问:
Is automorphism group of lie group a lie group?
Let $G$ be a lie group. Is its automorphism group $\text{Aut}(G)$ of smooth automorphisms then a lie group?

According to G. Hochschild is $\text{Aut}(G)$ when equipped with the compact-open topology a lie group if the component group $G/G_0$ is finitely generated. Here is link https://www.jstor.org/stable/1990752?seq=1#metadata_info_tab_contents

What if $G/G_0$ is not finitely generated? I have tried to find an example of lie group with non-finitely generated component group whose automorphism group is not a lie group when equipped with the compact-open topology, but haven't found one. I suspect a discrete countable group whose automorphism group is uncountable can be such an example.

答:
In practice I think it's very rare for people to consider Lie groups $G$ where $\pi_0(G)$ isn't just finite, but technically any countable discrete group is a (zero-dimensional) Lie group. So it suffices to find a (necessarily infinitely generated) countable discrete group $G$ whose automorphism group is not discrete when equipped with the compact-open topology. (You already know this but I'm clarifying for any other readers.)

We can take $G = \mathbb{Q}/\mathbb{Z}$ (with the discrete topology). It's a nice exercise to show that $\text{End}(G)$ is the profinite integers $\widehat{\mathbb{Z}}$ and so $\text{Aut}(G)$ is the group of units of the profinite integers, which is in particular uncountable. The compact-open topology here is the product topology on $G^G$; I don't understand this topology very well but I'd guess it doesn't have any uncountable discrete subspaces and hence that $\text{Aut}(G)$ is not discrete in it. The nicest possibility is that $\text{Aut}(G)$ would just have the profinite topology (which is certainly not discrete) but I don't know if that's true.


问:
Proving a relation R is not a function
I'm working on this basic discrete math question and struggling to understand the notation,

We have a relation $R$ on $Z^+$ defined as follows:

$mRn$ if and only if $m|n$

Explain why the relation R is not a function

Let $A$ and $B$ be nonempty sets. Then by definition, a relation $f$ from $A$ to $B$ is a "function" if each element $a\in A$ is related by $f$ to one and only one element of $b$.

Now the question provides the relation $m|n$, which provides us the information "$m$ divides $n$ if $n=km$ for some integer $k$".

How exactly do we related this simple statement to not being a function? I seems that this would be a function as $n=km$ for some integer $k$ appears to have one and only out output for each input.

答: $2 | 6$ and $2 | 12$. Hence $2R6$ and $2R12$. Meaning that $2$ would have at least two images if $R$ was a function. A contradiction with the definition of a function.

ContentLicense: CC BY-SA 4.0

Title: Proving a relation R is not a function


问:
Calculating percent reduction in variance after deductible

The amount of a claim that a car insurance company pays out follows an exponential distribution. By imposing a deductible of $d$, the insurance company reduced the expected claim payment by $10\%$. Calculate the percentage reduction on the variance of the claim payment.

This question was previously discussed here. Based on this discussion, I came up with this solution:

Let $X$ be the random variable for the claim amount before the deductible, d,  is applied and $Y$ be the random variable for the claim amount after the deductible, d,  is applied. Then, $Y = X-d$ if $d \le X <\infty$ and $Y = 0$ otherwise.

Now, $E[Y] = E[X] - 0.1 E[X] \implies E[Y] = 0.9E[X] = E[0.9X] \implies Y = 0.9X$.

So, $Var[Y] = Var[0.9X] = 0.9^2Var[X] = 0.81 Var[X]$

which means that the variance has been reduced by approximately $19\%$ but the correct answer is $1\%$. What have I done incorrectly?

答:
The law of total expectation gives
$$\begin{align}
\operatorname{E}[Y] &= \operatorname{E}[(X - d)_+] \\
&= \operatorname{E}[X - d \mid X > d]\Pr[X > d] + \operatorname{E}[0 \mid X \le d]\Pr[X \le d] \\
&= \operatorname{E}[X]\Pr[X > d].
\end{align}$$
Note the last equality uses the memoryless property.

Similarly,
$$\begin{align}
\operatorname{E}[Y^2] &= \operatorname{E}[(X-d)^2 \mid X > d]\Pr[X > d] \\
&= \operatorname{E}[X^2]\Pr[X > d].
\end{align}$$
Since we know for an exponential distribution with mean $\mu$ has survival $S_X(x) = \Pr[X > x] = e^{-x/\mu}$ and second moment $\operatorname{E}[X^2] = 2\mu^2$, it follows that $$\operatorname{E}[Y] = \mu e^{-d/\mu}, \quad \frac{\operatorname{E}[Y]}{\operatorname{E}[X]} = e^{-d/\mu} = 0.9,$$ and $$\operatorname{Var}[Y] = 2\mu^2 e^{-d/\mu} - (\mu e^{-d/\mu})^2, \quad \frac{\operatorname{Var}[Y]}{\operatorname{Var}[X]} = 2e^{-d/\mu} - (e^{-d/\mu})^2 = 2(0.9) - (0.9)^2 = 0.99$$ and there is a $1\%$ reduction in the variance.

No integration is needed.  If you did not recall the second moment of the exponential distribution, you can get it from the variance which is $\mu^2$ and first moment $\mu$:  $$\operatorname{E}[X^2] = \operatorname{Var}[X] + \operatorname{E}[X]^2 = \mu^2 + \mu^2 = 2\mu^2.$$


问:
$V_\omega$ is countable
Is there an easy way to prove this? I found a book that suggests the injection $h:V_\omega\to\omega$ defined by

$$h(\{x_1,x_2,\dots,x_n\})=2^{h(x_1)}+2^{h(x_2)}+\cdots+2^{h(x_n)},$$

but I hit some snags while proving that the function is well-defined. Are there any other clever ways to prove this fact? (Note: I am making a formal proof, so I need to be able to justify my steps.)

Edit: Let me see if I can make my difficulty more clear. Before I can start the proof by induction, I need a precise statement of the induction hypothesis, and this is what I am trying to hammer out. The hypothesis:

For each $n\in\omega$, there exists a unique bijection $h_n:V_n\to|V_n|$ such that $$h_n(x)=\sum_{m<|V_{n-1}|}\operatorname{if}(h_{n-1}^{-1}(m)\in x,2^m,0).$$
Moreover, the $h_n$ as defined satisfies $h_n\restriction |V_{n-1}|=h_{n-1}$.

Does this hypothesis contain enough to prove the claim? The reason for the odd way of stating the sum is that it avoids the issue in the linked question (we need to show that addition is a well-defined operation over sets), but as a consequence I have to prove that
the $h_n$ is a bijection and not just an injection, and I think I also need the restriction part in order to prove that the union $h=\bigcup_{n\in\omega}h_n$ is the required bijection.

答: If you just need to prove that $V_\omega$ is countable, and if you're allowed to use the axiom of choice, then I think you're working too hard.  Prove, by induction on natural numbers $n$ that $V_n$ is finite (because the power set of a finite set is finite).  Then $V_\omega=\bigcup_{n\in\omega}V_n$ is the union of countably many finite sets, hence countable.</p>

ContentLicense: CC BY-SA 3.0

Title: $V_\omega$ is countable


问:
Combining and simplifying equations
I've come across a derivation in a journal article where one of the steps is to substitute one expression into another, but I cannot simplify to their solution. The Expressions are:

Expression 1

$$
\frac{d\theta_{t}}{dt} = k_{a}q_{max}X\theta_{t}^2 -(k_{a}C_{0} + k_{a}q_{max}X+k_{d})\theta_{t} +k_{a}C_{0}
$$

Expression 2

$$
\theta_{e} = \frac{k_{a} (q_{max}X + C_{0})+k_{d} - \sqrt{\Delta}}{2k_{a}q_{max}X}
$$

In the article, it says that Expression 2 is substituted into Expression 1, to obtain Expression 3:

Expression 3

$$
\frac{d\theta_{t}}{dt} = k_{1}(\theta_{e}-\theta_{t})+k_{2}(\theta_{e}-\theta_{t})^2
$$

With the definitions

$$
k_{1}=\sqrt{\Delta}
$$
$$
k_{2} = k_{a}q_{max}X
$$

Unfortunately the authors omitted the steps in between, so to check I initially simplified Expression 1 and Expression 2 using the definitions:

$$
\frac{d\theta_{t}}{dt} = k_{2}\theta_{t}^2 -(k_{a}C_{0} + k_{2}+k_{d})\theta_{t} +k_{a}C_{0}
$$
$$
\theta_{e} = \frac{k_{2}+ k_{a}C_{0}+k_{d} - k_{1}}{2k_{2}}
$$

Rearranging the second expression I get:

$$
k_{a}C_{0} =2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$

Which I substituted into the first expression to get

$$
\frac{d\theta_{t}}{dt} = k_{2}\theta_{t}^2 -(2k_{2}\theta_{e}+k_{1})\theta_{t} +2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$

Multiplying out the brackets and completing the square, I get:

$$
\frac{d\theta_{t}}{dt} =k_{2}(\theta_{t}-\theta_{e})^2 - k_{2}\theta_{e}^2 -k_{1}\theta_{t}+2k_{2}\theta_{e}+k_{1}-k_{2}-k_{d}
$$

Which is where I become stuck. Is there a mathematical trick other than completing the square that I am missing which will allow me to simplify this expression to obtain Expression 3?

The paper that I have obtained this derivation from is by Lui and Shen, titled "From Langmuir Kinetics to First and Second Order Rate Equations for Adsorption"

Edit

In response to a comment, Delta is defined as:

$$
\Delta=k_{a}^2(C_{0}-q_{max}X)^2+2k_{a}k_{d}(C_{0}+q_{max}X)+k_{d}^2
$$

答:
Note that expression for $\theta_e$ looks suspiciously like a formula for a root of quadratic equation.  Specifically, it's the smaller of the two roots of the quadratic in the expression for $d\theta_t/dt$.  And the aptly named $\Delta$ is the quadratic's discriminant.

That's all we need to know.  Suppose $x_1$ and $x_2$ where $x_1 \le x_2$ are the two roots of
$$
ax^2 + bx + c
$$
Then the quadratic can be rewritten as
$$a(x-x_1)(x-x_2)$$

But, since $x_2-x_1 = \sqrt{\Delta}/a$ where $\Delta$ is the quadratic's discriminant, we can substitute $x_2 = x_1 + \sqrt{\Delta}/a$ to obtain

$$a(x_1-x)^2 + \sqrt{\Delta}(x_1-x)$$

Which is the expression you were trying to derive

Title: Combining and simplifying equations


问:
Solve the integral in the expression $x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt$
I'm following the course MIT 18.03. In [this problem][1], we evaluate the following expression containing an integral:
$$
x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt\tag{1}
$$
which finally gives:
$$
x(t) = \frac{2k \sin(t/2)+ 4k^2\cos(t/2)}{1+4k^2}+ce^{-kt}\tag{2}
$$
One possibility would be to make a substitution $u = \sin(x)$, but more than that I'm stuck

Edit (with the help of the comments)

So we can use the following substitutions:

\begin{align}
u &= \cos(t/2)\tag{3}\\
\implies u' &= 1/2\sin(t/2)\tag{4}\\
v'&= e^{kt}\tag{5}\\
\implies v &= \frac{1}{k} e^{kt}\tag{6}\\
\end{align}

Now using integration by part we can write

\begin{align}
\int v'u &= vu - \int vu'\tag{7}\\
&=\cos(t/2) \cdot \frac{1}{k}e^{kt} - \int \frac{1}{k}e^{kt}\frac{1}{2}\sin (t/2)\tag{8}\\
\end{align}
[1]: https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-i-first-order-differential-equations/solution-by-integrating-factors/MIT18_03SCF11_rec_04s5_sol.pdf

答:
Observe the following by Integration By Parts:
\begin{align*}
x(t) &= ke^{-kt} \int e^{kt}\cos\left ( \frac{t}{2} \right ) \, \mathrm{d}t\\
&= ke^{-kt} \left [2e^{kt}\sin\left ( \frac{t}{2} \right ) - 2k\int e^{kt}\sin\left ( \frac{t}{2} \right ) \, \mathrm{d}t \right ] && \text{using } u=e^{kt}, v'=\cos\left (\frac{t}{2} \right )\\
&= ke^{-kt} \left [2e^{kt}\sin\left ( \frac{t}{2} \right ) + 4ke^{kt}\cos\left ( \frac{t}{2} \right ) - 4k^2 \int e^{kt}\cos\left ( \frac{t}{2} \right ) \, \mathrm{d}t \right ] && \text{using } u=e^{kt}, v'=\sin\left (\frac{t}{2} \right )\\
&= 2k\sin\left ( \frac{t}{2} \right ) + 4k^2\cos\left ( \frac{t}{2} \right ) - 4k^2 x(t) + Ce^{-kt}\\
&= \frac{2k \sin\left ( \frac{t}{2} \right ) + 4k^2 \cos\left ( \frac{t}{2} \right )}{1 + 4k^2}+ ce^{-kt}
\end{align*}
and the result follows.


问:
Show $\frac52\cot\frac{4\pi}7+\frac34\cot\frac\pi7\sec\frac{2\pi}7+\cot\frac{2\pi}7\cos\frac\pi7=\sqrt7$
While solving a definite integral in different ways, I found that the trigonometric expression involving the heptagonal angles below simply reduces to $\sqrt7$ $$\frac52\cot\frac{4\pi}7+\frac34\cot\frac\pi7\sec\frac{2\pi}7+\cot\frac{2\pi}7\cos\frac\pi7$$

I verified it numerically, but could not resist trying to do the same by hand.

Unlike the familiar ones, such as $4\sin\dfrac{2\pi}7-\tan \dfrac\pi7 =\sqrt7$, I have not come across any resembling patterns anywhere and would like to find a clean path to derive it analytically.

答:
Notice that
$$ \cos(\pi/7)=-\cos(6\pi/7)\quad \text{ and }
\quad\cot(\pi/7)=-\cot(6\pi/7). \tag{1}$$ Define
$$ x:=\frac{2\pi}7,\; y:=\frac52\cot\frac{4\pi}7+
\frac34\cot\frac\pi7\sec\frac{2\pi}7
+\cot\frac{2\pi}7\cos\frac\pi7,\; z:=e^{ix}. \tag{2}$$
Now $$ y = \frac52 \cot(2x)-\frac34 \cot(3x)\sec(x)
-\cot(x)\cos(3x). \tag{3}$$
Check that $\,y^2-7\,$ is a rational
function in $\,z\,$ with a factor of $\,\sum_{n=0}^6 z^n\,$
in the numerator, but this $=0$ since $\,z\,$ is a 7th root of unity thus
$\,y=\sqrt{7}.$ I used Mathematica to do the rational function
expansion but some other Computer Algebra System could be used instead.
My Mathematica code is

```<code>y = 5/2 Cot[2x] - 3/4 Cot[3x]Sec[x] - Cot[x]Cos[3x];
Factor[TrigToExp[y^2-7] /. x->Log[z]/I]
</code>```

Body:
Notice that
$$ \cos(\pi/7)=-\cos(6\pi/7)\quad \text{ and }
\quad\cot(\pi/7)=-\cot(6\pi/7). \tag{1}$$ Define
$$ x:=\frac{2\pi}7,\; y:=\frac52\cot\frac{4\pi}7+
\frac34\cot\frac\pi7\sec\frac{2\pi}7
+\cot\frac{2\pi}7\cos\frac\pi7,\; z:=e^{ix}. \tag{2}$$
Now $$ y = \frac52 \cot(2x)-\frac34 \cot(3x)\sec(x)
-\cot(x)\cos(3x). \tag{3}$$
Check that $\,y^2-7\,$ is a rational
function in $\,z\,$ with a factor of $\,\sum_{n=0}^6 z^n\,$
in the numerator, but this $=0$ since $\,z\,$ is a 7th root of unity thus
$\,y=\sqrt{7}.$ I used Mathematica to do the rational function
expansion but some other Computer Algebra System could be used instead.
My Mathematica code is

y = 5/2 Cot[2x] - 3/4 Cot[3x]Sec[x] - Cot[x]Cos[3x];
Factor[TrigToExp[y^2-7] /. x->Log[z]/I]


问:
real mechanism behind addition, subtraction, multiplication, division
we all know the basic rules for operations of addition, subtraction, multiplication and division.

but what i don't know is why these rules (of addition, subtraction, multiplication and division) works.

as if we have been given algorithm to do these operations but not explained how they were derived.

if anyone can explain please expain .

答:
One thing you'll want to understand here is that you are only talking about 2 operations: addition and multiplication.

Subtraction is a way of abstracting the concept of additive inverses (e.g. $3 + (-3) = 0$), and division is really multiplicative inverses ($3 \times \frac{1}{3} = 1$).

I know this doesn't fully answer your question, but it does simplify it.


问:
There are 45 egg boxes in a store.
Twenty are Brand A, fifteen are brand B, and ten are Brand C. Brand A and C each have half green boxes and half yellow boxes. Brand B is all yellow.

If you have Brand B or C, what's the chance that you have a green box?

My work:

let $G$ be the event that the box is green, $B$ be the event that the box is from brand B, $C$ be the event that the box is from brand C.

I believe that we are trying to calculate $P(G \cap (B \cup C)) = P(G\cap B) \cup P(G\cap C)$ via the distributive law

So $P(G\cap B) = 0 $ since all of Brand B has yellow boxes. and $P(G\cap C) = P(C)P(G|C) = \cfrac{10}{45} \cdot \cfrac{1}{2} = \cfrac{1}{9}$

so $P(G \cap (B \cup C)) = 0 + \cfrac{1}{9} = \cfrac{1}{9}$ is the probability

But the book says that $1/5$ is the probability.

Can someone show me where I went wrong in my reasoning? Thank You!

答:
There are 10 boxes in C, of which 5 are green. None of the 15 in B are green.

If you select from either B or C, then you have 5 green boxes and 10 + 15 = 25 total items, so 5/25 or 1/5.

Title: There are 45 egg boxes in a store.


答:
You have computed the probability that you chose $B$ or $C$ and that you got a green box. You are given that you have $B$ or $C$.
$$
\overbrace{\quad\frac{15}{25}\quad}^{P(B|B\lor C)}\cdot\overbrace{\ \quad0\quad\ \vphantom{\frac11}}^{P(G|B)}+\overbrace{\quad\frac{10}{25}\quad}^{P(C|B\lor C)}\cdot\overbrace{\ \quad\frac12\quad\ }^{P(G|C)}=\overbrace{\ \quad\frac15\quad\ }^{P(G|B\lor C)}
$$

Title: There are 45 egg boxes in a store.

Body:
Twenty are Brand A, fifteen are brand B, and ten are Brand C. Brand A and C each have half green boxes and half yellow boxes. Brand B is all yellow.

If you have Brand B or C, what's the chance that you have a green box?

My work:

let $G$ be the event that the box is green, $B$ be the event that the box is from brand B, $C$ be the event that the box is from brand C.

I believe that we are trying to calculate $P(G \cap (B \cup C)) = P(G\cap B) \cup P(G\cap C)$ via the distributive law

So $P(G\cap B) = 0 $ since all of Brand B has yellow boxes. and $P(G\cap C) = P(C)P(G|C) = \cfrac{10}{45} \cdot \cfrac{1}{2} = \cfrac{1}{9}$

so $P(G \cap (B \cup C)) = 0 + \cfrac{1}{9} = \cfrac{1}{9}$ is the probability

But the book says that $1/5$ is the probability.

Can someone show me where I went wrong in my reasoning? Thank You!


问:
How to prove $\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) \, dx=\ln 2$
How to prove the integral
$$\begin{align}&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\=&\int_0^\infty\left (\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right)dx\\=&\ln 2\end{align}$$

I try to use integration by parts
$$\begin{align}
&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\
=&\int_0^\infty \int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\\
=&x\left.\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right|_0^\infty+\int_0^\infty \cos x \int_x^\infty\frac{\sin t}{t}dx+\int_0^\infty \sin x  \int_x^\infty\frac{\cos t}{t}dx
\end{align}$$
But the last two integrals are divergent.

答:
The Laplace transform is useful for providing integral representations in terms of non-oscillating functions:
$$\text{si}(x)=\int_{x}^{+\infty}\frac{\sin t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(\cos(x)+s\sin(x)\right)\,ds $$
$$\text{Ci}(x)=\int_{x}^{+\infty}\frac{\cos t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(s\cos(x)-\sin(x)\right)\,ds $$
This allows to apply Fubini's theorem:
$$\begin{eqnarray*} \int_{0}^{+\infty}\text{si}(x)\text{Ci}(x)\,dx &=& \iiint_{(0,+\infty)^3}\frac{e^{-(a+b)x}}{(1+a^2)(1+b^2)}(\cos x+a\sin x)(b\cos x-\sin x)\,dx\,da\,db\\&=&\iint_{(0,+\infty)^2}\frac{1}{(1+a^2)(1+b^2)}\left(1-\frac{a}{a+b}+\frac{ab-3}{4+(a+b)^2}\right)\,da\,db\end{eqnarray*} $$
and the last double integral is pretty simple to be computed:
$$ \iint_{(0,+\infty)^2}\frac{da\,db}{(1+a^2)(1+b^2)}=\left(\frac{\pi}{2}\right)^2 $$
$$ \iint_{(0,+\infty)^2}\frac{a}{(1+a^2)(1+b^2)(a+b)}\,da\,db=\iint_{(0,+\infty)^2}\frac{b}{(1+a^2)(1+b^2)(a+b)}\,da\,db =\frac{\pi^2}{8}$$
and the difficult part
$$ \iint_{(0,+\infty)^2}\frac{ab-3}{(1+a^2)(1+b^2)(4+(a+b)^2)}\,da\,db=\int_{0}^{+\infty}\frac{-\pi(a^2+1)+2(a^2-1)\arctan\frac{a}{2}+2a\log(a^2+4)}{4(a^2+1)^2}\,da$$
can be broken by integration by parts.


问:
Prove that $f$ is constant $\Leftrightarrow$ for all point pair $x,y \in \mathbb{R} $, $ | f(x) - f(y) | \leq (x-y)^2 $
Prove that $f$ is constant if and only if for all point pair $x,y \in \mathbb{R} $, $$ | f(x) - f(y) | \leq (x-y)^2 $$

My attempt:

It's a double implication so...

To right:

$f$ is constant $\Rightarrow \forall c \in \mathbb{R}, f(c) = k $, with $ k \in \mathbb{R} \Rightarrow \forall x,y \in \mathbb{R} , | f(x) - f(y) | = 0 \Rightarrow 0 \leq (x-y)^2  $

We observe that $(x-y)^2$ is the square of the difference of two real numbers, hence, is always positive.

To left:

We cant think $(x-y)^2$ as $|x-y|^2$, so:

$$ | f(x) - f(y) | \leq |x-y|^2$$
$$ \Bigg|\frac{f(x) - f(y)}{x-y}\Bigg| \leq |x-y| $$
$$ \lim_{x \to y}{\Bigg|\frac{f(x) - f(y)}{x-y}\Bigg|} \leq \lim_{x \to y}{|x-y|} $$
$$ |f'(y)| \leq 0 $$
We know that the module of a number is always greater or equal to $0$. So $ |f'(y)| = 0 $.
$$ |f'(y)| = 0 \Rightarrow f(y) \ \text{is constant} $$

But I'm not sure if I can get into the absolute value with the limit. The statement doesn't mention that $f$ is differentiable.

答: You proof works, the fact that $f$ is not supposed being differentiable is normal. Indeed, you showed that $\left|\frac{f(x)-f(y)}{x-y}\right|\leqslant |x-y|$ for all $x\neq y$, since $\lim\limits_{x\rightarrow y}|x-y|=0$, we also have $\lim\limits_{x\rightarrow y}\frac{f(x)-f(y)}{x-y}=0$. By doing this, you proved that $f$ is differentiable at $y$ and $f'(y)=0$, since $y$ is arbitrary, this means that $f'=0$ and therefore $f$ is constant.


答:
From part c: "... a consequence, you may recall, of two integrations by parts..."  So we expect we can integrate (or differentiate) the cosine piece twice to get a simple multiple of the integral with which we start.

By factoring out the $k$ from your integral, let
$$  I = \int \mathrm{e}^{kt} \cos(t/2) \,\mathrm{d}t  \text{.}  $$
Taking $u_1 = \cos(t/2)$ so $\mathrm{d}u_1 = (-1/2) \sin(t/2) \,\mathrm{d}t$ and $\mathrm{d}v_1 = \mathrm{e}^{kt} \,\mathrm{d}t$ so $v_1 = (1/k) \mathrm{e}^{kt}$, we obtain
$$  I = \frac{1}{k} \mathrm{e}^{kt} \cos(t/2) + \frac{1}{2k} \int \mathrm{e}^{kt} \sin(t/2) \,\mathrm{d}t  \text{.}  $$
Taking $u_2 = \sin(t/2)$ so $\mathrm{d}u_2 = (1/2) \cos(t/2) \,\mathrm{d}t$ and $\mathrm{d}v_2 = \mathrm{e}^{kt} \,\mathrm{d}t$ so $v_2 = (1/k)\mathrm{e}^{kt}$, we obtain
$$  I = \frac{1}{k} \mathrm{e}^{kt} \cos(t/2) + \frac{1}{2k} \left( \frac{1}{k} \mathrm{e}^{kt} \sin(t/2) - \frac{1}{2k} \int \mathrm{e}^{kt} \cos(t/2) \,\mathrm{d}t \right)  \text{.}  $$
Notice that the remaining integral is $I$.  So we have
\begin{align*}
I &= \frac{1}{k} \mathrm{e}^{kt} \cos(t/2) +  \frac{1}{2k^2} \mathrm{e}^{kt} \sin(t/2) - \frac{1}{4k^2} I + C \\
I + \frac{1}{4k^2} I &= \frac{1}{k^2} \mathrm{e}^{kt} \left( k\cos(t/2) + \frac{1}{2} \sin(t/2) \right) + C  \\
I &= \frac{1}{k\left(1 + \frac{1}{4k^2}\right)} \mathrm{e}^{kt} \left( k \cos(t/2) + \frac{1}{2} \sin(t/2) \right) + C  \\
    &= \frac{4k}{\left(4k^2 + 1\right)} \mathrm{e}^{kt} \left( k \cos(t/2) + \frac{1}{2} \sin(t/2) \right) + C
\end{align*}
... and you should be able to finish from here.

Title: Solve the integral in the expression $x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt$

Body:
I'm following the course MIT 18.03. In [this problem][1], we evaluate the following expression containing an integral:
$$
x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt\tag{1}
$$
which finally gives:
$$
x(t) = \frac{2k \sin(t/2)+ 4k^2\cos(t/2)}{1+4k^2}+ce^{-kt}\tag{2}
$$
One possibility would be to make a substitution $u = \sin(x)$, but more than that I'm stuck

Edit (with the help of the comments)

So we can use the following substitutions:

\begin{align}
u &= \cos(t/2)\tag{3}\\
\implies u' &= 1/2\sin(t/2)\tag{4}\\
v'&= e^{kt}\tag{5}\\
\implies v &= \frac{1}{k} e^{kt}\tag{6}\\
\end{align}

Now using integration by part we can write

\begin{align}
\int v'u &= vu - \int vu'\tag{7}\\
&=\cos(t/2) \cdot \frac{1}{k}e^{kt} - \int \frac{1}{k}e^{kt}\frac{1}{2}\sin (t/2)\tag{8}\\
\end{align}
[1]: https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-i-first-order-differential-equations/solution-by-integrating-factors/MIT18_03SCF11_rec_04s5_sol.pdf


问:
Calculate the limit of a sequence $\lim_{n\to \infty} \frac{I_{n}}{n^{n}}$

Calculate $$\lim_{n\to \infty} \frac{1}{n^{n}}\int_{0}^{n}x^n\arctan(\sqrt[n]{x})dx$$

My approach: I was trying to solve first the integral, so for example for $n=1$, we have that
$$\int x\arctan(x)dx=\frac{1}{2}x^{2}\arctan(x)-\frac{x}{2}+\frac{1}{2}\arctan(x)+c$$
and for $n=2$, we can see that $$\int x\arctan(\sqrt{x})dx=\frac{1}{6}(3(x^{2}-1)\arctan(\sqrt{x})-(x-3)\sqrt{x}+c$$
and for $n=3$, we have
$$\int x\arctan(\sqrt[3]{x})dx=\frac{1}{30}(-3x^{5/3}+15(x^{2}+1)\arctan(\sqrt[3]{x})+5x-15\sqrt[3]{x})+c$$

I was trying to see if by induction I could give a closed formula for the integral but the calculation of the integral does not show any pattern. How can I solve this limit?

答:
You can also use $t = \left(\frac{x}{n}\right)^{n+1}$

$$\implies \lim_{n\to\infty} \frac{n+1}{n}\int_0^1 \tan^{-1}\left(n^{\frac{1}{n}}t^{\frac{1}{n^2+n}}\right)dt \to \int_0^1 \tan^{-1}(1)dt = \frac{\pi}{4}$$

by dominated convergence since the sequence is dominated by $g(t)=\pi$.


问:
Proof that $V\odot W\cong W\odot V$ for vector spaces $V$ and $W$
I've read a proof that the algebraic tensor product of vector spaces is 'commutative' in the sense that $V\odot W\cong W\odot V$. Although I believe I have managed to prove this by showing that they both have same dimension, therefore are isomorphic, this is not a 'constructive' proof in the sense that it not provide an explicit example of a linear isomorphism.

In Appendix T of Wegge-Olsen's book $K$-Theory and $C^*$-Algebras, this is proved by showing that there is a linear isomorphism $\varphi:V\odot W\to W\odot V$ with $\varphi(v\otimes w)=w\otimes v$ for all $v\in V, w\in W$.

The only way I see of doing this is by using a result which is later on in the book i.e. proving injectivity via $$f\otimes e=f'\otimes e'\Rightarrow e=e',f=f'\Rightarrow e\otimes f=e'\otimes f',$$ where $e,e'\in E$, $f,f'\in F$ and $E,F$ are bases of $V,W$, respectively.

But that result uses commutativity, so this approach would be circular. Is there another way of doing this?

In other words, without using the result that when $\sum e_i\otimes f_i=0$ and $f_i$ lin. indep. implies $e_i=0$ for all $i$.

答:
You have constructed a linear map $\phi:V\otimes W \to W\otimes V$ with the property that for all $v\in V, w\in W$, we have $\phi(v\otimes w) = w\otimes v$. Similarly, construct a linear map $\psi:W\otimes V\to V\otimes W$ with the property that for all $v\in V, w\in W$, we have $\psi(w\otimes v)=v\otimes w$.

Then, it is easy to see that for such "pure tensors", $\phi\circ \psi$ and $\psi\circ \phi$ are the identity map (on appropriate spaces). But in general, an element of $V\otimes W$ is a finite sum of pure tensors, so $\psi\circ \phi = \text{id}_{V\otimes W}$. Similarly, an arbitrary element of $W\otimes V$ is a finite sum of pure tensors hence $\phi\circ \psi=\text{id}_{W\otimes V}$. This directly shows $\phi,\psi$ are inverse isomorphisms.

So, really, all we're using is that pure tensors span the whole tensor product, and that if a linear map is equal to identity when restricted to a spanning set, then it is equal to identity on the whole space. In general when trying to prove isomorphisms involving tensor products, I think it is much easier to construct the maps in both directions (and very often this is almost immediate by the universal property of tensor products) rather than proving injectivity/surjectivity directly, because often times we don't really understand the particular elements of the tensor product well.


问:
Non continuous function with directional derivatives zero in all directions
I was asked the following question:
A function f(x,y) has directional derivatives in all directions, and they are 0 in all directions. Does f have to be continous?

I thought I proved it by:

\begin{align}\lim_{(x,y)\to (0,0)}f(x,y)-f(0,0)&=\lim_{r\to0}f(r\cos\theta,r\sin\theta)-f(0,0)\\&=\lim_{r\to0}r\cdot (f(r\cos\theta,r\sin\theta)-f(0,0))/r\\&=\lim_{r\to0}r\cdot \lim_{r\to0}(f(r\cos\theta,r\sin\theta)-f(0,0))/r\end{align}

The right limit is the directional derivative in direction $\theta$ which is said to be zero, and the left limit is zero and we get:

$$\lim_{(x,y)\to (0,0)}f(x,y)=f(0,0)$$

Is there a mistake in this proof? Because the solution said that f does not have to be continuous.
Thank you!

答: No, it doesn't have to be continuous. Take, for instance$$f(x,y)=\begin{cases}1&\text{ if }y=x^2\text{ and }x\ne0\\0&\text{ otherwise.}\end{cases}$$It is discontinuous at $(0,0)$, but the directional derivatives at $(0,0)$ are all equal to $0$.


问:
A square appears randomly within a square of ten time its area. What is probability that the smaller square contains the larger square's center?
The alignment relative to the larger square can be anywhere between 0° to 90° (0° or 90° being identical orientation to the larger square, 45° being all the way skewed), and the probability distribution of the alignment is based on percentage of total possibility given the constraint guaranteeing that it will fit within the larger square.

<s>The largest discrete probability, for an alignment of 0°=90°, should be simple enough to figure-out‡. The smallest discrete probability, for alignment of 45°, is a bit trickier but straightforward enough.</s>Edit:The off-kilter version would be more likely to contain the center, since it has less possible places to exist and a larger proportion of them are the center. Intuitively this should correspond directly (though not necessarily uniformly) to what I am totally clueless about, which is:

how to determine the probability  distribution for between 0° to 45°,

‡ 2. how this continuum corresponds to the probability of the original problem (e.g. {1/17?} for 0° up to {2/9?} for 45° back down to {1/17?} for 90°, along finite part of some sort of curve function presumably),

and 3. how then to apply these equations formulated into the original problem to find the final probability that a square true-randomly generated wholly within a larger square with sides √(10)-times longer would surround (or contain exactly on an edge or corner) the center of the larger square.

‡ I haven't calculated the maximum or minimum ‘discrete’ probabilities (let alone any inbetween) yet. Those guesstimate values are placeholders. I am especially interested in the processes required to solve this, and appreciate any insight or clues.

答:
Just to solve the simpler problem to get an approximate answer to check another "exact" answer.  I don't think rotating the square will give an answer too far off ...

Suppose the larger square is $\sqrt{10}$ by $\sqrt{10}$ inches, and the smaller square is $1$ square inch.  Then the center of the smaller square will be uniform within a square with side of length $\sqrt{10}-1$.

The center of the smaller square then needs to be within a square of one square inch near the center square.  Thus

$$P = \frac{1}{(\sqrt{10}-1)^2}= \frac{1}{11-2\sqrt{10}} \approx 0.214.$$


问:
Evaluate integral with complex exponential. Inverse Fourier transform.
I'm trying to follow and understand a bigger problem in the book but at one step they simply write out the answer of this integral and take it as a given but it seems to be tricky to compute this integral. I have $g(\omega)=2a/((\omega^2+a^2)(\omega^2+b^2))$ where $a,b>0$ and I want to evaluate

$$\int\limits_{-\infty}^{\infty}g(\omega)e^{j\omega \tau} \ d\omega.$$

This is supposed to evaluate to

$$\frac{1}{(a^2-b^2)b}\left(ae^{-b|\tau|}-be^{-a|\tau|}\right).$$

Anyone who has a trick up their sleeve? I'm pretty sure it's possible to rewrite $g(\omega)$ in a way that allows us to use some neat trick.

答:
By partial fraction decomposition, we have $$g(\omega)=\frac{2a}{(\omega^2+a^2)(\omega^2+b^2)} = \frac{2a}{b^2 - a^2}(\frac{1}{\omega^2 + a^2} - \frac{1}{\omega^2 + b^2}) = \frac{2a}{b^2 - a^2}(\frac{1}{2ia}(\frac{1}{\omega-ia} - \frac{1}{\omega+ia}) - \frac{1}{2ib}(\frac{1}{\omega-ib} - \frac{1}{\omega+ib}))$$If we define $$\mathcal{F}\{x(t)\} = X(i\omega) = \int_{-\infty}^{+\infty} x(t)e^{-i\omega t}dt$$ The inverse transform is given by $$\mathcal{F}^{-1}\{X(i\omega)\} = x(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} X(i\omega)e^{i\omega t}d\omega$$ It's easy to show that if $x(t) = e^{-at}u(t), \ \ a\gt0$ then $X(i\omega) = \frac{1}{a+ i\omega}$. Using this result, $$\mathcal{F}^{-1}\{\frac{1}{\omega - ia}\} = ie^{-at}u(t) \\ \mathcal{F}^{-1}\{\frac{1}{\omega + ia}\} = -ie^{at}u(-t)$$We should put these facts together.

We are interested in $h(t) = 2\pi\mathcal{F}^{-1}\{g(\omega)\}$. Using linearity property, we have $$h(t) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{1}{2ia}(\mathcal{F}^{-1}\{\frac{1}{\omega-ia}\} - \mathcal{F}^{-1}\{\frac{1}{\omega+ia}\}) - \frac{1}{2ib}(\mathcal{F}^{-1}\{\frac{1}{\omega-ib}\} - \mathcal{F}^{-1}\{\frac{1}{\omega+ib}\})) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{1}{2ia}(ie^{-at}u(t) +ie^{at}u(-t)) - \frac{1}{2ib}(ie^{-bt}u(t) +ie^{bt}u(-t))) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{e^{-a|t|}}{2a} - \frac{e^{-b|t|}}{2b})$$

Also we can check the result using Mathematica. Entering

```<code>Assuming[a > 0 && b > 0,
InverseFourierTransform[(2 a/((w^2 + a^2) (w^2 + b^2))) , w , t ,
FourierParameters -> {1, -1}]]
</code>```

gives
$$\frac{1}{b(b^2 - a^2)}((b e^{a t} - a e^{b t})u(-t)
  +  (b e^{-a t} - a e^{-b t}) u(t))$$this can be simplified to $$\frac{1}{b(b^2 - a^2)}(be^{-a|t|} - ae^{-b|t|}) = \frac{1}{(b^2 - a^2)}(e^{-a|t|} - \frac{a}{b}e^{-b|t|})$$
which agrees with our previous result since $$h(t) = \frac{2\pi}{b^2 - a^2}(e^{-a|t|} - \frac{a}{b}e^{-b|t|})$$
Note that there are various ways for calculating $\mathcal{F}^{-1}\{\frac{1}{\omega^2 + a^2}\}$. See for example this and this.

Body:
By partial fraction decomposition, we have $$g(\omega)=\frac{2a}{(\omega^2+a^2)(\omega^2+b^2)} = \frac{2a}{b^2 - a^2}(\frac{1}{\omega^2 + a^2} - \frac{1}{\omega^2 + b^2}) = \frac{2a}{b^2 - a^2}(\frac{1}{2ia}(\frac{1}{\omega-ia} - \frac{1}{\omega+ia}) - \frac{1}{2ib}(\frac{1}{\omega-ib} - \frac{1}{\omega+ib}))$$If we define $$\mathcal{F}\{x(t)\} = X(i\omega) = \int_{-\infty}^{+\infty} x(t)e^{-i\omega t}dt$$ The inverse transform is given by $$\mathcal{F}^{-1}\{X(i\omega)\} = x(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} X(i\omega)e^{i\omega t}d\omega$$ It's easy to show that if $x(t) = e^{-at}u(t), \ \ a\gt0$ then $X(i\omega) = \frac{1}{a+ i\omega}$. Using this result, $$\mathcal{F}^{-1}\{\frac{1}{\omega - ia}\} = ie^{-at}u(t) \\ \mathcal{F}^{-1}\{\frac{1}{\omega + ia}\} = -ie^{at}u(-t)$$We should put these facts together.

We are interested in $h(t) = 2\pi\mathcal{F}^{-1}\{g(\omega)\}$. Using linearity property, we have $$h(t) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{1}{2ia}(\mathcal{F}^{-1}\{\frac{1}{\omega-ia}\} - \mathcal{F}^{-1}\{\frac{1}{\omega+ia}\}) - \frac{1}{2ib}(\mathcal{F}^{-1}\{\frac{1}{\omega-ib}\} - \mathcal{F}^{-1}\{\frac{1}{\omega+ib}\})) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{1}{2ia}(ie^{-at}u(t) +ie^{at}u(-t)) - \frac{1}{2ib}(ie^{-bt}u(t) +ie^{bt}u(-t))) = \frac{2a\times 2\pi}{b^2 - a^2}(\frac{e^{-a|t|}}{2a} - \frac{e^{-b|t|}}{2b})$$

Also we can check the result using Mathematica. Entering

Assuming[a > 0 && b > 0,
InverseFourierTransform[(2 a/((w^2 + a^2) (w^2 + b^2))) , w , t ,
FourierParameters -> {1, -1}]]

gives
$$\frac{1}{b(b^2 - a^2)}((b e^{a t} - a e^{b t})u(-t)
  +  (b e^{-a t} - a e^{-b t}) u(t))$$this can be simplified to $$\frac{1}{b(b^2 - a^2)}(be^{-a|t|} - ae^{-b|t|}) = \frac{1}{(b^2 - a^2)}(e^{-a|t|} - \frac{a}{b}e^{-b|t|})$$
which agrees with our previous result since $$h(t) = \frac{2\pi}{b^2 - a^2}(e^{-a|t|} - \frac{a}{b}e^{-b|t|})$$
Note that there are various ways for calculating $\mathcal{F}^{-1}\{\frac{1}{\omega^2 + a^2}\}$. See for example this and this.


问:
What is the best theoretical Big O for some ideal algorithm?
Big O notation has many commonly found real-world examples, like O(N), O(log N), O(N log N),... and so on.
Other than a constant Big-O like O(1), is there some theoretical golden "value" of O that some perfectly efficient algorithm could achieve, one that is faster than O(log N)? As in, an function with O([this expression]) could never be reduced to a less time-complex form?

Apologies if this is the wrong StackExchange for this, but this seems more a math question to me.
Thanks for any help.

答:
Between $O(\log n)$ and $O(1)$ are things like
$$
O(\log\log n)
\\
O(\sqrt{\log n})
$$
and lots of others.


问:
Dimension of $V = \{P \in \mathbb{R}_n [X] : P (X^2)-( X^n + 1) P (X) =0\}$

For $n \in \mathbb {N}$, consider the vector space $V$ defined by
$$
V = \{P \in \mathbb{R}_n [X] : P (X^2)-( X^n + 1) P (X)  =0\}
$$
What is the dimension of $V$?

I know that $\forall P\in V\setminus\{0\}$ $\deg(P)=n$. Will this help me solve the problem?

答:
Clearly $p(x)$ is not monomial, that is, exist second higest degree: $$p(x) = ax^n+bx^k+...$$ where $a\ne 0$ and $b\ne 0$. Obviously $k<n$. Then we get $$ax^{2n}+bx^{2k} +... = ax^{2n} +bx^{k+n} +ax^n+bx^k+...$$

By comparing the coefficients we get:

If $k>0$: Then $b=0$ or $2k=k+n$ i.e. $k=n$, a contradiction.

If $k=0$: Then $p(x) = ax^n+b$ so $$ax^{2n}+b = ax^{2n} +(a+b)x^{n} +b$$ so $b=-a$

So $V = \{a(x^n-1);a\in \mathbb{R}\} \implies \dim V =1$.


问:
Is it possible to ensure that the following characterisation of the function $a^x$ is well-defined?
Say we define $a^x$ in the obvious way for $x\in\mathbb{Q}$, with $a^{p/q}=\left(\sqrt[q]{a}\right)^p$. This definition can be extended to irrational numbers by requiring that $a^m<a^x<a^n$, where $m$ and $n$ are rational numbers such that $m<x<a$.* How can we prove that this definition is enough to narrow down the value of $a^x$ to just one number? Moreover, does this definition automatically imply that the function $x \mapsto a^x$ is continuous?

Ideally, I would look to see a solution that doesn't rely on other characterisations of the exponential function, e.g. when $a^x$ is defined as $\exp(x\log(a))$.

*This definition only applies when $a>1$. If $a<1$, then the inequalities are reversed.

答:
It's best to define $x\mapsto a^x$ on irrational $x$ by demanding continuity, rather than making inequalities' directions depend on whether $a>1$, with $1^x:=1$. In other words, $a^x:=\lim_{n\to\infty}a^{x_n}$ for any sequence $x_n$ in $\Bbb Q$ with $x=\lim_{n\to\infty}x_n$. Then you just need to check the choice of $x_n$ is irrelevant; this is equivalent to proving all null sequences (those with $x=0$) give $\lim_{n\to\infty}a^{x_n}=1$. Of course, once this is done, your inequality-based definition is trivially equivalent.

Edit, for requested elaboration:

Suppose $y_n,\,z_n$ are rational sequences with common limit $L$. Let $\delta_n:=y_n-z_n$, so $\lim_{n\to\infty}\delta_n=0$; we say $\{\delta_n\}_{n\ge0}$ is a null sequence. If we can prove this is sufficient to force $\lim_{n\to\infty}a^{\delta_n}=1$,$$\frac{\lim_{n\to\infty}a^{y_n}}{\lim_{n\to\infty}a^{z_n}}=\lim_{n\to\infty}a^{\delta_n}=1\implies\lim_{n\to\infty}a^{y_n}=\lim_{n\to\infty}a^{z_n},$$so $a^L$ is well-defined. (Well, we also need to check $\lim_{n\to\infty}a^{y_n},\,\lim_{n\to\infty}a^{z_n}$ exis, so the division is legal.) However, that null sequences satisfy $\lim_{n\to\infty}a^{\delta_n}=1$ is just the special case of the desired result where the exponents tend to $0$, so the general case reduces to that special case.


答:
(Assuming $a>1$), you can perhaps let
$$ a^x=\inf\{\,y\in\Bbb R\mid \forall p\in\Bbb Z\colon\forall q\in\Bbb N\colon qx>p\to y^q>a^p\,\}.$$

Title: Is it possible to ensure that the following characterisation of the function $a^x$ is well-defined?

Body:
Say we define $a^x$ in the obvious way for $x\in\mathbb{Q}$, with $a^{p/q}=\left(\sqrt[q]{a}\right)^p$. This definition can be extended to irrational numbers by requiring that $a^m<a^x<a^n$, where $m$ and $n$ are rational numbers such that $m<x<a$.* How can we prove that this definition is enough to narrow down the value of $a^x$ to just one number? Moreover, does this definition automatically imply that the function $x \mapsto a^x$ is continuous?

Ideally, I would look to see a solution that doesn't rely on other characterisations of the exponential function, e.g. when $a^x$ is defined as $\exp(x\log(a))$.

*This definition only applies when $a>1$. If $a<1$, then the inequalities are reversed.


问:
Giving an example of a strictly monotone function which is not injective.
The following question/task is from Taylor's, "Practical Foundations of Mathematics," page 175, ISBN 0 521 63107 6 hardback. I kind of answered it myself in typing this up, but I'll share it anyway so that others may benefit. This Approach0 search indicates that it is new to MSE.

The Details:

From page 103, ibid. . . .

Definition 2.6.1: A function $f:(X,<)\to(Y,\prec)$ between sets with binary relations is said to be strictly monotone if
$$\forall x, x'.\, x'<x\implies f(x')\prec f(x).$$

(NB: The lack of $\in$ here is used throughout most of the book so far; I can't, for the life of me, figure out why - the answer is buried deep and forgotten long ago.)

The Question:

Give an example of a strictly monotone function which is not injective.

My Attempt:

I want to show that there exists a strictly monotone $f:(X,<)\to(Y,\prec)$ such that there exist $a,a'\in X$ with $f(a)=f(a')$ but $a\neq a'$.

One idea is to picture a v-shaped lattice on three nodes as $(X,<)$, where $a$ and $a'$ occupy the top nodes; essentially: there exists a $b\in X$ with $b<a$ and $b<a'$, but $a\neq a'$ and are otherwise not comparable. Then I'll take $(Y,\prec)$ as the lattice with two nodes, say $y,y'\in Y$, where $$y=f(b)\prec y'=f(a)=f(a').$$

Now for all $x,x'\in X$, if $x<x'$, then $f(x)\prec f(x')$, so $f$ is indeed strictly monotone and, by construction, not injective.

Is this correct?

Please help :)

答:
Yes, this is a perfectly valid example. Good job!

Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.


问:
Help solving recurrence relation, $a_n = 3a_{n-1} + 4a_{n-2} - 12a_{n-3}$
This is in my homework, and I am not sure how to go about this, I've read the book but I can't seem to grasp what to do. Help?

$$a_n = 3a_{n-1} + 4a_{n-2} - 12a_{n-3}$$

where $a_0 = 2$, $a_1 = -1$, $a_2 = 3$

答:
Use the techniques in Wilf's "generatingfunctionology". Write:
$$
a_{n + 3} = 3 a_{n + 2} + 4 a_{n + 1} - 12 a_n
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
  = 3 \frac{A(z) - a_0 - a_1 z}{z^2} + 4 \frac{A(z) - a_0}{z} - 12 A(z)
$$
Solve for $A(z)$, split into partial fractions, and read off the coefficients.</p>


问:
The Hasse principle for bivariate cubic forms
In a talk he gave at the MSRI in 2011, Manjul Bhargava sketches a proof of why the Hasse principle holds for all bivariate cubic forms. It goes as follows:

Suppose the bivariate cubic forms $f$ has no root in $\mathbb{P}_{\mathbb{Q}}^1$.

Then $f$ factors over a cubic field.

By the Chebotarev density theorem, this implies that $f$ is irreducible mod $p$ for some prime $p$.

Thus $f$ has no root in $\mathbb{P}_{\mathbb{Q}_p}^1$ for some prime $p$.

I regret to say that I am having trouble following this proof. My main difficulties lie in understanding how the Chebotarev density theorem comes into it, seeing as I thought that to be a statement about the density of primes in non-Abelian extensions.

Also, step (2): Why does the non-existence of a root in $\mathbb{P}_{\mathbb{Q}}^1$ imply that $f$ must factor over some cubic field extension?

If a textbook proof of the above statement exists, I would much appreciate it if someone could point it out.

Lastly: He also mentions that the reason the Hasse principle holds for bivariate cubic forms is "basically the same as for bivariate quadratic forms". But surely, the Hasse principle for bivariate quadratic forms simply follows from the fact that any element $a$ in an algebraic number field $K$ is a square in $K$ if and only if it is a square in every completion of $K$, which is a much simpler argument, is it not?

Many thanks.

答:
Let $f(x,y) = ax^3 + bx^2y + cxy^2 + dy^3$ for $a,b,c,d \in \mathbf Q$. If $a$ or $d$ were $0$ then $f(1,0) = a$ would be $0$ or $f(0,1) = d$ would be $0$. Therefore the assumption that $f$ has no zeros in $\mathbf P^1(\mathbf Q)$ implies $a$ and $d$ are not $0$.

Now consider
$$
F(x) = f(x,1) = ax^3 + bx^2 + cx + d,
$$
which is cubic in $\mathbf Q[x]$ with no rational root: if $F(r) = 0$ for some $r \in \mathbf Q$ then $f(r,1) = 0$, but $f$ has no rational zero. A cubic in $\mathbf Q[x]$ with no root in $\mathbf Q$ is irreducible over $\mathbf Q$ (an accident of small-degree polynomials!), so $F(x)$ is irreducible over $\mathbf Q$. The following theorem explains why $F(x)$ has no root in some $\mathbf Q_p$.

Theorem. If $F(x) \in \mathbf Q[x]$ is irreducible over $\mathbf Q$ and has a root in $\mathbf Q_p$ for all $p$ then $F(x)$ is linear.

The same result and proof hold with $\mathbf Q$ replaced by an arbitrary number field (and the $\mathbf Q_p$'s replaced by all the nonarchimedean completions of that number field), but I stick to coefficients in $\mathbf Q$ to keep the notation simple.

Proof. Let $K = \mathbf Q(\alpha)$ where $F(\alpha) = 0$ and let $L$ be the Galois closure of $K$ over $\mathbf Q$.

The discriminant ${\rm disc}(F)$ is nonzero since $F(x)$ is irreducible over $\mathbf Q$.
Let $S$ be the primes $p$ that appear in the factorization of ${\rm disc}(F)$ or that ramify in $L$ or that appear in a coefficient of $F(x)$.
That is a finite set of primes. For $p \not\in S$, $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable since ${\rm disc}(F) \not\equiv 0 \bmod p$.

For $p \not\in S$, $F(x)$ has a root $r$ in $\mathbf Q_p$ by hypothesis, and the root is in $\mathbf Z_p$ since $F(x)$ has $p$-integral coefficients and its leading coefficient (in fact each coefficient) is a $p$-adic unit.
Therefore we can reduce the equation $F(r) = 0$ to $\overline{F}(\overline{r}) = 0$ in $\mathbf Z_p/(p) = \mathbf F_p$.  We have $K \cong \mathbf Q(r)$, so some prime ideal factor of $p$ in $K$ has residue field degree $1$: the embedding of $K$ into $\mathbf Q_p$ by mapping $\alpha$ to $r$ puts a non-archimedean absolute value $|\cdot|_{\mathfrak p}$ on $K$ with residue field $\mathbf F_p$, so $f(\mathfrak p|p) = 1$.

Now we bring in Frobenius elements. Set $G = {\rm Gal}(L/\mathbf Q)$ and $H = {\rm Gal}(L/K)$.  Since $p$ is unramified in $L$ (by $p \not\in S$), we can talk about a Frobenius element at a prime $\mathfrak P$ over $p$ in $L$.  Since $F(x)$ is $p$-integral and $F(x) \bmod p$ is separable, the permutation action of ${\rm Frob}(\mathfrak P|p)$ on the roots of $F(x)$ has a cycle structure that matches the degrees of the irreducible factors of $F(x) \bmod p$. (The actual choice of $\mathfrak P$ over $p$ doesn't matter, since all Frobenius elements at primes over $p$ are conjugate to each other in $G$ and thus permute the roots of $F(x)$ with the same cycle structure.)

We know $F(x) \bmod p$ has a root in $\mathbf F_p$, so it has a linear irreducible factor mod $p$.  Thus ${\rm Frob}(\mathfrak P|p)$ has a fixed point in its permutation action on the roots of $F(x)$: this Frobenius element fixes a root of $F(x)$ in $L$, so this Frobenius element fixes one of the subfields of $L$ conjugate to $K = \mathbf Q(\alpha)$.  The subgroup of $G$ corresponding to that subfield is conjugate to $H$.  Hence ${\rm Frob}(\mathfrak P|p) \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$.

Now we bring in the Chebotarev density theorem: every element of $G$ is a Frobenius element at infinitely many primes, in fact at a positive density of primes.  So each $g \in G$ is a Frobenius element at a prime ideal lying over a prime outside of $S$. The above argument proves $g \in \bigcup_{\sigma \in G} \sigma H\sigma^{-1}$.  Thus
$$
G = \bigcup_{\sigma \in G} \sigma H\sigma^{-1}.
$$

A finite group is never the union of conjugates of a proper subgroup, so $H = G$ and thus $K = \mathbf Q$.  Therefore $\mathbf Q(\alpha) = \mathbf Q$, so $F(x)$ has a root in $\mathbf Q$. Since $F(x)$ is irreducible over $\mathbf Q$ with a rational root, $F(x)$ is linear.  QED

In this proof, we did not need $F(x)$ to have a root in $\mathbf Q_p$ at every prime $p$. It is okay to say $F(x)$ has a root in $\mathbf Q_p$ outside a set $S$ of primes of density $0$, since Chebotarev implies that each $g \in G$ is a Frobenius at a prime outside $S$.  For example, if $F(x)$ has a root in $\mathbf Q_p$ for all but perhaps finitely many $p$ then we can still deduce that $F(x)$ has a root in $\mathbf Q$.

Therefore an irreducible $F(x) \in \mathbf Q[x]$ of degree greater than $1$ has no root in $\mathbf Q_p$ for infinitely many $p$. When $F(x)$ is cubic, not having a root in a field is equivalent to being irreducible over that field. Therefore a cubic irreducible in $\mathbf Q[x]$ has no root in $\mathbf Q_p$ for infinitely many $p$. Returning to the binary cubic form, $f(x,1)$ has no root in $\mathbf Q_p$ of infinitely many $p$, so for infinitely many $p$ we can't solve $f(x,y) = 0$ with $[x,y] \in \mathbf P^1(\mathbf Q_p)$ with $y \not= 0$. Of course we can't solve it with $y = 0$ either, since that would be the point $[x,0] = [1,0]$ and $f(1,0) = a \not= 0$.


问:
Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $
As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2)  +2t} + 1
$$ which simplifies to  $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?

答:
(This has a fairly obvious improvement
of the inequality.)

Here is a simple algebraic proof of the stronger inequalities
$x^2+y^2=1
\implies x^3+y^3
\ge \dfrac{x+y}{2} \ge \sqrt{xy}.$

More generally,
without the restriction on
$x^2+y^2$,
this gives
$x^3+y^3
\ge \dfrac{(x+y)(x^2+y^2)}{2}
$.

Proof that this inequality is stronger:

$xy \le (x^2+y^2)/2 = 1/2$
so
$1/\sqrt{xy} \ge \sqrt{2}$.
Therefore
$\sqrt{xy} = xy/\sqrt{xy}
\ge xy \sqrt{2}$.

Proof of the inequality.

$\begin{array}\\
x^3+y^3
&=(x+y)(x^2-xy+y^2)\\
&= (x+y) \dfrac{x^2+y^2+x^2-2xy+y^2}{2}\\
&= (x+y) \dfrac{1+(x-y)^2}{2}\\
&\ge \dfrac{x+y}{2}\\
&\ge \sqrt{xy}\\
\end{array}
$

In all these inequalities, there is equality when x=y and strict inequality otherwise.

Example.

If
$x=3/5, y=4/5$
then

$x^3+y^3 = (27+64)/125 =91/125=0.728,\\
\dfrac{x+y}{2} = 7/10 =0.7\\
\sqrt{xy} = \sqrt{12/25} = 2 \sqrt{3}/5 = 0.6928...,\\
xy \sqrt{2} = 0.6788...
$


问:
Is this a correct assertion regarding universal quantification?
Given an arbitrary number of universal quantifiers n and an arbitrary number of domain elements d:

$ \forall x_1, \forall x_2, \space ... , \forall x_n P(x_1, x_2, ..., x_n) $

Assertion: There are $d^n$ possible combinations for which the predicate P must be true, in order for the formula to be true.

Example:

$ \forall x \forall y P(x, y) $

Domain: {a, b}

n = 2, d = 2

$ 2^2 = 4 $

(a, a), (a, b), (b, a), (b, b)

答:
Yes. This is essentially asking for the $n$-fold cartesian product on $D$:

$D^n = \underbrace{D \times \ldots \times D}_{n \textrm{ times}} = \{\langle d_1, \ldots, d_n \rangle: d_i \in D\}$.

The number of possible such combinations is the cardinality of this cartesin product, which is given by $|D^n| = |D|^{|n|}$.


问:
Fuchsian groups from explicit representations corresponding to genus $2$ surfaces
I am learning hyperbolic geometry on my own. I have worked out the explicit description of genus $2$ surface as the upper half plane modulo group of Deck Transformation, that is, as follows:

Let $\Sigma_2$ is a genus $2$ Riemann surface. It can be thought of in the following way: it is the quotient space  $\mathbb{H}/\pi_1(\Sigma_2)$ where an element of $\pi_1(\Sigma_2)$  can be thought of as a biholomorphic map of upper half plane $\mathbb{H}$. So, for an explicit representation of $\pi_1(\Sigma_2)$, the fundamental group of the genus two surface, into $PSL(2,\mathbb {C})$, I have constructed $8$ Mobius transformations (mainly, I have constructed $4$ Mobius transformations, also I am considering their inverses) and those satisfy the defining relation of fundamental group of genus $2$ surface $\pi_1(\Sigma_2)$.

$\textbf{Question:}$ Can I get Fuchsian group of genus $2$ surface from this? Or, can I get  Fuchsian group of genus $g$ surface in general?

Also, I have heard that the explicit representation of $\pi_1(\Sigma_2)$, fundamental group of the genus two surface, into $PSL(2,\mathbb{C})$ so that the upper half plane, modulo the induced action of $\pi_{1}$, is a "$\textbf{hyperbolic structure}$" on the topological surface $\Sigma_{2}$. And this is somehow related to theory of Tiechmuller Space. I want to learn Tiechmuller theory. Kindly advise a road-map to get this.

Please help me. Thanking in advanced.

答:
For references to the Teichmuller theory, see for instance, here, both the accepted answer and comments. However, prior to this, I suggest reading something more introductory, such as Katok's book "Fuchsian Groups" or/and

Anderson, James W., Hyperbolic geometry, Springer Undergraduate Mathematics Series. London: Springer. ix, 230 p. (1999). ZBL0934.51012.

If all what you have is merely four elements $a_1, b_1, a_2, b_2$ of $PSL(2, {\mathbb R})$ satisfying the surface group relator
$$
[a_1,b_1][a_2,b_2]=1,
$$
this is far from enough for defining a Fuchsian group (whose quotient space will be a genus 2 surface $\Sigma_2$).

The missing conditions are:

i. $a_1, b_1, a_2, b_2$ generate a discrete subgroup $\Gamma$ of $PSL(2, {\mathbb R})$.

ii. $\Gamma$ is isomorphic to $\pi_1(\Sigma_2)$. Equivalently, $[a_1,b_1][a_2,b_2]=1$ is a defining relator of the group $\Gamma$ (informally, all other relators are its consequences).

In case you are wondering, the constructions described in the answers to your MO question here satisfy the missing conditions.


问:
Doubling Wager coin game. Expected Value and Optimal strategy. (hard)
The game:

You are player A, and your opponent is player B. We have a fair coin. The score, $s$ starts at $0$ and the wager, $w$ at \$$1$

We toss the coin and if heads the score adds $1$ and if tails we takeaway $1$. For example if we tossed  Heads Tails Tails Heads Heads Heads the score would move as follows: $0 \to 1 \to 0 \to -1 \to 0 \to 1 \to 2$

The game ends when the score is $\pm 2$ If the score is $+2$ then player A wins, and if $-2$ then player B wins. The looser pays the winner the wager.

Here is where your advantage and the strategy comes in. Before every toss of the coin player A gets to decide if they want  to double the wager or not.

Example:

First toss: Player A decides to double wager, $w = \$2$. A head is tossed, $s = 1$

Second toss: Player A decides to double the wager again, $w = \$4$. A tail is tossed, $s=0$

Third toss: Player A decides to not double the wager, $w = \$4$. A tail is tossed, $s=-1$

Fourth toss: Player A decides to not double the wager, $w = \$4$. A tail is tossed, $s=-2$

Player A now must pay Player B $\$4$

What is the optimal Strategy, with respect to maximising EV? I have found a way to make the expected value +2 but I believe there is a better strategy. I will explain through some of the reasoning I had with this problem:

First attempt, Strategy 1:

Let us say we double if we reach $+1$ for the first time. If we reach the score $+1$ our probability of winning has now risen to $\frac{3}{4}$. The probability that we reach $+1$ ever is $\frac{2}{3}$ (Simple Markov Chains) Let $S_1$  denote Strategy 1. And the event that we reach $s=1$ be denoted  $J$

Hence $\mathbb{E}[S_1] = \mathbb{P}(J^c)\mathbb{E}[S_1|J^c] + \mathbb{P}(J)\mathbb{E}[S_1|J] =\frac{1}{3} \cdot -\$1 + \frac{2}{3} \cdot ( \$2 \cdot \frac{3}{4} - \$2 \cdot \frac{1}{4}) = \$\frac{1}{3} $

Amazing! We now have reached a positive expected value of this game. Surely if the wager simply started at $2$ instead of $1$ i.e we are playing for twice us much then our expected value has indeed doubled! Hence if we modify $S_1$ slightly by doubling before the first toss, so the game is twice as valuable then our expected payoff of now $\frac{2}{3}$ You can check that the maths works out for this but it is intuitive. Expectation is linear.

Second attempt, Strategy 2:

Why only double the wager if we reach $s=1$ for the first time. Since we have this advantage we ought to double every single time we reach $s=1$. I.e if the score goes $0 \to 1 \to 0 \to 1$ why not double again? We showed it was profitable before. Let us define $S_2$ by doubling every single time we reach $s=1$. We will calculate its expected payoff by conditioning on the first 2 tosses, using recurrence. Instead of recording the tosses we show simply write the sequence of the score. i.e Heads Tails shall be represented as $(0,1,0)$

We have four possible states for the first two tosses:
$ \{ (0,1,2), (0,1,0) , (0,-1,-2) , (0,-1,0) \} $

Under the first we simply doubled once when we reached $s=+1$ and then won $\$2$

Under the second we doubled the value of the game and then are back at the beginning.

Under the second we simply lost $\$1$. We never doubled the game and lost.

Under the fourth nothing of significance has happened. The game is at $s=0$ and we have done nothing.

Hence: $\mathbb{E}[S_2] = \frac{1}{4}\cdot \$2 +  \frac{1}{4} \cdot 2\mathbb{E}[S_2] + \frac{1}{4} \cdot (-\$1) + \frac{1}{4} \cdot \mathbb{E}[S_2] $

Solving this yields $\mathbb{E}[S_2] = \$1$

Great so now we have made even more money! And like before if we choose to make the game twice as valuable, think of this as simply the starting wager is $\$2$. This happens if we choose to simply double before the first toss. Then we have a strategy, $S_2^{'}$, with expected payoff of $2$.

I have ran both of these strategies in python and the empirical values of EV match our "theoretical" answers.!

The problem

Now you might think I am just getting greedy but lets say now we double the wager ALWAYS, unless $s=-1.$ This means we double initially, we double if $s=1$ and we now also double on $s=0$. My reasoning is this: If you are player $A$ and I told you that you could pick any initial value for the wager, you should always want a larger value. Ultimately as long as Player $A$ is not "loosing" they should always want to have the game very expensive because of there innate advantage. We soundly justified doubling the game initially so why not always double on $s=0$ too ? Lets call $S_3$ this strategy. I cannot seem the calculate the expected value of $S_3$ Python simulations show nothing at all helpful, it oscillates from about -20000 to 20000 ?

Could anyone show a method for calculation $\mathbb{E}[S_3]$ ?

Thanks a lot!

Oskar

答: With your third strategy the expected value is undefined because it does not converge.  Let us assume there is some expected value $E$ when the stake on the first toss is $2$ because you doubled already.  If we get two heads we win $4$ because we double once more.  If we get two tails we lose $2$ because we don't double any more.  If we get tails and then heads we are back to a score of $0$, but we double again, so the expected value is $2E$.  If we get heads and then tails we are back to a score of $0$ but we have doubled twice, so the expected value is $4E$  The contribution of each round of two flips increases without bound, as the chance we return to $0$ is $\frac 12$ but the expected value from then is $\frac 32E$ (plus the bit that comes from terminating the game).


问:
Is the set of functions from $X$ to $\Bbb C$ with sup norm Banach?
Let $X$ be a nonempty set.

Define $F(X)= \{f:X\longrightarrow \Bbb C , f\text{ is a function}\}$ and $\|f\|=\sup|f|$. Then $(F(X),\|.\|)$ is a Banach space? If not why?

答: Hint: What if $f$ is unbounded?

Body: Hint: What if $f$ is unbounded?


问:
If $\lim\inf_{n\to\infty}(a_n+b_n)=\lim\inf_{n\to\infty}\ a_n+\lim\inf_{n\to\infty} b_n$ for any sequence$\{b_n\}$, does $\{a_n\}$ have to converge?

If $\underset{n\to\infty}{\lim\inf}(a_n+b_n)=\underset{n\to\infty}{\lim\inf}\ a_n+\underset{n\to\infty}{\lim\inf}\ b_n$ for any sequence $\{b_n\}$ in $\Bbb R$, does $\{a_n\}$ have to be convergent?

My attempt:

This statement is somewhat converse from the one proven here. I considered two sequences: $x_n=-a_n$ and $y_n=-b_n$. Then
$$-\underset{n\to\infty}{\lim\inf}(a_n+b_n)=\underset{n\to\infty}{\lim\sup}(-a_n-b_n)=\underset{n\to\infty}{\lim\sup}(x_n+y_n)$$

Instead of the definition $$\underset{n\to\infty}{\lim\inf}\ x_n=\lim_{k\to\infty}\underset{n\ge k}{\inf}\ x_n,$$
I used the fact if $\{x_n\}$ is bounded in $\Bbb R,\space \underset{n\to\infty}{\lim\inf}\ x_n$, as an accumulation point of $\{x_n\}$, is a limit of a convergent subsequence $\{x_{p_n}\}_n$ of $\{x_n\}$, i. e., $\lim\limits_{n\to\infty} x_{p_n}=\underset{n\to\infty}{\lim\inf}\ x_n$.

If the corresponding sequence $\{y_n\}$ is defined by $$y_n:=\begin{cases}\varepsilon,& n\in\{p_n\}\\0, &n\notin\{p_n\}\end{cases},$$ let $\varepsilon=\underset{n\to\infty}{\lim\sup}\ x_n-\underset{n\to\infty}{\lim\inf}\ x_n$. Then $\color{blue}{-\underset{n\to\infty}{\lim\inf}(a_n+b_n)\ge-\underset{n\to\infty}{\lim\inf}\ a_n (*)}$.

Since $x_{p_n}\overset{n\to\infty}{\to}\underset{n\to\infty}{\lim\inf}$,
$$(\forall\varepsilon>0)(\exists n_\varepsilon\in\{p_n\})(n\in\{p_n\}\land n>n_\varepsilon)\implies\underbrace{(|x_n-\underset{n\to\infty}{\lim\inf}\ x_n|<\varepsilon)}_{\implies x_n<\underset{n\to\infty}{\lim\inf}\ x_n+\varepsilon}$$

For $\varepsilon=\underset{n\to\infty}{\lim\sup}\ x_n-\underset{n\to\infty}{\lim\inf}\ x_n$, we then
obtain $x_n<\underset{n\to\infty}{\lim\sup}\ x_n\tag 1$.

On the other hand, $\forall n\in\Bbb N$ sufficiently large, it holds $x_n<\underset{n\to\infty}{\lim\sup}\ x_n\tag 2+\varepsilon,$ so $(1)\land(2)\implies x_n+y_n<\underset{n\to\infty}{\lim\sup} x_n+\varepsilon$, and it should hold $\underset{n\to\infty}{\lim\sup}(x_n+y_n)\le\underset{n\to\infty}{\lim\sup} x_n$, but I'm not sure how to precisely justify it. This is equivalent to $\color{blue}{-\underset{n\to\infty}{\lim\inf}(a_n+b_n)\le-\underset{n\to\infty}{\lim\inf}\ a_n(**)}$.

Finally, $(*)\land(**)\implies\underset{n\to\infty}{\lim\inf}(a_n+b_n)=\underset{n\to\infty}{\lim\inf}\ a_n\implies\underset{n\to\infty}{\lim\sup}\ y_n=\varepsilon=0\implies\{x_n\}\text{ is convergent}$.

May I ask if my arguments are valid and how to improve my proof if it makes any sense?

Thank you in advance!

答:
I think you are on the right track, but it becomes a bit confusing because $\epsilon$ is used both as the difference between $\limsup x_n$ and $\liminf x_n$ (which even might be infinite) and also as an arbitrary positive value.

I would argue as follows: Let $(a_n)$ be a sequence and assume that
$$-\infty \le I = \liminf_{n \to \infty} a_n < S = \limsup_{n\to \infty} a_n \le \infty\, .$$

Let $c$ be a real number with $I < c < S$, and $(a_{n_k})$ a subsequence of $(a_n)$ with $a_{n_k} \to S$. Then define the sequence $(b_n)$ as
$$
 b_n = \begin{cases}
 0 & \text{ if } n \in \{ n_1, n_2, n_3,\ldots \} \\
 c - a_n & \text{ otherwise.}
\end{cases}
$$
Then $a_{n_k} + b_{n_k} = a_{n_k} > c$ for all sufficiently large $k$, and $b_n \ge 0$ for all sufficiently large $n$. It follows that
$$
 I < c \le \liminf_{n \to \infty} (a_n + b_n) = \liminf_{n \to \infty} a_n + \liminf_{n \to \infty} b_n  = I + 0 = I
$$
which is impossible. It follows that necessarily $I=S$, which means that the sequence $(a_n)$ is convergent.


问:
Number of factors of irreducible polynomial of degree $mn$ over $\mathbb{F}_p$ in $\mathbb{F}_{p^n}[x].$
Given a polynomial $f(x)$ of degree $mn$ over $\mathbb{F}_p$, how can we calculate how many irreducible factors it has in $\mathbb{F}_{p^n}[x]$?

Since we have $\mathbb{F}_p\subset \mathbb{F}_{p^n}\subset\mathbb{F}_{p^{mn}},$ we must have a factor of degree $m$. But how can we continue this process? I am pretty sure $f(x)$ should factor into a product of polynomial of degree $m$, but don't see an immediate solution.

Body: $$f=\prod_{k=1}^{mn} (x-a^{p^k}) =\prod_{l=1}^n \prod_{k=1}^m (x-(a^{p^l})^{p^{kn}})$$


问:
Solve the integral in the expression $x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt$
I'm following the course MIT 18.03. In [this problem][1], we evaluate the following expression containing an integral:
$$
x(t) = e^{-kt} \int e^{kt} k \cos(t/2)dt\tag{1}
$$
which finally gives:
$$
x(t) = \frac{2k \sin(t/2)+ 4k^2\cos(t/2)}{1+4k^2}+ce^{-kt}\tag{2}
$$
One possibility would be to make a substitution $u = \sin(x)$, but more than that I'm stuck

Edit (with the help of the comments)

So we can use the following substitutions:

\begin{align}
u &= \cos(t/2)\tag{3}\\
\implies u' &= 1/2\sin(t/2)\tag{4}\\
v'&= e^{kt}\tag{5}\\
\implies v &= \frac{1}{k} e^{kt}\tag{6}\\
\end{align}

Now using integration by part we can write

\begin{align}
\int v'u &= vu - \int vu'\tag{7}\\
&=\cos(t/2) \cdot \frac{1}{k}e^{kt} - \int \frac{1}{k}e^{kt}\frac{1}{2}\sin (t/2)\tag{8}\\
\end{align}
[1]: https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-i-first-order-differential-equations/solution-by-integrating-factors/MIT18_03SCF11_rec_04s5_sol.pdf

答:
Integrating by parts twice is an obvious way to handle this integral.  Another approach is to write $\cos t/2= \frac{1}{2} (e^{it/2}+ e^{-it/2})$ so the integral becomes

$$\begin{align} k \int e^{kt} \cos t/2 \, dt = &\frac{k}{2} \int  \left(e^{t(k+i/2)}+ e^{t(k-i/2)}\right)\, dt \\= &\frac{k}{2} \left( \frac{e^{t(k+i/2)}}{k+i/2} + \frac{e^{t(k-i/2)}}{k-i/2}\right) + \text{const.}\\= &\frac{k^2e^{tk}}{k^2+1/4}\left(\cos t/2 + \frac{1}{2k} \sin t/2\right)+\text{const.}\end{align}$$
and
$$x(t) = \frac{k^2}{k^2+1/4}\left(\cos t/2 + \frac{1}{2k} \sin t/2\right)+ c \,e^{-kt},$$ where $c$ is a constant.


问:
Finding the minimum and the maximum when speaking of relations
I'm trying to figure out how to find the minimum and the maximum when speaking of relations. Consider $A=(2,3,\ldots,13)$. Also consider the following relation $R$:
$$
(x,y)\in R \Leftrightarrow x|y
$$
where $x,y\in A$. I need to find the minimum and the maximum of the relation. What does it mean? How do I find them?

答:
$
\newcommand{\p}{\preccurlyeq}
$
A "maximal" element of a set $A$ with partial order $\p$ is an $M \in A$ such that there is no $a \in A$ where both $M \p a$ and $M \ne a$. Likewise, a minimal element of $(A,\p)$ is an $m \in a$ such that no $a \in A$ satisfies $a \p m$ and $a \ne m$ together.

Consequently, $a \in A$ is not maximal if there exists $b \in A$, $b\ne a$, such that $a \p b$. Similarly, $a$ is not minimal if $b \p a$ but $b \ne a$.

So, run through the list invalidating candidates for maxima first:

$\not \exists x \in A$ such that $x \mid 2$ (except itself)

$\not \exists x \in A$ such that $x \mid 3$ (except itself)

$2 \mid 4$

$\not \exists x \in A$ such that $x \mid 5$ (except itself)

$3 \mid 6$

$\not \exists x \in A$ such that $x \mid 7$ (except itself)

$2 \mid 8$

$3 \mid 9$

$2 \mid 10$

$\not \exists x \in A$ such that $x \mid 11$ (except itself)

$2 \mid 12$

$\not \exists x \in A$ such that $x \mid 13$ (except itself)

Thus the maxima are the primes in $A$ -- which makes sense, for the divisibility relation.

We can similarly go through the list again and eliminate candidates for minima:

$2 \mid 4$

$3 \mid 6$

$4 \mid 8$

$5 \mid 10$

$6 \mid 12$

$7,8,\cdots,13$ only divide themselves in $A$

Thus the maxima are $7, 8,9,10,11,12$.

Note that there is no singular "greatest maximum"/"least minimum" for this relation. If there were, we would not have multiple maximal/minimal elements.


答: A famous example is given by the Union-Find problem, where the amortized time can be made $\Theta(\alpha(n))$ where $\alpha(n)$ is the Inverse Ackermann function, which is extremely slowly growing (slower than the iterated logarithm $\log^*(n)$). For all practical $n$, the function value does not exceed $5$ (it would take much much much more than the number of subsets of atoms in the universe to reach $5$).

Title: What is the best theoretical Big O for some ideal algorithm?

Body:
Big O notation has many commonly found real-world examples, like O(N), O(log N), O(N log N),... and so on.
Other than a constant Big-O like O(1), is there some theoretical golden "value" of O that some perfectly efficient algorithm could achieve, one that is faster than O(log N)? As in, an function with O([this expression]) could never be reduced to a less time-complex form?

Apologies if this is the wrong StackExchange for this, but this seems more a math question to me.
Thanks for any help.


答:
As stated, $\lVert f\rVert$ could be infinite, so it's not even a norm. What is true is that if $V$ is a Banach space (either real or complex), and $X$ a non-empty set, then the
set of bounded functions
\begin{align}
B(X,V):= \{f:X\to V| \, \, \text{$f$ a function with $\lVert f\rVert_{\infty}:=\sup_{x\in X}\lVert f(x)\rVert_V < \infty$}\}
\end{align}
is a Banach space with the norm $\lVert \cdot\rVert_{\infty}$. In particular you can take $V=\Bbb{C}$ to get a complex Banach space $B(X,\Bbb{C})$.

Title: Is the set of functions from $X$ to $\Bbb C$ with sup norm Banach?

Body:
Let $X$ be a nonempty set.

Define $F(X)= \{f:X\longrightarrow \Bbb C , f\text{ is a function}\}$ and $\|f\|=\sup|f|$. Then $(F(X),\|.\|)$ is a Banach space? If not why?


问:
explaining logarithmic and exponential differentiation without invoking e
Is there a way to explain the differentiation of logarithmic and exponential functions without invoking the assumption that e exists?  Is the existence of e necessary to differentiate these functions?</p>

答:
If we define $\log(x)$ as $\int_{1}^{x}\frac{1}{t} \, dt$, then once we have established that $\log$ is strictly increasing, $\exp$ can be defined as its inverse. Furthermore, $\log'(x)=1/x$ by the first part of the fundamental theorem of calculus. We can then use the formula for the derivative of an inverse function, which can be derived in the following way:
\begin{align}
y &= f^{-1}(x) \\
x &= f(y) \\
\frac{dx}{dy} &= f'(y) \\
\frac{dy}{dx} &= \frac{1}{f'\left(f^{-1}(x)\right)} \, .
\end{align}
Hence,
\begin{align}
\exp'(x) &= (\log^{-1})'(x) \\
&= \frac{1}{\frac{1}{\exp(x)}} \\
&= \exp(x) \, .
\end{align}
This approach makes no reference to the constant $e$, which is simply the value of $\exp(1)$. Indeed, one could argue that it is not $e$ which is of fundamental importance, but rather the exponential function.

All this being said, full appreciation of the exponential function can only be reached once we have established that $\exp(x+y)=\exp(x) \cdot \exp(y)$. This provides the motivation behind using the more familiar notation $e^x$ instead of $\exp(x)$. $a^x$ can then be defined as $e^{x\log(a)}$ for $a>0$.


答:
Here is another way to prove $\dim V = 1$. It is easy to find (guess) a solution $p(x)= x^n-1$ so $\dim V \geq 1$.

Now observe a transformation $\phi: V\to \mathbb {R}$ defined by $\phi (P):= P(2)$ which is clearly linear. Now let us show it is injective. Take $P\in {\rm ker}(\phi)$, then $\phi (P)=0$ so $P(2)=0$. But this means that $$P(4) = (2^n+1)P(2) =0$$
The same way we see (use formal induction) that $2^{2^n}$ is zero for $P$ for every $n\in \mathbb{N}$. But this means that $P(x)=0$ for all $x$ since it has infinite zeroes. So $\phi $ has trivial kernel and thus, by rank-nullity theorem $$\dim V = {\rm rang}(\phi) \leq  \dim \mathbb{R}=1$$

Conclusion $\dim V =1$.

Title: Dimension of $V = \{P \in \mathbb{R}_n [X] : P (X^2)-( X^n + 1) P (X) =0\}$

Body:
For $n \in \mathbb {N}$, consider the vector space $V$ defined by
$$
V = \{P \in \mathbb{R}_n [X] : P (X^2)-( X^n + 1) P (X)  =0\}
$$
What is the dimension of $V$?

I know that $\forall P\in V\setminus\{0\}$ $\deg(P)=n$. Will this help me solve the problem?


问:
Point-wise and Norm Convergence of Vectors in a finite dimensional space
I'm trying to prove the theorem, that states, that if I have a normed vector space with a finite dimension (so that each vector I can express as a linear combination $$ \vec{v}=a\vec{e}_1+b\vec{e}_2+c\vec{e}_3+\cdots+q\vec{e}_n $$ where the vectors $\{\vec{e_k}\}_{k=1}^{n} $ are linearly independent), and I have some norm defined there, and I have a sequence of vectors $$ \{\vec{v_k}\}_{k=1}^{\infty} $$ that converges in norm to some vector $\vec{V}$, then it also converges to it Point-wisely, that is, if
$$\vec{v}_k=a_k\vec{e}_1+b_k\vec{e}_2+c_k\vec{e}_3+\cdots+q_k\vec{e}_n, $$ and
$$\vec{V}=a\vec{e}_1+b\vec{e}_2+c\vec{e}_3+\cdots+q\vec{e}_n, $$ and
$$\lim_{k\rightarrow\infty} \lVert\vec{v}_k-\vec{V}\rVert=0,$$ then also
$$\begin{align*}
\lim_{k\rightarrow\infty} |a_k-a|&=0,\\
\lim_{k\rightarrow\infty} |b_k-b|&=0,\\
&\,\vdots\\
\lim_{k\rightarrow\infty} |q_k-q|&=0.
\end{align*}$$

Thanks.

答:
Apart from the given norm $\|\cdot\|$, we can define another norm by $$\|\sum_{j=1}^k \lambda_je_j\|_{\infty}=\max_{1\le j\le k} |\lambda_j|$$ where the $\lambda_j$ are scalars. (You should check that this is a well-defined norm.)
Any two norms on a finite-dimensional vector space are equivalent, so in particular there is a constant $C>0$ so that $\|x\|_{\infty}\le C\|x\|$ for all vectors $x$. This inequality shows that if $\|v_k-V\|\to 0$, then we must have $\|v_k-V\|_{\infty}\to 0$ as $k\to\infty$, which implies that each component of $v_k-V$ converges to $0$, or that each component of $v_k$ converges to the corresponding component of $V$.


问:
Can you find the resultant force between these two vectors?
Determine the magnitude of the resultant force on an object if force $A$ is pulling the object with $150$ lbs of force and force $B$ is pulling with $300$ lbs, and the angle between the two forces is $110^\circ$.</p>

答:
If we assume $f_A$ points in the positive $x$ direction and $f_B$ is located in quadrant II, we have,

$$f_A =
\left[ \begin{array}{c}
150\\
0 \end{array} \right]$$

and

$$f_B =
\left[ \begin{array}{c}
300\cos(110) \\
300\sin(110) \end{array} \right]$$

The resultant is

$$f_R = f_A + f_B =
\left[ \begin{array}{c}
150 + 300\cos(110) \\
300\sin(110) \end{array} \right]$$

then

$$\|f_R\| = \sqrt{(150 + 300\cos(110))^2 + (300\sin(110))^2}=285.86$$

Note:  Notice that $\cos(110)$ is negative.


答:
Another proof inspired by marty cohen.

By Power Mean Inequality,

$$\left(\frac{x^3+y^3}{2}\right)^{1/3} \ge \left( \frac{x^2+y^2}{2}\right)^{1/2}=2^{-\frac 12}\\
\implies x^3+y^3\ge 2\cdot 2^{-\frac 32}=\frac{1}{\sqrt 2} = \frac{x^2+y^2}{\sqrt 2}\ge \sqrt 2 xy.\blacksquare$$

Title: Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $

Body:
As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2)  +2t} + 1
$$ which simplifies to  $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?


问:
Given a relation, how do I find the smallest symmetric/transitive relation containing it, and the smallest relation with two equivalence classes?
$R$ is relation on $A = \{1,2,3,4\}$

$R = \{(1,2), (1,4),(4,1), (3,3)\}$

I need to find $S$, the smallest relation that contains $R$ that is

symmetric and transitive

$S$ is equivalence relation on $A$ so that $S$ has exactly two equivalence class

My answers are

$S = \{(1,2),(1,4),(4,1),(3,3),(2,1),(2,4),(1,1),(4,4)\}$

$S = \{(1,2),(2,1),(2,3),(3,4),(4,1),(1,4),(1,1),(2,2),(3,3),(4,4),(1,3),(2,4),(3,1),(4,2)\}$

but there are more than exactly two equivalence classes. How can I change that?

答:
Observe that, if we represent your initial relation $R$ as a directed graph (where $x$ points to $y$ iff $(x,y) \in R$), then we have this graph:

In graphs like these, certain properties can be intuitively and visually observed:

Reflexivity appears if every node points to itself.

Symmetry happens if, whenever $x$ points to $y$ (and thus a path exists), then $y$ points to $x$ (and thus a path to "backtrack" exists). Basically, there are no "stray," one-way edges - only closed loops.

Transitivity is visible when, if there exists paths $x$ to $y$ and $y$ to $z$, then there is a path $x$ to $z$. For distinct nodes, then this means that, whenever two sides of a triangle are filled in, the last one is too, though the direction depends on the direction of the other sides in question.

Your equivalence classes are all of the nodes in each of the disjoint graphs you have. (Two graphs are disjoint if, basically, one is entirely separate from another, totally isolated.)

For your first part, you want the symmetric and transitive closure. We can work each part in turn. To get symmetry, we need to close a loop with $2 \to 1$. We then need to use transitivity: $4 \to 1 \to 2$ requires $4 \to 2$; similarly, $2 \to 1 \to 4$ requires $2 \to 4$. The relation $\{(3,3)\}$ itself is transitive and symmetric, so there's nothing to change there. We thus get the following graph at this stage:

However, since $1 \to 2 \to 1$, $2 \to 4 \to 2$, and $4 \to 2 \to 4$, we need arrows from each of $1,2,4$ pointing to themselves as well by symmetry. Thus, the final result is

This relation is given by our original relation, plus the arrows we added. Thus, the $S$ you seek is

\begin{align*}
S
&= R \cup \{(2,1),(4,2),(2,4),(1,1),(2,2),(4,4)\} \\
&= \Big\{\underbrace{(1,2),(1,4),(4,1),(3,3)}_{\in R},(2,1),(2,4),(1,1),(4,4), \color{red}{(4,2),(2,2)}\Big\}
\end{align*}

In red are those you forgot.

To turn $R$ into an equivalence relation in the most minimal fashion possible, we can turn it into the reflexive/transitive/symmetric closure of itself. However, notice that the relation previously is, in fact, also reflexive. Thus, the relation given by that is also an equivalence relation.

The nodes of the disjoint graphs are $1,2,4$ and $3$, and thus each group is an equivalence class. That is,

$$A/S = \Big\{ \{1,2,4\},\{3\} \Big\}$$

is your quotient set - your set of equivalence classes.

...granted, this question is a bit old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Describing the plane given by a line and a vector
I'm studying lines and planes and am stumped by the following problem:

If $L$ is the line through the point $A = (3,2,1)$ and parallel to the
vector $v = (-2,1,3)$, what's the equation of the plane that contains
$L$ and the point $B = (-2,3,1)$?

I'm trying to describe the plane in terms of $A$, $v$, and $B$ and then find a normal vector. The line $L$ is the span of $v$ shifted by $A$, so it should be $L = \{A + a v : a \in \mathbb R\}$. Then the plane itself is formed by adding in $B$, so the plane is $H = \{A + av + bB : a,b \in \mathbb R\}$. This looks like the plane through the origin $\mathrm{span}\{v, B\}$ shifted by $A$, so I can describe the plane by getting a normal vector $\mathbf n$ to $\mathrm{span}\{v, B\}$ and then a vector $x$ is in the plane exactly when $\mathbf n^T(x - A) = 0$. I found $\mathbf n$ by picking something parallel to $v \times B$ (I got $\mathbf n = (2,1,1)$). But when I do this I get $2x + y + z = 9$ as the equation which my book says is wrong. But I can't see a flaw in my argument. Am I wrong about $H$ or $L$ or something else?

答:
The vector $\mathbf{n}=(a,b,c)$ is normal to the plane.  The plane contains a line parallel to $v$, and contains a line parallel to $B-A$.  Let $\mathbf{r}=(x,y,z)$ and $B=(x_0,y_0,z_0)$ be the point on the plane which is given:

So solve the system:

$$\mathbf{n} \cdot \mathbf{v} =0$$
$$\mathbf{n} \cdot (B-A) = 0$$

The equation of the plane is then

$$\mathbf{n}\cdot (\mathbf{r}-B)=0$$
or

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0.$$


答:
The plane $H$ may not be parallel to the vector $B$;

it is parallel to the vector $v$ and the vector from point $A$ to point $B$, which is $(-5,1,0)$.

Now can you use the cross-product to find a normal vector to it

and get the equation your book says for the plane?

Title: Describing the plane given by a line and a vector

Body:
I'm studying lines and planes and am stumped by the following problem:

If $L$ is the line through the point $A = (3,2,1)$ and parallel to the
vector $v = (-2,1,3)$, what's the equation of the plane that contains
$L$ and the point $B = (-2,3,1)$?

I'm trying to describe the plane in terms of $A$, $v$, and $B$ and then find a normal vector. The line $L$ is the span of $v$ shifted by $A$, so it should be $L = \{A + a v : a \in \mathbb R\}$. Then the plane itself is formed by adding in $B$, so the plane is $H = \{A + av + bB : a,b \in \mathbb R\}$. This looks like the plane through the origin $\mathrm{span}\{v, B\}$ shifted by $A$, so I can describe the plane by getting a normal vector $\mathbf n$ to $\mathrm{span}\{v, B\}$ and then a vector $x$ is in the plane exactly when $\mathbf n^T(x - A) = 0$. I found $\mathbf n$ by picking something parallel to $v \times B$ (I got $\mathbf n = (2,1,1)$). But when I do this I get $2x + y + z = 9$ as the equation which my book says is wrong. But I can't see a flaw in my argument. Am I wrong about $H$ or $L$ or something else?


问:
Let $p,q$ be prime numbers , Find all the solutions to the congruence $x^2\equiv (p+q)x \pmod{pq}$
Let $p,q$ be prime numbers , find all the solutions to the congruence $x^2\equiv (p+q)x \pmod{pq}$

Attempt:

$$x^2\equiv (p+q)x \pmod{pq} \Longrightarrow \cases {x^2\equiv (p+q)x \pmod{p}\\x^2\equiv (p+q)x \pmod{q}} $$

For $\pmod{p}$

$$x^2\equiv (p+q)x \pmod{p} \Longrightarrow x(x-(p+q))\equiv 0 \pmod{p}$$

Because $p$ is prime number then  we can write :

$$x\equiv 0 \pmod{p}\space\space\space(1)$$
and
$$x\equiv (p+q) \pmod{p}\Longrightarrow x\equiv q \pmod{p}\space\space\space(2)$$

For $\pmod{q}$

$$x^2\equiv (p+q)x \pmod{q} \Longrightarrow x(x-(p+q))\equiv 0 \pmod{q}$$

Because $q$ is prime number then  we can write :

$$x\equiv 0 \pmod{q}\space\space\space(3)$$
and
$$x\equiv (p+q) \pmod{q}\Longrightarrow x\equiv p \pmod{q}\space\space\space(4)$$

Now, we can say because $(1)$ and $(3)$ that $x\equiv 0 \pmod{pq}$.

There are more solutions using $(2)$ and $(4)$ but I cannot see a legal way to solve them.

Edit: note that $p,q$ can(not must) be equal to one another, which I think is another case for a solution.

答:
Your ‘and’ should be ‘or’.

Hint:

You just have to use the inverse isomorphism of the isomorphism of the Chinese remainder theorem, which relies on a Bézout's identity: $\enspace up+vq=1$, between $p$ and $q:$

The system of congruences $\begin{cases}x\equiv a\mod p,\\x\equiv b\mod q,\end{cases}$ has as solution
$\enspace x\equiv b up+a vq\mod pq.$


问:
Mathematicians conceived of black holes long before astronomers actually found any?How?
I read this statement in Lockhart's - "A Mathematician's Lament". But how could mathematicians  figure out something like black holes even before astronomers noticing any of them?</p>

答:
Without going into the details (which I'm not intimately familiar with), it was possible because physicists and mathematicians (such as Lorentz, Einstein and Minkowski) had developed a good mathematical model of gravitation.  Using this model - which is a system of partial differential equations - physicists and mathematicians were able to predict that gravitational singularities should form under certain conditions.  As a very (very) simplified example of what I mean, suppose that we have modeled some quantity with the simple ODE

$$
\dot{x}=x^2,\quad x(0)=1
$$  The solution to this equation is the function

$$
x(t)=\frac{1}{1-t}
$$

This function exhibits finite time blow-up, since when $t=1$, we have a singularity - a place where $x(t)$ becomes infinite.  If the ODE models our system accurately, we should be able to observe this blow-up with measurements.  The same is true of gravitational singularities - the model predicts a blow-up, so astronomers went looking for (and found) those singularities.


问:
partitioning a set into denumerable subsets

Let $A$ and $B$ be sets. If there exists an injection $f : A\to B,$ then write $|A|\leq |B|,$ where $|A|$ is the size of the set $A$. Let $A$ be a nonempty set. Show that $|\mathbb{N}|\leq |A|$ iff $A$ can be partitioned into denumerable sets. That is, there is a set of sets $\{K_i\}_{i\in I} \subseteq \mathcal{P}(A)$ so that

Each $K_i$ is denumerable

$K_i\cap K_j=\emptyset$ for $i\neq j$

$A = \cup_{i\in I} K_i$

Suppose $A$ can be partitioned into denumerable sets. Then $A = \cup_{i\in I} K_i$ for some index set $I$, where the $K_i$'s are disjoint and denumerable. In particular, one can find a bijection $f : \mathbb{N}\to K_{\alpha}, \alpha \in I.$ This is clearly an injection from $\mathbb{N}$ to $A$ (as $K_\alpha \subseteq A$), so $|\mathbb{N}|\leq |A|.$

Suppose there is an injection $f : \mathbb{N}\to A.$ Then $A$ is clearly infinite, since its size is at least the size of an infinite set. Also, $A$ has a denumerable subset. I'm not really sure how to show that the given family exists, though I think Zorn's lemma should be useful. However, I'm not really sure which family of sets to consider. For instance, would it be useful to consider the set of all denumerable subsets of $A$? And if so, how would I define the partial order?

答:
HINT: Let $\mathscr{P}$ be the family of all pairwise disjoint families of countably infinite subsets of $A$, partially ordered by inclusion. Use Zorn’s lemma to show that $\mathscr{P}$ has a maximal element $\mathscr{M}$. If $\bigcup\mathscr{M}=A$, we’re done. If not, let $F=A\setminus\bigcup\mathscr{M}$.

Show that $F$ is finite.

Then fix some $M\in\mathscr{M}$, let $M'=M\cup F$, and show that $(\mathscr{M}\setminus\{M\})\cup\{M'\}$ is a partition of $A$ into countably infinite subsets.


问:
Proof that $V\odot W\cong W\odot V$ for vector spaces $V$ and $W$
I've read a proof that the algebraic tensor product of vector spaces is 'commutative' in the sense that $V\odot W\cong W\odot V$. Although I believe I have managed to prove this by showing that they both have same dimension, therefore are isomorphic, this is not a 'constructive' proof in the sense that it not provide an explicit example of a linear isomorphism.

In Appendix T of Wegge-Olsen's book $K$-Theory and $C^*$-Algebras, this is proved by showing that there is a linear isomorphism $\varphi:V\odot W\to W\odot V$ with $\varphi(v\otimes w)=w\otimes v$ for all $v\in V, w\in W$.

The only way I see of doing this is by using a result which is later on in the book i.e. proving injectivity via $$f\otimes e=f'\otimes e'\Rightarrow e=e',f=f'\Rightarrow e\otimes f=e'\otimes f',$$ where $e,e'\in E$, $f,f'\in F$ and $E,F$ are bases of $V,W$, respectively.

But that result uses commutativity, so this approach would be circular. Is there another way of doing this?

In other words, without using the result that when $\sum e_i\otimes f_i=0$ and $f_i$ lin. indep. implies $e_i=0$ for all $i$.

答:
Here is a general proof (for tensor products of modules over commutative rings) using only the universal property.
$\DeclareMathOperator{\id}{id}$

Consider the bilinear maps $\phi_1\colon V\times W\to W\odot V$, $(v,w)\mapsto w\otimes v$ and $\phi_2\colon W\times V\to V\odot W$, $(w,v)\mapsto v\otimes w$.

By the universal property of the tensor products, you get $\tilde\phi_1\colon V\odot W\to W\odot V$ and $\tilde\phi_2\colon W\odot V\to V\odot W$. Now, it is easy to check that the maps in question satisfy $\tilde\phi_2\tilde\phi_1(v\otimes w)=v\otimes w$ and $\tilde\phi_1\tilde\phi_2(w\otimes v)=w\otimes v$ for any $v,w$.

On the other hand, by the universal property, there is a unique map $V\odot W\to V\odot W$ with $v\otimes w\mapsto v\otimes w$. Since identity clearly satisfies this, it follows that $\tilde\phi_2\tilde\phi_1=\id_{V\odot W}$. The same argument shows that $\tilde\phi_1\tilde\phi_2=\id_{W\odot V}$.

(Note that the conclusion does not make much sense in the non-commutative context - for $V\otimes_R W$ to make sense, we need $V$ to be a right $R$-module and $W$ to be a left $R$-module. The difference is that if $R$ is commutative, all left $R$-modules can be thought of as right $R$-modules with the same multiplication, which would, in general, violate mixed associativity.)


问:
Proving that for $x \ge 17$, $\prod\limits_{p < x\text{, prime, & }p \nmid (x^2+x)}p > x^2 + x$
Let:

$x$ be an integer

$p_n$ be the $n$th prime

$x\#$ be the primorial of $x$

$f(x) = x^2 + x$

$P(x) = \prod\limits_{p < x\text{, prime, & } p \nmid f(x)}p$

$m(p_n)$ be the minimum $P(i)$ where $p_n \le i < p_{n+1}$

Does it follow that $P(x) > f(x)$?

The challenge in the argument is to compare a strictly increasing function with a function that is not strictly increasing:

$P(x)$ is not strictly increasing.  In some cases $P(x) > P(x+1)$.

Example:  $P(4) = 3$ but $P(5) = 1$

$f(x)$ is strictly increasing since $x \ge 17$ and:

$$\frac{d}{dx}(x^2 + x) = 2x + 1$$

I handle this situation by leveraging Bertrand's Postulate and attempting to show:

$$m(p_n) > (2p_n - 1)^2 + 2p_n - 1 = (2p_n)^2 - 2p_n$$

Since $m(p_n) > \dfrac{p_n\#}{(p_n)^2}$ and $4(p_n)^2 > (2p_n - 1)^2 - 2p_n$, this will be demonstrated if:

$$p_n\# > 4(p_n)^4$$

(1)  Base Case: $p_n = 17$

$17\# = 510,510 > 4(17^4) = 334,084$

(2) Assume up to some $p_n \ge 17$, $p_n\# > 4(p_n)^4$

(3) Since $p_{n+1} \ge 19$, it follows that:

$$p_{n+1}\# \ge (19)(4(p_n)^4) > 4(2p_n)^4) > 4(p_{n+1})^4$$

I believe that the conclusion follows.  Please let me know if I made any mistakes in my reasoning or if there is a simpler way to make the same point.

答:
Another method, using properties of the first Chebyshev function (by $p$ we mean prime(s) everywhere below)
$$\vartheta (x)=\sum _{{p\leq x}}\log p$$
According to this paper, page 20
$${\vartheta (x)>0.985x}, \forall x\geq 11927$$
thus
$$\prod\limits_{p\leq x} p=e^{\vartheta (x)}\geq e^{0.985x} \tag{1}$$
It's not too difficult to show that
$$e^{0.985x} \geq \left(x^2+x\right)^2 \Rightarrow \frac{e^{0.985x}}{x^2+x}\geq x^2+x \tag{2}$$
Now
$$\prod\limits_{p\leq x} p=
\left(\prod\limits_{p\leq x \text{ and } p \nmid x^2+x} p \right)
\left(\prod\limits_{p\leq x \text{ and } p \mid x^2+x} p \right)$$
thus, from $(1)$
$$\prod\limits_{p\leq x \text{ and } p \nmid x^2+x} p \geq
\frac{e^{0.985x}}{\prod\limits_{p\leq x \text{ and } p \mid x^2+x} p} \tag{3}$$
Finally, because the product doesn't consider the multiplicity of the prime factors
$$\prod\limits_{p\leq x \text{ and } p \mid x^2+x} p \leq x^2+x \tag{4}$$
and combining $(2)$, $(3)$ and $(4)$
$$\prod\limits_{p\leq x \text{ and } p \nmid x^2+x} p \geq
\frac{e^{0.985x}}{x^2+x}\geq x^2+x$$
$\forall x\geq 11927$. The first $11926$ cases can be checked with a computer program.


问:
Conditional probability question about a r.v with geometric distribution
I'm going through my textbook for a measure theory and probability course I'm following at uni. I have encountered this problem that seems logical in nature, but I'm not sure I'm proficient enough at handling r.v's that I can solve it. The problem is,

For a r.v with geometric distribution and for $i,j>0$ show that,

$P(x>i+j\vert x>i)=P(x>j)$

It seems logical when putting the problem into words, but when I try to calculate it,

$P(x>i+j\vert x>i)=\frac{P(\{X>i+j\}\cap\{X>i\})}{P(X>i)}=\frac{P(\{X>i+j\})}{P(X>i)}$

and I hit a dead end. I have tried using their probabilities written as sums directly, but that doesn't seem to yield any result for me either.

答: $P(x>i)=\sum\limits_{k=i+1}^{\infty} p((1-p)^{k}=p\frac {(1-p)^{i+1} } {1-(1-p)}$ by the formula for a  geometric sum. This gives $P(x>i)=(1-p)^{i+1}$. Can you finish?


问:
Expected time spent in a node between visits to another node in an asymmetric random walk on integers
Recall that the first passage time to state $i$ is the random variable $T_i$ defined by
\begin{equation*}
T_i(\omega)=\mathrm{inf}\{n\geq1:X_n(\omega)=i\}
\end{equation*}
where inf$\emptyset=\infty$. Let $(X_n)_{n\geq0}$ be a simple random walk on $\mathbb{Z}$ with $p_{i,i-1}=q<p=p_{i,i+1}$. Find
\begin{equation*}
\gamma^0_1=\mathbb{E}_0(\sum_{n=0}^{T_0-1}1_{X_n=1}).
\end{equation*}

My efforts:

\begin{equation*}
\gamma^0_1=\sum^\infty_{n=0}\mathrm{Pr}(X_n=1,n+1\leq T_0).
\end{equation*}
\begin{equation*}
\mathrm{Pr}(X_{2k}=1,2k+1\leq T_0)=0.
\end{equation*}
\begin{equation*}
\mathrm{Pr}(X_{2k+1}=1,2k+2\leq T_0)=c_kp^{k+1}q^k.
\end{equation*}

We only need to determine $c_k$, which is a complicated counting problem. I know $c_0=1,c_1=1,c_2=2,c_3=4$. We only need to consider the walk to the right hand side of 0. In the case of $X_{2k+1}=1,2k+2\leq T_0$, the farthest place that the walker can reach is $k+1$. Of course the walker does not necessarily reach $k+1$. It can just repeat between some nodes, e.g., $k-1$ and $k$.

This is a transient random walk. So we cannot use the stationary distribution to calculate this.

The generating function method of calculating number of steps returning to 0 starting from 1 doesn't help either. Number of steps taken to return to 0 and number of visits to 1 are two very different things, although you must pass 1 so as to return to 0.

There seems to be no way to avoid counting. So my problem is: what is a smart way to count $c_k$?

答:
\begin{equation*}
\gamma^0_0=1.
\end{equation*}
\begin{equation*}
\gamma^0_i=q\gamma^0_{i+1}+p\gamma^0_{i-1},i\neq0.
\end{equation*}
Thus
\begin{equation*}
\gamma^0_i=A+(1-A)(\frac{p}{q})^i.
\end{equation*}
Consider the case of $i>0$. To visit $i+1$ and return to 0, the walker must visit $i$ at least twice. $\gamma^0_i$ is a non-increasing function of $i$. But $(p/q)^i$ is an increasing function. The best we can do is setting $A=1$ so that $\gamma^0_i=1$ for all $i>0$.

Consider the case to the left of 0. Still let $i>0$. We have
\begin{equation*}
\gamma^0_{-i}=A+(1-A)(\frac{q}{p})^i.
\end{equation*}
lim$_{i\rightarrow\infty}\gamma^0_{-i}=0$, so $A=0$ and $\gamma^0_{-i}=(q/p)^i$ for all $i>0$.


问:
Showing $5a + 2b \equiv 0 \pmod 7$ is symmetric, pictorially
I have a problem regarding the understanding of modulus.

So someone proved $5a + 2b \equiv 0\pmod 7$ is symmetric using the modulo circle,
and my brain cant comprehend how he meant that. I'm talking about circles like this:

I know there is a other way to prove it with $7 \mid 5a + 2b$

答:
The relation is defined on $\Bbb Z$ like this: $a \sim b$ if and only if $5a + 2b \equiv 0 \pmod 7$. On the assumption this holds, we wish to show $b \sim a$, i.e. $5b+2a \equiv 0 \pmod 7$.

Using the definition of modular congruence, then, $\exists k \in \Bbb Z$ such that

$$\frac{5a+2b-0}{7} = k \iff 5a + 2b = 7k$$

Subtract $7a$ and add $7b$ to both sides. Then you have

$$-2a - 5b = 7(k - a + b)$$

Now multiply both sides by $-1$:

$$2a + 5b = 7(-k+a-b)$$

Thus, there an $\ell \in \Bbb Z$ (namely $\ell = -k+a-b$) such that

$$\frac{5b+2a}{7} = \ell$$

ensuring that $5b + 2a \equiv 0 \pmod 7$, i.e. $b \sim a$. Thus, symmetry.

Now, with this in mind, consider the "circle" for equivalence modulo $7$:

If $5a + 2b \equiv 0 \pmod 7$, then the sum $5a+2b$ is in that topmost notch. Moreover, adding multiples of seven to it will make $5a+2b$ cycle back to that same place. So what we do, then, is add $-7a-7b= 7(-a-b) \in 7 \Bbb Z$ to it. This gives us $-2a-5b \equiv 0 \pmod 7$. However, in this notch of the diagram, if $x$ is in it, so is $-x$. (That is, $x \equiv -x \pmod 7$. I leave it to you to justify why.) Thus, $-(-2a-5b)= 2a +5b \equiv 0 \pmod 7$ as desired.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.