问:
Constructing a set of numbers so that all 3 element partitions have different subset sums
I am interested in constructing a set of $S$ of $24$ positive integers so that any partition into $8$ $3$-element subsets, $S_1,\dots,S_8$ has the property that the sums
$$\sum_{s \in S_i} s$$
are different for every $i =1,2,\dots,8$. I would like to do this using the smallest possible maximum value of any element in the set, that is minimize $$\max_{s \in S} s.$$ I imagine that this has been studied and I simply do not know the right name for the problem.

Note that what I want is different from the "subset sum" problem which says ALL subsets of our 24 integers have different sums, although I suppose this answers the question with a (most likely) suboptimal construction.

More generally, I'm curious what is known about such sets -- for $n$ integer, how do we construct a set of $n$ positive integers such that every partition into $k$-element subsets has the property that the $k$ element subsets all have a different sum.

答:
I tried the greedy approach I suggested in a comment, and it produced a maximum of $13796$, as opposed to the value $17404$ in the OEIS table mentioned in the comments.

I wrote this little python script:

```<code>from itertools import combinations

xs = [1,2,3]
twos = {3,4,5}
threes = {6}
nextX = 4
while len(xs) < 24:
    if len(xs) != 23:
        newTwos = {nextX + x for x in xs}
        if newTwos & twos:
            nextX += 1
            continue
    newThrees = {nextX + t for t in twos}
    if newThrees & threes:
        nextX += 1
        continue
    twos |= newTwos
    threes |= newThrees
    xs.append(nextX)
    nextX += 1
print(xs)

# audit
test3 = {sum(t) for t in combinations(xs,3)}
if len(test3)== 24*23*22//6:
    print('Passed audit')
else:
    print('Failed audit')
</code>```

We start with $\{1,2,3\}$ and repeatedly try adding the smallest integer not yet tested to the set.  We keep track of all the $2$-sums and $3$-sums.  When we test a new element we accept it if its addition would not create a duplicate $2$-sum or $3$-sum.  The $2$-sum test is necessary because if there were a duplicate  $2$-sum, the addition of an further element would necessarily create a duplicate $3$-sum.  This test is unnecessary when we add the $24$th elements to the set.

The script produced the output

```<code>[1, 2, 3, 5, 8, 14, 25, 45, 82, 140, 235, 388, 559, 839, 1286, 1582, 2221, 3144, 4071, 5795, 6872, 9204, 11524, 13796]
Passed audit
</code>```

Of course, there's no guarantee that the greedy algorithm produces the optimal answer.  The greedy algorithm is guaranteed to work only when we have an underlying matroid.  We might define a set to be independent if all of its $3$-subsets have distinct sums.  A subset of an independent set is independent, but we don't have the property that all maximal independent subsets have the same size.  $\{1,2,3,4\}$ and $\{1,2,3,5,8\}$ are both maximal independent subsets of $\{1,2,3,4,5,6,7,8\}$.  So, we should call a set independent if all of its $3$-subsets have distinct sums, and all of its $2$-subsets have distinct sums.  Now it doesn't seem so easy to produce a counterexample, though if I were forced to guess, I'd say it's still not a matroid.

Even if it did turn out to be a matroid, we'd only have found the minimum for a superset of $\{1,2,3\}$, so we'd have work left to do.

Body:
I tried the greedy approach I suggested in a comment, and it produced a maximum of $13796$, as opposed to the value $17404$ in the OEIS table mentioned in the comments.

I wrote this little python script:

from itertools import combinations

xs = [1,2,3]
twos = {3,4,5}
threes = {6}
nextX = 4
while len(xs) < 24:
    if len(xs) != 23:
        newTwos = {nextX + x for x in xs}
        if newTwos & twos:
            nextX += 1
            continue
    newThrees = {nextX + t for t in twos}
    if newThrees & threes:
        nextX += 1
        continue
    twos |= newTwos
    threes |= newThrees
    xs.append(nextX)
    nextX += 1
print(xs)

# audit
test3 = {sum(t) for t in combinations(xs,3)}
if len(test3)== 24*23*22//6:
    print('Passed audit')
else:
    print('Failed audit')

We start with $\{1,2,3\}$ and repeatedly try adding the smallest integer not yet tested to the set.  We keep track of all the $2$-sums and $3$-sums.  When we test a new element we accept it if its addition would not create a duplicate $2$-sum or $3$-sum.  The $2$-sum test is necessary because if there were a duplicate  $2$-sum, the addition of an further element would necessarily create a duplicate $3$-sum.  This test is unnecessary when we add the $24$th elements to the set.

The script produced the output

[1, 2, 3, 5, 8, 14, 25, 45, 82, 140, 235, 388, 559, 839, 1286, 1582, 2221, 3144, 4071, 5795, 6872, 9204, 11524, 13796]
Passed audit

Of course, there's no guarantee that the greedy algorithm produces the optimal answer.  The greedy algorithm is guaranteed to work only when we have an underlying matroid.  We might define a set to be independent if all of its $3$-subsets have distinct sums.  A subset of an independent set is independent, but we don't have the property that all maximal independent subsets have the same size.  $\{1,2,3,4\}$ and $\{1,2,3,5,8\}$ are both maximal independent subsets of $\{1,2,3,4,5,6,7,8\}$.  So, we should call a set independent if all of its $3$-subsets have distinct sums, and all of its $2$-subsets have distinct sums.  Now it doesn't seem so easy to produce a counterexample, though if I were forced to guess, I'd say it's still not a matroid.

Even if it did turn out to be a matroid, we'd only have found the minimum for a superset of $\{1,2,3\}$, so we'd have work left to do.


问:
Properties of Unbounded function
If $f$ is bounded on $A$ and $g$ is unbounded on $A$ then how can I prove that $f+g$ is unbounded on $A$?

Since $f$ is bounded then there exist $M>0$ such that $|f(x)| \le M\  \forall x\in A$ and $g$ is unbounded means $\exists M'>0$ such that $|g(x)| > M'\ \forall x\in A$.

Then what about $|(f+g) (x)|$?

答: As you wrote, since $f$ is bounded, there is some $M>0$ such that $(\forall x\in A):|f(x)|<M$. Since $g$ is unbounded, for any $n\in\Bbb N$, there is some $a_n\in A$ such that $|g(a_n)|\geqslant n$. So$$(\forall n\in\Bbb N):|(f+g)(a_n)|=|g(a_n)+f(a_n)|\geqslant n-M.$$So, $\lim_{n\to\infty}|(f+g)(a_n)|=\infty$, and therefore $f+g$ is unbounded.

ContentLicense: CC BY-SA 4.0

Title: Properties of Unbounded function


问:
Finding sum of sequence: $1, 2,2,3,3,3,4,4,4,4,5,5,5,5,5.......$
$1, 2,2,3,3,3,4,4,4,4,5,5,5,5,5.......n$ $(n$ times$)$

Approach:

Observation: For summation, let us consider a term $t_x$ as the sum of the terms which follow a general order.

Like:

$$t_1 = 1+2+3+4+5+6+.....n$$

$$t_2 = 2+3+4+5+6 .....n$$

$$t_3 = 3+4+5+6+.....n$$

$$.$$

$$.$$

$$.$$

$$t_{n-1} = n -1 + n$$

$$t_n = n$$

For complete summation, we need to add all the $t_x$ obtained from the above.

For this particular step, I am getting a really big and lengthy sigma, which I am not able to evaluate.

$$\sum_{n=1}^n  t_n$$

Now each term $t_n$ comprises of a summation of its very own.

$ $$\sum t_n $$= $$$\sum_{n=k}^n \cfrac{n(n+1)}{2}$$

where $k= 1,2,3,...n$

What to do next? After plugging in values for $k$, I am getting a really ugly expression. Please help.

P.S: This question might have been asked before, but I want to solve it through the above mentioned method. So please don't mark it as duplicate and close it.

Body: $$\sum\limits_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$$


问:
Prove that $A'\cap B=B\Longleftrightarrow A\subseteq B'$
Prove that $A'\cap B = B\Longleftrightarrow A\subseteq B'$

Attempt: First, let  $x\in A\subseteq B'$ . If  $x\in A'\cap B$, then   $x\in A'$  and  $x\in B$ ...

Conversely, assume  $x\in A \subseteq B'$ . Therefore,  $x\in A$ and  $x\in B'$, i.e.  $x\notin B$ ...

Question: Is it correct to let some $x$ be in each side of the proof to prove it because it seems to lead me to a dead end.
If not, please suggest a new way to prove this. Thank you!

答:
You're right that your arguments don't work, because they don't really go anywhere. You should be very clear about what you're supposed to prove, and what you're allowed to assume. Once you get that down pat, then usually the arguments have basically written themselves.

The first thing to notice is that you are proving an equivalence (which we can see from the $\iff$). This means that we will need two arguments: a $\implies$ argument and a $\impliedby$ argument (once you're more confident, it may be possible to combine these arguments into one, but don't run before you can walk).

Let's look at the $\implies$ argument. What you have to prove is:
$$A \subseteq B'.$$
You are allowed to assume that $A' \cap B = B$, but for now, it's best to focus on what you're trying to accomplish. How do you prove a subset relation like this? You should start by assuming that we have an element $x$ of $A$, and then logic our way into concluding that $x$ is an element of $B'$, which is to say, $x \notin B$.

I usually find it helpful to do this: to lay out what I'm trying to prove, filling in both the start ("Suppose $x \in A$...") and the end of my argument ("... therefore $x \notin B$, i.e. $x \in B'$"). This gives me a useful framework to fill in. Remember too that this is only the $\implies$ direction, and another argument is required for the $\impliedby$ direction.

So, let's suppose that $x \in A$. We are assuming that $A' \cap B = B$, and I want to use this fact to help us say something about $x$ with respect to $B$. So, my big idea is to take the complement of both sides of this equality. By De Morgan's laws:
$$B' = (A' \cap B)' = (A')' \cup B' = A \cup B'.$$
Now, since $x \in A$, we have $x \in A \cup B'$, since $A \subseteq A \cup B'$. But then, this set is $B'$, so $x \in B'$. And, this is what we wanted to conclude. In particular, this means that $A \subseteq B'$, as required (which finishes the $\implies$ direction).

I'll set up the other direction for you, and leave you to fill it in. This time, we wish to prove
$$A' \cap B = B.$$
To prove this set equality, you'll need to prove two things:

$A' \cap B \subseteq B$ (this should be easy to show),

$B \subseteq A' \cap B$ (this will be the majority of the argument).

To show the second point, your argument will start like "Suppose $x \in B$...", and should end like "... Therefore, $x \in A' \cap B$"). At some point, you should use the fact that $A \subseteq B'$, i.e. every point in $A$ is also in $B'$. Give it a go.


问:
Find the value of the expression: $\frac{1}{\sqrt{4}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{10}}+\cdots+\frac{1}{\sqrt{13}+\sqrt{16}}$

Find the value of the expression: $$\frac{1}{\sqrt{4}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{10}}+\cdots+\frac{1}{\sqrt{13}+\sqrt{16}}$$

After putting it into a calculator I worked out that it is equal to $\frac{2}{3}$. I tried to work it out on my own by saying $$\sum\limits_{k=1}^{5}\frac{1}{\sqrt{3k+1}+\sqrt{3k+4}},$$ must resemble somehow the telescopic identity: $$\frac{1}{m(m+1)}=\frac{1}{m}-\frac{1}{m+1},$$ but I couldn't find a way for it to become similar to it. I then said $$\sum\limits_{k=1}^{5}\frac{1}{\sqrt{3k+1}+\sqrt{3k+4}}=\sum\limits_{k=2}^{6}\frac{1}{\sqrt{3k-2}+\sqrt{3k+1}},$$ but that didn't bare crop either. Could you please explain to me how to solve this question?

Body:
Hint You had the nice idea with the telescopic sum :

$$\frac{1}{\sqrt{3k+1}+\sqrt{3k+4}}=\frac{\sqrt{3k+1}-\sqrt{3k+4}}{(\sqrt{3k+1}-\sqrt{3k+4})(\sqrt{3k+1}+\sqrt{3k+4})}=\frac{\sqrt{3k+1}-\sqrt{3k+4}}{3k+1-3k-4}=\frac{\sqrt{3k+1}-\sqrt{3k+4}}{-3}$$


问:
To find some particular solutions for Heat Equation type partial differential equation

Problem: If $~u(x,t)~$ is the solution of $$\dfrac{\partial u}{\partial t}=\dfrac{\partial^2 u}{\partial x^2},~~~~0<x<1,~~t>0\\u(x,0)=1+x+\sin(\pi x)\cos(\pi x)\\u(0,t)=1,~~u(1,t)=2$$then
$$1.~~~u\left(\dfrac 12,\dfrac 14\right)=\dfrac 32~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~2.~~~u\left(\dfrac 14,\dfrac 34\right)=\dfrac 54+\dfrac 12e^{-3\pi^2}\\3.~~~u\left(\dfrac 12,\dfrac 12\right)=\dfrac 32~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~4.~~~u\left(\dfrac 14,1\right)=\dfrac 54+\dfrac 12e^{-4\pi^2}$$

My approach:  We know that the general solution of the heat equation for one space variable
$$\dfrac{\partial u}{\partial t}=\alpha\dfrac{\partial^2 u}{\partial x^2},~~~~0\le x\le L,~~t\ge0\\u(x,0)=f(x),~~~~\forall~~x\in[0,L]\\u(0,t)=0=u(L,t),~~~~\forall~~t>0$$is $$u(x,t)=\sum_{n=1}^\infty D_n \sin\left(\dfrac{n\pi x}{L}\right)e^{-n^2\pi^2\alpha t/L^2}$$where$$D_n=\dfrac 2L\int_0^L f(x)\sin\left(\dfrac{n\pi x}{L}\right) dx~.$$ Clearly here $~L=1,~\alpha=1~$ and $~f(x)=1+x+\sin(\pi x)\cos(\pi x).$ But one can see that the boundary values are different from the Heat equation. So I stuck here. Please help to solve the problem.

答: Here's a hint: consider $v(x,t)=u(x,t)-(1+x)$.


问:
Epsilon-delta proof at infinity with square root
By using the $\varepsilon$-$\delta$-definition, how can I prove this limit:

$$\lim\limits_{x\to\infty}\frac{\sqrt{x}-5}{x+2}=0$$

I can't seem to simplify the fraction so that I can "solve for" $x$.

Body: Hint: $\frac{\sqrt{x}-5}{x+2}\leqslant \frac{1}{\sqrt{x}}$


问:
Convergence of the sequence $y_{n+1}=1-(n+1)y_n , y_0=(e-1)/e$
I would like to prove that the sequence $\{y_n\}_n$ defined recursively by $y_0=(e-1)/e$ and
$$y_{n+1}=1-(n+1)y_n , \quad n\geq 0$$
is convergent (if this is done then it clearly converges to $0$).

I tried to prove that $\{y_n\}_n$ is decreasing and bounded below by $0$ using mathematical induction, but the inequalities provided in the induction hypothesis are reversed and thus it's not possible to continue.

Could anyone please provide a hint/reference for this problem?

答:
I would find the general term of the sequence as follows :
Let $ n,k\in\mathbb{N}^{*} $, we have the following : \begin{aligned}x_{k}&=1-k x_{k-1}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{k}\frac{x_{k}}{k!}&=\frac{\left(-1\right)^{k}}{k!}+\left(-1\right)^{k-1}\frac{x_{k-1}}{\left(k-1\right)!}\\ \iff \sum_{k=1}^{n}{\left(\left(-1\right)^{k}\frac{x_{k}}{k!}-\left(-1\right)^{k-1}\frac{x_{k-1}}{\left(k-1\right)!}\right)}&=\sum_{k=1}^{n}{\frac{\left(-1\right)^{k}}{k!}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{n}\frac{x_{n}}{n!}-\frac{\mathrm{e-1}}{\mathrm{e}}&=\sum_{k=1}^{n}{\frac{\left(-1\right)^{k}}{k!}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{n}\frac{x_{n}}{n!}&=\sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{k!}}-\frac{1}{\mathrm{e}}\end{aligned}

Using the fact that $ \sum\limits_{k=0}^{n}{\frac{\left(-1\right)^{k}}{k!}}-\frac{1}{\mathrm{e}}=\frac{\left(-1\right)^{n}}{n!}\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x}$, we get that : $$\fbox{$\begin{array}{rcl} \left(\forall n\in\mathbb{N} \right),\ x_{n}=\displaystyle\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x}\end{array}$} $$

Now for the limit, we could do something like : $$ \left\vert\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x}\right\vert\leq\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{d}x}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0 $$

Thus $ \left(x_{n}\right)_{n} $ converges, and $ x_{n}\underset{n\to +\infty}{\longrightarrow}0 $.


答:
Assume that the expression of the sum is a polynomial in $n$, let $P(n)$.

Now if you set the last terms apart,

$$P(n)-P(n-1)=\underbrace{n+n+n\cdots n}_n=n^2.$$

This shows that the polynomial must be of degree $3$.

By direct computation, we have

$$P(0)=0,P(1)=1,P(2)=5,P(3)=14$$ and the Lagrangian interpolation polynomial by these four points is (after tedious direct computation or by identification*)

$$P(n)=\frac{n^3}3+\frac{n^2}2+\frac{n}6.$$

We can now confirm

$$P(n)-P(n-1)=\frac{3n^2-3n+1}3+\frac{2n-1}2+\frac16=n^2.$$

*With $P(n)=an^3+bn^2+cn+d$, we perform Gaussian elimination
$$d=0,\\a+b+c=1,\\8a+4b+2c=5,\\27a+9b+3c=14$$

$$6a+2b=3,\\24a+6b=11$$

$$6a=2$$
$$2b=1$$

$$c=1-\frac12-\frac13$$

Title: Finding sum of sequence: $1, 2,2,3,3,3,4,4,4,4,5,5,5,5,5.......$

Body:
$1, 2,2,3,3,3,4,4,4,4,5,5,5,5,5.......n$ $(n$ times$)$

Approach:

Observation: For summation, let us consider a term $t_x$ as the sum of the terms which follow a general order.

Like:

$$t_1 = 1+2+3+4+5+6+.....n$$

$$t_2 = 2+3+4+5+6 .....n$$

$$t_3 = 3+4+5+6+.....n$$

$$.$$

$$.$$

$$.$$

$$t_{n-1} = n -1 + n$$

$$t_n = n$$

For complete summation, we need to add all the $t_x$ obtained from the above.

For this particular step, I am getting a really big and lengthy sigma, which I am not able to evaluate.

$$\sum_{n=1}^n  t_n$$

Now each term $t_n$ comprises of a summation of its very own.

$ $$\sum t_n $$= $$$\sum_{n=k}^n \cfrac{n(n+1)}{2}$$

where $k= 1,2,3,...n$

What to do next? After plugging in values for $k$, I am getting a really ugly expression. Please help.

P.S: This question might have been asked before, but I want to solve it through the above mentioned method. So please don't mark it as duplicate and close it.


问:
Chebyshev Polynomials and the Hypergeometric Function
A problem related to Chebyshev Polynomials and the Hypergeometric Function involves transformation from one function to another. The task is to transform the Chebyshev polynomial into its correct scaled Hypergeometric Function. $$
T_{2n} (x) = ( - 1)^n {}_2F_1 \left( { - n,n;\tfrac{1}{2};x^2 } \right).
$$
Is there any way for me to make use of the relationship below?
$$
T_n (x) = {}_2F_1 \left( { - n,n;\tfrac{1}{2};\tfrac{{1 - x}}{2}} \right).
$$

答:
Let me carry out my suggestion.

Start HERE with the differential equation for ${}_2F_1$:

The solution of
$$
z(1-z)F''(z)+\left(\frac12 - z\right)F'(z)+n^2 F(z) = 0,\quad F(0)=1, F'(0)=-2n^2
\tag1$$
is
$$
F(z) = {}_2F_1\left(-n,n;\frac12;z\right) .
$$
Change variables $z=x^2$.  Then we have:  The solution of
$$
(1-x^2)F''(x)-xF'(x)+4n^2F(x) = 0,\quad F(0)=1,F'(0)=0
\tag2$$
is
$$
F(x) = {}_2F_1\left(-n,n;\frac12;x^2\right) .
$$

Next, HERE we find that the Chebyshev polynomial $T_m(x)$ satisfies
$$
(1-x^2)T_m''(x)-xT_m'(x)+m^2T_m(x)=0
\tag3$$
Put $m=2n$, then $(3)$ becomes $(2)$.  Also $T_{2n}(0)=(-1)^{n}, T_{2n}'(0)=0$.
Therefore
$$
T_{2n}(x) = (-1)^n\;{}_2F_1\left(-n,n;\frac12;x^2\right) .
$$

OwnerDisplayName: Lawrence Calixto


问:
Test Prep Question: Finding average rate of consumption
Here is a test prep question for a high school student:

Kimberly drives at a rate of $50$ miles per hour on her trip across the country. If it takes Kimberly $200$ gallons of gasoline to drive $500$ miles, how many minutes will it take her to use $x$ gallons of gasoline if she uses gasoline at the same average rate of consumption?

A) $20x$

B) $3/x$

C) $x/20$

D) $3x$

I get Answer D ($3x$) by cancelling the dimensions so I end up with minutes.

Can some please let me know if that seems right? And maybe another way to explain this problem.

Thanks!

答:
Your answer is correct.

In terms of easier wording, I can't think of a way. In my opinion, wordy math problems are less about mathematical ability, and more about reading comprehension. When working with weird quantities, the only way it to go by the units. Maybe I'm a moron in this respect, but during my bachelors in physics, as well as in my professional life, units are often the only way I know/knew what quantities I was working with.

I have probably solved your question in a very similar way as you:

We know $\frac{200}{500}$gl/mi, and we know $50$mi/h $=\frac{50}{60}$mi/min from the set-up.

We multiply to get gallons per minute (because we know we need this relationship from the question), in order to set up the right fraction value: $\frac{2}{5}*\frac{5}{6}*$gl/mi * mi/min$=\frac{1}{3}$gl/min.

We know that we need to multiply $x$ gallons by min/gl to get minutes, so we need to take the reciprocol of the above (gl/min -> min/gl), and we get the desired $3x$.


问:
Probability of rolling a 6 immediately after a 1 is rolled
Question

Ann and Bob take turns to roll a fair six-sided die. The winner is the first person to roll a six immediately after the other person has rolled a one. Ann will go first. Find the probability that Ann will win.

Answer

$\mathbb{P} (\mathrm {Ann\ wins}) = \frac {36} {73}$

I have thought long and hard about this question but I am unable to even start. I have tried considering cases, but got stuck along the way. For example, it is trivial to calculate the probability that Ann wins if there are only three rolls (which is the minimum number of rolls needed for Ann to win). However, the problem easily becomes very complicated when we consider five rolls and more.

The suggested solution by my professor uses first-step decomposition, but it is a new concept to me and I am struggling to understand it.

If anyone can provide a detailed and intuitive explanation as to how this problem should be solved, that will be greatly appreciated!

答:
You should solve it using Markov Chains by constructing the set of states varying from the absorbing state $s_{-2}$ (Bob wins) and the absorbing state $s_3$ (Alice wins). I denote the starting state $s_{\emptyset}$. Unless I made a silly mistake somewhere, you should get this set of recurrent equations:
$$
p_{\emptyset, 3} = \frac{5}{6}p_{1,3} + \frac{1}{6}p_{-1,3}\\
p_{1,3} = \frac{1}{6}p_{2,3} + \frac{5}{6}p_{\emptyset,3} \\
p_{2,3} = \frac{1}{6}p_{3,3} + \frac{4}{6}p_{\emptyset,3} + \frac{1}{6}p_{-1,3}\\
p_{-1,3} = \frac{1}{6}p_{-2,3} + \frac{1}{6}p_{2,3} + \frac{4}{6}p_{\emptyset,3}
$$

The boundary values are $p_{3,3}=1, p_{-2,3} = 0$. I'm pretty sure of this setup, but perhaps I made a typo somewhere during calculations, as I keep getting $p_{\emptyset,3}=\frac{18}{73}$.


问:
Solution to this specific differential equation $y'=f(y)$
Prove or disprove:

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function. Then the initial value problem
$y'=f(y)$ with $y(0)=1$ always has a solution $y\colon\mathbb{R}\to\mathbb{R}$.

Now, I think the crucial point is what we do with functions $f(y)$, which have roots. For all other functions, separation will work. Even for $f(y)=0$, we can just take $y=0$. Unfortunately, I dont know how to prove or disprove this.

答:
Extendability of solutions to the whole real line is not guaranteed even if $f$ satisfies the hypothesis of Cauchy-Lispchitz: namely, the one and only maximal solution to $$\begin{cases} y'=y^2\\ y(0)=1\end{cases}$$
is $y:(-\infty,1)\to\Bbb R,\quad y(t)=(1-t)^{-1}$.

Existence of solutions defined on some neighbourhood of $0$ is guaranteed by Peano's existence theorem. Since differentiable functions need not be locally Lipschitz, I don't think differentiability of $f$ can play a role in distinguishing this case from the general one, but I have been wrong before (in fact, I can't quite think right now of an example where $f$ is differentiable and the ODE has more than one solution).


问:
Mean value theorem for integration question
I have a question below. I would say $3$ with the formula for mean value theorem for integration but I don't see $3$ in the answers. What am I doing wrong?

The function $f(x)=x^2$ assumes the value of its integral average over the interval $[0,3]$.

a) at point $x_0=\sqrt 3 $

b) at point $x_0=\sqrt{5}/2 $

c) at point $x_0=\sqrt{3}/3 $

d) at point $x_0=3/2  $

e) never on the given interval

Body:
HINT:

$\int x^3dx$=$\frac{1}{3}x^3$+C,   $\frac{\int_0^{3} x^2dx}{3}$=3, $x_0$= ?


问:
Conversion of Riemann sum into definite integral

I was trying to do this but I dont know how to put convert it into a definite integral also I got confused about limits of the integral now im stuck.

答:
For an integral between zero and one you can take the partition
$$ 0 = 0 < \frac{1}{n} < \frac{2}{n} < \dots < \frac{n}{n} = 1.$$
Then the integral over $f(x)$ from zero to one becomes
$$ \int_0^1 f(x) dx = \lim_{n \to \infty} \sum_{k=0}^n \frac{1}{n} f\left(\frac{k}{n}\right).$$
Here first rewrite
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{(2n+5k)^2}{n^3} = \lim_{n \to \infty} \sum_{k=0}^n \frac{1}{n} \left(2+ 5 \frac{k}{n}\right)^2,$$
now you can read of $f(x)$ and evaluate the integral $\int_0^1 f(x) \,dx$ to find the value of the sum.


答:
This works well with states.

There are only $4$ active states:

Let $(A,0)$ denote the state in which it is $A's$ turn and the prior toss was not a $1$ (this includes the START state).

Let $(A,1)$ denote the state in which it is $A's$ turn and the prior toss was a $1$

Let $(B,0)$ denote the state in which it is $B's$ turn and the prior toss was not a $1$

Let $(B,1)$ denote the state in which it is $B's$ turn and the prior toss was  a $1$

For any state $S$ let $P(S)$ denote the probability that $A$ will eventually win, given that we are now in state $S$.    We see that $$P(A,0)=\frac 16P(B,1)+\frac 56P(B,0)$$
$$P(A,1)=\frac 16\times 1+\frac 16P(B,1)+\frac 46P(B,0)$$
$$P(B,0)=\frac 16P(A,1)+\frac 56P(A,0)$$
$$P(B,1)=\frac 16\times 0 + \frac 16P(A,1)+\frac 46P(A,0)$$

This system is easily solved and confirms the official result, here

Title: Probability of rolling a 6 immediately after a 1 is rolled

Body:
Question

Ann and Bob take turns to roll a fair six-sided die. The winner is the first person to roll a six immediately after the other person has rolled a one. Ann will go first. Find the probability that Ann will win.

Answer

$\mathbb{P} (\mathrm {Ann\ wins}) = \frac {36} {73}$

I have thought long and hard about this question but I am unable to even start. I have tried considering cases, but got stuck along the way. For example, it is trivial to calculate the probability that Ann wins if there are only three rolls (which is the minimum number of rolls needed for Ann to win). However, the problem easily becomes very complicated when we consider five rolls and more.

The suggested solution by my professor uses first-step decomposition, but it is a new concept to me and I am struggling to understand it.

If anyone can provide a detailed and intuitive explanation as to how this problem should be solved, that will be greatly appreciated!


问:
High school MCQ (geometry) about maximum of sine of an angle
No calculus is allowed in this question.

Also, it is not allowed to use any compound angle formulas such as

$\sin(A-B) =\sin A \cos B - \cos A \sin B$.

How to find the greatest value of $\sin \angle APB$?

I attempted by consider the difference between angles $\angle OPB$ and $\angle OPA$ but I do now know how to make use of the fact that $P$ makes the angle greatest.

答:
Recall from extended sine-rule,
$$\sin \angle APB = \frac{AB}{2R_{\triangle APB}}$$

Since $AB$ is fixed, to maximize $\sin \angle APB$, we have to minimize the circumradius $R_{\triangle APB}$. As the circumcenter $S$ lies on perpendicular bisector of $AB$, $R_{\triangle APB}$ will be least when $SP \perp OC$ as in the following diagram.

Using that $SPOD$ is a rectangle and $A$ is mid of $OB$,

$$\text{max of sin} \angle APB = \frac{1}{3}$$


问:
How to find one of the points of a triangle $3$d?
How to find the coordinates of $C$ as a function of $A, B, d_1, d_2$ and $d_3$?

$$f(A, B, d_1, d_2, d_3) = C$$

I need to find the coordinates of point C as a function of the distances d1, d2, d3.

A(0,0,0)

答:
C is not uniquely defined by the conditions you are mentioning simply because if we assume $C(x, y, z)$ then we have the following constraints

$$(x-A_x)^2+(y-A_y)^2+(z-A_z)^2={d_1}^2$$ and

$$(x-B_x)^2+(y-B_y)^2+(z-B_z)^2={d_2}^2$$

Both of them are spheres and hence they intersect in a circle. Thus locus of C is a circle of intersection of the above two spheres assuming that the initial constants are well chosen and they define a real triangle.


问:
Probability of more than 13 balls remaining in the box after some selections
Question

A box contains 20 balls, of which exactly 8 are red and 12 are green. Two balls are randomly selected from the box, without replacement. Then, any red balls selected are put back into the box, while any green balls selected are thrown away. Finally, 5 balls are randomly selected from the box, without replacement. Suppose that, following this procedure, all 5 balls are green. Find the probability that there remains more than 13 balls in the box.

My working

Observe that the only case when there are less than 14 balls in the box is when the two balls that are selected initially are both green, so we have three cases to consider here.

Case 1: Choose 2 red balls, put them back, then choose 5 green balls.

Case 2: Choose 1 red ball, then 1 green ball. Put the red ball back, throw the green one away and then choose 5 green balls.

Case 3: Choose 1 green ball, then 1 red ball. Put the red ball back, throw the green one away and then choose 5 green balls.

$\mathbb{P} (\mathrm {Case\ 1}) = \frac 8 {20} * \frac 7 {19} * \frac {12} {20} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} = \frac {231} {30685}$

$\mathbb{P} (\mathrm {Case\ 2}) = \frac 8 {20} * \frac {12} {19} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} * \frac 7 {15} = \frac {308} {30685}$

$\mathbb{P} (\mathrm {Case\ 3}) = \mathbb{P} (\mathrm {Case\ 2})$

$\therefore \mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {231} {30685} + 2 * \frac {308} {30685} = \frac {847} {30685}$

Answer

$\mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {154} {211}$

My answer is obviously off by a lot ): Am I missing cases, or is my thinking too naive etc.? I am not sure where I have gone wrong, so any intuitive explanations about my mistake(s) and also what the correct solution should be will be greatly appreciated!

P.S. My professor did provide a solution, but I am unable to understand his way of thinking, so I thought posting the problem here and asking for ideas would be better.

答:
I haven't checked your arithmetic, but it appears that you aren't computing the conditional probability.  Let $G$ be the event that all five balls are green, and let $T$ be the event that there are more than $13$ balls in the box.  We are asked for $$\Pr(T|G)=\frac{\Pr(T\cap G)}{\Pr(G)}$$

It looks to me like you've only computed the numerator.  You need to compute the probability that the first two balls are green and the last five are green, and add it to the numerator, in order to get the denominator.


问:
Trouble using Sylow's theorems

So, I found part (a) fairly easy, and also managed part (c) assuming (b) to be true.

Any help with (d) and especially (b) would really be appreciated.

My issue with part (b) is understanding which of Sylow's theorems to apply and how. I tried first to apply part (a) to find that, since Sylow $p$-subgroups by definition are non trivial (have $p$ as a factor of their order), then $$1<S\leq K \implies N_K(S) = N_G(S).$$ I was hoping doing this would allow me to somehow apply Sylow's second theorem as conjugation naturally comes up in the problem, but I can not see how I would go about using it.

答:
For (b) let $P_0$ be a Sylow $p$-subgroup of $K$, and let $P$ be a Sylow $p$-subgroup of $G$ which contains $P_0$. Then $Z(P)$ normalises $P_0$ so by (a), with $Q=P_0$, we have that $Z(P)\leqslant K$. Now $P$ normalises $Z(P)$, so by (a), with $Q=Z(P)$, we have that $P\leqslant K$.

For (d), there are $|G:N(K)|=|G:K|$ conjugates of $K$. These don't overlap except in the identity, so together they contain $1+|G:K| (|K|-1)$ elements. This leaves $|G:K|-1$ outside.

Title: Trouble using Sylow's theorems


答:
We want to prove that $$A^C \cap B = B \Longleftrightarrow A \subseteq B^C$$

$\Longrightarrow$

$$x\in A \Longrightarrow x \not \in A^C \Longrightarrow x\not\in A^C \cap B \Longrightarrow x\not\in B \Longrightarrow x\in B^C$$

$\Longleftarrow$

Let's do this by double containment, that is, $$A^C \cap B \subseteq B \quad \wedge \quad B \subseteq A^C \cap B$$ It is obvious that $A^C \cap B \subseteq B$, (it is basic property). The interesing one is $B \subseteq A^C \cap B$.

Let's supose, to arrive at a contradiction, that $B \not \subseteq A^C \cap B$, that is, it exists $x\in B$ such that $x \not \in A^C \cap B$ $$x \not \in A^C \cap B \Longrightarrow \neg (x \in  A^C \cap B) \Longrightarrow \neg (x \in  A^C \wedge x \in B) \Longrightarrow  x \not\in  A^C \vee  x \not\in B $$ So, we have that $$\exists x\in B \wedge(x \not\in  A^C \vee  x \not\in B)$$

Case $1)$. $x \in B$ and x $\not \in B$ which is clearly a contradiction.

Case $2)$. $x \in B$ and x $\not \in A^C$ which implies that $x \not\in B^C$ and x $\in A$ which is a contradiction to our hypothesis, because we have that $A\subseteq B^C$.

The contradiction comes of suposing that $B \not \subseteq A^C \cap B$, then $B \subseteq A^C \cap B$ and $A^C \cap B \subseteq B$ which implies that $A^C \cap B = B$.

Therefore, $$A^C \cap B = B \Longleftrightarrow A \subseteq B^C$$

Title: Prove that $A'\cap B=B\Longleftrightarrow A\subseteq B'$

Body:
Prove that $A'\cap B = B\Longleftrightarrow A\subseteq B'$

Attempt: First, let  $x\in A\subseteq B'$ . If  $x\in A'\cap B$, then   $x\in A'$  and  $x\in B$ ...

Conversely, assume  $x\in A \subseteq B'$ . Therefore,  $x\in A$ and  $x\in B'$, i.e.  $x\notin B$ ...

Question: Is it correct to let some $x$ be in each side of the proof to prove it because it seems to lead me to a dead end.
If not, please suggest a new way to prove this. Thank you!


问:
Has anyone studied this operator?
I've been studying a particular unary operator on the commutative ring $\mathbb{Z}/n\mathbb{Z}$. The operator is:

$\downarrow(x) = y\pmod{n}$ iff $n \equiv y \pmod{x}$, where $0< x,y \le n$.

The operator isn't distributive on addition or multiplication, so its use is probably fairly limited. I've proven a couple fairly trivial results, but I'm interested in seeing if there are more resources available.

One of the fairly trivial results:
If $n$ is composite and $n=pq$ and WLOG $p \le q$, then if $q\ge kp+k$ I can show that

$(\downarrow(q-k))\mid p$.

答: This is not really an operator on $\mathbb{Z}/n\mathbb{Z}$, it's an operator on the numbers $\{1,2,\ldots, n\}$.  The reason is that if $x\equiv x'\pmod{n}$, we would expect any such operator to satisfy $\downarrow (x)=\downarrow(x')$, but this one doesn't.</p>

ContentLicense: CC BY-SA 3.0

Title: Has anyone studied this operator?


问:
Do subrings of $\mathbb{H}$ containing 1 contain the conjugates of its elements?
Let $x\in\mathbb{H}$ which is the ring of real quaternions such that $$x=a+bi+cj+dk$$ where $a,b,c,d\in\mathbb{R}$. The quaternion conjugate of $x$ is $$x^* = a-bi-cj-dk.$$

I want to show (or find a counterexample) that if $R$ is a subring of $\mathbb{H}$ such that $1\in R$, then for any $x\in R$, $x^*\in R$.

I have tried some rings (finitely generated) but they either satisfy the property above or I can't show either. For example, I can't show whether $\pi -i$ is in the subring generated by $\{1,\pi+i\}$. But the property is easy to show when I included $j$ and $k$, that is, the property above is true for the subring generated by $\{1,\pi+i,j,k\}$.

Is property always satisfied? Any hint will be appreciated.

答:
Hint: If $\pi-i$ is in the subring generated by $\pi+i$, then so is $2i$.

Stronger hint:

 Now get a contradiction using the fact that $\pi+i$ is transcendental.

More details:

 Since $\pi+i$ is transcendental, the ring it generates is isomorphic to a polynomial ring $\mathbb{Z}[x]$.  This ring contains no square root of $-4$, so it cannot contain $2i$.


答:
I am only guessing on the "First Step Decomposition" technique you referred to in your question, because although I know a method suitable for this question, I do not know the name for the method.

Suppose the answer is $p$. Notice that the first roll has three types of outcomes:

Ann rolls 2, 3, 4, 5 or 6. The probability of Ann winning after that is $(1-p)$, since the game proceed with exactly the same rules, except Ann and Bob exchanges position.

Ann rolls a 1. In this case, let's suppose Bob's winning probability becomes $q$. Either Bob rolls a 6 (probability $1/6$) and wins the game, or

Bob rolls 2, 3, 4, or 5. The probability of Ann winning after this is $p$ again.

Bob rolls 1 too. The probability of Ann winning is also $q$.

And let's summarize. The total winning probabilities add up to $p$: $$p = \frac56 (1-p) + \frac16 (1-q).$$ Also there is a relation between $p$ and $q$: $$q = \frac16 + \frac46 (1-p) + \frac16 (1-q).$$

Can you see why these relations hold?

Now solve the equations. You get $p = \frac{36}{73}$.


问:
How to compute the time of upward movement of a Brownian motion
Suppose $B(t)$ is the standard Brownian motion. Does it make sense to compute the total time of upward movement
$$\int_0^t I(dB_\tau>0)d\tau$$
where $I$ is the characteristic function? What is a rigorous way to define it if it does make sense? Intuition dictates that the expectation of the above integral should be $\frac t2$. I am aware of Doob's upcrossing lemma. Perhaps we can use local time and Tanaka's formula. But I do not have a way to define my integral in similar terms.

答:
One answer to your question, in the negative, is this: With probability 1 the Brownian motion has no times of increase; that is, the probability that there exists a time $t>0$ and a $\delta>0$ such that $\sup_{s\in[t-\delta,t]}B_s=\inf_{s\in[t,t+\delta]}B_s$ is $0$.
This result if due to Dvoretzky, Erdos, & Kakutani ["Nonincrease everywhere of the Brownian motion process", (1961)]. A wonderful and quite brief proof was found thirty years later by K. Burdzy ["On nonincrease of Brownian motion" (1990)].


问:
Probability of drawing numbers in ascendin order
We have 10 uniquely numbered tickets from 0 to 9. We draw randomly 3 tickets, one at a time.

What is the probability of drawing the tickets in numerally ascending order?

By ascending order I mean any possible combination, where the numeral number of a ticket of the next draw is bigger than the last (i.e 2, 7, 9 or 3, 5, 8). Also it is to be noted that once a ticket is drawn, it will not be put back to be drawn again.

I gather that the actual number of the tickets, assuming that they all have an unique number, doesn't really matter. That said I would approach this by the permutation of the number of tickets to be drawn. In this case it would be

```<code>3! which would give us 3*2*1 = 6 = 1/6 = ~16,6 %
</code>```

Bonus: If it would be mandatory to draw consecutive numbers like 3, 4, 5 or 6, 7, 8, this would then be dependent of the total number tickets available. In this bonus example we only have 8 different variations of consecutive numbers to be drawn

```<code>0, 1, 2
1, 2, 3
2, 3, 4
3, 4, 5
4, 5, 6
5, 6, 7
6, 7, 8
7, 8, 9
</code>```

Which would give me

8 * 9 * 8 = 576 = 1/576

Am I right in this one?

答: The first one is right, but not the second.  There are $8$ admissible draws out of $10\cdot9\cdot8$ possible draws, so the probability is $\frac1{90}$.


问:
Hilbert transform of x (or a sawtooth)
This has a simple curiosity value, but I was converting the czt() (chirp z-transform) function from Octave into (wx)Maxima and, while testing it, I used a simple ramp, or $x$. The appearance of the transform is that of an "U". So I was wondering if there is a closed form formula for the response. If I use:

$$\dfrac{1}{\pi}\int{\dfrac{x}{t-x}\mathrm{d}x}=-\dfrac{t\log(x-t)+x}{\pi}$$

which is a logarithm on a slope; at best, it's only half the curve. There are other integral forms, I tried the difference version but that gives something very similar to the above. Then I realized that applying the FFT means the infinity $x$ becomes a modulo operation, or a sawtooth, and it's more fitting for a cosine sum: $\sum\frac{\cos{kx}}{k}$. I modified the formula above (and used a bit of hammer time) and got:

$$n\cdot\dfrac{\log(k+1)+\log(n-k+2)-7.33}{2.5},\quad k=1,2,...,n$$

which seems to come close to the actual curve (blue is the result of the Hilbert transform, green is the cosine sum, and red is the logarithm above):

$x$ is a discrete valued sequence of $n=100$ terms: $k+1,\;k=0,1,...,n-1$. OTOH, the cosine sum implies a $\sum{\frac{1}{x}}$ result, but that doesn't go negative by itself. Intuitively, I imagine this sum of cosines to be evaluated as a limit of $\sum{\frac{\cos{pk}}{k}}$ at each point $p$, and that would be the answer, but that seems like a brute-force approach, and I'm not even sure if it's the right way.

I am, most probably, overthinking it, or even diverging from the actual response (if there is one), so my question is: is there a continuous function that describes the Hilbert transform of (a periodic) $x$?

I've been fiddling with this. It looked as if the sum of cosines would converge to zero, since the cosine takes values between -1 and 1, while k goes to infinity, but I've been trying to evaluate the sum numerically with an imposed n and the results clearly converge towards a certain value. Here is the sum evaluated for x=0.9 and n=10000

:

The value at the end is 0.139365607458084, while the average of all the points is 0.139405249374687. This made me look at it from a different perspective: it's a sum of $\frac{\cos{kx}}{k}$, so it's a discrete evaluation of the integral of it, but results in $\frac{\sin{kx}}{k^2}$, which is nowhere near the result.

It looks like the only way to do it is to evaluate the limit of the sum of cosines at each point of the underlying DFT, which might avoid the spectral leakage, but it certainly won't give any continuous function. I hope the question doesn't reduce to "is this possible":.

答:
I found something, because the math seems to go the right way, but I can't quite match the responses (which most probably means it's wrong, but it's the only thing I've got so far).

Continuing from the sum of cosines, in the OP the integral used $x$ as the variable, but that is the value for the discrete point, and so $k$ should be used, instead, which leads to the cosine integral. Since the modulo transforms $x$ into a periodic function, and since the Hilbert transform is done using a Fourier transform, that implies sampling and, even if the formula I'm looking for is continuous, there is a spectrum folding, which is why there is a sum of logarithms in the OP with $x$ and $-x$. The same applies here:

$$\sum_{k=1}^N{\dfrac{\cos{kx}}{k}}=\int{\dfrac{\cos{kx}}{k}\mathrm{d}k}=-\dfrac{i\Gamma(0,-ikx)+i\Gamma(0,ikx)}{2}=\mathrm{Ci}(kx)\tag{1}$$

Maxima shows it as the incomplete $\Gamma$ function. Applying the spectrum mirroring is done relative to the length of the sequence, defined through $n$. With scaling it becomes:

$$-\dfrac{n}{\pi}\Biggl(\mathrm{Ci}\Bigl(\dfrac{x}{n}\Bigr)+\mathrm{Ci}\Bigl(\dfrac{n-x}{n}\Bigr)+1\Biggr)\tag{2}$$

Testing this seems to be a close match to the continuous sum of cosines ($n=10,100,1000$, skips $x[0]$ because of the $\mathrm{Ci}(0)$):

but there are apparent differences. If I grab the ol' trusty hammer, I end up with a correction value of $1.5$ for the argument of the function and $0.39$ as offset (differences between the traces):

But the offset is not the problem, the curve does not seem to fit exactly, and I don't know whether it's because I am using a finite amount of cosines for the sum, or if the formula is just plain wrong. Here are the differences for traces that are sampled at $\frac1n$, as opposed to $\frac{0.1}{n}$ until now:

It doesn't seem to be a sampling effect, since the curve for $n=100$ is quite similar for $n=1000$ (i.e. there are not more oscillations).

So I tried from a different angle. If the Hilbert transform is a sum of cosines and, thus, a cosine integral, then the original (modulo) $x$ signal should be a similarly made sine integral:

$$\begin{align}
\sum_{k=1}^N{\dfrac{\sin{kx}}{k}}&=\int{\dfrac{\sin{kx}}{x}\mathrm{d}k}=\dfrac{i\Gamma(0,-ikx)-i\Gamma(0,ikx)}{2}=\mathrm{Si}(kx)\tag{3} \\
&\Rightarrow -\dfrac{n}{2}\Biggl(\mathrm{Si}\Bigl(\dfrac{x}{n}\Bigr)-\mathrm{Si}\Bigl(\dfrac{n-x}{n}\Bigr)+1\Biggr)\tag{4}
\end{align}$$

where I had to use $\frac{n}{2}$ as the normalization, but it still doesn't match well the input, and I suppose it's to be expected that a continuously differentiable function should not be able to recreate a ramp. Here in blue is $x$, left - difference (same shape for $n=10$ as for $n=10000$):

So then, is this the way to go? Because the results are not very encouraging.

Title: Hilbert transform of x (or a sawtooth)


问:
$V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective.
Question : Let $\operatorname{Rad}(R)$ denote the Jacobson Radical of $R.$ Show that $V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective. Show that equality fails when $R=\mathbb{Z}$ and $V=\mathbb{Q}_\mathbb{Z}$

My Thoughts : I feel like this is a straightforward proof, but I am just having some issues as to where to start dealing with a free and projective module. Would it be best to consider ideals, or would it be best to try and appeal to Nakayama's Lemma ?  I would be appreciative for any help or insight ! Thank you.

答:
First the counterexample. The radical of $\mathbb{Q}$ as a $\mathbb{Z}$-module is $\mathbb{Q}$, because this module has no maximal submodule.

Indeed a quotient of $\mathbb{Q}$ is a divisible abelian group, whereas no simple abelian group is divisible.

Now the inclusion. You just need to show that $\DeclareMathOperator{\Rad}{Rad}\Rad(R)$ annihilates all simple modules, which is easy: if $S$ is a simple (right) $R$-module, then for $x\in S$, $x\ne0$, you have that $I=\{r\in R:xr=0\}$ is a maximal right ideal. Since $\Rad(R)$ is the intersection of all maximal right ideals, you conclude that $S\Rad(R)=\{0\}$.

If $V$ is any right $R$-module and $L$ a maximal submodule thereof, then $(V/L)\Rad(R)=\{0\}$, which amounts to saying that $V\Rad(R)+L=L$, that is, $V\Rad(R)\subseteq L$. Therefore $V\Rad(R)\subseteq\Rad(V)$ (the intersection of all maximal submodules).

For a free module $V=R^{(\Lambda)}$ (a direct sum of copies of $R$), you can consider, for each $\lambda\in\Lambda$ and each maximal right ideal $I$ of $R$, the submodule $L_{\lambda,I}=\pi_\lambda^{-1}(I)$, where $\pi_\lambda\colon V\to R$ is the projection on the component. Then $V/L_{\lambda,I}$ is isomorphic to $R/I$ and so is simple, which implies $L_{\lambda,I}$ is maximal.

If you take the intersection of all these maximal submodules you clearly get $(\Rad(R))^{\Lambda}$ and therefore $\Rad(V)\subseteq(\Rad(R))^{\Lambda}=V\Rad(R)$. The previously proved inclusion shows equality.

For a projective module $V$, prove that $\Rad(M\oplus N)=\Rad(M)\oplus\Rad(N)$ and apply the equality for free modules to $V\oplus V'$ a free module.


问:
Suppose $\mu_n \to \mu$ weakly for $n \to \infty$. What can we say about $\mu$?
Let $X$ is a metric space and $\mathcal{A}$ the canonical Borel sigma algebra on $X$. Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of probability measures on a measurable space $(X,\mathcal{A})$ and suppose $\mu_n \to \mu$ weakly for $n \to \infty$ i.e.
$$
\int_X f d\mu_n \underset{n \to \infty}{\longrightarrow} \int_X f d\mu
$$
for all $f \in C_b(X)$.

What can we say about $\mu$? Is it a probability measure? A finite measure?

Feel free to link to theorems etc.

答:
Provided $X$ is a Polish (i.e. complete and separable) metric space, $\mu$ is a probability measure. First, we test against the constant function $1$, explicitly

$$ 1 \equiv \int_X 1\,\mathrm{d}\mu_n \to \int_X 1\,\mathrm{d}\mu =\mu(X)$$

It remains to check that $\mu$ is a positive measure. By Hahn's decomposition theorem, write $X=X_+\cup X_-$ which is a disjoint union, with $\mu(E)\ge0$ for all $E\subseteq X_+$ and $\mu(E)\le0$ for all $E\subseteq X_-$. This means also $\mu=\mu_+-\mu_-$ where $\mu_+,\mu_-$ are the positive measures

$$\mu_\pm(E) := \mu(E\cap X_\pm) $$

Then, for any compact $K\subset X_-$ and $N\in\mathbb{N}_+$, consider the non-negative, bounded continuous function

$$ f_{K,N} := \max\{1-N\,\mathrm{dist}(x,K),0\} $$

By weak convergence,

$$ \int_Xf_{K,N}\,\mathrm{d}\mu = \lim_{n\to\infty}\int_Xf_{K,N}\,\mathrm{d}\mu_n\ge0$$

So

$$\int_Xf_{K,N}\,\mathrm{d}\mu_+ - \int_Xf_{K,N}\,\mathrm{d}\mu_-\ge0$$

Now take $N\to\infty$ and apply dominated convergence theorem,

$$-\mu_-(K) = \mu_+(K) - \mu_-(K) \ge0$$

Since $K$ is an arbitrary subset of $X_-$, by regularity of finite Borel measures on Polish spaces we see that $\mu_-$ is the zero measure. Hence $\mu=\mu_+$ is a positive measure.


问:
The subring test
This is how the Wikipedia article on subring defines the subring test

The subring test states that for any ring $R$, a nonempty subset of $R$ is a subring if it is closed under addition and multiplication, and contains the multiplicative identity of $R$.

When you follow the link for the subring test, it is stated as follows

In abstract algebra, the subring test is a theorem that states that for any ring, a nonempty subset of that ring is a subring if it is closed under multiplication and subtraction. Note that here that the terms ring and subring are used without requiring a multiplicative identity element.

My question is, is the first statement of subring test correct? This is also how a subring is "defined" in Atiyah-Macdonald. It seems incorrect to me as $\mathbb{R}_+$ satisfies those conditions and is not a subring unless I am missing something.

Looking at the responses I feel I should further clarify my question. Closure under subtraction and multiplication (with the added provision that the given subset contain the identity depending on how you define your rings), guarantees a subring, as in the second statement. I am comfortable with this statement as I know that closure under subtraction for a subset of a group (written additively) gives a subgroup. My question is whether the first statement is correct - is closure under addition and multiplication enough?

答:
Yes, you're right: the version of the subring test found in the wikipedia article on "subring" was faulty, whereas the article subring test has a correct statement.

I just edited the first wikipedia article to read as follows:

"The subring test states that for any ring R, a subset of R is a subring if it contains the multiplicative identity of R and is closed under subtraction and multiplication.'

I hope all will agree that this is an appropriate statement.

There is a slight discrepancy in that the article on subrings explicitly assumes we are working in the category of rings (in which we have a multiplicative identity which all the homomorphisms much respect), whereas the article on the subring test works in the category of rngs (i.e., there may not be a multiplicative identity and even if there is it need not be preserved by homomorphisms).  In the category of rngs, one should state the nonemptiness explicitly, whereas in the category of rings it is guaranteed by the presence of the multiplicative identity.

If anyone has further ideas for improving either of these two articles, please let me know.  Or rather, please go ahead and implement them -- be bold, as they say on that other site -- but it would be nice to come back here and tell us what you've done.

Title: The subring test


问:
Row operations do not change the dependency relationships among columns?
My book says this but doesn't explain it:

Row operations do not change the dependency relationships among columns.

Can someone explain this to me? Also what is a dependency relationship? Are they referring to linear dependence?

答:
Suppose a matrix A and one of its columns $v_k$ is a linear combination of the others, we can form an equation as such:

$$a_1v_1 + \ldots + a_{k-1}v_{k-1} + a_{k+1}v_{k+1} + \ldots + a_nv_n = a_kv_k$$

If the column vectors of A are linearly independent, $a_1, a_2, \ldots, a_n$ are all zero. This doesn't affect our proof.

Now we can rewrite the equation to:

$$A'x = b$$

where the columns of A' are $v_1, \ldots, v_{k-1}, v_{k+1}, \ldots, v_n$, and x = $[a_1, \ldots, a_{k-1}, a_{k+1}, \ldots, a_n]^T$, and $b = a_kv_k$.

Since we know that the equation still holds after row operations on the augmented matrix of A', thus they don't change the linear relationship of columns of A.


问:
Prove inequality bound for truncated random variables $X_{nk\delta}, X_{nk1}$
I need to prove a theorem from "The General Central Limit Theorem for Real and Banach valued random variables" (auth. E. Giné, A. Araujo) that gives necessary and sufficient conditions for weak convergence.

More specifically I need help proving the inequality below.

The inequality:

Notation etc.

$0 < \delta < 1 $

$ X_{nk1} = X_{nk} \ 1_{\{ X_{nk} \leq 1 \}} $

$X_{nk\delta} =    X_{nk} \ 1_{\{ X_{nk} \leq \delta \}}$

$\mu_{nk} = \mathcal{L}(X_{nk})$ $ \ \ \ $ (the distribution of $X_{nk}$)

My initial computation:

\begin{align*}
& \ \ \ \ \ \sum_{k=1}^{k_n} \bigg\vert \int _{\{ \vert x \vert \leq \delta \}} (x-E[X_{nk1}])^2 d\mathcal{L}(X_{nk1})(x)-\int _{\{ \vert x \vert \leq \delta \}} (x-E[X_{nk\delta}])^2 d\mathcal{L}(X_{nk\delta})(x)\bigg\vert \\[1em]
&= \sum_{k=1}^{k_n} \bigg\vert \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk1})(x) + \int _{\{ \vert x \vert \leq \delta \}} (E[X_{nk1}])^2 d\mathcal{L}(X_{nk1})(x) \\[1em]
& \quad \quad \quad - 2\int _{\{ \vert x \vert \leq \delta \}} xE[X_{nk1}] d\mathcal{L}(X_{nk1})(x)- \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk\delta})(x) \\[1em]
& \quad \quad \quad - \int _{\{ \vert x \vert \leq \delta \}} (E[X_{nk\delta}])^2 d\mathcal{L}(X_{nk\delta})(x) + 2\int _{\{ \vert x \vert \leq \delta \}} xE[X_{nk\delta}] d\mathcal{L}(X_{nk\delta})(x) \bigg\vert\\[1em]
&= \sum_{k=1}^{k_n} \bigg\vert \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)-2E[X_{nk1}]\int _{\{ \vert x \vert \leq \delta \}} x d\mathcal{L}(X_{nk})(x) + \mu_{nk}([-\delta,\delta])(E[X_{nk1}])^2 - \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)-\mu_{nk}([-\delta,\delta])(E[X_{nk\delta}])^2 + 2E[X_{nk\delta}]\int _{\{ \vert x \vert \leq \delta \}} x d\mathcal{L}(X_{nk})(x) \bigg\vert \\[1em]
\end{align*}

Thoughts: Now I can't quite see how to go from here. It also puzzles me quite a bit that I need the complements $[-1,1]^c,[-\delta,\delta]^c $ somehow. At some point we will probably use the triangle inequality and pull out max from the sums. Am I on the right track and if so can anyone help me finish the computation?

答:
I don't have a complete answer, but at least I have a comment that I think goes most of the way. A simple version of this problem is when $X_{nk}$ is symmetric and supported on $[-1,1]$, so that $E(X_{nk})= E(X_{nk1}) = E(X_{nk\delta})=0$, and $X_{nk} = X_{nk1}$ in which case the problem reduces to (when $k_n=1$)

$$\bigg\vert \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)-\int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk\delta})(x)\bigg\vert \le 2 \delta \mu_{nk}[-\delta, \delta]^c. $$

Intuitively $\int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk\delta})(x)>  \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)$, since the law of $X_{nk\delta}$ puts more mass on subsets of $[-\delta, \delta]$. Also, for any measurable subset $A$ of $[-\delta, \delta]$, $\mu_{nk\delta}(A) - \mu_{nk}(A) \le \mu_{nk}([-\delta,\delta]^c)$ (expand $\mu_{nk\delta}(A)$ as a sum of two terms depending on whether $X_{nk} \in [-\delta,\delta]$). Use this fact and the definition of the integral to get the bound in this simple case.


问:
Evaluate the following limit using Taylor
Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}$. Also $\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$ I am struggeling with finding the Taylor series for $\sin^2x^2$. So far I have this:
\begin{align*}
\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x.
\end{align*}
And by substituting $x$ with $2x$ in the Taylor series of $\cos x$ we get $\cos 2x=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}$.
This gives us
\begin{align*}
\sin^2 x&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}\\
&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}2^{2k}x^{2k}.
\end{align*}
And then, by substituting $x$ with $x^2$ we get
\begin{align*}
\sin^2 x^2 =\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}2^{2k}x^{4k}.
\end{align*}
Is this correct?

This would give:
\begin{align*}
&\lim_{x\to 0}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\\
&=\lim_{x\to 0}\frac{1+x^2+O(|x|^4)+2-\frac{2x^2}{2!}+O(|x|^4)-3}{\frac{1}{2}-\frac{1}{2}(1-\frac{4x^4}{3!}+O(|x|^8))}\\
&=\lim_{x\to 0}\frac{O(|x|^4)}{\frac{2x^4}{3!}-O(|x|^8)}\\
&=\lim_{x\to 0}\frac{1}{\frac{2}{6}-O(|x|^4)}\\
&=\frac{1}{\frac{1}{3}}\\
&=3
\end{align*}
Could someone tell me if this is correct? Thanks!

Body:
Note that $\sin x\le 1$ and $\cos \ge -1.$
$$\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge{e^{x^2}-5}.$$
So
$$\lim_{x\to\infty}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\ge \lim_{x\to\infty}{e^{x^2}-5}.\to\infty$$


问:
Exponential distribution with an upper bound
An exponential distribution is often used to model the lifespan of a component with a constant failure rate. What is a suitable distribution to use in the situation where a component has a constant failure rate $\lambda$ but that after a certain time $T$ if the part has not failed it is replaced anyway?

The question is whether there is a continuous distribution that approximates the following:
$$f(x;\lambda) = \begin{cases}
0 & x < 0,\\
\lambda  e^{-(\lambda x)} & x \ge 0 <T \\
1 & x\ge T.
\end{cases}$$

My best guess would be to use an appropriately scaled $Beta(\alpha,\beta)$ distribution.

I haven't worked it out fully but $\alpha$ and $\beta$ would obviously need to depend on $\lambda$ and $T$. For instance when $T >> \lambda$ the exponential distribution again would be a fair approximation.

Edited: added explicit definition of the distribution following 5201314's contribution.

答:
For a time to failure $X$, constant failure rate $\lambda$ and given that it is replaced after a certain time $T$ if it hasn't failed, $$F_X(x)=\begin{cases}
1-e^{-\lambda x}, & x<T\\
1, & x\ge T
\end{cases}$$


问:
Let $A$ and $B$ be complex $n\times n$ matrices such that $AB−BA$ is invertible and such that $A^2+B^2=c(AB−BA)$ for some rational number $c$.
Let $A$ and $B$ be complex $n\times n$ matrices such that $AB−BA$ is invertible and such that $A^2+B^2=c(AB−BA)$ for some rational number $c$. Prove $c\in\{−1,0,1\}$ and show that $n$ is a multiple of $4$ when $c\neq0$.

Please give some hints for this problem. I could not find a way to proceed. I had taken determinant on both sides. But it does not helping much

答: This can't be right. There are $n^2$ equations and $2 n^2$ unknown over complex numbers - there will be solution almost always.


问:
Should I be taking the partial derivative or the total derivative?
Say I have the following relationship
$$\frac{d}{dx}f(x) = y$$
where $x,y\in\mathbb{R}$, such that $f:\mathbb{R}\to\mathbb{R}$.  The function $f$ is Legendre Type, such that I can take its legendre transfrom (convex conjugate) and also yield the following relationship
$$
g(y) = f^{\ast}(y)
$$
So, in that first expression, in taking the derivative, I am finding the argument of the convex conjugate.

I now construct $f$ as a linear combination of basis functions, as follows

$$
f(a_1,\cdots,a_n,x) =\sum_ia_if_i(x)
$$
where $a_i ,x \in \mathbb{R}\;\forall i$ such that we now have $f:\mathbb{R}^{n+1} \to \mathbb{R}$ and $f_i:\mathbb{R} \to \mathbb{R}$.  Continuing with the Legendre transform relationships, where all $f_i$ are strictly convex and Legendre type, I therefore have
$$
g(a_1,\cdots,a_n,y) = \sum_ia_if^{\ast}\left(\frac{y_i}{a_i}\right)
$$

Again, I want to find $y$.  I am not sure if I should be taking the partial derivative with respect to $x$, holding all $a_i$ constant, or the total derivative with respect to $x$.

In other words, which of these is correct?
$$
\frac{\partial}{\partial x}f(a_i,\cdots,a_n,x) = \sum_ia_i\frac{\partial}{\partial x}f_i(x) = y
$$
or
\begin{align}
\frac{d}{dx}f(a_i,\cdots,a_n,x) &= \sum_ia_i\frac{\partial}{\partial x}f_i(x)\; + \;\sum_i\frac{\partial f}{\partial a_i}\frac{\partial a_i}{\partial x} \\
&=\sum_ia_i\frac{\partial}{\partial x}f_i(x)\; + \;\sum_if_i(x)\frac{\partial a_i}{\partial x}\\
&= y
\end{align}

Or perhaps neither?

Prompted by a comment, I should note that (at least some) $\frac{\partial a_i}{\partial x}$ are non-zero, so the two approaches are not equivalent.

答:
In calculating the form for g

$$
g(a_1,\cdots,a_n,y) = \sum_ia_i\frac{\partial}{\partial x}f_i(x) = y
$$

we hold the values of the variables $a_i$ constant, so it makes sense to do so in the derivative too.

Therefore, I think its sensible to take the partial derivative, holding $a_i$ constant.
This is also fairly simple to show numerically/empirically for a few choice examples.


问:
Choosing an answer randomly
You are asked a question and you don't have a clue about it. But, luckily you are given $K$ ($K \ge 4$) possible answers and only one is the right one. Because you choose your answer randomly you would like to increase your chance of answering the question right. You have two options:

Choose an answer randomly.

Choose an answer randomly, exclude it and then choose another answer randomly from the other $K-1$ given answers.

So the question, would you have better chance if you choose option 2 than option 1? And if it's better, how many times would be the best option, and stopping before it become a bad idea (excluding large part of the answers).

Let's say. $K=4$. Then you'll have $25$% odds choosing the right answer with option 1.
If you exclude that answer you have $33$% or $0$% odds of choosing the right answer.

答:
Intuitively, it should be clear that picking one, throwing it away, and picking another will give the same probability as just picking one.

We can explain it a little more using the notion of conditional probability:

Suppose you follow the procedure of picking one answer at random, throwing it away, and picking one of the remaining ones as your final answer.

The probability that you pick the correct answer on the first try is $1/4$ and the probability that you do not pick the correct answer on the first try is $3/4$.

The probability that you pick the correct answer on the second try assuming that you picked the correct answer on the first try is $0$. Thus the probability that you pick the correct answer on the first try and again on the second try is $0$.

The probability that you pick the correct answer on the second try assuming that you did not pick the correct answer on the first try is $1/3$.

Thus the probability that you do not pick the correct answer on the first try and you pick the correct answer on the second try is $\frac 3 4 \frac 1 3 = 1/4$.

Thus the total probability that you pick the correct answer on the second try is $0 + 1/4 = 1/4$, which is the same you'd have gotten otherwise.

Title: Choosing an answer randomly


问:
Given a system of ODE'S, find value of $I$ when $S=\frac{\gamma}{r}$
I've got the following system of equations:
$$\begin{matrix}
    \dot{S}(t) && = && -rSI-\lambda S\\
    \dot{I}(t) && = &&rSI-\gamma I
    \end{matrix}$$
where all constants are greater than $0$. Therefore
$$\frac{dI}{dS}=\frac{rSI-\gamma I}{-rSI - \lambda S}$$
I want to find the value of $I$ when $S=\frac{\gamma}{r}$. I think I will have to solve the equation above to do this, but I don't really know how to do this. Can someone help me?

Edit This is an extract of a SIR model. The $I$ coordinate I am looking for is the maximum number of infected people.

答:
$$\begin{matrix}
    \dot{S}(t)  =  \frac{dS}{dt} && = &&  -rSI-\lambda S\\
    \dot{I}(t) =  \frac{dI}{dt} && = &&rSI-\gamma I
    \end{matrix}$$
You want to find the value of $I$ when $S=\frac{\gamma}{r}$

$$\frac{dI}{dt}  = rSI-\gamma I = r\frac{\gamma}{r}I-\gamma I=0$$
$$I=\text{constant}$$
This seems strange. Is there a typo in the question ?

Even more strange:  $\quad S=\frac{\gamma}{r}\quad\implies\quad \frac{dS}{dt}=0$

$$\frac{dS}{dt} = (-rI-\lambda) S=0$$
$$I=-\frac{\lambda}{r}$$

Of course $\begin{cases} I=-\frac{\lambda}{r}\\ S=\frac{\gamma}{r} \end{cases}$ is a particular and trivial solution. But without initial condition what is its signifiance ? There is a lack of context.


答:
Neither the numerator nor the denominator has been correctly analysed.

The numerator is $O(x^4)$, but is asymptotic not to $x^4$ as your calculation incorrectly assumes, but $\tfrac12x^4+\tfrac{1}{12}x^4=\tfrac{7}{12}x^4$.

The denominator $\sin^2x^2=\tfrac12(1-\cos2x^2)$ has leading order term $\tfrac14(2x^2)^2=x^4$. You seem to have used the Taylor series for $\sin y$, rather than $\cos y$, with $y=2x^2$. An easier approach is $\sin^2x^2\sim(x^2)^2=x^4$.

So the correct limit is $\tfrac{7}{12}$, as @Atmos said.

Title: Evaluate the following limit using Taylor

Body:
Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \frac{x^{2k}}{k!}$. Also $\cos x=\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}x^{2k}$ I am struggeling with finding the Taylor series for $\sin^2x^2$. So far I have this:
\begin{align*}
\sin^2 x=\frac{1}{2}-\frac{1}{2}\cos 2x.
\end{align*}
And by substituting $x$ with $2x$ in the Taylor series of $\cos x$ we get $\cos 2x=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}$.
This gives us
\begin{align*}
\sin^2 x&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}(2x)^{2k}\\
&=\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}2^{2k}x^{2k}.
\end{align*}
And then, by substituting $x$ with $x^2$ we get
\begin{align*}
\sin^2 x^2 =\frac{1}{2}-\frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}2^{2k}x^{4k}.
\end{align*}
Is this correct?

This would give:
\begin{align*}
&\lim_{x\to 0}\frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}\\
&=\lim_{x\to 0}\frac{1+x^2+O(|x|^4)+2-\frac{2x^2}{2!}+O(|x|^4)-3}{\frac{1}{2}-\frac{1}{2}(1-\frac{4x^4}{3!}+O(|x|^8))}\\
&=\lim_{x\to 0}\frac{O(|x|^4)}{\frac{2x^4}{3!}-O(|x|^8)}\\
&=\lim_{x\to 0}\frac{1}{\frac{2}{6}-O(|x|^4)}\\
&=\frac{1}{\frac{1}{3}}\\
&=3
\end{align*}
Could someone tell me if this is correct? Thanks!


问:
Random points on a circle
Please assist with this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B$. Let $Z$ be the random variable denoting the distance of the point $B$ to its neighbour in the anti-clockwise direction.

Find the pdf of $Z$. (Well there's no measure theory for this problem, but I assume this pdf exists. Of course we can see for ourselves by computing the cdf $F_Z(z)= P(Z \le z)$ 1st and then hope the cdf is absolutely continuous.)

A. My model:

A.1. The circle is bijective with $[0,1)$, so let's call these 3 points $A,B,C$ s.t. they are iid $\sim \ Unif(0,1)$ (or $[0,1)$ or whatever).

A.2. (Not sure if any measure theory needed here, but they are all distinct $\mathbb P$-almost surely.)

A.3. Then $Z$'s image is also $(0,1)$ (or $[0,1)$ or whatever). (more details in part D.)

B. Answer (that I found online):

Edit: it says $Z$ is uniform, but I think this is wrong.

C. My intuition for the answer (but I would like to know precisely please):

Edit: Nvm. I think the answer is wrong.

For parts D,E,F,G: Edit: nvm this is wrong.

H. New section: What exactly is the formula of $Z$ in relation to $A,B,C$? Supposedly $f_Z(z)=2(1-z)1_{(0,1)}(z)$. But what exactly is $Z$? $Z=|A-C|$? $Z=\min\{?,?\}$

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
We can introduce an angular representation for $A, B, C$.

Since $A, B, C$ is considered iid. we pick $B$ as the origin for angles.

The distance between $A,B$ would be arc length $r\beta$, where $r$ is 1.
Similarly $C,B$ would have length $r \alpha$. As both A, C uniformly distributed. So should both $\alpha, \beta \sim U(0, 2\pi)$.

Now our question is simply asking for the distribution of this random variable $\gamma = min(\alpha, \beta)$.

$P(\gamma > z) = P(min(\alpha, \beta) > z) = P(\alpha > z)P(\beta>z) = (2\pi-z)^2 $

In the end you can change the coordinator back if you want :)

Title: Random points on a circle


问:
High school MCQ (geometry) about maximum of sine of an angle
No calculus is allowed in this question.

Also, it is not allowed to use any compound angle formulas such as

$\sin(A-B) =\sin A \cos B - \cos A \sin B$.

How to find the greatest value of $\sin \angle APB$?

I attempted by consider the difference between angles $\angle OPB$ and $\angle OPA$ but I do now know how to make use of the fact that $P$ makes the angle greatest.

答:
This is not a geometric solution and may not be the most recommended approach but meets the condition of not using calculus and no compound angle formula.

As $\angle APB \lt 90^0,$ maximizing $\sin \angle APB \ $ will maximize $\angle APB$.

Say $OA = AB = a, OP = k \cdot a \ $ ($k \gt 0)$

Equating the  area of $\triangle APB$ and $\triangle OAP$

$\frac{1}{2} AP \cdot BP \sin \angle APB = \frac{1}{2} OP \cdot OA$

$\sin \angle APB = \displaystyle \frac{k \cdot a^2}{\sqrt{a^2+k^2a^2} \cdot {\sqrt{4a^2+k^2a^2}}} = \frac{k}{\sqrt{k^4+5k^2+4}}$

$ = \displaystyle \frac{1}{\sqrt{k^2+\frac{4}{k^2}+5}}$

$\sin \angle APB$ is maximum when $k^2+\frac{4}{k^2}$ is minimum.

Now by A.M-G.M, $k^2+\frac{4}{k^2} \geq 4$

Hence $\sin \angle APB = \displaystyle \frac{1}{3}$.


问:
Let $G$ a nilpotent group such that $10$ divides $|G|$. It is true that $G$ has element of order $10$?
Let $G$ a nilpotent group such that $10$ divides $|G|$. It is true that $G$ has element of order $10$?

We know that $G$ can be expressed like  direct sum of Sylow's subgroups and that $G$ has a normal subgroup of order $10$ but how can we answer to this question?

答: Since $G$ is a direct product of its Sylow subgroups, let us call its $2-$ and $5-$ Sylow subgroups $P_2$ and $P_5$.  They are non-trivial since $|G|$ is divisible by $10$.  Taking a non-identity element of $P_2$ and raising to an appropriate power, we have $g\in P_2$ with $o(g)=2$ and similarly we have $h \in P_5$ with $o(h)=5$.  But since $G = P_2 \times P_5 \times \mathrm{Stuff} $, all elements of $P_2$ commute with all elements of $P_5$, so $g$ and $h$ commute.  Therefore, $o(gh)=10$, so the answer is affirmative and $gh$ is the required element.


问:
Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers

Find all $k \in \mathbb{R}$ such that both $\frac{12}{k+1}$ and $\frac{6}{k-1}$ are positive integers

Clearly $k$ is a rational number, so I let $k=\frac{a}{b}$ where $a, b \in \mathbb{Z}$ and $b \neq 0$, this gives $\frac{12b}{a+b}$ and $\frac{6b}{a-b}$ are positive integers. I tried different things here but couldn't make progress. I want to verify that $2, 3$ and $\frac{7}{5}$ are the only possible values for $k$. How to proceed from here? Thanks in advance.

答:
Starting from $\frac{6}{k-1}=n \in \mathbb N$, we can generate all possible rational $k=\frac{n+6}{n}$ without regard to the second condition.

Considering the second condition, $\frac{12}{k+1}=m \in \mathbb N$, we substitute to get $m=\frac{12}{\frac{n+6}{n}+1}=\frac{12n}{2n+6}=\frac{6n}{n+3}$. We need to find positive integers $n$ that yield positive integers $m$.

Note that $\frac{6n}{n+3}$ asymptotically  approaches $6$, so we only have to look at values of $n$ such that $\frac{6n}{n+3} \le 5$, which by algebraic rearrangement yields $n \le 15$. We find $m\in \mathbb N \Rightarrow n= 3,6,15$, which in turn yields $k=3,2,\frac{7}{5}$, which is exactly what you found by trial and error.


问:
Probability of more than 13 balls remaining in the box after some selections
Question

A box contains 20 balls, of which exactly 8 are red and 12 are green. Two balls are randomly selected from the box, without replacement. Then, any red balls selected are put back into the box, while any green balls selected are thrown away. Finally, 5 balls are randomly selected from the box, without replacement. Suppose that, following this procedure, all 5 balls are green. Find the probability that there remains more than 13 balls in the box.

My working

Observe that the only case when there are less than 14 balls in the box is when the two balls that are selected initially are both green, so we have three cases to consider here.

Case 1: Choose 2 red balls, put them back, then choose 5 green balls.

Case 2: Choose 1 red ball, then 1 green ball. Put the red ball back, throw the green one away and then choose 5 green balls.

Case 3: Choose 1 green ball, then 1 red ball. Put the red ball back, throw the green one away and then choose 5 green balls.

$\mathbb{P} (\mathrm {Case\ 1}) = \frac 8 {20} * \frac 7 {19} * \frac {12} {20} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} = \frac {231} {30685}$

$\mathbb{P} (\mathrm {Case\ 2}) = \frac 8 {20} * \frac {12} {19} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} * \frac 7 {15} = \frac {308} {30685}$

$\mathbb{P} (\mathrm {Case\ 3}) = \mathbb{P} (\mathrm {Case\ 2})$

$\therefore \mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {231} {30685} + 2 * \frac {308} {30685} = \frac {847} {30685}$

Answer

$\mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {154} {211}$

My answer is obviously off by a lot ): Am I missing cases, or is my thinking too naive etc.? I am not sure where I have gone wrong, so any intuitive explanations about my mistake(s) and also what the correct solution should be will be greatly appreciated!

P.S. My professor did provide a solution, but I am unable to understand his way of thinking, so I thought posting the problem here and asking for ideas would be better.

答:
So, based on another answer (it was more of a hint), it seems that I have simply forgotten to consider the conditional probability aspect of the problem.

We have indeed 3 cases that fit our criteria, but that is actually out of a total of 4 cases, the fourth being that the first two balls chosen are both green. I conveniently left the fourth case out of my original working and considering it will get us the correct answer, as shown below.

Case 1: Choose 2 red balls, put them back, then choose 5 green balls.

Case 2: Choose 1 red ball, then 1 green ball. Put the red ball back, throw the green one away and then choose 5 green balls.

Case 3: Choose 1 green ball, then 1 red ball. Put the red ball back, throw the green one away and then choose 5 green balls.

Case 4: Choose 2 green balls, throw them away and then choose 5 green balls, which is equivalent to simply choosing 7 green balls.

$\mathbb{P} (\mathrm {Case\ 1}) = \frac 8 {20} * \frac 7 {19} * \frac {12} {20} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} = \frac {231} {30685}$

$\mathbb{P} (\mathrm {Case\ 2}) = \frac 8 {20} * \frac {12} {19} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} * \frac 7 {15} = \frac {308} {30685}$

$\mathbb{P} (\mathrm {Case\ 3}) = \mathbb{P} (\mathrm {Case\ 2})$

$\mathbb{P} (\mathrm {Case\ 4}) = \frac {12} {20} * \frac {11} {19} * \frac {10} {18} * \frac 9 {17} * \frac 8 {16} * \frac 7 {15} * \frac 6 {14} = \frac {33} {3230}$

$\therefore \mathbb{P} (\mathrm {> 13\ balls\ left\ in\ the\ box}) = \frac {\mathbb{P} (\mathrm {Case\ 1}) + \mathbb{P} (\mathrm {Case\ 2}) + \mathbb{P} (\mathrm {Case\ 3})} {\mathbb{P} (\mathrm {Case\ 1}) + \mathbb{P} (\mathrm {Case\ 2}) + \mathbb{P} (\mathrm {Case\ 3}) + \mathbb{P} (\mathrm {Case\ 4})} = \frac {\frac {231} {30685} + 2 * \frac {308} {30685}} {\frac {231} {30685} + 2 * \frac {308} {30685} + \frac {33} {3230}} = \frac {154} {211}$


答:
To see why the second strategy is identical in outcome to that of the first strategy, note that after selecting randomly one answer, say randomly having selected $A$, out of four choices, so you the probability A is correct is $\dfrac 14.\;$

Alternatively, skip randomly chosen A (which has a $25$% chance of being correct, and move to the second answer, so if K = 4, and you exclude A, you have a $(1 - \frac 14) =\frac 34$ probability that one of the remaining choices (B, C, D) are correct, and $\;\frac 13\;$ probability that the choice randomly selected from those three remaining choices (B, C, D) is correct.

In all, with the second method, you have a $\dfrac 34 \times \dfrac 13 = \dfrac 14$ probability of selecting the correct answer using the second method...and so on:

If you exclude randomly selected A, then randomly selected C, then you'll have a $100 - 25 - 25 = 50$% chance that $B$ or $D$ is correct. Randomly selecting one of those will give you gives you a $50$% chance that if one of the two are correct, selecting it. But that means you have a $\frac 12 \times \frac 12 = \frac 14$ (25%) chance of selecting the correct answer if exluding two options randomly, and randomly selecting among the final two options.

And so too with excluding 3 gives you a 25% chance that the one not excluded, hence must be selected, is correct.

So with either method, you have a $25\%$ chance of selecting the correct answer for any given question.

Of course, if you study even just just a little bit before the exam, and attend class regularly, you can surely improve those odds, and not have to resort to "random selection." ;-)

Title: Choosing an answer randomly

Body:
You are asked a question and you don't have a clue about it. But, luckily you are given $K$ ($K \ge 4$) possible answers and only one is the right one. Because you choose your answer randomly you would like to increase your chance of answering the question right. You have two options:

Choose an answer randomly.

Choose an answer randomly, exclude it and then choose another answer randomly from the other $K-1$ given answers.

So the question, would you have better chance if you choose option 2 than option 1? And if it's better, how many times would be the best option, and stopping before it become a bad idea (excluding large part of the answers).

Let's say. $K=4$. Then you'll have $25$% odds choosing the right answer with option 1.
If you exclude that answer you have $33$% or $0$% odds of choosing the right answer.


答:
Use $\sin x \approx x$ in the denominator, it becomes $x^4$, next expand numerator complete up to $x^4$.Then use $e^{z}=1+z+z^2/2+O(z^3)$, $\cos z =1-z^2/2+z^4/24+O(z^6)$ when $|z|$ is very small
$$L=\lim_{x\to\infty} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}=\lim_{x\to 0} \frac{1+x^2+x^4/2+2(1-x^2/2+x^4/(24)-3}{x^4}$$.
$$L=\lim_{x \to 0} \frac{7x^4/12}{x^4}=\frac{7}{12}.$$


问:
Finding a matrix to force a part of another matrix to be with null trace
Assume I have a matrix $X$ whose size e.g., is $16$x$16$   and let's the matrix $x$ = $X_{(13:16,13:16)}$; it means it's the last square sub-matrix taken from the matrix $X$.  We know that $(x^H)^{-1}$$x^{-1} = z$ , where $(.)^H$ is the transpose of the matrix and $z$ is a scalar diagonal matrix having all its diagonal elements equal to e.g., $4$.

My question, I need to set a matrix $Y$ where $XY$ will make the trace of matrix $z$ null. it means for example a half of diagonal elements of matrix $z$ should be forced to be negative and the others positive. how can I select that matrix Y ?

答:
Here is how it can be solved in YALMIP (under MATLAB) as a nonlinear equation solve, which I am handling by specifying the nonlinear equations as constraints to a nonlinear optimization problem not having any objective function (to be minimized or maximized).

Let X21 denote the lower left 12 by 4 block of X.

Let X22 denote the lower right 4 by 4 block of X.

I will create "optimization" variables:

Y12, which is a 12 by 4 matrix variable consisting of the upper right 12 by 4 block of the matrix Y which you seek.

Y22, which is a 4 by 4 matrix variable consisting of the lower right 4 by 4 block of the matrix Y which you seek.

Let xy denote the lower right 4 by 4 block of $XY$, i.e., the analog of x as applied to $XY$. By use of block matrix multiplication, it can be seen that $xy$ depends only on the matrix blocks mentioned above. So, for instance, the upper left 12 by 12 blocks of X and Y are totally irrelevant to the problem at hand, as are the lower left 4 by 12 block of Y and upper 12 by 16 block of X.

Because $xy^{-1}$ can't be directly handled by YALMIP, I will create a variable xy_inverse. In order to make it meaningful as such, impose the constraint that xy*xy_inverse equals the Identity matrix.

Then the "z" corresponding to $xy$, is constrained to have trace equal to zero.

The solver is then called to try to find a solution. It may be that there is no solution, i.e., the problem is infeasible. In general, the problem need not be feasible.

In order to be sure the problem is infeasible, that determination must be made by a rigorous global optimization solver (not genetic algorithm and similar heuristic algorithms which are called global optimization). The solver I specified, bmibnb, is a rigorous global optimization solver included with YALMIP.

```<code>Y12 = sdpvar(12,4,'full');
Y22 = sdpvar(4,4,'full');
xy = X21*Y12+X22*Y22;
xy_inverse = sdpvar(4,4,'full');
Constraints = [xy*xy_inverse == eye(4),trace(xy_inverse'*xy_inverse) == 0];
optimize(Constraints,[],sdpsettings('solver','bmibnb'))
</code>```

If a feasible solution has been found, value(Y12) and value(Y22) contain values of Y12 and Y22 which meet the requirements.  The other blocks of Y can be anything you want and don't affect whether the requirements are met.

Body:
Here is how it can be solved in YALMIP (under MATLAB) as a nonlinear equation solve, which I am handling by specifying the nonlinear equations as constraints to a nonlinear optimization problem not having any objective function (to be minimized or maximized).

Let X21 denote the lower left 12 by 4 block of X.

Let X22 denote the lower right 4 by 4 block of X.

I will create "optimization" variables:

Y12, which is a 12 by 4 matrix variable consisting of the upper right 12 by 4 block of the matrix Y which you seek.

Y22, which is a 4 by 4 matrix variable consisting of the lower right 4 by 4 block of the matrix Y which you seek.

Let xy denote the lower right 4 by 4 block of $XY$, i.e., the analog of x as applied to $XY$. By use of block matrix multiplication, it can be seen that $xy$ depends only on the matrix blocks mentioned above. So, for instance, the upper left 12 by 12 blocks of X and Y are totally irrelevant to the problem at hand, as are the lower left 4 by 12 block of Y and upper 12 by 16 block of X.

Because $xy^{-1}$ can't be directly handled by YALMIP, I will create a variable xy_inverse. In order to make it meaningful as such, impose the constraint that xy*xy_inverse equals the Identity matrix.

Then the "z" corresponding to $xy$, is constrained to have trace equal to zero.

The solver is then called to try to find a solution. It may be that there is no solution, i.e., the problem is infeasible. In general, the problem need not be feasible.

In order to be sure the problem is infeasible, that determination must be made by a rigorous global optimization solver (not genetic algorithm and similar heuristic algorithms which are called global optimization). The solver I specified, bmibnb, is a rigorous global optimization solver included with YALMIP.

Y12 = sdpvar(12,4,'full');
Y22 = sdpvar(4,4,'full');
xy = X21*Y12+X22*Y22;
xy_inverse = sdpvar(4,4,'full');
Constraints = [xy*xy_inverse == eye(4),trace(xy_inverse'*xy_inverse) == 0];
optimize(Constraints,[],sdpsettings('solver','bmibnb'))

If a feasible solution has been found, value(Y12) and value(Y22) contain values of Y12 and Y22 which meet the requirements.  The other blocks of Y can be anything you want and don't affect whether the requirements are met.


答:
$$L=\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}$$

Lets try using our dear L'Hopital's rule to simplify the expression.

$$L=\lim_{x\to 0} \frac{2xe^{x^2}  -2\sin x }{2x\sin (2x^2)}$$

We are very familiar with these functions! So lets expand them since they're easier to deal with now.

$$L=\lim_{x\to 0} \frac{x(1+x^2+\cdots)  -(x-\frac{x^3}{6}+\cdots) }{x(2x^2-\frac{8x^6}{6})}=\lim_{x\to 0} \frac{\frac{7x^3}{6}+\cdots}{2x^3+\cdots}=\bbox[10px,#ffb]{\frac{7}{12}}$$


问:
Putting colored balls into colored urns

There are six balls and urns each. They are equally distributed in three colors: R, W, and B. Place the balls randomly in the six urns, arranged linearly. What is the probability that no ball is in an urn of the same color?

We can think of this as a matching problem, consider $X = \{r_1,r_2,b_1,b_2,w_1,w_2\}$. We want to find the number of bijections $f:X\to X$ such that:

$f(r_i) \ne r_j$ for $1\le i,j\le 2$

$f(b_i) \ne b_j$ for $1\le i,j\le 2$

$f(w_i) \ne w_j$ for $1\le i,j\le 2$

The total number of bijections is $6! = 720$. Define a few sets:

$R = \{f: X\to X \text{ is a bijection }: f(r_i)\ne r_j, 1\le i,j\le 2\}$

$B = \{f: X\to X \text{ is a bijection }: f(b_i)\ne b_j, 1\le i,j\le 2\}$

$W = \{f: X\to X \text{ is a bijection }: f(w_i)\ne w_j, 1\le i,j\le 2\}$

The answer we require then is $1 - \frac{N}{720}$, where $N = |(R\cap B\cap W)^c| = |R^c\cup B^c\cup W^c|$. Using inclusion-exclusion principle, we know that, $$|R^c\cup B^c\cup W^c| = |R^c| + |B^c| + |W^c| - |R^c\cap B^c| - |R^c\cap W^c| - |W^c\cap B^c| + |R^c \cap B^c\cap W^c|$$

Any other ways to approach the problem? Thanks a lot!

答:
In the answers above (and in the question) it is assumed that the balls and urns are distinct. If we let identically colored balls be identical (which is reasonable), we arrive at the same result nonetheless! I write this answer to merely demonstrate that. The urns are always distinct since their linear ordering distinguishes them.

The multiset $\mathcal B = \{R,R,B,B,W,W\}$ consists of balls, and the set $\mathcal U = \{R_1,R_2,B_1,B_2,W_1,W_2\}$ consists of the urns. The urns are assumed distinct (even though their colors may be identical) since they are arranged in a straight line - the linear ordering helps us distinguish between any two urns.

The total number of ways of putting balls into urns (no restrictions) would be the same as the total number of ways of arranging the balls in a straight line. In other words, $$\frac{6!}{(2!)^3} = 90$$
Now, let's find the total number of desirable arrangements!

Notation: $\{B_1,...,B_i\} \to \{U_1,...,U_i\}$ is used to replace the following statement: "The ball $B_j$ is put in envelope $U_j$, for all $1\le j\le i$".

Suppose $\{R,R\}\to \{B_1,B_2\}$. This uniquely determines the entire arrangement, as $\{B,B,W,W\} \to \{R_1,R_2,W_1,W_2\}$.

Suppose $\{R,R\}\to \{W_1,W_2\}$. This uniquely determines the entire arrangement, as $\{B,B,W,W\} \to \{R_1,R_2,B_1,B_2\}$.

Suppose $\{R,R\} \to \{B_1,W_1\}$. There are two possibilities for the remaining balls, namely $\{B,B,W,W\} \to \{R_1,W_2,R_2,B_2\}$ and $\{B,B,W,W\} \to \{R_2,W_2,R_1,B_2\}$. Simiarly, we get $3\times 2 = 6$ more favorable arrangements corresponding to $\{R_1,R_2\} \to \{B_2,W_2\}$, $\{R_1,R_2\} \to \{B_1,W_2\}$ and $\{R_1,R_2\} \to \{B_2,W_1\}$.

From (1), we have 1 arrangement. From (2), we have 1 arrangement. From (3), we have $4\times 2 = 8$ arrangements. The required probability is then
$$\frac{1+1+2\times 4}{90} = \frac{10}{90} = \frac{1}{9}$$

Title: Putting colored balls into colored urns


问:
Maximising Question
I need to maximise $P$ under these conditions. First, $$P=N[1.4W-0.31L]\,\text{ and }\,W\le19,\,7\le{L}\le11,\,N\in\{16,17\}.$$ I did this using elemantary methods. In order to maximise $P$, we need to maximise $[1.4W-0.31L]$, and I thought since $W$'s coefficient is positive and $L$'s coefficient is negative, to maximise we need to use maximum $W$ and minimum $L$, so $W=19$ and $L=7$, then we get $24,43$. Since we are looking for $\max\{P\}$, we need maximum $N$, and $N$ can be either $16$ or $17$ so $\max\{P\}=415,31$.

I have solved this question like above. However, my teacher asked me to write an essay and I reached this question and I need to do higher level mathematics like use of derivative. Can you suggest me another solution which involves more complicated mathematics?

答: Your solution is fine.  Yes, you can write things like $\frac {dP}{dW}=1.4N$ and argue that  since this is never zero the maximum must occur at one end of the interval, and since $N \gt 0$ it must be the high end, so $W=19$.  Really this is just showing off the fact that you can write the derivative and using the reasoning you already have.  I would clarify with your teacher whether the point of the essay is to expand the reasoning you give or to describe the process of using the derivative so that if you had a non-linear function you would find the optimum.  If the second, it would be better to pick a function that has its optimum in the range instead of at one end (or do one of each.)</p>

Title: Maximising Question


问:
Convolution of two Gaussians is a Gaussian (semi-group property)
I would like to show that
$$
g_{\sigma}(x, y) * g_{\tau}(x, y)=g_{\sqrt{\sigma^{2}+\tau^{2}}}\left(x, y\right)
$$
where $g_{\sigma}(x, y)=\frac{1}{2 \pi \sigma^{2}} e^{-\frac{x^{2}+y^{2}}{2 \sigma^{2}}}$. My attempt so far:

Since the Fourier transform of $g_{\sigma}$ is given by
$$
\hat{g}_{\sigma}(x, y) = e^{-(\alpha^2 + \beta^2)2\pi\sigma^2}
$$
However I simply cannot simplify/get the product
$$
\hat{g}_{\sigma}(x, y) \hat{g}_{\tau}(x, y)= e^{-(\alpha^2 + \beta^2)2\pi\sigma^2} e^{-(\alpha^2 + \beta^2)2\pi\tau^2}
$$
to $g_{\sqrt{\sigma^{2}+\tau^{2}}}\left(x, y\right)$  - am I on the wrong track here, or is it just some manipulation that needs to be done?

Body:
You are almost there. Use $$e^Ae^B=e^{A+B}$$
In your case $$e^{-(u^2+v^2)2\pi\sigma^2}e^{-(u^2+v^2)2\pi\tau^2}=e^{-(u^2+v^2)2\pi(\sigma^2+\tau^2)}=e^{-(u^2+v^2)2\pi(\sqrt{\sigma^2+\tau^2})^2}=\mathcal{F}\{G(x, y, \sqrt{\sigma^2+\tau^2})\}$$


问:
Expected value and variance from random variable
Given the following

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

$Y_2=α+βY_1+U \;where \; Y_1 \;and \;U\;is\;independent\;and\;U∼\mathcal{N}(0,v^2)$

Let $μ=350$ and $σ^2 =12365$

How would i calculate the expected value and variance from $Y_2$ ?
And how can i find the distribution of it?

答:
Pretty much by definition of expected value: $$E(Y_2) = E(\alpha + \beta Y_1 + U)\\ = E(\alpha) + E(\beta Y_1)+ E(U)\\ = \alpha + \beta E(Y_1) + E(U)$$

And variance:
$$var(Y_2) = E(Y_2^2) - E(Y_2)^2$$
We need to find $E(Y_2^2)$, which is

$$E\left[(\alpha + \beta Y_1 + U)^2\right]\\
= E\left[\alpha^2 + \alpha \beta Y_1 + \alpha U\\ + \alpha \beta Y_1 + \beta^2 Y_1^2 + \beta Y_1 U\\ + \alpha U + \beta Y_1 U + U^2\right].$$

The rest you can probably do with linearity and some arithmetic, I hope this helps.

Edit: Most of the work comes from linearity of the expect value function.


问:
How do I find the lateral surface area of an elliptical cone? (not a frustum)
The formula for the lateral surface area of a right circular cone is $\pi*r*l$ where $r$ is the circle's radius.

But ellipse has a semi-major axis as well as a semi-minor one.

Which one of them should be implemented in the formula and  how?

答:
If your cone is a section of a circular right cone, then a simple answer is possible.

Let $\alpha$ be the semi-aperture of the cone, $\beta$ the angle between the axis of the cone and the normal to the intersecting plane, $q$ the distance of the vertex from the intersection between the plane and the axis of the cone.

A parametric equation of the conic surface is then:
$$
x=t\sin\alpha\cos\phi,\quad
y=t\sin\alpha\sin\phi,\quad
z=t\cos\alpha,
$$
where $t$ is the distance from vertex $(0,0,0)$ and $\phi$ is the azimuthal angle about $z$-axis (which is the axis of the cone).

A suitable equation for the plane is simply $z=(\tan\beta) x+q$, and substituting here $x$ and $z$ from above one gets:
$$
t={q\over \cos\alpha-\tan\beta\sin\alpha\cos\phi},
$$
which is the distance from the vertex to a generic point $P$ on the ellipse at the base of the cone.

The area of the lateral surface is then:
$$
A=\int_0^{2\pi}\int_0^t r\sin\alpha\,dr\,d\phi=
{q^2\sin\alpha\over2\cos^2\alpha}
\int_0^{2\pi}{d\phi\over (1-\tan\alpha\tan\beta\cos\phi)^2}=
{\pi q^2\sin\alpha\over\cos^2\alpha}
{1\over(1-\tan^2\alpha\tan^2\beta)^{3/2}}.
$$


问:
Quadratic Equation Modulo an even composite
I am familiar with using the quadratic formula and Tonelli-Shanks with Hensel's Lifting Lemma to solve a quadratic equation, but how do I solve a quadratic equation in an even modulus? I can't use the quadratic formula because of the division by two. Would it be sufficient to multiply both sides of the equation (including modulus) by 2?

In particular, if I try to solve:

$x^2 + bx + c\equiv 0 \ (mod\ 2pq)$, where $p\equiv q\equiv 1\ (mod\ 2)$

I would end up with:

$\frac{-b \pm \sqrt{b^2-4c}}{2} \equiv 0\ (mod\ 2pq)$

$= -b \pm \sqrt{b^2-4c} \equiv 0\ (mod\ 4pq)$

does this work? Or am I terribly, terribly wrong?

edit:
To start with a simple one:
$x^2 + 5x + 6 \equiv 0\ (mod\ 28)$
the solutions are: $\{5,18,25,26\}$ I was able to find them using the quadratic formula without the division by 2 and then testing both possible solutions to the multiplicative inverse of 2 from the results there. However, I wasn't able to determine a pattern between finding those solutions and testing them versus checking the equation mod 56 either.

edit: corrected the solution set for the example.

答:
Here is a quick recipe:

Do it with one prime power factor of the modulus at a time.

For an odd prime power modulus you can use the quadratic formula and Hensel lifting.

When the modulus is a power of two, you can use this question and its answer.

In the end you can combine the solutions modulo different prime powers by an application of the Chinese Remainder Theorem.


Title: Prove $7^{71}>75^{32}$

Body:
My math teacher left two questions last week, prove (1) $6^9>10^7$ and (2) $7^{71}>75^{32}.$

I did the first question: \begin{align}\frac{6^9}{10^7}&=\frac{4}{5}\times\frac{27^3}{25^3}\\&=0.8\times1.08^3\\&>0.8\times(1+3\times0.08+3\times0.08^2)\\&>0.8\times(1+3\times0.086)\\&>0.8\times1.25=1.\end{align}

But I can't work out the second,  I calculated it out on my computer, $\frac{7^{71}}{75^{32}}=1.000000949\cdots$

Body:
$\color{green}{\textbf{Improved version.}}$

There is an alternative approach.

As it was shown in the my first answer,
$$R=\dfrac{75^{32}}{7^{71}}=\dfrac{5^2}{3^3}\,\dfrac{5^6}{3^7\cdot7}\,\left(\dfrac{3^3\cdot5^4}{7^5}\right)^{14}
 = \dfrac{625^2}{7\cdot 243^2}\,\left(\dfrac{16875}{16807}\right)^{14}.\tag1$$

Identity $(1)$ can be presented in the form of
$$R= \dfrac{625^2}{7\cdot 243^2}\,\left(\dfrac{225}{224}\right)^{14}
\,\left(\dfrac{2400}{2401}\right)^{14}
=\dfrac{625^2}{7\cdot 243^2}\,\left(\dfrac{225}{224}\right)^{14}
\,\left(1-\dfrac{1}{2401}\right)^{14}\tag{1.1}$$

$$<\dfrac{625^2}{7\cdot 243^2}\,\left(\dfrac{225}{224}\right)^{14}
\,\left(1-\dfrac{14}{2401}+\dfrac{91}{2401^2}\right),\tag{1.2}$$

$$R < \dfrac{35557}{37845}\,\left(\dfrac{225}{224}\right)^{14}.\tag2$$

Since
$$\dfrac{225^2}{224^2} = 1+\dfrac t7,\quad\text{where}\quad
t=\dfrac{449}{7168},\tag3$$
one can get
$$\left(1+\dfrac t7\right)^7 = 1 + \dfrac 77t + \dfrac{21}{7^2}t^2+\dfrac{35}{7^3}t^3 +\dfrac{35}{7^4}t^4+\dfrac{21}{7^5}t^5+\dfrac7{7^6}t^6+\dfrac1{7^7}t^7$$
$$< 1+t+\dfrac37t^2\left(1+\dfrac5{21}t+\dfrac {5^2}{21^2}t^2+\dots\right)
=1+t+\dfrac{9t^2}{21-5t} = \dfrac{21+16t+4t^2}{21-5t},$$
$$\dfrac{35557}{37845}\left(1+\dfrac t7\right)^7-1
< \dfrac{35557}{37845}\cdot\dfrac{21+16t+4t^2}{21-5t}-1 $$
$$=\dfrac{142228 t^2 + 758137 t - 48048}{37845(21-5t)},$$
wherein
$$(142228 t^2 + 758137 t - 48048)\bigg|_{\large\frac{449}{7168}}
= -\dfrac{7\,828\,635}{12\,845\,056} < 0.$$
Therefore,
$$\dfrac{35557}{37845}\left(1+\dfrac t7\right)^7 < 1$$
and $\;\color{brown}{\mathbf{R<1}}.\;$

Proved!


问:
A simple proof about finite subsets of a set
What is the simplest proof that for every cartesian square $A\times A$, which is a subset of $\Gamma$, the set $A$ is finite?

$$\Gamma = \left\{ ( x ; x) \hspace{.5em} | \hspace{.5em} x \in [ 0 ; 1] \right\}
\cup \left\{ ( x ; 0) \hspace{.5em} | \hspace{.5em} x \in [ 0 ; 1] \right\} \cup
\left\{ ( 0 ; x) \hspace{.5em} | \hspace{.5em} x \in [ 0 ; 1] \right\} .$$

The graph of $\Gamma$ follows:

答:
Suppose that $A$ contains more than one non-zero point: $y,z$. Without loss of generality $y<z$. So we have that $(y,z)$ is not of the form $(y,y)$ nor $(0,y)$ nor $(y,0)$. Therefore $A\times A$ is not a subset of $\Gamma$.

So it's not only finite, it has at most one non-zero point, so at most two points.


问:
Prove if $\operatorname{adj}(A) = O$ then $\operatorname{rank}A \leq n-2$
Prove if $\operatorname{adj}(A) =0$, then $\operatorname{rank}A \leq n-2$

My initial idea is to setup an induction. Base case $A \in M_{2×2}$ which obviously implies that $A = [O]_{2×2} \rightarrow \operatorname{rank}A \leq 0 = n-2$. However, I can't seem to reason how assuming it worked in the $M_{n\times n}$ case allows for the $M_{(n+1)×(n+1)}$ case to work....

答:
I don’t think that it require to use induction here. For avoid possible confusion, let’s just separate the case when $n=2$ and $n \neq 2$.

Proof. Let $n \in \Bbb{N}$ such that $n \geq 2$ and $A \in \mathcal{M}_{n}(\Bbb{K})$ and suppose that $\text{adj}(A) = \mathcal{O}_n$. If $n = 2$, then (as you well pointed out) $A = \mathcal{O}_2$. Since $\text{rank}\mathcal{O}_2 = 0$ and $0 \leq 0 = 2 - 2$, the proposition if true. Suppose now that, $n > 2$. Since, $\text{adj}A = \mathcal{O}_{n}$, we know that $\det{A} = 0$. So, $A$ is not invertible, which means that $\text{rank}A < n$, i.e., $\text{rank}A \leq n-1$. Suppose that $\text{rank}A = n-1$. This means that there is one row that is a linear combination of the remaking rows. Then, if we consider the matrix of size $n-1 \times n$ that we obtain by eliminating that row, we get a matrix that has rank equal to $n-1$. This means that we also have a column in this $n \times 1$ that is a linear combination of the remaining columns. So, the matrix of size $n-1 \times n-1$ that we obtain by further eliminating this column has rank equal to $n-1$. This means that this matrix is invertible. Hence, its determinant is not null. But this implies that one of the entries of $\text{adj}A$ is not zero, which is a contradiction. Therefore, the rank of $A$ cannot be $n-1$. Therefore, we conclude that $\text{rank}A \leq n-2$. $\square$