问:
Why is the integral test only for positive decreasing functions?
The way that the integral test is taught is that

If f is a continuous, positive and decreasing function where $f(n)=a_n$ on the interval [1,∞), then. the improper integral $\int_1^\infty f(x)dx$ and the infinite series $\sum_1^\infty a_n$. either both converge or both diverge.

However I am having quite a hard time building an intuition for this, say we have the infinite sum:
$$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}$$
And we compare it to the function
$$f(x)=\frac{\ln(x)}{x^3}$$

At the upper bound:
$$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \int^n_1\frac{\ln(x)}{x^3}dx$$
Solving by parts
$$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \left[\frac{-\ln(x)}{2x^2} \right]^n_1 +\int^n_1\frac{1}{2x^3}$$
$$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \frac{-\ln(n)}{2n^2} +\left[\frac{-1}{4x^2}\right]^n_1$$
$$\sum^\infty_{n=1}\frac{\ln(n)}{n^3}< \frac{-\ln(n)}{2n^2} -\frac{1}{4n^2} +\frac{1}{4}$$

Then by using L'Hôpital's rule
$$\lim \limits_{n  \to \infty} \frac{-\ln(n)}{2n^2}=\lim \limits_{n  \to \infty}\frac{-1}{4n^2}=0$$

Hence the integral converges to $\frac{1}{4}$ and so the sum also converges.

However looking at the graph of the function and by differentiating it it is apparent that the function only retains positive and decreasing for the interval for $x>e^{\frac{1}{3}}$

Yet the integral test seems to be still working properly over the greater interval of $x>1$ which contains both the increasing and decreasing parts of the function.

By drawing rectangles of unit width touching the function at the right corners it becomes apparent that for values $x>e^{\frac{1}{3}}$ the rectangles estimate the upper bound but for values less than this interval from $e^{\frac{1}{3}}>x>1$ they instead approximate the lower bound.

Why then does this test still apparently work and would this extend from $x=0$ to $x=\infty$ as well?

答: A series $\sum_{n=1}^\infty a_n$ converges if and only if, for some $N\in\Bbb N$, the series $\sum_{n=N}^\infty a_n$ converges. In other words, the first terms of the series do not matter when we are dealing with the proeblem of deciding whether a series converges or not. So, when we use the integral test, all that matters about $f$ is that, for some $a>0$, its restriction to $[a,\infty)$ is positive, decreasing and convergent to $0$. What happens to the left of that $a$ does not matter.


问:
Change of basis matrix from $\mathbb{R}^3$ to $\mathbb{R}^2$
Let $ f: \mathbb{R}^3 \rightarrow \mathbb{R}^2 $ be a linear application such that:
$$ \mathcal{M}(f; c.b.\mathbb{R}^3; c.b.\mathbb{R}^2) = \left[
 \begin{array}{lll}
  1 & 1 & 0 \\
  0 & 1 & 1
 \end{array}
\right]
$$
and considering the following basis

$ c.b.\mathbb{R}^3 = \{ (1,0,0), (0,1,0), (0,0,1)\} $

$ c.b.\mathbb{R}^2 = \{ (1,0), (0,1)\} $

$ \mathcal{B} = \{ (0,1,0), (1,0,1), (1,0,0)\} $

$ \mathcal{B}' = \{ (1,1), (1,0)\} $

using the change of basis matrix find:

a) $\mathcal{M}(f; \mathcal{B}; c.b.\mathbb{R}^2)$

b) $\mathcal{M}(f; c.b.\mathbb{R}^3; \mathcal{B}')$

c) $\mathcal{M}(f; \mathcal{B}; \mathcal{B}')$

I've already solved item a) by compting each vector in $\mathcal{B}$ with respect to  $c.b.\mathbb{R}^3$ and then multiplied this vector (on the rhs) by the given matrix.
So the solution to a) is the following:

$ (0,1,0) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 0 \land \alpha_2 = 1 \land \alpha_3 = 0$

$ (1,0,1) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 1 \land \alpha_2 = 0 \land \alpha_3 = 1$

$ (1,0,0) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 1 \land \alpha_2 = 0 \land \alpha_3 = 0$

$ \vec{e_1} =
 \left[ \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1
\end{array} \right]
%
\left[ \begin{array}{c}
0  \\
1 \\
0
\end{array} \right] =
\left[ \begin{array}{c}
1  \\
1
\end{array} \right]
 $

$ \vec{e_2} =
 \left[ \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1
\end{array} \right]
%
\left[ \begin{array}{c}
1  \\
0 \\
1
\end{array} \right] =
\left[ \begin{array}{c}
1  \\
1
\end{array} \right]
 $

$ \vec{e_3} =
 \left[ \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1
\end{array} \right]
%
\left[ \begin{array}{c}
1  \\
0 \\
0
\end{array} \right] =
\left[ \begin{array}{c}
1  \\
0
\end{array} \right]
 $

$$ \mathcal{M}(f; \mathcal{B}; c.b.\mathbb{R}^2) = \left[
 \begin{array}{lll}
  1 & 1 & 1 \\
  1 & 1 & 0
 \end{array}
\right]
$$

and I know that the solution for b) is

$$ \mathcal{M}(f; c.b.\mathbb{R}^3; \mathcal{B}') = \left[
 \begin{array}{lll}
  0 & 1 & 1 \\
  1 & 0 & -1
 \end{array}
\right]
$$

and solution for c) is

$$ \mathcal{M}(f; c.b.\mathbb{R}^3; c.b.\mathbb{R}^2) = \left[
 \begin{array}{lll}
  1 & 1 & 0 \\
  0 & 0 & 1
 \end{array}
\right]
$$
but I don't know how to do them. Any tips?

答:
There is a systematic way to do this. For such a problem, what you need is simply following the definitions. (It is very instructive to do it at least once to see what is really going on.)

Change the notations to make writing easier.

Let $\alpha=(e_1,e_2,e_3)$ be the standard basis of $\mathbb{R}^3$ and
$\alpha'=(e_1',e_2')$ the standard basis of $\mathbb{R}^2$.

Then the linear map $f$ is such that
$$
f(e_1)=e_1,\quad f(e_2)=e_1'+e_2',\quad f(e_3)=e_2'.
$$
This corresponds to the matrix on the second line of your post.

To find the matrix in (a), what you want is finding the following coefficients $b_{ij}$ where
$$
f(\beta_1)=b_{11}e_1'+b_{12}e_2'\\
f(\beta_2)=b_{21}e_1'+b_{22}e_2'\\
f(\beta_3)=b_{31}e_1'+b_{32}e_2'
$$

Here $\beta_1,\beta_2,\beta_3$ denote the three vectors of $\mathcal{B}$.

ContentLicense: CC BY-SA 4.0


问:
Cauchy sequence of positive terms, find it's limit.
Consider the sequence of positive terms, where $x_n$ it's difined by
\begin{equation}
x_1=3\ and\ x_{n+1}=\frac{3(1+x_n)}{3+x_n}.
\end{equation}
Find   $a=\underset{n\rightarrow \infty}\lim x_n$.

All that I have found is that exists $c=0.2$ such that
\begin{equation}
|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n|.
\end{equation}

I also tried to prove that
\begin{equation}
x_{k+1} \leq x_k,
\end{equation}
but I failed.

答:
$|x_{n+2}-x_{n+1}| \leq 0.2|x_{n+1}-x_n| \Rightarrow |x_{n+k}-x_{n}|\leq c^k |x_2-x_1|=c^k$

$s=\frac{1}{|c|}-1 \Rightarrow |c|=\frac{1}{1+s}$ and $Ns \geq \epsilon$

$|c|=\frac{1}{1+s} \Rightarrow |c|^k=(\frac{1}{1+s})^k=\frac{1}{(1+s)^k} \leq $
$\frac{1}{1+sk} < \frac{1}{sk} < \epsilon $

which means that $\lim_{n \to \infty}x_n=a$ exists

$\lim_{n \to \infty}x_n=a \Rightarrow \lim_{n \to \infty}x_{n+1}=a$

$x_{n+1}=\frac{3(1+x_n)}{3+x_n} \Rightarrow \lim_{n \to \infty}x_{n+1}=\lim_{n \to \infty}\frac{3(1+x_n)}{3+x_n}=\frac{3a+3}{3+a}=a$

$\frac{3a+3}{3+a}=a \Rightarrow a= \sqrt3$


问:
Basis for Product topology on $X \times Y$
Hi i am reading Topology by Munkres and there at the product topology i have one doubt. There it is written that the second condition is almost as easy, since the intersection of any two basis elements $U_1\times V_1$ and $U_2\times V_2$ is another basis element.My problem is that how does this prove the second condition? For reference i am writing the condition they have given for a collection to be called a basis.

If $X$ is a set, a basis for a topology on $X$ is a collection $\mathbb B$ of subsets of $X$ such that :

For each $x\in X$ there is at least one basis element B containing x.

If $x$ belongs to the intersection of two basis elements $B_1$ and $B_2$ , then there is a basis element $B_3$ containing $x$ such that $B_3\subset B_1\cap B_2$ .

After this they've defined the product topology as follows-

Let $X$ and $Y$ be topological spaces. The product topology on $X\times Y$ is the topology having as basis the collection $\mathbb B$ of all sets of the form $U\times V$ where $U$ is an open subset of $X$ and $V$ is an open subset of Y.

I am also attaching the screenshot of the section where i have highlighted the line in which i have the doubt. By the way i understand why the intersection of any two basis elements here is another basis element, i just can't understand how does this prove the second condition that they have given for a collection to be a basis.

答: In the condition we just pick $B_3= B_1 \cap B_2$ for any $x$ in it when $B_i = U_i \times V_i, i=1,2$. For open balls as a base in Euclidean metric on $\Bbb R^2$ we have work to do, because the intersection of two balls need not be a ball, but contains one around each point of their intersection. The product open squares are closed under intersections (as e.g. open intervals in $\Bbb R$) and then the proof is easier.


答:
Hint:

Simple – denote $P$ the change of basis matrix for$\mathbf R^3$, $X$ a column vector in the standard basis, $X'$ its column vector in basis $\mathcal B$, and $Q$ the change of basis matrix for $\mathbf R^2$,  $Y$ a column vector in the standard basis, $Y'$ its column vector in basis $\mathcal B'$.

We know  that $\;Y=QY'=\ M X= M PX'$, whence
$$Y'=(Q^{-1}MP)X'$$
which means th matrix you're seeking for is $Q^{-1}MP$.

Title: Change of basis matrix from $\mathbb{R}^3$ to $\mathbb{R}^2$

Body:
Let $ f: \mathbb{R}^3 \rightarrow \mathbb{R}^2 $ be a linear application such that:
$$ \mathcal{M}(f; c.b.\mathbb{R}^3; c.b.\mathbb{R}^2) = \left[
 \begin{array}{lll}
  1 & 1 & 0 \\
  0 & 1 & 1
 \end{array}
\right]
$$
and considering the following basis

$ c.b.\mathbb{R}^3 = \{ (1,0,0), (0,1,0), (0,0,1)\} $

$ c.b.\mathbb{R}^2 = \{ (1,0), (0,1)\} $

$ \mathcal{B} = \{ (0,1,0), (1,0,1), (1,0,0)\} $

$ \mathcal{B}' = \{ (1,1), (1,0)\} $

using the change of basis matrix find:

a) $\mathcal{M}(f; \mathcal{B}; c.b.\mathbb{R}^2)$

b) $\mathcal{M}(f; c.b.\mathbb{R}^3; \mathcal{B}')$

c) $\mathcal{M}(f; \mathcal{B}; \mathcal{B}')$

I've already solved item a) by compting each vector in $\mathcal{B}$ with respect to  $c.b.\mathbb{R}^3$ and then multiplied this vector (on the rhs) by the given matrix.
So the solution to a) is the following:

$ (0,1,0) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 0 \land \alpha_2 = 1 \land \alpha_3 = 0$

$ (1,0,1) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 1 \land \alpha_2 = 0 \land \alpha_3 = 1$

$ (1,0,0) = \alpha_1(1,0,0) + \alpha_2 (0,1,0) + \alpha_3 (0,0,1) \Leftrightarrow \alpha_1 = 1 \land \alpha_2 = 0 \land \alpha_3 = 0$

$ \vec{e_1} =
 \left[ \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1
\end{array} \right]
%
\left[ \begin{array}{c}
0  \\
1 \\
0
\end{array} \right] =
\left[ \begin{array}{c}
1  \\
1
\end{array} \right]
 $

$ \vec{e_2} =
 \left[ \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1
\end{array} \right]
%
\left[ \begin{array}{c}
1  \\
0 \\
1
\end{array} \right] =
\left[ \begin{array}{c}
1  \\
1
\end{array} \right]
 $

$ \vec{e_3} =
 \left[ \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 1
\end{array} \right]
%
\left[ \begin{array}{c}
1  \\
0 \\
0
\end{array} \right] =
\left[ \begin{array}{c}
1  \\
0
\end{array} \right]
 $

$$ \mathcal{M}(f; \mathcal{B}; c.b.\mathbb{R}^2) = \left[
 \begin{array}{lll}
  1 & 1 & 1 \\
  1 & 1 & 0
 \end{array}
\right]
$$

and I know that the solution for b) is

$$ \mathcal{M}(f; c.b.\mathbb{R}^3; \mathcal{B}') = \left[
 \begin{array}{lll}
  0 & 1 & 1 \\
  1 & 0 & -1
 \end{array}
\right]
$$

and solution for c) is

$$ \mathcal{M}(f; c.b.\mathbb{R}^3; c.b.\mathbb{R}^2) = \left[
 \begin{array}{lll}
  1 & 1 & 0 \\
  0 & 0 & 1
 \end{array}
\right]
$$
but I don't know how to do them. Any tips?


问:
Nth roots for positive elements of C*-algebra
There is a C*-algebra $A$ with a positive element $a$ which means $a=a^*$ and $\sigma_A{(a)}\in [0,+\infty)$, prove that $\forall n\in \mathbb{N}\exists b\in\{\text{positive}\}:b^n=a$

Of course this is the extension of the case in positive-definite matrices, but here there aren't eigenvectors! Is Gelfand-Naimark theorem (applied to a suitable sub-algebra) sufficient by simply considering root of a positive function? Or I can apply the taylor expansion of $\sqrt[n]{z}$ which has a large convergence disk?

答:
The function $t\longmapsto t^{1/n}$ is well-defined and continuous on $[0,\infty)$.

So you can just use the continuous functional calculus (which amount to having a C$^*$-isomorphism between $C^*(a)$ and $C(\sigma(a))$).


问:
Does the standard log normal distribution have a heavy right tail?
I read about heavy right tail and I saw that a distribution is said to have a heavy right tail if its tail probabilities vanish slower than any exponential
$$\forall t>0,\,\lim_{x\to\infty}e^{tx}P(X>x)=+\infty$$.

I also saw that the normal distribution does not have right tail .
Im not sure about the log normal distribution.

Does the standard log normal distribution have a heavy right tail?

答:
As I answered you here a distribution of a random variable $X$ is called heavy right tailed iff the moment generating function $M_X(t)$ is infinite for all $t>0$. So you either try to calculate the integrals
$$\mathbb E e^{tX}
$$
or you look here to read that the integrals do not converge.


问:
Math arithmetic
So I got this question in my exam yesterday, and only a few people from my school could solve this.

What's so hard about it? I have learned the simple arithmetic questions, but I don't really understand how to do this one.

formulas:

In the arithmetic the tenth organ is 4 times bigger than the  third organ.

The amount of all first 50 organs is 3725.
Find A1 and D (the first and the difference of the series).

How do I solve it?

答:
You write the equations for the 10th term, and 3rd term:

$T_n = a + (n-1)d$

$T_3 = a + (3-1)d = a + 2d$

$T_{10} = a + (10-1)d = a + 9d$

You are told that $T_{10}$ is 4 times $T_3$, or that: $T_{10} = 4T_{3}$

Which means:

$a + 9d = 4(a + 2d)$

And lastly, the sum all the first 50 terms is 3725, which means:

$S_n = \dfrac{n(2a+(n-1)d)}{2}$

$S_{50} = \dfrac{50(2a+(50-1)d)}{2} = 3725$

You should be able to solve those.

ContentLicense: CC BY-SA 3.0

Title: Math arithmetic


问:
Prove a graph is cyclic
I encountered the following problem:

Let $G(V, E)$ be a finite undirected graph with $2n$ vertices, in which every connected component has an even number of vertices. Given that exactly $n$ of the vertices of $G$ have a degree at least $3$, prove that $G$ is cyclic.

My solution to the problem goes as follows:

Let $C$ denote the set of all connected components $G_i(V_i, E_i)$ of $G$ and $|C| = k$.

Let's assume that $G$ is acyclic. That means that every $G_i$ is a tree. Then, by adding exactly $k-1$ edges to $G$, we can connect all connected components in such a way (root of one to another one's leaf, ordering them in a sequence that way) that we obtain a tree $G'(V', E')$, where $V'= V$ and $|E'| = |E| + k - 1$.

But since $G'$ is a tree, that means that $$|E'| = |V'| - 1 = |V| - 1 = 2n - 1$$

By the handshake lemma we also know that $$2|E'| = \sum_{v\in V'}\deg v \ge \underbrace{3n}_{\substack{\text{atleast $n$}\\ \text{with degree $\ge$ 3}}} + \underbrace{n}_{\substack{\text{G' is connected}\\\text{$\Rightarrow$ the remaining $n$ have $\deg\ge 1$}}}$$

Thus we obtain that $$2(2n-1) \ge 4n \Leftrightarrow -1 \ge 0$$
which is a contradiction, hence $G$ can't be acyclic and contains a cycle.

Now, what bothered me about that is that I never used the given fact, that every $G_i$ had an even number of vertices, so I felt like I was missing something huge or my solution is flawed in a fundamnetal way... and I really banged my head against the wall hard enough, but was nonetheless unable to figure out what was wrong. So my question would be: am I missing something or is the mentioned statement in the problem simply redundant? And if it is redundand, how would one go about proving it? I really got stuck somewhere about there.

Many thanks in advance!

答:
Here's something simple we can deduce from the extra hypothesis: because every connected component has an even number of vertices, there can be no vertices of degree $0$. This leads to a version of your argument that skips the "connect all the components" step.

That is, suppose by way of contradiction that $G$ is a union of $k$ trees. It's known that a forest with $2n$ total vertices and $k$ components has $2n - k$ edges. Therefore by Handshake we have
$$\sum \deg(v_i) = 2|E| = 4n - 2k.$$

Now, $n$ vertices have degree at least 3, and $n$ vertices have degree 1 or 2. Therefore
$$\sum \deg(v_i) \ge 3n + n = 4n.$$

Putting this together, we get
$$4n \le 4n - 2k \implies k \le 0,$$
a contradiction.

Remark: I don't see anything wrong with your argument, so maybe the inclusion of the hypothesis is just to make life a bit easier?

Title: Prove a graph is cyclic


问:
Rank of a matrix
If a $3 \times 3$ matrix has determinant zero, then is it possible that its rank could be $3$? I think it only could be $2$ or less. I am right or wrong? Please explain.</p>

答:
Just think about this image: A matrix can be viewed as a set of $n$ vectors of $n$ dimension (imagine the case of $n=3$). Thus the determinant is actually the volumn of the polyhedron spanned by these vectors (example: $(1,0,0)^T,(0,1,0)^T$ and $(0,0,1)^T$ form a unit cube), where the sign (plus or minus) indicates the orientation (right-handed system -- plus, and left-handed system otherwise).

Now you can understand many properties of determinants: Exchanging two columns (or rows) will flip the sign of determinant -- that's because the orientation has changed. Or, if one column (or row) becomes all zeroes, the determinant will also be reduced to zero -- that's because the polyhedron shrinks to a sheet, and thus occupies no space.

As for the rank, the rank of a matrix actually tells you what is the minimum dimension of the space holding all the vectors of the matrix. Take $n=3$ as an example again: If $r(A)=3$, it means, in order to contain all the column vectors of $A$, a space of at least 3 dimension is a must. As a result, they span a nonzero volumn. If $r(A)=2$, it means the vectors are in fact lying in a plane, or $r(A)=1$, lying within a line, either of which holds a zero volumn.

Title: Rank of a matrix


问:
Is it possible for a function to be analytic anywhere outside the circle of convergence of its power series expansion?
Is it possible for a function to be analytic anywhere outside the circle of convergence of its power series expansion? I'm referring to analytic fuctions of course (i.e. those with power series expansions but not necessarily analytic over the entire complex plane).</p>

答:
The question is a little confusing, but how about $f(z) = \dfrac{1}{1-z}$.

The power series of $f$ around $z=0$ is $\sum_{n=0}^\infty z^n$ and this converges on $|z|<1$. The function, however, is analytic on $\mathbb{C}\setminus \{1\}$.


问:
Best way to create sample set of numbers 1 to 5 to minimize expected value when betting on outcome of sample
You and I play the following game: You create a deck of as many cards as you want, and on every card you write one of the integers from $1$ to $5$, e.g. your deck could be just $\{1,2,3,4,5\}$ or $\{1,1,1,1,2,3,3,3\}$ or whatever you want, as long as it only consists of the integers from $1$ to $5$.
Then, I have a look at the deck, will select one number to bet on, and one random card will be chosen from your deck. If that number matches my bet, you pay me that that number.
How do you create a deck in order to minimize your losses, given that I play optimally?\

The best I figured out was $\{1,1,1,1,1,2,2,3,4,5\}$, giving you an expected loss of $\frac{1}{2}$ if I bet on $1$ or $5$ and less for the other numbers, but I'm quite sure that this is not optimal but also don't know how to proceed from there. Maybe there is a solution that gives me the same expected value no matter on which number I bet and is thus better for you?

答:
Suppose that a fraction $p_i$ of the cards are labelled $i$, for $1\leq i\leq 5$.

The crucial insight is that your best strategy will have
$$1p_1=2p_2=\dots=5p_5,$$
so that the opponent is indifferent to which card they guess. You can then see that the expected value is $\frac{60}{137}$.

Indeed, if we don't follow this distribution, then some $j$ would have $jp_j>\frac{60}{137}$, so the opponent could always guess card $j$ and have a higher EV.


问:
Why I am getting this slight deviation
I have solved this equation by two method

√3a = 50+a

Method 1

√3a-a =50

a(√3-1) = 50

a= 50/(√3-1)

a= 50/0.73 = 68.4

Method 2

a(√3-1) = 50

a = 50/(√3-1)

a =50(√3+1)/2.  [by rationalizing]

a =25 * 2.73

a = 68.25

Note : I have taken √3 = 1.73 in both of the above method
Still got different value of " a ".Can you please tell why this slight deviation has occured

答:
The error occurs because you are assuming that $\sqrt{3}=1.73$. This is incorrect: it should be $\sqrt{3} \approx 1.73$. Because you are only using the approximate value of $\sqrt{3}$, it is only natural that you get an approximate answer.

It is best to work through questions like these without rounding any figures until the very end. This is because small rounding errors can accumulate. For example, if I take $\sqrt{3}$ as $1.73$, then $1000\sqrt{3}$ would be approximated as $1730$. This is more than $2$ units away from the true answer of $1732.050\ldots$. Depending on when and where you round figures, the error that results could be marginal, or it could be much greater. In your example, the true answer is
$$
a=\frac{50}{\sqrt{3}-1} \approx 68.301 \, .
$$
Because you rounded early, both of your answers are a little off!

Title: Why I am getting this slight deviation


问:
Natural Deduction proof using basic rules only
I need some assistance solving what seems to be a very intuitive problem, but becomes tough when only using strict natural deduction and not assuming De Morgan laws.

Laws allowed: Implication, And, Or, MT, PBC, Copy Rule, Negation, Double Negation, Contradictions, law of excluded middle

I'm thinking it uses the law of excluded middle but I can't quite figure it out.

$$ \lnot(P \land \lnot Q), \; (\lnot P \to S) \land \lnot Q \;\;\; \text{premises} \tag{1} $$

$$ T \lor S \;\;\; \text{conclusion} \tag{2} $$

答:
Hint

$\lnot (P \land \lnot Q)$ --- premise

$(\lnot P \to S) \land \lnot Q$ --- premise

$\lnot Q$ --- from 2) by $(\land \text E)$

$P$ --- assumed [a]

$(P \land \lnot Q)$ --- from 4) and 3) by $(\land \text I)$

Contradiction !

and so on, deriving the sought conclusion: $T \lor S$.

The Natural Deduction rules needed, in addition to the $\land$-rules above, are $(\lnot \text I), (\to \text E)$ and $(\lor \text I)$.


问:
How do I find the infinite sum of z/(z-2)
To find the infinite sums, in general, the following approach seems to work:
$\sum_{0}^{\infty}z^{n+1} = z/(1-z)$, where $|z|<1$. This is just different version from the standard equation for power series:
$\sum_{0}^{\infty}a_nz^{n} = a_1/(1-z)$, with $|z|<1$.

When I do this for the following equation in the place of $z$: $z/(z+2)$, I get, as expected $z/2$. This is also what wolfram alpha gives me.

If I do this with $z/(z-2)$, I would expect to get $-x/2$ according to the same reasoning:
$\sum_{0}^{\infty}(z/(z-2))^{n+1} = \frac{z/(z-2)}{1-z/(z-2)} = -z/2$, for $|z/(z-2)|<1$.

Wolfram Alpha however, says that this is not correct. I do not understand what is going wrong. What is wrong about this approach?

Body:
For $|r|<1$,
$$
\sum_{n=0}^\infty r^n=\frac{1}{1-r}\tag{1}
$$

Let $r=z/(z-2)$. For $|z/(z-2)|<1$, you have
$$
\frac{1}{1-z/(z-2)}=\frac{z-2}{(z-2)-z}=\frac{z-2}{z-2-z}=-\frac12(z-2).
$$

For your edited version, you simply have
$$
\sum_{n=0}^\infty r^{n+1}=r\sum_{n=0}^\infty r^{n}=\frac{z}{z-2}\cdot(-\frac12)({z-2})=\frac{-z}{2}.
$$


问:
In the proof for intermediate value theorem, why can we assume that there is a point between some two values in the image set of $f$?
In these theorems, why can we assume that there is some $u$ between $f(a)$ and $f(b)$ for $a, b \in I$? What would happen if no such $u$ exists?

答:
In these theorems the function is continuous. If no such u would exist, then there's a "jump" in the function and thus it's not continuous.

Edit: As it seems I didn't "get" your question, I'd like to refer to this post and its answers, showing that there is an infinite number of real numbers between two distinct real numbers:

Prove that there are infinitely many numbers between two real numbers. (example from Hardy's book)


问:
Finding roots of polynomial $f(x)$ in $\mathbb{Z}_p = \mathbb{Z}/p\mathbb{Z}$, where $p$ is prime.
I am not quite sure how to approach such a problem. In particular, I am wondering how to find the roots of the following polynomial in $\mathbb{Z}_p = \mathbb{Z}/p\mathbb{Z}$:

$f(x) = x^{2p^2}- \overline{2}x^{p^2} - x^p + \overline{2} \in \mathbb{Z}_p[x]$.

Any explanation or advice would be much appreciated!

答:
$x^p \equiv x \bmod p \Rightarrow x^{2p^2}=(x^p)^{2p} \equiv x^{2p} \equiv x^2 \bmod{p}$

$x^p \equiv x \bmod p \Rightarrow x^{p^2} \equiv x \bmod p$

$x^{2p^2}- \overline{2}x^{p^2} - x^p + \overline{2} = x^2 -3x+2=(x-2)(x-1)=0$

$x \equiv 2$ or $x \equiv 1$


问:
Finding trigonometric integral (challenging)
Integrate:

$$\displaystyle\int \dfrac{(1+\cos^2A\times\cos(2A))^2}{\cos(2A)\times(2\cos^4A+\sin^2A)}dA$$

Again my working with trigonometric identities has not take me anywhere useful, that is I haven't been able to break free of fractions or nasty products. But again, I'm a beginner to calculus and I only know how to integrate the 6 primary trigonometric functions. Perhaps there is a good way to bypass the tedious mess?

Despite the most off-putting look, the teacher claims that the integral is not anywhere as nasty as one would expect. This, of course, is not a part of class work but a challenge she threw at us.

Thanks in advance and as usual, all advice welcome.

答:
Keep working with the trig identities! In this particular example, it might help to turn everything into powers of $\cos A$. You'll get some nice cancellations.

When you're done canceling, you should have some trig functions that will be easier to integrate.


问:
Solving $\ln(\lvert y \rvert)=\ln(\lvert 1+x \rvert)+ \ln(\lvert c \rvert)$ for $y$ where $c$ denotes an arbitrary constant
According to my book:
$$\ln(\lvert y \rvert)=\ln(\lvert 1+x \rvert)+ \ln(\lvert c \rvert)$$
$$\Rightarrow \ln(\lvert y \rvert)=\ln(\lvert c(1+x) \rvert)$$
$$\Rightarrow y=c(1+x)$$

I know that this is correct. However, I want to be able to understand the second step clearly. I want to make sure if this is why :
$$\ln(\lvert y \rvert)=\ln(\lvert c(1+x) \rvert) \Rightarrow \lvert y \rvert=\lvert c(1+x) \rvert$$
$$\Rightarrow y=± c(1+x) \Rightarrow y=C(1+x)$$
where $C=±c$ is just a substitution for the constant. Am I correct? Is this an often-used method that is considered rigorous even if we do not rename the constant?

Edit: $c$ is not a fixed constant, but an arbitrary one. To give some context, the equation to be solved is obtained from integration.

答:
Yes, you are correct. Though I would not write $C=\pm c$. Note that $c$ is a fixed (nonzero) constant. You cannot define $C$ with two values.

In general, what you have is that
$$
|y|=|kx|
$$
if and only if
$$
y=kx\quad \textrm{or}\quad  y=-kx\tag{1}
$$

(1) is sometimes written as $y=\pm kx$.

Notes 1.

If you get the identity when you solve an ODE and get two (sets of) solutions: $y=cx$ and $y=-cx$, you need to be careful if you want to combine them. In your case, the "arbitrary" constant $c$ is not zero. So you cannot conclude that $y=Cx$ where $C$ is "arbitrary". If you find further that $y=0$ is also a solution, then you can indeed claim that $y=Cx$, combining all the cases.

Notes 2.
Consider for instance the ODE you mentioned in the comment:
$$
\frac{dy}{dx}=\frac{y}{1+x}\tag{0}
$$
To solve it,
\begin{align}
\frac{dy}{y}&=\frac{dx}{1+x}\\
\ln|y|&=\ln|1+x|+\ln|c|&(c\ne 0)
\end{align}

(Remark. The reason why we can write $\ln|c|$ here is that the range of the function $x\mapsto \ln x$ ($x>0$) is the whole real line. So adding an "arbitrary" constant $c$ is the same as adding $\ln|c|$ for some arbitrary nonzero constant $c$.)

To go on,
\begin{align}
\ln|y|&=\ln|1+x|+\ln|c|&(c\ne 0)\\
\ln|y|&=\ln|c(1+x)|&(c\ne 0)\\
|y|&=|c(1+x)|&\textrm{(because $\ln x$ is injective)}
\end{align}

Now since $c$ is an arbitrary nonzero constant, you have following set of solutions,
$$
y=c(1+x),\quad c\ne 0.\tag{1}
$$

Since $c$ is "arbitrary", you don't need to add $\pm$ in front of it.

Observe that $y\equiv 0$ is also a solution to (0), and it is not of the form in (1).

So combining (1) and the zero solutions together, you get that the general solution for (0) is
$$
y=C(1+x)
$$
where $C$ is an arbitrary constant. (Note that $C$ is not required to be nonzero anymore.)


问:
How to determine h(x) in a polynomial partial fraction decompostion
Im susposed to do a partial fractional division;

$$ \frac{-2x^2 + 8x - 9} {(x-1) (x-3)^2}$$ Now I used the formula and this is what I got;

$$\frac {A}{(x-1)} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$

Now I've put the 3 fractions together,and got this

$$ \frac{A(x-3)^2+B(x-3)(x-1) +C(x-1)} {(x-1) (x-3)} $$

Now I looked at the starting function looked what my x^2 ,x terms were on the right on the left, solved it and got this

$$f(x) = \frac{-2} {4(x+1)} + \frac{-5} {4(x-3)} + \frac {3} {2(x-1)^2} $$

Now after looking at the formula I'd realise that I'm missing my h(x) the whole part if you will ( not sure how it is said in english) Now the h(x) doesnt exist when m < n but in my case m = n so I should also be having that? But how do I get that part? the formula says simply h(x) + the A B and C parts.How do I calculate h(x)?

EDIT: To clarify what h(x) means and what I was exactly trying to figure out.

$$f(x) = h(x) + \frac{r(x)} {q(x)}$$

Where m = deg r and n = deg q

答: Your approach is fine, but there is some mistake in your computations, since, in fact$$A=-\frac34,\ B=-\frac54\text{, and }C=-\frac32;$$therefore$$\frac{-2x^2+8x-9}{(x-1)(x-3)^2}=-\frac3{4(x-1)}-\frac5{4(x-3)}-\frac3{2(x-3)^2}.$$And, in your case, the degree of the numerator is $2$ and the degree of the denominator is $3$. Therefore, there is nothing else to be done.


问:
When will matrix multiplication become just "concatenation"?
Saw below entertaining matrix multiplication examples.

Obviously they are just coincidences. But I am curious when does below hold?

\begin{equation}
\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix} \times \begin{pmatrix}
a_2 & b_2 \\
c_2 & d_2
\end{pmatrix} = \begin{pmatrix}
\overline{a_1 a_2} & \overline{b_1 b_2} \\
\overline{c_1 c_2} & \overline{d_1 d_2}
\end{pmatrix}
\end{equation}

here $\overline{a_1 a_2}$ means gluing the integers together, not multiplications.. assuming we are working on integer matrices.

答:
Your question is quite interesting!

So, let's calculate: first, how can we write $\overline{a_1a_2}$? In fact, $a_1$ and $a_2$ are simply, as you said, "glued" together. So, we simply have $\overline{a_1a_2} = 10a_1+a_2$.

Now, let's re-write your equation using this result:

$$\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix} \times \begin{pmatrix}
a_2 & b_2 \\
c_2 & d_2
\end{pmatrix} = \begin{pmatrix}
\overline{a_1 a_2} & \overline{b_1 b_2} \\
\overline{c_1 c_2} & \overline{d_1 d_2}
\end{pmatrix} = \begin{pmatrix}
10a_1+a_2 & 10b_1+b_2 \\
10c_1+c_2 & 10d_1+d_2
\end{pmatrix}$$

Now, let's name this "equation" $(E)$ with every number being an integer.
Then, by calculating the product, we have
$$(E) \iff \begin{pmatrix} a_1a_2 + b_1c_2 & a_1b_2 + b_1d_2\\
c_1a_2 + d_1c_2 & c_1b_2 + d_1d_2 \end{pmatrix} = \begin{pmatrix}
10a_1+a_2 & 10b_1+b_2 \\
10c_1+c_2 & 10d_1+d_2
\end{pmatrix}$$

Finally, we are able to identify the coefficients and get a system of equations:
$$\boxed{(E) \iff \begin{cases} a_1a_2+b_1c_2 = 10a_1+a_2\\
c_1a_2+d_1c_2 = 10c_1+c_2 \\
a_1b_2+b_1d_2 = 10b_1+b_2\\
c_1b_2 + d_1d_2 = 10d_1+d_2 \end{cases}}$$

Now that we got this system, you can notice that there are only 4 equations for a total of 8 unknowns. So 4 of them will have to be fixed, then they'll give you the 4 other. Now it's up to you to find "groups" of integers that work!


问:
Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$
I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.

The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$

How am I supposed to solve this problem?

Body: If $t:=\tan x$ then $\frac{t(3-t^2)}{1-t^2}=\tfrac{2}{\sqrt{3}}$ so $t^3-\tfrac{2}{\sqrt{3}}t^2+3t+\tfrac{2}{\sqrt{3}}=0$. This has one real root, $t=\tfrac{2+33/a-a}{3\sqrt{3}}$ with $a:=\sqrt[3]{154+9\sqrt{443}}$. Now just calculate $\frac{1-t^2}{2}$.


问:
Number of additions in naive matrix multiplication
In the naive matrix multiplication algorithm where you have 3 loops, the total number of multiplications is $n^2(n)=n^3$ given that we have two matrices each with size $n\times n$. But a proof claims that the number of additions is $(n-1)$, so the total number of additions is $n^2(n-1)$. I don't understand how the number of additions is $(n-1)$ as I think it should be $n$ because whenever we multiply we add that element, so we should have the same number of multiplications $n$?

答: When you add $n$ terms there are only $n-1$ additions.  For example, $1+2+3$ has only two additions.  You are making a fencepost error.


答:
You made a mistake when you put three fractions together:
$$
\frac{A(x-3)^2+B(x-1)(x-3)+C(x-1)}{(x-1)(x-3)^\color{red}{2}}.
$$

Now you set
$$
A(x-3)^2+B(x-1)(x-3)+C(x-1)=-2x^2+8x-9
$$
to find $A,B,C$.

Set $x=1$: $4A=-2+8-9$.

Set $x=3$: $2C=-18+24-9$.

Set $x=2$: $A-B+C=-8+16-9$.

Now it should be very straightforward to go on.

Notes.

If you write
$$
f(x)=\frac{-2x^2+8x-9}{(x-1)(x-3)^2}
$$
then you can find polynomials $h,p,q$ so that
$$
f(x)=h(x)+\frac{p(x)}{q(x)}
$$
where the $\deg p<\deg q$.
This is the preliminary step for partial fraction decomposition.

But note that
$$
\deg (-2x^2+8x-9)<\deg (x-1)(x-3)^2
$$
you have $h(x)=0$, $p(x)=-2x^2+8x-9$ and $q(x)=(x-1)(x-3)^2$.

Title: How to determine h(x) in a polynomial partial fraction decompostion

Body:
Im susposed to do a partial fractional division;

$$ \frac{-2x^2 + 8x - 9} {(x-1) (x-3)^2}$$ Now I used the formula and this is what I got;

$$\frac {A}{(x-1)} + \frac{B}{x-3} + \frac{C}{(x-3)^2}$$

Now I've put the 3 fractions together,and got this

$$ \frac{A(x-3)^2+B(x-3)(x-1) +C(x-1)} {(x-1) (x-3)} $$

Now I looked at the starting function looked what my x^2 ,x terms were on the right on the left, solved it and got this

$$f(x) = \frac{-2} {4(x+1)} + \frac{-5} {4(x-3)} + \frac {3} {2(x-1)^2} $$

Now after looking at the formula I'd realise that I'm missing my h(x) the whole part if you will ( not sure how it is said in english) Now the h(x) doesnt exist when m < n but in my case m = n so I should also be having that? But how do I get that part? the formula says simply h(x) + the A B and C parts.How do I calculate h(x)?

EDIT: To clarify what h(x) means and what I was exactly trying to figure out.

$$f(x) = h(x) + \frac{r(x)} {q(x)}$$

Where m = deg r and n = deg q


问:
Pan Balance : Find Faulty Weight
Out of 9 equal weights, one is faulty (lighter). Find the least number of weighing required to identify a faulty weight using a two pan balance.

I came across this similar type of question posted here long back and I used the relation it mentioned :-

"In my module they have given some sort shortcut to evaluate the result which is for $n$ bags:
In first case if $3^a \lt n \lt 3^{a+1}$, then the minimum number of weighing required is $a+1$. In the second case, we need $2^a \lt n \lt 2^{a+1}$ which gives the minimum of $a+1$ weighing."

when I put 9 in the equation $3^a \lt 9 \lt 3^{a+1}$ and solve it, I get a<2 and a>1 which does not give a reasonable answer because it should be an integer answer right? Where I am going wrong?

答:
The answer is obviously two weighings for nine weights:

weigh three against three, revealing which batch of three has the light weight

from that batch weigh one against one,  revealing which one is the light weight

The expression should have been that you need at least $a$ weighings when $3^{a-1} \lt n \le 3^a$, i.e. $a=\lceil \log_3(n) \rceil$

It gets a little more complicated when you do not know whether the faulty weight is lighter or heavier than the others

Title: Pan Balance : Find Faulty Weight


问:
spherically symmetric configurations
$$\Delta S -S +S^3=0$$ How this Differential equation can be written in this form:

\begin{equation}
\frac{d^2S}{d\rho^2}+\frac{D-1}{\rho}\,\frac{dS}{d\rho}
-S+S^3=0
\end{equation}
Which is  spherically symmetric configurations. D= dimensions.
Details in the paper equations (21, 44)

答:
If $S$ is speherically symmetric, say $S(x) = S(\rho)$ with $\rho = |x|$, we have
\begin{align*}
  \Delta S &= \sum_i \partial_i^2S\\
              &= \sum_i \partial_i \bigl(\partial_\rho S\cdot \partial_i\rho\bigr)\\  &= \sum_i \left(\partial_\rho^2 S \cdot (\partial_i \rho)^2 +  \partial_\rho S \cdot \partial_i^2\rho\right)
\end{align*}
Now
$$ \partial_i\rho(x) = \frac{x_i}{\rho(x)} $$
and hence
$$ \partial_i^2\rho(x) = \frac{\rho(x) - x_i\frac{x_i}{\rho(x)}}{\rho^2(x)} = \frac{\rho^2(x) - x_i^2}{\rho^3(x)} $$
So $$\sum_i (\partial_i \rho)^2 = \frac{1}{\rho^2(x)}\sum_i x_i^2 = 1$$
and
$$ \sum_i \partial_i^2\rho = \frac{D\rho^2 - \sum_i x_i^2}{\rho^3} = \frac{D-1}{\rho} $$
and hence
$$ \Delta S = \partial_\rho^2 S + \frac{D-1}\rho \cdot \partial_\rho S $$</p>

Title: spherically symmetric configurations


问:
How to read $\frac{d^2y}{dx^2}$
Is it correct to say 'dee squared y over dee x squared'?

I am just trying to make sure because I have to teach this to someone.

I learnt that $f''(x)$ is pronounced 'f double prime of x' but I couldn't find how to pronounce Leibniz notation. Thanks!

答: You could read as: "dee two y over dee x two"</p>

Title: How to read $\frac{d^2y}{dx^2}$


答:
Call the two matrices (in blue and in green respectively) $A$ and $B$. Assuming for simplicity that we want $B$ to consist of single digits only, we can write the property we're looking for as a matrix equation: $AB = 10A + B$.

Moving everything to one side, we can add a fourth term to make things easier to factor: $AB - 10A - B + 10I = 10I$. This can be written as $(A - I)(B - 10I) = 10 I$. So, in terms of $B$, we have $$A = 10(B - 10 I)^{-1} + I.$$
This gives us a formula for $A$ no matter which $B$ we pick, though out of $10^4$ possibilities for $B$, my computer only found $429$ for which $A$ consists of nonnegative integers.

Title: When will matrix multiplication become just "concatenation"?

Body:
Saw below entertaining matrix multiplication examples.

Obviously they are just coincidences. But I am curious when does below hold?

\begin{equation}
\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix} \times \begin{pmatrix}
a_2 & b_2 \\
c_2 & d_2
\end{pmatrix} = \begin{pmatrix}
\overline{a_1 a_2} & \overline{b_1 b_2} \\
\overline{c_1 c_2} & \overline{d_1 d_2}
\end{pmatrix}
\end{equation}

here $\overline{a_1 a_2}$ means gluing the integers together, not multiplications.. assuming we are working on integer matrices.


答:
Let $A$ denote the first matrix in the product, and let $B$ denote the second.
I will consider only the case in which the entries of $B$ have one digit.

In this case, the "gluing" property of this matrix multiplication can be written as
$$
AB = 10A + B.
$$
Note that this equation can be rearranged into
$$
AB - 10 A - B + 10I = 10I \implies\\
(A - I)(B - 10 I)= 10 I,
$$
where $I$ denotes the identity matrix. Now, suppose that we select a matrix $B$ and we want a corresponding matrix $A$. We have
$$
(A - I)(B - 10 I) = 10 I \implies\\
A = 10(B - 10 I)^{-1} + I.
$$
Note that this equation only has a solution if $\det(B - 10 I) \neq 0$. Now, a question remains: how do we ensure that $10 (B - 10 I)^{-1}$ is an integer matrix? As it turns out, this will hold for a given integer matrix $B$ if and only if the determinant of $B - 10 I$ divides $10$.

In fact, we can generate pairs of matrices with non-negative integer entries that have the gluing property via the following steps:

Find a matrix $C$ whose diagonal entries satisfy $-10 \leq c_{ii} \leq -1$ and whose off-diagonal entries are a single digit positive number such that the determinant of $C$ is either $-1$, $-2$, $-5$, or $-10$.

Take $B = 10 I + C$ and $A = 10C^{-1} + I$.

For example, the matrix
$$
C = \pmatrix{-4 & 3\\7 &-4}
$$
has determinant $16 - 21 = -5$, which divides $10$. The corresponding matrix $B$ is
$$
B = 10 I + \pmatrix{-4 & 3\\7 &-4} = \pmatrix{6&3\\7&6}.
$$
The associated matrix $A$ is
$$
10C^{-1} + I = \frac{10}{-5} \cdot \pmatrix{-4 & -3\\-7 & -4} + \pmatrix{1&0\\0&1}
= \pmatrix{9 & 6\\14 & 9}.
$$
If we compute the product, we indeed find that
$$
\pmatrix{\color{blue}{9} & \color{blue}{6}\\
\color{blue}{14} & \color{blue}{9}} \cdot
\pmatrix{\color{green}{6} & \color{green}{3}\\
\color{green}{7} & \color{green}{6}} =
\pmatrix{\color{blue}{9}\color{green}{6} & \color{blue}{6}\color{green}{3}\\
\color{blue}{14}\color{green}{7} & \color{blue}{9}\color{green}{6}}.
$$

An interesting phenomenon: if $B$ is a single digit matrix for which there exists an $A,B$ pair with this "gluing" property, we will have $A = B$ if and only if $B$ is a "vampire matrix" (cf. my comment on the question), which holds if and only if $B$ has eigenvalues $0,11$, which holds if and only if $C = B - 10 I$ has eigenvalues $-10,1$, which holds if and only if $C$ has determinant $-10$ and trace $-9$.


问:
Is $\frac{\mathbb{Z}[\sqrt{10}]}{\langle4,\sqrt{10}\rangle}$ a field?
My answer is yes. I took the following route. Is it correct? $$\frac{\mathbb{Z}[\sqrt{10}]}{\langle4,\sqrt{10}\rangle}= \frac{\mathbb{Z}[\sqrt{10}]}{\langle2,\sqrt{10}\rangle}= \frac{\frac{\mathbb{Z}[X]}{\langle X^2-10\rangle}}{\langle2,\bar{X}\rangle}=\frac{\frac{\mathbb{Z}[X]}{X^2-10}}{\frac{\langle2,X,X^2-10\rangle}{X^2-10}}= \frac{\mathbb{Z}[X]}{\langle2,{X},X^2-10\rangle} = \frac{\frac{\mathbb{Z}[X]}{X}}{\langle2,\bar{X}^2-10\rangle}=\frac{\mathbb{Z}}{\langle2,10\rangle}=\frac{\mathbb{Z}}{\langle2\rangle}$$

Am I doing any wrong at any step? (Please assume I have tried using appropriate isomorphism theorems at appropriate steps.)

答:
In general
$$
\frac{\mathbb Z[\sqrt d]}{\langle k, \sqrt{d}\rangle} \cong \frac{\mathbb Z [X]}{\langle k, X, X^2-d\rangle} \cong \mathbb Z/\langle\gcd(k,d)\rangle,
$$
which is a field if and only if $\gcd(k,d)$ is prime. That's the case for  $\mathbb Z[\sqrt{10}]/\langle 4,\sqrt{10}\rangle$ with $\gcd(4,10)=2$ as well as $\mathbb Z[\sqrt{10}]/\langle 5,\sqrt{10}\rangle$ with $\gcd(5,10)=5$.


问:
A question about using the squeeze theorem?
If you have the inequalities not equal to, but the function is just strictly between two other functions, can you use the squeeze theorem?</p>

答:
Sorry for the late answer but just answering in case anyone else has the same doubt.

by looking at the question ,it seems that the op is wondering if $f(x) < g(x) < h(x)$ how can it be possible that the limits are equal as the functions are not allowed to be equal
The fact is $$f(x) < g(x) \implies \lim_{x\to a}f(x) \leq \lim_{x\to a}g(x) $$
this can be easily proved by taking h(x)=f(x)-g(x) and h(x)<0,now apply $\epsilon-\delta$ definition ,u cant rule out the possibility of $lim_{x\to a} h(x)=0$ and now u can use limit law to split f(x) and g(x) and hence proving the original statement
Several examples showing such scenario are provided in comments section.

Now coming to squeeze theorem,now u are possibly understanding that even if u have strict inequalities for the functions,it translates to having non strict inequalities for the limit and hence when the lower and upper bound for $\lim_{x \to a} g(x)$ are equal and it is allowed to take that value,it must take that value

alternatively, if u want to prove squeeze theorem for strict inequality from scratch ie. from $\epsilon -\delta $  definition , then proceed as shown in https://en.wikipedia.org/wiki/Squeeze_theorem#Proof and just remove equality wherever necessary


问:
Fewer degree-$3+$ nodes than leaf nodes in a tree
Considering this question, I was struck by the idea that:

For a graph $G$ that is a tree, the number of degree-$1$ nodes exceeds the number of nodes of degree $3$ or higher.

.. which would fairly directly solve that question.

The intuition is as follows: each degree-$3$ node adds a branch to the tree, which must also add a degree-$1$ node. Higher degree nodes add branches per extra degree. This intuition adds a numerical prediction:
$$ \sum_{k\in H}(d(k)-2) = |L|-2$$
where $H$ is the set of vertices in $G$ with degree $3$ or higher and $L$ is the set of degree-$1$ nodes, with two leaf nodes being allocated to a simple unbranched tree.

Can anyone produce or reference a more formal proof?

答:
A tree on $n$ vertices has $n-1$ edges. So if there are $a$ leaves and $b$ vertices of degree at least three, then by the handshaking lemma,
$$a+2(n-a-b)+3b\leq\sum \deg v=2(n-1)\implies a\geq b+2.$$


答: Yes, we read it this way. But it is actually the second derivative of the function $y$ which is a function of $x$.

Title: How to read $\frac{d^2y}{dx^2}$

Body:
Is it correct to say 'dee squared y over dee x squared'?

I am just trying to make sure because I have to teach this to someone.

I learnt that $f''(x)$ is pronounced 'f double prime of x' but I couldn't find how to pronounce Leibniz notation. Thanks!


答:
The expression for Laplacian operator in the $D$-dimensional spherical coordinates is
$$\Delta=\frac{\partial^2}{\partial\rho^2}+\frac{D-1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^2}\Delta_{S^{D-1}},$$
where $\Delta_{S^{D-1}}$ is a differential operator called Laplace-Beltrami operator on the $(D-1)$-dimensional sphere $S^{D-1}$. It depends only on the angular variables (in particular, all the derivatives are with respect to them). Therefore, if you are looking for rotationally symmetric solutions of your PDE, angular derivatives are all zero and you end up with the ODE you have written.

For example, as $D=2$, set $x=\rho\cos\varphi$, $y=\rho\sin\varphi$ and this angular part of Laplacian will be equal to
$$\Delta_{S^1}=\frac{\partial^2}{\partial\varphi^2}.$$

Similarly, for $D=3$, introducing the spherical coordinates
$$\begin{cases}x=\rho\cos\varphi\sin\theta,\\ y=\rho\sin\varphi\sin\theta,\\
z=\rho\cos\theta,\end{cases}$$
we have the expression
$$\Delta_{S^2}=\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\varphi^2}.$$

Title: spherically symmetric configurations

Body:
$$\Delta S -S +S^3=0$$ How this Differential equation can be written in this form:

\begin{equation}
\frac{d^2S}{d\rho^2}+\frac{D-1}{\rho}\,\frac{dS}{d\rho}
-S+S^3=0
\end{equation}
Which is  spherically symmetric configurations. D= dimensions.
Details in the paper equations (21, 44)


问:
Different definitions of covector
In the book An introduction to manifolds by Tu, Loring W, covector is defined as follows:

If $V$ and $W$ are real vector spaces, we denote by $\operatorname{Hom}(V, W)$ the vector space of all linear maps $f: V \rightarrow W$. Define the dual space $V^{\vee}$ of $V$ to be the vector space of all real-valued linear functions on $V$ :
$$
V^{\vee}=\operatorname{Hom}(V, \mathbb{R})
$$
The elements of $V^{\vee}$ are called covectors or $1$-covectors on $V$.

However, in the book Quick Introduction to Tensor Analysis by R. A. Sharipov, the covector is defined as follows:

Let's denote our hypothetical object by $\mathbf{a},$ and denote by $a_{1}, a_{2}, a_{3}$ that three numbers which represent this object in the basis $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$. By analogy with vectors we shall call them coordinates. But in contrast to vectors, we intentionally used lower indices when denoting them by $a_{1}, a_{2}, a_{3}$. Let's prescribe the following transformation rules to $a_{1}, a_{2}, a_{3}$ when we change $\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}$ to $\tilde{\mathbf{e}}_{1}, \tilde{\mathbf{e}}_{2}, \tilde{\mathbf{e}}_{3}$:
$$
\tilde{a}_{j} =\sum_{i=1}^{3} S_{j}^{i} a_{i} \tag{8.1}
$$
$$
a_{j} =\sum_{i=1}^{3} T_{j}^{i} \tilde{a}_{i}\tag{8.2}
$$
Here $S$ and $T$ are the same transition matrices as in case of the vectors in $(6.2)$ and $(6.5)$. Note that $(8.1)$ is sufficient, formula $(8.2)$ is derived from $(8.1)$.

DEFINITION 8.1. A geometric object a in each basis represented by a triple of coordinates $a_{1}, a_{2}, a_{3}$ and such that its coordinates obey transformation rules
$(8.1)$ and $(8.2)$ under a change of basis is called a covector.

My question is, are the two definitions equivalent? What's the relation between them? I cannot figure out the relation between "obeying transformation rules" and linear maps $f: V \rightarrow W$.

答:
The idea is that $V^\vee:=\operatorname{Hom}(V,\mathbf{R})$ is a vector space, and thus vectors in $V^\vee$ can have components with respect to a basis. If we have two different bases for $V^\vee$, then we will get two different sets of components for the same vector, and the way these components are related to each other is what we call a transformation law, or a change of basis, or something along those lines.

Let's look at the vector space $\mathbf{R}^3$. If we have two bases $E=\{e_1,e_2,e_3\}$ and $F=\{f_1,f_2,f_3\}$, we may express some vector $v\in\mathbf{R}^3$ as $v=a^1e_1+a^2e_2+a^3e_3$ or $b^1f_1+b^2f_2+b^3f_3$. How are the $a^i$'s and $b^i$'s related? Well, it's enough to know how $E$ and $F$ are related; so let $A$ be the matrix sending $e_i$ to $f_i$. That is, $f_i=Ae_i=A^1_ie_1+A^2_ie_2+A^3_ie_3$. Let's write $\epsilon^j$ for the map that takes a vector to its $j$-th component in the basis $E$, so $v=\epsilon^1(v)e_1+\epsilon^2(v)e_2+\epsilon^3(v)e_3$. Notice that $\epsilon^i$ is linear, and $\epsilon^j(Ae_i)=A^j_i$. We then compute
\begin{align*}
a^j=\epsilon^j(b^1f_1+b^2f_2+b^3f_3)
&=b^1\epsilon^j(f_1)+b^2\epsilon^j(f_2)+b^3\epsilon^j(f_3)\\
&=b^1\epsilon^j(Ae_1)+b^2\epsilon^j(Ae_2)+b^3\epsilon^j(Ae_3)\\
&=b^1A^j_1+b^2A^j_2+b^3A^j_3.
\end{align*}

So $a^j=\sum_ib^iA^j_i$. Roughly speaking, if $F$ is the old basis system and $E$ is the new one, this formula tells us that we must take the matrix $A$ sending new basis vectors $e_i$ to old ones $f_i$ and multiply it with the old components $b^i$ to get the new components $a^i$. As such, we say that the components of a vector transform contravariantly. (Later you'll find that the components of a covector transform covariantly. And a vector transforms covariantly as well, except people sometimes will say vector when they mean the components of the vector. And so they'll say a vector transforms contravariantly.)

So we can think of a vector as an abstract element of the vector space $\mathbf{R}^n$ subject to the vector space axioms, or as an equivalence class of $n$-tuples as subject to the constraints of transformation laws. Both ways end up imposing the same structure. Similarly for your question on covectors being linear maps from $V$ to $\mathbf{R}$. You should be working on a higher level of abstraction and thinking of covectors as elements of vector spaces, and thus admitting transformation laws with respect to different bases.

Later on you will see vector spaces, vectors and covectors generalized to tensors, and you will notice that mathematicians and physicists tend to treat them rather differently. To understand the interplay between the mathematician's abstract, coordinate-free, element of tensor space perspective and the coordinates, transformation laws perspective can be rather tricky. For that I recommend Michael Spivak's A Comprehensive Introduction to Differential Geometry, as well as section 7 of Keith Conrad's Tensor Products.

Title: Different definitions of covector


问:
Multiple choice question about the roots of a cubic equation
I am in class $7$th and studying high school maths. I encountered the following question. How can I solve it?

Well I have eliminated option A and B by using the concept that sum of roots in cubic equation is equal to $b/a$ and product of the roots is $d/a$. Here is my working:

$$x^2-ax^2-bx+c=0$$
Let the roots be $j$, $k$, and $l$. Then,
\begin{align}
j + k + l &= a \\
jkl &= -c \\
jk + kl + lj &= -b
\end{align}

答:
x1 + x2 + x3 =-a and x1x2x3 =-c

This means that

x1 + x2 + x3 >0
x1x2x3 <0

This is possible when two roots are greater than 0, and one less than 0


问:
Find the angle bisectors of a triangle with side $a$ and adjacent angles $\beta$ and $\gamma$
Find the angle bisectors of a triangle with side $a$ and adjacent angles $\beta$ and $\gamma$.

I am trying to solve the problem using the law of sines, because we have just studied it, but I can't seem to find a solution. Thank you in advance!

答:
Hint: find b and c by sine rule, then use following formula:

$$tan \frac{\gamma}2=\sqrt {\frac{(p-a)(p-b)}{p(p-c)}}$$

Where $p=\frac{a+b+c}2$


问:
How to integrate $\int_0^{\infty} \frac{x}{e^x+1} dx$
I want to integrate it explicitly. I looked up if this had an exact solution, and I got that it is $\frac{\pi^2}{12}$ here: https://www.integral-calculator.com/#expr=x%2F%28e%5Ex%2B1%29&lbound=0&ubound=inf

I thought it could be solved using geometric series, so I tried to turn my integral into something similar:
$$\int_0^{\infty} \frac{x}{e^x+1} dx= \int_0^{\infty} \frac{x}{e^x+1} \frac{e^x-1}{e^x-1} dx=\int_0^{\infty} \frac{x e^x }{e^{2x}-1}dx -\int_0^{\infty} \frac{x}{e^{2x}-1} dx= \int_0^{\infty} \frac{x e^x }{e^{2x}-1}dx -\frac{1}{4}\int_0^{\infty}  \frac{x}{e^{x}-1} dx=\int_0^{\infty} \frac{x e^x}{e^{2x}-1} - \frac{1}{4}\frac{\pi^2}{6}$$

Where the second term is easy to solve with geometric series, and it's also defined with the Riemann zeta:

$$\int_0^{\infty} \frac{x}{e^x-1}=\zeta(2)=\frac{\pi^2}{6}$$

I think I can use geometric series also with the other integral, I've tried to solve it this way and I got:
$$\int_0^{\infty}\frac{xe^x}{e^x-1}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{\left(n-\frac{1}{2}\right)^2} $$

And I don't know how to obtain a value from this infinite sum, although if I'm not wrong it should be $\pi^2/2$.

Anyone knows how to do this? Also, I think I might be looping the loop and there's some easier way to proceed from the beggining, so I'd appreciate another point of view to solve this.

Thanks.

Body:
Let $ n\in\mathbb{N} $, we have : \begin{aligned}\left\vert\sum_{k=1}^{n}{\left(-1\right)^{k-1}\int_{0}^{+\infty}{x\,\mathrm{e}^{-kx}\,\mathrm{d}x}}-\int_{0}^{+\infty}{\frac{x}{\mathrm{e}^{x}+1}\,\mathrm{d}x}\right\vert&=\left\vert\int_{0}^{+\infty}{x\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k-1}\,\mathrm{e}^{-kx}}\,\mathrm{d}x}\right\vert\\ &\leq\int_{0}^{+\infty}{x\left\vert\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k-1}\,\mathrm{e}^{-kx}}\right\vert\mathrm{d}x}\\ &\leq\int_{0}^{+\infty}{x\,\mathrm{e}^{-\left(n+1\right)x}\,\mathrm{d}x}=\frac{\Gamma\left(1\right)}{\left(n+1\right)^{2}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}

Which means that : $$ \sum_{n=1}^{+\infty}{\frac{\left(-1\right)^{n-1}}{n^{2}}}=\lim_{n\to +\infty}{\sum_{k=1}^{n}{\left(-1\right)^{k-1}\int_{0}^{+\infty}{x\,\mathrm{e}^{-kx}\,\mathrm{d}x}}}=\int_{0}^{+\infty}{\frac{x}{\mathrm{e}^{x}+1}\,\mathrm{d}x} $$

Thus : $$ \int_{0}^{+\infty}{\frac{x}{\mathrm{e}^{x}+1}\,\mathrm{d}x}=\eta\left(2\right)=\left(1-\frac{1}{2}\right)\zeta\left(2\right)=\frac{\pi^{2}}{12} $$


问:
Doubt on a question involving Euler's Theorem
I came across a question where I had to apply Euler's theorem on a function U(x,y) which is represented in terms of f(y/x), g(y/x) and another expression in terms of x and y. The question is as follows:

If $$u(x,y) =  \frac{(x^2+y^2)^n}{2n(2n-1)}  + x f(\frac{y}{x})+g(\frac{y}{x}),$$ where f and g are arbitrary functions, prove by using Euler’s theorem that

$$\Bigl(x\frac{∂}{∂x}+y\frac{∂}{∂y}\Bigr)^2u(x,y)=(x^2+y^2)^n.$$

I wrote the first expressions as $$x^{2n}h(\frac{y}{x}),$$
expanded the square term, and tried to get the result, but in vain. I'm not sure if this is the right method and I messed up the calculation.

Any inputs on how to proceed would be really helpful. Kindly refrain from providing the actual answer. Thank you!

Edit: Another input would be that the sum of two homogeneous functions need not be homogeneous. How do we then apply Euler's Theorem?

答:
Writing as an answer instead of a comment because use of pictures. Basically this is just a lot of algebra. Expand your operator as
$$x^2\partial^2_x+xy\partial_x\partial_y+y^2\partial^2_y$$
And take the partial derivatives. Not bothered to do the algebra by hand, I checked using Mathematica, and the result is correct:


问:
Help me understand Euclid's Second Postulate.
HI I'm using Greensberg book to learn Euclidean Geometry (with an aim to solve problems ranked D4-D6 lvl IMO wise)
I was unable to appreciate the postulate as stated in the book, ProofWiki's version of postulate was more intuitive

Since the book is going for general using notion of congruence as seen by the usage of the term in definition of circle of radius OA as set of all points P such that segment OP is congruent to segment OA

So by that logic, I can define CD as segment with any orientation and not necessarily parallel to AB .

So suppose CD is at obtuse slant to AB (>90 deg slope) then by the postulate I can produce segment BE congruent to CD: Then Here come the problem, it is intuitive to visualize B is between A and E even though the term is left undefined by the book

But in case where segment BE is congruent to segment CD , thus BE is obtuse to segment AB, then what does notion of between or betweeness imply in such case?

ProofWiki states : A straight line segment can be extended indefinitely to form a straight line.

But what I interpreted out of postulate as stated in the book(with help of intuition from ProofWiki):

A straight line segment can be extended indefinitely to form a zig-zag straight lines as desired.

Can someone shed light on where are I am going wrong with my interpretation?

答:
ProofWiki uses the "common" English translation of Euclid's Postulate 2:

To produce a finite straight line continuously in a straight line.

This means that the "extension" of the initial segment $AB$ must be done in such a way that the result is again a straight line (segment). Thus, "zig-zag"-ing is avoided.

Is Greenberg's version (page 15) more intuitive? Maybe.

Is it more precise? I think so.

You have to consider previous terms (page 12): "the second undefined term, 'line' is synonymous with 'straight line'" as well as "between" and "congruent".

From page 73 and on you will find the axioms regarding "between". The basic one is:

If point $B$ is between points $A$ and $C$, then $A,B,C$ are three distinct points all lying on the same line [emphasis added].

Thus, when Greenberg states his version of Postulate 2, he asserts that

"there exists a unique point $E$ such that $B$ is between $A$ and $E$"

and this amounts to saying that $A,B,E$ are on a straight line.


问:
Gradient of a scalar function of matrix
I referred to the following definition for gradient in matrices (so related to Frobenius norm in matrices)     Gradient and Hessian of a function with Matrix Variables.

Suppose I have $$g(X) = A(X-X_0)$$ where

$A, \, X, \, X_0 \in \mathbb{R}^{n\times n}$

$X$ is the variable

If $g(X)$ admits a scalar anti-derivative function, i.e., $\nabla_Xf(X)=g(X)$, where $f(X)\in \mathbb{R}$.

My question is what $f(X)$ should look like? (note: $A$ is a matrix, not a scalar)

If my question does not make sense, then why it does not make sense?  Thanks!

答: A sufficient conditions is to be able to write $A=B^T B$ then yes. Recall that $\nabla \frac{1}{2}\|B(x-x_0)\|^2=B^TB(x-x_0)$.

Title: Gradient of a scalar function of matrix


答:
Another way using areas :

$$[\triangle ABD] + [\triangle ADC] = [\triangle ABC]$$

In the usual notation, $BC=a, CA=b, AB=c$,

$$\frac{1}{2}c l_a \sin \frac{\alpha}{2} + \frac{1}{2}b l_a \sin \frac{\alpha}{2} = \frac{1}{2}b c \sin \alpha$$
$$\Rightarrow l_a=\frac{bc}{b+c}\frac{\sin \alpha}{\sin (\alpha/2)}$$
$$\therefore \boxed{l_a=\frac{2bc}{b+c}\cos (\alpha/2)}$$

Similar formulae for $l_b$, $l_c$.

Body:
Find the angle bisectors of a triangle with side $a$ and adjacent angles $\beta$ and $\gamma$.

I am trying to solve the problem using the law of sines, because we have just studied it, but I can't seem to find a solution. Thank you in advance!


问:
Rewriting $(e^b-1)(\frac{b/n}{e^{b/n}-1})$ and $(e^b-1)(e^{b/n})(\frac{b/n}{e^{b/n}-1})$ so that I can take the infimum and supremum
So I made an underapproximation and overapproximation for a sum and got that the underapprox is $(e^b-1)(\frac{b/n}{e^{b/n}-1})$ and the overapprox is $(e^b-1)(e^{b/n})(\frac{b/n}{e^{b/n}-1})$.
Now I want to rewrite those so that I can take the supremum of the underapprox and infimum of the overapprox. So for the underapprox I want to find something that is less than or equal to $(e^b-1)(\frac{b/n}{e^{b/n}-1})$ and for example we know that exp(x) $\ge$ 1+x but does that help me here? It would become $(b)(\frac{b/n}{e^{b/n}-1})$ I guess because I don't want to use it on the exp term in the denominator.

Ohh, and we also know that exp(x) $\le$ $\frac{1}{1-x}$

答:
As $n\to\infty$ we have

$(e^b-1)\left(\frac{b/n}{e^{b/n}-1}\right)< e^b-1$

and

$(e^b-1)(e^{b/n})\left(\frac{b/n}{e^{b/n}-1}\right)>e^b-1$

because $$\underset{x\to 0}{\text{lim}}\;\frac{x}{e^x-1}=1$$

Actually the sum is $$\int_0^b e^x \, dx=e^b-1$$


问:
Proving $abcd+3\geq a+b+c+d$

If $a,b,c,d$ are non negatives and $a^2+b^2+c^2+d^2=3$ prove that  $$abcd+3\ge a+b+c+d$$

The inequality is not as simple as it looks.The interesting part is that the equality occurs when $a=0,b=c=d=1$ upto permutation .(I don't know if there are further equality cases.)

I tried rewriting the inequality as $$a^2+a(bcd-1)+b^2+c^2+d^2-b-c-d\ge 0$$ As its a quadratic in $a$ it suffices to show  the discriminant$$\Delta_a={(bcd-1)}^2-4(b^2+c^2+d^2-b-c-d)\le 0$$ which unfortunately is wrong (when $a=\sqrt{3},b=c=d=0$)

P.S ;I am not aware of using Lagranges multipliers

Its from here

Update :An answer has been posted in the link above (AOPS) similar to Dr Mathva's answer

答:
Due to symmetry, assume wlog that $a\geqslant b\geqslant c\geqslant d\geqslant 0$. Since I will use this fact later, I am proving it before the cases:

Lemma: We have that $2\geqslant a+d$.

In fact, this follows from Cauchy-Schwarz:  $$2= 2\sqrt{\frac{a^2+b^2+c^2+d^2}{3}}\geqslant 2\sqrt{\frac{a^2+3d^2}{3}}=\sqrt{1+\frac13}\cdot \sqrt{a^2+3d^2}\geqslant a+d$$

We will now consider some cases depending on the position of the number $1$:

Case 1: $1\geqslant a\geqslant b\geqslant c\geqslant d\geqslant 0$. Observe that one may rewrite the inequality as $$3+abcd-(a+b+c+d)=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd) $$ Which is clearly positive.

Case 2: $a\geqslant 1\geqslant b\geqslant c\geqslant d\geqslant 0$. If $ab\geqslant 1$, we might proceed as in the next case. So we will work with $ab<1$. Thus \begin{align*}3+abcd-(a+b+c+d)&=(1-a)(1-b)+(1-c)(1-d)+(1-ab)(1-cd)\\&\geqslant \underbrace{(1-a)}_{<0}(1-b)+(1-c)(1-d)\\&\geqslant (1-a)(1-c)+(1-c)(1-d)\\&=(1-c)(2-(a+d))\geqslant 0  \end{align*} Where the last inequality follows from the lemma.

Case 3: $a\geqslant b\geqslant 1\geqslant c\geqslant d\geqslant 0$. This implies that \begin{align*}6+2abcd-2(a+b+c+d)&=a^2+b^2+c^2+d^2+3+2abcd-2(a+b+c+d)\\
&=(a-1)^2+(b-1)^2+(c+d-1)^2+2cd(ab-1)\geqslant 0\end{align*} Or, equivalently $3+abcd\geqslant a+b+c+d$.

Case 4: $a\geqslant b\geqslant c\geqslant 1\geqslant d\geqslant 0$. As @dezdichado noticed, this case is straightforward, since it forces directly $a=b=c=1, d=0$ due to the constraint $a^2+b^2+c^2+d^2=3$.

Case 5: $a\geqslant b\geqslant c\geqslant d\geqslant 1$. Clearly impossible, since this would yield $a^2+b^2+c^2+d^2\geqslant 4>3$.

Done!

Title: Proving $abcd+3\geq a+b+c+d$


问:
Automorphism of linear algebraic group $G$ preserves finite dimensional subspace of $k[G]$
I'm struggling to understand a claim in Springers "linear algebraic groups".

Suppose $G$ is a linear algebraic group and $\sigma$ an automorphism of $G$ as an algebraic group. How can I show the dual action of $\sigma$ on $k[G]$ stabilizes a finite dimensional subspace of generators?

答:
You can prove it using the following general fact (also written in Springer's book somewhere) :

An algebraic representation $G \times V \to V$ is the same as a comodule map $a : V \to V \otimes k[G]$ (where $k[G]$ is the ring of regular functions on $G$).

Now, if $v \in V$, we have $a(v) = \sum v_i \otimes f_i$ where each $f_i$ is a polynomial in $k[G]$. It follows from the comodule axiom that the subspace generated by the $v_i$ is $G$-stable and finite-dimensional. So actually $V$ is the union of finite-dimensional $G$-stable modules.


问:
Does every ring have a subring that is a field?
Well, that question arose at the very moment I saw two examples. The first one was about an integrity domain which has a subring that is a field (I don't remember the specific example) and the second one is:

Let $M = M_2(\mathbb{R})$ be the set of all $2 \times 2$ matrixes with entries in $\mathbb{R}$ and $1_M = \left[\begin{array}[cc] &1 & 0\\ 0 & 1\end{array}\right] $. The set $A = \left\{a \in \mathbb{R};\left[\begin{array}[cc] &a & 0\\ 0 & 0\end{array}\right]\right\} \subset M$ is a subring of $M$ with unity $1_A = \left[\begin{array}[cc] &1 & 0\\ 0 & 0\end{array}\right] \neq \left[\begin{array}[cc] &1 & 0\\ 0 & 1\end{array}\right] = 1_M$. Furthermore, since $A$ is isomorphic to $\mathbb{R}$, $A$ is a field.

So, there can be subrings with different unities than the ring they are subsets to. The question is, if a ring does not have unity and none of the other properties, there can be a subring that has unity?

Or if it is possible, can someone give me an example of a ring without unity that has a subring with unity?

And an example of a ring with no additional properties that has a subring that is a field?

Thanks in advance!

答: Let $R = \mathbb R \times 2\mathbb Z$.  Since $2\mathbb Z = \{ 2n : n \in \mathbb Z \}$ does not have a multiplicative identity, neither does $R$.  However, $R$ contains $\{ (x,0) : x \in \mathbb R\}$ as a subring which is a field, isomorphic to $\mathbb R$.


问:
Integrate $\int _0^1\frac{1}{\left(9+\:x^2\right)^{0.5}}\:dx$ using integration by parts and substitution.

Integrate:
$$\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx$$

Here's my attempt, using integration by parts in the first step:

\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\int \:-\frac{x^2}{\left(9+x^2\right)^{1.5}}dx\right]^1_0 \\
&= \left[\frac{x}{\left(9+x^2\right)^{\frac{1}{2}}}-\left(\frac{x}{\sqrt{9+x^2}}-\ln \left(\frac{1}{3} \left|x+\sqrt{9+x^2}\right|\right)\right)\right]^1_0 \\
&= \left[\ln \left(\frac{1}{3}\left|x+\sqrt{x^2+9}\right|\right)\right]^1_0 \\
&= \ln \left(\frac{1+\sqrt{10}}{3}\right)
\end{align}

I tried to solve this using substitution and I keep getting stuck at the same place. For the sake of understanding, I turned the definite integral into an indefinite one by remove the upper and lower bounds, since that's quite easy to do. I'm more concerned with the final result without the actual values. Here's my attempt:

\begin{align}
\int _0^1\frac{1}{(9+x^2)^{0.5}}\:dx &\stackrel{{u = \frac{x}{3}}, dx = 3\,du} = \int\frac{3}{\sqrt{9u^2+9}}\,\mathrm{d}u \\
&= \int \dfrac{1}{\sqrt{u^2+1}}\,\mathrm{d}u \\
&\stackrel{{u=\tan(v)} \to \mathrm{d}u=\sec^2(v)\,\mathrm{d}v} = \int \frac{\sec^2(v)}{\sqrt{\tan^2(v)+1}}\,\mathrm{d}v \\
&= \int \sec(v)\,\mathrm{d}v \\
&= \int \frac{\sec(v) (\tan(v)+\sec(v))}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \int \frac{\sec(v) \tan(v)+\sec^2(v)}{\tan(v)+\sec(v)}\,\mathrm{d}v \\
&= \cdots
\end{align}

Here is where I am stuck. I think that there should be another substitution here in order to get the final integral that can be solved, but I'm not sure which term to choose. I think it would have to be $w = w=\tan\left(v\right)+\sec\left(v\right)$, but I'm not $100 \%$ sure.

This really caught me off by surprise because it looks rather simple but boy is it tricky to do.

答:
$$I=\int_0^1\frac{dx}{\sqrt{9+x^2}}=\frac13\int_0^1\frac{dx}{\sqrt{1+(x/3)^2}}$$
now we know that $\cosh^2\theta-\sinh^2\theta=1\Rightarrow 1+\sinh^2\theta=\cosh^2\theta$ so we can make the substitution $x/3=\sinh u\Rightarrow dx=3\cosh u\,du$ so lets do that:
$$I=\frac13\int\limits_{\sinh^{-1}(0)}^{\sinh^{-1}(1/3)}\frac{3\cosh u}{\sqrt{1+\sinh^2u}}du=\int\limits_{\sinh^{-1}(0)}^{\sinh^{-1}(1/3)}du=\sinh^{-1}\left(\frac13\right)$$

using the fact that:
$$\sinh^{-1}(x)=\ln\left(x+\sqrt{x^2+1}\right)$$
we can say that:
$$I=\ln\left(\frac13+\sqrt{\frac19+1}\right)=\ln\left(\frac{1+\sqrt{10}}{3}\right)$$


问:
Is it possible for a function to be analytic anywhere outside the circle of convergence of its power series expansion?
Is it possible for a function to be analytic anywhere outside the circle of convergence of its power series expansion? I'm referring to analytic fuctions of course (i.e. those with power series expansions but not necessarily analytic over the entire complex plane).</p>

答:
If I understand you correctly, any meromorphic but not entire function satisfies your requirement.

You can also just stitch together different functions arbitrarily to make this true.


问:
Proving a relation between the ordinates of point of contact of a line on a circle
If the tangents at $(h,k)$ to the ellipse $x²/a² + y²/b² = 1$ , cuts the auxillary circle in the point whose ordinates are $p$ and $q$ , then show that $1/p + 1/q = 2/k$ .

I found this question while I was solving questions on ellipses.

These may help in understanding the question better:

Tangent : The line which touches a curve only at one point is called a tangent.

Auxillary circle : the circle passing through the vertices on the major axis of an ellipse and whose diameter is equal to major axis of the ellipse is called its auxillary circle.

Now, I did not understand how to proceed. I just tried to write the equation of the tangent at $(h,k)$ and then solve it with the equation of the circle. But I don't understand why I am only supposed to prove it using mathematical formulae and equations. I want to know whether there is a good way to prove it using geometry.

Use of trigonometry, differentiations, integrations are allowed wherever necessary.

答:
If the line through a point $C$ of an ellipse, perpendicular to the major axis, meets the auxiliary circle at $Q$, then the tangent to the ellipse at $C$ and the tangent to the circle at $Q$ meet the major axis at the same point $P$. This nice theorem is obvious if you think of the auxiliary circle as the ellipse stretched perpendicularly to the major axis, and allows to rephrase what you want to prove in a different way, with no connection to the ellipse:

Let $AB$ be a chord of a half-circle with diameter $RS$, line $AB$
meeting line $RS$ at $P$, and let line $PQ$ be tangent to the
half-circle at $Q$.
If $A'$, $B'$, $C'$ are the perpendicular projections of $A$, $B$, $Q$
on $RS$, and $C$ is the intersection between $QC'$ and $AB$, then:
$${1\over AA'}+{1\over BB'}={2\over CC'}.$$

To prove that, note first of all that we can rewrite the above equality as
$$
\tag{1}
{CC'\over AA'}+{CC'\over BB'}=2.
$$

On the other hand, if $O$ is the center of the circle, from similarity inside right triangle $PQO$ and power of point $P$ we have:
$$
\tag{2}
PQ^2=PC'\cdot PO=PA\cdot PB.
$$
But
$$
{AA'\over PA}={BB'\over PB}=\sin\alpha
\quad\text{and}\quad
{CC'\over PC'}=\tan\alpha,
$$
where $\alpha=\angle APR$ (see figure).
Hence equation $(2)$ can be rewritten as
$$
CC'\cdot PO={AA'\cdot BB'\over\sin\alpha\cos\alpha}
$$
whence we obtain, if $M$ is the midpoint of $AB$ and $M'$ its projection on $RS$:
$$
{CC'\over AA'}+{CC'\over BB'}=
{BB'+ AA'\over PO\sin\alpha\cos\alpha}=
{BB'+ AA'\over MM'}=2,
$$
as it was to be proved.


问:
2nd ODE Question Help
$$
\frac{d^2u}{dy^2} + \frac{V}{\rho} \frac{du}{dy} = 0
$$
conditions $u = 0$ at $y = 1$ and $u = U_0$ at $y= 0$

$V,\rho$ are constants

find solution for $u(y)$

My final answer was

$$u(y)=\frac{U_0}{1-\exp(-V/\rho)}-\frac{U_0}{(1-\exp(V/\rho))\exp(V/\rho)}$$

would anyone be able to clarify if i did this correctly please

答:
$$\frac{d^2u}{dy^2} + \frac{V}{\rho} \frac{du}{dy} = 0$$
$$\frac{du}{dy} + \frac{V}{\rho}u = C_1$$
$$(ue^{yV/{ \rho}})'=C_1e^{yV/{\rho}}$$
$$ue^{yV/{ \rho}}=Ce^{yV/{\rho}}+C_2$$
$$u(y)=C+C_2e^{-yV/{ \rho}}$$
Apply initial conditions. I think that you can simplify a little bit your answer with the exponential.

Title: 2nd ODE Question Help


问:
what are all the continuous functions $f(x)$ that is has a domain $x>0$ and $f(ab)=\frac{f(a^2)+f(b^2)}{2}$
what are all the continuous functions $f(x)$ that is has a domain $x>0$ and $f(ab)=\frac{f(a^2)+f(b^2)}{2}$.

this question was changed because the answer to the original problem what are all the continuous functions $f(x)$ that is has a domain $x>1$ and $f(ab)=f(a^2)+f(b^2)$ was trivial and User$8128$ pointed out

I think it would slightly more interesting to consider $f(ab) = \frac{f(a^2) +f(b^2)}{2}$ That way, there is a formal similarity to the inequality $ab \le \frac{a^2 + b^2}2$

答:
It's $\log(x)$ it follows the same structure as logs

$\log(ab)=\log(a^2)/2+\log(b^2)/2$

$\log(a)/2=\log(a^{0.5})$

$\log(a^2)/2=\log(a)$

$\log(ab)=\log(a)+\log(b)$

definition

$\log$ is in any base


问:
Aperiodic Markov Chain without self loops
If we have an irreducible Markov Chain in which any state has a self loop, it becomes aperiodic. This fact  made sense to me, but it got me wondering about two things:

Is it possible for an irreducible Markov Chain to exist which is aperiodic but does NOT have a self loop?

Is it possible to have a non irreducible Markov Chain which is aperiodic? (I'm leaning towards a yes on this one). In particular, a non irreducible Markov Chain without any self loops which is aperiodic.

It would be great if someone could intuitively or with examples explain the above queries.

TIA!

答:
Both of these are possible.

First, suppose we have a Markov chain with three states $A$, $B$, $C$ and a transition from any state to any other state (no loops). From state $A$ we can return back to $A$ in $2$ steps ($A \to B \to A$) or in $3$ steps ($A \to B \to C \to A$) and these have GCD $1$, so state $A$ is aperiodic; the same argument applies to other states.

Now take two copies of this Markov chain: states $A, B, C$ with transitions between any two of them, and three more states $A', B', C'$ with transitions between any two of them. This is still aperiodic for all the same reasons, but because we can't get from $\{A,B,C\}$ to $\{A', B', C'\}$, it's not irreducible.


问:
Cumulative distribution function (CDF) and distribution of a random variable
I would like to show that, given two real random variables $X$ and $Y$ over the probability space $(\Omega,\mathcal{A},P)$, and with the same CDF (namely $F_X=F_Y$), then they have the same distribution, namely $\mu_X(A):=P(X \in A)=P(Y \in A)=:\mu_Y(A)$ for all $A \in \mathcal{B}(\mathbb{R})$, where $\mathcal{B}(\mathbb{R})$ is the Borel sigma algebra over $\mathbb{R}$.

I'm pretty stuck from the start. Any hint would be appreciated! Thank you!

$ $

Post Scriptum:

I know that I have to put in my efforts, but I'm having big problems since the start.

The fact that $F_X=F_Y$ is obviously equivalent to the fact that $\forall \,x \in \mathbb{R} \quad \mu_X((-\infty,x])=\mu_Y((-\infty,x])$.

I'm learning this by myself, so maybe this is a deep result that is too hard to prove for me. I really don't know.

Again, I've searched a lot online but I didn't find any valid proof of this.

答:
If $F_X = F_Y$, i.e. $F_X(z) = F_Y(z)$ for all $z\in\mathbb R$, we have that $\mu_X(I_z) = \mu_Y(I_z)$ for all $z\in\mathbb R$, where $I_z = (-\infty,z]$.

The set $\mathcal I = \{I_z:z\in\mathbb R\}$ generates the Borel $\sigma$-algebra, i,e, $\sigma(\mathcal I) = \mathcal B(\mathbb R)$, and the set $\mathcal I$ is $\pi$-stable (i.e. $I,J\in\mathcal I$ implies that $I\cap J\in\mathcal I$).

The set $\mathcal D = \{A\in\mathcal B(\mathbb R) : \mu_X(A) = \mu_Y(A)\}$ is a $\lambda$-system (https://en.wikipedia.org/wiki/Dynkin_system), and it obviously holds that $\mathcal D\subseteq\mathcal B(\mathbb R) = \sigma(\mathcal I)$. By hypothesis, $\mathcal I\subseteq\mathcal D$ also. From the $\pi$-$\lambda$-theorem then follows that $\sigma(\mathcal I) \subseteq \mathcal D$. Thus, $\mathcal D = \mathcal B(\mathbb R)$, which means $\mu_X = \mu_Y$ for all $A\in\mathcal B(\mathbb R)$.


问:
Given two points in three dimensions, how can one find the direction of motion and angle of descent of a projectile?
Say you shoot a basketball and you have the data for the ball's x, y, and z coordinates (where z is height and peaks at around z=16) through its time in the air. That's the data that I have. I want to take the ball's x,y,z coordinates when a certain z is reached on its descent, and compute the direction of motion (in terms of x and y) and angle of descent (at what angle is the ball falling at this point? like how steep is the fall) using data from that point and the previous point.

In other words, I have two points taken from the descent of a projectile: $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$, and I want to compute the descent angle and the direction of motion.

I found the following formula for descent angle: $\theta=cos^{-1}(\frac{(x_1x_2+y_1y_2+z_1z_2}{\sqrt{x_1+y_1+z_1}\sqrt{x_1+y_2+z_1}})$. However, I'm getting outputs of less than 1 degree while I'd expect the descent to be pretty steep so I'm not sure if this formula is appropriate for this problem.

For direction of motion, I'm thinking maybe I should just calculate the slope of the points $(x_1,y_1)$ and $(x_2,y_2)$ such that $\text{dir}=\frac{y_2-y_1}{x_2-x_1}$, but again I'm not 100% sure, especially because the answer would not be in degrees.

I also have data for time at these points if that is needed.

答:
In an interval of time $\Delta t$ the ball will fall $\frac 12g\Delta t^2$ due to gravity.  If this is enough to matter to you, add this to the final $z$ measurement to get the direction of travel at the beginning of the interval.

Now we can construct a coordinate system where there is no motion along $y$.  $x$ is along the horizontal direction of travel.  $z$ is vertical with positive up.  Our measurements are $(x_1,z_1)$ and $(x_2,z_2)$.  You can convert your measurements by putting in $\sqrt {(x_2-x_1)^2+(y^2-y_1)^2}$ for our $x_2-x_1$.

The ball now travels $x_2-x_1$ horizontally and $z_2-z_1$ vertically.  The angle of travel is then $\arctan \frac{z_2-z_1}{x_2-x_1}$.  This is equivalent to the expression you are using.  Note that the output is in radians.  If you want degrees, you need to multiply by $\frac{180}{\pi}$


问:
The direct sum $\oplus$ versus the cartesian product $\times$
In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines $A^n := A\times A\times\ldots\times A$; however in, for example, the fundamental theorem of finitely generated abelian groups, we normally write that every such group is isomorphic to one of the form
$$
\mathbb{Z}^n \oplus \mathbb{Z}_{r_1} \oplus \cdots \oplus \mathbb{Z}_{r_t}
$$
where $\mathbb{Z}^n$ now means $\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$.

From an intuition perspective, and in the sense of sets, is this more or less the same as $\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}$? (Bear in mind I am normally using these ideas in relation to homology groups.)

答:
For a finite number of factors, the direct sum and direct product of abelian groups (and more generally, of $R$-modules) are equal.

However, when you have an infinite number of summands/factors, the two constructions are different. Explicitly, the direct product
$$\prod_{i\in I} A_i$$
is the collection of all functions $f\colon I\to \cup A_i$ with $f(i)\in A_i$ for each $i$ (you can think of them as the "tuples" indexed by $i$, with the $i$th coordinate being the value $f(i)$).

But the direct sum
$$\bigoplus_{i\in I} A_i$$
is the collection of all functions $f\colon I\to \cup A_i$ with:

$f(i)\in A_i$ for each $i$; and

$f(i) = 0$ for all except perhaps a finite number of $i$.

That is, the direct sum is the subgroup/submodule of the direct product that consists of the almost-null elements. When there are only finitely many coordinates, saying "all are zero except perhaps for a finite number of coordinates" is the same as saying nothing.

For categories where the coproduct is not the direct sum (for example, when dealing with not-necessarily-abelian groups), it used to be common to refer to the direct product as the "cartesian product", the "unrestricted direct product", or even the "complete direct product" or "complete direct sum" (e.g., Hungerford offers the latter as a parenthetical alternative in page 59); and to refer to the subgroup of almost-null elements as the "restricted direct product" or "weak direct product" (again, the latter is used in Hungerford).

In the category of abelian groups, the direct (cartesian) product is a product, in the categorical sense: $P=\prod\limits_{i\in I}A_i$ is a group, equipped with homomorphism $\pi_i\colon P\to A_i$ for each $i$ (the projections), with the universal property that for every abelian group $B$ with homomorphism $f_i\colon B\to A_i$ for each $i\in I$, there exists a unique $f\colon B\to P$ such that $f_i =\pi_i\circ f$ for each $i\in I$; the map $f$ is defined by letting $f(b)(i) = f_i(b)$ (remember that the elements of $P$ are maps $g\colon I\to\cup A_i$ with $g(i)\in A_i$ for each $i$; that's why we have $f(b)(i)$: $f(b)$ is a map from $I$ to $\cup A_i$, so $f(b)(i)$ is the map $f(b)$ evaluated at $i$).

The direct sum, on the other hand, is a coproduct in the categorical sense. $C=\oplus_{i\in I}A_i$ is a group, equipped with homomorphism $\iota_j\colon A_j\to C$ for each $j\in I$ (the map that sends $a\in A_j$ to the element that has $a$ in the $j$th coordinate and $0$s elsewhere), with the universal property that for every abelian group $B$ with homomorphism $g_i\colon A_i\to B$, there exists a unique homomorphism $g\colon C\to B$ such that $g_i = g\circ \iota_i$ for each $i$.

Because the set of homomorphisms from one abelian group to another form an abelian group (under pointwise addition) and composition is bilinear under this addition, then one can prove that any object which is a product for $A$ and $B$ is also a coproduct for $A$ and $B$ and conversely, any object which is a coproduct is also a product (Theorem 2 in Section VIII.2 of Categories for the Working Mathematician, by Saunders Mac Lane); in particular, finite products coincide with finite coproducts in the category of all abelian groups, which is why the direct product and direct sum agree when there are only finite many factors/summands.


问:
Integration of volumes of revolution: bisector surface

$y = (16-x^2)^{0.5}$ is rotated around the $x$-axis to give a sphere of radius $4$ units. Find the equation of the straight line that passes through $(-4,0)$, such that when also rotated around the $x$-axis, will create a surface that bisects the volume of the sphere.

The equation will obviously take the form $y=kx+4k$. I tried to solve for the volume bounded by the revolutions of the semicircle and the straight line being equal to exactly half the volume of the sphere. The result was a very complicated series of substitutions and expansions which yielded no valid solution (it was a $6^{\text{th}}$ degree polynomial with only $1$ positive solution of $2.27$ for $k$, which is quite obviously wrong).

My question is whether there is a simpler, perhaps more ingenious way of finding $k$.

Again, thanks in advance and all contributions welcome!

Body:
The volume of your sphere is $\,\displaystyle{\frac43\pi 4^3=\frac{256\pi}{3}}\,$ . You need a line $\,y=kx+4k\,$ ,  which intersects $\,y=\pm\sqrt{16-x^2}\,$ at

$$16-x^2=k^2x^2+8k^2x+16k^2\implies(k^2+1)x^2+8k^2x+16(k^2-1)=0\implies$$

$$\Delta=64k^4-64(k^4-1)=8^2\implies$$

$$x_{1,2}=\frac{-8k^2\pm8}{2(k^2+1)}=\begin{cases}\;\;\;\;\;\;-4\\{}\\-\frac{4(k^2-1)}{k^2+1}\end{cases}$$

Thus, we want

$$\pi\int\limits_{-4}^{-\frac{4(k^2-1)}{k^2+1}}(kx+4k)^2x\,dx=\left.\frac\pi{3k}(kx+4k)^3\right|_{-4}^{-\frac{4(k^2-1)}{k^2+1}}=\frac\pi{3k}\frac{8^3k^3}{(k^2+1)^3}=\frac{512k^2\pi}{3(k^2+1)^3}$$

And now we want

$$\frac{2^9k^2\pi}{3(k^2+1)^3}=\frac{2^7\pi}{3}\iff4k^2=(k^2+1)^3$$

and according to WA the resulting sixtic equation doesn't even have any real roots...


问:
Channel Decoding: Gaussian Channel as Time-Varying Binary Symmetric
I am reading Information Theory, Inference and Learning Algorithms by David MacKay. It is available free of charge online: http://www.inference.org.uk/mackay/itila/book.html (official site, not pirated copy).

My question is to solve Exercise 25.1. It is marked as one of the easier ones. I think I'm just missing some fundamental understanding.
Assistance would be enormously appreciated!

I quote the desired section, slightly paraphrased for abbreviation.
It is from Section 25.1 on Page 324 (Page 336 of my pdf).

A codeword $t$ is selected from a code with codewords which form a subset of $\{0,1\}^N$ and transmitted over a noisy channel and $y$ is received.
We have
$$ P(t \mid y) = P(y \mid t) P(t) / P(y) $$
by Bayes's theorem.
Assume that he channel is memoryless.
Thus
$$ P(y \mid t) = \prod_{n=1}^N P(y_n \mid t_n). $$

For example, if the channel is a Gaussian channel with transmissions $\pm x$ and additive noise of standard deviation $\sigma$, then the probability density of the received signal $y_n$ in the two cases $t_n \in \{0,1\}$ are given as follows:
$$
\begin{aligned}
P(y_n \mid t_n = 1) &= (2 \pi \sigma^2)^{-1/2} \exp\bigl( - (y_n - x)^2 / (2 \sigma^2) \bigr);
\\
P(y_n \mid t_n = 0) &= (2 \pi \sigma^2)^{-1/2} \exp\bigl( - (y_n + x)^2 / (2 \sigma^2) \bigr).
\end{aligned}
$$
[Here we are interpreting $t_n = 0$ as "send $-x$" and $t_n = 1$ as "send $+x$" (plus noise).]
From the point of view of decoding, all that matters is the likelihood ratio:
$$
\frac
{P(y_n \mid t_n = 1)}
{P(y_n \mid t_n = 0)}
=
\exp\biggl( \frac{2 x y_n}{\sigma^2} \biggr).
$$

Exercise. Show that from the point of view of decoding a Gaussian channel is equivalent to a time-varying binary symmetric channel with a known noise level $f_n$ which depends on $n$.

For further reference, Section 11.2 (Page 179 of book and 191 of my pdf) discusses inferring the input from a Gaussian channel. It's not so helpful for this, though.
I do not see what is time-varying about this.

PS Multiple different SE networks seemed appropriate for this: maths, cstheory, stats.
I picked this one but if a mod wants to move it, no objection from me.

答:
Arash presents one possibility. Here is another. It gives a completely different conclusion. MacKay's text appears ambiguous to which is intended.

Although it looks like a variable, here $y = (y_n)_{n \in [N]}$ is a single given realisation---the expression
$$
\frac
{P(y_n \mid t_n = 1)}
{P(y_n \mid t_n = 0)}
=
\exp\biggl( \frac{2 x y_n}{\sigma^2} \biggr)
$$
is not to be thought of as expressing a function of $y_n$.
Rather, let $q_n(b) := P(y_n \mid t_n = b)$---this is a function of the bit $b \in \{0,1\}$.
The likelihood ratio is saying that
$$
\frac{q_n(1)}{q_n(0)} = \exp( 2 x y_n / \sigma^2 ).
$$
Thus it is a binary symmetric channel with time-varying flip-probability.
Converting this into the flip parameter $f_n$ requires knowledge of the prior.
For example, if the prior is uniform, then I think
$$
\frac{1 - f_n}{f_n} = \frac{q_n(1)}{q_n(0)},
\quad\text{ie}\quad
f_n = \frac1{1 + q_n(1)/q_n(0)}.
$$
I could be mistaken with this final claim, however.


问:
Prove a graph is cyclic
I encountered the following problem:

Let $G(V, E)$ be a finite undirected graph with $2n$ vertices, in which every connected component has an even number of vertices. Given that exactly $n$ of the vertices of $G$ have a degree at least $3$, prove that $G$ is cyclic.

My solution to the problem goes as follows:

Let $C$ denote the set of all connected components $G_i(V_i, E_i)$ of $G$ and $|C| = k$.

Let's assume that $G$ is acyclic. That means that every $G_i$ is a tree. Then, by adding exactly $k-1$ edges to $G$, we can connect all connected components in such a way (root of one to another one's leaf, ordering them in a sequence that way) that we obtain a tree $G'(V', E')$, where $V'= V$ and $|E'| = |E| + k - 1$.

But since $G'$ is a tree, that means that $$|E'| = |V'| - 1 = |V| - 1 = 2n - 1$$

By the handshake lemma we also know that $$2|E'| = \sum_{v\in V'}\deg v \ge \underbrace{3n}_{\substack{\text{atleast $n$}\\ \text{with degree $\ge$ 3}}} + \underbrace{n}_{\substack{\text{G' is connected}\\\text{$\Rightarrow$ the remaining $n$ have $\deg\ge 1$}}}$$

Thus we obtain that $$2(2n-1) \ge 4n \Leftrightarrow -1 \ge 0$$
which is a contradiction, hence $G$ can't be acyclic and contains a cycle.

Now, what bothered me about that is that I never used the given fact, that every $G_i$ had an even number of vertices, so I felt like I was missing something huge or my solution is flawed in a fundamnetal way... and I really banged my head against the wall hard enough, but was nonetheless unable to figure out what was wrong. So my question would be: am I missing something or is the mentioned statement in the problem simply redundant? And if it is redundand, how would one go about proving it? I really got stuck somewhere about there.

Many thanks in advance!

答:
It seems even a little simpler than that, once we can show:

Lemma: Any tree has more leaf nodes than degree-3 and higher nodes.

Proof: Let a graph $G$ be a tree on $n$ vertices with $\ell$ leaf nodes and $h$ vertices of degree $3$ or more. Then $G$ has $n{-}1$ edges and by handshaking we know:

$2(n-1) = \sum d(v_i) \;  \geq \; \ell + 2(n-\ell-h) + 3h =  2n-\ell+h,\quad$ so $\ell-2\geq h$ as required.

So the fact that half the nodes are degree $3$ or more shows that not all components are trees; i.e. that $G$ contains cycles. We could join components into a tree but it isn't really needed.

(thanks to jlammy for a compact statement of the lemma proof)

Title: Prove a graph is cyclic


问:
Minimum Perimeter of Pentagon given area
A pentagon is made by mounting an isosceles triangle on top of a rectangle. What dimensions minimize perimeter $P$ for a given area $K$. This was asked in a test today.

Let $x$ and $y$ denote the sides of the rectangle, and let $z$ be the height of the triangle mounted on the side with length $y$.

The perimeter is $P=2x+y+2\sqrt{0.25y^2 + z^2}$ while the area is $K = xy + 0.5yz$.

My attempt at solving it was equating dP/DK to zero. The resulting expression was extremely messy and very hard to rewrite. Is there a better, more elegant way to solve this than by forcing the answer through repeated application of chain rule.

Any answer would be appriciated.

答:
The optimization problem is:
$$\min_{x,y,z} \left\{ 2x+y+2\sqrt{0.25y^2 + z^2} : xy + 0.5yz = K\right\}$$
The Lagrangian is:
$$L(x,y,z,\lambda) =2x+y+2\sqrt{0.25y^2 + z^2} + \lambda(xy+0.5yz-K)$$
The first order conditions are:
$$2+\lambda y =0 \quad (1)$$
$$1+\frac{0.5y}{\sqrt{0.25y^2+z^2}}+\lambda x + 0.5 \lambda z = 0 \quad (2)$$
$$\frac{2z}{\sqrt{0.25y^2+z^2}}+0.5 \lambda y = 0 \quad (3)$$
You can substitute (1) into (3) to get $z^2 = 4x^2 - 0.25y^2$. Substitute that into (2):
$$1+\frac{0.5y}{2x}+\lambda x + 0.5 \lambda \sqrt{4x^2 - 0.25y^2} = 0 \quad$$


问:
Symmetry of Plancherel measure (for $S_n$)
For each $n \geq 1$ consider the reverse lexicographical order on the set $P(n)$ of partitions of $n$. Example for $n=7$:

$$
\begin{pmatrix}
  \hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
  \hline
  7 &   6 &   5 &   5 &   4 &   4 &   4 &   3 &   3 &   3 &   3 &   2 &   2 &   2 &   1 \\
    0 &   1 &   2 &   1 &   3 &   2 &   1 &   3 &   2 &   2 &   1 &   2 &   2 &   1 &   1 \\
    0 &   0 &   0 &   1 &   0 &   1 &   1 &   1 &   2 &   1 &   1 &   2 &   1 &   1 &   1 \\
    0 &   0 &   0 &   0 &   0 &   0 &   1 &   0 &   0 &   1 &   1 &   1 &   1 &   1 &   1 \\
    0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   1 &   0 &   1 &   1 &   1 \\
    0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   1 &   1 \\
    0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   0 &   1 \\
   \hline
\end{pmatrix}
$$

Thus a partition of $n$, equivalently a Young diagram, is a column $j$ of the table above. For each $j$ let $Y_n(j)$ be the number of standard Young tableaux of shape $j$. Then, as shown by the figures below, the sequence $Y_n(1), \ldots, Y_{n}(T(n))$ is not symmetric when $n \geq 6$. Equivalently the Plancherel probability measure on $P(n)$ is not symmetric for this ordering.

However it seems that symmetry occurs for another ordering of $P(n)$. Is it true and if so what is the order making symmetry ?

EDIT

I think I'm wrong: for $n=8$ there's no possible symmetry (unless my graph is wrong). I'm going to delete this question soon.

EDIT 2

Sorry I have not deleted this post yet because I have in mind a variant of the question but I don't have the time to develop it now (shortly, I am under the impression that the bar chart is "almost symmetric" for $n \geq 8$.)

答:
Your question is related to the fact that the lexicographic ordering of partitions is not reversed by taking the transpose of all partitions. (Taking transpose does not affect the number of standard Young tableaux.) The important partial odering by dominance does have the property that transposition is an anti-isomorphism.

If you want a total ordering that is reversed by transposition, then your best bet is to extend the dominance ordering so as to preserve the anti-isomorphism property.
This means breaking ties for the incomparable pairs in the dominance ordering. The smallest such pairs are for $n=6$, namely $([4,1,1],[3,3])$ and the transposed pair $([3,1,1,1],[2,2,2])$. Since these pairs are separated in the dominance ordering by the partition $[3,2,1]$, an extension is possible. But incomparable pairs get more numerous quickly as $n$ grows. For $n=7$ one can still arrange things symmetrically around the unique self-dual partition $[4,1,1,1]$, but for $n=8$ there are two (necessarily incomparable) self-dual partitions $[4,2,1,1]$ and $[3,3,2]$. They have different numbers of tableaux ($90$ and $42$, respectively) which numbers then have odd multiplicity in the list of the $Y_n(j)$ (in fact both are unique), which means that for a symmetric arrangement both these numbers need to be at the central position in the list, which obviously that cannot.

This type of problem rapidly gets worse as $n$ increases.


问:
Trajectory of stationary stochastic process not in $\mathrm{L}^2$
Let $(X(t))_{t\in T}$ with $T\subseteq \mathbb{R}^n$ be a (strictly) stationary stochastic process. In the book 'Fourier Analysis and Stochastic Processes' Pierre Brémaud states on p.157 (Chapter 3.4.1) that the trajectory of $(X(t))_{t\in T}$ is almost surely not in $\mathrm{L}^2(\mathbb R)$ (he assumes $T\subseteq \mathbb R$). Unfortunately, he does not include a proof. I've got two questions regarding his statement:

Does someone know where I can find a proof of this statement or how to proof it?

Does this statement also hold for wide-sense stationary processes, i.e. are the trajectories of wide-sense stationary processes almost surely not in $\mathrm{L}^2(\mathbb R)$?

I would be happy if someone could help! Thanks a lot!

答:
Elaboration:

Stationary implies that $(X(t+s))_{s\ge 0}$ has the same distribution as $(X(s))_{s\ge 0}$, for each $t$. This is why $Z(t)$ and $Z(0)$ have the same distribution, being the same functional of $(X(t+s))_{s\ge 0}$ and  $(X(s))_{s\ge 0}$, respectively. [The definition of stationarity involves the finite-dimensional distributions, with the implication that probabilities of events in $\mathcal X:=\sigma\{X(t): t\in\Bbb R\}$ are also shift invariant. That is, if we use $X^t$ to denote the time-shifted process $(X(t+s))_{s\ge 0}$ then $P[B] = P[\{\omega: X^t(\omega)\in B\}]$ for all $B\in\mathcal X$.]

A stationary process is ergodic provided the $\sigma$-field of invariant events is trivial (all such events have probability 0 or 1). The event $\{\|X\|_2=+\infty\}$ is invariant, so if it has positive probability then it must have probability 1.
[Using the previous notation, an event $B$ is invariant provided $B=\{\omega: X^t(\omega)\in B\}$ for each $t$.]


问:
Gradient of Kronecker product in the form of $y^\top (X \otimes I_N)^{-1} y$
Consider the following scalar-valued function of a matrix $X$:

$$
f(X) = y^\top (X \otimes I_N)^{-1} y,
$$
where $y$ is a column vector, $X$ is positive definite and has dimension $L$, and $I_N$ is an identity matrix of dimension $N$.

The question is to compute the gradient of $f(X)$ over $X$. (This is a follow-up question of this question)

I did not find an explicit formula in matrix cookbook. Then I tried to look for the chain rule for matrix derivatives, but the chain rule in (17-19) of matrix cookbook is not explicitly stated in a full matrix form, I think.

答:
Use a colon as a convenient product notation for the trace
$$A:B = {\rm Tr}(A^TB) = B:A$$
and the well known relationships
$$\eqalign{
(X\otimes I)^{-1} &= (X^{-1}\otimes I) \\
{\rm vec}(ABC) &= (C^T\otimes A)\,{\rm vec}(B) \\
A:B &= {\rm vec}(A):{\rm vec}(B) \\
dX^{-1} &= -X^{-1}dX\,X^{-1}\\
{\rm vec}(Y) &= y \quad\iff\quad Y={\rm unvec}(y) \\
}$$
Then you can write the function and calculate its gradient as follows
$$\eqalign{
f &= y^T(X\otimes I)^{-1}y \\
 &= y:(X^{-1}\otimes I)\,y \\
 &= y:{\rm vec}(YX^{-T}) \\
 &= Y:YX^{-T} \\
 &= Y^TY:X^{-1} \\
\\
df &= -Y^TY:X^{-1}dX\,X^{-1} \\
 &= -X^{-T}Y^TYX^{-T}:dX \\
\\
\frac{\partial f}{\partial X} &= -X^{-T}Y^TYX^{-T} \\
}$$


问:
non-bijective morphism between indecomposable modules
I am trying to prove the following statement:

Let $A$ be a finite dimensional Algebra. Let $M$ and $N$ be indecomposable $A$-modules.
Suppose $f: M \rightarrow N$ is not an isomorphism. Then $f$ doesn't admit a left or right inverse.

I guess I don't know enough about indecomposable modules to prove this. Can anybody help?

答: If there's a $g$ such that $f \circ g$ is the identity, then $f$ is surjective. Since $f$ is not an isomorphism, it's not an injection, and hence $\operatorname{Ker}(f)$ is a nontrivial submodule of $M$. This yields a nontrivial decomposition of $M$ as $\operatorname{Ker}(f) \oplus \operatorname{Im}(g)$. To see this, notice that an element $m \in M$ can be written as $\big(m - g(f(m))\big) + g(f(m))$, and that $\operatorname{Ker}(f) \cap \operatorname{Im}(g) = \{0\}$. Such a decomposition is in contradiction with the indecomposability of $M$.