问:
Finding a limit using a definition
I want to prove, for $a,b$ positive numbers that $$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) = \cos a$$ by using the definition of a limit. In order to prove this, can I first prove by letting $a_n$ be the sequence which equals $1/n$, where $n = 1,2,3,\ldots$ then prove $a_n \to 0$ as $n$ approaches infinity, then apply this sequence $a_n$ to this equation like $$\lim_{n\to\infty} \cos \left( a + b\cdot a_n \right) = \cos \left( a + b \cdot \lim_{n\to\infty} a_n \right) = \cos(a)$$ as $n$ approaches infinity?

答:
A similar approach is to use Taylor expansions of sine and consine funtions:

$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$,    $\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$ to prove

$\lim_{x\to 0}\cos x=1$, $\lim_{x\to 0}sin x=0$

Then,

$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) $

=$ \lim_{n\to \infty} (\cos  a\cos\frac{b}{n}-\sin  a\sin\frac{b}{n})$

=$ \lim_{x\to 0} (\cos  a\cos x-\sin  a\sin x)$

=$ \cos  a  (\lim_{x\to 0}\cos x)-\sin  a( \lim_{x\to 0}sin x)$ (cosine function is continuous)

=$ \cos  a.1-\sin  a.0$

=$\cos a$

Title: Finding a limit using a definition


问:
Buying Same Combo in Multiple Lotteries
Odds of winning the jackpot in 6/49 lotto is 1 in 13,983,816. So imagine there were 13,983,816 countries and each country had its own lotto 6/49. If you bought one ticket in each country with the same numbers (eg. 1, 2, 3, 4, 5, 6) and all the lotteries made their draw on the same day, what are your odds of winning the jackpot in one of the countries?

The scenario above is hypothetical and the reason I ask is because in my country, we have two 6/49 lotteries (national and regional) that draw on the same day and I would like to know if buying the same combination of numbers for both lotteries on the same day actually improves chances of winning the jackpot in one of the two lotteries?

Please explain what the odds are for the hypothetical question and the odds for the real-life scenario of buying the same combo in the national and regional 6/49.

Thank you

答:
If the probability of winning in a single lottery is $p$, and you buy a ticket in $n$ different lotteries, the probability that you win in at least one of them is $1$ minus the probability that you lose in all of them or $$1-(1-p)^n$$  When $n=2$ this gives $$1-(1-p)^2=2p-p^2\approx 2p,$$ since $p^2$ is miniscule.  If you but two different tickets in one lottery, the probability that one of them will win is $2p$, effectively the same, but in theory, a tiny bit more.

For $n>2$, the same remark holds.  As a practical matter, you do just as well by buying $n$ different tickets in one lottery, but in theory, you do  just a tiny bit better.

ContentLicense: CC BY-SA 4.0

Title: Buying Same Combo in Multiple Lotteries


问:
Finding the sample size of students with the confidence level
Here is my question:
How large a sample of students would be needed in order to estimate the population prediction mean within ±2 with 85% confidence?
Given that the sample of wight students predictions was
72    83    78    65    69    77    81    71
that I have calculated the sample standard deviation(s) = 6.2335, the sample mean = 74.5 and the sample size is 8.

In my own trial, 85% refers to z=1.440. The question requires my calculation base on t(20,0.075) but I cannot find the 0.075 in the table. So I doubt whether my directions aren't correct.

Thanks a lot for help!

答:
Here is a simulation that serves as a 'reality check' on my answer in Comments that the necessary sample size
is around 20 or 22.

In R, the procedure t.test with parameter conf.lev=.85) will use data x show an 85% confidence interval for the population mean $\mu.$.

Suppose, for planning purposes, we use the sample mean and SD from your pilot
experiment with eight observations to be the population mean and SD.
Then we can simulate vector x of $n=22$ observations from $\mathsf{Norm}(\mu=74.5, \sigma=6.23).$ And use t.test to find the resulting 85% CI for $\mu.$ Finally, we can find the half-length of the CI to be the margin of error $E$ and see how close it is to $E = 2.$

```<code>set.seed(320)
x = rnorm(22, 74.5, 6.23)
CI = t.test(x, conf.lev = .85)$conf.int;  CI
[1] 73.21693 77.44468
attr(,"conf.level")
[1] 0.85

me = diff(CI)/2;  me
[1] 2.113874
</code>```

The result is a little larger than we hoped. But this is
only for one sample. If we do this for 10,000 samples and
average the results, we can get an idea whether $n = 22$
is, on average, sufficient to have a margin of error
of the desired size.

```<code>set.seed(2021)
me = replicate(10000,
                diff(t.test(rnorm(21,14.5,6.24),
                 conf.lev=.85)$conf.int)/2)
mean(me)
[1] 1.969126

hdr = "Simulated Margins of Error for n = 21"
hist(me, prob=T, col="skyblue2", main=hdr)
</code>```

So margins of error are often as small as 1.5 or as large as 2.3, but they are mostly near 2.0, as desired.

Body:
Here is a simulation that serves as a 'reality check' on my answer in Comments that the necessary sample size
is around 20 or 22.

In R, the procedure t.test with parameter conf.lev=.85) will use data x show an 85% confidence interval for the population mean $\mu.$.

Suppose, for planning purposes, we use the sample mean and SD from your pilot
experiment with eight observations to be the population mean and SD.
Then we can simulate vector x of $n=22$ observations from $\mathsf{Norm}(\mu=74.5, \sigma=6.23).$ And use t.test to find the resulting 85% CI for $\mu.$ Finally, we can find the half-length of the CI to be the margin of error $E$ and see how close it is to $E = 2.$

set.seed(320)
x = rnorm(22, 74.5, 6.23)
CI = t.test(x, conf.lev = .85)$conf.int;  CI
[1] 73.21693 77.44468
attr(,"conf.level")
[1] 0.85

me = diff(CI)/2;  me
[1] 2.113874

The result is a little larger than we hoped. But this is
only for one sample. If we do this for 10,000 samples and
average the results, we can get an idea whether $n = 22$
is, on average, sufficient to have a margin of error
of the desired size.

set.seed(2021)
me = replicate(10000,
                diff(t.test(rnorm(21,14.5,6.24),
                 conf.lev=.85)$conf.int)/2)
mean(me)
[1] 1.969126

hdr = "Simulated Margins of Error for n = 21"
hist(me, prob=T, col="skyblue2", main=hdr)

So margins of error are often as small as 1.5 or as large as 2.3, but they are mostly near 2.0, as desired.


问:
A inclusion property for relations
Suppose that $R$ and $S$ are two well-defined relations,
then
$$
(R \cap S)^{-1} \circ R^{-1} \subset (R \circ S)^{-1} ?
$$

Here, $M \circ N$ denotes the composition of the relations $M$ and $N$ and $L^{-1}$ the inverse of a relation $L$.

I try many counter-examples, but none works. For instance $A = \{ a,b\}, B = \{ a,d\}, C = \{ a,c\}$ and $D = \{ b,d\}$ for $R \subset A \times B$ and $S \subset C \times D$. Also, the identity $(M \circ N)^{-1} = N^{-1} \circ M^{-1}$ does not seem to help.

答:
I'm guessing you write $R \circ S$ for $S$ first and then $R$.

Note that the inverse of an intersection is the intersection of the inverses:
$$(R \cap S)^{-1} = R^{-1} \cap S^{-1}.$$
So the inclusion you are interested in is
$$(R^{-1} \cap S^{-1}) \circ R^{-1} \subset S^{-1} \circ R^{-1}$$
so this is not really about inverses and we can just think about:
$$(R  \cap S ) \circ R  \subset S  \circ R.$$
This does hold because for any $(x,z) \in (R  \cap S ) \circ R $, there must be
$y$ where
$(x,y) \in R$ and $(y,z) \in (R  \cap S)$. But then we have $(x,y) \in R$ and $(y,z) \in  S$, so that $(x,z) \in   S  \circ R $.

In a more general setting, if relations $R$ and $S$ satisfy $R \subset S$ and $T$ is a relation then $R \circ T \subset S \circ T$ and $T \circ R \subset T \circ S$.

Title: A inclusion property for relations


问:
Difficulty using Sylow theorems

I am really struggling to apply Sylow's theorems to various problems, so any help is much appreciated.

For part (a) here, since we are supposing that $H<G$, I tried to use an element $g$ of $G$ that is not in $H$ to find that $gH \neq Hg$ using that every element normalising $P$ is in $H$. I couldn't think of a way to do this.

I also tried to think of a way to prove it via the contrapositive (if $H$ is a proper normal subgroup of $G$, then $N_G(P)$ is not a subgroup of $H$. This also failed.

I am sure I should be using the Sylow theorems here but I can not tell how any of them can be applied in a useful way.

答: (a) Let $H$ be a normal in $G$. Then $H^x=H$ for every $x\in G$ and $P^x<H^x=H$. Since the subgroup $H$ has exactly one Sylow $p$-subgroup, it follows that $P^x=P$. So $x\in H$ and $H=G$. We have arrived at a contradiction.

Title: Difficulty using Sylow theorems


问:
Compute $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$
I haven't integrated in a while... so I'm a bit rusty. I have to integrate $\int_0^1\frac{x}{\sqrt{x(1-x)}}dx$. Should I apply some change of variables? If I take $u=x-1/2$, I get something of the form $\int\frac{1}{\sqrt{u^2+1/4}}du$ (along with another integral, which I can integrate), but I don't know how to deal with this one.

答:
Use the property $$ \int_0^a f(x) ~dx = \int_0^a f(a-x) ~dx $$

$$ \mathscr{I} =\int_0^1\frac{x}{\sqrt{x(1-x)}}~ dx ~= ~\int_0^1\frac{1-x}{\sqrt{(1-x)x}}~ dx $$

Add both the integrals
$$ 2 \mathscr{I} = \int_0^1\frac{x+1-x}{\sqrt{x(1-x)}}~ dx ~= \int_0^1\frac{1}{\sqrt{x(1-x)}}~ dx $$

Now, this integral can be easily evaluated by completing the square
$$ \mathscr{I} = \dfrac12 \int_0^1 \frac{1}{\sqrt{\frac14-(x-\frac12)^2}}~ dx  $$
$$ \mathscr{I} = \dfrac12 ~\sin^{-1} \left( \dfrac{x-\frac12}{\frac12}  \right)_0^1 = \color{red}{\dfrac{\pi}{2}} $$


问:
Optimization problem with Lagrangian multipliers
I'm trying to solve the following optimization problem:
\begin{equation}\end{equation} \begin{align} \text{argmax}_{\left\{x_i\right\},\left\{y_j\right\}} & \frac{1}{2}\left( \sum_{i=1}^{N} \log_2\left(1+\alpha_i x_i\right) + \sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) \right) \\[4pt]  \nonumber \text{subject to: }& \sum_{i=1}^N x_i + \sum_{j=1}^M y_j \leq A \\
& \sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) \leq B \end{align}

To this end, I tried to use the Lagrangian multipliers method. I defined the Lagrangian function
\begin{align}
    \mathcal{L}(x_1,\ldots,x_N,y_1,\ldots,y_M,\lambda_1,\lambda_2) &= \left( \sum_{i=1}^{N} \log_2\left(1+\alpha_i x_i\right) + \sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) \right) \\
    & - \lambda_1 \left(\sum_{i=1}^N x_i + \sum_{j=1}^M y_j - A\right) \\
    & -\lambda_2 \left(\sum_{j=1}^{M} \log_2\left(1+ \beta_j y_j\right) -B\right)
\end{align}
and I computed the partial derivatives of the Lagrangian function with respect to generic $x_i$ and $y_j$. Setting them to zero led to
\begin{align}
    & \frac{\partial \mathcal{L}}{\partial x_i} = \frac{\alpha_i}{\alpha_i x_i \log(2) + \log(2)} - \lambda_1 = 0, \\
    & \frac{\partial \mathcal{L}}{\partial y_j} = \frac{\beta_j}{\beta_j y_j \log(2) + \log(2)} - \lambda_1 -\lambda_2 \frac{\beta_j}{\beta_j y_j \log(2) + \log(2)} = 0,
\end{align}
which resulted in
\begin{align}
    & x_i = \frac{1}{\log(2) \lambda_1} - \frac{1}{\alpha_i}, \\
    & y_j = \frac{1-\lambda_2}{\log(2) \lambda_1} - \frac{1}{\beta_j}.
\end{align}
I don't know how to proceed from here onwards. Any help?

答:
Hint.

I will use $\ln$ instead to facilitate the notation issues. First you should transform your inequalities into equations with the introduction of some slack variables $s_1, s_2$ so the Lagrangian reads

\begin{align}
    \mathcal{L}(x_1,\ldots,x_N,y_1,\ldots,y_M,\lambda_1,\lambda_2,s_1,s_2) &= \frac 12\left( \sum_{i=1}^{N} \ln\left(1+\alpha_i x_i\right) + \sum_{j=1}^{M} \ln\left(1+ \beta_j y_j\right) \right) \\
    & - \lambda_1 \left(\sum_{i=1}^N x_i + \sum_{j=1}^M y_j - A+s_1^2\right) \\
    & -\lambda_2 \left(\sum_{j=1}^{M} \ln\left(1+ \beta_j y_j\right) -B+s_2^2\right)
\end{align}

now, the stationary points are the solutions for

\begin{align}
    & x_i = \frac{1}{2 \lambda_1} - \frac{1}{\alpha_i}, \\
    & y_j = \frac{1-2\lambda_2}{2\lambda_1} - \frac{1}{\beta_j},\\
    & \sum_{i=1}^N x_i + \sum_{j=1}^M y_j - A+s_1^2= 0,\\
    & \sum_{j=1}^{M} \ln\left(1+ \beta_j y_j\right) -B+s_2^2=0,\\
    & \lambda_1s_1 = 0,\\
    & \lambda_2 s_2 = 0,
\end{align}

We can proceed by substituting the values found to $x_i,y_j$ into the restrictions giving

\begin{align}
    &\frac{N}{2 \lambda_1} - \sum_{i=1}^N\frac{1}{\alpha_i}+M\frac{1-2\lambda_2}{2\lambda_1}- \sum_{j=1}^M\frac{1}{\beta_j}-A+s_1^2=0,\\
    &\left(\frac{1-2\lambda_2}{2\lambda_1}\right)^M\prod_{j=1}^M\beta_j=2^Me^{B-s_2^2}
\end{align}

Now using $\lambda_is_i=0$ we can (?) calculate $\lambda_i$ and finally to have the stationary points. The solutions obtained with $s_i = 0$ are solutions at the feasible region boundary.


答:
Let $A$ be a set, $R, S \subseteq A^2$ any two binary relations and $x,y \in A$. Then

\begin{align*}
x (R \cap S)^{-1} \circ R^{-1}y & \implies \exists z \in A \colon x R^{-1}z \wedge z (R \cap S)^{-1}y \\
& \implies \exists z \in A \colon zRx \wedge y R \cap S z \\
& \implies \exists z \in A \colon z R x \wedge y R z \wedge y S z \\
& \implies \exists z \in A \colon y S z \wedge z R x \\
& \implies y R \circ S x \\
& \implies x (R \circ S)^{-1}y.
\end{align*}

Therefore, $(R \cap S)^{-1} \circ R^{-1} \subseteq (R \circ S)^{-1}$. $\square$

Title: A inclusion property for relations

Body:
Suppose that $R$ and $S$ are two well-defined relations,
then
$$
(R \cap S)^{-1} \circ R^{-1} \subset (R \circ S)^{-1} ?
$$

Here, $M \circ N$ denotes the composition of the relations $M$ and $N$ and $L^{-1}$ the inverse of a relation $L$.

I try many counter-examples, but none works. For instance $A = \{ a,b\}, B = \{ a,d\}, C = \{ a,c\}$ and $D = \{ b,d\}$ for $R \subset A \times B$ and $S \subset C \times D$. Also, the identity $(M \circ N)^{-1} = N^{-1} \circ M^{-1}$ does not seem to help.


问:
A loss and gain problem
This is a very simple but confusing puzzle.

A customer buys goods worth $200$ rupees from a shop. The shopkeeper selling these goods makes zero profit from this purchase.

The lady gives him a $1000$ rupee note.

The shopkeeper has no change, so he goes next door to another shopkeeper to get change for the $1000$ rupee note. He keeps $200$ for himself and returns $800$ to the customer.

Later, the second shopkeeper from next door comes back with the $1000$ rupee note with a stamp on it saying "counterfeit" and takes his money back.

How much loss does the first shopkeeper face?

答: since shopkeeper selling good with zero profit so $200$rs of good selling is not his profit.so when he return 1000 to his neighbor shopkeeper he is in LOSS of $1000$rs. People thinks its answer will be 800.</p>

ContentLicense: CC BY-SA 3.0

Title: A loss and gain problem


问:
Function which is continuous exactly one point and differential exactly one point
I found a Function which is continuous exactly one point and differentiable exactly one point. And 2 point continuous and 2 point differentiable and more for infinite finding this type of function is possible?</p>

答:
It is analogous to a bunch of dominoes standing on a table. If one domino falls, then the domino right next to it will fall. However, if the first domino itself doesn't fall, then none of the dominoes will fall.

If you are able to prove that a statement holds for $n=k+1$ provided it holds for $n=k$ then that itself isn't enough to prove that the statement holds for say, all natural numbers. If it holds for $n=1$ then it will hold for $n=2$ and in turn, it will hold for $n=3$ and so on.

Take for example, the statement $$ \{x\} = 2 $$
Assuming that this statement holds for $x=k$ we can easily show that this also holds $x=k+1$ as $ \{k\} = \{k+1\} ~~ \forall ~~ k \in \mathbb{R} $

But as we all know, $ \{x\} = 2 $ has no real solutions. Despite proving that $ x=k \implies x=k+1$ it is still not an identity as there is no real $x$ that satisfies the equation. That is why proving the statement for $x=1$ is necessary if you want to prove it for all natural numbers.


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

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

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

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

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

答:
Another approach:

In figure OC=OB. As canbe seen $\APB$ is maximum when P is at distance $\frac 34 OC$. You can find that:

$AD=\frac{\sqrt 2}4 OB$

$AC=\frac{\sqrt 5}2 OB$

$\sin (\angle ABC)=\frac {AD}{AC}\approx 0.316$

$\sin (\angle APB)>0.316$

So option $\frac 13\approx 0.3333>0.316$ is corect.


问:
Calculating loan + interest recursively
I am trying to calculate a loan amount where the total principle depends on the payment amount.

Here's a simple example, leaving aside that a) no bank would actually do this, and b) my income is not specified:

I have $0 in the bank right now. I want to take out a loan for a $100 car, for which I will be making a regular annual payment, starting at the end of year 0 with an interest-only payment. My loan rate is 5%. In order to qualify for the loan, the bank wants to make sure that I have a balance equal to twice my annual payment amount throughout the life of the loan.

If I borrow exactly $100, my payment at the end of year 0 would be $5 ($100 x 5%). However, I need to borrow a little extra so that I have a $10 balance in the account (to satisfy the requirement that I maintain a balance of twice my annual payment). This means I actually need a loan of $110. However, a loan of $110 actually raises my payment amount to $5.50, which means I need to maintain a balance of $11 in my account.

This means that I actually need to borrow $111 to begin with, which raises my payment to $5.55, which means I actually need to borrow $111.10, etc.

I have worked it out to a summation that resembles the following, although I can't be certain this is accurate:

$$\sum_{n=1}^x (x-1-n)r^n$$

Where $r$ is the interest rate

I believe this would allow for me to iterate $x$ number of times over a formula that sums the additional principle needed for each incremental amount of interest. However, I'm not sure how to carry this out to infinity.

Ultimately, I need to plug this into Excel as part of a cash flow, but I wanted to think through the math first.

Any ideas?

答:
In the Wikipedia article you link to, cash on hand does not increase the DSCR.  Unless the cash generates income, borrowing more lowers the DSCR by the ratio of the new payment to the old payment.  If the cash can generate income greater than the old DSCR times the payment, borrowing will help the DSCR.  The new income enters into the Annual Net Income term.

If I inherit a bunch of money, but deposit it at zero interest, my DSCR doesn't change.  Paying down the loan in a way that lowers the payment would help.

Title: Calculating loan + interest recursively


问:
Global min/max verification
Take $x^3-3x$ for example, would $x=1$ be a local min as well as a global min if we consider the range of the domain to be all of $x \in \mathbb{R}$ as min/max must be at a finite $x$?

答: For $f(x) = x^3 - 3x$ the point $x = 1$ is a local minimizer, but not a global minimizer.  The reason for this is that in a small region around $x = 1$, $f(1) =-2$ is the smallest value, meaning it is a local min.  The reason it is not a global min is because there are points where the $y$-value is smaller than $-2$.  For example, $f(-3) = -18 < -2$, which means that $f(1) = -2$ cannot be the global min.  In this case, the function has no global min or global max, because we have $$\lim_{x\to-\infty}f(x) = -\infty \quad \text{and}\quad \lim_{x\to\infty}f(x) = \infty.$$

Title: Global min/max verification


答:
That's correct but you should mention why $a_n\to 0\implies \cos (a+ba_n)\to\cos a$. That is because of continuity of $\cos$ function. But let's pretend that we don't know about continuity of $\cos$ function. If $b=0$, we are done so assume $b\ne 0$.

$a_n\to 0\implies \forall \epsilon\gt 0, \exists N\in \mathbb N: n\gt N\implies|a_n|\lt \epsilon/|b|$

For $n\gt N$, we also have

$|\cos (a+ba_n)-\cos a|=2|\sin \frac{a+ba_n}{2}\sin\frac{ba_n}{2}|\le|2 (1)\sin \frac{ba_n}{2}|\le |ba_n|\lt \epsilon$.

This proves the result by definition.

Title: Finding a limit using a definition

Body: I want to prove, for $a,b$ positive numbers that $$\lim_{n\to \infty} \cos \left( a + \frac{b}{n}\right) = \cos a$$ by using the definition of a limit. In order to prove this, can I first prove by letting $a_n$ be the sequence which equals $1/n$, where $n = 1,2,3,\ldots$ then prove $a_n \to 0$ as $n$ approaches infinity, then apply this sequence $a_n$ to this equation like $$\lim_{n\to\infty} \cos \left( a + b\cdot a_n \right) = \cos \left( a + b \cdot \lim_{n\to\infty} a_n \right) = \cos(a)$$ as $n$ approaches infinity?


问:
Transformation matrix between carthesian to spherical coordinates to draw parallel circumferences
I've seen some algorithms with the following transformation matrix:

In such an algorithm, this matrix is used to draw some parallel circumferences on a sphere, i.e. the blue circumferences in the following figure:

Each curve is approximated as a set of points (so each circumference is approximated with a polygon). Each point is defined with the spherical coordinates $\rho, \phi, \theta$.

I want to find out some informations about this matrix. It seems to contain the classical transformation matrix from cartesian to spherical coordinates:

but it as another coloumn at right. Why has this been inserted in my matrix?

$X_c, Y_c, Z_c$ express the centre of the sphere.

答:
Too long for a comment.
The mapping $\mathbb R_+\times (-π, π)\times (0,π)\ni(r,\theta, \phi)\mapsto (x,y,z)$ is given by\begin{cases}
x=r \cos \theta\sin\phi\\
y=r \sin \theta\sin\phi\\
z=r\cos \phi
\end{cases}
and its differential is
$$
\begin{pmatrix}
\cos \theta\sin\phi& -r \sin \theta\sin\phi&r \cos \theta\cos\phi\\
\sin \theta\sin\phi&r \cos \theta\sin\phi&r \sin \theta\cos\phi\\
\cos \phi&0&-r \sin\phi
\end{pmatrix}.
$$
Permuting the second and third row, making $r=1$ you get
$$
\begin{pmatrix}
\cos \theta\sin\phi& \cos \theta\cos\phi& - \sin \theta\sin\phi\\
\sin \theta\sin\phi& \sin \theta\cos\phi& \cos \theta\sin\phi\\
\cos \phi&- \sin\phi&0
\end{pmatrix}.
$$
The longitude lines are defined by $\theta=\theta_0$ and the latitude lines by $\phi=\phi_0$ (the latitude is $\frac π2-\phi_0$, positive on the northern hemisphere, negative on the southern).
What was the use of your $4\times 4$ matrix?


问:
Proving Redei's Theorem
Can someone prove Redei's theorem that every tournament has a directed Hamiltonan path in simple english?

The articles I found on it are very abstract and the only video on the topic is in Hindi.

答:
A very relevant idea is the median order of a directed graph. It is a vertex ordering $v_1, v_2, \dots, v_n$ that maximizes the number of "forward edges": edges $(v_i, v_j)$ for which $i < j$. Since we are optimizing over $n!$ possible orderings, which is a finite set, a maximum must exist, though it is not always unique.

This is NP-hard to compute for general directed graphs. But it is a useful idea for proofs, where we don't have to worry about computing it! In particular, it immediately gives us the result you want here:

Theorem. $v_1, v_2, \dots, v_n$ is a median order of a tournament, then $(v_1, v_2, \dots, v_n)$ is a Hamiltonian path.

Proof. Suppose not: suppose there is an edge between $v_i$ and $v_{i+1}$ oriented $(v_{i+1}, v_i)$ rather than $(v_i, v_{i+1})$. Then swapping the two vertices $v_i$ and $v_{i+1}$ increases the number of forward edges by $1$: the edge between these two vertices is now a forward edge, and no other edges are affected. This contradicts the maximality of the median order we started with. $\square$

Title: Proving Redei's Theorem


问:
A basic intuition on a probability problem
Two players take turns shooting at a target, with each shot by player $i$ hitting the target with probability $p_i$, $i=1,2$. Shooting ends when two consecutive shots hit the target. Let $\mu_i$ denote the mean number of shots taken when player $i$ shoots first, $i=1,2$.

Now, I have calculated that $$\mu_1 - \mu_2 = \frac{p_2 - p_1}{p_1+p_2-p_1p_2-2}=\frac{q_2-q_1}{1+q_1q_2}$$

where $q_i=1-p_i, i=1,2 $

i.e. if the player with higher winning probability starts the game then the mean number of shots taken is higher than the case when the the player with lower winning probability starts the game. I don't understand the intuition behind this.

答:
Since the correctness of the OP's formula for $\mu_1-\mu_2$ has been questioned here is a roundup of the argument that leads to it:

Consider $\mu_i$ as the expected number of remaining shots when it is $P_i$'s turn and there is no immediately foregoing hit, and similarly let $\lambda_i$ be the expected number of remaining shots when it is $P_i$'s turn and there is an immediately foregoing hit. Then we have the following equations:
$$\eqalign{\mu_1&=1+p_1\lambda_2+(1-p_1) \mu_2 \cr
\mu_2&=1+p_2\lambda_1+(1-p_2) \mu_1 \cr
\lambda_1&=1+p_1\cdot 0+(1-p_1)\mu_2\cr
\lambda_2&=1+p_2\cdot 0+(1-p_2)\mu_1\ .\cr}$$
Solving this system for the $\lambda_i$, $\>\mu_i$ leads to
$$\mu_1-\mu_2={p_1-p_2\over 1+q_1q_2}\ ,$$
where $q_i=1-p_i$.

The fact that $p_1>p_2$ implies $\mu_1>\mu_2$ can intuitively be explained as follows: Since a hit by the weaker player is necessary to end the game, letting him begin gives him an extra chance to shoot and will, on the average, shorten the game.


问:
Finding a Basis of a Polynomial Space using Eigenvectors from a Linear Map
I'm working on a homework assignment and think I've got the answer, but wanted to double check that I'm interpreting it correctly. The problem and subproblems are as follows:

Let $T: P_2({\rm I\!R}) \to P_2({\rm I\!R})$ be a linear map defined by $$T(f(x)) = f(x)+(x+1)f'(x)$$ and let $\beta = \{1,x,x^2\}$ be a basis for $P_2({\rm I\!R})$

Determine the eigenvalues of $[T]_\beta$<br

For each eigenvalue of $[T]_\beta$, find the corresponding eigenvectors<br

If possible, find a basis for $P_2({\rm I\!R})$ consiting of eigenvectors

So I found the answers for 1 and 2, which was fairly straightforward once I created the transformation matrix of $T$ with respect to basis $\beta$, but 3 is giving me some issues. Given that $\beta$ is already the standard basis vectors for $P_2({\rm I\!R})$, couldn't I just use the eigenvectors found in SubProblem 2 to create said basis eigenvectors? If so, would it be as simple as just multiplying the basis $\beta$ by each eigenvector to attain the new eigenvector basis representation?

答: A basis of eigenvectors is $\{1,x+1,2x^2+4x+3\}$, which correspond to the eigenvalues $1$, $2$, and $3$ respectively. So, take $\alpha=\{1,x+1,2x^2+4x+3\}$, and $\alpha$ is a basis of $P_2(\Bbb R)$ which consists of eigenvectors.


问:
Two nodes consensus network and its phase plot
This is my first day to learn these stuff. I am confused about the following:

Suppose the underlying graph is just two nodes with edge value $1$.

So the adjacency matrix is:  $$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$$

with the degree matrix $$D =\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$

So the graph Laplacian: $L = D - A$. The consensus protocol should be $$\dot{x} = -Lx = \begin{bmatrix} x_2-x_1  \\ x_1-x_2 \end{bmatrix}$$

So let $x_0 = [0.25;-0.25]$, we have the following plot:

However, the phase plot is the following. Does this phase plot imply any consensus information? I cannot understand "consensus" from the following phase plot. What I mean is $x=0.25$ won't go to $0$ which contradicts the above plot. Please advise, thanks!

答: Your phase plot is correct. The states may not converge to zero since the Laplacian matrix has exactly 0 (zero) eigenvalue. According to your initial conditions, the states converge to the line $x=y$. Besides, consensus means that all states converge to the same value. It doesn't have to be zero.


问:
Converse to a result of convergence of $\lim\frac{1}{x}\sum_{n\leq x}a_n=k$
In this post it is proved that, if $(a_n)$ is a sequence in $\mathbb C$:

$$
\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}a_n=k\implies\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k.
$$

Is the converse to this statement true? In other words, is there a sequence $(a_n)$ with

$$
\lim_{x\to\infty}\frac{1}{\log x}\sum_{n\leq x}\frac{a_n}{n}=k
$$

but $\lim_{x\to\infty}\frac1x\sum_{n\leq x}a_n$ converges to some other limit, or diverges? Thanks.

答:
If $a_n=(-1)^n n$, then:

$\displaystyle \lim_{x \rightarrow +\infty} \dfrac{1}{\log x} \sum_{n\leqslant x} \dfrac{a_n}{n} = \lim_{x \rightarrow +\infty} \dfrac{1}{\log x} \sum_{n\leqslant x} (-1)^n = 0 $

$\displaystyle \lim_{x \rightarrow +\infty} \dfrac{1}{ x} \sum_{n\leqslant x}a_n$ doesn't exist.


问:
Approximate solutions to a transcendental equation of two variables
Say I have an equation of the form:
$$ y = A + B\sin (x) +C\sin (x+Dy) $$ on the domain $0<x<2\pi$.

I want to get $y$ as a function solely of $x$, i.e. remove the $y$-dependence from the R.H.S. This cannot be done analytically, but are there any useful tricks to analytically obtain approximate solutions? The trouble is that $x$ and $y$ can take on arbitrary values, i.e. I can't take them to be small and make a first-order approximation of $\sin$, for example.

答:
You can get $x$ solely in terms of $y$
$$y= A + B\sin(x) + C\sin(x+Dy)$$
$$y = A + B\sin(x) + C\sin(x)\cos(Dy) + C\sqrt{1-\sin^2 (x)}\sin(Dy)$$
$$z=\sin(x)$$
$$y = A + Bz + Cz\cos(Dy) + C\sqrt{1-z^2}\sin(Dy)$$
$$w=\cos(Dy)$$
$$y= A + Bz + Czw + C\sqrt{1-z^2}\sqrt{1-w^2}$$
Which yields:
$$z_{\pm}=\frac{\pm\sqrt{-C^2 \left(w^2-1\right) \left(-A^2+2 A y+B^2+2 B C w+C^2-y^2\right)}-A (B+C w)+B y+C w y}{B^2+2 B C w+C^2}$$


问:
Define term using Stirling Numbers
I am trying to solve the following exercise - quite unsuccessful yet.

Let a(m,n) be defined as
  $$ \sum\limits_{n=0}^m a(m,n) \prod\limits_{i=1}^n (x+i-1) = x^m $$
  Express a(m,n) using S(m,n) while S(m,n) are the Stirling numbers of the second kind which count the number of ways to partition a set of n elements into k nonempty subsets.

Hint: use the following identity : $$x^m = \sum\limits_{n=0}^m S(m,n) \cdot x \cdot (x-1) \cdots (x-n + 1) $$

First I rewrote the "hint"-identity as
$$ x^m = \sum\limits_{n=0}^m S(m,n) \prod\limits_{i=1}^n (x+1-i)$$

and got
$$ m = 0 \rightarrow a(0,0) = x^0 = S(0,0) $$
$$ m = 1 \rightarrow a(1,0) + a(1,1) \cdot x = S(1,0) + S(1,1) \cdot x $$
and m = 2
$$ a(2,0) + a(2,1) \cdot x + a(2,2) \cdot x \cdot (x+1) = S(2,0) \cdot x + S(2,1) \cdot x + S(2,2) \cdot x \cdot (x-1)$$
and both compared for m = 3
$$ \begin{array}{llll} a(3,0) & + a(3,1) \cdot x & + a(3,2) \cdot x \cdot (x+1) & + a(3,3) \cdot x \cdot (x+1) \cdot (x+2) \\ \underbrace{S(3,0) \cdot x}_{\text{always 0}} & +S(3,1) \cdot x & +S(3,2) \cdot x \cdot (x-1) & +S(3,3) \cdot x \cdot (x-1) \cdot (x-2) \end{array}
 $$

Replacing x with -x in the "hint"-identity as recommended by user9325 results in

$$ \begin{array}{llll} a(3,0) & + a(3,1) \cdot x & + a(3,2) \cdot x \cdot (x+1) & + a(3,3) \cdot x \cdot (x+1) \cdot (x+2) \\ S(3,0) & +S(3,1) \cdot (-x) & +S(3,2) \cdot (-x) \cdot (-x-1) & +S(3,3) \cdot (-x)(-x-1)(-x-2) \end{array} $$

Multiplying each summand of the already modified identity by $(-1)^{(n+1)}$ gets

$$ \begin{array}{llll} a(3,0) & + a(3,1) \cdot x & + a(3,2) \cdot x \cdot (x+1) & + a(3,3) \cdot x \cdot (x+1) \cdot (x+2) \\ S(3,0) & +S(3,1) \cdot x & +S(3,2) \cdot x \cdot (x+1) & +S(3,3) \cdot x\cdot(x+1)\cdot(x+2) \end{array} $$

Is this correct? How do I put this altogether?

答: The shortest way to find the answer is to replace $x$ by $-x$ in one of the identities and then compare them.</p>

Title: Define term using Stirling Numbers


问:
Vector spaces: Name for restricting scalar multiplication to a subfield
Say we have a complex vector space $V$? Do we have a name for the real vector space obtained by restricting the scalar product to $\mathbb{R}$? (I like to denote it as $\mathbb{V}_\mathbb{R}$, but I don't know if it has a standard name.)

More generally, say we have a field extension $\mathbb{L}\leq\mathbb{K}$. Consider the forgetful functor from the category of $\mathbb{K}$-vector spaces to the category of $\mathbb{L}$-vector spaces. Does this functor have a name? Do we have a name for the $\mathbb{L}$-vector space obtained from a $\mathbb{K}$-vector space?

答:
This functor is called "restriction of scalars." I general if we have a ring homomorphism $f:R\to S$ then this induces a functor $S\mathrm{-mod}\to R\mathrm{-mod}$ which sends an $S$-module $M$ to the $R$-module with the same underlying abelian group and the multiplication given by $r\cdot m:=f(r)m.$

There's also the "extension of scalars" functor going in the other direction which is just given by $-\otimes_R S$ where we use $f$ to consider $S$ as an $R$-module.


问:
Locally constant sheaves over an irreducible space is constant.
In Hartshorne's  Algebraic Geometry Chapter II Proposition 6.15: If $X$ is an integral scheme, the homomorphism $CaCl X \rightarrow Pic X$ is an isomorphism.

In the proof he wants to prove that $\mathcal L \otimes \mathcal K= \mathcal K$. It is clear that on an open cover $\{U_i\}$ $(\mathcal L \otimes \mathcal K)|_{U_i}\cong \mathcal K$.

From this he concludes that $\mathcal L \otimes \mathcal K \cong \mathcal K$, which follows from a general fact that if "$X$ is irreducible, a sheaf whose restriction to each open set of a covering of $X$ is constant, (*) to  is in fact a constant sheaf".

Can someone please give a  proof of the above fact that locally constant sheaves over an irreducible space is actually constant .

I guess Harthsorne wants to say that it is isomorphic to a constant sheaf.

答:
It suffices to show that for any such sheaf $\mathcal{F}$ and any global section $s\in\Gamma (X,\mathcal{F} )$, there exists an open cover $\{U_{\alpha}\}$ of $X$ and a set $A$ so that $s\vert_{U_{\alpha}}\in A$ for each $\alpha$. This is equivalent to saying that $\mathcal{F}$ is the sheaf associated to the constant presheaf $\underline{A}$.

Now since we have had an open cover $\{U_{\alpha}\}$ so that $\mathcal{F}\vert_{U_{\alpha}}$ is constant, there exists an open cover $\{V_{\gamma}\}$ of $X$ which is a refinement of $\{U_{\alpha}\}$ so that the restriction of $s$ to $V_{\gamma}$ lies in $A_{\alpha}$ if $V_{\gamma}\subseteq U_{\alpha}$. Since $X$ is irreducible, any two open subsets has non-empty intersection, and since $\mathcal{F}\vert_{U_{\alpha}}$ is constant for all $\alpha$, we have for $x\in U_{\alpha\beta }=U_{\alpha }\cap U_{\beta}$, $\mathcal{F}_x =A_{\alpha} =A_{\beta}$, hence the sets are all the same, and we have constructed such an open cover and find such a set, which implies that $\mathcal{F}$ is constant.


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

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

答:
This is user91684's repertoire. Since he/she has left this community, I will outline a possible approach using one of his/her favourite tricks (which for instance was applied in this answer to a similar question).

Let $A=X+iY$ and $B=iX+Y$ or equivalently, let $X=\frac12(A-iB)$ and $Y=\frac{1}{2i}(A+iB)$. The equality $A^2+B^2=c(AB-BA)$ can then be rewritten as
$$
(c-i)XY=(c+i)YX.
$$
Since $AB-BA=2(XY-YX)$ is invertible, neither $XY$ nor $YX$ is zero. Therefore $c\ne\pm i$. It follows that $XY=\omega YX$ where $\omega=\frac{c+i}{c-i}\ne 1$. Hence $AB-BA=2(XY-YX)=2(\omega-1)YX$ and $YX$ is invertible. Since $XY$ and $YX$ have identical spectra, the equality $XY=\omega YX$ implies that the spectrum of $XY$ is invariant when it is multiplied by $\omega$. It follows that $\omega$ is some primitive $m$-th root of unity and the spectrum of $XY$ is then a disjoint union of $m$-cycles of the form $\lambda\{1,\omega,\omega^2,\ldots,\omega^{m-1}\}$. Thus $m$ divides $n$. If you can show that the only values of $c$ such that $\omega=\frac{c+i}{c-i}$ is a primitive root of unity are $0,1$ and $-1$, then you are done because $m=4$ (and $\omega=\pm i$) when $c=\pm1$.

Remark. Two matrices $X$ and $Y$ that satisfies the equation $XY=\omega YX$ are said to be quasi-commutative or $\omega$-commutative. According to the aforementioned answer by user91684, there is an interesting result, discovered by Drazin, about quasi-commutativity. See section 5 of Holtz, Mehrmann and Schneider, Potter, Wielandt, and Drazin on the Matrix Equation $AB=\omega BA$: New Answers to Old Questions, Amer. Math. Monthly, 111:8 (2004), 655-667.


问:
help to understand the partial derivative of $f(x,y) = {1 + h(x) \over 1 + (g(y))^2}$
Can you please help me to understand how given this function:

$f(x,y) = {1 + h(x) \over 1 + (g(y))^2}$

Its partial derivative is:

$\frac{\partial f}{\partial x} = {1 \over 1 + (g(y))^2}(1 + h'(x))$

My initial attempt was:
$\frac{\partial f}{\partial x} = {1 \over 1 + (g(y))^2}(h'(x))$

答:
$$f(x,y)=\frac{1+h(x)}{1+(g(y))^2}=\underbrace{\frac{1}{1+(g(y))^2}}_{\text{independent of }x}+\underbrace{\frac{1}{1+(g(y))^2}}_{\text{independent of }x}\,h(x)$$

When taking the partial derivative with respect to $x$, you can treat expressions independent of $x$ as constants. Hence

$$f'_x(x,y)=0+\frac{1}{1+(g(y))^2}h'(x)$$

and your attempt is correct.


问:
What does "does not cancel on the interval" mean?
I have a question which asks me to consider the system
$$f(x)=a+bx+\frac{P(x)}{Q(x)},\qquad x\in I,$$
where $P$ and $Q$ are polynomials, and stipulates "$Q(x)$ does not cancel on the interval $I$ ".

What does this mean?

Does this mean $Q(x)\neq0$ on this interval, or that $Q(x)\neq P(x)$ on this interval, or perhaps something else? I have tried Googling the expression but couldn't find anything, and I haven't came across it before.

答: It means $Q(x) \neq 0$ for all $x\in I$.


问:
The set of all nilpotent elements of a weird binary operation.
On the set of integers $Z$ define the binary operations:

Addition: a#b=a+b+1;

Multiplication :a*b=a+b+ab;

For a, b in $Z$.

$Z$ is a ring. With the zero element= $-1$ and a unity= $0$.

Find the set of all nilpotent elements.

I could found only 1 element which is $-1$. How to find the rest, considering that doing the multiplication operation n times has no clear form, so we can solve the equation : $x^n=-1$.

I tried (using induction) to find the pattern of doing the multiplication n times. No clear pattern emerged.

答:
Take $a\in\Bbb Z\setminus\{-1\}$. Can we have $a*a=-1$? No, because$$a*a=-1\iff a^2+2a=-1\iff(a+1)^2=0,$$which is impossible, since $a\ne-1$.

Can you have $a*a*a=-1$? No, because$$a*a*a=a=-1\iff a^3+3a^2+3a=-1\iff(a+1)^3=0,$$which, again, is impossible, since $a\ne-1$.

And so on; you can prove by induction that$$\overbrace{a*a*\cdots*a}^{n\text{ times}}+1=(a+1)^n.$$


问:
Proving that a 3 regular graph has edge connectivity equal to vertex connectivity.
Was one of my homework problems in Graph theory. I got marked wrong by our teaching assistant on the solution below that I provided:

Note that any 3 regular graph can be constructed by drawing 2 cycles of 1/2 |V(G)| vertices, and connecting inner vertices with the outer ones. Thus, it is obvious that edge connectivity=vertex connectivity =3.

What's wrong with this proof?

答:
It is not true that any $3$-regular graph can be constructed in this way, and it is not true that any $3$-regular graph has vertex or edge connectivity $3$.

First of all, you can take two $3$-regular components, and get a $3$-regular graph that's not connected at all.

Other examples are also possible. Here's an example with connectivity $1$, and here's one with connectivity $2$. They are also shown below:

As a hint to get started, since you should already know that vertex connectivity is at most the edge connectivity, which is at most the minimum degree, you have only a few things to check:

If a $3$-regular graph has a vertex cut of order $1$, it also has an edge cut of size $1$.

If a $3$-regular graph has a vertex cut of order $2$, it also has an edge cut of size $2$.

Draw a picture of each of these, and see if you can spot the edge cut. (You'll have two cases in the second bullet point, since the two vertices in the vertex cut may or may not be adjacent.)


问:
Why is the interpretation of infinite union of sets as a limit so incorrect (or even dangerous)?
In some reputable notes on a Probability course at the IIT I read:

When the index set $\mathcal I$ is a finite set, say $\mathcal I = {1, 2, 3}$ the definition of union given above coincides with the “middle-school” understanding of unions, i.e., taking the union of sets one-at-a-time. For example, $\bigcup_{i=1}^3 A_i= A_1 \bigcup A_2 \bigcup A_3.$ However, this “one-by-one” interpretation completely breaks down when the index set $\mathcal I$ is infinite. For example when $\mathcal I = \mathbb N,$ the union $\bigcup_{i=1}^\infty A_i$ does not have any interpretation in terms of taking unions one by one, till infinity. After all, there is no $A_∞$ in the family $\{A_i, i ∈ \mathbb N\},$ and there is no notion of “limiting unions”. Thus, $\bigcup_{i=1}^\infty A_i$ should be interpreted just as Definition 1.3 says: it is the set of all elements contained in at least one of the $A_i, i ∈ \mathbb N.$ In order to avoid the (dangerous) temptation to interpret $\bigcup_{i=1}^\infty A_i$ as some sort of a limit of finite, “one-by-one” unions, a better notation would be to use $\bigcup_{i\in \mathbb N} A_i,$ instead of the potentially misleading but more commonly used notation $\bigcup_{i=1}^\infty A_i.$

What sort of misunderstanding would the approach of this union as a limit result in?

答:
The reason viewing infinite unions as limits is technically incorrect in this context is for the exact reason your quoted passage says: there is no definition of a "limit" when it comes to taking the union of sets in this context. When you think of, for example
$$
\lim_{x\to a}f(x)
$$
this has a definition. ${\forall\ \epsilon>0\ \exists\ \delta>0}$... yadda yadda yadda.

But when it comes to sets, we don't have any such notion. Nor do we need it. You can simply say that ${x \in \bigcup_{i \in \mathbb{N}}A_i}$ if and only if ${x \in A_i}$ for some ${i \in \mathbb{N}}$. That's it. That's all we need to define the infinite union.

EDIT: @Peter Morfe has corrected me in the comments of this post. There is such a notion defined, and it is useful in a lot of contexts. The main point still stands, though, that we do not need a notion of set limits to define at least a basic notion of the union and intersection of sets. In fact, for the union/intersection of uncountably infinitely many sets, the standard definition I referenced is the one you need to use.


问:
Prove inequality consisting of sum using mathematical induction
I wish to seek help for a question from the Finnish matriculation exam in mathematics. There are multiple methods to solve the exercise, but I am ONLY interested in solving it using mathematical induction. The question is as follows:

$$\text{let}\ n\ \text{be a positive integer}, n\ge1$$
$$\text{Show that:}$$
$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$

I have tried hard to solve this using induction, but I cannot get very far at all. Any help would be appreciated.

EDIT: What I have done so far.

Test with n = 1, which gives:

$$0\le1\le\frac{127}{7}, \text{which is true}$$

Assumption:

$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$

Induction step:

$$\frac{127}{7}\cdot\ n^7\le\sum_{k=n+1}^{2n+1}k^6\le\frac{127}{7}\cdot\ (n+1)^7$$

But I don't know how to prove this in terms of the assumption. I would assume that there is some way to remove the sum from this inequality. It seems to be possible to expand it using my CAS-calculator. But I just can't wrap my head around it.

I hope this enough explanation of what I have done so far.

(I might have used the wrong mathematical terms here, since English is not my first language)

答:
Thanks to the help of @maxmilgram, I've managed to figure out the solution.

SOLUTION:

Test with n = 1, which gives:

$$0\le1\le\frac{127}{7}, \text{which is true}$$

Assumption:

$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$

Induction step:

$$\frac{127}{7}\cdot\ n^7\le\sum_{k=n+1}^{2n+1}k^6\le\frac{127}{7}\cdot\ (n+1)^7$$

And since:

$$\sum_{k=n+1}^{2n+1}k^6=\sum_{k=n}^{2n-1}k^6-n^6+(2n)^6+(2n+1)^6$$

we can write the induction step in terms of the assumption:

$$\frac{127}{7}n^7\le\sum_{k=n}^{2n-1}k^6-n^6+\left(2n\right)^6+\left(2n+1\right)^6\le\frac{127}{7}\left(n+7\right)^7$$

$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \sum _{k=n}^{2n-1}k^6\le \frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6$$

In other words, if we could prove that:

$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \frac{127}{7}\left(n-1\right)^7$$ and

$$\frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\ge \frac{127}{7}n^7$$

we would know that the induction step is indeed true. After some simplification we end up with:

$$-573n^5+395n^4-795n^3+321n^2-139n+\frac{120}{7}\le 0, \text{which is true since } n\ge 1$$

$$189n^5+395n^4+475n^3+321n^2+115n+\frac{120}{7}\ge 0, \text{which is true since } n \ge 1$$

In other words, we have shown that the induction step is true if the induction assumption also is true. Therefore, the statement is true for all positive integers n.


问:
Formal logic question - why is this invalid?
I typed this into a logic calculator, and it comes out as invalid:

\begin{align*}
\forall x, & E(x) \implies I(x) \vee A(x) \\
\forall x, & A(x) \implies \neg E(x) \\
\forall x, & E(x) \implies \neg A(x)\\
\exists x, & E(x) \implies I(x)\\
\exists x, & E(x)\\
E(x)⟹I(x)
\end{align*}

Can someone help me explain why it's invalid?

https://www.umsu.de/trees/#((((((%E2%88%80y((Ex%E2%86%92(Ix%E2%88%A8Ax))))%E2%88%A7(%E2%88%80y((Ax%E2%86%92(%C2%ACEx)))))%E2%88%A7(%E2%88%80y((Ex%E2%86%92(%C2%ACAx)))))%E2%88%A7(%C2%AC(%E2%88%80y((%C2%AC(Ex%E2%86%92Ix))))))%E2%88%A7(%C2%AC(%E2%88%80y((%C2%ACEx)))))%E2%88%A7(Ex%E2%86%92Ix))

答:
The third line is the contrapositive of the second line.  The first and third lines together imply $$\forall x,E(x)\implies (I(x)\lor A(x))\land \lnot A(x)\implies I(x)$$  which is the conclusion.  The fourth and fifth lines are unnecessary, but they don't contradict anything in the first three lines, so the argument seems perfectly valid to me.

I don't know anything about the calculator.


问:
Find a regular homotopy
Firstly, we define a regular homotopy between regular closed curves as a continuous map $F: I \times I \to \mathbb{R}^n$ satisfying the following conditions:

(i) for each fixed $u \in I$, the map $\gamma_{u}(t)=F(u,t)$ is a regular closed curve.

(ii) All derivatives $\gamma_{u}^{(k)}(t)$ (with respect to $t$) depend continuously on $u$.

So here's the question:

Let $\gamma : [0, 2\pi] \to \mathbb{R}^2$ be given by $\gamma(t)=((1+2 \cos(t) ) \cos(t) , (1+2 \cos(t) ) \sin(t))$. By finding
  a suitable regular homotopy compute the rotation index of the curve. You also need to justify the regularity of your homotopy.

Now, I know that if 2 curves are regularly homotopic, then they have the same rotation index, and to check regularity I must check that the derivative never vanishes.  But I am a little unsure of how to find a regular homotopy.

My first thoughts involved splitting up the function into $$\gamma(t)=(\cos(t) , \sin(t) ) +(2\cos^{2}(t), 2\cos(t) \sin(t) ).$$
And then perhaps trying $$F(u,t) =(\cos(t) , \sin(t) )+(1-u)(2\cos^{2}(t), 2\cos(t) \sin(t) ) \ ?$$
Or maybe $$F(u,t) =(1-u)(\cos(t) , \sin(t) )+(2\cos^{2}(t), 2\cos(t) \sin(t) ) \ ?$$

All help is greatly appreciated.

答:
OK, so you've settled on $F(u,t) = (\cos t,\sin t) + 2(1-u)(\cos^2t,\cos t\sin t)$. To me it is evident that this is a smooth (infinitely differentiable) function of $t$ and $u$. What you really need to check is that $\dfrac{\partial F}{\partial t} \ne 0$ for every value of $t$ and $u$. This is the regularity condition for the curve $\gamma_u$. It might help to write
$$\frac{\partial F}{\partial t} (u,t)= (-\sin t,\cos t) + 2(1-u)(-\sin 2t,\cos 2t)$$
and think a bit. Are these two vectors ever parallel?

Here's a picture of $\gamma$ for you, so you can verify that the rotation index is actually $2$.

For the homotopy you had, to turn a circle into this curve, one must introduce a cusp or kink when the second loop is born (thinking in reverse time here):

Title: Find a regular homotopy


问:
Covariance of two discrete variables
Cards with numbers $1, 2, 3, 4, 5, 6$ are put into a bag (6 cards). The monkey randomly pulls out a card. If the monkey gets an even number, she gets $0$ peanuts, if she gets an odd number, she gets $1$ peanut. Let $X$ be the number she picks (from $1$ to $6$), and $Y$ - the number of peanuts she gets ($0$ or $1$). I need to find the covariance of $X$ and $Y$.

My solution:

For $X$ and $Y$ their joint PMF looks like this:

Then I try to apply the definition of covariance: $$cov\left(X,Y\right)=\mathbf{E}\left(\left(X-\mathbf{E}X\right)\left(Y-\mathbf{E}Y\right)\right)=\mathbf{E}\left(XY\right)-\mathbf{E}X\mathbf{E}Y$$

I know how to find $\mathbf{E}X$ and $\mathbf{E}Y$, but I'm having trouble finding $\mathbf{E}(XY)$

The way I see it: $1*1*1/6 + 3*1*1/6+5*1*1/6=9/6$

But the answer $9/6-3.5*0.5$, which is $\mathbf{E}\left(XY\right)-\mathbf{E}X\mathbf{E}Y$ seems wrong.

Where is my mistake?

Title: Covariance of two discrete variables

Body:
$$\mathbb{E}(X)=3.5$$

$$\mathbb{E}(Y)=\frac{3}{6}$$

$$\mathbb{E}(XY)=\frac{1}{6}+3\cdot\frac{1}{6}+5\cdot\frac{1}{6}=\frac{9}{6}$$

Thus

$$\mathbb{Cov}(X,Y)=\frac{9}{6}-3.5\cdot\frac{3}{6}$$


问:
Harvard MIT math question
Let P be a point selected uniformly at random in the cube $[0, 1]^3$. The plane parallel to x+y+z = 0
passing through P intersects the cube in a two-dimensional region R. Let t be the expected value of
the perimeter of R. If $t^2$ can be written as $\frac{a}{b}$, where a and b are relatively prime positive integers,compute 100a + b

Let me know if the solution below is right?

Let $x_1,y_1,z_1$ be any point P in the cube.

The plane that passes through the point parallel to x+y+z 0

is ($x+y+z=x_1+y_1+z_1 = t$)

This plane cuts the x axis at say (x,0,0) then x = t and the point is (t,0,0)

This plane cuts the y axis at say (0,y,0) then y = t and the point is (0,t,0)

This plane cuts the z axis at say (0,0,z) then z = t and the point is (0,0,t)

The region of intersection is an equilateral triangle whose perimeter is given to be $P = 3\sqrt{2t^2} = 3\sqrt{2}t$

Was it a fluke that i hit the correct answer or is the rationale correct?

$E(P) = 3\sqrt{2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{\sqrt{3}-x_1-y_1} (x_1+y_1+z_1)dz_1dy_1dx_1 = \frac{11}{2\sqrt{2}}$.  After extraction the answer is $12108$

答:
If you take a unit cube and try to cut it along a plane $x + y + z = r$, then you will notice that the two-dimensional region can be two different shapes.

More precisely, if $r \leq 1$ or $r \geq 2$ (which are symmetric via $x, y, z \mapsto 1 - x, 1 - y, 1 - z$), then the region $R$ is a triangle with side length $\sqrt 2 r$ or $\sqrt 2 (3 - r)$, respectively. Thus $R$ has perimeter $3\sqrt 2 r$ and area $\frac {\sqrt 3} 2 r^2$ when $r \leq 1$ (and symmetrically when $r \geq 2$).

If $1 < r < 2$, then the region $R$ will be a hexagon, whose angles are all $120^\circ$ and whose side lengths are $\sqrt 2(r - 1)$ and $\sqrt 2(2 - r)$, alternatively. In this case, $R$ has perimeter $3\sqrt 2$ and area $\frac{\sqrt 3}2 (-2r^2 + 6r - 3)$.

Since the point $P$ is chosen uniformly in the cube, the probability density of the plane being $x + y + z = r$ is proportional to the area of $R$. Therefore the expected value of the perimeter is calculated as
$$t = \left(2\int_0^1 3\sqrt 2 r \cdot \frac {\sqrt 3} 2 r^2 dr + \int_1^2 3\sqrt 2 \cdot \frac{\sqrt 3}2 (-2r^2 + 6r - 3)dr\right)/\left(2\int_0^1 \frac {\sqrt 3} 2 r^2 dr + \int_1^2 \frac{\sqrt 3}2 (-2r^2 + 6r - 3)dr\right) = \frac {11} 4 \sqrt 2.$$ Therefore $t^2 = \frac{121}{8}$ and the final answer is $12108$.

Title: Harvard MIT math question


问:
Fractional exponentiation of functions
I run into a simple question about the fractional exponentiation of a function. Suppose that $f(x)$ is a real valued function which is always positive definite on its domain. Is it possible to obtain a negative value by a fractional exponentiation of the function $f(x)$? i.e., the fractional exponentiation ${\left( {f(x)} \right)^{\frac{m}{n}}}$ (where $m$ and $n$ are some integers) could lead to a negative result? if yes,  when $m$ be an even integer, can we still obtain a negative result?

Of course, I know that this question is reminiscent of the fractional exponentiation of real numbers in calculus. Can we simply apply our knowledge about fractional exponentiation of real numbers to functions?

Thank you very much in advance.

答: You are only exponentiating the function values when it comes down to it, and these are real numbers, so of course what you learned in calculus applies. No, you'll never get negative values.</p>

Title: Fractional exponentiation of functions


答:
Your question is not specific to functions, but essentially to exponentiation of positive reals. Which always yields a positive result.

Anyway, as pointed by @Riemann'sPointyNose, there can be a confusion with the  roots of a real, which can to some extent be negative (for a rational with even denominator).

Title: Fractional exponentiation of functions

Body:
I run into a simple question about the fractional exponentiation of a function. Suppose that $f(x)$ is a real valued function which is always positive definite on its domain. Is it possible to obtain a negative value by a fractional exponentiation of the function $f(x)$? i.e., the fractional exponentiation ${\left( {f(x)} \right)^{\frac{m}{n}}}$ (where $m$ and $n$ are some integers) could lead to a negative result? if yes,  when $m$ be an even integer, can we still obtain a negative result?

Of course, I know that this question is reminiscent of the fractional exponentiation of real numbers in calculus. Can we simply apply our knowledge about fractional exponentiation of real numbers to functions?

Thank you very much in advance.


答:
It depends how you define fractional exponentiation.

For example, we typically view ${x^{\frac{1}{2}}}$ with being synonymous with "${\sqrt{x}}$" (which is always positive); but perhaps a better, more complete view is to think of ${x^{\frac{1}{2}}}$ as being a set of values satisfying the fact that their square gives ${x}$:
$$
x^{\frac{1}{2}} = \{a\ |\ a^2 = a\times a = x\}
$$
in this way, ${\sqrt{x}}$ (which always gives a positive value) is just one of the solutions in ${x^{\frac{1}{2}}}$. In fact, we have that
$$
\sqrt{x},-\sqrt{x} \in x^{\frac{1}{2}}
$$
So just take ${\left(f(x)\right)^{\frac{1}{2}}}$ as the negative solution and you are set.

I WANT TO MAKE IT CLEAR THOUGH: if you are treating fractional exponentiation as being synonymous with only taking the positive roots, then no, you will obviously never get a negative result. Hopefully this helps.

Title: Fractional exponentiation of functions


问:
Resource Recommendations for Epistemic Logic
I wish to learn about Epistemic Modal Logic in a way that's rigorous yet easy$\color{red}{^1}$ to understand. Could you recommend books, courses, or any other relevant resources I could look into?

I have a background in Set Theory, Propositional Logic, First-order Logic, and Basic Modal Logic (I am familiar with the basic modal language, models, model constructions, frames, completeness, etc. My primary reference for this has been Blackburn's Modal Logic - the first four chapters). Happy to provide any other detail necessary.

$\color{red}{1.}$ By "easy", I mean something suited for undergraduate math majors. This isn't a hard requirement though - I am open to all kinds of recommendations!

P.S. Since this question is really specific to Epistemic Logic, in all probability it is not a duplicate.

答:
Since you mention that you are OK with the first four chapters of Blackburn's et al. Modal Logic, then, technically, you already know Epistemic Logic (EL), which is essentially S5. A classic textbook on EL, which includes various notions of knowledge, completeness and complexity results, etc. is Reasoning about knowledge by Fagin, Halpern, Moses, and Vardi.

For continuation, I would recommend Dynamic Epistemic Logic (DEL) by Hans van Ditmarsch, Wiebe van der Hoek, and Barteld Kooi. It fits the description of being 'easy yet rigorous' perfectly.

For an overview of topics in EL and relatively recent developments, Handbook of Epistemic Logic is a nice option.

But, tbh, if you tackle DEL, then going straight to papers is the best.

Edit: also, for gentle introduction, I would suggest the corresponding chapters of Modal logic for open minds by Johan van Benthem.


问:
Dividing a rectangle in 3 equal parts half at time?
Here's an interesting puzzle that has got me thinking:

You need to divide a rectangle into 3 equal parts along it's length. However you can only divide things into half. You don't have any other tools at your disposal - a saw/cutter and a scale that can accurately tell you the half of a full length. How can this be done?

Clarification:

Assume a piece of wood:

```<code>Length: (16x)      ----------------
Step 1: (2 halves) --------|--------
Step 2: (4 quarts) ----|----|----|---- (1/3rd is somewhere between the first and second cut)
</code>```

And so on till you are close to $1/3$ with some $\pm\epsilon$. How can I mathematically frame and solve this problem and say after how many divisions will I be close to the 1/3 mark?

答: Since $\frac 13=\sum_{i=1}^\infty \frac 1{4^i}$ the closest ones are  $\frac 14, \frac 5{16}, \frac {21}{64},\frac {85}{256},$ where each numerator is four times the previous plus one.</p>


问:
Function which is continuous exactly one point and differential exactly one point
I found a Function which is continuous exactly one point and differentiable exactly one point. And 2 point continuous and 2 point differentiable and more for infinite finding this type of function is possible?</p>

答:
To be really precise when you prove it for one particular case, that is all that there is to it.

Buy, wait, you say? We want to prove that the statement is true for $k+1$ if it is true for $k$. And you are right. However, what you are proving is that your single value proof, for example, for $1$ is valid for all other chained statements too.

You are technically saying that through the filter of the used statement there is no difference between $1$ and all other integers. Which is to say that you proof for $1$ bears sufficient strength to cover all other integers.

Although it is almost always much more difficult to justify $k$ to $k+1$ step, the proof for one single value is more essential for induction. Or to say if you prove $k$ to $k+1$ step you have nothing unless all $k$ are the same, in nature, as any one single value.


问:
Why do you use this form for solving linear ODE instead of the other?
My known Method of solving the homogenous part of a linear ODE is given by separating the variables.
E.g:

\begin{equation*}
\begin{split}
y'=4y &\implies \frac{dy}{dx}=4y\\
&\implies \int\frac{1}{4y}dy =\int1dx\\
&\implies \frac{1}{4}\ln|y| =x+c, ~~c\in\mathbb{R}\\
&\implies y =e^{4x+C}, ~~C\in\mathbb{R},~~C=4c
\end{split}
\end{equation*}

Why is the method for solving the homogenous part mostly given per definition by using \begin{equation*}y=Ce^{4x},~ C\in\mathbb{R}\end{equation*}

instead? And how could you use the first method for $y_h$ ?
Thanks in advance

答:
Note that
$$y=e^{4x+C}=e^Ce^{4x}=\tilde{C}e^{4x}$$

so both forms are equivalent.

Personally, I (and most people, I believe) will find that $Ce^{4x}$ is clearer and cleaner.


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

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

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

Let $μ=350$ and $σ^2 =12365$

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

答:
Because of independence:

$m_Y=E(Y_2)=\alpha +\beta \mu$
$\sigma^2_Y=var(Y_2)=var(\alpha)+var(\beta Y_1)+var (U)$

$=\beta \sigma^2+ v^2$

$Y_2$ ~ $N(m_Y,\sigma^2_Y)$.

You need values for $\alpha$, $\beta$ and $v^2$.


问:
If $f \leq g \leq h$, $\lim_{x \to a} f = L$ and $\lim_{x \to a} h = L$, then $\lim_{x \to a} g = L$
I don't know if this property is right:

If $f \leq g \leq h$, $\forall x\in\mathbb{R}$ and $\lim_{x \to a} f = L$, $\lim_{x \to a} h = L$, then $\lim_{x \to a} g = L$.

My intention on these question is a generalization, I mean, let $$f, g, h: A\subseteq\mathbb{R^n} \longrightarrow \mathbb{R}$$ such that $$f\leq g\leq h, \forall \bar{x}\in A$$ and $$\lim_{\bar{x} \to \bar{x_0}} f = \lim_{\bar{x} \to \bar{x_0}} h = L$$ then $$\lim_{\bar{x} \to \bar{x_0}} g = L$$

答:
This is known as the squeeze theorem. It follows by
$$
L = \lim_{x\to x_0} f(x) \leq \liminf_{x\to x_0} g(x)\leq \limsup_ {x\to x_0} g(x) \leq \lim_ {x\to x_0} h(x)=L
$$.

Edit: add missing $\leq$


答:
Let $(X,d_{X})$ be a metric space and $f:X\to\mathbb{R}$, $g:X\to\mathbb{R}$ and $h:X\to\mathbb{R}$ be functions.

Moreover, let $E\subseteq X$ such that $a\in X$ is an adherent point of $E$ and $L\in\mathbb{R}$.

Also, let us assume that $f$ and $h$ converges to $L$ as $x$ approaches $a$ along $E$ s.t. $f\leq g\leq h$ for every $x\in X$.

Then for every $\varepsilon > 0$, there corresponds $\delta^{1}_{\varepsilon} > 0$  s.t. for every $x\in E$ one has that
\begin{align*}
d_{X}(x,a) < \delta^{1}_{\varepsilon} \Rightarrow |f(x) - L| < \varepsilon \Rightarrow L - \varepsilon < f(x) < L + \varepsilon
\end{align*}

Similarly, for the same $\varepsilon > 0$, there corresponds a $\delta^{2}_{\varepsilon} > 0$ s.t. for every $x\in E$ one has that
\begin{align*}
d_{X}(x,a) < \delta^{2}_{\varepsilon} \Rightarrow |h(x) - L| < \varepsilon \Rightarrow L - \varepsilon < h(x) < L + \varepsilon
\end{align*}

Hence for every $\varepsilon > 0$, there corresponds a $\delta = \min\{\delta^{1}_{\varepsilon},\delta^{2}_{\varepsilon}\}$ s.t. for every $x\in E$ we have that
\begin{align*}
d_{X}(x,a) < \delta \Rightarrow L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon \Rightarrow |g(x) - L| < \varepsilon
\end{align*}

and we are done.

Hopefully this helps!

Title: If $f \leq g \leq h$, $\lim_{x \to a} f = L$ and $\lim_{x \to a} h = L$, then $\lim_{x \to a} g = L$

Body:
I don't know if this property is right:

If $f \leq g \leq h$, $\forall x\in\mathbb{R}$ and $\lim_{x \to a} f = L$, $\lim_{x \to a} h = L$, then $\lim_{x \to a} g = L$.

My intention on these question is a generalization, I mean, let $$f, g, h: A\subseteq\mathbb{R^n} \longrightarrow \mathbb{R}$$ such that $$f\leq g\leq h, \forall \bar{x}\in A$$ and $$\lim_{\bar{x} \to \bar{x_0}} f = \lim_{\bar{x} \to \bar{x_0}} h = L$$ then $$\lim_{\bar{x} \to \bar{x_0}} g = L$$


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

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

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

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

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

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

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

Which would give me

8 * 9 * 8 = 576 = 1/576

Am I right in this one?

答:
The sample space in this case will be all possible draws.
$\\ n(S) = ^{10}P_3$
Let, the event of getting three consecutive increasing numbers be $E$.
Than, we can find $n(E) $ by selecting 3 numbers from 0-9. As there is only one way to arrange them, we don't need to permute them.
Thus,
$n(E) =^{10}C_3$

Now, as we have to select three cards in all the cases in the sample space, we can directly say,

$$P(E) = \frac{n(E) }{n(S) }
\\ P(E) = \frac{1}{3! }
\\ P(E) = \frac{1}{6}$$

A more intuitive way to think about it is that, suppose you draw any three cards, there will be 3! ways in which you could have taken them out but only one of them would have satisfied the condition provided.

Hope this is helpful!!!


问:
Prove that $\mathbf{R} = \{ (x,y) \in \mathbb{N^2} : x \ge y \}$ is an equivalence relation

For the binary relation R over $\mathbb{N}$ described below,

establish whether R is a reflexive, symmetric or transitive relation.

$\mathbf{R} = \{ (x,y) \in \mathbb{N^2} : x \ge y \}$

How can I prove that?

my try:

$$
\text{reflexive}: x \ge x \ \text{(true)}\\
\text{symmetric}: x \ge y \iff y\ge x \ \text{(??That's not true)} \\
\text{transitive}: x \ge y, \ y \ge z \implies x \ge z \ \text{(true)}
$$

symmetric propriety doesn't seem true if $x \neq y$.

答: It is not an equivalence relation because, as you see in the comments, symmetry fails (just pick the point $(2,1)$). The $\ge$ relation on, say, the natural numbers is, instead, satisfying the antisymmetry property.


问:
Diagonalizable binary matrices over $\mathbb{Z}_4$
I'm trying to figure out the following question:

Are symmetric, binary $n\times n$ matrices with zeros on the diagonal, are diagonalizable over $\mathbb{Z}_4$?

I know that it isn't true that $\textit{any}$ binary matrix is invertible over $\mathbb{Z}_2$, but I'm not sure about what happens in $\mathbb{Z}_4$, in particular in the case where the diagonal is all-zero.

Thanks!

Body:
<a href="https://pdf.sciencedirectassets.com/272332/1-s2.0-S0021869312X00255/1-s2.0-S0021869312005601/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEEIaCXVzLWVhc3QtMSJGMEQCIHKB5y6GrgEVVNxYQiE9W5zog7%2FjNUGkBQ8DTHaaQAp5AiBQp5sMwkooOhnTWdH1LdZWyWhFrTqwj9RJSz%2B78CZ1Oyq0Awh7EAMaDDA1OTAwMzU0Njg2NSIM5YShAf%2Fp0i1AkBYWKpEDtRgmhb482bAqItxUu8hOgTduW%2Fn6ig7RF5CM2qeSZ3FvJYRyv3GcyZJLEtCr%2BXrhrewtZAAErGEGpQ38zLaAEkqDbhyKgljELWIIEGepTOGguBq6rQiqHcBDZfXTCsp3oFau2v8hBndUpp%2BrLiiJYdfvRgX%2FB3glLw383mG7kOeuet2BgZIcKAE5v9DSoDjAMzpX%2B%2F5SsVG2CRDMALTFbltSPG6p29z1G9r2uEGyWA%2Buy1MIAWuLis5a9aeTltmi4rCxfckZbJ9PnG22t9KvteFABPza2m8QC8rsaXf0wmqw3dz0RtSoUzZ%2FGcwXo68ESsECYMUQewaxV6XtboyOtObYkKvi5ucnDwfLNbp7K1hpUb%2FvH5HBNt04fnbF4hzElMtIFr8rwaoIAgLubCAr8GakvC1N4sL2hUMCoA%2BifYbCuVX5aeBtjuLCa2Uew6kLfBykkUBd9TCqf0YvMGFqCI%2BnBeqIkrrUD2sUQLZt%2B0TVtQLHv8Ee%2FGG4fLR9mvemFHuILGhqgkwvd2MQ%2BYpBqQcw6eHYggY67AE79rblcK6msy46gOMnhWfHt34eQ%2Fa0hlAAt2e0Yo2pnFDW3WOROvQqyEDldNBUZE1JaEgorvsAFCS4KOaeQwaAom0iH%2FKa8JXdrGisQ8abpew7%2BGnLtkD3WWfWOOK3CJHrYsP%2B4q41P%2BIntvXaMmWT9Lk4nQIg4G8JqTLFcwxnhKk%2FSqvIx30ZBm%2FVMuY17d1c4xPljw9qL0iIoVe6ZXNxw9ACyJdvrEOUFb2MmUBQQerFiXdYUTnxr8c3AXuUMnpBNFKoZr4eRFHspzQ7XflTCkIZVHtXlsEHbz7j%2BsaO3SIiCCxwNcfCpeBe5Q%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20210320T181337Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTY4ZHJBYH3%2F20210320%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=6f11872c8b5d12591c67b208c95246ab705a0215febf8dea143889a2fc45e00e&hash=a81bcc269e1ce9da78f5e16d78d0624b46b2b9926bb4825b915f49ab85d8d7c6&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=S0021869312005601&tid=spdf-34d222f9-3571-48f1-8971-4efb43917f75&sid=883da4c35a07c54cd7290065b0826d4ff0f3gxrqa&type=client" rel="nofollow noreferrer">This paper sounds relevant, but I don't know anything about matrices over commutative rings.  I don't understand the second sentence of the abstract!

I tried to post this has a comment, but the link to the paper wouldn't fit.


问:
Curvature for a curve with constant angle between position vector and tangent vector
Let $\alpha(s)$ be a plane Frenet curve such that the angle $\theta$ between $\alpha(s)$ and its tangent vector $e_1(s)$ is constant. Show that either $\kappa = 0$ or

$$\kappa(s) = \frac{1}{\cot \theta + c}$$

where $c$ is a constant.

Hint: It holds that

$$\big< \alpha, e_1 \big> = |\alpha| \cdot \cos \theta$$

as well as

$$\big< \alpha, e_2 \big> = \pm |\alpha| \cdot \sin \theta$$

where $\dot{e}_1(s) = \kappa(s) e_2(s)$. Differentiate the first equation ...

--- ${}$ --- ${}$ --- ${}$ --- ${}$ --- ${}$ --- ${}$ ---

I tried solving this by using the hint given but I always ended up with a term containing $| \alpha |$, which I cannot get rid of. I already considered that maybe $| \alpha|$ is the constant $c$ and I therefore do not even have to get rid of it. But then again I cannot assume $|\alpha|$ to be constant a-priori although It may be (as for e.g. $\theta = \pi/2$).

I humbly ask for your help.

答:
This problem is just wrong. The only planar curves with constant curvature are circles, and only when the circle is centered at the origin will the angle between $\alpha$ and $e_1$ be constant (as you surmised, $\pi/2$). So $\cot\theta$ must be $0$ and $\kappa=1/c$.

I tried solving this the way I usually approach all such problems. I wrote the arclength parametrized curve $\alpha = \lambda  e_1 + \mu e_2$ for some functions $\lambda$ and $\mu$, differentiated and applied the Frenet equations. The hypothesis in the problem tells us that $\lambda = \pm\cot\theta \mu$ (or $\mu=0$, which leads to $\kappa=0$). Assuming $\kappa\ne 0$, we get the equations
\begin{alignat*}{3}
\kappa\cot\theta&\mu + &\mu' &= 0 \\
-\kappa&\mu + \cot\theta&\mu' &= 1,
\end{alignat*}
and so  $\mu = \dfrac{-1}{\kappa\csc^2\theta}=\dfrac{-\sin^2\theta}{\kappa}$ and $\mu' = \dfrac{\kappa\cot\theta}{\kappa\csc^2\theta} = \cos\theta\sin\theta$. This tells us that $\mu(s)=(\cos\theta\sin\theta)s + c$, and so
$$\kappa(s) = \frac{-\sin^2\theta}{\mu(s)} = \frac{-\sin^2\theta}{(\cos\theta\sin\theta) s + c},$$
very different from the statement of the problem. By the way, since $\kappa>0$, we should probably have opted for the negative sign choice.


问:
A loss and gain problem
This is a very simple but confusing puzzle.

A customer buys goods worth $200$ rupees from a shop. The shopkeeper selling these goods makes zero profit from this purchase.

The lady gives him a $1000$ rupee note.

The shopkeeper has no change, so he goes next door to another shopkeeper to get change for the $1000$ rupee note. He keeps $200$ for himself and returns $800$ to the customer.

Later, the second shopkeeper from next door comes back with the $1000$ rupee note with a stamp on it saying "counterfeit" and takes his money back.

How much loss does the first shopkeeper face?

答: The shopkeeper lost 200 worth of goods, plus 800 worth of money, which totals to 1000. </p>

Title: A loss and gain problem


答:
You don't have to take $n=1$ unless $n$ is the first case it is true.  You could take $n = 3562$ if we want to prove it is true for all numbers equal and beyond $3562$ but it might not be true for numbers below $3562$.

Induction has two key components.

The induction step. You show that if the statement is ever true, then it will be true for the next one as well.... the implication is that then it will be true for the one after that and by bootstrapping the one after that and after than and so on forever.

....but.....

You have to also show that the statement is true somewhere.  If the statement is never true it doesn't matter if the induction step holds.  If you don't have a first case, you don't have to have a second case or any cases after that.

So your base case is simply stating: "We show the statement is true for some first case." And the induction step is "Every time it is true it is true for the next case so once we have a first case, all the cases after that must be true".

....

Here is an induction proof that doesn't start with $n =1$.  It starts with $n = 5$.

Statement: For all integers $n \ge 5; n^2 \ge 5n$.

Base case:  $n= 5$ then $5^2 = 25 = 5\cdot 5$.  Done.

Induction step:  $n=k$ assume $k^2 \ge 5k$ then $(k+1)^2 = k^2 + 2k + 1\ge 5k + 7k + 1 = 12k + 1 = 5k + 7k +1$.  As $k\ge 1$ then $7k + 1 > 5$ so $5k + 7k + 1 > 5k + k = 5(k+1)$ so $(k+1)^2 \ge 5(k+1)$. So the induction step is done.

So $n^2 \ge 5n$ for all integer $n \ge 5$.

Why did we choose base case of $n =5$ and not $n=1$.  Because it's not true for $n =1,2,3,4$.  $n =5$ is the first case it is true. But once it is true then it is true for all later cases.

.....

And here is a case where if we don't have a base case the whole thing fails.

Statement:  Every natural number is irrational.  (Obviously false)

Pf:

Base case: .... well, we'll just ignore that........

Induction step: Assume: $n = k$ is irrational.  So there are no integers $a,b$ so that $a-b$ is an integer, and $\frac {a-b}b =k$. That is impossible if $k$ is irrational.

And if there are no integers $a,b$ so that $\frac {a-b}b = k$ then there are no integers $a,b$ so that $k+ 1 = \frac {a-b}b + 1 = \frac {a-b}b+\frac bb = \frac {a-b+b}{b} =\frac ab$.  That is impossible.  SO $k +1$ is irrational.

So that is our induction step. If $k$ is irrational then $k + 1$ is irrational.

There is NOTHING wrong with that and it is absolutely true.  It will work for any irrational.  If $k$ is irrational then $k + 1$ is irrational.

So why doesn't this prove that all natural numbers are irrational? We have the induction step?

Well, now all we need to complete the step is a base case.

Now to prove our (obviously false) result we just need a base case where we have an natural number that is irrational to start us off.  Once we have a first case then all others fall into place.

So $n=1$ then ... well $1 =\frac 11$ is rational.  Um.... $n=2$????? Then $2 =\frac 21$. ... rational.  Um.... $n=3562$???? then $3562 =\frac {3562}1$  .... rational.

If we can't find a first case our proof doesn't work.

Title: Function which is continuous exactly one point and differential exactly one point

Body: I found a Function which is continuous exactly one point and differentiable exactly one point. And 2 point continuous and 2 point differentiable and more for infinite finding this type of function is possible?</p>


问:
Why is the interpretation of infinite union of sets as a limit so incorrect (or even dangerous)?
In some reputable notes on a Probability course at the IIT I read:

When the index set $\mathcal I$ is a finite set, say $\mathcal I = {1, 2, 3}$ the definition of union given above coincides with the “middle-school” understanding of unions, i.e., taking the union of sets one-at-a-time. For example, $\bigcup_{i=1}^3 A_i= A_1 \bigcup A_2 \bigcup A_3.$ However, this “one-by-one” interpretation completely breaks down when the index set $\mathcal I$ is infinite. For example when $\mathcal I = \mathbb N,$ the union $\bigcup_{i=1}^\infty A_i$ does not have any interpretation in terms of taking unions one by one, till infinity. After all, there is no $A_∞$ in the family $\{A_i, i ∈ \mathbb N\},$ and there is no notion of “limiting unions”. Thus, $\bigcup_{i=1}^\infty A_i$ should be interpreted just as Definition 1.3 says: it is the set of all elements contained in at least one of the $A_i, i ∈ \mathbb N.$ In order to avoid the (dangerous) temptation to interpret $\bigcup_{i=1}^\infty A_i$ as some sort of a limit of finite, “one-by-one” unions, a better notation would be to use $\bigcup_{i\in \mathbb N} A_i,$ instead of the potentially misleading but more commonly used notation $\bigcup_{i=1}^\infty A_i.$

What sort of misunderstanding would the approach of this union as a limit result in?

答:
You're really looking for $\limsup$ and $\liminf$ of sets. Let $(A_n)_{n\geq 0}$ be a countable family of sets. Then $$\limsup_{n\to \infty}A_n = \bigcap_{n\geq 0} \bigcup_{k \geq n} A_k \qquad \mbox{and}\qquad \liminf_{n\to \infty} A_n = \bigcup_{n\geq 0} \bigcap_{k \geq n}A_k.$$If these two sets coincide, then we can write $\lim_{n\to \infty} A_n$.

Example: if $A_n=[-n,n]$ for all $n$, then $\lim_{n\to \infty}[-n,n]=\Bbb R$, as you would expect.

In general, if $(A_n)_{n\geq 0}$ in non-increasing or non-decreasing, the limit exists and it equals their union or intersection.

The bottom line is that in the same way that the limit of a numerical real sequence exists if and only if both the limsup and liminf exist and are equal, this is what you have to be careful with when dealing with sets.


问:
Distribution of continous random variable
Given the following

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

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

Let $μ=350$ and $σ^2 =12365$

How do i find the distribution of $Y_2$?

答:
The CDF of random variable $Y_2$ is given by
$$
F_{Y_2}(y_2)\equiv P(Y_2\leq y_2)
$$
Therefore,
$$
P(Y_2\leq y_2)=Pr(\alpha+\beta Y_1+U\leq y_2)=Pr(\beta Y_1+U\leq y_2-\alpha)=F_{\beta Y_1+U}(y_2-\alpha).
$$
The sum of normal distributions is normal. Therefore, $\beta Y_1+U\sim N(\beta \mu,\beta^2\sigma^2+v^2)$.


问:
Evaluating certain limits of integrals
I'm dealing with the following limits:

$$ 1.) \lim_{n \rightarrow \infty} \int_{-1}^{1} e^{\frac{x^2}{n}} dx  $$

$$ 2.) \lim_{n \rightarrow \infty} \int_{1}^{3} \frac{nx^2 +3}{x^3+nx} dx  $$

$$ 3.) \lim_{n \rightarrow \infty} \int_{0}^{\frac{\pi}{2}} (\sin{\frac{x}{n}}+\cos{\frac{x}{n}})^{\frac{1}{2}} dx  $$

I assumed that all the sequence of functions all converge uniformly so that the limit and the integral can be interchanged. So, the first integral would be $2$ as $e^{\frac{x^2}{n}} \rightarrow 0 $ as $n \rightarrow \infty $. Similarly, the other two integrals would be $4$ and $\frac{\pi}{2}$ respectively. Am I correct in making these assumptions? Are my results correct?

答:
You can interchange limit and integral with the help of the DCT. Choose for

$g(x)=e^{x^2}$

$g(x)=(nx^2+3)/n$

$g(x)=\sqrt{2}$

These functions are all integrable in their given domain. So yes it is possible to interchange limit and integral in this three cases.

Title: Evaluating certain limits of integrals


问:
Tree graph intuition + mathematical notation

Let $T$ be a tree with at least two vertices. Consider a path of maximal length in $T$. Prove that the start and end point of this path have degree 1

by contradiction : If we let the start and end point of this path NOT have degree 1, there would be another vertex after the end/beginning making the path longer, so it can only be maximal if it is an end point.

Is this a good intuition for this? I wouldnt even know where to start with the mathematical notation..

答:
Your idea is good, but it is incomplete. What if the other vertex is actually on the path?
This is how I would write it (without using almost any notation):

Let $w$ be any vertex of $T$ and $P$ be a path of maximal length in T going through $w$. [Note that $P$ exists as T is connected and has at least two vertices, so the set of all paths going though $w$ is not empty and finite (I am assuming $T$ itself is finite)].

Now let $a$ be the starting point of $P$. I show that the only vertex adjacent to $a$ in $T$ is the vertex adjacent to $a$ on the path $P$ (then this would mean that $a$ has degree 1). Suppose for a contradiction there is a different vertex, say $b$, adjacent to $a$. If $b$ is not on $P$, then we can extend $P$ to include $b$, contradicting the maximality of $P$. If $b$ is on $P$, then the path from $a$ to $b$ in $P$ and the edge from $a$ to $b$ would form a cycle, contradicting that $T$ is acyclic. So $a$ has degree 1. Similarly for the endpoint.


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

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

答:
A probabilistic answer: $g_{\sigma}$ is the probability density of a two-dimensional Gaussian random vector $X_{\sigma}$ with mean zero and covariance matrix $\Sigma_{\sigma} = \sigma^{2} \text{Id}$.  In general, if $f_{1}$ and $f_{2}$ are two probability densities, then $f_{1} * f_{2}$ is the probability density of the sum of two independent random variables, one with density $f_{1}$ and the other, density $f_{2}$.  Hence $g_{\sigma} * g_{\tau}$ is the probability density of $X_{\sigma} + X_{\tau}$ if $X_{\sigma}$ and $X_{\tau}$ are taken to be independent.

Now the sum of two independent Gaussian vectors is itself a Gaussian vector and hence is characterized by its mean and covariance.  The mean of $X_{\sigma} + X_{\tau}$ is zero and the covariance is $\sigma^{2} + \tau^{2}$ by independence.  Thus,
\begin{equation*}
g_{\sigma} * g_{\tau} = g_{\sqrt{\sigma^{2} + \tau^{2}}}.
\end{equation*}


问:
GCD and LCM number theory problem
I am having some trouble finishing a solution to this problem, so I thought I might ask you all for help.

The problem goes as follows:
Find all natural numbers $a,b>0$ for which holds:

$$lcm(x,y)-gcd(x,y)=\frac{xy}{2021}$$

The problem is annoted with the hint, that 2021=43*47.

I have tried using the fact, that the $gcd=d$ from which follows $x=a*d$ and $y=b*d$. The $lcm(x,y)$ therefore must be $lcm(x,y)=a*b*d$.
Given this and the fact that the product of two numbers $x,y$ and their $gcd$ and $lcm$ is the same, meaning
$$xy=lcm(x,y)*gcd(x,y)=abd*d$$ I arrive at the equation:

$$(abd)-d=(abd^2)/2021$$
which equals
$$2021(ab-1)=abd$$

Where do I go from here? Any help or correction of mistakes that I may have made is greatly appreciated!

答:
Hint

Let $(x,y)=d$ and WLOG $x/a=y/b=d\implies(a,b)=1$

$$abd-d=\dfrac{abd^2}{2021}$$

$$\iff\dfrac{2021(ab-1)}{ab}=d$$

But $2021=43\cdot47$

So, $a,b\in[1,43,47,2021]$

Can you take it home from here?

Title: GCD and LCM number theory problem


问:
Is there an irreducible Markov chain admitting a non-constant bounded harmonic function?
I am asked the following question: Does there exist an irreducible Markov chain which admits a non-constant bounded harmonic function? Either give an example or prove that such examples do not exist.

So far I showed that an irreducible recurrent Markov $(X_n)_{n \in \mathbb{N}}$ does not admit such a function. Namely, suppose $\varphi$ is such a function. Define for $y$ in the state space $\tau_y := inf\{n \geq 1 \mid X_n = y\}$. Then by recurrence $\tau_y$ is finite a.s. and by optional stopping time theorem it follows that $(\varphi(X_{n \wedge \tau_y}))_{n \in \mathbb{N}}$ is a martingale. Therefore for $X_0 = x$ we have
\begin{align*}
\varphi(x) = \varphi(X_0) = \mathbb{E}_x[\varphi(X_{n \wedge \tau_y})] \rightarrow \mathbb{E}_x[\varphi(X_{\tau_y})] = \varphi(y)
\end{align*}
which shows that $\varphi$ must be constant.

How about the transient case?

答:
A typical argument for proving that harmonic functions are constant is to consider the state with the maximum value of the function, and argue from there that adjacent states must also have same value, and so forth.

This fails for infinite Markov chains, because then the maximum value may never be achieved. So that gives us a place to go with finding an example: we want an infinite state space and a bounded harmonic function on that space that approaches but does not reach its minimum or maximum.

So let's suppose that the state space of our Markov chain is $\mathbb Z$, and the function $\varphi$ is something without a minimum or maximum. (Let's assume it's strictly increasing; for the natural "shape" of a random walk on $\mathbb Z$, local minima or maxima would be just as tricky as global ones.) $\varphi(n) = \frac{2^n}{2^n+1}$ and $\varphi(n) = \arctan(n)$ are two easy-to-define examples of this form.

We can make this harmonic by defining our random walk appropriately. For each $n$, define $p_{n,n+1} = \frac{\varphi(n) - \varphi(n-1)}{\varphi(n+1) - \varphi(n-1)}$ and $p_{n,n-1} = \frac{\varphi(n+1) - \varphi(n)}{\varphi(n+1) - \varphi(n-1)}$. This will produce valid transition probabilities whenever $\varphi$ is strictly increasing. For $\varphi(n) = \frac{2^n}{2^n+1}$, it actually gives relatively nice formulas: $p_{n,n+1} = \frac23 (\frac{2^{n+1}+1}{2^{n+1}+2})$ and $p_{n,n-1} = \frac13(\frac{2^n+2}{2^n+1})$.

As a result:

$\varphi$ is harmonic because $p_{n,n+1} \varphi(n+1) + p_{n,n-1} \varphi(n-1)$ simplifies to $\varphi(n)$.

The random walk is irreducible because $p_{n,n+1}$ and $p_{n,n-1}$ are strictly positive for all $n$.


问:
The Lie algebra of the generators of the projective transformation is isomorphic to the Lie algebra of traceless matrices.
The general projective transformation of the $x$-$y$ plane is given by
$$\tilde{x}=\frac{a_1x+a_2 y+a_3}{a_7x+a_8y+a_9},\quad\tilde{y}=\frac{a_4x+a_5y+a_6}{a_7x+a_8y+a_9}$$
for some constants $a_i\in\mathbb{R}$. From this, we find the infinitesimal generators
$$\begin{array}{lll}
X_1=x\partial_x,& X_2=y\partial_x,& X_3=\partial_x, \\
X_4=x\partial_y,& X_5=y\partial_y,& X_6=\partial_y, \\
X_7=-x^2\partial_x-xy\partial_y,& X_8=-xy\partial_x-y^2\partial_y,& X_9=-x\partial_x-y\partial_y
\end{array}
$$
corresponding respectively to each parameter $a_i$. With the commutator
$$[X_i,X_j]=X_iX_j-X_jX_i,$$
the vector space generated by $\{X_1,\ldots,X_9\}$ forms a Lie algebra. Let us denote this vector space by $\mathfrak{g}$. A basis for that space could be $\{X_1,\ldots,X_8\}$ since $X_9=-X_1-X_5$.

I want to show that it is isomorphic to the Lie algebra
$$\mathfrak{sl}(3,\mathbb{R}):=\{A\in\mathbf{M}_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$$
with commutator $[A,B]=AB-BA$, by explicitly giving an isomorphism
$$f:(\mathfrak{sl}(3,\mathbb{R}),[\cdot,\cdot])\longrightarrow(\mathfrak{g},[\cdot,\cdot])$$

I made a few attempts that all failed. Since $X_1+X_5+X_9=0$, we can see a sort of correspondence with traceless matrices, but I don't see how to use that.

EDIT: To get the generators $X_i$, we differentiate with respect to each $a_i$ both $\tilde{x}$ and $\tilde{y}$ and evaluate at the identity transformation, that is at $a_1=1,a_5=1,a_9=1$ and $a_2=a_3=a_4=a_6=a_7=a_8=0$. For example, to get $X_2$, we have
$$\frac{d\tilde{x}}{da_2}=\frac{y}{a_7x+a_8y+a_9},\quad\frac{d\tilde{y}}{da_2}=0$$
which evaluates to $y$ and $0$ respectively so that $X_2=y\partial_x+0\partial_y=y\partial_x$.

答: Rough idea: Consider the projective plane $PR^2$. A protective transformation on $R^2$ corresponds to a projective tranformation on $PR^2$. Since $Aut(PR^2)= \{T\in M_{3\times 3}| \det T=1\}$, you get $\mathfrak{aut}(PR^2)=\{t\in M_{3\times 3}| \operatorname{tr}(t)=0\}$.</p>


问:
Given a point, a line and a circle; how can I find a point on the line having the following property?
First of all: I searched with different search engines and on StackExchange, but did not find a solution to my problem. This may be because its specific description is kind of long. For the same reason, I had to choose an unspecific title to this question due to the character limit.

Now, here is my problem:
Given

A point $D$

A line $l$

A circle $K$ (with center $C$ and radius $r$)

all in $\mathbb{R}^2$

I want to find a point $G$ such that:

$G$ is incident to $l$

Let the intersection of the line $GC$ with $K$ be $H$. It should hold that $G$ has the same distance to $D$ as to $H$, i.e. $\left\lVert\vec{GD}\right\rVert = \left\lVert\vec{GH}\right\rVert$

So far I tried to find some property that this point has that allows me to construct it, but haven't found one yet. I am also very inexperienced and have little to no knowledge in geometry, which might be why this is hard to me.

Edit (clarifications):

$D$ and $C$ may be assumed to lie on the same side of $l$ (I guess this can even be assumed w.l.o.g., as we could otherwise just mirror the point $D$).

$H$ is the intersection point that is further away from $G$ than $C$.

$D$ can be assumed to be exterior $K$.

$l$ can be assumed to intersect $K$ at two different points and to not pass through $C$.

I preferably look for geometric solutions or algebraic solutions in closed form.

答:
Sketch for an algebraic solution.

Translate, rotate and scale your picture so that the circle is the unit circle, the line $\ell$ is horizontal at known height $y=h$. Then $D$ will have known coordinates $(c,d)$. The distance from $D$ to $\ell$ is $d-h$. Suppose $G=(x,h)$. Then the line from $G$ through the center of the circle has meets the circle at the two points $\theta$ for which $b/a = \tan(\theta)$. Moreover, $G = ((\tan \theta)/h, h)$.

Equate the squares of the two distances you care about:
$$
(c- ((\tan \theta) / h)^2 + (d- h)^2= (a-\cos \theta)^2 + ( b - \sin \theta)^2
$$

That's sufficient information to solve for the angle $\theta$. You may encounter algebraic problems that will reflect the fact that the ambiguities in the picture suggest cases with no solutions and multiple solutions.


问:
Examples of Axiom of Choice used in introductory-level undergradute math
All of the applications of AoC I've encountered have been in upper level undergraduate or graduate math courses. Are there any basic results from courses like Calc I-III which (unbeknownst to students) rely on AoC? </p>

答: Again, more upper-level undergrad, or intro grad level: "in a commutative ring, every proper ideal is contained in a maximal ideal" is equivalent to AC.</p>


问:
Compute the lengths of $\gamma,\sigma$ and $\eta$ with respect to $g$.
Consider r the manifold $M = \{(x^1,x^2) \in \mathbb{R}^2 | x^2 > 0\}$ equipped with the Riemannian metric

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

a. Sketch the following three curves in $M$:

$\gamma (t)=(1-2t,1), \ t\in [0,1]$

$\sigma (t)=(\sqrt{2} \cos(t),\sqrt{2} \sin(t)), \ t\in [\frac{\pi}{4},\frac{3\pi}{4}]$

$\eta (t)= \begin{cases}
        (1,1+t) & t\in [0,1]\\
        (1-2(t-1),2) & t \in [1,2] \\
        (-1,2-(t-2)) & t\in [2,3]
        \end{cases}
        $

Here is what I got

1

b. Compute the lengths of $\gamma,\sigma$ and $\eta$ with respect to $g$. Which curve is the shortest? Which is the longest?

I am not sure how to calculate the length in manifolds. Can anyone calculate the length of one of the curves and I will do the others. I know that I have to use the formula

$$\mathcal{L}_g (\gamma)=\int_0^1 ||\dot \gamma(t)||_g dt$$

but I am not sure how to do this with respect to $g$, especially for $\eta$ since it contains $3$ curves.

Any help would be appreciated. Thanks.

答:
Write $g|_p$ for the metric evaluated at the point $p = (p^1,p^2) \in M$, so that $$g|_p =\frac 1 {[p^2]^2} \left(dx^1|_p \otimes dx^1|_p + dx^2|_p \otimes dx^2|_p\right).$$ (Here $dx^j|_p$ is the standard differential form in the $j$-th direction on $M$, evaluated at $p$.) For any curve $\alpha : [a,b] \to M$ you should calculate
$$\begin{split}
\mathcal L_g (\alpha) &= \int_a^b \sqrt{g|_{\alpha(t)}(\dot\alpha(t),\dot\alpha(t))} dt =\int_a^b \sqrt{g_{ij}({\alpha(t)})\ \dot\alpha^i(t)\ \dot\alpha^j(t)} dt \\
&=\int_a^b \sqrt{\frac{1}{[\alpha^2(t)]^2}\left([\dot\alpha^1(t)]^2 + [\dot\alpha^2(t)]^2\right)} dt.
\end{split}$$
Also keep in mind that integrals are additive, so if $\alpha$ is defined piecewise, you can always split the integral up into those pieces.

For a specific calculation, see here.


问:
Dimension of $K[x]/x^{2}$ as a vector space
Let $K$ be a field, and $R=K[x]$ the polynomial ring over K. Let $J$ be the ideal generated by $X^{2}$ Show that $R/J$ is a K-space. What is its dimension? </p>

<p>I know that the torsion submodule of $R/J$ is trivial, but I don't know if it $R/J$ is finitely generated, so I don't know if it has a basis. Hints? </p>

答:
Here is further intuition. As others have noted, any coset must have the form $\overline{a + bx} := a + bx + x^2K[x]$ for some $a + bx \in K[x]$. Addition is straightforward, and ring multiplication looks like

$$ \left(\overline{a + bx}\right) \left( \overline{c + dx} \right) = \overline{ac + (ad + bc)x + bdx^2}= \overline{ac + (ad + bc)x}\text.$$

So as a ring, $K[x]/(x^2)$ is just like $2 \times 2$ upper-triangular matrices over $K$ with constant diagonal:

$$\begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \begin{pmatrix} c & d \\ 0 & c \end{pmatrix} = \begin{pmatrix} ac & ad + bc \\ 0 & ac \end{pmatrix} \text.$$

The scalar multiplication in $K[x]/(x^2)$ behaves the same as in this matrix ring. It's clear that a basis is $\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} $ and the identity matrix.

In general, if $R$ is a commutative ring and $f(x)$ is a nonconstant monic polynomial in $R[x]$, then every element of $R[x]/(f)$ can be written uniquely as a linear combination of $(\overline{1}, \overline{x}, \overline{x}^2, \dots, \overline{x}^{\deg f - 1})$, with coefficients coming from the embedding of $R$ in $R[x]/(f)$.


问:
fixed point of a continuous map on a projective space

Let $f:S^{2n} \rightarrow S^{2n}$ be a continuous map. Show that

there exists $x \in S^{2n}$, such that $f(x) =x$ or $f(x) = -x$;

any continous map $g: \mathbb R P^{2n} \rightarrow \mathbb RP^{2n}$ has a fixed point;

I think 1 implies 2. So, how can I find an $x$ in 1?

Thanks for the help.

答:
Hint: If $f(x)\neq -x$ for all $x$, show $f$ is homotopic to the identity, $I:S^{2n}\to S^{2n}$.  If $f(x)\neq x$ for all $x$, show $f$ is homotopic to $-I$. Therefore, if $f(x)\neq \pm x$ for all $x$, then $I$ and $-I$ are homotopic.

Presumably, you know that in $S^{2n}$ that $I$ and $-I$ cannot be homotopic.


问:
How to write an equation where both independent variables and dependent variables are log transformed in a multiple regression?
How to write the multiple regression model when both the dependent variable and independent variables are log-transformed?

I know that without any log transformation the linear regression model would be written as

But now I have transformed both my dependent variables and independent variable with log. So is correct to write as

Or since I am transforming both sides of question so can I write it as

答:
I am afraid that all your transformed forms are false.

Starting from

$$y=\beta_0+\beta_1(x_1)+\beta_2(x_2)+...$$

you can write :$$\ln(y)=\ln\bigg(\beta_0+\beta_1(x_1)+\beta_2(x_2)+...\bigg)$$
This is not equivalent to $\beta_0+\beta_1(\ln(x_1))+\beta_2(\ln(x_2))+...$

Also you can change of variables :
$$X_1=e^{x_1}\quad;\quad X_2=e^{x_2}\quad,...$$
$$y=\beta_0+\beta_1\ln(X_1)+\beta_2\ln(X_2)+...$$
And with the change of :
$$Y=e^y$$
$$\ln(Y)=\beta_0+\beta_1\ln(X_1)+\beta_2\ln(X_2)+...$$
Sorry if I misunderstood your question.


问:
Josephus Problem recurrence relation
I am going through the Josephus problem and I am not able to understand the recurrence relation.

The solution says that for $2n$ people, the actual number of person in the $2n$-d round can be mapped using

$$J(2n)=2\cdot J(n)-1 \qquad\rm{eq.1}$$

and for 2n+1 people the numbering in the 2nd round can be mapped using

$$J(2n+1) = 2 \times J(n)+ 1 \qquad\rm{eq.2}$$

I understand the R.H.S of these $2$ equations.

In case of $2n$ people, after the first round the actual number of the person at position $1$ is $1 = 2 \times J(1)-1$, actual number of person $2$ is $3 =2 \times J(2)- 1$.

Same goes for the $2n+1$ equation too, actual number of person $1$ after round $1$ is $3 = 2 \times J(1)+ 1$.

What I do not understand is L.H.S, what does $J(2n)$ and $J(2n+1)$ mean here?

Why $2n$ and $2n+1$?

If I substitute '$1$' in eq.1, we get $J(2 \times 1)=2 \times J(1) - 1 \implies J(2)=1$; which is not correct.

What am I missing here?

Also how is it that using these equations I can calculate the position of the person that survives? Since these equations only represent the position of the person.

Thanks

答:
You’ve misunderstood the function $J$. $J(n)$ is the position number of the last survivor when we start with a circle of $n$ people. It is not the number of people left after the first round.

If $n=1$, the last survivor is of course person $1$, so $J(1)=1$. If $n=2$, the last survivor is again person $1$, so $J(2)=1=2J(1)-1$. If $n=4$, the first round kills off persons $2$ and $4$, and the second round kills off person $3$, so person $1$ is yet again the final survivor, and $J(4)=1=2J(2)-1$. If $n=3$, however, the first round kills off person $2$, the second round kills off person $1$, and person $3$ is the last survivor, so $J(3)=3=2J(1)+1$.

You can use the recurrences $J(2n)=2J(n)-1$ and $J(2n+1)=2J(n)+1$ and the initial condition $J(1)=1$ to calculate the final survivor for any $n$ simply by working backwards. If we start with $n=21$, for instance, we have

$$\begin{align*}
J(21)&=J(2\cdot 10+1)\\
&=2J(10)+1\\
&=2J(2\cdot 5)+1\\
&=2\big(2J(5)-1\big)+1\\
&=4J(5)-1\\
&=4J(2\cdot 2+1)-1\\
&=4\big(2J(2)+1\big)-1\\
&=8J(2)+3\\
&=8J(2\cdot 1)+3\\
&=8\big(2J(1)-1\big)+3\\
&=16J(1)-5\\
&=16\cdot 1-5\\
&=11\,.
\end{align*}$$

Title: Josephus Problem recurrence relation