问:
Is the following polynomial reducible over Q?
I am looking at an exercise saying that "Demonstrate that x4-22x2+1 is reducible over Q. I have the solution manual and it solves like the following:

If x4-22x2+1 is reducible over Z, then it factors in Z[x], and must therefore
either have a linear factor in Z[x] or factor into two quadratics in Z[x]. The only possibilites for a
linear factor are x ± 1, and clearly neither 1 nor -1 is a zero of the polynomial, so a linear factor is
impossible. Suppose
x4-22x2+1 = (x2 + ax + b)(x2 + cx + d).
Equating coefficients, we see that
x3 coefficient : 0 = a + c
x2 coefficient : −22 = ac + b + d
x coefficient : 0 = bc + ad
constant term : 1 = bd so either b = d = 1 or b = d = −1.
Suppose b = d = 1. Then −22 = ac + 1 + 1 so ac = −24. Because a + c = 0, we have a = −c, so
−c2 = −24 which is impossible for an integer c. Similarly, if b = d = −1, we deduce that −c2 = −20,
which is also impossible. Thus the polynomial is irreducible.

My question is:

1)What does it mean "over Q"? What is the difference between saying over Z and over Q?

2)Why do we factor it as (x2 + ...)(x2 + ...)? Can't it be like (x3 + ...)(x + ...). Also, why don't we factor it like (ax2 + ...)(bx2 + ...), i mean how do we know that the head coefficients are 1?Can someone help with this?

Thanks

答:
1) Over $\mathbb{Q}$ means that factorizations can have rational coefficients. If one were to factor this polynomial over $\mathbb{Z}$, we would only admit factorizations in which the product polynomials have integer coefficients.

2) There is a theorem which says that a polynomial $P(x) \in \mathbb{Q}[x]$ has a root $\alpha \in \mathbb{Q}$ if and only if $P(x) = (x- \alpha)Q(x)$ for some $Q(x) \in \mathbb{Q}[x]$. Furthermore, by the rational root theorem, the polynomial $x^{4} - 22x + 1$ only has possible roots $\pm 1$, which are not roots, as your solution states. Thus, there are no linear factorizations of $x^{4} - 22x + 1$, so the only possible remaining factorization is into a product of quadratics.


问:
prove that an entire function is bounded
Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a entire function and suppose that $\lim\limits_{z\rightarrow\infty}f(z)=0$. Does it impliy that $f$ is bounded? If so, how can I prove it?

答:
The neighborhoods of $\infty$ on the Riemann sphere are the exteriors of closed disks.

To say that $\lim_{z\to\infty}f(z)=0$ is to say that $\forall \varepsilon>0,$ $\exists N$ such that $\mid z\mid>N$ implies $\mid f(z)\mid <\varepsilon$.  But $\mid z \mid \leq N$ is compact, so $f$ is bounded on this region too, and so bounded on $\mathbb{C}$.

ContentLicense: CC BY-SA 4.0


问:
Number of inequivalent one-dimensional representations of a group of order $p^n$
This is from an old comprehensive exam:

Prove that any group with $p^n$
elements has a nontrivial center
and use this to prove that any group with $p^n$
elements has at
least $p$ inequivalent one-dimensional complex representations.
Here $p$ is a prime number.

The part about having a nontrivial center is easily done with the class equation, but I'm having trouble with the part about $p$ inequivalent one-dim representations. The number of one dimensional representations is given by the order of the abelianization of the group: $$G/[G,G]$$
If the group is order $p$ or $p^2$ I can show that this holds, given that groups of order $p$ are cyclic, and therefore abelian, so the commutator subgroup is trivial. I also know that groups of order $p^2$ are abelian, so the commutator subgroup is again trivial.

But besides these cases I'm struggling to see how having a nontrivial center will yield the result. Any help would be appreciated!

答: Mark already indicated that $|G:G'|\geq p$ and that is the proof, but if you want to construct linear representations, you might throw in a bit more sophistication for example if you have a character table at your disposal. Let $H \lt G$ with $|G:H|=p$ (such a maximal subgroup exists). Look at $1_H^G$, the principal character of $H$ induced to $G$. By Frobenius' Reciprocity $[1_H^G,1_G]_G=[1_H,1_H]_H=1$, so $1_H^G=1_G + \sum_{\chi \in Irr(G)-\{1_G\}} a_{\chi}\chi$, with $a_{\chi}$ non-negative integers. Comparing degrees: $1_H^G(1)=|G:H|=p=1 + \sum_{\chi \in Irr(G)-\{1_G\}} a_{\chi}\chi(1)$. And we conclude that if $a_{\chi} \neq 0$ then $\chi(1) \lt p-1$. Since $\chi(1) \mid |G|$, we must have $\chi(1)=1$ for all $\chi$ and hence $1_H^G=1_G + \chi_2 + \cdots \chi_p$, with all $\chi_i$'s linear. And so here are your at least $p$ linear representations!


问:
$A^TPA-P$ is negative definite, does it imply that $APA^T-P$ is negative definite
If $P$ is positive definite, and if $A^TPA-P$ is negative definite, does it imply that $APA^T-P$ is negative definite?

答:
No. Take this example $$P=\begin{pmatrix}1&0\\0&2\end{pmatrix},\quad A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$

Then $$A^TPA=\begin{pmatrix}0&0\\0&1\end{pmatrix}<P$$
but $$APA^T=\begin{pmatrix}2&0\\0&0\end{pmatrix}\not\le P$$


问:
Convergent series proof
Suppose that $\{x_m\}$ is a sequence of positive, non-zero numbers such that $\sum_{n=1}^{\infty} {x_n}$ diverges, and let $S_n = x_1 + x_2 + ... + x_n$. If for all $n \geq 2$, $\frac{x_n}{S_n^2} \leq \frac{S_n - S_{n-1}}{S_nS_{n-1}}$, show that $\sum \frac{x_n}{S_n^2}$ converges.

My attempt:

Notice that $\frac{x_n}{S_n^2} \leq \frac{S_n - S_{n-1}}{S_nS_{n-1}} \leq S_n - S_{n-1} = x_n$.

The Cauchy criterion is satisfied meaning that $\sum \frac{x_n}{S_n^2}$ converges.

I strongly believe that my attempt is wrong but I'm not sure how to go about the proof. Any hints or suggestions are appreciated.

答:
HINT

$$\frac{S_n-S_{n-1}}{S_nS_{n-1}}=\frac{1}{S_{n-1}}-\frac{1}{S_n}$$

The RHS gives a telescoping series.

Title: Convergent series proof


问:
Convergence of a series containing arithmetic mean.
I want to ask the following problem:

Let $a_n$ be a sequence of positive numbers such that $\sum a^2_n$ converges. For $b_n = \dfrac{a_1 + a_2 + \cdots + a_n }{n}$, does $\sum b^2_n$ converges?

At first, I tried to find a counterexample by computing the case where $a_n = 1/n^{1/2 + \epsilon} $, but sadly that made $\sum b^2_n$ convergent. Also, I tried to prove that the sequence converges by using C-S inequality, limit comparison test, etc... but all ideas failed.

Please give me your advice or ideas for this problem. Thank you in advance.

答: Try applying Hardy's inequality. Your case is $p = 2$ and you have $\sum b_n^2 \leq 4 \sum a_n^2$, but you already know that $\sum a_n^2$ converges. Can you work with the definition of series convergence from here?


答:
$1.$  Being irreducible 'over $\Bbb Q$' means that it can not be factored as the product of polynomials (with degree greater than $0$) with all coefficients in $\Bbb Q$. The difference between irreducibility over $\Bbb Z$ and over $\Bbb Q$ is none due to (the second) Gauss's lemma.

$2.$ If it could be factored as the product of a polynomial of degree $1$ times something else, then it would have a root, namely the root of that same polynomial of degree $1$.

ContentLicense: CC BY-SA 3.0

Title: Is the following polynomial reducible over Q?

Body:
I am looking at an exercise saying that "Demonstrate that x4-22x2+1 is reducible over Q. I have the solution manual and it solves like the following:

If x4-22x2+1 is reducible over Z, then it factors in Z[x], and must therefore
either have a linear factor in Z[x] or factor into two quadratics in Z[x]. The only possibilites for a
linear factor are x ± 1, and clearly neither 1 nor -1 is a zero of the polynomial, so a linear factor is
impossible. Suppose
x4-22x2+1 = (x2 + ax + b)(x2 + cx + d).
Equating coefficients, we see that
x3 coefficient : 0 = a + c
x2 coefficient : −22 = ac + b + d
x coefficient : 0 = bc + ad
constant term : 1 = bd so either b = d = 1 or b = d = −1.
Suppose b = d = 1. Then −22 = ac + 1 + 1 so ac = −24. Because a + c = 0, we have a = −c, so
−c2 = −24 which is impossible for an integer c. Similarly, if b = d = −1, we deduce that −c2 = −20,
which is also impossible. Thus the polynomial is irreducible.

My question is:

1)What does it mean "over Q"? What is the difference between saying over Z and over Q?

2)Why do we factor it as (x2 + ...)(x2 + ...)? Can't it be like (x3 + ...)(x + ...). Also, why don't we factor it like (ax2 + ...)(bx2 + ...), i mean how do we know that the head coefficients are 1?Can someone help with this?

Thanks


问:
Binomial series expansion of this equation
I want to expand this seris in ascending power of x, up to and including the terms $x^3$. Give each coefficient as a fraction in its simplest form

$\frac{1}{(2+5x)^3}, |x| < \frac{2}{5}$

What I have tried:

$(2+5x)^{-3}$

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

Then simplifying the expression:

$(2+5x)^{-3} = 2-15x+30x^2-50x^3$

Although, The question asks for the coefficient in fractions. Perhaps I approached this incorrectly?

Following the advice from the comments:

$(2+5x)^{-3}=2(1+\frac{5x}{2})^{-3}= 2[1+(-3)(\frac{5x}{2})+\frac{-3(-3-1)}{2!}\frac{5x^2}{2} + \frac{-3(-3-1)(-3-2)}{3!}\frac{5x^3}{2}]$

this is simplified:

$2(1+\frac{5x}{2})^{-3}= 2[1+(\frac{-15x}{2})+\frac{30x^2}{2} + \frac{-50x^3}{2}]$

Is this form giving the coeffcients as a fraction in simplest form, or must I replace $x$?

答:
The binomial series is
$$(1+z)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} z^k = 1 + \alpha z + \frac{\alpha(\alpha-1)}{2!} z^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!} z^3 + \cdots$$ so we need to first write $$(2+5x)^{-3} = (2(1 + \tfrac{5}{2}x))^{-3} = 2^{-3} (1 + \tfrac{5}{2}x)^{-3}.$$  You forgot that the exponent $-3$ must also distribute over the factor.

Then the binomial series can be applied to the second factor $(1 + \tfrac{5}{2}x)^{-3}$ with the obvious choice $z = \tfrac{5}{2}x$, $\alpha = -3$, to yield
$$(1 + \tfrac{5}{2}x)^{-3} = 1 + (-3)(\tfrac{5}{2}x) + \frac{(-3)(-4)}{2} (\tfrac{5}{2}x)^2 + \frac{(-3)(-4)(-5)}{6} (\tfrac{5}{2}x)^3 + \cdots.$$

Your lack of the ellipses "$\cdots$" makes the equation incorrect.  You also again did not distribute the exponent for each term in the series, instead incorrectly writing $\tfrac{5}{2} x^2$ and $\tfrac{5}{2} x^3$ rather than $\tfrac{25}{4} x^2$ and $\tfrac{125}{8} x^3$.  The final result is

$$(2+5x)^{-3} = \frac{1}{8} - \frac{15}{16} x + \frac{75}{16} x^2 - \frac{625}{32} x^3 + \cdots.$$  The radius of convergence for this series expansion is $|x| < 2/5$.


问:
Dirichlet series for $\zeta(s) \zeta(2s)\zeta(3s) \zeta(6s)^{-1}$
Let $\kappa(n)$ be defined, for $n=p_1^{a_1}\dots p_k^{a_k}$, by $\kappa(n)=a_1\cdot a_2\dots \cdot a_k$. Prove that
$$
\sum_{n=1}^{\infty}\frac{\kappa(n)}{n^s}=\frac{\zeta(s)\cdot \zeta(2s) \cdot \zeta(3s)}{\zeta(6s)}.
$$

I would like to prove this statement, however I'm not sure what to do with the $\zeta(n)$ function. Are there any useful properties I can use?

答: @vitamind's approach notes$$(\kappa\ast1)(p^k)=\sum_{j=0}^kj=\tfrac12k(k+1)$$(except of course $(\kappa\ast1)(p^0)=1$) so, since $\sum_{k\ge0}\tfrac12k(k+1)z^k=\tfrac{z}{(1-z)^3}$,$$\begin{align}\sum_{n\ge1}(\kappa\ast1)(n)n^{-s}&=\prod_{p\in\Bbb P}\left(1+\frac{p^{-s}}{(1-p^{-s})^3}\right)\\\implies\sum_n\kappa(n)n^{-s}&=\prod_p\frac{1-p^{-s}+p^{-2s}}{(1-p^{-s})^2}.\end{align}$$Now just verify$$\frac{1-p^{-s}+p^{-2s}}{(1-p^{-s})^2}=\frac{1-p^{-6s}}{(1-p^{-s})(1-p^{-2s})(1-p^{-3s})}.$$Another approach lacking Dirichlet convolution uses$$1+\sum_{k\ge1}kz^k=\frac{1-z+z^2}{(1-z)^2}$$with $z:=p^{-s}$, after which the calculation is the same as above. @pisco's deleted answer used this.


问:
Can dimension of manifolds be understood similarly to dimension of schemes?
I’m only beginning to learn about schemes, but I know that at least in some cases, the dimension of a scheme (or variety) is 1 less than the length of the longest chain of irreducible closed subsets.

For (let’s say real, smooth) manifolds, this definition doesn’t work: all closed sets are reducible unless they are points. But what if we replace “irreducible closed sets” with “connected compact (c.c.) submanifolds (without boundary)”? It seems that c.c. manifolds are somewhat similar to irreducible closed sets in that they cannot be written as finite unions of other c.c. (proper)submanifolds not equal to themselves (I think). Does this idea lead anywhere interesting / is there a source that discusses dimension from this perspective (or can the validity thereof be disproven)?

The obvious follow up is: given a smooth manifold X, does the ideal of smooth, real valued functions on X which vanish on a given c.c. submanifold Y have any “nice” properties analogous to prime ideals (of course in the ring of smooth functions on X, this ideal itself is unfortunately not actually prime... but maybe something comparable but weaker? Or maybe this works better in the real analytic case?)

答:
I'm able to answer some of this question now.

You don't need to restrict to compact, connected submanifolds. You can just look at closed submanifolds. Everything I'll be discussing will be in the smooth case.

Define a submanifold to be m-reducible if it can be written as a finite union of two closed, proper submanifolds. We see that this matches the typical topological definition of reducibility, but that we're only considering decomposition into submanifolds, not arbitrary closed sets. Then, we claim that the dimension of a manifold M is equal to the length of the longest proper chain of m-irreducible closed submanifolds, $M_1 \subsetneq ...\subsetneq M_n$. Call this number $\ell(M)$. We must show that $\ell(M) \leq dim(M)$ and $\ell(M) \geq dim(M)$.

We show that $\ell(M) \geq dim(M)$ first. Consider a manifold $M$ with dimension $n$. Take $M_n$ to be a connected component of $M$. This is clearly a closed submanifold of dimension $n$. Then, $M_n$ has a subset that is isomorphic to $\mathbb{R}^n$ by definition. But $\mathbb{R}^n$ contains $\mathbb{S}^{n-1}$ as a closed submanifold, so $M_n$ contains an isomorphic copy thereof. By minor abuse of notation, we write $\mathbb{S}^{n-1} \subsetneq M_n$, and we define $M_{n-1} = \mathbb{S}^{n-1}$. From here, we can just proceed by building a "chain of spheres", noting that the $n-1$ sphere has an $\mathbb{R}^{n-1}$ patch, and thus contains an $n-2$ sphere. So, a chain of length n is: $\mathbb{S}^1 \subsetneq ... \subsetneq \mathbb{S}^{n-1} \subsetneq M_n$. Are all of these m-irreducible? They are all connected, closed submanifolds. By invariance of domain, if they are m-reducible, their constituent closed pieces must open subsets. But this would then render them disconnected, which they are not. So, they are all m-irreducible, and we're done.

Now we want to show that $\ell(M) \leq dim(M)$. Suppose not: suppose there was some chain of closed submanifolds of $M$ longer than $n$. Clearly no manifold can have higher dimension than $n$, so (by the always helpful pigeonhole principle) a chain longer than n must have two closed submanifolds of M, call them $X$ and $Y$, with $X \subsetneq Y$ but $dim(X) = dim(Y)$ (note we haven't proven that they are at consecutive positions in the chain, it doesn't matter - all that matters is that one is contained in the other). Now, again by invariance of domain, the inclusion map is open, and so $X$ is open in $Y$. But $X$ is closed in $M$ by assumption, and so also closed in $Y$. Thus, $X$ is either all of $Y$ (which is disallowed), or it disconnects $Y$. But $Y$ being disconnected means that it is m-reducible ($X$ and $X^c$ are proper closed submanifolds), which is also a contradiction. So we cannot construct a chain longer than $n$. This fully completes the proof - the chain of m-irreducible submanifolds definition of dimension matches the traditional definition.

I can also give some partial information about the ideals $I(X) \subset \mathcal{C}^{\infty}(M)$ where $X$ is a submanifold (where $I(X)$ is the set of all maps which vanish on all of $X$). We've proven that the maximum chain length for chains of m-irreducible closed submanifolds is $n$, but it's well known that $M$ is not Noetherian - for example, if we take $M = \mathbb{R}$, then any closed interval can be written as a union of two smaller closed intervals. The only irreducible closed sets are points. However, closed intervals are not submanifolds - they are submanifolds with boundary. A key part of our construction was disallowing the use of submanifolds with boundary for decompositions of closed submanifolds. We know that the $I(X)$ are not prime if $X$ is not a point. But can we tell the difference between the ideals generated by manifolds, and those generated by manifolds with boundary? Yes: from Dominic Joyce's "Algebraic Geometry over $\mathcal{C}^\infty$-rings" (available at https://arxiv.org/pdf/1001.0023.pdf), Proposition 3.1, we can see that the ring of smooth functions on a manifold (without boundary) is finitely presented, while the ring of smooth functions on a manifold with boundary is not. Recall that for a submanifold $X$ of $M$, we have $\mathcal{C}^\infty (X) = \mathcal{C}^\infty (M) / I(X)$, so this implicitly gives us information about the ideal.


问:
Check for values of $x$ that are not a basis for $V$ with basis $B$
I have a vector space $V$ with basis $B = \{v_1, v_2, v_3\}$.

I have a set $Z = \{v_1 + v_2, v_2 + v_3, v_3 + av_1\}$.

What value of $a$ makes $Z$ not a basis for $V$?

I know that $Z$ needs to be linearly independent to be a basis. But I'm lost on what to do after this...

答:
Hint: Evaluate the determinant:

$\det\left(\begin{array}{ccc}
1 & 0 & a\\
1 & 1 & 0\\
0 & 1 & 1\\
\end{array}\right)$.

The vectors are given by the columns of the matrix.

$\det = 1 + a$ and so $\det$ is zero (i.e., vectors are linearly dependent and so not a basis) if $a=-1$ as in the other answer.


问:
Prove that, there does not exist any $A ⊆ R$ satisfying $m(A ∩ I) = α l(I)$, for all intervals $I ⊂ R$
Let $0 < α < 1$. Prove that, there does not exist any $A ⊆ R$ satisfying $m(A ∩ I) = α l(I)$, for all intervals $I ⊂ R$. Here, $m$ denotes the Lebesgue measure on $R$ and $l$ denotes the length of the interval.

I have been asked to prove this, but I have found a counterexample instead.

$I=(1,3)$ and $α=\frac{1}{2}$, I have found an $A=(2,4)$ such that $m(A ∩ I) = 1 = α l(I)= \frac{1}{2}*2$

I have also been asked a follow up question:
Does there exist $A ⊆ R$ satisfying
$m(A ∩ I) = l(I)$, for all intervals $I ⊂ R?$ and I think a similar example can be constructed. Am I going wrong somewhere?

答:
If such set existed then it would not have zero measure.

Assume that such set $A$ exists with $m^*(A)>0$

Let $\epsilon>0$

There exists a covering $A \subseteq \bigcup_{n=1}^{\infty}$ such that $\sum_{n=1}^{\infty}l(I_n)<m^*(A)+\epsilon$

Now $$0<m^*(A) \leq \sum_n m^*(A \cap I_n)=a\sum_nl(I_n)<a(m^*(A)+\epsilon)$$

Choosing $\epsilon<\frac{1-a}{a}m^*(A)$, we arrive to contradiction.


问:
Finding the volume of a solid through 2$\pi$ radians
With reference to this question

Curve $C$ with the equation:
$$y = (2x-1)^{\frac{3}{4}}, \quad x \ge\frac{1}{2}$$

The finite region $S$, is bounded by the curve $C$, the $x$-axis. The $y$-axis and the line $y = 8$. This region is rotated through $2\pi$ radians about the $x$-axis.

Find the exact value of the volume of the solid generated.

Here's what I have tried:
$\mathbf{k}$ or $\mathbf{x}$ was found to be at $\frac{17}{2}$

$$\int\limits^{17/2}_{1/2}(2x-1)^{3/4}dx=\frac{256}{7}$$

I'm wondering if I approached this correctly?

答:
Curve $C$ is given by $y = (2x-1)^{\frac{3}{4}}$ for $x \ge\frac{1}{2}$.

Note that the region S, is bounded by the curve C, the x-axis, the y-axis and the line $y = 8$. This region is rotated through $2\pi$ about the x-axis.

a) If we find the volume using shell method, it is one integral.

$\displaystyle y = (2x-1)^{\frac{3}{4}} \implies x = \frac{y^{4/3} + 1}{2}$

So the integral can be set up as,

$V = \displaystyle \int_0^8 \int_0^{(1 + y^{4/3})/2} 2 \pi \ y \ dx \ dy = \frac{1696 \pi}{5}$

b) If we find volume using washer method, we will have to split the integral into two, which is between $0 \leq x \leq \frac{1}{2}$ and between $\frac{1}{2} \leq x \leq \frac{17}{2}$. The volume for the first part will be a cylinder of radius $8 (0 \leq y \leq 8)$ whereas the second part will have radius as $(2x-1)^{\frac{3}{4}} \leq y \leq 8$.

$V = \displaystyle \int_{1/2}^{17/2} \int_{(2x-1)^{3/4}}^8 2\pi \ y \ dy \ dx + \int_0^{1/2} \int_0^8 2\pi \ y \ dy \ dx$


问:
Let be $A=$ $\begin{pmatrix} 2 & 1 \\ 8 & 3 \end{pmatrix}$ and $H=SL_n(\mathbb{R})$ show that $HA=AH$
Let be $A=$ $\begin{pmatrix}
2 & 1 \\
8 & 3
\end{pmatrix}$

and $H=SL_n(\mathbb{R})$ show that $HA=AH$

Well i take $x\in H$ then $det(x)\neq 0$

and $xA=Ax'$ because $A\in GL_n(R)$ and $Sl_n(R)$ is normal in $GL_n(R)$ is right?

答:
This follows from   multiplicativity of the determinant. $HA$ is exactly the set $D$ of all 2 by 2 matrices with determinant $-2$. Indeed,
clearly $HA \subset D$ and if $M \in D$ then $H^{-1}D \in SL_2$. The same argument applies to $AH$.


问:
Using the word or in proofs
When using the word "or" in proofs, what if one of the statements is true and provides the correct justification that finishes the proof, and the other thing you state is either false or true but not always true? Would this mean that the proof is incorrect? For example if you arrive at a contradiction and say this is false because otherwise $p$ or $q$ would happen leading to a contradiction, but $p$ is the statement that actually leads to the correct contradiction, whereas $q$ is not necessarily true, would this mean that the proof is incorrect?

I would think it would be incorrect because the statements $P \implies Q \vee R$ is logically equivalent to $P \wedge \sim Q \implies R$.

答:
You are asking two different questions.

In a direct proof:  Prove $P$ or $Q$.  Pf: Bunch of stuff...... therefore $P$ is true.  And if $P$ is true then $P$ or $Q$ is true. SO $P$ or $Q$.

That's an utterly valid and correct proof.

In a proof: by contradiction:  Prove "$P$ or $Q$" is not true.   Pf:  If we assume $P$ we get a contradiction $R$ that we know is not true.  So as $P \implies (P\lor Q)$.  And $P \implies R$.  We have a contradiction is completely invalid. and wrong.

It's wrong because $P \implies (P\lor Q)$.  But $(P\lor Q) \not \implies  P$.  So we haven't proven anything and we have not reached the contradiction yet.  $Q\lor P$ was a contradiction but $P$ is not the only way to get $Q\lor P$.

On the other hand if $(P\lor Q)$ does lead to a contradiction we can conclude that $P$ is false, because $P$ would imply $(P\lor Q)$ which implies the contradiction.  That is valid.

.....

But you seem to be arguing that if assuming $p$ or $q$ leads to a contradiction but it's because $p$ and not $q$ that lead to the contradiction that that wouldn't be valid.  That's not logical because $p$ only can't lead to a contradiction because $p$ need not be true for $p$ and $q$ to be true.

If I assume "Either The moon is made of blue cheese or $7$ is prime"  I won't be able to reach a contradiction based on "The moon is made of blue cheese" alone.

To have "Either the moon is made of blue cheese or $7$ is prime" to lead to a contradiction we must have all three of these statements lead to a contradiction:

The moon is made of blue cheese and $7$ is prime must lead to a contradiction (it could because the moon is not made of blue cheese)

The moon is made of blue cheese and $7$ is not prime must lead to a contradiction (it could because the moon is not made of blue cheese and $7$ is prime).

The moon is not made of blue cheese and $7$ is prime must lead to a contradiction (IMPOSSIBLE.... The moon isn't made of blue cheese and $7$ is prime so this can't ever lead to a contradiction!)

So "The moon is made of blue cheese or 7 is prime" can't lead to a contradiction because it IS true.

Title: Using the word or in proofs


问:
The product of the $p−1$ integers between any two non-consecutive multiples of $p$
What is the meaning of this statement by Wilson'theorem:

The product of the $p−1$ integers between any two consecutive multiples
of $p$ is congruent to $−1$ modulo $p$, where $p$ is a prime.
That's:
$$(n+1)(n+2)(n+3)(n+4)\equiv-1\pmod 5$$
When we see that by example, then we write: for $n=10$, $p=5$
$$(10+1)(10+2)(10+3)(10+4)\equiv-1\pmod 5$$

My question is: where is the two consecutive multiples of $5$? Can anyone help me and give me an example of this statement: The product of the $p−1$ integers between any two non-consecutive multiples of $p$, where $p=5$?

答:
The multiples of $p$ are:

$$\{\ldots, -2p, -p, 0, p, 2p, 3p, \ldots, np, (n+1)p, \ldots\}$$ and if we pick any two consecutive multiples of $p$, they would have the form

$$np, (n+1)p$$ for some integer $n$.  Between any two such consecutive multiples are $p-1$ non-multiples
$$np + 1, np + 2, \ldots, np + (p-1).$$  Notice the last term is equal to $(n+1)p-1$.  The product of these is what is being considered in Wilson's theorem.

The reason why this interval must be chosen is because, if a set of $p-1$ consecutive integers is chosen in such a way that one of those integers is a multiple of $p$, then their product is a multiple of $p$, hence is $0$ modulo $p$, not $-1$.


答:
Consecutive multiples of $5$ look like $5n$ and $5(n+1)$, so $10$ and $15$, as Slugger stated in the comments.

There are thus $5(n+1)-5n-1 = 5-1 = 4$ numbers between them, namely
$$5n+1, 5n+2, 5n+3, 5n+4$$.

Your statement about non consecutive multiples doesn't really make sense, as there would be more than $p-1=4$ numbers between them. In addition, it would include a multiple of $5$ between them so the product would be $0\bmod 5$

Body:
What is the meaning of this statement by Wilson'theorem:

The product of the $p−1$ integers between any two consecutive multiples
of $p$ is congruent to $−1$ modulo $p$, where $p$ is a prime.
That's:
$$(n+1)(n+2)(n+3)(n+4)\equiv-1\pmod 5$$
When we see that by example, then we write: for $n=10$, $p=5$
$$(10+1)(10+2)(10+3)(10+4)\equiv-1\pmod 5$$

My question is: where is the two consecutive multiples of $5$? Can anyone help me and give me an example of this statement: The product of the $p−1$ integers between any two non-consecutive multiples of $p$, where $p=5$?


问:
Rolling two dice and picking the higest
So a common mechanic in many dice games such as Warhammer or Dungeons and Dragons is that to provide and advantage on a dice roll you roll twice and pick the highest result.  I'm curious from a combinatorics perspective how this works?  It obviously makes intuitive sense like if you roll twice you get two chances to pass a test but is it possible to quantify this a little better?  Like if I am testing on a d20 (a 20 sided die for those who don't know) and I need to roll over a 10 how much does rolling twice actually increase my chances of success?</p>

答: Given two rolls of a $20$ sided dice, let $P(x)$ be the probability distribution of the maximum of the two rolls. Note that $P(x\geq y)$ for a chosen $y$ is $1-P(D_{20}<y)^2=1-\frac{y-1}{20}^2$. And, $P(x=y)=P(x\geq y)-P(x\geq y+1)$.

Title: Rolling two dice and picking the higest


答:
There are not p-1 integers between two non-consecutive multiples if p, but at least 2p-1. For example 9 integers between 110 and 120 which are non-consecutive multiples of 5.

Since there is at least one multiple of p between two non-consecutive multiples of p (that’s what makes them non consecutive, the fact that there is another one in between), the product is a multiple of p, therefore 0 modulo p.

Title: The product of the $p−1$ integers between any two non-consecutive multiples of $p$


问:
Kernel, Basis and Column Space
If a matrix $A$ has row reduced form

$A=$$\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$

then the kernel of the linear map $T : \mathbb{R}^4 → \mathbb{R}^3$ defined by $T(x) = Ax$ has basis $$(−3,1,0,0), (−3, 0, 1, 1).$$

How do I prove or disprove the claim regarding the kernel?

Any help is appreciated guys. Also, how can I show that columns 2 and 4 forms a basis for the column space?

答:
The reduced matrix is $A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$ and if we want to find the generator of the Kernel we have to solve the system $A\underline x=\underline0$:
$$A=\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}\underline x=\underline 0\implies \begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \end{pmatrix}\begin{pmatrix}x\\y\\z\\w \end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\implies$$
$$\begin{cases}x+3y+3w=0\\z-w=0 \end{cases}\implies z=w=t\in\mathbb R,y=s\in\mathbb R,x=-3t-3s.$$
So we can state that $\operatorname{Ker}(F)=\Big\langle\begin{pmatrix}-3\\1\\0\\0 \end{pmatrix},\begin{pmatrix}-3\\0\\1\\1 \end{pmatrix} \Big\rangle$.

Title: Kernel, Basis and Column Space


问:
Proving a result from the Fourier Energy Theorem
So I don't know what the result is called as it is unstated in the problem but it is a proof following from the Fourier Energy theorem.
Prove:
$$\int_{-\infty}^{\infty}f'(x)f'^*(x)dx = \frac{1}{2\pi}\int_{-\infty}^{\infty}\omega^2\hat{f}(\omega)\hat{f}^*(\omega)d\omega$$

using:
$$ \int_{-\infty}^{\infty}(f(x))^2dx  = \frac{1}{2\pi}\int_{-\infty}^\infty\|\hat{f}(\omega)\|^2 $$

What I have done so far is replace $\|\hat{f}(\omega)\|^2 $ with $ \hat{f}(\omega)\hat{f}^*(\omega)$
to get

$$ \int_{-\infty}^{\infty}f(x)f(x)dx  = \frac{1}{2\pi}\int_{-\infty}^\infty\hat{f}(\omega)\hat{f}^*(\omega) $$

now my idea was to use the Fourier transform property:
$\mathcal{F}\left(\frac{d^nf}{dx^n}\right)  = (i\omega)^n\hat{f}(\omega)$
but was confused as to how this would work, any ideas?

答:
you nearly got it. If you use in your formula $f'$ instead of $f$ you can use $\mathcal F(f')(\omega)=i\omega f(\omega)$.

So we are using the Parval identiy on $f'$ to get:
$$\int_{-\infty}^\infty f'(x)f'^{*}(x)dx=\tfrac 1 {2\pi}\int_{-\infty}^\infty\widehat{ f'}(\omega)\widehat {f'}^*(\omega)d\omega$$
Now we can use the Fourier poperty and get with $\widehat{f'}(\omega)=i\omega\hat f(\omega)$:
$$\tfrac 1{2\pi}\int_{-\infty}^\infty\underbrace{(i\omega)\hat f(\omega)}_{=\widehat{f'}(\omega)}\underbrace{(i\omega)^*\hat f^*(\omega)}_{=\widehat{f'}^*(\omega)}d\omega=\tfrac 1 {2\pi}\int_{-\infty}^\infty\underbrace{-i^2}_{=1}\omega^2\hat f(\omega)\hat f^*(\omega)d\omega$$


问:
sub-algebra of inductive limit
Let $A_{i}$ be sequence of C* algebras, $\alpha_{i}:A_{i}\rightarrow A_{i+1}$ be sequence of *-homomorphisms. Consider some sub-algebra $B$ of its inductive limit $A$. Denote the canonical homomorphism be $\beta_{i}:A_{i}\rightarrow A$. Are $\beta_{i}^{-1}(B)$ a sub-algebra of $A_{i}$?

I know that the kernal of $\beta_{i}$ form an ideal in $A_{i}$, but I don't know if it works on preimage of other elements.

答:
Yes, and this has nothing to do with inductive limits. If $\alpha\colon A_1\to A_2$ is an algebra homomorphism and $B$ a subalgebra of $A_2$, then $\alpha^{-1}(B)$ is a subalgebra of $A_1$. I assume you know this result for vector spaces and linear maps, so I'll just prove that $\alpha^{-1}(B)$ is closed under taking products.

If $x,y\in \alpha^{-1}(B)$, then $\alpha(x),\alpha(y)\in B$ by definition and therefore $\alpha(xy)=\alpha(x)\alpha(y)\in B$ since $B$ is an algebra. Thus $xy\in \alpha^{-1}(B)$.

The same holds for (unital) $\ast$-homomorphisms and (unital) $\ast$-subalgebras by a similar argument.

Title: sub-algebra of inductive limit


问:
Choosing desired self-conjugate closed-loop pole locations
If i am given:

```<code>A = [1 -2; 0.5 -1];
B = [2; 2];
C = [-1 1];
D = 0;
\zeta = 0.707;
\omega_0 = 2 rad/s;
</code>```

I need to design a state controller for the system such that the closed-loop system becomes two complex poles. I want to use the place function in matlab but how do i find p, the desired self-conjugate closed loop pole locations?

答:
We have with null initial conditions

$$
sX(s)=AX(s)+BU(s)\Rightarrow X(s) = (sI-A)^{-1}B U(s)
$$

but $Y(s) = C X(s)$ and calling the controller $K(s)$ we have

$$
U(s) = K(s)(Y_r(s)-Y(s))
$$

then in closed loop

$$
Y(s) =  C(sI-A)^{-1}B K(s)(Y_r(s)-Y(s))
$$

or

$$
\frac{Y(s)}{Y_r(s)} = \frac{C(sI-A)^{-1}B K(s)}{1+C(sI-A)^{-1}B K(s)} = \frac{K(s)}{K(s)+s^2}=\frac{\omega^2}{s^2+2\zeta\omega s + \omega^2}
$$

now solving for $K(s)$ we obtain

$$
K(s) = \frac{\omega^2s}{s+2\zeta\omega}
$$


问:
Solution of an IVP cannot be extended beyond an Interval.
Consider the following  differential equation

$\dfrac{dy}{dx} = 1 + y^2$ , $y(0) = 1$

I need to show the solution cannot be extended beyond
$(-3\pi/4, \pi/4)$

By uniqueness and existence theorem I can get an interval which guarantees a unique solution, but I don't really understand how should I use this to prove the above claim ?

A complete solution will be helpful here, since I am really struggling with these kind of problems.

答:
In fact, you can integrate this differential equation as $y(x)= \tan(x+\frac{\pi}{4})$.

From there, you can prove the desired result.


答:
$$\frac{dy}{1+y^2}=dx$$
$$x+C=\arctan y;\;y(0)=1\to C=\arctan 1=\frac{\pi}{4}$$
$$x +\frac{\pi}{4}=\arctan y$$
RHS range is $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

thus for the LHS must be $-\frac{\pi}{2}<x+\frac{\pi}{4}<\frac{\pi}{2}$

and the solution $y=\tan \left(x+\pi/4 \right)$ is valid only when $-\frac{3}{4}  \pi < x< \frac{\pi }{4}$

Title: Solution of an IVP cannot be extended beyond an Interval.

Body:
Consider the following  differential equation

$\dfrac{dy}{dx} = 1 + y^2$ , $y(0) = 1$

I need to show the solution cannot be extended beyond
$(-3\pi/4, \pi/4)$

By uniqueness and existence theorem I can get an interval which guarantees a unique solution, but I don't really understand how should I use this to prove the above claim ?

A complete solution will be helpful here, since I am really struggling with these kind of problems.


问:
Is $\mathbb{Z}[\frac{1}{p}]$ a lattice in $\mathbb{R} \times \mathbb{Q}_p$?
So we have that $\overline{\mathbb{Z}[\frac{1}{p}]} = \mathbb{R}$ that $\mathbb{Z}[\frac{1}{p}]$ is dense in $\mathbb{R}$.  How can it be that $\mathbb{Z}[\frac{1}{p}]$ under the diagonal map $t \mapsto (t,t)$ is now a lattice in $\mathbb{R} \times \mathbb{Q}_p$ ?

First we have the diagonal embedding and then we have the inclusion map:
$$ \mathbb{Z}[\tfrac{1}{p}] \to \mathbb{Z}[\tfrac{1}{p}] \times \mathbb{Z}[\tfrac{1}{p}] \to \mathbb{R} \times \mathbb{Q}_p $$
So we have that $\mathbb{Z}[\frac{1}{p}] \subseteq \mathbb{R}$ in one metric and $\mathbb{Z}[\frac{1}{p}] \subseteq \mathbb{Q}_p$ in the $p$-adic metric.

The product of two metric spaces is could be metric space, there is certainly a product topology.  Here one metric Euclidean and the other is non-Archimedian.

答:
$\Bbb{R\times Q_p}$ with the norm $\|(a,b)\|=|a|_\infty+|b|_p$,

$\iota(c)=(c,c)$,

$\iota(\Bbb{Z}[p^{-1}])$ is a discrete subgroup and $[0,1)\times\Bbb{Z}_p$ is a fundamental domain of $\Bbb{R\times Q_p}/\iota(\Bbb{Z}[p^{-1}])$ which is compact.


答:
Rather than thinking of the chance of success, think of the chance of failure.  If you take the best roll out of a sequence of multiple rolls, and the outcome of each roll is independent of the outcome of all previous rolls, then the only way to fail is if each roll is a failure.  If any given roll has probability $q$ of failure, then the probability that every roll fails is $q^n$, where $n$ is the number of rolls performed.  If $0 < q < 1$, then $q^n < q$ for $n > 1$.

So for instance, if you have to roll at least a $10$ on a $20$-sided die, then the probability of success on a single roll is $10/20 = 1/2$, hence the probability of failure is $q = 1/2$.  If you get to take the maximum of $n = 2$ rolls, then that means the chance of failing both rolls is $q^2 = 1/4 < 1/2$.  If you get to roll $10$ times, then the chance of failing all of your rolls is $q^{10} = 1/1024$.  Again, this works because by taking the maximum value rolled, you are basically saying that every roll has to fail in order for you to fail overall.

Title: Rolling two dice and picking the higest

Body: So a common mechanic in many dice games such as Warhammer or Dungeons and Dragons is that to provide and advantage on a dice roll you roll twice and pick the highest result.  I'm curious from a combinatorics perspective how this works?  It obviously makes intuitive sense like if you roll twice you get two chances to pass a test but is it possible to quantify this a little better?  Like if I am testing on a d20 (a 20 sided die for those who don't know) and I need to roll over a 10 how much does rolling twice actually increase my chances of success?</p>


问:
Prove $\sin \left(1/n\right)$ tends to $0$ as $n$ tends to infinity.
I'm sure there is an easy solution to this but my mind has gone blank! Any help on proving that $\sin( \frac{1}{n})\longrightarrow0$ as $n\longrightarrow\infty$ would be much appreciated.

This question was set on a course before continuity was introduced, just using basic sequence facts. I should have phrased it as:

Find an $N$ such that for all $n > N$ $|\sin(1/n)| < \varepsilon$

答:
Hint: $\sin$ is a continuous function.

Edit: Since $\dfrac{1}{n}\longrightarrow 0$ the statement below is true: $$(\forall \delta>0)(\exists M\in \Bbb N)(\forall n\in \Bbb N)\left(n\ge M\implies \left\vert \dfrac{1}{n}\right\vert<\delta\right)$$

Using $M$ from above and the fact that $\left \vert \sin\left(\dfrac{1}{n}\right)\right \vert<\left \vert \dfrac{1}{n}\right \vert$ for all $n\in \Bbb N$ try to find a suitable $N$.


问:
Quadratic residues of Fermat primes
Given a Fermat prime $p=2^{2^k}+1$, I can show that every quadratic nonresidue mod $p$ is also a primitive root for $p$, by showing that there are $(p-1)/2$ primitive roots.

However, I am wondering how to show the converse implication: given a prime $p$, to show that if every quadratic nonresidue is a primitive root, then $p$ is necessarily a Fermat prime. Any hints would be appreciated.

答:
Consider an odd prime $p$; this has a primitive root $g$. Now $g^i$ is also a primitive root iff $\gcd(i,p{-}1)=1$ so there are $\phi(p{-}1)$ primitive roots. Since $p{-}1$ is even, $\phi(p{-}1)\leq(p{-}1)/2$ and the only case when $\phi(p{-}1)=(p{-}1)/2$ will be when $p{-}1$ has no odd factors, so $p{-}1$ is a power of $2$ and $p{-}1= 2^m$.

Then since $2^m+1$ is prime, $m$ must also be a power of $2$ as required.

Title: Quadratic residues of Fermat primes


问:
1990 BMO1- Maximising/minimising $PC/PD$ where $P$ is a point on the line $AB$ and $ABCD$ is a square
Recently, I've asked for help from my teacher, as well as gathering help from other people. At the moment, what troubles me is how to even start to find the maximum and minimum points without a calculator, other than that there is some symmetry in the question, so if $PC/PD$ is a maximum where point $P$ is one side of the line, then if point $P$ was on the opposite side of the line $AB$, then it would be the minimum instead. As well as this, given that this was once an Olympiad question, I haven't found any simple solutions to the problem without the use of a computer or calculator to maximise and minimise it. From some research, the only other source of help I had found were hints from The Mathematical Olympiad Handbook- an Introduction to Problem Solving based on the First 32 British Mathematical Olympiads 1965-1996

If anyone has solutions to the question without using computation, it would be much appreciated. Preferably, simplify it and show each step, as this is meant to be a question aimed at students, and I just want to see a solution in full and see the process.

Here is the question written here in full:

(1990 British Mathematical Olympiad Round 1 - Question 2)

$ABCD$ is a square and $P$ is a point on the line $AB.$ Find the maximum and minimum values of the ratio $PC/PD$, showing that these occur for the points $P$ given by $$AP \times BP=AB^2$$

答:
This can be answered in a geometrical way, if you remember that the locus of points $P$ in a given ratio $r$ to two points $C$ and $D$ is a circle.
If $a=CD$, then this circle has a radius given by
$$
R={ar\over1-r^2}
$$
(I'm supposing $r<1$, in the other case just reflect everything about the perpendicular bisector of $CD$)
and its center $O$ lies on line $CD$, at a distance
$$
x={ar^2\over1-r^2}
$$
from $C$.
When $r$ is minimum then $R$ is minimum too. Hence
point $P$ on line $AB$ corresponding to a minimum of $r$ is the tangency point when that circle is tangent to $AB$, that is when $R=a$.
This is equivalent to
$$
{r\over1-r^2}=1
$$
which also entails
$$
x(a+x)=a^2,
$$
as it was to be proved.


问:
Normal form of a vector field in $\mathbb {R}^4$.
EDIT (In responce to xpaul's answer): I'm looking for the exact normal form, not the one up to $O(|x^5|)$

Those are two analogous problems, the first one of which I have already accounted for.

Find the normal form of the vector fields:

a) Solved.

b) $$\dot x_1=x_2$$
$$\dot x_2=-x_1 $$
$$\dot x_3=\sqrt 2 x_4+x_1^3 $$
$$\dot x_4=-\sqrt 2 x_3+x_3x_4^2 $$

Solution:

The matrix of the linearised system is
$\left( \begin{array}{cccc}
0 & 1 & 0 &  0 \\
-1 & 0 & 0 & 0 \\
0 & 0 & 0 & \sqrt 2 \\
0 & 0 & -\sqrt 2 & 0 \\
\end{array} \right) $.

The eigenvalues are
$\{\pm i\sqrt 2, \pm i\}$.
Therefore, we have infinitely many resonances
$\lambda_i=\lambda_i+k(\lambda_1+\lambda_2)+p(\lambda_3+\lambda_4), (i=\overline{1,4}), (k+p\geq1)$. There are no other resonances.

Furthermore, by Poincare-Dulac's theorem, I can map the vector field to its linear part up to resonant monomes, but how should I proceed?

EDIT:

The resonant monomes are then
$$x_i\dot{}x_1^k\dot{}x_2^k\dot{}x_3^p\dot{}x_4^p\dot{}e_i$$
where $e_i$ is the unit vector along the i-th coordinate.

Then by Poincare-Dulac's theorem, the vector field can be "reduced" to:
$$\dot x_1=x_2 + \sum_{k+p\geq 1} c_{1kp}x_1^{k+1}\dot{}x_2^k\dot{}x_3^p\dot{}x_4^p$$

$$\dot x_2=-x_1 + \sum_{k+p\geq 1} c_{2kp}x_1^k\dot{}x_2^{k+1}\dot{}x_3^p\dot{}x_4^p$$

$$\dot x_3=\sqrt 2 x_4 + \sum_{k+p\geq 1} c_{3kp}x_1^k\dot{}x_2^k\dot{}x_3^{p+1}\dot{}x_4^p$$

$$\dot x_4=-\sqrt 2 x_3+ \sum_{k+p\geq 1} c_{4kp}x_1^k\dot{}x_2^k\dot{}x_3^p\dot{}x_4^{p+1}$$

Is that the correct answer? Is there a complexification of the system that leads to this answer, or gives it in a more pleasing form?

EDIT 2: Is it unreasonable to consider the normal form in order to study the dynamics of this system. What would be a good approach to find the stationary points and examine their stability?

答: This question is so-called "double Hopf bifurcation" and you can obtain the normal form from http://www.scholarpedia.org/article/Hopf-Hopf_bifurcation.</p>


问:
Let $x'=x \ln(1+x^2)$ with $x(0)=1$. why the maximal solutions are defined in whole line?
Let  $x'=x \ln(1+x^2)$ with $x(0)=1$. Why the maximal solutions are defined in whole line?

Comment: Any idea is welcome. The existence theorem guarantees the solution, however why is it defined in $\mathbb{R}$?
The function seems to be monotone increasing and positive.

答:
Here's a hint: The solution is given implicitly by
$$
\int_1^{x(t)} \frac{dx}{x \ln(1+x^2)} = t
.
$$
Can you show that $\int_1^0 (\cdots) = -\infty$ and $\int_1^\infty (\cdots) = +\infty$?


问:
Choosing a text for a First Course in Topology
Which is a better textbook - Dugundji or Munkres? I'm concerned with clarity of exposition and explanation of motivation, etc.</p>

答:
Topology For Beginners by Steve Warner
It starts with Set theory and goes from there. It ends with Algebraic topology


问:
Is the following polynomial reducible over Q?
I am looking at an exercise saying that "Demonstrate that x4-22x2+1 is reducible over Q. I have the solution manual and it solves like the following:

If x4-22x2+1 is reducible over Z, then it factors in Z[x], and must therefore
either have a linear factor in Z[x] or factor into two quadratics in Z[x]. The only possibilites for a
linear factor are x ± 1, and clearly neither 1 nor -1 is a zero of the polynomial, so a linear factor is
impossible. Suppose
x4-22x2+1 = (x2 + ax + b)(x2 + cx + d).
Equating coefficients, we see that
x3 coefficient : 0 = a + c
x2 coefficient : −22 = ac + b + d
x coefficient : 0 = bc + ad
constant term : 1 = bd so either b = d = 1 or b = d = −1.
Suppose b = d = 1. Then −22 = ac + 1 + 1 so ac = −24. Because a + c = 0, we have a = −c, so
−c2 = −24 which is impossible for an integer c. Similarly, if b = d = −1, we deduce that −c2 = −20,
which is also impossible. Thus the polynomial is irreducible.

My question is:

1)What does it mean "over Q"? What is the difference between saying over Z and over Q?

2)Why do we factor it as (x2 + ...)(x2 + ...)? Can't it be like (x3 + ...)(x + ...). Also, why don't we factor it like (ax2 + ...)(bx2 + ...), i mean how do we know that the head coefficients are 1?Can someone help with this?

Thanks

答:
As to the difference between "over $\mathbb{Q}$" and "over $\mathbb{Z}$, there is a simple answer, and a more complicated one. The simple answer is that the question asks about $\mathbb{Q}$. And if we are curious about $\mathbb{Z}$, it is clear that irreducibility over $\mathbb{Q}$ implies irreducibility over $\mathbb{Z}$.

In fact one can prove that the implication goes the other way: irreducibility over $\mathbb{Z}$ implies irreducibility over $\mathbb{Q}$. But this is not immediately obvious.

For the other question, the solution quoted did deal with the possibility $(x^3+\cdots)(x+\cdots)$. By the Rational Roots Theorem, the only possible rational root candidates for $x^4-22x^2+1=0$ are $\pm 1$. Neither works. So there is no degree $1$ polynomial with rational coefficients that divides $x^4-22x^2+1$.


问:
Almost, but not quite a trapping region
I'm looking for a term for something that almost fits the definition of a trapping region. According to wikipedia, a set is a trapping region for a dynamical system if every trajectory which starts within the set will move to the region's interior, and will remain there as the system evolves.

I've proved that a closed set I'm working with almost fits this definition: trajectories which start in the set remain in the set as the system evolves, but they may remain on the boundary of the set, rather than moving to the interior. Is there a term for this? Is the term trapping region sometimes used in this way?

答:
It's called a forward invariant set or positively invariant set.

(The condition is that solutions starting in the set stay in the set for all $t \ge 0$. This includes actually being defined for all $t\ge 0$, in contrast to blowing up after finite time; if the set is compact, this detail is nothing to worry about.)

Title: Almost, but not quite a trapping region


问:
Step by step method for $\int\frac{x^2+1}{x^2-1}\,dx$
$$\int\frac{x^2+1}{x^2-1}\,dx$$

What is the step-by-step way of solving this integral problem?
I tried using substitution which was $x^2-1=t^2$, but end up with an even more complicated equation.
Substituting trigonometric functions for example :$x^2=\sec^2(a)$ so that $x^2-1$ part becomes $\tan^2(a)$ did not help either.

答:
$$\begin{align}
\int\frac{x^2+1}{x^2-1}\,\mathrm dx&=\int\frac{x^2-1+2}{x^2-1}\,\mathrm dx\\[1ex]
&=\int\left(1+\frac2{x^2-1}\right)\,\mathrm dx
\end{align}$$

To integrate the second term, you can split it into partial fractions to get the expression mentioned in the comments, or make a substitution $x=\sec t$ and $\mathrm dx=\sec t\tan t\,\mathrm dt$. Then

$$\begin{align}
\int\left(1+\frac2{x^2-1}\right)\,\mathrm dx&=x+2\int\frac{\sec t\tan t}{\sec^2t-1}\,\mathrm dt\\[1ex]
&=x+2\int\csc t\,\mathrm dt\\[1ex]
&=x-2\ln|\csc t+\cot t|+C\\[1ex]
&=x-2\ln\left|\csc(\sec^{-1}x)+\cot(\sec^{-1}x)\right|+C\\[1ex]
&=x-2\ln\left|\frac x{\sqrt{x^2-1}}+\frac1{\sqrt{x^2-1}}\right|+C
\end{align}$$

With some massaging you can get this in a simpler form that immediately agrees with the partial-fraction-decomposition result:

$$x+\ln|x-1|-\ln|x+1|+C$$


问:
Convergence of Subsequences in Gromov's Theorem
A key theorem in metric geometry is Gromov's Compactness Theorem, which describes when a sequence of metric spaces has a Gromov-Hausdorff converging subsequence.

I'm not sure I follow how this works: does this mean that the sequence of metric spaces itself has to have a Gromov-Hausdorff limit in order for it to contain a Gromov-Hausdorff converging subsequence?  Also, what kinds of examples are there of converging subsequences in sequences of metric spaces apart from repeatedly rescaling a metric space?

答:
First of all, I highly recommend

Burago, D.; Burago, Yu.; Ivanov, S., A course in metric geometry, Graduate Studies in Mathematics. 33. Providence, RI: American Mathematical Society (AMS). xiv, 415 p. (2001). ZBL0981.51016.

specifically, Chapter 7, for a treatment of the GH-distance and convergence. If you want an example of a sequence of metric spaces which does not GH-converge but contains a convergent subsequence, the simplest example I know is:

For $n$ odd, take the metric space $X_n$ consisting of a single point.

For $n$ even  take the metric space $X_n$ consisting of two points $\{x, y\}$ such that $d(x,y)=1$.

Then the resulting sequence of metric spaces $(X_n)$ will not GH-converge, but will contains two convergent subsequences (one is $(X_{2k-1})$ and the other is $(X_{2k})$).

If you want more interesting examples (not obtained by a rescaling):

a. Take the sequence of metric spaces $X_n$ which are ellipses in the plane
$$
\{(x,y): a_n x^2+ y^2=1\}, a_n= 1+\frac{1}{n}.
$$
(Take the restriction of the planar distance function to define the metric on these ellipses.) The limit of this sequence will be the unit circle.

b. Take the sequence of Euclidean rectangles of the shape $a_n\times b_n$, where $a_n\to 1, b_n\to 0$. This sequence will converge to the unit interval.

c. Take your favorite compact connected Riemannian manifold  $(M,g)$ (say, a round sphere) and a sequence of finite subsets  $X_n\subset  M$ which form $1/n$-nets in $M$ (i.e. for every $n$, each point of $M$ is within distance $1/n$ from some point of $X_n$). Equip $X_n$ with the restriction of  the Riemannian distance function $d_g$ on $M$. Then the sequence $(X_n)$ will converge  to the metric space $(M, d_g)$.

d. Consider the standard (ternary) Cantor set $C\subset {\mathbb R}$, obtained by inductively removing "middle thirds" from the unit interval $I$. Let $C_n$ denote the $n$-th leftover subset of $I$ (the subset left in $I$ after removing middle thirds $n$ times), see here. Equip $C$ and $C_n$ with the restriction of the standard distance on the real line. Then $(C_n)$ will converge to $C$.


问:
the kernel of surjective homomorphism between topological groups is closed
Suppose $\varphi : G\to H$ is a surjective homomorphism of topological groups, then $K=ker\varphi$ is a closed normal subgroup of $G$. If $G$ is compact, then $\varphi' : G/K\to H$ is an open map.

It is straightforward to me that $K$ is a normal subgroup, could some tell me why $K$ is closed and why if $G$ is compact then $\varphi'$ is open?

Thanks!

答:
Since $\ker\varphi=\varphi^{-1}\bigl(\{e_H\}\bigr)$, since $\varphi$ is continuous and since $\{e_H\}$ is a closed set, $\ker\varphi$ is a closed set.

If $G$ is compact, then $\varphi'$ is a continuous bijection with compact domain. Therefore, it is a homeomorphism (it maps closed sets into closed sets); in particular, it is an open map.


问:
What is the point of giving a tensor identity in normal coordinates?
I have been confused about this for some time. For example, the curvature tensor in general for the Levi-Civita connection of $g$ in the local frame $\{\partial_i\}$ induced by the coordinates $\{x^i\}$ is

$R_{ijk}^l = \partial_i \Gamma_{jk}^l - \partial_j \Gamma_{ik}^l + \Gamma_{jk}^m \Gamma_{im}^l - \Gamma_{ik}^m\Gamma_{jm}^l$

and if you take the partials of the Christoffel symbols and grind through you can ultimately get

$R_{ijk}^l = \frac{1}{2}g^{ls}(\partial_i \partial_k g_{js} - \partial_i \partial_s g_{jk} -\partial_j \partial_k g_{is} + \partial_j\partial_s g_{ik}) + g^{lm}g_{st}(\Gamma_{jm}^s\Gamma_{ik}^t - \Gamma_{im}^s\Gamma_{jk}^t)$

which is fine. But in every book I read everyone just says "at a point $p$ in normal coordinates we have..."

$R_{ijk}^l = \frac{1}{2}g^{ls}(\partial_i \partial_k g_{js} - \partial_i \partial_s g_{jk} -\partial_j \partial_k g_{is} + \partial_j\partial_s g_{ik})$

which of course follows right away from what I already have, but what is the point of this second formula at all? Are they trying to tell me that if I change coordinates back to the coordinate frame in the latter formula that I'm supposed to get the former? I've been trying to do this with no success, am I failing because I'm making some computational mistake or because my logic is wrong altogether?

Since $R$ is a tensor its value at a point $p$ does not depend on the coordinates:

Does this mean that if $\tilde\partial_j$ is the frame induced by the coordinates $\{y^j\}$ then $g(R(\partial_i, \partial_j) \partial_k, \partial_l))=\tilde{g}(\tilde{R}(\tilde\partial_i, \tilde\partial_j)\tilde\partial_k,\tilde\partial_l)$ as real numbers at every point $p$ in the overlap of the two charts? Does this mean that if for some reason I wanted to integrate the component $R_{123}^4$ over some compact $M$ I could just integrate its formula in normal coordinates?

If the questions seem vague its because I really am confused by this. Thanks for whatever clarification you can give.

答:
As object is said to be coordinate independent when it can be defined without direct reference to coordinate charts. For instance, the Riemannian curvature $R$ can be viewed as a multilinear map between tangent spaces
$$
R|_{p}:T_pM\times T_pM\times T_pM\to T_pM \\
(X_p,Y_p,Z_p)\mapsto R(X_p,Y_p)Z_p
$$
And similarly for other tensors. This does not mean that the components $R^{a}{}_{bcd}$ will be independent of the choice of coordintes (though they will be related by a simple transformation rule). A component such as $R^4{}_{123}$ has no meaning unless we fix a coordinate chart. Index expressions remain useful because they hold in any coordinate chart, even though the values of the individual components may change.

Generally speaking, coordinates are most often used as an intermediate step in computing coordinate-independent relationships between coordinate-independent objects. In this context, one can choose an arbitrary coordinate chart, or one which is convenient for the task at hand.

For instance, if we want to prove the first Bianchi identity
$$
R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0
$$
we can work one point at a time by choosing an arbitrary point $p$ and showing it holds at $p$. By choosing normal coordinates centered at $p$, we can do the computation in a setting where the components of $R$ are particularly simple, since half of the terms vanish.


问:
Slick proof the determinant is an irreducible polynomial
A polynomial $p$ over a field $k$ is called irreducible if $p=fg$ for polynomials $f,g$ implies $f$ or $g$ are constant. One can consider the determinant of an $n\times n$ matrix to be a polynomial in $n^2$ variables. Does anyone know of a slick way to prove this polynomial is irreducible?

It feels like this should follow quite easily from basic properties of the determinant or an induction argument, but I cannot think of a nice proof. One consequence of this fact is that $GL_n$ is the complement of a hypersurface in $M_{n}$. Thanks.

答:
Here is another algebraic argument which may be of interest, using induction on the size of the matrix:

Let $A_{ij}$ be the minor we get when we delete row $i$ and column $j$. If we expand along the first row of the matrix, we get
$$\det A=(\det A_{11})x_{11}-(\det A_{12})x_{12}+\cdots\pm(\det A_{1n})x_{1n}.$$
None of the determinants $\det A_{1i}$ contains any variable $x_{1i}$, so this is a linear polynomial in the variables $x_{1i}$. Therefore it can only factor if the coefficients of the $x_{1i}$ have a common factor. But these coefficients are determinants of size $n-1$ so by induction they are all irreducible; since they are not scalar multiples of each other they have no (non-constant) common factor.


问:
natural language proof assistant
I was wondering whether there has been any attempt to create a proof assistant that you write in it, in english,

I mean you write your proof the usual way in TeX(maybe use a 'simpler english') then instead of sending it to a journal to have it verified you use the proof assistant to have it verified for you.

There are programming languages like inform7 in which you program in english. I think what is needed is a set of macro's to turn the tex into, lets say, something that Coq can verify. Is there any such thing out there?

Do you think if this happens casual mathematicians will use it? or are there deeper problems that people don't use them now??

答:
I believe the RISC (Research Institute for Symbolic Computation) group at Castle Hagenberg in Austria have been involved in such things for many years.

See:  www.theorema.org

Based on the OP's comments below, it appears that they are not seeking an automated 'theorem prover', but an automated 'proof verifier'. If so, there is already a thread on this at:

How do proof verifiers work?

Title: natural language proof assistant


问:
Decomposition of Field extensions and Transcendence basis
Let $K \subset L$ be arbitrary extension of fields and assume (for the first time) it has a finite
transcendence degree $n:= \operatorname{Trdeg}_K(L)$.
Let $\{a_1,...,a_n\}$ it's transcendence basis. I'm interested in
canonical ways how the extension $K \subset L$ can be decomposed und where
I should be more careful.

At first, it seems that it is always possible to decompose it in two
ways since I not see what could fail there:

T1) as $K \subset K(a_1,...,a_n) \subset L$ where
$K \subset K(a_1,...,a_n)$ is pure transcendental and
$K(a_1,...,a_n) \subset L$ algebraic

T2) as  $K \subset L_{alg} \subset L$ where
$K_{alg} := \{a \in L \vert a \text{ algebraic } \}$ is the algebraic
closure of $K$ in $L$ such that
$K \subset L_{alg}$ is tautologically algebraic and
$L_{alg} \subset L$ pure transcendental

First of all du these two decompositions always exist and work or
should one be more carefully at this point? Do we obtain always
same two decompositions if we assume $\operatorname{Trdeg}_K(L)
= \infty$ if we replace $\{a_1,...,a_n\}$ by infinite trans basis
$\{a_1, a_2, ...\}$?

If we have now always such two compositions we can how study the algebraic
part separately, therefore assume from now that $K \subset L$ is an
algebraic extension. Can we always decompose it as:

A1) as $K \subset K_s \subset L$ where $K_s$ is the separable closure
of $K$ in $L$. It's  a standard exercise in algebra that such
$K_s \subset L$ always exist and
$K \subset K_s$ is again tautologically a separable extension
and $K_s \subset L$ pure inseparable

A2) about this hypothetical one I'm not sure: does there exist a decomposition
$K \subset K_i \subset L$ where $K_i$ might be the 'inseparable
closure' of $K$ in $L$ and $K \subset K_i$ is pure inseparable ext and
$K_i \subset L$ separable.

About A2) I not sure if such decomposition exist because I haven't found
any textbook refering to 'inseparable closure'. On the other hand it
seems to be always possible to define $K_i$ as the subfield
$K_i := \{a \in L \vert a \text{ inseparable } \}$ where we recall that
an element $a \in L$ is inseparable over $K$ if it's minimal polynomial
$f_a \in K[t]$ has the form $t^{p^m}-c$ where $p >0$ is the characteristic
of $K$.

i know that if $L/K$ is normal and $G= \operatorname{Aut}(L)$ then
one can choose $K_{G}= L^{\text{Fix(G)}}$ as the the subfield of elements
fixed by $G$ and check that $K_{G}$ 'plays the role' of my
'inseparable closure'. What is going wrong with $K_i$ I defined
if $L/K$ is not normal?

Can the $K_i$ from my definition replaced in case of not normal ext by
by another itermediate field $K \subset M \subset L$ having the property
$K \subset M$ pure inseparable and $M \subset L$ separable.

In summary: If $L/K$ is arbitrary field extension do decompositions
$T1$ and $T2$ always exist and if $L/K$ algebraically what is going wrong with
$A2$ and my $K_i$? (since I assume that construction $A1$ always exist).

答:
$L/K(a_1,\ldots,a_n)$ is algebraic and $K(a_1,\ldots,a_n)/K$ is purely transcendental.

In general there is no other canonical (*) decomposition as shown with

$\Bbb{Q}(x,\sqrt{x^3+x})/\Bbb{Q}$ (an elliptic curve) whose field of constants is $\Bbb{Q}$.

(*) the decomposition depends on a choice of transcendental basis so it is not really canonical, it is interesting to check if there are some natural conditions giving a particular one. Minimizing $[L:K(a_1,\ldots,a_n)]$ is natural.

For your second question, the non-separable normal algebraic extensions decompose in the two ways:  $K(L^{p^\infty})$ means $\bigcap_{n\ge 1} K(L^{p^n})$ where $p$ is the finite characteristic, then $L/K(L^{p^\infty})$ is purely inseparable and $K(L^{p^\infty})/K$ is separable, while with $H$ the $K$-embeddings $L\to \overline{L}$ and $L^H$ the subfield fixed by all the embeddings then $L/L^H$ is separable and $L^H/K$ is purely inseparable.

When $L/K$ is not normal then $L/L^H$ doesn't have to be separable as shown there but I didn't know it before so I  don't have much to say about it.


问:
Is there an analytical solution to this probabilistic problem?
I am a big fan of MMORPGs, and this one, Cabal Online, has a particularly punishing upgrade system.

Items can be upgraded to increase their stats, ranging from level 0 to level 15 (for simplicity).

Upgrading costs cores (can be interpreted as currency).

Attempting an upgrade can result can result in the item's level going up (+1), remaining the same (0), or going down (-1 or -2).

The probabilities of these are summarized in the table below:

The question now is:
What is the expected number of cores to upgrade an item from level 0 to level 15?

I initially tried approaching this using a binomial distribution approach, as it involved successes and trials, and failures, but realized that the trials change depending on the item grade and are not independent.

I then tried drawing it as a probability tree of sorts, but ended up with infinite recursion.

I realize that my current knowledge is not adequate to tackle such complex problems, analytically unfortunately.

However, I have managed to arrive at a value of 22,000 using a simulation that involved multiple nested for loops.

I'm now curious about whether there is a way to analytically calculate the number of cores required.

Body:
An analytic solution is possible by the method of Markov chains, but we will need some mathematical software such as Matlab, Octave, or Mathematica.

Let's number the stages from $1$, corresponding to zero weapons, to $16$, corresponding to 15 weapons, and let $\mathbf{P}$ be the $16$ by $16$ matrix of one step transition probabilities; i.e., $\mathbf{P}_{i,j}$ is the probability of moving from stage $i$ to stage $j$. So from the table in the problem statement,

$$\mathbf{P} = \left(
\begin{array}{cccccccccccccccc}
 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0.1 & 0.9 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0.2 & 0 & 0.8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0.3 & 0 & 0.7 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0.4 & 0 & 0.6 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0.27 & 0.21 & 0 & 0.52 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0.36 & 0.2 & 0 & 0.44 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0.45 & 0.19 & 0 & 0.36 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.48 & 0.16 & 0 & 0.36 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.48 & 0.16 & 0 & 0.36 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.48 & 0.16 & 0 & 0.36 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.5 & 0.14 & 0 & 0.36 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.5 & 0.14 & 0 & 0.36 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0.5 & 0.14 & 0 & 0.36 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{array}
\right)$$
Let's also number the steps of the game starting at $0$, so at step $0$ the player is in stage $1$, and let $\pi_n$ be the vector of probabilities of being in each stage at step $n$, written as a $1$ by $16$ matrix, so
$$\pi_0 = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]$$
Given these definitions, the probabilities at step $n$ are given by
$$\pi_n = \pi_0 \mathbf{P}^n$$
It remains to work in the cost in number of cores associated with each stage. Let
$$c = (1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,0)^T$$
Then the expected cost of being at step $n$, disregarding the cost required by previous steps, is
$$x_n = \pi_0 \mathbf{P}^n c$$

We want to know the expected total cost,
$$\sum_{n=0}^{\infty} x_n = \sum_{n=0}^{\infty} \pi_0 \mathbf{P}^n c = \pi_0 \left( \sum_{n=0}^{\infty}  \mathbf{P}^n \right) c = \pi_0 (I-\mathbf{P})^{-1} c$$

Using some mathematical software to invert $I-\mathbf{P}$ and perform the matrix multiplications involved, we find that the expected number of cores required to reach stage $16$ is
$$\pi_0 (I-\mathbf{P})^{-1} c = \boxed{24120.6}$$


问:
Simplify this statistical average
I have a quantity $\tau$ given by:
$$
\frac{1}{\tau} = \frac{1}{\tau_1}+\frac{1}{\tau_2}+\frac{1}{\tau_3}
$$

where $\tau_1$, $\tau_2$ and $\tau_3$ are some constituent quantities. Now these $\tau$'s are function of a variable $x$, but the ensemble averages of $\tau_i$ are known, given by:

$$
\frac{\langle \tau_i^2\rangle}{\langle \tau_i\rangle^2} = \alpha_i
$$

Is there a way to express or simplify the ensemble average of total $\tau$ in terms of $\alpha_i$'s:
$$
\frac{\langle \tau^2\rangle}{\langle \tau\rangle^2}
$$

If not, what would be the necessary information needed w.r.t. the actual distribution of $\tau_i$ with $x$.

答:
First things first, as we know no relations between the $\tau_i$'s a priori, it's best to split this up via linearity of expectation.

Second, there is not a way in terms of your $\alpha_i$'s. This can be shown by considering the exponential distribution, where the $\alpha_i$'s will always be $2$, but the expectation of the inverse varies.

As for what would be proper information,I believe the only real way is to find $\mathbb{E}(1/X_i)$. One trick people I've seen used to compute this is to write it as
$$\mathbb{E}(1/X_i) = \mathbb{E}\left(\int_0^\infty e^{-tX_i}dt \right) = \int_0^\infty \mathbb{E}(e^{-tX_i})dt$$
So maybe that will help. Best of luck.

Title: Simplify this statistical average


问:
Pontryagin classes of a tensor product of bundles
This question is related to " How to calculate characteristic classes of tensor products? " but interested in the Pontryagin classes instead of $c_1$.

Specifically, given two real vector bundles $E$, $F$ over the same base space $B$ (paracompact, or whatever you need for your answer), is there a known formula for the total Pontryagin class $p(E\otimes F)$ in terms of $p(E)$ and $p(F)$?  If not, is there at least a formula for classes with small degree, analogous to the formula for $c_1$ of a tensor product of complex line bundles?

The particular example I have in mind is where $E=F$ is the realification of the tautological complex line bundle $\eta$ over $\mathbb{CP}^n$, where $n\geq 4$ so that $\eta \otimes_\mathbb{R} \eta$ has potentially non-zero $p_1$ and $p_2$.

I've checked the usual references (Milnor-Stasheff, various Hatchers) but they don't seem to tackle this particular problem.  Maybe there is an argument using Chern-Weil theory and a formula for the curvature form of a tensor product of bundles with connection.

答:
As Qiaochu Yuan alluded to in the comments, we have $(E\otimes_{\mathbb{R}} F)\otimes_{\mathbb{R}}\mathbb{C} \cong E_{\mathbb{C}}\otimes_{\mathbb{C}}F_{\mathbb{C}}$ where $E_{\mathbb{C}} = E\otimes_{\mathbb{R}}\mathbb{C}$ and $F_{\mathbb{C}} = F\otimes_{\mathbb{R}}\mathbb{C}$, so

$$p_i(E\otimes_{\mathbb{R}}F) = (-1)^ic_{2i}((E\otimes_{\mathbb{R}}F)\otimes_{\mathbb{R}}\mathbb{C}) = (-1)^ic_{2i}(E_{\mathbb{C}}\otimes_{\mathbb{C}}F_{\mathbb{C}}).$$

Therefore, you can obtain a formula for the $i^{\text{th}}$ Pontryagin class of a tensor product (of real vector bundles) if you can find a formula for the $2i^{\text{th}}$ Chern class of a tensor product (of complex vector bundles). This can be achieved using the multiplicativity of the Chern character, as is done in this answer.

Example: For real vector bundles $E$ and $F$ of ranks $r$ and $s$ respectively, we have

\begin{align*}
p_1(E\otimes_{\mathbb{R}}F) &= -c_2(E_{\mathbb{C}}\otimes_{\mathbb{C}}F_{\mathbb{C}})\\
&= -rc_2(F_{\mathbb{C}}) - sc_2(E_{\mathbb{C}}) - \binom{r}{2}c_1(F_{\mathbb{C}})^2 - \binom{s}{2}c_1(E)^2 - (rs-1)c_1(E_{\mathbb{C}})c_1(F_{\mathbb{C}})\\
&= rp_1(F) + sp_1(E) - \binom{r}{2}c_1(F_{\mathbb{C}})^2 - \binom{s}{2}c_1(E)^2 - (rs-1)c_1(E_{\mathbb{C}})c_1(F_{\mathbb{C}}).
\end{align*}

In particular, modulo $2$-torsion, we have $p_1(E\otimes_{\mathbb{R}}F) = rp_1(F) + sp_1(E)$.


问:
Can you solve a trig equation with a variable both inside a trig function and outside one?
I have the equation:

$$d=\frac{t}{2}-\frac{sin(t)}{4}$$

I'm completely failing at how to get this in terms of $t$

I only care about it for values of $0<t<2\pi $

I've seen the graph so I know there's an inverse of it but I'm struggling on getting there

答:
I do not think it is possible. If we look at a simpler version of the problem,  $$\sin(t) = t,$$ we could maybe divide both sides by $t$ to get $$\frac{\sin(t)}{t} = 1.$$ Then we could take the inverse sinc function(if it exists) on both sides. If sinc inverse exists, then I think that it should be restricted in its domain in order to be a function.

However with a problem like this, I do not think it is possible to isolate $t$ or to get the inverse sinc function.


问:
$E = A^C \cap B^C$, is my answer good?
Could anyone tell me if my answer is correct regarding this set(complement) question. As to confirm myself that I understand it or not.

$U = \{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}$.

$A = \{2,4,6,8,10,12,14\}$,
$B = \{1,2,3,4,5\}$.

Answer i got :

$E = A^C \cap B^C$.

$E = \{1,3,5,7,9,11,13,15\} \cap \{6,7,8,9,10,11,12,13,14,15\}$.

$E = \{9,11,13,15\}.$

Can anyone tell me if my answer is good or not?

Title: $E = A^C \cap B^C$, is my answer good?

Body:
The general idea is correct, but you made some small mistakes.

$E = A^c \cap B^c$

$E=\{1,3,5,7, \mathbf{9},11,13,15\} \cap \{6,7,8,9,10,11,12,13,14,15\}$

$E=\{\mathbf{7},9,11,13,15\}$


问:
Show that $G\left(n,p\right)e^{-\frac{1}{12np\left(1-p\right)}}<{n\choose pn} < G\left(n, p\right)$
The problem:

Using the inequalities

$\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{\left(\frac{1}{12n}-\frac{1}{360n^{3}}\right)} < n! < \sqrt{2\pi n}\left(\frac{n}{e}\right)^ne^{\frac{1}{12n}}$

show that, for a constant $p$ such that $0<p<1$, where $pn$ is an integer,

$G\left(n, p\right)=\frac{2^{nH\left(p\right)}}{\sqrt{2\pi n p\left(1-p\right)}}$, and $H\left(p\right)$ is the binary entropy function, show that

$G\left(n,p\right)e^{-\frac{1}{12np\left(1-p\right)}}<{n\choose pn} < G\left(n, p\right)$

My work so far:

First I have noted that

${n\choose pn} = \frac{n!}{\left(pn\right)!\left(\left(1-p\right)n\right)!}$

$ \approx \frac{\sqrt{2\pi n} \left( \frac{n}{e} \right)^{n}}{\left(pn\right)!\left(\left(1-p\right)n\right)!}$

$\approx \frac{ \sqrt{2\pi \left(1-p\right)n}\left( \frac{\left(1-p\right)n}{e}\right)^{\left(1-p\right)n}}{2\pi \sqrt{\left(1-2p\right)}\left(\frac{pn}{e}\right)^{pn} \left(\frac{\left(1-2p\right)n}{e}\right)^{\left(1-2p\right)n}} e^\left({\frac{1}{12\left(1-p\right)n}}-\frac{1}{12pn+1}-\frac{1}{12\left(1-2p\right)n+1}\right)$

$\leq 2^{nH\left(p\right)} \frac{e^{\frac{1}{12n}}}{\sqrt{2\pi np\left(1-p\right)}}$

and from here I am not sure where to go. Any hints or help would be greatly appreciated.

答:
In inequality $\text{(4d)}$ of this answer, it is shown that
$$
\frac1{12n+1}\le\log\left(\frac{n!\,e^n}{\sqrt{2\pi n}\,n^n}\right)\le\frac1{12n}\tag1
$$
Assuming that
$$
H(p)=-p\log_2(p)-(1-p)\log_2(1-p)\tag2
$$
and using
$$
G(n,p)=\frac{2^{nH(p)}}{\sqrt{2\pi np(1-p)}}\tag3
$$
$(1)$ implies
$$
\scriptsize\frac1{12}\overbrace{\left[\frac1{n+\frac1{12}}-\frac1{np}-\frac1{n(1-p)}\right]}^{\large\frac1{n+1/12}-\frac1{np(1-p)}}
\le\log\left(\frac1{G(n,p)}\binom{n}{pn}\right)
\le\frac1{12}\overbrace{\left[\frac1{n}-\frac1{np+\frac1{12}}-\frac1{n(1-p)+\frac1{12}}\right]}^{\large\le\frac1n-\frac1{p(1-p)(n-1/6)+1/12}}\tag4
$$
Note that the upper bound is negative when $n\ge1$ since $0\le p(1-p)\le\frac14$.

Therefore,
$$
e^{\frac1{n+1/12}-\frac1{np(1-p)}}\le\frac1{G(n,p)}\binom{n}{pn}\le e^{\frac1n-\frac1{p(1-p)(n-1/6)+1/12}}\tag5
$$
which is stronger than the requested inequality.


问:
What does integral and rational mean in the following text?
The following text is from a textbook about college algebra:

A term is integral and rational in certain literals (letters which represent numbers) if the term consists of (a) positive integer powers of the variables multiplied by a factor not containing any variable, or (b) no variables at all.

For example, the terms $ 6x^2y^3, 5y^4, 7, 4x, {\sqrt 3x^3y^6} $ are integral and rational in the variables present.
However, $3{\sqrt x}$ is not rational in x, ${\frac{4}{x}}$ is not integral in x.

答:
It means what it literally says.

If the variables (we don't care at all about the constants) are to a positive whole number powers (possibly $1$), or not any variable at all, then we say the term is "integral" and "rational".  If the variables are either to positive whole number powers, not any variable at all, or in the denominator as a power to a whole positive number, it is called "rational".  All "integral" expressions are "rational" but not all rational expressions are integral.  If the expression contains a variable in a denominator (and it's not being canceled out but an equally large or larger power in the numerator) it is "rational" but not "integral".

Anything else (say the varable is under a square root sing or is itself an exponent) is not rational.

This is an abstraction.  The origin referes to  $5, 10, 7\times 3, 8^2 \times 3$ are all integers (and rational) while $\frac 15, \frac {2\times 7}{9}, \frac {51^3}{19^7}$ are rational and $\sqrt {43}, etc.$ are neither.  However in this case we are talking about unknown variables and we don't really care what they actually are; we are concerned with how they are manipulated and represented.

.....

And we don't care AT ALL about the constant factors.... AT ALL.


问:
Prove $\sin \left(1/n\right)$ tends to $0$ as $n$ tends to infinity.
I'm sure there is an easy solution to this but my mind has gone blank! Any help on proving that $\sin( \frac{1}{n})\longrightarrow0$ as $n\longrightarrow\infty$ would be much appreciated.

This question was set on a course before continuity was introduced, just using basic sequence facts. I should have phrased it as:

Find an $N$ such that for all $n > N$ $|\sin(1/n)| < \varepsilon$

答:
Because $\sin$ is continuous, we can take:

$$\lim\limits_{n\to\infty}\sin\left(\frac{1}{n}\right)=\sin\left(\lim\limits_{n\to\infty}\frac{1}{n}\right)=\sin{(0)}=0$$

To answer the new question:

For each $\epsilon>0$ we need to find how big $n$ (for $N>0$, $n>N$) must be to ensure that $|\sin\left(\frac{1}{n}\right)-0|<\epsilon$. Thus we have:

$$|\sin\left(\frac{1}{n}\right)|<\epsilon$$

Let's only consider positive instances of $\sin1/n$ as negative ones will always be less than $\epsilon$. Then:

$$\frac{1}{\arcsin\epsilon}<n$$

So we put $N=1/\arcsin\epsilon$.

Then, formally:

Let $\epsilon>0$ and let $N=1/\arcsin\epsilon$. Then $n>N$ implies $n>1/\arcsin\epsilon$, hence $\arcsin\epsilon>\frac{1}{n}$, hence $\epsilon>\sin1/n$, hence $|\sin1/n|<\epsilon$. Finally, we have $|\sin1/n-0|<\epsilon$ which proves that the limit is 0.


问:
When is $a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n$
I know that: $$a\equiv b \pmod n \implies a^2 \equiv b^2 \pmod n $$ But this is not an equivalence relation so we cannot say that if:
$$a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n $$

But are there some restrictions that makes this proposition true?

答: Hint:take $a \equiv -b \pmod{n}$

Body: Hint:take $a \equiv -b \pmod{n}$


问:
finding the coordinate that lies in the vector and through it
I'm trying to find the coordinates of A, where the line $l_1$ is given by this equation:

$$\begin{pmatrix}8 \\ 1 \\ -3 \end{pmatrix} +\mathbf{\mu} \begin{pmatrix}-5 \\ 4 \\ 3 \end{pmatrix} $$

Where $\mu$ is a scalar parameter = 1. The point A lies on the equation provided above.

Find the coordinates of A.

My approach:

Given $\mu$ = 1; Then is A = $8  -5\mu=3?$

Furthermore, The point P has position vector:

$$\begin{pmatrix}  1  \\ 5 \\ 2 \end{pmatrix} $$

The line $l_2$ passes through the point P and is parallel to the line $l_1$, whats the vector equation for line $l_2$?

Do I have to do this?

$$\begin{pmatrix}8 \\ 1 \\ -3 \end{pmatrix} +\mathbf{\mu} \begin{pmatrix}1 \\ 5 \\ 2 \end{pmatrix} $$

Given that $\mu$ = 1; then

$$\begin{pmatrix}  10 \\ 11 \\ 1 \end{pmatrix} \implies \begin{pmatrix}10 \\ 11 \\ 1 \end{pmatrix} +\mathbf{\mu} \begin{pmatrix}1 \\ 5 \\ 2 \end{pmatrix} $$

Then If I'm to find the exact distance AP, giving an answer in the form $k\sqrt{2}$, where is to be determined how to I proceed?

And if the acute angle between AP and $l_2$ is $\theta$, how to find $\cos\theta?$

答:
When $\mu=1$ the point is
$$A=\begin{pmatrix}8 \\ 1 \\ -3 \end{pmatrix} +1\cdot\begin{pmatrix}-5 \\ 4 \\ 3 \end{pmatrix} =\begin{pmatrix}3 \\ 5 \\ 0 \end{pmatrix} $$
The line $l_2\parallel l_1$ is simply
$$\begin{pmatrix}1 \\ 5 \\ 2 \end{pmatrix} +\mathbf{\lambda} \begin{pmatrix}-5 \\ 4 \\ 3 \end{pmatrix} $$
$AP=\sqrt{(3-1)^2+(5-5)^2+(0-2)^2}=2\sqrt{2}$

$\vec{AP}=(3,5,0)-(1,5,2)=(2,0,-2)$ and $|\vec{AP}|=\sqrt{2^2+0^2+(-2)^2}=2\sqrt{2}$

The direction vector of $l_2$ is $\vec{u}=(-5,4,3)$, and $|\vec{u}|=\sqrt{(-5)^2+4^2+3^2}=5\sqrt 2$

The cosine of the angle between $AP$ and $l_2$ is
$$\cos\theta=\frac{\vec{AP}\cdot \vec{u}}{|\vec{AP}||\vec{u}|}=\frac{2\cdot 3+0\cdot 5+(-2)\cdot 0}{(2\sqrt 2)(5\sqrt 2)}=\frac{3}{10}$$

edit

Given $AP=PE$ and $E$ is on line $l_2$ means that we intersect the line $l_2$ with the sphere with center $(1,5,2)$ and radius $2\sqrt 2$, whose equation is $$(x-1)^2+(y-5)^2+(z-2)^2=8$$
while parametric equation of line $l_2$ is
$$l_2:(1-5\lambda,5+4\lambda,2+3\lambda)$$
substitute these last ones into the equation od the sphere and get
$$(3 \lambda +2-2)^2+(4 \lambda +5-5)^2+(-5 \lambda +1-1)^2=8$$
which gives
$$50 \lambda ^2=8\to \lambda=\pm\frac{2}{5}$$
if $\lambda=\frac{2}{5}$ then point $E$ has coordinates $E_1=\left(-1,\frac{33}{5},\frac{16}{5}\right)$

if $\lambda=-\frac{2}{5}$ then point $E$ has coordinates $E_2=\left(3,\frac{17}{5},\frac{4}{5}\right)$

The triangle is isosceles, so the area is $\frac12\cdot AE_2\cdot PM$ where $M$ is the midpoint of $AE_2$. Coordinates of $M$ are the mean (average) of the coordinates of $A$ and $E_2$.

$M=(3,4.2,0.4)$ and $$PM=\sqrt{\left(1-3\right)^2+\left(5-4.2\right)^2+\left(2-0.4\right)^2}=\frac{6}{\sqrt{5}}$$
$$AE_2=PE_2=\frac{4}{\sqrt{5}}$$
$$Area_{APE_2}=Area_{APE_1}=\frac12\cdot\frac{6}{\sqrt{5}}\cdot \frac{4}{\sqrt{5}}=2.4$$


问:
find the best fit between a long vector and a matrix
I have a long vector $a \in \mathbb{R}^{m \times 1}$ representing spatial information, and a matrix $C = \mathbb{R}^{m \times n}$ representing spatio-temporal information, how to find a good fit of the row vector $b \in \mathbb{R}^{1 \times n}$ (represents temporal information) so that $ab \approx C$?

One easy way seems to be to iterate each column of $C$ and find the ratio between its norm and norm(a), and use the ratio as entries for $b$, however, it treats each column of $C$ independently, so it's a pity not utilizing the correlations between the columns. I wonder if there's some standard way of dealing with this?

答:
You have a linear equation
$$ab^T = C \quad\iff\quad ba^T = C^T$$
You can solve for either vector in terms of the Moore-Penrose inverse of the other vector. In the case of vectors, there is an exact formula for the pseudoinverse:  $\quad v^+=\frac{v^T}{v^Tv}$

Therefore
$$\eqalign{
a^T &= b^+C^T \quad&\implies\quad &a = \frac{Cb}{b^Tb} \\
b^T &= a^+C   \quad&\implies\quad &b = \frac{C^Ta}{a^Ta} \\
}$$
NB: You wrote $b$ as a row vector, whereas I used column vectors in these formulas.


答:
The "least squares" approximation for $b$ is given by
$$
b = \frac{a^\top C}{a^\top a},
$$
where $M^\top$ denotes the transpose of $M$.

Title: find the best fit between a long vector and a matrix

Body:
I have a long vector $a \in \mathbb{R}^{m \times 1}$ representing spatial information, and a matrix $C = \mathbb{R}^{m \times n}$ representing spatio-temporal information, how to find a good fit of the row vector $b \in \mathbb{R}^{1 \times n}$ (represents temporal information) so that $ab \approx C$?

One easy way seems to be to iterate each column of $C$ and find the ratio between its norm and norm(a), and use the ratio as entries for $b$, however, it treats each column of $C$ independently, so it's a pity not utilizing the correlations between the columns. I wonder if there's some standard way of dealing with this?


问:
Is the following polynomial reducible over Q?
I am looking at an exercise saying that "Demonstrate that x4-22x2+1 is reducible over Q. I have the solution manual and it solves like the following:

If x4-22x2+1 is reducible over Z, then it factors in Z[x], and must therefore
either have a linear factor in Z[x] or factor into two quadratics in Z[x]. The only possibilites for a
linear factor are x ± 1, and clearly neither 1 nor -1 is a zero of the polynomial, so a linear factor is
impossible. Suppose
x4-22x2+1 = (x2 + ax + b)(x2 + cx + d).
Equating coefficients, we see that
x3 coefficient : 0 = a + c
x2 coefficient : −22 = ac + b + d
x coefficient : 0 = bc + ad
constant term : 1 = bd so either b = d = 1 or b = d = −1.
Suppose b = d = 1. Then −22 = ac + 1 + 1 so ac = −24. Because a + c = 0, we have a = −c, so
−c2 = −24 which is impossible for an integer c. Similarly, if b = d = −1, we deduce that −c2 = −20,
which is also impossible. Thus the polynomial is irreducible.

My question is:

1)What does it mean "over Q"? What is the difference between saying over Z and over Q?

2)Why do we factor it as (x2 + ...)(x2 + ...)? Can't it be like (x3 + ...)(x + ...). Also, why don't we factor it like (ax2 + ...)(bx2 + ...), i mean how do we know that the head coefficients are 1?Can someone help with this?

Thanks

答:
Over $\mathbb Q$ means that rational coefficients are allowed, not just integers. In fact one can show that they are not needed - but that has to be proved. Suppose $p(x)q(x)=x^4+22x^2+1$ is a rational factorisation. Multiply by $n$ to clear fractions, so that $P(x)Q(x)=n(x^4+22x^2+1)$ is an equation with integer coefficients. Now suppose $p$ is a prime factor of $n$. Then, starting with the highest power of $x$ and the coefficients of the factorisation, it is possible to prove that $p$ is a factor of every coefficient of either $P(x)$ or $Q(x)$. One then cancels the $p$ from both sides. Eventually, every factor of $n$ is cancelled, and we are left with an integer factorisation of the original equation.

If $p(x)=(ax^3+bx^2+cx+d)(x+e)$ then $p(-e)=0$ and $e$ is a root of the equation.


问:
Why care about ultrafilters (in operator algebras)?
I recently learned about filters and ultrafilters and the notion of convergence with respect to an ultrafilter in a topological space. I have to say, at first I could not appreciate their potential. I think the two central theorems are the following:

Proposition: Let $X$ be a set. Then any $S\subset\mathcal{P}(X)$ having the finite intersection property is included in some ultrafilter.

Proposition: Let $X$ be a topological space. Then $X$ is compact if-f every ultrafilter converges to at least one point in $X$. Also, $X$ is Hausdorff if-f every ultrafilter converges at most to one point in $X$.

I realized that ultrafilters are a powerful tool when I saw a very simple proof of Tychonoff's theorem using ultrafilters. I always thought that Tychonoff's theorem had a difficult proof and I was amazed to see such a simple proof using ultrafilters.

I would like to ask the following question: how are ultrafilters used in operator algebras? What are some interesting/ standard applications they have in this field? What books or papers would you recommend to someone that wants to understand the use of ultrafilters in operator algebras?

Bonus question: What about topology, in general? To what cause would someone use ultrafilters besides proving that a space is compact or Hausdorff? Are there any interesting constructions based on ultrafilters?

答:
Here is an application of ultra-filters that is very much in the spirit
of operator algebras.  The ultra-filters in question are not quite on
topological spaces, but instead, in the context of meet
semi-lattices.

For example, if   $X$ is a topological space, then the set of all subsets
of $X$,  denoted $\mathscr P(X)$, is a meet semi-lattice relative to the
order of inclusion and the usual ultra-filters  in $X$ correspond
precisely to ultra-filters in the meet semi-lattice $\mathscr P(X)$.

With these disclaimers, here we go:  many fundamental examples of
C$^*$-algebras  possess a set $S$ of generators consisting of partial
isometries.   Moreover the set $S$ is often closed under
multiplication and adjoint, in which case it forms an inverse
semigroup.

Since the algebraic structure of inverse semigroups is way easier to
understand as  compared to C$^*$-algebras, the question arises as to what
extent the given C$^*$-algebra may be reconstructed entirely in terms of
the generating  inverse semigroup $S$.

As it turns out the answer is generally negative because one can easily
obtain examples of non-isomorphic C$^*$-algebras generated by isomorphic
inverse semigroups of partial isometries.

Nevertheless, there is a construction of a specific C$^*$-algebra from
any given
inverse semigroup $S$, called the tight C$^*$-algebra, which is able to
recover some of the most important C$^*$-algebra in applications, such as
Cuntz-algebras and a host of other examples.

The contruction of the tight C$^*$-algebra can be made from a certain
dynamical system on the closure of the space of all ultra-filters of the
idempotent semi-lattice of the given inverse semigroup.


问:
Expressing functions $n\to n$ as evaluation of polynomials mod $n$
Consider $f:n\to n$ where $n:=\{0,1,2,\ldots,n-1\}$. We can associate to $f$ a polynomial $P_f$ of degree $<n$ by defining $P_f(x)=f(0)x^0+f(1)x^1+\cdots +f(n-1)x^{n-1}$. This gives another way to evaluate $f$: we have the literal function evaluation $x\mapsto f(x)$, and the "fancy" evaluation $x\mapsto \sum_{k<n}f(k)x^k$. I wish to understand a connection between these two evaluations.

Question: For $f:n\to n$, is there some polynomial $P(x)$  which evaluates to $P(x)\equiv f(x)$ (mod $n$) for all $x\in \mathbb{N}$?

What I've done: Assuming $n$ is prime, we only need to consider polynomials of degree $<n$ since by Fermat's Little Theorem $x^n\equiv x$. My idea was to find polynomials $i_k(x)$ which evaluate to the indicator function of the singleton $\{k\}$ so that $i_k(x)\equiv \begin{cases}1&x=k\\0&\text{else}\end{cases}$

Once we have these, every function can be written as a linear combination of the $i_k$ by $\left(\sum_{k<n}f(k)\cdot i_k\right)(x)\equiv f(x)$. To find these $i_k$, I used linear algebra to go in reverse. We start with the $n\times n $ matrix of the powers of $k<n$ reduced mod $n$. That is, $A_{i,j}=i^j$. For example, for $n=3$ this matrix is $A_3:=\begin{bmatrix}1&0&0\\1&1&1\\1&-1&1\end{bmatrix}$. For $n=4$, we have $A_4:=\begin{bmatrix}1&0&0&0\\1&1&1&1\\1&2&0&0\\1&-1&1&-1\end{bmatrix}$ and for $n=5$ it is $A_5:=\begin{bmatrix}1&0&0&0&0\\1&1&1&1&1\\1&2&-1&-2&1\\1&-2&-1&2&1\\1&-1&1&-1&1\end{bmatrix}$.

If the inverse of $A_n$ exists, it gives us the conversion matrix. In the case $n=3$ we have $A_3^{-1}=\begin{bmatrix}1&0&0\\0&-1&1\\-1&-1&-1\end{bmatrix}$. Consider for example the function $0\mapsto 1,1\mapsto 2, 2\mapsto 0$. We compute $\begin{bmatrix}1&0&0\\0&-1&1\\-1&-1&-1\end{bmatrix}\begin{bmatrix}1\\2\\0\end{bmatrix}\equiv\begin{bmatrix}1\\1\\0\end{bmatrix}$ so our desired polynomial is $x^0+x^1$. Indeed, this evaluates correctly mod 3.

For $n$ prime, this matrix seems to always be invertible (at least up to 11 according to wolfram). Moreover, the leftmost column of the inverse matrix is always $\begin{bmatrix}1\\0\\\vdots\\0\\-1\end{bmatrix}$ so the indicator $i_0(x)$ is the polynomial $1-x^{p-1}$. This corresponds to the corollary of Fermat's Theorem that $p\not\mid x\Rightarrow x^{p-1}\equiv 1$. Also, the bottom row of the inverse matrix always seems to be $\begin{bmatrix}-1&-1&\cdots&-1&-1\end{bmatrix}$. I have not been able to prove that $A_p$ is always invertible for $p$ prime.

For $n$ composite the $A_n$ seems never to be invertible, though I haven't been able to prove this either. However a proof of this wouldn't even answer the original question. Fermat's Theorem doesn't hold for composites, so we might theoretically find a suitable polynomial of degree $\geq n$.

I am also wondering about potential applications of this idea. If the symmetric group $S_p$ can be represented using polynomials in this way, group multiplication would correspond to composition of the polynomials (mod $p$). For example, we would have that $S_3\cong \{x,x+1,x+2,2x,2x+1,2x+2\}$. We don't typically think of $S_3$ as being linear polynomials (mod 3) under composition, but apparently (up to isomorphism) it is ! We could take this idea even further: for every $n$, there's always a least prime $p\geq n$ so we could potentially pick an embedding $S_n\leq S_p$. Then by Cayley's Theorem we could represent any finite group as composing polynomials (mod $p$) for some prime. Perhaps representing groups  this way would give new insights into group theory? I would also appreciate any other ideas for potential applications. Thank you!

答:
The answer is yes. First, note that

$$(x+n)^r=\sum_{k=0}^r\binom{r}{k}x^{r-k}n^k=x^r+\sum_{k=1}^r\binom{r}{k}x^{r-k}n^r\equiv x^r\ (\text{mod }n)$$

This then implies that for any polynomial, we have that

$$P(x+n)=\sum_{i=0}^r a_i(x+n)^i\equiv \sum_{i=0}^r a_ix^i=P(x)\ (\text{mod }n)$$

Thus, it is sufficient to find $P(x)$ such that

$$P(x)\equiv f(x)\ (\text{mod }n)$$

for $x\in \{0,1,...,n-1\}$. But the problem becomes trivial at this point. We simply need to find the interpolating polynomial such that

$$P(0)=f(0)$$

$$P(1)=f(1)$$

$$\vdots$$

$$P(n-1)=f(n-1)$$

This is a well known problem and algorithms to find such a $P(x)$ can be found in many places (for example here).


问:
Conjugate of negative normalized entropy
Given$$f(x) = \sum_{i=1}^n x_i \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right), \qquad dom f = R^n_{++}$$
Find $f^*(y)$

$$f^*(y) =
\sup_x ( y^T x - f(x) ) =$$
$$
\sup_x \left( y^T x - \sum_{i=1}^n x_i \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right) \right) =$$
$$
\sup_x \sum_{i=1}^n x_i \left(y_i - \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right) \right) =$$
$$
\sup_x \sum_{i=1}^n x_i \left(\log e^{y_i} - \log \left( \frac{x_i}{\sum_{j=1}^n x_j} \right) \right) =$$
$$
\sup_x \sum_{i=1}^n x_i \log \left( \frac{e^{y_i}\sum_{j=1}^n x_j}{x_i} \right)$$
Unfortunately, I can't come up what to do next. This is problem 3.41 from Convex Optimization by Boyd and Vandenberghe, and there is an answer to this problem available:

$$
\begin{equation*}
f^*(y) =
 \begin{cases}
   0 & \sum_{i=1}^n e^{y_i} \leq 1 \\
   +\infty & \text{otherwise}
 \end{cases}
\end{equation*}$$

答:
The derivation is straightforward if you break it down in small steps. The first step is decomposing $f$:

$$f(x) = \left( 1^Tx \right) g\left( \frac{x}{1^Tx}\right) \text{ with } g(x)=\sum_{i=1}^n g_i(x) \text{ and } g_i(x) = x_i \log(x_i).$$

We start by using that the conjugate of $x\log x$ is $e^{v-1}$, so:
$$g_i^*(v) = \begin{cases} e^{v_i-1} & \text{if } v_j =0 \;\forall j \neq i \\ \infty & \text{otherwise.} \end{cases}$$
Therefore, $g^*(v) = \sum_{i=1}^n e^{v_i-1}$. The largest step is computing the conjugate of the perspective:
\begin{align}
f^*(v) &= \sup_{x} \left\{x^Tv- \left(1^Tx\right) g\left( \frac{x}{1^Tx}\right)\right\} \\
&= \sup_{x,y\geq 0} \inf_z \left\{x^Tv- y g\left( \frac{x}{y}\right) + z(y-1^Tx)\right\} \\
&= \inf_z \sup_{x,y\geq 0} \left\{x^Tv- y g\left( \frac{x}{y}\right) + z(y-1^Tx)\right\} \\
&= \inf_z \sup_{y\geq 0} \left\{zy + y \sup_x \left\{(x/y)^T(v-z1)- g\left( \frac{x}{y}\right)\right\} \right\} \\
&= \inf_z \sup_{y\geq 0} \left\{zy + y g^*(v-z1) \right\} \\
&= \inf_z \sup_{y\geq 0} \left\{ zy + y \sum_{i=1}^n e^{v_i-z-1} \right\} \\
&= \inf_z \begin{cases}0 & \text{if } \sum_{i=1}^n e^{v_i-z-1} + z \leq 0 \\ \infty & \text{otherwise.}\end{cases}
\end{align}
The last step is to find when there exists a $z$ such that $h(z) := \sum_{i=1}^n e^{v_i-z-1} + z \leq 0$, which is equivalent to solving $\min_z h(z) \leq 0$. Setting the derivative of $h$ to $0$ we obtain $\sum_{i=1}^n e^{v_i-z^*-1} = 1$ or $z^* = \log(\sum_{i=1}^n e^{v_i})-1$, so  $f^*(v) = 0$ if $h(z^*)=\log(\sum_{i=1}^n e^{v_i}) \leq 0$.


答:
If you mean restrictions solely on $n$, then the only $n$ for which this is true is $n = 1,2$.

If $n \geq 3$, then there is always an element $a$ with $a \neq -a$.

Taking $b := -a$,
$a^2 = b^2$ but $a \neq b$.

In general, when $n$ is not prime, there can be more that just these two elements which square to the same thing: for example in $\mathbb{Z} / 8 \mathbb{Z}$, each of $1,3,5,7$ square to $1$!

Title: When is $a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n$

Body:
I know that: $$a\equiv b \pmod n \implies a^2 \equiv b^2 \pmod n $$ But this is not an equivalence relation so we cannot say that if:
$$a^2 \equiv b^2 \pmod n \implies a\equiv b \pmod n $$

But are there some restrictions that makes this proposition true?


问:
An Idea to solve a limit involve integral
Suppose $f(x)$ is a continuous function over $[0,1]$,and $0<a<b$ find the below limit $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=?$$
My work is below

$\int_{at}^{bt}\frac{f(x)}{x}=\int_{at}^{bt}f(x)d(\ln(x))$

Using the mean value theorem, and $f\in c^1[0,1]$ we have $$\exists c:a<c<b \ |f(ct)=\frac{\int_{at}^{bt}f(x)d(\ln(x))}{\ln(bt)-\ln(at)}$$ so $$\lim_{t\to 0^+}{\int_{at}^{bt}\frac{f(x)}{x}}=\lim_{t\to 0^+}{\int_{at}^{bt}f(x)d(\ln(x))}=\\
\lim_{t\to 0^+}f(ct)ln(\frac{bt}{at})=f(0)\ln(\frac ba)
$$
My questions:

1:Is my working true?

2:Is there another Idea or method to solve the problem?

Thanks in advance

答:
A couple things. First the assumption only says $f$ is continuous, not that $f\in C^1[0,1]$, so mean value theorem doesn't work.

Second, I believe the bounds in the integrals should be $\ln(at)$ and $\ln(bt)$ when you've switched to $d(\ln(x))$. Though this could just be some convention I am unaware of.

A similar idea is to use inequalities to bound it i.e.
$$\int_{at}^{bt} \frac{f(x)}{x} dx \leq \sup_{at \leq x \leq bt} f(x) \int_{at}^{bt}\frac{1}{x}dx = \ln(b/a)\sup_{at\leq x\leq bt}f(x) $$
and
$$\int_{at}^{bt} \frac{f(x)}{x} dx \geq \inf_{at \leq x \leq bt} f(x) \int_{at}^{bt}\frac{1}{x}dx = \ln(b/a)\inf_{at\leq x\leq bt}f(x) $$
Since we are taking $t\to 0$ and $f$ is continuous, both the supremum and infimum will become $f(0)$ in the limit.


问:
triangle free and connected graph
Let G be a connected and triangle-free graph with 82 vertices,
Prove that alpha(G)>=10
Where alpha(G) is the maximal independent set of G.

I've shown that if the maximal independent set is on 9 vertices ,
Then any other vertex we add to this set, it will be connected to at least one of them.
Then if we part the 73 vertices left to which one they're connected (if it is two different vertices choose one randomally)
We get one vertex of the independent set connected to atleast 9. If more then 9 we can find a triangle,
If 9 they create a new anti clique of 9 vertices. I don't know how to continue from here.

Thank you.

答:
There is another solution for this kind of exercises:

We can use Brook's theorm- since the graph isn't clique and the graph isn't an odd cycle, $\chi(G) \leq \Delta(G)$.

Now there are two cases:

Case 1: there is a vertex $x$ , s.t  , $deg(x)>  9$ : In that case, since there are no triangulars, the neighbers of $x$ form an indep set of order $\geq 10$.

Case 2: $\Delta(G) \leq 9$: Using Brook's thm, we can colour G with 9 colours. By piegonhole theorm, there is colour class with 10 vertices- hence an independent set with the demanded order.

Title: triangle free and connected graph


问:
The category $\operatorname{Fun}(\sf{C},\sf{A})$ is abelian is $\sf{A}$ is abelian and $\sf{C}$ is small. Why the set-theoretic condition?
Proposition 1.4.4 in F. Borceux Handbook of Categorical Algebra II says that if $\sf{A}$ is an abelian category and $\sf{C}$ is a small category, then the category of functors $\operatorname{Fun}(\sf{C},\sf{A})$ is abelian.

I have two questions:

Is this true without demanding $\sf{C}$ to be small? The only place where Borceux proof seem to demand this condition is where he uses that in this case limits and colimits are computed pointwise. But I think I know how to prove that products, kernels and cokernels are computed pointwise without demanding $\sf{C}$ to be small.

If the theorem is true without this condition, then why aren't limits and colimits computed pointwise in functor categories without this assumption?

答: If $\mathcal{C}$ is not small, the functor category $\operatorname{Fun}(\mathcal{C},\mathcal{A})$ will not be locally small in general. The morphisms in this category are natural transformations, whose components are indexed in $\operatorname{Ob}\mathcal{C}$, which is a proper class/large set if $\mathcal{C}$ is not small. Thus $\operatorname{Fun}(\mathcal{C},\mathcal{A})$ cannot be $\mathsf{Ab}$-enriched and hence cannot be abelian.


答:
The issue is when will $n|(a^2 - b^2) = (a-b)(a+b)$ imply that $n|(a-b)$.  An intermediate result of number theory is that $r|st$ and $r$ relatively prime to $t$ implies that $r|s$.

Applying that to your query, a sufficient condition is if $n$ is relatively prime to $(a+b)$.  The word sufficient is emphasized to indicate that individual examples are certainly feasible where $n$ is not relatively prime to $(a+b)$, but $n$ still divides $(a-b)$.