问:
Rewriting a system of two equations found in The Chemical Basis of Morphogenesis (A. Turing, 1952)
In reading The Chemical Basis of Morphogenesis
 by A. Turing, I am unable to follow a small section of his working. On page $47$, Turing states that

\begin{align*} x_r&=\sum_{s=0}^{N-1} \exp\left(\frac{2\pi
irs}{N}\right)\xi_s, \\ y_r&=\sum_{s=0}^{N-1} \exp\left(\frac{2\pi
irs}{N}\right)\eta_s, \end{align*}

can be written as

\begin{align*} \xi_r&=\frac{1}{N}\sum_{s=1}^{N} \exp\left(-\frac{2\pi
irs}{N}\right)x_s, \\ \eta_r&=\frac{1}{N}\sum_{s=1}^{N} \exp\left(-\frac{2\pi irs}{N}\right)y_s. \end{align*}

This does not seem obvious to me. It is given that this can be shown by using the equations
\begin{align}
\sum_{s=1}^N \exp\left(\frac{2\pi irs}{N}\right)&=0 \ \ \text{if} \ \ 0<r<N, \\
&=N \ \ \text{if} \ \ r=0 \ \ \text{or} \ \ r=N.
\end{align}
However, I do not see how this helps. I hint would be most helpful.

答:
Put the sums for $x_s$ and $y_s$
in the sums for $\xi$ and $\eta$
and then reverse the order of summation.

Use those last sums to make almost everything go to zero except the desired terms.


问:
Is it possible to evaluate the integral $I=\int_0^\infty \frac{\sqrt{x}\arctan(x)}{1+x^2}dx$ with residue theorem?
let's consider the following integral

$$I=\int_{0}^{+\infty}{\frac{\sqrt{x}\arctan{x}}{1+x^{2}}\,\mathrm{d}x}$$
I know exactly how to evaluate it with sneaky methods and obtained $I=\frac{\pi\sqrt{2}}{8}(\pi+2\ln{\left(2\right)})$, but would it be possible to evaluate it with contour integration (Residue theorem)? Of course I have done some simple contour integrations before for example a half circle to evaluate $J=\int_{0}^{+\infty}{\frac{\mathrm{d}x}{1+x^{2}}}$, but for some reason I cannot come up with ideas for this.

Thanks in advance.

In case it is possible, I would like to get hints.

答:
We will evaluate a more general integral $$I(a)=\int_0^\infty\frac{\sqrt{x}\arctan{x}}{a^2+x^2}dx$$
where initially we consider $a>1$.

We will do it due to two reasons:

We will get more general expression for integral;

We want to avoid integration around branch points which coincide with poles (at $x=+-i$). This would be more complicated technically and would not bring any additional value. Because our integrand does not have any peculiarity at $a=1$ the solution can be expanded to any $a$ aftermath.

We will also use the explicite form for $\arctan$: $\arctan(x)=-\frac{i}{2}\log\frac{i-x}{i+x}$. The branch of logarithm is chosen in such a way that we get $\arctan(x)$ on the right half of axis $X$ (on both banks of the cut).
We will integrate along the following closed contour (special keyhole):

We made two cuts on the complex plane: from $0$ to $\infty$ of the axis $X$ and from $-i$ to $+i$.
We go along the upped bank of the cut on $X$, then around a big circle $R$ in the positive direction (counter clockwise), then along the lower bank of the cut on $X$ in the opposite direction and around the cut from  $-i$ to $+i$.  Logarithm gets an additional factor $+2\pi{i}$ on the left side of the vertical cut when we go around either $x=+i$ or $x=-i$ in the positive direction.

$$\oint_c=\oint_c\frac{\sqrt{x}(-\frac{i}{2}\log\frac{i-x}{i+x})}{a^2+x^2}dx=2I(a)+I_{C1}+I_R$$  where $I(a)$ is the required integral and $I_{C1}$ and $I_R$ are the integrals around the vertical cut and along the big circle correspondingly. While integrating along the contour we have to go along small circles $r$ around points $x=0$, $x=i$ anf $x=-i$. All these integrals $\to0$ at $r\to0$ (and $I_R\to0$ at $R\to\infty$), so further we will not take them into consideration. But we have to care about the additional factors (the integrand phases) which we get when go around these points.

Inside the closed contour the integrand is a single-valued function, therefore $\oint_C=2\pi{i}\sum{Res}\frac{\sqrt{x}(-\frac{i}{2}\log\frac{i-x}{i+x})}{a^2+x^2}$ (residuals at $x=ia$ and $x=-ia$).
Evaluation of residuals in not complicated - we just have to bear in mind that (due to the chosen branch of logarithm) logarithm will get an additional factor $\pi{i}$ at $x=a\exp(\frac{\pi{i}}{2})$ (i.e. $x=ia$) and $x=a\exp(\frac{3\pi{i}}{2})$ (i.e. $x=-ia$).

We get:
$\oint_C=2\pi{i}\sum{Res}=2\pi{i}\left((-\frac{i}{2})\frac{\sqrt{a}\exp(\frac{\pi{i}}{4})(\log\frac{a-1}{a+1}+\pi{i})}{2ia}+(-\frac{i}{2})\frac{\sqrt{a}\exp(\frac{3\pi{i}}{4})(\log\frac{a+1}{a-1}+\pi{i})}{-2ia}\right)$
$$2\pi{i}\sum{Res}=\frac{\pi}{\sqrt{2a}}\left(\log\frac{a-1}{a+1}+\pi\right)$$

$I_{C1}=\frac{-i}{2}\left(\int_o^{-i}\frac{\sqrt{x}\log\frac{i-x}{i+x}}{a^2+x^2}+\int_{-i}^{0}\frac{\sqrt{x}(\log\frac{i-x}{i+x}+2\pi{i})}{a^2+x^2}+\int_{0}^{i}\frac{\sqrt{x}(\log\frac{i-x}{i+x}+2\pi{i})}{a^2+x^2}+\int_i^0\frac{\sqrt{x}\log\frac{i-x}{i+x}}{a^2+x^2}\right)$. The integrals of logarithms cancel each other; the remaining part is:

$I_{C1}=\pi\left(\int_0^1\frac{i\sqrt{t}\exp(\frac{\pi{i}}{4})}{a^2-t^2}+\int_1^0\frac{-i\sqrt{t}\exp(\frac{3\pi{i}}{4})}{a^2-t^2}\right)=-\sqrt2\pi\int_0^1\frac{\sqrt{t}}{a^2-t^2}dt$.

After easy manipulations the integral can be reduced to
$I_{C1}=-\sqrt2\pi\int_0^1\left(\frac{1}{\sqrt{a}-x}+\frac{1}{\sqrt{a}+x}\right)\frac{dx}{2\sqrt{a}}+\sqrt2\pi\int_0^1\frac{dx}{a+x^2}=-\frac{\pi}{\sqrt{2a}}\log\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\pi\sqrt2}{\sqrt a}\arctan{\frac{1}{\sqrt{a}}}$

Finally we get $2I(a)=2\pi{i}\sum{Res}-I_{C1}=\frac{\pi}{\sqrt{2a}}\log\frac{(\sqrt{a}+1)^2}{a+1}+\frac{{\pi}^2}{\sqrt{2a}}-\frac{\pi\sqrt2}{\sqrt a}\arctan\frac{1}{\sqrt{a}}$
$$I(a)=\frac{\pi}{2\sqrt{2a}}\log\frac{(\sqrt{a}+1)^2}{a+1}+\frac{{\pi}^2}{2\sqrt{2a}}-\frac{\pi}{\sqrt{2a}}\arctan\frac{1}{\sqrt{a}}$$ $$=\frac{\pi}{\sqrt{2a}}\Bigl(\log\frac{\sqrt{a}+1}{\sqrt{a+1}}+\arctan\sqrt a\Bigr)$$

At $a=1$ we get the desired result:

$$I(1)=\frac{\pi}{2\sqrt{2}}\log2+\frac{{\pi}^2}{4\sqrt{2}}$$

ContentLicense: CC BY-SA 4.0


问:
Question about star-finite collection in general topology
According to this link (which is related to the definition of star-finite collection),

A star-finite collection of subsets of a set is a collection of subsets such that every member of the collection intersects only finitely many members.

Again, in the "weaker condition" section, they claim that

If we have a start-finite open collection (in other words, a star-finite collection comprising of open sets), then it is a locally finite collection.

If the given star-finite open collection, say $\mathcal U$, is a covering of the whole space, say $X$, then it is trivially true, because for each $x \in X$, we can find a suitable open set in $\mathcal U$ that contains $x$. But my question is, what about the case when $\mathcal U$ is not a cover of $X$? Does above holds for that case too?

Any comments are appreciated.

答: It need not. Let $X$ be a hedgehog space of infinite spininess, obtained by identifying the $0$ points of the disjoint union of infinitely many copies of $[0,1]$, and let $\mathscr{U}$ be the collection of copies of $(0,1]$; then $\mathscr{U}$ is a pairwise disjoint family of open sets, hence certainly star-finite, but every open nbhd of the common $0$ point meets every member of $\mathscr{U}$. The most that we can say in general is that $\mathscr{U}$ is a locally finite cover of its union.


问:
Example of ring such that the nil radical is prime and 0 is not
I was just trying to think about an example of a ring that is not a domain and the nilradical is prime, however I could not find anyone.

Thanks in advance.

答: $\mathbb{Z} / 4 \mathbb{Z}$ works.</p>

ContentLicense: CC BY-SA 3.0

Body: $\mathbb{Z} / 4 \mathbb{Z}$ works.


问:
Evaluating the isomorphism $\wedge^{2}V\cong\mathfrak{so}(2n)$ on basis elements.
Im back with another question about section 20 of Fulton and Harris (a little earlier than I had hoped).

Let $V$ be an even dimensional complex vector space with a quadratic form $Q$ and an orthonormal basis $\{e_{1},\cdots,e_{2n}\}$. We have the following isomorphism of Lie algebras:
$$
\phi:\wedge^{2}V\to\mathfrak{so}(2n)\subseteq\text{End}(V)\text{ , }\ a\wedge b\mapsto\phi_{a\wedge b}
$$
where
$$
\phi_{a\wedge b}(v)=2(\langle b,v \rangle a-\langle a, v\rangle b)
$$

Fulton and harris claim that this isomorphism maps the standard basis elements $\frac{1}{2}e_{i}\wedge e_{n+j}$ of $\wedge^{2}V$ to $E_{i,j}-E_{n+j,n+i}$, where the $E_{i,j}$ are $2n\times 2n$ matrices with the $i$-th $j$-th entry $1$ and all other entries $0$.

This should be an easy enough calculation, but it is not working out. I proceed as follows:

$$
\phi_{\frac{1}{2}e_{i}\wedge e_{n+j}}(e_{k})=\langle e_{n+j},e_{k}\rangle e_{i}-\langle e_{i}, e_{k}\rangle e_{n+j}
=
\delta_{n+j,k}e_{i}-\delta_{i,k}e_{n+j}
$$

While on the other hand:

$$
(E_{i,j}-E_{n+j,n+i})e_{k}
=
\delta_{j,k}e_{i}
-
\delta_{n+i,k}e_{n+j}
$$

These two expressions are clearly different, and so either I have made an elementary mistake somewhere, or I have a fundamental misunderstanding of something. I would really appreciate it if somebody could point out my mistake, or help to reconcile these two expressions.

答:
I have managed to resolve my issue.

In the above calculation I had been using the standard bilinear form $\langle x,y \rangle=x^{T}y$ which is associated with the quadratic from given by the identity matrix. On the other hand Fulton and Harris were using the bilinear form given by $\langle x,y \rangle=x^{T}Sy$, where $S$ is the block anti-diagonal matrix $S=\begin{pmatrix}0&\mathbb{1}_{n}\\\mathbb{1}_{n}&0\end{pmatrix}$.

Both these bilinear forms are symmetric and non-degenerate and so (at least over $\mathbb{C}$), they define isomorphic copies of the special orthogonal group and the associated Lie algebra.

Despite both forms leading to isomorphic Lie algebras, the elements look different. With the first convention the elements are antisymmetric matrices, whereas for the second convention they are matrices $A$ such that $A^{T}S+SA=0$. In particular, the given $H_{i}$ form a Cartan subalgebra in the second convention, but do not even belong to $\mathfrak{so}(2n)$ in the first convention.

Using the second convention we have that for $1\leq i,j\leq n$, $\langle e_{i},e_{j}\rangle=\langle e_{n+i},e_{n+j}\rangle=0$ and $\langle e_{i},e_{n+j}\rangle=\langle e_{n+i},e_{j}\rangle=\delta_{ij}$, which leads to the desired result.


问:
Small Detail in Folland Prop.2.12

2.12 Proposition. Let $(X,\mathcal{M},\mu)$ be a measure space and let $(X,\overline{\mathcal{M}},\overline{\mu})$ be its completion. If $f$ is an $\overline{\mathcal{M}}$-measurable function on $X$, there is an $\mathcal{M}$-measurable function $g$ such that $f=g$ $\overline{\mu}$-almost everywhere.

Proof. This is obvious from the definition of $\overline{\mu}$, if $f=\chi_E$ where $E\in\overline{\mathcal{M}}$, and hence if $f$ is an $\overline{\mathcal{M}}$-measurable simple function. For the general case, choose a sequence $\{\phi_n\}$ of $\overline{\mathcal{M}}$-measurable simple functions that converge pointwise to $f$ according to Theorem 2.10, and for each $n$ let $\psi_n$, be an $\mathcal{M}$-measurable simple function with $\psi_n=\phi_n$ except on a set $E_n\in\overline{\mathcal{M}}$ with $\overline{\mu}(E_n)=0$. Choose $N\in\mathcal{M}$ such that $\mu(N)=0$ and $N\supset\bigcup_1^\infty E_n$, and set $g=\lim \chi_{X\setminus N}\psi_n$ Then $g$ is $\mathcal{M}$-measurable by Corollary 2.9, and $g=f$ on $N^c$. $\square$

Just wondering for the characteristic function case, why does the statement hold.

I see that by definition of completion, we have $E= A \cup Z$, where $A \in M$ and $Z \subset N$  for some null set $N \in M$.

Let $g= \chi_ A$. I'm not sure why we have $f=g$ almost everywhere.

Thanks!

答:
The question is:

Consider $E\in \overline{\mathcal{M}}$ and $f= \chi_E$.  Then we have that
$E= A \cup Z$, where $A \in \mathcal{M}$ and $Z \subseteq N$  for some null set $N \in \mathcal{M}$.
Let $g= \chi_ A$. How to prove that  $f=g$ $\bar{\mu}-$almost everywhere.

Let $[f \neq g] =\{x \in X: f(x) \neq g(x)\}$. Note that $[f \neq g] \subseteq Z \subseteq N$. Since  $N \in \mathcal{M}$ and $\mu(N)=0$, any subset $C$ of $N$ is in $\overline{\mathcal{M}}$ and $\bar{\mu}(C)=0$ . Since $[f \neq g] \subseteq N$, we have that $[f \neq g] \in \overline{\mathcal{M}}$ and $\bar{\mu}([f \neq g])=0$. So  $f=g$ $\bar{\mu}-$almost everywhere.

Title: Small Detail in Folland Prop.2.12


问:
Evaluate $\int u(1+u)^{\tfrac32}du$
I want to evaluate:$$\int u(1+u)^{\tfrac32}du$$

I tried to write the expression in different ways: $\int(u^{\tfrac23}+u^{\tfrac53})^{\tfrac32}du$ but it didn't help. I also wrote the expression as:

$$\int u(u+1)\sqrt{u+1}\; du\quad\quad\text{or}\quad\quad \int(u^2+u)\sqrt{u+1}\;du$$
But it didn't help too.

答:
Adding and subtracting 1 in the $u$ factor gives

\begin{align}
\int u(1+u)^{\tfrac32}du&=\int (u+1)(u+1)^{3/2}\,du -\int (u+1)^{3/2}\,du\\
&=\int (u+1)^{5/2}\,du -\int (u+1)^{3/2}\,du
\end{align}

Can you continue from this?

Title: Evaluate $\int u(1+u)^{\tfrac32}du$


问:
what are the last 2 digits of $n^n$ without computing the formula
Computing this exponential is costly for number like $n=10^{400}$

Would rather find a way to find the answer without any exponentiation.

答:
As another response has indicated, calculating (for example)
$(1337)^{(1337)} \pmod{100}$ is facilitated by noting that for any
$n \in \{0,1,2,\cdots, 99\}$ and any $r,k \in \mathbb{Z^+},$
we have that

$$n^r \equiv (n + 100k)^r \pmod{100}.$$

This means that computing

$$(1337)^{(1337)} \pmod{100}\tag1$$ is reduced to computing

$$(37)^{(1337)}\pmod{100}.\tag2$$

Expression (2) above can be resolved
via Euler's Theorem.

It is known that $\varphi(100) = \varphi(4) \times \varphi(25) = 40.$

Given any $r \in \mathbb{Z^+}$, let $f(r)$ denote the unique element in
$k \in\{0,1,\cdots,39\}$ such that $r \equiv k\pmod{40}.$

Then Euler's theorem implies that
for any $a$ relatively prime to $(100)$
and any $r \in \mathbb{Z^+}, a^r \equiv a^{[f(r)]} \pmod{100}.$

Since (for example) $(37)$ is relatively prime to $(100)$ and $f(1337) = 17$,
you then have that

$$(1337)^{(1337)} \equiv (37)^{(1337)} \equiv (37)^{(17)} \pmod{100}. \tag3$$

So, the problem reduces to handling the $(60)$ elements in
$\{0,1,2,\cdots, 99\}$ that are not relatively prime to $100$.

$$\{a^b \equiv a^c\pmod{25} ~\text{and}~ a^b \equiv a^c\pmod{4}\}
\implies \{a^b \equiv a^c \pmod{100}\}.$$

Similar to the definition of $f(r),$ let $g(r)$ denote the residue of
$r \pmod{\varphi[25] = 20}$.

Also, let $h(r)$ denote the residue of $r \pmod{\varphi[4] = 2}$.

Then, the $(40)$ elements $n \in \{0,1,2,\cdots, 100\}$
that are relatively prime to $5$ but not relatively prime to $2$ can be handled
as follows:

For $r \geq 2, n^r \equiv 0\pmod{4}$ and $n^r \equiv n^{g(r)} \pmod{25}.$

In a similar fashion, the 10 elements $n \in \{0,1,2, \cdots, 100\}$
that are relatively prime to $2$ but not relatively prime to $5$
can be handled as follows:

For $r \geq 2, n^r \equiv 0\pmod{25}$ and $n^r \equiv n^{h(r)} \pmod{4}.$

Now, still unresolved are the $(10)$ elements $n \in \{0,1,\cdots, 99\}$ that
are multiples of $(10)$.  As one might guess, these are handled as follows:

For $r \geq 2, n^r \equiv 0\pmod{100}$.

Addendum

In accordance with a comment by Jyrki Lahtonen, who indicated (I have no way of confirming without doing further research - no thanks) that the
Carmichael number of $100$ is $(20)$ [rather than $\varphi(100) = 40$],
see also The Carmichael function.


问:
Representation matrix as for Basis B
Does it hold that \begin{equation*}\mathcal{M}_{\mathcal{B}}^{\mathcal{B}}(\phi_a)=\left (\gamma_{\mathcal{B}}\left (\phi_a(b_1)\right )\mid \gamma_{\mathcal{B}}\left (\phi_a(b_2)\right )\mid \gamma_{\mathcal{B}}\left (\phi_a(b_3)\right )\right )\end{equation*} where $\phi_a$ is a linear function and $\gamma_B(v_i) $ is the coefficient vector of linear combination of vector $v_i$ in respect to basis B ?

答:
Let me expand on my comment. If V is a finite dimensional k-vector space with a basis $\beta = \{b_1,...,b_n\}$, then for $L\in \mathcal{L}(V,V)$ (a linear map on V),  $\mathcal{M}_{\beta}^{\beta}(L)$ is defined to be the unique matrix A s.t.

$A\gamma_{\beta}(v) = \gamma_{\beta}(L(v))$ $\forall v\in V$.  (1)

Write $A:= \mathcal{M}_{\beta}^{\beta}(L) = (A^1...A^n)$ which gives $AE^j = A^j$ (where $\{E^1,...,E^n\}$ is the standard basis of $k^n$). So that we can compute the $j^{th}$ column of A using (1). In particular, taking $v := E^j$ in (1) we have

$AE^j = \mathcal{M}_{\beta}^{\beta}(L)E^j = \mathcal{M}_{\beta}^{\beta}(L)\gamma_{\beta}(b_j) = \gamma_{\beta}(L(b_j))$.

(so taking $n = 3$, $L = \phi_a$, and $\beta = \mathcal{B}$, we have your equation; technically the equation (1) is defined more generally, namely $\mathcal{M}_{\beta '}^{\beta}(L)$, Lang's Linear algebra (not the introduction) is a good source for more details on that).

Title: Representation matrix as for Basis B


问:
General Addition Rule for Probability extended to 4 events?
I just started statistics and need to use the general addition rule. I know what it looks like for $3$ events:

$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) - (2 * P(A \cap B \cap C)) + P(A'\cup B' \cup C').$$

But I'm confused to exactly how it extends to 4 events?

Thanks.

答:
The general formula is

$P\left( \bigcup\limits_{i=1}^{n} A_i\right) = \sum\limits_{i=1}^{n} P(A_i) - \sum\limits_{i, j : i < j} P(A_i \cap A_j) + \sum\limits_{i, j, k : i < j < k} P(A_i \cap A_j \cap A_k) - ... + (-1)^{n-1} P\left(\bigcap\limits_{i=1}^{n} A_i\right)$

When $n = 4$, we get

$P\left( \bigcup\limits_{i=1}^{4} A_i\right) = \sum\limits_{i=1}^{4} P(A_i) - \sum\limits_{i, j : i < j} P(A_i \cap A_j) + \sum\limits_{i, j, k : i < j < k} P(A_i \cap A_j \cap A_k) -  P\left(\bigcap\limits_{i=1}^{4} A_i\right)$

If we further assume the events $\{A_i\}$ to be exchangeable, we get

$P\left( \bigcup\limits_{i=1}^{n} A_i\right) = n \times P(A_1) - {n \choose 2} \times P\left(\bigcap\limits_{i=1}^{2} A_i\right) +  {n \choose 3} \times P\left(\bigcap\limits_{i=1}^{3} A_i\right) - ... + (-1)^{n-1} P\left(\bigcap\limits_{i=1}^{n} A_i\right)$

For $n = 4$ with exchangeable events, this gives

$P\left( \bigcup\limits_{i=1}^{4} A_i\right) = 4 \times P(A_1) - 6 \times P\left(\bigcap\limits_{i=1}^{2} A_i\right) + 3 \times P\left(\bigcap\limits_{i=1}^{3} A_i\right) - P\left(\bigcap\limits_{i=1}^{4} A_i\right)$


问:
Approximate Lebesgue measure functions by simple functions
Suppose $f$ is a Lebesgue measurable function.

I know that if $f$ is non-negative then we can approximate $f$ by a sequence of simple functions.

But I came across the statement " every Lebesgue measurable function is equal almost everywhere to a limit of simple Lebesgue measurable functions".

I'm not sure why this holds if $f$ is negative.

Thanks!

答:
We have that $f  = f^+ - f^-$ where $f^+ = \max{f, 0}$ and $f^-=\max{-f,0}$. Note that since $f$ is measurable, we have that $f^+$ and $f^-$ are measurable.

Let $\{h_n\}_n$ be a sequence of simple function converging almost everywhere to $f^+$ and  let $\{k_n\}_n$ be a sequence of simple function converging almost everywhere to $f^-$. It folows that $\{h_n-k_n\}_n$  is a a sequence of simple function converging almost everywhere to $f^+- f^-$, that means, converging almost everywhere to $f$.


问:
Variance and expectations of $20S^2/144$ following chi-squared distribution
Not sure how the coefficient in this chi-squared problem affects the expectation and variance?  I have written out my answer so far but I'm not sure how the coefficient's might affect the answer.  Can anyone help me understand if and what I am doing wrong?

$\sigma^2 = 12; \mu= 170; n= 21$

Using $(n-1)S^2/\sigma^2$ I have that the p.d.f. follows a chi-squared distribution of $20S^2/144$ with 20 degrees of freedom.

$E(20S^2/144) = \sigma^2 = 12^2 = 144$

$Var(20S^2/144) = 2n = 2 * 21 = 42$

答:
Set

$$Y=\frac{(n-1)S^2}{\sigma^2}\sim \chi_{20}^2$$

You have that

$$\mathbb{E}[Y]=n=20$$

and

$$\mathbb{V}[Y]=2n=40$$

First proof (sketch)

just observe that

$$\chi_{n}^2=Gamma\Bigg[\frac{n}{2};\frac{1}{2}\Bigg]$$

and you can easy calculate mean and variance with the integral. Here $1/2$ is the "rate parameter"

Second proof:

Observe also that

$$Y=\frac{(n-1)S^2}{\sigma^2}=\frac{20S^2}{12}$$

Thus

$$\mathbb{E}\Bigg[\frac{20S^2}{12}\Bigg]=\frac{20}{12}\mathbb{E}[S^2]=\frac{20}{12}\sigma^2=20$$

and similarly for its variance


答:
Integrate by parts

$$\int u(1+u)^{\tfrac32}du=\frac25\int u \>d((1+u)^{\frac52})
$$

Title: Evaluate $\int u(1+u)^{\tfrac32}du$

Body:
I want to evaluate:$$\int u(1+u)^{\tfrac32}du$$

I tried to write the expression in different ways: $\int(u^{\tfrac23}+u^{\tfrac53})^{\tfrac32}du$ but it didn't help. I also wrote the expression as:

$$\int u(u+1)\sqrt{u+1}\; du\quad\quad\text{or}\quad\quad \int(u^2+u)\sqrt{u+1}\;du$$
But it didn't help too.


问:
Can we find $\int\sqrt{1+a^2\sin^2ax}\,dx$?
I do not think trigonometic subsitutions work here, $\displaystyle\int\sqrt{1+a^2\sin^2ax}\,dx$ where $a$ is a real number. (clear when $a=0$)

I was wondering if there is a way to find the antiderivative or not.

答:
You can find an antiderivative, but it is not in terms of elementary functions.

Hint 1: Use elliptic integrals, specifically use the one defined as:

$$E(\varphi | k^2):= \int_0^{\varphi} \sqrt{1-k^2\sin^2(\theta)} d\theta $$

This is known as the incomplete elliptic integral of the second kind.

Hint 2: To use the above hint, make a substitution of the form $u=ax$ in your integral. This is to get it into the form so that you can apply the elliptic integral. So your integrand will be:

$$\cfrac{\sqrt{1+a^2\sin^2(u)}}{a} $$

Do you think you can take it from there?

Edit: Just to be clear, the integral defined above by $E(\varphi|k^2)$ is also an indefinite integral (just like the one you want to solve). The variable $\theta$ is just a dummy variable.


问:
$P^2=P$, $Q^2=Q$, $im P\cap im Q=\{0\}$. Show $V=\ker P\oplus \ker Q$.
Let $P,Q$ be linear transformations from $n$ dimensional vector space $V$ to $V$ such that $P^2=P$, $Q^2=Q$, $im P\cap im Q=\{0\}$. Show $V=\ker P\oplus \ker Q$.

It sounds easy geometrically. But how to give the proof? In particular, how to show $\ker P\cap \ker Q=\{0\}$?

答:
I think a reasonable assumption is that $\text{im}P\oplus \text{im}Q=V$ and $PQ=QP.$ If so, then since $\ker P\oplus\text{im}P=V=\ker Q\oplus \text{im}Q$ since they are projections, $\dim\ker P + \dim\ker Q =n.$ Furthermore, if $v\in\ker P\cap\ker Q,$ then we can write $v=Pa+Qb,$ and hence
$$0=Pv=Pa+PQb$$
and
$$0 = Qv = QPa+Qb$$
so
$$Qb=-QPa=QPQb=PQb,$$
i.e., $Qb\in\text{im} P.$ Thus $Qb\in\text{im}P\cap \text{im}Q=\{0\}.$ Likewise we can show $Pa=0.$ Thus $v=0$

I am not sure if the condition $PQ=QP$ is necessary.

Original answer:

It seems that you need other conditions to assure your conclusion.

Suppose $V=\mathbb R^3,$ $P$ the orthogonal projection onto the $x$-axis, and $Q$ the orthogonal projection onto the $y$-axis (where we view the coordinate $(x,y,z)\in\mathbb R^3$). Then their images have trivial intersection, but their kernel doesn't (since $(0,0,1)$ is in both of their kernels).


问:
Is it possible to construct a $p$-adic numbers out of $\mathbb{Z}[[x]]$ like we can for the real numbers?
I saw this question earlier today about the $p$-adic numbers generally and am trying to figure out how to visualize them or construct an algebraic entity with the right structure.

There's a straightforward way to build the real numbers out of formal power series in $\mathbb{Z}[[x]]$. Is there a way to do a similar thing for the $p$-adic numbers?

The intuition is that we are considering generalizations of infinite decimals $123456.123123\cdots$, but we only have a single decimal place to the left of the decimal point, and we allow any integer to appear in any decimal place.

I'm defining a notion of normal form to capture the fact that some-but-not-all formal power series converge when evaluated at $\frac{1}{10}$, without circularly assuming that those sums converge to a real number. The normal form for a given real isn't unique, $1.3$ and $1.299999\cdots$ are the same number. Note that in the example below $a_0$ is unconstrained.

$$ \sum_{i=0}^{\infty} a_ix^i \;\; \text{is in normal form} \;\; \textit{if and only if} \;\; \text{$0 \le a_i \le 9$ for all $i \ge 1$} $$

Let's define a relation $R$ as follows

$$ \left(\left(\sum a_i x^i \right), \left(\sum b_i x^i\right) \right) \in R \\ \textit{if and only if} \\ \text{there exists a $k$ such that $10a_k + a_{k+1} = 10b_k + b_{k+1}$} \\ \text{and all indices besides $k$ and $k+1$ are the same. } $$

Let $\simeq$ be the transitive, reflexive, symmetric closure of $R$.

If we consider only the formal power series in $\mathbb{Z}[[X]]$ that are equivalent to a power series in normal form, and then take equivalence classes mod $\simeq$, then we have a field isomorphic to the real numbers $\mathbb{R}$.

We can also define orderings on this structure.

$$ \sum a_i x^i \;\; \text{is non-negative} \;\; \textit{if} \;\;  \text{$\sum a_i x^i$ is in normal form and $a_0 \ge 0$} $$

$$ \sum a_i x^i \;\; \text{is non-negative} \;\; \textit{if and only if} \;\;  \text{$\sum a_i x^i$ is equivalent to a non-negative series in normal form} $$

And then the definition of $\sum a_i x^i \le \sum b_i x ^i$.

$$ \sum a_i x^i \le \sum b_i x ^i \;\; \textit{if and only if} \;\; \sum (b_i-a_i)x^i \;\; \text{is non-negative} $$

Is there a way to do something similar for the $p$-adic numbers?

答:
"Decimal expansion" happens to be the/a correct way to view $p$-adics, and power series in particular are a good way in. Intuitively, one can think about $\mathbb Z_p$ as series which look like
$$\sum_{k=0}^\infty a_kp^k,$$
where $a_k\in[0,p)\cap\mathbb Z.$ The $a_k,$ then, are more or less digits in base-$p.$ This should look familiar: if you put $p=\frac1{10}$ into these series, you will recover $\mathbb R.$

However, the caveat here is that we have allowed this to be an infinite series when it certainly does not converge in any meaningful way in $\mathbb R.$ This is not a problem: we don't live in $\mathbb R$ anymore because we live in $\mathbb Z_p,$ and convergence is different in this universe.

Let's be a bit more formal. As suggested, let's take two power series $a(x),b(x)\in\mathbb Z[[x]]$ and ask when they should give the same element of $\mathbb Z_p.$ For concreteness, we'll write
$$a(x)=\sum_{k=0}^\infty a_kx^k,\qquad b(x)=\sum_{k=0}^\infty b_kx^k.$$
To test equality, what we want to do is test$\pmod{p^N}$ for increasingly large $N.$ That is, if
$$\sum_{k=0}^\infty a_kp^k\equiv\sum_{k=0}^\infty b_kp^k\pmod{p^N}$$
as $N\to\infty,$ then we'll have the corresponding elements of $\mathbb Z_p$ be equal. However, these series are only defined formally, so how should we check$\pmod{p^N}$? Well, there's no reason to check elements past $k\ge N$ because they don't add any modular information. So we might as well check if
$$\sum_{k=0}^Na_kp^k\equiv\sum_{k=0}^Nb_kp^k\pmod{p^N}$$
as $N\to\infty.$ Now that we have finite sums, this notion of equality potentially makes more sense and can be made rigorous.

Anyways, the point is that, yes, we can think about $\mathbb Z_p$ as power series in $p.$ This can even be made rigorous, for $\mathbb Z_p\cong\mathbb Z[[x]]/(x-p).$


答:
Use $u=ax$, which means $dx=\dfrac{1}{a}\,\mathrm{d}u$
$$\int\sqrt{1+a^2\sin^2ax}\,dx=\class{steps-node}{\cssId{steps-node-1}{\dfrac{1}{a}}}
{\displaystyle\int}\sqrt{a^2\sin^2\left(u\right)+1}\,\mathrm{d}u$$

Solving for
$${\displaystyle\int}\sqrt{a^2\sin^2\left(u\right)+1}\,\mathrm{d}u=\operatorname{E}\left(u\,\middle|\,-a^2\right)$$

And plugging them in
$$=\dfrac{\operatorname{E}\left(ax\,\middle|\,-a^2\right)}{a}+C$$

Title: Can we find $\int\sqrt{1+a^2\sin^2ax}\,dx$?

Body:
I do not think trigonometic subsitutions work here, $\displaystyle\int\sqrt{1+a^2\sin^2ax}\,dx$ where $a$ is a real number. (clear when $a=0$)

I was wondering if there is a way to find the antiderivative or not.


问:
Making precise Dehn filling
Dehn surgery along a knot is a well-known construction: choose a regular neighbourhood $N(K)$ of a knot $K \subset S^3$, let $X_K := S^3 - N(K)$ and choose an essential simple closed curve $\alpha$ on the torus $\partial X_K$.

Then surgery along $K$ is the pushout $X_K \cup_\varphi S^1 \times D^2$. In the literature it is common to find that the attaching map $\varphi : S^1 \times \partial D^2 \to \partial X_K$ is completely determined by mapping the meridian $* \times \partial D^2$ to $\alpha$. How to make precise that such assignment extends to the required map $\varphi$?

答:
One way to see that the image of the meridian determines how to attach the entire torus is by first gluing in a thickened disk, then a 3-ball (which gives the entire torus).

More precisely, first attach a $D^2 \times I$ to $X_K$ via a homeomorphism $f: (\partial D^2 \times I) \to N(\alpha)$, where $N(\alpha)$ is a regular neighborhood of $\alpha$ in $\partial X_K$. This step is just identifying two cylinders.

We then have a 3-manifold $X_K \cup_f (D^2 \times I)$ which has an $S^2$ boundary. Attaching a $D^3$ via a homeomorphism of the $S^2$ boundaries gives a closed 3-manifold.

Here the $D^3$ and the $D^2 \times I$ combined form the total $S^1 \times D^2$ you are gluing in.

As a final note, the resulting 3-manifold does not depend (up to homeomorphism) on the choices of homeomorphisms for gluing, for example since the mapping class groups of $S^1$ and $S^2$ are trivial. See also this question.

Title: Making precise Dehn filling


问:
JEE Main 2019:finding $\lim_{x \to \infty} \frac{f(x)}{x}$

let $f$ be a differentiable function such that $$f'(x)=7-\frac{3}{4} \frac{f(x)}{x}\tag 1$$ then $\lim_{x \to \infty} \frac{f(x)}{x}=?$

This came in JEE Main 2019.Although I got the answer $4$ which matches with the answer key,I( think my solution is not complete.

Approach:Assuming the limit exists we have by L-Hospitals rule $$\lim_{x \to \infty} \frac{f(x)}{x}=\lim_{x\to \infty}f'(x)=L \tag{say}$$ Then By taking $\lim_{x\to \infty}$ on both sides of equation $(1)$ we have $$L=7-\frac{3}{4}L \implies L=4$$

Question:Although this approach has quickly given the answer the problem is that I have assumed that the Limit exists which need not be the case.I am aware of the method of actually solving the differential equation and then calculating the limit as done here but I was wondering if my method could be actually tweaked  to show that the limit exists or any other method without solving the differential equation.

答:
In order to apply L'Hopital's rule, you must show that $f(x) \to \infty$ as $x \to \infty$, thus satisfying the criteria that $f(x)/x$ is an indeterminate form.  This is the main gap in the solution.

The question about the existence of the limit in general is more subtle, since even if L'Hopital's rule may be applied to an indeterminate form, the resulting limit may not be finite; e.g., $$\lim_{x \to \infty} \frac{e^x}{x^2}$$ has an indeterminate form of the same type but obviously does not have a finite limit.  That said, what your work does show is that if the limit is well-defined and finite, it must equal the value you obtained.


问:
$a_n$ is the number of sequences with length $n$ over ${1,2,3}$
$a_n$ is the number of sequences with length $n$ over ${1,2,3}$ such as there ain't no two number '1' next to each other.

Find a formula for $a_n$.

I thought about splitting it to one case where number in the last place was one, so the following must be 2 or 3 and a case where the three options are available. I'm having hard time putting it into formula and in general approaching those kind of questions.

I looked into other questions up here and still had hard time putting the concepts into this one.

答:
Let's call a sequence of $1$'s, $2$'s, and $3$'s "valid" if it doesn't have two consecutive $1$'s. Every valid sequence of length-$n$ fits one of these cases:

A) [valid sequence of length $n-1$], $2$

B) [valid sequence of length $n-1$], $3$

C) [valid sequence of length $n-2$], $2$, $1$

D) [valid sequence of length $n-2$], $3$, $1$

By considering these cases, can you find the number of valid sequences of length-$n$ in terms of the number of valid sequences of lengths $n-1$ and $n-2$? i.e. find $a_n$ in terms of $a_{n-1}$ and $a_{n-2}$.

After doing that, you'll have a recursive formula for $a_n$. From there, you just need to come up with some initial values and solve the 2nd order linear recursion.


问:
Show $2^{n-1} (z^n+1)-(z+1)^n=0$ for $n \geq 2$ implies $|z|=1$.
I've been trying to prove the following question:

For $n \geq 2$, if $z \in \mathbb{C}$ solves $2^{n-1} (z^n+1)-(z+1)^n=0$, then $|z|=1$.

I rewrite the equation as $\frac{z^n+1}{2}=\left(\frac{z+1}{2}\right)^n$. I don't know how to proceed from here. Any help would be appreciated! Thanks!

答:
I can't think of an elementary proof, but we can use the general criteria of Cohn, discovered in 1922. This characterizes the class of polynomials whose roots lie only on the unit circle.

$\mathrm{ Cohn \ Theorem}$ : Suppose that $p(z) = \sum a_kz^k$ is a complex polynomial of degree $n$. Then all the roots of $p$ lie on the unit circle, if and only if :

There is a $\mu \in \mathbb C$, $|\mu |=1$ so that $a_{n-k} = \mu\overline{a_{k}}$ for all $0 \leq k \leq n$. Such a polynomial is called self-inversive.

Any root $z_0$ of $p'(z)$ (where $p'$ is the derivative of $p$) satisfies $|z_0| \leq 1$.

Actually Cohn's theorem is a little more general than this, but I'll leave the generalization for future discussion.

Proof via Cohn's theorem : Let $p(z) = 2^{n-1}(1+z^n) - (1+z)^n$.One can write the condition for self-inversiveness as : there exists a $|\mu|=1$ so that $p(z) = \mu z^n  \overline{p}(\frac 1z)$, where $\bar{p}$ is the polynomial $p$ with each coefficient conjugated. However, our $p$ has real coefficients.

You can compute $z^n \bar{p}(\frac 1z)$ and check that $\mu = 1$ works. When $\mu=1$ works, the polynomial is referred to as a reciprocal polynomial, because its coefficients read the same back-to-front, which allows for some nice parametrization.

But now we get to $p'(z) = n2^{n-1}z^{n-1} - n(z+1)^{n-1}$. We are finding roots of $p'$. At a root, we must have $2^{n-1}z^{n-1} = n(z+1)^{n-1}$, and (ignoring $z
=0$ which doesn't work) rearranging and taking the absolute value, we get :$$
2^{n-1} = \left|1+\frac 1z \right|^{n-1} \implies \left|1+ \frac 1z\right| = 2
$$

and therefore if $|z| > 1$ we can't have $\left|1+ \frac 1z\right| = 2$ by the triangle inequality. The result for $p(z)$ follows by Cohn's theorem.


问:
Error for a Taylor Series
So in estimating $(1.6)^{4/5}$ using the first 3 terms of the Taylor Series for function $f(x) = (1+x)^{4/5}$ we want to use the remainder theorem to obtain the error estimate.

$|R_2(0.6)| ≤ $Max $0 ≤ z ≤ 0.6$          $|\frac{f'''(z)}{3!}(0.6)^3|$

Now $f'''(x) = \frac{24}{125(1+x)^{11/5}}$ I assume you would sub in the value for $z$ in place of $x$ before applying to the formula, but how do you calculate it if $z$ can take on multiple values?

答:
I think you were trying to write

$$\lvert R_2(0.6)\rvert \leq \max_{0 \leq z \leq 0.6}
\left\lvert\frac{f'''(z)}{3!}(0.6)^3\right\rvert.$$

The point of the "max" is to choose the $z$ that makes the expression the largest.

If $f'''(z) = \frac{24}{125(1+z)^{11/5}},$ which value in the range
$0 \leq z \leq 0.6$ produces the largest value of $f'''(z)$?
That's the value to use.

Title: Error for a Taylor Series


问:
Sum equal to the product
I know that $2 + 2 =  2 \cdot 2$ and $1 + 2 + 3 = 1 \cdot 2 \cdot 3.$

My question: Are there other positive integers with sum equal to the product? (The number 1 can not appear more than once among them.)

I am therefore excluding the infinity of solutions obtained by repeating 1, such as:

$$1 + 1 + 2 + 4 = 1 \cdot 1 \cdot 2 \cdot 4$$

$$1 + 1 + 1 + 2 + 5 = 1 \cdot 1 \cdot 1 \cdot 2 \cdot 5$$

答:
First we shall look at 2 numbers.

Consider $a,b \in \mathbb{N}$.  We have $ab=a+b$ $$\implies ab-a=b$$ $$\implies a(b-1)=b$$
$$\therefore b-1|b$$ $$\implies b=2$$ $$\implies a=2$$ And this is the only solution for two numbers.

Now let us look at $n$ numbers, $a_1,a_2,a_3, \dots a_n \in \mathbb{N}$. WlOG $a_1 \leq a_2 \leq \cdots \leq a_n$

We have $a_1a_2a_3 \cdots a_{n-1}a_n=a_1+a_2+a_3+ \cdots + a_n$. Clearly the way to optimize the situation (i.e minimizing the L.H.S while maximizing the R.H.S) is to put $a_1=1$ and $a_2,a_3, \cdots a_n=2$

Then $2^{n-1}=2n-1$ $$\implies2^{n-2}<n-1$$ $$\implies n\leq3$$

So now we look at $n=3$ (we have already done $n=2$).

If none of the terms is equal to one the most likely case is 2,2,2 which doesn't work so none can work. So $a_1=1$ then we have $$a_2a_3=1+a_2+a_3$$ $$\implies a_2a_3-a_2=a_3+1$$ $$\implies a_2(a_3-1)=a_3+1$$ $$\implies a_3-1 | a_3+1$$ $$\implies a_3=2,3$$ Which you can sub in and you find the only solution to be $a_1=1,a_2=2,a_3=3$.

Title: Sum equal to the product


问:
How to solve a system of 4 coupled non-linear differential equations?
The system of equations I have is:
$$\frac{dx_1}{dt} = k_2 \left( \frac{y_2 - y_1}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$
$$\frac{dy_1}{dt} = k_2 \left( \frac{x_1 - x_2}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$

$$\frac{dx_2}{dt} = k_1 \left( \frac{y_1 - y_2}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$
$$\frac{dy_2}{dt} = k_1 \left( \frac{x_2 - x_1}{(x_1-x_2)^2 + (y_1-y_2)^2} \right)$$

where $k_1,k_2$ are constants. How would I go about solving this for $x_1(t),y_1(t),x_2(t),y_2(t)$?

For some context, these differentials equations determine how 2 vortex velocity fields of arbitrary circulation ($k_1, k_2$) interact with each other. The x-y time derivatives represent the component-wise speed of each vortex due to the other vortex.

My attempt:

So far, I multiplied the first and second equation by $(x_1−x_2)$ and $(y_1−y_2)$ and added them to get $(x_1−x_2)(dx_1/dt)+(y_1−y_2)(dy_1/dt)=0$. Doing the same thing for the third and fourth equations, I got $(x_1−x_2)(dx_2/dt)+(y_1−y_2)(dy_2/dt)=0$. Don't know where to go from here.

答:
Your system can be written more compact in complex numbers as
$$
\dot z_1=ik_1\frac{z_1-z_2}{|z_1-z_2|^2}\\
\dot z_2=ik_2\frac{z_2-z_1}{|z_1-z_2|^2}\\
$$
It follows immediately that $k_2z_1+k_1z_2=const.$ As the right side only depends on differences, the system is translation invariant, thus one can shift it so that the constant is zero. Now set $z_1=k_1w$, $z_2=-k_2w$ so that the dynamic reduces to one equation
$$
\dot w=i\frac{w}{(k_1-k_2)|w|^2}
$$
Write in polar coordinates $w(t)=r(t)e^{i\varphi(t)}$ so that
$$
\dot r+ir\dot φ=\frac{i}{r(k_1-k_2)}
$$
which implies $\dot r=0$, the motion of $w$ is circular, and $\dot\phi=\frac1{r^2(k_1-k_2)}$ is the constant angular velocity.

From here on one can substitute back to find the full solution.


问:
Complex Analysis boundedness and limits
True or false. If $f: \mathbb{C}\to\mathbb{C}$ is bounded, then $\lim_{z\to 0} f(z)$ exists.

R If f is bounded means that there is some M∈R such that ∀z ∈C holds that |f(z)|≤M.

My Answer

Would $\frac{z}{|z|+1}$ be an example of such a function? Bounded above by 1 and below by -1.

答: False. Take the function that is $1$ on the real axis and $0$ elsewhere. </p>

Title: Complex Analysis boundedness and limits


问:
Permutation Problem: Select m objects from n ones of r types by sampling without replacement

Problem Description

If there are $n$ objects with:

$n_1$ indistinguishable objects of type 1,

$n_2$ indistinguishable objects of type 2,

...

$n_r$ indistinguishable objects of type r

s.t. $$n_1 + n_2 + ... + n_r = n$$
How many possible arrangements are there for selecting $m$ objects from them by sampling without replacement, where orders of the same type of objects do not matter while orders of different types matter.

Background

I thought of this question when I was reviewing the counting chapter of the book Discrete Mathematics and Its Applications. When it came to the generalized problems of permutations and combinations, I came up with and got confused of this question which I thought belonged to the box-ball problems. But I could not find a solution in the book, neither in Google. No doubt that this question is interesting and valuable.

Progress

The only idea about the solution is the case of $m=n$, and the solution should be
$$\frac{n!}{n_1!n_2!...n_r!}$$

答:
The general open-form solution could be of the form $\sum_{\begin{matrix}k_1,k_2,...,k_r\text{ such that }\\k_1+k_2+...+k_r=m\\k_1\in[0,min\{n_1,m\}]\\k_2\in[0,min\{n_2,m\}]\\...\\k_r\in[0,min\{n_r,m\}]\\k_i integers\end{matrix}}\left(\begin{matrix}m\\k_1k_2...k_r\end{matrix}\right)$.

Writing out each $n_i$ in a grid like format, it remains to find all possible combinations of values that sum to m given differing lengths of $n_1,n_2,...,n_r$, a seemingly challenging task.

$\begin{matrix}0 & 1 & 2 & 3 & \textbf{4} &... & n_1\\0&1&\textbf2&3&4&...&...&n_2\\...\\0&1&2&\textbf{$n_r$}\end{matrix}$

I have found it easy to begin small and consider the case with $\sum_{i=1}^rn_i=m$. In this case, You must pick all the things, and there is only one possibility. Then the answer is the multinomial coefficient with the corresponding values of $k_1=n_1,...,k_r=n_r$.

Now consider when $\sum_{i=1}^rn_i=m+1$. Looking at the hastily drawn image above, there is one degree of freedom that must be distributed across the r rows. There are r ways to do this. Now you must select all r combinations of $k_1,...,k_r$ such that they sum to m, plug that into the multinomial coefficient, and sum them up.

Consider when $\sum_{i=1}^rn_i=m+2$. Depending on r, the answer is different. $\begin{cases}r=1&r\\r\ge2&r+{r\choose2}\end{cases}$. When you've got only one type of item, you must pick all of them. But if you've got at least 2 types of things, you can distribute the 2 degrees of freedom to the same item (r possibilities) or distribute 1 each to 2 items (r choose 2). Note that this does not include actually finding the combinations of the $k_i$, this only tells you how many combinations there are.

Consider $\sum_{i=1}^rn_i=m+3$. Again, there are cases depending on r. $\begin{cases}r=1&r\\r=2&r+2{r\choose2}\\r\ge3&r+2{r\choose2}+{r\choose3}\end{cases}$. Notice the number of cases is how much smaller m is than n.

For $\sum_{i=1}^rn_i=m+4$, $\begin{cases}r=1&r\\r=2&r+2{r\choose2}+{r\choose2}\\r=3&r+2{r\choose2}+{r\choose2}+\left(\begin{matrix}3\\2,1\end{matrix}\right){r\choose3}\\r\ge4&r+2{r\choose2}+{r\choose2}+\left(\begin{matrix}3\\2,1\end{matrix}\right){r\choose3}+{r\choose4}\end{cases}$. Notice that a multinomial coefficient has appeared. I believe there will be at most one multinomial coefficient per "term" if you proceed this way, i.e. you won't have to worry about multinomial coefficients multiplying multinomial coefficients etc.

It has occurred to me that you can also start the other way, with $m=0$ and proceed upward.


问:
Polynomial division over different fields
Suppose that we have irreducible polynomials over a field, $f,g\in F[x]$, and that $g$ divides $f$ in $L[x]$ where $L$ is a field containing splitting fields of $f$ and $g$ over $F$. Can we say that $g$ divides $f$ in $F[x]$?

I've got this question from the following proof: "If there is finite group $G\subseteq \text{Aut }E$ satisfying $F=\text{Inv}(G)$ then $E/F$ is a Galois' extension."

Considerations:

$f\in F[x]$ irreducible;

$r$ such that $f(r)=0$;

$\mathcal{O}_r=\left\{\sigma(r):\: \sigma\in G\right\}=\left\{r_1,...,r_m\right\}$;

$g=\prod (x-r_i)$.

The author concludes that $g$ divides $f$ in $F[x]$ immediately after showing that $g\in F[x]$. But I don't consider that so trivial. I proved that it's true because if $f=gh$ for some $h\in L[x]$ then $\widehat{\sigma}(h)=h$ for any $\sigma\in G$ where $\widehat{\sigma}\:\colon E[x]\to E[x]$ is the unique ring homomorphism mapping $a\mapsto \sigma(a)$ and $x\mapsto x$, hence $h\in F[x]$.

Is there any forward way?

答:
One way is the following. Remember polynomial division. It works over any field $K$.

So if $a(x),b(x)\in K[x]$, $b(x)\neq0$, then there exists unique polynomials $q(x),r(x)\in K[x]$ such that

$a(x)=q(x)b(x)+r(x)$, and

$\deg r(x)<\deg b(x)$.

At the point you reached you know that when $K=L$ you have $r(x)=0$. Possibly I should temporarily write $r_L(x)=0$. Because $F\subseteq L$, whichever solution $(q_F(x),r_F(x))$ worked over $F$, will also work over $L$. Therefore $r_F(x)$ must be equal to $r_L(x)$, and the claim follows.

A consequence of this is that Euclid's algorithm runs with exactly the same intermediate steps in $F[x]$ and $L[x]$ for any pair of inputs $f(x),g(x)\in F[x]$. In other words, $\gcd(f(x),g(x))$ means the same thing in both $F[x]$ and $L[x]$.


问:
Measurability from $\mathbb R$ to $\mathbb R^n$?
I was following the definitions from Royden, where the measurability is given as

We say $A \subseteq \mathbb R$ is measurable if $\forall X \subseteq \mathbb R$
$$m^*(X) = m^*(X \cap A) + m^*(X \setminus A)$$

Similarly,

$A\subseteq \mathbb R^n$ is said to be measurable in $\mathbb R^n$ if for all $X\subseteq \mathbb R^n$, $$m^\ast(X)=m^\ast(A\cap X)+m^\ast(X\backslash A).$$

An open rectangle in $\mathbb R^n$ is a set of the form $R=(a_1,b_1)\times\dots\times (a_n,b_n)$. The volumn of the open rectangle is given as $\ell(R)=(b_1-a_1)(b_2-a_2)\cdots(b_n-a_n)$.

We then define Lebesgue outer measure on $\mathbb R^n$ by $m^\ast: P(\mathbb R^n)\rightarrow [0,\infty)\cup\lbrace \infty\rbrace$, $$m^\ast(A)=\inf\left\lbrace \sum_{i=1}^\infty \ell(R_i): A\subseteq \bigcup_{i=1}^\infty R_i\right\rbrace,$$

I am wondering how can we show something in $\mathbb R^n$ is measurable if each of its component is measurable in $\mathbb R$.

That is:

Let $A=A_1\times A_2\times\cdots \times A_n\subseteq \mathbb R^n$, where each $A_i\subseteq \mathbb R$ is measurable. Prove that $A$ is measurable.

答: Hint: Consider sets $B_i=\mathbb{R}\times \ldots \times A_i \times \mathbb{R}\times\ldots\times\mathbb{R}$ so that if you project $B_i$ onto its $i^{th}$ component you get $A_i$ and if you project onto any other component you get $\mathbb{R}$. Then observe that $A = \bigcap_i B_i$ and that the countable intersection of measurable sets is measurable.


问:
change of random variable in calculating expectations: $\mathbb{E}_x [f(X)] = \mathbb{E}_z [f(g(Z))] $?
Let $X$ be some function $g$ of a random variable $Z$, such that $X=g(Z)$.
Does the following hold in general? If so, how should I give a formal proof?
$$\mathbb{E}_x [f(X)] = \mathbb{E}_z [f(g(Z))] $$

For example, in showing the linearity of expectation, it's common to let $X=aZ+b$, then write $\mathbb{E}_x[X]=\mathbb{E}_z[aZ+b] = \int_{dz}(az+b) f(z) = a \mathbb{E}_z[Z] + b $, where I've always treated expectation $\mathbb{E}$ as an operator that averages w.r.t. the distribution of whatever random quantity in the argument.

答:
I believe I've finally learned enough measure theoretic probability to answer this question :)

It really just comes down to the formal definition of the expectation of a random variable, and the transformation ("change of measure") theorem.

The transformation theorem (see e.g., Schilling text theorem 15.1) basically says that given an "original" measure space $(\Omega, \mathcal{A}, \mu)$ and a measurable map $T: \Omega \to \Gamma$,  integrating a measurable function $u: \Gamma \to \bar{\mathbb{R}}$ w.r.t. the image measure $T(\mu) = \mu \circ T^{-1}$ is the same as integrating the composite function $u \circ T$ w.r.t. the "original" measure $\mu$,
i.e.,
$$\int_{\Gamma} u \text{ d} T(\mu) = \int_\mathcal{\Omega} u \circ T \text{ d} \mu(z) $$

So to justify the identity "$\mathbb{E}_x[X]=\mathbb{E}_z[aZ+b]$ if $X = g(Z) = aZ + b$" that I previously asked about, we just need to apply the above theorem to change the measure with respect to which we integrate from the probability law $\mathbb{P}$ of the underlying sample space $\Omega$, to the distribution $\mathbb{P}_Z:=Z(\mathbb{P})$ of $Z$, and finally to the distribution $\mathbb{P}_X:= g(Z(\mathbb{P})) = g(\mathbb{P}_Z)$ of $X:=g \circ Z$:

$$\mathbb{E}[X]=\mathbb{E}[g \circ Z]:= \int_\mathcal{\Omega} g \circ Z \text{ d} \mathbb{P} = \int g \text{ d} \mathbb{P}_Z =  \int x \text{ d} \mathbb{P}_X(x) $$

(the $u$ function in the last change of measure step is just the identity function, $u(x) = x$, and we consider $g: z \to x$ to be a random variable with $ \mathbb{E}[g]: = \int g \text{ d} \mathbb{P}_Z$).

In my example, $g(z) = az + b$, and suppose $\mathbb{P}_Z$ has density $f_Z$, $\mathbb{P}_X$ has density $f_X$, then

$$\mathbb{E}[X] := \int g \circ Z \text{ d} \mathbb{P} = \int (az + b )\text{ d} \mathbb{P}_Z (z) = \int (az + b) f_Z(z) \text{ d}z  = \int x f_X(x) \text{ d} x $$
as desired.


答:
Without additional assumptions, this doesn't hold: Consider

$f(z) = \begin{cases} 1\text{, if }\Re(z) \ge 0 \\ 0 \text{, if } \Re(z) < 0 \end{cases}$

Title: Complex Analysis boundedness and limits

Body:
True or false. If $f: \mathbb{C}\to\mathbb{C}$ is bounded, then $\lim_{z\to 0} f(z)$ exists.

R If f is bounded means that there is some M∈R such that ∀z ∈C holds that |f(z)|≤M.

My Answer

Would $\frac{z}{|z|+1}$ be an example of such a function? Bounded above by 1 and below by -1.


问:
Second solution to $ \sin(x) - \sqrt{2} \cos(x) = 1$?
I think I'm missing a solution to the following problem:

Solutions to the equation $ \sin(x) - \sqrt{2} \cos(x) = 1$ are....

I get $x = 1.571 \pm 2 n \pi$. This matches the answer in the book I'm using, Engineering Mathematics, 5th Edition, by Stroud.

However, shouldn't there be 2 answers per period? Unlike their previous example, $ 4 \sin(x) - 3 \cos(x) = 5$, this solution isn't at the peak amplitude. I would think there is an additional solution as the function decreases after the peak?

I don't know how to find it if that's the case, or maybe I'm turned all around and thinking about this wrong. What am I missing? Thanks for your help.

答:
You are correct, there is another solution per period. I found it as follows. Start by squaring both sides to get
$$
(\sin(x)-\sqrt{2}\cos(x))^2 = \cos^2(x) - 2 \sqrt{2} \sin(x) \cos(x) + 1 = 1^2 = 1
$$
Then subtracting 1 from both sides, and dividing by $\cos^2(x)$ (you already found the solution where $\cos(x) = 0$) we have
$$
1 - 2 \sqrt{2} \tan(x) = 0.
$$
Solving then gives $\tan(x) = \sqrt{2}/4$, from which we find the solutions $\arctan(\sqrt{2}/4) + \pi n$. Checking these back in the original equation (we squared so we may have extra incorrect solutions!) we see that these solutions work exactly when $n$ is odd. That is, $x = \arctan(\sqrt{2}/4) + (2n+1)\pi$.


问:
General Inequality Problem.
The question goes like this :

Let $a,b,c,d$  be positive reals  and  given that  $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$

My approach goes like this:

I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$

Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$

Although I am not sure if its the correct method, I tried reducing the powers.

Then, I got:

$6(a+b)(a^2+b^2-ab) -a^2-b^2 \geqslant c^2+d^2 - 6(c+d)(c^2+d^2-cd) +\frac{1}{8} $

I tried substituting $c,d$ in terms of $a,b$ but the calculation is lengthier than I expected. Please check if my reasoning is correct, and help me solve this in a shorter method. Thanks!

答:
Use Tangent Line Method
$$6x^3-x^2 \geqslant \frac{5x-1}{8}, \quad \forall x > 0.$$

Title: General Inequality Problem.


问:
If $n$ is even, then $P(x)$ cannot have an inverse
Here is the problem:

Let $P(x)$ be a polynomial of degree $n$. If $n$ is even, prove that $P(x)$ cannot have an inverse over $\Bbb R$.

I am not exactly sure how to proceed from here. But I did start off a little bit. Since $n$ is an even, then there is some value $k$ such that $$n = 2k$$ But from here, I am not exactly sure how to proceed.

Some help is appreciated!

答: Assuem w.l.o.g. that the leading coefficient is positive. Then $P(x) \to \infty$ as $ x\to \pm \infty$. But any injective continuous function is strictly monotonic. These two facts contradict each other.


问:
Existance of limit for value of $a$
Consider the limit

$$ \lim_{x \to \infty} x^a \left ( \left(x+\frac{1}{x}\right)^\frac{2}{7} - x^\frac{2}{7} \right ) $$

Does there exist a value for $a$ for which the limit exists, and it's value is different from $0$ and $\infty$. I was able to confirm through trial that for $a = \frac{12}{7} $ the condition listed is satisfied.

I'm not sure how to approach this problem, I've tried to use LHopital's but that didn't work and got very messy quickly.

Cheers

Title: Existance of limit for value of $a$

Body:
Answer is $12/7$. In fact, write
$$
(x+1/x)^{2/7}-x^{2/7}=x^{2/7}[(1+x^{-2})^{2/7}-1].
$$
In general when $c>0$ we have $\lim_{z\to 0}\frac{(1+z)^c-1}{z}=c$ by l'Hosptal. So
$$
x^{\frac{12}{7}}[(x+1/x)^{2/7}-x^{2/7}]=x^{\frac{12}{7}}x^{2/7}\frac{(1+x^{-2})^{2/7}-1}{x^{-2}}x^{-2}=\frac{(1+x^{-2})^{2/7}-1}{x^{-2}}\to \frac 27.
$$


Title: Second solution to $ \sin(x) - \sqrt{2} \cos(x) = 1$?

Body:
I think I'm missing a solution to the following problem:

Solutions to the equation $ \sin(x) - \sqrt{2} \cos(x) = 1$ are....

I get $x = 1.571 \pm 2 n \pi$. This matches the answer in the book I'm using, Engineering Mathematics, 5th Edition, by Stroud.

However, shouldn't there be 2 answers per period? Unlike their previous example, $ 4 \sin(x) - 3 \cos(x) = 5$, this solution isn't at the peak amplitude. I would think there is an additional solution as the function decreases after the peak?

I don't know how to find it if that's the case, or maybe I'm turned all around and thinking about this wrong. What am I missing? Thanks for your help.

Body:
Let $\tan(x/2)=t$. then
$$\sin x-\sqrt{2} \cos x=1 \implies \frac{2t}{1+t^2}-\sqrt{2} \frac{1-t^2}{1+t^2}=1$$
$$\implies (\sqrt{2}-1)t^2+2t-(\sqrt{2}+1)=0 \implies t=1,-(3+2\sqrt{2})$$ So this equation has two branches of solutions:
$$x/2=\pi/4+n\pi \implies x=\pi/2+2n\pi,n0,1,2,...$$ and
$$x/2=-\tan^{-1}[(3+2\sqrt{2}]+m\pi\implies x=-2\tan^{-1}(3+2\sqrt{2})+2m\pi, m=0,1,2,..$$
$$\implies x\approx -0.89 \pi+2m\pi$$
In $[0,2\pi]$ two roots are $1.57, 3.48,$ when $n=0,m=1.$ See  the Fig. below:


问:
Showing a biconditional statement about function lim sups in $\Bbb R^n$, and codifying the intuition into a proof
$
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\BB}{\mathcal{B}}
\newcommand{\ve}{\varepsilon}
\newcommand{\para}[1]{\left( #1 \right)}
\newcommand{\set}[1]{\left\{ #1 \right\} }
$The Statement: Let $f : E \subseteq \R^n \to \R$. The claim to prove:

$$\beta = \limsup_{x \to x_0} f(x) \iff \text{conditions (i) & (ii) below hold}$$

The conditions:

(i) $\exists \set{x_n}_{n \in \N} \subseteq \R^n$ a sequence such that $f(x_n) \to \beta$

(ii) $(\forall b > \beta)(\exists \delta > 0)(f(x) < b \text{ for } x \in B^* \cap E)$

Notable Definitions & Notations: I think the more relevant definitions to go over would be these two:

We denote the "punctured ball" centered at $x_0$ with radius $\delta$ by $B^*$, and more explicitly may also write
$$B^* := B^*(x_0;\delta) := B(x_0;\delta) \setminus \set{x_0}$$
(Some of you may be more familiar with the notation $B_\delta(x_0)$ to denote a ball of radius $\delta$ centered at $x_0$.) Essentially the punctured ball $B^*$ is the usual ball with the center removed.

We define the limit supremum of $f$ as $x \to x_0$ as so:

$$\limsup_{x \to x_0} f(x) = \lim_{\delta \to 0} \left( \sup_{x \in B^* \cap E} f(x) \right)$$

Context & Attempts: Ultimately this is a homework problem, so I would prefer to not have full proofs given out; moreso just nudges and such.

Now, in the senses of geometry, visuals, and intuition, I think I have some ideas as to what is going on, though not quite the whole picture.

In the forward direction: let $\beta$ be $\limsup f(x)$. Then that means two things. Firstly, we can find a sequence $x_n$ in the domain where that sequence under $f$, $f(x_n)$, converges to $\beta$. I think that helps get $\beta \le \sup f(x)$ on the relevant set. Moreover, whenever we have a number $b$ larger than $\beta$, then condition (ii) essentially constricts $b$ and $\beta$ in a sense. The closer $b$ and $\beta$, the smaller $\delta$ is, and eventually I think that will help us get to equality.

In the backwards direction: we know there is a sequence $\set{x_n}$ where $f(x_n) \to \beta$, and whenever $b > \beta$, $x \in B^*(x_0;\delta) \cap E$ have $f(x) < b$ (though where $\beta$ lies between the two is not known). I think this would give that $\sup f(x) \le b$ for the relevant set, and as $\delta \to 0$, $b$ and $\beta$ get closer and closer together until we somehow get equality.

My issues with these are more like ... trying to codify my intuitions. I have written down some things, but I'm not too certain they're on the right track.

For instance, for the forward direction:

Attempt at the Forward Direction:

Suppose $f : E \subseteq \R^n \to \R$ and $\beta := \displaystyle \limsup_{x \to x_0} f(x)$. Then by definition,
$$
\beta = \lim_{\delta \to 0} \para{ \sup_{x \in B^*(x_0;\delta) \cap E} f(x) }
$$
Let $\delta_k := 1/k$. Then define $\BB_k := B^*(x_0;\delta_k)$. Since $\delta_k \to 0$, it holds that
$$
\sup_{x \in \BB_k \cap E} f(x) \to \beta
$$
Thus, define $x_k$ by $x_k \in \BB_k \cap E \subseteq E$, and then $\set{x_k}_{k \in \N}$ is one such that $f(x_k) \to \beta$, satisfying condition (i).

Suppose $b > \beta$. To see condition (ii) holds, suppose otherwise, that $\forall \delta > 0$, $f(x) \ge b$ for $x \in B^* \cap E$. This would mean that
$$
\sup_{x \in B^*(x_0;\delta) \cap E} f(x) \ge b > \beta
$$
for all $\delta > 0$ and in particular the case $\delta \to 0$. This would give that $ \displaystyle\limsup_{x \to x_0} f(x) = b$ instead from the definition, contradicting that $\beta$ is the lim sup. Thus, a contradiction is reached and condition (ii) holds.

Attempt at the Backwards Direction: (which is moreso scratch work that stalled hard)

We're going to suppose (i) & (ii) hold. Since $\exists \set{x_k}_{k \in \N} \subseteq E$ such that $f(x_k) \to \beta$, from definition, $\forall \ve > 0$, $\exists N \in \N$ such that, $\forall n \ge N$, $|f(x_n) - \beta| < \ve$. In particular, we can consider the elements of $\set{x_k}$ which are within a radius $\delta > 0$ of $x_0$ and are different from $x_0$; these elements define a subsequence $\set{x_{n_k}}_{k \in \N} \subseteq B^* \cap E$.

From the fact that limits are unique, and if a sequence $z_k \to L$, then all subsequences of $\set{z_k}_{k \in \N}$ converge to $L$ as well. Thus, $\set{x_{n_k}}_{k\in\N}$ is a sequence in $B^* \cap E$. If we take the limit as $\ve,\delta \to 0$, it then follows from these and the definition of supremum that
$$
\lim_{\delta \to 0} \para{ \sup_{B^* \cap E} f(x) } \le \beta
$$

Consider condition (ii) and that for a $b > \beta$, $\exists \delta > 0$ such that $f(x) < b$ for $x \in B^* \cap E$.

(That's where I get lost though...)

Questions & Concerns:

Is my intuition for what's going on correct?

If not, what's the correct intuition and the general approach I should use?

Is there anything remotely salvageable from my approaches?

If there are errors with my approach, what are they? How might I rectify them?

Thanks for any insight you can give me!

Edit: (thoughts for converse as of 1/27/2021)

I had a thought for the approach that seems to neatly tie conditions (i) and (ii) together.

Let $\set{b_k}_{k \in \N}$ be a monotone decreasing sequence with limit $\beta$, i.e. $b_k > \beta$ for every $k$, and $b_k \searrow \beta$. Define $\ve_k := b_k - \beta$.

Consider the preimage $f^{-1}(\beta,\ve_k)$. This may consist of a set of disjoint sets, so consider only the one $\mathcal{A_k}$ the point $x_0$ lies in. For each preimage determined by $\ve_k$, we can get a $\mathcal{A_k}$, containing a ball $B(x_0;\delta_k)$.

Thus, each $\ve_k$ determines a preimage, which determines a ball (a subset of the preimage), which determines a $\delta_k$.

Define a set of points $x_k$ by being in that preimage and different from $x_0$, i.e. take $x_k \in B^*(x_0;\delta_k)$.

Obviously, as $b_k \to \beta$, then $\delta_k \to 0$. Moreover, $x_k \to x_0$, and $f(x_k) \to \beta$.

For each $k \in \N$, we have that

$$\sup_{x \in B^*(x_0;\delta_k) \cap E} f(x) = b_k$$

and, as $\delta_k \to 0$, we have $b_k \to \beta$, giving the lim sup as desired.

(I'm sure there are details to iron out in this, but I feel there's a grain of truth at the solution in there somewhere...)

答:
We assume that in (i) $\{x_n\}$ is a sequence of points of $E\setminus\{x_0\}$, converging to $x_0$.

The statement is technical. To avoid to be lost in details, we shall follow a path, which is a suitable chosen auxiliary sequence. Namely, for each natural $n$ put $\beta_n=\sup_{x\in B^*(x_0,1/n)\cap E} f(x)\in \Bbb R\cup\{\pm\infty\}$. Clearly, a sequence $\{\beta_n\}$ is non-increasing, so there exists $\beta^*=\lim_{n\to\infty}\beta_n\in \Bbb R\cup\{\pm\infty\}$. Since the sequence $\{\beta_n\}$ is non-increasing and for each $\delta>0$ there exists natural $n$ such that $B^*(x_0,1/n)\subset B^*(x_0,\delta)$, $\beta^*=\lim\sup_{x \to x_0} f(x)$.

$(\Rightarrow)$ (i) I assume that $x_0$ is an accumulation point of the set $E$, that is $x_0\in\overline{E\setminus \{x_0\}}$, otherwise the conclusion can fail. The following case are possible:

1)) $\beta^*=-\infty$. Then for each natural $n$ we can pick $x_n\in B^*(x_0,1/n)\cap E$ such that $f(x_n)<-n$.

2)) $\beta^*=+\infty$. Then for each natural $n$ we can pick $x_n\in B^*(x_0,1/n)\cap E$ such that $f(x_n)>n$.

3)) $\beta^*\in\Bbb R$. Then for each natural $n$ we can pick $x_n\in B^*(x_0,1/n)\cap E$ such that $|f(x_n)-\beta^*|<1/n$.

(ii) Since $b>\beta^*$ and the sequence $\{\beta_n\}$ tends to $\beta^*$, there exists $n$ such that $\beta_n<b$. We can pick $\delta=1/n$.

$(\Leftarrow)$ Let $\beta<b\in \Bbb R\cup\{+\infty\}$. By (ii), $\beta_n\le b$ for all sufficiently big $n$, so $\beta^*\le b$, thus $\beta^*\le\beta$. Let $m$ be any number. Since $\{x_n\}$ is a sequence of points of $E\setminus\{x_0\}$, converging to $x_0$,
there exist a natural number $N$ such that $x_n\in B^*(x_0,1/m)$ for each $n\ge N$, so $f(x_n)\le\beta_m$. Therefore $\beta\le\beta_m$. Thus $\beta\le\beta^*$ and so $\beta=\beta^*$.


问:
$A$ is representable as a difference of two closed sets if and only if $\overline A \setminus A$ is closed.

Show that the set $A$ is representable as a difference of two closed sets if and only if $\overline A \setminus A$ is closed.

My try:

$(\Leftarrow)$ Suppose $\overline A \setminus A$ is closed. Since $\overline A$ is closed and  $A = \overline A \setminus (\overline A \setminus A)$, we see that $A$ is represented as a difference of two closed sets. $\square$

$(\Rightarrow)$ Suppose $A = B\setminus C$ where $B$ and $C$ are closed sets. We need to show that $\overline A \setminus A$ is closed. Since $\overline A = \overline {B\setminus C}$, we have $$\overline A \setminus A = (\overline {B\setminus C}) \setminus (B \setminus C) = \partial (B\setminus C)$$ Since the boundary is closed (?), $\overline A \setminus A$ is closed $\square$

Last arguments of mine seem to be vague not clear. How to solve the problem rigorously?

答:
Partial answer

Your equality $$(\overline {B\setminus C}) \setminus (B \setminus C) = \partial (B\setminus C)$$ is wrong.

For a set $S$ you have

$$\partial S = \overline S \setminus \mathring{S}.$$


答:
Proof of $\implies$:

$\overline A \setminus A =\overline  {(B\setminus C)} \cap (B^{c} \cup C)$. Since $\overline  {(B\setminus C)} \cap B^{c} $ is empty we get $\overline A \setminus A =\overline  {(B\setminus C)} \cap  C$ which is closed.

Body:
Show that the set $A$ is representable as a difference of two closed sets if and only if $\overline A \setminus A$ is closed.

My try:

$(\Leftarrow)$ Suppose $\overline A \setminus A$ is closed. Since $\overline A$ is closed and  $A = \overline A \setminus (\overline A \setminus A)$, we see that $A$ is represented as a difference of two closed sets. $\square$

$(\Rightarrow)$ Suppose $A = B\setminus C$ where $B$ and $C$ are closed sets. We need to show that $\overline A \setminus A$ is closed. Since $\overline A = \overline {B\setminus C}$, we have $$\overline A \setminus A = (\overline {B\setminus C}) \setminus (B \setminus C) = \partial (B\setminus C)$$ Since the boundary is closed (?), $\overline A \setminus A$ is closed $\square$

Last arguments of mine seem to be vague not clear. How to solve the problem rigorously?


答:
$\cfrac {3\sqrt{2x-1}- \cfrac{(3x+4)}{\sqrt{2x-1}}} {2x-1}$

Take $LCM$ of the numerator term.

$\cfrac{\cfrac {3\sqrt{2x-1}(\sqrt{2x-1})- {(3x+4)}}{\sqrt{2x-1}}}{2x-1}$

You get:

$\cfrac{\cfrac {3(2x-1)- {(3x+4)}}{\sqrt{2x-1}}}{2x-1}$

$\cfrac {\cfrac{6x-3- 3x -4}{\sqrt{2x-1}}}{2x-1}$

$\cfrac {\cfrac {3x-7}{\sqrt{2x-1}}}{2x-1}$

And then, the powers add up, by the addition rule of exponents

so the final answer comes out to be :

${\cfrac {3x-7}{{(2x-1)^\frac{3}{2}}}}$

Title: How does $\frac{3\sqrt{2x-1}-\frac{3x+4}{\sqrt{2x-1}}}{2x-1}$ simplify to $\frac{3x-7}{(2x-1)^{3/2}}$?

Body: How does $$\frac{3\sqrt{2x-1}-\dfrac{3x+4}{\sqrt{2x-1}}}{2x-1}$$ simplify to $$\frac{3x-7}{(2x-1)^{3/2}}$$


问:
order of a class-preserving automorphism of $G$ have all its prime factor divides $|G|$
Let $\alpha$ be an automorphism of finite group $G$, if $\alpha$ maps any element $g$ to its conjugate $x_ggx_g^{-1}$ ($x_g$ could depend on g, so $\alpha$ is not necessarily inner), then the order of $\alpha$ has its prime factors all divide $|G|$.

These automorphisms are called class-preserving automorphisms denoted $\mathrm{Aut}_c(G)$. source [1] says

"One of the basic facts about class-preserving automorphisms is that primes
dividing the order of $\mathrm{Aut}_c(G)$ also divide the order of $G$", which easily leads to the above result, however I can't find its proof anywhere.

[1] Class-Preserving Automorphisms of Finite Groups,
Martin Hertweck, doi:10.1006/jabr.2001.8760

答: Let $\alpha$ be a class-preserving automorphism of $G$. WLOG we can assume that $\alpha$ has order $p$, a prime number. So assume $p$ does not divide $|G|$. Since $\alpha$ acts on a conjugacy class, the orbits are of length $p$ or $1$, but they cannot be all $p$ since then $p$ would divide the cardinality of the class and hence (this cardinality is an index of $G$) $p$ would divide $|G|$. So for each class $K_i$ we have a fixed-point $x_i$ $(i=1, \cdots , k$, where $k=k(G)$ the number of conjugacy classes of $G$.) Put $H=\langle x_1, x_2 , \cdots , x_k \rangle$. Then $\bigcup_{g \in G}H^g=G$. But this can only happen (this is a classical argument, see for example here) when $G=H$, implying $\alpha$ acts trivially on $G$, that is $o(\alpha)=p=1$, a contradiction.


问:
Computationally simple way to find N26 direction
Assume we have an integral 3D grid $\mathbb{Z}^3$ and a set of line segments $V = \{((0, 0, 0), (x, y, z)): x, y, z \in \mathbb{Z} \land (x, y, z) \ne (0, 0, 0)\}$ from its origin to another grid point. I would like to find the closest (by Euclidian metric or by an angle between $v$ and $\vec{0p}$) single point $p \in N_{26}$, where $N_{26} = \{(x, y, z): x, y, z \in \{-1, 0, 1\} \land (x, y, z) \ne (0, 0, 0)\}$, for any $v \in V$. This is equivalent to finding a primary N26 direction of $v$. If there is a technical term for it, I would like to know.

I would like the method to be computationally efficient, perhaps avoiding using trigonometric functions. It doesn't have to be perfectly accurate: a few percent error is fine if it simplifies calculations.

答:
Without loss of generality assume the variable endpoint of $v$ is $(x,y,z)$ where $x\ge y\ge z\ge0$. There are only three points to test – $(1,0,0)$, $(1,1,0)$ and $(1,1,1)$.

Knowing that $u\cdot v=|u|\cdot|v|\cos\theta$, compute the three values
$$v\cdot(1,0,0)=x$$
$$\frac{v\cdot(1,1,0)}{\sqrt2}=\frac{x+y}{\sqrt2}$$
$$\frac{v\cdot(1,1,1)}{\sqrt3}=\frac{x+y+z}{\sqrt3}$$
The vector that gives the largest result (for $\cos$ is decreasing on $[0,\pi]$) is the correct primary $N_{26}$ direction.


问:
I am confused about strong induction.
I am confused strong induction.....
For example,

We consider $p(n)$:$12|(n^4-n^2$).

Clearly, $p(1)$ is true.
In strong induction, assume $p(1),p(2),,...,p(k)$ is true.
We will show that $p(k-5) \Rightarrow p(k+1)$.

But, since $k-5\geq1$, we have $k \geq6$. Then in this case, do we have basis step $p(1),p(2),...,p(6)$ ? and then do we also have to check these are all true ?

What I am confused about is as follows:

At first, we already assumed $p(1),p(2),...,p(k)$ are all true. Nevertheless, do we necessarily have to check that basis step $p(1),...,p(6)$ are all true ?

答: If you are able to prove that $p(k-5)\implies p(k+1)$, then you will have to prove that $p(1)$, $p(2)$, $p(3)$, $p(4)$, $p(5)$, and $p(6)$ hold. After that, you shall have proved that $(\forall n\in\Bbb N):p(n)$ by strong induction. Or you can prove by induction that the statement $p(n)$ when $n$ is of the form $6k$, then when $n$ is of the form $6k+1$ and so on until $6k+5$.

Title: I am confused about strong induction.


问:
How do you take the multiplicative inverse of a p-adic number?
I am reading the wiki page for p-adic numbers and it states that they are a field extension of the rationals so each member has to have a modular multiplicative inverse.
So how would I take the inverse of, say, 35 in the ring of 2-adic numbers?</p>

答:
With Mathematica, the following command will provide you an answer for 20 digits :

BaseForm[PowerMod[35, -1, 2^20], 2]

= 10001010111110001011 (base 2)


答:
$$\sin x-\sqrt2\cos x=\sqrt3\left(\frac1{\sqrt3}\sin x-\sqrt{\frac{2}3}\cos x\right)=\sqrt3\sin\left(x-\arccos\frac1{\sqrt3}\right)$$

Therefore $$\sin x-\sqrt2\cos x=1\iff\sin\left(x-\arccos\frac1{\sqrt3}\right)=\frac1{\sqrt3}$$

In $[0,2\pi)$, this has solutions $$x=\arcsin\frac1{\sqrt3}+\arccos \frac1{\sqrt3}\lor x=\pi-\arcsin\frac1{\sqrt3}+\arccos\frac1{\sqrt3}$$

which are $\frac\pi2$ and $\approx3.481$ respectively. So you're right, one is missing.


问:
Looking for analytical proof that this function given as a power series is constant.
Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$

Is there an analytical way to prove that $P_t(x)$ is a continuous function on $x\in(0,1)$ and:

$P_t(x)=1$ for all $x:\ 0\le 1-x\le\frac1t$,

$P_t(x)<1$ for all $x:\ \frac1t< 1-x\le1$.

I proved this for trivial cases $t=1,2$ but did not find a way to deal with general $t$.

The behavior of the function for $t=1,2,3,4$ is demonstrated below:

Body:
Too long for comments.

Assuming that $t$ is a positive integer, there is no problem to get explicit formulae up to $t=3$.

For $t \geq 4$ appear hypergeometric function and the result write
$$P_t(x)=\frac x {1-x} \big[Q_t(x)-1\big]$$
$$Q_4(x)=\,
   _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{2}{3},\frac{4}{3};\frac{4^4}{3^3} (1-x) x^3\right)$$
$$Q_5(x)=\,
   _4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{2}{4},\frac{3}{4},\frac{5}{4};\frac{5^5}{4^4} (1-x) x^4\right)$$
$$Q_6(x)=\,
   _5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5};\frac{6^6} {5^5}(1-x) x^5\right)$$
$$Q_7(x)=\,
   _6F_5\left(\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7};\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},\frac{7}{6};\frac{7^7}{6^6} (1-x) x^6\right)$$

For $x=x_*=1-\frac 1t$, $Q_t(x_*)=\frac t{t-1}$ and then $P_t(x_*)=1$ as you observed.


答:
You are correct that you would need 6 base cases for an induction proof using the inductive step $p(k-5) \to p(k+1)$. The assumption that $p(1),p(2),\dots,p(k)$ are true is not really related to this however. In fact, you don't need that many assumptions, just assuming $p(k-5)$ will do.

In general you may need more than one assumption in the inductive step, but the number of base cases needed is the distance to the 'earliest' assumption you needed, regardless of how many assumptions you have.

Title: I am confused about strong induction.

Body:
I am confused strong induction.....
For example,

We consider $p(n)$:$12|(n^4-n^2$).

Clearly, $p(1)$ is true.
In strong induction, assume $p(1),p(2),,...,p(k)$ is true.
We will show that $p(k-5) \Rightarrow p(k+1)$.

But, since $k-5\geq1$, we have $k \geq6$. Then in this case, do we have basis step $p(1),p(2),...,p(6)$ ? and then do we also have to check these are all true ?

What I am confused about is as follows:

At first, we already assumed $p(1),p(2),...,p(k)$ are all true. Nevertheless, do we necessarily have to check that basis step $p(1),...,p(6)$ are all true ?


答:
The first half is of course fine. The problem with the second half is that

$$\operatorname{cl}(B\setminus C)\setminus(B\setminus C)$$

is not necessarily the boundary of $B\setminus C$. In the real line, for instance, take $B=[0,2]$ and $C=[0,1]$; then $B\setminus C=(1,2]$ and $\operatorname{cl}(B\setminus C)=[1,2]$, so

$$\operatorname{cl}(B\setminus C)\setminus(B\setminus C)=\{1\}\,,$$

but $\operatorname{bdry}(B\setminus C)=\{1,2\}$.

Clearly $A\subseteq B$, and $B$ is closed, so $\operatorname{cl}A\subseteq B$, and therefore

$$(\operatorname{cl}A)\setminus C\subseteq B\setminus C=A\,.$$

On the other hand, $A\cap C=\varnothing$, so $A\subseteq(\operatorname{cl}A)\setminus C$. Thus, $A=(\operatorname{cl}A)\setminus C$.

To complete the argument, show that $(\operatorname{cl}A)\setminus A=C\cap\operatorname{cl}A$, which is of course closed.

Title: $A$ is representable as a difference of two closed sets if and only if $\overline A \setminus A$ is closed.


问:
How to express $\ln\left(\frac{1}{1+j}\right)$ in rectangular form?
Trying to understand how to express this complex number $\ln\left(\frac{1}{1+j}\right)$ into rectangular form.

I thought to use $x = r\cos(\theta)$ and  $ y = r\sin(\theta)$, but I am pretty sure that is wrong.

答:
$$\ln\frac1{1+i}=-\ln(1+i)=-\ln\sqrt2e^{i\pi/4}=-(\ln\sqrt2+i\pi/4)=\ln\frac1{\sqrt2}-i\pi/4$$
with the usual caveat about $\ln$'s multivaluedness.


答: If by $\ln$, you mean the main branch of the complex logarithm (remember, the complex logarithm is a fairly complicated topic), then use the fact that $$\log z = \log|z| + i\cdot \mathrm{Arg}(z).$$

Title: How to express $\ln\left(\frac{1}{1+j}\right)$ in rectangular form?

Body:
Trying to understand how to express this complex number $\ln\left(\frac{1}{1+j}\right)$ into rectangular form.

I thought to use $x = r\cos(\theta)$ and  $ y = r\sin(\theta)$, but I am pretty sure that is wrong.


问:
Solutions of $x\frac{\partial \psi}{\partial x} + y \frac{\partial \psi}{\partial y} + \psi = f(x)e^{-2\pi i y}$?
I stumbled across the following PDE for a function $\psi(x, y)$:
$$
x\frac{\partial \psi}{\partial x} + y \frac{\partial \psi}{\partial y} + \psi = f(x)e^{-2\pi i y}
$$
where $f(z)$ is some arbitrary function. I was wondering if anyone knows anything about equations of this kind or how to solve them. Separation of variables didn't seem to work, so I'm fresh out of ideas.</p>

答:
Maple says
$$
\psi(x,y) = \frac{1}{x}\left(\int_a^xf(t)\exp\left(\frac{-2\pi i y t}{x}\right)\;dt+F\left(\frac{y}{x}\right)\right)
$$
where $F$ is an arbitrary function.</p>


问:
Please help me understand this geometry problem
ABCD is a quadrilateral; P,Q,R and S are the points of trisection of sides AB,BC,CD and DA respectively and are adjacent to A and C; prove that PQRS is a parallelogram.

In the above question it is mentioned that "P,Q,R,S" are the points of trisection.What does this statement mean?

As far as I know points of trisections are the points which divides the line segment into three equal parts and for this, there must exist two points on that segment.But in the above question there exists only one point on a  given line segment

For example in question it is provided that P is the point of trisection of segment AB.

Where am I wrong?

答:
As stated in several comments, e.g., achille hui's comment, the problem is indicating that when choosing among any of the mentioned $2$ possible internal trisection points, always choose the ones closest (which would also be the ones most "adjacent") to the points $A$ and $C$. This is shown in the diagram below. Also, the green $w$, $x$, $y$ and $z$ represent the lengths between each set of respective trisection points, and I've added the quadrilateral diagonals in red.

Next, note

$$\frac{|DR|}{|DC|} = \frac{2y}{3y} = \frac{2}{3} = \frac{2z}{3z} = \frac{|DS|}{|DA|} \tag{1}\label{eq1A}$$

Since $\angle RDS$ is a common angle, we have by SAS (i.e., side-angle-side, with this being the second bullet point in the triangle similarity section) that $\triangle RDS \sim \triangle CDA$ (i.e., the triangles are similar). Thus, $\angle DRS = \angle DCA$ and $\angle DSR = \angle DAC$. This means $SR \parallel AC$ (due to the corresponding angles of the transversal being congruent, which is point $3$ in the Condition for parallelism section). Repeating these arguments with the corresponding triangles on the other side of $AC$ shows $\triangle PBQ \sim \triangle ABC$ and $PQ \parallel AC$. Since $SR$ and $PQ$ are each parallel to $AC$, this means $SR \parallel PQ$.

Repeating what was done above with the angles and triangles, but with using the diagonal $BD$ instead, results in $PS \parallel QR$. This means both sets of opposite sides of $PQRS$ are parallel to each other, thus making it a parallelogram (this is the first bullet point in the Characterizations section, but note I could have instead (or in addition to) used the similar triangle properties to get that $\left| SR \right| = \frac{2}{3}\left| AC \right| = \left| PQ \right|$ and $\left| PS \right| = \frac{1}{3}\left| BD \right| = \left| QR \right|$, meaning the opposite sides' lengths are the same, thus matching the second bullet point).


答:
Tangent line detailed:

You can rewrite your inequality as $$f(a)+f(b)+f(c)+f(d)\geq {1\over 8}$$ where $f(x)= 6x^3-x^2$. Now we see that we have equality case if $a=b=c=d={1\over 4}$ so it is reasonable to calculate tangent line for $f$ at $x= {1\over 4}$ which is $g(x)={5x-1\over 8}$.

If you draw a graph for $f$ and $g$ you see that $f$ is all the time over $g$ for $x\in[0,1]$, so $$f(x)\geq g(x)$$ This, of course, you should verify with some calculations. So $$f(a)+f(b)+f(c)+f(d)\geq g(a)+g(b)+g(c)+g(d)= {1\over 8}$$

Title: General Inequality Problem.

Body:
The question goes like this :

Let $a,b,c,d$  be positive reals  and  given that  $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$

My approach goes like this:

I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$

Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$

Although I am not sure if its the correct method, I tried reducing the powers.

Then, I got:

$6(a+b)(a^2+b^2-ab) -a^2-b^2 \geqslant c^2+d^2 - 6(c+d)(c^2+d^2-cd) +\frac{1}{8} $

I tried substituting $c,d$ in terms of $a,b$ but the calculation is lengthier than I expected. Please check if my reasoning is correct, and help me solve this in a shorter method. Thanks!


答:
$$\ln\Bigl(\frac{1}{1+i}\Bigr)=-\ln\Bigl(1+i\Bigr)= -\ln\Bigl(\sqrt{2}\Bigl(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\Bigr)\Bigr)$$

$$-\ln\Bigl(\sqrt{2}\Bigl(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\Bigr)\Bigr)=-\ln\Bigl(\sqrt{2}\Bigl(\cos\frac{\pi}{4}+{i}\sin\frac{\pi}{4}\Bigr)\Bigr)$$
$r(\cos\theta +i\sin\theta)$ can be written as $re^{i\theta}$

$$-\ln\Bigl(\sqrt{2}\Bigl(\cos\frac{\pi}{4}+{i}\sin\frac{\pi}{4}\Bigr)\Bigr)=-\ln\Bigl(\sqrt{2}\Bigl(e^{i\frac{\pi}{4}}\Bigr)\Bigr)=-\ln\sqrt{2} -\ln{e^{i\frac{\pi}{4}}}$$

$$= \ln\frac{1}{\sqrt{2}} -i\frac{\pi}{4}$$


Title: General Inequality Problem.

Body:
By Chebyshev's inequality,

$\frac{a^2+b^2+c^2+d^2}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a+b+c+d}{4}$

$\implies a^2+b^2+c^2+d^2 \geq \frac{1}{4}$

Similarly,

$\frac{a^3+b^3+c^3+d^3}{4} \geq \frac{a+b+c+d}{4} \cdot \frac{a^2+b^2+c^2+d^2}{4}$

$\implies 4(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2$

$6(a^3+b^3+c^3+d^3) \geq a^2+b^2+c^2+d^2+ \frac{a^2+b^2+c^2+d^2}{2} \geq a^2+b^2+c^2+d^2+ \frac{1}{8}$


问:
Prove function is always zero.
Suppose $f$ is a function defined on $\mathbb{R}^2$. If for any square ABCD,

$f(A)+f(B)+f(C)+f(D)=0$.

How to prove that $f\equiv0$? Thanks!

答:
We take two squares A, B, C, D and C, D, G, H which share the C-D side. Then, it follows that $f(A)+f(B)=f(G)+f(H)$.

Now, A, G, H, B is a rectangle. Take another rectangle A, I, J, G which shares the A-G side. It follows that $f(I)+f(A)=f(J)+f(G)$.

We get $f(B)-f(I)=f(H)-f(J)$, subtracting the previous two equations. But then since B, I, G, H is square, we also have $f(B)+f(I)+f(H)+f(J)=0$. Together we get $f(I)+f(H)=0$.

This means, the sum of two diagonals of a square is always zero.

Next, inscribe a square E, F, G, H inside another A, B, C, D using the middle points of the edges of the latter. G-H is a diagonal, as is F-H. Therefore, $f(G)=f(F)$.

Similarly, we can get $f(G)=f(F)=f(E)=f(H)=0$, since the sum of all is 0.

Thus $f$ is identically zero since the choice of vertices was arbitrary.

Title: Prove function is always zero.


问:
Equation of a line passing through a point and forming a triangle with the axes
How can I find the equation of a line that;

is passing through the point (8, 6) and

is forming a triangle of area 12 with the axes

?

So I tried to start using $A = |{\frac{mn}{2}}|$ and got that $m\cdot n$ is either -24 or 24. What now?

答:
Equation of lines that passes through $(8,6)$ is $y=ax+6-8a$
and with axes that has common points $(0,6-8a),(8-\frac{6}{a},0)$. Then the surface of right triangle is
$$S=\frac{(8-\frac{6}{a})(6-8a)}{2}=12$$

after simplification we get equation

$$16a^2-30a+9=0$$
with solutions
$$a_1=6,a_2=24$$
there exists two lines with such proppertie
$$l_1:y=6x-42$$
$$l_2:y=24x-186$$


问:
Time Complexity for $T(n)=T(\sqrt{n})+\sqrt{n}$
Can you help me figure out the time complexity of $T(n)=T(\sqrt{n})+\sqrt{n}$

答: Try the Ansatz $T(n)\propto n^p$ so $n^p\sim n^{p/2}+n^{1/2}$. While $p=\tfrac12$ works, if $p<\tfrac12$ then $n^p\in o(n^{p/2}+n^{1/2})=o(n^{1/2})$, whereas if $p>\tfrac12$ then $n^{p/2}+n^{1/2}\sim n^{\max(1/2,\,p/2)}\in o(n^p)$.


答:
It could be interesting to make the problem more generat
$$\lim_{x \to \infty} x^a \left ( \left(x+\frac{1}{x}\right)^b - x^b \right )$$
$$ \left(x+\frac{1}{x}\right)^b - x^b =x^b\left(\left(1+\frac{1}{x^2}\right)^b-1\right)=x^b \left(\frac{b}{x^2}+\frac{(b-1) b}{2
   x^4}+O\left(\frac{1}{x^6}\right)\right)$$

So, for large values of $x$
$$x^a \left ( \left(x+\frac{1}{x}\right)^b - x^b \right )\sim b x^{a+b-2}$$ which, for a finite limit (which will be $b$), requiers $a=2-b$

Title: Existance of limit for value of $a$

Body:
Consider the limit

$$ \lim_{x \to \infty} x^a \left ( \left(x+\frac{1}{x}\right)^\frac{2}{7} - x^\frac{2}{7} \right ) $$

Does there exist a value for $a$ for which the limit exists, and it's value is different from $0$ and $\infty$. I was able to confirm through trial that for $a = \frac{12}{7} $ the condition listed is satisfied.

I'm not sure how to approach this problem, I've tried to use LHopital's but that didn't work and got very messy quickly.

Cheers


问:
In how many ways can a team of $4$ boys and $4$ girls be selected out of $6$ boys and $5$ girls if a particular boy and girl refuse to work together?

In how many ways can a team of $4$ boys and $4$ girls be selected out of $6$ boys and $5$ girls if a particular boy and a particular girl refuse to work together.

My attempt:

Case 1: That particular boy is not considered. So, 4 boys to be selected out of 5 boys. 4 girls to be selected out of 5 girls. So, $\dbinom{5}{4} \cdot \dbinom{5}{4}=25$

Case 2: That particular girl is not considered. So, 4 boys to be selected out of 6 boys. And 4 girls to be selected out of 4 girls. So, $\dbinom{6}{4}\cdot\dbinom{4}{4}=15$

Case 3: That particular boy and girl are not considered. So, 4 boys out of 5 and 4 girls out of 4. So, $\dbinom{5}{4}\cdot\dbinom{4}{4}=5$

So, total cases: $45$.

Answer is given as $35$.

What's my mistake?

答:
Total cases are 40 and refused cases five so 40 -5 = 35
Or      no of cases by all students = 75 and
No of cases with unplayed students=40 so answer
Is 35</p>


问:
For which $a \in \mathbb R$ does the line $y=a$ cut the graph of a sine function?
I really dont get this question. Dont I need to know what $a$ is?

答:
Here is a more expanded version of the same problem. Maybe that helps you understand it better:

Does the line $y=0$ cut the graph of a sine function? What about $y=2$? What about $y=-\frac12$? These are three possible lines, but there are many more. How can you tell, just by looking at the right-hand side of such an expression, whether the line it describes intersects the sine graph?


问:
Proving that multiplication acts linearly on the complex plane as the composition of a dilation and a rotation.
Primary purpose of this problem: To prove that the multiplication by $z_0 \neq 0$ acts linearly on the complex plane as the composition of a dilation and a rotation.

I have been asked to prove that there exists $r_0>0$ and $\theta_0 \in \mathbb{R}$ such that the matrix $A_{z{_0}}$ can be decomposed as:

$$A_{z{_0}}=\begin {bmatrix} r_0 & 0 \\ 0 & r_0 \end{bmatrix} \begin {bmatrix}\cos(\theta_0) & -\sin(\theta_0)\\ \sin(\theta_0) & \cos(\theta_0) \end{bmatrix} =: D_{r_{0}}R_{\theta_{0}}$$ where $D_{r_{0}}$ represents dilation and $R_{\theta_{0}}$ represents rotation.

I have the basic intuition regarding the concepts in this problem. The idea of multiplying these two matrices allows us to get combination of Dilation and Rotation, however, I am not sure exactly how to "prove" this.

A few relations that I have come up with:

The matrix expansion of $A_{z_0}=\begin {bmatrix} r\cos(\theta_0) -r\sin(\theta_0)\\ r\sin(\theta_0) + r\cos(\theta_0) \end{bmatrix}$

Let $z=x+iy$. In polar, this is represented by $z=re^{i\theta}=r \left(\cos(\theta)+i(\sin(\theta)\right)$. I was also able to identify that the rotation matrix comes from: $\cos(\theta)+i(\sin(\theta))$

If we use Matrices to represent Euler: $z=re^{i\theta}=\left(\cos(\theta)+i(\sin(\theta)\right)$, we will have $$z=\begin {bmatrix} \cos(\theta_0) & -\sin(\theta_0)\\ \sin(\theta_0) & \cos(\theta_0) \end{bmatrix}
\begin {bmatrix} r_0 \\ r_0 \end{bmatrix}=\begin {bmatrix} r\cos(\theta_0) -r\sin(\theta_0)\\ r\sin(\theta_0) + r\cos(\theta_0) \end{bmatrix}$$

However, I need some help understanding how to actually prove what the $r_0>0$ and $\theta_0 \in \mathbb{R}$ is.

答: If $z_0\in\Bbb C\setminus\{0\}$, then $z=\rho_0\bigl(\cos(\theta_0)+i\sin(\theta_0)\bigr)$, for some $\rho_0>0$ and some $\theta_0\in\Bbb R$. So, if $z\in\Bbb C$, and if you write $z$ as $a+bi$ ($a,b\in\Bbb R$), then$$z_0\times z=\rho_0\bigl(\cos(\theta_0)a-\sin(\theta_0)b+\bigl(\cos(\theta_0)b+\sin(\theta_0)a\bigr)i\bigr)$$and therefore multiplication by $z_0$ the composition of $w\mapsto \rho_0 w$ (a dilation)  with the rotation whose matrix with respect to the basis $\{1,i\}$ is $\left[\begin{smallmatrix}\cos(\theta_0)&-\sin(\theta_0)\\\sin(\theta_0)&\cos(\theta_0)\end{smallmatrix}\right]$.


问:
How can I "see" that calculus works for multidimensional problems?
Let's say I have some function f(x) = x^2 + b. I can see what's going on, I can count the slope geometrically even without knowing the rules of derivatives.

When I need to minimize some cost function for linear problem (linear regression with gradient descent), I just need to picture 3d, and I can "see", and be quite confident why and how it works.

How can I "see" or get intuition that calculus works for multidimensional problems? Let's say I have a problem with many variables like:

f(area_km, bedrooms, ...) = theta + theta1 * area_km + theta2 * bedrooms etc

If I want to apply gradient descent, I know I need to calculate partial derivatives and multiply it with a learning rate etc. It works. But it's kinda magical that it works.

I am sorry this is a silly question, I am just beginning.

答: When I think about a function $f: \mathbb R^n \to \mathbb R$, I can only visualize the cases where $n = 2$ or $n = 3$ (or $n = 1$, of course). I like to imagine that $f$ tells us the temperature at each point in space. A mosquito who likes cool temperatures is flying around. Which way should it move? The gradient vector at its current location points in the direction in which the temperature is increasing most rapidly. So the mosquito should move a little bit in the opposite direction, which is the direction in which the temperature is decreasing most rapidly. Then the mosquito should recompute the gradient of $f$ at its new location, and move a bit again in the negative gradient direction. And so on. This is gradient descent.


问:
Is the the function $f(k) = \frac{k\ln 2}{2\ln k}$ strictly increasing at $k \ge 10$
Is the the function $f(k) = \frac{k\ln 2}{2\ln k}$ strictly increasing at $k \ge 10$ where $\ln$ is the natural log.

I believe that the answer is yes.

Here's my thinking:

$\dfrac{d}{dk}\left(\frac{k\ln 2}{2\ln k}\right) = \dfrac{\ln 2(\ln k - 1)}{2\ln^2(k)}$

The conclusion follows since for $k\ge 10$, $\dfrac{\ln 2(\ln k - 1)}{2\ln^2(k)} > 0$

Would I also be right in concluding that since $\dfrac{\ln 2(\ln k - 1)}{2\ln^2(k)} < 1$, it follows that the rate of increase decreases as $k$ increases?

答:
A different approach:

Note that $$f(k)=\frac{k\ln2}{2\ln k}=\frac{\ln2^k}{\ln k^2}$$ and for $k \geq10$, clearly $\ln(2^k)$ and $\ln(k^2)$ are both positive and greater than $1$. Moreover, $\ln$ is an increasing function and $2^k> k^2$ for $k\geq 10$ which implies $f(k)$ is strictly increasing for $k\geq 10$.


问:
What is the expected payment that you will have to make the company?
The claim amount for medical expenses insurance is modeled with an exponential random variable with mean 20. The insurance company applies a deductible of 5. The payment function of the company is the amount of the claim.
minus the deductible, when this amount is positive and zero otherwise. What is the expected payment that you will have to make
the company?

Let $R$ be the random variable that represents the claim amount then
$$f_R(r)=\frac{1}{20}e^{-\frac{1}{20}r}\mathbb{I}_{[0,\infty)}(r)$$
Then let $P$ be the payment function of the firm then
$$P(r)=\left\{ \begin{array}{lcc} \frac{1}{20}e^{-\frac{1}{20}r}-5 \hspace{1cm} r<-40ln(10)\\0 \hspace{1cm} In\hspace{.2cm} another\hspace{.2cm} case \end{array}\right.$$

It is what I have tried but I suspect that it is wrong, can someone help me?

答:
Here $P = (R-5)\cdot\mathsf 1_{(5,\infty)}(R)$. It follows that
$$
\mathbb P(P=0) = \mathbb P(R<5) = \int_0^5 \frac1{20} e^{-\frac1{20} r}\ \mathsf dr = 1-e^{-\frac14},
$$
and for $r>0$
$$
\mathbb P(P>r) = \mathbb P(R>r+5) = \int_{r+5}^\infty\frac1{20} e^{-\frac1{20} r}\ \mathsf dr = e^{-\frac{5+r}{20}}.
$$
From this we may compute
$$
\mathbb E[P] = \int_0^\infty \mathbb P(P>r)\ \mathsf dr = \int_0^\infty e^{-\frac{5+r}{20}}\ \mathsf dr = 20 e^{\frac14}.
$$


问:
Finitely generated flat modules that are not projective
Over left noetherian rings and over semiperfect rings, every finitely generated flat module is projective. What are some examples of finitely generated flat modules that are not projective?

Compare to our question f.g. flat not free where all the answers are f.g. projective not free.

答:
Over a von Neumann regular ring, every right module (and every left module) is flat. Let $V$ be an countable dimensional $F$ vector space, and let $R$ be the ring of endomorphisms of that vector space. It's known that $R$ is a von Neumann regular ring with exactly three ideals.

The nontrivial ideal $I$ consists of the endomorphisms with finite dimensional image. Then $R/I$ is flat but it cannot be projective. If it were projective, then $I$ would be a summand of $R$... but it is not, because it's an essential ideal.

A second example over any non-Artinian VNR ring:  you can take $R/E$ for any maximal essential right ideal $E$ to get a nonprojective, simple flat module. The reasons are very much the same, since a proper essential right ideal can't be a summand of the ring.

You can even make a commutative version: take an infinite direct product of fields $\prod F_i$ (this is von Neumann regular). The ideal $I=\oplus F_i$ is an essential ideal, and $R/I$ is flat, nonprojective. (This one also has the added benefit of supplying examples of ideals which are projective but not free. Any summand of the ring will do, since the ring has IBN. The argument at the other post can be carried out again.)

Incidentally, Puninski and Rothmaler have written a nifty paper investigating which rings have all f.g. flat modules projective.


问:
Evaluate the limit $\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$
Evaluate the limit$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$$

I used L'Hopital's Rule

$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx=\lim_{n\to\infty}\frac{\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx}{\frac{1}{\sqrt{n}}}=\lim_{n\to\infty}\frac{e^{-1}\frac{1/n}{\sin (1/n)}.\frac{-1}{n^2}}{-\frac{1}{2n\sqrt{n}}}=0$$

Is it correct

答:
You did not differentiate the numerator correctly. For  a simpler proof use the inequality $\sin x \geq \frac 2 {\pi} x$ for $0 <x <\frac {\pi} 2$. You will see very easily that the limit is $0$.

[ $\int_0^{1/n}e^{-ny}dy=\frac {1-e^{-1}} n$].


答:
We have

$$0 \le \sqrt{n}\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx \le \sqrt{n}\int_{0}^{\frac{1}{n}}\frac{x}{\sin x}dx.$$

By the mean value theorem there is $t_n \in (0,1/n)$ such that

$$\int_{0}^{\frac{1}{n}}\frac{x}{\sin x}= \frac{1}{n} \cdot \frac{t_n}{\sin t_n}.$$

Hence

$$0 \le \sqrt{n}\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx \le \sqrt{n}\int_{0}^{\frac{1}{n}}\frac{x}{\sin x}dx \le \frac{1}{\sqrt{n}}\cdot \frac{t_n}{\sin t_n}$$

This gives that the limit in question $=0.$

Title: Evaluate the limit $\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$

Body:
Evaluate the limit$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx$$

I used L'Hopital's Rule

$$\lim_{n\to\infty}\sqrt{n} \int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx=\lim_{n\to\infty}\frac{\int_{0}^{\frac{1}{n}}e^{-nx}\frac{x}{\sin x}dx}{\frac{1}{\sqrt{n}}}=\lim_{n\to\infty}\frac{e^{-1}\frac{1/n}{\sin (1/n)}.\frac{-1}{n^2}}{-\frac{1}{2n\sqrt{n}}}=0$$

Is it correct


问:
Pointwise convergence of indicators
Let $X$ be a compact metric space and $\{h_n\}_{n=1}^{\infty}$ be a uniformly convergence sequence of continuous functions on $X$ converging to $h:X\rightarrow \mathbb{R}$.  Let $A_n=[0,\delta_n]$ be a nested sequence of closed subsets of $\mathbb{R}$ for which $[0]=\cap_{n=1}^{\infty} A_n$.  Suppose also that $A\subseteq h_n^{-1}[A_n]$ for each $n$.

Then, can we guarantee that
$$
I_{A_n}\circ h_n \mbox{ converges point-wise to } I_{A}\circ h?
$$

** What I have:**  $I_{A_n}$ converges point-wise $I_A$, pretty much by definition.  But where do we go from here?

答: False. Let  $X=[0,1]$ , $h_n(x)=3\delta_n x$ and $h(x)=0$ for all $x$. Then $I_A(h(\frac 1 2))=1$ but $I_A(h_n(\frac 1 2))=0$ for all $n$.

Title: Pointwise convergence of indicators


问:
Assume that G is of order 4 (not a cyclic group) find all its subgroups.
I know that the cyclic group it contains at least one element that generates the group itself</p>

答:
The only noncyclic group of order 4 is the Klein-Four group $V = \{e,a,b,c\}$,
where $e$ is the unit element and $a^2=b^2=c^2=e$.

The nontrivial subgroups are $\{e,a\}$, $\{e,b\}$, and $\{e,c\}$.


问:
Probabilities and rates in a CTMC
I have a CTMC with the following transition rates for the states $a,b,c$:
$$Q= \begin{bmatrix}
-2 &  3& 3\\
 2&  -4& 0\\
 0& 1 & -3
\end{bmatrix}$$
where those are the exponential rates per hour.

Now, if I'm in state $c$, what is the probability that after 2 hours I'll be in state $a$?

I feel like the solution would use the formula $P(t+s)=e^{(t+s)Q}$, but I cannot seem to find a way to impose the condition that I'm starting from the $c$ state. (No actual results needed, just the theory behind it).

Bonus question: the problem said that the transition rate from state $a$ to state $b$ was 2, but there was also a process involved that could make the transition "restart" that was a Poisson process with rate 3. For example I'm building a sand castle with a rate of 2 and waves are destroying it with a rate of 3.

Shouldn't the actual transition rate be $2*\frac{2}{5}=\frac{4}{5}$?

答:
Recall the Kolmogorov backward equation $P'(t) = QP(t)$ with initial condition $P'(0)=Q$. This has the unique solution $P(t) = e^{Qt}$, where
$$
e^{Qt} = \sum_{n=0}^\infty \frac{t^nQ^n}{n!}.
$$
In this case we have
$$
\left(
\begin{array}{ccc}
 \frac{2 e^{-5 t}}{5}+\frac{3}{5} & \frac{1}{10} \left(e^{-5 t} \left(5 e^t-8\right)+3\right) & \frac{1}{10} e^{-5 t} \left(-5 e^t+e^{5 t}+4\right) \\
 \frac{3}{5}-\frac{3 e^{-5 t}}{5} & \frac{1}{10} \left(e^{-5 t} \left(12-5 e^t\right)+3\right) & \frac{1}{10} e^{-5 t} \left(5 e^t+e^{5 t}-6\right) \\
 \frac{3}{5}-\frac{3 e^{-5 t}}{5} & \frac{3}{10} e^{-5 t} \left(-5 e^t+e^{5 t}+4\right) & \frac{1}{10} e^{-5 t} \left(15 e^t+e^{5 t}-6\right) \\
\end{array}
\right),
$$
and hence the probability in question is
$$
P(2)_{ca} = \frac35\left(1- e^{-10}\right).
$$

(Note that I have transposed the matrix $Q$, as the generator marix for a continuous-time Markov chain is usually written so that the rows sum to zero, as opposed to the columns.)

Title: Probabilities and rates in a CTMC


问:
How do you determine the points of inflection for $f(x) = \frac{e^x}{1+e^x}$?
$$f(x) =\dfrac{e^x}{1+e^x}$$
I know we can find points of inflection using the second derivative test. The second derivative for the function above is $$f''(x) = \dfrac{e^x(1-e^x)}{(e^x+1)^3}$$ I have found one critical point for the second derivative which is $0$. I then determined that the function is concave up from $(-\infty,0)$ and concave down from $(0,\infty)$. I am now asked to find the points of inflection. How would I determine the exact points from where the function switches from concave up to concave down?</p>

答:
You've found the inflection point by identifying the value of $x$ at which the graph shifts from concave up, to concave down. Plus, it matches the solution to the $f''(x) = 0$.

That gives you an inflection point at $\left(0, \frac 12\right)$,