问:
Having trouble finding the fault in my logic for this coding question(it's heavily math related)
So I have been doing some coding questions to practice my coding and I ran into this problem: http://www.usaco.org/index.php?page=viewproblem2&cpid=989

The general gist of the problem is that there is a person named Bessie is racing a length of $K\ge1$ meters. She starts at $0$ meters per second at the $0$ second mark, and then every second after that, she can either increase her speed by $1$ meter per second, stay at the same speed, or decrease her speed by $1$ meters per second. Her speed can not drop below zero(so if Bessie is at $0$ meters per second, she can not decrease her speed.).

Bessie wants to finish the race in an integral amount of seconds, either finishing at the $K$ meter mark or past the $K$ meter mark. But Bessie also doesn't want to finish the race too quick. At the instant when she finishes $K$ meters, she wants to be traveling at a speed of no more than $X\ge1$ meters per second. Bessie wants to know the minimum speed required to finish $K$ meters, given $K$ and $X$.

The logic I use to solve this problem only works for the first 4 test cases, and I'm sure it's not because of an coding error. So my logic is as follows:

Before we do anything, we first have to test whether or not a speed of $X$ meters per second can be reached, as the following solution assumes that $X$ meters per second is reachable.

To do so, we first note that the quickest way to get to $X$ meters per second is to increase the speed by $1$ each second for $X$ seconds. We then note that if after increasing $X-1$ times, if the distance covered is $<K$ meters, then it is guaranteed that $X$ meters per second is reachable. But if after increasing $X-1$ times the distance covered is $\ge K$, then we know that $X$ meters per second is unobtainable.

To calculate the distance covered after $X-1$ increases, we can calculate the following sum:$$1+2+\cdots+(X-2)+(X-1)$$which can be represented as$$\frac{X(X-1)}2$$. We want to test whether or not$$\frac{X(X-1)}2\ge K$$. If this inequality is false, then go to the solution under the gray line. If this inequality is true, then we know $X$ meters per second is unobtainable, and therefore we need to calculate how many increases are required to surpass $K$ meters. We will call this amount $n$. To find the value of $n$ that will cause the distance to go over $K$ meters, we first need to find the formula of the distance covered after $n$ increases. That can be represented with the sum $$1+2+3+\cdots+n=\frac{n(n+1)}2$$. So then we set this sum to be $<K$, then use this inequality to maximize $n$:$$\frac{n(n+1)}2<K\\\frac{n^2+n}2<K\\n^2+n<2K\\(n+1/2)^2-1/4<2K\\n+1/2<\sqrt{2K+1/4}\\n<\frac{\sqrt{8K+1}-1}2$$So the value of $n$ would be:$$n=\left\lceil\frac{\sqrt{8K+1}-1}2\right\rceil$$(without the ceiling function we would be calculating the largest amount of increases that doesn't surpass $K$ meters, instead of actually passing $K$ meters)

First we want to find the maximum speed in which Bessie can go at. Let's say that this maximum speed is $m$ and the target speed(the speed we want to have at $K$ meters) be $X$ meters per second(as stated in the problem). We can find the maximum speed by allowing Bessie to increase its speed every second until it reaches $m$, then immediately start decreasing her speed until she hits $X$ meters per second. We then know that the total distance traveled after this increase and decrease is(which I will denote as $d$): $$d=\underbrace{1+2+3+\cdots+m}_{\text{increasing speed}}+\underbrace{(m-1)+(m-2)+\cdots+(X+1)+X}_{\text{decreasing speed}}$$. We can then find the formula for this sum to be: $$d=m^2-\frac{X(X-1)}2$$. This sum has to be $\le K$(or else we can't decrease enough in time), so we have the following inequality:$$m^2-\frac{X(X-1)}2\le K$$. $K$ and $X$ are already given as inputs in the problem, so we just have to isolate $m$. We get that: $$m\le \sqrt{K+\frac{X(X-1)}2}$$(positive square root). To get the highest $m$, we just need to take the floor of the RHS so:$$m=\left\lfloor\sqrt{K+\frac{X(X-1)}2}\right\rfloor$$. Then if $d$ is $<K$, we need to find out the remaining distance that we need to cover. That is easy to calculate: $K-d$. From this we can calculate how many seconds we need to stay at $m$ meters per second(if we stay at a speed $<m$, we can always stay at a higher speed to potentially reach $K$ meters quicker. Not too sure about this logic though). Each second we stay at $m$ meters per second adds an extra $m$ meters to our distance. So we need to divide $K-d$ by $m$ to see how many times we need to add $m$ to $d$ to reach $K$(I will denote this as $s$). So we get that we need to stay$$s=\left\lceil\frac{K-d}m\right\rceil$$seconds at $m$ meters per second to pass $K$ meters. Then we need to calculate the amount of seconds that passed for traveling $d$ meters using the method stated. To calculate this, we need to count how many terms we added together in the sum. So we need to find the length of this list:$$1,2,3\dots,m,(m-1),(m-2),\cdots,(X-1),X$$This can be calculated with the following formula: $$2m-X$$So finally we calculate $$2m-X+s$$ to obtain the final answer.

The problem is this only works for the first 4 test cases, and this strategy only works for certain values of $K$ and $X$ past test case 4, and is really close to the actual answer for other values(yes, I downloaded the test data), so I'm actually not too sure where I went wrong here. If you guys want the code I can also put it here, but this is more of a math problem, so I decided not to put the code for now.

答:
I checked based on the example you gave. Yes your earlier solution did not handle the case when you crossed the finish line before you could attain $X$. On your new logic to handle that case, you need to modify it a bit. Also there are a few other problems as I suspected on boundary conditions.

A) Problem with the newly added logic -

Take an example of $K = 10^9, X = 44721$

$\frac{X(X-1)}{2} = 999961560,$ which is less than $K$. So you would go to your original logic. That will give you a max speed of $44720$ and you will stay at that speed for $(k-d)$ distance to cross the finish line. But this is not optimal. You can attain the speed of $44721$ before the finish line and cross at that speed. (Or you fix your max speed logic that currently considers floor function that I have called out later, it would work too).

So you should check whether $\frac{X(X+1)} {2} \geq K$ and your formula for $n$ works.

B) On your original logic -

i) you are not attaining max speed every single time. Take example of $K = 22, X = 5$. Your logic will get to max speed of $5$ but you can actually attain $6$. Another example is $K = 27, X = 4$ or $K = 31, X = 3$.

So please fix your logic to calculate $d$ as per (ii). Then to find maximum speed,

$m$ is either a floor or a ceiling (Round function does not work for all cases either. I checked further.). There are two ways to fix -

Take $m$ as floor like you are doing right now and then do a test with $m + 1$ to confirm whether $m$ or $m+1)

or

Take the floor to find $m$ the way you do right now. Recalculate $d$ and check whether $(K - d + X - 2m) \geq 2$. If so, you can attain speed of $m + 1$ and you will still be able to get back to $X$ and cross the finish line.

ii) It is not clear how you calculate $d$. As per details in your questions,

You calculate $d=m^2-\frac{X(X-1)}2$. Then go on to calculate

$m=\left\lfloor\sqrt{K+\frac{X(X-1)}2}\right\rfloor$ but I do not see you refreshing $d$ based on the final value of $m$ you choose (which is a floor function). You should do that.

iii) Last point - calculation of $s$. While the final answer $2m - X + s$ comes correct. It needs a bit more clarification as there may be two different speeds (adjacent) at which you go for more than a second. Take example of $K = 29, X = 3$. How long will you go at speed of $m = 5$ beyond one initial second? It should be one more time (till distance is $20$). Then you go twice at a speed of $4$. Then you reduce to $3$. Your logic does not fully clarify that though the final answer works.


问:
Show $\log 15$ is irrational and that $\log 3$+ $\log 5$ is also irrational.
I am trying to show that $\log 15$ and $\log 3$ + $\log 5$ is irrational.

For $\log 15$ I feel like I have no issues showing this is irrational.

By contradiction assume $\log 15$ is rational then

$\log 15 = \frac{m}{n}$ where $m,n\in\mathbb{N}$ and $n\not= 0$

$15 = 10^{\frac{m}{n}}\Rightarrow 15^n = 10^m \Rightarrow 3^n5^n = 5^m2^m$

The left side of the equation has an odd multiple of $5$ but the right side has an even multiple of $5$ and we have arrived at a contradiction.

But for  $\log 3$ + $\log 5$  I do not even know where to begin, I am assuming I cannot use the fact that   $\log 15 = \log 3$ + $\log 5$ as that would be trivial. Should I show they are individually irrational? But that leads me to my next thought as the sum of two irrational numbers is not always irrational (even though in this case I know the sum is irrational). And then would that lead me to breaking it into a case by case where I assume one is rational and the other is not, etc.  I am still an undergrad so I do not know much advanced number theory so I would appreciate a more intuitive approach if possible.

答:
Where does the restriction end? Without using the log identity directly, I can still assume $\log 3 + \log 5 = \frac mn$ and then

$$10^{\log 3+\log 5} = 10^{m/n} \implies 10^{\log 3}\cdot 10^{\log 5} = 10^{m/n} \implies 3\cdot 5 = 10^{m/n}$$

ContentLicense: CC BY-SA 4.0


问:
Characterisation of the orthogonal projection
In an euclidean vector space $E$ how can I prove that a projection $p$ is orthogonal if and only if $$||p(x)||\leq ||x||\quad \forall x\in E$$

I can prove that if $p$ is an orthogonal projection then we have the inequality but I'm stuck in the other implication. Any suggestion is welcome.

答:
A property of all projections is that $P^2=P$.

Let $v=Pu-u$. Then
$$
Pv=P(Pu-u)=P^2u-Pu=0
$$
Next, by assumption
$$
\begin{align}
\|Pu\|^2
&=\|P(Pu+tv)\|^2\\
&\le\|Pu+tv\|^2\\
&=\|Pu\|^2+2t\langle v,Pu\rangle+t^2\|v\|^2
\end{align}
$$
The only way that this can be true for all $t$, is if $\langle v,Pu\rangle=0$. That is,
$$
\langle Pu-u,Pu\rangle=0
$$
Therefore, $P$ is orthogonal.

ContentLicense: CC BY-SA 3.0


问:
A ball around of a generic point has a positive ergodic measure
Let $T:X \to X$ be a continuous function on the compact metric space $X$. We say that $\mu$ is $T-$ invarint if $\mu(T^{-1}(A))=\mu(A) \hspace{0.2cm} \forall A \in \beta.$

We denote by $\mathcal{M}(X,T)$ the set of $T-$invariant Borel probability measures.

For $\mu \in \mathcal{M}(X, T),$ the set $G_{\mu}$ of $\mu-$generic points is defined by
$$
G_{\mu}:=\left\{x \in X: \frac{1}{n} \sum_{j=0}^{n-1} \delta_{T^{j} x} \rightarrow \mu \text { in the weak }^{*} \text { topology as } n \rightarrow \infty\right\}
$$
where $\delta_{y}$ denotes the probability measure whose support is the single point $y$.

Let $\mu$ be an ergodic measure. By Birkhoff ergodic theorem, $\mu(G_{\mu})=1$.

$\textit{Question:}$ Let $\mu$ be an ergodic measure and $x$ be a $\mu-$generic point.  Why is $\mu(B(x, \frac{1}{h}))>0$ for every $h \in \mathbb{N}$, where $B(x, \frac{1}{h})$ is the ball of radius $\frac{1}{h}$ centered at $x.$?

I think one should consider $\chi_{B(x, 1/h)}$ in Birkhoff ergodic theorem, but I don't know how to get the sum is positive.

答: This claim seems false to me. Consider $X=\{0,1\}$ with $T(0)=T(1)=1$. Then $0$ is a generic point for the measure $\delta_1$, but is not contained in the support of the measure.


问:
How to approach proof of "For an integer n, if $n > 7$, then $n^2-8n+12$ is composite"?
For the first part of this question, I was asked to find the either/or version and the contrapositive of this statement, which I found as follows:

i) either $n \leq 7$, or $n^2-8n+12$ is composite

ii) if $n^2-8n+12$ is not composite, then $n \leq 7$

Then we are asked to prove the statement.

I've examined all of our class notes and found mention of proof via factorising "to be covered in more detail later" (but no detail later included - perhaps it is considered too obvious?).

As this is such a simple question, judging by the two marks available for it, I'm concerned that asking another student will cause them to inadvertently tell me the answer.

I am finding this question hard to google around because findable simple proof examples do not seem to include quadratic expressions.

What would the right steps be to approach a proof like this? And/or is there an online resource that shows some similar examples with working/notes? I don't mind handing in an incorrect answer if I'm able to make an attempt, but I'm just so uncertain of the right starting point that I cannot make a start.

答:
$$n^2-8n+12=(n-4)^2-2^2=(n-2)(n-6)$$

Thus, if $n>7$, then both of the factors $n-2$ and $n-6$ are greater than $1$ and $n^2-8n+12=(n-2)(n-6)$ is a nontrivial factorization of $n^2-8n+12$.


答:
$$n^2-8n+12=(n-2)(n-6)$$
Now if $n>7$ then both factors are at least $2$, which means $n^2-8n-12$ is composite. This settles the matter.

Title: How to approach proof of "For an integer n, if $n > 7$, then $n^2-8n+12$ is composite"?

Body:
For the first part of this question, I was asked to find the either/or version and the contrapositive of this statement, which I found as follows:

i) either $n \leq 7$, or $n^2-8n+12$ is composite

ii) if $n^2-8n+12$ is not composite, then $n \leq 7$

Then we are asked to prove the statement.

I've examined all of our class notes and found mention of proof via factorising "to be covered in more detail later" (but no detail later included - perhaps it is considered too obvious?).

As this is such a simple question, judging by the two marks available for it, I'm concerned that asking another student will cause them to inadvertently tell me the answer.

I am finding this question hard to google around because findable simple proof examples do not seem to include quadratic expressions.

What would the right steps be to approach a proof like this? And/or is there an online resource that shows some similar examples with working/notes? I don't mind handing in an incorrect answer if I'm able to make an attempt, but I'm just so uncertain of the right starting point that I cannot make a start.


问:
About the sub-mean value property
Let $\Omega$ be an open subset of $\mathbb{R}^N$. We say that a function $u\in C(\Omega)$ has the (surface) sub-mean value property in $\Omega$ if (1) $u(x)\leq\frac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)} u(y)dSy$ for any ball $\bar{B}_r(x)\subseteq\Omega$.
We say that a function $u\in C(\Omega)$ has the (volumetric) sub-mean value property in $\Omega$ if (2) $u(x)\leq \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y)dy$ for any ball $\bar{B}_r(x)\subseteq\Omega$.
A function that satisfies the condition (1) is called subharmonic on $\Omega$. I know that this is equivalent to (if $u\in C^2(\Omega)$) $\Delta u\geq 0$, and that (1) implies (2), but what i wish to know is if (2) implies (1). I think that is not the case but i cannot find a counterexample (The analogue version in which there is the equalities are equivalent).

答:
I am not sure about the case $u \in C(\Omega)$ but if $u \in C^2(\Omega)$, then (2) does indeed implies (1). So, in all what follows, let us assume that $u \in C^2(\Omega)$.

Since you already know that $\Delta u \geq 0$ implies (2), I assume that you also know the implication
$$\Delta u \leq 0\phantom{a}\text{on }\Omega \phantom{aaa}\Longrightarrow\phantom{aaa}u(x) \geq \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y) d y\phantom{aa}\text{for all }\overline{B_r(x)} \subseteq \Omega$$
This follows from "$\Delta u \geq 0$ $\Rightarrow$ (2)" by multiplication with $-1$.

Now assume (2) holds. It is sufficient to show $\Delta u \geq 0$ since you already know that this is equivalent to (1).

Let's prove $\Delta u \geq 0$ by contradiction: Assume this is false. Then, by continuity of $\Delta u$, there is an open $\Omega_0 \subseteq \Omega$ such that $\Delta u < 0$ on $\Omega_0$. By my remarks above, this implies
$$u(x) \geq \frac{1}{|B_r(x)|} \int_{B_r(x)} u(y) d y\phantom{aa}\text{for all }\overline{B_r(x)} \subseteq \Omega_0$$
If we combine this with (2), we get
$$u(x) = \frac{1}{|B_r(x)|}\int_{B_r(x)} u(y) d y\phantom{aa}\text{for all }\overline{B_r(x)}\subseteq\Omega_0$$
But this means $\Delta u =0$ on $\Omega_0$ which contradicts our hypotheses.

Title: About the sub-mean value property


问:
Find the limit of the function without L'Hôpital's rule
I have a problem, such as:

$$\lim_{x \to 0} \left(\frac{\cos(ax)}{\cos(bx)}\right)^\frac{1}{x^2}$$

How do I solve this problem without using L'Hôpital's rule or small-o?

Thanks!

Body:
$$\lim_{x\to0}\cos(ax)^{1/x^2}=\lim_{t\to0}\cos(t)^{a^2/t^2}=\lim_{t\to0}(\cos(t)^{1/t^2})^{a^2}=\left(\lim_{t\to0}(\cos(t)^{1/t^2}\right)^{a^2}$$ and the answer is of the form

$$c^{a^2-b^2}.$$

We can find $c$ with

$$(\cos(t))^{1/t^2}=\left(\left(1-2\sin^2\left(\frac t2\right)\right)^{1/\sin^2(\frac t2)}\right)^{\sin^2(\frac t2)/4(\frac t2)^2}\to (e^{-2})^{1/4}.$$


问:
Unique nearest point in epsilon neighborhood of compact real manifold?
I have to proof the following assertion:

Let $X$ be a compact submanifold of $\mathbb{R}^n$ and $\mathcal{U}^\varepsilon=\{p\in\mathbb{R}^n\;:\; |p-q|<\varepsilon \text{ for some }q\in X\}$. Then there is $\varepsilon>0$ such that every point $p\in\mathcal{U}^\varepsilon$ has a unique nearest point $\pi(x)\in X$.

But is this true at all?

Consider $f(x)=x\cdot\sin(1/x)$ for $x\ne0$ and $f(0)=0$. Since $f$ is continuous, $X=\{(x,f(x)\;|\;x\in[0,1]\}$ is a compact submanifold of $\mathbb{R}^2$. Let $x_1 > x_2 > \ldots$ be all roots of $f$ in $(0,1]$. For every point $\big(\frac{x_{i+1}+x_i}{2},0\big)\in\mathbb{R}^2$ both $(x_i,0)$ and $(x_{i+1},0)$ are nearest points in $X$. And since $x_{i+1}-x_i$ tends to zero, we have such points in every $\mathcal{U}^\varepsilon$.

What is wrong with my argument?

Thanks in advance!

答: Your homework problem is about smooth submanifolds.


问:
Find gcd of $a = 170 + 32\sqrt{3}$ and $b = 61 + 35\sqrt{3}.$ Then find $f,g \in \mathbb{Z}[\sqrt{3}]$ such that $af + bg = d$ using norm function.
I want to solve that question:

Find a greatest common divisor $d$ of $a = 170 + 32\sqrt{3}$ and $b = 61 + 35\sqrt{3}.$ Then find $f,g \in \mathbb{Z}[\sqrt{3}]$ such that $af + bg = d.$

Note that we defined the norm function as follows:  Define the norm function $N:\mathbb{Q}[\sqrt{3}] \rightarrow \mathbb{Q}$ by $N(a_{1} + a_{2}\sqrt{3}) = \operatorname{det} \phi (a_{1} + a_{2}\sqrt{3}) = a_{1}^2 - 3 a_{2}^2.$ And we showed that $N(xy) = N(x)N(y)$ for all $x,y \in \mathbb{Q}[\sqrt{3}].$

My questions are:

a-I found those questions here:

1- Efficiently computing GCDs in $\mathbb{Z}[(1+\sqrt{-19})/2]$ and

2- GCD of two elements in $\mathbb Z \left[\frac{1 + \sqrt{-11}}{2}\right]$

but still I do not understand the general procedure of finding gcd using norm function, could anyone explain that for me please?

b- I know that to solve this part of the question "find $f,g \in \mathbb{Z}[\sqrt{3}]$ such that $af + bg = d.$" in case of $\mathbb{Z}$ we usually take the calculations we did in the first part (using the division algorithm )in reverse direction i.e. from last step to first step, But I do not know how we usually do that in case of  $\mathbb{Z}[\sqrt{3}].$ Could anyone clarify that for me please?

My trial so far:

Since $N(a) = 25828 > N(b) = 46 = 2 \times 23$ and since $N(a)$ is not divisible by $23,$ then their common divisor is $2.$ then we will calculate $$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23},$$ But this is not in $\mathbb{Z}[\sqrt{3}].$

答:
Here is a partial answer.

Since $d$ divides both $a$ and $b$, $N(d)$ divides both $N(a)$ and $N(b)$ and so $N(d)$ divides $\gcd(N(a),N(b))=2$.

Therefore, $N(d) \in \{ \pm 1, \pm 2 \}$. Since neither $-1$ nor $2$ are norms, $N(d) \in \{ 1, -2 \}$.

Let's try $N(d)=-2=(1+\sqrt3)(1-\sqrt3)$ and so we compute
$$
\frac{170+32\sqrt3}{1+\sqrt3} = -37 + 69 \sqrt3,
\qquad
\frac{61+35\sqrt3}{1+\sqrt3} = 22 + 13 \sqrt3
$$
Thus, $d \sim 1+\sqrt3$ because $2$ does not divide $a$ and $b$.

For the second part, it suffices to solve $ (-37 + 69 \sqrt3)f + (22 + 13 \sqrt3)g=1$ .


问:
Binomial Distribution problem $n$ and $p$ unknown
$$X \sim B(n, p)$$

$$P(X=2)= \frac{80}{243} \quad \quad \quad P(X=5)= \frac{12}{729}$$

Find $n$ and $p$.

Why is $n=6$ and $p=\frac{1}{3}$?
We cannot use simultaneous equations. However, I think its something to do with $243 = 3^5$ but I am not too sure.

答:
I don't think there is a unique solution of (n,p).
put n equal to a any integer greater than 5, you will find a particular p as a solution.</p>


问:
Transformation of random variable?
i know that if we have a random variable having standard normal distribution (mean = 0 and var = 1) then its square is chi-square distribution with 1 degree of freedom. My question is " does the same thing applies if the random variable is having normal distribution (mean = non-zero and var = other than 1).</p>

答:
Let $\phi$ be a test function. We compute
$$ \mathbb{E}[\phi(X^2)]$$
for $X \sim \mathcal{N}(\mu,\sigma^2)$. This gives
\begin{equation}
\mathbb{E}[\phi(X^2)] = \int_{-\infty}^{\infty} \phi(x^2) \frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sqrt{2 \pi \sigma^2}} dx \\
\qquad \qquad \qquad = 2 \int_0^{\infty} \phi(y) \frac{e^{-\frac{(\sqrt{y}-\mu)^2}{2\sigma^2}}}{2 \sqrt{y} \sqrt{2 \pi \sigma^2}} dy
\end{equation}
where I split the integral into $\mathbb{R}_-$ and $\mathbb{R}_+$ and perform the change of variable $y = x^2$ in both integrals.
Since this is valid for all test function, we can identify the density of $X^2$ for $X \sim \mathcal{N}(\mu,\sigma^2)$. It is given, on $\mathbb{R}_+$ by
$$ f(y) = \frac{1}{\sqrt{y 2\pi \sigma^2}} e^{-\frac{(\sqrt{y}-\mu)^2}{2\sigma^2}}. $$
Notice that for $\mu = 0$, $\sigma^2 = 1$ you recover indeed the $\chi^2_1$ distribution.

Title: Transformation of random variable?


问:
Equivalence of definitions of tame ramification
There are three different characterisations of a tamely ramified extension $L/K$ of a local field $K$ and I don't understand why they are equivalent.

$\bullet$ $p \nmid e$ where $p = \rm{char}(k)$, $k$ is the residue field of $K$ and $e$ is the ramification index.

$\bullet$ $G_1 = \{1\}$ where $G_1$ is the ramification group of $L/K$.

$\bullet$ $L/K$ has conductor $1$, where the conductor is the least integer $\frak f$ such that $1+\frak{m}^f \subseteq \rm{Ker}(\phi_{L/K})$. $\frak m$ is the unique maximal ideal of $K$ and $\phi_{L/K}: K^\times \longrightarrow \rm{Gal}(L/K)$ is the local Artin map.

The equivalence of the first two definitions has already been discussed on stackexchange but as far as I know, no proof has been provided.

Hints are very welcome too. Thank you.

答:
$L$ is a finite abelian extension of local field $K$. I will briefly describe the upper numbering to prove that conductor $1$ implies $G_1=\{1\} $.

Let $g_i=\# G_i $ so $g_0 \geq g_1 \geq g_2 \geq \cdots \geq1 $. Define a piecewise linear and continuous function $\phi : [0, \infty) \rightarrow [0, \infty) $ such that $\phi(0)=0  $ and it has slope $g_i/g_0 $ in $(i,i+1)$. You can see that $\phi$ is strictly increasing and thus invertible.

For $v\in \mathbb R_{\geq0 }$ define the upper numbering as  $G^v=G_{\lceil \phi ^{-1} v \rceil}$ where $\lceil . \rceil $ is ceiling function.

The image of $U^{m}_K $ under local reciprocity map is $\theta_{L/K}(U^m_{K})=G^m  $.

Thus $G^1=\{1\}  $   when the conductor is 1.
We say $u$ is a jump in uppper numbering if $G^u \neq G^{u+\epsilon} $ for all $\epsilon >0 $.

Theorem (Hasse - Arf): $L/K$ abelian extension of local fields. The jumps in upper numbering are at integers.

Thus $G^1 =G^v $ for all $v \in (0,1] $. If we let $v = g_1/g_0 $ then $$\{1\}=G^1 =G^v=G_{\lceil \phi ^{-1}(g_1/g_0) \rceil} = G_1$$ which is what we wanted.

The proofs of the theorems I have used above, I believe, are given in Serre's article on Local Class Field Theory in Cassels- Frohlich Algebraic Number Theory.


问:
Prove existence of $n$ for an infinite product
I've been studying a paper in which the author says:

Fix $n$ such that $m^n \prod_{j=1}^n \frac{j}{j+\delta} > 1$, where $1<m<\infty$, and $\delta >0$.

I seem not to be able to show why such $n$ must exist. I tried rewriting it this way:

$m^n \prod_{j=1}^n \frac{j}{j+\delta} =m^n \prod_{j=1}^n (1-\frac{\delta}{j+\delta})= m^n \exp\bigg[\sum_{j=1}^n \log(1-\frac{\delta}{j+\delta})\bigg] \geq m^n\bigg[1+\sum_{j=1}^n \log(1-\frac{\delta}{j+\delta})\bigg]$

but after that I'm stuck. In one part the author writes

$m^n \prod_{j=1}^n (1-\frac{\delta}{j+\delta}) \geq m^n e^{c_0} \exp\bigg[-\delta\sum_{j=1}^n \frac{1}{j+\delta}\bigg]$, where $c_0$ is some constant,

but then I still don't know how this guarantees that there exists $n$ such that $m^n \prod_{j=1}^n \frac{j}{j+\delta} > 1$. Does anyone have any idea on how to prove it?

答: Find an $N$ such that $\frac{mN}{N+\delta}>1+\eta$ for some $\eta>0$. Then take $M$ such that $(1+\eta)^M>\left(\prod_{j=1}^N\frac{j}{j+\delta}\right)^{-1}$. Then take $n=N+M$.


问:
simplify this formula
I'm attempting to simplify this formula for when r is equal to
$$
\frac{119}{2\sin(θ/2)+2(π-θ)/2}
$$
the formula is:

$$A = \frac{\pi r^2}{2} - \frac{1}{2} (θ-\sinθ) r^2$$

I've tried but i'm confused particularly about the fact that you end up with a fraction in a fraction. Overall, my aim is to make θ the only variable so i can find the maximum by finding its derivative and making the equation equal 0.

Ps. sorry if my brackets are 100% correct :)

答:
I can do the differentiation for you. Let's do it in small steps and using the chain rule. First we have
$$
r(\theta) = \frac{119}{2\sin(θ/2)+\pi-\theta}  = \frac{119}{g(\theta)}
$$
I re-named the denominator as $g(\theta)$ to simplify the notation. The derivative of $r$ with respect to $\theta$ is now
$$
r'(\theta) = -\frac{g'(\theta)}{g(\theta)^2}
$$
The derivative of $g$ is relatively easy to compute, it's just
$$
g(\theta) = 2\sin \frac{\theta}{2} + \pi - \theta \quad \Rightarrow \quad
g'(\theta) = \cos \frac{\theta}{2} -1
$$
Therefore, we have the intermediate result
$$
r'(\theta) = -119 \frac{\cos \frac{\theta}{2} -1 }{ \left( 2\sin \frac{\theta}{2} + \pi - \theta \right)^2}
$$
(It looks like the whole derivative expression will be quite complicated!)
Now we can calculate the derivative of $A$ with respect to $\theta$, using the chain rule and product rule!
$$A = \underbrace{\frac{\pi r(\theta)^2}{2}}_{=B(\theta)} + \underbrace{\frac{-1}{2} (θ-\sinθ)}_{=C(\theta)} \underbrace{r(\theta)^2}_{=D(\theta)} = B(\theta)+C(\theta)D(\theta)$$
$$
\begin{split}
A'(\theta) &= B'(\theta) &+ C'(\theta)\cdot D(\theta) &+ C(\theta) \cdot D'(\theta)\\
&= r'(\theta)\cdot \pi r(\theta) &+ \frac{-1}{2}(1-\cos \theta)\cdot r (\theta)^2
&+ \frac{-1}{2} (θ-\sinθ)\cdot 2 r (\theta)
\end{split}
$$
We can still take $r(\theta)$ as a common factor to get
$$
A'(\theta) = \left[ \pi r'(\theta)  - \frac{1}{2}(1-\cos \theta)r (\theta)
- θ+\sinθ) \right] r(\theta)
$$
Now you just need to plug in the expressions for $r(\theta)$ and $r'(\theta)$ ...
$$
A'(\theta) = \left[ -119 \pi \frac{\cos \frac{\theta}{2} -1 }{ \left( 2\sin \frac{\theta}{2} + \pi - \theta \right)^2}  - \frac{1}{2}(1-\cos \theta)\frac{119}{2\sin(θ/2)+\pi-\theta}
- θ+\sinθ) \right] \frac{119}{2\sin(θ/2)+\pi-\theta}
$$

Title: simplify this formula


问:
$E \xi_n^2 \le c $ and $\xi_n \to \xi$ almost surely. Prove that $E \xi $ is finite and $E \xi_n \to E \xi.$

Suppose $\{\xi_n\}$ is a sequence of random variables on a probability space such that $E \xi_n^2 \le c $ for some constant $c.$ Assume that $\xi_n \to \xi $ almost surely as $n \to \infty$. Prove that $E \xi $ is finite and $E \xi_n \to E \xi.$

This question has been asked before, but I don't want to take the path outlined there. I'm trying to use just the theorems from my class notes: Egorov, Lebesgue Dominated Convergence, Levi Monotone Convergence, Fatou Lemma.

Attempt: By Fatou's Lemma, $E \liminf\limits_{n \to \infty} \xi_n^2 \le \liminf\limits_{n \to \infty} E \xi_n^2 = c.$ Since $\xi^2 = \lim\limits_{n \to \infty} \xi_n^2 = \liminf\limits_{n \to \infty} \xi_n^2$ almost surely, I'm tempted to conclude that $E \xi^2 \le c \, (*),$ which would solve the problem.

Assuming $(*)$ is true for the moment, we get $E |\xi| \le \sqrt{c}$ by Cauchy-Schwarz, so $E \xi$ is finite and all that remains to show is $E|\xi_n - \xi| \to 0.$ Fix $\epsilon > 0.$ By Egorov's Theorem, there exists $A$ such that $\xi_n \to \xi$ uniformly on $A$ and $P(A) \ge 1 - \frac{\epsilon}{4\sqrt{c}}.$ Then $E|\xi_n - \xi| = \int_A |\xi_n - \xi| dP + \int_{\Omega \setminus A} |\xi_n - \xi| dP.$ Using uniform convergence, there exists $N$ such that the first integral is $\le \sup\limits_{\omega \in A} |\xi_n - \xi| dP \le \epsilon/2$ for all $n \ge N.$ The second integral is at most $\frac{\epsilon}{4\sqrt{c}} (E|\xi_n|+E|\xi|) \le \epsilon/2.$ Thus, $E|\xi_n - \xi| \le \epsilon$ for all $n \ge N,$ which implies $E|\xi_n - \xi| \to 0.$

Now how do I show $(*)$? Let $A = \{w : \xi_n(\omega) \not\to \xi(\omega)\}$ be the set of measure zero on which convergence fails. What we showed is that $\int_{\Omega \setminus A} \xi^2 dP \le c.$ If we show that $\int_A \xi^2 dP = 0,$ that would finish things off. I believe this follows from the convention that $\infty \cdot 0 = 0,$ and should still apply even though we only know a priori that $\xi$ is measurable, not integrable.

Anyways, it seems I ended up answering my own question, so I just want to know if my proof is correct.

Edit: Inspired by this question, I found a fix. Let $B = \Omega \setminus A.$ Cauchy-Schwarz on the indicator function (earlier I was using $E1 =1,$ which is too weak) gives $\int_B |\xi_n - \xi| dP = E(|\xi_n - \xi| I_B) \le (E I_B E (\xi_n-\xi)^2)^{1/2} \le (P(B) \cdot 2c)^{1/2},$ which really is bounded now.

答:
There is a problem with your proof: You said $\int_{\Omega \setminus A} |\xi_n-\xi|dP \leq \frac {\epsilon} {4\sqrt c }(E|\xi_n|+E|\xi_i|)$. This is not correct. (It is not true that $\int_E |f| d\mu \leq P(E)\int |f|d\mu$). It is hard to get  any control over $E|\xi_n I_{\Omega \setminus A}|$? I think this where the idea of uniform integrbility comes in. I doubt if this approach can be fixed.

If $A$ has measure $0$ then $\int |f| d\mu=0$ for any measurable function $f$. This is an easy consequence of the way integrals are defined: first verify it for simple functions and then take limits. Thus any measurable function is integrable over a set of measure $0$ and its integral is $0$.


问:
Merging polynomials together
Given the (monic) polynomials $P$ and $Q$, we can split them over an appropriate field into linear factors to write $P(x)=(x-a_1)\dotsb(x-a_m)$ and $Q(x)=(x-b_1)\dotsb(x-b_n)$, and then form the polynomial
$$ P(x-b_1)\dotsb P(x-b_n) = Q(x-a_1)\dotsb Q(x-a_m) . $$
Is there any standard notation or name for this new polynomial? Have its properties been studied?

答:
I am going to give an answer to the question asked in the comments

What are alternative ways to define the polynomial in question without factoring the original P and Q

Let $T_x(y)=Q(x-y)$ be understood as a polynomial in $F[x][y]$, where $F$ is a field. I claim that

The resultant $\operatorname{Res}(P(y),T_x(y),y)$ with respect to $y$ is equal to $\prod_{i=1}^mQ(x-a_i)$.

Proof. Let $R$ be an integral domain and $f$ and $g$ monic polynomials in $R[y]$ of degrees $m$ and $n$, and with roots $\lambda_i$ and $\mu_j$ in any algebraically closed field containing $R$, respectively. Their resultant with respect to $y$ is
$$\operatorname{Res}(f,g,y)=\prod_{i=1}^m g(\lambda_i)=(-1)^{mn}\prod_{j=1}^n f(\mu_j).$$
Now apply the above for $f(y):=P(y)$ and $g(y):=T_x(y)=Q(x-y)$. The first equality gives
$$\operatorname{Res}(P(y),T_x(y),y)=\prod_{i=1}^m T_x(a_i)=\prod_{i=1}^mQ(x-a_i).$$

Corollary. The polynomial $\prod_{j=1}^nP(x-b_j)=\prod_{i=1}^mQ(x-a_i)$ can be computed without factoring $P$ or $Q$.

Proof. Compute the Sylvester determinant for $P(y)$ and $Q(x-y)$ with respect to $y$ (for the relevant definitions see e.g. the wiki articles for Resultant or Sylvester matrix).

Title: Merging polynomials together


问:
slowness of growth of polynomials
How to show that for any $d\in\mathbb{N}$ and $\delta>0$ there exists $\varepsilon=\varepsilon(d,\delta)>0$ such that is $f$ is a polynomial of degree $d$, and
$$|f(t)|\leq 1,\quad t\in[0,1],$$ then $$|f(t)|<1+\delta,\quad t\in[0,1+\varepsilon].$$
 (This is an exercise from "Introduction to Ratner Theorems..." by D.W. Morris). It is immediate for $d=0,1,2$, but I am probably lacking some trick to approach the general case. Any help is greatly appreciated.</p>

答:
Claim: For every $d\in\mathbb{N}$, there exists $M_d>0$, such that for $f(x)=\sum_{i=0}^da_ix^i$, if $|f(t)|\le 1$ when $t\in[0,1]$, then $|a_i|\le M_d$, $i=0,\dots d$.

Proof of Claim:
Given $f(x)=\sum_{i=0}^da_ix^i$ and given $b_0,\dots,b_d\in[-1,1]$, consider the following system of linear equations with unknowns $a_0,\dots,a_d$:
$$f\big(\frac{i}{d}\big)=b_i,\quad i=0,\dots,d.\tag{1}$$
Note that the coefficient matrix of $(1)$ is a non-singular Vandermonde matrix whose entries are independent of $b_0,\dots,b_d\in[-1,1]$. It follows that $(1)$ has a unique solution $(a_0,a_1,\dots,a_d)$ and there exists  $M_d>0$, only dependent on $d$, such that $|a_i|\le M_d$, $i=0,\dots d$, which proves the claim. $\quad\square$

The statement follows from the claim immediately. If $f(x)=\sum_{i=0}^da_ix^i$ satisfies that $|f(t)|\le 1$ when $t\in[0,1]$, then $|a_i|\le M_d$, $i=0,\dots d$, and for every $\delta>0$, there exists $\varepsilon=\min(1,\frac{\delta}{2^dM_d})$, such that when $t\in[1,1+\epsilon]$,

$$|f(t)|\le |f(1)|+|f(t)-f(1)|\le 1+\sum_{i=0}^d|a_i|(t^i-1)\le M_d\varepsilon\sum_{i=1}^d 2^{i-1}\le 1+2^dM_d\varepsilon\le 1+\delta.$$

Edit: An alternative way(indirect but a little more general) to prove the claim is to use the fact that all norms on finite-dimensional vector spaces are equivalent. The claim follows from comparing the two norms of $\mathbb{R}^{d+1}$ below:
$$\|(a_0,\dots,a_d)\|:=\max_{0\le i\le d}|a_i|~,\quad \|(a_0,\dots,a_d)\|':=\max_{0\le t\le 1}|\sum_{i=0}^d a_it^i|.$$

Title: slowness of growth of polynomials


问:
How to solve this nonlinear vector equation?
Let $Z \in \mathbb{R}^n$, $\alpha \in \mathbb{R}^n$, $\gamma > 0$. Now I would like to find a solution for $\alpha$ of the following equation,

\begin{align}
Z - \alpha = \gamma\frac{\alpha}{\|\alpha\|_{2}} .
\end{align}

I am totally unfamiliar with such an equation. So I already get stuck here. Any help is really appreciated (or sources/references).

Thanks,

p.s apologies for the limited (just one) tags. I was not sure which tags would be appropriate.

答:
$$Z - \alpha = \gamma\frac{\alpha}{\|\alpha\|_{2}}  \Longleftrightarrow Z = \alpha \left( 1 + \gamma\frac{1}{\|\alpha\|_2}\right) \quad \quad (*)$$

Taking the $||.||_2$, you get that $$||Z||_2 = ||\alpha||_2 \left( 1 + \gamma\frac{1}{\|\alpha\|_2}\right) = ||\alpha||_2 + \gamma $$

so you must have $||\alpha||_2=||Z||_2-\gamma$. So if $||Z||_2 \leq \gamma$, the equation has no solution. If $||Z||_2 > \gamma$, then $(*)$ gives
$$Z = \alpha \left( 1 + \gamma\frac{1}{||Z||_2-\gamma}\right), \quad \text{so} \quad \alpha = \frac{1}{\left( 1 + \gamma\frac{1}{||Z||_2-\gamma}\right)}Z =\frac{||Z||_2-\gamma}{||Z||_2}Z $$

(if $Z \neq 0$) and you can check that this is indeed a solution.


问:
Bounding sum of row norms of an orthogonal matrix
Let $A$ be an $n\times k$ real matrix such that $A'A=I_k$. Then if $b_j$ denotes the $j$th column of $A$ we have $\sum_{j=1}^k || b_j || = k$ where $ || \cdot||$ denotes the Frobenius norm.

Question: Can I bound the sum of row norms:

$$\sum_{i=1}^n || a_i ||$$

where $a_i$ denotes the $i$th row of $A$? I feel like I should be able to bound it by $k$ as well, but somehow I can't show it.

Edit: In the comments user1551 provided a counterexample for the $k$ bound. Hence my question is what is the best bound I can put on this sum?

Thanks a lot for your help.

答:
It isn't true that $\sum_i\|a_{i\ast}\|\le k$. E.g. let
$$
A=\frac12\pmatrix{1&1&1\\ 1&1&-1\\ 1&-1&1\\ 1&-1&-1}.
$$
It is straightforward to verify that $A^TA=I_3$, but $\sum_i\|a_{i\ast}\|=2\sqrt{3}>3$.

However, as $\sum_i\|a_{i\ast}\|^2=\|A\|_F^2=\operatorname{tr}(A^TA)=k$, by Cauchy-Schwarz inequality we have
$$
\sum_{i=1}^n\|a_{i\ast}\|\le\sqrt{\left(\sum_{i=1}^n\|a_{i\ast}\|^2\right)\left(\sum_{i=1}^n1^2\right)}=\sqrt{kn}.
$$
Tie occurs when the norms of all rows of $A$ are the same, as shown by the above example.


问:
An odd combinatorics formula
I ran into a formula during my TA session with an intro discrete course, that I am having a hard time proving.
$$
\displaystyle
\prod_{i=0}^{n-1}{{n^2 - in} \choose n} = \frac{(n^2)!}{(n!)^n}
$$
Does anyone have an idea of how to prove this?

Title: An odd combinatorics formula

Body:
Hint:
\begin{eqnarray*}
& & \prod_{i=0}^{n-1}{{n^2 - in} \choose n} \\&=& \color{blue}{\binom{n^2}{n} } &\color{red}{\binom{n^2-n}{n}} &\cdots \binom{n}{n} \\ &=& \color{blue}{\frac{(n^2)(n^2-1)\cdots (n^2-n+1)}{(n!)}} &\color{red}{\frac{(n^2-n)(n^2-n-1)\cdots (n^2-2n+1)}{(n!)}}& \cdots \frac{n!}{n!} \\.
\end{eqnarray*}


答:
let $\mathbb{R}^n=\alpha\mathbb{R}\bigoplus\left(\alpha\mathbb{R}\right)^{\perp} $

You have : $Z=z_{\alpha}u_{\alpha}+z_{\perp}u_{\perp} $ and $\alpha=\alpha u_{\alpha} $

So you get $z_{\perp}=0 $ and :
$$z_{\alpha}=\alpha+\gamma \\\alpha=\left(|| Z||-\gamma\right)\frac{Z}{||Z||} $$

(which is also equivalent to the result of the other answer)

Title: How to solve this nonlinear vector equation?

Body:
Let $Z \in \mathbb{R}^n$, $\alpha \in \mathbb{R}^n$, $\gamma > 0$. Now I would like to find a solution for $\alpha$ of the following equation,

\begin{align}
Z - \alpha = \gamma\frac{\alpha}{\|\alpha\|_{2}} .
\end{align}

I am totally unfamiliar with such an equation. So I already get stuck here. Any help is really appreciated (or sources/references).

Thanks,

p.s apologies for the limited (just one) tags. I was not sure which tags would be appropriate.


问:
How to solve a system of ordinary differential equation squared?
Say we are given a system of differential equations
\begin{cases} \frac{d^2x}{dt^2}=w\frac{dy}{dt} \\ \frac{d^2y}{dt^2}=-w\frac{dx}{dt}  \\ \frac{d^2z}{dt^2}=0\end{cases}
The teacher told us to use  $ u= \frac{dx}{dt}+i \frac{dy}{dt} $

I think it looks like that
$$\left[ \begin{array}{c} x' \\ y'\\ z' \end{array} \right] = \begin{bmatrix} w & 0 & 0\\ 0 & -w & 0 \\ 0&0&0 \end{bmatrix} \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$$

I guess we have to use $u$ to get the correct A(3x3) but how ?

答:
If $w$ is a constant then you can integrate the second DE:
$$y''=-wx'$$
$$\implies y'=-wx+c_1$$
And plug this in the first DE:
$$x''=wy'$$
$$x''=w(-wx+c_1)$$
$$x''+w^2x=c_1w$$
And solve the diferential equation.
$$ \implies x(t)=A\cos (wt)+B\sin(wt)+\dfrac {c_1}w$$
For $y$ we have:
$$y'(t)=-wx+c_1$$
$$ y'(t)=-w(A\cos (wt)+B\sin(wt))$$
$$ \implies y(t)=-A\sin (wt)+B\cos(wt)+c_2$$
The third DE can be integrated easily too.
$$\frac{d^2z}{dt^2}=0 \implies z(t)=k_1t+k_2$$

$$\begin{cases} \dfrac{d^2x}{dt^2}=w\dfrac{dy}{dt} \\ \dfrac{d^2y}{dt^2}=-w\dfrac{dx}{dt} \end{cases}$$
You can also use the substitution given by your teacher. For the first two differential equations you have :
$$u= \frac{dx}{dt}+i \frac{dy}{dt}$$
$$u'= \frac{d^2x}{dt^2}+i \frac{d^2y}{dt^2}$$
Multiply by $i$ the second differntial equation. Sum the differential equations. So that the system becomes:
$$u'=w\frac{dy}{dt}-iw \frac{dx}{dt}$$
$$u'=-i^2w\frac{dy}{dt}-iw \frac{dx}{dt}$$
$$u'=-iw\left(i\frac{dy}{dt}+ \frac{dx}{dt}\right)$$
$$u'(t)=-iwu(t)$$
$$u(t)=e^{-iwt}C$$
Back to $x(t),y(t)$
$$\pmatrix {x'(t) \\ y'(t)}=\pmatrix {\cos(wt) & \sin(wt) \\ -\sin(wt) & \cos(wt)}\pmatrix {C_1 \\ C_2}$$


问:
Show that the following product equals 1 (involves trig)
How can I show that:
$$\prod_{k=1}^{n}\left ( 1+2\cos\frac{2\pi .3^{k}}{3^{n}+1} \right )=1$$
Could you please explain to me how to approach this problem?
Thank you.</p>

答:
Hint:

$$1+2 \cos{\theta} = \frac{\sin{(3 \theta/2)}}{\sin{(\theta/2})}$$

I get as a result for the product:

$$\frac{\sin{\left ( \frac{\pi \cdot 3^{n}}{3^n+1}\right)}}{\sin{\left( \frac{\pi}{3^n+1}\right)}} = 1$$

The way to get this result is to let $\theta_0 = 2 \pi/(3^n+1)$.  The product is then

$$\frac{\sin{(3 \theta_0/2)}}{\sin{(\theta_0/2)}} \frac{\sin{(3^2 \theta_0/2)}}{\sin{(3 \theta_0/2)}} \cdots \frac{\sin{(3^n \theta_0/2)}}{\sin{(3^{n-1} \theta_0/2)}}$$

The product telescopes, so we are left with just the last term in the numerator and the first term in the denominator.  The result follows.


问:
Prove that in a real inner product space $V$ an endomorphism $f$ is skew-Hermitian if and only if $\langle v,f(\vec v)\rangle=0$ for all $v\in V$

Definition

Let be $V$ and $U$ real vector spaces equipped with an inner product. So given a linear transformation $f:V\rightarrow U$ a function $f^*:U\rightarrow V$ is called the adjoint of $f$ if
$$
\langle\vec u,f(\vec v)\rangle=\langle f^*(\vec u),\vec v\rangle
$$
for all $\vec v\in V$ and for all $\vec u\in U$. In particular an endomorphism $f\in\mathscr L(V,V)$ is called Hermitian if $f=f^*$ and skew-Hermitian if $f=-f^*$.

Clearly if $f$ is skew-Hermitian then
$$
\langle v,f(\vec v)\rangle=\langle f^*(\vec v),v\rangle=-\langle f(\vec v),v\rangle=-\langle\vec v,f(\vec v)\rangle
$$
for all $\vec v\in V$ and so $\langle v,f(\vec v)\rangle=0$ for all $v\in V$. But unfortunately I don't be able to proce that if $f$ is such that
$\langle v,f(\vec v)\rangle=0$ for all $v\in V$ then it is skew-Hermitian. So could someone hep me, please?

答:
A common trick is to select two vectors $u$ and $v$ and apply the hypothesis to $u+v$. Using linearity you have
$$0 = \langle u+v,f(u+v) \rangle = \langle u,f(u) \rangle + \langle u,f(v) \rangle + \langle v,f(u) \rangle + \langle v,f(v) \rangle = \langle u,f(v) \rangle + \langle v,f(u) \rangle$$
so that
$$\langle u,f(v) \rangle = \langle -f(u),v\rangle.$$


答:
Clearly $f $ is uniformly continuous on closed intervals, So given $ \delta >0 $ there exists $ \epsilon'(d,\delta)>0 $ such that if $ |y-t| \leq \epsilon' $ then for all such $y$ we have
$$ |f(y)-f(t)|< \delta $$
Take $ \epsilon = \min\{ \epsilon',1\} $, then for all $ t \in [1,1+\epsilon] $ we have $ 0 \leq 1-\epsilon \leq t-\epsilon \leq 1 $ and thus $ t-\epsilon \in [0,1] $. So we can choose $y \in [t-\epsilon,1]\subseteq [0,1] $ which clearly satisfies $ |y-t|\leq \epsilon $. Hence we have for all $ t \in [1,1+\epsilon] $
$$ |f(t)|\leq |f(y)|+|f(y)-f(t)| < 1 + \delta $$
For $ t\in [0,1] $ we trivially have $ |f(t)| \leq 1 < 1 + \delta $. So $ |f(t)| < 1+ \delta $ for all $ t \in [0,1+\epsilon ] $</p>

Title: slowness of growth of polynomials

Body:
How to show that for any $d\in\mathbb{N}$ and $\delta>0$ there exists $\varepsilon=\varepsilon(d,\delta)>0$ such that is $f$ is a polynomial of degree $d$, and
$$|f(t)|\leq 1,\quad t\in[0,1],$$ then $$|f(t)|<1+\delta,\quad t\in[0,1+\varepsilon].$$
 (This is an exercise from "Introduction to Ratner Theorems..." by D.W. Morris). It is immediate for $d=0,1,2$, but I am probably lacking some trick to approach the general case. Any help is greatly appreciated.</p>


问:
Probability of Getting 3 aces in 13 cards
There are 52 cards in a poker card set. If the cards are being given to 4 players, W, X, Y, and Z. Each of them will get 13 cards. What is the probability of a player Y getting exactly 3 aces?

Initially, I get the probability using
$$
P=\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}=\frac{1}{5525}
$$

I tried using the other way

$$
P =\frac{{{13}\choose{1}} {{12}\choose{10}}{{4}\choose{3}} {{4}\choose{1}}^{10}}{{52}\choose{13}} = 5.667 \times10^{-3}
$$

Can someone please explain to me why these 2 ways have different probabilities and which way is the most direct way to tackle this kind of situation?

答:
Player $Y$ can receive $13$ cards in
$$\binom{52}{13}$$
ways.  For player $Y$ to receive exactly three aces, player $Y$ must receives three of the four aces and $10$ of the remaining $48$ cards in the deck.  Hence, the number of favorable cases is
$$\binom{4}{3}\binom{48}{10}$$
Therefore, the probability that player $Y$ receives exactly three aces is
$$\frac{\dbinom{4}{3}\dbinom{48}{10}}{\dbinom{52}{13}}$$

You have calculated the probabilities of other events.

Your first approach is the probability that three aces are selected when three cards are selected from the deck.  Your second approach gives the probability that three cards of one rank are selected and one card each is drawn from ten of the other twelve ranks.  If you wanted to find the probability that three aces are selected and one card each is drawn from ten of the other twelve ranks, you would drop the factor of $13$ since there is only one way to choose the rank ace.


问:
Online statistics course using python for teaching
I am interested in data science and data analysis. I have been reading been reading on self-study but when it comes to statistics, I feel I need a course from an expert to help me grasp some of the statistics concepts and particularly when using python to analyse them. can someone point me to a statistics course on edx or coursera, or even anywhere, that uses python? I keep seeing R and know nothing about R during my search.</p>

答:
I think it's best (that's how I did it) to learn the concepts and then try to translate it to Python yourself. So then you would need two resources: a good resource to learn the statistical concepts step by step ánd a good resource to translate it to Python from scratch.

These are the two resources I advise respectively:

The youtube channel 'StatQuest' by Josh Starmer.

The site by Jason Brownlee on how to build algorithms from scratch (focused on statistical concepts for machine learning): https://machinelearningmastery.com/

You can find so many courses online covering Python and data science, so you can't expect a complete answer to your question (I also think your question belongs more to the cross validated page or something similar). I suggest you browse down the web and look at several options. I did this as well and these two resources hereabove, gave me the highest increase in knowledge. So it doesn't mean these are the only two you need!


问:
Find $a$ and $b$ such that $|ax+by|+|bx+ay|=|x|+|y|$ holds for any $x,y$; Same problem with [...] instead of |...|.
Found these 2 problems in an old notebook of mine . Tried to solve them , but i'm stuck on both of them. They both require me to find a and b real numbers for which the equation is true no matter what value x and y take (x,y real numbers too)

1.) $|ax+by|+|bx+ay|=|x|+|y|$

2.) $[ax+by]+[bx+ay]=[x]+[y]$

Note : [] refers to the integer part .

答:
For the first problem.

If $y=x=1$ then   $|a+b|=1$.

If $y=-x=1$ then  $|a-b|=1$.

Case 1: $a+b=1$ $\iff$ $b=1-a$. Then

$$1=|a-b|=|a-(1-a)|=|2a-1| \iff \begin{cases}2a-1=1&\iff a=1 \\ 2a-1=-1 &\iff a=0\end{cases}.$$

Case 2: $a+b=-1$ $\iff$ $-b=1+a$. Then

$$1=|a-b|=|2a+1|\iff \begin{cases}2a+1=1&\iff a=0\\ 2a+1=-1&\iff a=-1.\end{cases}$$

Finally. For $x=0,\, y=1$ you'll get  $|a|+|b|=1$.

If $a=\pm1$, then $1+|b|=1$, whence $b=0$.

If $a=0$, then $|b|=1$ and $b=\pm1$.

Now you can CHECK by hand that $a=0, b=\pm1$ and vice versa are solutions to your problem.

So they are the only solutions.


问:
The Mittag-Leffler condition and $\varprojlim^1$
Recall that an inverse system of abelian groups $$\cdots \rightarrow G_2 \stackrel{\alpha_2}{\rightarrow} G_1 \stackrel{\alpha_1}{\rightarrow} G_0$$ is said to satisfy the Mittag-Leffler condition if, for each $i$, there exists a number $N$ such that the image of $G_{i+n} \rightarrow G_i$ is invariant for all $n \geq N$ (the map is obviously the composition $\alpha_{i+n} \circ \cdots \circ \alpha_{i+1}$).

I'm trying to show that, given a system that satisfies this condition, $\varprojlim^1G_i = 0$. Recall that $\varprojlim^1 G_i$ is defined to be the cokernel of the map $$\delta \colon \prod_i G_i \rightarrow \prod_i G_i$$ given by $(\ldots, g_i, \ldots) \mapsto (\ldots, g_i - \alpha_{i+1}(g_{i+1}), \ldots)$.

I've spent a half-hour trying to show that $\delta$ is surjective, but I haven't gotten very far with it. I don't actually need this result right now - all the morphisms in the inverse system i'm consider are surjective, and it's trivial to show that $\varprojlim^1$ dies in this case - so I'm just going to leave it and move on for now, but does anybody have a proof they could share? I feel as if it only needs a little thought brought to bear on it, but thought is a rare commodity on Friday afternoons... $\ddot \smile$

答:
I don't know if there is a direct proof, but one can prove things indirectly as follows:

If one has a short exact sequence $0\to \mathcal{A}\to \mathcal{B}\to \mathcal{C}\to 0$ of inverse sequences, then the snake lemma yields an exact sequence

$$ 0 \to {\displaystyle\lim_{\longleftarrow}}\mathcal{A} \to {\displaystyle\lim_{\longleftarrow}}\mathcal{B} \to{\displaystyle\lim_{\longleftarrow}}\mathcal{C} \to{\displaystyle\lim_{\longleftarrow}}^1\mathcal{A} \to{\displaystyle\lim_{\longleftarrow}}^1\mathcal{B} \to{\displaystyle\lim_{\longleftarrow}}^1\mathcal{C} \to 0$$

You know that ${\displaystyle\lim_{\longleftarrow}}^1\mathcal{A}$ vanishes when the maps are surjective.  The next step is to show that it vanishes when $\mathcal{A}$ satisfies the trivial ML-condition, that is, when the images are eventually all $0$.  Finally, if $B_i$ is the eventual image inside of $A_i$, we can look at the short exact sequence

$$0\to \mathcal{B}\to \mathcal{A}\to \mathcal{A}/\mathcal{B}\to 0$$

The first term has surjective maps and the last satisfies trivial ML.


问:
Minimum number of distances determined by $n$ points on a plane
Set $S$ is the set of the distances between any two dots in a panel. Prove that given $202$ dots in a panel, the set will have at least $11$ members.

I tried using pigeonhole principle. Every three dots can form a equilateral triangle but adding a fourth dot will definitely produce another distance. But I don't know how to proceed.

答:
This concerns a famous result of P.Erdos, later generalized by others. We in fact have :

If $f(n)$ denotes the minimum number of distances determined by $n$ points on a plane, then $$
\sqrt{n - \frac 34} - \frac12 \leq f(n)
$$

Proof of the lower bound : The $n$ points can be enclosed in a convex shape $P$, called the convex hull of the points. This can be proved using linear algebra, for example, and can be shown in this case to be a convex polygon which has vertices made up from these points.

Let $P_1$ be any one of the points lying on $P$ as well. Consider the distances given by $P_1P_i, i=2,3,...,n$, and let $K = |\{|P_1P_i|, i=2,3,...,n\}|$.

If $N$ is the maximum number of times a distance occurs, we must have $KN \geq n-1$, obviously i.e. $f(n) \geq \frac{n-1}{N}$.

Furthermore, if a distance $r$ occurs $N$ times, then these points occuring $n$ times must lie in the same semicircle of radius $r$ around $P_1$, since these points lie on one side of $P_1$ because $P_1$ is on the convex hull.

It is easy to see that these points among themselves have at least $N-1$ different distances (draw a diagram and see this). Therefore $f(n) \geq N-1$.

Thus, for any $N$ we have $f(n) \geq \max\{N-1 , \frac{n-1}{N}\}$. It is easy to minimize this over all $n>N >0$ and obtain the expression given.

It is even easier to obtain a rough upper bound, but remarkably there are better lower bounds than the above. One of these is $Cn^{\frac 57}$ for a fixed constant $C>0$, which I believe is the best.


问:
If $E(Y\mid X) = a + bX$, what is $E(XY)$?
If $X$ and $Y$ are dependent random variables, and $E(Y\mid X) = a + bX$ for some constants $a$, $b$.

What is $E(XY)$?

Here is what I have tried:

$Y E(Y\mid X) = aY + bXY$

$\implies  E(YE(Y|X)) = aE(Y) + bE(XY) = a\mu_y + bE(XY)$

$\implies  E(XY) = (E(YE(Y|X)) - a\mu_y) / b$

But I have no idea how to deal with $E(YE(Y\mid X))$.

Very Thanks!

答:
Me neither. I recommend using the law of total expectation (essentially, multiply by $X$ and not by $Y$):

$$E(XY)=E(E(XY\vert X))=E(XE(Y\vert X))=E(aX+bX^2)=aE(X)+bE(X^2)$$


答: You should multiply by $X$ instead of $Y$. We have $XE(Y|X)=aX+bX^{2}$. But $XE(Y|X)=E(XY|X)$ so we get $E(XY|X)=aX+bX^{2}$. Taking expectation on both sides we get $EXY=aEX+bEX^{2}$.

Title: If $E(Y\mid X) = a + bX$, what is $E(XY)$?

Body:
If $X$ and $Y$ are dependent random variables, and $E(Y\mid X) = a + bX$ for some constants $a$, $b$.

What is $E(XY)$?

Here is what I have tried:

$Y E(Y\mid X) = aY + bXY$

$\implies  E(YE(Y|X)) = aE(Y) + bE(XY) = a\mu_y + bE(XY)$

$\implies  E(XY) = (E(YE(Y|X)) - a\mu_y) / b$

But I have no idea how to deal with $E(YE(Y\mid X))$.

Very Thanks!


问:
Lemma right inverse is also left inverse
I read about a Lemma that says, for any Group $\langle G, \circ\rangle$ with $\forall a \in G $ the right inverse is also the left inverse. Without explicitly being said that the group is an abelian group the lemma shows:

$a^{-1} \circ a = a^{-1} \circ a \mathbb{1} = a^{-1} \circ a \circ a^{-1} \circ (a^{-1})^{-1} = \mathbb{1}$

I don't fully understand the logic behind that. In the calculation it's already assumed that $a^{-1} \circ a = \mathbb{1}$ and therefor the right inverse is also the left inverse. I do trust my book but I would like to be able to understand how the calculation can be understood.

答: They are not assuming that $a^{-1}\circ a=1$, they are starting with the LHS, eventually solving it and proving to be equal to identity. First they postmultiply by identity, then write $1=a^{-1}\circ (a^{-1})^{-1}$, where $(a^{-1})^{-1}$ is right inverse of $a^{-1}$. Then they use associativity to multiply the inner ones to get $a\circ a^{-1}=1$, since $a^{-1}$ is right inverse. Then they multiply the outer elements in the same way.


问:
Minimization problem of $e^x$ and straight line.

Given the curve $C$: $y=e^{x}$ and a point $P$: $(x_0,y_0)$ with $y_0 > e^{x_{0}}$, a straight line $L$ of slope $k$ goes through the point $P$ and intersects $C$ at two points $A$ and $B$, try to find out: $(1).$  $k$ that minimizes the distance of $A$ and $B$, $(2).$ k that minimizes the area of triangle OAB ($O$ is the origin $(0,0)$).

Edit:

I've tried the Lagrange multiplier method and only solved part of the problem(the minimal triangle when P on yAxis, in this case, $k=y_0$). However, the steps are quite complicated (for high school students). I'd like to know if there is some elementary methods to approach this problem. Thanks!

Edit 2:

(add the image)

I took the following Steps:

as shown in the chart, points P,Q,A,B are colinear. then we got the constraits of variables a,b and q, which are: $g(a, b) = 0, g(a,q) = 0, g(b,q) = 0$.

the distance between A and B is given by $f(a,b) = sqrt((b-a)^2+(e^b-e^a)^2)$

the area of the triangle OAB is given by $f(a,b,q)=1/2 * q * (b-a)$

${\mathcal {L}}(a,b,\lambda) = f(a,b)+\lambda*g(a,b)$

solve equations: $\nabla _{a,b,\lambda }{\mathcal {L}}(a,b,\lambda )=0$

The last step was the part that is very complicated.

答:
From the constraint $$\dfrac{e^b-y_0}{b-x_0} = \dfrac{y_0-e^a}{x_0-a}$$ it is not difficult to solve it for $b$. This gives
$$b(a)=x_0+\frac{y_0 (x_0-a)}{e^a-y_0}-W\Big[\frac{x_0-a}{e^a-y_0}\exp\left(\frac{e^a x_0-a y_0}{e^a-y_0} \right)\Big]$$ The minimum of $$f(a) = \big[a-b(a)\big]^2+\big[e^a-e^{b(a)}\big]^2$$ will happen at a value of $a$ a bit larger than $\log(y)$.

We do not need optimization since the problem is now to find the root of $f'(a)=0$.

$$f'(a)=(a-b(a)) \left(1-b'(a)\right)+\left(e^a-e^{b(a)}\right) \left(e^a-e^{b(a)}   b'(a)\right)$$ The edit gives the formal expression of $b'(a)$.

Now, using a very poor starting point $a_0=3$, Newton iterates are
$$\left(
\begin{array}{cc}
n & a_n \\
 0 & 3.000000000 \\
 1 & 2.518474855 \\
 2 & 2.048112714 \\
 3 & 1.597475428 \\
 4 & 1.194726874 \\
 5 & 1.004001060 \\
 6 & 1.029256086 \\
 7 & 1.031630200 \\
 8 & 1.031646684
\end{array}
\right)$$

If we want a better starting point, compute $f'(a)$ for $a=k \log(y)$ with $k >1$ until $f'(a)> 0$. For the worked example, this would give
$$\left(
\begin{array}{cc}
k & f'(k \log(y)) \\
 1.1 & -324.676 \\
 1.2 & -55.3709 \\
 1.3 & -18.4489 \\
 1.4 & -5.89996 \\
 1.5 & +0.61317
\end{array}
\right)$$

So, our starting guess $a_0=1.5\log(2) \sim 1.03972$ which is more than decent. Now, Newton iterates will converge without any overshoot since $f'(a_0)$ is positive and $f'''(a)>0$ (by Darboux theorem).
$$\left(
\begin{array}{cc}
 n & a_n \\
 0 & 1.039720771 \\
 1 & 1.031465818 \\
 2 & 1.031646591 \\
 3 & 1.031646684
\end{array}
\right)$$

In short, the problem seems to be simple from a numerical point of view. Beside the root finder algorithm (using numerical derivatives), the only requirement is a routine for the computation of Lambert function (there are many which are available).

Edit

For the formal calculation of $b'(a)$ defining three intermediate functions of $a$
$$u=\frac{x-a}{e^a-y} \qquad \qquad v=\frac{e^a x-a y}{e^a-y}\qquad \qquad w=W(u\, e^v)$$ to make
$$b(a)=x+y\,u-w\implies b'(a)=y u'-w'$$
$$u'=\frac{e^a (a-x-1)+y}{\left(e^a-y\right)^2}\qquad \qquad v'=y\, u'\qquad \qquad w'=\frac w {1+w}\left(\frac {u'}u +v'\right)$$


问:
What is the set containing only integers congruent to 89 modulo 17?
What is the set containing only integers congruent to 89 modulo 17</p>

答: Since $89$ is congruent modulo $17$ to $4$, it is just $\{4,\ 4 \pm 17,\ 4\pm2\cdot17,\ 4\pm3\cdot17,\ldots\}$.</p>


问:
simplify this formula
I'm attempting to simplify this formula for when r is equal to
$$
\frac{119}{2\sin(θ/2)+2(π-θ)/2}
$$
the formula is:

$$A = \frac{\pi r^2}{2} - \frac{1}{2} (θ-\sinθ) r^2$$

I've tried but i'm confused particularly about the fact that you end up with a fraction in a fraction. Overall, my aim is to make θ the only variable so i can find the maximum by finding its derivative and making the equation equal 0.

Ps. sorry if my brackets are 100% correct :)

答:
To make things slightly less messy, let's write $\phi=\pi-\theta.$  Then $$r=\frac{119}{\sin\left(\frac{\pi-\phi}{2}\right)+\phi}=\frac{119}{\cos\frac\phi2+\phi}$$ and $$A=\frac{r^2}2(\phi+\sin \phi)$$
Now $$\frac{dr}{d\phi}=
\frac{-119}{\left(\cos\frac\phi2+\phi\right)^2}\left(1-\sin \frac\phi2\right)=r\frac{\sin\frac\phi2-1}{\phi+2\cos\frac\phi2}$$ abd by implicit differentiation, $$\begin{align}
\frac{dA}{d\phi}&=r(\phi+\sin\phi)\frac{dr}{d\phi}+\frac{r^2}{2}(1+\cos\phi)\\
&=r^2(\phi+\sin\phi)\frac{\sin\frac\phi2-1}{\phi+2\cos\frac\phi2}+\frac{r^2}{2}(1+\cos\phi)\\
\end{align}$$
Since $r^2\neq0$, $$\frac{dA}{d\phi=0}\iff(\phi+\sin\phi)\frac{\sin\frac\phi2-1}{\phi+2\cos\frac\phi2}+\frac12(1+\cos\phi)=0$$ and clearing denominators, this last is equivalent to
$$(\phi+\sin\phi)\left(\sin\frac\phi2-1\right)+\frac12(1+\cos\phi)\left(\phi+2\cos\frac\phi2\right)=0$$

WolframAlpha finds many solutions to this equation, including one at $\phi=\pi,$ or $\theta=0,$  You haven't given any bounds on $\theta$ in the question, so I don't know how to carry on from here.

In the above, I have ignored the possibility that the derivatives don't exist, which will happen if $\phi+\cos\frac\phi2=0$.  WolframAlpha finds one solution to this equation at $$\phi\approx-1.47817$$ or $\theta=\pi+1.47817$.  If this is in the range you are interested in, you also have to consider the value of $A$ at this point.  Also, if $\theta$ is restricted to a finite interval, don't forget to check the endpoints.

Title: simplify this formula


答:
Maybe it's clearer if we use other names instead of $a^{-1}$.

Assume that for $a$ we have $b$ and $c$ such that $b\circ a=1$ and $a\circ c=1$. Then $b=b\circ1= b\circ (a\circ c)=(b\circ a)\circ c=1\circ c=c$, so $b=c$ and we can call them $a^{-1}$

Title: Lemma right inverse is also left inverse

Body:
I read about a Lemma that says, for any Group $\langle G, \circ\rangle$ with $\forall a \in G $ the right inverse is also the left inverse. Without explicitly being said that the group is an abelian group the lemma shows:

$a^{-1} \circ a = a^{-1} \circ a \mathbb{1} = a^{-1} \circ a \circ a^{-1} \circ (a^{-1})^{-1} = \mathbb{1}$

I don't fully understand the logic behind that. In the calculation it's already assumed that $a^{-1} \circ a = \mathbb{1}$ and therefor the right inverse is also the left inverse. I do trust my book but I would like to be able to understand how the calculation can be understood.


问:
Clearify I understood the question (best aproximation)
Define the $p$ norm for $1<p<\infty$ $$
\|x\|_{p}=\left(\sum_{i=1}^{n}\left|x_{i}\right|^{p}\right)^{1 / p}
$$

Find all the best approximations of $(1, 1)$ by vectors in span {$(1, −1)$}with $||\cdot||_p$ on $\mathbb {C^2}$

First of all why they say approximations? I thought approximation is a unique value.
second do I need to solve this equation :$$<(1,1) - \alpha (1,-1),(1,1)>=0$$
$$<(1,1),(1,1)> - \alpha <(1,1),(1,-1)>=0$$
and from here just open it according to the definition of $||\cdot||_p$?

答:
Let
$$e= (1,1)- \alpha(1,-1)$$
be the difference or error between the vector $(1,1)$ with a vector $\alpha(1,-1)\in \mbox{Span} \{(1,-1)\}$. For best approximations the length of the error should be minimimum. That is you have to find $\alpha$ such that
$$\|e\|_p = \|(1,1)-\alpha(1,-1)\|_p=\bigg((1-\alpha)^p +(1+\alpha)^p\bigg)^{\frac{1}{p}},$$ is minimum.


问:
Expand: $e^{\sin z}$ at $z = 0$
I need hints to find the Laurent expansion of $\displaystyle e^{\sin z}$ at $z = 0$ in a simpler way. I am getting double series which I can't simplify.

ADDED:: Can it be simpler that the below? I also need to find the radius of convergence.
$$\Large e^{\sin z} = e^{\sum_{k = 0}^\infty \frac{(-1)^nz^{2n+1}}{(2n+1)!}} = \sum_{n = 0}^\infty \frac{\small{B(1, -1/3!, \dots , (-1)^n/(2n+1)!)} z^n}{n!}$$

答:
Exponential formula.

Supose
$$ f(x)=a_1 x+{a_2 \over 2}x^2+{a_3 \over 6}x^3+\cdots+{a_n \over n!}x^n+\cdots.$$

Then

$$\exp f(x)=e^{f(x)}=\sum_{n=0}^\infty {b_n \over n!}x^n,$$

where

$$b_n=\sum_{\pi=\left\{\,S_1,\,\dots,\,S_k\,\right\}} a_{\left|S_1\right|}\cdots a_{\left|S_k\right|},$$

and the index $\pi$ runs through the list of all partitions $\{S_1,\ldots,S_k\}$ of the set $\{1,\ldots,n\}$. (When $k=0$ the product is empty and so equals $1$.)

We have $a_1=-a_3=a_5=-a_7=\cdots=1$ and $a_2=a_4=a_6=\cdots=0$.

So suppose we want $b_4$.  There are $15$ partitions of $\{1,2,3,4\}$: one corresponding to $1+1+1+1$, six corresponding to $2+1+1$, three corresponding to $2+2$, four corresponding to $3+1$, and one corresponding to $4$.  In this case we can discard the ones containing even numbers, so we need only consider $1+1+1+1$ and $3+1$.  We get
$$
b_4 = a_1^4 + 4a_3 a_1 = 1 - 4 = -3.
$$
Similarly,
$$
\begin{align}
b_0 & & & =1 \\[6pt]
b_1 & = a_1 & & = 1 \\[6pt]
b_2 & = a_1^2 & & = 1 \\[6pt]
b_3 & = a_1^3 + a_3 & & = 1-1 = 0  \\[6pt]
b_4 & = a_1^4 + 4a_3 a_1 & & = 1 - 4 = -3 \\[6pt]
b_5 & = a_1^5 + 10a_3 a_1^2 + a_5 & & = 1 - 10 + 1 = -8 \\[6pt]
b_6 & = a_1^6 + 20 a_3 a_1^3 + 10 a_3^2 + 6a_5 a_1 & & = 6 - 20 + 10 + 1 = -3 \\[6pt]
b_7 & = a_1^7 + 35 a_3 a_1^4 + 70 a_3^2 a_1 + 21a_5 a_1^2 + a_7 & & = 1 - 35 + 70 + 21 - 1 = 56
\end{align}
$$
And so on.

The radius of convergence is $\infty$, because this is a composition of two entire functions, and is therefore an entire function.

Title: Expand: $e^{\sin z}$ at $z = 0$


问:
Is this quadratic function monotonically increasing?

Given vector $x'$ and positive definite matrix $W$, let $$ f(x) := (x-x')^TW(x-x') $$ Show that $f$ is monotonically increasing, i.e. if $y-x' > x-x'$ component-wise, then $f(y) > f(x)$.

I do not know how to approach this. I do not know how to use the positive definiteness of $W$.

答:
This is not true if $y \geq x$ is taken to be a componentwise condition -- at least not without further restrictions
A simple counterexample is with $x' = 0$,
$$
W = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, \\~\\
x = (-1, -1), \\~\\
y = (0, 0),
$$
then $y \geq x$ and $x^T W x = 2$ but $y^T W y = 0$.

Or in case you are worried about the use of vectors with negative entries in the previous example, here is one in which $x, y \geq 0$. Take $ W = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$, $x = (1, 0)$ and $y = (4/3, 1)$.


问:
If there a formal definition for the sum of two irrationals is rational if it is only the additive inverse plus some rational?
If $x$ and $y$ are two irrational numbers and $x + y$ is a rational number, I know one of them has to be the additive inverse with/without some rational term I believe. I know it is not always the case that the sum of two irrational numbers is irrational.

for example:

$(\sqrt2 -1) + (-\sqrt2 +4) = 3$

But I can only seem to come up with examples that are of this nature and that seems to be the general consensus. But is there a formal definition stating it has to be of this form or is there a counterexample I am ignorantly unaware of.

then if $x+y\in\mathbb{Q}$ would it be true that $x-y$ would always just produce a multiple of that irrational plus some rational implying that if $x$ and $y$ are irrational and $x+y\in\mathbb{Q}$ then $x-y\not\in\mathbb{Q}$

答:
Yes, but it is trivial.  Say $x,y$ are two rational numbers, but $x+y = r$ is rational.  Then the sum is of the form
$$
x + y = (x+0) + (-x+r) = r
$$


问:
Finding conditions when an initial value problem has unique solution
For each $\alpha>0$ find all pairs $(x_0,y_0)\in\mathbb R^2$ such that the following IVP has a unique solution in a neighbourhood of $(x_0,y_0)$
$$y'=y^\alpha,\  y(x_0)=y_0$$
I know about the Picard-Lindelof theorem which asserts uniqueness but it is only a sufficient condition. The question ask to find all such pairs.

答:
The function $f(y):=y^\alpha$ is not locally Lipschitz in $y=0$ for $\alpha \in (0,1)$. If we choose $\alpha$ in this interval and $y_0\neq 0$ then we can apply the Picard-Lindelöf theorem.
The same theorem applies when $\alpha \geq 1$ since $y^\alpha$ is locally Lipschitz for all $\alpha \geq 1$ in all $y\in \mathbb{R}$.

When we lose the Lipschitz continuity we can construct $2$ solutions to prove non-uniqueness. For example if $\alpha=1/2$ and $y(1)=0$ we have the solutions $y\equiv 0$ and $y=\frac{(x-1)^2}{4}$.   We can easily generalize this example to all $\alpha \in (0,1)$ and $y(x_0)=0$.


问:
Proving that a group is closed under multiplication by scalars
I need to prove that the group $U = \{A \in M_n(\mathbb{R})\mid AB=BA\}$ is closed under multiplication by scalars.

So I let $\alpha \in \mathbb{R}$ and let $A \in U$, which means I need to prove that $\alpha A \in U$.

$(\alpha A)B = \alpha(AB) = \alpha(BA)$

But that does not prove what I need, I need to prove that $B(\alpha A) \in U$.

Where do I go from here?

答: It seems that the $U$ is a subspace of $V=M_{n\times n}(\mathbb R)$. So you are on a right path. In fact, if $A,C\in U$ then $$(aA+bC)=(aA)B+(bC)B=a(AB)+b(CB)=...=B(aA)+B(bC)=B(aA+bC)$$. Note that I amasuuming $B$ is a certain fixed matrix in vector space $V$</p>


答:
You would not look for a formal definition but for a theorem, and also let's not talk about consensus. (Namely, an argument in math is either right or wrong, regardless on how many people are "for" or "against" it.)

So if $\alpha, \beta$ are irrational and $\alpha+\beta=q$ - rational, then $\beta=-\alpha+q$ - i.e. indeed $\beta$ is the inverse of $\alpha$ plus a rational constant, as you have claimed. Then $\alpha-\beta=2\alpha-q$, which must then be irrational. (Because, otherwise you would add $q$ and conclude that $2\alpha$ would be rational, and then you would halve it and conclude that $\alpha$ would be rational - contradiction!)

Title: If there a formal definition for the sum of two irrationals is rational if it is only the additive inverse plus some rational?

Body:
If $x$ and $y$ are two irrational numbers and $x + y$ is a rational number, I know one of them has to be the additive inverse with/without some rational term I believe. I know it is not always the case that the sum of two irrational numbers is irrational.

for example:

$(\sqrt2 -1) + (-\sqrt2 +4) = 3$

But I can only seem to come up with examples that are of this nature and that seems to be the general consensus. But is there a formal definition stating it has to be of this form or is there a counterexample I am ignorantly unaware of.

then if $x+y\in\mathbb{Q}$ would it be true that $x-y$ would always just produce a multiple of that irrational plus some rational implying that if $x$ and $y$ are irrational and $x+y\in\mathbb{Q}$ then $x-y\not\in\mathbb{Q}$


问:
Number of toss to reject it is a fair coin
You toss the coin n times, and you have observed 60% of times, it is head.

How large n needs to be in order to achieve 95% confidence that it is not a fair coin?

=======

Attempt: Basically use Bionmial distribution, and but I have no idea how to take account of the 60% number into my calculation.

答:
The question implies that you are to use the number of heads tossed as the test statistic for a null-hypothesis significance test of the hypothesis $\ H_0\ $ that the coin being tossed is fair, and find the smallest value of $\ n\ $ for which $\ 0.6n\ $ lies in a critical region of significance $5\%$ (or, equivalently, outside the complementary confidence interval of $95\%$).

Unless you have a strong reason to believe that the coin could only be biassed towards heads (which is not suggested by anything in the wording of the question), you should be using a two-tailed test, for which the confidence interval would have the form $\ \left[\lfloor0.4n\rfloor+1, \lceil0.6n\rceil-1\right]\ $.  So if $\ N_h\ $ is the number of heads tossed, you have to find the smallest value of $\ n\ $ such that
\begin{align}
P\left(N_h\in \left[\lfloor0.4n\rfloor+1, \lceil0.6n\rceil-1\right]\big|H_0\right)&=\frac{1}{2^n}\sum_{k=\lfloor0.4n\rfloor+1}^{\lceil0.6n\rceil-1}{n\choose k}\\
&\ge0.95
\end{align}
Under the null hypthesis, the mean of $\ N_h\ $ is $\ \frac{n}{2}\ $, and its variance is $\ \frac{n}{4}\ $, so $\ 0.6n\ $ is $\ 0.2\sqrt{n}\ $ standard deviations above the mean, and for large enough $\ n\ $ the probability that $\ N_h\ $ lies in the given interval can be well approximated by $\ \mathcal{N}(0,1)\big(0.2\sqrt{n}\big)-$$\mathcal{N}\big(0,1\big)(-0.2\sqrt{n})\ $, or, equivalently, by $\ 2\mathcal{N}(0,1)\big(0.2\sqrt{n}\big)-1\ $. So if $\ u\ $ is the (unique) positive number such that $\ \mathcal{N}(0,1)(u)=0.975\ $, then $\ n=\left(\frac{u}{0.2}\right)^2\ $ will be a reasonable approximation for the value you require.


问:
Solution (non-zero) for $\frac{dy}{dx} = |y|^{\alpha}, y(0) = 0$
I am having difficulty finding an approach for the following problem:

Problem: "Take an arbitrary $\alpha \in (0,1)$.  Construct a non-zero solution for

$$\frac{dy}{dx} = |y|^{\alpha}, y(0) = 0$$

and then show that there are infinitely many solutions.

Hint: First find a solution in the following form:

\begin{equation*}
    y(x) =
    \begin{cases}
    0, \quad &\text{if } x \leq 0\\
Cx^\beta &\text{otherwise}
    \end{cases}
\end{equation*}

with C and $\beta > 0$ to be determined."

My attempt: I solved the ODE for $y > 0$ and then $y < 0$, but didn't know what to do from there.

\begin{align}
    &\text{When } y > 0 \Rightarrow y(x) = ((1 - \alpha)x)^{-(1-\alpha)}\\
    &\text{When } y < 0 \Rightarrow y(x) = -((1 - \alpha)x)^{-(1-\alpha)}
\end{align}

But I don't know where to go from here, or if this is the right first step at all. Any help would be appreciated. Thank you!

答:
If you plug $y=C x^{\beta}$ in the ODE we obtain conditions on $C$ and $\beta$.
We get
$$
C\beta x^{\beta-1} = C^\alpha x^{\alpha \beta}.
$$
Hence we need $\beta-1=\alpha \beta$ and $C=C^\alpha$, i.e. $C\equiv 1$ (otherwise we get the trivial solution) and $\beta=\frac{1}{1-\alpha}\in (1,\infty)$.


答:
For the second problem.

For $y=0$ we have $[ax]+[bx]=[x]$.

For $x=y$ we have $[ax+bx]+[bx+ax] = [x]+[x]$, so $[(a+b)x]=[x]$.

In particular, for $x=1$ we have $[a+b] = 1$.

If $a+b \notin \mathbb Z$, write $a+b = [a+b] + \epsilon$.

Then for $N\in\mathbb N$ and $N > \frac 1 \epsilon$, $[N]=[(a+b)N] = [[a+b]N+\epsilon N]\ge N + 1$, a contradiction.

This gives $a+b=1$.

We have $[a]+[b]=[1]=1=a+b$. Since $[a]\ge a$ and equality hold iff $a\in\mathbb Z$, we have $a,b \in\mathbb Z$.

WLOG suppose $a>b$ and $a \ne 1$. Then $a > 1$.

We once again have $[ax]+[(1-a)x] =[x]$.

Take $x = \frac1{2a}$. Then $RHS =0$. $LHS=[\frac12]+[\frac{1-a}{2a}]=0+(-1) = -1$, which is a contradiction.

Hence $a=1, b=0$. Without the WLOG, $a=0, b=1$ works as well.

Title: Find $a$ and $b$ such that $|ax+by|+|bx+ay|=|x|+|y|$ holds for any $x,y$; Same problem with [...] instead of |...|.

Body:
Found these 2 problems in an old notebook of mine . Tried to solve them , but i'm stuck on both of them. They both require me to find a and b real numbers for which the equation is true no matter what value x and y take (x,y real numbers too)

1.) $|ax+by|+|bx+ay|=|x|+|y|$

2.) $[ax+by]+[bx+ay]=[x]+[y]$

Note : [] refers to the integer part .


问:
Chain Rule for multivariable functions to one variable functions
I'm trying to understand some questions about the chain rule of multivariable and how it relates with one variable functions.

Ok, so, imagine that I have two functions $f: U \subset E \rightarrow \mathbb{R}$ and $\alpha: (-\epsilon, \epsilon) \subset \mathbb{R} \rightarrow E$, where $E$ is a normed vector space.

Then, by the chain rule, the derivative of $f \circ \alpha$ in some point $a \in U$ is:

$(f \circ \alpha)'(a) = Df_{\alpha(a)} \cdot D\alpha_a$.

The point that I don't get is that $(f \circ \alpha)'(a)$ is a number, but $Df_{\alpha(a)} \cdot D\alpha_a$. is a linear application. So, I'm a little bit confused, can the number $(f \circ \alpha)'(a)$ be interpreted in some way by a linear application?

Thanks!

答:
The derivative of a function $f:\mathbb R\to\mathbb R$ at $x\in\mathbb R$ is a number $c\in\mathbb R$ characterized by the fact that
$$f(x+h)=f(x)+ch+o(h^2).$$
The term $ch$ is essentially a linear map $h\mapsto ch$, which is entirely characterized by the number $c$, which is the derivative of $f$. That's how the derivative of a 1d function should be interpreted as a linear map. More conceptually, the tangent function $t(x+h)=f(x)+f'(x)h$ is a linear map (affine linear, to be exact) as a function of $h$, determined by the derivative. The actually linear part $f'(x)h$ is the linear map we're talking about.

For higher dimensions, this works exactly the same: If $\mathrm Df_x$ is the derivative of $f$ in $x$, then
$$f(x+h)=f(x)+\mathrm Df_x(h)+o(\vert h\vert^2).$$
Just like in the 1d case.


问:
Is every subset of a product a product of subsets?
Is every subset of a product a product of subsets ?

i.e Let $E$ and $F$ two non empty sets and we define the Cartesian product $E \times F$.

Now given a non empty subset $A$ of $E\times F$, can we write $A$ as the product of two subsets of $E$ and $F$: i.e is there $E_1 \subset E$ and $F_1 \subset F$ such that
$$A=E_1 \times F_1$$

My idea is that this statement is false, and some counter-example I thought of is
$$\{(x,y) \in \mathbb{R}^2, \,\, x^2+y^2=1\}$$
$$\{(x,1/x), \,\, x\in \mathbb{R}^*\}$$
But I could not find a way to prove we can't write these two sets as product of two subsets of $\mathbb{R}.$

答:
Assuming the product is $A\times A$, the subset of $A\times A$, $B$ is a product of subset of $A$ if $B$ is some form of "rectangle". That is to say, for any $x,y,d,e\in A$, $(d,x)\in B$ and $(e,y)\in B$ implies $(d,y)\in B$ and $(e,x)\in B$.

If the condition above is supposed, then for any pair of elements $C=(x,y)\times (d,e)$, always have the conclusion: $C$ is a subset of $A\times A$ and $C$ is a cartesian product. Also, let $x,y,d,e$ go through all the elements of $B$, so that is proved.

It should also be correct for cartesian of 2 different sets.


问:
Finding delta for epsilon-delta definition of a piece wise function
I am trying to find the necessary $\delta$ that satisfies the $\epsilon - \delta$ definition of continuity of the following function at $x = 0$:

$$f(x) = \begin{cases} x & \text{ for } x \ge 0,\\
x^2 & \text{ for } x < 0.\end{cases}$$

My work:

right limit:

$\lim\limits_{x \to 0+}x$ $\implies \forall \epsilon$ $\exists \delta $  ($|x|<\delta$ $\implies$ $|f(x)|< \epsilon)$

clearly we can let $\delta = \epsilon$ as $f(x) = x$

For the left limit:

$\lim\limits_{x \to 0-}x^2$ $\implies \forall \epsilon$ $\exists \delta $  ($|x|<\delta$ $\implies$ $|f(x)|< \epsilon)$

here we let $\delta = \sqrt\epsilon$ as $f(x) = x^2$

My problem is that which $\delta$ do I choose?

答:
Once you've found one value $\delta_1 > 0$ and used it to prove a statement of the form

$|x| < \delta_1 \implies P(x)$

then you can replace $\delta_1$ by any value $\delta$ such that $0 < \delta \le \delta_1$ and it will still be true that

$|x| < \delta \implies P(x)$

Using that, suppose you have two values $\delta_1 > 0$ and $\delta_2 > 0$, and you have proved two statements like this:

$|x| < \delta_1 \implies P(x)$

$|x| < \delta_2 \implies Q(x)$

It follows that you can replace each of $\delta_1$ and $\delta_2$ by the same value $\delta = \min\{\delta_1,\delta_2\}$, and then you'll know its true that

$|x| < \delta \implies P(x) \quad\text{and}\quad Q(x)$

In your problem, you have $\delta_1 = \epsilon$ and $\delta_2 = \sqrt{\epsilon}$, and so you use $\delta = \min\{\epsilon,\sqrt{\epsilon}\}$.


问:
Probability in Minesweeper
Suppose I click a random tile during a Minesweeper game. It is a 1. During each time I click an adjacent square, what are the chances of hitting a mine? How would this change if it were a 2 or another number?</p>

答:
In the absence of any other information (which is rarely the case in Minesweeper), clicking a $1$ implies a $\frac{1}{8}$ chance that any of the adjacent $8$ squares is a mine. Now suppose you click an adjacent cell and it reveals a $1$. At this point, you have something like:

$$
\begin{array}
& \star & \star & \star & \star \\
\star & 1 & 1 & \star\\
\star & \star & \star & \star
\end{array}
$$

Suppose that mines are distributed uniformly at random. There are $10$ unknown squares. Suppose further that the number of mines and number of cells are both much larger than $10$. Let's start counting!

First, keep in mind that there are $2^{10}$ possible mine placements in the cells marked with stars (each is either a mine or not a mine).

But there are much fewer mine placements among the $10$ cells that result in $1\ 1$ being displayed on the two middle cells:

There are $4$ one-mine configurations (corresponding to placing a mine in one of the four centermost cells).

There are $9$ two-mine configurations (corresponding to placing one mine in the left-most column, and independently placing one mine in the rightmost column).

We have enumerated all $13$ configurations.

We can tabulate the mine probabilities per-square as follows:
$$
\begin{array}
& \frac{3}{13} & \frac{1}{13} & \frac{1}{13} & \frac{3}{13} \\
\frac{3}{13} & \star & \star & \frac{3}{13}\\
\frac{3}{13} & \frac{1}{13} & \frac{1}{13} & \frac{3}{13}
\end{array}
$$

In addition, by summing up these probabilities we can conclude that the expected total number of mines among the $10$ squares is $\frac{22}{13}$.

This is the kind of analysis used in probabilistic Minesweeper solvers. In general, many configurations must be analyzed separately so your original question is a little bit ill-posed.

Title: Probability in Minesweeper


问:
linear functional definition
I am unable to understand
what is meant by 'unique'  in the theorem on linear functionals given in Axler's  Linear Algebra Done Right, edition 3.

Theorem 6.45 Suppose $\phi$ is a linear functional on vector space $V$. Then there is a unique vector $v$ in $V$ such that
$\phi(u)= \langle u,v\rangle$ for every $u$ in $V$.

Does it mean  a fixed vector $v$ or a special ( unique)  vector $v$ ?
Taking functional as  product of a row vector $(v)$ and a column vector $(u)$  of same dimension , one can take any fixed $v$ and get a scalar that is linear with respect to $u$. Is it correct ?

答: For a given functional $\phi$, there is only one such vector $v$.  Another way of putting it is that all linear functionals are of the form $$\phi(u_1,u_2,\dots,u_n)=v_1u_1+v_2u_2+\cdots+v_nu_n$$ for some constants $v_1,v_2,\dots,v_n.$  Then $v=(v_1,v_2,\dots,v_n).$

Title: linear functional definition


问:
Find the volume of the region D in spherical coordinate
Question: Use spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere $x^2+y^2+z^2=4$  but outside the cylinder $x^2+y^2=1$

So question is pretty similar to my previous with a slight variation of the region. This is my answer:

$$\int _0^{2\pi }\int _{\frac{5\pi }{6}}^{\frac{\pi }{6}}\int _1^{\cos \phi \ }\rho \ ^2\sin \phi \ d\rho \, d\phi \ d\theta \ \ $$

Can anyone please please correct me? If my answer is correct then I'll remove this post, otherwise if there is any mistake, I'd be more than happy if someone can correct me!
Thanks in advance

答:
Its almost correct! The only problem is the bounds for $\rho$. Since you want to describe
$\rho$ in terms of the angle $\phi$ you can proceed as follows:

Fix two angles $\theta\in [0,2\pi),\ $$\phi\in [\pi/6,5\pi/6]$. Then for this two fixed angles take the unique point $(\rho,\theta,\phi)\equiv (x,y,z)$ which lies in the cylinder $x^2+y^2=1$. You want to calculate the radius $\rho$ of this point. Using the relations of $(x,y,z)$ in terms of $(\rho,\theta,\phi)$ and the equation $x^2+y^2=1$ we get
\begin{align}
(\rho \cos\theta \sin\phi)^2&+(\rho \sin\theta \sin \phi)^2=1\\
&\implies \rho^2\sin^2\phi=1\\
&\implies \rho=\frac{1}{|sin\phi|}
\end{align}
and since $\phi\in [\pi/6,5\pi/6]$ we eventually end up with $\rho=1/\sin\phi=\csc \phi$. Now , from that point and since the cylinder is inside the sphere of radius $2$ you move till you hit the sphere (for fixed $\theta,\phi$). This will happen when $\rho$ becomes $2$. So, the bounds for $\rho$ must be $\csc \phi\leq \rho\leq 2$.

Hence, the volume is
$$\int_{0}^{2\pi}\int_{\pi/6}^{5\pi/6}\int_{\csc\phi}^{2}\rho^2\sin\phi\, d\rho d\phi d\theta$$


问:
Where can I learn more about commutative hyperoperations?
I just learned about commutative hyperoperations, and they look interesting. However, the wikipedia page doesn't link to more information.

Is there an article or book where I can learn more? I'm especially interested in whether this is a "natural" sequence of definitions.

Note that the expression $k^{\log_k(a)+\log_k(b)}, k>0$ is independent of $k$, since it always equals $ab$. On the other hand, the expression $k^{\log_k(a)\log_k(b)}$ is dependent on $k$, and this makes me wonder whether there isn't a "better", more "natural" sequence out there.

答:
Try to check the chapter $1$ (from Pag $9$) of this  book (New Mathematical Objects-C.A.Rubtsov) :

Here the autor creates a kind of generalization of the Albert Bennet's Hyperoperations. But what he does is much more general. He creates a procedure that he calls $ω$-reflection using a function of connection $f$.

In the book the only function he uses is $f(x):=k^x$ where $k\gt 1$ is called in the book "factor of image".

Then he defines an infinite hierarchy of "reflexive binary operations" that are homomorphic via $f$:

$x\circ_iy=x+y$ if $i=1$

$f(x)\circ_{i+1}f(y)=f(x\circ_iy)$

and we have that

$x\circ_{i+1}y=f(f^{\circ-1}(x)\circ_if^{\circ-1}(y))$

The first chapter the autor puts more attention on the homomorphic operation $\circ_{3}$ that is denoted by $\odot$ in the book and called "reflexive multiplication". Since he uses the exponentiation as function of connection, $\circ_{3}$ is isomophic to the mutiplication, then commutative and associative.

$k^x \odot k^y = k^{a \times b}$

In this chapter the autor builds an infinite numbers of what he calls reflexive functions, reflexive (homomorphic) binary operations , reflexive algebras (that are homomorphics via f) and other mathematical objects in this way:

let $F$ a bijection $F:\Bbb R \rightarrow \Bbb R$ (function of connection)

he calls $f'$ the reflexive image of the function of $f$ via $F$ if we have

$f'\circ F=F \circ f$

If you want to go deeper, on Tetration forum there is a topic where the possible propeties/evaluation of reflexive binary operations with non-integer indexes $i\le 2$ using tetration is discussed: Rational Operators

More informations:

$1$ The first autor is C.A.Rubtsov that with G.F.Romerio worked on the
Hyperoperations topic:

(Ackermann's function and new  arithematical operations-Rubtsov,
Romerio-2004)

$2$ About Albert Bennet here a pdf about his commutative Hyperoperations:

(Note on an Operation of the Thrid Grade- Albert A.Bennet )

if the link is broken try these

```<code>https://web.archive.org/web/20210715091833/http://www.geocities.ws/rubcov/english/09.htm

https://web.archive.org/web/20210715091855/http://www.geocities.ws/rubcov/english/10.htm
</code>```

Body:
Try to check the chapter $1$ (from Pag $9$) of this  book (New Mathematical Objects-C.A.Rubtsov) :

Here the autor creates a kind of generalization of the Albert Bennet's Hyperoperations. But what he does is much more general. He creates a procedure that he calls $ω$-reflection using a function of connection $f$.

In the book the only function he uses is $f(x):=k^x$ where $k\gt 1$ is called in the book "factor of image".

Then he defines an infinite hierarchy of "reflexive binary operations" that are homomorphic via $f$:

$x\circ_iy=x+y$ if $i=1$

$f(x)\circ_{i+1}f(y)=f(x\circ_iy)$

and we have that

$x\circ_{i+1}y=f(f^{\circ-1}(x)\circ_if^{\circ-1}(y))$

The first chapter the autor puts more attention on the homomorphic operation $\circ_{3}$ that is denoted by $\odot$ in the book and called "reflexive multiplication". Since he uses the exponentiation as function of connection, $\circ_{3}$ is isomophic to the mutiplication, then commutative and associative.

$k^x \odot k^y = k^{a \times b}$

In this chapter the autor builds an infinite numbers of what he calls reflexive functions, reflexive (homomorphic) binary operations , reflexive algebras (that are homomorphics via f) and other mathematical objects in this way:

let $F$ a bijection $F:\Bbb R \rightarrow \Bbb R$ (function of connection)

he calls $f'$ the reflexive image of the function of $f$ via $F$ if we have

$f'\circ F=F \circ f$

If you want to go deeper, on Tetration forum there is a topic where the possible propeties/evaluation of reflexive binary operations with non-integer indexes $i\le 2$ using tetration is discussed: Rational Operators

More informations:

$1$ The first autor is C.A.Rubtsov that with G.F.Romerio worked on the
Hyperoperations topic:

(Ackermann's function and new  arithematical operations-Rubtsov,
Romerio-2004)

$2$ About Albert Bennet here a pdf about his commutative Hyperoperations:

(Note on an Operation of the Thrid Grade- Albert A.Bennet )

if the link is broken try these

https://web.archive.org/web/20210715091833/http://www.geocities.ws/rubcov/english/09.htm

https://web.archive.org/web/20210715091855/http://www.geocities.ws/rubcov/english/10.htm


问:
Definition of the Second Order Partial Derivative from First Principles
I was recently introduced to the wave equation and in the derivation of said equation they used the definition of the second order partial derivative for the final result. The source I used for the definition basically said that the following equation:

is the equivalent of:

And mentioned that this is because the partial second order derivative of u w.r.t x is defined as such.

I presumed that this meant if derived from first principles (taking lim h->0) for the definied formula one gets the second order partial derivative. However, I could not find an online resource which derives such a formula. I have tried to attempt this myself, however, I do not understand where the h squared term comes from.

If someone knows of the derivation, if they could kindly direct me to it, that would be much appreciated. Thanks.

答:
I think the insight can be seen just by seeing single variable derivatives:

$$ f' = \lim_{h \to 0} \frac{ f(x+h) - f(x) }{h}$$

$$ f'' = \lim_{h \to 0} \frac{ f'(x+h) - f'(x) }{h}= \lim_{h \to 0} \frac{ \frac{ f(x+2h) - f(x+h) }{h} - \frac{ f(x+h) - f(x) }{h} }{h}$$

Or,

$$ f'' =  \lim_{h \to 0} \frac{ f(x+2h) + f(x) - 2 f(x+h)}{h^2}$$


问:
How do you write ${r \over (r+1)!}$ as a difference?
We are asked to evaluate the following sum

$$\sum_{r=1}^\infty {r\over (r+1)!}$$

I thought it best to write the expression out as a difference and use the method of differences to find the sum.

However, how would one go about converting it into a difference? I lucked upon the solution, as it turns out:

$$
{r\over (r+1)!} \equiv {1 \over r!} - {1\over (r+1)!}
$$

But it was simply a result of me playing about and was not rigorous at all. What would've been a sure-fire method to reach this? More generally how should one approach expression like these when looking to convert them into differences, similar to how converting some expressions like ${1\over 4r^2-1 }$ into partial fractions can ease the solution?

答:
There is a standard trick that involves adding and subtracting the same number (often $1$) to make things line up. In this case
$$
\frac{r}{(r+1)!}=\frac{r+1-1}{(r+1)!}\\
=\frac{r+1}{(r+1)!}-\frac{1}{(r+1)!}\\
=\frac{1}{r!}-\frac{1}{(r+1)!}
$$
How would one discover this? Well, we would really like to have $r+1$ in the numerator, because that would cancel very nicely with the denominator. So we make it so that we have $r+1$ in the numerator, by adding $1$, and then subtracting $1$. The rest is standard arithmetic manipulations.


问:
Find two independent series solutions
I'm trying to find two independent series solutions, expanded about x = 0, that satisfy:

$$ g''+2xg'+4g=0 $$

so far I have gotten the indicial equation and found $r=0$ and $r=1$. Then I substitute the derivatives by the respective sigma notation and found $a_n=\frac{-2}{n+r-1}$.

Now, if $r=0$ then $a_n=\frac{-2}{n-1}$.

And if $r=1$ then $a_n=\frac{-2}{n}$.

I tried from here to collect coefficients and find the respective power series but I do not understand how there 2 independent series solutions.

Also, on the problem it's given the general solution that I'm trying to approach which is:

$$g(x)=Axe^{-x^2}+B\Sigma^{\infty}_{n=0}\frac{(-4)^nn!}{(2n)!}x^{2n}$$ with A and B arbitrary constants.

Any suggestions?

Thanks in advance!

答:
If you plug $\sum_{n=0}^\infty x^n$ in your equation we get
$$
\sum_{n=2} a_n n (n-1)x^{n-2}+2\sum_{n=1}^\infty a_n n x^n +\sum_{n=0} 4 a_n x^n=0.
$$
The left-hand side can be written as a unique series
$$
\sum_{n=0}^\infty  \underbrace{[a_{n+2} (n+2)(n+1)+a_n(2n+4)}_{=:b_n} x^n=0.
$$
Hence we need all the coefficients $b_n =0$ and this gives a recurrent relation $f(a_n)=a_{n+2}$.
If we choose $a_1$ we find all the odd terms ($a_{2k+1}$) and if we choose $a_2$ we determine all the even terms $a_{2k}$.

You can look here to understand why these solutions are independent.

If you Taylor expand the general solution given by the text you get your 2 solutions (observe that $xe^{-x^2}$ has only odd terms in the expansion and the second one only the odd ones, hence $A,B$ corresponds to a rescale of $a_1,a_0$ respectively).

Title: Find two independent series solutions


问:
Is my proof correct? (minimal distance between compact sets)
I'm working out the following problem form Ahlfors' Complex Analysis text:

"Let $X$ and $Y$ be compact sets in a complete metric space $(S,d)$. Prove that there exist $x \in X,y \in Y$ such that $d(x,y)$ is a minimum."

My attempt:

Define $E:=\{d(x,y): x \in X,y \in Y \}$. We will prove that $E \subset \mathbb R$ is compact.

Firstly we will prove that $E$ is bounded:

$X$ is compact, and therefore it is bounded. That is, there exist $x_0 \in X,r_1>0$ such that $d(x,x_0)<r_1$ for all $x \in X$. $Y$ is also compact, and similarly there exist $y_0 \in Y,r_2>0$ such that $d(y,y_0)<r_2$ for all $y \in Y$.

Now, for any $d(x,y) \in E$, we have $$d(x,y) \leq d(x,x_0)+d(x_0,y_0)+d(y,y_0)<r_1+d(x_0,y_0)+r_2 =:M.$$ This proves that $E$ is bounded.

And now we will prove that $E$ is also closed:

Let $a_n=d(x_n,y_n)$ be sequence in $E$, which is convergent in $\mathbb R$. We will prove that its limit $a:=\lim_{n \to \infty} a_n$ is in $E$.

$(x_n)$ is a sequence in the compact set $X$, therefore it admits a convergent subsequence $(x_{n_k}) \to \bar{x}$. $(y_{n_k})$ is a sequence in the compact set $Y$, and therefore it admits a convergent subsequence $(y_{n_{k_l}}) \to \bar{y}$. The sequence $(x_{n_{k_l}},y_{n_{k_l}})_l$ converges to $(\bar{x},\bar{y})$ in the product space, and the continuity of the metric gives $a=\lim_{l \to \infty} a_{n_{k_l}}=\lim_{l \to \infty} d(x_{n_{k_l}},y_{n_{k_l}})=d(\bar{x},\bar{y})$. This shows that $a \in E$ as required.

In summary I have shown that $E$ is compact, and from a well-known theorem it has a minimum.

Is this proof OK? I think that the completeness of $S$ is unnecessary.

Thank you.

答:
The boundedness of $E$ isn't really the problem. We only need $E$ to be bounded below, which is the case : it is the set of the values of a positive function.

The main part is to prove that $E$ is closed. And the fact that a bounded closed subset of $\mathbb{R}$ contains its minimum isn't really a famous theorem : it's only the fact that the infimum of a set is a limit point of it, and therefore contained in it if the set is closed.


问:
Prove $<x, f'(x) \geq pf(x)$
Let f: A →ℝ, where A ⊆ ℝn is an open. We will say that f is homogeneous
of grade p over A if

f (tx) = t^(p).f(x)

for each t ∈ ℝ and x ∈ A for which    tx ∈ A. If such a function is differentiable
at x ∈ A, prove that

```<code><x, f'(x) > = pf(x)
</code>```

答:
I think there are many things you need to have more clarifications in your statement. However, in my opinion, the solution does not vary much.

Roughly speaking, we have:

$f(x+tx)= f(x)+ < \nabla f(x), tx>+ o( tx)$ when $t \longrightarrow 0$

which combines with homogenous fact given, we deduce:

$ \left[ \frac{(t+1)^p-1}{t} \right]f(x) = < \nabla f(x), x> +o(x)$

Hence, we're done.

Surely, this one is not a rigorous solution but i think the necessary elements are all presented.

Title: Prove $<x, f'(x) \geq pf(x)$


问:
decide for what $n \in \mathbb{N}: 3^n \equiv n \mod 13$ is valid.
It is not enough to try every option mod 13
because

3^1≡1 (mod 13) not true

3^2≡2 (mod 13) not true

3^3≡3 (mod 13) not true

3^4≡4 (mod 13) not true

3^5≡5 (mod 13) not true

3^6≡6 (mod 13) not true

3^7≡7 (mod 13) not true

3^8≡8 (mod 13) not true

3^9≡9 (mod 13) not true

3^10≡10 (mod 13) not true

3^11≡11 (mod 13) not true

3^12≡12 (mod 13) not true

3^13≡0 (mod 13) not true (a^(prime number)≡a(mod (prime number)

I dont know how to find those numbers, there is no repetition...

答:
Hint:

In the multiplicative group $(\mathbf Z/13\bf Z)^{\times}$, the element  $3$ has order $3$, hence $3^n\equiv3^{n\bmod 3}\mod 13$, and there are three cases

If $n\equiv 0\mod 3$, $3^n\equiv 1\mod 13$;

If $n\equiv 1\mod 3$,  $3^n\equiv 3\mod 13$;

If $n\equiv 2\mod 3$,  $3^n\equiv 9\mod 13$.

Therefore, we have to solve the systems of congruences
$$ \begin{cases}
n\equiv a\mod 3, \\n\equiv b\mod 13,
\end{cases}\qquad (a,b)\in\bigl\{(0,1),(1,3),(2,9)\bigr\}.$$
You can use  the inverse of the isomorphism of the Chinese remainder theorem:
\begin{align}
\mathbf Z/39\mathbf Z&\longrightarrow\mathbf Z/3\mathbf Z \times\mathbf Z/13\bf Z\\
n\bmod 39&\longmapsto (n\bmod 3, n\bmod 13)
\end{align}
This isomorphism uses a Bézout's relation: $3u+13v=1$ and is defined thus
\begin{align}
\mathbf Z/3\mathbf Z \times\mathbf Z/13\bf Z&\longrightarrow \mathbf Z/39\mathbf Z,\\
(a\bmod 3, b\bmod 13)&\longmapsto 3ub+3va\bmod 39.
\end{align}


问:
What is the $\lim _{n \to \infty} \sum_{k=0}^{n} \frac{\sqrt k }{n^{\frac32}}$?
What is $$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right)$$
Wolframalpha tells me its 2/3 but I don't understand why.

What I do know is that:
$$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right) < \lim _{n\to \infty }\left(\frac{n\sqrt{n}\:}{n^{\frac32}}\right) =1$$

答: The key-word is "Riemann sum" :$$\frac{\sum _{k=1}^n\sqrt{k}}{n^{1.5}}=\frac{1}{n}\sum _{k=1}^n\sqrt{\frac{k}{n}} \rightarrow_{n \rightarrow +\infty} \int_0^1 \sqrt{x} \ dx = \frac{2}{3}$$


问:
Use eigenvalues to find formula for recurrence relation
(1) Sequence $ \{a_n\} $ is defined as $ a_{0} = a_1 = a_2 = 1  $
and

$ a_n = 7a_{n-2} + 6a_{n-3}, for. \  n \geq 3$.

Find a general formula for $ a_n $

I have only solved recurrence relations for n >= 2 so far in my class, I am not sure as to how I should extend it to n >= 3?

Edit: P =

(2) $ C = \begin{pmatrix} 2 &2 & 0 \\ -1 & -1 & 0 \\ 2 & 2 & 3  \end{pmatrix}\ $. Diagonalise the matrix in 3 ways so that the diagonal matrices $ D_1, D_2, D_3 $ are different.
I found it's eigenvalues to be

$ \lambda_1 = 0, $ corresponding vector is $ (-1,1,0) $.

$ \lambda_2 = 1, $ corresponding vector is $ (-2,1,1) $

$ \lambda_3 = 3, $ corresponding vector is $ (0,0,1) $

Is it correct to say that $ D_1 =  \begin{pmatrix} 3 &0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0  \end{pmatrix}\ $. $ D_2 =  \begin{pmatrix} 1 &0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0  \end{pmatrix}\ $ $ D_1 =  \begin{pmatrix} 3 &0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1  \end{pmatrix}\ $?

答:
Hint for (1): We have
$$
\begin{pmatrix} a_n \\ a_{n-1} \\ a_{n-2} \end{pmatrix}
=
\begin{pmatrix} 0 &7 & 6 \\ 1 & 0 & 0 \\ 0 & 1 & 0  \end{pmatrix}
\begin{pmatrix} a_{n-1} \\ a_{n-2} \\ a_{n-3} \end{pmatrix}
$$
whose characteristic polynomial is $x^3 - 7 x - 6$ (no surprises here).

Finally, $x^3 - 7 x - 6=(x + 1) (x + 2) (x - 3)$.