问:
Is a spanning tree *minimal* if any cycle resulting from adding an edge $e$ contains no edges with weight greater than $w(e)$?
Let $T$ be a spanning tree on some graph $G$. Adding any edge $e$ from $G$ to $T$ would result in a cycle. Consider the following two statements:

$T$ is a minimum spanning tree on $G$.

The cycle resulting from adding any $e$ from $G$ to $T$ will not have any edges with weight greater than $e$.

It is easy to see that 1 implies 2, since if 2 were false, then you could add $e$ in and remove the edge with higher weight, resulting in a tree with lower weight than $T$, a contradiction.

However, does 2 imply 1? I'm guessing yes, though it's unclear to me how to prove the impossibility of "locally minimum" spanning trees, for which 2 is true but the tree is still not a global minimum.

There's a theorem that's somewhat related, but I'm not sure how to apply it to this question:

if $G$ has some cycle $c$, then
there exists some MST of G that does not include the edge $e'$ with the
largest weight on that cycle $c$.

答:
The answer is yes: the spanning tree is guaranteed to be minimum.

The following proof paraphrases another answer.

Define a "local MST" $T$ as a spanning tree with property 2 in the question. Let $w(T)$ be the total weight of all edges in tree $T$.

Lemma: For any two local MSTs $T$ and $T'$, we have $w(T)$ = $w(T')$.

Proof:

If $T$ and $T'$ have the same edges, then we are done.

Otherwise, consider the set $S$ of edges that are in exactly one of $T$ and $T'$. Clearly $|S|$ must be even. Let $e$ be an edge with minimal weight in $S$, and without loss of generality let $e \in T$.

Adding $e$ to $T'$ would create a cycle, and that cycle must contain at least one other edge $e'$ in $S$, since $e$ existed in $T$ without causing a cycle.

By definition of $e$, we have $w(e) \leq w(e')$, but since $T'$ is a local MST, we have $w(e) \geq w(e')$, and thus $w(e) = w(e')$.

Thus, replacing $e'$ with $e$ in $T'$ keeps $w(T')$ invariant and removes both $e$ and $e'$ from $S$. Repeat the above until $S$ is empty and thus $T = T'$. Since $w(T')$ stayed constant the entire time, we must have had $w(T) = w(T')$ in the very beginning, QED.

Main result: A local MST is a true "global" minimum spanning tree.

Proof:

As discussed in the question, property 1 implies property 2 and thus an MST is a local MST. From the lemma, all local MSTs have the same weight, and thus the result follows.


问:
Show that the relation defined on an integral domain by "$x\sim y$" ("$x$ is associated with $y$") is a relation of equivalence.
I did the following demonstration: Would it be possible to give me a feedbak if i'm to wrong.

Be $A$ an integral domain and $x,y \in A$, then:

$x\sim y \Rightarrow$ $\exists u \in \mathfrak{U}(A)$ such that $y=xu \Rightarrow x \mid y$ and $y\mid x$. So there's $a,b \in A$ such that $xa=y$ and $yb=x$.

$1º$ case: $x=0.$

$x=0$ and $xa=y \Rightarrow y=0.$

Take $u=1 \in \mathfrak{U}(A)$ and then,
$0=0\cdot 1$, or $y=xu$.

$2º$ case: $a\neq 0$.

$xa=y$ and $yb=x$.

$\Rightarrow xab=x$

$\Rightarrow x(ab-1)=0$

$\Rightarrow ab=1$

$\Rightarrow a$ is invertible.

Taking $u=a \in \mathfrak{U}(A)$ and then $xu=y$.
By hypothesis, $y=xu$, $u \in \mathfrak{U}(A)$. It follows that $x\mid y.$ In addition, from $y=xu$, we have $y \cdot u^{-1}=x$ and then $ y\mid x$.

So, $x\sim y$.

$\square$

答:
I'm not sure exactly what you're trying to show in your exposition. To answer the question in the title, i.e. show that a relation ~ is an equivalence relation, we need to show three things, reflexivity, symmetry, and transitivity.

Reflexivity is clear, every element is associate to itself since $x = 1 \cdot x \forall x \in R$.

Symmetry is also easy since units are invertible, so if $x$~$y$ then there exists some unit $u$ such that $x = uy$ so $u^{-1}x=y$ i.e. $y$ ~ $x$.

Lastly we consider transitivity. Suppose $x$ ~ $y$ and $y$~$z$. Then there exist units $u,v$ such that $x = uy$ and $y = vz$. So we have that $x = uvz$. Since the units are closed under multiplication, $x$~$z$ as desired.

ContentLicense: CC BY-SA 4.0


问:
How to prove this beautiful statement on linear forms using induction?
Here is the statement :

Let $E$ a finite $\mathbb{K}$-vectorial space and $\phi_1, \ldots, \phi_p, \psi$ a collection of linear forms on $E$ such that : $\bigcap \limits_{i=1}^p\ker(\phi_i)\subset \ker(\psi)$. Then $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_p\}$.

The suggestion is to make an induction on $p\ge 1$ (I already know that this method is not the fastest way) :

For $p=1$ then $\ker(\phi_1)\subset \ker(\psi)$.

If $\phi_1 \equiv 0$ then $\ker(\phi_1)=E$ so $\psi \equiv 0$ and the statement is true.

If not, we have the fact that $\phi_1$ and $\psi$ are linearly dependent so $\psi \in \operatorname{span} \{\phi_1\}$ and the statement is also true.

Now we want to check the statement for $p+1$.

We suppose $\bigcap \limits_{i=1}^{p+1}\ker(\phi_i)\subset \ker(\psi)$ and we want to prove that  $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_{p+1}\}$.

If all the linear forms are null then $\psi \equiv 0$ and the statement is true.

If not, we have $\bigcap \limits_{i=1}^{p+1}\ker(\phi_i)\subset \bigcap \limits_{i=1}^p \ker(\phi_i) \subset \ker(\psi)$ so by the induction's hypothesis, $\psi \in \operatorname{span}\{\phi_1,\ldots, \phi_p\}$.

Moreover (not sure) $\ker(\phi_{p+1})\subset \ker(\psi)$ so $\psi \in \operatorname{span} \{\phi_{p+1}\}$.

Then I'm stuck...

Thanks in advance

答:
Take $p \in \mathbb Z^+$and suppose that for any $\mathbb K$-vector space $F$ and any $p$ linear forms $f_1,\dots,f_p$ on $F$ we have that $$\textstyle (\forall g \in F^*) \quad \bigcap_{i=1}^p \ker f_i \subseteq \ker g \ \ \Rightarrow \ \ g \in \operatorname{span}\{f_1,\dots,f_p\}.$$ Now, consider $p+1$ linear forms $\phi_1,\dots,\phi_{p+1}$ on $E$, and take $\psi \in E^*$ such that $\bigcap_{i=1}^{p+1} \ker \phi_i \subseteq \ker \psi$. Denote $\ker \phi_{p+1}$ by $K$ and observe that $\phi_1|_K,\dots,\phi_p|_K$ are $p$ linear forms on $K$ such that $\bigcap_{i=1}^p \ker \phi_i|_K \subseteq \ker \psi|_K$ because if $v \in \bigcap_{i=1}^p \ker \phi_i|_K \subseteq K$, then $v \in \bigcap_{i=1}^{p+1} \ker \phi_i \subseteq \ker \psi$ and so $0 = \psi(v) = \psi|_K(v)$.

Thus, by induction hypotheses there are $a_1,\dots,a_p \in \mathbb K$ such that $$\psi|_K = a_1\phi_1|_K + \cdots + a_p\phi_p|_K.$$ Finally, the linear form $j := \psi-(a_1\phi_1+\cdots+a_p\phi_p)$ on $E$ vanishes on $K$, and so $\ker \phi_{p+1} \subseteq \ker j$, which implies that $j$ is a scalar multiple of $\phi_{p+1}$, meaning that $\psi \in \operatorname{span}\{\phi_1,\dots,\phi_{p+1}\}$.


问:
Disk in $\mathbb{C}$ is homeomorphic to $\mathbb{C}$
This seems to be a really frequently used fact.

An open disk in $\mathbb{C}$ is homeomorphic to $\mathbb{C}$.

Does anyone know a nice way to construct the homeomorphism? I for some reason can't construct this, yet this is apparently fairly straightforward. I'm looking for a general construction method for these kinds of functions.

答:
If you can homeomorph $[0,1)$ to $[0,\infty)$ then you can rotate its graph to get a homeomorphism from the open unit disk with the 2D plane. That is, given $f:[0,1)\to[0,\infty)$ unbounded and strictly increasing (and $f(0)=0$), we can define a map $\overline{f}:D\to\mathbb{C}$ by $f(re^{i\theta})=f(r)e^{i\theta}$.

A nice example might be $f(x)=\tan(\frac{\pi}{2}x)$.

Kavi's example from the comments is $f(x)=x/(1-x)$.

An example I think is more geometrically interesting though is $f(x)=x/\sqrt{1-x^2}$. If we take a point $(x,y)$ on the unit circle in quadrant $I$ with $y\in[0,1)$ and say the line through $(0,0)$ and $(x,y)$ has slope $m$ then the function $f$ is defined by $f(y)=m$. This is a kind of stereographic projection.


问:
Comparing the value of $\int _0 ^{\frac{\pi}{2}} x^n \sin^nx\,dx$ for different $n$
Let $$I_n = \int _0 ^{\frac{\pi}{2}} x^n \sin^nx \, dx.$$

Could you tell me how to decide which one is true: $I_{2012}>I_{2013} $, or $I_{2013}>I_{2012}$?

答:
Denote $f(x)=x\sin x$ and for every real number $p>0$, denote
$$I_p:=\int_0^{\frac{\pi}{2}}\big(f(x)\big)^pdx.$$

Claim: $$I_n<I_{n+1},\quad\forall ~n\ge 2.\tag{1}$$

Proof 1(using Jensen's inequality without the evaluation of $I_2$)

Note that $f> 0$ on $(0,\frac{\pi}{2}]$, and
$$I_1=\int_0^{\frac{\pi}{2}}f(x)dx=-x\cos x\Big|_0^{\frac{\pi}{2}}+\int_0^{\frac{\pi}{2}}\cos xdx=1\tag{2}.$$
Then by Jensen's inequality, for every non-constant continuous function $g:[0,\frac{\pi}{2}]\to[0,\infty)$ and every strictly convex function $\varphi:[0,\infty)\to\mathbb{R}$, we have:
$$\varphi\big(\int_0^{\frac{\pi}{2}}g(x)f(x)dx\big)<\int_0^{\frac{\pi}{2}}\varphi(g(x))f(x)dx.\tag{3}$$
In particular, for any $p> 0$, $g=f^p\ge 0$ is continuous and not constant on $[0,\frac{\pi}{2}]$, and $\varphi(x)=x^{\frac{p+1}{p}}$ is strictly convex on $[0,\infty)$. Substituting them into $(3)$, we have:

$$\left(I_{p+1}\right)^{\frac{p+1}{p}}< I_{p+2}\iff \left(I_{p+1}\right)^{p+1}<\left(I_{p+2}\right)^p.\tag{4}$$
Since $(4)$ holds for every $p>0$, letting $p\to 0$ in the right hand side of $(4)$ and using $(2)$, we obtain that $I_2\ge 1$. Staring with $n=2$, an induction on $n\in\mathbb N$ using $(4)$ shows that
$$I_n\ge 1,\quad\forall~n\ge 2.\tag{5}$$
$(1)$ follows from $(4)$ and $(5)$ immediately. $\quad\square$

Proof 2(using Cauchy-Schwarz inequality with the evaluation of $I_2$)

For every $n\ge 2$, since $f^{\frac{n-1}{2}}$ and $f^{\frac{n+1}{2}}$ are linearly independent on $[0,\frac{\pi}{2}]$, by Cauchy-Schwarz inequality,
$$0<\left(I_n\right)^2=\big(\int_0^\frac{\pi}{2}f^{\frac{n-1}{2}}\cdot f^{\frac{n+1}{2}}dx\big)^2<\int_0^\frac{\pi}{2}f^{n-1}dx\cdot\int_0^\frac{\pi}{2}f^{n+1}dx=I_{n-1}\cdot I_{n+1}.$$
The inequality above implies that
$$\frac{I_{n}}{I_{n-1}}<\frac{I_{n+1}}{I_n},\quad\forall~n\ge 2.\tag{6}$$
Due to $(6)$, to prove $(1)$, it suffices to show that $I_2\ge I_1$.
A direct calculation shows that
$$I_2=\int_0^{\frac{\pi}{2}}x^2\cdot\frac{1-\cos 2x}{2}dx=\frac{\pi^3}{48}+\frac{1}{2}\int_0^{\frac{\pi}{2}}x\sin 2x dx=\frac{\pi^3}{48}+\frac{\pi}{8}>1=I_1,$$
which completes to proof.  $\quad\square$

ContentLicense: CC BY-SA 3.0


问:
Gradient of least-squares loss function
Given $m \times n$ matrix $\bf X$ and $m \times p$ matrix $\bf Y$, define the loss function in $n \times p$ matrix $\bf R$

$$ \operatorname{Loss} ({\bf R}) := \| \mathbf{X} \mathbf{R} - \mathbf{Y} \|_F^2 $$

where the square of the Frobenius norm of $n \times m$ matrix $\mathbf{A}$ is defined as

$$ \| \mathbf{A} \|_F^2 = \sum_{i=1}^n \sum_{j=1}^m a_{ij}^2 $$

I have to compute the gradient $\nabla_{{\bf R}} \operatorname{Loss}$. My source says:

$$ \nabla_{{\bf R}} \operatorname{Loss} ({\bf R}) = \dfrac2m \mathbf{X}^T (\mathbf{X} \mathbf{R} - \mathbf{Y} ) $$

but I am not sure how to obtain this result.

答:
Use a colon to denote the trace/Frobenius product
$$\eqalign{
A:B \;=\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij}
 \;=\; {\rm Tr}(A^TB)  \;=\; {\rm Tr}(AB^T) \\
}$$
then the derivative/gradient is fairly simple to calculate
$$\eqalign{
W &= XR-Y \\
dW &= X\,dR\\
Loss &= \big\|W\big\|^2_F \,\;=\; W:W \\
d\,Loss &= 2W:dW \;=\; 2W:X\,dR \;=\; 2X^TW\color{red}{:dR} \\
\frac{\partial\,Loss}{\color{red}{\partial R}} &= 2X^TW \;=\; 2X^T(XR-Y) \\
}$$
So the answer from your source is off by a factor of $\left(\frac 1m\right)$

Title: Gradient of least-squares loss function


问:
How to find the total spendings in a restaurant when it is given as a system of equations using exponents?
The problem is as follows:

Jenny goes to dinner to a well known restaurant in town. The number of times she goes to the restaurant is $\frac{a}{2}$ times and for each time she went, she spent $b^{2}$ hundreds of dollars. Find the money Jenny spent assuming the following system of equations was observed by Jenny.

$$\begin{cases}
3^a-2^{b^{2}}=77 \\
3^{\frac{a}{2}}-2^{\frac{b^2}{2}}=7 \\
\end{cases}$$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{600 dollars}\\
2.&\textrm{500 dollars}\\
3.&\textrm{400 dollars}\\
4.&\textrm{300 dollars}\\
\end{array}$

In this given situation the only thing which I can understand is as this:

$\textrm{Money spent}=\frac{a}{2}\cdot b^2\cdot 10^2$

But the thing is how to find $a$ and $b$, which sort of simplication or algebraic manipulation can be done to solve this system of equation.

Can someone help me?.

答:
Guide:

We have

$$3^a-2^{b^2}=77$$

and

$$3^{\frac{a}2}-2^{\frac{b^2}2}=7$$

Dividing the equations, we get

$$3^{\frac{a}2}+2^{\frac{b^2}2}=11$$

Now, solve a linear system equation to solve for $3^\frac{a}2$ and $2^\frac{b^2}2$.


问:
Laurent series of an integral with parameter
To find the Laurent series of function $f(a)$ at point $a=0$
$$
f(a)=\int^1_0 \frac{d x}{x^2+a^2}
$$
one can first do the integral
$$
f(a)=\frac{1}{a}\arctan(1/a)
$$
then expand $\arctan(1/a)$ and obtain
$$
f(a)\approx \frac{\pi}{2a}-1+O(a^2).
$$
However if one expands the integrand before integration,
one would get divergent integral.

My question is: is there any method helping us expand the function $f(a)$ before performing the integral?
The motivation arises from more complicated cases, when the functions can not be integrated out analytically.

==========================================

Updated: I think I find the solution to this integral. When I expand the function $f(a)$, I have to expand all the integral, not only the integrand
$$
\frac{1}{x^2+a^2}\approx \frac{1}{x^2}+O(a^2)
$$
which means that the upper limit of integral has also to be changed, i.e.
$$
f(a)\approx \int^1_a d x\left(\frac{1}{x^2}+O(a^2)\right)
$$
this gives us
$$
f(a)\approx \frac{13}{15 a}-1+O(a^2)
$$
Except the coefficient of $a^{-1}$, all the other orders are consistent.

答:
Step 1: Expanding the integrand with respect to $a$ round $a\sim 0$
$$
\frac{1}{x^2+a^2}=\sum_{n=0}^\infty (-1)^n\frac{ a^{2n}}{x^{2n+2}}
$$
Step 2: Deforming the integral
$$
\int^\infty_0 dx \to \int^\infty_{\epsilon a} dx
$$
Step 3: Performing the integral
$$
\sum_{n=0}^\infty (-1)^n a^{2n}\int^\infty_{\epsilon a} \frac{dx}{x^{2n+2}}
=\sum_{n=0}^\infty (-1)^n \frac{a^{2n}}{2n+1}
+\frac{1}{a}\sum_{n=0}^\infty  \frac{(-1)^n}{(2n+1)\epsilon^{2n+1}}
$$
The second term is nothing else but ${\rm arccot}(\epsilon)/a$.

Step 4: Taking the limit $\epsilon \to 0$

$${\rm arccot}(\epsilon)/a\to \pi/(2a)$$
and
$$f(a) = \frac{\pi}{2a}+\sum_{n=0}^\infty (-1)^n \frac{a^{2n}}{2n+1}$$


问:
Graph theory - a separation property of a normal tree
I'm a novice in this new field of mathematics and would like to ask you for a more concrete explanation underlying the single-line explanation on the textbook.

After a brief introduction of the definition of a normal tree in a graph, these two lemma immediately follows. Author emphasizes the usefulness of a normal tree as an examining tool for the structure of a graph: it projects its separation properties onto the graph.

I understand the first lemma but the very last part of the second one is bothering me a lot. How can we use the result in (i) to prove that there are no other minimal elements? As I understand, it at least needs to precede that the given component C := G-S has a normal tree in it.

Of course there is a theorem that states the existence of normal (spanning) trees in finite connected graphs, but the theorem hasn't been introduced since. So I conjecture that author herein assumes a quite straightforward proof of a weaker theorem. Would anyone help me with this?

*Because I'm not familiar with this field, I'm worried that I might be missing some essential understanding. That's why I'm trying to avoid those theorems covered later on the textbook.

refer to the pg.16 for the book

答: Suppose that $C$ is a component of $G-S$ and that $x$ and $y$ are minimal vertices of $C$. $C$ is connected, so there is an $x$-$y$ path in $C$. By (i) we know that this path contains a vertex $v\in\lceil x\rceil\cap\lceil y\rceil$. Then $v\le x$ and $v\in C$, so $v=x$, since $x$ is minimal in $C$. Similarly, $v\le y$, so $v=y$. But then $x=v=y$, and $C$ must have a unique minimal vertex.


问:
Differentiability implies Lipschitz continuity

Let $f:[0,1]\to\mathbb{R}$ be a continuous function and suppose $f$ is differentiable at $x_0\in [0,1]$. Is it true that there exists $L>0$ such that $\lvert f(x)-f(x_0)\lvert\leq L\lvert x-x_0\lvert$?

I know that local continuously differentiable implies local Lipschitz continuity. Is this still true in the case given above?

答: From differentiability at $x_0$, you will find an $L_1$ such that $| f(x)-f(x_0) |\leq L_1| x-x_0|$ for $|x-x_0| < \delta$. Since $f$ is continuous on a compact set, $|f(x)|_{\infty} < \infty$. This will give you an $L_2$ such that $| f(x)-f(x_0) |\leq L_2| x-x_0|$ for $|x-x_0| \geq \delta$. Take $L = \max \{L_1, L_2\}$.</p>


问:
How to show equally likeliness if it's not given in the premise of the hat problem
Given question below is a slightly modified cliche example for probability.

The Hat Problem. Suppose that 4 people throw their hats in a box and then each picks one hat at random. (Each hat can be picked by only one person.) What is the expected value of X, the number of people that get back their own hat?

For n = 4, we create a sample space whether people gets his/her hat or not. After we create sample space through permutations, we derive our probability mass function by using discrete uniform probability law. Then we can solve it for:

$E[X] = \sum\limits_{x} x*p_{x}(x)$

Since we assume all permutations are equally likely, is this solution can be accepted as valid? How to show equally likeliness for permutations?

If the solution doesn't valid, what is the proper way to solve without knowing premise of equally likeliness for permutations?

答:
If you want to compute the expectation, you can ask first what the probability that one person gets his hat. This probability is $1/n$.
Then, using linearity of expectation, the expected number of people who got their hats is $n \cdot (1/n) = 1$


问:
Summation of Cosine Series
Find the summation of the series $\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} $

My approach is as follow

$\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)}  \Rightarrow \sum\limits_{k = 0}^n {\left( {1 - \cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \right)} $

$ \Rightarrow \sum\limits_{k = 0}^n {\left( 1 \right)}  - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)}  \Rightarrow \left( {n + 1} \right) - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} $

$\sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)}  = \sum\limits_{k = 0}^n {\cos \left( {\frac{{2\pi }}{{n + 2}} + \frac{2}{{n + 2}}k} \right)} ;a = \frac{{2\pi }}{{n + 2}};d = \frac{2}{{n + 2}}$

From the website I got the following formula but the summation of the series is from $0$ to $n-1$.

$\sum\limits_{k = 0}^{n - 1} {\cos \left( {a + kd} \right)}  = \frac{{\sin \left( {\frac{{nd}}{2}} \right)}}{{\sin \left( {\frac{d}{2}} \right)}} \times \cos \left( {a + \frac{{\left( {n - 1} \right)d}}{2}} \right)$

Where as in the question it is from $0$ to $n$, a total of $n+1$ terms. How do I proceed

答:
Move the last term of the sum out of the sum.  The remaining sum will match your reference.
$$\sum_{k=0}^n f(k) = \left( \sum_{k=0}^{n-1} f(k) \right) + f(n)  \text{.}$$

Title: Summation of Cosine Series


问:
A couple has two children. At least one is a boy. What is the probability that the couple has one child of each sex?
I assume that the question assumed that the probability of having a boy is 50% and the probability of having a girl is 50%

So I'm confused because this seems like a really easy probability question. If they already have a boy. The probability that they have a girl is just $.5$ isn't it?

But the book says that the probability is $2/3$

This seems completely illogical to me because then that would assume that the boy to girl birth ratio is $1$ to $2$ which doesn't seem right at all.

What am I misinterpreting here? Is the book wrong?

答:
By careful of what is said and what it means.

The question said "at least one" was a boy; not "a specific child" is a boy.

There are four equally likely possibilities.  They have two boys, they have two girls, they have a boy and a girl, or they have a girl and a boy.

If we know they have at least one boy that rules out having two girls.  That leaves three equally likely scenarios.  They have two boys, they have a boy and a girl, or they have a girl and a boy.

Now had they question been, the couple have two children and one of them is a boy, what is the probably the other one is a girl, you'd be correct.  But that wasn't the question.  The question was (could be paraphrased as) the couple of two children, given that one or the other, or maybe both are boys, what is the probability that they are one (either one) is a boy and the other is a girl.


答:
$\dfrac{\sin\frac{(n+1)d}{2}}{\sin \frac{d}{2}} \cdot \cos(a + \dfrac{nd}{2})$

For summation from $0$ to $n$

That will  work

A very basic proof if you want would be substitute n = t -1 and then using the summation formula for cosine.

Proof of summation formula of cosine and sine up to n-1 terms

Title: Summation of Cosine Series

Body:
Find the summation of the series $\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} $

My approach is as follow

$\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)}  \Rightarrow \sum\limits_{k = 0}^n {\left( {1 - \cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \right)} $

$ \Rightarrow \sum\limits_{k = 0}^n {\left( 1 \right)}  - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)}  \Rightarrow \left( {n + 1} \right) - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} $

$\sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)}  = \sum\limits_{k = 0}^n {\cos \left( {\frac{{2\pi }}{{n + 2}} + \frac{2}{{n + 2}}k} \right)} ;a = \frac{{2\pi }}{{n + 2}};d = \frac{2}{{n + 2}}$

From the website I got the following formula but the summation of the series is from $0$ to $n-1$.

$\sum\limits_{k = 0}^{n - 1} {\cos \left( {a + kd} \right)}  = \frac{{\sin \left( {\frac{{nd}}{2}} \right)}}{{\sin \left( {\frac{d}{2}} \right)}} \times \cos \left( {a + \frac{{\left( {n - 1} \right)d}}{2}} \right)$

Where as in the question it is from $0$ to $n$, a total of $n+1$ terms. How do I proceed


问:
Convergence of $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$
I want to show that $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$ is convergent.

I tried to use the limit convergence test by comparing with $\frac{1}{x^2}$ but that doesn't work. Any hints on how to proceed with this?

答:
The problem is at the endpoint $1$.
You want to compare with $\int_1^2 \frac{1}{\sqrt{x-1}}\; dx$.
Possibly helpful is
$$  x^4-1 = (x-1)(x^3 + x^2 + x + 1)$$


问:
How to prove $A\cap B \subseteq \overline{A \triangle B}$
An exercise asks me to prove the following:
$$A\cap B \subseteq \overline{A \triangle B}$$

Unlike most other exercises, this one implies a symmetric difference, of which I am unfamiliar in this kind of proofs. There was little I could do, here:

The statement can be rewritten as the following:
$$A\cap B \subseteq \overline{(A-B)\cap (B-A)}$$
$$A\cap B \subseteq \overline{(A-B)}\cap \overline{(B-A)}$$
$$A\cap B \subseteq (\overline{A} - \overline{B}) \cap (\overline{B} - \overline{A})$$
I rewrote it because the symmetric difference doesn't seem "primitive" enough for me to operate with. Then my proof begins:
$$x \in A \cap B \implies x\in A \land x \in B$$
$$\implies x \in (A \cap B) \land x \in (B \cap A)$$
$$\implies (x \in A \land x \in B) \land (x \in B \land x \in A)$$
And then, I got stuck. I don't see how can $(x \in A \land x \in B) \land (x \in B \land x \in A)$ become what I needed at all.

答:
A straightforward calculational proof, which uses the following definition of $\triangle$: $$x \in A \triangle B \;\equiv\; x \in A \not\equiv x \in B$$ starts at the most complex side (here: the right hand side) of the statement in question, and goes like this:
$$
\begin{align}
& x \in \overline{A \triangle B} \\
\equiv & \;\;\;\;\;\text{"definition of $\overline{\phantom\square}$; the above definition of $\triangle$"} \\
& \lnot(x \in A \:\not\equiv\: x \in B) \\
\equiv & \;\;\;\;\;\text{"logic: simplify"} \\
& x \in A \:\equiv\: x \in B \\
\Leftarrow & \;\;\;\;\;\text{"logic: weakening -- suggested by the shape of the left hand side ($A \cap B$)"} \\
& x \in A \:\land\: x \in B \\
\equiv & \;\;\;\;\;\text{"definition of $\cap$"} \\
& x \in A \cap B \\
\end{align}
$$
By the definition of $\subseteq$ this proves the original statement.</p>


答: You can set $x^2=sect$ and perform this substitution. When you do that, you in up with an integral of the form$\int\frac{dt}{\sqrt{cost}}$. The key is in the upper and lower bound. For both $x=1$ and $x=2$ you can conclude that on the corresponding $t$ interval, the function is continuous. in other words, it is no longer an improper integral, the integral has an answer (though through elementary techniques we can' figure that out) thus convergent.

Title: Convergence of $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$

Body:
I want to show that $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$ is convergent.

I tried to use the limit convergence test by comparing with $\frac{1}{x^2}$ but that doesn't work. Any hints on how to proceed with this?


问:
Area of triangle with incircle tangent to semicircle whose diameter is on one triangle side
Triangle $\triangle ABC$ with incenter $I$ and inradius $r$ has the following property: the incircle is tangent to the semicircle with diameter $AB$ and is within the semicircle. Find the area of $\triangle ABC$ as a function of the side $AB$ and the inradius $r$.

So far I've noticed that $\angle AEF=\angle BEF=\frac{\pi}{4}$ but I don't think that that's useful in any way.

答:
$\triangle ABC = \frac{1}{2} AB \times h$

Let's get onto finding $h$ and then we are done.

We know $R = \frac{AB}{2} \,( R$ being the radius of the semicircle).

$AB = h \cot A + h \cot B \implies h = \frac{AB}{\cot A + \cot B}$

Now, $\cot \frac{A}{2} = \frac{AF}{IF} = \frac{R+x}{r}$

Similarly, $\cot \frac{B}{2} = \frac{R - x}{r}$

Using the identity $\, \cot 2 \alpha  = \frac{1}{2} (\cot \alpha - \tan \alpha), \,$ we have,

$2 \cot A = \frac{(R+x)^2 - r^2}{r(R+x)} \,, \, 2 \cot B = \frac{(R-x)^2 - r^2}{r(R-x)}$

$ 2(\cot A + \cot B) = \frac{(R^2-x^2)(R+x) + (R^2-x^2)(R -x) - 2 r^2R}{r(R^2-x^2)}$

Also, $R^2 - x^2 = R^2 - ((R-r)^2 - r^2) = 2Rr$

So, $ 2(\cot A + \cot B) = \frac{4R - 2r}{2r} = \frac{AB - r}{r}$

And hence $\frac{h}{2} = \frac{AB \, r}{AB - r} \, , \,$ Area $\triangle ABC = \frac{AB^2 \, r}{AB - r}$


答:
You can use the fact that $\frac{1}{\sqrt{x^4-1}} < \frac{1}{\sqrt{x-1}}$ for all $x>1$. This is because $x^4-1>x-1$ for all $x>1$.

Therefore, we have $\int_1^2 \frac{1}{\sqrt{x^4-1}} < \int_1^2 \frac{1}{\sqrt{x-1}} = 2\sqrt{x-1} |_1^2 = 2$


问:
If every convergent subsequence converges to the same limit, does the whole sequence converge to that limit?
$(E,d)$ be a metric space and $(x_n)_{n\in\mathbb N}\subseteq E$ be relatively compact.

Assume that there is a $x\in E$ such that every convergent subsequence converges to $x$. Does it already follow that $(x_n)_{n\in\mathbb N}$ is convergent and converges to $x$?

The convergence of $(x_n)_{n\in\mathbb N}$ to $x$ is equivalent to the assertion that every aribtrary subsequence is convergent and converges to $x$; but we only know that every subsequence, for which we a priori assume that it is convergent, converges to $x$.

答:
What about the following example in the interval (0,2): $1,\frac{1}{2},1,\frac{1}{3},1,\frac{1}{4},1,\frac{1}{5},1,\frac{1}{6},1,\frac{1}{7},\dots$

Every convergent subsequence converges to 1, but the entire sequence does not converge


问:
Can I use 2 sample t-test in the same distribution
I need to check if there is a significant difference in recorded level values between the two decades. I have a data that includes both of the decades.

I decided to use a t-test and was thinking using the two-sample t test but the definition of 2-sample t tests are that they are from 2 different samples and are independent which is not true in my case.

Can I assume they are 2 different samples or should I use something else?

答:
You are interested in seeing if there is significant difference in recorded level values between the two decades.

If you are certain that the two distributions are independent, go for the 'two sample t-test.
(two sample t-test requires independent sample)

If the two distributions are dependent, 'paired t-test' would be more appropriate.
(paired t-test requires dependent sample)

I'd like to start with what are dependent or independent samples are and which tests to use for each case.

Dependent samples are such data that are 'related' to each other.
For example, let's say you are interested in the drug effect on blood pressure(bp) level and would like to compare the bp level measured before taking the drug and bp level measured after taking the drug.
'BP level BEFORE taking drug' gives you some information about 'BP level AFTER taking drug', which means BP level is dependent on the existence of drug in each patient's body(same to say that the bp level of the 'Group Before taking drug' is dependent on bp level of the 'Group After taking drug'). In other words, two groups are related in regards to the bp level.
Obviously, we have the same patient for each pair of Before/After BP level measurement, this would give us the idea about the effect of the drug, which again, Before group will give us information about After group about drug effects on BP level and vice versa.

Thus,'paired t-test' would be appropriate for the dependent sample above. You are looking at the paired observation(BP Before and After taking the drug) and testing to see if there's any significant mean difference between these paired dependent samples.
The null hypothesis would be "paired difference(Before - After) is significantly different from 0"

(You are most likely to use paired t-test when the same subjects are measured on the two different things - like same patient's bp is measured before/after the drug. However, there are exceptions too. If we are looking at the twins and the difference of heights between twins it's the two different individuals that we're looking at. However, again, since they share the gene, diets, etc, they are dependent samples. I'm brining up this example to say that study subjects that are NOT the same individual can also be the dependent samples, depending on how the one group of sample is related to another group of the sample.)

Moving onto the independent samples. independent samples of first group does not give such information to the second group. Let's say we're interested in measuring the difference of height in people in France and Nepal. Two populations are independent of each other because people in France does not really give information about people in Nepal.

In this case, two sample t-test will be appropriate.

Now, in your case of recorded 'level values between the two decades'. I cannot tell much about which one you should use because I do not have much information about the data.

However, if you are certain that the first decade somehow gives information about the second decade or if the first decade is related to the second decade, use the paired t-test.

However, if you find that the first decade is totally independent to the second decade, please use the two sample t-test. It really depends on the description of the data and I do not have much information of each decade so please look into how the data was collected and interpreted to determine if the two decades are related or not.

Thank you.


问:
Show that the relation defined on an integral domain by "$x\sim y$" ("$x$ is associated with $y$") is a relation of equivalence.
I did the following demonstration: Would it be possible to give me a feedbak if i'm to wrong.

Be $A$ an integral domain and $x,y \in A$, then:

$x\sim y \Rightarrow$ $\exists u \in \mathfrak{U}(A)$ such that $y=xu \Rightarrow x \mid y$ and $y\mid x$. So there's $a,b \in A$ such that $xa=y$ and $yb=x$.

$1º$ case: $x=0.$

$x=0$ and $xa=y \Rightarrow y=0.$

Take $u=1 \in \mathfrak{U}(A)$ and then,
$0=0\cdot 1$, or $y=xu$.

$2º$ case: $a\neq 0$.

$xa=y$ and $yb=x$.

$\Rightarrow xab=x$

$\Rightarrow x(ab-1)=0$

$\Rightarrow ab=1$

$\Rightarrow a$ is invertible.

Taking $u=a \in \mathfrak{U}(A)$ and then $xu=y$.
By hypothesis, $y=xu$, $u \in \mathfrak{U}(A)$. It follows that $x\mid y.$ In addition, from $y=xu$, we have $y \cdot u^{-1}=x$ and then $ y\mid x$.

So, $x\sim y$.

$\square$

答:
This holds not only for the unit group but for any subgroup $G$ of the multiplicative group, i.e.

suppose that $\,G\subset A^{\times}\,$ satisfies $\,\color{#c00}1\in G\,$ and $\,g,h\in G\,\Rightarrow\, \color{#08f}{gh}\in G,\,$ and $\, \color{#0a0}{g^{-1}}\in G$.

Note $\ x\approx y \!\overset{\rm def\!\!}\iff\! x = g\, y\ \ {\rm for\ some}\ \ g\in G\ $ is an equivalence relation, $ $  since

$\qquad\ \ \ \  \approx\,$ is $\ \ \rm\color{#c00}{reflexive}\,\ \ $ by $\  x = \color{#c00}1x\,\Rightarrow\, x\approx x$

$\qquad\ \ \ \  \approx\,$ is $\,\rm\color{#0a0}{symmetric}\,$ by $\ x\approx y\,\Rightarrow\, x = g y\,\Rightarrow\, y = \color{#0a0}{g^{-1}}x\,\Rightarrow\,y\approx x$

$\qquad\ \ \ \ \approx\,$ is $\ \rm\color{#08f}{transitive}\,\ $  by $\,x\approx y\approx z\,\Rightarrow\, y = hz,\,  x = g y = g(hz) = (\color{#08f}{gh})z\,\Rightarrow\, x\approx z$

Remark $\ G x = \{ gx\, :\, g\in G\}\ $ is called the $G$-orbit of $x.\, $ It is a basic concept in group theory.

A quick way to recognize such group structure is by the subgroup test, i.e. a nonempty $G\subset H$ of a group $H$ forms a group $\iff$ it is closed under division, i.e. $\, g,h\in g\,\Rightarrow g/h = gh^{-1}\in G,\,$ which is clear for units.

When studying divisibility it is often useful to ignore units by factoring out the unit group, e.g. see the divisibility group and  its associated characterizations of UFDs and gcd, valuation and Reisz domains mentioned there.


问:
find out the value of $\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$
If $(x-3)^2+(y-5)^2+(z-4)^2=0$,then find out the value of $$\dfrac {x^2}{9}+\dfrac {y^2}{25}+\dfrac {z^2}{16}$$

just give hint to start solution.

Body: Hint: $(x-3)^3\geq0,(y-5)^2\geq0,(z-4)^2\geq0$.


答: Let the original sequence be $\sigma$, and let $K=\operatorname{cl}\{x_n:n\in\Bbb N\}$; by hypothesis $K$ is compact. If $\sigma$ does not converge to $x$, there are an $\epsilon>0$ and a subsequence $\sigma_1=\langle x_{n_k}:k\in\Bbb N\rangle$ such that $d(x_{n_k},x)\ge\epsilon$ for all $k\in\Bbb N$. Clearly $\sigma_1$ is a sequence in the compact set $K$, so it has a convergent subsequence $\sigma_2=\langle x_{n_{k_i}}:i\in\Bbb N\rangle$. But $\sigma_2$ is also a subsequence of $\sigma$, so by hypothesis it converges to $x$, contradicting the fact that $d(x_{n_{k_i}},x)\ge \epsilon$ for each $i\in\Bbb N$. Thus, $\sigma$ must converge to $x$.

Title: If every convergent subsequence converges to the same limit, does the whole sequence converge to that limit?

Body:
$(E,d)$ be a metric space and $(x_n)_{n\in\mathbb N}\subseteq E$ be relatively compact.

Assume that there is a $x\in E$ such that every convergent subsequence converges to $x$. Does it already follow that $(x_n)_{n\in\mathbb N}$ is convergent and converges to $x$?

The convergence of $(x_n)_{n\in\mathbb N}$ to $x$ is equivalent to the assertion that every aribtrary subsequence is convergent and converges to $x$; but we only know that every subsequence, for which we a priori assume that it is convergent, converges to $x$.


问:
Can two events in probability be both non-disjoint AND non-independent? Confused about interpretation of conditional probabilities?
I am currently reviewing conditional probability in my textbook and have stumbled across this:

The formula for conditional probability is this: Where $A$ and $B$ are two events where $P(B) > 0$ (naturally)

$$P(A \mid B) = \frac{P(A\cap B)}{P(B)}$$

so naturally: $P(A\cap B) = P(B)\cdot P(A\mid B)$

and if $A$ and $B$ are independent, $P(A \mid B) = P(A)$

so $P(A\cap B) = P(B) \cdot P(A)$ which matches the definition of independence.

What I wonder is this. Say that $P(B)$ and $P(A \mid B)$ are non-zero probabilities and $P(A \mid B) \neq P(A)$ so $P(A\cap B) \neq 0$ and $A$ and $B$ are non-disjoint AND not independent events. Is this legal? Have I broken a law of probability? The textbook doesn't specify. If it's legal can someone give me an example in real life of two events that are non-disjoint and not independent?

Also can someone answer my second question?

Does the RHS of $P(A\cap B) = P(B)\cdot P(A\mid B)$ mean that event $B$ occurs first and THEN event $A$ occurs in which event $B$ occurring first affected the probability of event $A$ also occurring?

OR

Does the RHS mean that event $B$ and $A$ occur simultaneously? (If it's this second interpretation, can someone explain how events $A$ and $B$ occurring simultaneously affects event $A$'s probability of occurring?)

答: It not only is not illegal but it is the typical situation. For example: The event that a man is more than six feet tall and the event that he weighs more than 200 pounds are not independent and not disjoint.</p>


问:
What’s a good (fairly advanced) book on probability, cumulants and CLT please?
Can anyone recommend a book covering probability, cumulants and a proof of the CLT (ideally including a proof of Levy’s continuity theorem).

I studied Math undergrad and PhD many years ago, but don’t work in the field.

I’ve read Yuval Filmus‘s note on the CLT and the Wiki pages.  Cumulants make intuitive sense, but e.g. I can’t see a proof of Levy’s continuity theorem online.

As a bonus, if there’s any version of the CLT for correlated random variable size , that would be interesting.

Thanks

答:
I would recommend Introduction to mathematical statistics by Robert V. Hogg, Joseph Mckean, Allen T. Craig 7th Edition.
(I'm not sponsored)

The first chapter gives you pretty detailed information about the probability and distributions. I also recommend you to look at the second and the third chapter which talks about multivariate distribution and some special distributions.

If you go to the Index page of the book, you can look for the CLT. The book that I currently own indicates that page 170,220,225,249,313,317,323-324, 340, 342, 352,370, 448, 456, 463, 541, 552, 570, 605, and 626 have information related to CLT.

I was struggling a lot with statistics, not coming from the math or stat background and this book really helped me a lot.

Thanks.


问:
Real Analysis Qualifying Exam Sample Question
Studying for the RA Quals, when my best friend and I came across this problem in Chapter 4 of Wheeler's, "Measure and Integral an Introduction to Real Analysis," #15 pg 62. We're just curious if anybody could/would tell us how to approach it.

Let $\{f_{k}\}$ be a sequence of measurable functions defined on a measurable $E$ with $|E|<+\infty$. If $|f_{k}(x)|\leq M_{x}<+\infty$ for all $k$ for each $x\in E$, show that given $\epsilon >0$, there is a closed $F\subset E$ and a finite $M$ such that $|E-F|<\epsilon$ and $|f_{k}(x)|\leq M$ for all $k$ and all $x\in F$.

答:
Without loss of generality, we may assume $M_x = \sup_k \lvert f_k(x) \rvert$. For each positive integer $n$, let $S_n = \{x \in E : M_x \leq n\}$. Since each $M_x$ is finite, $\bigcup_{n \geq 1} S_n = E$. Also, for each $n$, we have that
$$S_n = \bigcap_k \{x \in E : \lvert f_k(x) \rvert \leq n\}$$
is measurable (since it is a countable intersection of measurable sets). Let $\varepsilon > 0$ be arbitrary. Since the sequence $\lvert \bigcup_{n=1}^m S_n \rvert$ increases monotonically to $\lvert E \rvert < \infty$, there is some $M > 0$ such that $F_0 := \bigcup_{n=1}^M S_n$ satisfies $\lvert E - F_0 \rvert < \varepsilon/2$, and moreover $\lvert f_k(x) \rvert \leq M$ for all $k$ and all $x \in F_0$.

The only issue is that $F_0$ might not be closed! Here, we use the regularity of the Lebesgue measure to produce an open set $O \supseteq F_0$ and a closed set $F \subseteq F_0$ such that $\lvert O - F \rvert < \varepsilon/2$. Now, since $F$ is a subset of $F_0$, we have
$$\lvert E - F \rvert = \lvert E - F_0 \rvert + \lvert F_0 - F\rvert < \frac{\varepsilon}{2} + \lvert O - F \rvert < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$
and $\lvert f_k(x) \rvert \leq M$ for all $k$ and all $x \in F$.


问:
what's the coefficient inverse polynomial?
The polynomial $(1+x)^n$ has a form in coefficient $\sum_{i=0}^{n} w_ix^i$,  and i'm interested in this polynomail $\sum_{i=0}^n \frac{1}{w_i}x^i$, and is there any inverse presentation in the form of $\prod_{i=0}^{n} (1+k_ix)$ ?

答:
Even in the case $n = 2$, this is hopeless.

$$  (1+x)^2 = 1+2x+x^2  \text{.}  $$

The polynomial
$$  P(x) = 1 + \frac{1}{2} x + x^2  $$
is not of the desired form.  In detail,
$$  P(x) = \left( x + \frac{1}{4} + \frac{\mathrm{i}}{4} \sqrt{15} \right)\left( x + \frac{1}{4} - \frac{\mathrm{i}}{4} \sqrt{15} \right)  $$
and if we force
$$  1+k_1 x = x + \frac{1}{4} + \frac{\mathrm{i}}{4} \sqrt{15}  \text{,}  $$
then
$$  k_1 = \frac{-3 + \mathrm{i}\sqrt{15} + 4x}{4x}  \text{,}  $$
which is not independent of $x$ (for instance, its values at $x=1$ and $x=2$ are different).


问:
In what formal system is Godel's Incompleteness Theorem (or similar statements of undecidability) proven?
For example, consider the proof using Rosser's trick as shown on wikipedia.

https://en.wikipedia.org/wiki/Rosser%27s_trick#The_Rosser_sentence

That proof isn't inside the arithmetical theory T, but in some higher level meta-theory. So if I wanted to do meta-reasoning about Godel's proof (or Rosser's etc), I would need a system that could refer to or even encode statements like "For any fixed arithmetical theory $T$, let Proof$_T(x,y)$..." In what formal system would the statement "For any fixed arithmetical theory $T$..." exist? It obviously wouldn't be in $T$ itself since that would be tantamount to a proof that $T$ is consistent that is inside $T$, in which case $T$ would be inconsistent by Godel's second incompleteness theorem, thus making $T$ useless as a formal system. Is it second order logic? ZFC?

答:
You write

It obviously wouldn't be in $T$ itself since that would be tantamount to a proof that $T$ is consistent that is inside $T$.

This is a common misconception. The incompleteness theorem, appropriately phrased, can be proved in (first-order) $\mathsf{PA}$ or indeed much less. Here's the precise statement of the theorem:

Suppose $T$ is a computably axiomatizable consistent theory which interprets Robinson arithmetic. Then $T$ is incomplete.

Note that consistency is folded into the hypothesis. In particular $\mathsf{PA}$ can prove

If $\mathsf{PA}$ is consistent, then it is incomplete

(since it's easy for $\mathsf{PA}$ to prove that $\mathsf{PA}$ is computably axiomatizable and interprets Robinson arithmetic), but this doesn't contradict the inability of $\mathsf{PA}$ to prove its own consistency.

That said, there is a "fundamentally meta" aspect to the claim that the incompleteness theorem can be proved in $\mathsf{PA}$.

What $\mathsf{PA}$ actually proves is a sentence $(*)$ about numbers, which we claim is a faithful rephrasing of "Every computably axiomatizable consistent theory interpreting Robinson arithmetic is incomplete" in some sense. But where is that claim made and evaluated?

Basically, on some level we have to commit to the position that a particular formal system successfully describes (at least to a certain extent) the behavior of formal systems themselves. This can feel like a bit of a rabbit hole, but remember: we do need to start with something.


问:
Is there any software which does Ring Computations?
Are there any software that is able to compute the following problems:

Verifying if or not a subset of a ring is an ideal.

Generating all ideals of a given finite ring.

Finding maximal multiplicative closed subset of a finite ring, etc.

I found this question previously posted, but this deals with Groups.

答:
For one thing, you can work with rings in GAP! Documentation

If this isn't powerful enough, maybe look into Sage or Magma. These are general-purpose computer algebra systems that have some built-in support for rings and such.

Macaulay2 and Singular are also popular software tools, but I think they're more focused towards commutative algebra and algebraic geometry (I've never used them, though -- perhaps they do have what you're looking for).


问:
Finding $PR + RQ$ given $PQ = 1$
Suppose we have triangle $ABC$ such that $\angle B$'s external angle bisector intersects $AC$'s extension at $P$ and $\angle A$'s external angle bisector intersects $BC$'s extension at $Q.$ Then, let the intersection of $BP$ and $AQ$ be $R.$ If the circumcircle of $\triangle PQC$ passes through $R$ and $PQ = 1,$ find the value of $PR + RQ.$

Due to the fact that $RPQC$ is cyclic, I got $$\angle RQP = \angle ACR = \angle RCB = \angle RPQ.$$ Therefore, I found that $PR = RQ.$ However, I was unsure how to proceed from here. Can somebody help please?

答:
Here is a picture:

Since you have shown that $RQ = RP$, we just need to find $\angle QRP$ (the green angle).

All same-colored angles are equal since they relate to the external angle bisectors. Now consider:

\begin{align}
\angle QRP &= \angle QCP\\
\text{In }\triangle ABC, \ \angle ACQ &= (180^\circ - 2\angle ABR) + (180^\circ - 2\angle BAR)\\
\text{In }\triangle ABR, \ \angle ARP &= \angle ABR + \angle BAR
\end{align}

This gives $\angle QRP = \angle ABR + \angle BAR = 120^\circ$, and the rest of the question follows.

Title: Finding $PR + RQ$ given $PQ = 1$


问:
How to find the total spendings in a restaurant when it is given as a system of equations using exponents?
The problem is as follows:

Jenny goes to dinner to a well known restaurant in town. The number of times she goes to the restaurant is $\frac{a}{2}$ times and for each time she went, she spent $b^{2}$ hundreds of dollars. Find the money Jenny spent assuming the following system of equations was observed by Jenny.

$$\begin{cases}
3^a-2^{b^{2}}=77 \\
3^{\frac{a}{2}}-2^{\frac{b^2}{2}}=7 \\
\end{cases}$$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{600 dollars}\\
2.&\textrm{500 dollars}\\
3.&\textrm{400 dollars}\\
4.&\textrm{300 dollars}\\
\end{array}$

In this given situation the only thing which I can understand is as this:

$\textrm{Money spent}=\frac{a}{2}\cdot b^2\cdot 10^2$

But the thing is how to find $a$ and $b$, which sort of simplication or algebraic manipulation can be done to solve this system of equation.

Can someone help me?.

答:
Let $x = 3^a$ and $y = 2^{b^2}$

Therefore, $x   = 77+y$   (as $x-y = 77$)

and, $\sqrt{x} = \sqrt{y} + 7$  (as $\sqrt{x} - \sqrt{y} = 7$)

Squaring the second and subtractiing from the first we have,

$0 =28-14\sqrt{y}$

Therefore, $y = 4$ and $x=81$

So $a =4$ and $b^2 =2$

The answer is 400


问:
A question about undecidable sentences with purely mathematical contents
I'm struggling with the idea that the continuum hypothesis does indeed have a purely mathemathical/set theoretical meaning, but is neither provable nor disprovable in ZFC (according to Gödel and Cohen). Moreover, CH is not self-referential, like the Gödel and Rosser sentences are.

So why isn't it considered an example of a non-metamathematical undecidable sentence, so that we consider the Paris-Harrington sentence to be the first one with purely mathematical content? Where is my misunderstading?

答:
The notion of a "purely mathematical" statement is highly subjective. It's certainly intended to rule out sentences which arise via "encoding" techniques a la Godel/Rosser/Kleene/etc., but it may be taken further. In particular, I think reasonable people (who grant in the first place that "purely mathematical" is an adjectival phrase worth having) could disagree over whether $\mathsf{CH}$ is indeed purely mathematical. The strictest approach would be something like:

A purely mathematical statement is one which is significant outside of mathematical logic,

and one might argue that $\mathsf{CH}$ should not be considered purely mathematical according to this criterion on the grounds that, while it does occasionally crop up outside of logic, it does so very rarely and generally indicates that a certain question isn't really the "right" one. Moreover, theorems like Shoenfield absoluteness show that in a precise sense $\mathsf{CH}$ can't have consequences for "concrete" mathematics.

By contrast, this same hypothetical one might argue, Paris-Harrington is very obviously "just" a problem of pure combinatorics. In fact, the only thing connecting it with logic is its unprovability analysis - it's not a sentence from logic, it's a sentence about which logic has a lot to say.

Of course all of this is highly subjective, and quite frankly I don't put much stock in the notion to begin with. But this is what's behind the attitude that $\mathsf{PH}$ has a distinguished status as a natural undecidable (in $\mathsf{PA}$ rather than $\mathsf{ZFC}$, however) sentence which $\mathsf{CH}$ lacks, which is what the OP asked.


问:
Evaluating $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$
Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?

I've calculated that the recurrence relation for this integral is:

$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$

So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.

Can this be useful here?

I would appreciate all your help.

答:
Following the Ishan Banerjee comment let $t=x^n$ hence $dx=\frac{1}{n}t^{\frac{1}{n}-1}dt$ and then
$$ (2n+1) \int_0 ^{1} x^n e^x dx=\frac{2n+1}{n}\int_0^1t^{1/n}e^{t^{1/n}}dt\to2e$$
by using the dominated convergence theorem.</p>


问:
Convergence of $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$
I want to show that $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$ is convergent.

I tried to use the limit convergence test by comparing with $\frac{1}{x^2}$ but that doesn't work. Any hints on how to proceed with this?

Body:
Check that the integral $$I=\int_{a}^{b} \frac{dx}{(x-a)^p}$$ is finte if $p<1$. So
$$J=\int_{1}^{2} \frac{dx}{(x-1)^{1/2} (x+1)^{1/2} \sqrt{x^2+1}},$$
converges as $p=1/2<1$


问:
Two groups overlapping by a fixed number
A travel agency offers the chance to its customers to win vouchers for 2 free excursions during the Covid crisis. Out of the x total customers, y will be randomly chosen to participate in the first excursion and z in the second. In how many of all possible draws we will have a total of w customers who will participate in both excursions?

I am not sure about the exact interpretation: Let's say we have 25 customers and we randomly choose 8 for the first and 7 for the second. If 3 of them are in both excursions (that is, if the 2 groups have 3 people in common), then we have a total of 3 people who will participate in both excursions.
The first draw gives us $xCy$ and the second $xCz$. Obviously it must be $w \leq y+z$.

Can anyone help? I am not good at combinatorics!
(This is a puzzle just for challenging between friends - not an assignment or anything.)

答:
There are $3$ distinct groups: $(y-w)$ for only excursion $1$, $w$ for both excursions, and $(z-w)$ for only excursion $2$

Use permutations with repetitions to get  $\,\dfrac{x!}{(y-w)!\,(w)!\,(z-w)!}$

The above assumes that all $x$ people get at least one excursion. If this is not true, there is a fourth group of $u$ people who don't get any excursion, and the formula becomes$\,\dfrac{x!}{(y-w)!\,(w)!\,(z-w)!(u)!}$


问:
Mass of an element in an Alloy
An alloy is prepared by mixing metals A, B, C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of metals A, B, C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg, of the metal C is

I tried solving this question but I was not able to. Also the solutions that I looked up for on the web are not well-explained. Any help will be much appreciated.

答:
Say their weights are $5x, 2x, 6x$ for unit volume.

In alloy of volume $14$ units, we have volume of $A$ as $3$ units, of $B$ as $4$ units and of $C$ as $7$ units.

Multiplying by their weights of unit volume, we have weight of $15x, 8x, 42x$ for metals $A, B$ and $C$ respectively. Summing, the total weight of the alloy is $65x$.

Now weight of metal $C$ in $65x$ unit weight of alloy is $42x$, so weight of metal $C$ in $130$ kg of alloy

$ = \frac{42x}{65x} \times 130 = 84 \,$ kg.

Title: Mass of an element in an Alloy


问:
Analytic function with derivative zero
Theorem :- Let $f(z) =u(x,y) +iv(x, y)$  If, $f'(z)=0$ everywhere in a domain $D$ then $f(z)$ must be constant throughout D.
Here

In this theorem, by using cauchy reimann equations, we get $u_x =v_y=0$ and $u_y=-v_x=0$

Now I am not getting why it does not suffice to show that f is constant.

Please help me with this, (I am thinking that both partial derivatives are zero than $u(x, y)$ is constant similarly $v(x, y)$)
I don't understand what I am missing.

答:
As pointed out by our OP user772744 in the text of the question
itself,

$f'(z) = 0 \tag 1$

for

$z \in D \tag 2$

implies

$u_x = u_y = v_x = v_y = 0 \tag 3$

in $D$ as well; that is

$\nabla u = \nabla v = 0 \tag 4$

in $D$.

Now, since $D$ is a connected set in $\Bbb C$, it is path connected; see this wikipedia entry.  And since $D$ is a path connected set in $\Bbb C$, it is $C^1$ path connected, see this question and its answers; that is, any two points

$z_1, z_2 \in D \tag 5$

may be joined by a $C^1$ path

$\gamma:I = [0,1] \to D \tag 6$

with

$\gamma(0) = z_1, \gamma(1) = z_2; \tag7$

such a path $\gamma(t)$ may of course be viewed in either of two ways, as

$\gamma:I \to \Bbb C, \tag{7.1}$,

with

$\gamma(t) = x(t) + iy(t), \tag{7.2}$

or as

$\gamma:I \to \Bbb R^2, \tag{7.3}$

with

$\gamma(t) = (x(t), y(t)); \tag{7.4}$

for any such points $z_1$, $z_2$ and such a real path $\gamma(t)$ as in (7.4) we have the real integral formula

$u(z_2) - u(z_1) = u \circ \gamma(1) - u \circ \gamma(0) = \displaystyle \int_{z_1}^{z_2} \dfrac{d(u \circ \gamma)(t)}{dt} \; dt$
$= \displaystyle \int_{z_1}^{z_2} \nabla u \cdot \gamma'(t) \; dt = \int_{z_1}^{z_2} 0 \cdot \gamma'(t) \; dt = 0, \tag 8$

where

$\gamma'(t) = (x'(t), y'(t)), \tag{8.1}$

yielding

$u(z_2) = u(z_1); \tag 9$

likewise,

$v(z_2) = v(z_1); \tag{10}$

that is, $u$ and $v$ are constant in $D$; hence, $f(z)$ is constant as well. $OE\Delta$.

Note Added in Edit, Sunday 27 December 2020, 6:35 PM PST:  The above demonstration takes its cue from our OP user772744's observation that (1) implies (3); that is, it reduces the problem to  integrals of the form (8) which compute $u(z_2) - u(z_1)$ and $v(z_2) - v(z_1)$ separately. It is of course possible to work directly from (1):

$f(z_2) - f(z_1) = \displaystyle \int_{z_1}^{z2} f'(z) \; dz = \int_{z_1}^{z2} 0 \;dz = 0, \tag{11}$

where now $\gamma(t)$ is of the form (7.2), so that

$\gamma'(t) = x'(t) + iy'(t), \tag{11.1}$

and

$dz = \gamma'(t) \; dt. \tag{12}$

End of Note.

Title: Analytic function with derivative zero


问:
A suitable $\sigma$-algebra for this probability distribution over $\mathbb N$?
Let $\Omega= \mathbb N\backslash \{0\}$. Let the function $S(A)$ be defined:
$$S(A)= \begin{cases}
\sum_{i\in A} \frac {1}{2^i}, &\text {if $A \ne \emptyset$ and the sum is defined.}\\
0, &\text {if $A=\emptyset$.}
\end {cases}$$

The probability distribution I'm interested in is: $\forall i\in \Omega:P(\{i\})=\frac {1}{2^i}$. Let $\mathcal F$ be the set of subsets $A$ of $\Omega$ such that $S(A)$ is defined. I don't think $\mathcal F$ is a $\sigma$-algebra. What is a $\sigma$-algebra that is commonly used for this distribution?

答:
The question was: "What is a σ-algebra that is commonly used for this distribution?"

The answer is: $2^{\Omega}$.

Probability spaces of such type a called "Discrete probability spaces".


问:
Finishing the proof that $\textrm{Spec}K[X,Y]\setminus\{(X,Y)\}$ is not an affine scheme
If $K$ is a field, $\mathbb A^2_K=\textrm{Spec}K[X,Y]$  and $U=\mathbb A^2_K\setminus\{(X,Y)\}$, I want to prove that $(U,\mathscr O_{\mathbb A^2_K|U})$ is not an affine scheme. I know that this topic has been handled many times, but I want to point a specific passage in the language of schemes.

I have proved that $\mathscr O_{\mathbb A^2_K|U}(U)=\mathscr O_{\mathbb A^2_K}(\mathbb A^2_K)=K[X,Y]$. Now suppose that $(U,\mathscr O_{\mathbb A^2_K|U})$ is an affine scheme, there is a ring $R$ such that  $(U,\mathscr O_{\mathbb A^2_K|U})\cong (\textrm{Spec }R,\mathscr O_{\textrm{Spec}R})$; by the equivalence between the opposite category of rings and the category of affine schemes we have that $R\cong \mathscr O_{\mathbb A^2_K|U}(U)=K[X,Y]$. The conclusion is that $(U,\mathscr O_{\mathbb A^2_K|U})\cong (\mathbb A^2_K,\mathscr O_{\mathbb A^2_K})$ but I don't understand where is the contraddiction. Why $\mathbb A^2_K$ and $U$ shouldn't be homeomorphic (look here for a similar question)? Even if they are homeomorphic (I don't think so) why there shouldn't exist an isomorphism as locally ringed spaces?

The fact that the open embedding is not an isomorphism doesn't imply that there is no isomorphisms between the two schemes.

答:
You've misunderstood the argument. Let's recall the fundamental theorem of affine schemes:

Theorem.

For all commutative rings $A$, there is a natural isomorphism $A \cong \Gamma (\operatorname{Spec} A, \mathscr{O}_{\operatorname{Spec} A})$.

For all locally ringed spaces $(X, \mathscr{O}_X)$ and all commutative rings $A$, there is a natural bijection between morphisms $(X, \mathscr{O}_X) \to \operatorname{Spec} A$ and ring homomorphisms $A \to \Gamma (X, \mathscr{O}_X)$.

Corollary. The functor $\operatorname{Spec} : \mathbf{CRing}^{\mathrm{op}} \to \mathbf{Sch}$ is fully faithful and has a left adjoint.

Now, let's apply this to the problem at hand.

Lemma. If $X$ and $Y$ are affine schemes, then a morphism $f : X \to Y$ is an isomorphism if and only if the ring homomorphism $f^* : \Gamma (Y, \mathscr{O}_Y) \to \Gamma(X, \mathscr{O}_X)$ is an isomorphism.

Proof. If $f : X \to Y$ is an isomorphism, then $f^* : \Gamma (Y, \mathscr{O}_Y) \to \Gamma(X, \mathscr{O}_X)$ must also be an isomorphism, by functoriality. Conversely, if $f^*$ is an isomorphism, say with inverse $g^* : \Gamma (X, \mathscr{O}_X) \to \Gamma(Y, \mathscr{O}_Y)$, then the fundamental theorem implies there is a unique scheme morphism $g : Y \to X$ inducing $g^*$, and moreover $g$ must be a two-sided inverse for $f$.　◼

Corollary. If $X$ is a scheme, $f : X \to \operatorname{Spec} A$ is a morphism but not an isomorphism, and $f^* : A \to \Gamma (X, \mathscr{O}_X)$ is an isomorphism, then $X$ is not affine.

Proof. This is the contrapositive of the above lemma.　◼

Of course, in the case of interest, $X = U$, $A = K[X, Y]$, and $f : X \to \operatorname{Spec} A$ is the inclusion. This is certainly not an isomorphism: it is not even a bijection on points!


问:
A family that is not equicontinuous in the unit disc

How to show that if  a family of (complex-valued) functions in the unit disc $D$ is equicontinuous  in every compact subset of $D$
then it is equicontinuous
in $D$?

What is an example of a family of (complex-valued) functions in the unit disc $D$ that is not uniformly equicontinuous
in $D$ but uniformly  equicontinuous  in every compact subset of $D$?

答:
There is no such example -- in other words, if a set of functions is equicontinuous on every compact subset of $D$, then it is equicontinuous on $D$.

Proof. Let $F$ be a set of functions $D \to \mathbb{C}$ such that $F$ is equicontinuous on every compact subset of $D$. Now let $x \in D$ be arbitrary. Let $t > 0$ be such that $\overline{B_t(x)} \subseteq D$. Since $\overline{B_t(x)}$ is compact, $F$ is equicontinuous on $\overline{B_t(x)}$.

In particular, for all $\varepsilon > 0$, there exists some $\delta > 0$ such that for all $f \in F$ and all $y \in \overline{B_t(x)} \cap B_\delta(x)$, we have $\lvert f(y) - f(x) \rvert < \varepsilon$. Without loss of generality, we may suppose $\delta < t$ so that $\overline{B_t(x)} \cap B_\delta(x) = B_\delta(x) = D \cap B_\delta(x)$. But now we have shown precisely that $F$ is equicontinuous at $x \in D$! Since $x$ was arbitrary, we conclude that $F$ is equicontinuous on $D$. $\square$

Here's an alternate question: is there a set of functions which is uniformly equicontinuous on every compact subset of $D$, but not uniformly equicontinuous on $D$? Now the answer is "yes", but not for a very interesting reason.

Let $f : D \to \mathbb{C}$ be the function $f(z) = (1-\lvert z \rvert)^{-2}$. This function is continuous but not uniformly continuous (exercise!). Thus, $F = \{f\}$ is equicontinuous but not uniformly equicontinuous. However, continuity implies uniform continuity on compact domains. Thus, $F$ is uniformly equicontinuous on all compact subsets of $D$. This is an "uninteresting" example because it doesn't highlight anything interesting about equicontinuity: the set $F$ was a singleton, so (uniform) equicontinuity of $F$ is the same as (uniform) continuity of $f$!


答:
The only problem could be at the lower bound.

But, by Taylor
$$ \frac 1{\sqrt{x^4 - 1}}=\frac{1}{2 \sqrt{x-1}}-\frac{3 \sqrt{x-1}}{8}+O\left((x-1)^{3/2}\right)$$ Then, no problem, and you will get a number (the answer is quite complicated).

But, if you continue the expansion
$$ \frac 1{\sqrt{x^4 - 1}}=\frac{1}{2 \sqrt{x-1}}+\sum_{n=0}^\infty(-1)^n a_n  (x-1)^{n+\frac 12}$$ the first coeffiicients form the sequence
$$\left\{\frac{3}{8},\frac{11}{64},\frac{7}{256},\frac{141}{4096},\frac{627}{16
   384},\frac{2465}{131072},\frac{81}{524288},\frac{142019}{16777216},\cdots\right\}$$ You will have
$$\int_1^t \frac 1{\sqrt{x^4 - 1}}=\sqrt{t-1}+2\sum_{n=1}^\infty(-1)^n \frac{ a_n }{2 n+3}(t-1)^{n+\frac{3}{2}}$$

Using the above coefficients and $t=2$, the approximation will give
$$\frac{22479314089}{27808235520}\approx 0.8084$$ while the excat value is $0.8078$


问:
Prove that $T(W^{\bot}) \subset W^{\bot}$ where $T$ is unitary operator and $T(W) \subset W$
Lets say $T$ is unitary operator and $T(W)\subset W$, where $W$ is a linear subspace of finite dimensional space $V$.

Prove that $T(W^{\bot}) \subset W^{\bot}$, where $W^{\bot}$ is a subspace orthogonal to subspace $W$.

I was trying to do it like this.
Let $w^{\bot} \in W^{\bot}$ we have to show that
$\langle w, Tw^{\bot} \rangle = 0$ where $w \in W$.

But I don't where do I have to use the fact that $T$ is unitary.

答:
We know if $W$ is subspace of inner product space $V$, then if $$T(W) \subset W$$ then,
$$T^*(W^{\bot}) \subset  W^{\bot} $$
So, for any $w\in W$ and any $w^{\bot} \in W^{\bot}$, we have,
$$ \langle w,T^*(w^{\bot} )\rangle=0$$
$$\langle Tw,w^{\bot}\rangle=0$$
$$\langle TTw,Tw^{\bot}\rangle=0$$
And we know $T$ is unitary operator and $T(W) \subset W$ , so $T$ must be an isomorphism from $V$ to $V$, so,
$$TTw=w'$$
for some $w' \in W$
So, $$T(W^{\bot}) \subset W^{\bot}$$
Hence, the proof!


答:
Let $I_n=\int\limits_0^1x^n\mathrm e^x\mathrm dx$. By integration by parts, $(n+1)I_n=\left.x^{n+1}\mathrm e^x\right|_0^1-I_{n+1}=\mathrm e-I_{n+1}$. Now, $0\leqslant I_{n+1}\leqslant I_n$ hence $(n+1)I_n\leqslant\mathrm e\leqslant(n+2)I_n$.

This is enough to show that
$$
\left(2-\frac3n\right)\cdot\mathrm e\leqslant(2n+1)I_n\leqslant2\mathrm e,
$$
hence
$(2n+1)I_n\to2\mathrm e$.

Title: Evaluating $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$

Body:
Could you help me evaluate $\lim _{n \rightarrow \infty} (2n+1) \int_0 ^{1} x^n e^x dx$?

I've calculated that the recurrence relation for this integral is:

$\int_0 ^{1} x^n e^x dx = x^ne^x | ^{1} _{0} - n \cdot \int_0 ^{1} x^{n-1} e^x dx$

So if we let $I_n = \int_0 ^{1} x^n e^x \ dx$, we get $I_n = \left.x^ne^x \,\right |^1 _0 - n \cdot I_{n-1}$.

Can this be useful here?

I would appreciate all your help.


问:
Prove that $a ( n ) = b( n + 2)$

let $a(n)$ denotes the number of ways of expressing the positive integer $n$ as an ordered sum of 1's and 2's. Let $b(n)$ denote the number of ways of expressing n as an ordered sum of integers greater than 1. prove that  $a(n) = b(n+2)$. for $n=1,2,3...$

My approach:
$a(1) = 1$ (only) , $b(3) = 3$

$a(2) = 2$, because $2=1+1=2$ ,and  $b(4)=2$, because $4=2+2=4$

$a(3) = 3$, because $3=1+1+1=1+2=2+1$ and, $b(5) = 3$ , because $5=2+3=3+2=5$

$a(4)= 5$, because $4=1+1+1+1=2+2=1+1+2=1+2+1=2+1+1$ and, $b(6)=5$, because $6=3+3=2+4=4+2=2+2+2=6$

$a(5)=8$, because $5=1+1+1+1+1=2+1+1+1=1+2+1+1=1+1+2+1=1+1+1+2=2+2+1=2+1+2=1+2+2$ and, $b(7)=8$, because
$7=3+2+2=2+3+2=2+2+3=3+4=4+3=2+5=5+2=7$

By this way i am able to show that  $a(n)=b(n+2)$. but is there any general method for this problem. I mean any recursion relation which  i can  understand.

Background:-This problem is from pathfinder for Olympiad mathematics.

答:
Hint: Observe that the sequence $1, 2, 3, 5, 8, \ldots $ looks like the Fibonacci sequence, offset by 1

Hint: Show that $a_n$ is indeed the Fibonacci sequence (offset by 1), because $ a_{n+2} = a_{n+1 } + a_{n} $, with initial conditions $ a_1 = 1, a_2 = 2$.

Hint: Show that $b_n, n \geq 2$ is indeed the Fibonacci sequence, because $b_{n} = b_{n-2} + b_{n-3} + b_{n-4} + \ldots + b_3 + b_2 + 1$, and $ b_2 = 1, b_3 = 1$.

Hence, $ a_n, n \geq 1 $ and $ b_n, n \geq 3$ are the Fibonacci sequence offset by 1.

Title: Prove that $a ( n ) = b( n + 2)$


问:
Restricting to affine charts
this is a fairly basic question that I'm struggling with. I'm trying to solve this problem:

Let $C$ be an affine or projective plane curve over $\mathbb{C}$, and let $p$ be a smooth point of $C$.
Show that $p$ has an open neighbourhood in $C$ that is homeomorphic to $\mathbb{C}$.

In the affine case, this seems to be a direct application of the inverse function theorem, which I state here:

$Q$ is a non-constant polynomial in $\mathbb{C}[x, y],$ and let $D$ be the affine plane curve defined by $Q=0 .$ Let $(a, b)$ be a point in $D$ such that $\partial_{y} Q(a, b) \neq 0 .$ Then there exists an open neighbourhood $U$ of $(a, b)$ in $D$ such that the projection
$$
\psi: U \rightarrow \mathbb{C},(x, y) \mapsto x
$$
is a homeomorphism onto an open subset $V$ of $\mathbb{C},$ and such that the composition of $\psi^{-1}: V \rightarrow U$ with the projection $\mathbb{C}^{2} \rightarrow \mathbb{C},(x, y) \mapsto y$ is holomorphic.

So from the above set $V$, I can find an open disk around point $p$ identified to be in $V$ (i.e. $\psi(p)$), and this is homeomorphic to $\mathbb{C}$.

But in the case of the projective plane curve, the only explanation that I'm given is that "we can intersect $C$ with an affine chart containing $p$ to reduce to the affine case". I can't justify this line. My attempt would be something like:

$V$ is open in $\mathbb{C}$ $\iff$ $\exists$ $W \subset \mathbb{P}^2_\mathbb{C}$ such that $W \cap C = V$, and $W \cap U_i$ is open in $U_i$ for $i = 0,1,2$ $\iff$... something appropriate to justify the above step.

Could anyone help with this? Thanks. $U_i$ is defined as in the conventional sense, such that $U_i$ form an open cover of $\mathbb{P}^2_\mathbb{C}$.

答:
Since the affine charts $U_i$ cover $\mathbb{P}^2$, there is some $i$ such that $p \in U_i$. Let $f : \mathbb{C}^2 \to U_i$ be the canonical homeomorphism. Now "pull back" $C$ to a plane curve $C' \subseteq \mathbb{C}^2$ (what polynomial should this be given by?). Then $C'$ is an affine plane curve such that $f(C') = U_i \cap C$.

We have $f^{-1}(p) \in C'$, and we can check that $f^{-1}(p)$ is a smooth point of $C'$. Thus, there is an open neighborhood $V \subseteq \mathbb{C}^2$ of $f^{-1}(p)$ such that $V \cap C' \cong \mathbb{C}$. Finally, $f(V) \subseteq \mathbb{P}^2$ is an open neighborhood of $p$ such that $$f(V) \cap C = f(V \cap C') \cong V \cap C' \cong \mathbb{C},$$ as desired.

Title: Restricting to affine charts


问:
Given a linear operator $T:\mathbb R^3 \to \mathbb R^3$ with $(x_1,x_2,x_3) \mapsto (3x_1,x_1-x_2,2x_1+x_2+x_3)$. Show that $(T^2-I)(T-3I)=0$
Given a linear operator $T:\mathbb R^3 \to \mathbb R^3$ with $(x_1,x_2,x_3) \mapsto (3x_1,x_1-x_2,2x_1+x_2+x_3)$.

Show that $$(T^2-I)(T-3I)=0$$

The linear operator $T$ is a function such that $$
\begin{bmatrix}
x_1  \\
x_2\\
x_3 \\
\end{bmatrix}
\quad
 \mapsto P\begin{bmatrix}
x_1  \\
x_2\\
x_3 \\
\end{bmatrix}
\quad$$
Where $$P=
\begin{bmatrix}
3& 0 & 0 \\
1 & -1 & 0\\
2 & 1 & 1\\
\end{bmatrix}
\quad
$$

Moreover $$T^2(x_1,x_2,x_3)=T(3x_1,x_1-x_2,2x_1+x_2+x_3)=(9x_1,2x_1+x_2,9x_1+x_3)$$
$$I(x_1,x_2,x_3)=(x_1,x_2,x_3)$$
$$T^2(x_1,x_2,x_3)-I(x_1,x_2,x_3)=(8x_1,2x_1,9x_1)$$
$$T(x_1,x_2,x_3)-3I(x_1,x_2,x_3)=(0,x_1-4x_2,2x_1+x_2-2x_3)$$

But I don't understand what does $(T^2-I)(T-3I)$ means and that's why I can't show the desired result,besides is the matrix $P$ helpful here?

答:
$(T-3I)(a,b,c)=(0,a-4b, 2a+b-2c)$. Call this vector $(x_1,x_2,x_3)$ and apply $T^{2}-I$ to it. From your computaion of $T^{2}-I$ is is clear that $(T^{2}-I)(T-3I) (a,b,c)=((8)(0), (2)(0),(9)(0))=0$. Since this is true for all $(a,b,c)$ we conclude that  $(T^{2}-I)(T-3I)=0$.

If you want to use $P$ you have to compute $P^{2}$ and $P^{3}$ and show that $(P^{2}-I)(P-3I)=P^{2}-3P^{2}-P+3I=0$ but that is a lengthy computation.


问:
Area of triangle with incircle tangent to semicircle whose diameter is on one triangle side
Triangle $\triangle ABC$ with incenter $I$ and inradius $r$ has the following property: the incircle is tangent to the semicircle with diameter $AB$ and is within the semicircle. Find the area of $\triangle ABC$ as a function of the side $AB$ and the inradius $r$.

So far I've noticed that $\angle AEF=\angle BEF=\frac{\pi}{4}$ but I don't think that that's useful in any way.

答:
Let the midpoint of $AB$ be $O$. Draw $OE$ and $IF$. Label the sides $BC$, $CA$ and $AB$ as $a$, $b$ and $c$ respectively. Let the semiperimeter of $\triangle ABC$ be $s$. Points $O$, $I$ and $E$ will be collinear.

$OI=OE-IE=\frac{c}{2}-r$

$OF=OB-FB=\frac{c}{2}-\left(s-b\right)$

Applying Pythagorean theorem on $\triangle OIF$ gives,

$OF^2 + IF^2= OI^2$

$\Rightarrow  \{\frac{c}{2}-\left(s-b\right)\}^2 + r^2=(\frac{c}{2}-r)^2$

On simplification, this leads to, $r=\frac{\left(s-b\right)\left(s-a\right)}{c}$

But, $r= \frac{\Delta}{s}=\sqrt{\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}$

$\Rightarrow  r^2=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}$

$\Rightarrow  r^2=\frac{cr\left(s-c\right)}{s}$

$\Rightarrow    r=\frac{c\left(s-c\right)}{s}$

$\Rightarrow    rs=c\left(s-c\right)$

Hence, $\Delta=rs=c\left(s-c\right)$

$rs=c\left(s-c\right)$   $\Rightarrow  s=\frac{c^2}{c-r}$   $\Rightarrow  \left(s-c\right)=\frac{cr}{c-r}$

Thus, $\Delta=rs=c\left(s-c\right)=c\cdot\left(\frac{cr}{c-r}\right)=\boxed {\frac{c^2r}{c-r}}$


问:
Calculate the number of different words, where $0$ appears an even number of times.
Let the set of words with an even length $n$ from the alphabet: $\{ 0,1,2\}$.

Calculate the number of different words, where $0$ appears an even number of times.

For example, for $n=6$ , the words $121212,001212,000000$ are allowed,but the word $100011$ is not allowed.

My idea is:

$$\sum_{k=0}^{\frac{n-2}{2}} \binom{n}{2k} \cdot \binom{n-2k}{2}$$

Is this correct?

答:
The recurrence relation for this problem will be
$$A_n = A_{n-1} + 3^{n-1} ; n>1 $$

$$A_1 = 2$$

Solve this recurrence equation using homogeneous and particular solution method.
You will get the answer: $$(3^n + 1)/2$$


问:
Does $\int_{-\infty}^\infty \frac{1}{x-a}dx$ converge?
I don't think this integral converges:

$$\int_{-\infty}^\infty \frac{1}{x-a}dx\qquad x,a\in\mathbb R$$

The integrand is unbounded.

I keep getting undefined natural logarithms and non-existent limits (infinites) whenever I try to work this integral around.

Can you confirm or prove otherwise?

Thank you!

答:
Yes, the integrand is unbounded, but it doesn't follow from that that the integral diverges. In fact, the integral $\int_0^1\frac{\mathrm dx}{\sqrt x}$ converges, although the integrand is unbounded too in this case.

On the other hand, by definition,$$\int_{-\infty}^\infty\frac{\mathrm dx}{x-a}=\int_{-\infty}^a\frac{\mathrm dx}{x-a}+\int_a^\infty\frac{\mathrm dx}{x-a}$$and$$\int_a^\infty\frac{\mathrm dx}{x-a}=\int_a^b\frac{\mathrm dx}{x-a}+\int_b^\infty\frac{\mathrm dx}{x-a}$$for some $b>a$ (the choice of $b$ is irrelevant). But\begin{align}\int_a^b\frac{\mathrm dx}{x-a}&=\lim_{y\to a}\int_y^b\frac{\mathrm dx}{x-a}\\&=\lim_{y\to a}\bigl(\log(b-a)-\log(y-a)\bigr)\\&=\infty\end{align}and\begin{align}\int_b^\infty\frac{\mathrm dx}{x-a}&=\lim_{y\to\infty}\int_b^y\frac{\mathrm dx}{x-a}\\&=\lim_{y\to\infty}\bigl(\log(y-a)-\log(b-a)\bigr)\\&=\infty\end{align}too.


问:
How to estimate the money given by a kid when the goal is to purchase a computer game?
The problem is as follows:

Mike and Roger intend to purchase a computer game which has become popular. Assuming Roger gives $P$ times the amount given by Mike and the cost of the computer game is $60$ dollars. Use the following information to find the amount given by Mike.

$$P=\left[\sqrt[\leftroot{-1}\uproot{2}\scriptstyle a+b]{\frac{\sqrt[a]{2^{b+c}}\cdot\sqrt[b]{2^{a+c}}\cdot \sqrt[c]{2^{a+b}}}{\sqrt[a]{2^{b-c}}\cdot\sqrt[b]{2^{c-a}}\cdot \sqrt[c]{2^{a-b}}}}\qquad\right]^{\frac{c}{2}}$$

and $\frac{b-c}{b}=\frac{c^2}{a^2}$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{15 dollars}\\
2.&\textrm{12 dollars}\\
3.&\textrm{20 dollars}\\
4.&\textrm{10 dollars}\\
\end{array}$

In my attempt to solve this problem I did the following:

Please note that for brevity and displaying purposes I'm preserving the root as it is indicated above.

$P=\left[\sqrt[\leftroot{-1}\uproot{2}\scriptstyle a+b]{\frac{\sqrt[a]{2^{b+c}}\cdot\sqrt[b]{2^{a+c}}\cdot \sqrt[c]{2^{a+b}}}{\sqrt[a]{2^{b-c}}\cdot\sqrt[b]{2^{c-a}}\cdot \sqrt[c]{2^{a-b}}}}\qquad\right]^{\frac{c}{2}}$

$P=\left[\sqrt[\leftroot{-1}\uproot{2}\scriptstyle a+b]{\frac{2^{\frac{b+c}{a}}\cdot 2^{\frac{b+c}{b}} \cdot 2^{\frac{a+c}{c}}}{2^{\frac{b-c}{a}}\cdot 2^{\frac{c-a}{b}}\cdot 2^{\frac{a-b}{c}}}}\qquad\right]^{\frac{c}{2}}$

$P=\left(\left(\frac{2^{\frac{bc(b+c)+ac(a+c)+ab(a+b)}{abc}}}{2^{\frac{bc(b-c)+ca(c-a)+ab(a-b)}{abc}}}\right)^{\frac{1}{a+b}}\right)^{\frac{c}{2}}$

$P=\left(\left(2^{\frac{bc(b+c)+ac(a+c)+ab(a+b)-bc(b-c)-ca(c-a)-ab(a-b))}{abc}}\right)^{\frac{1}{a+b}}\right)^{\frac{c}{2}}$

The issue here is what's happening in the numerator of the exponent of $2$:

$bc(b+c)+ac(a+c)+ab(a+b)-bc(b-c)-ca(c-a)-ab(a-b))=2bc^2+2ac^2+2ab^2$

From the initial condition:

$a^2b-a^2c=c^2b$

$2(a^2b-a^2c)+2ac^2+2ab^2=2a^2b+2ab^2=2ab(a+b)$

Then:

$\frac{2ab(a+b)}{abc}=\frac{2(a+b)}{c}$

Then this is:

$P=\left(\left(2^{\frac{2(a+b)}{c}}\right)^{\frac{1}{a+b}}\right)^{\frac{c}{2}}$

$P=2$

Hence it seems that all such fuss was to say $P=2$ hence:

Since it mentions Roger is $P$ times the money given by Mike and all of this equates to $60$ dollars.

$R=PM$

$R=2M$

$2M+M=60$

$3M=60$

$M=20$

Thus I believe that the money given by Mike is $\textrm{20 usd}$?. But is this the only way to approach this problem?.

答:
$$P=\left[\sqrt[\leftroot{-1}\uproot{2}\scriptstyle a+b]{\frac{\sqrt[a]{2^{b+c}}\cdot\sqrt[b]{2^{a+c}}\cdot \sqrt[c]{2^{a+b}}}{\sqrt[a]{2^{b-c}}\cdot\sqrt[b]{2^{c-a}}\cdot \sqrt[c]{2^{a-b}}}}\qquad\right]^{\frac{c}{2}}$$

I would simplify
$$\frac{\sqrt[a]{2^{b+c}}}{\sqrt[a]{2^{b-c}}} = \sqrt[a]{2^{2c}}$$
$$\frac{\sqrt[b]{2^{a+c}}}{\sqrt[b]{2^{c-a}}} = \sqrt[b]{2^{2a}}$$
$$\frac{\sqrt[c]{2^{a+b}}}{\sqrt[c]{2^{a-b}}} = \sqrt[c]{2^{2b}}$$

This simplifies $P$ to
$$P = {\left(\sqrt[a+b]{\sqrt[a]{2^{2c}}\cdot \sqrt[b]{2^{2a}}\cdot \sqrt[c]{2^{2b}}}\right)}^{c/2} =
{\left(\sqrt[a+b]{\sqrt[a]{2^{c}}\cdot \sqrt[b]{2^{a}}\cdot \sqrt[c]{2^{b}}}\right)}^{c}$$

Then I would turn this into exponential notation:
$${\left(\sqrt[a+b]{\sqrt[a]{2^{c}}\cdot \sqrt[b]{2^{a}}\cdot \sqrt[c]{2^{b}}}\right)}^{c}
= {\left( 2^{c/a}\cdot 2^{a/b} \cdot 2^{b/c} \right)}^{\frac{c}{a+b}}
= 2^{(c/a+a/b+b/c)\cdot \frac{c}{a+b}}$$
and then simplify the exponent to
$$\frac{bc^{2}+a^{2}c + ab^{2}}{abc}\cdot \frac{c}{a+b} = \frac{bc^{2}+a^{2}c+ab^{2}}{ab(a+b)}.$$

Then your equation $\frac{b-c}{b}=\frac{c^2}{a^2}$ tells you
$$bc^{2} = a^{2}(b-c)$$
so
$$\frac{bc^{2}+a^{2}c+ab^{2}}{ab(a+b)} = \frac{a^{2}(b-c)+a^{2}c+ab^{2}}{ab(a+b)}
= \frac{ab(a+b)}{ab(a+b)} = 1,$$
and so $P = 2^{1} = 2$.

I don't know if this is any easier than the version you had. I don't think it's supposed to be simple.


问:
A related rate problem of a trough with a trapezoidal cross section
From the MIT exercise book (2E-7):

A trough is filled with water at a rate of $1$ cubic meter per second. The trough has a trapezoidal cross section with the lower base of length half a meter and one meter sides opening outwards at an angle of $45$ degrees from the base. The length of the trough is $4$ meters. What is the rate at which the water level $h$ is rising when $h$ is one half meter?

There is a solution here, but omits a lot of details that I couldn't follow.

My main question is how to come up with the volume of the trough.

答:
It is not the best diagram but hopefully explains better. From the first diagram,

$BCFE$ is the base of the trough. $ABCD$ and $EFHG$ are standing vertical to the base.

$ABEG$ and $CDHF$ are walls that are making $135^0 ( = 90^0 + 45^0)$ to the base.

From the second diagram which is the cross section, if the water has filled till height $h$, $AM = ND = h,$ as the angle is $\frac{\pi}{4}$.

The volume of water, $V$ = length of trough  $\times$ area of cross section

$ = 4 \times Area \, (\triangle ABM + \square MBCN + \triangle CDN)$

$ = 4 \times (\frac{1}{2}h^2 + \frac{1}{2} h + \frac{1}{2}h^2) = 4h^2 + 2h$

$\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} = (8h + 2)\frac{dh}{dt}$

Also note when the trough is completely full, $AB = 1 \,$ so $h_{max} = \frac{1}{\sqrt2}$.


问:
Find a non-normal operator that is similar to a normal operator
Find a non-normal operator that is similar to a normal operator

My Solution:

For non-normal operator consider, $$T=\begin{bmatrix} 1&1\\0&1\end{bmatrix} .$$ and For normal operator consider, $$I=\begin{bmatrix} 1&0\\0&1\end{bmatrix} .$$

Can we say that: they are similar?

答:
Let we take,
$$A=\begin{bmatrix} 2&3\\0&5\end{bmatrix} $$
and
$$B=\begin{bmatrix} 2&0\\0&5\end{bmatrix} $$
Here, $B$ is self-adjoint matrix, so it is normal also!
And $A$ is non normal matrix.
Clearly, $A$ and $B$ are similar.


问:
Determinants of matrices defined by the minimum/maximum indices of their entries
$
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
$
The Problem: Consider the matrices $A_n := \left(a^{(n)}_{i,j}\right)_{1 \le i,j \le n},B_n := \left(b^{(n)}_{i,j}\right)_{1 \le i,j \le n} \in M_{n \times n}(\Bbb R)$ with entries defined as follows:

$$a^{(n)}_{i,j} := \min \{i,j\} \qquad b^{(n)}_{i,j} := \max \{i,j\}$$

(Examples of these matrices follow momentarily, to show how the structure of each "evolves" and emerges as $n$ increases.)

What are $\det(A_n)$ and $\det(B_n)$ in the general $n \times n$ case, for each $n \in \Bbb Z^+$?

Background/Context: I happened to have a question similar to this on my linear algebra final a few weeks ago. (In full disclosure: final grades have already been determined and all that, so this isn't any longer an "active" test question.) However, that problem used a matrix similar to $A_n$, without specific numbers, and only the $4 \times 4$ case.

I then recently found a video by Dr. Peyam which considered the $5 \times 5$ cases for both $A_n$ and $B_n$, and hinted that these possibly came from a Putnam exam. I'm not sure if it was, though. I looked back through all of the Putnam problems back to $1985$ and didn't see this exact one, though one was similar: $A2$ from $2014$, where $A_n$ is instead defined by $a^{(n)}_{i,j} = 1/\min\{i,j\}$. (Maybe I'll look at that some other time.)

I imagine the definition for $B_n$ is just considered a natural generalization of this problem. In any event, I couldn't find this problem handled in its generality on MSE, so I figured I'd try to figure it out, and post my own solution.

Of course, if you have your own solutions, they're greatly welcomed! Each can provide their own insights.

Small Cases for the Matrices: To help show the structure of the matrices, $A_n$ and $B_n$ look like, in the cases for $n=1,2,\cdots,6$, as follows. Determinants are included, but calculated by Wolfram.

\begin{alignat*}{3}
&\boxed{n=1} &&\qquad A_1 = \m{
1  }  &&\qquad \det(A_1) = 1\\
& &&\qquad B_1 = \m{
1 }
&&\qquad \det(B_1) = 1\\
&\boxed{n=2} &&\qquad A_2 = \m{
1 & 1 \\
1 & 2  }
&&\qquad \det(A_2) = 1\\
& &&\qquad B_2 = \m{
1 & 2    \\
2 & 2   }
&&\qquad \det(B_2) = -2\\
&\boxed{n=3} &&\qquad A_3 = \m{
1 & 1 & 1 \\
1 & 2 & 2 \\
1 & 2 & 3 }
&&\qquad \det(A_3) = 1\\
& &&\qquad B_3 = \m{
1 & 2 & 3   \\
2 & 2 & 3  \\
3 & 3 & 3  }
&&\qquad \det(B_3) = 3\\
&\boxed{n=4} &&\qquad A_4 = \m{
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4 }
&&\qquad \det(A_4) = 1\\
& &&\qquad B_4 = \m{
1 & 2 & 3 & 4  \\
2 & 2 & 3 & 4 \\
3 & 3 & 3 & 4   \\
4 & 4 & 4 & 4  }
&&\qquad \det(B_4) = -4\\
&\boxed{n=5} &&\qquad A_5 = \m{
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 & 3 \\
1 & 2 & 3 & 4 & 4 \\
1 & 2 & 3 & 4 & 5 }
&&\qquad \det(A_5) = 1\\
& &&\qquad B_5 = \m{
1 & 2 & 3 & 4 & 5  \\
2 & 2 & 3 & 4 & 5 \\
3 & 3 & 3 & 4 & 5  \\
4 & 4 & 4 & 4 & 5 \\
5 & 5 & 5 & 5 & 5 }
&&\qquad \det(B_5) = 5\\
&\boxed{n=6} &&\qquad A_6 = \m{
1 & 1 & 1 & 1 & 1 & 1\\
1 & 2 & 2 & 2 & 2 & 2\\
1 & 2 & 3 & 3 & 3 & 3\\
1 & 2 & 3 & 4 & 4 & 4\\
1 & 2 & 3 & 4 & 5 & 5\\
1 & 2 & 3 & 4 & 5 & 6}
&&\qquad \det(A_6) = 1\\
& &&\qquad B_6 = \m{
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 2 & 3 & 4 & 5 & 6\\
3 & 3 & 3 & 4 & 5 & 6 \\
4 & 4 & 4 & 4 & 5 & 6 \\
5 & 5 & 5 & 5 & 5 & 6 \\
6 & 6 & 6 & 6 & 6 & 6}
&&\qquad \det(B_6) = -6
\end{alignat*}

One reasonably conjectures that $\det(A_n) = 1$ for all $n$ and $\det(B_n) = (-1)^{n-1} n$ for all $n$.

答:
The inductive approach (IE How I first dealt with them when I saw them)

Using notation from Eevee Trainer.

$
\newcommand{\RR}{\mathcal{R}}
\newcommand{\m}[1]{\left( \begin{matrix} #1 \end{matrix} \right)}
\newcommand{\mdet}[1]{\left| \begin{matrix} #1 \end{matrix} \right|}
$
Finding $\det(A_n)$:

I will pursue this one by simple row reduction: adding multiples of one row to another will preserve the determinant. So, for now, note the form of $A_n$:

$$A_n =
\m{
1 & 1 & 1 & \cdots & 1 \\
1 & 2 & 2 & \cdots & 2 \\
1 & 2 & 3 & \cdots & 3 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 2 & 3 & \cdots & n}$$

Throughout the entirety of this answer, let $\RR_i$ denote the $i$th row of the matrix, at whatever stage in the row reduction process we're at. We'll let $\sim$ denote equivalence by row reduction. At this point, subtract $\RR_1$ from each of $\RR_k$ for $k \ge 2$. We see that

$$A_n  \sim
\m{
1 & 1 & 1 & \cdots & 1 \\
0 & 1 & 1 & \cdots & 1 \\
0 & 1 & 2 & \cdots & 2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 1 & 2 & \cdots & n-1}$$

Now, expanding along the first column, we see that $\det(A_n) = 1 \times \det(A_{n-1})$.

Since $ \det (A_1) = 1$, hence $\det (A_n) = 1$.

Finding $\det(B_n)$:

I will similarly utilize row reduction for this, as I did for $A_n$. In general, $B_n$ looks like this:

$$B_n = \m{
1 & 2 & 3 & \cdots & n \\
2 & 2 & 3 & \cdots & n \\
3 & 3 & 3 & \cdots & n \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n & n & \cdots & n}$$

At this point, subtract $\RR_n$ from each of $\RR_k$ for $k \leq n-1$. We see that

$$B_n = \m{
1-n & 2-n & 3-n & \cdots & 0 \\
2-n & 2-n & 3-n & \cdots & 0 \\
3-n & 3-n & 3-n & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n & n & \cdots & n}$$

Expanding along the last column, $ \det B_n = n \times \det C_{n-1}$, where $C_n$ is the $(n-1) \times (n-1)$ minor.

Multiplying each of the columns by $ (-1)$, we get that

$$\det(C_n) = ( -1)^{n-1}   \det \m{
n-1 & n-2 & n-3 & \cdots & 1 \\
n-2 & n-2 & n-3 & \cdots & 1 \\
n-3 & n-3 & n-3 & \cdots &1 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 1 & 1 & \cdots & 1}$$

Note that the matrix on the RHS is just $A_{n-1}$, though with the rows and columns completely flipped. Hence, by accounting for the flip, the determinant is $ (-1)^{2(n-2) } \det(A_n) = 1 $.

Hence, $ \det (B_n) = n \times \det (C_{n-1}) = n \times (-1)^{n-1} \times 1 $.


问:
Question on binomial theorem: Find the term independent of $x$ in the expansion of $(2+x)\left(2x+\frac{1}{x}\right)^5$

Find the term independent of $x$ in the expansion of $$(2+x)\left(2x+\frac{1}{x}\right)^5$$

I am only able to do this with one binomial, but when there are two I do not totally understand the process to do this.

答:
First, rewrite the fifth power as a sum $$(2+x)\left(2x+\frac 1x\right)^5 = (2+x) \sum_{k=0}^5 {5\choose k} (2x)^\color{red}{5-k}\left(\frac 1x\right)^\color{blue}{k}$$
General idea: Note that to have a free (of $x$) term when multiplied by $(\color{orange}2+\color{purple}x)$, we must have a constant term (when multiplied by $\color{orange}2$ yields a free term) or term with $\frac{1}{x}$ (when multiplied by $\color{purple}x$ yields a free term) in the sum.

Let's see if there is a constant term in the sum. To check this, we equate the red and blue powers since $x$'s should cancel each other:
$$5-k = k \implies k = 2.5 \qquad \color{red}{\sf X}$$
So, there is no such term since $k$ was not an integer.

Let's see if there is a term  with $\frac{1}{x}$ in the sum. To check this, we set the $\color{red}{\text{red}} = \color{blue}{\text{blue}} - 1$ since after cancelling $x$'s, one $x$ should "survive" in the denominator:
$$5-k = k - 1 \implies k = 3 \qquad \color{gree}{\checkmark}$$
Yes! There is such term when $k=3$:
$${5\choose k} (2x)^{5-k}\left(\frac 1x\right)^k\Bigg|_{k=3} = {5\choose 3} \frac 4x$$

Finally, when we multiply the sum and $(2+x)$ our constant (free) term will be
$$x \cdot {5\choose 3} \frac 4x = 4 {5\choose 3} = 40$$


问:
Find $\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]$ where $N$ is Poisson random variable with mean $m$
Let $N$ be a Poisson random variable with the mean parameter $m$.
We are interested in finding the following limit
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right].
\end{align}

Things that I tried:
First, I found an upper bound by using Jensen's inequality
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]\le  \lim_{m \to \infty} m  \log\left( \frac{ E \left[ N  \right]+\frac{1}{2}}{m+1} \right)= \lim_{m \to \infty} m  \log\left( \frac{ m+\frac{1}{2}}{m+1} \right)=-\frac{1}{2}.
\end{align}

For the lower bound I tried to use the CLT argument from a related question in here:
\begin{align}
m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] P[ N \ge \frac{m}{k} ]+ m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N < \frac{m}{k} \right] P[ N < \frac{m}{k} ]
\end{align}
for some $k>0$.  Via the CLT it can be shown that $P[ N < \frac{m}{k} ] \to 0$ for all $0<k <1$, and we have that
\begin{align}
\lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] .
\end{align}
Now if we use the lower bound at this point, we get
\begin{align*}
 \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right]  \ge   \lim_{m\to \infty}m  \log\left( \frac{ \frac{m}{k}+\frac{1}{2}}{m+1} \right) =-\infty,
\end{align*}
for all $k \in (0,1)$.

答:
We will show that $lim_m m E\ln \frac{N+ \frac12}{m+1} = -1$.

We have $$m E\ln \frac{N+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + Em\ln \frac{m+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + m \ln (1 - \frac{0.5}{m+1})$$
and thus
$$lim_m m E\ln \frac{N+ \frac12}{m+1} = \lim_m E\ln \frac{N+ \frac12}{m+\frac12}  -\frac12.$$

It's sufficient to show that $\lim_m E\ln \frac{N+ \frac12}{m+\frac12} \to -\frac{1}2$.

Condsider $m \ge 100$, $1+Y_i \sim Pois(1)$, $\xi_m \sim Pois(m - [m])$. Thus $\sum_{i=1}^{[m]} (Y_i + 1) + \xi_m = N$ in distribution.

Fix $x$. $$P(\frac{\xi_m}{\sqrt{m}} \ge x) \le \frac{E \frac{\xi_m}{\sqrt{m}} }{x} \to 0, m \to \infty$$
thus $ \frac{\xi_m}{\sqrt{m}}  \to 0$ in probability.
It follows from CLT that $\frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}} \to N(0,1)$. Hence
$$\frac{N-m}{\sqrt{m}} = \frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}}\frac{\sqrt{[m]}}{\sqrt{m}}   +  \frac{\xi_m}{\sqrt{m}} \to N(0,1).$$
Put $$\eta_m = \frac{m}{m+\frac12}\cdot \frac{N-m}{\sqrt{m}} = \frac{\sqrt{m}(N-m)}{m+\frac12}$$
Then $\eta_m \to N(0,1)$  and $\eta_m \ge \frac{m}{m+\frac12}\frac{-m}{\sqrt{m}} \ge -\sqrt{m} + \frac{1}{3\sqrt{m}}$.
Further,
$$\lim_m E\ln \frac{N+ \frac12}{m+\frac12} = E m \ln(1 + \frac{\eta_m}{\sqrt{m}}).$$
Put $g(x) = \ln(1+x) - (x - \frac{x^2}2)$. Then
$$ E m \ln(1 + \frac{\eta_m}{\sqrt{m}}) = E m(  \frac{\eta_m}{\sqrt{m}} - \frac12 ( \frac{\eta_m}{\sqrt{m}})^2 + g( \frac{\eta_m}{\sqrt{m}}) ) = - \frac12 + Emg( \frac{\eta_m}{\sqrt{m}})  $$
and it's sufficient to show that $Emg( \frac{\eta_m}{\sqrt{m}})  \to 0$.

Put $A = [-\sqrt{m}+ \frac{1}{3\sqrt{m}}, - m^{\frac{3}{8}}], B= [-m^{\frac{3}{8}}, m^{\frac{3}{8}}], C = [m^{\frac{3}{8}}, \infty)$. Then

$$|Emg( \frac{\eta_m}{\sqrt{m}})| \le Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le $$
$$\le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|P(\eta_m  \in A)  + m \max_{x \in B} |g(\frac{x}{\sqrt{m}})|P(\eta_m  \in B) + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C}$$
As $m\ge 100$, we know that  $\max_{x \in A} |g(\frac{x}{\sqrt{m}})| = |\ln(\frac{1}{3m})| \le 2 \ln m $.
Further, by Taylor formula
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$, $0 \le \theta \le 1$. Thus
$g(x) = \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$
and $\max_{x \in B} |g(\frac{x}{\sqrt{m}})| =  O(\frac{1}{ m^{\frac98 }  })$.

We have $|g(x)| \le 2x^2$ for $x > 0$.
Thus $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le E m 2 (\frac{\eta_m}{\sqrt{m}})^2 I_C = 2 E \eta_m^2 I_C$.
Further there will be heuristic arguments.
$2 E \eta_m^2 I_C \approx 2 E N^2(0,1) I_{(N(0,1) > m^{\frac38})} \to 0$ hence $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \to 0$.

$P(\eta_m \in A) \approx P(N(0,1) \in A) \le P(N(0,1) \le -m^{\frac38})$ and thus
$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A}  \le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|  P(N(0,1) \le -m^{\frac38}) \to 0$.

It's sufficient to show that$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \to 0$.

But $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \le m \max_{x \in B} |g(\frac{x}{\sqrt{m}})| = m  O(\frac{1}{ m^{\frac98 }  }) = o(1)$, q.e.d.


问:
Why is the Riemann zeta function defined with limits when 0 < s < 1?
The book Introduction to Analytic Number theory by Tom M. Apostol defines Riemann zeta function like this:

$\zeta(s)$ denotes the Riemann zeta function which is defined by the equation $$ \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \text{ if } s \gt 1, $$
and by the equation $$ \zeta(s) = \lim_{x \to \infty} \left( \sum_{n \le x}\frac{1}{n^s} - \frac{x^{1-s}}{1 - s} \right) \text{ if } 0 < s < 1.$$

Why is the Riemann zeta function defined differently for $ 0 < s < 1 $? What would go wrong if we were to use the same definition as that for $ s > 1 $ for the case $ 0 < s < 1 $ too?

What is the intuition behind the second definition? Why was that specific formulation chosen instead of something else?

答:
I could be wrong about this, but I don't think the second formula is a good definition, pedagogically, because in order to motivate it you need to already have $\zeta(s)$ defined for $\mathrm{Re}(s)<1$ in the first place.

Here is the canonical story of how $\zeta(s)$ goes:

Let's start with $\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$. This is convergent for $\mathrm{Re}(s)>1$ and divergent otherwise. However, the Dirichlet eta function $\eta(s)=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}$ converges for $\mathrm{Re}(s)>0$ and it is not hard to see that we can write $\eta(s)=(1-2^{1-s})\zeta(s)$ and we can use this to analytically continue $\zeta(s)$ so it is defined also for values $\mathrm{Re}(s)>0$ (except $s=1$, where it has a pole).

If you're unfamiliar with analytic continuation, consider the geometric series $(1-z)^{-1}=\sum_{n=0}^\infty z^n$ for a complex variable $z$. The function $(1-z)^{-1}$ is defined for all $z\ne1$ but the series only converges hence is only defined for $|z|<1$. So the function defined by the series can be "continued" beyond its original domain to a function $(1-z)^{-1}$ defined on all of $\mathbb{C}\setminus\{1\}$.

There is a really cool reason we ought to multiply $\zeta(s)$ by another function to "complete" it, which gives the function $\xi(s)=\binom{s}{2}\pi^{-s/2}\Gamma(s/2)\zeta(s)$. Interchanging summation and integration entailed respectively in defining the zeta and gamma function in this formula shows $\xi(s)$ is a Mellin transform of a theta function. The theta function has a functional equation deriving from the Poisson summation formula, which in turn implies the functional equation $\xi(s)=\xi(1-s)$, and we can use this to extend the domain of $\zeta(s)$ to all $s\ne 1$.

[Here is the cool reason BTW. The local factors of the Euler product factorization $\prod(1-p^{-s})^{-1}$ are in fact $p$-adic Mellin transforms of Gaussian functions (i.e. fixed points of the $p$-adic Fourier transform). Note the Mellin transform is actually the multiplicative version of the Fourier transform. The corresponding local factor for the real numbers is the Mellin transform of the usual Gaussian function, which gives us the stuff in front of $\zeta(s)$ in the definition of $\xi(s)$. This is a bumper sticker version of Tate's thesis.]

Now, we know that $\sum\frac{1}{n^s}$ only converges to $\zeta(s)$ for $\mathrm{Re}(s)>1$, so it seems like a reasonable question to ask how the partial sums grow (asymptotically) with $n$ otherwise. Note $H_{n,s}=\sum_{k=1}^n k^{-s}$ is called a generalized harmonic number. For $\mathrm{Re}(s)>0$ we can proceed by multiplying and dividing by $n^s$ - that is,

$$H_{n,s}=n^{1-s}\left(\frac{1}{n}\sum_{k=1}^n (k/n)^{-s}\right)\sim n^{1-s}\int_0^1 x^{-s}\,\mathrm{d}x=\frac{n^{1-s}}{1-s}. $$

In fact, the Euler-Maclaurin formula can be used to derive a full asymptotic series:

$$ H_{n,s}=\zeta(s)-\sum_{k=-1}^{\infty} \frac{B_{k+1}}{(k+1)!} (s)_k n^{-s-k}. $$

This series is divergent, but for a particular value of $s$ if we lop off all summands that $\to0$ with $n$ we get a finite sum. In particular, this shows your formula

$$ \zeta(s) = \lim_{n\to\infty} \left( H_{n,s}-\frac{n^{1-s}}{1-s} \right). $$


问:
Given triangle ABC and its inscribed circle O, AO = 3, BO = 4, CO = 5, find the perimeter of ABC
As title,

I could only figure out that

$$a^2 + r^2 = 9$$
$$b^2 + r^2 = 16$$
$$c^2 + r^2 = 25$$
$$r^2 = \frac{abc}{a+b+c}$$

But couldn't get how to derive $2(a+b+c)$.

答:
I doubt if it can be solved by hand. By Mathematica, we have
$$r=\frac{1}{360} \left(\sqrt[3]{245083391+21600 i \sqrt{314509827}}+\frac{591361}{\sqrt[3]{245083391+21600 i \sqrt{314509827}}}-769\right)$$

Computing its minimal polynomial, it is a root of $120 r^3+769 r^2-3600=0$

And thus it is very complicated to represent the result in radicals.

The numerical solutions is $a= 2.32148,b= 3.51984,c= 4.62485,r= 1.9002$.