问:
Probability of breast cancer
I'm having trouble with a probability problem I've been trying to solve for a while. It's about the accuracy of breast cancer testing. Relevant probabilities are listed below, where:

"$\text{cancer}$" is the event "has breast cancer".

"$+$" is the event "tests positive for breast cancer".

$P(\text{cancer}) = \frac{12}{1000}$

$P(+|\text{cancer}) = \frac{11}{12}$

$P(+) = \frac{31}{1000}$

$P(\text{cancer}|+) = 0.355$

This last line is a result from a previous problem. The next part involves updating the probability of having cancer, but I'm having trouble figuring out what the answer is.

In the next part of the question, there's a woman who has tested positive and her doctor says she's part of a population for which there's a 40% chance of breast cancer.

I need to find the probability that the woman has cancer.

I am confused by this update to the cancer probability, but i will assume that this means $P(\text{cancer})$ has changed.

I also assume this means I need to find a new value for $P(\text{cancer}|+)$, but I'm not getting this right.

$P(+ | \text{cancer}) = \frac{11}{12} = \frac{P(\text{cancer} | +)\cdot P(+)}{P(\text{cancer})} = \frac{P(\text{cancer} | +) \cdot \frac{31}{1000}}{0.40}$

$P(\text{cancer} | +) = \frac{11}{12} \cdot 0.40 \cdot \frac{1000}{31} = 11.828$

The result can't be correct because it's way over 1.

How can i fix this? Thank you in advance for any insight.

答:
By Bayes' Theorem:

$$P(cancer|+) = \frac{P(+|cancer) P(cancer)}{P(+)}$$

Where $$P(+) = P(+|cancer)P(cancer)+ P(+|no-cancer)P(no-cancer)$$
$$P(+|cancer) = \frac{11}{12}$$
$$P(cancer) = \frac{4}{10}$$
Therefore,
$$P(cancer|+) = {\frac{11}{12} \cdot \frac{4}{10} \over \frac{11}{12} \cdot \frac{4}{10} + \frac{6}{10} \cdot P(+|no-cancer)}$$

To find the rate of false positives for the test, P(+|no-cancer), we can use the information from the general population, that $P(+)= \frac{31}{1000}$ and that $P(cancer) = \frac{12}{1000}$.
Then, $$ \frac{31}{1000} = \frac{11}{12} \cdot \frac{12}{1000} + P(+|no-cancer) \cdot \frac{988}{1000} $$
Rearrange to get $$ P(+|no-cancer) = \frac{5}{247}$$

Plug this back into the previous equation to get $$P(cancer|+) = \frac{2717}{2807} \approx 0.968$$

Title: Probability of breast cancer


问:
Introduction to cluster algebras
I am doing an undergraduate research project and the topic I have been assigned is cluster algebras (an introduction). I was reviewing some bibliography and videos on the internet and the truth is I don't know where to start.

I have taken the following courses: algebra (not Galois theory), analysis, topology.
I have notions of quivers and their representations.

If you could give me an outline or guidelines to start reading it would be very helpful. As well as bibliography. Thanks!

答:
There is a book in progress available by Fomin, Zelevinsky, and Williams. You can find the first 6 chapters here, here, and here. This gives enough material for an introduction, I think.

You can also look at a particular motivating example: total positivity and it's tests. A good survey of very slick results can be found here. I strongly recommend trying a few of the examples, and thinking through these proofs. They helped me a lot.

Finally, there are a collection of lecture notes and old courses here. I have not reviewed them myself, so the quality of links and material may vary wildly.

ContentLicense: CC BY-SA 4.0

Title: Introduction to cluster algebras


问:
Proof that $\mathbb Q$ is not cyclic.
I tried to show that $\mathbb Q$ with addition is not a cyclic group. Here is the proof:

If possible assume that there is $a\in \mathbb Q$ such that $\mathbb Q=\langle a \rangle$. Note that $a\neq 0$.
For $1\in \mathbb Q$ we must have $1=ma$ for some $m\in \mathbb Z$.

Now for each $n\in \mathbb N$ we must have $\frac {1}{n}=m_n a$  for some $m_n\in \mathbb Z$. Then $1=nm_n a$ for each $n\in \mathbb N$.
Then $ma=nm_n a$.
Then we get $nm_n-m=0$. So $n$ must divide $m$ for each $n\in \mathbb N$. Then $m$ must be a zero which is not possible.
So our supposition is wrong. Hence $\mathbb Q$ is not cyclic.

Is this correct?

答: You can show this more quickly in the following way: We must have $m_{1}a=a/2$ for some $m_{1} \in \mathbb{Z}$ so that $2m_{1}a=a \implies 2m_{1} = 1 \implies m_{1} = \frac{1}{2}$ but $\frac{1}{2} \notin \mathbb{Z}$.

Title: Proof that $\mathbb Q$ is not cyclic.


问:
Problem on Hahn-Banach and separation theorem
A problem on Hahn-Banach theorem and convex separation theorem:

Let $X$ be a real normed linear space. $E$ a convex subset of $X$ with nonempty interior, $F$ a linear subspace, with $(\operatorname{Int} E)\cap F=\emptyset$. Then there exists a bounded linear functional $f\in X^*, f\not\equiv0$, such that $f(E)\leq0,f(F)=0$.

I think the following corollary of Hahn-Banach theorem may be somewhat useful, but I can't figure out how to combine it with the convex separation theorem.

Let $A$ be a linear subspace of normed linear space $X$, $x_0\in X$, $d(x_0,A)>0$, then there exists $f\in X^*$ s.t. $f(A)=0,f(x_0)=d(x_0,A)$ and $||f||=1$.

Thanks for any help or suggestions.

答:
The statement you want to prove is the classic first version of the Hahn-Banach Separation Theorem .

Usually one has available the Hahn-Banach Theorem, in the form

If $M\subset X$ is a subspace, $\varphi:M\to\mathbb R$ is linear, and $\mu$ is a real seminorm on $X$ (subadditive, $\mu(\lambda x)=\lambda\,\mu(x)$ for $\lambda\geq0$) with $|\varphi(x)|\leq\mu(x)$ for all $x\in M$, then there exists $\tilde\varphi:X\to\mathbb R$, linear, with $\tilde\varphi|_M=\varphi$ and $|\tilde\varphi(x)|\leq\mu(x)$ for all $x$.

Since the interior of a convex set is convex, we may assume that $E$ is open (as the inequality will hold for any limit point). Fix $e'\in E$, $f'\in F$; The condition $E\cap F=\varnothing$ guarantees $e'\ne f'$. Then the set $Z=E-F+f'-e'$ is convex, open and $0\in E-F+f'-e'$. If $\mu$ is its Minkowski Functional, the conditions on $Z$ (open convex neighbourhood of $0$) guarantee that
$$\tag1
Z=\{x\in X:\ \mu(x)<1\}.
$$

On the subspace $M=\mathbb R(f'-e')$, define $\varphi:M\to\mathbb R$ by $\varphi(\lambda(f'-e'))=\lambda$. Then $\varphi$ is linear. As $f'-e'\not\in Z$, we have $\mu(f'-e')\geq1$. So
$$
|\varphi(\lambda(f'-e'))|=|\lambda|\leq|\lambda|\,\mu(f'-e')=\mu(\lambda(f'-e')).
$$
That is, $|\varphi(x)|\leq\mu(x)$ for all $x\in M$. By the Hahn-Banach Theorem, there exists $\tilde\varphi:X\to\mathbb R$, linear, with $|\tilde\varphi(x)|\leq\mu(x)$ for all $x\in X$.

Now, for any $e\in E$, $f\in F$,
$$
\tilde\varphi(e)-\tilde\varphi(f)+1=\tilde\varphi(e-f+f'-e)<1
$$
by $(1)$. Thus
$$\tag2
\tilde\varphi(e)<\tilde\varphi(f),\qquad e\in E,\ f\in F.
$$
Let $c=\sup\{\tilde\varphi(e):\ e\in E\}$. Then $c\leq\tilde\varphi(f)$ for all $f\in F$. Using that $F$ is a subspace, we get $c\leq t\tilde\varphi(f)$ for all $t\in\mathbb R$. This forces $\tilde\varphi(f)=0$ for all $f\in F$. Thus
$$\tag3
\tilde\varphi(f)=0,\ \tilde\varphi(e)\leq0,\qquad f\in F,\ e\in E.
$$
Finally,  limit points of $E$ are also limit points of $Z$, and so they satisfy $\mu(x)\leq1$ as the Minkowski Functional is continuous. So the argument above still works, just with $\leq$ instead of $<$.


答:
Yes, the proof is correct.

There is another way: write $a=\frac pq$ with $\gcd(p,q)=1$. Then $\frac p{2q}\in\mathbb Q$ implies $\frac p{2q}=\frac{np}q$ for some $n\in\mathbb Z$, which means $2\mid1$, which is impossible.

There is yet another fun way:

Consider a surjective group homomorphism $f:A\rightarrow \mathbb Z$. For any such group homomorphism there is a group homomorphism $g:\mathbb Z\rightarrow A$ such that $f\circ g=id_{\mathbb Z}$.

But this question's answer constructs examples of surjective homomorphisms $f:A\rightarrow \mathbb Q$ such that no $g:\mathbb Q\rightarrow A$ can exist so that $f\circ g=id_{\mathbb Q}$. This means $\mathbb Q$ is not isomorphic to $\mathbb Z$.

Hope this helps.

Title: Proof that $\mathbb Q$ is not cyclic.

Body:
I tried to show that $\mathbb Q$ with addition is not a cyclic group. Here is the proof:

If possible assume that there is $a\in \mathbb Q$ such that $\mathbb Q=\langle a \rangle$. Note that $a\neq 0$.
For $1\in \mathbb Q$ we must have $1=ma$ for some $m\in \mathbb Z$.

Now for each $n\in \mathbb N$ we must have $\frac {1}{n}=m_n a$  for some $m_n\in \mathbb Z$. Then $1=nm_n a$ for each $n\in \mathbb N$.
Then $ma=nm_n a$.
Then we get $nm_n-m=0$. So $n$ must divide $m$ for each $n\in \mathbb N$. Then $m$ must be a zero which is not possible.
So our supposition is wrong. Hence $\mathbb Q$ is not cyclic.

Is this correct?


问:
Probability that the machines are broken down
I have a problem of probability as below:

A company has 4 machines with the probability of being broken down in a day is $0.01$, $0.05$, $0.1$ and $0.1$ respectively.

a. In one random day with one-machine observation,what is the probability that machine is broken down?

b. When $2$ machines are tested, there is exactly one broken machine. What is the probability that the broken one is the first one?

For the first question, I have done like this:

Call $A$ is the probability one machine is chosen, $P(A) = \frac{1}{4}$

Call $H_{i}$ is the probability that the $i^{\text{th}}$ machine is broken $\rightarrow$ $P(A.H_{i})$ is the probability exactly one $i^{\text{th}}$ machine is broken.

I just think that the answer wil be the sum of all $P(H_i)$ but I still really confused here. I really need your help to check for me

答:
Your answer to part A is correct, namely

$(1/4) [ .01 + .05 + .1 + .1].$

Part B should be solved with intuition rather than math.

Suppose that there were two pieces of paper, each with a number.  One of the numbers is 0, and the other number is 1.  You choose one of the two pieces of paper at random.  You are then handed the piece of paper, and you look at it.  What are the chances that you chose the paper with the larger number?

Now, vary the problem.  Instead of handing you the piece of paper that you chose, a mediator first looks at that piece of paper, and then looks at the 2nd piece of paper.  Then he announces whether you chose the paper with the larger number.

Now vary the problem again.  This time the mediator looks at the other piece of paper first.  Obviously, the answer to all 3 questions is (1/2).

By the exact same intuition, the answer to part B is (1/2).


问:
How to show that $E([X|Y])^{2} = E[X^{2}|Y]$?
I have some isssue with this equation:
$$
\mathrm{var}[X|Y] = E([X|Y])^{2} - (E[X|Y])^{2}
$$
How to show that
$
E([X|Y])^{2} = E[X^{2}|Y]
$?

答:
I have some isssue with this equation: $$var[X|Y] = E([X|Y])^{2} - (E[X|Y])^{2} $$

The conditional variance of $X$ measured against $Y$ is by definition:

$$\begin{align}\mathsf {Var}(X\mid Y)&=\mathsf E((X-\mathsf E(X\mid Y))^2\mid Y)\\&=\mathsf E(X^2-2X\,\mathsf E(X\mid Y)+\mathsf E(X\mid Y)^2\mid Y)\\&=\mathsf E(X^2\mid Y)-2\,\mathsf E(X\,\mathsf E(X\mid Y)\mid Y)+\mathsf E(\mathsf E(X\mid Y)^2\mid Y)\\&=\mathsf E(X^2\mid Y)-2\,\mathsf E(X\mid Y)^2+\mathsf E(X\mid Y)^2\\&=\mathsf E(X^2\mid Y)-\mathsf E(X\mid Y)^2\end{align}$$

How to show that $E([X|Y])^{2} = E[X^{2}|Y]$?

You generally do not, as it requires $\mathsf{Var}(X\mid Y)=0$, which is to say that $X$ is completely determined by $Y$ (ie there is no unexplained variance).


问:
Finding the x-intercept of a cubic function
How would one go about solving for x in this cubic equation?

$f(x) = x^3 + x - 4$

I don't think factoring works in this situation:

$f(x) = x^2(x + 1) - 4$

and I don't know how to find the value of $x$ where f(x) = 0.

答:
For solving cubic equations, Cardano's method is not the $\alpha$ and $\omega$.

If you follow the (simple) steps described here, you will immediately notice that there is only one real root since $\Delta=-436$.

So, use the hyperbolic method and obtain
$$x=\frac{2}{\sqrt{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(6 \sqrt{3}\right)\right)$$ which is a bit nicer than
$$x=\frac{\sqrt[3]{18+\sqrt{327}}}{3^{2/3}}-\frac{1}{\sqrt[3]{3
   \left(18+\sqrt{327}\right)}}$$


问:
Subset Question Dealing with Subsets With One element or Less
How many subsets of a set with $100$ elements have more than one element?

I know that a set, $S$, has $100$ elements, $|S| = 100$, and that all the subsets from $S$ is $2^{100}$, but how do I know how many subsets have one element? I know one is the empty set.

答:
Total subsets $-$ singleton sets $-$ empty set:

$2^{100} - 100 - 1$.


问:
Find $\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}$

Find $$\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}.$$

This is a recent exam question, which I couldn't figure out in the exam. My guess is it doesn't exist.

答:
For small $|x|$ we have $\sin(x)\approx x$.
Hence
\[  i \cdot \sin\frac{1}{i} \approx 1  \]
for big values of $i$.
Hence
\[ \frac{\sum_{i=1}^n i\cdot \sin\frac{1}{i} }{n}\approx \frac{n}{n}=1\]</p>

ContentLicense: CC BY-SA 3.0


问:
Prove identity involving $\operatorname{lcm}(x, y, z)$
I received this problem to prove that:
$$
\operatorname{lcm}(x, y, z) = \frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))}
$$
Knowing that: $$\operatorname{lcm}(x, y) = \frac{xy}{\operatorname{gcd}(x, y)}$$

I started saying that:
\begin{align*}
\operatorname{lcm}(x, y, z) &= \operatorname{lcm}(\operatorname{lcm}(x, y), z)\\
&= \frac{z\cdot\operatorname{lcm}(x, y)}{\operatorname{gcd}(z, \operatorname{lcm}(x, y))}\\
&=\operatorname{lcm}(x, y, z)\\
&= \frac{z\cdot\operatorname{lcm}(x, y)}{\operatorname{gcd}(z, \operatorname{lcm}(x, y))}\\
&= \frac{xyz}{\operatorname{gcd}(x, y)\cdot\operatorname{gcd}(z,\operatorname{lcm}(x, y))}
\end{align*}

And I got stuck, how do I prove that it equals:

$$
\frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))}
$$

答:
Lemma: For any nonnegative integers $a_i, b_i, c_i$, we have
$$\mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\} + \mathrm{min}\{a_i, b_i, c_i\} +\mathrm{max}\{a_i, b_i, c_i\} = a_i + b_i + c_i.$$

Proof: Without loss of generality, we can impose an arbitrary ordering, say $a_i \leq b_i \leq c_i$. Then $$\mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\} + \mathrm{min}\{a_i, b_i, c_i\} +\mathrm{max}\{a_i, b_i, c_i\} = \mathrm{min\{b_i, c_i, c_i\} + a_i + c_i = b_i + a_i + c_i = a_i + b_i + c_i}. \square$$

Now, we have
$$\max\{a_i, b_i, c_i\} = a_i + b_i + c_i - \mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\} - \mathrm{min}\{a_i, b_i, c_i\}$$
so that
\begin{align*}
\mathrm{lcm}(x, y, z) = \prod_{p}p_i^{\max\{a_i, b_i, c_i\}} &= \frac{\prod_p p_i^{a_i
 + b_i + c_i}}{\prod_p p_i^{\mathrm{min}\{\mathrm{max}\{a_i, b_i\},\mathrm{max\{b_i, c_i}\},\mathrm{max\{c_i, a_i}\}\}} \prod_{p} p_i^{\mathrm{min}\{a_i, b_i, c_i\}}} \\ &= \frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))}
\end{align*}
as desired. $\blacksquare$


问:
Give the context-free grammar that generates the pushdown automaton language
The pushdown automaton :

From Here i figured out that having an idea of the language is a good first step so i came to the conclusion that the language represented by this pushdown automaton is something that looks like this :
$a^n b(b^*aa^*bb^*)^*a^n :n \in\mathbb{N}$ (correct me if i am wrong)

But from there i am kinda lost on how to approach the problem to figure out the CFG, in class were not really told a precise technique to do this so i would like to know how someone would proceed to solve this problem.

Thanks a lot.

EDIT my attempt: could someone validate if it's good or what i am missing thanks

$S \Rightarrow aSa$

$S \Rightarrow B$

$S \Rightarrow bB$

$B \Rightarrow aC$

$C \Rightarrow \epsilon$

$C \Rightarrow bB$

$B \Rightarrow b$

答:
We have shown (link in comment)  that the language of this PDA is L = { a^n b (a* b)* a^n , n ≥ 0 } , now let's build the grammer ,

S --> aSa | bT

T --> AbT | ε

Α --> aA | ε

The first rule generates a^n b T a^n accounting for n = 0 , T generates (a* b)* , note how A generates a* , Ab is the same as a* b , and adding T , AbT allows for repeating ( you can form AbAbT , AbAbAbT and so on , or use T --> ε ) which is analogous to the *

As for your grammer , comparing it to the language you provided ( which is not the language for the PDA ) , it doesn't describe the language correctly , it doesn't also describe the correct language of the PDA

If we use the rules
S --> aSa , then S --> B , we arrive at aBa , now use B --> ε ,and you get the string aa , which doesn't belong to the language you provided or that of the PDA ( note how the languages require at least one b to be in any string )


问:
Addition of variance
I have a question about the addition of variance and mean, and I can't seem to have the same answer for variance

Assume A is normally distributed with a mean of 10 mm and a standard deviation of 0.1 mm.
B and C is normally distributed with a mean of 2 mm and a standard deviation of 0.05 mm.
Assume all Random variable are indepedent.
And D=A-B-C
Find the variance and the mean of D

Method 1

E(D)=E(A-B-C)=E(A)-E(B)-E(C)=10-2-2=6

V(D)=V(A-B-C)=C1^2V(A)+C2^2V(A)+C2^2V(A)=1^2(0.1)^2+(-1)^2*(0.05)^2+(-1)^2*(0.05)^2=0.015

Method 2

Since B=C

E(D)=E(A-2B)=E(A)-2*E(B)=10-(4)=6

V(D)=V(A-2B)=C1^2V(B)+C2^2V(B)=1^2*(0.1)^2+(-2)^2*(0.05)=0.02

So the mean gives the same value, however the variance in both cases gives me two different values.I don't understand why? Am I doing something wrong?

Thank you

答:
By independence, $Var(A-B-C) = VarA + Var(-B) + Var(-C) = Var(A) + Var(B) + Var(C)$. This is correct.

Your method 2 is the incorrect one. The problem is that you assume $B = C$ just because they are identically distributed. This is not the case and it is a mistake that quite a lot of people make when learning probability.

By hypothesis you have that $A$, $B$ and $C$ are independent. But if $B=C$ then they cease to be independent. Let me illustrate.

By independence, we have (this is the correct calculation)

$$
P(B  >0|C<0) = P(B>0)
$$

But if $B=C$ we'd have

$$
P(B  >0|C<0) = P(B>0|B<0) = P(\emptyset) = 0
$$

Title: Addition of variance


问:
Birthday problem with large $n, d$ values
In the Birthday problem, the formulas

$${\displaystyle {\begin{aligned}p(n;d)&={\begin{cases}1-\displaystyle \prod _{k=1}^{n-1}\left(1-{\frac {k}{d}}\right)&n\leq d\\1&n>d\end{cases}}&\approx 1-e^{-{\frac {n(n-1)}{2d}}}&\approx 1-\left({\frac {d-1}{d}}\right)^{\frac {n(n-1)}{2}}\end{aligned}}}$$

work well for $d = 365$ and $n=23$, and gives the usual estimation that if you have 23 people in the same room, the probability to have at least two people born the same day is $\geq 50 \%$.

Question: what formula is available for $p(n; d)$ with more precise error terms?

Concrete application:

I'm using random 5-alphanumeric-character identifiers for an inventory of objects.

Example: V4QH7, WYJ9X, LK6H4, etc.

If I have $n = 10,000$ objects, what is the probability that at least 2 objects have the same ID?

Note: the last formula (the one after the one with exponential function above) gives Error, numeric exception: overflow in Maple when I take $d=(26+10)^5=60,466,176$ and $n=10,000$.

The formula with exp gives $p \approx 3.8 \%$ but since no error term is given, I don't know if this is accurate.

Edit: Mistake: $p \approx 3.8 \%$ was obtained when I took $d=33^6$ (6-alphanumeric characters with a few letters removed for easier identification: I vs 1, etc.). With $d=36^5$, we get $56.3 \%$ probability of having a collision with the exp formula above, which is in accordance with the accepted answer.

答:
This can be done pretty straightforwardly as follows: we want to upper bound

$$- \sum_{k=1}^{n-1} \log \left( 1 - \frac{k}{d} \right)$$

and we can do this by applying an upper bound for $f(x) = - \log \left( 1 - x \right)$. On $x \in [0, a]$, Taylor's theorem with remainder gives that

$$|f(x) - x| \le \frac{f''(\xi)}{2} x^2$$

where $\xi$ maximizes $f''(x)$ for $x \in [0, a]$. We have $f'(x) = \frac{1}{1 - x}$ and $f''(x) = \frac{1}{(1 - x)^2}$, which is maximized when $x = a$. This gives that

$$- \log (1 - x) \le x + \frac{x^2}{2(1 - a)^2}, x \in [0, a]$$

and setting $a = \frac{n}{d}$ gives

$$- \sum_{k=1}^{n-1} \log \left( 1 - \frac{k}{d} \right) \le \sum_{k=1}^{n-1} \left( \frac{k}{d} + \frac{k^2}{2 (d^2 - n^2)} \right) \le \frac{1}{d} {n \choose 2} + \frac{n^3}{6(d^2 - n^2)}.$$

On the other hand, by convexity we have $-\log (1 - x) \ge x$ which gives a lower bound

$$- \sum_{k=1}^{n-1} \log \left( 1 - \frac{k}{d} \right) \ge \frac{1}{d} {n \choose 2}.$$

Exponentiating both of these gives

$$\boxed{ 1 - \exp \left( - \frac{1}{d} {n \choose 2} \right) \le p(n, d) \le 1 - \exp \left( - \frac{1}{d} {n \choose 2} - \frac{n^3}{6(d^2 - n^2)} \right) }.$$

Plugging in $n = 10^4, d = 36^5$ gives

$$\boxed{ 0.5625\color{red}{6} \ldots \le p(n, d) \le 0.5625 \color{red}{8} \ldots }$$

which disagrees with your answer. Heuristically these estimates tell us to expect a non-negligible probability of a collision once $d \approx {n \choose 2}$ (equivalently, once the expected number of collisions, which is exactly $\frac{1}{d} {n \choose 2}$ and which is an upper bound on the probability, is approximately $1$), and we have ${n \choose 2} \approx 5 \times 10^7$ and $d \approx 6 \times 10^7$, so if you were given these numbers by someone else that was probably deliberate.


问:
Is it an abuse of notation to omit the leading zero in a decimal less than 1?
Is it acceptable to write $.001$ rather than $0.001$ when using decimal notation?

Are there contexts in which omitting the leading zero is acceptable, and other situations in which it is not?

答: It is not an abuse of notation as long it is clear, though I would assume most people would prefer to write it as $0.001$. Of course, you could also write it as $00.001$ and some else might prefer to write it as $000.001$. In general, it is preferred to write it as $0.001$, especially if it is used in a sentence since . could be confused with period .. Also, when we write a number say $.1$, it might be possible that someone might miss the .infront of the $1$ and might read it as $1$. Whereas, if we precede the number with a $0$, then we give the reader a heads-up that since we have a zero infront, watch out for a number less than $1$.


问:
Finding all prime numbers such that $8p^4-3003$ is also prime

Find all the prime numbers $p$  such that $8p^4-3003$ is also a prime

I was trying to solve the previous problem but I haven't been able to. I was thinking about using Fermat's Little Theorem but I'm not sure. Any help is welcome.

答:
Hint:

If $5\nmid p$, then, by Fermat's Little Theorem, $p^4\equiv1\pmod5$, so $8p^4-3003\equiv \,?\pmod5$.


问:
Proving that solutions to linear systems form a vector space
I've been given to systems of equations and I have to prove whether their solutions are or aren't vector spaces, but I'm confused on how to do it.
The first system:
$$
x+y+z=0\\
x+y+2z=1
$$
The second system is:
$$x+y+z=0\\
x+y+2z=0$$

I think that for the second system once you find the solution, you're solving for $Ax=b$, but if I have two solutions $x1$ and $x2 A(x1+x2)$ doesn't equal $b$ so it shouldn't be a vector space. For the, first system since it's homogenous I believe it is a vector space.

答:
As you mentioned in the comment, in order to have a vector space one needs to be able to add two vectors and get a vector in the same set, or scale a vector to yield a vector in the same set. In the first system subtract the first equation from the second, to get $z=1$. Then the first equation becomes $$x+y+1=0$$or $$y=-x-1$$
So the solution in a vector form is $$\vec v=(x,-x-1,1)$$
If you scale this vector, say by a factor of $2$, you need to be able to write it in the same form:
$$2\vec v=(2x,-2x-2,2)$$
Notice that the last term of $2\vec v$ is $2$, and $2\ne 1$. So $2\vec v$ is not a solution of the first system. So the solutions of the first system do not form a vector space.

Can you do that for the second system?


问:
If-then statement problems

Your guarantee is good only if you bought your CD player less than 90 days ago.

If your guarantee is good, then you must have bought your CD player less than 90 days ago.

Can we also write the if-then statement of the sentence of this link like the following:

if  you have bought a CD player less than 90 days ago then the guarantee
is good

If I am wrong then please correct me.

Edit: Thanks for the answers

答:
These kinds of unclear situations can largely be avoided by phrasing the sentences with “it follows” instead of “if”.

From the fact that your guarantee is good, it follows that you must have bought the CD player <90 days ago.

(Note that the first version can not directly be phrased in this version.)

So that mean indeed the correct direction is guarantee → time-since-buy.

$$
  \text{valid}(\text{guarantee}) \Longrightarrow t_\mathrm{buy} <90\:\mathrm{d}
$$

The other direction is not true, as witnessed by the real-world situation that the warrantee is also void if you smash the CD player on the floor – then $t_\mathrm{buy} <90\:\mathrm{d}$ would still be fulfilled, but you could not conclude that the guarantee is still valid.

What does work is, turning around the implication arrow whilst also negating the statements.

From the fact that you have not bought the player <90d ago, it follows that the guarantee is void.

$$
  \neg(t_\mathrm{buy} <90\:\mathrm{d}) \Longrightarrow
 \neg(\text{valid}(\text{guarantee}))
$$

Title: If-then statement problems


问:
Is there a bounded continuous function $f$ s.t. a certain integral does not converge uniformly to zero?
The following question was the reason for question Is there a bounded function $f$ with $f'$ unbounded and $f''$ bounded?, which had a negative answer.
So I'd like to post my original problem:

I'm trying to find a continuous bounded function $f:\mathbb{R}\to\mathbb{R}$
such that

$$\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi t}}e^{-(y-x)^2/2t}f(y)dy -f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-y^2/2}(f(x+\sqrt{t}y)-f(x))dy$$

does NOT uniformly converge to $0$ for $t\to 0+$. I'm not completely sure that there is actually one. But I'm fairly certain.
I was playing around with functions like $\text{sin}(e^{2x})$. But so far without any luck. Thanks for any comments.

答:
Here is an example.

Let me write $p_t(x) = \frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}$ for short.  Let $f$ be a continuous function such that for $n \ge 2$, centered around $x=n$, there is a triangular bump of height $1$ and width $2/n^2$.  Elsewhere it is 0.  Here is a very crude picture.

Notice that the bump centered at $n$ has area $1/n^2$, so $\int_{\mathbb{R}} f(x) dx = \sum_{n=2}^\infty \frac{1}{n^2} < \infty$.  Also, since I left out $n=1$, each bump has width less than $1/2$ and so is supported in $[n-\frac{1}{2}, n+\frac{1}{2}]$.

Note $f(n) = 1$ for each $n$.  Let's estimate $\int_{\mathbb{R}} f(x) p_t(n-x)dx$.  First let's consider the part of the integral outside of  $[n-\frac{1}{2}, n+\frac{1}{2}]$, where we have $p_t(n-x) \le p_t(1/2) = \frac{1}{\sqrt{2 \pi t}} e^{-1/8t}$.  Thus we can bound this part of the integral by $p_t(1/2) \int_{\mathbb{R}} f(x) dx$.  This goes to $0$ as $t \to 0$, so take $t$ small enough that this term is less than, say, $1/4$.  (Note this part is independent of $n$.)

Now let's consider the integral over $[n-\frac{1}{2}, n+\frac{1}{2}]$ which contains a single bump.  Since $p_t(x)$ attains its maximum at $x=0$, we have $p_t(x) \le p_t(0) = (2 \pi t)^{-1/2}$.  Since the bump is actually supported in $[n-\frac{1}{n^2}, n+\frac{1}{n^2}]$ and $f \le 1$, this part of the integral is bounded by $\frac{2}{n^2} p_t(0) = \frac{2}{n^2 \sqrt{2 \pi t}}$.  We can take $n$ so large that this is also less than $1/4$.

Thus we have shown that for any sufficiently small $t$, there exists $n$ such that $\int_{\mathbb{R}} f(x) p_t(x-n)dx \le 1/2$ whereas $f(n) = 1$.  So we do not have uniform convergence.

The Wikipedia page cited by Theo Buehler is incorrect.  It cites "general facts about approximation to the identity"; however, the usual condition for $f * \psi_k \to f$ uniformly is that $f$ be uniformly continuous.  Of course, my $f$ is not uniformly continuous.


问:
How to you model a game when the rules you are given are wrong?
The title-question is the main focus of my confusion but I have several supporting questions at the end that try to clarify.

To illustrate my question, I will base it off of a game I played a while back during a team-building exercise at a university camp. I can't remember exactly how the rules were phrased. Basically all the students were blindfolded and led to a place where we were given a singular, long-taught rope to hold.

These were the constraints that were given to us or that could be inferred:

We were told that when the whistle blew we could begin moving.

We could move anywhere in space as long as we were touching the rope at any given moment (it was possible to pass other people with some talking and physical co-ordination).

If we thought we'd found the end of the rope we could raise our hand and some authority would tell us whether we had actually found it. There was no limit to how often we could do this.

If we found the end of the rope, we won.

We could ask to surrender; the game ended for us and we left with our blindfolds off.

It might not be the best formulation, but I have decided to model it as a one-player game with two actions, namely: surrender or continue searching for the end of the rope. Let $k$ be a quotient representing how much patience and/or resolve we have. And let $t$ be how much time has elapsed in minutes. Given the following decision matrix.

$$\begin{array}{c|c|}
\hline
Surrender & 0\\
Continue & 15-(t/k)\\
\hline
\end{array}$$

As time went on in the game, I heard as people asked if they had found the rope and being told that they hadn't. Soon, people with lower patience $k$ started surrendering. Everything fit the model of the rules explained so far.

The problem arose when I reached a knot in the rope. I recognised the knot from my starting point and realised that I had come full-circuit. The implication of this was that the theoretical win-condition stated at the beginning was un-achievable.

Much later, after the game, the facilitator told us that the solution was to surrender. The game was supposedly designed to show us how self-dependent we try to be. Before that, however, after I had realised that the actual rules were different to the way I had perceived them, I tried to formulate a new game with my new information. Since I wasn't satisfied to end with a solution that gave me the minimum utility.

Therefore, considering I was effectively playing against the patience of the facilitators, I adapted it as minimally as I could to a two-player game like the utility matrix bellow (where $j$ is the patience and/or resolve of the facilitator).

$$\begin{array}{c|c|c|}
& Surrender & Continue\\
\hline
Surrender & 0,0 & 0,5\\
Continue & 5,0 & 15-(t/k),15-(t/j)\\
\hline
\end{array}$$

The logic was that if I could motivate myself to have $k > j$, then I would have to win. Which, if this game was the measure, I eventually did, to the annoyance of the facilitator.

So what I want to know is, what was the actual game?

At what point did the rules change (if at all)?

If they did change, was my adaptation from the original one more reasonable or did the facilitator's "admission" about lying about the win condition not remove credence from their new win condition?

Is there a better way to model the two games (together maybe)?

This all borders on philosophy because it is unclear who determines the rules but I basically want to know what the "mathiest" way to talk about it is.

If there is any way this question can be clearer, feel free to edit it or suggest a change. I am just a hobbyist mathematician.

答:
What you have sounds like a game with the students (or a single student) and the facilitator, but it's a game of incomplete information. In general incomplete information games things players might not have common knowledge of are:

Payoffs

Who the other players are

What moves are possible

How outcome depends on the action

What the opponent knows and what they know you know...

What game is exactly being played

Using an incomplete information formulation the rules of the game never changed just the information available to pick an optimal strategy.

For modeling the game you describe though I don't think there is much to be done. With incomplete information the best rational strategy is the search for the end unit you get back to the start and then give up. If played with the missing information given then best rational strategy would likely be to give up immediately, as your time is likely worth more then out waiting a facilitator for no greater reward.

Probably a modification to the game to make it "better" to analyze would be to assume the players are in a larger game, in that the students would be told that with a random draw with some probability they will be given a rope with no end or a rope with an end, and that may give something more amenable to modeling decisions on both sides.


答:
It's enough to do it one prime at a time.

Let $p$ be a prime that divides one or more of $x,y$ or $z$, and let $k_x$ be the highest power so that $p^{k_x}|x$ and $k_y$ the highest power so that  $p^{k_y}|y$ and so on (note: some of these may be equal to $0$).

By symmetry we can assume $k_x \le k_y \le k_z$.  If that seems too glib, just relabel that $w = $ the $x,y$ or $z$ where $p$ divides by the lowest power, $u=$ the $x,y$ or $z$ where $p$ divides by the middle power, and $v = x,y$ or $z$ where $P$ divides by the highest.

The highest power of $p$ that divides $lcm(x,y,z)=lcm(w,u,v)$ is $p^{\max(k_x,k_y,k_z)} = p^{k_v}$.

The highest power of  $p$ that divide the RHS of the equation id $p^{(k_x+ k_y + k_z)- [\min(k_x,k_y,k_z) + \min(\max(k_x,k_y),\max(k_x,k_z),\max(k_y,k_z)]}=$

$p^{(k_w+ k_u + k_v)- [\min(k_w,k_u,k_v) + \min(\max(k_w,k_u),\max(k_w,k_v),\max(k_u,k_v)]}=$

$p^{(k_w+ k_u + k_v)- [k_w + \min(k_u,k_v,k_v)]}=$

$p^{(k_w+ k_u + k_v)- [k_w + k_u]}=$

$p^{k_v}=$

$p^{\max(k_x,k_y,k_z)}$.

So for every $p$ that divides any of $x,y,z$ the power of $p$ that divides both sides is exactly the same!

So do that for every prime that divides $x,y,$ or $z$ and the result follows.

Title: Prove identity involving $\operatorname{lcm}(x, y, z)$

Body:
I received this problem to prove that:
$$
\operatorname{lcm}(x, y, z) = \frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))}
$$
Knowing that: $$\operatorname{lcm}(x, y) = \frac{xy}{\operatorname{gcd}(x, y)}$$

I started saying that:
\begin{align*}
\operatorname{lcm}(x, y, z) &= \operatorname{lcm}(\operatorname{lcm}(x, y), z)\\
&= \frac{z\cdot\operatorname{lcm}(x, y)}{\operatorname{gcd}(z, \operatorname{lcm}(x, y))}\\
&=\operatorname{lcm}(x, y, z)\\
&= \frac{z\cdot\operatorname{lcm}(x, y)}{\operatorname{gcd}(z, \operatorname{lcm}(x, y))}\\
&= \frac{xyz}{\operatorname{gcd}(x, y)\cdot\operatorname{gcd}(z,\operatorname{lcm}(x, y))}
\end{align*}

And I got stuck, how do I prove that it equals:

$$
\frac{xyz}{\operatorname{gcd}(x, y, z)\cdot\operatorname{gcd}(\operatorname{lcm}(x, y), \operatorname{lcm}(x, z), \operatorname{lcm}(y, z))}
$$


问:
Proving that $G$-action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr)$ is a homomorphism.
I am using the definition of group action of $G$ on $X$ which says that any group homomorphism from $G$ to $\operatorname{Aut(X)}$ is a group action.

And I am trying to prove $(d)$ in the following question (after proving $(a), (b), (c)$)

Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\varphi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$

$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\varphi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$

$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$

$(c)$ Given $\varphi \in \operatorname{Aut_{I}(R)},$ show that $\varphi(I) = I$ and $\bar{\varphi} \in \operatorname{Aut(R/I)}.$\ Hint: Use part $(b).$

$(d)$ Let $G$ be a group which acts on $R$ by ring automorphisms and restricts to $I.$ Show that the $G-$action on $R$ induces a  $G-$action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$

My trial:

I was given  hint here Show that the $G$-action on $R$ induces a $G$-action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$ to prove that the given action ( $G-$action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr).$) is a homomorphism which I do not know exactly how to do as the hint mentioned $\bar{\varphi_{g}}.$ Also,  I do not know if I must show that this action is a group homomorphism or ring homomorphism, could anyone clarify this for me please?

Also, I was told that the meaning of the question is:

It means that G is a subgroup of the group Aut(R) and g(I)=I for each g in I. Elements of Aut(R) are bijective ring homomorphisms f:R-->R, which are called automorphisms.

Here is my trial to show that it is a ring homomorphism:

(a) Showing that it preserves addition:

${}^{(g_{1} + g_{2})} \pi(r) = \pi({}^{(g_{1} + g_{2})}r) = {}^{(g_{1} + g_{2})}r + I = {}^{g_{1}}r + {}^{g_{2}}r + I =  ({}^{g_{1}}r + I) + ({}^{g_{2}}r + I) = {}^{g_{1}} \pi(r) + {}^{g_{2}} \pi(r)$

I am not sure from my third equality here, specifically the first $+$ sign in it, if it is correct or not, could anyone check it for me please?

(b) Showing that it preserves multiplication:

${}^{(g_{1}g_{2})} \pi(r) = \pi({}^{(g_{1}g_{2})}r) = {}^{(g_{1}g_{2})}r + I = {}^{g_{1}}(^{g_{2}}r) + I =  {}^{g_{1}}(^{g_{2}}r + I)  = {}^{g_{1}} \pi({}^{g_{2}}r)$

Where the fourth equality because I know that $g(I) = I \forall g \in G.$ And after the last step I do not know how to complete, could anyone show me the rest of steps or the correct ones please?

(c) for showing that the identity of $G$ is mapped to the identity of $\operatorname{Aut(R/I)}$ I do not know how should I prove that, any help will be appreciated in that proof.

Also I have a question: what is the relation between $G-$action on $R/I$ defined by ${}^g \pi(r) = \pi({}^gr)$ and the function $\bar{\varphi_{g}}.$ Is it just a name for the homomorphism between $G$ and $\operatorname{Aut(R/I)}$? vould anyone explain that to me please?

答: For a) and b): Why are you using two different operations ($g_1+g_2$ and $g_1g_2$) for elements of G? For c): The action (denoted by ".") of the group's identity, say e, on elements $\pi(r)$ of $R/I$ is given by $e.\pi(r)=\pi(e.r)=\pi(r)$. Finally, the action $\cdot:G\times R/I\rightarrow R/I$, given by $(g,\pi(r))\mapsto g.\pi(r)$ corresponds with a homomorphism $\overline{\varphi}:G\rightarrow Aut(R/I)$ given by $g\mapsto \overline{\varphi}_g:R/I\rightarrow R/I$, where $\overline{\varphi}_g(\pi(r)):=\pi(\varphi_g(r))$, for the homomorphism $\varphi:G\rightarrow Aut(R)$ which induces $\overline{\varphi}.$


问:
Evaluate $\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$, for integrable $f$
Let $f : [0, 1] \to \mathbb{R}$ be Riemann integrable. Evaluate
$$\lim_{n \to \infty} \int_{0}^1 x^{n+1} f(x) \ dx$$
I have tried applying by parts by assuming
$$\\ \int_{0}^1 f(x) \ dx = F(1)-F(0)$$
but it doesn't seem to get me anywhere. Do I have to use partitions and if so how do I go about it?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\lim_{n \to \infty}\,
\int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x}:\
{\Large ?}}$.

As $\ds{n \to \infty}$, the main contribution to the integral comes from $\ds{x \lesssim 1}$. So, I'll make the change $\ds{x \mapsto 1 - x}$ such that the main contribution comes from $\ds{x \gtrsim 0}$. Namely,
\begin{align}
&\bbox[5px,#ffd]{\lim_{n \to \infty}\,
\int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\,
\int_{0}^{1}\pars{1 - x}^{n + 1}
\,\on{f}\pars{1 - x}\,\dd x
\\[5mm] = &\
\lim_{n \to \infty}\,
\int_{0}^{1}\expo{\pars{n + 1}\ln\pars{1 - x}}
\,\,\,\on{f}\pars{1 - x}\,\dd x
\\[5mm] = &\
\lim_{n \to \infty}\,
\int_{0}^{\infty}\expo{-\pars{n + 1}x}
\,\,\,\on{f}\pars{1 - 0}\,\dd x
\end{align}

where I used the
Laplace's Method.

\begin{align}
&\mbox{Then,}
\\ &\
\bbox[5px,#ffd]{\lim_{n \to \infty}\,
\int_{0}^{1}x^{n + 1}\,\on{f}\pars{x}\,\dd x} =
\lim_{n \to \infty}\,\,{\on{f}\pars{1} \over n + 1} =
\bbx{\large 0} \\ &
\end{align}


问:
Vector Analysis- Curves in Space

I have attached both the problem and approach to the solution. But I am confused how to proceed. Kindly help in providing the simplified version of the solution.

PS- I tried pasting the problem and solution from a word file, but I don't understand why the editor is not accepting my text. That's why I have attached the images.

答: The approach to problem solution is attached. Kindly help in providing the simplified version of the solution.

Title: Vector Analysis- Curves in Space


问:
simple calculation with Christoffel symbols on Poincare half-plane
Equip $H=\{(x,y):y>0, x,y \in \mathbb{R}\}$ with the metric $$ds^2=\frac{dx^2+dy^2}{y^2}.$$
(https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model). I want to show that the sectional curvature
$$ K(\partial_i,\partial_j) = \frac{\langle R(\partial_i,\partial_j)\partial_j,\partial_i\rangle}{\det(g)}=\frac{R_{ijji}}{\det(g)}$$
is $-1$. I know this is a simple calculation, but for some reason I'm off by a sign and it's driving me nuts. Here $g$ is the metric in matrix form:

$$g=(g_{ij})=
\begin{pmatrix}
\frac{1}{y^2} & 0 \\
0 & \frac{1}{y^2}
\end{pmatrix}.$$

Let $\{ \partial_1=\partial/ \partial x,\partial_2=\partial/ \partial y\}$ be the coordinate basis, and just consider the computation of $K(\partial_1,\partial_2).$ So I just need to compute $R_{1221}=g_{1m}R^m_{\ 221}=g_{11}R^1_{\ 221}$, where

$$R^i_{jkl}=\partial_k \Gamma^i_{jl}-\partial_l \Gamma^i_{jk}+\Gamma^p_{jl}\Gamma^i_{pk}-\Gamma^p_{jk}\Gamma^i_{pl}.$$

I already have the Christoffel symbols, namely

\begin{equation*}
\Gamma^1_{12}=\Gamma^1_{21}=-\frac{1}{y},\\
\Gamma^2_{12}=\Gamma^2_{21}=0,\\
\Gamma^1_{11}=\Gamma^1_{22}=0,\\
\Gamma^2_{11}=\frac{1}{y},\\
\Gamma^2_{22}=-\frac{1}{y}.
\end{equation*}

So
\begin{align*}
R^1_{\ 221} &= \partial_2 \Gamma^1_{21}-\partial_1 \Gamma^1_{22}+\Gamma^p_{21}\Gamma^1_{p2}-\Gamma^p_{22}\Gamma^1_{p1}\\
&= \partial_2 \left( -\frac{1}{y}\right)+\Gamma^1_{21}\Gamma^1_{12}-\Gamma^2_{22}\Gamma^1_{21}\\
&= \frac{1}{y^2}+\left(-\frac{1}{y}\right)^2-\left(-\frac{1}{y}\right)^2\\
&= \frac{1}{y^2}.
\end{align*}

But this gives $R_{1221}=1/y^4$, and hence $K(\partial_1,\partial_2)=1$.

Please tell me what I'm doing wrong here or what formula is wrong.

答:
The sectional curvature for a Riemannian manifold, $(M,g)$ with respect to an orthonormal basis for $P=\text{span}_{\mathbb{R}}(e_1,e_2)\subset T_pM$ at a point $p\in M$ is given by:
$$K(P)=\langle R(e_1,e_2)e_2,e_1\rangle.$$
Recall that the components of the Riemann curvature tensor take the form: $$R_{\kappa\lambda\mu\nu}=\langle R(e_\mu,e_\nu),e_\lambda,e_\kappa\rangle.$$
So, $\langle R(e_1,e_2)e_2,e_1\rangle=R_{1212}.$ You calculated $R_{1221}$, so if you use the symmetries of the Riemann curvature tensor in its last two indices you find the missing negative sign.


问:
If the square of an irrational number r is irrational, can it be equal to a + br, where both a and b are rational
I am trying to evaluate such a statement:

$$\forall r \in \mathbb{R} \setminus \mathbb{Q}: \ ({r^{2}} \notin \mathbb{Q} \implies \forall a \in \mathbb{Q} \ \forall b \in \mathbb{Q}: {r^{2}} \neq a + br)$$

but it seems to exceed my skills. Could you please help?

答:
The following is a counterexample:

Set $r = \sqrt{2} +1 \in \mathbb{I}$.

Then $r^{2} = (\sqrt{2} +1)^2=1+2\sqrt{2}+2=1+2(\sqrt{2}+1) = 1+2r.$


问:
Non-homogeneous Poisson process first occurrence distribution
I am trying to do the following problem in order to prepare for my Stochastic Processes exam:

Problem statement:
Consider $N(t)$ to be a non homogeneous Poisson process with rate $\lambda(t)$ over the interval [0, a].
Let $T_{1}$ denote the time for the first event to occur. Find the distribution of $T_{1}$ given that $N(a) = 1$. Here, $N(t)$ denotes the number of occurrences up until time t.

->What have i tried so far? I started by investigating the following: $$P(T_{1} > t) = P(N(T_{1})=0)$$

Which, since we have a non homogeneous Poisson process, turns out to be:

$$\frac{\left[\int_{0}^{T_{1}} \lambda(s) d s)\right]^{0}e^{-\int_{0}^{T_{1}} \lambda(s)ds}}{0 {\displaystyle !\,}} = e^{-\int_{0}^{T_{1}} \lambda(s)ds}$$

Finally: $$P\left(N\left(T_{1}\right)=0\right) = e^{-\int_{0}^{T_{1}} \lambda(s) d s}$$

Now, i tried to use the remaining information to finish the problem. However, i got stuck. I don't know how to proceed in regards to conditioning the variables (I dont know if that is indeed the path to solve the problem). I tried doing:

$$P\left(N\left(T_{1}\right)=0 \mid N(a)=1\right) = \frac{P\left(N\left(T_{1}\right)=0, N(a)=1\right)}{P(N(a)=1)}$$

Now, i tried a few things, but i dont know if that is indeed correct. Knowing that $N(T) = 0$ and $N(a) = 1$, one can infer that $T \le a$, otherwise $N(T)$ would not equal zero, as one event would have ocurred. This information can be translated into: $$P(N(a)=1) = e^{-\int_{a}^{T_{1}} \lambda(s) d s}$$

Now, if i could expand the below term: $$P\left(N\left(T_{1}\right)=0, N(a)=1\right)$$ into something useful, i could start manipulating some terms. What do you guys think of my approach? Any help is appreciated.

Thanks in advance!

答:
On your approach

You have done all the right things initially. $P(N(T_1) = 0) = e^{-\int_0^{T_1} \lambda(s)ds}$ is correct. The conditional formula $P(N(T_1) = 0 | N(a) = 1) = \frac{P(N(T_1) = 0, N(a) = 1)}{P(N(a) = 1)}$ is correct (because the denominator of RHS is not zero). Furthermore, this is the way to solve the problem, so you are going in the right direction.

Now, $P(N(a) = 1)$ is incorrectly written. Note that $N(a)$ is a Poisson random variable with parameter $\int_0^a \lambda(s)ds$. Therefore, we have :
$$
P(N(a) = 1) = \frac{\left(\int_0^a \lambda(s)ds\right)^1e^{-\int_0^a \lambda(s)ds}}{1!} =  \left(e^{-\int_0^a \lambda(s)ds}\right)\int_0^a \lambda(s)ds
$$

Answer

Now, how do we handle the numerator? Recall the independence of increments of the Poisson process. This means, in particular, that $N(T_1)$ and $N(a) - N(T_1)$ are independent for any $T_1$.

Now, it is easy to see that $$P(N(T_1) =0, N(a)= 1) = P(N(T_1) = 0,N(a) - N(T_1) =1) = P(N(T_1)= 0)P(N(a) - N(T_1) = 1)$$

thanks to independence of the two random variables. Finally, each of these involves a Poisson random variable, so we should be through here. One part you've already done :
$$
P(N(T_1) = 0) = e^{-\int_0^{T_1} \lambda(s)ds}
$$

The other part is easy as well :
$$
P(N(a) - N(T_1) = 1) = \frac{\left(\int_{T_1}^a \lambda(s)ds\right)^1e^{-\int_{T_1}^a \lambda(s)ds}}{1!} = \left(\int_{T_1}^a \lambda(s)ds\right)e^{-\int_{T_1}^a \lambda(s)ds}
$$

Multiplying these two, and combining the integrals sitting on top of the exponentials (basically, $\int_a^c f + \int_c^b f = \int_a^b f$ for any $a,b,c$ and function $f$), gives us :
$$
P(N(T_1) = 0 , N(a) = 1) = \left(\int_{T_1}^a \lambda(s)ds\right)e^{-\int_{0}^a \lambda(s)ds}
$$

Dividing by $P(N(a) = 1)$ and doing some change of variables to get back to our original situation, gives an answer that feels right, in addition to being right of course!

$$\bbox[yellow, 2pt, border:2px solid red]{
P(T_1 > t | N(a) = 1) = \frac{\int_{t}^a \lambda(s)ds}{\int_{0}^a \lambda(s)ds}}
$$


问:
Find $\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}$

Find $$\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}.$$

This is a recent exam question, which I couldn't figure out in the exam. My guess is it doesn't exist.

答:
Start by writing it in summation form. That is,

$$
\lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n}
$$
Now, $\sin(1/i)<1/i$ for $i\in\mathbb{N}$. Furthermore, $\sin(1/i)>\frac1i-\frac1{6i^3}$ for $i\in\mathbb{N}$.
Now, we squeeze the result by noting that
$$
\lim_{n\to\infty} \sum_{i=1}^n \frac1n = 1
$$
and
$$
\lim_{n\to\infty} \sum_{i=1}^n \frac{1-\frac1{6i^2}}n = \lim_{n\to\infty} \sum_{i=1}^n \frac1n-\frac1{6i^2n} = 1
$$
Therefore, we have that
$$
\lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n}=1
$$


问:
Deriving a new distribution from repeated samples
it's been many moons since I've been in a math class, so I'm not sure how to formulate the problem precisely. I'll describe it informally instead.

This is related to response times in a computer system. Let's say I have some system, foo, that I can make requests to and I can measure its response time. I've collected enough data to be able to  describe the TP50 and TP90 latencies. Now, let's say I'm building a new system, bar, and for every request that bar needs to service, I need to make two requests to foo. The calls to foo are done in parallel and let's say they are independent of each other. bar needs to wait for them both to complete. How can I calculate TP50 of bar? What if instead of 2 calls, I need to make n calls?

Through a little toy simulation, I get the TP50 of bar is ~TP70 of foo, assuming bar makes two calls. I'd like to understand how to think about this mathematically though.

Thanks for any help!

答:
Unfortunately, knowing only the $50^{\rm th}$ and $90^{\rm th}$ percentiles of the latency distribution for the first system is inadequate information to estimate the $50^{\rm th}$ percentile of the second system.

For instance, suppose the first system's latency is exponentially distributed with mean $1$ time units (say, seconds); i.e., the random response time $T$ has cumulative distribution function $$\Pr[T \le t] = 1 - e^{-t}, \quad t > 0.$$  Then the "TP50" is the $50^{\rm th}$ percentile $m_{0.5}$ of this distribution, namely $$\Pr[T \le m_{0.5}] = 1/2$$ which implies $$m_{0.5} = \log 2 \approx 0.693147.$$  This means that $50\%$ of all requests to the first system will be served in at most $693$ milliseconds.  Similarly, $m_{0.9} = \log 10 \approx 2.30259$ seconds.

Now under this distributional assumption, we can establish that the maximum response time of $n$ requests to this first system has the cumulative distribution function $$\Pr[T_{max} \le t] = (1 - e^{-t})^n, \quad t > 0.$$  The $50^{\rm th}$ and $90^{\rm th}$ percentiles are $$m_{0.5} = - \log (1 - (1/2)^{1/n}), \\ m_{0.9} = - \log (1 - (9/10)^{1/n})$$ and for $n = 2$, these become $m_{0.5} \approx 1.22795$ and $m_{0.9} \approx 2.96974$.

But this applies only to exponentially distributed latencies.  Suppose I instead choose to model the latency as normally distributed, in such a way that matches the median and $90^{\rm th}$ percentiles above; i.e., $$\mu = m_{0.5} = \log 2,$$ and we need to numerically solve for the variance $\sigma^2$ such that the $90^{\rm th}$ percentile equals $\log 10$.  This gives us $$\sigma^2 \approx 1.57716.$$  Then the maximum of $n = 2$ requests does not have a nice closed form, but again with a computer, we can solve for the percentiles.  These are $m_{0.5} \approx 1.37753$ and $m_{0.9} \approx 2.74297$, different from the exponential model.

If the use of a normal distribution is objectionable, then we could use a Gamma distribution with suitable shape parameter.  But the point that I have shown mathematically is that even when the $50^{\rm th}$ and $90^{\rm th}$ percentiles are fixed, I could choose two different distributional assumptions and get different results, illustrating that knowing these values is not in itself sufficient to determine the TP50 of the second system.  Consequently, you must provide additional information or assumptions about the latency of the first system with which to properly model the behavior of the second.


问:
Name for nearly linear (line-shaped) subgraph of a graph
Given a graph, it's straightforward to define a "linear" or line-shaped subgraph: a connected subgraph with two ends which are nodes of degree one, and intermediate nodes of degree two. And maybe a single point as a degenerate variant.

Now add new edges to nodes outside of the subgraph to it. The result is "nearly" a "linear" graph, maybe one could call it 1-linear: If one removes those edges again, the result is a "linear" graph.

And so on for 2-linear etc.

Is this a known concept? Is there some generally accepted terminology for this?

Examples, subgraphs only:

```<code>1 -- 2 -- 3 -- 4
</code>```

is linear.

```<code>A         B
|         |
1 -- 2 -- 3 -- 4 -- E
|    |
C    D
</code>```

is "1-linear", if you remove edges to A..E, you end up with a linear "spine".

答:
And with the link to Wikipedia provided by @perpetuallyconfused I can answer my own question:

There is indeed the name "linear graph" or "path graph", and what I called "1-linear" is a Caterpillar Tree. What I called "2-linear" seems to be a Lobster Tree or Lobster Graph.

There are also generalizations called K-tree, but they look a bit different than what I had in mind.

Though in the application where this came up the extended graphs are not necessarily trees, but at least this points me in the right direction.


问:
Find minimum value of $|z|+|z-e^{ix}|+|z-5e^{ix}|+|z-7e^{ix}|+|z-32e^{ix}|$ where $z$ is a complex number

Find minimum value of $$|z|+|z-e^{ix}|+|z-5e^{ix}|+|z-7e^{ix}|+|z-32e^{ix}|$$ where $z$ is a complex number.

we notice that we have to minimze the sum of distances from the points in the argand plane  $(0,e^{ix},5e^{ix},7e^{ix},32e^{ix})$.We also see these points are on a straight line which makes an angle $x$ with real axix. .We now rotate the plane by $x$ radians anticlockwise so that the points lie on the real line.Let the coordinates of $z$  after rotation be $(h,k)$.Intuitively this does not change the minimum.So now we have to minimise: $$\sqrt{h^2+k^2}+\sqrt{{(h-1)}^2+k^2}+\sqrt{{(h-5)}^2+k^2}+\sqrt{{h-7)}^2+k^2}+\sqrt{{(h-32)}^2+k^2}$$  which happens when
$k=0$.We minimize  $$f(h)=|h|+|h-1|+|h-5|+|h-7|+|h-32|$$
But $f(h)$ becomes  a lenear function for each of the intervals $h\in [-\infty,0],[0,1],[1,5],[5,7],[7,32],[32,\infty]$.But a lenear function achieves its maximum and minimum at the endpoints of interval.

We seek:$$\min(f(0),f(1),f(5),f(7),f(32))=f(5)=38$$

Is it good?
I am looking for alternative proofs

答:
You can use only the triangle inequality:

$$|z| + |z-32 e^{ix}| \ge |32 e^{ix}| = 32 \\
|z-e^{ix}| + |z - 7 e^{ix}| \ge |7 e^{ix} - e^{ix}| = 6 \\
|z- 5e^{ix}| \ge 0$$

Adding up we get : $f(z) \ge 32 + 6 + 0 = 38$.  We have equality for $z = 5e^{ix}$  ( an easy but important check). Conversely, if we have equality then necessarily $|z- 5 e^{ix}| = 0$ so $z = 5 e^{ix}$.

Note: $x$ is a parameter, $z$ is the variable.


问:
To find $h=\max_{a,b}(\min\{a, f(a,b)\}), a\geq 0, b\geq 0$
How to find $$h=\max_{a,b}(\min\{a, f(a,b)\}), a\geq 0, b\geq 0$$ where $f$ is some fixed non negative continuous function. My teacher told me put $a=f(a,b)$ and then solve $b$ in terms of $a$. But I am not getting the point. Usually this type of $h$ is needed to
find interval of existence of unique solution by Existence and uniqueness theorem of Picard for ordinary differential equation. I have confusion when $a<f(a,b)$ or $a>f(a,b)$, then how can we put $a=f(a,b)$? Please clear how to find this $h$ in easy way and how by putting $a=f(a,b)$ is working?

答:
we are given $$h= \max_{a,b} \min(a,f(a,b)); a \ge 0, b \ge 0$$
The $\min$ function either selects its first or second argument (or they are equal in which case it doesn't matter which argument is selected.)

case 1 (arg 1 selected):
$$ h= \max_{a,b} a; f(a,b) \ge a \ge 0, b \ge 0$$

case 2 (arg 2 is selected):
$$ h= \max_{a,b} f(a,b); a \ge f(a,b), a \ge 0, b \ge 0$$

In the first case, we can increase $a$ until $a=f(a,b)$, but no further, and so $h=a$ when $a=f(a,b)$
In the second case, $f(a,b)$ might be able to be increased until $f(a,b)=a$, but no further, in which case $h=f(a,b)$ when $f(a,b)=a$. However, we have to consider the possibility that $\forall a,b. f(a,b) \lt a$. There are again two cases. In the first case, $f$ may take on any value and still have $f(a,b)<a$. For example, if $f$ were defined such that $f(a,b)=a-1$. In that case, the "maximum" value, if it is said to exist, would be $h=f(a,b)=\infty$, and since $f(a,b)<a$ then $a=\infty$. We might write this as $f(\infty,b)=\infty$

The final case is that $f(a,b)$ reaches some finite maximal value $k$ for any $a,b$ while $\forall a,b. a\gt f(a,b) \ge 0$. An example of such a function would be $$f(a,b)=\min(k,\max(0,a-1)$$

It is clear that in that case
$h=k$ and the solution is not related to setting $f(a,b)=a$


问:
Interior of $A=\{(a_n)_{n \geq 1} \in l_1 : |a_n| \leq |\frac{1}{2^n}| \}$
Let $A=\{(a_n)_{n \geq 1} \in l_1 : |a_n| \leq |\frac{1}{2^n}| \}$, where $l_1$ is the space of absolutely summable real sequences. Show that the interior of $A$ is empty in $(l_1,d_1)$.

How to prove that for any $(a_n) \in A$ there is some distance $r>0$ such that
$$(x_n) \in A,~d_1((a_n),(x_n))=\sum_{n=1}^\infty|a_n-x_n|<r~\Rightarrow x_n=a_n$$

答: If $B((a_n);\varepsilon)\subset A$, you can construct a $(b_n)$, with $b_n=a_n$ for all $n$ except for one $N$, which you choose considering that $|b_N|> \frac{1}{2^N}$ and $|a_N-b_N|<\varepsilon.$


问:
Closed form of a recurrence relation using generating functions
It's been awhile since I have done this. The sequence is $\displaystyle a_n = a_{n-1} + 5~a_{n-2}$ with $a_{0}=0$ and $a_{1}=1$. </p>

<p>I found the generating function to be $\displaystyle G(x) = \frac{x}{(1-x-5x^2)}$ but I am lost on where to go from here. </p>

答:
Factor the denominator. Slightly unpleasant, since the roots are not rational, but doable. Then use partial fractions to express the generating function in the form $\frac{A}{1-ax}+\frac{B}{1-bx}$.

It is easy to write down the power series for $\frac{1}{1-ax}$ and $\frac{1}{1-bx}$, by using the fact that $\frac{1}{1-t}=1+t+t^2+t^3+\cdots$.

Then you can get an explicit expression for the coefficient of $x^n$ in each of the terms, and hence in the sum.

If you run into trouble, I can give more detail. I got the following partial fractions representation:
$$\frac{1}{\sqrt{21}}\left(\frac{1}{1-\frac{1}{2}(1+\sqrt{21})x}-\frac{1}{1-\frac{1}{2}(1-\sqrt{21})x}                             \right) .$$

Remark: There are (to me) easier ways for this particular problem than the generating functions approach. Characteristic equations, though they involve roughly similar manipulations, feel easier.


问:
Converting Piecewise function into a single expression in MATLAB
I have a piecewise function. I want to turn this into a single expression. How can this be accomplished IN MATLAB and what  is the  general procedure?

$$f(x,t) =
\begin{cases}
20e^{-4x^2} & \text{if $0 \leq t \leq T/2,$} \\
0 & \text{otherwise.}
\end{cases}$$

答:
You can write $$f(x,t)=20e^{-4x^2}\Theta(T/2-t), t\in (0,\infty)$$
The Heaviside step function is defined as: $\Theta(z\ge 0)=1, \Theta(z<0)=0.$


问:
How to solve this equation with matrices
can you please give me some hints to solve the following? I really don't know how to start.
$$X^2= \begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix}.$$
I tried to express this matrix as $4\cdot I + \begin{pmatrix}
2 & 2 \\ 3 & 3
\end{pmatrix}$ And somehow solve it, but I really have no clue. Please some help.

答:
One simple but laborious way of doing it is as follows

Let $$X = \begin{bmatrix}a & b \\c & d\end{bmatrix}$$

Now, using the Caley-Hamilton Theorem, we have

$$X^2 - (a+d)X + |X|I = 0$$

Now, from the equation, we have $|X| = \pm 6$

Hence, we will have two sets of solutions each corresponding to one of these

$$|X| = 6$$

$$\implies X = \frac{1}{a+d}\begin{bmatrix}12 & 2 \\3 & 13\end{bmatrix}$$

Using the equivalence of the two representations of $X$, we have

$$a =  \frac{12}{5} \\ b =  \frac{2}{5} \\ c = \frac{3}{5} \\ d = \frac{13}{5}$$

Another solution to this is when each of the entries is negative. Similarly complete it for $|X| = -6$ to get the full set of solutions


问:
Covariant derivative of a covector field along a curve
During my special relativity course we defined the covariant derivative of a vector field $ \ \xi^\mu (\tau)$ along a curve $x^\mu(\tau)$ as:

$$ \frac{\mathrm{D}\xi ^\mu}{\mathrm{D} \tau} = \frac{\mathrm{d} \xi ^\mu}{\mathrm{d} \tau} + \Gamma^{\mu}_{\alpha \beta}  \, \xi^\alpha \frac{\mathrm{d} x^\beta}{\mathrm{d} \tau} $$

The justification given for this definition was straightforward: the equation of motion for a free particle in general coordinates is

$$\frac{\mathrm{D}u ^\mu}{\mathrm{D} \tau}=0,$$

where $u^\mu$ is the four-velocity of the particle. So, we didn't have to learn a lot about differential geometry for that course. Now, what if I want to compute the covariant derivative of a covector, like $u_\mu$? I've been searching around here and most of what I find doesn't refer to covariant derivatives along a curve. And what does deal with that, is written in ways I honestly don't understand (I don't have a lot of background in differential geometry). From what I've learned from searching about this topic, the following would be incorrect:

$$ \frac{\mathrm{D}\xi _\mu}{\mathrm{D} \tau} = \frac{\mathrm{d} \xi _\mu}{\mathrm{d} \tau} + \Gamma_{\mu \alpha \beta}  \, \xi_\alpha \frac{\mathrm{d} x^\beta}{\mathrm{d} \tau} $$

And, even though I haven't seen it written exactly like this, I would guess the correct expression is:

$$ \frac{\mathrm{D}\xi _\mu}{\mathrm{D} \tau} = \frac{\mathrm{d} \xi _\mu}{\mathrm{d} \tau} - \Gamma^\beta _{ \alpha \mu}  \, \xi_\beta \frac{\mathrm{d} x^\alpha}{\mathrm{d} \tau} $$

I would like to know if this is correct, and most importantly, why. Hopefully it doesn't require too much background?

答: Recall that the covariant derivative of a scalar reduces to the ordinary derivative: $\nabla_Xf=X(f).$ If $X$ is the coordinate basis, then $\nabla_\mu f=\partial_\mu f.$ Calculate the covariant derivative of the following scalar: $\langle Y,\omega\rangle=Y^i\omega_i$, which is the contraction of a vector and a covector. You will need to use the product rule. Lastly, let $X$ be a vector field defined along the curve of interest.


问:
Random Variable Transformation With Dirac delta function
Consider  $f_X(x) = 1/2+x$ where $0 <x\leq 1$. We want to find the pdf of $Y=X^{-1}$ by utilizing the following transformation.

$$f_Y(y) = \int_0^1 f_X(x) \delta (y - g(x)) dx$$

Where $u =g(X) = Y$.

From my understanding we can represent $x$ as $x=g^{-1}(g(x))$ and then we can write $dx$ as $dx = \frac{d(g(x))}{dx}dx \implies dx=-x^2du$. Since $x \in (0,1]$, we can conclude that $y \in [1,\infty)$. So we can rewrite our expression as:

$$f_Y(y) = \int_1^\infty f_X(g^{-1}(g(x))) \delta (y - g(x)) (-\frac{1}{y^2})du$$
However, after this point I don't really know how to continue. Yes I can factor out a negative sign, but i just don't see what my next step is. Any ideas?

答:
\begin{align}
&\int_{0}^{1}\operatorname{f}_{X}\left(x\right)
\,\delta\left(y - {1 \over x}\right)\mathrm{d}x =
\int_{0}^{1}\operatorname{f}_{X}\left(x\right)
\,{\delta\left(x - 1/y\right) \over \left\vert\,{ 1/x^{2}\,}\right\vert}\mathrm{d}x
\\[5mm] = &\
\left.\vphantom{\Large A}
\left[0 <  {1 \over y} < 1\right]
x^{2}\operatorname{f}_{X}\left(x\right)
\,\right\vert_{\, x\ =\ 1/y}\,\,\,\,\, =
\,\,\,\,\,
\bbox[10px,border:1px groove navy]{\left[y > 1\right]\,{\operatorname{f}_{X}\left(\,{1/y}\,\right) \over y^{2}}}
\\ &
\end{align}


答:
We have
$$
\begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix} = QDQ^{-1}$$
where
$$
Q = \begin{pmatrix} -1 & 2 \\ 1 & 3 \end{pmatrix},\  D = \begin{pmatrix} 4& 0 \\ 0 & 9 \end{pmatrix}.
$$
It follows that we may take any of the four matrices
$$
X = Q\begin{pmatrix} \pm 2 & 0 \\ 0 & \pm 3 \end{pmatrix} Q^{-1}.
$$

Title: How to solve this equation with matrices

Body:
can you please give me some hints to solve the following? I really don't know how to start.
$$X^2= \begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix}.$$
I tried to express this matrix as $4\cdot I + \begin{pmatrix}
2 & 2 \\ 3 & 3
\end{pmatrix}$ And somehow solve it, but I really have no clue. Please some help.


问:
A transpose of a matrix and its application
Since we know that a matrix can be a tool to describe and change a space,What is exactly a transpose used for and what can we get when we use a  transpose to describe a space?Is there a intuitive explanation?</p>

答:
Can be used to find the point in a line closer to a given position outside the line in the plane.

If we consider $f(x)=Ax$ the equation  of a line with matrix column $A$ and $b$ is a given point then one seeks for $Ax-b$ to be minimum and that is by finding $x_0$ such that it must satisfy $Ax_0\approx b$, so by multiplication with $A^{\top}$ to get
$$A^{\top}Ax_0\approx A^{\top}b$$
this implies
$$x_0\approx (A^{\top}A)^{-1}A^{\top}b,$$
that is the instant where $Ax_0$ is closer to $b$.

This is the same to calculate the estimate $x_0\approx\dfrac{A\cdot b}{A\cdot A}$.


问:
In an either or statement in logic, is "either TRUE STATEMENT or NONSENSE STATEMENT" a true or nonsense statement?
I apologise for having no education in logic, so I am clueless on where to start with this issue. I will explain my predicament in full:

I recently loosely adapted the barber's paradox into a logic puzzle of sorts. This is the formulated puzzle:

In a town, everybody must shave their beards. All the people living there either shave their own beard or have the barber shave their beard. There is a person in this town named Ed. Does Ed shave his beard or have his beard shaved by the barber?

The answer which I considered to be correct is the following:

Consider the barber. It is given that everyone in the town either shaves their own beard or has the barber shave their beard. Were the barber to shave his own beard, he would shave his own beard and have the barber shave his beard. This is a contradiction of the previous 'either or' statement, and thus the barber cannot exist. Therefore everyone in the town shaves their own beard. So Ed shaves his own beard.

When I posed this problem to a friend, they responded by saying that the statement "All the people living there either ... or have the barber shave their beard" implies the existence of one barber. If no barber exists, the statement is nonsense and no implications can be made from it.

So, my question is essentially: In a town with no barber, is it true to say that "everyone either shaves their own beard or has the barber shave their beard" or is this just nonsense, and the puzzle solution thus invalid? Furthermore, could this puzzle be improved by writing "any barber" as opposed to "the barber", as this wording does not necessitate the existance of a barber?

答:
If there is no barber, the statement "everyone either shaves their own beard or has the barber shave their beard" is equivalent to the statement "everyone shaves their own beard".  It is true unless somebody has a non-barber (like their friend) shave their beard or somebody doesn't shave their beard at all.  It is certainly not nonsense.

When I have seen the problem it was specified that there was exactly one barber in the town.  It is important to get the details right on problems like this.  Even with that added on, we can't know about Ed.  It is important to have the puzzles ask about the barber, not about a different person.  Then the statement should be that the barber shaves all those who do not shave themselves.  Now you have a contradiction, but only with these changes.


问:
Show that this sequence of functions converges uniformly to $e^{-x}$
Define

$$
f_n(x)=
\begin{cases}
\left(1-\frac xn\right)^n & 0\le x<n \\
0 & x\ge n
\end{cases}
$$

I am able to show that $f_n(x)\to e^{-x}$ pointwise on $[0,+\infty)$, and I wonder if my justification for uniform convergence is valid:

By properties of logarithm, I transformed the sequence into

$$
\left(1-\frac xn\right)^n=\exp\left\{n\log\left(1-\frac xn\right)\right\}
$$

Employing the fact that for sufficiently large $n$

$$
\left|n\log\left(1-\frac xn\right)+x\right|=\left|n\left[-\frac xn+\mathcal O\left(1\over n^2\right)\right]+x\right|=\mathcal O\left(\frac1n\right)
$$

we have $n\log\left(1-\frac xn\right)\to-x$ uniformly, hence for each $\delta>0$ there exists an $N$ such that for all $n>N$:

$$
|\log f_n(x)+x|<\delta
$$

By the uniform continuity of the exponential function, for all $\varepsilon>0,x>0,y>0$, we have $|e^{-x}-e^{-y}|<\varepsilon$ whenever $|x-y|<\delta$. As a result let $y$ be $-\log f_n(x)$, so  for all $\varepsilon>0,x>0$ there exists an $N$ such that for all $n>N$

$$
|f_n(x)-e^{-x}|<\varepsilon
$$

I wonder if this proof is correct or please point out my mistake. Thank you all!

答:
Your proof is not valid. Your estimates are valid only for fixed $x$ and there is no uniformity. In particular $n \log (1-\frac x n) $ does not tend to $-x$ uniformly: If $|n log (1-\frac x n) +x| <\epsilon $ for $n \geq n_0$ and for all $x <n$ you get a contradiction by letting $x \to n$.

Hint for  a valid proof: Choose $T$ such that $x>T$ implies $e^{-x} <\epsilon$. Note that $0 \leq f_n(x)\leq e^{-x} <\epsilon$ for all $n$ if $x>T$. So it is enough to consider $\sup_{x \leq T} |f_n(x)-e^{-x}|$ and here your approach can be used.


问:
Prove or disprove the following statement about sequences
Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.

Prove or disprove (by providing a counterexample) each of the following statements:

$(i)$ Let $\{x_n\}$ be a sequence such that $\underset{\rm n}{\rm lim}$ $x_n = 0$, then  $\underset{\rm n}{\rm lim}$ $ \frac{1}{|x_n|} = +\infty$

$(ii)$ Let $\{y_n\}$ be a sequence such that $\underset{\rm n}{\rm lim}$ $y_n = 0$, then either  $\underset{\rm n}{\rm lim}$ $ \frac{1}{y_n} = +\infty$ or $\underset{\rm n}{\rm lim}$ $ \frac{1}{y_n} = -\infty$

Answer:

$(i)$ Let $M \in \mathbb{R}, M>0$, we need to show that $N \in \mathbb{N}$ such that $\frac{1}{|x_n|}>M, \hspace{5pt} \forall n \ge N$. As $\underset{\rm n}{\rm lim}$ $x_n = 0$ for $\zeta = \frac{1}{n}>0, \hspace{5pt} \exists n \in \mathbb{N}$ such that

$|x_n|<\frac{1}{M} \hspace{5pt} \forall n \ge N$ $\Rightarrow$ $\frac{1}{|x_n|}>M \hspace{5pt} \forall n \ge N$ $\Rightarrow$ $\underset{\rm n}{\rm lim}$ $ \frac{1}{|x_n|} = +\infty$. (Hence statement $(i)$ is true).

$(ii)$ Using a counterexample to disprove the statement, let $y_n = \frac{(-1)^n}{n}.$ Then $\underset{n\rightarrow \infty}{\rm lim} y_n =0. $ And $\underset{n\rightarrow \infty}{\rm lim} y_n$ is neither $+ \infty$ nor $- \infty$ (since $\underset{n\rightarrow \infty}{\rm lim} y_n$ does not exist). Here $y_n$ is an oscillating sequence. Therefore, the statement is false.

答: For (ii) you have $y_n = \frac{(-1)^n}{n}$ which satisifies the hypothesis, but then $\ \frac{1}{|y_n|} = \frac{1}{\bigg|\frac{(-1)^n}{n}\bigg|} = n$ so that $\underset{n\rightarrow \infty}{\rm lim} \frac{1}{|y_n|} = +\infty$. Your chosen $y_{n}$ is not a counterexample.


问:
Does $3-4+1$ equal $0$ or $-2$.. or maybe $2$?
Does $3 - 4 + 1 = 0$ or $3 - 4 + 1 = -2$?

Makes sense that $(3 - 4) + 1 = 0$ and $3 - (4 + 1) = -2$, but what if there are no parenthesis?

Also, if I have $4$ apples and I add $1$ more apple, then I have $5$ apples, but if I eat $3$ apples, then I have $2$ left.

Any ideas?

答:
Without parenthesis, the equation would remain as your first case:

$$ (3 − 4) + 1 = 3 - 4 + 1 $$

According to the MDAS rules, from the methods of addition and subtraction, the equation will start off with any of the beginning methods. Afterwards, the equation will do all those methods (depending if subtraction or addition goes first from the equation). Then it will go to the next operation if its there.

So it is still in fact $ 0 $ without groupings.


答:
Outside of a minor typo, your analysis for part (i) is valid (and accurate).  Typo :

$\exists n \in \mathbb{N}$ such that...

should be

$\exists N \in \mathbb{N}$ such that

Your conclusion for part (ii) is wrong and your counter example is invalid.

In part (ii), the assertion concerns $\frac{1}{|y_n|}$ while your counter example is incorrectly focusing on $y_n$.

More to the point, if you inspect the assertions of part (i) and (ii) closely, you will see that part (ii) requires no work.

This is because, by the wording of parts (i) and (ii), it is impossible for part (ii) to be false if part (i) is true.

Further, you have already (validly) proven that part (i) is true.

Therefore, part (ii) must be true.

As an irrelevancy, looking more closely at the trailing possibility in part (ii), you will see that it is nonsensical because both the numerator and denominator must be non-negative.  Therefore, how can the fraction be going to $-\infty?$

Addendum

Reaction to OP's editing of the query to remove the absolute value sign in the denominators, re part (ii).

Yes, this editing permits part (ii) to be false, re $y_n$ can oscillate between positive and negative.

Title: Prove or disprove the following statement about sequences

Body:
Below the following question is my solution but I am not sure whether the proofs are correct and rigorously exhaustive so, if present, kindly point out any errors.

Prove or disprove (by providing a counterexample) each of the following statements:

$(i)$ Let $\{x_n\}$ be a sequence such that $\underset{\rm n}{\rm lim}$ $x_n = 0$, then  $\underset{\rm n}{\rm lim}$ $ \frac{1}{|x_n|} = +\infty$

$(ii)$ Let $\{y_n\}$ be a sequence such that $\underset{\rm n}{\rm lim}$ $y_n = 0$, then either  $\underset{\rm n}{\rm lim}$ $ \frac{1}{y_n} = +\infty$ or $\underset{\rm n}{\rm lim}$ $ \frac{1}{y_n} = -\infty$

Answer:

$(i)$ Let $M \in \mathbb{R}, M>0$, we need to show that $N \in \mathbb{N}$ such that $\frac{1}{|x_n|}>M, \hspace{5pt} \forall n \ge N$. As $\underset{\rm n}{\rm lim}$ $x_n = 0$ for $\zeta = \frac{1}{n}>0, \hspace{5pt} \exists n \in \mathbb{N}$ such that

$|x_n|<\frac{1}{M} \hspace{5pt} \forall n \ge N$ $\Rightarrow$ $\frac{1}{|x_n|}>M \hspace{5pt} \forall n \ge N$ $\Rightarrow$ $\underset{\rm n}{\rm lim}$ $ \frac{1}{|x_n|} = +\infty$. (Hence statement $(i)$ is true).

$(ii)$ Using a counterexample to disprove the statement, let $y_n = \frac{(-1)^n}{n}.$ Then $\underset{n\rightarrow \infty}{\rm lim} y_n =0. $ And $\underset{n\rightarrow \infty}{\rm lim} y_n$ is neither $+ \infty$ nor $- \infty$ (since $\underset{n\rightarrow \infty}{\rm lim} y_n$ does not exist). Here $y_n$ is an oscillating sequence. Therefore, the statement is false.


问:
Partitioning of square lattice into equivalence classes under congruence relation
Let $m,n$ be positive integers. Consider the set $S^{\small(m)}_n=([1,n]\cap\mathbb N)^m$ having $n^m$ points in the Euclidean space $\mathbb R^m$ arranged in a square (cubic, etc) lattice.

Two subsets of $S^{\small(m)}_n$ are congruent if one can be transformed into the other by a combination of rigid motions: translations, rotations, and reflections. We can see that the congruence is an equivalence relation. Partition the powerset $\mathcal P(S^{\small(m)}_n)$ into equivalence classes under this relation. Let $a^{\small(m)}_n$ be the number of resulting equivalence classes.

We can see that for the linear case $a^{\small(1)}_n=2^{n-1}+1.$

Can we find similar formulae for $a^{\small(2)}_n$ or $a^{\small(3)}_n$, or even a general case $a^{\small(m)}_n$? Or can we at least find a few initial terms of these sequences?

答:
The case $a_n^{(2)}$ is given by sequence A054247 in the OEIS, which starts $1, 2, 6, 102, 8548,\ldots$.

I believe that $a_2^{(3)}=22$. As a subset of the $27$ vertices of a $3\times 3\times 3$ cube can be taken to at most $48$ subsets via symmetries of the cube, $a_3^{(3)}$ is at least $\left\lceil\frac{2^{27}}{48}\right\rceil=2796203$. If my calculation for $a_2^{(3)}$ is correct, then the sequence $a_n^{(3)}$ is not in the OEIS, as it contains no increasing sequences which start with $1,2,22$ and continue to a sufficiently large third term.

In general, the lower bound $2^{n^m}/(2^m\cdot m!)$ will, for fixed $m$, be asymptotically tight as $n$ grows large. This is because a randomly-chosen subset of a large $m$-dimensional cube will typically have no symmetries, and so will have an orbit of full size. One can see this in the $m=2$ case: the lower bound of $2^{n^2}/8$ is already at $99.99\%$ of the true value when $n=6$.


问:
Proving $X=Y$ almost surely
Given the three conditions:
$X_n \overset{a.s.}{\longrightarrow} Y_n$, $X_n \overset{a.s.}{\longrightarrow} X$, $Y_n \overset{L^1}{\longrightarrow} Y$, where $X_n, Y_n, X, Y$ are random variables,

Can we conclude that $X=Y$ almost surely? I think we can conclude this, but I don't have mathematically rigorous proof for this. If the conclusion can't be made, what is the counter example to this?

Thank you,

答:
$X_n \to Y_n$ as is meaningless. I will interpret this as $X_n -Y_n \to 0$ a.s.

$Y_n \to Y$ in $L^{1}$ implies that there is  a subsequence $(Y_{n_k})$ which converges to $Y$ a.s. Now $X =\lim X_{n_k}=\lim (X_{n_k}-Y_{n_k}) +Y_{n_k}=Y$ a.s.

Title: Proving $X=Y$ almost surely


问:
What will be a circle look like considering this distance function?
I am working on some exercises in the book Geometry: A Metric Approach with Models by R.S. Millman. He defines the following map: $$d_S(P,Q):\mathbb R^2\times\mathbb R^2\to\mathbb R\\\ d_S(P,Q)=\max\{~|x_2-x_1|,~|y_2-y_1|~\}$$ where in $P(x_1,y_1),~Q(x_2,y_2)$ are points in $\mathbb R^2$ and then wants the reader to show that $d_S$ is a distance function on $\mathbb R^2$.

I was thinking to myself of what would be a circle looked like regarding this function in Cartesian plane? I fixed a point, for example the origin, as a centre and then probed the case with a given positive constant $r$ as a radius. I just got two horizon $|y|=r$ or two vertical $|x|=r$ lines. the results didn't have good taste. Any help? Was I right? Thanks

答: Yes, the "circle" in this case, looks like a square.</p>


问:
How two line segments are not equal if function of one line is ($f(x)=x$) and of another ($g(x)=cx))$
Line $c=g(x)=cx$;

Line $b=f(x)=x$

If for every point $(x,0)$ on line $(b)$ ,there exists a point  $(x,cx)$ on line $(c)$ and vice versa,then how two given  line segments are not equal (considering this reason).

答:
Consider points $A(x_A, y_A) $ and $B(x_B, y_B)$:

If these points are on line b then:

b: $y=x$; $m=\frac{y_B-y_A}{x_B-x_A}=1$

And the length of AB is:

$AB=\sqrt{(x_B-x_A)^2+(y_B+y_A)^2}=(x_B-x_A)\sqrt{1+1=2}$

The gradient of line c; (y=cx) is $c$, so it can have one intersection with line b; (y=x), let this intersection is point A. Then point B will have different coordinates to when it was located on line b.Let mark this point as B', we have, $x_B=x_B'$ due to statement also:

$\frac{y_B'-y_A}{x_B-x_A}=c$

Then the length of AB' is:

$AB'=\sqrt{(x_B-x_A)^2+(y_B'+y_A)^2}=(x_B-x_A)\sqrt{1+c}$


答:
Yes. What you obtained is right. In general, the $p$-metric, where $p \geq 1$, is defined as
$$d_S(P,Q) = \left(\vert x_2 - x_1 \vert^p + \vert y_2 - y_1 \vert^p\right)^{1/p}$$
Now it is easy to prove that as $p \to \infty$, we get the metric $\max\{|x_2-x_1|,|y_2-y_1|\}$. Now you can use a plotting software and see how unit ball looks like for different $p$'s. You will find the for $p=1$, it is a diamond and as you increase $p$ you encompass "more" and "more" region in the plane. For $p=2$, you get the "usual" circle and for $p \to \infty$, you get the "unit square".

The diagram below shows the unit ball for different $p$'s. The inner most diamond is the case for $p=1$, which is also called as the taxi-cab metric. As $p$ increases, the "area" enclosed by the unit ball slowly increases. The outer unit square is the metric you get when you let $p \to \infty$, i.e., you get the metric $\max\{|x_2-x_1|,|y_2-y_1|\}$.

$\hskip0.5in$

Title: What will be a circle look like considering this distance function?

Body:
I am working on some exercises in the book Geometry: A Metric Approach with Models by R.S. Millman. He defines the following map: $$d_S(P,Q):\mathbb R^2\times\mathbb R^2\to\mathbb R\\\ d_S(P,Q)=\max\{~|x_2-x_1|,~|y_2-y_1|~\}$$ where in $P(x_1,y_1),~Q(x_2,y_2)$ are points in $\mathbb R^2$ and then wants the reader to show that $d_S$ is a distance function on $\mathbb R^2$.

I was thinking to myself of what would be a circle looked like regarding this function in Cartesian plane? I fixed a point, for example the origin, as a centre and then probed the case with a given positive constant $r$ as a radius. I just got two horizon $|y|=r$ or two vertical $|x|=r$ lines. the results didn't have good taste. Any help? Was I right? Thanks


问:
Find location of a point on an oval shape
I have an oval shape (let's assume the center is (0,0)), and I have two laser pointers (at known locations) pointing to the two different objects (marked as X). One object is on the perimeter, the other somewhere inside the oval area.

I have two questions:

With a sinlge laser pointer, given the angle of pointing and its (x,y) location, can we calculate the location of the object?

Now with two laser pointers, their angle of poining and locations, can we calculate the location of the object?

Please assume whatever variables needed, like the equation for the oval geometry. I just want to see what are minimum information needed to localize an object on the area or perimeter.

答:
We assume that, after reading @achille hui’s comment, you have realized that we need two laser pointers to locate an object, which is not on the perimeter of the oval (we call it Case 1). Furthermore, that comment emphasizes that we need only one laser pointer to pinpoint an object on the perimeter of the oval, which we call Case 2. The answer given below takes its cue from the aforementioned comment.

Here are the equations for the Case 1, i.e. spotting an object at an arbitrary position using two laser guns, locations of which are known. We use the extended sketch shown in $\mathrm{Fig.\space 1}$ to derive the following equations. Coordinates of the guns at the top (i.e. $A$) and bottom (i.e. $B$) are assumed as $\left(x_A, y_A\right)$ and $\left(x_B, y_B\right)$ respectively. Furthermore, for brevity, let $AD=m$ and $BD=n$.
$$s=x_A-x_B \quad \mathrm{and}\quad h=y_A-y_B$$
$$m\sin\left(\alpha\right)-n\sin\left(\beta\right)=s=\left(x_A-x_B\right)$$
$$m\cos\left(\alpha\right)+n\cos\left(\beta\right)=h=\left(y_A-y_B\right)$$

When we solved these two simultaneous equations, we get
$$n=\frac{\left(y_A-y_B\right) \sin\left(\alpha\right)-\left(x_A-x_B \right) \cos\left(\alpha\right)}{\sin\left(\alpha\right) \cos\left(\beta\right)+ \cos\left(\alpha\right)\sin\left(\beta\right)}=\frac{\left(y_A-y_B\right) \sin\left(\alpha\right)-\left(x_A-x_B \right) \cos\left(\alpha\right)}{ \sin\left(\alpha+\beta\right)}.$$

Let the coordinates of the object $D$ equal $\left(x_D, y_D\right)$, Then we have
$$x_D=x_B-n\sin\left(\beta\right) = x_B-\frac{\left(y_A-y_B\right) \sin\left(\alpha\right)-\left(x_A-x_B \right) \cos\left(\alpha\right)}{ \sin\left(\alpha+\beta\right)} \sin\left(\beta\right)$$
$$y_D=y_B+n\cos\left(\beta\right)=y_B+\frac{\left(y_A-y_B\right) \sin\left(\alpha\right)-\left(x_A-x_B \right) \cos\left(\alpha\right)}{ \sin\left(\alpha+\beta\right)}\cos\left(\beta\right)$$

Now we illustrate the Case 2. There are several types of ovals. As far as we know, only Cassini oval can be described by a single equation. To describe other types of ovals, we usually need two or more equations. For instance, the ovals called Moss’ egg and Thom’s egg need four equations each. Only Cassini’s curves with $a\gt \sqrt{2}c \gt 0$ (see $\mathrm{Fig.\space 3}$) and ellipses resemble the figure given in your sketch. Equations similar to ones given for Case 1 can be developed for the case of ellipse without much ado. However, deriving equations for Cassini oval takes some doing. To make matters worse, solving the equations is not possible without resorting to numerical methods.

The equation of Cassini oval (see $\mathrm{Fig.\space 3}$) with $a\gt \sqrt{2}c$ in polar coordinates is
$$r^2=c^2 \cos\left(2\phi\right)+\sqrt{a^4-c^4\sin^2\left(2\phi\right)} \quad \mathrm{where}\quad a\gt \sqrt{2}c \gt 0. \tag{1}$$

Now, please pay our attention to $\mathrm{Fig.\space 2}$.
$$\tan\left(\alpha\right)=\frac{CE+OF}{AF+OE}=\frac{x_A-\left(\sqrt{ c^2 \cos \left(2\phi\right)+\sqrt{a^4-c^4\sin^2\left(2\phi\right)}}\right)\cos\left(\phi\right)}{y_A-\left(\sqrt{ c^2 \cos\left(2\phi\right)+\sqrt{a^4-c^4\sin^2\left(2\phi\right)}}\right)\sin\left(\phi\right)} \tag{2}$$

Once we determine the value of $\phi$ by solving equation (2), we can find the value of $r$ using equation (1). Then, the Cartesian coordinates of the object at $C$ follows from
$$ x_C=CE=r\sin\left(\phi\right)\quad \mathrm{and}\quad y_C=OE=r\cos\left(\phi\right).$$

We also give here, for your reference, the set of equations needed to locate an object lying on the perimeter of an ellipse. The equation that describes the ellipse, whose semimajor axis and semiminor axis are $a$ and $b$ respectively can be express in parametric form as
$$x=a\cos\left(t\right)\quad \mathrm{and}\quad y=b\sin\left(t\right). \tag{3}$$

The equation equivalent to (1) is
$$\tan\left(\alpha\right)=\frac{x_A+ a\cos\left(t\right)}{y_A+ b\sin\left(t\right)}.$$

From this we can obtain the following quadratic equation in $\sin\left(t\right)$, which, unlike equation (2), can be solved straightforward.
$$\left(a^2+b^2\tan^2 \left(\alpha\right)\right) \sin^2 \left(t\right)+2b\left(y_A\tan\left(\alpha\right)-x_A\right)\tan \left(\alpha\right)\sin\left(t\right)+\left(y_A\tan\left(\alpha\right)-x_A\right)^2-a^2=0$$

Once we solved this equation to determine $t$, we can use it in (3) to calculate the coordinates of $C$.


问:
Solving for $d$ from $a=r^2\arccos{\frac{d}{2r}}-\frac{d}{2}\sqrt{r^2-\left(\frac{d}{2}\right)^2}$
I would like to solve this equation for d with a given r value. Is there a way to write in explicit form? $a$ is of course a known value, I am a little rusty on trigonometry.

$$a  = r^2\arccos{\frac{d}{2r}}-\frac{d}{2}\sqrt{r^2-\left(\frac{d}{2}\right)^2}
$$

Thank you very much.

EDIT: corrected the equation

答:
Let $d = 2r \cos \theta$

$$a = r^2\theta - r\cos \theta r\sin \theta$$

$$a = r^2(\theta - \frac{1}{2}\sin(2\theta))$$

Hence the equation becomes

$$2\theta - \sin(2\theta) = \frac{2a}{r^2}$$

Generally, equations of this form cannot be solved analytically, but it's fairly easy to graph and get an approximate value depending on the values of $a, r$. Since it is a monotone function, there is exactly one solution to this

The graph of $y = x - \sin x$ oscillates between the graphs of $y=x+1$ and $y = x - 1$, touching them at maxima/minima points of $\sin x$


问:
Every Element of $\mathbb F_p$ is a Square in $\mathbb F_{p^2}$ if $p \equiv 3 \mod 4$
How can we show that every element of $\mathbb F_p$ is a square in $\mathbb F_{p^2}$ if $p \equiv 3 \mod 4$ is a prime?

This comes from the first paragraph of Koblitz's "Introduction to Elliptic Curves and Modular Forms" on page 61, self-study.

答:
You actually don't need the condition that $p\equiv 3 \text{(mod }4)$, only that $p$ is odd. Let $F=\mathbb{F}_{p^2}$; clearly it suffices to show that the units of $\mathbb{F}_p\subset F$ are all squares.

Note that $F^\times$ is cyclic; indeed, recall that a finite abelian group is cyclic if and only if there is at most one subgroup of order $k$ for any $k\in\mathbb{N}$. Hence suppose there are subgroups $E_1,E_2\leqslant F^\times$ with $|E_1|=k=|E_2|$. The polynomial $x^k-1\in F[x]$ has $E_1\cup E_2$ as roots, but also has at most $k$ roots in $F$ since $F$ is a field, so we must have $E_1=E_2$, and hence $F^\times$ is indeed cyclic. Note that we can use the same argument to show that $\mathbb{F}_p^\times$ is cyclic.

Thus let $\theta\in F^\times$ a generator for $F^\times$, and let $\alpha\in\mathbb{F}_p^\times$ a generator for $\mathbb{F}_p^\times\subset F^\times$. Note that it suffices to show that $\alpha$ is a perfect square, for every element of $\mathbb{F}_p^\times$ is of the form $\alpha^k$ for some $k$. Now, since $\theta$ is a generator for all of $F^\times$, $\alpha=\theta^m$ for some $m\in\mathbb{N}$. Thus we just need to show that $m$ is even.

But now recall $|F^\times|=p^2-1$, so the order of $\theta$ is $p^2-1$, and $|\mathbb{F}_p^\times|=p-1$, so the order of $\alpha$ is $p-1$. We have $1=\alpha^{p-1}=\theta^{m(p-1)}$, so $p^2-1=(p+1)(p-1)$ divides $m(p-1)$.

This means that $p+1$ divides $m$, and so since $p+1$ is even by hypothesis we are done. Indeed, we have that $\alpha=(\theta^{m/2})^2$ is a perfect square, as desired.


问:
Find the required gcd

Question: Let $\gcd(a,n)=d_1\gt 1$ and $\gcd(b,n)=d_2\gt 1$, given that $\gcd(d_1,d_2)=1$, find $\gcd(a+b,n)$.

My try
Since $\gcd(a,n)=d_1$ so $a=pd_1$ and since $\gcd(b,n)=d_2$ so $b=qd_2$.

My guess is since the greatest common divisors of $a,b$ have gcd =1, so $\gcd(a+b,n)=1$.
But I cant prove it in a compact and well defined fashion

Can someone please help me out?

The question boils down to this:

How is $\gcd(a+b,n)$ related to $\gcd(a,n)$ and $\gcd(b,n)$?

答: As counter-examples to $\gcd(a+b,n)=1$, let $n=ab(a+b)$, for example $30$.

Title: Find the required gcd


答:
Let $$X=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies X^2=\begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc\end{bmatrix}=\begin{bmatrix} 6 & 2\\ 3 & 7 \end{bmatrix}.$$
$$\implies a^2+bc=6~~~(1), b(a+d)=2~~~(2), c(a+d)=3~~~(3), d^2+bc=7~~~(4)$$
From (2) and (3) $b/c=2/3$, let $b=2k, c=3k$. From (1) and (4), we get $a^2-d^2=-1 \implies (a-d)(a+d)=-1$. Again from (2) and (3) we get $a+d=1/k, a-d=-1/k \implies a=(1/k-k)/2, d=(1/k+k)/2$. Finally,inserting $a,b,c$ in (1), we get
$$25k^4-26k^2+1=0 \implies k=\pm 1,\frac{\pm 1
}{5}$$
For $k\pm 1$, we get
$$X=\pm \begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix}~~~(1)$$

For $k=\pm\frac{1}{5}$, we get
$$X=\pm \frac{1}{5} \begin{bmatrix} 12 & 2 \\ 3 & 13 \end{bmatrix}~~~(2)$$

This sheer brute force method gives 4 matrices. one can check their correctness by squaring them.


问:
Confusion on Combinatorics Problem
How many ways are there to put 6 employees in 4 identical offices if at least one employee
must be assigned to each office?

So my thinking was put a person in each office and the last office gets the extra two people, therefore the number of ways is $\binom{6}{3}$.

The other set up is putting two people in the first two offices and the last two we put an extra person in there. So the number of wasys is $\binom{6}{2} \cdot \binom{4}{2}$, but this is incorrect. The right answer is: $\binom{6}{2} \cdot \binom{4}{2}\cdot\frac{1}{2}$.

Why is the $\frac{1}{2}$here?

答:
Why is the (1/2) here.

Because of overcounting.

Person-1, Person-2 in one office, Person-3, Person-4 in another office is counted twice.

Once when Person-1, Person-2 are the first two selected out of the 6, and once when Person-3, Person-4 are the first two selected out of the 6.

Remember, the offices are construed to be identical.

Title: Confusion on Combinatorics Problem


答:
The correct result is squares. For simplicity, take a circle centered at $O=(0,0)$. For $P=(x,y)$,
$$d_S(P,0)=\max\{|x|,|y|\}=r$$
if and only if $x=\pm r$ and $|y|\leq r$ or $y\pm r$ and $|x|\leq r$.

For a nice picture from Mathematica, I ran

```
center = {0, 0}; ContourPlot[ Max[Abs[x - center[[1]]], Abs[y - center[[2]]]],
{x, -2, 2}, {y, -2, 2}]
```

which produces

Body:
The correct result is squares. For simplicity, take a circle centered at $O=(0,0)$. For $P=(x,y)$,
$$d_S(P,0)=\max\{|x|,|y|\}=r$$
if and only if $x=\pm r$ and $|y|\leq r$ or $y\pm r$ and $|x|\leq r$.

For a nice picture from Mathematica, I ran

center = {0, 0}; ContourPlot[ Max[Abs[x - center[[1]]], Abs[y - center[[2]]]],
{x, -2, 2}, {y, -2, 2}]

which produces


问:
Let $f(x)=x+2x^2\sin{1/x}$ for $x\neq0$ and $f(0)=0$. Show that $f'(0)=1$ but $f$ is not monotonic in any interval around $0$.
Let $f(x)=x+2x^2\sin{1/x}$ for $x\neq0$ and $f(0)=0$. Show that $f'(0)=1$ but $f$ is not monotonic in any interval around $0$.

I have shown the first part. How to show its monotonicity part.

答: Consider $f'(\frac  1{2n \pi})$ and $f'(\frac  1{(4n+1) \pi /2})$. Check that for large $n$ the first one is negative and the  second one is positive and any interval of the type $(-\epsilon,\epsilon)$ contains these points for large $n$. If $f$ is  monotonic in $(-\epsilon,\epsilon)$  the $f'$ would not change sign. This proves that $f$ is not  monotonic in $(-\epsilon,\epsilon)$ for any $-\epsilon>0$.


答:
$$f(x) = \begin{cases}x+2x^2\sin\frac{1}{x}, \ \ x\neq 0, \\0, \ \ x = 0\end{cases}$$
$$f'(x) = \begin{cases}1+4x\sin\frac{1}{x} - 2\cos\frac{1}{x}, \ \ x \in U^\circ_\varepsilon(0)  \\ 1,  \ \ x = 0\end{cases}$$

So, if $f$ is monotonic, say increasing, on an interval containing $0$ then you must have $f'(x) \ge 0, \ \ \forall x \in U_\varepsilon(0)$, i.e
$$1+4x\sin\frac{1}{x} - 2\cos\frac{1}{x} \ge 0,  \ \ \forall x \in U_\varepsilon(0)$$
However, you can always find a point $c$ such that $f'(c) < 0$, see this similar post.

The converse ($f$ - decreasing) is similar.

Body:
Let $f(x)=x+2x^2\sin{1/x}$ for $x\neq0$ and $f(0)=0$. Show that $f'(0)=1$ but $f$ is not monotonic in any interval around $0$.

I have shown the first part. How to show its monotonicity part.


问:
Stars and Bars Question Dealing with Number of Solutions
How many solutions are there to the equation $x+y+z = 9$ if $x, y, \text{and }z$ are nonnegative with the restriction $x \le y\le z$?

For this problem I have so far the idea of splitting $x$ into cases. Like if $x=0$, then $y$ ranges from $(0-4) \text{ while $z$ ranges from } z= 9 - y$. But how do I find the number of solutions using this idea?

答:
We know that the total number of solutions to the equation is given by $\binom{11}{9} = 55$. Out of these $55$ solutions, there are certain sets of solutions, in which a number appears more than once.

There are $3$ permutations for the elements of the solution sets having $2$ identical numbers, which are $(0,0,9), (1,1,7), (2,2,5),(1,4,4)$ and there is only $1$ permutation for the set having all $3$ numbers identical, i.e. $(3,3,3)$. i.e. there are a total of $3.4 + 1 = 13$ solutions sets in these $55$ having at least two elements identical. It follows that $55-13 = 42$ solution sets have all elements distinct.

Now, remember that the elements solution set having distinct elements can be permuted in $3! = 6$ ways. But we only want the one case, in which they are arranged in the increasing order. Hence the cases in which the distinct solutions are in increasing order is given by  $42/6 = 7$.

Also, as discussed above, there are also some solution sets having $2$ elements identical, and also all $3$ identical, which are $(0,0,9), (1,1,7), (2,2,5),(1,4,4), (3,3,3)$. We don't want to count their permutations. Hence there are $5$ of these cases.

Hence the total number of solutions satisfying your condition are $\fbox{7+5 = 12}$.


问:
Prove or disprove if $T$ be idempotent t-norm and $a\leq b$ then $T(a,b)=a$ for all $a,b\in[0,1]$?
Let $T : [0, 1]\times [0, 1] \rightarrow [0, 1]$. A $t$-norm is a function $T$ with properties:

$ T (x, 1) = x$

If $y\leq z$ then $T(x,y)\leq T(x,z)$

$T (x, y) = T (y, x) $

$T (x, T (y, z)) = T (T (x, y), z) $

$T$ said to be a idempotent t-norm if for all $x\in[0,1]$, $T(x,x)=x$.

Prove or disprove:

Let $T$ be idempotent t-norm. If $a\leq b$ then $T(a,b)=a$ (for all
$a,b\in[0,1])$?

If we take example $T(x,y)=\min(x,y)$, obviously if $a\leq b$ then $T(a,b)=a$. Generally, if $T$ be idempotent t-norm and $a\leq b$ then prove or disprove $T(a,b)=a$?

I don't have an idea to prove it.

答:
After thinking a long time, I have an idea to prove it.

Let $a,b\in[0,1]$. Then,
$$
a=T(a,a)\leq T(a,b)\leq T(a,1)=a.
$$
So, I can conclude $T(a,b)=a$.


问:
3x3 Rotation matrix from various angles
I am getting 3 angles from another system that I need to convert into a 3x3 rotation matrix.

Here is the diagram:

P is the point where all angles are 0.

A is the tilt angle limited to the angles 0 to 90 degrees.

B is the angle that A is applied in. Note B does not rotate the object itself. If A's value is zero then this angle does nothing. Range is 0 to 180 anticlockwise and 0 to -180 clockwise.

C is the rotation around P. Same ranges as B. This rotation is applied first.

In short, rotate object around P by C, then tilt by A in the direction of B.

Let me know if you need more info.

Edit:
I'll start off with what I have got already and that is the C rotation. Pretty easy for that one as it is just the rotation around the Z axis.

$$\begin{pmatrix}
\cos(C)&-\sin(C)&0\\
\sin(C)&\cos(C)&0\\
0&0&1\end{pmatrix}$$

That works for my purposes but I am unsure of how to convert the other angles to a matrix so I can multiply the two together.

答:
After J. M.'s comment it occurred to me that I can make B + 90 degrees the arbitrary axis and spin in the amount of A. So I can use that to get an axis-angle rotation and then convert that to a matrix.

$$\begin{pmatrix}
xxt+c&xyt-zs&xzt+ys\\
yxt+zs&yyt+c&yzt-xs\\
zxt-ys&zyt+xs&zzt+c\end{pmatrix}$$

Where:

$$\begin{align*}
x &= \cos(B + 90°)\\
y &= \sin(B + 90°)\\
z &= 0\\
s &= \sin(A)\\
c &= \cos(A)\\
t &= 1 - c
\end{align*}$$

Multiplying this matrix with the C matrix in the question above give the correct rotation matrix.

Title: 3x3 Rotation matrix from various angles