问:
Conclude that $A^TA=I$, so $A^{-1}=A^T$.
A matrix $A$ is said to be orthogonal if for every vector $\textbf{x}$ we have

$||A\textbf{x}||_2=||\textbf{x}||_2$.

Show that such a matrix will necessarily have the property that

$\left=\left<\textbf{x},\textbf{y}\right>$.

From here, conclude that $A^TA=I$, so $A^{-1}=A^T$.

$\textbf{Hint:}$ Make use of the identity $\left<\textbf{x},\textbf{y}\right>=\frac{1}{4}||\textbf{x}+\textbf{y}||^2_2 - \frac{1}{4}||\textbf{x}-\textbf{y}||^2_2$.

I am not sure how to even begin this question. If someone could at the least provide definitions of the notation. How do you read the first line? Is it read, "The norm of the matrix A with vector x in $\mathbb{R}^2$"? Then what does the hint say? Why is there a 2 as a superscript and subscript?

答:
Hint

$$\left<Ax,Ay\right>=\left<x,y\right>
\text{ }\text{ }\Rightarrow\text{ }\text{ }
\left=\left<x,y\right>
\text{ }\text{ }\Rightarrow\text{ }\text{ }
\left<(A^TA-I)x,y\right>=0$$

Body:
Hint

$$\left<Ax,Ay\right>=\left<x,y\right>
\text{ }\text{ }\Rightarrow\text{ }\text{ }
\left<A^TAx,y\right>=\left<x,y\right>
\text{ }\text{ }\Rightarrow\text{ }\text{ }
\left<(A^TA-I)x,y\right>=0$$


问:
Finding the limit, multiplication by the conjugate
I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$

I tried and tried... friends of mine tried as well and we don't know how to get out of:

$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$

(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)

How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.

答: Multiply both numerator and denominator by $2+\sqrt{3+x}$, simplify, cancel out. I get $-\frac{1}{4}$</p>

ContentLicense: CC BY-SA 3.0


问:
When do I need to take the Memorylessness property in consideration when calculating Exponential distribution?
The exersice is:
The waiting time in a line for a person who arrives to get vaccinated is exponentially distributed with the expected value of 20 minutes. John is waiting on the line for 15 minutes already to get the vaccine and he is not yet been called. What is the probability that his turn to get the vaccine will happen in the next 10 minutes?

Do I need to take the Memorylessness property in consideration here or is it enough to calculate F(25)-F(15)? That is, to use the cumulative distribution function of the exponential distribution for the values 25 and 15 and substract it? Is there a rule of thumb for this?

答:
You can calculate the conditional probability $\dfrac{F(10+15)-F(15)}{1-F(15)}$ in general

but the memoryless property of the exponential distribution tells you this is $F(10)$ as a shortcut

ContentLicense: CC BY-SA 4.0


答:
To try to decipher this for you:

Here $A \in \mathbb{R}^{n \times n}$ is some matrix and $x,y \in \mathbb{R}^n$ are vectors. Then $Ax = y \in \mathbb{R}^n$ and
$$
\|Ax\|_2 = \|y\|_2 = \sqrt{\sum_{k=1}^n y_k^2}
$$
where $y = (y_1, y_2, \ldots, y_n)$.

So the first line claims that $A$ has the property that
$$
\|Ax\|_2 = \|x\|_2.
$$

Title: Conclude that $A^TA=I$, so $A^{-1}=A^T$.

Body:
A matrix $A$ is said to be orthogonal if for every vector $\textbf{x}$ we have

$||A\textbf{x}||_2=||\textbf{x}||_2$.

Show that such a matrix will necessarily have the property that

$\left=\left<\textbf{x},\textbf{y}\right>$.

From here, conclude that $A^TA=I$, so $A^{-1}=A^T$.

$\textbf{Hint:}$ Make use of the identity $\left<\textbf{x},\textbf{y}\right>=\frac{1}{4}||\textbf{x}+\textbf{y}||^2_2 - \frac{1}{4}||\textbf{x}-\textbf{y}||^2_2$.

I am not sure how to even begin this question. If someone could at the least provide definitions of the notation. How do you read the first line? Is it read, "The norm of the matrix A with vector x in $\mathbb{R}^2$"? Then what does the hint say? Why is there a 2 as a superscript and subscript?


问:
Relationship between the parity of the number of forests in a graph and Eulerianness
Given a connected multigraph $G$, consider the number of forests in it – subsets of the edge set of $G$ that form no cycle, empty set included. This is a specific value $T_G(2,1)$ of the Tutte polynomial; I played around with some graphs using this relation and saw the following pattern:

The number of forests is odd iff the graph is Eulerian.

For example, any simple tree with $m\ge1$ edges is not Eulerian and all its edge subsets are forests, hence $2^m$ possible forests, an even number. The octahedral graph has $1083$ forests and the $6$-regular Shrikhande graph $187888382977$. The dipole graph with $m$ edges has $m+1$ forests.

If the above observation is always true, how can I prove it? It seems that I would need to use the deletion-contraction recurrence satisfied by the count of forests (and spanning trees, and all other graph invariants encompassed by the Tutte polynomial), but all I get from this is that removing an edge from Eulerian graph always gives a non-Eulerian graph, and contracting always gives an Eulerian graph.

答:
(...) but all I get from this is that removing an edge from Eulerian graph always gives a non-Eulerian graph, and contracting always gives an Eulerian graph.

But isn't this just what you need? Let $f(G)$ denote the number of forests of $G$ and let us prove the statement for arbitrary (not necessarily connected) multigraphs by induction on the number $m+n$ of vertices plus edges. If $m=0$, then the graph is Eulerian and has $1$ forest, so the statement holds. Now if $G$ is Eulerian and $e$ is an edge of $G$, $f(G) = f(G-e) + f(G / e)$ is a sum of an odd and an even number, so it is odd. If $G$ is non-Eulerian then either $G$ is a single edge, in which case the number of forests is $2$, as needed, or you can find an edge of $G$ such that neither $G-e$ nor $G/e$ is Eulerian: if there is a vertex with even degree, take any edge incident to it, and otherwise just take any edge. In this case $f(G)$ is a sum of two odd numbers, so it is even.


问:
What are the bases of a function space (Hilbert space)?
1. Motivation

I was learning about the Hilbert space and function spaces, i.e., roughly speaking, infinite-dimensional vector spaces.

Let's now think about ordinary 3D Euclidean vectors. A vector $\vec{x}$ may be given by

$$ \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}} \vec{x} = \mycolv{1\\3\\2}$$

and this is equivalent to saying that $\vec{x} = 1 \hat{i} + 3 \hat{j} + 2 \hat{k}$. So even if we think of this vector as an ordered triple, $(1, 3, 2)$, then a mathematical structure behind is saying that a vector can be demonstrated in the form of a linear combination of basis vectors.

2. My understanding of function spaces

As someone studying engineering and hasn't been exposed to rigorous mathematical proofs of linear algebra, I understood, through quite an intuitive approach, why function spaces are infinite-dimensional vector spaces. I considered the inner product defined, for example, of functions $\phi: R \rightarrow C$ and $\psi: R \rightarrow C$.

$$\langle \phi | \psi \rangle = \int \phi^*(x) \psi(x) \ dx$$

Intuitively speaking this is about adding up all the differentials $\phi^*(x) \psi(x) \ dx$, which is analogous to the scalar product in Euclidean space, so I thought each and every function values corresponding to individual (though an infinite number of) elements of the domain should all be the components of vectors $\phi(x)$ and $\psi(x)$. (And thus both function spaces should be of an infinite dimension.) So, like the below:

$$ \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}} |\psi(x) \rangle = \mycolv{...\\\psi(a-2\epsilon) \\ \psi(a-\epsilon) \\\psi(a)\\\psi(a + \epsilon) \\ \psi(a+2 \epsilon) \\ ...}$$

for some $a \in R$.

3. Question

It's fine up to here. But, there should at least be an infinite number of linearly independent basis vectors that make up the actual infinite-dimensional column vector, i.e., $|\psi(x)\rangle$. As if $\hat{i}, \hat{j}, \hat{k}$ corresponded respectively to the coefficients $1, 3, 2$, there should be basis vectors that correspond to each of $\psi(a), \psi(a+\epsilon), \psi(a-\epsilon), $ and so on. What are they?

答:
Thank Ethan and md2perpe for your answers. I found a good YouTube video that pretty much explains all points I was confusing. So it would be good if anyone else is also agonising about the basis of function spaces. I will introduce the key ideas of the video.

We know that a Dirac delta function, which is informally defined as

$$\delta(x) = 0 \ when \ x = 0$$
$$\delta(x) = \infty \ when \ x \neq 0$$

has a peak at $x=0$ and is zero elsewhere.

So we can make it have a peak at $x=a$ by using $\delta(x-a)$ instead. And it satisfies the following property:

$$\int_{-\infty}^{\infty} f(x) \delta (x-a) dx = f(a)$$

So a function can be expressed in terms of a linear combination of the appropriate basis vectors $\delta(x-a)$, that is roughly

$$f(x) = \sum_{i=-\infty}^{\infty} f(x_i) \delta (x-x_i)$$

However, this is for the case of functions whose domain is the integer set, and we can extend this idea appropriately to ordinary continuous functions by changing the sum to the integral.

$$f(x) = \int_{-\infty}^{\infty} f(a) \delta(x-a) \ da$$

Notice that within the integral there is $f(a)$ which corresponds to $f(x_i)$ and the integration is done with respect to the variable $a$.

There are other ways of forming basis for a function, such as Taylor series which uses monomial basis ($x^0, x^1, x^2, x^3, ...$) and Fourier series which uses trigonometric/exponential basis ($e^{ikx}, e^{-ikx}, e^{2ikx}, e^{-2ikx}, ...$).


问:
Given tangents from A and C, find minimum length of AC
Let $ H $ be the intersection point of the heights of the acute-angled triangle $ ABC $. From points $ A $ and $ C $, tangents $ AK $ and $ CT $ are drawn to the circle drawn on the segment $ BH $ as on the diameter. Let $ 15 $ and $ 19 $ be the lengths of these tangents. What is the square of smallest possible $ AC $ side length?

It can be easy for me if it is with value. As It is an acute triangle, can't find the way. Please help me with a solution to go with this type in the future.

My Simple Answer:

$AK^2 + AC^2 = CT^2$

$15^2 + AC^2 = 19^2$

$AC^2 = 361 - 225$

$AC^2 = 136$

答:
Let the feet of perpendiculars from point $A$, $B$ and $C$ on the respective opposite sides be $D$, $E$ and $F$ respectively. The circle with diameter $BH$ will pass through points $D$ and $F$. Let the tangents from $A$ and $C$ to this circle be $AS$ and $CT$ respectively.

Considering the power of points $A$ and $C$ with respect to this circle gives,

$AS^{2}=AF\cdot AB$

$CT^{2}=CD\cdot BC$

Now, observe that, $CD\cdot BC=CE\cdot AC$ because $\triangle ACD\sim \triangle BCE$. Similarly, $AF\cdot AB=AE\cdot AC$ because $\triangle CAF\sim \triangle BAE$.

$\Rightarrow AS^{2}=AF\cdot AB=AE\cdot AC$ and similarly $CT^{2}=CE\cdot AC$.

Adding these two equations gives,
$AC^{2}=\boxed {AS^{2}+CT^{2}}$

Now plugging in the lengths of the tangents will give the value of $AC^{2}$.


问:
Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.

My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.

There are $2$ cases to consider:

Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:

\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) +  abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}

Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?

答:
Unfortunately, I wasn't able to pick up from where you left off.

Starting from the following inequality:
$$a(b+c-a) \leq b(a+c-b) \leq c(a+b-c) $$
which is true given your initial WLOG $a\geq b\geq c$.

And note that $\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c}$.

You can use Rearrangement Inequality so that:

\begin{align}\frac{1}{a} a(b+c-a) + \frac{1}{b} b(a+c-b) + \frac{1}{c} c(a+b-c) & \geq \frac{1}{c} a(b+c-a) + \frac{1}{a} b(a+c-b) + \frac{1}{b} c(a+b-c) \\ a+b+c &\geq \frac{a(b-a)}{c} + a + \frac{b(c-b)}{a} + b + \frac{c(a-c)}{b} + c \\ 0 &\geq \frac{a(b-a)}{c} + \frac{b(c-b)}{a}  + \frac{c(a-c)}{b} \\ \frac{a(a-b)}{c} + \frac{b(b-c)}{a} + \frac{c(c-a)}{b} & \geq 0 \\ a^2b(a-b) + b^2c(b-c)+c^2a(c-a) &\geq 0\end{align}

Hope it's correct.


问:
Quick Integration with Vectors Question
So I was given the following prompt:

The instantaneous rate of change of the vector-valued function $g(t)$ is given by $r(t)=\langle\sqrt{t^2+1}, \sin(t^2)\rangle$. If $g(7)=\langle\sqrt{2},\pi\rangle$, find $g(0)$.

I guess I'm a bit confused about the integration that would be involved with this problem. I understand that I'm going to have to integrate both given functions, which I would then plug $0$ in to, but I'm a bit confused about what this integration would look like. Some clarification would be appreciated!

Edit: I understand what the integral formula would look like here, it's the actual integration that's confusing to me, I'm not coming up with a normal integral when I integrate.

Title: Quick Integration with Vectors Question

Body:
You have

$$g(7) - g(0) = \int_0^7 r(t) dt.$$ hence

$$g(0)  = g(7) - \int_0^7 r(t) dt = \left(\sqrt 2 - \int_0^7 \sqrt{t^2+1} dt,\pi - \int_0^7\sin t^2 dt\right)$$


答: Multiplying by the conjugate does indeed work.  You just forgot to carry the negative sign throughout.  After multiplying by the conjugate, the correct expression is $\frac{1-x}{(x-1)(2+\sqrt{3+x})}$</p>

Title: Finding the limit, multiplication by the conjugate

Body:
I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$

I tried and tried... friends of mine tried as well and we don't know how to get out of:

$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$

(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)

How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.


问:
Is this identity trivial? $n^n=n!+\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^n$
If $m$ and $n$ are two positive integers such that $m\ge n$, then it is a standard counting exercise that the number of surjective functions between the sets $\{1, 2, ..., m\}$ and $\{1, 2, ..., n\}$ is $\displaystyle n^m-\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^m$.

Now, if we let $m=n$ in this formula, we get that the number of permutations of the set $\{1, 2,...,n\}$ equals $ \displaystyle \sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^n$. But we also know that the number of the permutations equals $n!$. Hence, by double counting, we get the identity I mentioned in the title: $$ n^n=n!+\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^n.$$
Now, I haven't seen this identity before. When we found this in class, we suspected that it is just trivially true, but no one was able to actually show that it is so. I have come across this identity again today while I was revising for my final exam and I couldn't help but wonder whether this is actually trivially true or if there is something more to it. I have tried using the Binomial Theorem, but things only started to look messier. So, I am looking forward to seeing some clever manipulations which may render this trivial (I doubt that it is some remarkable identity, but who knows, maybe I am wrong).

答:
It is not quite trivial, but your identity
$$ n^n=n!+\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^n $$
can be easily simplified to
$$ n!=\sum_{k=0}^n(-1)^k\binom{n}{k}(n-k)^n =
\sum_{k=0}^n(-1)^{n-k}\binom{n}{k}k^n $$
which is well known in combinatorics and proved in
several ways including inclusion-exclusion.
The OEIS sequence A000142
for the factorial sequence has this formula
$$ \sum_{i=0}^n (-1)^i(n-i)^n \binom{n}{i} = n!. $$
Also see
MSE question 1512743
"An identity for the factorial function".


问:
Find the set of 'fair' groupings from non-square binary matrix and cost vector
Given an m by n binary matrix $P$ and a 'cost' vector $v_1$ I want to find the set of $P$ matrices where the product $Pv_1=v_2$ results in all the elements of $v_2$ being within a certain lower and upper bound (i.e. $\pm$ 20% of the sum of the elements of $v_1$/m).

This is for a fair allocation problem where 1000 items are being distributed into 20 groups (each item being allocated only once). The costs of the items are not normally distributed so the groups will necessarily be of different sizes to achieve a set of fair costs. I am currently using iterative methods but I was wondering if there was a neater way of approaching this.

Many thanks

答:
Thanks to the edited tags on this question I now see it is a variation of the 'fair allocation problem', and there is a good resource here on the topic: http://recherche.noiraudes.net/resources/papers/comsoc-chapter12.pdf

Often it's just a case of finding out what your problem's name is :-)


问:
For a point K inside a triangle show an equality
Let ABC be a triangle whose heights are $h_a,h_b$ and $h_c$. Let $K$ be any point inside the triangle, and $d_a,d_b$ and $d_c$ the distances of $K$ from the sides $a,b$ and $c$, respectively. Show that $$\dfrac{d_a}{h_a}+\dfrac{d_b}{h_b}+\dfrac{d_c}{h_c}=1.$$

We can draw a line from $K$ to each of $A, B,$ and $C$, forming three triangles $KAB, KBC,$ and $KCA$. We know $$S_{KBC}+S_{KAC}+S_{KAB}=S_{ABC}$$ or $$\dfrac{ad_a}{2}+\dfrac{bd_b}{2}+\dfrac{cd_c}{2}=\dfrac{ah_a}{2}=\dfrac{bh_b}{2}=\dfrac{ch_c}{2}.$$ If we multiply the last by 2, we get $$ad_a+bd_b+cd_c=ah_a=bh_b=ch_c.$$ I don't see anything else. Any help would be appreciated. (I would love to hear your thoughts on the problem)

答:
Now from your last equation, call

$M=ah_a=bh_b=ch_c$.

Then you have that $\frac{ad_a+bd_b+cd_c}{M}=1$, that is, $\frac{ad_a}{M}+\frac{bd_b}{M}+\frac{cd_c}{M}=1$, that is, $\frac{ad_a}{ah_a}+\frac{bd_b}{bh_b}+\frac{cd_c}{ch_c}=1$, which is what you want.


问:
Meromorphic matrix
I am starting to study some documents for my thesis, maybe this question is basic, but I don´t know precisely what is.

What is a meromorphic matrix?.

Where i can read about of  the precisely definition and/or examples?

could you recommend me some book?

thanks.

答: Some context might be helpful, but I think the most reasonable assumption is that this refers to a matrix whose entries are meromorphic functions.</p>

Title: Meromorphic matrix


答:
I'm not sure if this counts as "trivial" (and this is at heart identical to your surjections argument) but just note that you are counting the number of functions $\{1,\dots,n\}\to\{1,\dots,n\}$ in two ways.

The LHS is just $n^n$ because each of $1,\dots,n$ is mapped to one of $n$ possible values.

The RHS is inclusion-exclusion, where your sets $A_i$ are the set of all functions that don't include $i\in\{1,\dots,n\}$ in their image. The number of functions that don't hit (at least) $k\geq1$ points is $\binom {n}{k}(n-k)^n$, as you choose $k$ "illegal" points, and then each of $1,\dots,n$ map to any of the other $n-k$ "legal" points.

The $n!$ term is to account for the number of functions that hit every point, i.e. the permutations.

Title: Is this identity trivial? $n^n=n!+\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^n$

Body:
If $m$ and $n$ are two positive integers such that $m\ge n$, then it is a standard counting exercise that the number of surjective functions between the sets $\{1, 2, ..., m\}$ and $\{1, 2, ..., n\}$ is $\displaystyle n^m-\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^m$.

Now, if we let $m=n$ in this formula, we get that the number of permutations of the set $\{1, 2,...,n\}$ equals $ \displaystyle \sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^n$. But we also know that the number of the permutations equals $n!$. Hence, by double counting, we get the identity I mentioned in the title: $$ n^n=n!+\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)^n.$$
Now, I haven't seen this identity before. When we found this in class, we suspected that it is just trivially true, but no one was able to actually show that it is so. I have come across this identity again today while I was revising for my final exam and I couldn't help but wonder whether this is actually trivially true or if there is something more to it. I have tried using the Binomial Theorem, but things only started to look messier. So, I am looking forward to seeing some clever manipulations which may render this trivial (I doubt that it is some remarkable identity, but who knows, maybe I am wrong).


问:
Representing sums of matrix algebras as group rings
Let $A = M_{n_1}(\mathbb R) \oplus M_{n_2}(\mathbb R) \oplus ... \oplus M_{n_m}(\mathbb R)$ be a direct sum of real matrix algebras. Under what conditions does there exist a group ring $\mathbb R[G]$ which is isomorphic to $A$?

I know every group ring is isomorphic to some such $A$ by Wedderburn's theorem, and I want to determine the extent to which the converse holds. I know that $A$ must have a $\mathbb R$-summand, corresponding to the linear span of $\sum_{g\in G} g \in \mathbb R[G]$. Is this condition sufficient as well? If not, is there a nice criterion, at least for small numbers of summands?

答:
This is in fact not what Artin-Wedderburn says over $\mathbb{R}$. Artin-Wedderburn says that $\mathbb{R}[G]$ is a product of matrix algebras over (finite-dimensional) real division algebras. Over $\mathbb{R}$ there are three such algebras, namely $\mathbb{R}, \mathbb{C}, \mathbb{H}$.

Which of these algebras appear can be computed from the character table of $G$ (see Frobenius-Schur indicator). In order for $\mathbb{R}$ to be the only division algebra that occurs, the Frobenius-Schur indicator of every irreducible representation of $G$ over $\mathbb{C}$ must be equal to $1$. In particular, every irreducible representation of $G$ over $\mathbb{C}$ must be self-dual, which is equivalent to the character table being real, which is in turn equivalent to every element of $G$ being conjugate to its inverse (ambivalent groups).

If the above condition holds, then the $n_i$ are precisely the dimensions of the real irreducible representations of $G$. There are various conditions that these dimensions need to satisfy (for example, the dimensions of the complex irreducible representations must divide $|G|$, so I think for each $i$ either $n_i$ or $\frac{n_i}{2}$ must divide $|G|$, or something like that) and I am doubtful that anything simple can be said in general.

Note that we must have $|G| = \sum n_i^2$, so if this number is particularly simple (e.g. prime) then you can hope to classify all of the finite groups of order $|G|$ and compute their character tables.

Here are some things that can be said for very small values of $m$.

$m = 1$: Every group has the trivial representation, so one of the $n_i$ must be equal to $1$. So in the case of one summand, the only possibility is $A = \mathbb{R}$ and $G$ is the trivial group.

$m = 2$: We must have $A = \mathbb{R} \times M_n(\mathbb{R})$ for some $n$ and $G$ must be a finite group with $|G| = n^2 + 1$ such that $G$ has only one nontrivial real irreducible representation. This is a very difficult constraint to satisfy: it implies that $G$ has an irreducible complex representation of order either $n$ or $\frac{n}{2}$, hence either $n$ or $\frac{n}{2}$ divides $n^2 + 1$. The only $n$ with this property are $n = 1, 2$. In the second case $n^2 + 1 = 5$ and the cyclic group of order $5$ does not work, so we can only have $G = C_2$ and $A = \mathbb{R} \times \mathbb{R}$.

Alternatively, it's known that the number of real irreducible representations of $G$ is the number of "real conjugacy classes" of $G$, where real conjugacy is the equivalence relation generated by conjugacy and inverses. Since elements of different order can't be real conjugate, the number of real conjugacy classes is at least one more than the number of prime factors of $|G|$, hence if this number is equal to $2$ then $|G|$ must be prime and in fact must be equal to $2$ as well.


问:
Banach spaces admitting a coarser topology for which the closed unit ball is compact.
Let $X$ be a Banach space and let $B$ be its closed unit ball.

It is well known that $B$  is compact in the weak topology provided $B$ is reflexive.   Otherwise,   if $X$ is at least
a  dual space, then $B$ is compact in the weak* topology.

These examples show that, in many situations, there exists a Hausdorff locally convex topology on $X$, coarser than the
norm topology,  relative to which
$B$ is compact.

For $X=c_0$, I cannot think of such a topology and I doubt it exists.  The reason is I feel the sequence $\{x_n\}_n$ given
by
$$
  x_n= (1,1,\ldots, 1,0,0, 0\ldots)
  $$
($n$ ones) cannot possibly have a cluster point in any sensible topology.

Question.  Given a Banach space $X$,  is there always  a Hausdorff locally convex topology on $X$, coarser than the
norm topology, relative to which the
closed unit ball is compact? Can one characterize the spaces for which this is true?
What if $X=c_0$?

答: It is an old result of K.F. Ng [On a theorem of Dixmier, Math. Scand. 29 (1971), 279–280 (1972)] that a Banach space $X$ whose unit ball $B$ is compact in some coarser locally convex topology $\tau$ is a dual Banach space. Namely, it is the dual of $Y=\{f\in X^*: f|_B$ is $\tau$-continuous$\}$ endowed with the dual norm.


问:
tensoring with a field preserves exactness?

Let $A$ be a local ring and $M$ be a $A$-module. Then for given exact sequence $0\to B\to C\to D\to 0$ where $B,C,D$ are all $A$-modules, $0\to \mathcal{k}\otimes B\to\mathcal{k}\otimes C\to\mathcal{k}\otimes D\to 0$ is exact where $\mathcal{k}$ is a residue field of $A$.

I want to prove this by showing $f:B\to C$ is injective then $1\otimes f:\mathcal{k}\otimes B\to\mathcal{k}\otimes C$ is injective. Could you give any hint?

And if for general commutative ring with unity $A$, if $k$ is a field which has $A$-module structure, then can we say $k$ is flat module?

Any comment will be appreciated thanks.

Edit:

The original question is from exercise of Introduction to commutative algebra by Atiyah-Macdonald. In the above image, they say what I wrote above without verification. I wonder why it's true

答:
I don't think the first claim is true as stated. Typically, the residue field $k$ is not flat over the local ring $A$. For instance, if we take $A$ to be the local ring $\mathbb{Z}_p$, and we consider the short exact sequence $0 \to (p) \to \mathbb{Z}_p \to \mathbb{F}_p \to 0$, then if we tensor with the residue field, the map $(p) \otimes \mathbb{F}_p \to \mathbb{F}_p$ will be the zero map.

In the situation of the exercise, things are fine, but this is because it is the $F$'s which are flat over $A$. I suppose that there's a coordinatewise way to see that $N \otimes k \to F \otimes k$ is exact, but here's a more roundabout approach. If you expand the a priori just right-exact sequence through Tor groups you find the sequence
$$\cdots \to \operatorname{Tor}^1_A(F,k) \to \operatorname{Tor}^1_A(F,k) \to N \otimes k \to F \otimes k \to F \otimes k \to 0.$$
As $F$ is flat over $A$, $\operatorname{Tor}^1_A(F,k)$ vanishes and the desired result follows.


问:
Conjecture: For almost all $n$, the sum of $n \bmod k$ for $k<n$ is unique.
It looks to me like $$f(n):= \sum_{k=1}^{n}{(n\bmod k)}$$ is almost unique to $n$, with the only systematic exceptions being $(2^m-1, 2^m)$ pairs, which share a value for any $m \in \mathbb N$. In other words, the set $\{f(n): n\in\mathbb N\land n\neq 2^m
\ \forall m\in\mathbb N\}$ shouldn't have many duplicate values.

As discussed in the answers, a handful of counterexamples have been found, but they seem more sparse than one would expect, and nobody has yet suggested why that might be, or what's special about the counterexamples found, apart from a moderate preponderance of small factors in the mean of the pair.

Although there's not much data to go on, it also looks like counterexamples may come in the form $(n,n+2i)$, or possibly even $(n,n+2^i)$, and I would be interested if anyone can find a case outside of those. The latter form would also include the $(2^m-1,2^m)$ exceptions, which seems elegant.

Also, if someone could set me straight on this: My understanding is that the term 'almost all' in this context means approaching $0$ density in the limit; so if one could show that the exceptions aren't necessarily finite, but they are less common than powers of two, would that be sufficient?

答:
This is just a comment on the counterexample in the OP's answer, noting a relatively quick way to verify that $f(38183)=f(38185)$ without computing each to equal $258840858$.

A formula at the OEIS entry for the OP's sequence says

$$f(n)=n^2-\sum_{k=1}^n\sigma(k)$$

where $\sigma(k)$ is the sum of the divisors of $k$. To show that $f(m)=f(n)$ with $m\lt n$, then, it suffices to show that

$$\sum_{k=m+1}^n\sigma(k)=n^2-m^2=(n-m)(n+m)$$

For $m=38183$ and $n=38185$, we have $(n-m)(n+m)=2\cdot76368=152736$ while

$$\begin{align}
\sigma(38184)+\sigma(38185)
&=\sigma(2^3\cdot3\cdot37\cdot43)+\sigma(5\cdot7\cdot1091)\\
&=15\cdot4\cdot38\cdot44+6\cdot8\cdot1092\\
&=100320+52416\\
&=152736
\end{align}$$

Remark: The trickiest part of the verification is the factorization into primes (in particular, recognizing $1091$ as a prime number).


问:
Stokes theorem to find flux
Find flux of field $(z,x,y)$ on curve $x^2+y^2=1, z=1$ oriented positive looking from positive part of z axis using Stokes formula. So this should come to calculating $3\iint_S dS$ or $3\int\limits_{0}^{2\pi}\int\limits_{0}^{1}drd\phi$. Is this correct? I think we should use plus in front of integral because it is oriented positively.

答:
The Stokes' Theorem states that $\displaystyle \int_C \vec{F} \cdot d\vec{r} = \iint_{S}(\nabla \times \vec{F}) \cdot d\vec S$

The surface integral of the curl of a vector field is equal to the line integral of the vector field over the smooth closed boundary curve $C$ which is the bound of the surface.

In this case, boundary curve is $x^2+y^2 = 1$ at $z=1$. We can parametrize the curve as $\vec{r}(t) = (\cos t, \sin t, 1), 0 \leq t \leq 2\pi$.

$\vec{r'}(t) = (-\sin t, \cos t, 0)$

$\vec {F} = (z,x,y) = (1, \cos t, \sin t)$

and then the line integral.

Now if you are not applying Stokes' Theorem and are doing a double integral, you need to find a surface whose boundary curve is $x^2+y^2 = 1$ at $z=1$. It can be any simple surface, let's take a disc itself that is given by $x^2+y^2 = 1$ in plane $z=1$. The normal surface to the plane is $(0,0,1)$ based on the orientation.

$curl(\vec{F}) = \nabla \times \vec{F} = (1,1,1)$

$(\nabla \times \vec{F}) \cdot \hat{n} = 1$ which is a constant so double integral equals to nothing but the surface area of the disc which is $\pi$. Check the result to see both sides (integrals) are same.

Title: Stokes theorem to find flux


答:
Note that the identity can be rearranged to
\begin{eqnarray*}
n!= \sum_{k=0}^{n-1} (-1)^k \binom{n}{k} (n-k)^n =n^n-\binom{n}{1} (n-1)^n+ \cdots.
\end{eqnarray*}
Now condiser maps from $[n]$ to$[n]$, the LHS is permutations of $[n]$.

The RHS counts the permutations using inclusion-exclusion principle.


答:
Observe that, $\frac {d_{a}}{h_{a}}=\frac{\frac12\times a\times d_{a}}{\frac12\times a\times h_{a}}=\frac{Area(KBC)}{Area(ABC)}$.

Similarly, $\frac{d_{b}}{h_{b}}=\frac{Area(KCA)}{Area(ABC)}$ and $\frac{d_{c}}{h_{c}}=\frac{Area(KAB)}{Area(ABC)}$.

Adding these three ratios will give the desired result.

Title: For a point K inside a triangle show an equality

Body:
Let ABC be a triangle whose heights are $h_a,h_b$ and $h_c$. Let $K$ be any point inside the triangle, and $d_a,d_b$ and $d_c$ the distances of $K$ from the sides $a,b$ and $c$, respectively. Show that $$\dfrac{d_a}{h_a}+\dfrac{d_b}{h_b}+\dfrac{d_c}{h_c}=1.$$

We can draw a line from $K$ to each of $A, B,$ and $C$, forming three triangles $KAB, KBC,$ and $KCA$. We know $$S_{KBC}+S_{KAC}+S_{KAB}=S_{ABC}$$ or $$\dfrac{ad_a}{2}+\dfrac{bd_b}{2}+\dfrac{cd_c}{2}=\dfrac{ah_a}{2}=\dfrac{bh_b}{2}=\dfrac{ch_c}{2}.$$ If we multiply the last by 2, we get $$ad_a+bd_b+cd_c=ah_a=bh_b=ch_c.$$ I don't see anything else. Any help would be appreciated. (I would love to hear your thoughts on the problem)


问:
Is the Kantorovich duality only for 1-Wasserstein?
I’m reading this paper here, and the authors make the following assertion:

“Unfortunately, their algorithm cannot be readily applied to any other OT cost, because the famous Kantorovich duality holds only for W1”

So, they claim that the Kantorovich Duality is valid only for the 1-Wasserstein distance. I think they are actually speaking of the Kantorovich-Rubinstein duality. But is it really only valid for the 1-Wasserstein?

答:
Yes, they are referring to the Kantorovich-Rubinstein duality. The paper they reference involves using neural networks to approximate the Lipschitz potentials.

In general (with suitable assumptions on $c$, especially for powers of the euclidean norm) you have the duality
$$ \inf_{\pi \in \Pi(\mu,\nu)} \int_{X \times Y} c(x,y) d \pi = \sup_{\substack{\varphi \in C_0(X) \\ \varphi(x) + \varphi^c(y) \leq c(x,y)} }\int_X \varphi d\mu + \int_Y \varphi^c d\nu $$
where
$$ \varphi^c(y) = \inf_x c(x,y) - \varphi(x) $$
for a reference you can see Santambrogio's Optimal Transport for Applied Mathematicians - Chapter 1.2.

In the case of $c(x,y) = d(x,y)$, this reduces to lipschitz functions since $\varphi^c = - \varphi$.


问:
Self intersection of degree $d$ plane curves in $\mathbb P^3$
Let $C$ be a nonsingular degree $d$ plane curve  in $\mathbb P^3$ over complex numbers. My interest is to calculate self-intersection of $C$. Does the following calculation make sense?

From adjunction formula we know that if $C$ is a non-singular curve of genus $g$ on $\mathbb P^2$ and if $K$ is the canonical divisor of $\mathbb P^2$, then $2g-2=C.(C+K)$. We know that $g= \frac{1}{2}(d-1)(d-2)$ and as $K \cong \mathcal O_{\mathbb P^2}(-3)$ we have, $(d-1)(d-2) -2 =C^2 -3d $. Which means $C^2=d^2$ .

But I am nowhere using that the entire phenomenon is happening inside $\mathbb P^3$. Am I missing something very obvious?

Any help from anyone is welcome.

答: The intersection of two curves in $\mathbb{P}^3$ (in the sense of intersection theory) is always $0$.  In general, the intersection of a codimension $d$ cycle and a codimension $e$ cycle has codimension $d+e$.  In the case of curves in $\mathbb{P}^3$, they are codimension $2$, so their intersection has codimension $4$, i.e. it is a cycle of dimension $-1$, which just means it is automatically $0$.


问:
Family of geometric shapes closed under division
The family of rectangles has the following nice properties:

Every rectangle $R$ can be divided to two disjoint parts, $R_1 \cup  R_2 = R$, such that both $R_1$ and $R_2$ are rectangles (i.e. belong to the same family).

Moreover, for every number $p > 0$, it is possible to cut $R$ to two disjoint rectangles $R_1 \cup  R_2 = R$, such that the ratio of their areas is $p$.

What other non-trivial families of geometric shapes have one or two of these properties?

Triangles have properties 1 and 2 - a triangle can always be divided to two triangles using a ray from one of its vertices.

Right-angled triangles have property 1, because they can be divided using a ray from the 90-degree vertex that is perpendicular to the opposite side, and the result will be two right-angled triangles. However, they don't have property 2 because the ratio between the two parts cannot be changed.

Squares have none of these properties, as a square cannot be divided to two squares. Ditto for circles.

L-shapes seem to have properties 1 and 2, as well as rings.

This raises several questions:

Is it possible to give a complete characterization of such families (i.e. find all families with the above properties)?

Given a family of shapes, how can we find a minimal containing family that has the above properties? (e.g. for squares, a possible extension is the rectangles; another extension is the L-shapes).

(terminology) Is there a formal term for these two properties? If not, what could be a good term? Maybe "closed under division" / "closed under arbitrary proportion division"?

MOTIVATION: When designing algorithms for geometric shapes, if the shapes have property 1 (and possibly property 2), then it is possible to design a recursive algorithm. Otherwise, a recursive algorithm is not possible and we must think of other algorithms.

EDIT: A concept closely related to my property #1 is rep-tile. A rep-tile is a shape that can be divided to two or more smaller copies of itself. This concept was recently generalized to a rep-tile set by Sallows.  Specifically, all rep-2 shapes and all rep-tile sets with 2 elements have property #1. But property #1 is more general, as it does not require that the parts will be copies of the original (for example, a rectangle can be divided into two rectangles that are rectangles, although they are not smaller copies of the original).

Property #2 is still open - I haven't found anything similar.

答:
This is not a complete answer, but here are some thoughts that may be useful.

These families are closed under unions; if families $A$ and $B$ have this property, so does $A\cup B$. (Infinite unions work as well.) This suggests to me that a  "nice" characterization of these families may be unrealistic; it would have to include things like "shapes which are either a rectangle or a triangle", which is not a very natural category.

Perhaps you could hope to focus on "prime" families, for which no nonempty strict subset of them is a valid family, but a problem is that there exist families that have no prime subsets; see some of the cases below.

Here are several more families not listed in your post. I'll assume boundedness in all cases, and only specify shapes up to similarity rather than congruence (otherwise we could have examples like "every subset of the following region").

All convex shapes.

Circular sectors of central angle at most $\theta$, for any $0<\theta\le2\pi$.

All parallelograms with an angle of $\theta$ for any $0<\theta<\pi$, and hence also "all parallelograms". Rectangles are the case $\theta=\pi/2$.

Triangles whose smallest angle is at most $\theta$, for any $0<\theta<\pi$. Also "strictly less than $\theta$", for which the case $\theta=60^\circ$ gives us the family of non-equilateral triangles.

Obtuse triangles.

Scalene triangles. (Not positive about this one.)

All triangles which are neither equilateral nor right triangles. (Just excluding right triangles doesn't work!)

Trapezoids.

Quadrilaterals.

Non-convex quadrilaterals.

Strictly convex quadrilaterals (all internal angles less than $180^\circ$).

Convex polygons with at most $n$ sides, for all $n\ge3$ (and including $n=\infty$).

Non-triangular convex polygons with at most $n$ sides, for all $n\ge4$ (and including $n=\infty$).

Convex shapes whose boundaries are a combination of line segments and circular arcs.

Shapes with a line of reflectional symmetry.

For any two families $A$ and $B$, the set of disjoint unions of a member of $A$ and a member of $B$. (Similar constructions abound for more than two base families, "two $A$s and three $B$s", etc.)

Some non-examples, and their counterexamples where not obvious:

Acute triangles.

Isosceles triangles.

Rhombuses.

Strictly convex pentagons; in fact, no strictly convex pentagon is the union of two others! This applies to strictly convex $n$-gons for all $n\ge 5$, as well.

All triangles whose angles are not $(50^\circ,60^\circ,70^\circ)$. (An equilateral triangle needs to use one of these to satisfy property $2$.)

All triangles without $10^\circ$ angles. (Again, an equilateral triangle fails here.)

The variety of these examples seems like further evidence to me that a complete classification is probably not feasible.


答:
$$\int \left(\sqrt{t^2+1},\sin \left(t^2\right)\right) \, dt=\left(a+\frac{1}{2} \left(t \sqrt{t^2+1}+\text{arcsinh} (t)\right),b+\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} t\right)\right)$$
Where $S(x)$ is the Fresnel integral.

To get the two constant we substitute $(\sqrt 2,\pi)$ so we get
$$\small g(t)=\left(\frac{1}{2} \left(t \sqrt{t^2+1}+\text{arcsinh} (t)\right)+\sqrt{2}+\frac{1}{2} \left(-35 \sqrt{2}-\text{arcsinh} (7)\right),\sqrt{\frac{\pi }{2}} S\left(\sqrt{\frac{2}{\pi }} t\right)-\sqrt{\frac{\pi }{2}} S\left(7 \sqrt{\frac{2}{\pi }}\right)+\pi \right)$$

Plugging $t=0$ we have
$$g(0)=\left(\sqrt{2}+\frac{1}{2} \left(-35 \sqrt{2}-\text{arcsinh}(7)\right),\pi -\sqrt{\frac{\pi }{2}} S\left(7 \sqrt{\frac{2}{\pi }}\right)\right)\approx (-24.6566,2.53571)$$

Title: Quick Integration with Vectors Question

Body:
So I was given the following prompt:

The instantaneous rate of change of the vector-valued function $g(t)$ is given by $r(t)=\langle\sqrt{t^2+1}, \sin(t^2)\rangle$. If $g(7)=\langle\sqrt{2},\pi\rangle$, find $g(0)$.

I guess I'm a bit confused about the integration that would be involved with this problem. I understand that I'm going to have to integrate both given functions, which I would then plug $0$ in to, but I'm a bit confused about what this integration would look like. Some clarification would be appreciated!

Edit: I understand what the integral formula would look like here, it's the actual integration that's confusing to me, I'm not coming up with a normal integral when I integrate.


问:
Positive functionals on the disc algebra
I don't know how to work out 10(d) in Rudin's functional analysis -

Let $U$ be the open unit radius disc in the complex plane and let $A := \{ f \in C\left(\bar{U}\right) : f|_U \text{ is holomorphic} \}$ be equipped with the sup norm as well as the involution $f^*(z):= \overline{f(\bar{z})}$ (Not a $C^*$ algebra).

Given a positive Borel measure $\mu$ on $[-1,1]$, the functional $f\mapsto \int fd\mu$ is clearly positive.

Are there any other positive functionals on $A$?

Attempt: (You should probably ignore unless you have a neurosis about questions lacking effort) My guess would be to connect such positive functionals to functionals on the space $C[-1,1]$. The correspondingly defined involution here does turn the space into a $C^*$ algebra whence the positive functionals are determined by a Bochner-type theorem.

There is the restriction map $r:A \to C[-1,1]$ which is continuous, injective, has dense image (polynomials) and relates both involutions.
Let $\phi$ be a positive linear functional on $A$.
Given a $g\in C[-1,1]$, I'm tempted to take a sequence $f_n \in A$ with $r(f_n) \to g$ and use it to define a functional.
However the $f_n$ need not be Cauchy. E.g. if $g$ is a bump function, any such sequence of holomorphic functions $f_n$ cannot have a convergent sub-sequence? Have I messed up here? Is this approach useless?

Please help me.

答:
Given a positive linear functional $\varphi $ on $A$,  let $(\pi , H, \xi )$ be the associated GNS representation,  which means that
$\pi $ is a *-representation of $A$ on the Hilbert space $H$,  and $\xi $ is a cyclic vector for $\pi $, such that
$$
  \varphi (f) = \langle \pi (f)\xi , \xi \rangle , \quad \forall f \in A.
  $$

Setting $f_1(z)=z$, observe that $f_1^*(z) = \overline {f_1(\bar z)}= z = f_1(z)$, which means that $f_1$ is self-adjoint.  It
follows that the operator
$
  T:= \pi (f_1)
  $
is self-adjoint, and clearly
$$
  \|T\| = \|\pi (f_1)\| \leq  \|f_1\| = 1.
  $$

We then have that $\sigma (T)\subseteq [-1,1]$, while
the functional calculus for $T$ is a *-homomorphism
$$
  \gamma :C(\sigma (T)) \to  B(H),
  $$
sending the function $f_1$ (more precisely its restriction to $\sigma (T)$) to $T$.

The functional
$$
  \psi : g\in   C(\sigma (T)) \mapsto  \langle \gamma (f)\xi , \xi \rangle \in  \mathbb C
  $$
is clearly positive, so by Riesz there exists a positive Borel measure $\mu $ on $\sigma (T)$ such that
$$
  \psi (g) = \int_{\sigma (T)}g(t)\, d\mu (t), \quad \forall g \in C(\sigma (T)).
  $$

On the other hand,
if we let $\rho$ be the *-homomorphism defined by
$$
  \rho :f\in A\mapsto f|_{\sigma (T)}\in  C(\sigma (T)),
  $$
we get  that $\pi (f_1) = T = \gamma \big (\rho (f_1)\big )$.  Observing that $A$ is generated by $f_1$ as  a  unital Banach algebra,
we conclude that $\pi = \gamma \circ \rho $, whence for all $f$ in $A$ we have
$$
  \varphi (f) = \langle \pi (f)\xi , \xi \rangle  = \langle \gamma \rho (f)\xi , \xi \rangle  = \psi (\rho (f)) = \int_{\sigma (T)}f(t)\, d\mu (t).
  $$

Recalling that $\sigma (T)\subseteq [-1, 1]$,  we may view $\mu $ as a measure on $[-1, 1]$ (whose support is contained in $\sigma (T)$), in
which case we may write the above as
$$
  \varphi (f) = \int_{-1}^1f(t)\, d\mu (t).
  $$


问:
Dimensions of cycles and boundaries in a full simplex
$\newcommand\rk{\operatorname{rk}}$Let $\Delta_n$ denote the full $n$-simplex $\{n,\dotsc,0\}$.  It is clear that there are $\binom{n+1}{d+1}$ many $d$-simplices, since a $d$-simplex corresponds to a subset of $\Delta_n$ of $d+1$ elements.  This means that the cochains $C_d(\Delta_n)$ are free of rank $\binom{n+1}{d+1}$.

I would like to know the ranks of the cycles $Z_d(\Delta_n)$ and the boundaries $B_d(\Delta_n)$.  Certainly we have $\rk C_d(\Delta_n) = \rk B_{d-1}(\Delta_n) + \rk Z_d(\Delta_n)$, and since $\Delta_n$ is contractible, we get $\rk Z_{d}(\Delta_n) = \rk B_d(\Delta_n)$.

I have read that $\rk Z_d(\Delta_n) = \binom{n}{d+1}$; the above formula then reflects the recursion formula for binomial coefficients.  However, I struggle to see how the rank of $\rk Z_d(\Delta_n)$ is obtained.

答:
As mentioned in the comments, the correct equation is $\text{rk } Z_d(\Delta^n)=\binom{n}{d+1}$.

Well, using the two equations you gave, you can prove that $\text{rk } Z_d(\Delta^n)=\binom{n}{d+1}$ by induction on $d$. For simplicity, let $z_d=\text{rk }Z_d(\Delta^n)$, and $b_d=\text{rk } B_d(\Delta^n)$.
\begin{align}
z_d
&=\binom{n+1}{d+1}-b_{d-1}\tag{rank-nullity}\\
&=\binom{n+1}{d+1}-z_{d-1}\tag{contractible}\\
&=\binom{n+1}{d+1}-\binom{n}{d}\tag{induction hypothesis}\\
&=\binom{n}{d+1}
\end{align}


问:
Perpendiculars on a line segment
Two points A and B are given. Find the set of feet of the perpendiculars dropped from the point A onto all possivle straight lines passing through the point B.</p>

答:
It seems a perpendicular from $A$ to a line through $B$ can be any length between $0$ and $|\overline{AB}|$, that is, if $L$ is the set of all possible lengths we have

$$L=\{l\mid0\le l\le|\overline{AB}|\}$$

The minimum occurs when the line is through $A$, and the maximum occurs when the line is perpendicular to the segment $\overline{AB}$.  All values in between occur as the line rotates about $B$.

Title: Perpendiculars on a line segment


问:
Conclude that $A^TA=I$, so $A^{-1}=A^T$.
A matrix $A$ is said to be orthogonal if for every vector $\textbf{x}$ we have

$||A\textbf{x}||_2=||\textbf{x}||_2$.

Show that such a matrix will necessarily have the property that

$\left=\left<\textbf{x},\textbf{y}\right>$.

From here, conclude that $A^TA=I$, so $A^{-1}=A^T$.

$\textbf{Hint:}$ Make use of the identity $\left<\textbf{x},\textbf{y}\right>=\frac{1}{4}||\textbf{x}+\textbf{y}||^2_2 - \frac{1}{4}||\textbf{x}-\textbf{y}||^2_2$.

I am not sure how to even begin this question. If someone could at the least provide definitions of the notation. How do you read the first line? Is it read, "The norm of the matrix A with vector x in $\mathbb{R}^2$"? Then what does the hint say? Why is there a 2 as a superscript and subscript?

答:
The 2 in the superscript means squared. The 2 in the subscript means the L-2 norm, aka $\Vert x \Vert_2=\sqrt {\sum_{i=1}^n x_i^2}$. Usually the subscript is dropped when you know it’s L-2 because it’s most common. Other norms include 1 norm, infinity norm, etc. Notice that after squaring the norm there’s no overarching square root.

So
$$
\begin{split}
(Ax)^T(Ay)
 &= 1/4(Ax+Ay)^T(Ax+Ay)-1/4(Ax-Ay)^T(Ax-Ay)\\
 &= 1/4(A(x+y))^T(A(x+y))-1/4(A(x-y))^T(A(x-y)) \\
 &= 1/4(x+y)^T(x+y)-1/4(x-y)^T(x-y) \\
 &= x^Ty
\end{split}
$$

Using the notation you’re given,
$$
\begin{split}
\langle Ax, Ay\rangle
 &= 1/4\Vert Ax+Ay\Vert ^2-1/4\Vert Ax-Ay \Vert^2 \\
 &= 1/4\Vert A(x+y)\Vert^2 -1/4\Vert A(x-y) \Vert^2 \\
 &= 1/4\Vert x+y\Vert^2 -1/4\Vert x-y\Vert^2 \\
 &= \langle x, y\rangle \ldots
\end{split}
$$

For the second part, You showed already that $(Ax)^T(Ay)=x^Ty$. But $(Ax)^T(Ay)=x^TA^TAy$. Looking at both these, It’s sandwiched in between, so it must be that $A^TA=I$.

Because $AA^{-1}=I$, A^T is the inverse. Learn the moves involving matrix distribution, inner products, norms squared, and transposing a matrix-vector product and you’re good to go.


问:
What's wrong with this part of ratio test

Prove that if $\sum_{n=1}^\infty a_n$ is a series and $\lim_{n\to\infty} \frac{|a_{n+1}|}{|a_n|}>1$, then $\sum_{n=1}^\infty a_n$ diverges.

Attempt: Suppose  $r= \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|},r>1$. Then  there exists $N \in\mathbb{N}$ such that $n \geq N \implies |\frac{|a_{n+1}|}{|a_n|} -r|<r-1=\epsilon$. Thus, $n \geq N \implies \frac{|a_{n+1}|}{|a_n|}>1$. Thus $n \geq N \implies |a_{n+1}|>|a_n|$. So that $\sum a_n$ diverges by the $n'th$ term test.

Comment:According to my professor, there are minor issues with this proof?Can anyone explain where the fault is?

答: You proved correctly that $|a_{n+1}|>|a_n|$ (if $n\gg1$). In order to apply the $n$th term test, you should now say that a strictly increasing sequence of non-negative numbers cannot have limit equal to $0$ (and perhaps that you could justify why). So, you don't have $\lim_{n\to\infty}|a_n|=0$, and this is the same thing as asserting that you don't have $\lim_{n\to\infty}a_n=0$. And now the $n$th term test says that the series diverges.


问:
Show that in a group $G$ there is exactly one element $g$ for which $g^2 = g$?

Show that in a group $G$ there is exactly one element $g$ for which $g^2 = g$?

What are they essentially asking here? This seems to be a bit general question since its considering all sets that form groups? For the group $(\Bbb{R} \setminus \{0\}, \cdot)$ this would correspond to $1^2=1=e$? Any elaboration/hints on how to show this would be welcome...

答: $$g^2=g \Longrightarrow g^{-1}(g^2)=g^{-1}g \Longrightarrow g=e$$


问:
One formula for function
Let $f(x)$ is a function such that $f(x)=x$ if $x\ge0$ and $f(x)=-x$ if $x<0$ then we can write $f(x)=|x|$.

How about $g(x)$ such that $g(x)=x^3$ if $x\ge0$ and $g(x)=x^2$ if $x<0$? Can we write $g(x)$ use one formula?

答:
Yes we can! Use $g(x)=x^{2.5+\frac{sgn(x)}{2}}$

(where $sgn(x)$ is the sign function)

Title: One formula for function


问:
Linear equation AYB + Y = C
I have the following system of equations:

$AYB + Y = C$ with $Y\in \mathbb{R}^{m\times n}$, $A\in \mathbb{R}^{m\times m}, B\in\mathbb{R}^{n\times n}, C\in \mathbb{R}^{m\times n}$.

I know that $A$ and $B$ are symmetric and semi-definited positive matrix. I want to obtain $Y$. Do you know a way about do it? Thanks in advance!

答:
With vectorization, we have
$$
AYB + Y = C \implies\\
\operatorname{vec}(AYB + Y) = \operatorname{vec}(C) \implies\\
(B^T \otimes A + I)\operatorname{vec}(Y) = \operatorname{vec}(C) \implies\\
\operatorname{vec}(Y) = (B^T \otimes A + I)^{-1} \operatorname{vec}(C).
$$
In the above, $\otimes$ denotes the Kronecker product. Because $A$ and $B$ are positive semidefinite, the matrix $B^T \otimes A$ is positive semidefinite which means that $(B^T \otimes A + I)$ is necessarily invertible.
The matrix $Y$ can then be attained by "unvectorizing", i.e. stacking the entries of the vector into an $m \times n$ matrix.

Title: Linear equation AYB + Y = C


问:
Absolute Convergence in Probability
$X_{1},X_{2},...$ are random variables, with $\sum_{n=1}^{\infty}\mathbb{E}[\left | X_{n} \right |]<\infty$. My goal is to prove that $\sum_{n=1}^{\infty}X_{n}$ converges absolutely with probability 1.

So, I have to prove that

$$\mathbb{P}(\left \{ \omega \in \Omega:lim_{n\rightarrow \infty}\sum_{j=1}^{n}\left | X_{j} \right |=L, \ L<\infty \right \})=1$$

First, I define  $B_{n,k}=\left \{ \omega \in \Omega:\left | \sum_{j=1}^{n}\left | X_{j} \right |-L\leq \frac{1}{k} \right | \right \}$

And then I have to prove that the following has probability 1.

$$\mathbb{P}(\cap_{k=1}^{\infty}\cup_{m=1}^{\infty}\cap_{n=m}^{\infty}B_{n,k})=1$$

From decreasing monotonicity property we have that

$$=lim_{k\rightarrow \infty}\mathbb{P}(\cup_{m=1}^{\infty}\cap_{n=m}^{\infty}B_{n,k})$$

And then it seems that I can use the Borel-Cantelli Lemma , but how do I show that $\sum_{n=1}^{\infty}\mathbb{P}(B_{n,k})<\infty$, i.e. $\sum_{n=1}^{\infty}\mathbb{P}(\left \{ \omega \in \Omega:\left | \sum_{j=1}^{n}\left | X_{j} \right |-L \right |\leq \frac{1}{k} \right \})<\infty$.

I could use Chebyshev inequality but $L$ is not the mean value of $\sum_{j=1}^{n}\left | X_{j} \right |$.

答:
By Fubini's theorem,
$$
\mathbb E\left[\sum_{n=1}^{+\infty}\vert X_n\vert\right]=\sum_{n=1}^{+\infty}\mathbb E[\vert X_n\vert]<+\infty.
$$

Hence the random variable $\sum_{n=1}^{+\infty}\vert X_n\vert$ is integrable and therefore almost surely finite, which means that with probability $1$, $\sum_{n\ge1}\vert X_n\vert$ is absolutely convergent.

Title: Absolute Convergence in Probability


问:
The number of vertices that the traversing algorithm visited with at least d sons
Given a directed tree $T$ with $n$ leaves, each vertex has at least $2$ sons.  Let's define $d(v)$ as the number of outgoing edges of vertex $v$ or the number of its sons.

A traversing algorithm is defined as follows:
Starting from the tree root and for each step, the next vertex is chosen uniformly from the current vertex's sons, with probability $\frac{1}{d(v)}$.

Let $Z_d$ be the number of vertices that the algorithm visited with at least $d$ sons.

We would like to prove that:
$$E(Z_d) \leq \frac{\log(n)}{\log(d)}$$

I notice that $n + 1 \leq |v| \leq 2n + 1$. The single root case is the lower bound, and if the tree is a chain with leaves is the upper bound.

Therefore, if we define $K$ as the number of vertices with at least $d$ sons we can say that
$$K \cdot d \leq |V| - 1 \leq 2n + 1 - 1 = 2n$$
$$K \leq \frac{2n}{d}$$

答:
Let the root have $s$ sons and proceed by induction on the height of the tree.

BASE CASE

Suppose the tree has height $1$. If $s<d$ then $E_d=0$ and there is nothing to prove. If $s\ge d$ then $n=s$ and $E_d=1 \le \frac{\log n}{\log d}. $

THE INDUCTIVE STEP

Suppose  $s\ge d$ sons and let the $i$th son be the root for a tree with $n_i$ leaves. If the $i$th son is chosen initially, then the expected number of vertices visited which have at least $d$ sons is less than or equal to $1+\frac{\log(n_i)}{\log(d)}$.

By the AM-GM inequality, $$\log{(n_1n_2...n_s)}^{1/s}\le \log \big (\frac{{n}}{s}\big ). $$

The sons are each chosen with probability $\frac{1}{s}$ and so the required expected value is at most $$ 1+ \frac{1}{s}\sum  \frac{\log{n_i}}{\log d}=\frac{\log d+\log{(n_1n_2...n_s)^{1/s}}}{\log d} \le \frac{\log \big (\frac{{nd}}{s}\big )}{\log d} \le \frac{\log n}{\log d}. $$

If $d<s$ then the above sum is reduced and so the inequality is still true.


问:
Topological entropy of the geodesic flow for manifold with non-negative constant sectional curvature and norms of Jacobi fields
I am trying to see why it's true that if we have a compact Riemannian manifold with non-negative constant sectional curvature $K$ then it's geodesic flow $\phi_t:TM\rightarrow TM$ has topology entropy $0$, i.e, $h_{top}(\phi)=0$.

Now to do this I am trying to see why we have that it's Lyapunov exponents are identically zero and the using Ruelle's Inequality I will get the desired result. So basically we need to compute $\limsup_{n\rightarrow \pm \infty}\frac{1}{n}\log \langle \langle d_{\theta}\phi_n(\xi),d_{\theta}\phi_n(\xi)\rangle\rangle= \limsup_{n\rightarrow \pm \infty}\frac{1}{n}\log (\langle J_{\xi}(n), J_\xi(n)\rangle +\langle \dot J_{\xi}(n),\dot J_{\xi}(n)\rangle$ and check that this is zero , where $J_{\xi}(t)$ is a jacobi field.

Now my initial idea for this would be to decompose the jacobi field into the tangential and normal jacobi fields and do the computations for this fields. Now for manifold of constant non-negative sectional curvature we know that if $J$ is a jacobi field with $J(0)=0$ then $J(t)=as_c(t)E(t)$ where $s_c(t)=R\sin(\frac{t}{R})$ if $K>0$ and $s_c(t)=t$ if $K=0$ and $E$ is any parallel unit normal vector field along $\gamma$. Now for these it's easy to see that the previous limit will be going to zero. But how can I deal with the case that $J(0)\neq 0$? I know that we have general expressions for the jacobi field but I don't know if there is anything that can help me here .

Basically my question is what happens to the norm of a normal  jacobi field $J$ such that $J(0)\neq 0$, is it possible to know anything about this ?

Any help regarding this is appreciated.

答:
First of all, I am sorry I could not find a complete reference for this fact: I think with some effort one can find something similar in Petersen, Riemannian geometry, Lee, Riemannian geometry, an introduction to curvature or Gallot, Hulin Lafontaine, Riemannian Geometry. But I could recover the result that follows ; there may be some typo. This is quite long but really is elementary.

Suppose $J$ is a Jacobi field along a geodesic $\gamma : I \to (M,g)$. We fixe some notations: we denote by $\sec$ the sectionnal curvature of $(M,g)$ and by $R_{\gamma}$ the tensor along $\gamma$ defined by $R_{\gamma}X = R(\gamma',X)\gamma'$. It is equivalent to say $g\left(R_{\gamma}X,X \right) = {\|X\|_g}^2\sec(\gamma',X)$. Recall the Jacobi equation is
$$
J'' +R_\gamma J = 0
$$
The function $t \in I \mapsto \|J(t)\|^2$ is smooth, and we have
\begin{align}
\dfrac{\mathrm{d}^2}{\mathrm{d}t^2}\left\|J\right\|^2 &= \dfrac{\mathrm{d}}{\mathrm{d}t} 2g\left(J',J\right)\\
&= 2g\left(J'',J \right) + 2g\left(J',J'\right) \\
&=-2g\left(R_\gamma J, J\right) + 2\|J'\|^2
\end{align}

Moreover, the left hand side can be computed another way for $t$ such that $J(t) \neq 0$:
\begin{align}
\dfrac{\mathrm{d}^2}{\mathrm{d}t^2}\left\|J\right\|^2 &= \dfrac{\mathrm{d}}{\mathrm{d}t} 2\|J\|\|J\|' \\
&= 2\left( {\|J\|'}^2 + \|J\|\|J\|'' \right)
\end{align}
and this tells us that for $t$ such that $J(t) \neq 0$, then
$$
2\left( {\|J\|'}^2 + \|J\|\|J\|'' \right) = -2g\left(R_\gamma J, J\right) + 2\|J'\|^2
$$
from which we deduce
$$
\left\|J \right\|'' = -\frac{g\left(R_{\gamma}J,J \right)}{\|J\|} = -\sec(\gamma',J)\|J\|
$$
Suppose $M$ has positive sectionnal curvature bounded from below along $\gamma$, say $\sec \geqslant \kappa^2 >0$, and define on $I$ $f(t) = \|J(0)\|\cos(\kappa t) + \|J(0)\|'\frac{\sin(\kappa t)}{\kappa}$. It is solution to the second order ODE $y'' = -\kappa^2 y$.

Now appears the trick: consider $g(t) = f(t) \|J(t)\|' - f'(t)\|J(t)\|$. Then on an interval containing $0$ on which $f(t) \geqslant 0$ and $J(t) \neq 0$ we have
\begin{align}
g' &= f\left(\|J\|'' + \kappa^2 \|J\| \right) \\
&= f \left(\|J\|'' + \sec(\gamma',J)\|J\| - \sec(\gamma',J)\|J\| +\kappa^2\|J\| \right) \\
&= f \left(\kappa^2 - \sec(\gamma',J) \right)\|J\| \geqslant 0
\end{align}
Hence, $g$ is non-decreasing and $g(0) = 0$. This shows that
$$
\forall t \text{ as above},~ g(t) \geqslant g(0) = 0
$$
which turns out to be
$$
\forall t \text{ as above},~ \frac{\|J\|'}{\|J\|} \leqslant \frac{f'}{f}
$$
integrating gives
$$
\ln \frac{\|J\|}{\|J(0)\|} \leqslant \ln \frac{f}{f(0)}
$$
and as $f(0) = \|J(0)\|$, we can deduce that for all $t$ such that all the above works
$$
0 \leqslant \|J(t)\| \leqslant f(t)
$$
Now, you can deduce vanishing properties of $J$ thanks to this inequality.

Also, the exact same study in case the sectionnal curvature is bounded from above $\sec \leqslant -\kappa^2 <0$ shows that
$$
\forall t,~ \|J(t)\| \geqslant \|J(0)\|\cosh (\kappa t) + \|J(0)\|' \frac{\sinh(\kappa t)}{\kappa}
$$

Comment: if $J(0) = 0$ one can adapt this proof to fix this. Also, $\|J\|$ may not be differentiable at $0$ if $J(0)=0$, but if the curvature is of constant sign, then above calculations imply that $\|J\|$ is concave or convex, thus differentiable from the right at zero and we are done.

Another comment:  if $\kappa \geqslant 0$ without having a lower bound, we cannot say much, because the euclidean case shows that $J$ may not vanish for positive $t$.


问:
Finding the limit, multiplication by the conjugate
I need to find $$\lim_{x\to 1} \frac{2-\sqrt{3+x}}{x-1}$$

I tried and tried... friends of mine tried as well and we don't know how to get out of:

$$\lim_{x\to 1} \frac{x+1}{(x-1)(2+\sqrt{3+x})}$$

(this is what we get after multiplying by the conjugate of $2 + \sqrt{3+x}$)

How to proceed? Maybe some hints, we really tried to figure it out, it may happen to be simple (probably, actually) but I'm not able to see it. Also, I know the answer is $-\frac{1}{4}$ and when using l'Hôpital's rule I am able to get the correct answer from it.

答:
You had the right idea: the issue is in your simplification of the numerator:

$$\begin{align}
(2 - \sqrt{3 + x})(2 + \sqrt{3 + x}) & = 2^2 - \left(\sqrt{(3 + x)}\right)^2 \\ \\
& = 4 - (3 + x) \\ \\
& = 4 - 3 - x \\ \\
& = 1 - x = -(x - 1)
\end{align}$$

That gives you $$\begin{align} \lim_{x \to 1} \frac{-(x - 1)}{(x - 1)(2 + \sqrt{3 + x}} &  = \lim_{x\to 1} \frac{-1}{2 + \sqrt{3 + x}}
& = -\frac 14
\end{align}$$


问:
How to take the derivative of a matrix with respect to itself?
Could someone please explain how to take the derivative of matrix with respect to itself?

$$\frac{\partial \textbf{X}}{\partial \textbf{X}}$$

where $\textbf{X}$ is an M x N matrix

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$A matrix/matrix
gradient produces a 4th order tensor, which is easily evaluated in
index notation
$$\eqalign{
{\cal E} &= \p{X}{X} \quad\implies\quad
{\cal E}_{ijk\ell} &= \p{X_{ij}}{X_{k\ell}} &= \delta_{ik}\delta_{j\ell} \\
}$$
where the Kronecker delta symbol is defined as
$$\eqalign{
\delta_{ik} &=
\begin{cases}
{\tt1}\quad {\rm if}\;i=k \\
0\quad {\rm otherwise}
\end{cases}
}$$
In words:   If $X_{ij}$ and $X_{k\ell}$ refer to different elements then the derivative is $0$, otherwise it's $\tt1$.

This is analogous to the vector/vector derivative which produces the identity matrix
$$\eqalign{
I &= \p{x}{x} \quad\implies\quad
{I}_{ij} &= \p{x_i}{x_j} &= \delta_{ij} \\
}$$


答:
A variant: the simplification rule is valid in groups, so
$$g^2(=gg)=g(=ge)\implies g=e.$$

Body:
Show that in a group $G$ there is exactly one element $g$ for which $g^2 = g$?

What are they essentially asking here? This seems to be a bit general question since its considering all sets that form groups? For the group $(\Bbb{R} \setminus \{0\}, \cdot)$ this would correspond to $1^2=1=e$? Any elaboration/hints on how to show this would be welcome...


答:
It's asking you to show that, no matter the element $g$ in your group $G$, if $g^2=g$, then $g=e$ (since $e^2=e$). So let $g\in G$ such that $g^2=g$. Multiply on, say, the left, by $g^{-1}$. Then

$$\begin{align}
e&=g^{-1}g\\
&=g^{-1}g^2\\
&=(g^{-1}g)g\\
&=eg\\
&=g.
\end{align}$$

Title: Show that in a group $G$ there is exactly one element $g$ for which $g^2 = g$?


问:
Understanding: If two finite measures agree on a $\pi$-system then they agree on the $\sigma$-algebra generated by the $\pi$-system
In my course notes there is this theorem that says if two finite measures agree on a $\pi$-system then they agree on the whole of the $\sigma$-algebra generated by the $\pi$-system.

Theorem:

Let $\Omega$ be a set, $\mathcal{A} \subseteq 2^\Omega$ be a $\pi$-system on $\Omega$ and $\sigma(\mathcal{A}) := \mathcal{F}$ be the sigma algebra generated by $\mathcal{A}$. Let $\mu_1,\mu_2: \mathcal{F} \rightarrow \bar{\mathbb{R}}$ be two measures on $\mathcal{F}$ with $\mu_1(\Omega) = \mu_2(\Omega) < \infty$. If $\mu_1(A) = \mu_2(A)$ for all $A \in \mathcal{A}$ then $\mu_1(A) = \mu_2(A)$ for all $A \in \mathcal{F}$.

The hints given to prove it are to define a set: $$\mathcal{D} = \{A \in \mathcal{F}: \mu_1(A) = \mu_2(A)   \}$$

and then show that $\mathcal{D}$ is a $\lambda$-system containing $\mathcal{A}$ and use the $\pi-\lambda$ theorem to conclude $\mathcal{D} = \mathcal{F}$

I have managed to complete all of this but I am having trouble understanding how this proves this result. I understand $\mathcal{D}$ to be the set of all sets where the two measures agree on $\mathcal{F}$ and then by showing $\mathcal{D} = \mathcal{F}$ we are showing that the sets where the measures agree are all the sets in $\mathcal{F}$. How does this make a conclusion related to the original $\pi$-system.

答: You want to prove that if $\mu_1(A)=\mu_2(A)$ for all $A\in\mathcal A$, then $\mu_1(A)=\mu_2(A)$ for all $A\in\mathcal F$. You can rephrase it as: show that if $\mathcal A\subset\mathcal D$, then $\mathcal F=\mathcal D$. The latter is true by the $\pi$-$\lambda$ theorem.


问:
weak star convergent net is uniformly convergent on weakly compact sets?
Let $X$ be a Banach space and $X^\ast$ its (topological)dual space.

Let $x_\alpha^\ast,x^\ast\in B_{X^\ast}$, where $B_{X^\ast}$ denotes the closed unit ball of $X^\ast$, such that $x_\alpha^\ast\to x^\ast$ in the weak star topology, and $A\subset X$ is weakly compact. Does $x_\alpha^\ast$ converge to $x^\ast$ uniformly on $A$?

Under what condition(s) could this statement be true?

答: The answer is no in general.  If $X$ is reflexive, then $B_X$ is weakly compact and clearly $x_\alpha^*\to x^\star$ uniformly on $A=B_X$ iff the convergence is in norm.


问:
Why is $U$ open if and only if $j^{-1}(U \cap A)= j^{-1}(U)$?
Suppose $(X, \tau)$ is a topological space and $A \subseteq X$. Suppose $A$ has the initial topology on $A$ with respect to the inclusion function $j: A \to X$.

Now suppose $U$ is open $X$ $\iff$ $j^{-1}(U)$ is open in   $A\iff j^{-1}(U \cap A)= j^{-1}(U)$ is open in $A$

Why we have the second equivalence? The set $U=U \cap A$ are not equal.. Has someone a solution to this question?

Body: $j^{-1}(U\cap A)=j^{-1}(U)\cap j^{-1}(A)=j^{-1}(U)\cap A=j^{-1}(U)$.


问:
thrown n dice and using IRV sum of face that do not appear on top
Consider n dice are thrown simultaneously. Use indicator random variables to compute
the expected value of the sum of the faces that do not appear on top. E.g., if 4 appears on top
then the sum is 1+2+3+5+6=17, for 1 die. If there are 6 dice then what probability? and also with n dice</p>

答:
Reading this as we roll all $n$ of our dice, we look at the collection of faces, we add the values of all numbers from $\{1,2,\dots,6\}$ which did not appear as a face for any of the dice.  So, for example, having rolled the eight dice 1 1 3 3 3 4 6 6 we are missing faces 2 and 5 so the sum would have been $7$.  Meanwhile having rolled just two dice and gotten 5 5 we are missing every value except $5$ and so our sum would have been $16$, etc...

Now... our random variable can be calculated in the following way:  Begin with a running total of $0$.  If a $1$ is not present as the upper face on any of the $n$ thrown dice, then add $1$ to our total.  Similarly, if a $2$ is not present as the upper face on any of the $n$ thrown dice, then add $2$ to our total.  Continue in this fashion for $3$, and on, if a value is not present as the upper face on any die add that value to our running total.

Recognize that these "If this then add that" statements can be written with indicator random variables and our random variable can be written as a linear combination of those indicator random variables.

Letting $X$ be our random variable and $X_i$ be the indicator random variable $X_i=\begin{cases}1&\text{if }i\text{ was not among the die results shown}\\0&\text{otherwise}\end{cases}$, we have:

 $$X = X_1+2X_2+3X_3+\dots+6X_6$$

Now... we can look at $\Pr(X_1=1)$ and $\Pr(X_2=1)$ etc...

First, notice that $\Pr(X_1=1)=\Pr(X_2=1)=\dots=\Pr(X_6=1)$.  Next, recall what $\Pr(X_1=1)$ means.  It means that in our $n$ throws of the dice, no $1$'s occurred.  That means that in all $n$ throws of the dice, one of the five results different than $1$ occurred.

 $$\Pr(X_i=1)=\dfrac{5^n}{6^n}$$

Finally, remember that for indicator random variables we have that $E[X_i]=\Pr(X_i=1)$ and remember that expected value is linear so $E[\alpha A+\beta B] = \alpha E[A]+\beta E[B]$ for any scalars $\alpha,\beta$ and random variables $A,B$.  Complete the necessary algebra and arithmetic and reach your final conclusion.


问:
Solve $\sqrt{3}\tan\theta=2\sin\theta$
I am trying to solve $\sqrt{3}\tan\theta=2\sin\theta$ on the interval $[-\pi,\pi]$.

$$\sqrt{3}\tan\theta=2\sin\theta \Rightarrow \sqrt{3}=\frac{2\sin\theta}{\tan\theta}$$

$$\Rightarrow \sqrt{3}=2\sin\theta \cdot \frac{\cos\theta}{\sin\theta} \Rightarrow 3 = 4\cos^2\theta$$

I get $\displaystyle \cos \theta = \pm{\frac{\sqrt{3}}{2}}$; the cosine of $30^{\circ}$ and $150^{\circ}$ so arrived at the solutions $\displaystyle -\frac{5}{6}\pi,-\frac{1}{6}\pi,\frac{1}{6}\pi,\frac{5}{6}\pi$.

Looking in the back of the book (and checking with Wolfram), the answer is $\displaystyle -\pi,-\frac{1}{6}\pi,0,\frac{1}{6}\pi,\pi$.

Where am I going wrong please?

答:
When you cancelled the $\sin \theta$ factors, you lost some solutions. This is similar to starting with $x^2 = x$, cancelling an $x$ and concluding that $x=1$; the obvious, $x=0$ solution has been lost. The remedy for this is to factor instead. In the example I provided, that would amount to writing $x^2 - x = 0$, then factoring to get $x(x-1)=0$ then setting each factor equal to zero, as usual.

You also gained solutions when you squared. This is similar to starting with $x=1$, then squaring to get $x^2 = 1$, so $x= \pm 1$. Notice that in your problem, squaring was superfluous since a few steps later you took a square root that "undid" that squaring anyway.

Title: Solve $\sqrt{3}\tan\theta=2\sin\theta$


问:
a case where contraction of a principal ideal is principal
Let $K$ be a field and $R_1,\cdots,R_n$ DVRs of $K$ with $m_i$ the maximal ideal of $R_i$. Define $A=\cap R_i$. Then $A$ is semilocal with maximal ideals $p_i=m_i \cap A$. Also, $A_{p_i} = R_i$.

Question: How can we see that $p_i$ is principal?

My effort: certainly $m_i$ is principal and in view of the isomorphism $A_{p_i} \cong R_i$ we have $p_i A_{p_i} = m_i = x_i A_{p_i}$ and we can take $x_i$ to be in $A$. Then $x_i$ is inside $p_i$. Now for every $y \in p_i$ we have $y s = a x_i$ with $s$ not inside $p_i$. That's as far as i got.

Reference: Theorem 12.2 in Matsumura.
Note: Matsumura's argument is unclear to me.

答:
Note: This is edited to correct a blunder pointed out by YACP in comments below.

Let $J := p_1 \cap \cdots \cap p_n$; this is the Jacobson radical of $A$.
Note that the inclusion $J p_1 \subset p_1^2 \cap p_2 \cap \cdots \cap p_n$
is an equality (check after localizing at each $p_i$).  Thus the Chinese Remainder Theorem shows that $p_1/J p_1 \cong p_1 /p_1^2 \times A/p_2 \times \ldots \times A/p_n,$ which is $\cong m_1/m_1^2 \times A/p_2 \times \ldots \times A/p_n$,
and hence is a cyclic module. (If $C_i$ is a collection of cyclic modules over the rings $A_i$, then $\prod C_i$ is cyclic over $\prod A_i$ (a cyclic generator
is given by taking a product of cyclic generators).  By Nakayama, we see that $p_1$ itself is cyclic,
i.e. a principal ideal.  The same argument applies to each $p_i$.

Some more details added at OP's request: $p_1$, $p_2$, etc. are distinct maximal
ideals, thus each pair $p_i, p_j$ ($i \neq j$) generates the unit ideal.  Then $p_1^2, p_2,\ldots, p_n$ have the same property, and so CRT gives $A/Jp_1 = A/(p_1^2 \cap p_2 \cap \cdots \cap p_n) = A/p_1^2 \times A/p_2 \times \cdots A/p_n.$

Thus the image of $p$ in  $A/ J p_1,$ which is equal to $p_1/J p_1$,
is equal to the ideal generated by $p_1$ in the product.  In each $A/p_i$ ($i > 1$) we have $p_1$ generates the unit ideal, and so we see that $p_1 / J p_1$
equals $p_1/p_1^2 \times A/p_2\times \cdots \times A/p_n$, as claimed.

A variation of this is as follows: a Dedekind domain with only finitely many prime ideals is necessarily a PID, by the same Chinese Remainder Theorem argument.  Note that this gives a proof of the infinitude of primes: if it were false,
then by general Dedekind domain theory, each ring of algebraic integers
would again only have finitely many prime ideals, and hence would be a PID.  But e.g. $\mathbb Z[\sqrt{-5}]$ is easily checked to not be a PID, and so there must actually be infinitely many primes!  (This proof is due to Larry Washington, see here.)


问:
Is sylvester's criterion can indicate a positive semi-defined matrix?
for example i want to know if the matrix:

$\begin{pmatrix}a&2&0\\ 2&a+1&0\\ 0&0&a\end{pmatrix}$ is positive semi-defined $∀ m_i\le 0$

is it possible to use sylvester's criterion and demand

$a\le 0$

$a^2+a\le 0$

$a^2\left(a+1\right)-4a\le 0$

and find $a=\frac{-1+\sqrt{17}}{2},\:a=\frac{-1-\sqrt{17}}{2}$

or criterion works only to positive defined matrix?

(i know that if a matrix is positive defined is also positive semi-defined,i'm just trying to figure out the principle)

答:
In your case, the matrix is block-diagonal. Thus, it sufficient to check whether each of the submatrices
$$
\pmatrix{a & 2\\ 2 & a + 1}, \quad a
$$
are positive semidefinite. More generally, the naive generalization of Sylvester's criterion is not sufficient to ensure that a matrix is positive semidefinite. For example, the matrix
$$
\pmatrix{1&0&0\\0&0&0\\0&0&-1}
$$
has non-negative leading principal minors but is not positive semidefinite.


问:
Why the matrices for integral operator is so different from its approximation?
When we try to invert the derivative operator, we get a matrix with infinite elements:

$\frac1{D+s}=\left(
\begin{array}{ccccccc}
 \frac{1}{s} & -\frac{1}{s^2} & \frac{2}{s^3} & -\frac{6}{s^4} & \frac{24}{s^5} & -\frac{120}{s^6} & . \\
 0 & \frac{1}{s} & -\frac{2}{s^2} & \frac{6}{s^3} & -\frac{24}{s^4} & \frac{120}{s^5} & . \\
 0 & 0 & \frac{1}{s} & -\frac{3}{s^2} & \frac{12}{s^3} & -\frac{60}{s^4} & . \\
 0 & 0 & 0 & \frac{1}{s} & -\frac{4}{s^2} & \frac{20}{s^3} & . \\
 0 & 0 & 0 & 0 & \frac{1}{s} & -\frac{5}{s^2} & . \\
 0 & 0 & 0 & 0 & 0 & \frac{1}{s} & . \\
 . & . & . & . & . & . & . \\
\end{array}
\right)$

Yet, the antiderivative has a matrix:

$\frac 1D=\left(
\begin{array}{ccccccc}
 0 & 0 & 0 & 0 & 0 & 0 & . \\
 1 & 0 & 0 & 0 & 0 & 0 & . \\
 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & . \\
 0 & 0 & \frac{1}{3} & 0 & 0 & 0 & . \\
 0 & 0 & 0 & \frac{1}{4} & 0 & 0 & . \\
 0 & 0 & 0 & 0 & \frac{1}{5} & 0 & . \\
 . & . & . & . & . & . & . \\
\end{array}
\right) $

These two matrices do not look similar and the second one is not a limit case of the first. Why it is so?

Can the second one somehow be obtained from the first one?
The $\ln (D+s) $ operator matrix also has infinite elements as $s\to0$:

$\ln (D+s)=\left(
\begin{array}{ccccccc}
 \ln s & \frac{1}{s} & -\frac{1}{s^2} & \frac{2}{s^3} & -\frac{6}{s^4} & \frac{24}{s^5} & . \\
 0 & \ln s & \frac{2}{s} & -\frac{3}{s^2} & \frac{8}{s^3} & -\frac{30}{s^4} & . \\
 0 & 0 & \ln s & \frac{3}{s} & -\frac{6}{s^2} & \frac{20}{s^3} & . \\
 0 & 0 & 0 & \ln s & \frac{4}{s} & -\frac{10}{s^2} & . \\
 0 & 0 & 0 & 0 & \ln s & \frac{5}{s} & . \\
 0 & 0 & 0 & 0 & 0 & \ln s & . \\
 . & . & . & . & . & . & . \\
\end{array}
\right)$

Notice that the elements of $\ln (D+s)$ are antiderivatives of $\frac1{D+s}$ Can we somehow oftain a matrix with finite elements from this, similar to the first example?

答:
When you say
$$
\frac{1}{D} 1 = f
$$
it seems that you want the solution of the differential equation
$$
1 = f'(x)
\tag1$$
with solution $f(x) = x + C$.  You drop the $C$ and you get the first column of your  $1/D$ matrix.  Why do you drop the $C$?

Next, when you say
$$
\frac{1}{D+s} 1 = f
$$
you want the solution of the differential equation
$$
1 = f'(x) + sf(x)
\tag2$$
with solution
$$
\frac{1}{s} + C e^{-sx}
$$
You drop the second term and get the first column of your $1/(D+s)$ matrix.

Maybe it is better so set things up where we do not just drop unwanted terms, but instead specify a way to choose one of the solutions.  For example, what if we specify $f(0)=0$?  Then the solution of $(2)$ is
$$
f_s(x) = \frac{1-e^{-sx}}{s}
\tag3$$
and the solution of $(1)$ is
$$
f_0(x) = x
\tag4$$
and we do have
$$
\lim_{s\to 0} f_s = f_0 .
$$


问:
How do I find the bifurcation points of the quadratic map?
I am exploring the quadratic map $x_{n+1} = x_n^2 + c$. If we plot the value of $x$ after many iterations on the vertical axis against the input $c$ value on the horizontal axis we get a classic bifurcation diagram. Going from right to left, the diagram starts with a single stable curve at $\frac{1}{4}$. At $-\frac{3}{4}$ it splits into two branches, and then at $-\frac{5}{4}$ it splits into four branches. This period-doubling happens again and again, over increasingly narrow intervals, until the plot dissolves into chaos a little beyond $-1.401$. The ratio of the width of one stable region to the next tends towards the Feigenbaum constant of approximately 4.669.

I am trying to find the exact values of these bifurcation points from period-4 onward, or at least a method of approximating them. Wikipedia gives approximate values of $c$ for some of the bifurcation points, to eight significant figures for periods 8 through 256 and thirteen significant figures for periods 512 and 1024, plus eleven significant figures for the limit as the period goes to infinity. It does not explain how these values are calculated, however, and I haven't been able to find that information elsewhere.

How do I calculate these bifurcation points myself? My math education extends through Calc 1; if this requires math beyond that, I'd appreciate as much detail as you can provide so I can follow along.

I know that the bifurcations happen for values of $c$ such that there exist points where $f^{2n}(x) - x$ and the first derivative of that expression are both zero. For example, $c = -0.75$ at the first bifurcation point, where $f^2(x) - x = x^4 + 2cx^2 - x + c + c^2$ and its derivative $x^3 + 4cx - 1$ both equal zero for $x = -0.5$. For the second bifurcation point at $c = -1.25$, $f^4(x) - x$ and the first derivative both equal zero for $x = \frac{-1 ± \sqrt{2}}{2}$.

I can't go far with that, though, because the function $f^{2n}(x) - x$ gets exponentially longer as $n$ increases, to the point that the software I have available refuses to work with $n ≥ 8$.

答:
The contraction factor $f_c'(x_1)f_c'(x_1)···f_c'(x_{2^n})$ along the cycle $x_1,x_2,...,x_{2^n}$ changes, from right to left, from $1$ to $-1$. There at the point of the bifurcation the cycle doubles to $x_1,x_2,...,x_{2^n},x_1,x_2,...,x_{2^n}$, and the new cycle starts again with $(-1)^2=1$. Along the old path the $2^n$ cycle continues, but is now unstable. Close-by a stable $2^{n+1}$ cycle branches off, locally looking like a pitch-fork bifurcation.

So you get two equations to solve to determine the two variables $x,c$ to get to a bifurcation point, $$F_{2^n}(x,c)=f_c^{2^n}(x)=x$$ and $$\partial_xF_{2^n}(x,c)=\prod_{k=0}^{2^n-1} f_c'(f_c^k(x))=-1$$

This behavior can be explored by plotting the function $x\mapsto f_c^{n}(x)-x$ for several values of $c$ close to one line of fixed points. From the graph the 4-cycle around $c=-1.3$ has a fixed point close to $0$.

One can see that as the slope at the root in the 4-cycle map becomes steeper in falling direction, the graph in the 8-cycle map develops a cubic looking shape, which is typical for this type of bifurcation.

It now becomes helpful to compute the Taylor polynomial of at least 3rd degree $f_c^N(x_0+h)=a_0+a_1h+a_2h^2+a_3h^3+O(h^4)$ to represent this pattern in an accessible form.

It can be used to find better root approximation by using the roots of the Taylor polynomial for $f_c^N(x)-x$, that is $a_0-x_0+(a_1-1)h+a_2h^2+a_3h^3$, as increments.

If $x_0$ is actually a cycle point, $a_0=x_0$ and $a_1$ determines the stability, the cycle is stable if $|a_1|<1$. In constructing a map from $c$ to $a_1$, one can determine the points where $a_1=-1$, the bifurcation point for the doubling to a $(2N)$-cycle, with standard root-seeking methods.

For the above plots the coefficients at the fixed points of the $4$-cycle are
$$\small\begin{array}{c|l|l}
c&\text{4-cycle coefficients}&\text{8-cycle coefficients}\\
\hline
-1.300&[ 0.01943,  0.18054,  4.62595, -1.02845]&[ 0.01943,  0.03260,  0.98597,  7.53528]\\
-1.325&[-0.02352, -0.24406,  5.15920,  1.20675]&[-0.02352,  0.05957, -0.95184, -13.30452]\\
-1.350&[-0.05994, -0.67882,  5.48437,  3.01329]&[-0.05994,  0.46079, -1.19572, -43.82344]\\
-1.375&[-0.09219, -1.12390,  5.69108,  4.59867]&[-0.09219,  1.26316,  0.79252, -84.49985]\\
-1.400&[-0.12145, -1.57948,  5.82748,  6.09527]&[-0.12145,  2.49476,  5.33379, -140.92247]\\
\end{array}$$

Simply applying the root-finding idea works for the initial doubling cycles up to the $32\to64$ transition. After that the remaining space becomes so narrow and the iteration map too oscillating that strange defects in my implementation lead to paradox results, for instance graphing the leading coefficient function shows graphs that do not take negative values.

$$\small\begin{array}{c|l|l}
N & \text{bifurcation parameter $c$ for $N\to2N$ cycles}\\
\hline
1 & -0.750000000000 \\
2 & -1.250000000000 \\
4 & -1.368098939391 \\
8 & -1.394046156601 \\
16 & -1.399631238874 \\
32 & -1.400828742371 \\
\end{array}$$


问:
Perpendiculars on a line segment
Two points A and B are given. Find the set of feet of the perpendiculars dropped from the point A onto all possivle straight lines passing through the point B.</p>

答:
Let $O$ be the midpoint of $AB$ and let $AO=R$ . If $l$ is any line passing through $B$ and $C$ is the leg of the perpendicular of $A$ on $l$ then the triangle $ABC$ is right at $C$ and hence $CO=AO=BO$.

Thus $C$ is on the circle with center at $O$ and radius $R$.
This shows that the locus is a subset of this circle.

Moreover, if $C'$ is any point on this circle excepting $A$ and $B$, since $AB$ is a diameter then $<C'=90^0$ which shows that $C$ is a point on the locus.

If $C'=A$ then this is the case $l=AB$ and if $C'=B$ then this is the case $l \perp AB$.

This shows that the locus is the circle with diameter $AB$.

Title: Perpendiculars on a line segment


问:
Partial derivatives in a function with "linked" variables
Let $x,y,z$ be three variables, $\phi=\phi(x,y,z)$ be a function of these variables and $c_1,c_2$ be two constants. If we are given that

$$x=c_1\left(\frac{y}{z}\right)^{1/3} \tag{1}$$
$$y=c_1^{-3}x^3z \tag{1'}$$
$$\phi(x,y,z)=c_2zx^4 \tag{2}$$

and we want to find $\frac{\partial\phi(x,y,z)}{\partial z}$, then we get that

$$\frac{\partial\phi(x,y,z)}{\partial z}=\frac{\partial (c_2zx^4)}{\partial z}=c_2x^4+c_2z\frac{\partial (x^4)}{\partial z} \tag{3}$$

But, to find $\frac{\partial (x^4)}{\partial z}$, we would need to use $(1)$ to write the dependence of $x$ on $z$. However, by doing that, we would need to find $\frac{\partial (y^{3/4})}{\partial z}$, which would depend on $z$ by $(1')$, so we would have the same problem ad infinitum.

How are partial derivatives calculated in case the variables are linked, as in this example?

答:
It should just be $$\frac{\partial\phi(x,y,z)}{\partial z} = c_2x^4.$$  We can see this from the definition of partial derivative:

\begin{align}
\frac{\partial\phi(x,y,z)}{\partial z} &= \lim_{h\to 0}\frac{\phi(x,y,z+h) - \phi(x,y,z)}{h}\\
&= \lim_{h\to 0}\frac{c_{2}(z+h)x^{4} - c_{2}zx^{4}}{h}\\
&= \lim_{h\to 0}\frac{c_{2}hx^{4}}{h}\\
&= c_{2}x^{4}.
\end{align}


问:
I am trying to figure out the critical points of a multivariable function $\ f(x,y)=(x-2)(y-2)(x+y-2)$
After multiplying it out I get: $\ f(x,y)=x^2y+xy^2-6xy-2x^2-2y^2+8x+8y-8$

Where I would take the partial derivative

with respect to x and set to zero: $\ y^2+2xy-6y-4x+8=0$,

with respect to y and set to zero: $\ x^2+2xy-6x-4y+8=0$

This is where I am stuck and unable to proceed. Any tips would be helpful.

答:
Let me try to "cheat" a little: introduce the variables
$$
u = x - 2, \; v = y - 2.
$$
Then the function becomes
$$
F(u, v) = f(x, y) = u v (u + v + 2).
$$
The partial derivatives are:
$$
F_{u} = v ( (u + v + 2) + u) = v ( 2u + v + 2), \quad
F_{v} = u ( (u + v + 2) + v) = u ( u + 2v + 2).
$$
Equating $F_{u}$ to zero gives:
$$
v = 0 \mbox{ or } v = - 2u - 2.
$$
Analogously, equating $F_{v}$ to zero gives:
$$
u = 0 \mbox{ or } u = - 2v - 2;
$$
At the outset, this seems to give four possibilities.  Three of them are:
$$
(u = 0, v = 0); \quad (u = -2, v = 0); \quad (u = 0, v = -2).
$$
The last one is the intersection of the lines
$$
u = - 2v - 2, \quad v = - 2u - 2.
$$
Having found the critical points in the $uv$-coordinates, one can convert to the $xy$-coordinates.


问:
Show that a finite domain is a division ring
Let $R$ be a finite ring. Show that the following are equivalent:

i. $R$ is a division ring.

ii. $R$ is nontrivial and if $r$,$s \in R$, with $rs=0$, then either $r=0$ or $s=0$.

$\textbf{NOTE:}$ A commutative ring that satisfies (ii) is called an integral domain or sometimes just a domain. The problem tells us that finite domains are fields.

I am not sure if I have one direction done properly. The other direction stumps me.

$\textbf{Proof:}$ (i) $\implies$ (ii) Suppose $R$ is a division ring. Then every nonzero element is invertible.
Let $r,s \in R$ where $r \neq 0$. Suppose $rs=0$. Then $r^{-1}(rs)=s \implies s=r^{-1}0=0 \implies s=0$.

$\therefore$ $rs=0 \implies r=0 \,\, \text{or} \,\, s=0$.

(ii) $\implies$ (i) Suppose $R$ is a finite ring that is nontrivial. Let $r,s \in R$ with $rs=0$ and either $r=0$ or $s=0$. [We want to show that $R$ is a division ring. So we want to show that every nonzero element is a unit. I know that a ring is a division ring if and only if the set of units denoted $U(R)=R-\{0\}$, where $R-\{0\}=R^\times$ is the multiplicative group of $R$.] How do I being to prove this direction?

答: (ii) $\Rightarrow$ (i): Let $r$ be a nonzero element of $R$.  You want to show that $r$ has a multiplicative inverse.  What can you say about the elements $r, r^2, r^3$ etc.?


答:
Since both $x^2-6x-4y$ and $y^2-6y-4x$ are equal to $-2xy-8$, you have $x^2-6x-4y=y^2-6y-4x$, or $x^2-y^2=2(x-y)$. Therefore, $x=y$ or $x+y=2$.

If $x=y$, your first equation becomes $3y^2-10y+8=0$, which has two solutions: $\frac43$ and $2$. So, this provides two critical points: $\left(\frac43,\frac43\right)$ and $(2,2)$.

And, if $x=2-y$, your first equation becomes $2y-y^2=0$, which has two solutions: $0$ and $2$. So, this provides another two solutions: $(2,0)$, and $(0,2)$.

Title: I am trying to figure out the critical points of a multivariable function $\ f(x,y)=(x-2)(y-2)(x+y-2)$

Body:
After multiplying it out I get: $\ f(x,y)=x^2y+xy^2-6xy-2x^2-2y^2+8x+8y-8$

Where I would take the partial derivative

with respect to x and set to zero: $\ y^2+2xy-6y-4x+8=0$,

with respect to y and set to zero: $\ x^2+2xy-6x-4y+8=0$

This is where I am stuck and unable to proceed. Any tips would be helpful.


问:
How many other means exist out there? How do they relate to each other?
I am aware of the Quadratic Mean, Arithmetic Mean, Geometric Mean, and Harmonic Mean, and they are related by the inequality chain $QM \ge AM \ge GM \ge HM$. For two numbers $a,b$, the quadratic mean is equal to $\sqrt{\frac{a^2+b^2}2}$, the arithmetic mean equals $\frac{a+b}2$, the geometric mean equals $\sqrt{ab}$, the harmonic mean equals $\frac{2}{\frac1a+\frac1b}$. Some means I made up: Logarithmic mean ($ln(e^a+e^b)$) and Trigonometric Mean: $arcsin(sin(a)+sin(b))$. So, are their any other such means?

答:
$M_r(a)=(\frac{1}{n}\sum a^r)^{\frac{1}{r}}$ gives a whole class of means. In the formula $a$ stands for a sequence of numbers you are taking the mean of, and $r$ is the 'order' if you like, of the mean. So $$\lim_{r\rightarrow \infty}M_r(a)=\max a$$ and $M_{-\infty}=\min a$. The less trivial connection is a connection with the geometric means$$\lim_{r\rightarrow0}M_r(a)=G(a)$$ Moreover the means can be generalized a bit to have different weights $$M_r(a)=(\sum \omega a^r)^{\frac{1}{r}}$$ where the $\omega$ are non-negative and $\sum \omega=1$.

However you define the mean, it needs to satisfy homogeneity $M(ka)=kM(a)$ and $$\min a\leq M(a)\leq\max a$$ these are like the 'axioms' for means and are similar to those of group theory, topological space etc.


问:
Application of maths in economics
What are the branches of maths where we can see undoubtful connections with economics? Where can we use mathematical methods or models and apply them to analyze economic concepts? </p>

答:
Very late answer, but I figured I give it a shot, from the perspective of an economics graduate student.

There are clear connections between econ and math but the statement needs some qualification: Economics is about solving economics problem in the economics paradigm. How much math is involved in the solution depends on the subfield and the economist in question. The mathematical sophistication across the econ community varies greatly. Some economists are mathematicians in their own right, while others (probably most) don't know much math at all. So, yes, at its most rigorous level, economics is highly mathematical but that's small subcommunity in econ.

Having given that caveat, mathematics used, when they are, in the three main econ subfields, along with some references:

Microeconomics: Consumer theory is about how an individual consumer makes his decision. The mathematical context is convex analysis. See Theory of Value by Debreu (that level of rigor, standard in math, will never been seen in an econ course.) Game theory---used in analysing conflict. See Game Theory by Fudenberg and Tirole.

Macroeconomics: Real and (basic) functional analysis. The main mathematical tool used in macro is dynamic programming, whose rigorous formulation requires analysis. See Recursive Methods in Economics by Stokey and Lucas, written at the level of Rudin.

Econometrics: This is basically statistics applied to econ. I don't know a PhD level text that would be considered rigorous by a mathematician. I'd refer to journal papers published in Econometrica.

Hope this helps.

Title: Application of maths in economics


答:
Hint:

Let $r\in R,\;r\ne 0$. Consider the endomorphism of the underlying additive group:
$\begin{aligned}[t]
m_r:R&\longrightarrow R\\
s&\longmapsto rs
\end{aligned}$

Show that $m_r$ is injective. As $R$ is finite, injective $\longleftrightarrow$ surjective $\longleftrightarrow$ bijective, and consequently, $1$ is attained.

Title: Show that a finite domain is a division ring

Body:
Let $R$ be a finite ring. Show that the following are equivalent:

i. $R$ is a division ring.

ii. $R$ is nontrivial and if $r$,$s \in R$, with $rs=0$, then either $r=0$ or $s=0$.

$\textbf{NOTE:}$ A commutative ring that satisfies (ii) is called an integral domain or sometimes just a domain. The problem tells us that finite domains are fields.

I am not sure if I have one direction done properly. The other direction stumps me.

$\textbf{Proof:}$ (i) $\implies$ (ii) Suppose $R$ is a division ring. Then every nonzero element is invertible.
Let $r,s \in R$ where $r \neq 0$. Suppose $rs=0$. Then $r^{-1}(rs)=s \implies s=r^{-1}0=0 \implies s=0$.

$\therefore$ $rs=0 \implies r=0 \,\, \text{or} \,\, s=0$.

(ii) $\implies$ (i) Suppose $R$ is a finite ring that is nontrivial. Let $r,s \in R$ with $rs=0$ and either $r=0$ or $s=0$. [We want to show that $R$ is a division ring. So we want to show that every nonzero element is a unit. I know that a ring is a division ring if and only if the set of units denoted $U(R)=R-\{0\}$, where $R-\{0\}=R^\times$ is the multiplicative group of $R$.] How do I being to prove this direction?


问:
Is it possible that individually, pieces of evidence increase the chance of guilt but together they decrease the chance of guilt?
Let G be the event that a certain individual is guilty of a certain robbery. In gathering evidence, it is learned that an event $E_1$ occurred, and a little later it is also learned that another event $E_2$ also occurred.
Is it possible that individually, these pieces of evidence increase the chance of guilt (so $P(G|E1)$ $> P(G)$ and $P(G|E2) > P(G)$), but together they decrease the chance of guilt (so $P(G|E1, E2)$ $ < P(G)$)?

This is the example in the solution:

Suppose that the crime was
committed between 1 pm and 3 pm on a certain day. Let E1 be the event
that the suspect was at a nearby coffeeshop from 1 pm to 2 pm that
day, and let E2 be the event that the suspect was at the nearby
coffeeshop from 2 pm to 3 pm that day. Then P(G|E1) > P(G), P(G|E2) >
P(G) (assuming that being in the vicinity helps show that the suspect had the opportunity to commit the crime), yet
P(G|E1, E2) < P(G) (as being in the coffeeshop from 1 pm to 3 pm gives
the suspect an alibi for the full time).

I've been trying to think of another example but I can't think of any. Can someone provide another example? Moreover, in terms of general problem-solving heuristics, I find myself terrible at coming up with examples/counter-examples. My only strategy is to look for simple/extreme cases, but there's no simple/extreme cases for this question.

Thank you very much!

答:
Assume that Person-1 only wears the shoes that he owns, and that he only wears a left shoe on his left foot and a right shoe on his right foot.

Assume that both tangerine shoes and red-polka-dot shoes are very unusual.  Assume that Person-1 owns both a left tangerine shoe and a left red-polka-dot shoe, but does not own any right shoes that are tangerine or red-polka-dot.

One witness says that at least one of the shoes worn by the perpetrator was tangerine, but the witness unsure whether shoe was on left foot or right foot.

2nd witness says the same, except with respect to red-polka-dot shoe.  So police conclude that perpetrator owns a tangerine-left and red-polka-dot-right pair of shoes, or vice-versa.

Each witness individually casts legitimate suspicion on Person-1, but collectively, they exonerate him.


问:
If $Ax=b$ has infinitely many solutions, prove there exists $c$ such that $Ax=c$ has no solutions
Where $A\in M_n(F)$, and $b,c\in F^n$.

What is the approach? Thanks.

答: As the dimension of the column spaced is less than $n$: what conclusion can you draw?


答:
Hint :

$$Ax=b \text{ has infinitely many solutions } \Longrightarrow A \text{ is not invertible } \Longrightarrow \text{rank}(A) < n$$

Title: If $Ax=b$ has infinitely many solutions, prove there exists $c$ such that $Ax=c$ has no solutions

Body:
Where $A\in M_n(F)$, and $b,c\in F^n$.

What is the approach? Thanks.


答: If $Ax =b$ has infinitely many solutions, then $A$ has rank less than $n$ and so its column space has at most dimensions $n-1$. So there is the matrix $D$ such that $DAD^{-1}$ is the identity with the zero column at the end. So you can pick $\tilde{c}$ to be $e_n$ (the $nth$ standard basis vector). Now does $DAD^{-1}x = \tilde{c}$ have any solutions?


问:
Why do people use "K" to represent a complete graph?
Why do people use the letter "K", rather than "C", to represent a complete graph? Does it come from German "komplett"?</p>

答:
My understanding is that Harary introduced the notation $K_5$ and $K_{3,3}$ for the graphs appearing in Kuratowski's theorem, and the choice of $K$ as symbol was in honour of Kuratowski.

I must admit that I do not have this first hand, but it was certainly the accepted explanation as I was "growing up" graph theoretically.

Edit: The story is in print on page 259 of Doug West's "Introduction to Graph Theory".


答:
Let $R=\{r_1=1,r_2,\dots,r_k\}$ be the distinct elements of $R.$

Assume $r\in R$ does no have an inverse. Then the set $rR=\{rr_1,rr_2,\dots,rr_k\}$ must have fewer than $k$ elements, since $1$ is not in $rR.$ So, by the pigeonhole principle, there must be some $i,j$ with $i\neq j$ Such that $rr_i=rr_j.$

But then $r(r_i-r_j)=0.$. So either $r=0$ or $r_i-r_j=0.$

But $i\neq j,$ so $r_i\neq r_j$ and thus by (ii) $r=0.$


答: Let's see $Ax=b$ has infinitely many solutions implies the columns of $A$ are dependent, so it does not give a basis for the entire vector space $F^n$. If the $n$ column vectors of $A$ does not form a basis, this means some vectors $c$ in your vector space is not expressible as $Ax=c$. The linear map $A: F^n\rightarrow F^n$ is not surjective.


答: If $Ax = b$ has infinitely many solutions, it means that the nullspace of $A$ is not the zero subspace, but contains at least one nonzero vector, hence has dimension 1 or higher.  Thus, by the Rank-Nullity Theorem, the rank of $A$ is less than $n$.  It follows that $A$ fails to be onto.  This completes the proof.


问:
What is the boundary of a plane excluding a line.
```<code>f(x,y) = y/x^2
</code>```

Here, we can see that the x = 0 is not in the function domain, so what should be the domain's boundary?

答: The boundary is the entire line $x=0$, since any ball centred on the line will always contain both points inside and outside the domain.


答:
While DMcMor's answer gives the rigorous reason why the partial derivative is $c_2x^4$, let me give you a more 'intuitive' treatment of how to compute partial derivatives quickly.

How I think about it when i have to compute something like $\frac{\partial}{\partial z}c_2x^4z$, is that I treat it as as if it was a function of $z$ only (since I'm interested in the partial derivative in $z$-direction), and treat the other variables (here $x^4$) as if they were constants. Then, computing the partial derivative here is nothing more than computing $f'(z)$, where $f(z) = a\cdot z$, which is of course just $f'(z) = a$. And this constant $a$ happens to be $a = c_2x^4$ in this case.

As I said, this is not as rigorous as using the definition, but there are situations in which you have to compute a lot of partial derivatives so it pays off to have a quick method.

Title: Partial derivatives in a function with "linked" variables

Body:
Let $x,y,z$ be three variables, $\phi=\phi(x,y,z)$ be a function of these variables and $c_1,c_2$ be two constants. If we are given that

$$x=c_1\left(\frac{y}{z}\right)^{1/3} \tag{1}$$
$$y=c_1^{-3}x^3z \tag{1'}$$
$$\phi(x,y,z)=c_2zx^4 \tag{2}$$

and we want to find $\frac{\partial\phi(x,y,z)}{\partial z}$, then we get that

$$\frac{\partial\phi(x,y,z)}{\partial z}=\frac{\partial (c_2zx^4)}{\partial z}=c_2x^4+c_2z\frac{\partial (x^4)}{\partial z} \tag{3}$$

But, to find $\frac{\partial (x^4)}{\partial z}$, we would need to use $(1)$ to write the dependence of $x$ on $z$. However, by doing that, we would need to find $\frac{\partial (y^{3/4})}{\partial z}$, which would depend on $z$ by $(1')$, so we would have the same problem ad infinitum.

How are partial derivatives calculated in case the variables are linked, as in this example?