问:
How to prove this property of the Wronskian?
I couldn't understand why it says the linear system has a nontrivial solution since the number of rows is :$1$(the original function）$+(n-1)$ (take the derivative $n-1$ times)$=n$. Also there are $n$ columns or unknowns. Why the system is not an independent one?

答:
I use [1] as the basis of my proof. As in [1], I employ a proof by contradiction [2].

Step 0. Define the proposition to be proved

By $P$, I denote the proposition to be proved, which is

$P := $ Given $n$ functions $f_1(x), f_2(x), \ldots, f_n(x)$ that are elements of the set $C^{n-1}{(I)}$ [3]  (i.e., the space of continuous functions that have continuous first $n-1$ derivatives in the interval $I$), if the Wronskian is not equal to zero for any value $c$ in the interval $I$, then the $n$ functions are linearly independent in the interval $I$ [4].

Step 1. The proposition to be proved, $P$, is assumed to be false

I assume that given $n$ functions $f_1(x), f_2(x), \ldots, f_n(x)$ that are elements of the set $C^{n-1}{(I)}$, if the Wronskian is not equal to zero for any value $c$ in the interval $I$, then the $n$ functions are  linearly dependent in the interval $I$.

Step 2. I derive a contradiction

I know that a sequence of functions $f_{1}{(x)},f_{2}{(x)},\dots ,f_{n}{(x)}$  from a function space are said to be linearly dependent, if there exist constant scalars $a_{1},a_{2},\dots ,a_{n}$, which are not all zero, such that
$$a_{1}\,f_{1}{(x)}+a_{2}\,f_{2}{(x)}+\cdots +a_{k}\,f_{k}{(x)}= 0_f,\quad\text{for all}~x\in  I ;\tag{1}$$
where $ 0_f$ denotes the zero function [5].  Next, I take the derivative of Equation $1$ $n-1$ times. I obtain the following system of $n$ equations in $n$ unknowns:
$$
\begin{bmatrix}
f_1{(x)} &f_2{(x)} & \cdots & f_n{(x)}
\\
f^{'}_1{(x)} &f^{'}_2{(x)} & \cdots & f^{'}_n{(x)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(x)} &f^{(n-1)}_2{(x)} & \cdots & f^{(n-1)}_n{(x)}
\end{bmatrix}
\,
\begin{bmatrix}
a_1\\a_2\\\vdots\\a_n
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\\vdots\\0
\end{bmatrix},\quad\text{for all}~x\in  I.
$$
This may be re-written in terms of any constant $c\in I$.
$$
\begin{bmatrix}
f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)}
\\
f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)}
\end{bmatrix}
\,
\begin{bmatrix}
a_1\\a_2\\\vdots\\a_n
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\\vdots\\0
\end{bmatrix}\qquad~\text{for any}~c\in I.
$$

According to my proposition, I have
$$
\begin{vmatrix}
f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)}
\\
f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)}
\end{vmatrix}
\neq 0\qquad~\text{for any}~c\in I.
$$
I can therefore invert the square matrix of coefficients. I obtain a solution for my unknowns (i.e., $a_1, a_2,\ldots, a_n$). I obtain that
$$
\begin{bmatrix}
a_1\\a_2\\\vdots\\a_n
\end{bmatrix}
=
\begin{bmatrix}
f_1{(c)} &f_2{(c)} & \cdots & f_n{(c)}
\\
f^{'}_1{(c)} &f^{'}_2{(c)} & \cdots & f^{'}_n{(c)}
\\
\vdots & \vdots & \vdots & \vdots
\\
f^{(n-1)}_1{(c)} &f^{(n-1)}_2{(c)} & \cdots & f^{(n-1)}_n{(c)}
\end{bmatrix}^{-1}
\begin{bmatrix}
0\\0\\\vdots\\0
\end{bmatrix}\qquad~\text{for any}~c\in I.
$$
Upon matrix mulitplication, I find
$$a_1 = a_2=\cdots = a_n=0.$$

These functions were said to be linearly dependent since there existed scalars $a_1, a_2, \ldots, a_n$ that are not all zero. Yet, I just arrived at the contradictory statement that the scalars $a_1, a_2, \ldots, a_n$ are all zero.

Step 3. Conclusion

Since the statements that, the scalars $a_1, a_2, \ldots, a_n$ are not all zero, and the statement that, the scalars $a_1, a_2, \ldots, a_n$ are all zero, cannot both be true, the assumption that $P$ is false must be wrong.  I conclude that $P$ must be true.

BIBLIOGRAPHY

[1] https://math.vanderbilt.edu/sapirmv/msapir/prwronskian.html

[2] https://en.wikipedia.org/wiki/Proof_by_contradiction#:~:text=In%20logic%20and%20mathematics%2C%20proof,false%20leads%20to%20a%20contradiction.

[3] https://en.wikipedia.org/wiki/Function_space

[4] https://mathworld.wolfram.com/LinearlyDependentFunctions.html

[5] https://en.wikipedia.org/wiki/Linear_independence#Definition


问:
general solution of equation and relation
I am interested in learning the below question in some elementary way. Please discuss this problem and help me to get mind free state.

How to get solutions for $x^2 - 10y^2$ = $1$? I would like to learn some easy method. Also, I have seen in some text book $x= 19$ and $y = 6$ is one of the solution and general solution can be prepared by using $x = 19$ and  $y = 6$ in terms of $n$. where $n$ is some positive integer. Please explain this general solution and relation with $x = 19$ and $y =6$.

Advance thanks to all.

答:
We add a few comments. First we should mention the Brahmagupta Identity, which for this case is
$$(s^2-10t^2)(u^2-10v^2)=(st+10uv)^2-10(sv+tu)^2.\tag{$1$}$$
This identity can be verified by expanding both sides.

From this identity, if we have a representation of $m$ in the form $x^2-10y^2$, and also a representation of $n$ of that form, then the identity gives us a representation of $mn$.

$(1.)$ In particular, let $s=u=3$ and $t=v=1$. Then $s^2-10t^2=-1$. Thus by the Brahmagupta Identity, we have
$$(s^2-10t^2)(s^2-10t^2)=(9+10)^2-10(3+3)=(-1)(-1)=1.$$
Thus the solution $(19,6)$ can be thought of as coming from the simpler "almost" solution $(3,1)$.

$(2.)$ Let $a_1=19$, $b_1=6$. For $n\ge 1$, define $a_{n+1}, b_{n+1}$ by
$$a_{n+1}=19a_n+6b_n,\qquad\text{and}\qquad b_{n+1} =6a_n+ 19b_n.\tag{$2$}$$
The recurrences $(2)$ give an easy way to produce more solutions of our equation $x^2-10y^2=1$. For example, $(a_2,b_2)=(721,228)$.

It turns out that all positive solutions come from the recurrence $(2)$.

$(3.)$ You asked how one obtains $(19,6)$, the fundamental solution. There is a mechanical and relatively quick way of doing it, using the continued fraction expansion of $\sqrt{10}$. Computations can be arranged so that they are pure integer computations, meaning we do not need to know anything about the decimal expansion of $\sqrt{10}$.

There is a very nice book on the Pell equation by Ed Barbeau. Many number theory texts discuss the Pell equation in detail.

ContentLicense: CC BY-SA 3.0


问:
in non-separable Hilbert spaces, is the adjoint of a $p$-Schatten class operator again $p$-Schatten class with the same norm?
If $T \colon U \to V$ is an operator of Schatten class $p$ between two separable Hilbert spaces $U$ and $V$, it follows directly from the fact that the singular values of an operator and its adjoint agree that the adjoint $T^{\ast}$ is of Schatten class $p$ as well and $\lVert T \rVert_p = \lVert T^{\ast} \rVert_p$.

Does the situation change if the Hilbert spaces are non-separable? If I interpret the Wikipedia page correctly, the Schatten norms are no longer defined in terms of singular values but instead of traces of powers of the absolute value.

答:
Every Schatten  class operator is compact, and every compact operator lives on  separable spaces.

By this I mean that there are closed, separable subspaces $U_0\subseteq U$, and $V_0\subseteq V$, such
that

$T=0$ on $U_0^\perp$,  and

$\text{Ran}(T)\subseteq V_0$.

Thus, if we define $T_0$ to be the restriction of $T$ to $U_0$, then
$$
  T=T_0\oplus 0 :  U_0\oplus U_0^\perp \to   V_0\oplus V_0^\perp.
  $$

Using this,  one may easily generalize  all of the machinery developed to deal with  operators on
separable spaces to non-separable ones.

In particular,  yes,   the singular values of a $p$-Schatten class  operator and its adjoint agree, and hence
$\|T^*\|_p = \|T\|_p$.

ContentLicense: CC BY-SA 4.0


问:
Order $25$ people in $4$ circles, $3$ circles with $6$ people and $1$ circle with $7$ people - the order between the circles isn't important
I am learning to my discrete math test and I tried to answer this question.

I know how to count the options with importance of the order between the circles, using the multiplication principle.

I count the options to put $6$ of $25$ people in the first circle - $p(25,6)$ and options to order them in the circle which is $5!$

Then count the options to put $19$ of $25$ people in the second circle - $p(19,6)$ and the options to order them in the circle which is $5!$

Then count the options to put $13$ of $25$ people in the second circle - $p(13,6)$ and the options to order them in the circle which is $5!$

Then count the options to put $7$ of $7$ people in the second circle - $1$ and the options to order them in the circle which is $6!$

Then I multiply all of them and I get that the number of options to order the people in the circles is: $(25, 6) \times (19, 6) \times (13, 6) \times (5!)^3 \times 6!$

Now I need to use diverse principle - I know that for every group of circles there are $4!$ different ways to order the circles, so if I am dividing the count with importance of the order by $4!$ I get the answer.

The problem is that in the solution they say that I need to diverse it by $3!$ because I already know that the circle with $7$ people will be last in the order and i am having struggles to understand the idea.

The question is as I write in the header, so I hope I was clear enough that people can help me get the idea in the solution.

答:
Rather than dividing by symmetry, perhaps a more direct argument will make more sense.

First, let us choose the seven people who sit at the seven-person-table.  There is no ambiguity as to which table is the seven-person-table, it is the only table with seven chairs.  We choose those seven people in $\binom{25}{7}$ ways.

Once having chosen those people, we look at them and one of those people will be youngest.  Let them sit at the seven-person-table in whatever chair they like, it doesn't matter to us since we aren't keeping track of that information.  From there, take the remaining six people and arrange them around the remaining seats of the table.  We do keep track of this information since we care about how the people are seated with relation to each other (though we don't care how they are seated with relation to the door to the room).  This can be done in $6!$ ways.

From here, and this is where my proposed solution differs from yours, among the remaining people there will be some youngest person and that person will eventually be seated somewhere.  Go ahead and let them choose somewhere to sit.  It doesn't bother us which of the remaining tables they sit at, they all look the same to us.  It doesn't matter which of the seats they sit at either since we only care how people are seated relative to one another, not how the table is situated around the room or where the chair is relative to the entrance etc...

Now that the youngest has chosen their seat, choose five more people to sit at the table with them and then choose how they are arranged around the table relative to that youngest person.  This can be done in $\binom{17}{5}\cdot 5!$ ways.

Now, among the remaining $12$ people, again, there will be a youngest remaining.  Let them sit in one of the remaining available seats at one of the remaining available tables, again it matters not which.  Choose the $5$ other people to sit with them and arrange around the table in $\binom{11}{5}\cdot 5!$ ways.  Do this one final time for the final set of people in $\binom{5}{5}5!$ ways.

This gives a grand total of:

$$\binom{25}{6}6!\cdot \binom{17}{5}5!\cdot \binom{11}{5}5!\cdot \binom{5}{5}5!$$

As for your division by symmetry approach and dividing by $3!$ instead of by $4!$, the point being that when choosing which table they sat at, as alluded to in the teacher's solution we can correctly distinguish which of the tables is the seven-person-table by the fact that it is the only table with seven people at it... this even if the labels of the table were hidden from us.  The three six-person-tables however if left unlabeled we would be unable to distinguish so our initial approach to counting which had these tables labeled as the "first" "second" and "third" had counted each discernable outcome six times, once for each different arrangement of the labels on the tables.


问:
Answer to an algebra question

The parabola $y=x^2-3bx-5$, in the $x-y$ plane, has its vertex at
$(3,4)$. What is the value of $b$?

I can't answer it. There is no value for $b$ such that, when multiplied by 3, gives 5 and also makes the equation equal 4 when $x$ is 3 (to get the vertex). I'm pretty stumped all around. The problem is from a private SAT prep company. I would appreciate any help.

答:
It is easiest to see whether or not there is a solution to this problem using graphing.

https://www.desmos.com/calculator/x7cwulidgt

Hint: manipulate the value of $b$ to convince yourself of whether or not there is some value of $b$ for which the parabola with the general equation $y=x^2+3bx-5$ can have a vertex of $P(3,4)$.

Title: Answer to an algebra question


问:
Similar triangles question
If I have a right triangle with sides $a$.$b$, and $c$ with $a$ being the hypotenuse and another right triangle with sides $d$, $e$, and $f$ with $d$ being the hypotenuse and $d$ has a length $x$ times that of $a$ with $x \in \mathbb R$, is it necessary for $e$ and $f$ to have a length $x$ times that of $b$ and $c$ respectively?

EDIT: The corresponding non-right angles of both triangles are the same.

答:
EDIT: This answer is now incorrect since the OP changed his question.

This is a good way of visualizing the failure of your claim:

Imagine the point C moving along the circle from A to the north pole. This gives you a continuum of non-similar right triangles with a given hypotenuse (in this case the diameter).

Title: Similar triangles question


问:
To calculate type I error of hypothesis testing on a discrete random variable
Suppose X is a random variable with $P(X=k)=(1-p)^kp$ for $k\in{0,1,2,...}$ and some $p\in(0,1)$. For the hypothesis testing problem $H_0:p=1/2$ and $H_1:p\neq 1/2$.

Consider the test "Reject $H_0$ if $X\leq A$ or if $X\geq B$", where $A<B$ are given positive integers. Express the type-I error of this test in terms of A and B.

My thoughts: The probability we need is P(reject $H_0|H_0$ is true). So, p=1/2. Now the probabilities at each k will be as follows:

$\frac{1}{2}\;\frac{1}{2^2}\;\frac{1}{2^3}\;\cdots\frac{1}{2^{A+1}}\;[\cdots\text{Here $H_0$ cannot be rejected}\cdots]\frac{1}{2^{B+1}}\;\cdots$

I need the probability of $1$-above region. How do I get an expression from this series ?

Edit: The given answer is $1+2^{-B}-2^{-A-1}$. I don't understand how.

答:
See here.

The sum of the first terms is $a+a r+a r^2+...+a r^n$ where $a=\frac{1}2$, $r=\frac{1}2$ and $n=A$.

The sum of the second set of terms is also a geometric series.

Use the formula to find the two sums and add them together.

Alternative:

Call the sum of all the terms $x$.

Multiple $x$ by $-\frac{1}2$, which is the sum of all those terms multiplied by $-\frac{1}2$.

Add the original sum and the new sum of terms. Almost all the terms are cancelled out.

Figure out which terms are not cancelled out. Then, you are left with an equation:

$x-\frac{1}2 x=\text{whatever terms are not cancelled out}$

Solve that for $x$.


问:
Splitting up a conjugacy class with respect to a subgroup
Suppose we have a conjugacy class $\mathcal{K}$ of elements of a group $G$, and a subgroup $H \leq G$.

How can we efficiently (computationally) partition the elements of $\mathcal{K}$ according to whether they are conjugates by elements of $H$? If there is a way to do this in Magma using essentially built-in intrinsics, that is ideal, otherwise I am interested in knowing algorithmic ideas.

This is similar to this question:

Conjugacy class of $G$ splitting as Conjugacy class of $H\unlhd G$

But my question differs in that:

I am not assuming that $H \unlhd G$

I am not assuming that $\mathcal{K} \subseteq H$

I am interested in doing this computationally in a "nice" way.

One obvious solution (mentioned in the answer to the linked question) is to compute the permutation action of $H$ acting on $\mathcal{K}$, and computing the orbits. I don't think this is ideal when the groups are large, but if this is the most practical way that could be an acceptable answer.

答:
The orbits of $H$ on the orbit $\cal K$ of $G$ are in bijection with the double cosets ${\rm Stab}_G(\kappa)\backslash G/H$, orbit representatives can be obtained by conjugating $\kappa$ with representatives of the double cosets.

The reason this is better (also in practice) than calculating orbits directly, is that the double coset computations can use other subgroups of $G$ to compute orbits on smaller domains and use imprimitivity of an action.

Since the question was tagged [gap], and you said you were looking for a nice solution, here is an example how you could do it in GAP. (I also think the GAP implementation for double cosets is better than the one in Magma, but having written it I am more than biased.)

```<code>gap> g:=SimpleGroup("HS");;Size(g);
44352000
gap> c:=ConjugacyClass(g,g.2);;Size(c);
88704
gap> s:=SylowSubgroup(g,2);;Size(s);
512
gap> dc:=DoubleCosets(g,Centralizer(c),s);;Length(dc);
219
gap> reps:=List(dc,x->Representative(c)^Representative(x));;
gap> Sum(reps,x->Index(s,Centralizer(s,x))); # verify
88704
</code>```

Body:
The orbits of $H$ on the orbit $\cal K$ of $G$ are in bijection with the double cosets ${\rm Stab}_G(\kappa)\backslash G/H$, orbit representatives can be obtained by conjugating $\kappa$ with representatives of the double cosets.

The reason this is better (also in practice) than calculating orbits directly, is that the double coset computations can use other subgroups of $G$ to compute orbits on smaller domains and use imprimitivity of an action.

Since the question was tagged [gap], and you said you were looking for a nice solution, here is an example how you could do it in GAP. (I also think the GAP implementation for double cosets is better than the one in Magma, but having written it I am more than biased.)

gap> g:=SimpleGroup("HS");;Size(g);
44352000
gap> c:=ConjugacyClass(g,g.2);;Size(c);
88704
gap> s:=SylowSubgroup(g,2);;Size(s);
512
gap> dc:=DoubleCosets(g,Centralizer(c),s);;Length(dc);
219
gap> reps:=List(dc,x->Representative(c)^Representative(x));;
gap> Sum(reps,x->Index(s,Centralizer(s,x))); # verify
88704


问:
Finding the explicit formula for the given recurrence relation
This is the recurrence relation:$ a_n=4-a_{n-1}, a_2=-3$.I noticed that the result for even index of $a$ is always -3 and the result for odd index of $a$ is 7,But I don't know where to go with it.

Body:
$$a_n-2=4-a_{n-1}-2=(-1)(a_{n-1}-2)$$
$\{a_n-2\}$ is a geometric sequence so
$$a_n-2=(-1)^{n-2}(a_2-2)=5(-1)^{n-1} \implies a_n = 2+5(-1)^{n-1}.$$


答:
Why do you need to divide by $3!$ and not by $4!$?

Note all the circles of size $6$ can be interchanged with one another, while the circle of size $7$ is "special" - cannot be swapped with any other circle. Your solution before the last division assumed you knew which of the circles of "size" $6$ was "first", which was "second" and which was "third". Now, as we don't care about the order of those three circles, we divide by the number of permutations of those circles, i.e. $3!$.

To illustrate further: in a hypothetical similar problem, where $26$ people would be split into two circles of size $7$ and two circles of size $6$, you would, as the last step, need to divide by $2!\cdot 2! = 4$, because the two circles of size $7$ can be swapped, and two circles of size $6$ can be swapped, but a circle of size $6$ cannot be swapped for a circle of size $7$.

Title: Order $25$ people in $4$ circles, $3$ circles with $6$ people and $1$ circle with $7$ people - the order between the circles isn't important

Body:
I am learning to my discrete math test and I tried to answer this question.

I know how to count the options with importance of the order between the circles, using the multiplication principle.

I count the options to put $6$ of $25$ people in the first circle - $p(25,6)$ and options to order them in the circle which is $5!$

Then count the options to put $19$ of $25$ people in the second circle - $p(19,6)$ and the options to order them in the circle which is $5!$

Then count the options to put $13$ of $25$ people in the second circle - $p(13,6)$ and the options to order them in the circle which is $5!$

Then count the options to put $7$ of $7$ people in the second circle - $1$ and the options to order them in the circle which is $6!$

Then I multiply all of them and I get that the number of options to order the people in the circles is: $(25, 6) \times (19, 6) \times (13, 6) \times (5!)^3 \times 6!$

Now I need to use diverse principle - I know that for every group of circles there are $4!$ different ways to order the circles, so if I am dividing the count with importance of the order by $4!$ I get the answer.

The problem is that in the solution they say that I need to diverse it by $3!$ because I already know that the circle with $7$ people will be last in the order and i am having struggles to understand the idea.

The question is as I write in the header, so I hope I was clear enough that people can help me get the idea in the solution.


问:
Understanding Bayes' Theorem
I worked through some examples of Bayes' Theorem and now was reading the proof.

Bayes' Theorem states the following:

Suppose that the sample space S is partitioned into disjoint subsets $B_1, B_2,...,B_n$.  That is, $S = B_1 \cup B_2 \cup \cdots \cup B_n$, $\Pr(B_i) > 0$ $\forall i=1,2,...,n$ and $B_i \cap B_j = \varnothing$ $\forall i\ne j$.  Then for an event A,

$\Pr(B_j \mid A)=\cfrac{B_j \cap A}{\Pr(A)}=\cfrac{\Pr(B_j) \cdot \Pr(A \mid B_j)}{\sum\limits_{i=1}^{n}\Pr(B_i) \cdot \Pr(A \mid B_i)}\tag{1}$

The numerator is just from definition of conditional probability in multiplicative form.

For the denominator, I read the following:

$A= A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)\tag{2}$

Now this is what I don't understand:

The sets $A \cup B_i$ are disjoint because the sets $B_1, B_2, ..., B_n$ form a partition.$\tag{$\clubsuit$}$

I don't see how that is inferred or why that is the case.  What does B forming a partition have anything to do with  it being disjoint with A.  Can someone please explain this conceptually or via an example?

I worked one example where you had 3 coolers and in each cooler you had either root beer or soda.  So the first node would be which cooler you would choose and the second nodes would be whether you choose root beer or soda.  But I don't see why these would be disjoint.  If anything, I would say they weren't disjoint because each cooler contains both types of drinks.

Thank you in advance! :)

答:
You can easily show this using set theoretic arguement. </p>

<p>$A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)$</p>

Title: Understanding Bayes' Theorem

Body:
You can easily show this using set theoretic arguement.

$A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)$


问:
How do you prove Osborn's rule?
Given a trigonometric identity written in terms of sine and cosine, it is possible to write down the corresponding hyperbolic identity using Osborn's rule:

Replace $\cos$ with $\cosh$

Replace $\sin$ with $\sinh$

However, negate the product of two $\sinh$ terms. For example,
$$
\cos^2x+\sin^2x \equiv 1 \implies \cosh^2x-\sinh^2x=1 \, .
$$
I tried proving this rule using the following two identities
\begin{align}
\cos ix &\equiv \cosh x \\
\sin ix &\equiv i\sinh x \, .
\end{align}
Given an arbitrary trigonometric identity of the form
$$
f(\cos z,\sin z) \equiv g(\cos z, \sin z)
$$
then letting $z=ix$ we see that
\begin{align}
f(\cos ix,\sin ix) &\equiv g(\cos ix, \sin ix) \\
f(\cosh x,i\sinh x) &\equiv g(\cosh x, i\sinh x) \\
\end{align}
From here, it seems the rule makes sense because the product of two $i\sinh x$ terms gives us $\color{red}{-}\sinh x$. But I'm not quite sure how to formalise this argument. Moreover, an identity such as
$$
\sin 2z \equiv 2\sin z \cos z
$$
would correspond to
$$
i\sinh x \equiv 2i\sinh x \cosh x \, .
$$
Here, we could divide through by $i$ and the expected hyperbolic identity pops out, but I'm unsure if this 'divide by $i$' trick always works, since there may be other terms that don't contain an $i$ in them. To give a slightly contrived example,
$$
\sin x + \cos 2x \equiv \sin x + \cos^2 x - \sin^2x
$$
corresponds to
$$
i\sinh x + \cosh 2x \equiv i\sinh x + \cosh^2 x + \sinh^2 x \, ,
$$
and here, dividing by $i$ does not help. So I think I've understood the crux of how to prove Osborn's rule, but there are a few loose ends that I need to sort out.

答:
You could proceed in the following way:$$
\sin x + \cos 2x \equiv \sin x + \cos^2 x - \sin^2x
$$
corresponds to
$$
i\sinh x + \cosh 2x \equiv i\sinh x + \cosh^2 x + \sinh^2 x \, ,$$

consequently, by equalling the imaginary parts and the real ones, we get that,

$$\begin{cases}
\sinh x\equiv\sinh x\\
\cosh 2x\equiv\cosh^2 x+\sinh^2 x
\end{cases}$$

and, by adding both sides,

$$\sinh x + \cosh 2x \equiv \sinh x + \cosh^2 x + \sinh^2 x\,.$$

Title: How do you prove Osborn's rule?


问:
Find the area using integrals

What is the area defined by the curves $y = \sin x$ and $y = \cos x$ for $0 < x < \frac{\pi}{4}$?

I think it's $$\int_{0}^{\frac{\pi}{4}} [\cos x - \sin x]\, \mathbb{d}x$$ but I'm not sure if my step is right or not.

答: Yes, your integral is correct.  If you have two curves above the $x$-axis on an interval $[a,b]$, then the area between them is given by $$\int_{a}^{b}(\text{top curve} - \text{bottom curve})\,dx.$$

Title: Find the area using integrals


问:
need help with Laplace transform of $t^2 \cos(\omega t)$
I looked up the Laplace transform table on tutorial.math.lamar.edu, and it says that $$t^n f(t) = (-1)^n F^n(s)$$

In my head that should equal

$$-1^2 (s/s^2 + \omega^2)^2$$

I put the question into Symbolab to check myself and it says that the answer is
$$\frac{-(2s(-s^2+3\omega^2)}{(s^2+\omega^2)^3}$$

According to symbolab you have to take the derivative of the $(\mathcal{L}f)(s)$.Any guidance is much appreciated.

答:
Well, as you stated we know that:

$$\mathscr{L}_x\left[x^\text{n}\cdot\text{f}\left(x\right)\right]_{\left(\text{s}\right)}=\left(-1\right)^\text{n}\cdot\frac{\text{d}^\text{n}}{\text{ds}^\text{n}}\left(\mathscr{L}_x\left[\text{f}\left(x\right)\right]_{\left(\text{s}\right)}\right)\tag1$$

And we also know that:

$$\mathscr{L}_x\left[\cos\left(\omega x\right)\right]_{\left(\text{s}\right)}=\frac{\text{s}}{\text{s}^2+\omega^2}\tag2$$

So, for your problem we get:

$$\mathscr{L}_x\left[x^2\cos\left(\omega x\right)\right]_{\left(\text{s}\right)}=\left(-1\right)^2\cdot\frac{\partial^2}{\partial\text{s}^2}\left(\frac{\text{s}}{\text{s}^2+\omega^2}\right)=\frac{2\text{s}\left(\text{s}^2-3\omega^2\right)}{\left(\text{s}^2+\omega^2\right)^3}\tag3$$


问:
Proof of partition of natural numbers
I have a set $A_n$ = $ \{2^i(2n-1): i \in \mathbb{N} \cup {0} \}$

It is said that the set  $P = $ $\{A1, A2, A3,...\}$ partitions the natural numbers.

I am attempting to solve this by the definition of a partition, I can see that no $An = \emptyset$, but I can't see how the union of all the $An$'s would give $\mathbb{N}$ can anyone help?

答: A good way to get some intuition is to take some numbers and see which set they belong to.  For example, $12=2^2\cdot 3 \in A_3$  You divide out all the factors of $2$ and are left with an odd number $N$.  You let $N=2n-1, n=\frac 12(N+1)$ and that is the set $A_n$ it belongs to.

Title: Proof of partition of natural numbers


答: Consider $A_1$. This is the set $\{2^i\}$, which contains the elements $1,2,4,8,16,\cdots$. Now consider $A_2$, which is $2^i\times3$ which contains the elements $3,6,12,24,\cdots$. Do something similar for $A_4$ and so on.  You should see that the least element of $A_{k}$ is always the $k$th odd number, which is $2k-1$ , and so you can generate every odd element of $\mathbb{N}$ in such a way. By the definition of the set, we can multiply every element by $2$, which makes every even number. Now you need only formalise this proof.

Title: Proof of partition of natural numbers

Body:
I have a set $A_n$ = $ \{2^i(2n-1): i \in \mathbb{N} \cup {0} \}$

It is said that the set  $P = $ $\{A1, A2, A3,...\}$ partitions the natural numbers.

I am attempting to solve this by the definition of a partition, I can see that no $An = \emptyset$, but I can't see how the union of all the $An$'s would give $\mathbb{N}$ can anyone help?


问:
Equality in Jensen's inequality for strictly convex functions
Jensen's inequality

Let $\phi : \mathbb{R} \rightarrow \mathbb{R}$ be a convex function and $X$ be a random variable. Then
$$\phi(E[X]) \leq E[\phi(X)],$$ if $E[X]$ and $E[\phi(X)]$ exist.

Exercise
Let $\phi : \mathbb{R} \rightarrow \mathbb{R}$ be a strictly convex function, that is,
$$\phi(x) \geq ax+ b \forall x \in \mathbb{R} (1) $$

Then if
$$\phi(E[X]) = E[\phi(X)]$$

$\Rightarrow X = c$ almost everywhere, where $c$ is c a constant.

Question 1

I have found that a function $f:X \rightarrow \mathbb{R}$ is called strictly convex iff
$$f(tx_1 + (1-t)x_2) < tf(x_1) + (1-t)f(x_2) \forall t \in (0,1), \forall x_1,x_2 \in X$$

Why does the exercise mention $(1)$? Is it an equivalent definition?

Question 2

How should I approach this exercise?

答:
Equation $(1)$ requires a little more context. A function $\phi:\mathbb R\to\mathbb R$ is convex if and only if, for all $x_0\in\mathbb R$, $\phi$ has a subderivative at $x_0$, i.e. there exists a number $c\in\mathbb R$ such that

$$ \phi(x) - \phi(x_0) \ge c(x-x_0) \qquad\qquad\qquad(*)$$

for all $x\in\mathbb R$. Moreover, $\phi$ is strictly convex if and only if equation $(*)$ is a strict inequality for $x\neq x_0$.

Translating this into the language of equation $(1)$, given $x_0\in\mathbb R$, there exists $a,b\in\mathbb R$ such that:

$\phi(x_0) = ax_0+b$, and

$\phi(x) > ax + b$ for all $x\neq x_0$.

(Specifically, one can take $a=c$ and $b=\phi(x_0)-cx_0$.) This can be taken as an equivalent definition of strict convexity.

To use this to solve your exercise, let $x_0=E[X]$. Then we have

$$\phi(X) \ge aX + b, \qquad\qquad\qquad(\dagger)$$

and so taking expectation, we find

$$E[\phi(X)] \ge aE[X] + b = ax_0 + b = \phi(x_0) = \phi(E[X]).$$

This proves Jensen's inequality. Moreover, since $\phi$ is strictly convex, we know that either $X=x_0$ almost surely, or the inequality in $(\dagger)$ is strict with positive probability. In the latter case, this of course implies that the inequality above is also strict, completing the proof.


答:
To concur with you, and demonstrate visually, I graphed both functions from $-\pi$ to $\pi$:

Between $x= 0$ and $x=\pi/4$, we see that $y=\cos x > y= \sin x$.  So your set up is spot on.  By noting $$\int_{o\to {\pi/4}} (\cos x- \sin x)\,\mathrm dx,$$

we find the area below $y = \cos x$ and subtract the area below $y=\sin x$, giving us the area between the two curves, on that interval.

Title: Find the area using integrals

Body:
What is the area defined by the curves $y = \sin x$ and $y = \cos x$ for $0 < x < \frac{\pi}{4}$?

I think it's $$\int_{0}^{\frac{\pi}{4}} [\cos x - \sin x]\, \mathbb{d}x$$ but I'm not sure if my step is right or not.


问:
Find subsequence with even and odd indices $(a_{2k})_{k}$ and $(a_{2k-1})_{k}$
I have the sequence $(a_{n})_{n}$ with first term $a_{1}=2$ and recursive relation $$a_{n+1}=\frac{2}{1+a_{n}}$$
So I have to find subsequence with even and odd indices $(a_{2k})_{k}$ and $(a_{2k-1})_{k}$

Any help?

答:
Well, "to find" means an explicit expression, doesn't it? So you just prove
$$a_n=\frac{2^{n+1}-2\,(-1)^n}{2^{n+1}+(-1)^n}$$ by induction (that's almost trivial, given the recursive relation), and then, you get the subsequences, using $(-1)^{2k}=1$ and $(-1)^{2k+1}=-1$.

This may seem a bit ad hoc, but it's in the spirit of the question (no context, no source, no motivation,...).


问:
if S and T are Hilbert-Schmidt between two Hilbert spaces U and V, is $\operatorname{trace}_U(S^{\ast}T) = \operatorname{trace}_V(T^{\ast}S)$?
Let $U$ and $V$ be two real Hilbert spaces and $S$, $T$ two Hilbert-Schmidt operators between $U$ and $V$. Denote their adjoints by $S^{\ast}$ and $T^{\ast}$, both viewed after identification of $U$ and $V$ with their respective duals as Hilbert-Schmidt operators between $V$ and $U$. Is it true that
$$\operatorname{trace}_U(S^{\ast}T) = \operatorname{trace}_V(T^{\ast}S)$$
holds (at least when the Hilbert spaces are separable)? Formulated differently, is the set of Hilbert-Schmidt operators between different Hilbert spaces still a Hilbert space when equipped with the inner product $\langle S,T \rangle=\operatorname{trace}_U(S^{\ast}T)$?

答:
Assuming both $U$ and $V$ are separable,  infinite dimensional Hilbert spaces,  it is well known that there exists an isometry
$\varphi :V\to U$.

Considering the map
$$
  \Phi: T\in  \mathscr B(U, V) \mapsto  \varphi T \in  \mathscr B(U, U),
  $$
notice that $\Phi$ restricts to a bijective linear map from  $L_2(U, V)$,  the space of Hilbert-Schmidt operators, onto   $L_2(U, U)$.

Moreover,
for every $S, T\in L_2(U, V)$,   one has that
$$
  \text{trace}_U(S^*T) =
  \text{trace}_U(S^*\varphi ^*\varphi T) =
  \text{trace}_U\big (\Phi(S)^*\Phi(T)\big ),
  \tag 1
  $$
so $\Phi$ is compatible with the usual inner-product on $L_2(U, U)$ and the proposed inner-product on
$L_2(U, V)$.

This answers the last question posed by the OP, namely that the set of Hilbert-Schmidt operators between
different Hilbert spaces is still a Hilbert space with the indicated inner-product.

However,
the relation
$$
  \text{trace}_U(S^{\ast}T) = \text{trace}_V(T^{\ast}S),
  \tag 2
  $$
is not true, as it compares  a map that is conjugate-linear on $S$, with one that is  linear on $S$.
Nevertheless,
we may use (1) and the fundamental property of the trace to write
$$
  \text{trace}_U(S^*T) =
  \text{trace}_U\big (\Phi(S)^*\Phi(T)\big ) =
  \text{trace}_U\big (\Phi(T)\Phi(S)^*\big ) = $$$$ =
  \text{trace}_U(\phi TS^*\varphi ^*) =
  \text{trace}_V(TS^*),
  $$
where the last step is a consequence of the unitary invariance of the trace.


问:
$x^{2}+y^{2}=3(2023)^{z}+77$: to prove that if $4 \mid x^{2}+y^{2} \implies 4\mid x^{2},y^{2}$
This is a math Olympiad problem, and actually, I was in it.
$x^{2}+y^{2}=3(2023)^{z}+77$ and $x,y,z \in \mathbb{N}$.

Prove that if $4 \mid x^{2}+y^{2} \implies 4\mid x^{2},y^{2}$
then find all $(x,y,z)$ solutions to the equation.

My attempt :

A claim:
$3(2023)^{z}+77$ Is divisible by $4$ iff $z$ is even.

But I can't prove this claim.
Do not use any modular arithmetic

答:
The original question was with 4 The only thing you need is $4|(x^2+y^2)$. If $4$ divides $x^2+y^2$ then $4$ must divide both $x$ and $y$. Indeed: Then as all odd squares are $1 \mod 4$, if either $x$ or $y$ are odd then $(x^2+y^2) \mod 4$ is either $1$ or $2$. So both $x$ and $y$ must be even i.e., multiples of $2$. Which gives $x^2$ and $y^2$ are both multiples of $2^2=4$.

To see mod 7, this follows from the fact that $-1$ is not a square mod 7. i.e., the only way $x^2+y^2$ could divide $7$ without both $x^2,y^2$ dividing 7 would be if $y^2 \mod 7 = (-1)×x^2\mod 7$ which would imply that $-1$ is a square in $\mathbb{Z}/7\mathbb{Z}$, which it is not. This can be easily checked directly.


问:
Example of a locally Noetherian space $X$ and a point $x$ such that $\dim_xX=\infty$
There is a well-known example that a Noetherian ring $A$ with infinite Krull dimension due to Nagata, which gives a Noetherian space $\operatorname{Spec}A$ with infinite Krull dimension.

Given the definition $\dim_xX=\inf_{x\in U}\dim U$ where $U$ takes over all open neighborhood of $x$, then does there exist a locally Noetherian space $X$(not necessary a scheme) with a point $x\in X$ such that $\dim_xX=\infty$?

Edit. User leoli1 has given an example, could someone give an example of scheme?

答:
With your definition of $\dim_x X$, Nagata's example satisfies $\dim_x X=\infty$ at every point.  Indeed, every nonempty open subset of $X$ still contains all but finitely many of the maximal ideals (since any element of the ring is only contained in finitely many of the maximal ideals), so it contains arbitrarily long chains of prime ideals.

Let me also mention that you can get a scheme (though not a locally Noetherian scheme) whose underlying topological space is the space $\mathbb{N}\cup\{\infty\}$ described in leoli1's answer.  For instance, you can just take a valuation ring with appropriate value group.  Consider the ordered ring $\mathbb{Z}[t]$ where $t$ is infinitely large.  Let $A$ be a valuation ring with $\mathbb{Z}[t]$ as its value group.  The prime ideals of $A$ then correspond to the initial segments of the positive part of $\mathbb{Z}[t]$ which are closed under addition.  For each $n$, there is such a segment consisting of elements which are less than some integer multiple of $t^n$, and the only other such segments are $\{0\}$ and the entire positive part of $\mathbb{Z}[t]$.  So the set of prime ideals in $A$ under reverse inclusion is order-isomorphic to $\mathbb{N}\cup\{\infty\}$.  It is the easy to see that the Zariski topology coincides with the topology described in leoli1's answer.


问:
Given the function $f(x) =xe^{ax}$
Show that if $f(x) =xe^{ax}$, where $a>0$, the absolute minimum value of $f$ is $-\frac{1}{a e}$.

I'm unsure how to approach this question. I've tried several things including plugging in the critical point of the first derivative.

答: Let $f(x)=xe^{ax}$ with $a>0$. Then, $f'(x)=axe^{ax}+e^{ax}=e^{ax}(1+ax)$ and it follows that the only critical point is at $x=-\frac{1}{a}$ (why?). In particular, $f'(0)>0$ and $f'(-\frac{2}{a})<0$ so this cricital point is at least a local minimum. But note that f is decreasing always to the left of $-\frac{1}{a}$ and increasing always to the right of this point. What does this then imply about the local minimum $x=-\frac{1}{a}$?

Title: Given the function $f(x) =xe^{ax}$


问:
How to find inverse function of function with parameter?
I'm trying to find the inverse function of this function:
$$
f_a(x) = \frac{x^a}{x^a + (1-x)^a}
$$

$a$ - parameter

$x$ - variable

Is it possible to find if there is a parameter here?

Hope for your help! Thanks in advance!

答:
On the domain you mentioned $(0,1)$, I think you just need to invert the parameter $a$ itself, and it works.
$$f^{-1}_a(x) = \frac{x^\frac{1}{a}}{x^\frac{1}{a} + (1-x)^\frac{1}{a}}$$


问:
How to define linear equations in an introductory linear algebra class
I believe there is an issue of clarification with respect to the definition of linear equations in many linear algebra texts.

Here is a typical one ``A linear equation in the $n$ variables $x_1,x_2, ..., x_n$ is an equation that can be written in the form $a_1x_1+a_2x_2+ ... +a_nx_n=b$ where the coefficients $a_1,a_2,...,  a_n$ and the constant term $b$ are constants.''

No problem with the definition so far except that the text proceeds to state that $x^2-y^3=5$ is not linear because "it contains powers".  There is no further explanation. Of course a well-prepared student would say: $(x+y+5)^2-(x+y)^2=35$ appears to contain powers and is a linear equation, how do you know that $x^2-y^3=5$ cannot be simplified?

I want to say the issue is with "[it] can be written" in the definition. That some operational explanation should be added.

Here are my two questions:

(a) Which introductory textbooks go deeper in the issue?

(b) How do you approach the issue? (Again this is introductory linear algebra, students come in with just calculus I as prerequisite).

答:
This is a good point. Most textbooks I have used begin with the definition you supply, although many of them would not make assertions about $x^3 - y^3 = 5$ that aren't proved, unless they made clear that a proof might come later, or made clear that they were just glossing over details to make the essential point that the world of linear equations is not the world of all multivariable equations.

Your example $(x+y+5)^2 - (x+y)^2 = 35$ makes a point that is very well taken (i.e., there is some reduction of powers that takes place that isn't visually apparent, but requires a moment or two of thought). Formally speaking, the assertion that something can or cannot be put in a certain form requires proof, and the definition would seem to require that, and formal discussion of these issues can be thorny and alienating for beginners.  (For example, it isn't a priori clear that a polynomial of degree $n$ can't be identically equal, as a function, to a polynomial of lower degree, and I have never seen a calculus textbook take the small amount of time it would require to explain this, although they all use the result, at least implicitly.  See my answer to https://math.stackexchange.com/a/98370/18076 for more on that.)

For beginners I think it's best just to tell people to focus on linear equations and not worry about proving that things are not linear.  (One way of proving it later in your specific example would be to derive geometric properties of solution sets of linear equations and show that the solution set to $x^3 - y^3 = 5$ does not have these properties.  Of course, that gets into finding solutions to a polynomial system, and doesn't handle the case of nonlinear systems whose solution sets do coincide with solution sets to systems of linear equations.  You can see how this could go down a long and winding road.)

One simplification might be to qualify "can be written" by restricting what you get to use when you rewrite, e.g., addition, subtraction, and scalar multiplication only.  And make clear that this is a pedagogical expedient.  I think the focus of "can be written" is simply that everybody wants something like $y = 2x + 5$ to satisfy the definition even though the variables aren't all on one side.

And you're not going to lose a lot of generality there, because linear systems that somehow masquerade as non-linear systems aren't that common in practice.  By the time students get to anything like that, they will be able to handle it.


问:
How to prove the continuity of the following Homeomorphism?
Let $f$ be a function from the one-point compactification of $\mathbb{R}$ to the circle defined as follows

$t\mapsto (x,y) = \begin{cases} \phantom{\lim\limits_{t\to\infty}} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t\ne\infty, \\[10pt]
\lim\limits_{t\to\infty} \left( \dfrac{1-t^2}{1+t^2}, \dfrac{2t}{1+t^2} \right) & \text{if }t=\infty.
\end{cases}$

I'm having trouble proving continuity and that it's open. Can you help me?

N/A = I tried many things, but none had a sense or took to somewhere. I actually need help...

答:
Continuity is a local concept, so the only thing to check here is that $f$ is continuous at $\infty$ (continuity at all other points is obvious). Note that $f(\infty) = (-1,0)$. By definition of the topology in the one-point compactification, it suffices to show that if $t \to \pm \infty$, then $f(t) \to (1,0)$, but this holds pretty much from the way $f$ is defined.

Once you have shown that $f$ is continuous, verify that it is bijective. Then you are done, because a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism, as continuity of the inverse follows from $f$ being closed.

Alternatively, you can exhibit a formula for the inverse, which will be manifestly continuous. For any point $p$ in the circle not equal to $(-1,0)$, consider the ray starting at $(-1,0)$ and passing through $p$. This ray will cross the $y$-axis at a single point -- the $y$-coordinate of such point is $f^{-1}(p)$, and then you set $f^{-1}(-1,0) = \infty$.


问:
Finding a Function That Approaches Another Function
One of my math professors gave me the following challenge. It isn't graded, it's just for fun.

Consider the function:

\begin{equation*}
f_n(x)=x+3^3x^3+5^3x^5+...+(2n-1)^3x^{2n-1},~x \in (0, 1).
\end{equation*}

I want to find which of the following functions $f_n$ is getting close to as $n$ gets larger:

$ \displaystyle a)\frac{x(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle b) \frac{x(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

$\displaystyle c) \frac{x^2(x+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$ $\displaystyle d) \frac{x^2(x^2+1)(x^4+22x^2+1)}{(x-1)^4(x+1)^4}$

Based on some tests i ran in mathematica by giving $n$ and $x$ values, it looks like $b)$ is the answer, but I am not sure. Can anyone confirm or deny this, and show how one might find the right answer, either with pen and paper or by using mathematica or maple or some other software?

答:
Here's yet another way to verify that it's (b). Let Mathematica do partial fractions decomposition:

```<code>Apart[x(x^2+1)(x^4+22x^2+1)/(x^2-1)^4]
</code>```

This gives $\frac{1}{2 (x+1)}-\frac{7}{2(x+1)^2}+\frac{6}{(x+1)^3}-\frac{3}{(x+1)^4}+\frac{1}{2 (x-1)}+\frac{7}{2(x-1)^2}+\frac{6}{(x-1)^3}+\frac{3}{(x-1)^4}$, which can be expanded into power series using the geometric series and its derivatives.

Of course this is ugly, and requires you to know the answer in advance. Moron's answer is much better. ;-)

Body:
Here's yet another way to verify that it's (b). Let Mathematica do partial fractions decomposition:

Apart[x(x^2+1)(x^4+22x^2+1)/(x^2-1)^4]

This gives $\frac{1}{2 (x+1)}-\frac{7}{2(x+1)^2}+\frac{6}{(x+1)^3}-\frac{3}{(x+1)^4}+\frac{1}{2 (x-1)}+\frac{7}{2(x-1)^2}+\frac{6}{(x-1)^3}+\frac{3}{(x-1)^4}$, which can be expanded into power series using the geometric series and its derivatives.

Of course this is ugly, and requires you to know the answer in advance. Moron's answer is much better. ;-)

ContentLicense: CC BY-SA 2.5


问:
Computing the gradient of Cross Entropy Loss
The categorical cross entropy loss is expressed as:

$$L(y,t) = -\sum_{k=1}^{K}t_k\ln{y_k}$$

where $t$ is a one-hot encoded vector. $y_k$ is the softmax function defined as:

$$y_k = \frac{e^{z_k}}{\sum_{j=1}^{K}e^{z_j}}$$

I want to compute the gradient, $\nabla_z$, of the loss function with respect to the input of the output node. What I know: I understand how to compute the partial derivative of L with respect to a selected node (say, $z_k$). This yields the following expression:

$$\frac{\partial L}{\partial z_k} = y_k - t_k$$

But I am not sure how to generalize this to the entire vector, $z$. In essence, I know how to compute $\frac{\partial L}{\partial z_k}$ when $k = j$ and $k \neq j$, but I don't know how to calculate the gradient, $\nabla_z$.

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\D{{\rm Diag}}\def\o{{\tt1}}$Given
an independent vector $z$, define the variables
$$\eqalign{
p &= \exp(z)
 \quad&\implies\quad P=\D(p)
 \quad&\implies &\quad dp&=p\odot dz &= P\,dz
\\
 &&&&\o^Tdp&=p^Tdz
\\
y &= \frac{p}{\o^Tp}
 \quad&\implies\quad Y=\D(y)
 \quad&\implies\; &\;Ydz&=\frac{P\,dz}{\o^Tp}&=\frac{dp}{\o^Tp}
\\
}$$
Now calculate the differential of the elementwise softmax function
$$\eqalign{
dy &= \frac{(\o^Tp)dp-p(\o^Tdp)}{(\o^Tp)^2} = \Big(Y-yy^T\Big)\,dz \\
}$$
and substitute it into the differential of the loss function
$$\eqalign{
{\cal L}  &= -t^T \log(y) \\
d{\cal L} &= -t^T\Big(d\log(y)\Big) \\
 &= -t^T\Big(Y^{-1}dy\Big) \\
 &= -t^TY^{-1}\Big(Y-yy^T\Big)\,dz \\
 &= -t^T\Big(I-\o y^T\Big)\,dz \\
 &= \Big(y(t^T\o) - t\Big)^T\,dz \\
 &= \Big(y - t\Big)^T\,dz \\
\p{\cal L}{z} &= y-t \\
}$$
Taking the components of this vector-valued gradient recovers your solution.


问:
How to show that the parabola $\{(x, a x^2+b x+c) \mid x\in [0,1]\}$ is a manifold.
I have not had any geometry courses yet but have had one topology course where did not learn about manifolds. Unfortunately i am in a situation where i need to show that for $a,b,c\in\mathbb{Q}$ the set $\mathcal{M}=\{(x, a x^2+b x+c)\mid x\in[0,1]\}$ is a sub-manifold of $\mathbb{R}^n$.

I have read that the idea is to show that for any point $x\in\mathcal{M}$ we can find an open set $U_x$ in $\mathcal{M}$ (open w.r.t. $\mathcal{M}$ i believe) with $x\in U_x$, a function $\varphi_x :U_x\to \varphi_x(U_x)\subset\mathbb{R}^n$ where $\varphi_x(U_x)$ is open.

My guess is that i need to find a way to map open sets in $\mathcal{M}$ to open sets on the line $[0,1]$ since that probably is a manifold, but i can not seem to find anywhere where it is all explained in a way that i understand.

答:
This $\mathcal{M}$ is going to be a manifold with boundary, and $\partial\mathcal{M} = \{(0,c), (1, a+b+c)\}$. You have a global chart $\varphi: \mathcal{M} \to [0,1]$, which is the projection into the first factor. Namely, $\varphi = \pi|_{\mathcal{M}}$, where $\pi\colon \Bbb [0,1]\times \Bbb R \to \Bbb [0,1]$ is given by $\pi(x,y) = x$. Then $\varphi$ is continuous because it is the restriction of a continuous map. The inverse is $\varphi^{-1}\colon [0,1] \to \mathcal{M}$ given by $\varphi^{-1}(x) = (x,ax^2+bx+c)$.

Replacing $[0,1]$ with any open interval would give a manifold without boundary instead. The fact that $a$, $b$ and $c$ are rational is irrelevant. Same strategy works for showing that if $f\colon \Bbb R^n \to \Bbb R^k$ is continuous, then the graph of $f$ is homeomorphic to $\Bbb R^n$, and thus it is a topological manifold (if $f$ is smooth instead, then a smooth manifold).


问:
Euler's Totient with primes?
I know how to solve for phi if a number has prime factorization, but how do you do it with a prime number? for eg, 13
.</p>

答:
Since $ \phi (n) $ denotes the number of integers x in the range $ 1 \leq x \leq n $ such that x and n are coprime and the definition of a prime number $p$ is that it's coprime with every $ x $ in the range of $ 1,...,p-1$.

So $ \phi (p) = p-1$, if p is a prime.

Title: Euler's Totient with primes?


答:
If you want to use your formula, you likely know that $$\phi(p^n) = p^n (1-1/p) = p^n-p^{n-1}$$
Thus $\phi(p) = \phi(p^1) = p^1-p^0=p-1$.

However, it is much easier to use the other definition, the $\#$ of relatively prime numbers to $p$ in the range $[1,p-1]$. All of these are relatively prime as $p$ is prime, and so $\phi(p) = p-1$

Title: Euler's Totient with primes?

Body:
I know how to solve for phi if a number has prime factorization, but how do you do it with a prime number? for eg, 13
.</p>


答:
If you want to give a basic idea to the students, I suggest you telling them this definition:

"If a function in x and y donot contain any powers on x and y terms, then the function is a linear function."

Now, the answer to your 1st question:

One of the books I trust is Introduction to linear algebra by Gilbert Strang, because I have myself read from this book.

Now since you mention that students know calculus, I would approach the problem the following way:

"If the slope (obtained by differentiation of the equation with respect to x) is a constant, then the equation is a linear one"

For example, we want to know whether x³-y³=9 is linear or not, then we do the following calculation:$$d(x^3)/dx - d(y^3)/dx = 0$$
$$3x^2 - 3y^2(dy/dx) = 0$$
$$dy/dx = x^2/y^2$$
Which is not constant as it depends on x and y.

Thus, $x^3 - y^3 =9$ is not a linear function.

Point to remember:

$dy/dx$ is called the slope of an equation and is constant for all linear equations.

Title: How to define linear equations in an introductory linear algebra class

Body:
I believe there is an issue of clarification with respect to the definition of linear equations in many linear algebra texts.

Here is a typical one ``A linear equation in the $n$ variables $x_1,x_2, ..., x_n$ is an equation that can be written in the form $a_1x_1+a_2x_2+ ... +a_nx_n=b$ where the coefficients $a_1,a_2,...,  a_n$ and the constant term $b$ are constants.''

No problem with the definition so far except that the text proceeds to state that $x^2-y^3=5$ is not linear because "it contains powers".  There is no further explanation. Of course a well-prepared student would say: $(x+y+5)^2-(x+y)^2=35$ appears to contain powers and is a linear equation, how do you know that $x^2-y^3=5$ cannot be simplified?

I want to say the issue is with "[it] can be written" in the definition. That some operational explanation should be added.

Here are my two questions:

(a) Which introductory textbooks go deeper in the issue?

(b) How do you approach the issue? (Again this is introductory linear algebra, students come in with just calculus I as prerequisite).


问:
USA MO 1980's inequality
Let $x;y;z$ be real numbers and $x;y;z \in [0;1]$

Prove: $\dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$

The official solution uses calculus, so I can't understand it.
I hope someone have a "simpler solution" (for middle-school student). Thank you.

答:
Assume $x+y+z>0$, else the inequality is easy. Do some rearranging:
\begin{align*}(1-x)(1-y)(1-z)\leq1-\sum\frac{x}{y+z+1}&=\sum\left(\frac{x}{x+y+z}-\frac{x}{y+z+1}\right) \\
&= \frac{1}{x+y+z}\sum\frac{x(1-x)}{y+z+1}
\end{align*}
Multiplying through by $x+y+z>0$, and moving everything to one side gives
$$\sum x(1-x)\left[\frac{1}{y+z+1}-(1-y)(1-z)\right]\geq0.$$
Now it's just AM-GM:
$$1=\frac{(1-y)+(1-z)+(y+z+1)}{3}\geq\sqrt[3]{(1-y)(1-z)(y+z+1)},$$
so $\frac{1}{y+z+1}\geq(1-y)(1-z)$, which implies the result.

Title: USA MO 1980's inequality


问:
Electrical networks, calculating voltage of a vertex
Given a finite graph $G$, and three vertices $v,s,t$, I am wondering about the probability that simple random walk starting at $v$ will reach $s$ before it reaches $t$. I know this is equal to $\phi(v)$, where $\phi$ is the unique harmonic function on $G$ with boundary $\{s,t\}$ such that $\phi(s) = 1,\phi(t) = 0$. Alternatively, if treat $G$ as an electric network, where current flows into $s$ and out from $t$, we get that $\phi(v)$ is the voltage at $v$.

Is a way to interpret $\phi(v)$ in terms of spanning tree counts? In practice, what are efficient ways to calculate/approximate $\phi(v)$ with a program?

答:
Let $G*$ denote $G$ with an additional edge $e$ connecting $s$ to $t$.  Denote by $A$ the collection of spanning trees in $G*$ that contain $e$. Then the voltage $\phi(v)$ is the fraction of  trees in $A$ that contain a path from $v$ to $s$ that does not include $e$. This follows from the property of the uniform spanning tree in $G$ with $s,t$ identified,  that the random path from $v$ to the conjoined vertex $\{s,t\}$ in the tree is obtained by a loop-erased walk; This result is due to Pemantle, extended by Wilson; see an exposition and detailed references in Chapter 4 of [1].

Let $n$  denote the number of vertices in $G$. While $\phi(v)$ can be estimated via random sampling, and also by relaxation methods the most efficient methods rely on $\phi(\cdot)$ being the unique solution of $n-2$ equations in $n-2$ unknowns. These equations reflect the harmonicity of $\phi$ at all nodes different from $s,t$. There are many linear systems solvers available online. The equations that arise here have a special nature, they are often sparse and specialized algorithms have been developed for them.
See e.g. [2] for a masterful 2010 survey and [3] for a more recent algorithm.

[1] Lyons, Russell, and Yuval Peres. Probability on trees and networks. Vol. 42. Cambridge University Press, 2017. Available from

https://rdlyons.pages.iu.edu/prbtree/

[2] Spielman, Daniel A. "Algorithms, graph theory, and linear equations in Laplacian matrices." In Proceedings of the International Congress of Mathematicians 2010 (ICM 2010) (In 4 Volumes) Vol. I: Plenary Lectures and Ceremonies Vols. II–IV: Invited Lectures, pp. 2698-2722. 2010.

[3] Kyng, R., Lee, Y.T., Peng, R., Sachdeva, S. and Spielman, D.A., 2016, June. Sparsified cholesky and multigrid solvers for connection laplacians. In Proceedings of the forty-eighth annual ACM symposium on Theory of Computing (pp. 842-850).


问:
Why we need complete space for the interchange of limits to be valid?
I am learning Real analysis using the book Analysis 2 by Terence Tao. I'm confused with the condition on which we can interchange the limit.

Concerning the below proposition, my question is that: Why we need the space Y to be complete ? What will happen if the space Y is not complete ?

Thank you for your help!

答:
Consider the following counterexample where $Y$ is not complete.

Let $X = [0,1]$ and $Y = (0,2)$. Let $E = (0,1)$. Consider the sequence $f^{(n)}:E\to Y$ with $f^{(n)}(x) := x+1/n$ and $f(x) := x$. Let $x_0 = 0$.

Then

$\lim_{x\to 0}f^{(n)}(x)=\frac1n$ for each $n$;

$f^{(n)}$ converges to $f$ uniformly on $E$.

But the limit of the sequence $(\frac1n)_{n=1}^\infty$ does not exist in $Y$.

In the conclusion of the proposition, it is said that the limit of the sequence
$$
(\lim_{x\to x_0;x\in E}f^{(n)}(x))_{n=1}^\infty
$$
exists, which is not an assumption.

One way to prove it is to show that it is a Cauchy sequence using the following estimate (for $x\in E$ near $x_0$ and large $n,m$):
$$
d_Y(g(x_0;n),g(x_0;m))
=d_Y(g(x_0;n),f^{(n)}(x))+d_Y(f^{(n)}(x),f^{(m)}(x))+d_Y(f^{(m)}(x),g(x_0;m))
$$
where
$
g(x_0;n):=\lim_{x\to x_0;x\in E}f^{(n)}(x)
$.

This is where you need completeness of $Y$. If $Y$ is not complete, then the limit cannot be guaranteed to exist.


答:
$3\equiv -1\mod(4)$

$2023\equiv -1\mod (4)$

$77\equiv 1\mod(4)$

$\Rightarrow$ $3(2023)^z+77\equiv(-1)(+1)+1\equiv 0\mod(4)$

If z is even.

For second question:

$2023=7\times 17^2\equiv 0\mod 7$

$\Rightarrow 3(2023)^z+77\equiv 0 \mod(7)$

Title: $x^{2}+y^{2}=3(2023)^{z}+77$: to prove that if $4 \mid x^{2}+y^{2} \implies 4\mid x^{2},y^{2}$

Body:
This is a math Olympiad problem, and actually, I was in it.
$x^{2}+y^{2}=3(2023)^{z}+77$ and $x,y,z \in \mathbb{N}$.

Prove that if $4 \mid x^{2}+y^{2} \implies 4\mid x^{2},y^{2}$
then find all $(x,y,z)$ solutions to the equation.

My attempt :

A claim:
$3(2023)^{z}+77$ Is divisible by $4$ iff $z$ is even.

But I can't prove this claim.
Do not use any modular arithmetic


问:
Is the multiplication operator on $\ell^p$ compact?
Let $p\in[1,\infty]$, $a\in\ell^\infty$ and $$T:\ell^p\to\ell^p\;,\;\;\;x\mapsto ax.$$ Note that $\left\|T\right\|=\left\|a\right\|_{\ell^\infty}$. It's easy to see that if $a\in\ell^1$, then $T$ is nuclear and hence compact.

Can we more generally conclude that $T$ is compact?

Assume $p<\infty$. The intuitive approach would be to define $$T_n:\ell^p\to\ell^p\;,\;\;\;x\mapsto(a_1x_1,\ldots,a_nx_n,0,\ldots)$$ for $n\in\mathbb N$ and to show that $\left\|T_n-T\right\|\to0$. Now, $$\left\|(T_n-T)x\right\|^p=\sum_{i=n+1}^\infty|a_ix_i|^p\tag1$$ for all $x\in\ell^p$ and $n\in\mathbb N$. The right-hand side of $(1)$ clearly converges to $0$ as $n\to\infty$, but this only yields convergence in the strong operator topology.

On the other hand, if $(a_n)_{n\in\mathbb N}$ would be nonincreasing and $a_n\to0$ (e.g. $a_n=\frac1n$), then $$\left\|(T_n-T)x\right\|^p\le|a_{n+1}|^p\left\|x\right\|_{\ell^p}^p\tag2$$ for all $x\in\ell^p$ and $n\in\mathbb N$ and we could immediately conclude $\left\|T_n-T\right\|\to0$.

EDIT: Maybe the general result can be obtained from the convergence in strong operator topology (which holds by $(1)$) by noting that $\ell^p$ has the approximation property; but I'm not sure whether this is true.

答:
First of all, you cannot conclude that an operator is compact if it is the strong limit of finite rank operators, for example in any Hilbert space the identity operator is the strong limit of a net of finite-rank projections, but of course it is compact if and only if the dimension of our space is finite.

As Timur Bakiev says, the multiplication operator $T:\ell^p\to\ell^p$ is compact if and only if the sequence of $\ell^\infty$ that acts on $\ell^p$ belongs to $c_0$.

You already showed the converse: In your notation, $$\|(T_n-T)x\|^p=\sum_{k=n+1}^\infty|a_k|^p\cdot|x_k|^p\leq\sup_{k\geq n+1}|a_k|^p\cdot\|x\|^p$$
so $\|T_n-T\|\leq\sup_{k\geq n+1}|a_k|\to0$.

Conversely, suppose that $T$ is compact. We will use the following lemma from real analysis: if $(t_n)$ is a sequence of real numbers and every subsequence of $(t_n)$ has a further subsequence converging to $t$, then $(t_n)$ converges to $t$. See this post for example: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.

By definition of compactness, if $(x_n)\subset\ell^p$ is a bounded sequence, $(Tx_n)$ has a convergent subsequence in $\ell^p$. Fix indexes $n_1<n_2<n_3<\dots$ and consider the elements $e_k\in\ell^p$ having $0$ everywhere and $1$ in the $n_k$ position. Then $\|e_k\|=1$, so this sequence is bounded and therefore $(Te_k)$ has a convergent subsequence in $\ell^p$. To simplify my notation, I will assume that $n_1=1, n_2=2,\dots$ but the method below can be applied to arbitrary subsequences too, you can verify this yourself easily.

So as I said, let $e_n$ be the sequence having $0$ everywhere and $1$ in the $n$-th position, so $\|e_n\|=1$ for all $n$. Note that $Te_n$ is the sequence that has $0$ everywhere and $a_n$ in the $n$-th position, so $\|Te_n\|_p=|a_n|$. Now $(Te_n)$ has a convergent subsequence, say $(Te_{n_k})$ and say that $Te_{n_k}\to y\in\ell^p$, thus $\|Te_{n_k}-y\|_p^p\to0$. If $y=(y_i)$, then
$$\|Te_{n_k}-y\|_p^p=\sum_{i=1}^{n_k-1}|y_i|^p+|a_{n_k}-y_{n_k}|^p+\sum_{i=n_k+1}^\infty|y_i|^p$$
Fix an integer $m$. For $\varepsilon>0$ choose $k>m$ so large so that $\|Te_{n_k}-y\|_p^p<\varepsilon$.

Then $|y_m|^p\leq\sum_{i\neq n_k}|y_i|^p+|a_{n_k}-y_{n_k}|^p=\|Te_{n_k}-y\|_p^p<\varepsilon$ and as $\varepsilon>0$ was arbitrary, we conclude that $y_m=0$, so $y=0$ since $m$ was arbitrary before fixing. We conclude that $|a_{n_k}|\to0$.

Now as I said, this can be done for any subsequence (this is why I fixed $n_1<n_2<\dots$ in the beginning, nevermind that I eventually went wLOG to the case $n_1=1,n_2=2\dots$), so the lemma will yield that $|a_n|\to0$, i.e. $a_n\in c_0$.


答:
"how do you know that $x^2 - y^3 = 5$ cannot be simplified"

For a finite system with finitely many variables, one can take derivatives of the left-hand side.  Namely, hold $y$ constant, differentiate with respect to $x$ twice, and collect the like powers--this is now a little easier to do, and that's is the only merit of this method.  If you don't get a zero, the polynomial is nonlinear.  Similarly for $y$: hold $x$ constant, differentiate with respect to $y$ twice.  Do we get a zero?  If not, the polynomial is nonlinear.  For the polynomial to be linear, it is necessary and sufficient that each of these second-derivative tests (for each variable involved) gives us zero.

Trying the above test on $(x+y+5)^2-(x+y)^2=35$, we see that the second derivative w.r.t. $x$ on the left gives us $0$, and the same for $y$.

Generally, though, the definition of a concept does not always give an immediate recipe for testing each object as to whether it satisfies the definition.  If you've seen limits in calculus, you probably noticed that the definition of a limit gives no recipe for determining whether a limit exists, and if it does, what its value is.

The issue of "linear equations" gets a little easier to deal with once the concept of a linear transformation is introduced.  Unfortunately, many courses try to teach "linear equations" before introducing linear transformations.

An equation is linear if it is (or can be) written in the form
$$Ax = b,$$
where $A$ is a linear transformation, $x$ is a sought vector in the domain of $A$, and $b$ is the given vector in the range of $A$.

This is similar to what you quoted from your text.  However, it makes it easier to see that the linearity of the equation can be tested by taking two solutions, $x_{1}, x_{2}$ and seeing whether their difference is a solution of the homogeneous version
$$
Ax = \vec{0}
$$
of the original equation (think of this difference as a finite difference that approximates the first of the aforementioned derivatives).  So, we are a little better off.  If $A$ is given in a complicated, but explicit form, we can always consider numerical tests of linearity.  Plus, this latter, coordinate-free, definition echoes the Superposition Principle, which pervades physics and engineering.

Just out of curiosity, I looked into Kurosh's Higher Algebra (a classic in its own time and a champion of clarity).  It jumps into systems of linear equations from the get-go and doesn't seem too concerned with polynomials that, upon collecting the like powers, turn out linear.  I haven't seen any textbook on linear algebra sweat this too much; perhaps, experience shows that testing linearity is usually not that much of difficulty, probably because linearity is so scarce.


答:
Show $ \ \dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$

WLOG, $x \geq y \geq z$

Then $ \ \dfrac{y}{z+x+1} \leq \dfrac{y}{y+z+1} \ $ and $ \ \dfrac{z}{x+y+1} \leq \dfrac{z}{y+z+1}$

So LHS $\leq \dfrac{x+y+z}{y+z+1}$ and hence it suffices to show,

$(1-x)(1-y)(1-z) \leq 1 - \dfrac{x+y+z}{y+z+1} = \dfrac{1-x}{y+z+1}$

or it suffices to show $(1-y)(1-z) \leq \dfrac{1}{(1+y)(1+z)} \ $

[as $ \ \dfrac{1}{(1+y)(1+z)} \leq \dfrac{1}{y+z+1}$]

or to show that $(1-y^2)(1-z^2) \leq 1 \ $ which is obviously true.

Title: USA MO 1980's inequality

Body:
Let $x;y;z$ be real numbers and $x;y;z \in [0;1]$

Prove: $\dfrac{x}{y+z+1}+\dfrac{y}{z+x+1}+\dfrac{z}{x+y+1} \leq 1-(1-x)(1-y)(1-z)$

The official solution uses calculus, so I can't understand it.
I hope someone have a "simpler solution" (for middle-school student). Thank you.


问:
A group of order $150$ has at least $4$ conjugacy classes made by elements of order a power of $5$?
Let be $G$ a group of order $150$ and $g \in G\,\,$ s.t. $|g|=25\,$:

Prove that $\exists \,\,K\lhd G\,\,$s.t. $K \simeq C_5$. $\,\,$ Say whether $\exists \,\,h \in G\,\,$ s.t. $|h| =15$.$\,\,$ Prove that if $\,\,\exists\,\, t\in G/K$ s.t. $\,\,|t|=15$ then $G/K$ has only one $5$-Sylow; in this case how many elements of order $15$ does $G$ have? $\,$Prove that $G$ has at least $4$ conjugacy classes made by elements of order a power of $5$.

I am struggling with the last part of this exercise, any help will be greatly appreciated. My solution:

We know, by hypothesis, that $\exists \,\,g\in G$ s.t. $|g|=25\,\,$ and this means that $C_{25} \leqslant G\,$, where $\langle g \rangle = C_{25}\,$, which is abelian because $25$ is the square of a prime, but it is also a $5$-Sylow of $G$, because $150=2 \cdot 3 \cdot 5^2$, hence, for the Sylow's Theorems, the conjugacy action of $G$ on $C_{25}$ is transitive, which implies $C_{25} \lhd G\,\,$. Observing that $g^5 \in C_{25}$ and that $|g^5|=5$ in $C_{25}$, we have $C_5 \simeq \langle g^5 \rangle \lhd C_{25}\,\, \Rightarrow \,\, C_5 \lhd G$.

$G$ has an element $h$ of order $15 \iff C_{15}\leqslant G$. Let be $P \in{\rm Sylow}_3(G)$, now $|P| = 3$ which is a prime, therefore $P \simeq C_3$. We have seen that $C_5 \lhd G$, thus $H := C_3C_5 \leqslant G$. From this: $C_3 \cap C_5 = \{1\}$, $|C_3C_5| = 15 = |H|$ and $C_3, C_5 \lhd H$ because they are the only $3$-Sylow and the only $5$-Sylow in $H$; hence $H \simeq C_3 \times C_5 \simeq C_{15} \leqslant G$. $h$ is the generator of $C_{15}$.

By hypothesis $C_{15} \leqslant G/K$, with $K \simeq C_{25}$. Now $|G/K|=30 =3 \cdot 2 \cdot 5$ and $[G/K : C_{15}]=2$, which is the minimum prime that divides the order of $G/K$, that implies $C_{15} \lhd G/K\,\,$. We can observe that $C_{15}$ has only one $5$-Sylow $Q$, hence $Q \lhd C_{15} \lhd G/K$, therefore $Q \lhd G/K$, which means that $Q$ is the only $5$-Sylow of $G/K$.

Any idea about the last two requests? Thank you.

答:
This is just a hint for the final question. Since the order of the group is not divisible by $4$ (the order of ${\rm Aut}(C_5)$), there must be at least two conjugacy classes of elements of order $5$ that lie in $K$.

Similarly, if $g$ has order $25$, then by considering $\langle g \rangle/K$, there must be at least two conjugacy classes of elements of order $25$ that lie in $\langle g \rangle$.


问:
How can I find the inverse of $y = \frac{x}{1-x^2}$?
In example $1.5$ of Cracking the GRE Subject Test, the authors make the following calculation in one step with no additional commentary:

we interchange $x$ and $y$ and solve for $y$:
\begin{align}
\vdots\\
xy^2 + y-x &= 0\\
y &= \frac{-1 \pm \sqrt{1+4x^2}}{2x}
\end{align}

What technique allows such a breezy solution?

答:
Using the quadratic formula for $ay^2 + by +c = 0$

$$y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Here $a = x$, $b = 1$, $c = -1$


问:
Are these two events mutually exclusive?

Let's consider event A and B where $\Pr(A)\neq 0$,$B=\emptyset$.

Are these events mutually exclusive since $\Pr(A\cap B)=\Pr(A\cap \emptyset)=0?$

What if event A = event B and $\Pr(A)=\Pr(B)=0$ but $A,B \neq \emptyset$

Are these events also mutually exclusive since $\Pr(A\cap B)=0?$

But for the 2nd part, $A\cap B =A=B\neq \emptyset$, so not mutually exclusive?

答: If two events are mutually exclusive then it is impossible for both to happen. This is stronger than saying there is probability $0$ that both happen. So the correct condition is $A\cap B=\emptyset$, not $\Pr(A\cap B)=0$. In your first case they are mutually exclusive and in the second they are not.


问:
Can the concept of field extensions be applied equally well to UFDs?
In a nutshell, a field extension is where you take a polynomial $p(x)$ that is irreducible in some field $F$, then define $\alpha$ as a root of $p$, then add $\alpha$ to $F$, then add the minimum number of additional elements to $f$ to maintain its field structure.

Can this still be done if $F$ is merely a UFD rather than a field?  Does something fundamental break down in the construction of a "UFD extension" that goes through in a field extension?  If not, why do people talk about field extensions when "UFD extension" seems more general?

答: A simple example of a non-UFD extension of a UFD is $\, \Bbb Z \subset R =\Bbb Z[x^2,x^3] \subset \Bbb Z[x].\,$ By $\,x\not\in R\,$ both $\,x^2,\, x^3$ are irreducible, but not prime. Indeed,  the factorization $\, (x^2)^3\! = (x^3)^2$ is not unique.</p>


答:
Mutual exclusivity is defined to mean that the intersection of the events is empty.

A consequence of the definition is that if two events are mutually exclusive then the probability of their intersection is zero.  The reverse however is not true.  It is possible for two events to have the probability of their intersection be zero while not being mutually exclusive.

"Are these events mutually exclusive since $\Pr(A\cap B)=0$?"  Not for that reason, no.  The first will be mutually exclusive since every event is mutually exclusive to the empty event.  The second will not be mutually exclusive since we are told $A\neq \emptyset$ and that $A=B$, hence $A\cap B\neq \emptyset$ as you correctly state in your post.

Title: Are these two events mutually exclusive?

Body:
Let's consider event A and B where $\Pr(A)\neq 0$,$B=\emptyset$.

Are these events mutually exclusive since $\Pr(A\cap B)=\Pr(A\cap \emptyset)=0?$

What if event A = event B and $\Pr(A)=\Pr(B)=0$ but $A,B \neq \emptyset$

Are these events also mutually exclusive since $\Pr(A\cap B)=0?$

But for the 2nd part, $A\cap B =A=B\neq \emptyset$, so not mutually exclusive?


问:
Integrating $\int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx$ with $x=a \sin^2 u$ - limits query
I had to evaluate the integral $$\int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx$$

I used the substitution $x=a \sin^2 u$ for this and got the correct answer of $\frac{\pi a^2}{8}$

However, part of my method involved the changing of the limits.

For example, when $x=0$, I reduce to $\sin u = 0$, which has multiple solutions $u=0, \pi, 2\pi, ...$. I chose $u=0$ to replace $x=0$, but why would it be incorrect to choose the others?

Similarly with $x=a$, I get $\sin^2 u = 1$.  I happened to choose $\sin u = 1$ but again why would it be incorrect to pick $\sin u = -1$ and use $u = -\frac{\pi}{2}$? It ends up giving the negative of the correct answer...

答:
You can choose whatever solution you want, provided that you choose the other one according to the first one. This is due to the periodicity of trigonometric function.

For example, your substitution reads $x = a\sin^2(u)$, which means a change in the first extrema as

$$\sin^2(u) = 0 \longrightarrow x = 0$$
And consequently the second extrema needs to be chosen with the "principal" solution:
$$\sin^2(u) = 1 \longrightarrow x = \frac{\pi}{2}$$

Say you want to choose the second solution, in the first extrema that is,

$$\sin^2(u) = 0 \longrightarrow x = \pi$$
allowed of course since it fits.

The other extrema then needs to be chosen in a shifted way according to the first solution that is,

$$\sin^2(u) = 1 \longrightarrow x = \frac{\pi}{2} \color{red}{ + \pi}$$

If you choose $x = 2\pi$ as a solution for the first extrema, then the second one shall be $x = \frac{\pi}{2} \color{red}{+ 2\pi}$

Indeed we have

$$\int_0^{\pi/2} \sin^2(u)\cos^2(u)\ \text{d}u = \frac{\pi }{16}$$

as well as

$$\int_{\pi}^{\pi/2 + \pi} \sin^2(u)\cos^2(u)\ \text{d}u = \frac{\pi }{16}$$

as well as

$$\int_{2\pi}^{\pi/2 + 2\pi} \sin^2(u)\cos^2(u)\ \text{d}u = \frac{\pi }{16}$$

and in general

$$\int_{0 + k\pi}^{\pi/2 + k\pi} \sin^2(u)\cos^2(u)\ \text{d}u = \frac{\pi }{16} ~~~~~~~~~~~ k\in\mathbb{Z}$$


Title: How can I find the inverse of $y = \frac{x}{1-x^2}$?

Body:
In example $1.5$ of Cracking the GRE Subject Test, the authors make the following calculation in one step with no additional commentary:

we interchange $x$ and $y$ and solve for $y$:
\begin{align}
\vdots\\
xy^2 + y-x &= 0\\
y &= \frac{-1 \pm \sqrt{1+4x^2}}{2x}
\end{align}

What technique allows such a breezy solution?

Body:
$$y = \frac{x}{1-x^2}$$

Replace $x$ with $y$

$$x = \frac{y}{1-y^2}$$

and solve $y,x=\frac{y}{1-y^2}$

$$-\frac{1\pm\sqrt{4x^2+1}}{2x}$$


问:
Why is $\mathbb{Z}[\alpha ]$ not finitely generated as $\mathbb{Z}$-module?

Assume that $\alpha \in \mathbb{C}$ is an algebraic number which is not an algebraic integer. My question is why $\mathbb{Z}[\alpha]$ is not finitely generated as  $\mathbb{Z}$-module.

Clearly there exists $ a_i \in \mathbb{Z}$, which are not all zero and $n>0$ such that $a_n\alpha^n+a_{n-1}\alpha^{n-1}+\dots + a_1\alpha + a_0=0$. Shouldn't this imply that $\mathbb{Z}[\alpha]$ is finitely generated? The motivation for this question is that it may be shown that $\alpha \in \mathbb{C}$ is an algebraic integer if and only if $\mathbb{Z}[\alpha ]$ is finitely generated as a $\mathbb{Z}$-module. Why does the coefficient $a_n$ has to be one for the module to be finitely generated? And, to check that I haven't misunderstood the terminology, $\mathbb{Z}[\alpha ]$ is equal to all polynomials in $\alpha$ over $\mathbb{Z}$, right?

答: What you want is to be able to express $\alpha^n$ and all other higher powers of $\alpha$ as a $\mathbb{Z}$-linear combination of the powers from $0$ to $n-1$. But if $a_n$ is not $\pm 1$, you will not be able to do this.</p>


答:
You may choose any of the intervals
$$u\in \cdots (-\frac\pi2,0), (0,\frac\pi2),(\frac\pi2,\pi)\cdots$$
Take $u\in (-\frac\pi2,0)$ under $ x=\sin^2 u$, for the example,

$$\int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx
= a^2\int^{-\frac\pi2}_0(-\sin u)\cos u\>d(\sin^2u)=\frac\pi8a^2
$$

All the interval choices result in the same result. In practice, though, the choice $(0,\frac\pi2)$ is the most straightforward.

Title: Integrating $\int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx$ with $x=a \sin^2 u$ - limits query

Body:
I had to evaluate the integral $$\int_0^{a} x^{\frac{1}{2}} (a-x)^{\frac{1}{2}} dx$$

I used the substitution $x=a \sin^2 u$ for this and got the correct answer of $\frac{\pi a^2}{8}$

However, part of my method involved the changing of the limits.

For example, when $x=0$, I reduce to $\sin u = 0$, which has multiple solutions $u=0, \pi, 2\pi, ...$. I chose $u=0$ to replace $x=0$, but why would it be incorrect to choose the others?

Similarly with $x=a$, I get $\sin^2 u = 1$.  I happened to choose $\sin u = 1$ but again why would it be incorrect to pick $\sin u = -1$ and use $u = -\frac{\pi}{2}$? It ends up giving the negative of the correct answer...


问:
Prove whether a language is finite or infinite.
Consider the language :

$L = \{w \mid w \in \{0, 1\}^∗, w \text{ has three times as many }1\text{'s as }0\text{’s}\}.$

Say whether $L$ is finite or infinite, and prove that you are correct.

I am assuming that the language described above is infinite. But, I am not sure how to set up the proof. What type of proof should I use? Contradiction, Inductive?
Any help or tips appreciated.

答: A subset of your language is $\{ (0 1 1 1)^n \colon n \ge 0 \}$ (each zero is followed by three ones), and that one is infinite. Thus your language is infinite.


问:
How to calculate this determinant (Finding characteristic polynomial).
I am trying to calculate this determinant in order to find a characteristic polynomial of a matrix, however I couldn't see any operation that I can do to make my life easier calculating this, so I tried to calculate the determinant according to column $3$ since it has a zero, but it was such a long journey and in the end I got a polynomial with degree of $3$, so I also needed to find its roots (eigen values) once again and it took alot of time.

So I'm wondering if there's any easy way to calculate this determinant since most likely I just couldn't see the right operation to do to calculate it and took the long road.

Here's the determinant:
\begin{vmatrix} x-2 &    -3    & -2  \\ 1 & x-3 & 0 \\ -1 & -1 & x+1\end{vmatrix}

Any feedback and tips are really appreciated, thanks in advance.

答:
Hint: There a simple method, you have just to do:
\begin{align}
(x-2)\begin{vmatrix}
x-3 & 0 \\
-1 & x+1
\end{vmatrix}
+
(-1)(-3)\begin{vmatrix}
1 & 0  \\
-1 & x+1
\end{vmatrix}
+
(-2)\begin{vmatrix}
1 & x-3  \\
-1 & -1
\end{vmatrix}
\end{align}

Can you continue from here?


问:
Push-through rule for square root of specific projection matrix
Is the following a valid identity?

$$
(I - XX^T)^{1/2} X = X (I - X^T X)^{1/2}
$$

Without the square root, it is quite easy to see that

$$
(I - XX^T) X = X (I - X^T X)
$$

holds but I have trouble verifying that the upper equation holds.

答:
Making use of the SVD $X = U \Sigma V^T$, one arrives at

$$
(I - X X^T)^{1/2} X = U (I - \Sigma^2)^{1/2} \Sigma V
$$

and

$$
X (I - X^T X)^{1/2} = U \Sigma (I - \Sigma^2)^{1/2} V
$$

where the diagonal entries may be swapped, $\Sigma (I - \Sigma^2)^{1/2} = (I - \Sigma^2)^{1/2} \Sigma$. Hence the identity holds!


问:
Is $\tau(x,y)\leq\sqrt{n}d(x,y)$ where $d$ is the Euclidean metric, and $\tau$ the taxicab metric?
Suppose $d$ is the Euclidean metric on $\mathbb{R}^n$, $d(x,y)=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}$, and $\tau(x,y)=\sum_{i=1}^n|x_i-y_i|$ is the taxicab metric on $\mathbb{R}^n$.

When showing this two metrics are equivalent, it is not hard to bound $\tau$ by
$$
\tau(x,y)\leq n d(x,y).
$$

Is it possible to tighten the bound to $\tau(x,y)\leq\sqrt{n}d(x,y)$?

答:
I think the counterexample forgot to take the square root of $d$. Instead, this follows from the Cauchy-Schwarz inequality.

If $x,y$ are in $\mathbb{R}^n$,
$$
\tau(x,y)=\sum_{i=1}^n|x_i-y_i|=\left|\sum_{i=1}^n|x_i-y_i|\right|\leq\sqrt{\sum_{i=1}^n (x_i-y_i)^2}\sqrt{\sum_{i=1}^n 1^2}=\sqrt{n}d(x,y).
$$


问:
Why does a helix have a radius of curvature of $(r^2+h^2)/r$?
I can follow the math to calculate the radius of curvature of a helix, but what has me confused is there is no $z-$component to the normal vector, so the normal vector points at the $z$ axis at all times.  Therefore, I would think the radius of curvature would just be $r$.  If the radius of curvature is $(r^2+h^2)/r$, which is larger than $r$, then that means the center of rotation is further away from a moving point than the $z-$axis, and yet the moving point seems to rotate about the $z-$axis, which is only a distance of $r$ away.

答:
The radius of curvature is larger than the radius of the cylinder because the helix is climbing as it winds. That means the best matching circle will be larger than the circle of the cylinder.

If you imagine a helix that climbs very steeply you can see that it's locally nearly a straight line, so tangent to a very large circle.


问:
Similarity by SSA (side-side-angle) in obtuse triangles
Source: Challenge and Thrills of Pre-College Mathematics, Page 74, Problem 54:

"In two obtuse triangles, an acute angle of the one is equal to an angle of the other, and sides about the other acute angles are proportional. Prove that the triangles are similar."

I tried this first by cosine rule, and the question reduced to:

Assume $\triangle ABC$ and $\triangle DEF$ with $\angle A$ and $\angle D$ being obtuse, $\angle B=\angle E$ and $AC/BC=DF/EF$. We may take $DF=k\cdot AC$, $EF=k\cdot BC$ and apply $\cos B= \cos E$, obtaining $DE=k\cdot AB$ and the problem is solved. EDIT: Even this however is difficult to prove, as pointed out by cosmo5 in his answer.

What is interesting and challenging is to prove the same without the cosine law (probably the expected solution, as cosine law doesn't seem to help), as the law hasn't been introduced in the book until later chapters, while this problem is taken from chapter 3.

My attempt:

Fix $\triangle ABC$ as well as points $E$ and $F$ of $\triangle DEF$. Assuming $DF=k\cdot AC$, $EF=k\cdot BC$, the locus of $D$ will be a circle with centre $F$ and radius $k\cdot AC$.

After this however, I couldn't go further with the same concept. Either a hint or a solution related to my thought process, or any other thought process for that matter, would be greatly appreciated.

答: Hint: note that there are at most two points on ray $BA$ which is of a certain distance from point $C$ (because you are intersecting a ray with a circle).


问:
Is $GCD(2^n,(x^2+1)^{n-1})=1$
Let $x>1,n>2$ be positive integers. I come across the following problem:

Is $$GCD(2^n,(x^2+1)^{n-1})=1$$

or $2^n$ is relatively prime to $(x^2+1)^{n-1}$ ?

Is $$GCD(x^{2n},(x^2+1)^{n-1})=1$$ ?

NB: $GCD$ is the greatest common divisor.

Title: Is $GCD(2^n,(x^2+1)^{n-1})=1$

Body:
If $x=3$, then
$$ (x^2+1)^{n-1}=10^{n-1} $$
and hence
$$ \gcd(2ⁿ,(x²+1)ⁿ⁻¹)=2^{n-1}\neq1.$$
The 2nd one is correct.


问:
Convergence of the sum of columns/rows of a matrix
Let $B$ be a real invertible $N\times N$ matrix, and $O$ the $N\times N$ matrix with all the entries equal to 1.
Define the $\epsilon$ dependent matrix $A(\epsilon)$ as
$$
A(\epsilon) = O + \epsilon B,\quad \epsilon\in\Bbb R.
$$

$A(\epsilon)$ is invertible for each $\epsilon\neq0$, so we can let $C(\epsilon)$ be the inverse of $A(\epsilon)$ for all $\epsilon\neq0$.
As $\epsilon\to0$ the matrix $A(\epsilon)$ converges to $O$, so clearly $C(\epsilon)$ diverges as $\epsilon\to0$.  Anyways, it happens that both the sum of the rows and the sum of the columns of $C(\epsilon)$ converge as $\epsilon\to0$.  I have clear numerical evidence of this fact, but I don't know how to formally prove it.  Can you help me?

答:
Notice that $O = ee^T$, where $e$ is the all ones vector. As a consequence, if you suppose everything is invertible, you can use the Sherman-Morrison formula and get
$$
(O + \epsilon B)^{-1}e = (\epsilon B)^{-1}e - \frac{(\epsilon B)^{-1}ee^T(\epsilon B)^{-1}}{1+e^T(\epsilon B)^{-1}e}e
\\=
\frac 1{1+e^T(\epsilon B)^{-1}e}
(\epsilon B)^{-1}e \\
=\frac 1{\epsilon+e^TB^{-1}e}
B^{-1}e
$$
This quantity explodes only if $e^TB^{-1}e = 0$, so to get a counterexample, take for example
$$
B = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}
$$
and you will get sum of row/columns of the order $O(1/\epsilon)$.


问:
How to find inverse function of function with parameter?
I'm trying to find the inverse function of this function:
$$
f_a(x) = \frac{x^a}{x^a + (1-x)^a}
$$

$a$ - parameter

$x$ - variable

Is it possible to find if there is a parameter here?

Hope for your help! Thanks in advance!

答:
$$\require{cancel}
f_a\circ f_b(x)=\dfrac{\left(\dfrac{x^b}{\cancel{x^b+(1-x)^b}}\right)^a}{\left(\dfrac{x^b}{\cancel{x^b+(1-x)^b}}\right)^a+\left(1-\dfrac{x^b}{\cancel{x^b+(1-x)^b}}\right)^a}\\
=\dfrac{x^{ab}}{x^{ab}+(\cancel{x^b}+(1-x)^b-\cancel{x^b})^a}\\
=\dfrac{x^{ab}}{x^{ab}+(1-x)^{ab}}
=f_{ab}(x)$$

So we have $\quad f_a\circ f_b = f_{ab}$

You can verify that since $f_1(x)=\dfrac{x}{x+(1-x)}=x$ then $f_a^{-1}=f_{\frac 1a}$


问:
Understanding Bayes' Theorem
I worked through some examples of Bayes' Theorem and now was reading the proof.

Bayes' Theorem states the following:

Suppose that the sample space S is partitioned into disjoint subsets $B_1, B_2,...,B_n$.  That is, $S = B_1 \cup B_2 \cup \cdots \cup B_n$, $\Pr(B_i) > 0$ $\forall i=1,2,...,n$ and $B_i \cap B_j = \varnothing$ $\forall i\ne j$.  Then for an event A,

$\Pr(B_j \mid A)=\cfrac{B_j \cap A}{\Pr(A)}=\cfrac{\Pr(B_j) \cdot \Pr(A \mid B_j)}{\sum\limits_{i=1}^{n}\Pr(B_i) \cdot \Pr(A \mid B_i)}\tag{1}$

The numerator is just from definition of conditional probability in multiplicative form.

For the denominator, I read the following:

$A= A \cap S= A \cap (B_1 \cup B_2 \cup \cdots \cup B_n)=(A \cap B_1) \cup (A\cap B_2) \cup \cdots \cup(A \cap B_n)\tag{2}$

Now this is what I don't understand:

The sets $A \cup B_i$ are disjoint because the sets $B_1, B_2, ..., B_n$ form a partition.$\tag{$\clubsuit$}$

I don't see how that is inferred or why that is the case.  What does B forming a partition have anything to do with  it being disjoint with A.  Can someone please explain this conceptually or via an example?

I worked one example where you had 3 coolers and in each cooler you had either root beer or soda.  So the first node would be which cooler you would choose and the second nodes would be whether you choose root beer or soda.  But I don't see why these would be disjoint.  If anything, I would say they weren't disjoint because each cooler contains both types of drinks.

Thank you in advance! :)

答: It is the sets $A\cap B_i$ that are pairwise disjoint, and that is precisely what you need to calculate the probability in the denominator.</p>

Title: Understanding Bayes' Theorem


问:
A function with a condition
This question was asked for one of our internal tests (high school)

$\mathit f(x):$  $ $$\Bbb R  \mapsto \Bbb R $ is a continuous function and has the property $\;\mathit f(f(x)) = 1-x \,\,$$\,\forall \,\, x \in [0,1] $ and $\mathit J= \int _0 ^1 f(x) dx$, then find the value of $\mathit 1/J.$

Will someone provide an example of such a function.

NB: I do not require a solution, please provide me only an example of such a function.

答:
Hints:

Show that $f(x)$ must be bijective and, therefore, strictly increasing or decreasing.

Show that $f(f(x))$ is strictly increasing no matter if $f(x)$ is increasing or decreasing.

Observe a contradiction since $1-x$ is strictly decreasing. So there is no such function.

Title: A function with a condition


问:
Rank of projection error
Let $X_1$,$X_2$ be $n\times k$ real matrices with rank $k$. Let $S_1,S_2$ denote the column spaces of $X_1,X_2$ respectively. I am trying to show that the rank of $MX_1 $, with $M:=I_n-X_2(X_2'X_2)^{-1}X'_2$, equals the dimension of $S_1 \cap S_2^\perp$.

I know that the matrix above is the error arising from the projection of each column of $X_1$ on $S_2$, but don't know where to go from there. I notice that we have $X_2'MX_1=0$.

Any help on this is greatly appreciated.

答: I don't think what you're saying is true. Define $X_1,X_2 \in \mathbb{R}^{2 \times 1}$ by $$X_1=(1,0)^T$$ $$X_2=(1,1)^T$$ Then $S_1$ is the $x-$ axis, $S_2$ is the line $y=x$, and $S_2^{\perp}$ is the line $y=-x$. Notice $S_1 \cap S_2^{\perp}=\{(0,0)^T\}$ has dimension $0$. On the other hand, your $2\times 1$ matrix $MX_1=(0.5,-0.5)^T$ has rank $1$.

Title: Rank of projection error


问:
What do you call a "chunk" of a circle
Like, what do you call a third of a circle, or a quarter of a circle</p>

答: A circular sector. The larger 'chunk' is called the major sector and the smaller chunk the minor sector.

Title: What do you call a "chunk" of a circle


问:
Can the order of adjacent and same quantifiers be mixed?
So can we mix say $\forall x \forall y$ to $\forall y \forall x$ or $\exists y \exists x$ to $\exists x \exists y$ for any possible predicate? And in general, could you mix the order of quantifiers up to a differing quantifier, so $\forall x \exists y \exists z$ could be changed to $\forall x \exists z \exists y$?

答:
Yes, quantifiers of the same type are commutative:

$$\forall u \forall v \phi \equiv \forall v \forall u \phi \text{ and }\exists u \exists v \phi \equiv \exists v \exists u \phi$$

for all variables $u, v$ and formulas $\phi$.

If two subformulas are logically equivalent, so are all formulas they occur in, so $\forall x \exists y \exists z \phi \equiv \forall x \exists z \exists y \phi$ also.

However, $\forall$ and $\exists$ are not commutative with each other: $$\exists v \forall u \phi \vDash \forall u \exists v \phi$$ but $$\forall u \exists v \phi \not \vDash \exists v \forall u \phi$$ i.e., whenever $\exists v \forall u \phi$ is true so is $\forall u \exists v \phi$, but not vice versa.

The proof comes down to the fact the variable assignments and quantification over them can be swapped on the meta level:

First observe that for a variable assignment $v$ and two objects $a, b$, $$v'' = v[x \mapsto a][y \mapsto b] = v[y \mapsto b][x \mapsto a] = v'$$ -- for two distinct variables $x$ and $y$, it doesn't matter in which order we compute the assignment modifications. Then:

$\newcommand{\A}{\mathfrak{A}}
\begin{align*}
& \A \models_v \forall x \forall y \phi\\
\iff & \text{for all $x$-variants $v'$ of $v$ : } \A \models_{v'} \forall y \phi\\
\iff & \text{for all $x$-variants $v'$ of $v$ and all $y$-variants $v''$ of $v'$: } \A \models_{v''} \phi\\
\iff & \text{for all  $y$-variants $v''$ of $v'$ and all $x$-variants $v'$ of $v$: } \A \models_{v'} \phi\\
\iff & \text{for all $y$-variants $v''$ of $v'$ : } \A \models_{v''} \forall x \phi\\
\iff & \A \models_v \forall y \forall x \phi
\end{align*}$

The middle step is justified because we can swap the two "all"s in the meta language (English) -- becaue that's just how the meaning of "all" works.

Analogous for the commutativity of existence.


答:
The standard proof that $\mathbb Z[\alpha]$ is finitely generated if $\alpha$ is an algebraic integer uses the fact that if
$\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+a_0=0$ then $\alpha^n=-(a_{n-1}\alpha^{n-1}+\cdots+a_0)$, so for any $m\geq n$ we have
$$\alpha^m=-(a_{n-1}\alpha^{m-1}+\cdots+a_0\alpha^{m-n})$$
i.e. $\alpha^m$ is in the module generated by $1,\alpha,\ldots,\alpha^{m-1}$. By induction this shows us that $\alpha^m$ is in the module generated by $1,\alpha,\cdots,\alpha^{n-1}$, hence $\mathbb Z[\alpha]$ is generated by these elements.

If $a_n$ is not a unit this all breaks down, because we no longer have a way to write $\alpha^n$ in terms of $a_{n-1}\alpha^{n-1}+\cdots+a_0$.

Title: Why is $\mathbb{Z}[\alpha ]$ not finitely generated as $\mathbb{Z}$-module?

Body:
Assume that $\alpha \in \mathbb{C}$ is an algebraic number which is not an algebraic integer. My question is why $\mathbb{Z}[\alpha]$ is not finitely generated as  $\mathbb{Z}$-module.

Clearly there exists $ a_i \in \mathbb{Z}$, which are not all zero and $n>0$ such that $a_n\alpha^n+a_{n-1}\alpha^{n-1}+\dots + a_1\alpha + a_0=0$. Shouldn't this imply that $\mathbb{Z}[\alpha]$ is finitely generated? The motivation for this question is that it may be shown that $\alpha \in \mathbb{C}$ is an algebraic integer if and only if $\mathbb{Z}[\alpha ]$ is finitely generated as a $\mathbb{Z}$-module. Why does the coefficient $a_n$ has to be one for the module to be finitely generated? And, to check that I haven't misunderstood the terminology, $\mathbb{Z}[\alpha ]$ is equal to all polynomials in $\alpha$ over $\mathbb{Z}$, right?


答:
$$ \textbf{Another way to solve that kind of integrals using Laplace transform} $$

$ \ $

For $ \alpha,\beta\in\mathbb{R}_{+}^{*} $, let's define a function $ f_{\alpha,\beta}:x\mapsto\int_{0}^{x}{y^{\alpha - 1}\left(x-y\right)^{\beta - 1}\,\mathrm{d}y} $, since $ f $ is a locally integrable function, we're allowed to talk about its Laplace transform which will then be : $$ \mathcal{L}\left\lbrace f_{\alpha,\beta}\right\rbrace=\mathcal{L}\left\lbrace g_{\alpha}*g_{\beta}\right\rbrace =\mathcal{L}\left\lbrace g_{\alpha}\right\rbrace\times\mathcal{L}\left\lbrace g_{\beta}\right\rbrace $$

Where $ g_{y} $, for $ y\in\mathbb{R}_{+}^{*} $, is the function defined as $ g_{y}:x\mapsto x^{y-1} $.

We know that, for any $ y\in\mathbb{R}_{+}^{*} $, and for all $ s\in\mathbb{C} $, $ \mathcal{L}\left\lbrace g_{y}\right\rbrace\left(s\right)=\frac{\Gamma\left(y\right)}{s^{y}} $. Hence, for all $ s\in\mathbb{C} $ : $$ \mathcal{L}\left\lbrace f_{\alpha,\beta}\right\rbrace\left(s\right)=\frac{\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}{s^{\alpha +\beta}}=\frac{\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}{\Gamma\left(\alpha +\beta\right)}\mathcal{L}\left\lbrace g_{\alpha +\beta}\right\rbrace\left(s\right) $$

Thus, for all $ x\in\mathbb{R}_{+}^{*} $, we have : $$ f_{\alpha,\beta}\left(x\right)=\frac{\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}{\Gamma\left(\alpha +\beta\right)}g_{\alpha +\beta}\left(x\right) $$

$$\text{(Because when dealing with continuous functions, $ \mathcal{L} $ is injective)}$$

Meaning that for all $ x\in\mathbb{R}_{+}^{*} $, $$ \int_{0}^{x}{y^{\alpha -1}\left(x-y\right)^{\beta -1}\,\mathrm{d}y}=\frac{\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}{\Gamma\left(\alpha +\beta\right)}x^{\alpha +\beta -1} $$

That holds for $ x = 1 $, and gives indeed one of the main properties (if not the definition) of the $ \beta $ function.


问:
Lagrange's method
Suppose that I am asked to determine the maximum and the minimum of a continuous and differentiable function $$ f(x, y, z)$$
with a constraint $$ g(x, y, z)=0.$$

I have some trouble understanding the method, even more in this case when it's harder to picture what the function $f$ looks like.

So far I have proven that when $f$ is restricted to those points for which $g(x,y,z)=0$, if it has a local maximum/minimum, the gradient of $f$ is parallel to the gradient of $g$ at that point, i.e.
$$
\nabla f-\lambda \nabla g=0.
$$
My question is: is it enough just to consider the critical points, i.e. those $(x, y, z)$ for which the gradient of Lagrange function
$$
L \left(x, y, z, \lambda \right)=f(x, y, z)-\lambda g(x, y, z)
$$
is zero, in order to find the maximum and minimum of $f$? Many examples of optimization exercises I have seen only do that.

答:
Yes, this is enough (assuming everything is differentiable and nice).

That's because if we take the gradient of the function $L$, the first three components together become $$\nabla f(x,y,z)-\lambda\nabla g(x,y,z)$$and the final and forth component (i.e. $\frac{\partial L}{\partial\lambda}$) becomes $g(x,y,z)$. We have a solution iff both of these are zero. Which is to say, iff the gradient of $L$ is zero.

Title: Lagrange's method


问:
Maximally compact Cartan subalgebra of $SO(p,q)$ and its adjoint representation
This is a somewhat technical question so please bear with me. However, apparently my understanding leads to a contradiction, so I must be missing something basic, and I would be grateful to gain better understanding and clarify the situation.

Consider the real semisimple Lie group $G = SO(2p,2q+1)$ say with $1<p\le q$. Data about this group can be found on p. 699 of Knapp's book. The absolute root system has type $B_{p+q}$ with absolute rank $n = p+q$. A maximal compact subgroup is $K= SO(2p)\times SO(2q+1)$. Thus $K$ has type $D_p \times B_q$ as a compact semisimple Lie group.

Take $T = (S_1)^n$ for maximal torus, consisting of copies of $S^1$ arranged in the "obvious" way as $2 \times 2$ blocks along the diagonal. The Lie algebra $\mathfrak{h} = \mathrm{Lie}(T)$ is a maximally compact Cartan subalgebra of $\mathfrak{g} = \mathrm{Lie}(G)$.

The associated Vogan diagram is that of $B_{n} = B_{p+q}$ with a trivial automorphism. All simple roots are imaginary (i.e. take imaginary values on $\mathfrak{h}$). There is  a single "painted" root $\alpha_{p} = e_{p} - e_{p+1}$. This means that  $\alpha_p$ is non-compact, while all other simple roots $\alpha_i$ with $i \neq p$ are compact. This makes sense - after all, the simple roots $\alpha_{p+1},\ldots,\alpha_{p+q}$ are a simple system for the $B_q$ factor. However, the $D_p$ factor is "missing" its simple root $e_{p-1} + e_{p}$, which is listed in Knapp under "simple roots for $\mathfrak{l} = \mathrm{Lie}(K)$".

I now come to my question. It has to do with what happens precisely when the non-compact ("painted") root $\alpha_p$ is being restricted to $T$ for the adjoint representation of $G$.

One the one hand, $\alpha_{p|T}$ is the sum of the algebraically integral form $e_p$ on the $D_p$ factor, with the negative root $-e_{p+1} = -(e_{p+1} - e_{p+2}) - \cdots  - (e_{p+q-1} - e_{p+q}) - e_{p+q}$ on the $B_{q}$ factor. So, if we replace $G$ with its adjoint $\mathrm{Ad}(G) \le \mathrm{GL}(\mathfrak{g})$ and so $T$ with its image under the adjoint $\mathrm{Ad}(T) \le \mathrm{GL}(\mathfrak{g})$, the integral form $e_p$ is not analytically integral  (regarded on the $D_p$ factor, since for an adjoint semisimple compact Lie group, the set of analytically integral forms coincides with $\mathbb{Z}$-span of roots, but $e_p$ does not belong to the $\mathbb{Z}$-span of roots w.r.t. $D_p$).

On the other hand, the restriction of the adjoint representation $\mathrm{Ad} : G \to \mathrm{GL}(\mathfrak{g})$ to the maximal torus $T$ still "remembers" the non-compact roots, so these should give rise to analytically integral forms on $\mathfrak{h} = \mathrm{Lie}(T)$. However, as noted above, $\alpha_p$ is not analytically integral.

I realize something must be flawed in the above picture. Presumably, the same confusion can arise more generally (not just with $SO(2p,2q+1)$ of course), but let's stick to this special case.

Thanks.

答:
I think I figured this out:

The group $G = \mathrm{SO}(2p,2q+1)$ is adjoint (i.e. has trivial center), so that $\mathrm{Ad}(G) = G$.

Its maximal compact subgroup $K = \mathrm{SO}(2p) \times \mathrm{SO}(2q+1)$ has non-trivial center.

Therefore the image of $K$ with respect to adjoint representation $\mathrm{Ad} : G \to \mathrm{GL}(\mathfrak{g})$ satisfies $\mathrm{Ad}(K) = K$ and is not adjoint.

Because of this, we should not expect the analytic form $\alpha_{p|T}$ to be a $\mathbb{Z}$-combination of roots.

This resolves the contradiction.