}$ is uncountable. Is $\mathcal{A}$ an atomless $\sigma$-algebra?

I always thought that it is indeed the case because only countable intersections are allowed in $\mathcal{A}$. Therefore, since I have a supply of uncountably many generators, there exist no minimal non-empty elements of $\mathcal{A}$. Is that correct? (I suddenly have a doubt.)

Also, the cardinality of $\mathcal{A}$ is at least $\mathfrak{c}=2^{\aleph_0}$. I am under the impression that the cardinality of $\mathcal{A}$ is $\mathfrak{c}$ if and only if $\mathcal{A}$ is atomic and strictly larger than $\mathfrak{c}$ if and only if $\mathcal{A}$ is atomless. Is that correct?

答:
No. Take any uncountable set $X$ and look at the $\sigma$-algebra on $X$ generated by the singletons, that is $\{\{x\}\mid x\in X\}$ which has the same cardinality as $X$, which by assumption is uncountable.

Easily, this is exactly $\{A\subseteq X:\min\{|A|,|X\setminus A|\}\leq\aleph_0\}$. I leave you to convince yourself why this is an atomic $\sigma$-algebra.


问:
Surjectivity of $f:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$ via universal covering of $\mathbb{S}^{1}$
Let $f:\mathbb{S}^{1}\longrightarrow\mathbb{S}^{1}$ be given by $f(x,y) = (2xy,x^{2}-y^{2})$. Prove that $f$ is a closed map and $f$ is surjective. Use the universal covering $\phi:\mathbb{R}\longrightarrow\mathbb{S}^{1}, \phi(x)=(\cos(2\pi x),\sin(2\pi x))$ to prove surjectivity.

For the first part, note that $\mathbb{S}^{1}$ is a compact space, so every closed subset is compact. But $f$ is continuous, so it maps compact sets to compact sets, so $f$ is closed. For the second part, I don't know how to proceed. Thanks for any help.

答: If $(x,y)\in\Bbb S^1$, then $(x,y)=(\sin\alpha,\cos\alpha)$, for some $\alpha\in\Bbb R$. But then$$(x,y)=\left(2\sin\left(\frac\alpha2\right)\cos\left(\frac\alpha2\right),\cos^2\left(\frac\alpha2\right)-\sin^2\left(\frac\alpha2\right)\right),$$and therefore$$(x,y)=f\left(\sin\left(\frac\alpha2\right),\cos\left(\frac\alpha2\right)\right).$$


问:
Why in the affine space can not we use Grassmann formula?
For example, in space three-dimensional affine space generated by two skew lines is all the space three-dimensional, since they are not coplanar.
  For this reason it is not worth the Grassmann formula, which in this case would say that the space generated by the two straight lines has dimension 1 +1-0.

The affine geometry is intermediate between the geometry of vector spaces and the projective subspaces in a vector space are forced to pass through the origin. The affine space is then built in order to remedy this lack unnatural, but in doing so you lose the Grassmann formula, and many problems will lengthen the list of cases to consider: two straight lines can be accidents, coplanar, skew ... The projective space eliminates re-adding phenomena of parallelism of "new points at infinity", without restoring a "vantage point", and so here is Grassmann.

Could you help me understand this concept? In a strict and intuitive way.

答: The dimension will be the same in affine and projective geometry if you use the same definition of "generated", except that additional general position assumptions could be needed in the affine case.   If you use different notions of generation for the two geometries then of course the answer could vary.</p>