问:
Let $R$ and $R'$ be two discrete valuation rings of $K$. Is $R=R'$?
Let $K$ be a finite extension of the $p$-adic field $\mathbb{Q}_p$. Assume $R$ and $R'$ be two discrete valuation rings of $K$ such that $R \subset R'$.

Lemma: Is $R =R'$  or $R \cong R'$ ?

Since $R$, $R'$ are discrete valuation ring,then both has unique maximal ideals $m$ and $m'$ respectively  such that $m \subset m'$.

If we show that $m' \subset m$, then $m=m'$ will imply $R=R'$.

Am I right here?

Now, let $0 \neq x \in m'$, then either $x \in R$ or $x ^{-1} \in R$. If $x^{-1} \in R$, then $x^{-1} \in R'$, which implies $x$ is invertible in $R'$ and hence $x \notin m'$, a contradiction.  Thus $m' \subset m$ and we already have $m \subset m'$ as $R \subset R'$. Thus $m=m'$, which forces $R=R'$.

$-----------------------------$

My main question:

Consider the Cyclotomic extension $\mathbb{Q}_p(\zeta_p)$ of $\mathbb{Q}_p$. We know that both $\mathbb{Z}_p[\zeta_p]$ and $\mathbb{Z}_p[\pi], \ \pi=\sqrt[(p-1)]{-p}$ are discrete valuation rings of $\mathbb{Q}_p(\zeta_p)$.

I want to show $\mathbb{Z}_p[\zeta_p]=\mathbb{Z}_p[\pi]$.

Clearly  $\mathbb{Z}_p[\zeta_p]\subset \mathbb{Z}_p[\pi]$ because $\zeta_p \in \mathbb{Z}_p[\pi]$ as $\pi=\sqrt[(p-1)]{-p}$.

Thus by above Lemma, we have $\mathbb{Z}_p[\zeta_p]=\mathbb{Z}_p[\pi]$.

Am I correct ?

Thanks for help

答:
Lemma 1: Let $R$ be a discrete valuation ring (with maximal ideal $m =(\pi)$), and let $(R', m')$ be any local ring such that $R \subseteq R'$. Then either $m \subseteq m'$ or $R'$ contains the field of fractions of $R$. In particular, if $R, R'$ are valuation rings of the same fraction field $K$, then $m \subseteq m'$ (unless $R' = K$ which you tacitly exclude.)

Proof: If $\pi \notin m'$, then $\pi \in (R')^\times$ and hence $R'$ contains $R[\pi^{-1}] = Frac(R)$.

Note that if we drop the condition that the valuation of $R$ is discrete, the result is not necessarily true even for two valuation rings of the same fraction field $K$:

(Counter-)Example 1: Let $K$ be the field of Laurent series in two variables $x,y$ as described in this answer. Let $R$ be the valuation ring for the (rank two) valuation described there (called $A$ in that answer). Note that $R$ consists of those series where the coefficients of $x^r y^s$ are zero if either $r$ is negative, or $r = 0$ and $s$ is negative.

Now consider the following bigger subring $R'$ of $K$: It consists of those series where the coefficients of $x^r y^s$ are zero if $r$ is negative. Note that $R \subsetneq R' \subsetneq K$: e.g. $y^{-1} \in R' \setminus R$. In fact $R'$ is the valuation ring to the rank one, discrete valuation $v'$ which just measures the degree of $x$ in $K$ viewed as $E((x))$, where $E:=\mathbb C((y))$ is the ground field and residue field, all its elements having trivial valuation $v'(e)=0$.

Now $y \in m$ but $y \notin m'$.

ADDED: Also look through the "similar but different", but maybe psychologically easier, example here.

Lemma 2: If $R$ is a valuation ring of the field $K$, and $R'$ is any proper subring of $K$ containing $R$, then $R'$ is local, and its maximal ideal $m'$ is contained in the maximal ideal $m$ of $R$.

Proof: Exactly as outlined in your question: Let $x$ be any non-zero non-unit in $R'$. Since $x$ is no unit in $R'$, we cannot have $x^{-1} \in R \subseteq R'$. So by $R$ being a valuation ring, we have $x \in R$, and again by assumption cannot have $x \in R^\times$, so $x\in m$.
This means that all non-units of $R$ must be in $m$. But then the sum of two non-units in $R'$ is a non-unit, and it's clear that a non-unit multplied by any element of $R'$ is still a non-unit, i.e. the non-units in $R'$ form an ideal, which then must be the unique maximal ideal of $R'$, and we showed all its elements must bei in $m$.

Note that (Counter-)Example 1 also gives an example of this where the inclusion $m' \subset m$ is proper.

Lemma 3: In the situation of lemma 2, assume $m=m'$. Then $R=R'$.

Proof: It suffices to show $R'^\times \subseteq R$. But if some unit $x$ of $R'$ were not in $R$, then by $R$ being a valuation ring we would have $x^{-1} \in R$. Now obviously $x^{-1} \notin m = m'$ (else it couldn't be a unit in $R'$), so $x^{-1} \in R^\times$, but then $x \in R^\times$ all along.

Corollary ("Yes" to your question): If $R \subseteq R'$ are two proper valuation rings of the field $K$, and $R$ is a discrete valuation ring, then $R=R'$.

Proof: Combine lemmas 1,2,3.

However, all of this is not the main problem in your main question. The difficult part in showing that

$$\mathbb{Z}_p[\zeta_p]=\mathbb{Z}_p[\pi]$$

is to show any inclusion $\pi \in \mathbb{Q}_p(\zeta_p)$ or $\zeta_p \in \mathbb{Q}_p[\pi]$ to begin with. That exact problem has been solved on this site both with Hensel's Lemma and Krasner's Lemma, and without anything of that price category, you will not be able to get any argument started here. Once one has one of those inclusions, one could continue with the general machinery above, but one might as well conclude with basic dimension/rank arguments and/or properties of extensions of complete DVRs, because when dealing with $p$-adics one has to use those all the time anyway.


问:
inner product space and polynomial
Let $V = \mathrm{span}\{1,x,x^2,x^3\}$ be a real inner product space with the
inner product defined by
$$
\langle f,g\rangle =\int\limits_{-1}^{1} fg
$$

Check that $T(f) = f(0)$ is a linear functional on this space and find the element of $V$ which represents it (corresponds to).

So how should I do it? From where to begin?

I am guessing the first part is easy because I have to prove that

$$T(f+g)=(f+g)(0)=f(0)+g(0)=T(f)+T(g)$$

and that

$$T(af)=af(0)=aT(f)$$

but what about this vector that I need to find?

I know that it has something to do with Gram-Schmidt but I do not know how.

答:
As you said, in order to prove that $T[f] = f(0)$ is a linear functional you just need to check the two conditions of linearity, ie.
$$T[f + g] = (f + g)(0) = f(0) + g(0) = T[f] + T[g]$$ and
$$T[af] = af(0) = aT[f]\qquad \forall a \in K = \mathbb{R}\ \text{or}\ \mathbb{C}.$$

Now you are asked to find an element representing a linear functional, therefore I suppose you are asked to use the Riesz Representation Theorem. This is what I am talking about: http://en.wikipedia.org/wiki/Riesz_representation_theorem.

Roughly speaking, you are asked to look for the vector $g$ satisfying the equality
$$Tf = (f,g).$$

Before doing this just a couple of observations:

$1)$ I suppressed the notation [] as is usually done with linear operators.

$2)$ I am assuming that the field is $\mathbb{R}$ only because there is no conjugacy in the integral defining the inner product.

$3)$ The existence of such a $g$ is given by the above Theorem by Riesz. (It is also unique!)

Finally the problem is just a matter of calculations: let $$g = a x^3 + bx^2 + cx + d.$$ It is determined if we determine its coefficients. To do this just remember that the integral of an odd function over an interval symmetric with respect to the origin equals $0$ and it is two times the integral on the positive subinterval if the function is even. In formulas, if $f(x) = -f(-x)$ then $\int_{-1}^1f(x)dx = 0$, while if $f(x) = f(-x)$ we have $\int_{-1}^1f(x)dx = 2\int_0^1f(x)dx$.

Now let's compute $\int_{-1}^1f(x)g(x)dx$ for some (non-random) $f$.
If f$(x) = 1$ then we impose
$$1 = f(0) = Tf = \int_{-1}^1(a x^3 + bx^2 + cx + d)dx = 2\int_0^1(bx^2 + d)dx = \frac{2}{3}b + 2d.$$
If $f(x) = x^2$ then we impose
$$0 = f(0) = Tf = \int_{-1}^1(ax^5 + bx^4 + cx^3 + dx^2)dx = 2\int_0^1(bx^4 + dx^2)dx = \frac{2}{5}b + \frac{2}{3}d$$
If can now find $b$, $d$ by solving the system
$$1 = \frac{2}{3}b + 2d$$
$$0 = \frac{2}{5}b + \frac{2}{3}d.$$

You now you should be able to continue by yourself, and you should also have an idea of what could be nice choises for $f$ to find $a$ and $c$.

I hope this helps!

ContentLicense: CC BY-SA 3.0

Title: inner product space and polynomial


问:
ith Center of Product is Product of ith Centers
The Lemmas:

-Lem. 1: Let $k\in\mathbb{N}$. Let $G_{1}$, ..., $G_{k}$ be groups. Then $$Z(G_{1}\times...\times G_{k})=Z(G_{1})\times...\times Z(G_{k}).$$

-Lem. 2: Let $k\in\mathbb{N}$. For $1\leq j\leq k$ let $G_{j}$ be a group and $N_{j}\trianglelefteq G_{j}$. Then \begin{align*}&i.)\;N_{1}\times...\times N_{k}\trianglelefteq G_{1}\times...\times G_{k};\\&ii.)\;(G_{1}\times...\times G_{k})/(N_{1}\times...\times N_{k})\cong (G_{1}/N_{1})\times...\times(G_{k}/N_{k}).\end{align*}

The Problem:

Let $k\in \mathbb{N}$. Let $G=G_{1}\times ...\times G_{k}$ where the $G_{j}$ are groups. For each $G_{j}$, let $N_{j}\trianglelefteq G_{j}$. Then for $i\geq 0$ we have $$Z_{i}(G_{1}\times...\times G_{k})=Z_{i}(G_{1})\times...\times Z_{i}(G_{k}).$$

My Attempt at Solution:

We proceed by induction on $i\geq 0$.

For the Base Case, $i=0$ is the trivial case. Note the $i=1$ case is given by Lem. 1 above.

So let $i\geq 0$. Suppose $Z_{i}(G_{1}\times...\times G_{k})=Z_{i}(G_{1})\times...\times Z_{i}(G_{k}).$ Now observe \begin{align*}Z_{i+1}(G_{1}\times...\times G_{k})/Z_{i}(G_{1}\times...\times G_{k}) &= Z((G_{1}\times...\times G_{k})/Z_{i}(G_{1}\times...\times G_{k}))\\&=Z((G_{1}\times...\times G_{k})/(Z_{i}(G_{1})\times...\times Z_{i}(G_{k})))\quad(\text{using induction hypothesis})\\&- - -\end{align*}

Comments:

... and this is where I'm stuck. My gut instinct was to use Lem. 2 to say $Z((G_{1}\times...\times G_{k})/(Z_{i}(G_{1})\times...\times Z_{i}(G_{k})))\cong Z(G_{1}/Z_{i}(G_{1})\times...\times G_{k}/Z_{i}(G_{k})),$ but I think this is wrong for two reasons: 1.) the problem asks to show equality, not isomorphism; and 2.) going through Dummit & Foote, we have not yet proven each $Z_{i}(G)\trianglelefteq G$ (this fact is a result of the very first exercise of Sec. 6.1 but the result I am trying to prove was presented mid-section by our professor). Also, no where in the proof I started above did I use each $N_{j}\trianglelefteq G_{j}$ which is concerning.

Any help, hints, or comments would be greatly appreciated!

答:
The right and simplest approach is to do this elementswise. Let me do it for $G_1 \times G_2$ and proving $Z_2(G_1 \times G_2)=Z_2(G_1) \times Z_2(G_2)$ and I leave it to you to apply induction on the number of factors and the members of the upper central series.

For all $g_1 \in G_1, g_2 \in G_2$ we have $$(x_1,x_2) \in Z_2(G_1 \times G_2) \iff \\ [(x_1,x_2),(g_1,g_2)] \in Z(G_1 \times G_2) \iff \\ [(x_1,x_2),(g_1,g_2)] \in Z(G_1) \times Z(G_2) \iff \\
([x_1,g_1]),[x_2,g_2]) \in Z(G_1) \times Z(G_2) \iff \\
[x_1,g_1] \in Z(G_1) \text { and }  [x_2,g_2] \in Z(G_2) \iff \\
x_1 \in Z_2(G_1) \text { and } x_2 \in Z_2(G_2) \iff \\
(x_1,x_2) \in Z_2(G_1) \times Z_2(G_2).$$
Can you finish?

ContentLicense: CC BY-SA 4.0


问:
Impossible integral? $\int_0^{\pi/2}\frac{\sin^2(x)}{\sqrt{x^2+(\pi/2-x)^2}}$
I have absolutely no idea how to solve:$$\int_0^{\pi/2}\frac{\sin^2(x)}{\sqrt{x^2+(\pi/2-x)^2}}$$I think I am supposed to use trig sub, but I can't seem to put it in the right form?

Body: By the King property$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx=\tfrac12\int_a^b[f(x)+f(a+b-x)]dx,$$your integral is$$\frac12\int_0^{\pi/2}\frac{\sin^2x+\cos^2x}{\sqrt{x^2+(\pi/2-x)^2}}dx=\frac12\int_0^{\pi/2}\frac{dx}{\sqrt{2x^2-\pi x+\pi^2/4}}=\frac{\operatorname{arsinh}1}{\sqrt{2}}.$$


问:
How to find a the partial derivative of a function that includes functions as inputs
Take $f(x,g(x)) = x \cdot g(x)$. where $g(x) = x^2$

The solution we're going to want to find is $3x^2$ but I'm having a hard time finding the general way of writing it...

I looked here

And I thoght it that would be doing something like this
$D_x [f(x,g(x))] = D_xf(x) + D_xg(x) = g(x) + 2x = x^2 + 2x$

But that's wrong... I know it's a simple question but for some reason I'm struggling with it.

答:
Let's write $\phi(x) = f(x, g(x))$, with $f(x, y) = xy$ and $g(x) = x^2$.

Then $\phi'(x) = \frac{\partial f}{\partial x}(x, g(x)) + \frac{\partial f}{\partial y}(x, g(x)) g'(x)$.


问:
When is there an adjunction between two categories?
Call two categories $C$ and $D$ adjivalent if there is an adjunction
$$F\dashv U,\quad F\colon C\to D,\quad U\colon D\to C$$
between $C$ and $D$.

Can one say something about when two categories are adjivalent? In particular, are there pairs of categories which aren't adjivalent?

答:
If $C$ and $D$ are groupoids then an adjunction must be an equivalence (because the unit and the counit must be invertible), so groupoids have an adjunction between them iff they're equivalent. In particular one-object groupoids $BG, BH$ corresponding to two non-isomorphic groups $G, H$ do not have an adjunction between them.

By the way, I don't recommend calling this condition "-valent"; to me that suffix suggests an equivalence relation, which this isn't. For example, any category $C$ admits a unique functor $C \to 1$ to the terminal category, which has a left adjoint iff $C$ has an initial object and a right adjoint iff $C$ has a terminal object. And there are categories with an initial object but not a terminal object and vice versa. So "adjivalence" is not symmetric; it has a "handedness" depending on which direction you ask for a left adjoint and which direction you ask for a right adjoint.


问:
Inequality $||DG(t)||\leq c||G(t)||$ implies $G\equiv 0$?
Continuous and differentiable function $G:[0,\infty)\to \mathbb{R}^n$ satisfies $G(0) = 0\in\mathbb{R}^n$ and there exists $c>0$ such that for all $t>0$ we have
$$||DG(t)||\leq c||G(t)||$$
Does it mean that $G\equiv 0$?

Remark:
$G(t) = (g_1(t),g_2(t),...,g_n(t))$, $DG(t)$ is the derivative of $G$ and $||\circ||$ is euclidean norm.

If $n=1$ then this statement is true, but every proof I've seen relied on Mean Value Theorem. As of yet I was unable to generalize it to higher dimensions or provide counter examples.

答:
Without loss of generality, $c=1.$ First suppose that $0\le x\le 1$ and let $M=\sup \{G(t):t\in [0,1]\}.$ Now then

$G(x)=\int_0^xG'(t)dt\Rightarrow |G(x)|\le \int_0^x|G'(t)|dt \le \int_0^x|G(t)|dt.$

By the mean-value inequality$^1$, there is an $0<\alpha<t$ such that

$|G(t)|\le t|G'(\alpha)|$ so $|G(x)|\le \frac{x^2}{2}|G'(\alpha)|\le \frac{x^2}{2}|G(\alpha)|\le M\frac{x^2}{2}.$

But this means that $|G(x)|\le M\frac{x^2}{2}$ and so inductively, that $|G(x)|\le M\frac{x^n}{n!}$ for each integer $n.$

We conclude that $G=0$ in $[0,1].$ Now define $G_1(t)=G(t+1).$ Then, $G_1$ satisfies the same hypothesis as $G$ and so $G_1=0$ on $[0,1]$ or what is the same thing, $G=0$ on $[1,2].$ Now, another induction proves the claim.

$^1$ This is the mean-value inequality for vector-valued functions. Rudin proves it this way: define $z:=G(b)-G(a)$ and $\varphi(t)=\langle z,G(t)\rangle$ and apply the single-variable mean-value theorem to find a $\alpha\in (a,b)$ such that $\varphi(b)-\varphi(a)=\varphi'(\alpha)=\langle z,G'(\alpha)\rangle.$ But we also have $\varphi(b)-\varphi(a)=|z|^2$ so the result follows by the Schwarz inequality.


问:
What is the largest finite number you can make using no more than 6 characters?
(I am new to the Math Stack Exchange community so tell me if this question is not allowed, however, I checked on Meta first.)

(Also, I don't really know which tags I should assign this question, so please offer tag suggestions if you don't agree with the tags I tagged this question with)

Sorry for the many edits to this question, but I needed to clarify what I was asking.

I have done some research and several google searches but I have not found the answer to this question:

What is the largest finite number you can make using no more than 6 characters? Characters include numbers and operations, etc. Rules: Parenthesis do not count as characters, and letters are not allowed. You can use the following operations: Basic Arithmetic, Exponents, Factorials, and Tetration (although you may also use Knuth's up-arrow notation).

Clarification Edit:
If you are using exponents - keep in mind this rule:
9^9 (9 to the power of 9) would be 3 characters. However, 9^99 (9 to the power of 99) would be 4 characters.

The following are some guesses that I have come up with, but I'm not confident in any of them

((((9!)!)!)!)!

((((9⁹)!)!)!)

$9^{9^{9{9{{}}}}}$

If I am not clear enough with the rules regarding what counts as a character please ask.

Any help is appreciated!

Thank you in advance!

Edit

Gerry Myerson has (quiet repeatedly) asked me to include the following links for reference: This link and this link.

答:
I need 50 of reputation to write a comment, but sadly enough I got no reputation because I am new.. So I am putting my comment here.

Using standard notations, the biggest number you can possibly write is $((((9!)!)!)!)!$

Then if you are allowed to, you can use notations for very large numbers like these: https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation https://en.wikipedia.org/wiki/Conway_chained_arrow_notation
https://en.wikipedia.org/wiki/Steinhaus-Moser_notation


问:
Projection of linear combination of real numbers
Suppose a real number $\alpha$ is a rational linear combination of two linearly independent real numbers, such as $\pi$ and $\sqrt 3$. For instance, say $$\alpha = \tfrac12\pi + 5\sqrt 3.$$

By a simple linear algebra argument ($\{\pi,\sqrt3\}$ is a $\mathbb Q$-basis), it's clear that the coefficients of $\alpha$ are uniquely determined.

Question: Given $\alpha\in\operatorname{span}_\mathbb Q(\{\pi,\sqrt 3\})$, how do we determine the coefficients of $\pi$ and $\sqrt 3$?

It feels like there is enough information, but at the same time, I imagine that we can find different coefficients which give us numbers arbitrarily close to $\alpha$, so it is important that we have an exact expression for $\alpha$.

(In other words, if we only know that $\alpha\approx 10.23105$, say, I imagine it would be hopeless to expect that we can determine the coefficients, even if we restrict our view to the module $\operatorname{span}_{\mathbb Z}(\{\pi,\sqrt 3\})$.)

The reason I ask this question is because for each $n$, the integral $$I(n)=\int_0^{\pi/3}\sin^{2n}x\,dx$$ is a number of the form $a\pi+b\sqrt 3$. It's easy to obtain the recurrence relation
$$I(n) = \begin{cases} \hfil\frac\pi3\hfil & \text{if $n=0$}\\[3pt]
(1-\tfrac1{2n})\,I(n-1)-\frac1{4n}\big(\tfrac{\sqrt3}2\big)^{2n-1} & \text{otherwise}
\end{cases}$$
for $I(n)$, but I want to be able to compute the coefficients of $\pi$ and $\sqrt 3$ separately (in exact form) using a computer program.

I appreciate any help with solving either of the two problems, I'm assuming a solution to one necessarily sheds light on the other, that's why I haven't posted these as two separate questions.

答:
Here's easy way to determine them:
let $t \in [0..1]$, we define
$$ a_0 = t * \alpha $$
$$ a_1 = (1-t) * \alpha $$
Basically we have $a_0 + a_1 = \alpha$.

Now, after we have $a_0$ and $a_1$, we can calculate co-efficients:
$$c_0 = \frac{a_0}{\pi}$$
$$c_1 = \frac{a_1}{\sqrt{3}}$$.

I have a feeling that the $t$ cannot be calculated from the $\alpha$, given that they are both located in same dimension, basically you need some criteria how $\alpha \in R$ can be divided to two separate dimensions. Thus your end result is large number of possible alternative coefficients. Every choice of $t$ will have different coefficients. Limiting $c_0,c_1 \in R$ to $Q$ sounds kinda difficult.


问:
Haar measure- uniqueness
I'm new to measure theory and topological groups, and I thought about a question. Let $G$ be a locally compact group, equipped with the Borel $\sigma$-algebra, there is exist one measure $\mu$ (up to positive scalar multiplication) such that $\mu$ is left (or right)-translation-invariant: $\mu (gS)=\mu (S)$ for every $ g\in G$ and all Borel sets $ S\subset G$ and $\mu(S)>0$.

I tried to think on a counter-example when $G$ is not locally compact. Is this measure even exist when $G$ is not locally compact? If it is, is there a counter-example to the uniqueness of Haar measure? (Or there is $G$ such that Haar measure exist and unique and $G$ is not locally compact?)
And if it does not exist, are there conditions on $G$ (avoiding locally compact) such that Haar measure exists?

Thanks in advance.

答:
The MO page here addresses your question (answer and comments).

By the way, your characterization of Haar measure is missing regularity conditions. Of course the main technical property of Haar measure that we care about is its invariance under left (or right) multiplication, but you need further properties to define a Haar measure. For example, according to your definition the counting measure on every topological group is a Haar measure!  It is a common "error" to describe Haar measure in such a way that counting measure fits the description even though counting measure is not a Haar measure on $G$ when $G$ is not discrete.

What you left out are the following two conditions on $\mu$:

(i) for each Borel subset $A \subset G$, $\mu(A) = \inf \mu(U)$ where $U$ runs over open subsets of $G$ containing $A$,

(ii) for each open or $\sigma$-finite subset $A \subset G$, $\mu(A) = \sup \mu(K)$ where $K$ runs over compact subsets of $G$ contained in $A$.

These conditions taken together are called $\sigma$-regularity, and a Haar measure on $G$ is a $\sigma$-regular Borel measure on $G$ fitting the conditions you gave.  Such conditions should be found in a proof that Haar measure is unique up to a scaling factor. Something like these conditions had better be used in a proof of uniqueness, because without them counting measure would be a Haar measure for $\mathbf R$ with its usual topology, which is false.

A measure that satisfies (i) and (ii) for all Borel subsets $A \subset G$ is called regular. Regularity is a stronger condition than $\sigma$-regularity and Haar measure on some locally compact groups is not regular.  For example, let $G = S^1 \times \mathbf R_d$ where $\mathbf R_d$ is $\mathbf R$ with the discrete topology. Topologically this is like a cylinder of "independent circles".
Counting measure is a Haar measure on $\mathbf R_d$ and on $S^1$ there is a standard normalized Haar measure $d\theta/2\pi$. The product of these is a Haar measure on $G$ and it is not regular: the subset $A = \{1\} \times [0,1]$ of $G$ is closed (and not open), so $A$ is a Borel subset of $G$. Check $\mu(K) = 0$ for all compact $K \subset A$ and $\mu(U) = \infty$ for all open $U \supset A$. Thus condition (ii) for a Haar measure does not apply to $A$ (which is neither open nor $\sigma$-finite, so there is no contradiction).

Title: Haar measure- uniqueness


问:
OLS/LAD on linear/quadratic programming type problem
I have a list of equations that define halfspaces (3 equation list example below)

$x_1 + x_2 + x_3 + x_5 < x_4 + x_6 + x_7 \\ x_4 + x_6 + x_7 < x_8 + x_9 + x_{21} + x_{43} + x_1 \\ x_4 + x_6 + x_7 < x_1 + x_2 + x_3 + x_5$

As shown, sometimes this system will not be solvable (I've shown a simple case; in n-dim, it's obvious how this may happen in a more complex way). In this case, I want to find the values of all $x_n$ that has the lowest euclidean distance from being viable.

I'm not entirely sure if I want euclidean distance or squared euclidean distance., but my first thoughts jump to OLS, LAD, and linear programming. However, I've always seen OLS/LAD done with a bunch of real number valued points, not to solve a linear programming system.

I also got informed that this may be reformulateable as a classic nonlinear (specifically quadratic) programming problem, although I'm unfamiliar with the space as a whole.

Any ideas on how to best find the values of this system.?

答:
If your problem is defined by $g_j(x) \le 0, j \in J$, you want to minimize $\sum_{j \in J} ||\max(0, g_j(x))||$ for a given norm.

If you take the 1-norm, you get

$\min_x \sum_{j \in J} \max(0, g_j(x))$

which is not differentiable. However you can reformulate it into a smooth problem by introducing the variables $s_j = \max(0, g_j(x))$:

$\min_{x, s} \sum_{j \in J} s_j$

s.t. $s_j \ge 0, s_j \ge g_j(x)$

which is a linear problem.


问:
Why is the term "point" not defined in topology? What are the elements of the open sets of the topology? What are the UR-elements of the open sets?
I looked in several books on topology: Munkres, Schaum's, etc. and cannot find a definition for the word "point". Why is the term "point" not defined in topology? What are the elements of the open sets of the topology?  What are the UR-elements of the open sets?</p>

答:
Any set $X$ can be given a topology. For instance, $\wp(X)$ is a topology on $X$. So is $\{\varnothing,X\}$. So is $\{X\setminus F:F\subseteq X\text{ is finite}\}$. The elements of $X$ can be anything: ordinal numbers, ordered $n$-tuples of real numbers, functions from $\Bbb R^2$ to $\Bbb R$, sequences of sets of complex numbers — you name it.

When we construct a specific space, we specify the underlying set, and the elements of that set become the points of the space.


答:
The point-set topology is the basic or general topology its really the study of the most basic ideas like when can a topology on a space be derived from a metric.

The ideas of continuity, dimension, compactness, and connectedness rely on the ideas of interior, closure, and boundary, so are not really defined in terms of points.

http://pi.math.cornell.edu/~hatcher/Top/TopNotes.pdf has some good notes if you want to read on.

Body: I looked in several books on topology: Munkres, Schaum's, etc. and cannot find a definition for the word "point". Why is the term "point" not defined in topology? What are the elements of the open sets of the topology?  What are the UR-elements of the open sets?</p>


问:
How to find out if this function is characteristic
I need to find out if $\phi$ is a characteristic function

$$
\phi(t_1,t_2) = (p\cos(t_1+t_2) + 1 - p)^n, \quad 0 < p < 1, n \in \mathbb{N_{>0}}
$$

What I've tried:

Instead I'm trying to prove that $p\cos(t_1+t_2) + 1 - p$ is characteristic because if it is then it's n-th power also is characteristic.

I've tried converting $\cos$ to trig form and then doing Fourier transform to get p.d.f. and got
$$
p \pi \delta_{x-1} \delta_{y-1} +
p \pi \delta_{x+1} \delta_{y+1} +
2 \pi \delta_{x} \delta_{y} +
2 p \pi \delta_{x} \delta_{y}
$$
but I'm not sure if this is a p.d.f.

Also, $\phi$ looks like a characteristic function of $Bin(n,p)$:
$$
{\displaystyle \!\,(1-p+pe^{it})^{n}}
$$
but I don't know if there is any connection between them.

答:
You have the right hunch.

You can check the characteristic function of the following random variable:
$$X=(B_1+...+B_n,B_1+...B_n)$$
where $B,B_1,...,B_n$ are iid random variables, B equals 0 with probability 1-p, B equals 1 with probability p/2, -1 with probability p/2


问:
Is there a method to find the possible exponents of an integer?
For example,

$16$ is $2^4$ and $4^2$

so the possible exponents are $2$ and $4$.

However, what if the number is larger, say

$634964616$

I'm writing a C program to compute possible exponents.

Thanks.

答:
We can see that $64=2^6$ so combinations $2^6, 4^3, 8^2$ are also valid.

For $634964616$ you can divide by $2$ three times before you get an odd number $79370577$ and $\sqrt{79370577}\approx 8909$ so the maximum tests you would have to make are $\frac{8908}{2}=4454$  which is the number of odd numbers $3\le d\le 8989$. In the end you will find

$$634964616 = 2^3×3^3×11×13×61×337$$
In any case, you never have to text more than $\frac{\sqrt{x}}{2}$ times to get all factors. If you keep track of factor exponents in an array, you can compare which have what numbers of exponents and figure out the "sought" exponents from there.

If you had a number that was a power of just one other, then the routine would be easier still. For instance, dividing $10077696$ by $2$ $9$ times, you get $19683$ and if you divide that by $3$ $9$ times you get the number $1$. So $10077696=6^9=216^3$.


答:
I submit $9\uparrow\uparrow\uparrow\uparrow 9$. (Wikipedia)

But it's possible something like the busy beaver function (wikipedia again) could be even more effective -- and at any rate once one starts talking about numbers this large it's effectively impossible to determine which is the largest anyway.

Body:
(I am new to the Math Stack Exchange community so tell me if this question is not allowed, however, I checked on Meta first.)

(Also, I don't really know which tags I should assign this question, so please offer tag suggestions if you don't agree with the tags I tagged this question with)

Sorry for the many edits to this question, but I needed to clarify what I was asking.

I have done some research and several google searches but I have not found the answer to this question:

What is the largest finite number you can make using no more than 6 characters? Characters include numbers and operations, etc. Rules: Parenthesis do not count as characters, and letters are not allowed. You can use the following operations: Basic Arithmetic, Exponents, Factorials, and Tetration (although you may also use Knuth's up-arrow notation).

Clarification Edit:
If you are using exponents - keep in mind this rule:
9^9 (9 to the power of 9) would be 3 characters. However, 9^99 (9 to the power of 99) would be 4 characters.

The following are some guesses that I have come up with, but I'm not confident in any of them

((((9!)!)!)!)!

((((9⁹)!)!)!)

$9^{9^{9{9{{}}}}}$

If I am not clear enough with the rules regarding what counts as a character please ask.

Any help is appreciated!

Thank you in advance!

Edit

Gerry Myerson has (quiet repeatedly) asked me to include the following links for reference: This link and this link.


问:
Proof of Möbius function on subset poset
Prove that Möbius function of subset poset of $[n]$ is following, given $A,B \subseteq [n]$.

$\mu (A,B)=\left\{
\begin{array}{l} (-1)^{|B|-|A|},& \text{if}\,\, A \subseteq B\\
0,& \text{otherwise}\end{array} \right.$

I searched for the proof and in one paper they use induction. $A = B$ this is obvious. Assume this holds up to $|B|= |A| + n, A\subset B$. Then for $|B|=n+1$

$\mu (A,B)=- \sideset{}{_{A \subseteq C \subset B} (-1)^{|C|-|A|}} \sum = (-1)^{|B|-|A|}$

But I do not understand the last step. How

$\sideset{}{_{A \subseteq C \subset B} (-1)^{|C|-|A|}} \sum = (-1)^{|B|-|A|}$

Can anyone explain this or point me to a more detailed proof? Thanks

答:
The most important step is to bring the term $(-1)^{|B|-|A|}$ into the summation, so that instead of showing $-\sum_{A \subseteq C \subset B} (-1)^{|C|-|A|} = (-1)^{|B|-|A|}$ one will show the equivalent
$
\sum_{A \subseteq C \subseteq B} (-1)^{|C|-|A|} =0
$ (under the assumption $A\subset B$).

Changing variables to $C'=C\setminus A$ this now follows easily
$$
  \sum_{A \subseteq C \subseteq B} (-1)^{|C|-|A|}
 =\sum_{C' \subseteq B\setminus A} (-1)^{|C'|} = 0
 \qquad\text{since $B\setminus A\neq\emptyset$.}
$$
The latter equality can be justified by recognising $\sum_{T\subseteq S}(-1)^{|T|}$ as the expansion of $\prod_{s\in S}(1-1)=0$ for the non-empty set $S=B\setminus A~$ (if you find this too cryptic, replace the terms $-1$ in each factor of the product by a separate variable $x_s$ which are then all set to $-1$; you may of course also use binomial coefficients).

A more advanced way to understand this Möbius function is to recognise the poset of subsets of $[n]$ as a product poset of $n$ copies of a $2$-element totally ordered set (namely the two subsets of a singleton $\{i\}$, one such copy for each $i\in[n]$). The Möbius function will then be the product of the Möbius functions of those basic orderings, as I've explained in this answer. Since always $\mu(X,Y)=0$ unless $X\subseteq Y$ and $\mu(X,X)=1$, the only interesting value of Möbius function $\mu_{\{i\}}$ of the subsets of $\{i\}$ is $\mu_{\{i\}}(\emptyset,\{i\})=-1$. The expression for the Möbius function of the subsets of $[n]$ then is
$$
  \mu(A,B)=\prod_{i\in[n]}\mu_{\{i\}}(A\cap\{i\},B\cap\{i\})
 =\begin{cases}0&\text{if $A\not\subseteq B$,}\\
          (-1)^{|B\setminus A|}&\text{if $A\subseteq B$.}\end{cases}
$$


问:
What do you call (and how do you solve) this kind of optimization problem?
I have a system of equations like:

\begin{cases}
X_1 = a_{11} x_1 + a_{12} x_2 + a_{13} x_3\\\\
X_2 = a_{21} x_1 + a_{22} x_2 + a_{23} x_3\\\\
X_3 = a_{31} x_1 + a_{32} x_2 + a_{33} x_3
\end{cases}

Then I have a "target" such as

$$T = t_1 x_1 + t_2 x_2 + t_3 x_3 $$

I want to choose coefficients $c_1, c_2, c_3$ such that $c_1 X_1 + c_2 X_2 + c_3 X_3$ is as close to $T$ as possible. By "as close to" I mean a least squares distance, so minimizing

$$ ((c_1 a_{11} + c_2 a_{21} + c_3 a_{31}) - t_1)^2 +  ((c_1 a_{12} + c_2 a_{22} + c_3 a_{32}) - t_2)^2 + ((c_1 a_{13} + c_2 a_{23} + c_3 a_{33}) - t_3)^2 $$

My end goal is actually to do this kind of problem in R--if you're able to show me that, amazing, but I'm really just trying to start by figuring out what this kind of problem is called! Right now I don't know what terms to look up or google....

Addendum: in case my notation is unclear, an example with numbers

\begin{cases}
X_1 = 2x_1 + 3x_2 + 4x_3\\\\
X_2 = 3x_1 - 2x_2 + 9x_3\\\\
X_3 = 5x_1 + 4x_2 + 4x_3
\end{cases}

The target is $T = 24x_1 + 12.5x_2 + 36x_3$.

In this case there is a perfect solution $c_1 = 1.5$, $c_2 = 2$, $c3=3$, but in general there won't be a perfect solution and I want to find the closest solution.

答:
It feels to me that the $x_i$ are more distracting than useful. You are using polynomial notation for what I'd rather consider a vector in this case. So you have $X_1,X_2,X_3,T\in\mathbb R^3$ and want $c_i$ which minimize $\lVert c_1X_1+c_2X_2+c_3X_3−T\rVert^2$.

In this formulation I'd call it a linear least squares problem because you are trying to find the last sum of squares residue, and each coordinate is linear in the parameters you control. Looking at Wikipedia more closely, you also find that this is an ordinary least squares problem.

To solve this, you multiply the matrix equation you'd get for the exact case by the transpose of the matrix:

$$A\cdot c\approx t\qquad\rightarrow\qquad A^T\cdot A\cdot c=A^T\cdot t$$

Personally I can't remember which matrix is the original one and which is the transpose, but if you imagine a situation where you are fitting more coordinates in $t$ than you have parameters in $c$ then $A$ needs to have more rows than columns for the equation to even make sense. That will tell you which direction is which.

This should be easy to reproduce in R. The crossprod function might be of some use.

Come to think of it, if $A$ is square, as in your question, and you don't have a perfect solution, then $A$ must be singular. Which translates back into a linear dependency: one of your polynomials $X_i$ can be expressed as a linear combination of the other two. You can't get a uniquely defined best solution from that. I'd recommend dropping polynomials $X_i$ until you have a basis, i.e. until they are all linearly independent.


问:
Struggling to understand proof, there is no simple group of order $525$
Prove there is no simple group of order $525.$

$N(H)$ is the normalizer of a set $H$.

Proof:Let $L_7$ be a Sylow $7$ subgroup of $G$. It follows that $|N(L_7)|=35$ by the sylow theorems. Let $L$ be a subgroup of $N(L_7)$ of order $5$. Since $N(L_7)$ is cyclic $N(L) \geq N(L_7)$( I do not understand this part of the proof), so that $35$ divides $|N(L)|$. But L is contained in a sylow $5$ subgroup which is abelian, thus $25$ divides $|N(L)|$ as well. It follows that $175$ divides $|N(L)|$, now the index theorem gives a contradiction.

I am having a lot of confusion proving nonabelian groups of specific orders are not simple, because all of the new facts appearing about normalizers, which I have never learnt before. How should I better learn how to derive facts about normalizers.

In particular in this proof, I know why $|N(L_7)|=35$, I do not need help on that, but why does this line hold?

Let $L$ be a subgroup of order $5$. Since $N(L_7)$ is cyclic $N(L) \geq N(L_7)$ so that $N(L_7)$ is a subgroup of $N(L)$?

答: Maybe the way to start is to prove that any group of order 35 is cyclic. Then, since $N(L_7)$ is cyclic, any subgroup of $N(L_7)$ is a normal subgroup as well. From this and from the definition of the normalizer of a subgroup, you should be able to conclude that $N(L_7)$ is contained in $N(L)$.


问:
Show that an even map on the circle $\mathbb S^1$ has even degree **without** any homology
I want to show that an continuous even map, i.e., it satisfies $f(z)=f(-z)$, on the circle $\mathbb S^1$ has even degree without any homology techniques.

I know the multiplicative property of degrees, i.e., $\deg(f \circ g ) =\deg(f) \deg(g)$. It is trivial that if $f$ is an even map and $p_2$ is the multiplication map $z\mapsto z^2$, then $f\circ A=f$, $p_2\circ A= p_2$ where $A$ is the antipodal map. I also have a continuous map $g$ which satisfies $\deg(g)=\deg(f)$ and $p_2\circ f = g \circ p_2$. Given these information, how can I proceed?

Any help is appreciated.

答:
Define $g:S^1\rightarrow S^1$ to be the function
$$g(z)=f(\sqrt z).$$
The square root is defined up sign, but the fact that $f$ is even guarantees that $g$ is well-defined. In fact $g$ is continuous. For the covering projection $p=\exp(2\pi it(-)):\mathbb{R}\rightarrow S^1$ is a quotient map, and $g\circ p$ is the continuous map
$$\mathbb{R}\ni t\mapsto f(\exp(\pi it))\in S^1.$$
Now if $p_2:S^1\rightarrow S^1$ is the map $p_2(z)=z^2$, then $g\circ p_2=f$. Thus
$$\deg(f)=\deg(g\circ p_2)=\deg(g)\cdot \deg(p_2)=2\cdot \deg(g).$$


问:
Find all linear transformations $\mathbb{R}^3 \rightarrow \mathbb{R}^3$ that satisfy the following:
I am currently taking a course on linear algebra and we are studying linear transformations. In an exercise, we are asked the following:

In $\mathbb{R}^3$ let $(e_1, e_2, e_3)$ with $e_1 = (1,0,0)$, $e_2 = (0,1,0)$ and $e_3 = (0,0,1)$, be a basis. Find all matrices of the linear transformations
\begin{equation*}
f: \mathbb{R}^3 \rightarrow \mathbb{R}^3
\end{equation*}
that satisfy

$Ker(f) = \langle e_1 + e_2 \rangle$.

$f(e_3) = e_2 - e_3$.

$f(v) \in \langle v \rangle$, being $v = 2e_1 + e_2 + e_3$.

Which of them verify $Im(f^2) \neq Im(f)$?

I don't know how to find $ALL$ of them or how to impose that they satisfy that and then find them.

答:
$\mathcal{B} = \{e_1 + e_2, e_3, 2e_1 + e_2 + e_3\}$ is a basis. You know the values of $f$ on that basis:

\begin{align}
f(e_1 + e_2) &= 0 \\
f(e_3) &= e_2 - e_3 \tag{$*$}\label{equations} \\
f(2e_1 + e_2 + e_3) &= t(2e_1 + e_2 + e_3)
\end{align}

(for some $t \in \mathbb{R}$). Remember: the linear transformation $f$ is determined uniquely by its values on some basis.

So if $\mathcal{S} = \{e_1,e_2,e_3\}$ denotes the standard basis, then

$$ [f]^{\mathcal{S}}_\mathcal{B} = \begin{bmatrix} 0 & 0 & 2t \\ 0 & 1 & t \\ 0 & -1 & t\end{bmatrix}. $$

Use this to find $[f]^\mathcal{S}_\mathcal{S}$. You can figure out what the image is directly from the equations \eqref{equations}.


答:
You seem to be assuming you will have a lot of choices.  You won't.

$634964616$ is divisible by $2^3$ but not by $2^4$.  So $634964616 = 2^3K$ where $K$ is odd.  If $634964616 = a^d$ then then $2^3K =a^d$ which means $2^{\frac 3d}K^{\frac 1d} = a$ is an integer.  But $2^{\frac 3d}$ is only rational if $d|3$ so $d = 1$ or $3$.

So we have $634964616 = 634964616^1$ or $634964616 = (2\sqrt[3]{K})^3$ are the only two options and the second is only an option if $K$ is a perfect cube.  Which it aint.

So you can do this but figuring out the power of $2$ factors, then $3$ and so on.  It will filter down really quickly$.

Title: Is there a method to find the possible exponents of an integer?

Body:
For example,

$16$ is $2^4$ and $4^2$

so the possible exponents are $2$ and $4$.

However, what if the number is larger, say

$634964616$

I'm writing a C program to compute possible exponents.

Thanks.


问:
Show that $f(z)=\sum_{k=1}^{n}|f_{k}(z)|$ cannot achieve a strictly local maximum inside a connected open set $D$.
Suppose that we have $f_{1},\cdots, f_{n}$ holomorphic in a connected open set $D$. For $$f(z):=|f_{1}(z)|+|f_{2}(z)|+\cdot+|f_{n}(z)|\ \ \text{for}\ \ z\in D.$$ I want to show that $f(z)$ cannot achieve a strict local maximum inside $D$.

I know that this must have something to do with the maximum modulus principle, especially we have some connected open set, this will enable us to directly conclude that a function $g(z)$ is constant all over $D$ if it is holomorphic in $D$ and $|g|$ achieves a local maximum at $z_{0}\in D$.

However, I cannot connect this theorem to $f(z)$ because even thought $f_{k}(z)$'s are holomorphic, the summation of moduli of them may not be holomorphic.

I know a way to show that

Let $D$ be a disc with radius $R$ and center $z_{0}$. Suppose $f_{k}(z)$'s are holomorphic on $D$ and continuous on $\partial D$ such that $f(z_{0})\geq z$ for all $z\in D\cup C$, then $f$ is constant in $D\cup C$.

I can prove this claim using Cauchy's Integral formula, but it is long. This lemma can certainly conclude the proof quickly, but I realized that I did not really use the "connectedness" of $D$.

Is there any way for me to drive a shorter proof using maximum modulus principle? Thank you!

Edit 1:

The post is duplicated, but I'd still like to post a relatively complete proof.

Thanks to Conrad, the following is a relatively complete proof:

Suppose $\omega$ is the local maximum of $f(z):=\sum_{k=1}^{n}|f_{k}(z)|$. Note that as $f_{k}(\omega)$ is a complex number, unless it is zero, we can represent (uniquely) as $$f_{k}(\omega)=|f_{k}(\omega)|e^{it_{k}},\ \ t_{k}\in [0,2\pi).$$ Take $\alpha_{k}:=e^{-itk}$, and if $f_{k}(\omega)=0$ for some $k$, we tae $\alpha_{k}:=1$.

Then, we have $$f_{k}(\omega)=\alpha_{k}^{-1}|f_{k}(\omega)|\ \text{with}\ |\alpha_{k}|=1.$$

Now, define $g(z):=\sum_{k=1}^{n}\alpha_{k}f_{k}(z).$ Then, we have \begin{align*}
|g(z)|\leq \sum_{k=1}^{n}|\alpha_{k}||f_{k}(z)|=\sum_{k=1}^{n}|f_{k}(z)|&=f(z)\\
&\leq \sum_{k=1}^{n}|f_{k}(\omega)|=f(\omega)\\
&=\sum_{k=1}^{n}f_{k}(\omega)\alpha_{k}=g(\omega)\leq|g(\omega)|\ \ \ (*).
\end{align*}

Hence, $\omega$ is also the point at which $|g|$ achieves local max. As $g$ is holomorphic and $D$ is connected open, by the global maximum modulus principle, we conclude that $g(z)=C$ for some constant $C$ and for all $z\in D.$

But then, $(*)$ implies that $$|C|\leq C\leq |C|\implies C=|C|\implies C\geq 0.$$

So we have $g(z)=g(\omega)\geq 0$. But then (*) implies that $$g(\omega)=g(z)=|g(z)|\leq f(z)\leq f(\omega)\leq g(\omega),$$ and thus we must have $$g(\omega)=g(z)=f(z)=f(\omega)=C\geq 0\ \ \text{for all}\ \ z\in D,$$ and thus the local maximum of $f$ cannot be strict.

I really appreciate the patient and inspiring help from Conrad.

答:
Thanks to Conrad's patient and inspiring help, I wrote the proof, as follows:

Suppose $\omega$ is the point at which local maximum of $f(z):=\sum_{k=1}^{n}|f_{k}(z)|$ is achieved. Note that as $f_{k}(\omega)$ is a complex number, unless it is zero, we can represent (uniquely) as $$f_{k}(\omega)=|f_{k}(\omega)|e^{it_{k}},\ \ t_{k}\in [0,2\pi).$$ Take $\alpha_{k}:=e^{-itk}$, and if $f_{k}(\omega)=0$ for some $k$, we tae $\alpha_{k}:=1$.

Then, we have $$f_{k}(\omega)=\alpha_{k}^{-1}|f_{k}(\omega)|\ \text{with}\ |\alpha_{k}|=1.$$

Now, define $g(z):=\sum_{k=1}^{n}\alpha_{k}f_{k}(z).$ Then, we have \begin{align*}
|g(z)|\leq \sum_{k=1}^{n}|\alpha_{k}||f_{k}(z)|=\sum_{k=1}^{n}|f_{k}(z)|&=f(z)\\
&\leq \sum_{k=1}^{n}|f_{k}(\omega)|=f(\omega)\\
&=\sum_{k=1}^{n}f_{k}(\omega)\alpha_{k}=g(\omega)\leq|g(\omega)|\ \ \ (*).
\end{align*}

Hence, $\omega$ is also the point at which $|g|$ achieves local max. As $g$ is holomorphic and $D$ is connected open, by the global maximum modulus principle, we conclude that $g(z)=C$ for some constant $C$ and for all $z\in D.$

But then, $(*)$ implies that $$|C|\leq C\leq |C|\implies C=|C|\implies C\geq 0.$$

So we have $g(z)=g(\omega)\geq 0$. But then (*) implies that $$g(\omega)=g(z)=|g(z)|\leq f(z)\leq f(\omega)\leq g(\omega),$$ and thus we must have $$g(\omega)=g(z)=f(z)=f(\omega)=C\geq 0\ \ \text{for all}\ \ z\in D,$$ and thus the local maximum of $f$ cannot be strict.


问:
Finding Exponents of Generating Function $A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $
I have the following generating function:

$$A(x)=\prod\limits_{k=1}^{\infty}\frac{1-x^{6k}}{(1-x^{2k})(1-x^{3k})} $$

Among multiple of $2$, we will get a multiple of 6 whenever we take a multiple that is a multiple of 3: possible multiples: $3m, 3m+1, 3m+2$, where $3m$ cancels with the numerator.

So, we will have $1$ left in the numerator. What is left are factors in which the exponent is either $2\times(3m+1) $, $2\times(3m+2)$ or a multiple of $3$ ($3\times(2m+1), 3\times(3m+2)$).

Questions:

Why $3m$ cancels with the numerator?

Why what is left are factors in which the exponent is either $2\times(3m+1) $, $2\times(3m+2)$ or a multiple of $3$: $3\times(2m+1), 3\times(3m+2)$.

答:
We obtain
\begin{align*}
\color{blue}{A(x)}&=\prod_{k=1}^\infty \frac{1-x^{6k}}{\left(1-x^{2k}\right)\left(1-x^{3k}\right)}\\
&=\prod_{k=1}^\infty \frac{1+x^{3k}}{1-x^{2k}}\tag{1}\\
&=\prod_{k=1}^\infty \frac{1-x^k+x^{2k}}{1-x^k}\tag{2}\\
&\,\,\color{blue}{=\prod_{k=1}^\infty \left(1+\frac{x^{2k}}{1-x^k}\right)}\tag{3}
\end{align*}

Comment:

In (1) we cancel $1-x^{3k}$.

In (2) we cancel $1+x^k$.

Using the coefficient of operator $[x^N]$ to denote the coefficient of $x^N$ of a series, we derive from (3)
\begin{align*}
\color{blue}{[x^N]A(x)}&\color{blue}{=[x^N]\prod_{k=1}^\infty \left(1+\frac{x^{2k}}{1-x^k}\right)}\\
&=[x^N]\left(1+x^2+x^3+x^4+\cdots\right)\\
&\qquad\qquad\cdot\left(1+x^4+x^6+x^8+\cdots\right)\\
&\qquad\qquad\cdot\left(1+x^6+x^9+x^{12}+\cdots\right)\\
&\qquad\qquad\cdot\ \cdots
\end{align*}
which is the number of partitions of $N$ where each part of $N$ occurs at least twice.


问:
For real scalars $a,b$ and a nonzero vector $v$ in a vector space, prove that if $av=bv$ then $a=b$.
Is it possible to prove this using the axioms of a vector space? I tried playing around with the $a(v+w)=av+aw$ and $(a+b)v=av+bv$ axioms, but to no avail because I always got back to where I started. i.e. if I replaced the $bv$ with $av$ in the second axiom, I get $(a+b)v=av+av$ then $av+bv=av=av$ and then $av=bv$.

答:
I use a single axiom to define vector space:

A vector space is an abelian group $V$ together with an action by a field $F$.

Then from $0v+0v=(0+0)v=0v$, we conclude $0v=0$ for all $v\in V$.
Likewise, $a0=a(0+0)=a0+a0$ implies $a0=0$ for all $a\in F$.

Now if $av=bv$ and $a\ne b$, we thus have
$$ v=1v=(a-b)^{-1}(a-b)v=(a-b)^{-1}(av-bv)=(a-b)^{-1}0=0.$$


问:
Please verify this Proportion Hypothesis Test (Statistics)
Questions:

$120$ voters, $39$ identify with radical movement.

does this provide sufficient evidence that more than $25\%$ of voters identify with radical movement?

use $p$-value approach.

My Steps:

(0) Check Requirements $(120)(.25)>10$ and $(120)(.75)>10$ -- passes requirements.

(1) Hypothesis will be using a right tailed test. $H_0 : p = .25; H_1 : p > .25$

(2) No alpha (level of significance) is given, so default to $0.05$.

(3) $Z_t (\text{Test Statistic}) = ((.325-.25)/(\sqrt{(.25)(.75)/120} = 1.897$

(4) $P(Z \ge 1.897) \ge 1 - (\text{about}).9713 = .0287$. Reject $H_o$ if $p$-value is less than alpha.

(5) At the $.05$ level of significance, there is sufficient evidence to support that more than $25\%$ of voters identify with the radical movement.

Does that all look correct? Thank you :-)

答: Yes, the steps all look correct.</p>

Body: Yes, the steps all look correct.


问:
$M$-Test for Uniform Convergence Proof (Complex Analysis)
I am reading Bak's Complex Analysis for my undergraduate complex analysis class, and I am a little confused on the line of reasoning in the proof that is given for the $M$-Test for uniform convergence. Here is their statement of the $M$-Test:

Suppose $f_k$ is continuous in $D$, $k=1,2,\dots$. If $|f_k(z)|\leq M_k$ throughout $D$ and if $\sum_{k=1}^\infty M_k$ converges, then $\sum_{k=1}^\infty f_k(z)$ converges to a function $f$ which is continuous.

And then the proof that they give is this:

The convergence of $\sum_{k=1}^\infty f_k(z)$ is immediate. Moreover, for each $\epsilon>0$, we can choose $N$ so that
$$\left|f(z)-\sum_{k=1}^\infty f_k(z)\right|=\left|\sum_{n+1}^\infty f_k(z)\right|\leq M_{n+1}+M_{n+2}+\dots<\epsilon$$
for $n\geq N$. Hence the convergence is uniform and $f$ is continuous.

So the first equality makes sense, and the $\leq$ makes sense since we are assuming that $f_k\leq M_k$ for all $k$. What confuses me is how we can say that $M_{n+1}+M_{n+2}+\dots<\epsilon$. I think it comes from the definition of convergence, but it's not immediately clear to me, so any insight would be appreciated.

答: Since the series $\sum_{k=1}^\infty M_k$ converges,$$\lim_{n\to\infty}\sum_{k=n+1}^\infty M_k=\lim_{n\to\infty}\left(\sum_{k=1}^\infty M_k-\sum_{k=1}^nM_k\right)=0.$$So, given $\varepsilon>0$, we have $\sum_{k=n+1}^\infty M_k<\varepsilon$ if $n$ is large enough.


问:
Epicycloid is periodic if and only if $R/r$ is rational
Let $R$ be the radius of the fixed circumference and $r$ the radius of the circumference that rolls without slipping around the fixed circumference. The general equation is $$f(t)= (R+r)e^{it}-re^{it\frac{R+r}{r}}$$
($t$ is measured form the centre of the circumference of radius $R$). Prove that epicycloid is periodic if and only if $\frac{R}{r}\in\mathbb{Q}$.

Assume epicycloid starts at $t=0$.

Let epicycloid be periodic. Then it would need to circle the fixed circumference a whole number $k$ to get back to the start. That is, it would travel an angle of $ 2 \pi k $:
\begin{equation}
\begin{split}
& f(0)=f(2\pi k)\\
& \iff (R+r)e^{i0}-re^{i0\frac{R+r}{r}}=(R+r)e^{i2\pi k}-re^{i2\pi k\frac{R+r}{r}}\\
& \iff R=R+re^{i2\pi k}-re^{i2\pi k\frac{R+r}{r}}\\
& \iff re^{i2\pi k}=re^{i2\pi k\frac{R+r}{r}}\\
& \iff e^{i2\pi k}=e^{i2\pi k\frac{R+r}{r}}\;\;\text{as}\; r\neq 0\\
& \iff 2\pi k\frac{R+r}{r}=2\pi k+2\pi k_{1} \;\;\text{with}\; k_{1}\in\mathbb{Z}\\
& \iff k\frac{R+r}{r}=k+k_{1}\\
& \iff \frac{R+r}{r}=1+\frac{k_{1}}{k}\\
& \iff \frac{R}{r}=\frac{k_{1}}{k}\in\mathbb{Q} \;\;\text{as}\;   k,k_{1}\in\mathbb{Z}\\
\end{split}
\end{equation}

I don´t know how to prove:$\frac{R}{r}\in\mathbb{Q}\Longrightarrow $ epicycloid is periodic.

$e^{it}$ is periodic with period $2\pi$. Now $e^{it\frac{R+r}{r}}$ has period $2\pi\frac{r}{r+R}$ or maybe I can write $e^{it\frac{R+r}{r}}=e^{it\frac{R}{r}}e^{it}$ and now $e^{it}$ has period $2\pi$ and $e^{it\frac{R}{r}}$ has period $2\pi\frac{r}{R}$.

I don't know how to sum up those ideas to conclude that the function $f$ is periodic.

答:
The implication you are missing comes simply from the relationship between the lengths of the circles in play. Because the periodicity of the epicycloid occurs only when the number of turns required by an arbitrary point of the minor circumference of radius r, say $C_r$ is finite to reach its starting point. In other words if $\dfrac rR$ is rational then after  an integer number of complete turns of $Cr$ on the greater circumference $C_R$, the arbitrary point in $C_r$ reachs its initial position. (Sorry for bad English).

Thus you need that the quotient of the lengths of $C_r$ and $C_R$ be rational to have periodicity and this quotient is $$\dfrac{2\pi r}{2\pi R}=\frac rR\text { be rational }$$
In the attached figure the number of turns in a case is equal to $5$ but in the other case it is infinite.


问:
For which real $s$ does $\sum _{n=1}^{\infty} \frac{1}{n^s}$ converge?
For which values of real $s$ does the following converge?

$$
\sum _{n=1}^{\infty} \frac{1}{n^s}
$$

I have read extensively that the answer is $s>1$ but I can't find a simple proof suitable for students not trained to university level mathematics

答:
Here is one way forward that relies only on the Generalized Binomial Theorem and
straightforward arithmetic.

For $s>1$, clearly we have

$$\begin{align}
n^{1-s}&=\sum_{k=1}^n \left(k^{1-s}-(k-1)^{1-s}\right)\\\\
&=\sum_{k=1}^n k^{1-s} \left(1-\left(1-\frac1k\right)^{1-s}\right)\\\\
\end{align}$$

Now, the Generalized Binomial Theorem reveals $\left(1-\frac1k\right)^{1-s}=1-\frac{1-s}{k}+O\left(\frac1{k^2}\right)$ so that

$$n^{1-s}=\sum_{k=1}^n \frac{1-s}{k^s}+O\left(\frac1{k^{s+1}}\right)\tag1$$

Rearranging $(1)$ yields

$$\sum_{k=1}^n \frac{1}{k^s}=\frac{1}{(1-s)n^{s-1}}+\sum_{k=1}^n O\left(\frac1{k^{s+1}}\right)\tag2$$

As $n\to \infty$, the first term on the right-hand side of $(2)$ goes to zero while the second term is a convergent series since $s+1>2$.  Therefore, the series on the left-hand side of $(1)$ must converge.  And we are done.


问:
Can someone help me figure out this complex number problem?
I don't understand how to do this.

If $i^2 = -1$, $(4 + i)^2$ = ?

答:
$$(4+i)^2 = (4+i)\cdot (4+i)$$

Now... recall your "FOIL" (firsts, outers, inners, lasts) technique of how to multiply two sums together... that is, $(a+b)(c+d) = ac+ad+bc+bd$.  The same works here even though these are complex numbers and not integers like you might have been used to working with up until now.

$$(4+i)\cdot (4+i)=4\cdot 4 + 4\cdot i + i\cdot 4 + i\cdot i$$

Now, you just use that $i^2=-1$ to simplify the last term and then combine like-terms to finish:

 $$4\cdot 4 + 4\cdot i + i\cdot 4 + i\cdot i = 16 + 4i + 4i -1 = 15 + 8i$$

As an aside, "FOIL" was how we were taught originally, but it is merely a double-application of the distributivity property of multiplication over addition, that $x(y+z) = xy + xz$ and that $(x+y)z=xz+yz$.

Here, treating $(a+b)$ as a term by itself, we have $(a+b)(c+d) = (a+b)c + (a+b)d$ and then using distributivity again from the other side $ = ac+bc+ad+bd$


问:
If sent the same message m to Alice and Bob, how someone who follow the channel can find m ?
Alice has public key (n,ea) and Bob has public key (n,eb) with gcd(ea,eb)=1. If sent the same message m to Alice and Bob, how someone who follow the channel can find m ? </p>

答: The eavesdropper gets $r=m^{e_a}\mod n$ and $s=m^{e_b}\pmod n$. Since $\gcd(e_a,e_b)=1$, eavesdropper can find $x,y$ with $e_ax+e_by=1$ (right?). Eavesdropper calculates $r^xs^y$. Can you take it from there?</p>


问:
A measure of connectedness in graph theory
From Wikipedia:

A complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge.

A tree is an undirected graph in which any two vertices are connected by exactly one path.

To me, these two types of graphs seem like some kind of opposites in terms of connectivity. I have been searching online for a graph property that reflects this difference, like Wiener index, circuit rank, strength, etc.

I couldn't find a property that measures the thing I desire: something like an averaged number that defines a node's connectivity to more than 1 node. A normalized value that should be minimum for a tree e.g. ~0 and maximum for a complete graph e.g. ~1.

Does such a measure exist in literature?

答:
Some relevant metrics are described here. First, there is the connectivity, which describes the number of vertices you need to remove to make the graph disconnected. In the case of a tree with 3 or more vertices, this is 1. In the case of a complete graph, it is V. And in a disconnected graph it's 0, so it's easy to normalize. A similar property holds if you replace the number of vertices to remove with the number of edges to remove.

A metric which is a bit more of a global property is the average connectivity, given by computing the minimum number of edges between any two vertices $\kappa(u, v)$ that you need to remove to cause the graph to be disconnected and averages over all pairs: $$\frac{\sum_{u\neq v\in V}\kappa(u,v)}{\binom{|V|}{2}}.$$
The first source also defines an analogous vertex connectivity, though I don't understand their definition for vertex connectivity when $u$ and $v$ are adjacent.

Another idea I think could be interesting, but I don't know that anyone has ever really researched is the expected number of edges/vertices that need to be removed to disconnect the graph, if you choose randomly.


问:
How many axioms does ZFC have?
ZFC doesn't have finitely many axioms because it has some axiom schemas. Let's look at Axiom Schema of Specification:

$$
\forall X \exists Y \forall a \quad (a \in Y \iff a \in X \land P(a))
$$

This specifies a different axiom for each predicate $P$. Since $P$ is essentially a function from $X$ to boolean values, taken literally as $X$ is an arbitrary set, this axiom schema specifies uncontainably many axioms. That is, the class of axioms this axiom schema specifies is not a set.

Still, I think this axiom schema actually specifies countably infinite axioms, because we can construct an $X$ and a $P$ only in finitely long "text" with finitely many "characters". Is this the case?

答:
In the schema that you've written $P$ needs to be a formula of the language of set theory.  The language of set theory is finite, and formulas are finitely long, so there are only countably many $P$ available to consider.

In other words, not every function from $X$ to $\{T, F\}$ is eligible to be $P$ in the schema.

Title: How many axioms does ZFC have?


问:
Is there an element in $^* \Bbb N$ is Dedekind-infinite?
One definition of a finite set is that it can be injected into an initial segment of $ \Bbb N$, thus any $n$ in $\Bbb N$ is finite.

Accordingly, if it's legitmate to define every element in $^* \Bbb N$  as its own intial segment, then every element $^* \Bbb N$ is finite.

What about Dedekind-infinity? Can we define an Dedekind-infinite element in $^* \Bbb N$ by constructing an injection into its proper subset?

Added: I'm not sure whether the construction of $^* \Bbb N$ is compatible with $\bf ZFC$ in the same way as $\Bbb N$ is, which is probably not, I guess. But how?

答:
There is a problem with this idea because we use $\in$ to construct $\omega$ which is a surrogate for $\Bbb N$ in $\sf ZFC$. This is fine because both $\in$ and $\Bbb N$ are well-founded.

But $^*\Bbb N$ is not well-founded. The non-standard integers must have a decreasing sequence. So we cannot use $\in$ to model them in a model of $\sf ZFC$. Of course, we can have non-standard models of $\sf ZFC$ whose $\omega$ is actually $^*\Bbb N$, but that's besides the point. Inside these models, the set which they think is $\omega$ is well-founded (internally!).

You could perhaps use $\subseteq$ to define the order $^*\Bbb N$ somehow and use $\omega$ for the initial segment of the standard integers. In that case it shouldn't be hard to note that the standard integers correspond to Dedekind-finiteness, and the non-standard integers are Dedekind-infinite sets.

Why? Because any non-standard integer would have to have infinitely many smaller integers, which correspond to infinitely many subsets, which means that it is a Dedekind-infinite set (remember we assume $\sf AC$).

To be fair, though, I don't see what good comes from that definition.


答:
A "predicate" in this context is not an arbitrary Boolean-valued function on $X$.  Instead, $P$ is restricted to predicates that can be written down as formulas in the language of set theory.  Such a formula is a finite string of characters from a finite alphabet, so there are only countably many of them.

(Technical point: the predicates used in the axiom schemas are allowed to have parameters; i.e., they are allowed to refer to arbitrary individual sets.  However, this doesn't change the fact that there are only countably many relevant predicates, because you handle the parameters by representing them as a variable with an extra quantifier.  For instance, Separation for a formula $P(x,y)$ with two free variables says that $$\forall b\forall X \exists Y \forall a \quad (a \in Y \iff (a \in X \land P(a,b)))$$ where the variable $b$ is serving as a parameter (and the outer $\forall b$ means you can substitute any set for it.))

Title: How many axioms does ZFC have?

Body:
ZFC doesn't have finitely many axioms because it has some axiom schemas. Let's look at Axiom Schema of Specification:

$$
\forall X \exists Y \forall a \quad (a \in Y \iff a \in X \land P(a))
$$

This specifies a different axiom for each predicate $P$. Since $P$ is essentially a function from $X$ to boolean values, taken literally as $X$ is an arbitrary set, this axiom schema specifies uncontainably many axioms. That is, the class of axioms this axiom schema specifies is not a set.

Still, I think this axiom schema actually specifies countably infinite axioms, because we can construct an $X$ and a $P$ only in finitely long "text" with finitely many "characters". Is this the case?


问:
An abelian, characteristically simple group is divisible (supposedly).
This is Exercise 4.13 of Roman's "Fundamentals of Group Theory: An Advanced Approach". According to this search and Approach0, it is new to MSE.

The Details:

On page 69 of Roman's book, we have

Definition: Let $G$ be a group. A subgroup $H$ of $G$ is characteristic in $G$ if it is invariant under all automorphisms of $G$.

Write this as $H\sqsubseteq G$.

On page 121 ibid., there is the

Definition: A nontrivial group $G$ is characteristically simple if it has no proper characteristic subgroups.

On page 145, the exercise in question starts out as follows:

An abelian group $A$ (written additively) is divisible if for any $a\in A$ and any positive integer $n$, there is a $b\in A$ for which $nb=a$.

The Question:

Prove that a characteristically simple abelian group $A$ is divisible.

Thoughts:

Let $A$ be a characteristically simple abelian group.

Such a group is classified in the first $\blacktriangleright$ of this answer by @YCor. No justification is given. (I would appreciate a reference to this end, please.)

Let $\varphi \in {\rm Aut}(A)$. Then, for any $B\sqsubseteq A$, either $B$ is trivial or $B=A$, and $\varphi(B)=B$ by definition.

Let $a\in A$. I aim to show that there for all $n\in \Bbb N$, there is some $b\in A$ with $nb=a$.

(I have nothing nontrivial to add.)

Please help :)

答:
Every abelian group $A$ has a canonical collection of characteric subgroups $nA = \{ na : a \in A \}, n \in \mathbb{N}$ given by the $n^{th}$ powers. If $A$ is characteristically simple, these subgroups must each be either the entire group $A$ or the trivial subgroup $0$. So we split into two cases:

If $nA = 0$ for some $n \ge 2$ then we must have $pA = 0$ for some prime divisor $p \mid n$ (otherwise, if $pA = A$ for all $p \mid n$ then we would have $nA = A$), so $A$ is an $\mathbb{F}_p$-vector space. I'll leave it as an exercise to check that every $\mathbb{F}_p$-vector space is characteristically simple (the basic idea is that the automorphism group acts transitively on subspaces of a fixed dimension, so no proper subgroups are characteristic). In particular, the exercise is false as stated.

If $nA = A$ for all $n$ then $A$ is divisible. This is probably what the author had in mind.

In the divisible case we can keep going with a second source of characteristic subgroups: the $n$-torsion subgroups $\sqrt[n]{A} = \{ a \in A : na = 0 \}$ (this is not standard notation; I think $A[n]$ might be standard, but that notation is already too overloaded for my taste) are characteristic, so again must satisfy either $\sqrt[n]{A} = A$ or $\sqrt[n]{A} = 0$. If $\sqrt[n]{A} = A$ then every element is $n$-torsion so $nA = 0$ and $A$ can't be divisible. So we conclude that $\sqrt[n]{A} = 0$ for every $n$, hence that $A$ is torsion-free. Torsion-free divisible abelian groups are precisely vector spaces over $\mathbb{Q}$ (divisibility allows us to define division and torsion-freeness guarantees that the result is unique), and it's again an exercise to show that $\mathbb{Q}$-vector spaces are characteristically simple.

This proves YCor's classification. He singles out the zero abelian group because it's both the zero $\mathbb{F}_p$-vector space for any $p$ and the zero $\mathbb{Q}$-vector space; it is the unique group which satisfies $A = 0$ so it's an edge case for every case of this argument. In particular it's the unique group which is both torsion-free and torsion!

Note that $nA = \text{im} \left( A \xrightarrow{n} A \right)$ while $\sqrt[n]{A} = \text{ker} \left( A \xrightarrow{n} A \right)$. So their characteristic nature comes from an even stronger property of the multiplication-by-$n$ endomorphism on an abelian group, which is that it's natural (commutes with arbitrary homomorphisms of abelian groups); formally, it's a natural transformation from the identity functor $\text{id}_{\text{Ab}} : \text{Ab} \to \text{Ab}$ to itself. It's a nice exercise to show that every such natural transformation has this form; in other words, $\text{End}(\text{id}_{\text{Ab}}) \cong \mathbb{Z}$ (as rings, even).

This naturality property shows that $nA$ and $\sqrt[n]{A}$ have a property stronger than being characteristic: they are natural subgroups, in that arbitrary homomorphisms $f : A \to B$ induce homomorphisms $nA \to nB$ and $\sqrt[n]{A} \to \sqrt[n]{B}$. Formally, they are both subfunctors of the identity functor, and as subfunctors can can be defined as the kernel resp. the image of multiplication-by-$n$ as a natural transformation.


问:
How to find the value $k$ of the line $x=k$ that Bisects the Area Under any Real Curve
As the title states; that is the problem. I am aware of the 1st Moment - however for $f(x) = x$ there is a small deviation from $1/2$ the area which is proportional to the range specified in the integral. For all similair functions this is also true.

答: Let $a, b$ be the lower bounds of $x$ for the region that you want to bisect. You then have to solve $$\int_a^kf(x)dx = \int_k^bf(x)dx$$ Let the integral of $f(x)$ be $F(x)$. Then this is $$F(k)-F(a)=F(b)-F(k) \to F(k) = \frac{F(a)+F(b)}{2}$$ Assuming $f(x) > 0$ means that $F(x)$ is monotonically increasing, so $$k = F^{-1}\left(\frac{F(a)+F(b)}{2}\right)$$


问:
How do I use Ito's formula to solve this $ dX = \frac{1}{3} X^{1/3} dt + X^{2/3} dB $
The title says it all. I tried many functions for Ito's formula but failed. $f(B,t)= B^{3}$ gives something close but that is not it. I'll appreciate any help.

答:
It makes sense to look for solutions in the form $X_t = Y_t^3$ so our equation simplifies to $dX_t = \frac 13 Y_t dt + Y_t^2 dB_t$.  We should also have that $dY_t = a_t dt + b_t dB_t$.  Now applying Ito to find $dY_t^3$ gives

\begin{align*}
dY_t^3 &= 3 Y_t^2 dY_t + 3 Y_t d \langle Y,Y \rangle_t = 3 Y_t^2 a_t dt + 3 Y_t^2 b_t dB_t + 3 Y_t b_t^2 dt \\
dX_t &= \frac 13 Y_t dt + Y_t^2 dB_t.
\end{align*}

Comparing the $dB_t$ term gives that $b_t = \frac 13$, and then comparing the $dt$ terms gives $\frac 13 Y_t = 3Y_t^2 a_t + 3 Y_tb_t = 3Y_t^2 a_t + \frac 13 Y_t$ so we can simply set $a_t = 0$.  Hence our solution is $Y_t = \frac 13 B_t$ and $X_t = \left(\frac 13 B_t\right)^3$.


问:
Mistake in a study book?
Here is the exercise from a Pinter's "A book of abstract algebra" from a chapter dealing with permutations on a finite set:

Let $\alpha$ and $\beta$ be cycles, not neccessarily disjoint. Prove that, if $\alpha^2=\beta^2$, then $\alpha=\beta$.

I think I've found counterexample: $(2345)^2 = (24)(35) = (2543)^2$. Am I right?

答: I have the 1990 edition of Pinter's Book of Abstract Algebra.  Chapter 8, "Permutations of a Finite Set" problem D6 apparently corrected the question to limit the proof to cycles of odd length.</p>

Title: Mistake in a study book?


问:
Can you use the set membership symbol ∈ to indicate subsequence membership in a larger set?
It's common practice to use ∈ to denote set membership, such as 2 ∈ A where A is a set:
A=(1,2,3,4,5). This just states 2 is an element of set A.

However, is it improper to use ∈ in the context of a subset, namely to state that a contiguous subset is included in a larger set? For example a subset:

X = (1,2,3)

A = (1,2,3,4,5)

Would it be incorrect to say X ∈ A?

答:
It's not too improper if you clearly state what you mean in your context by the $\in$ symbol. But viewed differently, yes, it is improper because $\in$ has a well established standard meaning.

So I think it's way more appropriate to assign a new symbol for your purpose, and just define its meaning rigorously and use it.

You didn't define this really rigorously since sets have no order, you probably meant sequences here.


问:
Linear Transformation Question Help
Question: Let $T: R^3$ -> $R^2$ be the linear transformation defined by $T(x)=Ax$ where
$$A =\begin{bmatrix}1 & -5 & -3\\-3&15&9\end{bmatrix}$$
If $\begin{bmatrix}2 \\b\end{bmatrix}$ is in the range of $T$, find the value of "b".

My attemped solution: I understand the basic concepts behind problems like this where $T(x)$ represents the range of $T$ and that $T(x)$ in $R^m$ is the image of $x$ under $T$. However, I'm not very sure how to solve this problem given that a vector with variable $b$ and we have to solve for the value(s) of b to see if it's in the range of $T$. Generally when I check if a column vector is in the range of $T$, I would put the column vector as the augumenting matrix for $A$ and row reduce the matrix to check if there are any correct solutions. But in this case it doesn't seem to work.

I'd much appreciate it if someone could show me how this is done. Thanks in advance!

答: The question is: if $\left[\begin{smallmatrix}2\\b\end{smallmatrix}\right]$ is of the form $A.\left[\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right]$ for some vector $\left[\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right]$, then what is $b$? Note that$$A.\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}x-5y-3z\\-3x+15y+9z\end{bmatrix}.\tag1$$Clearly, there are vectors of the form $\left[\begin{smallmatrix}2\\b\end{smallmatrix}\right]$ in the range of $T$; take $x=2$ and $y=z=0$, for instance. But the second line of the RHS of $(1)$ is the first one time $-3$. So, the only $b$ such that $\left[\begin{smallmatrix}2\\b\end{smallmatrix}\right]$ is in the range is $\left[\begin{smallmatrix}2\\-6\end{smallmatrix}\right]$, that is, $b$ must be $-6$.

Title: Linear Transformation Question Help


问:
The working of $\dim(\ker(X))$ in a square matrix
Does a square matrix $X \in\Bbb R^{n\times n}$ always follow that $\dim(\ker(X)) = \dim(\ker(X^T))$?

I know it works for a sizable amount, but I am not sure if it works for all of them, is there a way to prove this either way?

答:
Yes, and there are many ways to prove it. The easiest is probably considering that $$\dim\ker(X)=n-\operatorname{rank}(X),$$
so we just have to prove that $\operatorname{rank}(X)=\operatorname{rank}(X^T)$; but that follows immediately from the fact that the rank of a matrix $X$ is the maximal number of linearly independent rows (or columns) of $X$, since $X$ and $X^T$ have the same rows (and columns), only "switched".


答:
$$\begin{align*}
\begin{bmatrix}1&-5&-3\\-3&15&9\end{bmatrix} \mathbf x &= \begin{bmatrix}2\\b\end{bmatrix}\\
\begin{bmatrix}1&-5&-3\\0&0&0\end{bmatrix} \mathbf x &= \begin{bmatrix}2\\b+6\end{bmatrix} & R_2 \to R_2+3R_1
\end{align*}$$

If the matrix equation is consistent, then $b+6 = 0$, so $b = -6$.

For completeness, there are some $\mathbf x$s that satisfy $A\mathbf x = \begin{bmatrix}2\\-6\end{bmatrix}$, e.g. for $y,z\in\mathbb R$,

$$\begin{align*}
\mathbf x &= \begin{bmatrix}2+5y+3z\\y\\z\end{bmatrix}\\
&= \begin{bmatrix}2\\0\\0\end{bmatrix} + y \begin{bmatrix}5\\1\\0\end{bmatrix}+z\begin{bmatrix}3\\0\\1\end{bmatrix}
\end{align*}$$

Title: Linear Transformation Question Help

Body:
Question: Let $T: R^3$ -> $R^2$ be the linear transformation defined by $T(x)=Ax$ where
$$A =\begin{bmatrix}1 & -5 & -3\\-3&15&9\end{bmatrix}$$
If $\begin{bmatrix}2 \\b\end{bmatrix}$ is in the range of $T$, find the value of "b".

My attemped solution: I understand the basic concepts behind problems like this where $T(x)$ represents the range of $T$ and that $T(x)$ in $R^m$ is the image of $x$ under $T$. However, I'm not very sure how to solve this problem given that a vector with variable $b$ and we have to solve for the value(s) of b to see if it's in the range of $T$. Generally when I check if a column vector is in the range of $T$, I would put the column vector as the augumenting matrix for $A$ and row reduce the matrix to check if there are any correct solutions. But in this case it doesn't seem to work.

I'd much appreciate it if someone could show me how this is done. Thanks in advance!


问:
What is the probability? (illustrated example)
edit: Here is a short background for my question. I'm looking for the probability that at least two operators has made the same choice, thereby "matching". I've provided a sketch of the problem and this is my thought process and conclusions drawn from the sketch:

An operator is hired to operate a light diode. Each particular hour he can decide to switch it on or off. The likelihood of the operator deciding to switch it on is 50%.

A new operator is hired by the same rules to operate another light diode, the probability of at least two diodes being switched on at a particular hour is $\frac14, 25$%

Yet another third operator is hired, the probability of at least two diodes being switched on is now up to $\frac48,50$%

I have three questions:

Q1: What is the probability of at least two diodes being switched on at the same time when 4 operators are hired?

Q2: How many operators do you need to hire for the probability to be at least 90%?

Q3: If the probability that the operator switch on the diode is not $\frac12$ but instead $\frac1{16}$, how would this be accounted for in the equation?

edit: To find the number of possible states i guessed that if there are n operators, the possible states are $n^2$. This results in 2 operators having 4 possible states, 3 operators having 9 states and 4 operators having 16 states. However, when sketching it out it is clear that 3 operators only has 8 states. So something is wrong (I found out in the answer provided below that I mistakenly switched position of the base and the exponent, the correct way to calculate the states are $2^n$, not $n^2$)

答: Given $n$ operators, there are $2^n$ different ways the diodes can be lit, each with probability $(1/2)^n$. In how many of these scenarios are fewer than two diodes lit? There's one scenario in which no diodes are lit, and $n$ scenarios in which exactly one diode is lit. Subtracting these cases out gives probability $1 - \frac{n+1}{2^n}$ that at least two diodes are lit when there are $n$ operators. Plug in $n=4$ to solve Q1. Experiment with your calculator for different $n$ to solve Q2.


问:
Polynomial map on sphere is smooth
This is from Problem 2 from Milnor's "Topology from a differentiable viewpoint".

Show that every complex polynomial of degree $n$ gives rise to a smooth map from the Gauss sphere $S^2$ to itself of degree $n$.

My candidate for the desired map $f$ is the natural one: We can project $S^2\backslash\{N\}$ ($N$ is the North pole) onto the plane via the stereographic projection from $N$, do the polynomial map, then take the inverse projection. Then define $N\mapsto N$. It is clear that $f$ is smooth everywhere away from $N$.

To show that $f$ is smooth at $N$, I tried to consider the other chart of this map, by projecting $S^2\backslash \{S\}$ from the South pole $S$. If we decompose the polynomial map $P:\mathbb{C}\to \mathbb{C}$ to $(Q,R):\mathbb{R}^2\to \mathbb{R}^2$, then the map $f$ on this chart is
$$(x,y)\mapsto \begin{cases} \left(\frac{Q(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})}{Q(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})^2+R(\frac{x}{x^2+y^2},\frac{x}{x^2+y^2})^2},\frac{R(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})}{Q(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})^2+R(\frac{x}{x^2+y^2},\frac{x}{x^2+y^2})^2}\right) &\text{ if }(x,y)\neq (0,0)\\
(0,0) &\text{ if }(x,y)=(0,0)\end{cases}.$$
Maybe we can discuss the derivatives at $(0,0)$, but I can't see a clean argument in this way. I also think that there maybe something about an analytic continuation to $(0,0)$ since the function defined on $\mathbb{R}^2\setminus (0,0)$ is rational, but I am not familiar with this type of argument.

Any help to find a smooth map/show that my map is smooth at $(0,0)$ is greatly appreciated!

答:
The resulting rational function is much easier to analyze in complex form, where the transition map $\tau$ between the two steriographic projections has the form $\tau(z)=z/|z|^2$, or, more simply $\tau(z)=1/\overline{z}$, where $\overline{z}$ is the complex conjugate of $z$. Let $P(z)=p_0+p_1z+\cdots+p_dz^d$ be the polynomial of interest (with $p_d\neq 0$), and $Q=\tau\circ P\circ\tau$. We then have
$$
Q(z)=\frac{1}{\overline{P(1/\overline{z})}}=\frac{1}{\sum_{n=0}^d\overline{p}_n/z^n}=\frac{z^d}{\sum_{n=1}^d\overline{p}_nz^{d-n}}
$$
We see that the resulting function is a complex rational, and the constant term in the denominator is $\bar{p}_d\neq 0$. Clearly, then, there is no issue extending $Q$ smoothly to $0$.

Title: Polynomial map on sphere is smooth


问:
Is this a typo? $(t_0,x_0)\in \mathbb R^{n+1}$ or $(x_0,t_0)\in \mathbb R^{n+1}$?
Is this a typo?

From Ordinary Differential Equations and Dynamical Systems by Gerald Teschl (free copy), page 36, equation $2.10$ says:

Initial value problem
\begin{align}
\dot x &= f(t,x) \tag 1\\
x(t_0) &= x_0 \tag 2
\end{align}
We suppose $f\in C(U,\mathbb R^n)$, where $U$ is an open subset of $\mathbb R^{n+1}$ and $(t_0,x_0)\in U$.

Question:

Isn't the order wrong in $(t_0,x_0)$ and $f(t,x)$ due to $\mathbb R^{n+1}$?

For simplicity I here use $\mathbb R^{n+1}$ instead of the subset $U$. So, from $(t_0,x_0)\in \mathbb R^{n+1}$ we have that $t_0 \in \mathbb R^n$ and $x_0 \in \mathbb R$, but in $(1)$ the variable is $t$? If $t\in \mathbb R$, isn't also $t_0 \in \mathbb R$?

I.e. shouldn't it be
\begin{align}
\dot x &= f(x,t) \tag 3\\
(x_0,t_0) &\in U \tag 4\\
U &\subset \mathbb R^{n+1} \tag 5
\end{align}

Alternatively $\mathbb R \times \mathbb R^n =\mathbb R^{1+n}$ for $(1)-(2)$.

答:
It sounds like you are viewing $\mathbb{R}^{n+1}$ as $\mathbb{R}^{n}\times\mathbb{R}$.

But $\mathbb{R}^{n+1}$ is also $\mathbb{R}\times \mathbb{R}^{n}$.

The integer $n+1$ is the same as the integer $1+n$.


问:
Covering map on all neighbourhoods is covering map
Let $(X, \mathcal{T}_X), (Y, \mathcal{T}_Y)$ be topological spaces, and let $f : Y → X$ be a continuous map. Suppose that every $x ∈ X$ is contained in an open neighbourhood $U_x$ which has the property that $f|_{f^{-1}U_x} : f^{-1}U_x → U_x$ is a covering map.

That is, every $U_x$ admits an open covering $\mathcal{U}_x $ such that for all $U ∈ \mathcal{U}_x$, there is a set $\mathcal{W} ⊆ \mathcal{T}_Y$ of pairwise disjoint opens such that $U = \bigsqcup\mathcal{W} = \bigsqcup_{W ∈ \mathcal{W}} W$ and for all $W ∈ \mathcal{W}$,  $f|_W : W \tilde{→} U$ is a homeomorphism.

Consider the set $\mathcal{Z} = \bigcup_{x ∈ X}\mathcal{U}_x$. This is clearly an open covering of $X$. I want this $\mathcal{Z}$ to be an open covering that satisfies the conditions for $f$ to be a covering map of $X$. That is, for every $Z ∈ \mathcal{Z}$, I want to construct a set $\mathcal{V} ⊆ \mathcal{T}_Y$ of pairwise disjoint opens, such that $f^{-1}Z = \bigsqcup \mathcal{V}$ and for all $V ∈ \mathcal{V}$, $f|_V : V \tilde{→} Z$ is a homeomorphism.

Can this set $\mathcal{Z}$ do the job? If so, I am still not sure whether I'm basically there, or have actually not yet seen any key insight to prove this.

答:
Your approach works, but it is too complicated. You have to show that $f$ is surjective and each $x \in X$ has an open neigborhood $V_x$ which is evenly covered by $f$. You know that $f \mid_{f^{-1}(U_x)} : f^{-1}(U_x) \to U_x$ is a covering map for each $x \in X$. Each $f \mid_{f^{-1}(U_x)}$ is surjective, thus $x \in f(Y)$ for all $x$. This means that $f$ is surjective.

Moreover, there exists an open neigborhood $V_x$ of $x$ in  $U_x$ such that $V_x$ is evenly covered by $f \mid_{f^{-1}(U_x)}$. Since $U_x$ is open in $X$, also $V_x$ is open in $X$. Moreover, $(f \mid_{f^{-1}(U_x)})^{-1}(V_x)$ is the disjoint union of open $W_\alpha \subset (f \mid_{f^{-1}(U_x)})^{-1}(V_x)$ which are mapped by $f \mid_{f^{-1}(U_x)}$ homeomorphically onto $V_x$. Hence they are mapped by $f$ itself homeomorphically onto $V_x$. But clearly $f^{-1}(V_x) = (f \mid_{f^{-1}(U_x)})^{-1}(V_x)$, thus $V_x$ is evenly covered by $f$.


答: It is incorrect, full stop. If you really mean that $X$ and $A$ are just sets, the correct notation is $X\subseteq A$ (or, if you want to make it clear that $X$ is a proper subset of $A$, $X\subsetneqq A$ or the like). If $X$ and $A$ are sequences, or an ordered triple and an ordered $5$-tuple, there is no standard notation; if you need to talk about this kind of relationship, you should define a suitable notation. Here, for instance, you might for instance say that $X=A[1,3]$. Or you might define $\preceq$ to mean subsequence of and write $X\preceq A$.

Title: Can you use the set membership symbol ∈ to indicate subsequence membership in a larger set?

Body:
It's common practice to use ∈ to denote set membership, such as 2 ∈ A where A is a set:
A=(1,2,3,4,5). This just states 2 is an element of set A.

However, is it improper to use ∈ in the context of a subset, namely to state that a contiguous subset is included in a larger set? For example a subset:

X = (1,2,3)

A = (1,2,3,4,5)

Would it be incorrect to say X ∈ A?


问:
Are all non-squares generators modulo some prime $p$?
If $q\in\mathbb{Z}^+$ is not a perfect square, does there always exist an odd prime $p$ such that $q$ is a generator of $\mathbb{Z}/p\mathbb{Z}^\times$?  Can we find always find infinitely many such $p$?</p>

答: This is a conjecture of Artin. Hooley proved the conjecture is true for all nonsquare $q > 1$ if the generalized Riemann hypothesis is true for zeta-functions of number fields. Later work by Hooley, following work by R. Murty and Gupta, established the conjecture unconditionally (i.e., with no GRH assumptions) for all prime $q$ with at most 2 exceptions (and nobody expects there really are any exceptions at all). For example, the conjecture is provably true for at least one of the choices $q = 2$, 3, or 5 (and surely is true for all three!) but we can't pin down a specific one of those three values for which the conjecture is definitely true. To this day the conjecture is not proved for any specific value of $q$.


问:
What does a proof in an internal logic actually look like?
The nLab has a lot of nice things to say about how you can use the internal logic of various kinds of categories to prove interesting statements using more or less ordinary mathematical reasoning. However, I can't find a single example on the nLab of what such a proof actually looks like. (The nLab has a frustrating lack of examples in general.)

Can anyone supply me with some examples? I'd be particularly interested in the following kinds of examples:

I've heard in a topos one can internalize the real numbers and, in the topos $\text{Sh}(X)$ of sheaves on a topological space, this reproduces the sheaf of continuous real-valued functions $X \to \mathbb{R}$. Moreover, one can internalize "finitely generated projective $\mathbb{R}$-module" and in $\text{Sh}(X)$ this reproduces real vector bundles on $X$. What can you prove about vector bundles this way?

I'd also like to see examples of what you can do in the internal logic of Cartesian closed categories.

This MO question is related but it doesn't really satisfy my curiosity.

答:
Joyal and Tierney's 1984 monograph, An extension of the Galois theory of Grothendieck, is an example of a substantial piece of mathematics written using informal reasoning in internal logic. The main result is the following:

Theorem. Every open surjection of toposes is an effective descent morphism. In particular, every Grothendieck topos is equivalent to the category of equivariant sheaves on a localic groupoid.

If you look at Chapter I, you will find that reads just like any other mathematical text, save for the avoidance of classical logic and certain kinds of set-theoretic operations. The main difficulty with using internal logic is the interpretation of the conclusions – this requires much care! For example, the proposition scheme
$$(\forall b : B . \exists a : A . f(a) = b) \to (\forall h : B^T  . \exists g : A^T . h = g \circ f)$$
where $B$ is fixed but $f : A \to B$ and $T$ are allowed to vary, says that "if $f$ is surjective, then for any $h : T \to B$, there exists $g : T \to B$ such that $h = g \circ f$", or in short, "$B$ is a projective object"... but in the canonical semantics, it is neither necessary nor sufficient that $B$ be projective for the statement to hold! That is because what the formula actually means is the following,

If $f : A \to B$ is an epimorphism, then $f^T : A^T \to B^T$ is also an epimorphism.

whereas $B$ being projective is the statement below:

If $f : A \to B$ is an epimorphism, then $\mathrm{Hom}(T, f) : \mathrm{Hom}(T, A) \to \mathrm{Hom}(T, B)$ is a split surjection of sets.

If the topos in question has a projective terminal object, then the first statement (internal projectivity) implies the second, and if the topos is well-pointed, then the second statement implies the first.

So much for projective objects. What about finitely generated modules? Again there are subtleties, but the most straightforward way to formulate it is to take a mixed approach. Let $R$ be an internal ring. Then $M$ is a finitely-generated $R$-module if there exist global elements $m_1, \ldots, m_n$ (i.e. morphisms $1 \to M$) such that
$$\forall m : M . \exists r_1 : R . \cdots . \exists r_n : R . m = r_1 m_1 + \cdots + r_n m_n$$
holds in the internal logic. This amounts to saying that the evident homomorphism $R^{\oplus n} \to M$ is an epimorphism, which is what we want. It is tempting to formulate the whole statement internally, but this cannot work: at best one will obtain an internal characterisation of modules that are locally finitely generated.

Perhaps I should give a positive example. I'm afraid I can't think of anything interesting, so I'll opt for something simple instead. It is well-known that a two-sided unit element of a magma is unique if it exists. This is also true for internal magmas in any topos, and the proof is exactly the same (so long as it is formulated directly). More explicitly:

Let $M$ be a magma. Suppose $u$ is a left unit element in $M$ and $v$ is a right unit element in $M$. Then, $u = u v = v$.

Formally, we are deducing that
$$\forall u : M. \forall v : M. (\forall m : M. u m = m) \land (\forall m : M. m v = m) \to (u = v)$$
which means that, for all $u : S \to M$ and $v : T \to M$, if $\mu \circ (u \times \mathrm{id}_M) = \pi_2$ and $\mu \circ (\mathrm{id}_M \times v) = \pi_1$, then $u \circ \pi_1 = v \circ \pi_2$ as morphisms $S \times T \to M$.

Now, suppose we have an internal magma $M$ for which
$$\exists u : M . \forall m : M. (u m = m) \land (m u = m)$$
holds in the internal logic, i.e. there exists a morphism $u : T \to M$ satisfying the relevant equations, such that the unique morphism $T \to 1$ is an epimorphism. (The latter is the true content of the quantifier $\exists$.) We wish to show that $M$ has a global unit element, i.e. a morphism $e : 1 \to M$ satisfying the obvious equations. Applying the above result in the case $u = v$, we deduce that $u$ must factor through the coequaliser of $\pi_1, \pi_2 : T \times T \to T$. But this coequaliser computes the coimage of the unique morphism $T \to 1$, and we assumed $T \to 1$ is an epimorphism, so $T \to 1$ is itself the coequaliser of $\pi_1$ and $\pi_2$. Thus $u$ factors through $1$ (in a unique way), yielding the required $e : 1 \to M$.

Of course, the above paragraph takes place in the external logic, but this is unavoidable: there is no way to formulate the existence of a global element in the internal logic. I suppose the point is that, once you have built up a stock of these metatheorems that interpret statements in the internal logic, you can then prove various results using internal logic if so desired.


问:
For which real $s$ does $\sum _{n=1}^{\infty} \frac{1}{n^s}$ converge?
For which values of real $s$ does the following converge?

$$
\sum _{n=1}^{\infty} \frac{1}{n^s}
$$

I have read extensively that the answer is $s>1$ but I can't find a simple proof suitable for students not trained to university level mathematics

Body:
Fix $s>1$. For $k=2,3,\dots$, one has $2^k-1=3,7,15,\dots$ and

\begin{align*}
  \sum_{n=1}^{2^k-1}\frac{1}{n^s} & = 1 +
  %%
 \Big(\frac{1}{2^s}+ \frac{1}{3^s}\Big) +
  %%
 \Big(\frac{1}{4^s}+ \frac{1}{5^s} + \frac{1}{6^s}+ \frac{1}{7^s}\Big) +
  %%
  %%
 \cdots + \Big(\frac{1}{(2^{k-1})^s}+ \cdots + \frac{1}{(2^{k}-1)^s}\Big) \\
  %%
   & < 1 +
  %%
 \Big(\frac{1}{2^s}+ \frac{1}{2^s}\Big) +
  %%
 \Big(\frac{1}{4^s}+ \frac{1}{4^s} + \frac{1}{4^s}+ \frac{1}{4^s}\Big) +
  %%
  %%
 \cdots + \Big(\frac{1}{(2^{k-1})^s}+ \cdots + \frac{1}{(2^{k-1})^s}\Big) \\
  %%
   & = 1 + \frac{2}{2^s} +\frac{4}{4^s}+
  %%
 \cdots + \frac{2^{k-1}}{{2^{k-1}}^s}
  %%
    = 1 + 2^{1-s}+ \big(2^{1-s}\big)^2 +
  %%
 \cdots +  \big(2^{1-s}\big)^{k-1} \\
  %%
   &  = \frac{1 - \big(2^{1-s}\big)^{k}}{ 1-2^{1-s}}
  %%
 < \frac{1}{ 1-2^{1-s}}<\infty. \\
  %%
  %%
\end{align*}

Fix $s\leq 1$. For $k=2,3,\dots$, one has $2^k=4,8,16,\dots$ and

\begin{align*}
  \sum_{n=1}^{2^k}\frac{1}{n^s} & = 1 + \frac12+
 \Big(\frac{1}{3^s}+ \frac{1}{4^s}\Big) +
  %%
 \Big(\frac{1}{5^s}+ \frac{1}{6^s} + \frac{1}{7^s}+ \frac{1}{8^s}\Big) +
  %%
  %%
 \cdots + \Big(\frac{1}{(2^{k-1}+1)^s}+ \cdots + \frac{1}{(2^{k})^s}\Big) \\
  %%
   & > 1 + \frac12+
  %%
 \Big(\frac{1}{4^s}+ \frac{1}{4^s}\Big) +
  %%
 \Big(\frac{1}{8^s}+ \frac{1}{8^s} + \frac{1}{8^s}+ \frac{1}{8^s}\Big) +
  %%
  %%
 \cdots + \Big(\frac{1}{(2^{k})^s}+ \cdots + \frac{1}{(2^{k})^s}\Big) \\
  %%
   & \geq 1 + \frac12+
  %%
 \Big(\frac{1}{4}+ \frac{1}{4}\Big) +
  %%
 \Big(\frac{1}{8}+ \frac{1}{8} + \frac{1}{8}+ \frac{1}{8}\Big) +
  %%
  %%
 \cdots + \Big(\frac{1}{2^{k}}+ \cdots + \frac{1}{2^{k}}\Big) \\
  %%
   & = 1 + \frac12 +\frac12 + \cdots + \frac12
  %%
 = 1 + \frac{k}{2} \rightarrow \infty
  %%
\end{align*}


问:
Uniform convergence: If not Weierstrass M-test, then how?
Not so long ago, I've tried to learn the concept of uniform convergence and I must admit, my understanding is quite unpolished and based only on epsilon-delta definition.

About the problem itself, this is what I have to investigate:

$$\sum_{n=0}^\infty \frac{(-1)^n(x^{2n+1})}{(2n+1)!}$$

And I did it already with [-1,1] interval:

$$\frac{|(-1)^n(x^{2n+1})|}{|(2n+1)!|} = \frac{(x^{2n+1})}{(2n+1)!}$$

$x^{2n+1}$ is at most 1 on [-1,1], so:

$$\frac{(x^{2n+1})}{(2n+1)!} \le \frac{1}{(2n+1)!}$$

Which is then found convergent by ratio test:

$$\frac{1 * (2n+1)!}{(2n+3)! * 1} = \frac{1}{(2n+2)(2n+3)}$$

So, I'm actually confused with what I should do to investigate the uniform convergence on the interval (0, +$\infty$), as the same trick obviously won't do there. I've also noticed that $\sum_{n=0}^\infty \frac{(-1)^n(x^{2n+1})}{(2n+1)!} = \sin x$ by Tailor series, and I am sure that I've missed something really simple, yet important.

答: If it converges uniformly on $(0,\infty)$ then the general term would tend to $0$ uniformly. So there exist $N$ such that $|\frac {(-1)^{n}x^{2n+1)}} {(2n+1)!}| <1$ for all $n >N$ and for all $x$. You get a contradiction to this by letting $x \to \infty$.


问:
Field Extensions: Prove $F(a,b)^* = F(a)^* \cap F(b)^*$
Let $F$ be a field,
Let $K$ be a finite extension of $F$,
Let $a, b$ be members of $K$.

Prove $F(a,b)^* = F(a)^* \cap  F(b)^*$

From Pinter's book of abstract algebra chapter 32 problem G2

I'm assuming the asterisk means that zero is not included in the sets.  Not sure on that but it seems to be the usage earlier in this book.

But with or without the identity elements, since $F \subseteq F(a) \subseteq F(a,b)$ and $F \subseteq F(b) \subseteq F(a,b)$, how is "$F(a,b)^* = F(a)^* \cap  F(b)^*$" necessarily true?  Is this a typo, should it be union?  What do the asterisks have to do with it?

I wish I had more to go on, but this is it.

答:
I looked up Pinter's book. A couple paragraphs above Theorem 2 (same chapter) he defines $I^* = \operatorname{Gal}(K : I)$ (where $I$ is an intermediate field extension). He calls this the "fixer" of $I$ but I would like to point out that most people call this the "isotropy group" or maybe "stabilizer group" instead.

With this in mind, the exercise should not be too difficult. Basically it just comes down to the facts that $\sigma \in F(a)^*$ if and only if $\sigma(a) = a$ and $\sigma \in F(a,b)^*$ if and only if $\sigma(a) = a, \sigma(b) = b$.


答:
According to the comments, for an intermediate extension $L$, $F\leq L\leq K$, the notation means $$L^* = \mathrm{Gal}(K/L) = \{\sigma\in \mathrm{Aut}(K)\mid \sigma(x)=x\text{ for all }x\in L\}.$$ That is, the automorphisms of $K$ that fix $L$ pointwise.

(The notation often implies that the extension is Galois, but many authors do not require it for the definition; e.g., Lang’s Algebra calls the automorphism group as above “the Galois group”, and has a separate notation to denote the special case when the extension is Galois; in the comments it is stated that there is an assumption the extension will be Galois, so see remark below the horizontal line for a more immediate proof in that situation.)

With that understanding, the equality should be obvious: what are the automorphisms of $K$ that fix $F(a,b)$? Well, they must fix all of $F$, they must fix $a$, and they must fix $b$. Thus, they must fix both $F(a)$ and $F(b)$. Conversely, if an automorphism fixes both $F(a)$ and $F(b)$, then it fixes their compositum, $F(a)F(b) = F(a,b)$, which gives the equality.

If you know the extension is Galois then the equality is even more immediate: the Fundamental Theorem of Galois Theory establishes an inclusion-reversing correspondence between subgroups of $\mathrm{Gal}(K/F)$ and intermediate fields given precisely by the ${}^*$ operator. Since $F(a,b)$ is the smallest field that contains both $F(a)$ and $F(b)$, it follows that $F(a,b)^*$ must be the largest subgroup that is contained in both $F(a)^*$ and $F(b)^*$, i.e., the intersection.

Title: Field Extensions: Prove $F(a,b)^* = F(a)^* \cap F(b)^*$

Body:
Let $F$ be a field,
Let $K$ be a finite extension of $F$,
Let $a, b$ be members of $K$.

Prove $F(a,b)^* = F(a)^* \cap  F(b)^*$

From Pinter's book of abstract algebra chapter 32 problem G2

I'm assuming the asterisk means that zero is not included in the sets.  Not sure on that but it seems to be the usage earlier in this book.

But with or without the identity elements, since $F \subseteq F(a) \subseteq F(a,b)$ and $F \subseteq F(b) \subseteq F(a,b)$, how is "$F(a,b)^* = F(a)^* \cap  F(b)^*$" necessarily true?  Is this a typo, should it be union?  What do the asterisks have to do with it?

I wish I had more to go on, but this is it.


问:
Show y is odd in the equation $y^3 =x^2 +2$
Found a question which asks you to find all integer solutions $y^3 =x^2 +2$ .    In order to do so, it first asks you to show that y is odd.  How would one go about showing this?

Many Thanks

答: Suppose not. Then $x^2=y^3-2$, hence $x$, is even. Therefore, $2=y^3-x^2$ is divisible by $2^2$, which is a contradiction.</p>


问:
Why can the map between fundamental groups be extended to the whole polygon?
Let $M$ be a surface with genus $g$. Suppose there is an injection $\pi_1(M) \to \pi_1(N)$, where $N^2$ is compact and oriented. We can view $M$ as a $4g$-sided polygon with appropriate identification and defining a boundary map via the inclusion $\pi_1(M) \to \pi_1(N)$. I am wondering why this allows me to extend the map $\phi:M \to N$. The hint I am given is to use contractibility, but I am not very sure how to use this hint.

答:
Let's call

$P$ your $4g$-gon;

$f:P → M$ the quotient map;

$\hat\gamma:[0,1] → \partial P$, a simple loop circumnavigating $\partial P$;

$\gamma= f\circ \hat\gamma$, the image onto $M^1$; and

$a_1,b_1,\ldots,a_g,b_g:S^1→ M^1$ the loops generating $\pi_1(M)$, such that
$$\gamma = [a_1,b_1][a_2,b_2]\cdots[a_g,b_g].$$

Let $\phi_\ast$ denote your homomorphism, and $\phi$ the map to be derived.

Choosing representatives of the images of the generators $\phi_\ast(a_i),\ \phi_\ast (b_i)$ gives a map $M^1→N$ as you mention: $$\phi:\bigcup a_i\cup b_i → \bigcup \phi(a_i)\cup \phi(b_i) \tag(1)$$ The fact that $\phi_\ast$ is a homomorphism then tells us that

$$[\phi(a_1),\phi(b_1)][\phi(a_2),\phi(b_2)]\cdots[\phi(a_g),\phi(a_g)]$$

is nullhomotopic in $N$, since it represents the class

$$\phi_\ast([a_1,b_1][a_2,b_2]\cdots[a_g,b_g]) = \phi_\ast (1_{\pi_1(M)}) = 1_{\pi_1(N)}.$$

Let $F: S^1 × [0,1] → N$ denote the nullhomotopy. I.e.:

$F$ is continuous.

$F(\cdot, 1) = [\phi(a_1),\phi(b_1)][\phi(a_2),\phi(b_2)]\cdots[\phi(a_g),\phi(a_g)]$.

$F(\cdot, 0)$ is constant.

By the last bullet point, $F$ passes to the quotient to give a map

$$F: \frac{S^1 × [0,1]}{S^1 × \{0\}} → N,$$

Theis new domain of $F$ is, up to homeomorphism your polygon (a.k.a. disk): i.e. there exists a homeomorphism

$$\Psi:P → \frac{S^1 × [0,1]}{S^1 × \{0\}}.$$

One just has to show that $\Psi$ can be chosen so that $F\circ\Psi: P \to N$ passes again to the quotient $f$ to give a continuous map $M → N$.

This is a condition only on the restriction of $\Psi$ from boundary to boundary: $\partial P → S^1 × \{1\}$. Since $\gamma$ is simple, we can take

$$\Psi^{-1}(x,1) = \gamma(x)$$

and this does the trick nicely—from the second point about $F$, it follows that $\Psi \circ F$ restricted to $\partial P$ is equal to the function $\phi$ as in $(1)$, and so easily passes to the quotient, as required!