问:
Find a matrix such that $(1,1,1)\in C(A)$ and $N(A)=span(1,1,1,1)$
Find a matrix such  that $(1,1,1)\in C(A)$ and $N(A)=span(1,1,1,1)$.

From the dimensions, I know that $A\in\mathbb{R}^{3\times4}$, and a column should be $[1~1~1]^T$.

Also, it has to satisfy $A[\alpha~\alpha~\alpha~\alpha]^T=[0~0~0]^T$.

So the matrix would have 3 pivots and one free variable. However, I'm struggling trying to get the coefficients so the nullspace of A is that span, any hints or advice on how to get it?

答:
See you want $(1,1,1)$ in the column space of A, so write $(1,1,1)$ as the first column and you want $(1,1,1,1)$ in the null space so make sure each row sum adds upto 0$. For example, the following matrix works.

$\begin{pmatrix}
1 & 1& 1&- 3\\
1 & 1& 1&- 3\\
1 & 1& 1&- 3
\end{pmatrix}$
Here,  $A(1,0,0,0)^t= (1,1,1)^t$ and $A(1,1,1,1)^t= (0,0,0)^t$


问:
Parity of spherical harmonics
I would like to proof $Y_{\ell m}(-\mathbf{r}) = (-1)^\ell\, Y_{\ell m}(\mathbf{r})$. In this formula, $Y_{\ell m}$ are the spherical harmonics given by
\begin{equation}
Y_{\ell m}(\theta, \varphi) = \sqrt{\frac{2\ell + 1}{4\pi}\frac{(\ell-|m|)!}{(\ell+|m|)!}}\, P_\ell^m(\cos\theta)e^{\mathrm{i}m\varphi}.
\end{equation}
For the associated Legendre polynomials, I follow the convention
\begin{equation}
P_\ell^m(x) = \frac{(-1)^m}{2^\ell \ell!} (1-x^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}x^{\ell+m}}(x^2-1)^\ell.
\end{equation}

I already figured out that $\mathbf{r}\to -\mathbf{r}$ corresponds to $\theta\to \pi-\theta$ and $\varphi\to \pi+\varphi$ in spherical coordinates. This yields
\begin{align}
\cos\theta&\to -\cos\theta\\
e^{\mathrm{i}m\varphi}&\to (-1)^m\, e^{\mathrm{i}m\varphi},
\end{align}
but I do not see how $P_\ell^m(-x) = (-1)^{\ell+m}\, P_\ell^m(x)$. To me, it seems like it should not change anything as $(-x)^2=x^2$.

答:
Remember you are also taking the derivative, so you must apply the change rule, e.g, call $z = -x$

\begin{eqnarray}
P_\ell^m(z) &=& P_\ell^m(-x) \\
&=& \frac{(-1)^m}{2^\ell \ell!} (1-z^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}z^{\ell+m}}(z^2-1)^\ell  \\
&=& \frac{(-1)^m}{2^\ell \ell!} (1-(-x)^2)^{\frac{m}{2}}\frac{\mathrm{d}^{\ell+m}}{\mathrm{d}(-x)^{\ell+m}}((-x)^2-1) \\
&=& (-1)^{\ell + m} P_\ell^m(x)
\end{eqnarray}

ContentLicense: CC BY-SA 4.0

Title: Parity of spherical harmonics


问:
Show that this map is well-defined on cohomology
I'm following this paper, trying to understand Corollary 5.

Essentially, there is a Lie Bracket $[,]:C^*\otimes C^*\to C^*$ on a cochain complex $C^*$ such that the the differential $d$ of the complex is a derivation of the bracket, i.e.

$$d[x,y]=[dx,y]+(-1)^{|x|}[x,dy].$$

The author says that from this it is easy to see that this map takes cocycles to cocycles and is independent of the choice of representative of the cohomology classes. I can see that if $dx=0=dy$ (both $x$ and $y$ are cocycles) then $d[x,y]=0$ by the above identity, so the bracket takes cocycles to cocycles. My problem comes when trying to show that the value on different representatives of the same cohomology class is the same on cohomology.

I can compute the bracket of another representative using linearity

$[x+da,y+db]=[x,y]+[x,db]+[da,y]+[da,db]$

If I assume that $[x,da]=[x,0]=0$, then I would be done, but I think $[x,da]=[x,0]$ is part of what I want to prove. Otherwise I would need to show that $[x+da,y+db]-[x,y]$ is a coboundary. But this difference is

$[x,db]+[da,y]+[da,db]$

and I don't see how this is $d(something)$.

答:
I found out how to describe the $[x,db]+[da,y]+[da,db]$ as a coboundary.

It is $d((-1)^{|x|}[x,b]+[a,y]+[a,db])$. Here's the computation using the derivation property, note that $x$ and $y$ are cocycles, so they vanish when $d$ acts on them

$$d((-1)^{|x|}[x,b]+[a,y]+[a,db])=(-1)^{|x|}[dx,b]+[x,db]+[da,y]+(-1)^{|a|}[a,dy]+[da,db]+(-1)^{|a|}[a,ddb]=$$

$$[x,db]+ [da,y]+[da,db]$$

as I wanted.

P.S.: mathematicians, if something is easy, do it or at least indicate it, this is easy once you see it, but I don't think it's so obvious.


问:
Show that if $X$ is $T_3 \implies$ $X/ \sim$ is Hausdorff
Let $X$ be an Hausdorff space and $C \subset X$ closed.Let's define an equivalence relation $\sim$ such that two elements $x,y \in X$ are in relation $x \sim y \iff x=y$ or $x,y \in C$.This equivalence relation induces a quotient space $X/C$.

$1)$Show that $X/\sim$ is $T_1$;

$2)$Show that if $X$ is $T_3 \implies$ $X/ \sim$ is Hausdorff;

For the first part I noticed that $U$ is open in the quotien topology iff $U$ is an open saturated in $X$ topolgy.

So, let's consider $\pi: X \to X/\sim$ and we observe that $\forall x \in C \subset X$, $\pi^{-1}(\pi (x))=\pi^{-1}([x])=C$, while if $y \in X-C$, $\pi^{-1}(\pi (y))=\pi^{-1}([y])=y \implies \forall A\subset X-C$, $ \pi^{-1}(\pi (A))=\pi^{-1}([A])=A$.

As consequence, $[x] \neq [y]\iff x\neq y$ in $X-C$ or if $x \in C, y \in X-C$.

$X-C \ni x\neq y \in X-C \implies (\square)$ $\exists U,V\in \mathcal{T}_X:x\in U,y \in V$, with $U \cap V= \emptyset$ and $U,V \subset X-C$. This implies that $[y]\notin\pi(U)\ni [x]$ and $[x]\notin\pi(V)\ni [y]$.

If $C \ni x\neq y \in X-C$. I considered two open sets $A,B\subset X$ such that $C \subset A \ni x$ and $y \in B \subset X-C$ such that $x \notin B$ and $y \notin A$ (Hausdorff $\implies$ $T_1$). Note that $A$ is saturated since $\pi^{-1}(\pi(A))=A$, because all the fibres of classes that identify $C$ give $C$, and the elements $[z]$
in $\pi(A-C)$ give $z$, then $\pi(A)$ is open in the quotient topology. So, as I wrote in $(\square)$,  $X/\sim$ is $T_1$.

The part "$2)$" seems too similar to part "$1)$"... Am I missing something?

Thank you in advance.

答:
You’ve made the first part much harder than necessary. For convenience let $Y=X/\sim$, let $[x]\in Y$, and let $U=Y\setminus\{[x]\}$. If $x\in X\setminus C$, then $\pi^{-1}[U]=X\setminus\{x\}$, which is open in $X$, so $U$ is open in $Y$, and $\{[x]\}$ is closed. If $x\in C$, then $\pi^{-1}[U]=X\setminus C$, which is open in $X$, so $U$ is again open in $Y$, and $\{[x]\}$ is closed. Thus, each singleton $\{y\}$ in $Y$ is closed, so $Y$ is $T_1$. (If you’ve not already shown that a space is $T_1$ iff each singleton set in it is closed, this would be a good time to prove that useful fact.)

However, the extra effort isn’t really wasted, because the proof that $Y$ is Hausdorff is, as you suspect, very similar to what you’ve already done. If $x$ and $y$ are distinct points of $X\setminus C$, they have disjoint open nbhds disjoint from $C$, and $\pi$ maps those nbhds to disjoint open nbhds of $\pi(x)$ and $\pi(y)$. Regularity is needed only to separate $[x]$ and $[y]$ when one of $x$ and $y$ is in $C$.


问:
Show that there exists a set $E \in \mathbb{R} \times \mathbb{R}$ such that the cross sections are borel sets, but $E \notin B\otimes B$
From Axler's MIRA

Let $B$ denote the $\sigma$-algebra of Borel subsets of $\mathbb{R}$. Show that there exists a set $E \subset \mathbb{R} \times \mathbb{R}$ such that $[E]_a \in B$ and $[E]^a \in B$ for every $a \in \mathbb{R}$, but $E \notin B \otimes B$.

Here $[E]_a$ is the cross section with a fixed second coordinate and $[E]^a$ is the cross section with a fixed first coordinate. Currently my approach to this problem is to use a non-measurable set such as the vitali set $V$ so that $V \times V \notin B \oplus B$ and then having the projection onto an axis make it trivially a Borel subset, however everytime I try to write this up it ends up being wrong so I am wondering if I am even going in the right direction with this problem now.

edit: Here is what I wrote up so far btw

答:
We prove by assuming the Continuum Hypothesis. Let $\omega_1=\{\alpha\mid\alpha<\omega_1\}$ be the first uncountable ordinal. Assume that the Continuum Hypothesis holds, i.e., $\omega_1 = c := 2^\omega$. Choose a bijection $\theta:c\rightarrow\mathbb{R}.$ Denote $x_{\alpha}=\theta(\alpha)$,
$\alpha<c$. Define $E=\{(x_{\alpha},x_{\beta})\mid\alpha<\beta<c\}$.
We go to show that all $x$-sections and $y$-sections of $E$ are
Borel. Let $t\in\mathbb{R}$ be fixed, then there exists a unique
$\alpha_{0}<c$ such that $t=x_{\alpha_{0}}$.

Observe that
\begin{eqnarray*}
E_{t} & = & \{x\in\mathbb{R}\mid(x,t)\in E\}\\
 & = & \{x_{\alpha}\mid\alpha<\alpha_{0}\}
\end{eqnarray*}
which is countable, so $E_{t}$ is a Borel set.

Also,
\begin{eqnarray*}
E^{t} & = & \{y\in\mathbb{R}\mid(t,y)\in E\}\\
 & = & \{x_{\beta}\mid\alpha_{0}<\beta<c\}
\end{eqnarray*}
which is co-countable (i.e., its complement is countable. For, the
complement $(E^{t})^{c}=\{x_{\beta}\mid\beta\leq\alpha_{0}\}$ is
countable), so $E^{t}$ is also a Borel set.

Since the Lebesgue measure $\lambda$ on $\mathbb{R}$ is $\sigma$-finite,
Fubini-Tonelli Theorem is applicable. We prove by contradiction that
$E\notin\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$. Suppose
the contrary that $E\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}),$
then
$$
\int\left(\int1_{E}(x,y)dx\right)dy=\int\left(\int1_{E}(x,y)dy\right)dx.
$$
For each $y$, $1_{E}(x,y)=1_{E_{y}}(x)$, so $\int1_{E}(x,y)dx=\lambda(E_{y})=0$
because $E_{y}$ is a countable set. Hence, $\int\left(\int1_{E}(x,y)dx\right)dy=\int0dy=0$.
On the other hand, for each $x$, $1_{E}(x,y)=1_{E^{x}}(y)$, so $\int1_{E}(x,y)dy=\lambda(E^{x})=\infty$
because $E^{x}$ is co-countable. It follows that $\int\left(\int1_{E}(x,y)dy\right)dx=\int\infty dx=\infty$.
We arrive a contradiction.


问:
Convergence and Lebesgue Integration
I came across this question in a textbook on introductory Lebesgue Integration. I have been teaching myself this material but was unsure of how to do the following question:

Let $(g_n)$ be a sequence of functions that is uniformly bounded and converges pointwise to $g$ almost everywhere. Show that $(g_n)$ converges in the mean to $g$.

I am working with the following definition of convergence in the mean:

$$\lim \int_I{|f_n −f|\ d\mu} = 0$$

Any help would be greatly appreciated. Thank you.

答:
It is true when $I$ is bounded, either by dominated convergence or Egoroff's theorem.

However, when $I=\Bbb R_+$, it may be not true, for example if $f_n=\chi_{(n,n+1)}$ ($|f_n|\leqslant 1$ for all $n$, and $f_n\to 0$ almost everywhere but $\int_I|f_n-f|=1$).

ContentLicense: CC BY-SA 3.0

Title: Convergence and Lebesgue Integration


问:
Resolution of $2\mathbf{\hat{i}}+3\mathbf{\hat{j}}$ along $\mathbf{\hat{i}}+\mathbf{\hat{j}}$ and $\mathbf{\hat{i}}-\mathbf{\hat{j}}$
I was solving this problem:

Resolve $2\mathbf{\hat{i}}+3\mathbf{\hat{j}}$ along $\mathbf{\hat{i}}+\mathbf{\hat{j}}$ and $\mathbf{\hat{i}}-\mathbf{\hat{j}}$.

I proceeded as below:

Let $\mathbf{A}$, $\mathbf{a}$ and $\mathbf{b}$ be the given vectors in the same order. Let
$$\mathbf{A}=\lambda\mathbf{a}+\mu\mathbf{b}$$
Putting the values and rearranging,
$$2\mathbf{\hat{i}}+3\mathbf{\hat{j}}= (\lambda+\mu)\mathbf{\hat{i}}+(\lambda-\mu)\mathbf{\hat{j}}$$
Then, noting that two vectors are equal iff their magnitudes as well as directions are equal, first we equate the directions and taking $\tan$ of both sides.
$$\frac 3 2 = \frac{\lambda-\mu}{\lambda+\mu}$$
Thus, $\lambda = -5\mu$.

Now, taking magnitudes and squaring,
$$13 = 2(\lambda^2+\mu^2)$$
Using our previous equation and doing some work, we get
$$\lambda=\pm \frac{5}{\sqrt{2}}\textrm{ and } \mu = \mp \frac{1}{\sqrt{2}}$$
However, I find no way of choosing which sign.

The book says the answer is $\frac{5}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}$.

I know there is the way to take components by using dot product. But since the book has not introduced it yet, I wonder if there's a way I showed above.

Edit: As Jeea pointed out, the correct answers are
$$\lambda=\pm \frac{5}{2}\textrm{ and } \mu = \mp \frac{1}{2}$$
However, this worsens the situation, this time even the answers are different, let alone sign.

Body:
Corresponding unit vectors: $1/√2 (i+j),$ and $1/√2(i-j)$;

$2i+3j=$

$a(1/√2)(i+j)+b(1/√2)(i-j);$

$a+b=2√2; a-b=3√2;$

$a=(5√2)/2=5/√2;$

$b=√2/2=-1/√2.$


问:
Verification of answer of Maximum Modulus Principle
This question is from Textbook Ponnusamy Silvermann page 283 , section application of Cauchy theorem.

Find maximum and minimum values of (a) $|5+2i z^2| $ on |z|$\leq$1.
(b) f(z) =$ e^z$ on |z|=r

Attempt:  maximum and minimum modulus theorem is to be used.

For (b) maximum at $e^r$ and minimum at $e^{-r}$ . Is it correct?

For (a) I am confused on how to use the principle. Maximum and minimum both will be at boundary. But how to find those values in this case.

Kindly help for part (a). Hope I am right in(b) .

If wrong kindly tell.

答: Observe that $|5+2iz^2| \leq 5+2|z|^2 \leq 7 $ for $|z| =1$ which is achieved at $z= \sqrt{-i}$ which lies on the boundary as $|z|^2= |-i|=1  \implies |z|=1.$ For minimum, use reverse triangle inequality and observe that $ |5+2iz^2| \geq 3$ on $|z|=1$ which is achieved at $z=i$.


问:
How do I draw a log log or semi log graph in a normal graph?
Say I have a function $y=f(x)$. Now how can I change the function to $y=g(x)$ so that when $y=g(x)$ is plotted in a normal graph it looks like the semi log or log log graph of the old $y=f(x)$.

答:
For a semi-log graph ( log in the $x$ axis and linear in the $y$ axis) replace $x$ with $10^x$, i,e,. plot $y=f(10^x)$. The $x$ value in the horizontal axis should correspond to $10^x$ in the semi-log graph (like $1, 2, 3, \cdots$ goes to $10, 100, 1000, \cdots$).

For the case of log in the $y$ axis and linear in the $x$ axis, keep $x$ as it is but instead of $y=f(x)$ plot $y=\log(f(x))$.

For log-log graph do both, and plot $y=\log(f(10^x))$.

Here $\log$ is 10-based.


问:
How do I find the bounds of integration and integral for a triple integral?

So I'm gonna replace my stuff with spherical coordinates with the following equations:

$$x^2+y^2+z^2\:=\:\rho \:cos^2\left(\theta \right)+sin^2\left(\theta \right)sin^2\left(\phi \right)+cos^2\left(\phi \right)$$
And when I simplify:

$$\rho ^2=x^2+y^2+z^2,\:\rho ^2=25,\:\rho \:=5$$

Then since it's a ball I think:
$$0\le \theta \le 2\pi $$

And then $$z = \frac{5}{2}$$

so:

$$(5)cos{\phi} = \frac{5}{2}   ,   \phi = \frac{\pi}{3}$$

And so my integral is:

$$\int _{0\:}^{2\pi \:}\:\int _{0\:}^{\frac{\pi }{3}\:}\:\int _{\frac{1}{cos\left(\phi \right)}}^{\:5}\:\:\:\rho \:\:cos^2\left(\theta \:\right)+sin^2\left(\theta \:\right)sin^2\left(\phi \:\right)+cos^2\left(\phi \:\right)d\rho \:d\phi \:d\theta \:$$

But this doesn't feel right at all. What's my problem?

答:
The limits of integration are almost correct, but your integrand is wrong (and lacking parentheses). You should have
$$\rho\cos\phi\cdot \rho^{-3} \cdot \underbrace{\rho^2\sin\phi\,d\rho\,d\phi\,d\theta}_{dV}.$$
Now, the lower limit for $\rho$ is wrong, as you should have the equation $\rho\cos\phi = 5/2$, so $\rho=\frac52\sec\phi$. (Think about why what you did makes no sense.)


问:
Find image in a complex plane
Consider map $z \to w=f(z)= \sqrt{z} $ $\sqrt{1}=1$; and compute the image f(D) of $D = \{z:1/\sqrt{3}|Im(z)|<Re(z)<1\} $

I know that $f: D \to f(D)$ is bijective

答:
Here is the best I could do using Wolfram Mathematica.

Hope it can be useful

Title: Find image in a complex plane


答: Since $13=2(26\mu^2)$, I think you miscounted the powers of $2$. (The book probably instead worked with the unit vectors $(\mathbf{\hat{i}}\pm\mathbf{\hat{j}})/\sqrt{2}$.) The easiest simultaneous solution of $\lambda+\mu=2,\,\lambda-\mu=3$ averages the equations to get $\lambda=\tfrac52$, so $\mu=2-\lambda=-\tfrac12$.

Title: Resolution of $2\mathbf{\hat{i}}+3\mathbf{\hat{j}}$ along $\mathbf{\hat{i}}+\mathbf{\hat{j}}$ and $\mathbf{\hat{i}}-\mathbf{\hat{j}}$

Body:
I was solving this problem:

Resolve $2\mathbf{\hat{i}}+3\mathbf{\hat{j}}$ along $\mathbf{\hat{i}}+\mathbf{\hat{j}}$ and $\mathbf{\hat{i}}-\mathbf{\hat{j}}$.

I proceeded as below:

Let $\mathbf{A}$, $\mathbf{a}$ and $\mathbf{b}$ be the given vectors in the same order. Let
$$\mathbf{A}=\lambda\mathbf{a}+\mu\mathbf{b}$$
Putting the values and rearranging,
$$2\mathbf{\hat{i}}+3\mathbf{\hat{j}}= (\lambda+\mu)\mathbf{\hat{i}}+(\lambda-\mu)\mathbf{\hat{j}}$$
Then, noting that two vectors are equal iff their magnitudes as well as directions are equal, first we equate the directions and taking $\tan$ of both sides.
$$\frac 3 2 = \frac{\lambda-\mu}{\lambda+\mu}$$
Thus, $\lambda = -5\mu$.

Now, taking magnitudes and squaring,
$$13 = 2(\lambda^2+\mu^2)$$
Using our previous equation and doing some work, we get
$$\lambda=\pm \frac{5}{\sqrt{2}}\textrm{ and } \mu = \mp \frac{1}{\sqrt{2}}$$
However, I find no way of choosing which sign.

The book says the answer is $\frac{5}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}$.

I know there is the way to take components by using dot product. But since the book has not introduced it yet, I wonder if there's a way I showed above.

Edit: As Jeea pointed out, the correct answers are
$$\lambda=\pm \frac{5}{2}\textrm{ and } \mu = \mp \frac{1}{2}$$
However, this worsens the situation, this time even the answers are different, let alone sign.


问:
Lakatos on continuity and invariance to rotation
On page 159, note 21, of Cauchy and the continuum, Imre Lakatos writes: "The modern definition of continuity [the $\epsilon-\delta$ definition] is strongly counter-intuitive, e.g. it is not invariant to rotation".

I have two questions:

Does Lakatos mean that there is a continuous function $f$ such that ROTATION$(f)$ is not continuous?

What are rotation and invariance to rotation in this context? (Lakatos does not explain the terms and a search on the internet did not clarify the issue.)

Thanks very much.

答:
If the author did not further specify the meaning, then, in general,
there is no way to determine the meaning intended. However, it is
possible to make educated guesses. Given a function $\,f(x)\,$ on an
interval, then the set of points $\,(x,f(x))\,$ forms a parametrized curve. This curve
can be rotated. What the author probably meant is that there are some
examples where the original curve is continuous, and a rotated version
also comes from a function, but the rotated version is not continuous
at some points. This all has to be made explicit, of course,
but that is my guess.


问:
Sum of two squares in a $\Bbb Z/p\Bbb Z$
I need to show that every element in $\Bbb Z/p\Bbb Z$ can be written as a sum of two squares. The case $p=2$ is trivial and $0$ is always $0^2 + 0^2$. So all I have to do is show that every element of $(\Bbb Z/p\Bbb Z)^\times$ (the group of units) can be expressed as a sum of two squares. The question hints that I should consider the set of elements of $(\Bbb Z/p\Bbb Z)^\times$ expressible as a sum of two squares.

Associativity is trivial, the identity element $1 = 1^2 + 0^2$ exists.
Closure holds by the identity $(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$.
I don't know how to show that inverses exist.

答:
The following is another version of the argument by Jyrki Lahtonen. Beside the numbers $1,2,\dots, p-1$ write QR for quadratic residue and NR for quadratic non-residue. So we write QR beside $1$. </p>

<p>Note that we write each of QR and NR $\frac{p-1}{2}$ times. So at some time, QR is followed by NR. This means there is a NR $c$ which is congruent to $a^2+1^2$ modulo $p$. But as $x$ ranges over the numbers from $1$ to $p-1$, $x^2 c$ ranges over the non-residues. So the non-residues can be written as $(xa)^2+x^2$.  </p>

Body:
The following is another version of the argument by Jyrki Lahtonen. Beside the numbers $1,2,\dots, p-1$ write QR for quadratic residue and NR for quadratic non-residue. So we write QR beside $1$.

Note that we write each of QR and NR $\frac{p-1}{2}$ times. So at some time, QR is followed by NR. This means there is a NR $c$ which is congruent to $a^2+1^2$ modulo $p$. But as $x$ ranges over the numbers from $1$ to $p-1$, $x^2 c$ ranges over the non-residues. So the non-residues can be written as $(xa)^2+x^2$.


问:
Determine all three-digit positive integers *abc*
Determine all three-digit positive integers abc (a is hundreds b tens and c ones) such that 8abc = 3cba.

Here is my how I have started:

8 (100a + 10b + c) = 3(100c + 10b + a)

800a + 80b + 8c = 300c + 30b + 2a

797a + 50b - 292c = 0

50b = 292c - 797a

b = (292c - 797a)/50

Since abc and cba both need to be even we know that a must be even but not c because factor 292 makes the product even.

Then I started to test different values for a and c, it didn't take so long because a can be larger than 2, so the only values I tested were 1 and 2 for a. For c I tested all values up to 7 (then I found the answer) and it was kind of easy because you don't need to do the whole calculation since if the unit digit of the product of 292 multiplied by c minus 4 and 7 is not 0 we know that the difference is not divisible with 50.

So the only answer I found is c = 7  and a = 2 so b = 9

But I can't prove it algebraically or write a better solution for this problem, I don't think testing is the best method for solving it...

答:
$$797a+50b=292c$$
Here we see that $a$ must be even, and if $a\ge4$ then the LHS is too big for the RHS to match in magnitude (remember that $0\le c\le 9$). Hence $a=2$:
$$1594+50b=292c$$
Now we see that $c$ must be at least $6$ so that the RHS can match the LHS in magnitude, and must be $2$ or $7$ to match in the last digit (the LHS's last digit must be $4$). Hence $c=7$, from which we get $b=9$ and the unique solution as $\overline{abc}=297$.


问:
In how many ways can 15 identical math books be distributed to six students?
question: In how many ways can 15 identical math books be distributed to six students?
i try do this: P(15,6) could be this the answer or i should check the question, i find out other? help please</p>

答: The problem is not solved by a permutation because the books are identical. This is a stars-and-bars problem, with $15$ stars (the books) and $5$ bars creating six spaces for the six different students. The answer is therefore $\binom{15+5}5=\binom{20}5$.


问:
Prove $12^n>7^n+5^n $ for $ n\geq2$ by induction.

Prove the inequality $$12^n>7^n+5^n$$ holds true for $ n\geq2$ by induction.

I'm having difficulties understanding how to prove inequalities by induction.

Body:
Recurrence :

$12^n>7^n+5^n$

For $ n=2 $

$12^2>7^2+5^2$

$\Rightarrow $ $ 144>74$ that is true

Suppose :$12^n>7^n+5^n$

For $n+1$

Let's show $12^{n+1}>7^{n+1}+5^{n+1}$

We know:$12^n>7^n+5^n$

$\Rightarrow $ $12^{n+1}>7^n(5+7)+5^n(5+7)$

$\Rightarrow $ $12^{n+1}>7^{n+1}+5× 7^n+5^{n+1}+7×5^n$

$\Rightarrow $ $12^{n+1}>7^{n+1}+5^{n+1}+5 ×7^n+7×5^n>7^{n+1}+5^{n+1}$

So:
$12^{n+1}>7^{n+1}+5^{n+1}$

Finally

$12^n>7^n+5^n$


问:
Showing there always exists a sequence of numbers in an arbitrary list that sums to a certain value
Suppose that the integers 9, 10, $\dots$ , 20 are listed in some arbitrary order $(a_1, a_2, \dots)$. Prove that $\exists i, 0 \leq i \leq 10 $ and $a_i + a_{i+1} + a_{i+2} \geq 44$.

This is saying that no matter how you arrange the numbers, in the first 10, you can get three in a row that sum to 44 or more. I assume I would use a prove by contradiction to solve this, but I'm not sure where to start.

答:
Let $s_1$ be the sum of the first three numbers in the list, $s_2$ the sum of the next three, $s_3$ the sum of the next three, and $s_4$ the sum of the last three. Clearly

$$s_1+s_2+s_3+s_4=\sum_{k=9}^{20}k=174\,.$$

The average of these four numbers is $43.5$, so ... ?


问:
Implications of independence for two dichotomous random variables
I am trying to understand the relationship among the following:
$P(A | B)$, $P(A | \lnot B)$, and $P(A)$

A is independent of B iff:
$P(A | B) = P(A)$

But, if A is independent of B, then does this necessarily imply that:
$P(A | B) = P(A | \lnot B)$

And, if the A and B are not independent, does that necessarily imply that:
$P(A | B) \neq P(A | \lnot B)$

If the above all does hold, does it fail to always hold as soon as we add another outcome possibility to either random variable (i.e., A or B is not dichotomous).

答:
Yes, if $A$ and $B$ are independent then $P(A|B) = P(A|\lnot B)$. To see this, write $P(A)$ by conditioning on $B$:
$$P(A) = P(A|B)P(B) + P(A|\lnot B)P(\lnot B) = P(A|B)P(B) + P(A|\lnot B)(1-P(B))$$
Since $A$ is independent of $B$, it is independent of $\lnot B$ (exercise) so it follows that $P(A|\lnot B) = P(A)$ and we have
$$0 = P(A|B)P(B)- P(A|\lnot B)P(B)$$
Thus $P(A|B) = P(A|\lnot B)$.

When $A$ and $B$ are not independent then we must have $P(A|B) \neq P(A|\lnot B)$. Notice that this is exactly the converse of the previous statement, so we can simply show that $P(A|B) = P(A|\lnot B)$ implies that $A$ and $B$ are independent.
$$\frac{P(A\cap B)}{P(B)} = \frac{P(A\cap \lnot B)}{P(\lnot B)} = \frac{P(A) - P(A\cap B)}{1 - P(B)}\implies \frac{P(A\cap B)}{P(B)} - P(A\cap B) + P(A\cap B) = P(A).$$
Thus $P(A|B) = P(A)$ and $A$ and $B$ are independent.

I am not sure what you are asking about in the in last part. A random variable can take a range of values, but the probabilities in-terms of an event i.e. the event $X = x$ or the random variable $X$ takes on a certain value $x$. An event is either true or false so the above analysis still holds.


问:
Show that $z^5-1=(z-1)(z^2+2z\cos{\pi \over 5}+1)(z^2-2z\cos{2\pi \over 5}+1)$
Show that

$ \quad \quad z^5-1=(z-1)(z^2+2z\cos{\dfrac{\pi}{5}}+1)(z^2-2z\cos{\dfrac{2\pi}{5}}+1)$

and deduce from this the closed formulas for $\cos{\dfrac{\pi}{5}}$ and $\cos{\dfrac{2\pi}{5}}$.

Here's what I did:

I assumed that it was correct, and using the fact that $z^5-1=(z-1)(z^4+z^3+z^2+z+1)$, I said that:

$ \quad z^4+z^3+z^2+z+1=(z^2+2z\cos{\dfrac{\pi}{5}}+1)(z^2-2z\cos{\dfrac{2\pi}{5}}+1)$

Letting $a=\cos{\dfrac{\pi}{5}}$ and $b=\cos{\dfrac{2\pi}{5}}$

$
\quad z^4+z^3+z^2+z+1=z^4+(2a-2b)z^3+(2-4ab)z^2+(2a-2b)z+1$

Equating the coefficients:

$\quad a-b= {1 \over 2} \quad$ and $ \quad ab={1 \over 4}$

Simultaneously solving, and knowing that $0<a<1$ and $0<b<1$:

$\cos{\dfrac{\pi}{5}}=\dfrac{1}{4} + \dfrac{\sqrt{5}}{4} \quad$ and $\quad \cos{\dfrac{2\pi}{5}}=-\dfrac{1}{4} + \dfrac{\sqrt{5}}{4}$

Then I said that, if I can prove the cosine formulas above in a different way, I have indirectly shown that the relationship shown in the question was true.

So, since $\cos{\pi}= \cos{(5 × (\frac{\pi}{5}))} =-1$ and with a little trig indentity,

$\quad 16{\cos^5{\pi \over 5}}-20{\cos^3{\pi \over 5}}+5{\cos{\pi \over 5}}+1=0$

With substitution $x=\cos{\pi \over 5}$:

$16x^5-20x^3+5x+1=0$, a polynomial that has roots at $x=-1$, $x= \frac{1}{4} - \frac{\sqrt{5}}{4}$, and $x=\frac{1}{4}+\frac{\sqrt{5}}{4}$.

The only possible solution is $\cos{\pi \over 5}=\frac{1}{4}+\frac{\sqrt{5}}{4}$, and the double angle formula shows that $ \cos{\frac{2\pi}{5}}=-\frac{1}{4} + \frac{\sqrt{5}}{4}$

My question, is there a simpler or more straightforward way of proving this?

答:
Actually, if you want to solve this question in a quicker method, mostly using mental math and picturisation, I'd recommend keeping the circle diagram in mind. In the Argand plane, the 5 roots of unity are at equally spaced points on the circle of radius $1$, and start at $(1,0)$.
Going anticlockwise, the points are $1$, $α$, $α^2$, $α^3$ and $α^4$.
Using basic vectors, we can prove that

$α^2+α^3=-2cos(π/5)$.

Similarly,

$α+α^4=2cos(2π/5)$.

Now you can disintegrte the quadratic equations given to give;

$RHS=(z-1)(z-α)(z-α^2)(z-α^3)(z-α^4)$

Hence proved.

You can check this image for the mind-image I referred to, for more clarity. https://en.wikipedia.org/wiki/Root_of_unity#/media/File:One5Root.svg


问:
Determine whether the region $\text{Im}(\frac{z+1}{z-1}) \leq 3$ is open, closed or neither.

Determine whether the region $\text{Im}\left(\frac{z+1}{z-1}\right) \leq 3$ is open, closed or neither.

My attempt and the memo for the question is below.

My answer is somewhat similar to the memo's except that I didn't have the $z \neq 1+0i$ condition.
Is this additional $z \neq 1+0i$ part correct?
While I was working on the solution it didn't seem immediately evident to me, that that condition was necessary.
If the memo is correct, do you have any tips on how I can spot such conditions in similar questions in the future?
Thank you.

Memo answer
My answer

答:
See what happens when you substitute $z= 1+ 0i = 1\;$ into your inequality.

Then $$\text{Im}\left(\frac{z+1}{z-1}\right) = \text{Im}\left(\frac 20\right).$$

You can't do that, because $\frac 20$ is undefined; so we must forbid $z = 1 + 0i = 1$.


问:
Kind of Gronwall Inequality
Does somebody knows if it is possible to obtain an inequality (like for Gronwall inequality) on $f$ if $f$ verify
$$ f(t) \leq A+\int_0^{2t} g(s)f(s) ds $$.
Where $f$ and $g$ are as smooth as necessary and nonnegative.

Thanks.

答: See the first result of this work. I think that this inequality is called of Gronwall-Bellman (or Gronwall-Bellman-Bihari) inequality.</p>

Title: Kind of Gronwall Inequality


问:
characteristic function properties
In lecture, we had the following corollary (without proof, unfortunately):
If $ A \in (0,2) $ and $X$ is a random variable (real-valued) with the following characteristic:

$$ \mathbb P(X > x) = \mathbb P(X < -x) = \frac{x^{-A}}{2} \text{ for any } x \in [1, \infty).$$

Then, the following is valid as $t \to 0$:

$1-\phi(t)\sim C|t|^A $, where $\phi$ is the characteristic function and $\displaystyle C = \int_0^\infty \frac{1-\cos(u)}{u^{A+1}} \, du$.

Why is this correct? Can somebody show me?

答:
Note that $X$ is symmetric hence $1-\phi(t)=E[1-\cos(tX)]$, that is,
$$
1-\phi(t)=\int_1^{+\infty}(1-\cos(tx))\frac{A\mathrm dx}{x^{A+1}}=A|t|^A\int_{|t|}^{+\infty}\frac{1-\cos(u)}{u^{A+1}}\mathrm du,
$$
thanks to the change of variable $u=|t|x$. The last integral converges when $t\to0$ because $A\lt2$ and at $+\infty$ because $A\gt0$ hence the result holds with
$$
C=A\int_{0}^{+\infty}\frac{1-\cos(u)}{u^{A+1}}\mathrm du.
$$</p>

Title: characteristic function properties


问:
Why does the Big O notation of summation give log
I’m not understanding the following line in a proof

I believe it should be $O(h^2\psi(2h))$ since $\sum_{j=1}^{h}\frac{1}{j}\leqslant h\cdot1=h$

答:
This is but a harmonic series summation:
$$\sum_{j=1}^h\frac{2h\psi(2h)}j=2h\psi(2h)\sum_{j=1}^h\frac1j\in2h\psi(2h)\cdot O(\log h)\sim O(h\psi(2h)\log h)$$


问:
General form of a matrix within the support of another matrix
Suppose I have a Hermitian matrix which (after diagonalizing) is expressed as $$A = \sum_i a_i v_i v_i^*,$$ where $a_i$ are the eigenvalues, $v_i$ are the eigenvectors expressed as a column vector and $v_i^*$ is the transpose conjugate of $v_i$ i.e. a row vector.

I am given some other matrix $B$ (it is okay to assume that this is also Hermitian if needed) whose support is contained by the support of $A$. Here the support is defined as

$$\text{supp}(A) = \{u\in V| \langle u, v\rangle = 0, v\in \text{ker}(A)\}$$

What is the most general way to express $B$ if $\text{supp}(B)\subseteq \text{supp}(A)$?

答:
Without loss of generality, suppose that $a_1,\dots,a_r$ are the non-zero eigenvalues of $A$. The support of $A$ is spanned by $v_1,\dots,v_r$.

A Hermitian matrix $B$ will have its support contained in that of $A$ if and only if there exists a Hermitian $r \times r$ matrix $M$ for which $B = \sum_{i,j = 1}^r m_{ij} v_iv_j^*$. Equivalently, if $V$ is the matrix whose columns are $v_1,\dots,v_r$, then $M$ is such that $B = VMV^*$.

For the case that $B$ is not Hermitian, $B$ will have a support contained in that of $A$ if and only if the kernel of $B$ contains that of $A$, which occurs if and only if $B$ can be expressed in the form $B = MV^*$ for some $n \times r$ matrix $M$ (where $n$ is the size of $A$).

Further discussion for the Hermitian case: here's an explanation via block-matrix multiplication.

Let $v_1,\dots,v_{r},v_{r+1},\dots,v_n$ be an orthonormal basis of $\Bbb C^n$. As before, take $V$ to be the matrix with columns $v_1,\dots,v_r$, and take $W$ to be the matrix with columns $v_{r+1},\dots,v_{n}$. Note that $[V \ \ W]$ is a unitary matrix.

Because the kernel of $B$ contains the span of $v_{r+1},\dots,v_n$, we have $BW = 0$ and $W^*B = (BW)^* = 0$. With that, we note that
$$
[V \ \ W]^* B [V \ \ W] = \pmatrix{V^*BV & V^*BW\\ W^*BV & W^*BW} = \pmatrix{V^*BV & 0\\0 & 0}.
$$
Now, let $M = V^*BV$. Note that $M$ is Hermitian. We have
$$
B = (V\ \ W)\pmatrix{M & 0\\ 0 & 0} (V\ \ W)^* = VMV^*,
$$
which is what we wanted.


答: The notation $=O$ needs to be discouraged; it should be $\in O$. It's also important to note $o(f(h))\subseteq O(f(h))$, so very different looking arguments can go in the $O$. In this example,$$\sum_{j=1}^h\frac1j\sim\int_1^h\frac1JdJ=\ln h\in O(\ln h)\subseteq o(h)\subseteq O(h).$$

Title: Why does the Big O notation of summation give log

Body:
I’m not understanding the following line in a proof

I believe it should be $O(h^2\psi(2h))$ since $\sum_{j=1}^{h}\frac{1}{j}\leqslant h\cdot1=h$


问:
Why is the space of $L^p$ (Fourier) multipliers a Banach space? (Stein)
I am reading Stein's book 'Singular Integrals and Differentiability Properties of Functions', and he introduces the class $\mathcal{M}_p$ of $L^{\infty}$ multipliers $m$ such that the operator $T_m$ defined for $f\in{L^2}{\cap}{L^p}$ by $\widehat{(T_mf)}(x)=m(x)\hat{f}(x)$ is bounded on $L^p$, endowing the class with the operator norm of $T_m$.

He then states that this class is a Banach algebra, but I can't prove it. If $(m_n)_n$ is a sequence in $\mathcal{M}_p$, we can deduce that $T_{m_n}f\rightarrow{Tf}$ for some $T\in{\mathcal{B}}(L^p,L^p)$ as ${\mathcal{B}}(L^p,L^p)$ is a Banach space. But how can we show that $T$ corresponds to a Fourier multiplier? I think it would be enough to show that $T_{m_n}f\rightarrow{Tf}$ in $L^2$ also, to try and show that $(m_n)_n$ is also Cauchy in $L^{\infty}$ and so has a limit there, but I can't see how to prove this.
Any ideas greatly appreciated!

答:
Here's an argument that at least works for $1 < p < \infty$. We take for granted at the outset that $\mathcal{M}_2 = L^{\infty}$ and so is complete.

Step 1

First off, by a duality argument we can show that if $m \in \mathcal{M}_p$, then also $m \in \mathcal{M}_{p'}$, and that the norms are identical. (So these spaces are isomorphic.)

To see this, let $f$ and $g$ be Schwartz functions. Then using Plancherel's theorem, we obtain
$$
\int (T_m f) \overline{g} \, dx
= \int m \hat{f} \overline{\hat{g}} \, d\xi
= \int (T_m \tilde{g}) \overline{\tilde{f}} \, dx,
$$
where $\tilde{f}(x) = \overline{f(-x)}$. So
$$
| (T_m f, g) |
= | (T_m \tilde{g}, \tilde{f}) |
\leq \| T_m \tilde{g} \|_p \| \tilde{f} \|_{p'}
\leq \| m \|_{\mathcal{M}_p} \| g \|_p \| f \|_{p'}.
$$
This shows that $T_m$ is bounded on $L^{p'}$, with an operator norm not more than the operator norm on $L^p$. Inverting the roles of $p$ and $p'$ gives the equality of norms.

Step 2

Based on the above, we can extend $T_m$ to an operator on $L^p + L^{p'}$. Complex interpolation then gives us that $\log \| T_m \|_{L^q \to L^q}$ is a convex function of $1/q$, for $q$ between $p$ and $p'$. In particular, the equality of operator norms at the endpoints implies that $\| T_m \|_{L^q \to L^q} \leq \| m \|_{\mathcal{M}_p}$ for $q$ in this range.

Step 3

With these preliminaries set, suppose that $(m_n)_n$ is Cauchy in $\mathcal{M}_p \subseteq \mathcal{B}(L^p, L^p)$, with $p \leq 2$. By the norm comparison property in Step 2, it follows that the sequence is also Cauchy in $\mathcal{M}_q$ for all $q$ between $p$ and $p'$, and in particular for $\mathcal{M}_2$. So for each $q$ we can extract a limiting operator $T^{(q)} \in \mathcal{B}(L^q, L^q)$. Since $\mathcal{M}_2 = L^{\infty}$ is complete, we know that $T^{(2)}$ corresponds to a multiplier $m$.

Step 4

We now have a candidate multiplier, $m$, and it just remains to show that $T^{(p)}$ corresponds to $T_m$. For this, fix a (Schwartz, or $L^p \cap L^2$) function $f$. Then the sequence $(T_{m_n} f)_n$ converges in $L^p$ to $T^{(p)} f$. The corresponding convergence in $L^2$ to $T^{(2)} f = T_m f$ implies, via Hölder's inequaltiy, that if we restrict the $T_{m_n} f$ to a set $E$ of finite measure, then $T_{m_n} f |_E \to T_m f |_E$ in $L^p$. Piecing together the finite-measure sets, we see that the overall $L^p$ limit $T^{(p)} f$ must be identical to $T_m f$.

Step 5

Steps 3 and 4 show that $\mathcal{M}_p$ is complete for $1 < p < 2$. By the isomorphism indicated in Step 1, we also obtain completeness for $2 < p < \infty$.


问:
CNF formula for manipulating words
I am trying to create CNF formula for manipulation of a word.  word is a sequence of letters from a $\Sigma$ alphabet. A word is encoded by variables like $x_{i,a}$ which means that the letter $a$ is at position $i$.

So there is these two questions that I am interested in:

How to write a formula in CNF that expresses that the word does not contain any $b$ on it?

So far I have come up with thise but I am not sure if it is true:

$$\bigwedge_{i=1}^n \; (\bigvee_{b \in \Sigma}  \neg x_{i,b} )$$

How to write a formula (not necessarily in CNF) that expresses that the word has $abc$ on it in consecutive positions?

I have come up with this but again I am not sure:

$$\bigwedge_{i=1}^{n-2} \; ((x_{i,a} \rightarrow x_{i+1,b}) \rightarrow x_{i+2,c})$$

I would be appreciated if someone could help me.

答:
The first one is
$$\lnot \bigvee_{i=1}^n x_{i,b} = \bigwedge_{i=1}^n \lnot x_{i,b}$$

The second one is
$$\bigvee_{i=1}^{n-2} \left(x_{i,a} \land x_{i+1,b} \land x_{i+2,c}\right)$$

Title: CNF formula for manipulating words


问:
Zeno-like riddle with additional complication: Runner with dog running back and forth at different speeds
I came across this little riddle:

An amateur runner trains on a 1 250 m long track at a speed of $v_L = 2 m/s$. His dog runs ahead to the target, turns around and then runs back to him, then he runs back to the  target, back, etc. His speed is when running forth $v_H = 5 m/s$ and when running back (because of the headwind) $v_Z = 4 m/s$. Calculate the distance traveled by the dog.

It is little bit like Zeno's paradoxes but with the additional complication that here the dog is running back and forth at different speeds.

Could anyone please give some hints how to elegantly tackle the problem or references where a solution can be found (Perhaps this riddle even has a common name under which it can be found in the literature)? Thank you!

答:
This is a very famous problem, there is a hard way to solve it (formulate an infinite sum and calculate what it converges to) and a very easy way to do it (think like a physicist and apply the relation velocity = way per time). </p>

<p>I've heard the anecdote that a psychologist noticed that mathematicians always did it the hard way and needed several minutes to solve the puzzle, while physicists always did it the easy way and needed only seconds. When he asked John von Neumann to solve the puzzle, von Neumann answered within seconds. The psychologist asked "But you are a mathematician, you're supposed to do it the hard way and evaluate the sum!" von Neumann answered "That's what I did..." </p>

Body:
This is a very famous problem, there is a hard way to solve it (formulate an infinite sum and calculate what it converges to) and a very easy way to do it (think like a physicist and apply the relation velocity = way per time).

I've heard the anecdote that a psychologist noticed that mathematicians always did it the hard way and needed several minutes to solve the puzzle, while physicists always did it the easy way and needed only seconds. When he asked John von Neumann to solve the puzzle, von Neumann answered within seconds. The psychologist asked "But you are a mathematician, you're supposed to do it the hard way and evaluate the sum!" von Neumann answered "That's what I did..."


问:
Factorial of Infinite Cardinal
I have been thinking about the following problem:

Let $A$ be a set of cardinality $k$ and denote $\sum_A$ the set of all bijection from $A$ to $A$.

Also denote $k! = \mathrm{card}\left(\sum_A \right)$. Prove that $k!=2^k$.

My proof consists of finding a bijection $F:\sum_A\to P(A)$ which associates each bijection from the left to the set of its fixed points. Then the result would follow.
($P(A)$=the power set of $A$).

Since this proof seems quite easy I am afraid it is wrong. Can someone enlighten me?
Thank you very much!

答:
Hint:

Every on $A$ bijection is a function from $A$ to $A$.

Every function from $A$ to $A$ is a relation on $A\times A$.

A relation on $A\times A$ is just a subset of $A\times A$.

What's the cardinality of $A\times A$ when $A$ is infinite?

Title: Factorial of Infinite Cardinal


问:
Bartle 6.J. proof completion
Let $(X,S,\mu)$ be a measure space. Suppose $\mu(X)<\infty$. Put $E_n=\{$ $x\in X:$ $n-1\leq |f(x)|<n \}$

Then, $f\in \mathcal{L}^p$ if and only if $\sum_{n=1}^{\infty}n^p\mu(E_n)<\infty$.

My attempt:

Notice, $X=\bigcup_{n=1}^{\infty}E_n$. Since $(E_n)_n$ are pairwise disjoint, we deduce by a standard property of measures that $\mu(E_n)<\infty$ for each $n$. As a consequence,

$\int |f|^p d\mu=\sum_{n=1}^{\infty}\int_{E_n}|f|^p d\mu$

Now, observe that for each $n$,

$n^p\chi_{E_n} \geq |f|^p\chi_{E_n}\geq (n-1)^p\chi_{E_n}$

Upon taking integrals, we obtain:

$ \sum_{n=1}^{\infty}(n-1)^p \mu(E_n)\leq\int |f|^p d\mu \leq \sum_{n=1}^{\infty} n^p\mu(E_n)$

So, if $\sum_{n=1}^{\infty}n^p\mu(E_n)<\infty$ then $f\in \mathcal{L}^p$.

Now for the reverse, not that if $\mu(E_n)=0$ for each $n$ then we are done. If there is some $n$ such that $\mu(E_n)=0$  if $f\in \mathcal{L}^p$, then

$\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)<\infty$.

Now the question is how do we determine whether a series coneverges given another one converges? The answer is the $\textbf{limit comparison test}$:

```<code>Theorem: Let <span class="math-container">$\sum_{n=1}^{\infty}a_n$</span> and <span class="math-container">$\sum_{n=1}^{\infty}b_n$</span> be two series such that <span class="math-container">$a_n\geq 0$</span> and <span class="math-container">$b_n>0 0$</span> for all <span class="math-container">$n$</span>.

Put <span class="math-container">$c= \lim_{n\rightarrow \infty}\frac{a_n}{b_n}$</span>.

If <span class="math-container">$0<c<\infty$</span> then either both series converge or both series diverge.
</code>```

To invoke this theorem notice that if for all $n$, $\mu(E_n)=0$ then we are immediately done. So, we may assume that for some n , $\mu(E_n)>0$. If for all $n$, this is the case, then we are immediately done. If only finitely many satisfy this property, then we may disregard the ones with 0 measure. Thus, we may assume for each $n$, $\mu(E_n)>0$. In which case,

the limit comparison test above allows us to deduce that $c=1$. Since $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)$ converges, so must $\sum_{n=1}^{\infty}n^p \mu(E_n)$.

答:
Following your ideas, observe that

$$\sum_{n=1}^\infty (n-1)^p\mu(E_n)\le
\sum_{n=1}^\infty\int_{E_n} |f|^pd\mu \le \sum_{n=1}^\infty n^p\mu(E_n) $$

If $\sum n^p\mu(E_n)<+\infty$, then the integral is also finite and $f\in{\cal L}^p$.

Conversely, If the integral is convergent, then $\sum (n-1)^p\mu(E_n)<+\infty$ and this series have the same bahavior as yours. Indeed, by Ratio test
$$\lim_{n\to\infty}\frac{(n+1)^p\mu(E_n)}{n^p\mu(E_n)} =\lim (1+1/n)^p=1.$$

Title: Bartle 6.J. proof completion


问:
Rayleigh quotient of $q$ with respect to matrix A
my notes say (I'm self studying) that the Rayleigh quotient minimizes $Aq=pq$

I have no idea what that means. I did a quick search online and I found out that this quotient is a complex number $p$, $q^HAq = q^Hqp $

I have no idea how I shoud prove this. how are we supposed to show that p minimizes $Aq-pq$??

答:
Not entirely sure I understand your question but this is my guess about what you're asking.

For a matrix $A$ and a vector $q$, the rayleigh quotient $R_{A}(q) := \frac{q^{H}Aq}{q^{H}q}$ is bounded between the eigenvalues of $A$. Therefore, its minimum is the smallest eigenvalue $\lambda_{min}$, which is the smallest possible $p$ in the eigenvalue equation $Aq = pq$. So the minimum Rayleigh quotient is the smallest $p$ in the eigenvalue equation (which also corresponds to the eigenvector $q$ that minimises the Rayleigh quotient).

Additional clarification:

If $q$ is a linear combination of the eigenvalues of $A$ (i.e. $q = a_{1}u_{1} + ... a_{n}u_{n}$ where $\sum a_{i} = 1$ and the $u_{i}$ are the eigenvectors) and has norm of 1, clearly $R(q) = \frac{q^{H}Aq}{1} = q^{H}(\lambda_{1}a_{1}u_{1} + ... \lambda_{n}a_{n}u_{n}) = \sum_{i} \lambda_{i}a_{i}^{2}$.

To minimise this expession we just set the $a_{i}$ corresponding to the minimum eigenvalue to 1. Scaling of $q$ by some constant does not affect this.

Therefore the minimum of $R_{A}(q)$ is the minimum eigenvalue, which by definition is the minimum $p$ in $Aq = pq$.


问:
Are solutions to a second order differential equation unique in this case?
In my math book, a general solution to the differential equation $y'' + y = f(t)$ where $f(t)$ is continuous on $(-\infty, \infty)$ is given by $y = A\cos t + B\sin t + \int_0^t f(s)\sin(t-s)ds$.

Taking as an example $y'' + y = t$, I get the following result when I use undetermined coefficients: $y = A\cos t + B\sin t + t$

When I use the formula given above, I get $y = A\cos t + B\sin t + t - \sin t$

Both solutions also work when you plug them back into $y'' + y = t$, so is this to say that both solutions are valid and the only difference would be in the values of the constants?

答:
Short answer: yes, both are valid.

Long answer.  A function with undetermined coefficients is not a single function, but a family of functions. Two families (sets) are equal if every member  of the first family is also a member of the second, and the other way around. For example, the families {Jane, Joe, their son} and {Bobby, his Dad, his Mom} may be the same though the members are listed in different order and described in  different ways.

In your example:

Every member of the second family of functions is of the form  $y = A\cos t + B\sin t + t-\sin t$ with some numbers $A,B$. This function can also be written as  $y = A\cos t + (B-1)\sin t + t $. Therefore, it also belongs to the first family.

Every member of the first family is of the form $y = A\cos t + B\sin t + t$. This can be written as  $y = A\cos t + \tilde {B}\sin t + t-\sin t$ by choosing $\tilde B=B+1$. Therefore, $y$ belongs to the second family.

The first time one encounters this situation is in  integral Calculus, where $+C$ is used to describe families of antiderivatives. For example, one can write
$$
\int \frac{1}{x}\,dx = \ln |x| +C
$$ or $$
\int \frac{1}{x}\,dx =  \int \frac{2}{2x}\,dx \overset{\color{Red}{u=2x}}{=} \int \int \frac{1}{u}\,du = \ln|u|+C=\ln|2x|+C
$$

Both are the same family of functions.


问:
Logic if $x \bmod 3 ≠ 0$, then $3 \nmid x^2$

How would one go about writing out this proof?
Mod tends to confuse me, but my line of thought went to :

Perhaps proving by the contrapositive? Such as we would say: if 3|x^2, then x mod 3 = 0. But I am stuck as to where to go from here, or if contrapositive is even the way to go.

Body: If $x\neq 0\pmod{3}$, then $x=1,2\pmod{3}$. Thus, $x^{2} = 1^{2},2^{2} = 1\neq 0\pmod{3}$.


问:
Is the graph of the following type possible?
Let a simple undirected graph be denoted by $(V, E)$, where $V$ is the set of vertices and $E$ is the set of edges. Let each vertex of a graph be a natural number. Now, I want to define the graph's edge such that $(V_1\cup V_2, E)=(V_1, E)\mathbin{\pmb\cup} (V_2, E)$, where $\cup$ and $\pmb\cup$ are set union and graph union, respectively. How should I define the graph's edge such that the edge is preserved (or, above equality holds)?

The following is my failed attempt.

let the edge of each graph represents "divides", i. e.,  vertices $ a$ and $b$ are  connected by an edge if and only if $a$ divides $b$ or, $b$ divides $a$. In this case, it is found that the above equality doesn't hold.

答:
Using the formulation from the discussion in the comments above, I don't think this is possible unless $f(X)$ only has edges between nodes and themselves for all $X$ i.e. $\forall X, f(X)\subseteq \{\{a,a\}|a \in X\}$.

Here is why: assume $X=\{n_1, ..., n_k\}$ and $f(X)=\{\{n_a,n_b\}, ...\}$ where $n_a \neq n_b$. Now, let's create two sets, $X_a=X-\{n_b\}$ and $X_b=X-\{n_a\}$. Then, we know:

$\{n_a,n_b\} \notin f(X_a)$ since $n_b \notin X_a$

$\{n_a,n_b\} \notin f(X_b)$ since $n_a \notin X_b$

But

$\{n_a,n_b\} \in f(X)$

$X=X_a \cup X_b$

To put it in graph terms, take a graph $G(E,V)$ and take any edge ${a,b}$. Then you can create two new graphs which don't have that edge by $a$ into one graph and $b$ into the other graph.

In fact, we can specify $f$ even more. The only way to get the desired property is that for any $a \in X$, $\{a.a\} \in f(X)$ iff $\{a.a\} \in f(\{a\})$ i.e. the question of whether or not to include $\{a,a\}$ is independent of the other vertices in $X$. (You can prove this using a similar argument to the above).


问:
Assume that $G$ is abelian and $H \le G$. If $k=\{x \in G\mid x^2 \in H\}$ then prove that $k \le G$.

Assume that $G$ is abelian and $H \le G$. If $k=\{x \in G \mid x^2 \in H\}$ then prove that $k \le G$.

I tried to use this rule: Assume $k$ is a subset of $G$. If for each $x,y\in k$ there is $xy\in k$ and $x^{-1}\in k$ then $k$ is a subgroup of $G$.

I managed to prove the closed property of $k$, but couldn't prove the inversion property.

For the closed property I wrote:
for each $x,y \in k$ then $x^2, y^2\in H$, and since $H$ is a subgroup and has the closed property then $x^2y^2 \in H$ thus $(xy)^2 \in H$ and finally $xy \in k$.

Also, I don't know how the abelian property can be used here.

Thanks for your help!

答:
You are using abelian property of $G$  when you write $x^2 y^2 = xxyy\in H$ thus $(xy)(xy) = (xy)^2 \in H.$

In a nonabelian group $G'$, where commutativity does not hold, if $x, y \in G'$, then it is possible that $x^2y^2 = xxyy \neq xyxy = (xy)^2$.


问:
Existence of a prime
If $x$ is odd and natural and ${x^2}+2\equiv3\mod 4$, how can I show there exists a prime $p$ such that $p|x^2+2$ and $p\equiv3\mod 4$.</p>

答:
Let $n=x^2+2$. Note that $n$ is odd. If all the prime divisors of $n$ were of the shape $4k+1$, then $n$ also would be. For the product of any two numbers congruent to $1$ modulo $4$ is itself congruent to $1$ modulo $4$.

Remark: The $x^2+2$ stuff is just intended to distract you. Not nice!

Title: Existence of a prime


问:
What functor can make categories equivalent
I’m reading Jacobson’s book “Basic Algebra II”,and confused by the proof of the following theorem for long long time.

Let $F$ be a functor from $\mathcal{C}$ to $\mathcal{D}$.Then there exists a functor $G:\mathcal{D}\to \mathcal{C}$ such that $(F,G)$ is an equivalence iff $F$ is faithful, full and essentially surjective.

And my confusion is about the proof of the necessity. The author says that

$GF\cong 1_{\mathcal{C}}$ implies that the map $f\mapsto GF(f)$ of $Hom(A,B)$ onto $Hom(GF(A),GF(B))$ is bijective.
Similarly,$g\mapsto FG(g)$ is bijective.

The claim above is what I can accept. However, the following one borthered me for hours

the surjectivity of $g\mapsto FG(g)$ implies the surjectivity of $f\mapsto F(f)$.

I don’t know how can “imply”.

we know that $F$ is full iff for any given $A,B\in ob\mathcal{C}$, for any $g\in Hom_{\mathcal{D}}(F(A),F(B))$, there exists $f\in Hom_{\mathcal{C}}(A,B)$ such that $F(f)=g$. It seems that we want to use the surjectivity of $FG$. However, I don’t know how to ensure the appropriate domain of $FG$. It seems that we must have $A’,B’\in ob\mathcal{C}$ such that $G(A’)=A$,$G(B’)=B$. And then we can say that there must exist some $f’\in Hom(A’,B’)$ such that $FG(f’)=g$.

I thought this problem for hours and asked my classmate for help. Unfortunately,they also don’t know why.

I do feel like I'm going crazy, and I really wish someone would answer me in details.

Thank you very much

答:
We want to prove that $F$ is full and faithful, which is exactly the statement
$$\forall g\in\hom_{\mathcal D}(FA,\,FB)\,\exists! f\in\hom_{\mathcal C}(A,B):\,Ff=g$$
that you seem to understand already.

Then we have $Gg\in \hom_{\mathcal C}(GFA,\,GFB)$ which is in bijection with $\hom_{\mathcal C}(A,B)$ via the full and faithful functor $GF$.

Nevertheless, since it's an equivalence, $A'=FA$ and $B'=FB$ will be good candidates for what you seem to be missing.

Well, we can't guarantee mere equality $GA'=A$ and $GB'=B$, instead we only have (natural) isomorphisms $GFA\cong A$ and $GFB\cong B$, and composing by these given isomorphisms or their inverses readily proves the one-to-one correspondence
$$\hom_{\mathcal C}(A,B)\ \cong\ \hom_{\mathcal C}(GFA,\,GFB)\,.$$


问:
Calculate $ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{xy^{2}}{x^{4}+y^{2}} $
Im trying to calculate (or prove that the limit dosent exists)

$ \lim_{\left(x,y\right)\to\left(0,0\right)}\frac{xy^{2}}{x^{4}+y^{2}} $

Wolfram says the limit does not exists, but for every path I choose the limit ends up being 0.

If someone can tell if the limit exists or not, it would be helpful.

Thanks in advance

Body:
HINT

By $x^2=u$ and $y=v$ we obtain

$$\left|\frac{xy^{2}}{x^{4}+y^{2}}\right|= \frac{\sqrt u v^{2}}{u^{2}+v^{2}} $$


问:
Calculate the expected value with of a function with given variance
Question: Let $X$ be a random variable with $E(X) = 1$ and $Var(X) = 5$. Then $E((1 + X)^2)$ = ?

We know that for any discrete variable and function $f:R \to R$ the expected value is
$$E[f(X)]= \sum_{x \in X(\Omega)} f(x)\cdot P(X=x)$$.

So in our case:
$$E[(1 + X)^2]= \sum_{x \in X(\Omega)} (1 + x)^2\cdot P(X=x)$$.

But I can't to anything with that or?

Body: $$\mathbb{E}[X+1]^2=\mathbb{V}[X+1]+(\mathbb{E}[X+1])^2=\mathbb{V}[X]+(\mathbb{E}[X]+1)^2=5+2^2=9$$


答:
It is not necessary to assume $X$ takes on discrete values.

Recall that $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2.$$  Then $$\operatorname{E}[(1+X)^2] = \operatorname{E}[1+2X+X^2] = 1 + 2\operatorname{E}[X] + \operatorname{E}[X^2],$$ by linearity of expectation.  So if you know $\operatorname{E}[X] = 1$ and $\operatorname{Var}[X] = 5$, you can use this information to compute $\operatorname{E}[X^2]$, from which you can substitute back into the above equation to get the desired expectation.

Title: Calculate the expected value with of a function with given variance

Body:
Question: Let $X$ be a random variable with $E(X) = 1$ and $Var(X) = 5$. Then $E((1 + X)^2)$ = ?

We know that for any discrete variable and function $f:R \to R$ the expected value is
$$E[f(X)]= \sum_{x \in X(\Omega)} f(x)\cdot P(X=x)$$.

So in our case:
$$E[(1 + X)^2]= \sum_{x \in X(\Omega)} (1 + x)^2\cdot P(X=x)$$.

But I can't to anything with that or?


问:
How did it get from line 1 to line 2 to line 3 (Vector Cross Product)

How does it get from line 1 to line 2
can someone explain it in verybbasic terms
Thanks

答: The second $=$ expands the cross-product; like ordinary multiplication, it's linear in both arguments. The last $=$ uses antisymmetry $u\times v=-v\times u$, including the "diagonal" special case $u=v=a$, giving$$a\times a=-a\times a\implies 2a\times a=0\implies a\times a=0.$$


答:
Any number $n$ of the form $3 \pmod4$, must have a prime factor of the form $3 \pmod4$. If all the primes dividing $n$ were to be of the form $1 \pmod4$, then the product of these primes (including multiplicity) is again of the form $1 \pmod 4$, since $$\left(1 \pmod 4 \right) \times \left( 1 \pmod4 \right) \equiv (1 \times 1) \pmod4 \equiv 1 \pmod4$$

Additional note

We can in fact say something more, if we consider all the primes of the form $3 \pmod 4$ dividing $n$, say $q_1,q_2,\ldots,q_k$, and we write $n$ as
$$n = p_1^{a_1} p_2^{a_2} \cdots p_l^{a_l} q_1^{b_1} q_2^{b_2} \cdots  q_k^{b_k}$$ where $p_j \equiv 1 \pmod 4$, since $n \equiv 3 \pmod4$, we must have $\displaystyle \sum_{j=1}^k b_j \equiv 1 \pmod 2$.

Title: Existence of a prime

Body: If $x$ is odd and natural and ${x^2}+2\equiv3\mod 4$, how can I show there exists a prime $p$ such that $p|x^2+2$ and $p\equiv3\mod 4$.</p>


答:
More generally, if $\phi: G_1 \to G_2$ is a group homomorphism and $H_2 \le G_2$, then $\phi^{-1}(H_2) \le G_1$.

Apply this to $G_1=G_2=G$, $H_2=H$, and $\phi(x)=x^2$, which is a homomorphism because $G$ is abelian.

Title: Assume that $G$ is abelian and $H \le G$. If $k=\{x \in G\mid x^2 \in H\}$ then prove that $k \le G$.

Body:
Assume that $G$ is abelian and $H \le G$. If $k=\{x \in G \mid x^2 \in H\}$ then prove that $k \le G$.

I tried to use this rule: Assume $k$ is a subset of $G$. If for each $x,y\in k$ there is $xy\in k$ and $x^{-1}\in k$ then $k$ is a subgroup of $G$.

I managed to prove the closed property of $k$, but couldn't prove the inversion property.

For the closed property I wrote:
for each $x,y \in k$ then $x^2, y^2\in H$, and since $H$ is a subgroup and has the closed property then $x^2y^2 \in H$ thus $(xy)^2 \in H$ and finally $xy \in k$.

Also, I don't know how the abelian property can be used here.

Thanks for your help!


答:
Remember that for any vectors $u,v$, we have $u \times v = -v \times u$ and $v \times v = 0$. So,
$$
b \times c - a \times c - b \times a + a \times a =\\
b \times c - -(c \times a)  --(a \times b) + 0 =\\
b \times c + c \times a + a \times b.
$$

Title: How did it get from line 1 to line 2 to line 3 (Vector Cross Product)

Body:
How does it get from line 1 to line 2
can someone explain it in verybbasic terms
Thanks


问:
Prove $\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$
How to prove

$$\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$$

I got this result using Fourier representation of
$$|\log(\cos(\frac{x}{2}))|$$
and Parseval's identity. But I am wondering if there is a direct way to calculate this integral.

答:
With $t= \frac x2$

\begin{align}
I& = \int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx}\\
&= 4\int_0^{\frac\pi2}\ln^2 (\cos t) dt= 2\int_0^{\frac\pi2}(\ln^2 (\cos t)+ \ln^2 (\sin t)  )dt \\
&=\int_0^{\frac\pi2}\left( \ln^2 (\sin t\cos t) + \ln^2 \frac{\sin t}{\cos t} \right)dt
=J+K\tag1
\end{align}
where
\begin{align}
J &= \int_0^{\frac\pi2}\ln^2 (\sin t\cos t) dt
\overset{2t\to t}= \frac12 \int_0^{\pi}\ln^2 (\frac12\sin t)dt= \int_0^{\frac\pi2}\ln^2 (\frac12\sin t)dt\\
&= \int_0^{\frac\pi2}\left( \ln^2 2 -2\ln2 \ln\sin t +\ln^2(\sin t ) \right)dt\\
&= \frac\pi2\ln^22 -2\ln 2(-\frac\pi2\ln2) +\int_0^{\frac\pi2}\ln^2(\sin t )dt
= \frac{3\pi}2\ln^22 +\frac14I\\
K&=\int_0^{\frac\pi2} \ln^2 (\frac{\sin t}{\cos t}) dt\overset{u=\tan t}= \int^\infty_0 \frac{\ln^2 u} {1+u^2} du=\frac{\pi^3}8
\end{align}
Plug $J$ and $K$ into (1) to obtain

$$I = 2\pi\ln^22 +\frac{\pi^3}6$$


答:
I will use the one-step subgroup test.

Since $H\le G$, we have $e\in H$, so $e=e^2\in H$, which implies $e\in k$. Hence $k\neq\varnothing$.

Since $k=\{\color{red}{x\in G}\mid x^2\in H\}$, we have $k\subseteq G$.

Now consider $x,y\in k$. Then $x^2,y^2\in H$, so we have

$$(xy^{-1})^2=x^2y^{-2}$$

(since $G$ is abelian), but $H\le G$, so $H$ is closed (and closed under taking inverses), meaning $x^2y^{-2}\in H$, so $xy^{-1}\in k$.

Hence $k\le G$.


Title: Prove $\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$

Body:
How to prove

$$\int\limits_{-\pi}^{\pi}{\log^2{(\cos{\frac{x}{2})}}dx} = 2\pi\log^2{2} + \frac{\pi^3}{6}$$

I got this result using Fourier representation of
$$|\log(\cos(\frac{x}{2}))|$$
and Parseval's identity. But I am wondering if there is a direct way to calculate this integral.

Body:
Let us write $$I=\int_{-\pi}^{\pi} \ln^2(\cos(x/2)) dx=4\int_{0}^{\pi/2} \ln^2 \cos y dy$$ Let $\cos y=t$, then
$$I=4\int_{0}^{1} \frac{ln^2 t}{\sqrt{1-t^2}} dt =4\int_{0}^{1} \sum_{k=0}^{\infty} \ln^2 t ~ C_k ~(-1)^k~t^{2k} dt, C_k={-1/2,\choose k}$$ Let $t=e^{-u}$, then
$$I=4\int_{0}^{1}\sum_{k=0}^{\infty} (-1)^k C_k u^2 e^{-(2k+1)u} =8\sum_{k=0}^{\infty} \frac{(-1)^k ~ C_k}{(2k+1)^3}$$

I may get back.


问:
Examples of inner products defined on a matrix vector space
Is $\langle A,B \rangle = \operatorname{tr}(A^TB)$ the only inner product defined for $M(2\times 2)$?
If there are others, what are some examples?

答:
Let $\langle A,B \rangle = \operatorname{tr}(A^TB)$. For any invertible linear map $\Phi: M_{2 \times 2} \to M_{2 \times 2}$, the function
$$
(A,B) = \langle \Phi(A),\Phi(B) \rangle
$$
defines an inner product, and every inner product over $M_{2 \times 2}$ can be expressed in this fashion.  For example: for any invertible $2 \times 2$ matrices $P,Q$, defining $\Phi(X) = PXQ$ leads to
$$
(A,B) = \operatorname{tr}(Q^TA^TP^TPBQ) = \operatorname{tr}(A^T[P^TP]B[QQ^T]).
$$
Note also that adding any inner products together yields another inner product.


问:
Real Analysis Qualifying Exam Problem
I think this should be an easy question, and I believe the answer should be in the positive, but I am not sure how to start.  I would appreciate some help.  Thank you.

Suppose that $f_j$ is a sequence of functions in $L^1(\mathbb{R})$ satisfying

(i)  $\|f_j\|_\infty \leq 5$ and $f_j(x) = 0$ for $|x| \geq 10, j \in \mathbb{N}.$

(ii) the $f_j$ converge pointwise to $f \in L^1(\mathbb{R})$.

Does it follow that the $\hat{f_j}$ converge pointwise to $\hat{f}$?

答:
It's enough to show that $\lVert f_n-f\rVert_{L^1}\to 0$.

In order to do that, we can use Egoroff's theorem: fix $\varepsilon>0$, and $S_\varepsilon\subset [0,1]$ such that $[-10,10]\setminus S$ has a measure $<\varepsilon$ and there is uniform convergence of $f_n$ to $f$ on $S$. Since all the $f_n$ and $f$ are supported by $[-10,10]$, we get

$$\int_{\Bbb R}|f_n-f|d\lambda=\int_{[-10,10]}|f_n-f|d\lambda\leqslant 20\sup_{x\in S}|f_n(x)-f(x)|+10\varepsilon.$$

Title: Real Analysis Qualifying Exam Problem


问:
Tricker chain rule with partial derivative
$$\dfrac{\partial}{\partial x}\bigg(\frac{d}{dt} f(x(t)g(t))\bigg)$$

I begin substituting $y=xg$ and get

$$\dfrac{\partial}{\partial x}\bigg(\frac{df}{dy} (y(t))\cdot (x'(t)g(t)+x(t)g'(t))\bigg)$$

$$\dfrac{\partial}{\partial x\partial y}(y)\cdot (x'(t)g(t)+x(t)g'(t))+\dfrac{df}{dy}(y)\cdot(x''(t)g(t)+g'(t))$$

How could I improve this?

Thank you!

答:
Let $x=x(t)$ and assume that $x$ is invertible.  Then, from the chain rule

$$\begin{align}
\frac{d}{dx}\frac{dx}{dt}&=\frac{d x'(t)}{dx}\\\\
&=\frac{dx'(t)}{dt}\frac{dt}{dx}\\\\
&=\frac{x''(t)}{x'(t)}
\end{align}$$

Now, with $x(t)$ invertible, we can write

$$\begin{align}
\frac{d }{d x}\frac{df(x(t)g(t))}{dt}&=\frac{d}{dt}\frac{df(x(t)g(t))}{dt}\times\frac{dt}{dx}\\\\
&=\frac1{x'(t)}\frac{d^2f(x(t)g(t))}{dt^2}\end{align}$$

And the rest is left to the reader.


问:
Proving function f(z) to be constant
This question was asked in my complex analysis quiz and I was unable to solve it.

Suppose f(z) is analytic for $|z|\leq 1$ and $|f(z)|\geq 1$ for $ |z|\leq1$ . If f(0) =1 then show that f(z) is constant.

I am sorry but for this particular question I have no clue which theorem I should use. I am completely clueless.

Kindly tell about what theorem should I use. Rest I would like to work by myself.

答: Use minimum modulus principle which will imply that $|f(z)| =1$ only when $|z|=1$ otherwise $f$ is constant. Since $f(0)=1$. Hence, minimum of $|f(z)|$ is achieved at an interior point, therefore, $f$ is constant.

Title: Proving function f(z) to be constant


问:
How many ways are there to distribute eight different toys among four children if the first child gets at least two toys?
My work:

$e_1 + e_2 + e_3 + e_4 = 8$

let $e_1\geq2$

with no constraint on the other $e_i$'s

we want to find coefficient of $\cfrac{x^8}{8!}$

with this I've found the exponential generating function to be

$g(x) = (\cfrac{x^2}{2!} + \cfrac{x^3}{3!}+...)(1 +x+\cfrac{x^2}{2!}+...)^3 = (e^{x}-1-x)(e^x)^3 = e^{4x}-e^{3x}-xe^{3x}$

so first series would be $ \sum_{r=0}^{\infty} 4^r\cfrac{x^r}{r!}$ for $e^{4x}$

Second series would be -$ \sum_{r=0}^{\infty} 3^r\cfrac{x^r}{r!}$ for $-e^{3x}$

But I get stuck here

What would the third series be for $-xe^{3x}$?

I need to figure this third series out to figure out the coefficient

Any help would be appreciated here.

Also is my logic and methodology right for this question?

If it helps, I just need help with the $-x$ in front of $-xe^{3x}$

It's really throwing me off as I'm not sure how it would turn into a series.

答:
kludginess says hello

I don't know enough about generating functions to specifically
examine your work.  However, I do know that there is a strong
relationship between generating functions and stars and bars
analysis.  Stars and bars is discussed at https://brilliant.org/wiki/integer-equations-star-and-bars/.

I also know that stars and bars analysis, at least in its
simplified form, is inappropriate here, specifically
because the toys are deemed to be different.

The problem is actually ambiguous, depending on whether the second, third,
and fourth child must each get at least one toy.

$\underline{\textbf{Interpretation 1: Some children may get no toys}}$

This is easy.

$$4^8 - \left(8 \times 3^7\right) - 3^8.$$

$\underline{\textbf{Interpretation 2: Each child gets at least 1 toy}}$

There may be an elegant approach here that I am unaware of.  My kludgy approach
is to identify all of the possible satisfying toy distribution patterns and then
enumerate each pattern.

Patterns:

P-1 : 5 - 1 - 1 - 1 : # ways $= S_1$.

P-2 : 4 - 2 - 1 - 1 : # ways $= S_2$.

P-3 : 3 - 3 - 1 - 1 : # ways $= S_3$.

P-4 : 3 - 2 - 2 - 1 : # ways $= S_4$.

P-5 : 2 - 2 - 2 - 2 : # ways $= S_5$.

The final tally will be $S_1 + S_2 + S_3 + S_4 + S_5.$

P-1

1st child gets 5 toys, other 3 toys are permuted.

$$S_1 = \binom{8}{5} \times 3!.$$

P-2

Three ways of selecting which other child gets more than one toy.

Then, you examine the # of ways that 1st child gets 4 toys, and other child gets 2 toys.

Then, you multiply by 2, since the first child can switch roles with the other child here.

$$S_2 = 2 \times 3 \times \binom{8}{4} \times \binom{4}{2} \times 2!.$$

P-3

Analysis very similar to analysis of P-2.

However, since the first child and the odd man out child
will each get 3 toys, you must not apply the factor of 2,
which would correspond to their switching places.

$$S_3 = 3 \times \binom{8}{3} \times \binom{5}{3} \times 2!.$$

P-4

If the first child is to get three toys, with one of the children the odd man out,
then the first tally is

$$ S_{4a} = 3 \times \binom{8}{3} \times 5 \times \binom{4}{2}.$$

If the first child is to get two toys, then there are 3! ways of deciding which
of the other children get 3 toys, 2 toys, 1 toy.

With this decided:

$$ S_{4b} = 3! \times \binom{8}{2} \times \binom{6}{3} \times 3.$$

$$ S_4 = S_{4a} + S_{4b}.$$

P-5

The # of ways of first giving 2 toys to 1st child, then giving 2 toys to 2nd child, ...

$$ S_5 = \binom{8}{2} \times \binom{6}{2} \times \binom{4}{2}.$$


问:
Construct a 98% confidence interval for the population proportion of defective toys.
A toy company would like to estimate the proportions of toys produced by their younger workers. In a random sample of 300 toys, they found that 75 were defective. Construct a 98% confidence interval for the population proportion of the toys that are defective.

Can someone explain how I would go about this?

I recall that the formula for developing a confidence interval is (point estimate)$\pm$(critical value)(standard error).

Since we are working with a 98% confidence interval, I'm pretty sure that the critical value is 2.326.

Am I on the right track? How do I proceed?

答:
Yes, you are on the right track. The approximated confidence interval is

$$\large{\left[\hat p-z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}, \  \hat p+z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}\right]}$$

With $\alpha=1-0.98=0.02 \Rightarrow 1-\frac{\alpha}{2}=0.99$. And $z_{0.99}=2.326$. The other values are $\hat p=\frac{75}{300}=0.25$ and $n=300.$


问:
Find upper and lower bound given limit for submultiplicative sequence
I am working on an exercise and I am confused on the boundary situation.

Suppose $q_t,t\geq0$ satisfy $0 \leq q_{t} \leq 1$ and submultiplicative property, i.e.

$$q_{t+s}\leq q_s q_t$$
and we have ($\lambda$ is a constant, $\lambda \in (0, \infty)$)
$$\lim _{t \rightarrow \infty} \frac{\log (q_{t}e^{\lambda t})}{t}=0$$
then it is said that: there exists $0<c_{1}<c_{2}<\infty$ such that for all $t$,
$$
c_{1}  \leq q_{t}e^{\lambda t} \leq c_{2}
$$
I try to find $c_1,c_2$ but I fails, notice that $log(q_t)/t$ is decreasing as t increase. The situation as $t \rightarrow 0$ is unknown. And also if $\log (q_{t}e^{\lambda t})=\small o(t)$, then limit goes to zero. How can we know that $q_{t}e^{\lambda t} \leq c_{2} $ ?

Can anybody help?

答:
I can Prove one side of inequality:

From Wikipedia:

Theorem:  For every measurable subadditive function $f:(0, \infty) \rightarrow \mathbb{R},$ the limit $\lim _{t \rightarrow \infty} \frac{f(t)}{t}$ exists and is equal to $\inf _{t>0} \frac{f(t)}{t} .$

Hence

$$\lim _{t \rightarrow \infty} \frac{\log \left(q_{t} \right)}{t}=\inf _{t>0} \frac{\log (q_t)}{t}=-\lambda$$
So for any $t>0$,$\frac{\log (q_t)}{t}\geq-\lambda\Longrightarrow 1 \leq q_{t} e^{\lambda t}$. This prove the LHS.


问:
A question based on maximum modulus principle
This question is part of my assignment on complex analysis course and I was unable to solve it.

Find maximum and minimum values of $|\frac{z-\alpha}{1-\bar{\alpha} z} | $  on $|z|\leq 1$ . ( where $\alpha$ is such that |$\alpha$ |<1 and $\alpha$ is fixed.)

By maximum and minimum modulus principle it is clear that Maximum and minimum will be at |z|=1.

But I have no idea on how to find such valuevalues due to the complicated nature of function f.

So, it's my humble request to help me.

答:
The map given by $f(z)=\frac{z-\alpha}{1-\overline{\alpha}z}$ for $|\alpha|<1$ is an automorphism of unit disc i.e.  f maps unit disc onto unit disc, therefore, $|f(z)|=1$ for all $|z|=1$.
You can use the following link to read a detailed discussion:
Automorphisms of the unit disk


答:
Actually, the minimum is $0$, which is taken when $z=\alpha$.

On the other hand, if $|z|=1$, then$$\left|1-\overline\alpha z\right|=\left|\overline zz-\overline\alpha z\right|=\left|\overline z-\overline\alpha\right||z|=|z-\alpha|$$and therefore$$\left|\frac{z-\alpha}{1-\overline\alpha z}\right|=1$$and therefore the maximum is $1$.

Title: A question based on maximum modulus principle

Body:
This question is part of my assignment on complex analysis course and I was unable to solve it.

Find maximum and minimum values of $|\frac{z-\alpha}{1-\bar{\alpha} z} | $  on $|z|\leq 1$ . ( where $\alpha$ is such that |$\alpha$ |<1 and $\alpha$ is fixed.)

By maximum and minimum modulus principle it is clear that Maximum and minimum will be at |z|=1.

But I have no idea on how to find such valuevalues due to the complicated nature of function f.

So, it's my humble request to help me.


问:
Bartle 6.J. proof completion
Let $(X,S,\mu)$ be a measure space. Suppose $\mu(X)<\infty$. Put $E_n=\{$ $x\in X:$ $n-1\leq |f(x)|<n \}$

Then, $f\in \mathcal{L}^p$ if and only if $\sum_{n=1}^{\infty}n^p\mu(E_n)<\infty$.

My attempt:

Notice, $X=\bigcup_{n=1}^{\infty}E_n$. Since $(E_n)_n$ are pairwise disjoint, we deduce by a standard property of measures that $\mu(E_n)<\infty$ for each $n$. As a consequence,

$\int |f|^p d\mu=\sum_{n=1}^{\infty}\int_{E_n}|f|^p d\mu$

Now, observe that for each $n$,

$n^p\chi_{E_n} \geq |f|^p\chi_{E_n}\geq (n-1)^p\chi_{E_n}$

Upon taking integrals, we obtain:

$ \sum_{n=1}^{\infty}(n-1)^p \mu(E_n)\leq\int |f|^p d\mu \leq \sum_{n=1}^{\infty} n^p\mu(E_n)$

So, if $\sum_{n=1}^{\infty}n^p\mu(E_n)<\infty$ then $f\in \mathcal{L}^p$.

Now for the reverse, not that if $\mu(E_n)=0$ for each $n$ then we are done. If there is some $n$ such that $\mu(E_n)=0$  if $f\in \mathcal{L}^p$, then

$\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)<\infty$.

Now the question is how do we determine whether a series coneverges given another one converges? The answer is the $\textbf{limit comparison test}$:

```<code>Theorem: Let <span class="math-container">$\sum_{n=1}^{\infty}a_n$</span> and <span class="math-container">$\sum_{n=1}^{\infty}b_n$</span> be two series such that <span class="math-container">$a_n\geq 0$</span> and <span class="math-container">$b_n>0 0$</span> for all <span class="math-container">$n$</span>.

Put <span class="math-container">$c= \lim_{n\rightarrow \infty}\frac{a_n}{b_n}$</span>.

If <span class="math-container">$0<c<\infty$</span> then either both series converge or both series diverge.
</code>```

To invoke this theorem notice that if for all $n$, $\mu(E_n)=0$ then we are immediately done. So, we may assume that for some n , $\mu(E_n)>0$. If for all $n$, this is the case, then we are immediately done. If only finitely many satisfy this property, then we may disregard the ones with 0 measure. Thus, we may assume for each $n$, $\mu(E_n)>0$. In which case,

the limit comparison test above allows us to deduce that $c=1$. Since $\sum_{n=1}^{\infty}(n-1)^p \mu(E_n)$ converges, so must $\sum_{n=1}^{\infty}n^p \mu(E_n)$.

答:
For the converse notice that $n^p\chi_{E_n} \leq (|f|+1)^p\chi_{E_n}$. Since $f \in L_p$, $|f| \in L_p$. And since $\mu(X)<\infty$, $1 \in L_p$. So $|f|+1 \in L_p$ and
$$
\infty > \int (|f|+1)^p=\sum\limits_{n=1}^{\infty} \int_{E_n} (|f|+1)^p \geq \sum\limits_{n=1}^{\infty} n^p.
$$
This is another way of doing the converse. But @Tito Eliatron's way is totally right.

Title: Bartle 6.J. proof completion


问:
Proof: For all real numbers $x$ and $y$, if $\lfloor 2x\rfloor = \lfloor 2y \rfloor$ then $\lfloor x\rfloor = \lfloor y \rfloor$
Here is what I have so far:

The statement is true. Let $x$ and $y$ be real numbers and assume that $\lfloor 2x\rfloor = \lfloor  2y \rfloor=n$, where $n$ is an integer (we are trying to prove $\lfloor x\rfloor =\lfloor y\rfloor$). Then $n \le 2x < n + 1$ and $n \le 2y < n + 1$.

I'm not quite sure where to go from here. Any suggestions/guidance would be much appreciated!

答:
Since you want to prove something about $x$ and $y$, the natural thing is to divide those inequalities by $2$: $\frac{n}2\le x,y<\frac{n+1}2$. Now consider two cases, $n$ even and $n$ odd.

This works quite easily, but it’s even easier to prove the contrapositive: show that if $\lfloor x\rfloor\ne\lfloor y\rfloor$, then $\lfloor 2x\rfloor\ne\lfloor 2y\rfloor$.


问:
Necessary and sufficient condition for $f$ to be a closed mapping

Let $f:X\to Y$ be a mapping from the topological space $X$ to $Y$. Show that the existence of neighborhood $V_y$ of  $\forall y \in Y$ such that $f^{-1}(V_y)\subset U$ for any open $U: f^{-1}(y)\in U$ is necessary and sufficient condition for $f$ to be a closed mapping.

I'm fairly new to topology, so even the statement of the problem was quite tough to understand for me.

I started by trying to prove the sufficient condition:

Let $f:X\to Y$ be a mapping from the topological space $X$ to $Y$. Assume there exists a neighborhood $V_y$ of  $\forall y \in Y$ such that $f^{-1}(V_y)\subset U$ for any open $U: f^{-1}(y)\in U$. Now, we know that $f(y)=U$ and $V_y \subseteq f(U)$. This implies that $f(U)$ is an open set since $y$ lies in $f(U)$ with some neighborhood. Then, $f$ sends open sets to open sets $(\because U\overset f{\mapsto} f(U) \text{ is open to open})$.

I do not know if I am right so far, but even if I am, I do not see any use of that piece of observation.

For the second direction (assuming $f$ is closed), I do not have idea on how to use the definition of the closed mapping (if it's really needed).

Thanks in advance for your help!

答:
Assume that whenever $y\in Y$ and $U\subseteq X$ is an open set with $f^{-1}(y)\subseteq U$, then there exists an open set $V_y\subseteq Y$ with $y\in V_y$ such that $f^{-1}(V_y)\subseteq U$.

Suppose $C\subseteq X$ is closed. We will show that $f(C)$ is closed by showing that its complement $Y\setminus f(C)$ is open. So let $y\in Y\setminus f(C)$. Then $f^{-1}(y)\subseteq X\setminus C$. Note that $X\setminus C$ is open, so by assumption there is $V_y\subseteq Y$ such that $f^{-1}(V_y)\subseteq X\setminus C$. Necessary $V_y\subseteq Y\setminus f(C)$. We conclude that $Y\setminus f(C)$ is open, and so that $f$ is a closed map.

Now assume that $f:X\rightarrow Y$ is a closed map. Let $y\in Y$ and suppose that
$f^{-1}(y)\subseteq U$, where $U\subseteq X$ is open. Then $X\setminus U$ is closed, and thus also $f(X\setminus U)$, since $f$ is a closed map. Put $V_y=Y\setminus f(X\setminus U)$ and notice that $f^{-1}(V_y)\subseteq U$.


问:
Why does the Big O notation of summation give log
I’m not understanding the following line in a proof

I believe it should be $O(h^2\psi(2h))$ since $\sum_{j=1}^{h}\frac{1}{j}\leqslant h\cdot1=h$

答:
You found the correct statement

$$\sum_{j=1}^h \frac1j \in O(h).$$

The authors of the paper used the correct statement

$$\sum_{j=1}^h \frac1j \in O(\log h),$$

as explained in the answer by Parcly Taxel. Since $\log h \in o(h)$, the paper uses the sharper, "better" bound. Yours isn't wrong, it's just not as good as could be.

You can find details about why $\sum_{j=1}^h \frac1j \in O(\log h)$ in the wikipedia entry for the Harmonic numbers.


答:
To find the coefficient of $\frac{x^k}{k!}$ in $x e^{3x}$, I find it easier to find the coefficient of $x^k$ and then multiply by $k!$.

The coefficient of $x^k$ in $x e^{3x}$ is the coefficient of $x^{k-1}$ in $e^{3x}$, which is $\frac{3^{k-1}}{(k-1)!}$. So the coefficient of $\frac{x^k}{k!}$ is $k! \cdot \frac{3^{k-1}}{(k-1)!} = k 3^{k-1}$.

Therefore our final answer should be $4^k - 3^k - k 3^{k-1}$, where $k=8$ in your case. That's $4^8 - 3^8 - 8 \cdot 3^7 = 41\,479$, which I also checked by brute force:

```<code>Length@Select[Tuples[Range[4], 8], Count[#, 1] >= 2 &]
</code>```

gives $41\,479$ in Mathematica.

Title: How many ways are there to distribute eight different toys among four children if the first child gets at least two toys?

Body:
My work:

$e_1 + e_2 + e_3 + e_4 = 8$

let $e_1\geq2$

with no constraint on the other $e_i$'s

we want to find coefficient of $\cfrac{x^8}{8!}$

with this I've found the exponential generating function to be

$g(x) = (\cfrac{x^2}{2!} + \cfrac{x^3}{3!}+...)(1 +x+\cfrac{x^2}{2!}+...)^3 = (e^{x}-1-x)(e^x)^3 = e^{4x}-e^{3x}-xe^{3x}$

so first series would be $ \sum_{r=0}^{\infty} 4^r\cfrac{x^r}{r!}$ for $e^{4x}$

Second series would be -$ \sum_{r=0}^{\infty} 3^r\cfrac{x^r}{r!}$ for $-e^{3x}$

But I get stuck here

What would the third series be for $-xe^{3x}$?

I need to figure this third series out to figure out the coefficient

Any help would be appreciated here.

Also is my logic and methodology right for this question?

If it helps, I just need help with the $-x$ in front of $-xe^{3x}$

It's really throwing me off as I'm not sure how it would turn into a series.

Body:
To find the coefficient of $\frac{x^k}{k!}$ in $x e^{3x}$, I find it easier to find the coefficient of $x^k$ and then multiply by $k!$.

The coefficient of $x^k$ in $x e^{3x}$ is the coefficient of $x^{k-1}$ in $e^{3x}$, which is $\frac{3^{k-1}}{(k-1)!}$. So the coefficient of $\frac{x^k}{k!}$ is $k! \cdot \frac{3^{k-1}}{(k-1)!} = k 3^{k-1}$.

Therefore our final answer should be $4^k - 3^k - k 3^{k-1}$, where $k=8$ in your case. That's $4^8 - 3^8 - 8 \cdot 3^7 = 41\,479$, which I also checked by brute force:

Length@Select[Tuples[Range[4], 8], Count[#, 1] >= 2 &]

gives $41\,479$ in Mathematica.


问:
Probability of gaining more than $\$20$ in a two person die game

Jack and Jill gamble on a roll of a die, as follows. If the die shows $1$ or $2$ spots, Jack gives Jill $\$1$. If the die shows $5$ or $6$ spots, Jill gives Jack $\$1$. If the die shows $3$ or $4$ spots, no money changes hands. Suppose Jack and Jill play this game $400$ times. The chance that Jill’s net gain is more than $\$20$ is closest to:

a) $10\%$

b) $20\%$

c) $30\%$

d) $40\%$

Addition Rule: $\Pr(A\text{ or }B)= \Pr(A)+\Pr(B)$

$$\Pr(\text{Jack 1 or 2}, + \text{\$1 to Jill})=( 1/6)+(1/6)=2/6$$
$$\Pr(\text{Jill 5 or 6, + \$1 to Jack})=(1/6)+(1/6)=2/6$$
$$\Pr(\text{3 or 4; \$0 both Jack & Jill})= (1/6)+(1/6)=2/6$$

$n=400$

$\Pr(\text{net gain >\$20})$?

答:
In each game, with probability $1/3$ each, Jill wins $\$1$, loses $\$1$ or plays even. Letting $X_i$ denote the gain of Jill in some game $i$, we have

\begin{align}
E(X_i) &= \frac{1}{3}(1 + 0 + -1) = 0 \\
\mathrm{Var}(X_i) &= \frac{1}{3}\left(1^2 + 0^2 + (-1)^2\right) = \frac{2}{3}
\end{align}

Letting $Y = \sum_{i=1}^{400} X_i$ denote the overall profit of Jill in $n = 400$ games, we can compute the mean and variance of $Y$ as follows, using that the $X_i$ are i.i.d.:

\begin{align}
E(Y) &= E\left(\sum_{i=1}^{400} X_i\right) = \sum_{i=1}^{400} E(X_i) = 0 \\
\mathrm{Var}(Y) &= \mathrm{Var}\left(\sum_{i=1}^{400} X_i\right) = \sum_{i=1}^{400} \mathrm{Var}\left(X_i\right) = 400 \cdot \frac{2}{3} = \frac{800}{3}.
\end{align}

Using the Gaussian approximation, $Y$ can be approximated by a normally distributed random variable $Y'$ with the same mean and variance, i.e, with $Y'$ satisfying $Y' \sim \mathcal{N}(0, 800/3)$. So we finally get an approximation as

$$P(Y > 20) \approx P(Y' > 20) = P\left(Z > \frac{20}{\sqrt{800/3}}\right) \approx P(Z > 1.22474) \approx 0.11034.$$

(Here $Z$ denotes a standard normal random variable, i.e, $Z \sim \mathcal{N}(0,1)$.)

So the answer is closest to (a) $10\%$.


问:
Equivalent conditions on a finite Poset P.
Let $d \in P$. Show that the following two conditions on a finite poset $P$ are equivalent:

i. $P$ is the intersection of d linear orderings of $[n]$, where $|P| = n$,

ii. $P$ is isomorphic to a subposet of $\mathbb{N}^d$

Moreover, show that when $d=2$ the two conditions are also equivalent to:

iii. There exists a poset Q on [n] such that $s<t$ or $s>t$ in Q if and only if s
and t are incomparable in P.

So I know that I need to show $i \equiv ii$ and then either $i \equiv iii$ or $ii \equiv iii$, since the first two are already equivalent.

Here are my thoughts:

$\mathbb{N}^d$ is $(x_1,x_2,...,x_d)\leq (y_1,y_2,...,y_d)$ iff $x_i\leq y_i, \forall i\in [d]$

The intersection of $d$ linear orderings of $[n]$ is a new poset that $i\leq j$ iff $i\leq j$ in all $d$ linear orderings.

But I'm not sure how to show that $i. \equiv ii.$, so I can't even get started on showing the equivalence to $iii.$

Any insight, next steps, or  guides on how to show these equivalences would be appreciated.

答:
HINT: To show that (i) implies (ii), let $P$ be a partial order on $[n]$ such that $P=\bigcap_{k=1}^d\le_k$, where each $\le_k$ is a linear order on $[n]$. For each $k\in[d]$ there is a bijection $\varphi_k:[n]\to[n]$ such that for all $i,j\in[n]$, $\varphi_k(i)\le\varphi_k(j)$ iff $i\le_k j$. Let

$$\varphi:[n]\to[n]^d:i\mapsto\langle\varphi_1(i),\varphi_2(i),\ldots,\varphi_d(k)\rangle\,,$$

and show that $\varphi$ is an isomorphism. For the reverse implication you have to start with an isomorphism $\varphi$ and use it to define the linear orders $\le_k$ for $k\in[d]$; this is really just reversing the procedure that I just sketched.


问:
$\int_{\bigcup_{n=1}^{\infty}E_n}f=\sum_{n=1}^{\infty}\int_{E_n}f$ given $f$ positive and measurable
I'm learning about measure theory (specifically Lebesgue intregation) and need help with the following problem:

Let $f:\mathbb{R}\rightarrow[0,+\infty)$ be measurable and let $\{E_n\}$ be a collection of pairwise disjoint measurable sets. Prove that $\int_{\bigcup_{n=1}^{\infty}E_n}f=\sum_{n=1}^{\infty}\int_{E_n}f.$

For convenience I set $E=\bigcup_{n=1}^{\infty}E_n$.

This problem looks like an application of the monotone convergence theorem but I'm having a hard time applying it. I need to find a sequence of functions that is positive an nondecreasing but I don't know how to define it.

答:
Set $f_N=\sum_{n=1}^{N} f\chi_{E_n}$.

As $f\chi_{E_n}\geq 0$ for each $n$,  $f_1\leq f_2 \leq f_3 \leq f_4....$.

Now observe that $f_n \rightarrow f \chi_{E}$ as $n\rightarrow \infty$, where $E=\bigcup_{n=1}^{\infty}E_n$. Thus by MCT we obtain the following equality:

$\lim_{n} \int f_n d\mu= \int f\chi_{E} d\mu$.

Which is the conclusion you seek.


问:
When determining the equation of a plane given 3 points, why the election of the points to compute the normal vector matter?
Let's say I have three points $P_1 = (1, 2, -1)$, $P_2 = (-1, 1, 4)$ and $P_3 = (1, 3, -2)$ and I want to find the equation of the plane determined by it.

My reasoning is:

Given three points of a plane, there are three lines parallel to this
plane determined by the pairs of points ($P_1, P_2$), ($P_2, P_3$) and
($P_1$, $P_3$), so I need to find a normal vector $\vec{N}$ such that
it's perpendicular to two of those lines (and since all three lines
are in the same plane, then if a $\vec{N}$ is perpendicular to two
lines, then it must be perpendicular as well to the third line).

The standard way to compute $\vec{N}$ is by solving the equations system yielded by $\vec{N} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N} \cdot (\vec{P_2} - \vec{P_1}) = 0$, but if I happen to choose $\vec{N} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N} \cdot (\vec{P_3} - \vec{P_2}) = 0$ instead, although I get a $\vec{N} = (4, -3, 1)$ that is perpendicular to $\overline{P_{1}P_{3}}$ and $\overline{P_{2}P_{3}}$, it's not perpendicular to $\overline{P_{1}P_{2}}$ as I expected it to be. So at least the highlighted part of my reasoning above is false. But why? Why I seem to be forced to arbitrarily solve $\vec{N} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N} \cdot (\vec{P_2} - \vec{P_1}) = 0$ if the line determined by $P_2$ and $P_3$ is in the same plane as well (and thus its normal vector should be parallel to the normal vectors of the other two lines, which doesn't happen)?

Solving $\vec{N_1} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N_1} \cdot (\vec{P_2} - \vec{P_1}) = 0$ yields $\vec{N_1} = (2, 1, 1)$, which is the correct $\vec{N}$ of the plane.

Solving $\vec{N_2} \cdot (\vec{P_3} - \vec{P_1}) = \vec{N_2} \cdot (\vec{P_3} - \vec{P_2}) = 0$ yields $\vec{N_2} = (4, -3, 1)$.

But $\vec{N_1}$ and $\vec{N_2}$ are not parallel to each other, which is geometrically counterintuitive to me. What is going on?

答:
There is no flaw in the reasoning. The calculation is wrong.

The two normal vectors are not parallel.

$\vec N_2(4,-3,1)$ is not orthogonal to  $\vec{P_1P_3}=(0,1,-1)$ and $\vec{P_2P_3}=(2,2,-6)$ but it is orthogonal to $\vec{P_1P_2}=(-2,-1,5)$. Note that $\vec N_1(2,1,1)$ is orthogonal to all of them.

$\vec{N_1}(2,1,1)$ is the correct normal and you have run into an error while calculating $\vec{N_2}$.

$$\vec{N_2}\cdot\vec{P_1P_3}=0\implies y-z=0\\\vec{N_2}\cdot\vec{P_2P_3}=0\implies x+y-3z=0\\$$This gives $\vec{N_2}=z(2,1,1)$ parallel to $\vec{N_1}$.

Note that an alternative way to calculate the normal vector is to find $\vec{P_1P_3}\times\vec{P_2P_3}$ (or you can take the cross product of any two sides of the triangle $P_1P_2P_3$).


问:
Exact value of a sequence of coefficients.
Let $a_1=-1$ and $a_n=-1-\sum_{j=1}^{n-1}a_j[n/j]$, where $[x]$ is the integer part of $x$. Is it always true that $a_n\in\{-1,0,1\}$?

These coefficients are determined by the formal equality $\chi_{(0,1)}(x)=\sum_{n=1}^\infty a_n\rho(1/(nx))$, where $\rho(x)=x-[x]$, which is related to the characterization of the Riemann Hypothesis in terms of the approximation in $L^2(0,\infty)$ of the above characteristic function by the linear combinations of the sequence $\{\rho(1/(nx))\}_n$

答:
Let's identify your coefficients $a_j$:
Your equation can be rewritten as
$$\sum_{j=1}^n a_j\lfloor n/j\rfloor=-1,$$ since $\lfloor n/n\rfloor=1$
(I'll use the notation $\lfloor x\rfloor$ for the integer part of $x$, as it is more common nowadays than the $[x]$ used in my youth). For $n-1$ instead of $n$, this is
$$\sum_{j=1}^{n-1} a_j\lfloor (n-1)/j\rfloor=-1,$$ but can also be written
$$\sum_{j=1}^n a_j\lfloor (n-1)/j\rfloor=-1,$$ because $\lfloor (n-1)/n\rfloor=0$.
Subtraction of both equations gives
$$\sum_{j=1}^n a_j(\lfloor n/j\rfloor-\lfloor (n-1)/j\rfloor)=0.$$ Now it's easy to see that $\lfloor n/j\rfloor-\lfloor (n-1)/j\rfloor$ is $0$ if $n$ is not a multiple of $j$, and $1$ if it is. So our equation becomes $$\sum_{j|n}a_j=0.$$ This is exactly the equation satisfied by the Möbius function,
$$\sum_{j|n}\mu(j)=0,$$ just with the starting value $a_1=-1$ instead of $\mu(1)=1$. This means we have $a_j=-\mu(j)$ for all $j\ge1$.


问:
Identity Element of Subtraction (Solution Verification)
This is about an exercise from Norman L. Biggs Discrete Mathematics. One is asked to check which binary operations are valid when $*$ represents the usual $-$ of arithmetic. Now, Biggs says that $-$ has an identity value. But, using the definition:

G3: There is an element $e \in G$ such that $$e*x=x*e=x$$ for all $x \in G$. Now, $e-x=x \iff e=2x$ and $x-e=x \iff e=0$. As $2x≠0$ for all $x \in G$

I answered that subtraction did not have an identity element. What am I missing?

答:
You are not missing anything.

You are correct that there is no  identity element $\,e\,$ such that for all $\,x\,$ (in most sets), we have that $$e - x = x - e = x.$$

As Federica Maggioni states, "$0$" is a "right-identity" for subtraction, satisfying:

$$\; x - 0 = x,$$ with subtraction defined as we know it.

But this is not an identity in the sense that we've defined identity $e \in G$ for a binary operator on $G$: $$e * x = x*e = x$$ for all $x$ in a set $G$ on which $*$ is defined.


问:
Is every finite Galois extension of $\mathbb{Q}_p$ the completion of a number field at some valuation?
I've been wondering about this for a while, is it the case that if $K$ is a finite Galois extension of $\mathbb{Q}_p$, then there exists some field $L$ with $$\mathbb{Q} \subset L \subset K,$$ such that $[L:\mathbb{Q}]<\infty$ and $L\cdot \mathbb{Q}_p = K$.

I can see that this is true for $K$ unramified, because we can just attach the appropriate roots of unity to $\mathbb{Q}$ so it remains the case where $K$ totally ramifies. I guess it would be enough to show that there exists some $\alpha \in K$ which is algebraic over $\mathbb{Q}$ but not an element of $\mathbb{Q}$. I have tried to use Hensel's lemma but that fails spectacularly because the uniformizer is the root of some Eisenstein polynomial which reduces to be $x^n$.

答:
Yes, this is Hensel-Krasner lemma. Given a square-free polynomial $f\in \Bbb{Z}_p[x]_{monic}$, there is some $N$ such that for $n>N$, if $f(a_n) = 0\bmod (\pi^n), a_n\in O_K$ then $f$ has a root in $O_K$. The proof is the same as when $f$ is separable $\bmod p$ :

$$f(a_n+b)= f(a_n)+b f'(a_n)+O(b^2/p)$$
Normalize such that $v(\pi)=1$. For $N$ large enough we have $v(f'(a_n))\le m$.

With $b=-f(a_n)/f'(a_n)$, $a_{n+1}=a_n+b, 2(n-m)-v(p) > n$ ie. $N > 2m+v(p)$ we get that

$v(f(a_{n+1}))>v(f(a_n))$ and hence $\lim_{n\to \infty} f(a_n)=0$, by compactness $a_n$ has a subsequence converging in $O_K$, to a root of $f$.

Next $K= \Bbb{Q}_p(\pi+\zeta_{q-1})\cong \Bbb{Q}_p[x]/(g(x))$, take $f\in \Bbb{Z}[x]$ of same degree approximating $g$ with enough precision so that its $m$ is the same as $g$ and $f(\pi+\zeta_{q-1})=0\bmod (\pi^{2m+v(p)+2})$.

Take $a_{N+2} = \pi+\zeta_{q-1}$.

$f$ will have a root $a\in K$, which is $\equiv  \pi+\zeta_{q-1}\bmod (\pi^2)$, whence  $\Bbb{Q}_p(a)$ has the same residue field and uniformizer as $K$ ie. $K=\Bbb{Q}_p(a)$.


问:
Solving a functional equation with derivatives?
Am I doing this correctly?
Given
\begin{align}
f(x) = \frac{x^n}{1+x^n} - kf(x-1)
\end{align}
, then evaluating its the derivative
\begin{align}
\frac{df}{dx}=\frac{nx^{n-1}}{(1+x^n)^2}-k\frac{df}{dx}(x-1).
\end{align}
Let $t=x-1$ so $x=t+1$ and $dx=dt$. Thus,
\begin{align}
\frac{df}{dt} &= \frac{n(t+1)^{n-1}}{(1+(t+1)^n)^2} - k\frac{df}{dt}\\
\Leftrightarrow (k+1)\frac{df}{dt} &= \frac{n(t+1)^{n-1}}{(1+(t+1)^n)^2}\\
\Leftrightarrow \frac{df}{dt} &= \biggl(\frac{n}{k+1}\biggr)\frac{(t+1)^{n-1}}{(1+(t+1)^n)^2} \\
\Leftrightarrow \frac{df}{dt} &= \biggl(\frac{n}{k+1}\biggr)\frac{(x)^{n-1}}{(1+(x)^n)^2}\\
&= \biggl(\frac{1}{k+1}\biggr)\frac{nx^{n-1}}{(1+x^n)^2}\\
\Leftrightarrow f(x) &= \biggl(\frac{1}{k+1}\biggr)\frac{x^n}{1+x^n}
\end{align}

EDIT
As pointed out in the comments. My solution is incorrect. Can anyone please help me solve this?

答:
Formally a solution is given by
$$
f_0(x) = \sum_{j = 0}^\infty (-k)^j\frac{(x-j)^n}{1+(x-j)^n}
$$
One can expect convergence if $|k| < 1$. Mathematica can evaluate this for certain choices of $k$ and $n$. For example, if $k = 1/2, \, n = 2$, then
$$
f_0(x) = \frac{1}{6} \left(4+3 i \left(-\frac{1}{2}\right)^{x-i}
   \left(\left(-\frac{1}{2}\right)^{2 i}
   B_{-\frac{1}{2}}(-x-i,0)-B_{-\frac{1}{2}}(i-x,0)\right)\right)
$$
where $B$ is the Beta function. For $n = 1$, we obtain
$$
f_0(x) = \left(-k\right){}^{x+1} B_{-k}(-x,-1)-x \Phi
   \left(-k,1,-x-1\right)
$$
where $\Phi$ is the Lerch transcendent.
In other cases, the answer may be expressed in terms of hypergeometric functions.


问:
Convergence and the limit of the sequence $x_n=\frac {a_n}{b_n}$ where $(1+\sqrt3)^n=a_n+b_n\sqrt3, n \ge 1$ with $a_n,b_n$ integers
I have written the binomial expansion of $(1+\sqrt3)^n$ as

$$\binom n0 1 + \binom n1 \sqrt3 + \binom n2 3 +\binom n3 3\sqrt3 + ...$$

but I couldn't make any use of it. Can you give me a starting point to solve the problem?

Thanks for your effort and time in advance.

答:
It is not hard to see that $$(1-\sqrt3)^n=a_n-b_n\sqrt3$$

Now solve the system for $a_n$ and $b_n$...


问:
Second Order Finite Difference Scheme to Solve Initial Value Problem
I would appreciate help in a problem that I cannot figure out where am I doing the mistake. The mistake could be something I am unaware of so please excuse my lack of knowledge.

I am trying to solve the equation of motion,
$$M \ddot{x} + K x = F,$$
where $M$, $K$, and $F$ are variables independent of $x$, and $\ddot{x}=\frac{d^2x}{dt^2}$. Using a central difference finite scheme, I rewrote the equation as,

$$M \frac{ x_{i+1} - 2x_i + x_{i-1} }{ {\Delta t} ^2 } + Kx = F.$$

Then, I am solving this equation for $x_{i+1}$ at every time step $i$ as,

$$ x_{i+1} = \frac{{\Delta t}^2}{M} \cdot \left( F - K x + \frac{M}{{\Delta t}^2} \cdot 2 x_i - \frac{M}{{\Delta t}^2} \cdot x_{i-1} \right).$$

Now to solve this problem, one needs two initial conditions, to my knowledge, namely, $x(0)$ and $\dot{x}(0)$. At the first time step, I am setting $x_i$ to the $x(0)$ initial condition. Now I am stuck with two problems:

What value should I set $x_{i-1}$ at first time step?

How can I apply the initial condition $\dot{x}(0)$?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

In the firt step,lets move everything to the adimensional form:
\begin{align}
& x = {F \over k}\,X,\quad t = \root{M \over k}T
\\[2mm] &
\substack{\mbox{which yields, with}
\\[1mm]
\ds{\dot{a} \equiv \totald{a}{t},\quad b' = \totald{b}{T}}}
\quad
\left\{\begin{array}{l}
\ds{X'' + X = 1}
\\[2mm]
\ds{X\pars{0} = {k \over F}\,x\pars{0} \equiv X_{0}}
\\[2mm]
\ds{X'\pars{0} = {\root{kM} \over F}\,\dot{x}\pars{0} \equiv V_{0}}
\end{array}\right.
\end{align}

With $\ds{Y = X'}$, I'll have the couple of equations
\begin{align}
&\left.\begin{array}{rcl}
\ds{X'} & \ds{=} & \ds{Y}
\\
\ds{Y'} & \ds{=} & \ds{1 - X}
\end{array}\right\}\quad\mbox{with}\quad
\left\{\begin{array}{rcl}
\ds{X\pars{0}} & \ds{=} & \ds{X_{0}}
\\
\ds{Y\pars{0}} & \ds{=} & \ds{V_{0}}  \end{array}\right.
\end{align}
and the "discrete version"
$$
\left\{\begin{array}{rcl}
\ds{X_{n + 1}} & \ds{=} & \ds{X_{n} + Y_{n}\,\Delta T}
\\
\ds{Y_{n + 1}} & \ds{=} & \ds{Y_{n} + \pars{1 - X_{n}}\Delta T}
\end{array}\right.
$$

With $\ds{\color{red}{X_{0}} = \color{blue}{V_{0}} = 0}$, the following picture illustrates $\ds{10000}$ iterations of $\ds{\color{red}{X\pars{T}}}$ and $\ds{\color{blue}{Y\pars{T} = X'\pars{T}}}$ with $\ds{\Delta T = 0.005}$:


问:
Drawing numbers without replacement until you get your desired number.
If you have a bag of numbers $1-10$, and you draw (without replacement) until you get the number $1$, what is the probability that you get $1$ on the $i^\text{th}$ try?

I tried to approach this problem with: $P(i=1) = 1/10, P(i=2) = (9/10)(1/9), P(i=3) = (9/10)(8/9)(1/8)...,$ but this gave me $(1/10)$ for each $i$ value, which I am pretty sure is incorrect. How else should should I approach this problem?

答:
"There is only one way to pick the correct number first."  No.  You're not considering all the variations in subsequent number selections.

There are many ways to approach this classic problem to see the answer is $1/10$.  I prefer ones based on the logic of interchange symmetry:

Let the probability of choosing a $1$ on the first selection be $p$.  Can you see that the probability of choosing a $2$ on the first selection must also be $p$?  (How could it be different?)  And likewise for all $10$ numbers.  But SOME number will be chosen first, so the sum of all the $p$s must equal $1$.

QED.


问:
Random walk serial correlation
Given a model $$Y_t =b_0 + b_1 \cdot X_t + b_2 \cdot Z_t + e_t,$$ where the error term $e_t$ follows a random walk form of serial correlation $e_t = e_{t-1} + u_t$. Further assume $u_t$ has zero mean and a constant variance and zero covariance with $u_s$ where $s \neq t$.

How to transform a random walk serial correlation so that we can use the same data to estimate a model that satisfies Gauss-Markov assumptions? In other words, how can we transform the above model to get OLS estimates?

答:
How about first differencing? This yields
$$
Y_t-Y_{t-1}=(b_0-b_0)+b_1(X_t-X_{t-1})+b_2(Z_t-Z_{t-1})+((e_{t-1}+u_t)-e_{t-1})\\
\Delta Y_t=b_1\Delta X_t+b_2\Delta Z_t+u_t
$$
which you might find useful. </p>

Title: Random walk serial correlation