问:
Question about the differential function on a surface
My question is related to this Show that $\text{grad }f=\frac{f_uG-f_vF}{EG-F^2}\textbf{x}_u+\frac{f_vE-f_uF}{EG-F^2}\textbf{x}_v$

In both answers it's assumed that $df_{p}(X_u) = f_{u}$ and $df_{p}(X_v) = f_{v}$ but I don't know why this is true. I'm even confused about what $f_u$ and $f_v$ mean.

I know that $df_{p}(v)$ is such that $df_{p}(v)$ : $T_{p}S$ $\rightarrow$ $\mathbb{R}^3$ and that $\{X_u,X_v\}$ is a basis for $T_{p}S$ but I don't know why mapping basis vectors gives us $f_u$ and $f_v$

答: It's sort of by definition. Given a parametrization ${\bf x} = {\bf x}(u,v)$ for a surface $S$ and $f:S \to \Bbb R$, one writes $$\frac{\partial f}{\partial u}(q) \doteq \frac{\partial (f\circ {\bf x})}{\partial u}({\bf x}^{-1}(q)) $$for any $q$ in the image of ${\bf x}$. In other words, partial derivatives of a function defined on a surface only make sense once a coordinate system has been chosen. What happens is that people don't have the patience to be precise all the time, so they identify $f$ with $f\circ {\bf x}$ and then you have to struggle to understand what is going on. For a function $F:\Bbb R^n \to \Bbb R$, in general, one has $DF(p)(e_i) = (\partial F/\partial x_i)(p)$, so applying this to the local expression $f \circ {\bf x}$ gives that $${\rm d}f({\bf x}_u) = {\rm d}f({\rm d}{\bf x}(1,0)) = {\rm d}(f \circ {\bf x})(1,0) = \frac{\partial f}{\partial u} = f_u.$$Note that I'm supressing base points above. In any case, ${\rm grad}(f)\circ {\bf x}$ is a combination of ${\bf x}_u$ and ${\bf x}_v$, so write $${\rm grad}(f)\circ {\bf x} = A{\bf x}_u+B{\bf x}_v,$$for functions $A$ and $B$ to be found. Taking products with ${\bf x}_u$ and ${\bf x}_v$ gives $$f_u = EA + FB \quad \mbox{and}\quad f_v = FA + GB.$$So $$\begin{pmatrix}E & F \\ F & G \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} f_u \\ f_v \end{pmatrix}\implies \begin{pmatrix} A \\ B \end{pmatrix} = \begin{pmatrix} E & F \\ F & G  \end{pmatrix}^{-1}\begin{pmatrix} f_u \\ f_v \end{pmatrix}.$$This gives the expressions for $A$ and $B$.


问:
Proof by induction in sequences
My lecturer left this as an exercise and didn't go through it, I couldn't find it anywhere online so how is it done? Any help appreciated.

Original question: "Let $(x_n)$ be a real sequence satisfying that every subsequence of $(x_n)$ does not converge in $\mathbb{R}$. Prove that $|x_n|\rightarrow \infty$ as $n \rightarrow \infty$

Proof:

By contradiction, if $|x_n|$ does not tend to $\infty$ (i.e. negation of $|x_n| \rightarrow \infty$), then $\exists M > 0$, such that for all $N$ natural numbers, $\exists m > N$, such that $|x_m| \le M$ => $\exists$ subsequence $(x_{n_k})$ such that $|x_{n_k}| \le M$.

My main question is how to prove the last implication, it's been left as an exercise.

答:
We know that because  $x_{n} \not \to \infty$, when $n\to\infty$, that there exists a $M$ such that for any $N\in\mathbb{N}$ we can find a natural number $m>N$ such that $|x_{m}|\leq M$.

Lemma. $(x_{n}: n\in \mathbb{N})$ has a bounded sub-sequence $(x_{n_{k}} : k\in \mathbb{N})$.

I leave it to you to fill in the details of the following:

Proof sketch: Set $n_{1}$ to be $\min\{n\in \mathbb{N}: |x_{n}|\leq M\}$. This is made possible by the non-emptiness of the set $\{n\in \mathbb{N}:|x_{n}|\leq M\}$ and applying the Well Ordering Principle. Argue inductivly that $n_{k}:=\min\{n\in \mathbb{N}: |x_{n}|\leq M \ \text{and} \ n>n_{k-1}\}$ defines the index of the sought after sub-sequence $(x_{n_{k}}: k\in \mathbb{N})$. That is, prove that $n_{k}$ exists for all $k$ and conclude that each term in $(x_{n_{k}}: k\in \mathbb{N})$ bounded by $M$.

ContentLicense: CC BY-SA 4.0

Title: Proof by induction in sequences


问:
Finding an order of $n$ such that a product is exponentially small in $n$
I am looking for a function $f$ s.t. if $t=O(f(n))$, where $O$ stands for Big-O notation, then:

$$\frac{t}{n}\times\frac{t-1}{n}\times\cdots \times \frac{2}{n}\sim e^{-an} \tag{1}$$

i.e. when the LHS of $(1)$ is exponentially small in $n$?

Say $t=n/2$, so $t=O(n)$, then the LHS of $(1)\leq \left(\frac{1}{2}\right)^n$ so it is exponentially small in $n$. But I was wondering if I can do better than that like what if $t=O(\sqrt{n})$ or something like that in which case I don't know if the LHS of $(1)$ would still be exponentially small?

答:
You want
$$\frac{f(n)!}{n^{f(n)-1}}\sim e^{-an}.$$
Since
$$f(n)!\sim \sqrt{2\pi f(n)}\left(\frac{f(n)}e\right)^{f(n)},$$
so we want
$$n\sqrt{2\pi f(n)}\left(\frac{f(n)}{en}\right)^{f(n)}\sim e^{-an}.$$
Taking logs and writing $f(n)=f$ for simplicity, we have that
$$\log n+\frac{\sqrt{2\pi}}2+\frac{\log f}2+f\cdot(\log f-\log n-1)+an$$
is asymptotically constant.
Writing $g=f/n$ (since $f$ should be a bit smaller than $n$),
$$\frac{\sqrt{2\pi}}2+\frac{\log g+3\log n}2+gn(\log g-1)+an\to c$$
for some constant $c$. The asymptotically largest terms here are $gn\log g$ and $an$, so we want
$$-g\log g\sim a,$$
so $g$ should be (around) a constant. So, $f$ should be asymptotically some constant times $n$.


问:
Determine where the range of a function is positive
Consider the quadratic equation $y = -\dfrac{3}{4}x^2 + 4x - 4$, with roots at $x = \tfrac{4}{3}$ and $x = 4$. I'd like to find the values for $x$ where $y > 0$; that is where
$$
   -\dfrac{3}{4}x^2 + 4x - 4 > 0.
$$
Rewritting as
$$
(x - \tfrac{4}{3})(x - 4) > 0,
$$
the above inequality is true when

A) both $(x - \tfrac{4}{3})$ and $(x - 4)$ are positive

or

B) both $(x - \tfrac{4}{3})$ and $(x - 4)$ are negative.

Case A - both positive:

$x - \tfrac{4}{3} > 0$ => $x > 4/3$ and $x - 4 > 0$ => $x > 4$. This is true when $x > 4$.

Case B - both negative:

$x - \tfrac{4}{3} < 0$ => $x < 4/3$ and $x - 4 < 0$ => $x < 4$. This is true when $x < \tfrac{4}{3}$.

Therefore the solution is $(-\infty, \tfrac{4}{3}) \cup (4,\infty)$. Yet graphing this equation shows the positive range as $(4/3,4)$. The range $(-\infty, \tfrac{4}{3}) \cup (4,\infty)$ actually satisfies the equation $y = \dfrac{3}{4}x^2 - 4x + 4$.

Where have I run afoul?

Body: $$-\dfrac{3}{4}x^2 + 4x - 4 > 0$$ can't be written as$$(x-4/3)(x-4) > 0$$ but as $$-\frac34(x-4/3)(x-4) > 0$$which gives$$(x-4/3)(x-4)\color{red}<0$$


问:
Calculating missing values when the combined mean is given.
There are 150 workers in a factory running two shifts. The average wage of workers is Rs. 41.12. The average wage of worker in morning shift is Rs 43 and that of evening shift is Rs 35. Find out the number of workers in morning and that of evening shift.

I calculated the answer and it comes out to be 114.75 for the number of workers working in the morning shift. The book's answer says 94 and 56 workers working in morning and evening shifts respectively. Which is the correct one? And how to go about calculating it?

I used the standard formula. The one which is used to calculate the combined mean. And tried to calculate the value for the respective variables.

答:
Let's see:

$$\dfrac{114.75\times 43 + 35.25 \times 35}{150} = \dfrac{6168}{150} = 41.12$$
$$\dfrac{94\times 43 + 56 \times 35}{150} =\dfrac{6002}{150} \approx 40.01$$

Quite evidently you are correct (despite detractors who apparently didn't bother to check the math themselves).

Likely there is some typo in the problem statement.


问:
From matrix notation to element-wise
Let $B\in\mathbb R^{q\times p}, C\in\mathbb R^{q\times q}$ be two real matrices, and $C$ invertible and symmetric.

Define $A= B^\top C^{-1} B$.

Can we express the elements of $A$, $a_{i,j}$, using elements of $B$ and $C$ in general ?

答: @MostafaAyaz's objection remains, even with the constraints $\det C\ne0,\,C^\top=C$. You can write$$A_{ij}=B^\top_{ik}C^{-1}_{kl}B_{lj}=B_{ki}C^{-1}_{kl}B_{lj},$$but in general you have to do it with $C^{-1}$, not $C$. Even the special case $p=q,\,B=I_p$ requires you to somehow write components of $C^{-1}$ in terms of $C$, which you can't do, except verbosely. You certainly can't write something nice like $A_{ij}=B_{ki}B_{lj}X_{klmn}C_{mn}$.

Title: From matrix notation to element-wise


问:
A noetherian ring whose ideals are idempotent is artinian
I have to prove the folowing:

If $R$ is a Noetherian ring, and for every ideal $I$ of $R$ we have $I = I^{2}$, then $R$ is Artinian.

My first thought was to try to prove that the nilradical of $R$ is equal to the Jacobson, but I got nothing with that. Then I tried to prove that any prime ideal of $R$ is maximal, but I couldn't, so I got stuck.

Thank you for any help!

答: If $I=I^2$ and $I$ is finitely generated, then there exists $e\in I$, $e^2=e$ such that $I=(e)$; see here. Now take a prime ideal $\mathfrak p$ and look at $R/\mathfrak p$: any ideal of this ring has the form $I/\mathfrak p$ with $I$ ideal of $R$ containing $\mathfrak p$. We also have $I/\mathfrak p=(I/\mathfrak p)^2$ and $I/\mathfrak p$ is finitely generated, so $I/\mathfrak p$ is generated by an idempotent of $R/\mathfrak p$. Since $R/\mathfrak p$ is an integral domain its only idempotents are $0$ and $1$. We thus get that $I/\mathfrak p=(0)$ or $I/\mathfrak p=R/\mathfrak p$. This means that $R/\mathfrak p$ has only two ideal, that is, $R/\mathfrak p$ is a field.</p>

ContentLicense: CC BY-SA 3.0


问:
If $\lbrace y_{k}\rbrace_{k\in K}$ is a net such that $e^{ixy_{k}}\to 1$ pointwise, then $y_{k}\to 0$
Okay, so my brain doesn't seem to be working right now. I'm trying to prove that, given a net $\lbrace y_{k}\rbrace_{k\in K}$ in $\mathbb{R}$ such that $e^{ixy_{k}}$ converges pointwise to $1$ then $y_{k}$ necessarily goes to $0$. I know that if $y_{k}$ is bounded then there's a sufficiently large $n$ such that the whole net is contained in $[-n,n]$. Since this set is compact there must be a convergent subnet, say $y_{k_{j}}$. If we call its limit $y$ then $e^{ixy_{k_{i}}}\to e^{ixy}$ pointwise, so $e^{ixy_{k_{j}}}\to 1$ pointwise. Thus $e^{ixy}=1$ for all $x$. Then it must be the case that $y=0$. Otherwise we could take $x=\dfrac{\pi}{y}$ in $e^{ixy}=e^{i\pi}=-1$, which is a contradiction.

The only thing left to show is that $\lbrace y_{k}\rbrace_{k\in K}$ is bounded. If it wasn't we should be able to construct a subnet $\lbrace y_{k_{j}}\rbrace_{j\in J}$ (I'm pretty sure I can take it to be a sequence) that blows up to either $\infty$ or $-\infty$. In particular, we can assume without loss of generality that it blows up to $\infty$ (since the other case should be analogous). It seems intuitively true that this will lead to a contradiction, since plotting the real part of $e^{iax}$, $\cos(ax)$ for some $a\in\mathbb
{R}$ and letting $a\to\infty$ the resulting set of points appears to coincide (or at least be dense) in the band $(-\infty,\infty)\times[-1,1]$. This seems to imply that the set $(-\infty,\infty)\times\lbrace 1\rbrace$ is dense in this band, which is obviously false.

So, how should I go about proving that the original net $\lbrace y_{k}\rbrace_{k\in K}$ is bounded? Or am I approaching this in the wrong way? For context, I am trying to prove that $\mathbb{R}$ is homeomorphic to $\widehat{\mathbb{R}}$ (its dual group with the corresponding topology) via the identification $y\longmapsto e^{ixy}$. I already know this is a bijection (moreover, I know it defines a group isomorphism) and I know that mapping $y$ to $e^{ixy}$ is continuous. But I've been trying a lot of different approaches to show continuity in the other direction and nothing I've tried so far has worked. The idea I outlined in the post seems the most promising so far, since I know that convergence in the topology of $\widehat{\mathbb{R}}$ implies uniform convergence on compact sets (in particular, it implies pointwise convergence). All of that led me to the approach I described at the beginning.

For reference, I am following Chapter VII of John B. Conway's "A Course in Functional Analysis" (Second Edition). According to the author the fact that $\mathbb{R}$ and $\widehat{\mathbb{R}}$  are homeomorphic is easy to prove, but I've been thinking about this for two days straight and I've honestly ran out of ideas. Maybe I'm just too dumb to see it, I don't know. But I would really appreciate your help.

答:
A net $(y_k)_{k \in K}$ such that $e^{ixy_k} \to 1$ pointwise need not converge to $0$.

Take an arbitrary finite set $\{ x_{\nu} : 1 \leqslant \nu \leqslant n\} \subset \mathbb{R}$ and consider the point
$$\xi = (e^{ix_1}, \dotsc, e^{ix_n}) \in T^n\,.$$
Since $T^n$ is compact, the sequence $(\xi^m)_{m \in \mathbb{N}}$ has a limit point $\eta$. Then for every $\varepsilon > 0$ there are infinitely many $m \in \mathbb{N}$ with $\lvert e^{ix_{\nu}m} - \eta_{\nu}\rvert < \varepsilon$ for all $\nu$. Taking two such $m$, we see that there are arbitrarily large integers $r = m_1 - m_2$ such that $\lvert e^{ix_{\nu}r} - 1\rvert < 2\varepsilon$ for $1 \leqslant \nu \leqslant n$.

Thus if we take for $K$ the directed set $\mathscr{F} \times (0,1) \times \mathbb{N}$, where $\mathscr{F}$ is the set of finite subsets of $\mathbb{R}$, and the preorder is given by
$$(F_1, \varepsilon_1, m_1) \leqslant (F_2, \varepsilon_2, m_2) \iff F_1 \subseteq F_2 \land \varepsilon_2 \leqslant \varepsilon_1 \land m_1 \leqslant m_2$$
we obtain a net $(y_k)_{k \in K}$ such that $y_k \to +\infty$ and $e^{ixy_k} \to 1$ for all $x \in \mathbb{R}$ by setting
$$y_{F, \varepsilon, m} = \min \{ \mu \in \mathbb{N} : \mu \geqslant m \land \lvert e^{ix\mu} - 1\rvert \leqslant \varepsilon \text{ for all } x \in F\}\,.$$

The topology of pointwise convergence is too coarse. You need to use the uniform convergence on compact sets, and the fact that $\mathbb{R}$ is locally compact. Thus convergence in $\widehat{\mathbb{R}}$ implies (is) locally uniform convergence, and for $a > 0$ the set of $y \in \mathbb{R}$ such that $\lvert e^{ixy} - 1\rvert \leqslant \varepsilon$ holds for all $x \in [-a,a]$ is (for small enough $\varepsilon$) easy to determine.


问:
Summation indices in a proof of Incluson-Exclusion Principle
I rewrote (for easier time reading in the future) the proof of PIE given in my book as follows below and got tangled up in sum notation. I got questions about the $\color{purple}{\text{purple}}$ and $\color{brown}{\text{brown}}$  bits in the rewritten proof below. The expression $N_{\ge}(\emptyset)$ has no index $j$ so I think $\color{purple}{\displaystyle{\sum_{j=0}^nN_{\ge}(\emptyset)} = N_{\ge}(\emptyset)}$ because there's only one item to sum. Does that make sense? In the case of $\color{brown}{\displaystyle{N_{\ge}(J)\sum_{j=0}^n\left((-1)^m\binom nm\right) = N_{\ge}(J) \cdot 0}}$, we have $\color{brown}{\displaystyle{\sum_{j=0}^n\left((-1)^m\binom nm\right) = 0}}$, but the actual theorem says  $\displaystyle{\sum_{j=0}^n\left((-1)^j\binom nj\right) = 0}$ and so indices don't match in the  $\color{brown}{\text{brown bit}}$. How can I make the indices match up in the $\color{brown}{\text{brown bit}}$?

Here below is PIE as given in my book:

Definition

Statement of PIE

Proof

Here below is the rewritten proof:

Let $U$ be a universe of objects and let $P = \{p_1, p_2, p_3, \ldots, p_n\}$ be a set of properties that the objects may or may not have.

Let $N_=(J)$ be a set of all objects all of whose properties are in $J \subseteq P.$

Suppose $|J| = 2$. Then some of the $N_=(J)$s are $N_=(\{p_1, p_7\}), \ N_=(\{p_{n-1}, p_n\})$ etc. There are $\binom n2$ such $N_=(J)$s. Generally, if $|J| = j$, then there are $\binom njN_=(J)$ objects all of whose properties are in $J$ meaning $\color{red}{\displaystyle{\sum_{|J| = j}(-1)^jN_=(J) = (-1)^j\binom njN_=(J)}}$. Summing both sides of the expression in $\color{red}{\text{red}}$ above, we have $$\color{blue}{\sum_{j=0}^n\left(\sum_{|J| = j}(-1)^jN_=(J)\right) = \sum_{j=0}^n(-1)^j\binom njN_=(J)}$$

Since removing $N_=(J)$ elements also removes $N_{\ge}(J)$ elements, we can replace $N_=(J)$s in the expression $\color{blue}{\text{in blue}}$ above with $N_{\ge}(J)$s as follows: $$\color{green}{\sum_{j=0}^n\left(\sum_{|J| = j}(-1)^jN_{\ge}(J)\right) = \sum_{j=0}^n(-1)^j\binom njN_{\ge}(J)}$$

Note that the number of objects in $U$ with none of the properties is $N_{\ge}(\emptyset)$ meaning the number of times $N_{\ge}(\emptyset)$ includes each object in $U$ is $1$ when the object has none of the properties and $0$ when the object has at least one property. If we can say the same thing about the expression  $\color{green}{\text{in green}}$ above, then we can use it instead of $N_{\ge}(\emptyset)$.

Now assume an object in $U$ contains none of the properties in $P$. Then $|J| = 0$ meaning $\displaystyle{\sum_{j=0}^n\left(\sum_{|J| = 0}(-1)^0N_{\ge}(\emptyset)\right) = \color{\purple}{\sum_{j=0}^nN_{\ge}(\emptyset) = N_{\ge}(\emptyset)}}$. Since $N_{\ge}\emptyset$ includes each object in $U$ only $1$ time when the object has none of the properties, so does the expression  $\color{green}{\text{in green}}$ above.

Suppose an object in $U$ contains $m$ properties in $P$ where $1 \le m \le n$. Then $|J| = m.$ Now $\displaystyle{\sum_{j=0}^n\left(\sum_{|J| = m}(-1)^mN_{\ge}(J)\right) = \sum_{j=0}^n\left((-1)^m\binom nmN_{\ge}(J)\right) = \color{brown}{N_{\ge}(J)\sum_{j=0}^n\left((-1)^m\binom nm\right) = N_{\ge}(J) \cdot 0} = 0}$

Thus the expression  $\color{green}{\text{in green}}$ above doesn't even count an object with properties in $J$.

Edit (number of onto functions):

We consider functions $\{1, 2, 3, 4, \ldots,k\} \to \{A, B, C, D\}$. Let $A$ stand for a set of functions whose image does not contain $a$. Define $B, C, D$ similarly. Let $N_=(\emptyset)$ be the number of elements in a set of functions whose image does not miss any of $a, b, c, d$ and $N_{\ge}(\emptyset) -$ the number of elements in  set of all functions including $\emptyset.$ Let $N_{\ge}(m_a)$ be the number of elements in a set of functions whose image misses $a$ and also possibly $b, c, d$. In other words, $N_{\ge}(m_a) = |A|$ or $N_{\ge}(m_a) = |A \cap B|$ or $N_{\ge}(m_a) = |A \cap C \cap D|$ etc. Also, let $AB$ stand for $A \cap B.$

Now take a look at the given pics. Our goal is to count the elements in the square outside the union  of $A, B, C, D$ as shown in pic 1. To that end start by counting all the elements in the square as shown in pic 2, then remove the elements in the union of $A, B, C, D$ from the total number of elements. When we remove $A$, we remove $AB$ with it. Look at pic 3. When we remove $B$, we remove $AB$ again. Thus removing $A \cup B$ results in us removing $AB$ twice. Generally, removing $A \cup B \cup C \cup D$ results in us removing $AB, AC, AD, BC, BD, CD$ twice each meaning we need to add one copy of each of $AB, AC, AD, BC, BD, CD$ back as we meant to remove them only once. When we remove $A$, we remove $ABC$ with it as in pic 4. Similarly, removing $B$ also results in removal of $ABC$. This holds for $C$ as well meaning as we remove $A \cup B \cup C$ we also remove $ABC$ three times. But earlier we added $AB, BC, AC, AD$ back which resulted in us adding $ABC$ back four times. So, when  we remove $A \cup B \cup C \cup D$, one copy of $ABC$ is left behind which we need to get rid of. Similarly we need to remove one copy of each of $ACD, BCD, ABD.$ Now removing $A$ results in removal of $ABCD$ with it as in pic 5. Similarly, removing $B, C, D$ means we remove $ABCD$ three more times. Earlier, we added $AB, AC, AD, BC, BD, CD$ which added $ABCD$ back six times. Then we removed $ABC, BCD, ACD, ABD$ which removed $ABCD$ four times. Thus removing $A \cup B \cup C \cup D$ results in removing $ABCD$ twice meaning we need to add $ABCD$ back once. Algebraically,

$$N_=(\emptyset) = \\ N_{\ge}(\emptyset) \\ - (N_{\ge}(m_a) + N_{\ge}(m_b) + N_{\ge}(m_c) + N_{\ge}(m_d)) \\ + (N_{\ge}(m_am_b) + N_{\ge}(m_am_c) + N_{\ge}(m_am_d) + (N_{\ge}(m_bm_c) + N_{\ge}(m_bm_d) + N_{\ge}(m_cm_d)) \\ - (N_{\ge}(m_am_bm_c) + N_{\ge}(m_am_bm_d) + N_{\ge}(m_bm_cm_d) + N_{\ge}(m_am_bm_d)) \\ + N_{\ge}(m_am_bm_cm_d)$$

Now note, $$N_{\ge}(m_a), \ N_{\ge}(m_b), \ N_{\ge}(m_c), \ N_{\ge}(m_d) \ge (n - 1)^k \\ N_{\ge}(m_am_b), \ N_{\ge}(m_am_c), \ N_{\ge}(m_am_d), \ (N_{\ge}(m_bm_c), \ N_{\ge}(m_bm_d), N_{\ge}(m_cm_d \ge (n - 2)^k \\ N_{\ge}(m_am_bm_c), \ N_{\ge}(m_am_bm_d), \ N_{\ge}(m_bm_cm_d), \ N_{\ge}(m_am_bm_d) \ge (n - 3)^k \\ N_{\ge}(m_am_bm_cm_d) \ge (n - 4)^k \\ N_{\ge}(\emptyset) \ge (n - 0)^k$$

Note, $n = 4$, but we use either $n$ or $4$ depending on whichever is more instructive at the time. Thus, we have $$N_=(\emptyset) \ge \binom 40(n - 0)^k - \binom 41(n - 1)^k + \binom 42(n - 2)^k - \binom 43(n - 3)^k + \binom 44(n - 4)^k = \sum_{j=0}^4\binom 4j(-1)^j(n - j)^k$$

Finally, $N_= (m_a) = (n - 1)^k$. However, as we saw above (when working with overlapping circles) removing an object with $N_= (m_a)$ properties results in the removal of an object with $N_{\ge} (m_a)$ properties. Thus we can replace $N_{\ge} (m_a)$ with $N_= (m_a)$ meaning $$N_=(\emptyset) = \sum_{j=0}^4\binom 4j(-1)^j(n - j)^k$$

答:
I’ll take them in order. First,

$$\sum_{j=0}^nN_\ge(\varnothing)=(n+1)N_\ge(\varnothing)\,:$$

there are $n+1$ terms, one for each value of $j$ from $0$ through $n$, and each term is $N_\ge(\varnothing)$. However, this summation never actually occurs in the proof: your derivation of it further down is incorrect.

Your second question is also based on a misunderstanding of the argument; see below.

Now for the proof. $N_=(J)$ is not the set of all objects whose properties are in $J$: it is the number of objects that have precisely the properties in $J$.

You write:

Since removing $N_=(J)$ elements also removes $N_\ge(J)$ elements, ...

This is not true: $N_\ge(J)$ is the number of objects that have all of the properties in $J$ and possibly others as well and is generally greater than $N_=(J)$. It is the lefthand side of the green displayed expression that is the starting point for the argument; the righthand side is meaningless as written, since $J$ is undefined, and the blue displayed expression is irrelevant.

The number of objects in $U$ that have none of the properties is $N_=(\varnothing)$. It is not $N_\ge(\varnothing)$: that is the number of objects in $U$ that have at least $0$ of the properties, so it is in fact $|U|$, the number of objects in $U$. The point here is that an object with none of the properties is counted once in the $j=0$ term of the lefthand side of the green expression and $0$ times in each of the other terms of that sum, so altogether the sum counts it exactly once.

Now we consider an object in $U$ that has exactly $m\ge 1$ of the $n$ properties and want to know how many times it is counted by the sum

$$\sum_{j=0}^n\left(\sum_{\substack{J\subseteq P\\|J|=j}}(-1)^jN_\ge(J)\right)\,.\tag{0}$$

Let $I$ be the set of $m$ properties possessed by the object. If $J$ is any subset of $P$, the object is counted in $N_\ge(J)$ if and only if $J\subseteq I$, i.e., if and only if the object has all of the properties in $J$ (and possibly some others as well).

Clearly $J\nsubseteq I$ if $|J|>m$, so the object is not counted in $N_\ge(J)$. Thus, the object is not counted at all in any of the terms

$$\sum_{\substack{J\subseteq P\\|J|=j}}(-1)^jN_\ge(J)\tag{1}$$

with $j>m$: it affects only the terms with $j\le m$.

Now suppose that $j\le m$; how many times is the object counted in $(1)$? It is counted once in $N_\ge(J)$ for each $J$ of cardinality $j$ that is a subset of $I$. $I$ has $\binom{m}j$ subsets of cardinality $j$, and the object is counted once in each of them, so the object is counted $\binom{m}j$ times in the term $(1)$. That number is then multiplied by $(-1)^j$, so altogether the $j$ term $(1)$ contributes a total of $(-1)^j\binom{m}j$ to the sum $(0)$.

In short, the object contributes $0$ to the terms $(1)$ with $j>m$ and $(-1)^j\binom{m}j$ to the terms with $J\le m$, so altogether it contributes

$$\sum_{j=0}^m(-1)^j\binom{m}j=0$$

to the sum $(0)$.

Now put the pieces together: each object that has none of the properties in $P$ is counted once in the sum $(0)$, and each object that has at least one of those properties is counted a net total of $0$ times, so the sum $(0)$ is the number of objects in $U$ that have none of the properties in $P$.


问:
Normal subgroups of a group of order 12
Let $G$ be a group of order 12 and suppose $G$ does not have normal subgroups of order $4$. How to show that $G$ has a subgroup of order $6$?</p>

答: An alternative way to finish Peter Patzt's argument is as follows: Once you know that $G$ has a normal subgroup $N$ of order $3$, note that $G/N$ is of order $4$ and therefore abelian. Since normal subgroups of the quotient correspond to normal subgroups of $G$ that contain $N$, and every group of order $4$ contains a normal subgroup of order $2$, this normal subgroup of $G/N$ of order $2$ lifts to a normal subgroup $K$ of $G$ that contains $N$ with $[K:N]=2$. Since $|N|=3$, that means that $K=6$, so you get a normal subgroup of order $6$.</p>

Title: Normal subgroups of a group of order 12


问:
$E[X]$ finite iff $\sum\limits_{n} P(X>an)$ converges
Show that:

$$\sum\limits_{n \in N } P(X>an) < \infty\ \text{for some}\ a > 0 \Rightarrow E[X] < \infty \Rightarrow \sum\limits_{n \in N } P(X>an) < \infty\ \text{for every}\ a > 0.$$

We have an unknown probability space ($\Omega,\mathbb{F},P)$, where $\mathbb F$ is a sigma-algebra, and $P:\mathbb F \rightarrow [0,1]$ is a probability measure, $X$ is a $\mathbb{F}/\mathbb{B}(\mathbb{R})$-measurable random variable and $E[X]$ is the expectation of the variable $X$.

I've been working through past exam papers, and am stuck on this. Would appreciate a solution!

答:
We prove the first implication.

Let's take $a>0$ such that $\sum_{n \in N } P(X>an) < \infty$.

First of all, $P(X>an)=\sum_{k=n}^{\infty}P(X\in(an,a(n+1)])$. Hence, we write
$\infty>\sum_{n \in N } P(X>an) = \sum_{n \in N } \sum_{k=n}^{\infty} P(X\in(ak,a(k+1)])$. By Tonelli's theorem we can change the order of summation. We arrive at

$\sum_{k \in N } \sum_{n=0}^{k} P(X\in(ak,a(k+1)])=\sum_{k \in N }  (k+1) P(X\in(ak,a(k+1)])$.

If $d\mu(x)$ represents our probability measure (i.e. $P(X\in A)=\int_{x\in A} d\mu(x)$), then we can continue

$ \sum_{k \in N }  (k+1) P(X\in(ak,a(k+1)]) = \frac {1}{a}\sum_{k \in N }  a(k+1) \int_{(ak,a(k+1)]}d\mu$
$\ge \frac {1}{a}\sum_{k \in N }   \int_{x\in (ak,a(k+1)]}xd\mu(x)=\frac {1}{a}    \int_{x>0}xd\mu(x)\ge \frac {1}{a}E[X]$ by definition. In other words, $E[X]<\infty$. Note that there can still be $E[X]=-\infty$.

In order to prove the second implication, a little modification of this trick is enough.


问:
Residue of $\frac{1}{(1+\cos z)^2}$ for $z_0=\pi$
I'm trying to find the residue of $f(z)=\frac{1}{(1+\cos z)^2}$ around $z_0=\pi$. I know that $z=\pi$ is a pole of order 4, so I'm using the formula: $$\frac{1}{(4-1)!}\lim_{z\to\pi}\frac{d^3}{dz^3}\left[(z-\pi)^4\left(\frac{1}{(1+\cos z)^2}\right)\right]$$ But when I try to calculate the limit, it just gets worse. After several attempts, I'm stuck at: $$\frac{1}{6}\lim_{z\to 0}\left[\frac{z^4}{(1-\cos z)^2}\left(\frac{24}{z^2} - \frac{24\sin^3z}{(1-\cos z)^3} + \frac{2\sin z}{(1-\cos z)} + \frac{18\sin z\ \cos z}{(1-\cos z)^2} - \frac{24 \cos z}{z (1-\cos z)} + \frac{72\sin^2 z}{z(1-\cos z)^2} - \frac{72\sin z}{z^2(1-\cos z)}\right)\right]$$
Can anyone give me a hint?

答: The residue of $f$ at $\pi$ is the residue of $f(z+\pi)$ at $0$. But$$f(z+\pi)=\frac1{\left(1+\cos(z+\pi)\right)^2}=\frac1{\left(1-\cos(z)\right)^2},$$which is an even function. Therefore, its residue at $0$ is $0$.


问:
lemma from Milne
I am reading Milne book on algebraic geometry. I understood the proof presented below. I don't understand why use A[x] ? Why not proceed with the proof by using A[X]/(1 - hX)? x isn't defined anywhere.

答:
Milne's phrasing in the second sentence of the proof is a little bit informal with notation, but for clarity, you could insert the following sentence immediately before that:

Let $x$ be the class of $X$ in the quotient ring $A[X]/(1 - hX)$.

So, in particular, $A[x]$ isn't a polynomial ring over $A$ in one indeterminate $x$ (i.e., the ring over $A$ generated by an element $x$ with no relations other than those implied by the commutative ring axioms), but rather it's the ring over $A$ generated by an element $x$ that's subject to the relation $hx = 1$.

Title: lemma from Milne


问:
Visual understanding for the non-injectivity of $h(x_1,x_2) = (x_1^2+2x_1x_2+x_2^2, 2x_1+2x_2)$

Let $h(x_1,x_2) = (x_1^2+2x_1x_2+x_2^2, 2x_1+2x_2).$ Determine if $h$ is locally injective.

I know the function is not locally injective. Factoring one has $h(x_1,x_2) = ((x_1+x_2)^2, 2(x_1+x_2))$ from where it can be seen. However, I was trying to get some visual understanding of this and letting $x=(x_1+x_2)^2$ and $y=2x_1+2x_2$ I managed to get that $y = 2x_1+2(\sqrt{x}-x_1) = 2\sqrt{x}.$ Now this is just the square root function with a little shift upwards and looks as follows

And this is injective. Actually, it's even bijective, which seems to be in contradiction with the actual result that $h$ is not locally injective.

What might I be missing here?

答:
Your $2$-dimensional graph is wrong. You should be graphing $4x=y^2$. Just consider points with $y<0$ and you have a problem with writing $y=2\sqrt x$.

Nevertheless, the map $y\rightsquigarrow (4y^2,y)$ is an injective map. The problem comes from the fact that you have infinitely many values of $(x_1,x_2)$ with $x_1+x_2=y$ for any value of $y$. Just consider $x_2=c-x_1$. All the points $(x_1,c-x_1)$ map to the single point $(4c^2,2c)$.


问:
$\mathbb{Z}_p^*$ subgroup generators and element orders

Assume $\mathbb{Z}^*_p = \{1,2,\dots, p-1\}$ with operation of
multiplication modulo $p$, $p$ prime and let $g$ be a generator of
this group.

If $d$ divides $p - 1 $, find a $b \in \mathbb{Z}^*_p$ with $\text{ord}(b) = d$

How many elements of order $d$ exist in $\mathbb{Z}^*_p$ ?

How many generators has the cyclic subgroup generated by $b$ with $\text{ord}(b) = d$?

How many cyclic subgroups of order $d$ exist in $\mathbb{Z}^*_p$?

If you're given an element $h$, its order $d$ and a random element $a$, how can you check if $a$ belong in the subgroup generated by $h$
in polynomial time?

Solution:

$f: \mathbb{Z}_{p-1} \longrightarrow \mathbb{Z}^*_p$ with $f(x) = g^x \pmod{p}$ is an isomorphism. Also, in $\mathbb{Z}_{p-1}$ the element $\frac{p-1}{d}$ is of order $d$. Thus, $g^{\frac{p-1}{d}} \pmod{p}$ is of order $d$ in $\mathbb{Z}^*_p$.

Only one, since $f$ is an isomorphism and $\frac{p-1}{d}$ is the only element of order $d$ in $\mathbb{Z}_{p-1}$

This subgroup has order $d$ and thus $\varphi(d)$ generators (all the coprimes less than $d$)

Only one, with generator $\frac{p-1}{d}$

Check: $h^0 = a$ or $h^1 = a$ or $h^2 = a$ or $\dots$ or $h^{d-1} = a$. If one of the equalities is true, then $a$ belongs in the subgroup generated by $h$. In any other case, it doesn't.

Please, if you can, confirm my solution.

答: Note $\mathbb{Z}^*_p$ with $ p=5$, then $p-1=4$. Consider $4|p-1$, then there exist 2 elements of order $4$, namely $2,3$.


问:
translating a problem to a differential equaton
I've got a question like this:

```<code>The country's population has doubled in 50 years.
In a few years it will increase threefold,
assuming that the growth rate is proportional to the number of residents?
</code>```

in the context of learning to solve first order differential equations.
How can I translate it to a differential equation form?

答:
$p(t+50)=2p(t)$

$$p'(t)=kp(t)\to p(t)\to C e^{k t}$$
$$p(t+50)=2p(t)\to e^{k(50+t)}=2e^{k t}\to k=\frac{\log (2)}{50}\approx 0.013863$$
How many years will pass before population will be three times, means solving
$$Ce^{kt}=3Ce^{kt_0}\to t=t_0+79.25\text{ years}$$


问:
How to find all ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$
I am trying to find all homomorphisms from $\mathbb{Z}$  to    $\mathbb{Z}$. I am looking at a solution, but i do not understand it. Here it is:

Let  $\Phi\colon \mathbb{Z}\to\mathbb{Z}$  be a ring homomorphism. Because $1^2 = 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$. If $\Phi(1) = 1$ then $\Phi(n)= \Phi(n · 1)= n$, so $\Phi$ is the identity map of $\mathbb{Z}$ onto itself which is a homomorphism. If $\Phi(1)= 0$, then $\Phi(n)=\Phi(n · 1)0$, so  maps everthing onto $0$, which also yields a homomorphism.

I do not understand here what "Because $1^2= 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$.". All proof is built upon this and I do not understand what it means. Can someone help me with this?

Thank you

答: Since $1=1\cdot 1$, and then by definition of a homomorphism, $\Phi(1)=\Phi(1\cdot 1) = \Phi(1)\cdot \Phi(1)$.</p>


问:
Quotient group of normalizer N(K) with Sylow p-group
I am self studying from Pinter's "Abstract Algebra". Chapter 16, section M, question 5 has a question where we have to prove that no element of $N/K$ has order a power of $p$.

Let $K$ be a $p$-Sylow subgroup of $G$, and $N = N(K)$ the normalizer of $K$.

I use the correspondence theorem:

$$S \cong S^*/K$$

Where $S$ is the cyclic subgroup of $N/K$ generated by $Ka$.

$$S^* = \{ x \in N : Kx \in S \}$$

For any $x \in K$, then $Kx = K \in S$, so $x$ itself is in $S^*$. That means $K$ is a subgroup of $S^*$ but since $K$ is maximal, we conclude $S^* = K$ and for any $a \in N, Ka = K$.

Now my question is, does the correspondence theorem prove that $S = \{K\}$? If so we just proved no element of $N/K$ has order a power of $p$ (since it equals the whole group). If my logic is incorrect, then how should I approach this question?

Thanks

答:
It's much easier simply to note that since $K$ is a $p$-Sylow subgroup of $G$, it's necessarily also a $p$-Sylow subgroup of $N$, so $p$ does not divide $\vert N/K \vert$ and $N/K$ can't have any elements of order $p$.

If any group has an element $g$ of order $ap, a \in \Bbb N$, then $g^a$ has order $p$.  Since $N/K$ can't have any such elements and any power of $p$ is also a multiple of $p$, $N/K$ can't have any elements with order a power of $p$.


问:
Can any of sum of squares be realized as a sum of this specific form?
Let $x,y \in \mathbb{Z}$, and suppose that $x^2+y^2 \ge 4$.
Do there exist $a,b,c,d \in \mathbb{Z}$ such that
$
(a+d)^2+(b-c)^2=x^2+y^2
$
and $ad-bc=1$?

This question is motivated by an attempt to characterise the norms of matrices in $SL_2(\mathbb{Z})$: If $A =\begin{pmatrix} a & b \\\ c & d \end{pmatrix} \in SL_2(\mathbb{Z})$, then
$$
(a+d)^2+(b-c)^2=\|A\|^2+2,
$$
so I wonder whether $\|A\|^2+2$ can be any sum of squares.

答:
Let $x=1$ and $y=3$ so that $x^2+y^2=10$. Then if $\tbinom{a\ b}{c\ d}\in\operatorname{SL}_2(\Bbb{Z})$ is such that
$$(a+d)^2+(b-c)^2=x^2+y^2=10,\tag{1}$$
then $\{a+d,b-c\}=\{\pm 1,\pm 3\}$ for some for some choices of the signs. Then of course the matrices
$$\begin{pmatrix}-a&\hphantom{-}b\\\hphantom{-}c&-d\end{pmatrix},
\begin{pmatrix}\hphantom{-}a&-b\\-c&\hphantom{-}d\end{pmatrix},
\begin{pmatrix}\hphantom{-}b&\hphantom{-}a\\-d&-c\end{pmatrix}
\in\operatorname{SL}_2(\Bbb{Z}),$$
also satisfy $(1)$, so without loss of generality $a+d=1$ and $b-c=3$. Then
$$1=ad-bc=a(1-a)-b(b-3),$$
which shows that $b$ is an integral root of a quadratic with discriminant
$$\Delta=-4a^2+4a+5.$$
But this is never a perfect square; a contadiction! Hence no such matrix exists.


问:
Example of Diverging functions with converging integral
{Hey thanks for sparing some time I would very much appreciate your help on this problem :}

Give a example of a continuous function $f$ such that $f(x)\geq0$
$(1\leq{x}\leq\infty)$
and such that $$\int_1^\infty{f(x)dx}$$ converges but $$\sum_1^\infty{f(n)}$$  diverges.

I have just started studying analysis from R.R.Goldberg and find few of the problem a bit tricky  Will very much appreciate your help.

答:
Consider the function
$$f(x) = \begin{cases}
2x - 2(n - 1/2n) & \text{if} ~ x \in [n - 1/2n, n], n \in \mathbb{N}_0\\
-2x + 2(n + 1/2n) & \text{if} ~ x \in [n , n + 1/2n], n \in \mathbb{N}_0\\
0 &\text{otherwise}
\end{cases}$$
The integral converges because
$$\int_1^{\infty} f(x) = \sum_{n = 1}^\infty \int_{n - 1/2n}^{n + 1/2n} f(x) = \sum_{n = 1}^\infty \frac{1}{2n^2} < +\infty$$
and the associate serie
$$\sum_{n = 1}^\infty f(n) = \sum_{n = 1}^\infty \frac{1}{n} = +\infty. $$


问:
About a claim by Hintikka in Encycl. Britannica bearing on the limitations of First Order Logic
In the article History Of Logic, more precisely in the section dealing with " Logic After 1900" one can find this assertion :

"First-order logic is not capable of expressing all the concepts and
modes of reasoning used in mathematics; equinumerosity
(equicardinality) and infinity, for example, cannot be expressed by
its means. For this reason, the best-known work in 20th-century logic,
Principia Mathematica (1910–13), by Bertrand Russell and Alfred North
Whitehead, employed a version of higher-order logic."

Is it actually the case that equinumericity or infinity cannot be expressed using F.O.L. ?

That seems strange , since, in order to define equinumericity for example, one needs ( as far as I know) : the notion of a function, and the notion of bijectivity, whch apparently can be defined using FOL.

In the same way, in order to define infinity, one needs the proper subset notion and the concept of equinumericity; but it does not seem difficult to express the proper subset notion using FOL.

I guess that Hintikka is right in his asertion. So, what do I miss?

答:
There's a crucial subtlety here: in what context is our putative first-order definition taking place?

The obvious definitions of infinity, equinumerosity, etc. are indeed first-order formulas - in the language of set theory. That is, there is for example a first-order $\{\in\}$-formula $\psi(x,y)$ such that for all sets $a,b$, the sentence(-with-parameters) $\psi(a,b)$ is true in the universe of sets $V$ - that is, iff $V$ $\models$ $\psi(a,b)$.

However, note that we're leveraging the power of $V$ here, regardless of how simple the sets $a$ and $b$ are. What if we try to work more conservatively? This is where we get the impossibility results. For example, per Hintikka we have:

There is no sentence $\theta$ such that for every structure $\mathcal{M}$, we have $\mathcal{M}\models\theta$ iff $\mathcal M$ is infinite.

There is no sentence $\theta$ such that for every structure $\mathcal{M}$ with unary relations $A,B$ we have $\mathcal{M}\models\theta$ iff the sets $A^\mathcal{M}$ and $B^\mathcal{M}$ have the same cardinality.

There is no sentence $\theta$ such that for every structure $\mathcal{M}$, we have $\mathcal{M}\models\theta$ iff $\mathcal{M}$ is uncountable.

And so on. The most common tool for proving such limitative results is the compactness theorem, but the more technical downward Löwenheim–Skolem theorem also plays an important role.

(I've specified "downward" since the upward Löwenheim–Skolem theorem is really just a compactness corollary — this is a pet peeve of mine.)

And so forth. Note that here we're not sneaking in any "extra resources" beyond the structures in question themselves.


问:
How to calculate $\int \frac{dp}{p\sqrt{p^2 - c^2}}$
I'm trying to find a solution by using Euler-Lagrange, but I'm stuck with this integral.

$$\int \frac{dp}{p\sqrt{p^2 - c^2}}$$

I know I have to use a substitution, but I don't know how.

I'm stuck there and I don't even know if this is good.

$$\int \frac{dp}{p^2\sqrt{1 - \frac{c^2}{p^2}}}$$

答:
Set $p=\frac{c}{\sin (x)}\to dp=-c \cot (x) \csc (x)dx$

$x=\arcsin\left(\frac{c}{p}\right)$

Assuming $c>0$

$$\int \frac{dp}{p \sqrt{p^2-c^2}} =\int \frac{- \cot (x) \frac{c}{\sin x}}{\frac{c}{\sin (x)} \sqrt{\left(\frac{c}{\sin (x)}\right)^2-c^2}} \, dx=(*)\\
=\int \frac{-\cot x}{c\cot x}\,dx=-\frac{x}{c}+K=-\frac{1}{c}\arcsin\left(\frac{c}{p}\right)+K$$

$(*)$

$$\sqrt{\left(\frac{c}{\sin (x)}\right)^2-c^2}=\sqrt{\frac{c^2}{\sin^2 (x)}-c^2}=c\,\sqrt{\frac{1-\sin^2(x)}{\sin^2 (x)}}=c\,\sqrt{\frac{\cos^2(x)}{\sin^2 (x)}}=\\=c\,\sqrt{\cot^2(x)}=\cot(x)$$


答:
hint

Assume $ c>0 $ and consider
$$f(x)=\frac{1}{x\sqrt{x^2-c^2}}$$

the domain of $ f $ is
$$D=(-\infty,-c)\cup (c,+\infty)$$

$ f $ is continuous at $ D$, then it has antiderivative at each of the intervals of $ D$.

To find the antiderivative at $ (c,+\infty) $, we make the substitution
$$x=c\cosh(u)$$
with
$$x^2-c^2=c^2\sinh^2(u)$$

To find the antiderivative at $ (-\infty,-c)$ , you put $$x=-c\cosh(u)$$

Title: How to calculate $\int \frac{dp}{p\sqrt{p^2 - c^2}}$

Body:
I'm trying to find a solution by using Euler-Lagrange, but I'm stuck with this integral.

$$\int \frac{dp}{p\sqrt{p^2 - c^2}}$$

I know I have to use a substitution, but I don't know how.

I'm stuck there and I don't even know if this is good.

$$\int \frac{dp}{p^2\sqrt{1 - \frac{c^2}{p^2}}}$$


问:
How to represent the floor function using mathematical notation?
I'm curious as to how the floor function can be defined using mathematical notation.

What I mean by this, is, instead of a word-based explanation (i.e. "The closest integer that is not greater than x"), I'm curious to see the mathematical equivalent of the definition, if that is even possible.

For example, a word-based explanation of the factorial function would be "multiply the number by all integers below it", whereas a mathematical equivalent of the definition would be $n! = \prod^n_{i=1}i$.

So, in summary: how can I show what $\lfloor x \rfloor$ means without words?

答:
There exists exactly one integer $n$ such that $n \le x < n+1$. We define $\lfloor x \rfloor$ to be this $n$.

Or, if you really like symbols: $\forall x \in \mathbb{R} \ \exists! n \in \mathbb{Z} \ n \le x < n+1 \quad \lfloor x \rfloor := n$.


问:
Flaw in solving a PDE and Fourier Series

Consider the Laplacian equation in cylindrical coordinates with no theta $\left(~\theta~\right)$ dependence:
\begin{align}
&\mbox{So,}\quad\frac{1}{r} (r(u)_r)_r + (u)_{zz} = 0
\\[2mm]
&\mbox{subject to}\quad
\left\{\begin{array}{rcl}
u\left(r=R,z\right) & = & 1
\\
u\left(r=0,z\right) & = & \text{finite}
\\
u\left(r,z=\pm H\right) & = & 0
\end{array}\right.
\end{align}

Using separation of variables, $u(r,z) = A(r)B(z)$, we get
$$B''=-\lambda ^2 B \\
rA'' + A' - \lambda^2 r A = 0$$

Solving the $B(z)$ equation and imposing the boundary conditions, I get,
$$c_1 \cos(\lambda H) + c_2\sin(\lambda H) = 0 \\
c_1 \cos(\lambda H) - c_2\sin(\lambda H) = 0$$
I was told that this was incorrect, and I should be using $\sinh$ and $\cosh$ functions to this DE. My question is, why is that allowed? Aren't they non-periodic?

What am I doing wrong here, and how am I supposed to know when to use $\sin$ and $\cos$, and when to use $\sinh$ and $\cosh$ when solving PDEs?

答:
There is more than one representation of the solution in term of series and Bessel functions. I'll try to derive the one with $\sinh$ and $\cosh$.

First, let's make the $r = R$ conditions homogenous. Let $u = 1 - v$. The new unknown $v$ satisfies the same equation
$$
\frac{1}{r}\frac{\partial}{\partial r}(r v_r) + v_{zz} = 0,
$$
but the boundary conditions are
$$
v(r=R, z) = 0\\
|v(r=0, z)| < \infty\\
v(r, z=\pm H) = 1
$$
By separation of variables $v = Q(r) Z(z)$ we have
$$
Z'' = \lambda Z\\
(rQ')' + \lambda r Q = 0.
$$
Taking $\lambda < 0$ leads us to $\sin$ and $\cos$ for $Z$ (the path you've originally taken). There's noting wrong with it, but it leads to Bessel functions with imaginary arguments.

So let's assume $\lambda = \kappa^2 > 0$.
$$
Z'' = \kappa^2 Z\\
(rQ')' + \kappa^2 r Q = 0.
$$
The solutions are $Z(z) = A \cosh \kappa z + B \sinh \kappa z$ and $Q(r) = C J_0(\kappa r) + D Y_0(\kappa r)$. And since $Y_0(x)$ is infinite at $x = 0$ we need to state that $D = 0$.

The equation $Q(R) = 0$ is
$$
J_0(\kappa R) = 0.
$$
If $\mu_n$ are the roots of $J_0(\mu) = 0$ then
$$
\kappa_n = \frac{\mu_n}{R}.
$$
The roots of Bessel functions are quite common in such problems. Moreover (see Fourier-Bessel series, A&S 11.4.5),
$$
\int_0^R J_0(\kappa_n r) J_0(\kappa_m r) r dr = \delta_{nm} R^2 \frac{J_1(\mu_n)^2}{2}.
$$

So the solution should be sought as
$$
v(r,z) = \sum_{n=1}^\infty (A_n \cosh \kappa_n z + B_n \sinh \kappa_n z) J_0(\kappa_n r).
$$
Let's put $z = \pm H$:
$$
1 = v(r,\pm H) = \sum_{n=1}^\infty (A_n \cosh \kappa_n H \pm B_n \sinh \kappa_n H) J_0(\kappa_n r).
$$
Multiplying both sides with $J_0(\kappa_m r)$ and applying $\int_0^R \bullet rdr$ on the both sides gives:
$$
\int_0^R J_0(\kappa_m r) rdr = (A_m \cosh \kappa_m H \pm B_m \sinh \kappa_m H)
R^2 \frac{J_1(\mu_m)^2}{2}.
$$
The integral on the left is $R^2 \frac{J_1(\mu_m)}{\mu_m}$ (see A&S 11.3.20), so
$$
A_m \cosh \kappa_m H \pm B_m \sinh \kappa_m H = \frac{2}{\mu_m J_1(\mu_m)}.
$$
Clearly, $B_m = 0$ (we might guess this from $z$-symmetry) and
$$
A_m = \frac{2}{\mu_m J_1(\mu_m)}\frac{1}{\cosh \kappa_m H}.
$$

Finally, the answer is
$$
u = 1 - \sum_{n=1}^\infty \frac{2}{\mu_n J_1(\mu_n)} \frac{\cosh \kappa_n z}{\cosh \kappa_n H} J_0\left(\kappa_n r\right)
$$

I made a plot with 20 terms of Felix Marin's solution (red) and mine (blue):

The $\cos$-series has oscillations near $r = R$ while $\cosh$-series has oscillations near $z = \pm H$. These oscillations are in fact, Gibb's phenomenon and are due no approximation of constant $1$ with either Fourier series $\cos k_n z$ or by Fourier-Bessel series $J(\kappa_n r)$. Far from the oscillations the solutions are almost equal up to small difference (due to truncated infinite series).


问:
How to solve for a constant
I'm totally stuck at those equations:

$a+b=5$

     $ax+b=11$

     $a+bx=9$

I need to solve for $x$.

I know this is embarrassing but maybe someone can give me a clue...

答:
Add the second and third equation. We get
$$(a+b)x + (a+b) = 20$$
From the first equation, we have $a+b = 5$. Hence, we get $x = \ldots$

Since $a+b = 5$, we get $5x + 5 = 20 \implies 5x = 15 \implies x = 3$.

Title: How to solve for a constant


问:
Is a bounded real function defined on a singleton Riemann Integrable?
I know to be Riemann integral a real function $f$ must bounded be defined on a closed bounded interval. A singleton is technically a closed bounded interval so is this permitted?

答:
Yes. And note that a function from a singleton into $\Bbb R$ is automatically bounded.

On the other hand, the integral of such a function will alwyas be equal to $0$.


问:
Proving Positivity is Equivalent to $\langle v,v \rangle >0$ for some $v\in V$
Problem $3$ of section $6.A$ in the textbook Linear Algebra Done Right is as follows

Suppose $\mathbb{F}= \mathbb{R}$ and $V \not = \{0\}$. Replace the positivity condition (which
states that $\langle v,v \rangle > 0$  $\forall v\in V$, $v\not= 0$) in the definition of an inner product
$(6.3)$ with the condition that $\langle v,v \rangle >0$ for some $v\in V$. Show that this
change in the definition does not change the set of functions from $V \times V$
to $\mathbb{R}$ that are inner products on $V$.

I assumed that this intends to point out that when the field is the real numbers, the condition positivity is equivalent to having at least one non-zero vector whose inner product with itself is positive.

If that is true, then since we can immediately show that any scalar multiple of $v$ also satisfies the positivity condition, the next step would be to prove that positivity holds for any non-zero vector $u$ that is not in the span $v$. The information that I notice is given are $\langle v,v \rangle$ and conjugate symmetry is now become commutativity. SO I reasoned I must somehow form an expression using $\langle v,v\rangle >0$ to show that $\langle u, u \rangle> 0$.

One approach I tried was to expand the expression
\begin{align*}
\langle v,v\rangle &= \langle u+(v-u) , u + (v-u)\rangle \\
 &= \langle u,u\rangle + \langle u , v-u\rangle + \langle v-u, u \rangle + \langle v-u, v-u \rangle \\
&= \langle u,u \rangle + 2\langle u,v-u \rangle - \langle u, v-u\rangle + \langle v,v-u\rangle \\
&= \langle u,u\rangle - \langle u+v, u-v \rangle >0
\end{align*}

In another exercise we proved that if $u$ and $v$ have the same norm, then $\langle u+v, u-v \rangle = 0$. So if could argue that I can choose $u$ in this way, I should be done. But if I'm not mistaken, I can't do this since I don't know that the norm induced by the inner product is well-defined anymore.

I would appreciate a hint on how to solve this problem. Is the idea to use an expression equivalent to $\langle v,v \rangle$ and expand using the properties of inner products to eventually show $\langle u, u \rangle >0$?

答:
You need to show two things. First that the new condition implies the positivity condition in a real vector space and the converse.

To show that the new condition implies the positivity condition fix a $w \in V$ and then show that $\langle w,w \rangle \ \geq 0$.

You may find it useful to consider two cases: $w = \lambda v$ for some $v\in V$ and $w \neq \lambda v$ for all $v \in V$


答: You have one mistake that you haven't noticed. On the elements of order $d$ in $\mathbb{Z}_p^*$, you are right about the isomorphism between $(\mathbb{Z}_{p-1},+)$ and $\mathbb{Z}_p^*$, but you're missing there are more elements of order $d$ in $\mathbb{Z}_{p-1}$. I'll leave you with a hint: Consider $\frac{(d-1)(p-1)}{d}$, what is its order in $\mathbb{Z}_{p-1}$? After you answer this question, I wish you may have a greater insight in the problem.

Title: $\mathbb{Z}_p^*$ subgroup generators and element orders

Body:
Assume $\mathbb{Z}^*_p = \{1,2,\dots, p-1\}$ with operation of
multiplication modulo $p$, $p$ prime and let $g$ be a generator of
this group.

If $d$ divides $p - 1 $, find a $b \in \mathbb{Z}^*_p$ with $\text{ord}(b) = d$

How many elements of order $d$ exist in $\mathbb{Z}^*_p$ ?

How many generators has the cyclic subgroup generated by $b$ with $\text{ord}(b) = d$?

How many cyclic subgroups of order $d$ exist in $\mathbb{Z}^*_p$?

If you're given an element $h$, its order $d$ and a random element $a$, how can you check if $a$ belong in the subgroup generated by $h$
in polynomial time?

Solution:

$f: \mathbb{Z}_{p-1} \longrightarrow \mathbb{Z}^*_p$ with $f(x) = g^x \pmod{p}$ is an isomorphism. Also, in $\mathbb{Z}_{p-1}$ the element $\frac{p-1}{d}$ is of order $d$. Thus, $g^{\frac{p-1}{d}} \pmod{p}$ is of order $d$ in $\mathbb{Z}^*_p$.

Only one, since $f$ is an isomorphism and $\frac{p-1}{d}$ is the only element of order $d$ in $\mathbb{Z}_{p-1}$

This subgroup has order $d$ and thus $\varphi(d)$ generators (all the coprimes less than $d$)

Only one, with generator $\frac{p-1}{d}$

Check: $h^0 = a$ or $h^1 = a$ or $h^2 = a$ or $\dots$ or $h^{d-1} = a$. If one of the equalities is true, then $a$ belongs in the subgroup generated by $h$. In any other case, it doesn't.

Please, if you can, confirm my solution.


答:
Remember the basic properties of a ring homomorphism: $\Phi$ must be such that

$$\Phi(m+n)=\Phi(m)+\Phi(n)$$

and

$$\Phi(mn)=\Phi(m)\Phi(n)$$

for all $m,n\in\Bbb Z$. Since $1\cdot1=1$, we must have

$$\Phi(1)=\Phi(1\cdot1)=\Phi(1)\Phi(1)\;,$$

i.e., $\Phi(1)^2=\Phi(1)$. The only two integers with the property that $n^2=n$ are $0$ and $1$, so $\Phi(1)$ must be one of them.

Title: How to find all ring homomorphisms from $\mathbb{Z}$ to $\mathbb{Z}$

Body:
I am trying to find all homomorphisms from $\mathbb{Z}$  to    $\mathbb{Z}$. I am looking at a solution, but i do not understand it. Here it is:

Let  $\Phi\colon \mathbb{Z}\to\mathbb{Z}$  be a ring homomorphism. Because $1^2 = 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$. If $\Phi(1) = 1$ then $\Phi(n)= \Phi(n · 1)= n$, so $\Phi$ is the identity map of $\mathbb{Z}$ onto itself which is a homomorphism. If $\Phi(1)= 0$, then $\Phi(n)=\Phi(n · 1)0$, so  maps everthing onto $0$, which also yields a homomorphism.

I do not understand here what "Because $1^2= 1$, we see that $\Phi(1)$ must be an integer whose square is itself, namely either $0$ or $1$.". All proof is built upon this and I do not understand what it means. Can someone help me with this?

Thank you


问:
Axiomatization of physics
I'm studying mathematical physics and I often suffer from the fact that "big theories" in physics never have explicit axioms (quantum mechanics, QFT, GR, ...), or at least fundamental rigorous principles.

I am not talking about the fact that physics aims to describe the real world, but about the fact that the mathematical models used are often not well defined (at least to my actual knowledge).

Is this really the case? Is there some area of math that aims to axiomatise some physical models? Could you recommend some literature?

答:
In Mathematics, you start by axioms. then you choose good definitions and then you prove theorems. in general the result you get don't change with time.

In Physics, you start by Principles. Then you check the result you can predict by experience or test. In some cases, you have to change the principle if some experience give unexpected result.

In Philosophy, you start by Postulates. But every one has own postulates and own conclusions.

Title: Axiomatization of physics


问:
Alternate to Vieta's formula to find $a,b,c,d$ in this question.
Question :

Roots of the equation $$x^{4}+2 x^{3}-5x^{2}+7x+10=0$$ are $\alpha, \beta, \gamma, \delta$ and that of $x^{4}+a x^{3}+b x^{2}+c x+d=0$ be
$\alpha+\beta+\gamma, \alpha+\beta+\delta, \alpha+\gamma+\delta ; \beta+\gamma+\delta,$ then find the value of
$a+b-c-d$.

What I tried:

By applying Vieta's formula(In 2nd Equation):
$$\alpha+\beta+\gamma+\alpha+\beta+\delta+\alpha+\gamma+\delta+\beta+\gamma+\delta=-a$$
$$=3(\alpha+\beta+\gamma+\delta)=-a \tag1\label{eq1} $$
and By applying in Vieta's formula in 1st equation we get:
$$\alpha+\beta+\gamma+\delta=-2 \tag2\label{eq2}$$
From $\eqref{eq1}$ and $\eqref{eq2}$:

$$a=6$$
In similar way, I would find $b,c,d $ which is a very tedious and time consuming method(and it's not guaranteed that answer will come).

Is there any other way?

Hint

(Not complete answer)

Thanks

答: So the roots of the second polynomial $p(x)$ are $-2-\alpha$, $-2-\beta$,...so if we write it in a factor form we have $$p(x)=(x+2+\alpha)(x+2+\beta)...$$ so $$p(-x-2)= (x-\alpha)(x-\beta)...=q(x)$$ where $q(x)$ is a starting polynomial. So $$p(x) = q(-x-2)=...$$


答:
This is far from a complete answer, but it illustrates the basic fact that such axiomatization is very much a work in progress.

One Millennium Prize problem asks us to prove each compact simple gauge group is associated with a certain theory that would be of great interest to physicists. Reading that article, you'll find out about the Wightman axioms, which are one of several choices for axioms that such a theory ought to satisfy. We're not entirely sure which axioms are a sensible choice. If the problem is ever solved, it will likely involve saying, "Let's go with this list of axioms instead".

Why is physics in this situation? I'll only mention one contributing factor.

When a physical theory is shown with either empirical evidence, a thought experiment or a calculation to have a serious problem, physicists patch it to address this problem. Mathematicians aren't generally familiar with a need to do this. They arguably should be, but only in a historical sense; nowadays, "use these axioms and you'll be fine, and never mind the false starts before we got there" is a standard approach in undergraduate or higher mathematics education. When physicists have to "compromise" their axioms (if they even think of it in those terms; to be fair, mathematics itself largely didn't before about 1900, despite Euclid's influence), it's kind of like when Russell's paradox forced mathematicians to patch set theory. How to patch it has been the subject of some controversy ever since!

Mathematical theories often have equivalent choices for their axioms, such that which you'll use depends on what's convenient in your context, especially from the standpoint of pedagogy. In physics, the way problems historically evolve the subject compound this issue. Let's pretend it's 1800 for a second, so electromagnetism, special relativity, quantum mechanics etc. aren't an issue. Do you use Newtonian, Lagrangian, Hamiltonian or Jacobian mechanics? Horses for courses, but they're equivalent. As new physics is added, you often have to jump around between different choices of standpoint for further insights. This is an awkward analogy, but if a choice of axioms were like a basis of a vector space, it would be as if you had to rotate to another basis to get a now useful perspective.

Having said all that, I think you'll be interested to read about specific examples of axiomatisation, though I don't recommend specific texts. You'll want to read about the examples I've mentioned so far, Haag-Kastler and Mach.

Title: Axiomatization of physics

Body:
I'm studying mathematical physics and I often suffer from the fact that "big theories" in physics never have explicit axioms (quantum mechanics, QFT, GR, ...), or at least fundamental rigorous principles.

I am not talking about the fact that physics aims to describe the real world, but about the fact that the mathematical models used are often not well defined (at least to my actual knowledge).

Is this really the case? Is there some area of math that aims to axiomatise some physical models? Could you recommend some literature?


问:
Find the conditional density for $Z_1$ given $Z_2 = 0$
$X_1$ and $X_2$ are independent from Unif(-1,1) and have joint $f(x_1,x_2) = 1/4$ for $-1<x_1<1$ and $-1 < x_2 < 1$. Let $Z_1 = X_1$ and $Z_2 = X_2/(1+X_1^2)$. Find the conditional density of $Z_1$ given $Z_2 = 0$.

I have tried to get the joint density of $f(z_1,z_2) = 1/4*z_1^2 - 1/2*z_1z_2 + 1/4$, so I only need the $f(z_2 = 0)$ to complete this question. But I think the position $Z_2 = 0$ is not existed. How can I get the density for $Z_2$ at $Z_2 = 0$?

答:
But I think the position $Z_2 = 0 $ is not existed

Why? $Z_2 \in(-1;1)$ thus $f_{Z_2}(0)$ exists!

The joint density you calculated doesn't look right to me.

The Jacobian is $1+z_1^2$ thus

$$f_{Z_1Z_2}(z_1,z_2)=\frac{1+z_1^2}{4}$$

It is understood that you have to understand which is the bivariate support of $(Z_1,Z_2)$

It results to me the following

$$\mathbb{1}_{(-1;1)}(z_1)\cdot\mathbb{1}_{\Big(-\frac{1}{1+z_1^2};\frac{1}{1+z_1^2}\Big)}(z_2)$$

That is the area between these two curves

To get the density of $Z_2$ you have only to integrate in $Z_1$


问:
How to find a positive semidefinite matrix satisfying the following condition?
Consider such a problem:
$$\mathbf A\mathbf x=\mathbf b$$
where the vectors $\mathbf x$ and $\mathbf b$ are known.

I want to find a positive semidefinite matrix $\mathbf A$ satisfying the aforementioned equation.

What conditions do $\mathbf x$ and $\mathbf b$ need to satisfy?

Thanks a lot.

答:
The equation $Ax=b$ is solvable by a positive semidefinite $A$ if and only if
$$
b=0\ \text{ or }\ x^\ast b>0.\tag{$\ast$}
$$

For necessity, suppose $A$ is positive semidefinite and $Ax=b$. Then $x^\ast b=x^\ast Ax$ is nonnegative. If it is zero, then for every real scalar $c$, we have
$$
0\le(b-cx)^\ast A(b-cx)=b^\ast Ab - cb^\ast Ax - cx^\ast Ab + c^2x^\ast Ax = b^\ast Ab - 2c\|b\|^2.
$$
Hence $b$ must be zero.

For sufficiency, if $(\ast)$ is satisfied, we can take $A=0$ when $b=0$, or $A=\frac{bb^\ast}{b^\ast x}$ when $x^\ast b>0$.


问:
Protter-Morrey Problem 6.1.17
Protter-Morrey Problem 6.1.17

Let $S$ be a set and $d$ a function from $S \times S$ into $\mathbb{R}^1$ with the properties:

(i) $d(x,y) = 0$ if and only if $x = y$

(ii) $d(x,z) \le d(x,y) + d(z,y)$ for all $x$, $y$, $z$ $\in S$.

Show that $d$ is a metric and hence that $(S,d)$ is a metric space.

The only thing I have not figured out how to derive is that $d(x,y) = d(y,x)$ for all $x$, $y$ $\in S$.

Could someone give me some hint or guidance?

答:
Just pay attention to the order of points: use the triangle inequality twice to get $$d(x,y) \leq d(x,z)+d(y,z)\leq d(x,z) + d(y,x)+d(z,x)$$for all $x,y,z\in S$. Choose $z = x$ to get $d(x,y)\leq d(y,x)$ for all $x,y\in S$. Since $x$ and $y$ were arbitrary, switching their roles gives $d(y,x)\leq d(x,y)$ as well. Thus, we have symmetry.

Lastly, for positivity, we use the triangle inequality again: $$0 = d(x,x) \leq d(x,y)+d(x,y) = 2d(x,y)\implies d(x,y)\geq 0.$$

Title: Protter-Morrey Problem 6.1.17


问:
Unable to represent the function as a power series
Assuming I have this function below :

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

How do I represent it as a power series ?
The only solution that I know is when using the geometric series :
$\sum_{k=1}^\infty q^k = \frac{1}{1-q}$ but this is not possible in our case .

Body:
$$\frac{x-1}{x^2+1}=\frac{x}{x^2+1}-\frac{1}{x^2+1}$$
Integrate
$$\int\left(\frac{x}{x^2+1}-\frac{1}{x^2+1}\right)\,dx=\frac{1}{2}\log(1+x^2)+\arctan x+C$$
Series for these functions are known, we get
$$\frac{1}{2}\log(1+x^2)-\arctan x+C=\frac{x^{10}}{10}-\frac{x^9}{9}-\frac{x^8}{8}+\frac{x^7}{7}+\frac{x^6}{6}-\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}+\frac{x^2}{2}-x+O(x^{11})+C$$
Now differentiate all terms
$$\frac{x-1}{x^2+1}=-1 + x + x^2 - x^3 - x^4 + x^5 + x^6 - x^7 - x^8 + x^9+O(x^{10})$$
bonus

$$\frac{x-1}{x^2+1}=\sum _{n=0}^{\infty } a_n x^n;\quad a_n=3 n^2+n+3-4 \left\lfloor \frac{1}{4} \left(3 n^2+n+4\right)\right\rfloor $$


答: Try to take the sum of two last lines, then extract a factor $(a+b)$ on the left.</p>

Title: How to solve for a constant

Body:
I'm totally stuck at those equations:

$a+b=5$

     $ax+b=11$

     $a+bx=9$

I need to solve for $x$.

I know this is embarrassing but maybe someone can give me a clue...


问:
Concrete Mathmatics: Notation for sums of sums where inner sum depends on outer sum (equation 2.30)

Summary

In Concrete Mathematics (Graham, Knuth, Patashnik) equation 2.30 is as follows

$$
\sum_{j \in J} \sum_{k \in K(j)} a_{j,k} = \sum_{k \in K'} \sum_{j \in J'(k)}a_{j,k}
$$

Does the $K'$ mean then the complement of some set $K$? The book says further down

..., because we can let $J = K'$ be the set of all integers ...

If $K'$ is the set of all integers (along with $J$ if I understand correctly) then what is it the complement of? The non-integers?

答:
Note the sentence that immediately follows (2.30):

Here the sets $J,K(J),K'$, and $J'(k)$ must be related in such a way that $$[j\in J]\big[k\in K(j)\big]=[k\in K]\big[j\in J'(k)\big]\,.$$

In other words, we can reverse the order of summation by rewriting

$$\sum_{j\in J}\sum_{k\in K(j)}a_{j,k}$$

as

$$\sum_{k\in K'}\sum_{j\in J'(k)}a_{j,k}$$

whenever we can find $K'$ and $J'(k)$ for each $k\in K'$ such that

$$[j\in J]\big[k\in K(j)\big]=[k\in K]\big[j\in J'(k)\big]\,.$$

The sets $K'$ and $J'(k)$ for $k\in K'$ are not defined in (2.30): (2.30) simply describes in very general terms a kind of rearrangement of a double summation — reversing the order of summation — that is often very helpful, just as reversing the order of integration in a double iterated integral is often very helpful.

The authors then go on to illustrate a simple and rather common case,

$$\sum_{j=1}^n\sum_{k=j}^na_{j,k}=\sum_{1\le j\le k\le n}a_{j,k}=\sum_{k=1}^n\sum_{j=1}^ka_{j,k}\,,$$

analogous to

$$\int_0^1\int_x^1f(x,y)\,dy\,dx=\int_0^1\int_0^xf(x,y)\,dx\,dy\,.$$


问:
Is it the case that that $E(X_i) = 0 \implies \sum_{i=1}^n E|X_i| < \infty$?
Is it the case that that $E(X_i) = 0 \implies \sum_{i=1}^n E|X_i| < \infty$? If so, what's the proof? Here $X_1, X_2, ...$ are random variables.

I considered Jensen's inequality, i.e. $E|X_i|\geq |E(X_i)|$ but that doesn't help. Also, I "know" it's supposed to be true but I couldn't find a proof.

答: This is true because $\mathsf{E}|X_i|$ is finite for each $1\le i\le n$ (otherwise, either $\mathsf{E}X_i=\pm\infty$ or $\mathsf{E}X_i$ doesn't exist).


问:
How to find the value of L in the figure?
How do I find the value of L in the problem below?

答:
Let's call the angle between the red lines $2\beta$. Then the line connecting the reflection point to the origin is the bisector of the reflection angle. In the triangle origin, center of circle, reflection point, you write the law of sines:
$$\frac{\sin\alpha}R=\frac{\sin\beta}{c_x}$$
Then in the triangle, reflection point, center of circle, $(L,0)$ you have the angle at center $\alpha+\beta$, so the angle at $L$ is $180^\circ-(\alpha+\beta)-\beta$. Once again, write law of sines:
$$\frac{\sin(180^\circ-\alpha-2\beta)}R=\frac{\sin\beta}{L-c_x}$$


问:
Derivative equal to zero at a point
Currently playing around with functions that involve $\sin(\frac{1}{x}).$ And a question came up. Is there any function, which's derivative at a point is equal to $0,$ but this point is not local extrema or point of inflection?

答: Let $f:\mathbb{R}\setminus\{0\}\to\mathbb{R}$ such that $f(x)=\frac{(-1)^{\lfloor x \rfloor}}{x},$ where $\lfloor x \rfloor$ is the value of the floor function at $x.$

Title: Derivative equal to zero at a point


问:
Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$.

Prove that
  $$\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$$

I can't find counting interpretations for either of the sides. A hint of "if $S$ is a subset of $\{1, . . . , n\}$ and $S^\prime$ is its complement then $|S| + |S^\prime| = n$" was also given, but I still don't know how to begin.

答:
This is not really an answer to the question (very good ones have already been given), but to the more daunting challenge of finding a way to actually use the hint given in the question (in an interesting way).

The left hand side  $L=\sum_{i=0}^n\binom nii$ gives the sum of the sizes of all subsets of an $n$-element set, grouping together the $\binom ni$ subsets of size$~i$. Note that I've thrown in the empty set, which makes no difference for this sum, but makes the number of subsets summed over equal to $2^n$. Since taking the complement of all subsets again gives every subset exactly once (the operation is an involution), $L$ also gives the sum of the sizes of the complements of all subsets of an $n$-element set. But if a set has size $i$, then its complement has size $n-i$ (this is where the hint is used!) so this means that $L=\sum_{i=0}^n\binom ni(n-i)$.
Adding up the two summations gives
$$
 2L=\sum_{i=0}^n\binom ni(n+(n-i))=n\sum_{i=0}^n\binom ni=n2^n.
$$
Dividing both sides by$~2$ gives the desired equation $\sum_{i=0}^n\binom nii=L=n2^{n-1}$.


问:
Simple derivative of summation
I am a bit confused as to how the derivative of a summation works. I am trying to calculate the following:

$d/d{x_i} \sum_{i} x_i + cx_i$

Would the answer just be 1 + c then?

Any resources where I can learn more about this?
Thanks!

答: If we're more careful, we write the problem as$$\frac{\partial}{\partial x_i}\left(\sum_jx_j+cx_i\right)=\sum_j\delta_{ij}+c=1+c,$$with $\delta_{ij}$ the Kronecker delta.

Title: Simple derivative of summation


问:
I have difficulty counting
Take this example:

$$84 + 87 + 90 + 93 .. + 180 + 183$$

If we want to use Gauss' way of finding this sum, we have to find the number of elements in this. What I do is just take the difference of the highest and lowest sum (99) and divide by 3, but this yields 33 and that can't be correct because it needs to be an even number in order to make sense. Why does it count 34 elements instead of the 33 my (faulty) arithmetic gives me?

Title: I have difficulty counting

Body:
Every number in your series is divisible by 3, so we can write it as:

$$84+87+90+93..+180+183 = 3(28+29+...+60+61)$$

Know using Gauss formula we can get:

$$3(28+29+...+60+61) = 3(\frac{61*62}{2} - \frac{27*28}{2})$$

We can simplify this to:

$$3(\frac{34*62 + 27*62 - 27*28}{2}) = $$
$$3(\frac{34*62 + 27(62-28)}{2}) = $$
$$3(\frac{34*62 + 27*34}{2}) = $$
$$3(\frac{34(62+27)}{2}) = $$
$$3(\frac{34*89)}{2}) = $$
$$\frac{102*89}{2} = 4539$$


问:
The random variable $Y=F\left( X\right)$ is uniformly distributed on the interval $\left( 0,1\right)$
This is a problem from "Probability: An Introduction", by Grimmett & Welsh, Problem 5.8.5. I am having some troubles in prove it, because I don't know if it can go straight out of the definition.

Let $X$ be a random variable whose distribution function $F$ is a continuous function. Show that the random variable $Y=F\left( X\right)$ is uniformly distributed on the interval $\left( 0,1\right)$.

The book sugests to call $F^{-1}\left(y \right) = \sup\left\{x: F\left(x \right) = y\right\}$, but I cant proceed.

答:
$$F_Y(y)=\mathbb{P}[Y\leq y]=\mathbb{P}[F_X(x)\leq y]=\mathbb{P}[X\leq F_X^{-1}(y)]=F_X[F_X^{-1}(y)]=y$$

That's all!


问:
Proof explanation of an inverse function theorem

Let $I\neq\emptyset$ be an open interval and $f:I\to\mathbb{R}$ a continuous and injective function. If $f$ is differentiable at $x_0\in I$ and $f'(x_0)\neq 0$, then $f^{-1}$ is differentiable at $f'(x_0):=y_0$ and we have that $$ (f^{-1})'(y_0)=\frac{1}{f'(x_0)}=\frac{1}{f'(f^{-1}(y_0))}.$$

Proof:

Note that $f(I)$ is an interval, $f:I\to\mathbb{R}$ is an homomorphism, and $f^{-1}:J\to I$ is continuous.
Now, $f$ and $f^{-1}$ are strictly monotonic (by a a lemma we previously proved).

Let $(y_n)\in J\setminus\{y_0\}$ be any sequence such that $y_n\to y_0$, and let $x_n=f^{-1}(y_n)$.

The continuity and injectivity of $f^{-1}$ implies that $(x_n)$ is a sequence in $I\setminus\{x_0\}$ such that $x_n\to x_0$.
Since $f$ is differentiable at $x_0$ and $f^{-1}(x_0)\neq 0$, then we have $$\displaystyle\lim_{n\to\infty}\frac{f^{-1}(x_n)-f^{-1}(x_0)}{y_n-y_0}=\displaystyle\lim_{n\to\infty}\frac{x_n-x_0}{f(x_n)-f(x_0)}=\frac{1}{f'(x_0)}.$$

What I don't understand is how the continuity and injectivity of $f^{-1}$ implies that $(x_n)$ is a sequence in $I\setminus\{x_0\}$ such that $x_n\to x_0$, and how to go from $\displaystyle\lim_{n\to\infty}\frac{x_n-x_0}{f(x_n)-f(x_0)}$ to $\frac{1}{f'(x_0)}$. Any help is welcome

答:
Well, each $x_n = f^{-1}(y_n)$. So since $f^{-1} : J \to I$, we at least have $x_n \in I$. Also, each $y_n \neq y_0$ (because $y_n$ was chosen from $J \setminus \{y_0\}$). So, by definition of injectivity,
$$
x_n = f^{-1}(y_n) \neq f^{-1}(y_0) = x_0
$$ Thus, $x_n \neq x_0$ and so $x_n \in I \setminus \{x_0\}$. Finally, because $f^{-1}$ is continuous, it commutes with limits:
$$
x_0 = f^{-1}(y_0) = f^{-1}(\lim_{n \to \infty}y_n) = \lim_{n \to \infty}f^{-1}(y_n) = \lim_{n \to \infty}x_n
$$ So $x_n \to x_0$.

Next, let $L := f'(x_0)$ and let $D(x) := \frac{f(x) - f(x_0)}{x - x_0}$. This is a function $D : I \setminus \{x_0\} \to \mathbb{R}$. $x_0$ is not in the domain of $D$ but it is still an adherent point of that domain. So $\lim\limits_{x \to x_0; x \in I \setminus \{x_0\}}D(x)$ makes sense and, in fact, this limit is just $L$. In other words, $D(x)$ converges to $L$ at $x_0$ in $I \setminus \{x_0\}$. Or equivalently, by the sequential criterion for convergence:

If $x_n \to x_0$ where $x_n$ is in the domain of $D$ (i.e. $I \setminus \{x_0\}$), then $D(x_n) \to L$.

As a result, $\lim\limits_{n \to \infty} D(x_n) = L$. So, as long as $D(x_n) \neq 0$ and $L \neq 0$, we also have
$$
\lim_{n \to \infty} \frac{x_n - x_0}{f(x_n) - f(x_0)} = \lim_{n \to \infty} \frac{1}{D(x_n)} = \frac{1}{L} = \frac{1}{f'(x_0)}
$$ by elementary limit laws for sequences. But lo and behold, we indeed have $L = f'(x_0) \neq 0$ and $D(x_n) = \frac{f(x_n) - f(x_0)}{x_n - x_0} \neq 0$ because the numerator is non-zero: $f(x_n) = y_n \neq y_0 = f(x_0)$.


问:
Understanding the RSI Formula - $RSI = {\color {red}{100} - { \color {red}{100} \over {\color {red} 1} + {n_{up} \over n_{down}} }}$
RSI (Relative Strength Index) is an Oscillator Formula used in trading to detect overbought/oversold conditions. High values suggests overbought and low values suggest oversold conditions in the market.

While I understand what the denominator shown in the formula below does, I am having trouble understanding what the values highlighted in red does to the overall picture from a mathematical standpoint.

$$RSI = {\color {red}{100} -  { \color {red}{100} \over {\color {red} 1} + {n_{up} \over n_{down}}  }}$$

Any help is appreciated. Thanks.

答:
This is the percentage of moves which have been up rather than down.

The proportion up rather than down is clearly $\dfrac{n_{up}}{n_{up}+n_{down}}$ and gives something between $0$ and $1$.  Multiply it by $100$ if you want a percentage between $0$ and $100$ so $100\dfrac{n_{up}}{n_{up}+n_{down}}$.

The rest is manipulation: $$100\dfrac{n_{up}}{n_{up}+n_{down}} = 100-100\dfrac{n_{down}}{n_{down}+n_{up}} = 100-100\dfrac{1}{1+\frac{n_{up}}{n_{down}}}$$ though it is not obvious why anyone would do this.  Perhaps because technical analysis is little more than astrology.


问:
Solving an ODE that resulted from method of characteristics
I am trying to follow a development in which an ODE is solved.  The equation arises from using the method of characteristics to solve a PDE. We have $\dot{x}=v$. The ODE is
$$
m(x) v \frac{dv}{dx} = F(x) - v^{2} \frac{d}{dx}(m(x))
$$
with the substitution $\xi = v^{2}$ we have the following expression
$$
m(x) \frac{d \xi}{d x} + 2 \xi \frac{d}{dx}(m(x)) = 2 F(x)
$$
The solution should be

$$
C = \frac{m^{2}(x)}{2} v^{2} - \int m(x) F(x) dx
$$
But I cannot get to that solution. My first problem arises by not being able to identify what kind of ODE this is.
I hope you can help me get there.

Body:
$$m(x) \frac{d \xi}{d x} + 2 \xi \frac{d}{dx}(m(x)) = 2 F(x)$$
$$m(x) d \xi + 2 \xi d(m(x)) = 2 F(x)dx$$
Multiply by $m(x)$:
$$m^2(x) {d \xi} + 2m(x) \xi d(m(x)) = 2m(x) F(x)dx$$
$$m^2(x) {d \xi} +  \xi d(m^2(x)) = 2m(x) F(x)dx$$
$$d \xi m^2(x) = 2m(x)  F(x)dx$$
Integrate.
$$ \xi m^2(x) =C+ 2\int m(x)  F(x)dx$$
$$ C= (vm(x))^2 -2\int m(x)  F(x)dx$$


问:
How do I define Injective/Surjective functions in terms of sets and not the elements within them?
I am in a Proof writing class and we are currently on functions. I have a good understanding of what it means to be injective or surjective in terms of elements in the set but I am having trouble coming up with a formal definition that just generalizes the definitions in terms of sets using an iff statement.

So if we have a function  $f: A\rightarrow B$

Would my statement be considered a proper theorem?

If $G\subseteq B$ then  $f: A\rightarrow B$  is surjective if and only if  $G=B$ and $f(f^{-1}( B)) = B$.

My reasoning here being that if our function is surjective everything in B has to have some pre-image in A. I am not sure If I need to say anything about A.

For Injective I came up with a similar theorem:

If $H\subseteq A$ then $f: A\rightarrow B$ is injective if and only if $A=H$ and $f(f^{-1}(A))=A$

My reasoning again being we need everything in our domain to have some image in B. So our domain subset must take on the whole set A.

Your help and knowledge would be extremely appreciated!

答:
For the surjective definition, the $G$ is irrelevant. It just must verify that $f(f^{-1}(B))=B$, or simpler, just that $f(A)=B$.

For the injective definition, you can write it as:

$\forall H,N\subseteq A$, $f(H)=f(N) \implies H=N$.


问:
Exponential Sums
This is my first time on the Mathematics Stack Exchange. I am currently studying for my IIT entrance exam, and while studying and browsing the web, I came across an interesting problem.

Let $a,b,c,d \ge 0$. Find a closed form for the expression: $$\sum_{a+b+c+d = n} 2^{a+2b+3c+4d}$$ in terms of $n$.

At the bottom, it said that a possible hint might be that I should use generating functions. I've been working for the past couple of hours and can't seem to figure out what the generating function of this expression would be.

Any help would be greatly appreciated!

Thank you,
Rohit

答:
Hint: If you multiply out $$\dfrac{1}{1-2x} \cdot \dfrac{1}{1-2^2x} = \left(\sum_{a = 0}^{\infty}2^ax^a\right) \cdot \left(\sum_{b = 0}^{\infty}2^{2b}x^b\right),$$ the $x^n$ coefficient is the sum of $2^a \cdot 2^{2b} = 2^{a+2b}$ over all integers $a,b \ge 0$ such that $a+b = n$.

Can you extend this idea to your problem?

Title: Exponential Sums


问:
How do I show the the integral exists?
How do I prove the limit of Lebesgue integral exists and find its value: $\lim_{n \rightarrow \infty} \int_{0}^{\infty} f_n(x)\,dx. $ where $f_n(x)=e^{-x}\cos(x/n)$.

My attempt: I try to apply the Dominated convergence theorem:

i) $|f_{n}(x)|=|e^{-x}\cos(\frac{x}{n})|\leq e^{-x}$.

ii) $\lim_{n \rightarrow \infty} f_n(x)=\lim_{n \rightarrow \infty} e^{-x}\cos(\frac{x}{n})= e^{-x}$.

Then $f$ is integrable, and $\lim_{n \rightarrow \infty} \int_{0}^{\infty} f_n(x)\,dx=\lim_{n \rightarrow \infty} \int_{0}^{\infty} f(x)\,dx$.

$\lim_{n \rightarrow \infty} \int_{0}^{\infty} e^{-x}\cos(\frac{x}{n})\,dx= \lim_{n \rightarrow \infty} \int_{0}^{\infty} e^{-x}\, dx=1$.

Is this attempt correct? Can I say the integral exist because $f$ is integrable?

答:
You are correct. As a remark, you can solve this (and verify your solution) explicitly.

Let $I_n=\int_0^\infty f_n(x)\,dx=\int_0^\infty e^{-x}\cos(x/n)\,dx$. You can actually evaluate this using the bounce back technique. Let $u=\cos(x/n)$, $dv=e^{-x}dx \implies v=-e^{-x}$. Then
\begin{align}
I_n&=uv\Big|_{0}^\infty-\int_0^\infty v\,du \\
&=1-\frac{1}{n}\int_0^\infty \sin(x/n)e^{-x}\,dx \\
&\stackrel{IBP}{=}1-\frac{1}{n^2}\int_0^\infty\cos(x/n)e^{-x}\,dx \\
&=1-\frac{1}{n^2}I_n
\end{align}
This implies
$$
I_n=\frac{1}{1+1/n^2}.
$$

Title: How do I show the the integral exists?


问:
Prove that if P(E|F1) = P(E|F2) = P(E|G) for some F1, F2 ⊆ G, then P(E|F1 ∪ F2) = P(E|G).
Im struggling with proving the following identity any help would be great.</p>

答:
It is not obviously true.

Consider as a counterexample the following equally probable cases with indicators:

```<code>E  F1 F2 G Probability
0  1  0  1    0.2
1  1  0  1    0.1
1  1  1  1    0.1
1  0  1  1    0.1
0  0  1  1    0.2
1  0  0  1    0.1
1  0  0  0    0.1
0  0  0  0    0.1
</code>```

Then $F_1 \subseteq G$ and $F_2 \subseteq G$,

and $\mathbb P(E \mid F_1)=\frac{0.2}{0.4} = \frac12$, $\mathbb P(E \mid F_2)=\frac{0.2}{0.4} = \frac12$,  $\mathbb P(E \mid G)=\frac{0.4}{0.8} = \frac12$,

but $\mathbb P(E \mid F_1 \cup F_2)=\frac{0.3}{0.7} = \frac37 \not = \frac12$

Body:
It is not obviously true.

Consider as a counterexample the following equally probable cases with indicators:

E  F1 F2 G Probability
0  1  0  1    0.2
1  1  0  1    0.1
1  1  1  1    0.1
1  0  1  1    0.1
0  0  1  1    0.2
1  0  0  1    0.1
1  0  0  0    0.1
0  0  0  0    0.1

Then $F_1 \subseteq G$ and $F_2 \subseteq G$,

and $\mathbb P(E \mid F_1)=\frac{0.2}{0.4} = \frac12$, $\mathbb P(E \mid F_2)=\frac{0.2}{0.4} = \frac12$,  $\mathbb P(E \mid G)=\frac{0.4}{0.8} = \frac12$,

but $\mathbb P(E \mid F_1 \cup F_2)=\frac{0.3}{0.7} = \frac37 \not = \frac12$


问:
Formal counting of cube faces colorings wrt full symmetry
How many 3-colorings of cube's faces exist if we consider two colorings the same iff it's possible to rotate and/or mirror the cube such that one coloring goes to another?

I do know a similar question considering only the rotations was asked here. I do understand where the solution comes from: Burnside's lemma and the geometrical classification of cube rotations (by face, by edge, by vertex, etc). But I can't seem to construct a convenient geometrical classification of all the symmetries (is there really a sensible geometrical approach to this problem???)

I sense that the full symmetry group of a cube is $S_4 \times \mathbb Z_2$ since we can fix an arbitrary mirroring and combine it with all the rotations of the cube (which form a group isomorphic to $S_4$). I have heard of a solution which uses an embedding $S_4 \times \mathbb Z_2 \rightarrow S_6$ which "realizes all cube symmetries in permutations of cube's faces". I though have no idea how to build such an embedding and how can it help?

Basically I am looking for any approach to this problem which doesn't involve calculating $\text{Fix}(g)$ for all of the 48 cube symmetries $g$ separately and plugging it in Burnside's lemma (this is going to be on a test with limited time, so that is not acceptable).

答:
The rotation group permutes the four diagonals, whence $S_4$. The rotations and index two in the full subgroup, since reflection x reflection = rotation, and the central involution $-I_3$ commutes with everything so we find the full symmetry group is $S_4\times\mathbb{Z}_2$. Anyway, the conjugacy classes of rotations in $S_4$ are:

$1~$ - $~0^{\circ}$ rotation

$6~$ - $~90^{\circ}$ rotations around face axes

$8~$ - $120^{\circ}$ rotations around vertex axes

$3~$ - $180^{\circ}$ rotations around face axes

$6~$ - $180^{\circ}$ rotations around edge axes

It's also a good exercise to figure out the cycle types in $S_4$ corresponding to each item above.

The other possibilities are plane reflections and rotation-reflections (which are reflections across planes combined with rotation in said planes). Here's a list:

$3$ coordinate plane reflections

$3$ coordinate plane reflections + $180^{\circ}$ rotation

$6$ coordinate plane reflections + $90^{\circ}$ rotation

$6$ antipodal-edge plane reflections

$6$ antipodal-edge plane reflections + $180^{\circ}$ rotation

Finding $\mathrm{Fix}(g)$ for these $g$s should be doable.


答:
It's not very clear exactly what you are looking for, but the following are very useful "element-free" characterisations of injectivity and surjectivity:

$f : A \to B$  is injective iff it has a left inverse, i.e, iff there is a function $g: B  \to A$ such that $g \circ f = f$ (i.e., $g(f(x)) = x$ for all $x \in A$).

$f : A \to B$  is surjective iff it has a right inverse, i.e, iff there is a function $g: B \to A$ such that $f \circ g = f$ (i.e., $f(g(y)) = y$ for all $y \in B$).

Title: How do I define Injective/Surjective functions in terms of sets and not the elements within them?

Body:
I am in a Proof writing class and we are currently on functions. I have a good understanding of what it means to be injective or surjective in terms of elements in the set but I am having trouble coming up with a formal definition that just generalizes the definitions in terms of sets using an iff statement.

So if we have a function  $f: A\rightarrow B$

Would my statement be considered a proper theorem?

If $G\subseteq B$ then  $f: A\rightarrow B$  is surjective if and only if  $G=B$ and $f(f^{-1}( B)) = B$.

My reasoning here being that if our function is surjective everything in B has to have some pre-image in A. I am not sure If I need to say anything about A.

For Injective I came up with a similar theorem:

If $H\subseteq A$ then $f: A\rightarrow B$ is injective if and only if $A=H$ and $f(f^{-1}(A))=A$

My reasoning again being we need everything in our domain to have some image in B. So our domain subset must take on the whole set A.

Your help and knowledge would be extremely appreciated!


问:
Is the Oval ( based on Ptolemy inequality) known?
It has a property of enclosing quadrilaterals so the ratio of their diagonals product  and sum of their opposite sides pair  products is constant $(e<1)$. The curve is from a family defined by the Ptolemy Inequality

In order to rope in the Ptolemy Inequality Oval took three points on a unit generating circle radius $ a=1 $ and the fourth one outside the circle

$$(-1,0),(0,-1),(1,0),(x,y)$$

as particular vertices of a non-cyclic  quadrilateral. The ratio $e$ defines its equation.

$$ \dfrac{\sqrt 2 \sqrt{x^2+(1+y)^2}}{\sqrt{y^2+(x+1)^2} + \sqrt{y^2+(x-1)^2}} =e<1 \tag 1 $$

Special case  $e=1$ is the circle enclosing cyclic quadrilaterals that have the property given by Ptolemy theorem. A set of non-cyclic quadrilaterals can be inscribed in this oval shape. In this drawing $ e=0.95; $

Some shapes for Other $e$ values

Further simplification yields a fourth degree algebraic curve:

$$\left(-a^4-2 a^3 y+a^2 \left(2 \left(e^2-1\right) x^2-2 y^2\right)-2 a y \left(x^2+y^2\right)-\left(x^2+y^2\right)^2\right)+\frac{\left((a+y)^2+x^2\right)^2}{2 e^2}=0$$

答:
Not another answer. For comments/ achille hui with sketch:

```<code>h = 1.4; a = .5; b = 0.3; A = 1;
mirror = ParametricPlot[{A Cos[t], A Sin[t]}, {t, 0, 2 Pi}];
ell = ParametricPlot[{h + a Cos[t], b Sin[t]}, {t, 0, 2 Pi},
   PlotStyle -> {Thick, Purple}];
Oval = ParametricPlot[
   A^2 {h + a Cos[t], b Sin[t]}/((h + a Cos[t])^2 + (b Sin[t])^2), {t,
     0, 2 Pi}, PlotStyle -> {Thick, Red}];
Show[{ell, Oval, mirror}, PlotRange -> All, GridLines -> Automatic]
</code>```

Body:
Not another answer. For comments/ achille hui with sketch:

h = 1.4; a = .5; b = 0.3; A = 1;
mirror = ParametricPlot[{A Cos[t], A Sin[t]}, {t, 0, 2 Pi}];
ell = ParametricPlot[{h + a Cos[t], b Sin[t]}, {t, 0, 2 Pi},
   PlotStyle -> {Thick, Purple}];
Oval = ParametricPlot[
   A^2 {h + a Cos[t], b Sin[t]}/((h + a Cos[t])^2 + (b Sin[t])^2), {t,
     0, 2 Pi}, PlotStyle -> {Thick, Red}];
Show[{ell, Oval, mirror}, PlotRange -> All, GridLines -> Automatic]


问:
$\{a_1,...,a_n,b\}$ is discrete if and only if $t_j \in \mathbb{Q}$

Let $G= \mathbb{R}^{n}$ with the usual topology. If $\{a_1,...,a_n\} \subset \mathbb{R}^n$ is linearly independent and $b=\sum_{j=1}^n t_ja_j$ show that the subgroup generated by $\{a_1,...,a_n,b\}$ is discrete if and only if $t_j \in \mathbb{Q}$ for each $j=1,2,...,n$.

Let $G$ be a topological group and $ \Gamma $ a subgroup of $ G $. $ \Gamma $ is a discrete space if and only if $ \Gamma $ has an isolated point.

Let $\Gamma$ be the subspace generated by $\{a_1,...,a_n,b\}$ and suppose that $t_j \in \mathbb{Q}$, my idea is to show that $ b $ is an isolated point  and conclude with the previous result, it is possible to show that $ b $ is an isolated point?

Any help, I would appreciate it.

答:
Here's 95% of the details. You can check the other 5% yourself.

First, take the linear transformation $T$ which maps $a_i$ to the $i$-th standard basis vector. Then $T$ is a homeomorphism, and from that it follows that $\langle a_1, \dots, a_n, b \rangle$ is discrete, if and only if $\langle e_1, \dots, e_n, Tb \rangle$ is discrete.

Now, assume $Tb = (t_1, \dots, t_n) \in \mathbb{Q}^n$. If we write the numerators and denominators as $t_i = p_i/q_i$ then
$$\langle e_1, \dots, e_n, Tb \rangle \subseteq \left\langle \frac{e_1}{q_1},\dots,\frac{e_n}{q_n}, Tb \right\rangle \cong \langle e_1,\dots,e_n,(p_1,\dots,p_n)\rangle = \langle e_1,\dots,e_n\rangle = \mathbb{Z}^n.$$

Where for the isomorphism, we map $e_i \mapsto q_ie_i$.

Conversely, suppose that some $t_j \notin \mathbb{Q}$. Actually, for convenience's sake, let's suppose that $t_1,\dots,t_m \notin \mathbb{Q}$ and $t_{m + 1} + \dots + t_n \in \mathbb{Q}$. Since

$$\mathbb{Z}^n = \langle e_1,\dots,e_n \rangle \subset \langle e_1, \dots, e_n, Tb \rangle,$$
we can speak of the fractional part of a vector in this group. If $x$ is in this group, let us denote its fractional part by $\{x\}$.

Consider the sequence $x_k = \{k (Tb)\}$. In the $m+1$-st through $n$-th coordinates, $x_k$ is periodic. In the $1$-st through $m$-th coordinates, $x_k$ is aperiodic. Therefore the sequence $x_k$ represents infinitely many points in the unit cube $[0,1]^n$.

So now if we cover $[0,1]^n$ by finitely many balls of radii $\varepsilon$ then by the pigeonhole principle, one of these balls must contain at least two elements of our sequence. Thus $0 < |x_k - x_l| < \varepsilon$ and it follows that $\vec{0}$ is not an isolated point.

See also:

Why is this quotient space not Hausdorff?

Multiples of an irrational number forming a dense subset


问:
Diagonalising a $2 \times 2$ and $3 \times 3$ matrix

For each of the following matrices $A$, find an invertible matrix $P$ over $C$ such that
  $P^{-1}AP$ is upper triangular:

$$A = \begin{bmatrix}4 & 1\\-1 & 2\end{bmatrix} \quad \text{ and } \quad A = \begin{bmatrix}1 & 1 & 1\\1 & 1 & 1\\0 & 0 & 0\end{bmatrix}.$$

For the first matrix, I calculate the eigenvalue to be $3$, with algebraic multiplicity $2$.
However, I can only find one eigenvector for it, namely $\begin{bmatrix}1\\-1\end{bmatrix}$.

The second has two eigenvalues, $0$ (multiplicity $2$) and $2$ (multiplicity $1$)
But again, for the eigenvalue $0$, I can't find two linearly independent eigenvectors.

So, how do I proceed to find $P$ in these cases?

答:
For the first matrix, your work is all correct.

We need to find a generalized eigenvector.

One approach to this (did you learn why in class), is to setup:

$$[A - \lambda I]v_2 = v_1$$

I am going to write the eigenvector with the signs swapped for the first one.

We have $v_1 = (-1, 1)$, so we would get:

$$[A -\lambda I]v_2 = v_1 \rightarrow \begin{bmatrix}1 & 1\\-1 & -1\end{bmatrix}v_2 = \begin{bmatrix}-1\\1\end{bmatrix}$$

Solving this (use RREF), we get a second linearly independent eigenvector of:

$$v_2 = (-1, 0)$$

The matrix $P$ is formed as linear combination of the eigenvectors as $[v_1 | v_2]$, so we get:

$$P = [v_1 | v_2] = \begin{bmatrix}-1 & -1\\1 & 0\end{bmatrix}$$

This will lead to the Jordan Normal Form for the upper triangular matrix as:

$$J = P^{-1}AP = \begin{bmatrix}3& 1\\0 & 3\end{bmatrix}$$

Also note, there are other methods available and sometimes required over the method outlined above.

Can you use the method above for your second problem and see if it bears fruit?

For example, we find two eigenvalues of $\lambda_1 = 2 and \lambda_{2,3} = 0$. For the $\lambda_{2,3} eigenvalue, we will have a row-reduced-echelon-form of:

$$a+b+c = 0$$

We need two linearly independent choices and have a lot of free variables to choose from. For example, we can choose:

$$v_2 = (-1, 0, 1), v_3 = (-1, 1, 0)$$

For the $\lambda_1=2$ eigenvalue, we form:

$$[A-\lambda_1 I]v1 = 0$$.

After RREF that matrix, we arrive at: $v_1 = (1,1,0)$

Recall, you have to do this process over for EACH eigenvalue!

So, to summarize, we should arrive at (no generalized eigenvectors needed because of geometric and algebraic multiplicity):

$$\lambda_1 = 2, v_1 = (1, 1, 0)$$

$$\lambda_2 = 0, v_2 = (-1, 0, 1)$$

$$\lambda_3 = 0, v_3 = (-1, 1, 0)$$


问:
Find a homomorphism $\phi: \Bbb C \to \Bbb R^*$ such that $\phi(i) = 2$.

Find a homomorphism $\phi: \Bbb C \to \Bbb R^*$ such that $\phi(i) = 2$.

I can't think of anything here, I can only get $\phi(a+bi) = 2^b,$ but I'm pretty sure I can't just extract the imaginary part like that.

If it was $\mathbb C^*$ instead of $\mathbb C$ then I know I can use absolute value / magnitude somehow but in this case it wouldn't be a homomorphism.

答:
You can.

Let $x=a+ib,y=c+id\in\Bbb C$. Then

$$\begin{align}
\phi(x+y)&=\phi((a+c)+i(b+d))\\
&=2^{b+d}\\
&=2^b2^d\\
&=\phi(a+ib)\phi(c+id)\\
&=\phi(x)\phi(y).
\end{align}$$

Thus $\phi:\Bbb C\to \Bbb R^*$ is indeed a homomorphism.

Moreover, $\phi(i)=\phi(0+i1)=2^1=2$.


问:
Show that $x^4+x+1$ is a prime polynomial over $F_2$
To prove this, the only idea I had was to divide $x^4+x+1$ by all polynomials over $F_2$ and show that it cannot be evenly divided.

This seems tedious and I am sure there is a more elegant way to prove this. Can anyone give hints or guide me towards a proof?

答: Hint If a quartic polynomial factors over some field, then it has a linear factor and/or an irreducible quadratic factor. Over $F_2$ there are only two (strictly) linear polynomials and only one irreducible quadratic polynomial, namely $x^2 + x + 1$, and these facts together reduce the checking to just a few cases.


问:
Finding PMF of a zero-inflated Poisson random variable
A Zero-Inflated Poisson r.v. $X$ with parameter $p$ and $\lambda$ can be generated as follows. First flip a coin with probability of $p$ of Heads. Given that the coin lands Heads, $X = 0$. Given that the coin lands Tails, $X$ is distributed Pois($\lambda$). Note that if $X = 0$ occurs, there are two possible explanations: the coin could have landed Heads (in which case the zero is called a structural zero), or the coin could have landed Tails but the Poisson r.v. turned out to be zero anyway.
For example, if $X$ is the number of chicken sandwiches consumed by a random person in a week, then $X = 0$ for vegetarians (this is a structural zero), but a chicken-eater could still have $X = 0$ by chance (since they might not happen to eat any chicken sandwiches that week).

a) Find the PMF of a zero-inflated Poisson r.v. $X$.

b) Find the mean and variance of $X$.

Hint: Use the following facts from Calculus: $$\mathbb\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}=e^\lambda\,,$$
$$\mathbb\sum_{k=0}^{\infty}\frac{(k+1)\lambda^k}{k!}=e^\lambda+\lambda e^\lambda$$

For part A I got, $$P(X = 0) = p + (1 - p)\frac{\lambda^0}{0!}e^{-\lambda} = p + (1 - p)e^{-\lambda}$$ then for $k$ greater than or equal to $1$, $$P(X = k) = k(1 - p)\frac{\lambda^k}{k!}e^{-\lambda}$$

For part B I got, $$E(X^2) = E((1 - I)^2Y^2) = E((1 - I)^2)E(Y^2) = (1 - p)(\lambda + \lambda^2)$$ then $$E(Y^2) = Var(Y) + E(Y)^2 = (\lambda + \lambda^2)$$ and finally, $$Var(X) = E(X^2) - E(X)^2 = (1 - p)(\lambda + \lambda^2) - (1 - p)^2\lambda^2$$

My answers seem off to me but I am not exactly sure where I am off. Any help would be much appreciated.

答:
The answer you obtained for Part A is almost correct; you have an extra factor of $k$ when $X \ne 0$.  To see why, another way to write the zero-inflated Poisson model is to let $X = (1-B) N$, where $B \sim \operatorname{Bernoulli}(p)$, and $N \sim \operatorname{Poisson}(\lambda)$.  Then $$\Pr[X = x] = \Pr[X = x \mid B = 0]\Pr[B = 0] + \Pr[X = x \mid B = 1]\Pr[B = 1]$$ by the law of total probability.  This simplifies to $$\Pr[X = x] = \Pr[N = x](1-p) + \Pr[0 = x]p.$$  Then we have $$\Pr[X = x] = e^{-\lambda} \frac{\lambda^x}{x!} (1-p) + \mathbb 1(x=0)p$$ where $$\mathbb 1(x=0) = \begin{cases}1, & x = 0 \\ 0, & x \ne 0. \end{cases}$$  Writing this out explicitly, $$\Pr[X = x] = \begin{cases} e^{-\lambda} (1-p) + p, & x = 0 \\ e^{-\lambda}\frac{\lambda^x}{x!} (1-p), & x \ne 0. \end{cases}$$  When you add up the probability of all possible values of $X$, you should get $1$:  $$\begin{align}
\sum_{x=0}^\infty \Pr[X = x]
&= \Pr[X = 0] + \sum_{x=1}^\infty \Pr[X = x] \\
&= p + (1-p) \left( e^{-\lambda} + \sum_{x=1}^\infty e^{-\lambda} \frac{\lambda^x}{x!} \right) \\
&= p + (1-p)(1) \\
&= 1.
\end{align}$$

For Part B, I would use the laws of total expectation and total variance.  In particular:
$$\operatorname{E}[X] = \operatorname{E}[\operatorname{E}[(1-B)N \mid B]] = \operatorname{E}[(1-B)\operatorname{E}[N]] = \operatorname{E}[(1-B)\lambda] = \lambda \operatorname{E}[1-B] = \lambda(1-p),$$
and
$$\begin{align}
\operatorname{Var}[X]
&= \operatorname{E}[\operatorname{Var}[X \mid B]] + \operatorname{Var}[\operatorname{E}[X \mid B]] \\
&= \operatorname{E}[\operatorname{Var}[(1-B)N \mid B]] + \operatorname{Var}[\operatorname{E}[(1-B)N \mid B]] \\
&= \operatorname{E}[(1-B)^2 \operatorname{Var}[N]] + \operatorname{Var}[(1-B)\operatorname{E}[N]] \\
&= \operatorname{E}[(1-B)^2 \lambda] + \operatorname{Var}[(1-B)\lambda] \\
&= \lambda \operatorname{E}[1-2B+B^2] + \lambda^2 \operatorname{Var}[1-B] \\
&= \lambda (1 - 2p + p) + \lambda^2 \operatorname{Var}[B] \\
&= \lambda (1-p) + \lambda^2 p(1-p) \\
&= \lambda (1-p) (1 + \lambda p).
\end{align}$$


问:
What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?
Here is my candidate for
the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists.
I would be interested in seeing others.

$***$ Added after some comments:
I prove here by very elementary means that the limit exists.
Calling the limit "$e$" names it.
$***$

It only needs Bernoulli's inequality (BI)
in the form
$(1+x)^n \ge 1+nx$
for $x > -1$ and
$n$ a positive integer,
with equality only if
$x = 0$ or $n = 1$.
This is easily proved by induction:
It is true for $n=1$,
and $(1+x)^{n+1}
= (1+x)(1+x)^n
\ge (1+x)(1+nx)
= 1+(n+1)x+nx^2
\ge 1+(n+1)x
$.
(If $-1 < x < 0$,
if $1+nx \ge 0$,
the above proof goes through, and if
$1+nx < 0$,
$1+mx < 0$ for all $m \ge n$
so certainly $(1+x)^m > 1+mx$.)

This proof originally appeared in
N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563
and uses the
arithmetic-geometric mean inequality (AGMI)
in the form
$\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$
 (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.

Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.
We will prove that $a_n$ is an increasing sequence
 and $b_n$ is an decreasing sequence.
Since $a_n < b_n$, this implies,
for any positive integers $n$ and $m$ with $m < n$
 that $a_m < a_n < b_n < b_m$.

For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.
Since their sum is $n+2$
and their product is $(1+1/n)^n$,
by the AGMI,
$((n+2)/(n+1))^{n+1} > (1+1/n)^n$,
 or $(1+1/(n+1))^{n+1} > (1+1/n)^n$,
or $a_{n+1} > a_n$.
For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.
Since their sum is $n$
and their product is $(1-1/n)^n$,
by the AGMI,
$(n/(n+1))^{n+1} > (1-1/n)^n$
 or $(1+1/n)^{n+1} < (1+1/(n-1))^n$,
 or $b_n > b_{n+1}$.

Since $b_n-a_n
= (1+1/n)^{n+1} - (1+1/n)^n
= (1+1/n)^n(1/n)
=a_n/n
$
and every $a_n$ is less than any $b_n$
and $b_3 = (1+1/3)^4 = 256/81 < 4$,
$b_n-a_n < 4/n$,
so $b_n$ and $a_n$ converge to a common limit.

These proofs do not seem to be really elementary,
 since they use the AGMI.
However, they use a special form of the AGMI,
where all but one of the values are the same,
and this will now be shown to be implied by BI,
 and thus be truly elementary.

Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.
The AGMI for these values is
$(((n-1)u+v)/n)^n \ge u^{n-1}v$
with equality if and only if $u = v$.
We will now show that this is implied by BI:
$(((n-1)u+v)/n)^n \ge u^{n-1}v$
 is the same as
$(u+(v-u)/n)^n \ge u^n(v/u)$.
Dividing by $u^n$,
this is equivalent to
$(1+(v/u-1)/n)^n \ge v/u$.
By BI,
since $(v/u-1)/n > -1/n > -1$,
$(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$
with equality only if $n=1$ or $v/u-1 = 0$.
Thus BI implies this version of the AGMI.

答:
Could this be ?

$$\begin{align} \frac{d}{dx} \ln x &= \lim_{h \to 0}\; \frac{\ln(x + h) − \ln x}{h} \\
                   &= \lim_{h \to 0} \;\frac{\ln(1 + \frac{h}{x})}{h} \\
                   &= \lim_{h \to 0} \;\ln\left((1 + \frac{h}{x})^{\frac{1}{h}}\right) \\
\end{align}$$

let $h=\frac xc$

$$\begin{align} \frac{d}{dx} \ln x &= \lim_{c \to \infty} \;\ln\;\left((1 + \frac{1}{c})^{\frac{c}{x}}\right) \\
                                   &= \frac{1}{x} \;\lim_{c \to \infty}\; \ln\;\left((1 + \frac{1}{c})^{c}\right) \\
                                   &= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right)
\end{align}$$

Since $\frac{d}{dx} \ln x = 1/x$

$$\begin{align} \frac{1}{x} &= \frac{1}{x} \;\ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\
                          1 &= \ln\;\left(\lim_{c \to \infty}\;(1 + \frac{1}{c})^{c}\right) \\
                          e &= \lim_{c \to \infty}\;\left(1 + \frac{1}{c}\right)^{c}
\end{align}$$

That's not very rigorous but well...