问:
Proof of Strong Markov Property for SRW
In my course I have been given a proof that simple random walks satisfy the strong Markov property and I'm a bit stuck on a line in the proof. The theorem is stated as follows

Let $\tau$ be a stopping time with respect to the filtration $(\mathcal{G}_n)_{\in\mathbb{N}_0}$. Then on $\{\tau<\infty\}$ the process $(\hat S_n)_{\in\mathbb{N}_0}$, defined by $\hat S_n = S_{n+ \tau} - S_n$
is a SRW starting at the origin and independent of $\mathcal{G}_\tau := \{A \in \mathcal{F} : A \subset \{\tau \leq n\}\in \mathcal{G}_n \forall n \in \mathbb{N}_0 \}$.

In the proof given in my notes we arrive to the following equality which I am unsure why it holds and was wondering if anyone could point out why it holds.

$$\sum_{j=0}^\infty P(S_j-S_j = x_0 , \dots , S_{j+n} - S_j = x_n \cap A \cap \{\tau = j\})\\
=\sum_{j=0}^\infty P(S_0 = x_0 , \dots , S_n = x_n) P(A \cap \{\tau = j\})$$

I understand why the $S_i$'s change the way they do, this simply follows from the way they are defined, but what confuses me is why we can split the two events in the way we do. Since if $A$ was in the "past" of all the $S_i$'s it would follow by the Markov property for SRW but $A$ can't be in the past of $S_0$

Thanks in advance

答: There are two things happening. First of all, $A\cap \{\tau=j\}\in \mathcal{G}_j$, so it is independent of $(S_{j+1}-S_j,S_{j+2}-S_j,...,S_{n}-S_j)$ and this vector has the same distribution as $(S_1,S_2,...,S_n)$. Now, I did not include $S_0$ and, indeed, the above is not true unless $x_0=0$ and $S_0$ is $0$ almost surely, in which case you simply need to note that degenerate variables are independent of anything.

Title: Proof of Strong Markov Property for SRW


问:
Closed form of solutions of $xy-zv=±1$
Let us consider a real dynamical system $s′=h(s)$. In order to study the stability of the central manifold, we reformulate the problem as follows: I am asking if this equation
$$xy-zv=±1$$
has integer solutions $x,y,z,v$.

We can find some special cases, but I am interested in a closed-form.

答: By the Bezout identity, there are solutions iff $x,z$ are relative primes, and $y,v$ follow.

ContentLicense: CC BY-SA 4.0

Title: Closed form of solutions of $xy-zv=±1$


问:
Simple probability question on product rule of independent events
I would need to clarify a point on the intuition behind the product rule related to elementary probability:

Given a sample space Ω, we have that the probability of a conjunction of events $P(A,B) = P(A) \cdot P(B)$ if and only if $A$ and $B$ are statistically independent.

We can measure the probability of an event $E$ using set theory in the following way:
$$P(E) = \frac{|E|}{|\Omega|}$$

For instance, given a bag of words containing "table chair the a" and assuming they are all statistically independent from each other, $$P(\text{table,chair}) = \frac{|\text{event table}|}{4} \cdot \frac{|\text{event chair}|}{4}$$

What is the justification behind defining the probability of joint independent events as the product of the probability of the two simple events? Is there a way to formally prove the result using set theory, except deriving it from the case of the conditional probability?

答:
After having watched the some of the videos, I think it is somewhat more clear, but not entirely:
for instance, in the example I made of "table chair the a", I now get how, to measure set theoretically $P(the,a)$ [the probability that from a bag of words I extract the and then, from the same words, I extract a]. I have to measure the cardinality of the new event over a new sample spaces that has as simple events all the possibile results of the extraction of two words from that bag:

The number of possibile extraction of two words will be 16.
So, in a space $Ω$ of 16 possible outcomes, the probability of the outcome(simple event) "the-a" will be 1/16.

1/16 is equal, looking back at the sample space Ω of 4, to P(the) * P(a).

But how do I arrive to the general rule that tells me that, given an event A and an event B and a sample space $Ω$, then the probability of getting A and then B is equal to the probability of $|A|*|B|/Ω*Ω$?


问:
How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$?

How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$?

I have been trying to solve this but I can't use $$\det(A^{-1})=\det \Big(\frac{1}{\det(A)} \operatorname{adj}(A) \Big)$$ because $\det(A)=0$ and $\frac{1}{0}$ is not allowed.

答:
Hereafter, $A^*$ stands for adjoint, $|A|$ stands for the determinant of $A$.

Actually we have the conclusion that
$$
\operatorname{rank}(A^*) = \begin{cases}
    n   & \operatorname{rank}(A) = n   \\
    1 & \operatorname{rank}(A) = n-1 \\
    0   & \operatorname{rank}(A) = 0
  \end{cases}
$$

To prove the first one, notice that $A A^* = |A|E$ where $E$ is the identity matrix and $\operatorname{rank}(A)=n$ means $|A|\neq 0$. So $A^*$ is invertible and $\operatorname{rank}(A^*)=n$

To prove the second one, notice that $A A^* = |A|E = \mathbf{0}$. This means every column of $A^*$ is in the null space of $A$, which has dimension one. But the alternative definition of rank (see below) tells us $A^* \neq \mathbf{0}$, therefore $\operatorname{rank}(A^*)=1$.

To prove last one, simply use an alternative definition (or calculation) of rank of matrix: $\operatorname{rank}(A)$ is the size of its largest non-vanishing minor. See more at this wikipedia page under the subsection Determinantal rank – size of largest non-vanishing minor.

ContentLicense: CC BY-SA 3.0


问:
A question regarding moment generating function
The moment generating function of a real-valued random variable $X$ on a probability space $(\Omega,\mathscr A,P)$ is defined as $G_{X}(t)=E[e^{tx}]$ $(t \in \mathbb{R})$. Prove that $G_{\alpha X+\beta}(t)=e^{\beta t}G_{X}(\alpha t)$ for $\alpha, \beta \in \mathbb{R}$.

Attempt

$G_{X}(t)=E[e^{tx}]$, then

$$G_{\alpha X+\beta}(t)=E[e^{t(\alpha x+\beta)}]=E[e^{\alpha xt}e^{\beta t}]=E[e^{\alpha xt}]E[e^{\beta t}]=e^{\beta t}E[e^{\alpha xt}]=e^{\beta t}G_{X}(\alpha t).$$

Show that $G_{X+Y}=G_{X} \cdot G_{Y}$.

Attempt

$$G_{X+Y}(t)=E[e^{t(x+y)}]=E[e^{xt}e^{yt}]=E[e^{xt}]E[e^{yt}]=G_{X}(t)\cdot G_{Y}(t).$$

Questions

Are my attempts correct?

In general, does $E[XY]=E[X]\cdot E[Y]$? If yes, what conditions do the random variables $X, Y$ have to satisfy for $E[XY]=E[X]\cdot E[Y]$ to be true?

Suppose we have a random variable $X$, and $g$ is not a function of $x$, then does it follow that $E[g]=g$ for all $g$?

Thank you for your time.

答:
Your attempts are correct if you assume that $X$,$Y$ are independent in the proof of $G_{X+Y} = G_XG_Y$. Which btw you may, since that identity doesn't generally hold for non-independent $X$ and $Y$.

In general, if $X$ and $Y$ are independent, you have $\mathbb{E}\left(f(X)g(Y)\right) = \mathbb{E}f(X)\mathbb{E}g(Y)$. You can easily see this by expanding $\mathbb{E}$ into its definition. You get (assuming that the expectation exists, and that $\mu_X$ is the distribution of $X$ and $\mu_Y$ the distribution of $Y$) \begin{align}
  \mathbb{E}\left(f(X)g(Y)\right) &= \int f(x)g(y) \,d(\mu_X\times\mu_Y)(x,y) = \iint f(x)g(y) \,d\mu_X(x) \,d\mu_Y(y)\\
&= \int g(y) \int f\,d\mu_X \,d\mu_Y(y) = \int f\,d\mu_X \int g\,d\mu_Y \\
 &= \mathbb{E}f(X)\mathbb{E}g(Y) \text{.}
\end{align}

Your point (3) is correct too. The expected value of a constant $a$, is, well, $a$.


问:
If $f(nx)=f(x)^n$ then $f(x+y)=f(x)f(y)$
If $f:\mathbb{R}\to\mathbb{R}_{+}$ is injective monotone function and if, $f(nx)=f(x)^n$ for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$, then,
$$f(x+y)=f(x)f(y),\text{ for all }x,y\in\mathbb{R}$$

It is easy to check the equality if $y$ is a multiple of $x$.

I thought about proving the inequalities, $f(x+y)\le f(x)f(y)$ and $f(x+y)\ge f(x)f(y)$, but I couldn't get to them.

Assuming that $f$ is increasing and $x<y$, then $y=kx+r$, for some $k\in\mathbb{Z}$ and $0< r\le \lvert x \rvert$, so

$$f(x+y)=f(x+kx+r)=f(x(k+1)+r)\ge f(x(k+1))=f(x)^{k+1}=f(x)f(x)^k=f(x)f(kx)$$

That is, $f(x+y)\ge f(x)f(kx)$, but the problem is that $f(kx)<f(y)$ so I can't continue the demonstration.

答:
I would like to approach this a little bit differently - and possibly inferiorly as I will be looking only at the rational numbers so it could still be the case that the property does not hold for $x=\pi$ or some other irrational number - I will first proof that the only class of functions that satisfies the property
$$  f(nx) = f(x)^n \, \forall \,x \in \mathbb{Q}, n \in \mathbb{N} $$
is the class
$$f(x) = f(1)^x \,\forall \,x \in \mathbb{Q}$$

I will do this by first observing that
$$f\left(\frac{1}{n}\right) = \sqrt[n]{f(1)} \, \forall \, n \in \mathbb{N}$$

and combine that property with the original property to show that
$$ x = \frac{p}{q} \implies f(x) = \sqrt[q]{f(1)}^{p} = f(1)^x \, \forall \, p \in \mathbb{Z}, q \in \mathbb{N}$$
which would be equivalent to saying
$$ f(x) = f(1)^x \, \forall \, x \in \mathbb{Q}$$
since any rational number can be written as the quotient of an integer and a natural number

The first property will be proven by contradiction: let us assume that
$$\exists \, n \in \mathbb{N} : f\left(\frac{1}{n}\right) \neq \sqrt[n]{f(1)}$$
from that it would follow that
$$f(1) = f\left(\frac{n}{n}\right) = f\left(\frac{1}{n}\right)^n \neq \sqrt[n]{f(1)}^n = f(1)$$
This is a contraciction, therefore it must be so that
$$f\left(\frac{1}{n}\right) = \sqrt[n]{f(1)} \, \forall \, n \in \mathbb{N}$$

Now for our second step, let us pick two values $p \in \mathbb{Z}, q \in \mathbb{N}$ with $x = \frac{p}{q}$ making $x$ any rational number
$$ f(x) =  f\left(\frac{p}{q}\right) = f\left(\frac{1}{q}\right)^p = \sqrt[q]{f(1)}^{p} = f(1)^{p/q} = f(1)^x$$

Now, we have proven that $f(xn) = f(x)^n \implies f(x) = f(1)^x$ for any rational number $x$, because $f(1)$ is a constant we will give it a new name, $r$ to get that $f(x)$ must be of the form $r^x$. To this we can apply power rules $$f(x+y) = r^{x+y} = r^x r^y = f(x)f(y)$$

Therefore, we have shown that $$f(nx) = f(x)^n \, \forall x \in \mathbb{Q}, n \in \mathbb{Z} \implies f(x+y) = f(x)f(y) \, \forall x, y \in \mathbb{Q}$$


问:
OA + AB = AB vectors
OA= -2 2 OB 6 1 OC = 10 1
I drew on a diagram and I nave to use triangular law but I'm getting OA +AB = AB AND I DONT HOW TO ADD THIS BEACUAS IM TRYING TO GET AB</p>

答:
By definition of vector addition we have $$\overrightarrow{OA}+\overrightarrow{AB}=\overrightarrow{OB}$$
and therefore $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$.

Title: OA + AB = AB vectors


问:
Probability of an RNG?
For fun (relatively speaking), I'd like to do a little research on probabilities related to a pseudo-random number generator.

Specifically, how would I get started figuring out "what is the probability that a prng will generate the same random number twice?"

This is part of an investigation into a larger issue, but I'm not sure where to begin (or if my question is even relevant).

Edited to add:

After reading the comments and answers I see that my question isn't clear. As an example: the rng outputs a 5 What is the probability that another 5 will be the next output? How many "rolls" of the RNG must I make before I see another 5?

If we assume a good rng has uniform distribution, then is it 1 out of (2^32) (assuming 32bit integers)?

For a larger example, lets say I use the RNG to generate numbers that are used as an index into an array of characters. The array is 25 numbers long, so the output of the RNG is mod 25 I will generate a sequence of 5 characters.

How would I determine the probability of generating the same 5 character sequence?

If we assume that we have a uniform distribution of output numbers from the RNG, then the RNG's output shouldn't factor into the probability of generating the same character sequence.

However, is my assumption valid?

答:
This is an answer to the edited question, taking into account your comment that you're interested in "In a good PRNG, the same number will occur twice at lots of irregular intervals, and this won't have anything to do with periodicity for a very long time."

In a deterministic machine with finite state space, the output of every computation is necessarily finite or periodic. However, as J.M. has pointed out, there are good pseudo-random number generators with very long periods. For practical purposes, we can assume that a good PRNG has a uniform distribution, as you do in your new question. But then the question is no longer really about PRNGs; it's about random numbers. What you're asking for now is just the probability and (presumably expected) waiting time for drawing a certain number from a uniform distribution.

If the numbers are integers uniformly distributed between $0$ and $2^n-1$ (where $n$ is the number of bits), then the probability of a particular one of them occurring is $1$ in $2^n$. (You wrote $2^n-1$; there's no reason to subtract one; all of the $2^n$ numbers are equally probable.)

The expected waiting time $t$ has to satisfy the equation $t = p \cdot 1 + (1 - p) \cdot (1 + t)$, where $p$ is the probability $p=2^{-n}$ of getting the number you're waiting for. This is because with probability $p$ you have to wait just once, and with probability $1-p$ you wait once and then end up in the same situation, having to wait for the expected waiting time. Solving for $t$ yields $t=p^{-1}=2^n$. So for $32$-bit numbers you have to draw $2^{32}$ numbers on average to draw a specific number, and in particular to draw a specific number again that you'd already drawn.

Title: Probability of an RNG?


问:
Relation of the length of latus rectum with the distance between focus and vertex and distance to focus from directix
In this website, the following statement is written

Length of latus rectum is twice the distance between focus and vertex or four times the distance of focus from the directrix.

I thought of proving the statement for a parabola whose focus lies on the $y$ axis and this would prove the geometric idea for all other parabolas since all other are just shifted/stretched variations of the one in consideration. The proof is presented in the following section.

Set of points equidistance from line and point. For a parabole opening up like a U in the x-y plane, say it is parallel its line is $y=c$ and the point is $(a,b)$ then the distance from these two are the same.

$$ (y-c)^2 = (x-a)^2 + (y-b)^2 \tag{1}$$

Now, the for the points which are same vertical height as directrix, plug  $y=b$,

$$ a \pm (b-c) = x$$

Hence the distance between the two points:

$$ l= 2 (b-c)$$

"The length of latus rectum is twice the perpendicular distance between directrix and focus"

But the statement says that it is four times, so is it wrong or did I make mistake?

For the second part, we can find the maxima of $y$ is when $x=a$ by differentiation and hence,

$$ y-c = -(y-b)$$

$$ 2y = b-c$$

Hence, the vertex is given as:

$$ y = \frac{b-c}{2}$$

Now since focus at (a,b) the distance between vertex and focus :

$$ d = \frac{b+c}{2}$$

Which doesn't even seem to be related to original equation

答:
For the parabola $y^2=4ax$, the equation of Latus rectum is $x=a$ and end points of Lr are $(a,2a)$ and $(a,-2a)$ so $Lr=4a$. The vertex is $V(0,0)$ directrix is $x=-a$
The point of intersection of directrix and x-axis is $Z(-a,0)$ so $VZ=a$. Focus is $F(a,0)$ so $VF=a$. So finally $VF=VZ=Lr/4.$


问:
Why don't these samples come from a normal distribution?
I have two sets of 1000 $L_2$-normalized vectors (each being 512 elements long). I compute the similarities of each vector from one set with each vector in the other set using the inner product, which yields  a total of 1.000.000 similarity-computations between vectors. When i construct a histogram out of the resulting similarities i get the following image:

which looks very much like a normal distribution to me. When i manually construct a normal-distribution with $\mu$ and $\sigma$ of the empirical data i get the following plot:

```<code>from scipy import stats

mu = similarities.mean()
variance = similarities.var()
sigma = similarities.std()

x = np.linspace(mu - 5*sigma, mu + 5*sigma, 100)

sns.histplot(without_diag, stat='density');
plt.plot(x, stats.norm.pdf(x, mu, sigma), 'r', alpha=0.6);
</code>```

Which seems to fit just fine. I used numpy to perform the computation and seaborn to plot the histogram. I used scipy.stats.normaltest to test, if the samples come from a normal-distribution. Result:

```<code>>>> k2, p = stats.normaltest(without_diag)
>>> print(p)
3.2035057152116813e-221
</code>```

I read somewhere, that $p$ should be below $0.05$ in order to reject the null-hypothesis, that the data was sampled from a normal-distribution and $3\times 10^{-221}$ seems very much to be below that border. Does it have to do with the number of samples? Is there a name for this kind of distribution?

EDIT:

Here is the normal probability-plot suggested in the answer by @heropup:

答:
Rather than plotting a histogram, it may be more informative to create a normal probability plot.  This may show deviations for normality that might be hidden in a histogram; e.g. tails too heavy or sparse.

A caveat regarding statistical tests of non-normality is that, like many statistical tests, the power to reject becomes greater as the sample size increases.  With $n = 10^6$ this means that even a small deviation from normality may be detected by such a test.

That said, the resulting $p$-value you obtained does suggest deviation from normality; however, whether the extent of such deviation is meaningful--i.e., "effect size" as it pertains to measures of goodness-of-fit--is a separate question.


问:
A category which direct limits but no general colimits
I am looking for a (at best, real life) category that has direct limits, but no general small colimits, or a category that has inverse limits, but no general small limits. Are there any interesting examples that are not too obviously made to be an example for this?

I am asking this because I wonder why many lectures include the construction of direct/inverse limits as an exercise, instead of general small (co)limits. In particular, I wonder this for topology and algebra classes.

答:
I think this answer goes already into the direction that answers my second question.

When I wrote the assignment myself, I recognized that writing what a direct system is is much easier than writing what a general diagram is—the commutativity relations in a direct system are particularly easy, and you can effectively avoid talking about categories and functors. This is an advantage for lectures which typically haven't talked about categories and stuff yet when the exercise about the limits is assigned.


问:
Prove that $\lambda(\alpha A) = \alpha\lambda(A)$ where $\lambda$ is the Lebesgue measure and $A$ is a Borel-Set
Let $\mathscr{B}(\mathbb{R})$ be the Borel-Set over $\mathbb{R}$ and $\lambda: \mathscr{B}(\mathbb{R}) \to [0,\infty]$ the Lebesgue measure.

Prove: If $A \in \mathscr{B}(\mathbb{R})$, show that $\alpha A := \{\alpha x \; |\; x \in A\}\in \mathscr{B}(\mathbb{R})$ for $\alpha > 0$ and further show that $\lambda(\alpha A) = \alpha \lambda(A)$.

I know that by definition, $\mathscr{B}(\mathbb{R})$ is the smallest $\sigma$-Algebra that contains all open sets in $\mathbb{R}$. I'm confused how to derive $\alpha A \in \mathscr{B}(\mathbb{R})$ from $A \in \mathscr{B}(\mathbb{R})$. Any hints are greatly appreciated.

答:
It is clear that if $I\subset\Bbb R$ is an interval, then $\alpha I$ is an interval for every $\alpha>0$, and that ${\rm long\,}(\alpha I)=\alpha {\rm long\,}(I)$.

If $A\subset \bigcup I_n$ with $I_n$ intervals, then $\alpha A\subset \bigcup (\alpha I_n)$ and $\alpha I_n$ are intervals. So,
$$\lambda(\alpha A)\le \sum_n{\rm long\,}(\alpha I_n)=\alpha \sum_n{\rm long\,}(I_n).$$
Now, taking $\inf$ over all covers of $A$, and taking into account that $\alpha>0$, you get that
$$\lambda(\alpha A)\le \alpha \lambda (A).$$

For the reverse, $\alpha \lambda(A)=\alpha \lambda[(1/\alpha) (\alpha A)]\le\alpha (1/\alpha) \lambda(\alpha A)=\lambda(\alpha A)$.


问:
Newbie vector spaces question
So browsing the tasks our prof gave us to test our skills before the June finals, I've encountered something like this:

"Prove that the kernel and image are subspaces of the space V: $\ker(f) < V, \operatorname{im}(f) < V$, where $<$ means a subspace."

Is it just me or there's something wrong with the problem? I mean: I was rewriting the tasks from the blackboard by hand so I may have made a mistake but is a problem like this solvable or I messed up and should rather look for the task description from someone else? Cause for the time being, I don't see anything to prove here since we don't know what V is, right?

答: Let $f$ is a linear transformation from $V$ to $V$. So what is the kernel of $f$? Indeed, it is $$\ker(f)=\{v\in V\mid f(v)=0_V\}$$ It is obvious that $\ker(f)\subseteq V$. Now take $a,b\in F$ the field associated to $V$, and let $v,w\in\ker(f)$. We have $$f(av+bw)=f(av)+f(bw)=af(v)+bf(w)= 0+0=0$$ So the subset $\ker(f)$ has the eligibility of being a subspace of $V$. Go the same way for another subset.</p>

Title: Newbie vector spaces question


问:
Prove that for $n_1\neq n_2$ the measure of points in $(0,1)$ and that in their decimal representation has $1$ in $n_1,n_2$ is $\frac{1}{100}$
Prove that for $n_1\neq n_2$ the measure of points in  $(0,1)$ and that in their decimal representation has $1$ in $n_1,n_2$ is $\frac{1}{100}$ and it is measurable.

Any hint or intuition on how to do this?

答: How many posibilities do you have for position 1? And for position 2? and for position $n_1$? and for position $n_2$? and for any position DIFFERENT for $n_1$ and $n_2$?


问:
Interchange the order of integration and summation
I read some questions, where people usually discuss the interchange of definite integral and summation; for examples, here, here and here. But I would like to ask about the interchange of indefinite integral with summation, the reason is very practical. I am not sure if I read all the similar questions in SE, so if someone finds the answer somewhere else, please show me the link, and we could close this question.

The statement of question: I have a function $F(z)$, which is represented by an indefinite integral
$$
F(z)=\int d z\; f(z),
$$
and the integration can not be carried out analytically. I would like to study the zeros and singularities of $F(z)$ based on knowing the zeros and singularities of $f(z)$.
For example, if I know that $z=0$ is a singular point of $f(z)$, and $f(z)$ can just be expanded by Laurent series, say
$$
f(z)=\sum_{n=-\infty}^{\infty} a_n z^n.
$$
My question is in what conditions the following equality holds
$$
\int d z\;\sum_{n=-\infty}^{\infty} a_n z^n = \sum_{n=-\infty}^{\infty} \int d z\; a_n z^n+c,
$$
where $c$ is an arbitrary constant.

Generally the Laurent series is not necessary, the expansion could be in any other forms, say asymptotic expansion or fractional expansion.

Thanks in advance.

答:
You need compact convergence of the series, that is, convergence on compact sets. But we also need some more preliminaries:

First, I'm not going to talk about the indefinite integral, but just about antiderivatives, simply because we will need definite integrals down the line, and I think it will be less confusing if we reserve the integral sign for the definite ones. Second, the expression $\sum\int a_nz^n\mathrm dz$ is a bit icky, since the integrals all come with an arbitrary additive constant attached, which can make the sum diverge. We will have to specify which antiderivative we are choosing specifically. Third, we will only be considering connected domains, for simplicity's sake. You will soon see why. Fourth, interchanging sums and integrals or sums and antiderivatives is really about interchanging limits and integrals/antiderivatives, since infinite sums are just sequences/their limits written down in a particular way. So the real question is: If $f_n\to f$, does $F_n\to F$, where $F_n,F$ are suitably chosen antiderivatives of $f_n$ and $f$? And in what manner do they converge?

Now with these remarks out of the way, we can say the following (which is about sequences of functions, but series are sequences, so you can apply it to series exactly the same way):

Let $D\subseteq\mathbb C$ be a connected domain. Let $f_n:D\to\mathbb C$ be holomorphic for all $n\in\mathbb N$, and let $f_n$ converge to $f:D\to\mathbb C$ uniformly on compact subsets of $D$. Also, let all $f_n$ have an antiderivative on $D$.

Then the function
$$F:D\to \mathbb C,~z\mapsto\int_{z_0}^z f(w)\mathrm dw,$$
where the integral goes along any arbitrary path from a fixed $z_0\in D$ to $z$, is well defined and an antiderivative of $f$. We also have $F_n\to F$ uniformly on compact sets, where $F_n$ is an analogously defined antiderivative of $f_n$.

Proof:

First note that because $f_n\to f$ uniformly on compact subsets, $f$ is holomorphic. Also, since $D$ is connected and open, it is also path connected, so

a path from $z_0$ to $z$ is guaranteed to exist. And since the functions $f_n$ admit an antiderivative, the integral
$$F_n(z)=\int_{z_0}^z f_n(w)\mathrm dw$$
does not depend on the path and is thus well-defined. Due to uniform convergence of the integrand to $f$ on the arbitrarily chosen path (it's compact), the integral converges to
$F(z)=\int_{z_0}^z f(w)\mathrm dw,$ which thus also doesn't depend on the path and is then well-defined.
If we can show that the convergence is uniform on compact subsets, then $F$ is holomorphic and $F_n'\to F'$ uniformly on compact subsets, too. But since $F_n'=f_n$ and $f_n\to f$, we will then have $F'=f$, so $F$ is an antiderivative of $f$. So we show this uniform convergence on compact subsets:

Note that any compact subset of $D$ can be covered by a finite number of compact discs. And if a function converges uniformly on a finite number of sets, then it also converges uniformly on their union. So it is sufficient to show uniform convergence on compact discs. Let $\overline{U_r}(z_\ast)\subset D$ be such a disc with radius $r$ centered at $z_\ast$. On this disc we have

$$\begin{align}
\vert F(z)-F_n(z)\vert&=\left\vert\int_{z_0}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&=\left\vert\int_{z_0}^{z_\ast} f(w)-f_n(w)\mathrm dw+\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&=\left\vert F(z_\ast)-F_n(z_\ast)+\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&\leq\vert F(z_\ast)-F_n(z_\ast)\vert + \left\vert\int_{z_\ast}^z f(w)-f_n(w)\mathrm dw\right\vert\\
&\leq \vert F(z_\ast)-F_n(z_\ast)\vert + r\sup_{w\in\overline{U_r(z_\ast)}}\vert f(w)-f_n(w)\vert.
\end{align}$$

The last estimation gives a bound that doesn't depend on $z$ and goes to $0$ (the first term because $F_n\to F$ pointwise, the second because $f_n\to f$ uniformly on the compact disc). So $F_n\to F$ uniformly on the disc, and by the argument above, on any other compact set as well.


问:
Abelian category with all direct limits but without all colimits
As a follow-up to my previous question on the topic, I'd like to know if there is an abelian category that has all direct limits (or inverse limits, if dualising helps), but no  general colimits (limits).</p>

答:
An abelian category has all colimits if and only if it has all filtered colimits ("direct limit" is very confusing terminology and IMO should be avoided, because they aren't limits). This is because

an abelian category has finite coproducts and coequalizers,

arbitrary colimits can be expressed in terms of arbitrary coproducts and coequalizers, and

arbitrary coproducts are filtered colimits of finite coproducts.


问:
Puzzlement about a step in WW Rouse Ball's explanation of Hero's proof
I have a 4th edition copy of "A Short Account of the History of Mathematics" by W. W. Rouse Ball.  This work, written in the late 1800's (the 4th edition dates to the early 1900's) is a highly respected history, and contains accessible explanations of many historically important results.

In his explanation of how Hero derived the formula $\sqrt{s(s-a)(s-b)(s-c)}$ for the area of a triangle with sides $a, b, c$ and semi-perimeter $s$, on page 90 in chapter 4, Ball sets up a triangle $ABC$ with inscribed circle centered at $O$, touching side $AB$ at point $F$.  He then defines point K as the meeting of two lines:  The line passing through $C$ perpendicular to side $BC$, and the line passing through $O$ perpendicular to $BO$.  Then the following statement is made:

He then shews (sic) that the angle $OAF = $ angle  $CBK$; hence the triangles $OAF$ and $CBK$ are similar.

(Note that "shews" rhymes with "sews," not with "news.")  The formula then follows from a sequence of ratio manipulations.

I cannot prove or even convince myself that those two angles are equal.  I don't restrict the proof to classical geometry; I have tried the usual vector tricks (which for example easily demonstrate that if side $BC = \vec{u}$ and $BA = \vec{v}$ the angle bisectors meet at a point which differs from $B$ by $\frac{\vec{u}+\vec{v}}{3}$).  I have also tried using analytic geometry; the coordinates of $K$ get pretty messy and I was unable to "shew" that the two discussed angles are equal.

Could it be that this respected author simply erred?  I doubt it, because the subsequent ratio manipulations would then hint that Hero's formula does not work.

So I am looking for any simple demonstration that those two angles are equal.

答:
Let $\measuredangle CAB$, $\measuredangle ABC$, and $\measuredangle BCA$ be $2\alpha$, $2\beta$, and $2\omega$ respectively. For brevity, we also denote $\measuredangle CBK$ as $\phi$.

Since $O$ is the incenter, $OA$, $OB$, and $OC$ are the angle bisectors of $\measuredangle CAB$, $\measuredangle ABC$, and $\measuredangle BCA$ respectively

Since $\measuredangle KOB = \measuredangle KCB = 90^o$, $OBKC$ is a cyclic quadrilateral. Therefore,
$$\measuredangle BKO = \measuredangle BCO = \omega. $$

$\Delta KOB$ is a right angle triangle. Therefore,
$$\measuredangle BKO + \measuredangle OBK =\omega+\beta+\phi = 90^o\quad\rightarrow\quad \phi=90^o -\beta - \omega. \tag{1}$$

Now consider the triangle $ABC$. There we have,
$$2\left(\alpha+\beta+\omega\right)=180^o \quad\rightarrow\quad \alpha=90^o -\beta - \omega. \tag{2}$$

From (1) and (2), it is obvious that $\phi =\alpha$. Therefore,
$$\measuredangle CBK = \measuredangle OAF. $$


问:
Proving $\gcd(4k+2,k^2+k) = 2$ by direct proof
I came across this problem in a textbook (not an assignment) and was interested in how to solve it.

I imagine $\gcd(4k+2,k^2+k)$ can easily be proved to be an even integer because:

$4k+2 = 2(2k+1)$, which would be even because its $2$ times and integer, and

$k^2+k$ can be proved to be even by using $2$ cases, one where $k$ is even and is equal to $2n$ and the other for an odd $k$, where we can set it equal to $2n+1$, where $n \in \mathbb{Z}$.

From here, if we let $\gcd(4k+2,k^2+k)=x$, we can say if $x$ is the $\gcd$ of $(4k+2,k^2+k)$ then it must divide any linear combination of the 2. I'm stuck on this part, because there's so many directions this could go.

I'm thinking $x|(k^2+5k+2)$ or $x|(4k^2-2)$, but I can't find a way to prove the gcd isn't some even integer greater than $2$. Any insight is appreciated!

答:
Let $(4k+2,k^2+k)=d$. Then $d|4k+2-2(4(k^2+k)-k(4k+2))=2$.

$d=2$ since both numbers are even. The motivation is to play around until we get a constant (in this case, it is $2$).

It motivates us to multiply $4k+2$ by $k$ and $k^2+k$ by 4 and to subtract those numbers so we have something easier to handle (lower in degree): $4(k^2+k)-k(4k+2)=2k$. Seeing $4k+2$, we are motivated to multiply $2k$ by two and subtract from $4k+2$ to obtain $2$.


问:
Need help in finding the sum of the following infinite series.
I need to find the sum of the infinite series with the following general term
$\frac{2^{nlogn}}{n!}$

I initially started off by getting rid of the log in the the numerator, so that later became $\frac{n^{nlog2}}{n!}$

So I tried my best to split the terms so that the consecutive terms cancel out, but I couldn't find a way. Am I approaching this the wrong way?

答:
At least, you could show the convergence. Consider the more general case of
$$a_n=\frac{k^{n\,\log(n)}}{n!}\implies \log(a_n)=n\log(n)\log(k)-\log(n!)$$
Using Stirling approximation
$$\log(a_n)=n ((\log (k)-1) \log (n)+1)-\frac{1}{2} \log (2 \pi  n)-\frac{1}{12
   n}+O\left(\frac{1}{n^3}\right)$$ Use Taylor approximation to get
$$\log(a_{n+1})-\log(a_n)=((\log (k)-1) \log (n)+\log (k))+\frac{\log (k)-2}{2 n}+O\left(\frac{1}{n^2}\right)$$
$$\frac {a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n) }=k\,n^{\log (k)-1}\left(1+\frac{\log (k)-2}{2 n}+O\left(\frac{1}{n^2}\right) \right)$$ which tends to $0$ if $k<e$ which is the case for $k=2$.

Now, we know that $\sum_{n=2}^\infty a_n$ will converge to ... a number but, as already said in comments, I am quite skeptical about a possible closed form.

As also said in comments, the partial sums
$$S_p=\sum_{n=2}^p \frac{2^{n\,\log(n)}}{n!}$$ converge quite fast
$$\left(
\begin{array}{cc}
p & S_p \\
 10 & 19.337809138017622646 \\
 100 & 33.087403209274342050
\end{array}
\right)$$ This value with $20$ significant figures is obtained for $k=97$.

Inverse symbolic calculators do not recognize it.


问:
Does distinct sets mean disjoint sets?
If A and B are two distinct sets, does that mean A and B are disjoint? Or if A is the subset of B, can A and B be two distinct sets? I am confused here that whether distint and disjoint sets mean the same? I have to find the intersection of two sets. It could be empty set if these two sets are disjoint. And it could be any other set if they are not.
So please help me understand the difference between distinct and disjoint sets if there is any.

答:
Two sets are equal if and only if they both have EXACTLY the same elements.

Two sets are disjoint if they do NOT have commons elements.

So no, to be DIFFERENT/DISTINCT is not the same that being DIsJOiNT.

$A=\{0,1\}$ and $B=\{0\}$ are DIFFERENTS (DISTINCT) ($1\in A$ but $1\notin B$), but they are not disjoint: $A\cap B=\{0\}$ (in fact, $B\subset A$).

Moreover, $A$ and $C=\{0,2\}$ are also differents (distinct), not disjoint $A\cap C=\{0\}$ and neither $C\subset A$ nor $A\subset C$.

$A$ and $D=\{2,3\}$ are differents (distinct) and disjoint.

Title: Does distinct sets mean disjoint sets?


问:
A generalization of the theorem on natural irrationalities?
Here is the statement of problem 26 of chapter 5 from Morandi's Field and Galois Theory:

Let $K/F$ be a normal extension and let $L/F$ be an algebraic extension. If either $K/F$ or $L/F$ is separable, show that $[KL:L] = [K:K \cap L]$ Give an example to show that this can be false without the separability hypothesis.

Let's assume that the extensions are finite. If $K/F$ is separable, then it is Galois, and the theorem on natural irrationalities applies (the theorem states that $\mathrm{Gal}(KL/L) \cong \mathrm{Gal}(K/K\cap L)$. If we only assume $L/F$ to be separable, then I am stuck. I tried using the primitive element theorem, so that $L = F(a)$ for some $a \in L$ and then show that $[K(a):F(a)] = [K:K\cap F(a)]$, but I can't get anywhere with this.

I'm also a bit surprised by this problem; the theorem on natural irrationalities is fairly well-known, but everywhere I have seen it is stated with the assumption that $K/F$ is Galois. I was hoping that someone could provide some insight on how to do this, or a source for a proof that doesn't use the assumption that $K/F$ is Galois.

Edit: A tiny bit of progress. Let $L/F$ be separable. If $\min(F,a)$ has a root in $K$, then it splits over $K$ by normality of $K/F$. Therefore, $a \in K$, and $K(a) = K$ and $K \cap F(a) = F(a)$. The degree formula follows trivially in this case. We can then assume that the none of the distinct roots of $\min(F,a)$ lie in $K$.

答:
Assume that $L/F$ is separable, and note that$[KL:L]=[K:K\cap L]\Leftrightarrow [KL:K]=[L:K\cap L]$.

Let $L'$ be the normal closure of $L$. Because $L/F$ is separable, then $L'/F$ is separable, therefore is Galois. By natural irrationalities, $[KL':K]=[L':K\cap L']$

and
\begin{alignat*}{1}
[KL':K]&=&[L':K\cap L']\\
\Leftrightarrow [KL:K] &=& \frac{[L':L][L:K\cap L]}{[KL':KL][K\cap L':K\cap L]}\tag*{(1)}
\end{alignat*}

By natural irrationalities
\begin{alignat*}{1}
[KL':KL]&=&[L':KL\cap L']\tag{2}\\
[KL\cap L':L]&=&[(K\cap L')L:L]\\
&=&[K\cap L':K\cap L'\cap L]\\
&=&[K\cap L':K\cap L]\tag{3}
\end{alignat*}

Then we have

\begin{alignat*}{1}
[KL':KL][K\cap L':K\cap L]&=[L':KL\cap L'][KL\cap L':L]\tag*{by (2)(3)}\\
&=[L':L]\\
\Rightarrow 1&=\frac{[L':L]}{[KL':KL][K\cap L':K\cap L]}\\
\Rightarrow [KL:K] &= [L:K\cap L]\tag*{by (1)}\\
\Leftrightarrow [KL:L]&=[K:K\cap L]\tag*{QED}
\end{alignat*}


问:
Difference between two quaternions
I am currently writing up a program that needs to know the difference in degrees for each axis between two quaternions.

Now, suppose I have a starting quaternion Qs and I need to compute at each step the difference between my current orientation represented by the quaternion Qc.
I do that: Q = Qc^-1 * Qs

Now I would like to get values from the quaternion Q that would represent for each axis the number of degrees that differ from Qc to Qs. Is that possible?

Edit: If that's not clear here are other explanations
I know I can get the angle value from a quaternion by doing 2 * cos-1(q.w). Now I would like to get an idea on how to project this angle onto each axis. The expected output is something like:

From Qs to Qc there is a rotation of 30 degrees on the x axis, 20 degrees on the y axis and 0 degrees on the z axis

答:
Perhaps worth mentionning here, the angular difference between two rotations represented by unit norm quaternion is

$$d\left(\mathbf{q}_1, \mathbf{q}_2\right) = 2 \arccos\left(\mathbf{q}_1^T\mathbf{q}_2\right)$$

Title: Difference between two quaternions


问:
Find the volume inside the curve $x^2+4z^2+8y=16$ and on the positive side of $xz$-plane.
Find the volume inside the curve $x^2+4z^2+8y=16$ and on the positive side of $xz$-plane.

What will be the volume of this as I'm getting wrong answer?

As per me the volume is this.

Please help me as I'm unable to get it.

答:
Correct integral - $$v = 4\int_0^2 \int_0^\sqrt{16-4z^2} 1/8(16-x^2 +4z^2) \,dx\,dz$$

Correct Answer is $$ {8pi} $$

below is the correct solution of the given question -


问:
How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved
How are can the above recurrence relation be solved?

I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$

And then here:

$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$

And we are given that these: $T(0) = 2, T(1) = 3$

So, for $n = 0$ and $n = 1$ we get two equations but we need 3 more, yet we don't have any more constants.

答:
You can also solve it the following way:

Step 1: Find one particular $f(n)$ so that

$$f(n)=f(n-1)+f(n-2)+3n+1 \,.$$

If you look fora  polynomial $f$, the highest power doesn't cancel so you need to look for some $f(n)=an+b$. Find it.

Step 2: Let $g(n)=T(n)-f(n)$. Then $g(n)$ satisfies the homogeneous recurrence

$$g(n)=g(n-1)+g(n-2) \,.$$

Solve it.


问:
Brownian Motion Hitting Time of functions $y=t$ and $y=t^2$
Let $W(t)$ be a standard Brownian motion. Let $y(t)=t$. Define the following stopping time:

$T_h = \inf \{t>0: W(t)=y(t) \}$

What would be the probability:

$$\mathbb{P}(T_h\leq k)$$ for some constant $k>0$?

Now defining $y(t)=t^2$, what would be the probability?

Is there any literature on these types of problems? (I of course know how to compute the probability for the function $y(t)=constant$, but I haven't found any material for non-constant functions).

答:
Hint

On can prove the following proposition

If $(W_s)_s$ is a Brownian motion on a probability space $(\Omega ,\mathcal F, \mathbb P)$, then
$$
\mathbb P\left\{W_t\in \mathrm d x,\sup_{0\leq s\leq t}W_s\in \mathrm d y\right\}=\frac{2(2y-x)}{\sqrt{2\pi t^3}}\exp\left\{-\frac{(2y-x)^2}{2t}\right\}\boldsymbol 1_{(-\infty ,y]}(x)\,\mathrm d x\,\mathrm d y.
\label{1}\tag{1}$$

Recall Girsanov theorem

Let $(W_t)$ a Brownian motion and let $\theta \in L^2([0,T])$. Under the measure $$\mathbb Q(\mathrm d \omega )=\exp\left\{\int_0^T \theta _s\,\mathrm d W_s-\frac{1}{2}\int_0^T \theta _s^2\,\mathrm d s\right\}\mathbb P(\mathrm d \omega ),$$
the process $$\tilde W_t:=W_t-\int_0^t\theta _s\,\mathrm d s,\quad t\in [0,T],$$ is a Brownian motion.

By Girsanov theorem, under the measure $$\mathbb Q(\mathrm d \omega )=\exp\left\{W_t-\frac{1}{2}t\right\}\mathbb P(\mathrm d \omega )$$
the process
$$\tilde W_t:=W_t-t,$$
is a Brownian motion. Then

\begin{align*}
\mathbb P\{T_h>t\}&=\mathbb P\left\{\sup_{0\leq s\leq t}(W_s-s)<0\right\}\\
&=\mathbb E_{\mathbb Q}\left[\boldsymbol 1_{\left\{\sup_{0\leq s\leq t}\tilde W_s>0\right\}}\exp\left\{-\tilde W_t-\frac{3}{2}t\right\}\right].
\end{align*}
where $\mathbb E_{\mathbb Q}$ denote the expectation under the measure $\mathbb Q$. Using \eqref{1} allow you to conclude.

For $y(t)=t^2$, same idea using $$\mathbb Q(\mathrm d \omega )=\exp\left\{2\int_0^ts\,\mathrm d W_s-\frac{2}{3}t^3\right\}\mathbb P(\mathrm d \omega ),$$
and the process $$\tilde W_t=W_t-t^2.$$ But at the end, I'm not so sure that you can find a nice closed form as the previous case.


问:
Find the flux integral of the vector field $F=(0,0,-1)$ through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$

A uniform fluid flowing vertically downward (heavy rain) is described by the vector field $F=(0,0,-1)$. Find the total flow through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$.

b)Now consider $F=(-\frac{\sqrt{2}}{2},0,\frac{-\sqrt{2}}{2})$ and Find the total flow through the cone $z=\sqrt{x^2+y^2}$, $x^2+y^2 \leq 1$

Attempt a)
We shall use the expresion
$$\int {F dS}=\int \int_{D}F \cdot ( T_{u} \times T_{v})=\int \int_D (-Pg_x-Qg_y+R)dA$$
Notice that the region $S$ alredy was given $g(x,y)=z(x,y)=\sqrt{x^2+y^2}$
$$g_x=\frac{x}{\sqrt{x^2+y^2}}$$
$$g_y=\frac{y}{\sqrt{x^2+y^2}}$$
doing the substitution on the integral
$$\iint_{D}(-0g_x-0g_y+(-1))dA$$

Since $x^2+y^2 \leq 1$ we shall use polar cordinates

$$x=r \cos \theta $$
$$y=r \sin \theta$$
where $r\in [0,1]$ and $\theta \in [0,2\pi]$
Finally doing the integration
$$ \int_{0}^{2 \pi} \int_{0}^{1}-rdrd \theta=-\pi $$

Now for $b)

We shall use the expresion
$$\int {F dS}=\int \int_{D}F \cdot ( T_{u} \times T_{v})=\int \int_D (-Pg_x-Qg_y+R)dA$$
Notice that the region $S$ alredy was given $g(x,y)=z(x,y)=\sqrt{x^2+y^2}$
$$g_x=\frac{x}{\sqrt{x^2+y^2}}$$
$$g_y=\frac{y}{\sqrt{x^2+y^2}}$$
doing the substitution on the integral
$$\iint_{D}(\frac{\sqrt{2}}{2}g_x-0g_y+(\frac{-\sqrt{2}}{2}))dA$$
$$\frac{\sqrt{2}}{2} \iint_{D}(\frac{x}{\sqrt{x^2+y^2}}-1)dA$$
Usinng polar coordinates we get
$$\int_{0}^{2 \pi} \int_{0}^{1} (cos \theta-1)r dr d\theta=-\frac{\pi}{\sqrt{2}}$$
Are my answers right or probably i do a mistake applying the formula.

答:
Please note your normal vector is pointing upward. As per question, You need normal vector pointing downward through the cone surface. So the signs of your final answer will change otherwise it looks fine.

From cross product, you get $(-\frac{x}{\sqrt{x^2+y^2}}, -\frac{y}{\sqrt{x^2+y^2}}, 1)$.

Please see the third coordinate is positive that gives us upward direction but given we want downward direction, you should take the normal vector $(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}, -1)$.

Now please do the same calculations and you will get opposite sign in both your answers.

By the way, instead of converting to polar coordinates at the time of integration, you could also parametrize the cone as below -

$T(r, \theta) = (r \cos \theta, r \sin \theta, r) \,$ $(0 \leq r \leq 1, 0 \leq \theta \leq 2\pi)$

So you get $T_{\theta} \times T_{r} = \, (r \cos \theta, r \sin \theta, -r)$

Integral becomes $\iint_D F.(T_{\theta} \times T_{r}) \, dr \, d\theta$


问:
Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$
I need to compute a limit:

$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$

I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.

$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$

I've calculated several values of this function, and it seems to have a limit of $1$.

答:
Hint: Using the fact that $\frac {\sin x} x \to 1$ as $x \to 0$ verify that $ x\ln (\frac 1  2 \sqrt  x) <x\ln (2\sin\sqrt x+\sqrt x \sin (\frac 1 x)) <x \ln (3\sqrt x)$. Conclude that the limit is $e^{0}=1$.

[$x \ln x \to 0$ as $x \to 0+$]


问:
In a group $G$, if $a \in G$ is the only element such that its order is 2, prove that $G = C(a)$
(i) As $C(a) \leq G $, $x \in C(a) \implies x \in G$. Hence $C(a) \subset G$.

(ii) How to prove $G \subset C(a)$?

I tried to start with $a^2 = e$ and manipulate this to get $ax = xa$, but it doesn't work.

答:
Fun fact: in general, if $p$ is the smallest prime dividing $|G|$ and $N \unlhd G$ with $|N|=p$, then $N \subseteq Z(G)$.

Proof note that $G=N_G(N)$ and $G/C_G(N)$ can be embedded in Aut$(N) \cong C_{p-1}$ ($G$ acts by conjugation on $N$ with kernel $C_G(N)$). Since $p$ is minimal this implies $G/C_G(N)=\overline{1}$, whence $G=C_G(N)$, being equivalent to $N \subseteq Z(G)$.

Note that the uniqueness of your element $a$, implies $\{1,a\}$ is a normal subgroup of $G$.


问:
Explicit description of the elements of a local system
If I have a topological space $X$ with a universal cover $p:\tilde{X}\to X$ and the fundamental group $G=\pi_1(X)$, and (say) $A$ is some $G$-module and $A_{\tilde {X}}$ is the constant sheaf defined on this universal cover for $A$, then (as far as I can tell) this is supposed to be enough information to define a local system $\mathcal{A}$, which is a sheaf on on $X$ itself. But I can't make sense of any of the explicit definitions of that I can find for what that sheaf actually is. By which I mean, exactly what the elements of the image of some given open set under the sheaf are in terms of the constant sheaf on the universal cover.

As an example, and with some notation translating, Wikipedia's description of a local system seems to say that for any open subset $U$ of $X$, $\mathcal{A}(U)=\lbrace a\in A_{\tilde{X}}(p^{-1}(U)):g \circ a=g\cdot a \rbrace \text{ for every }g\in G$, but what does it mean to "compose" an element of $g$ with an element of $a$ here? I understand that $g$ can be thought of as just the deck transformation map $\tilde{X}\to \tilde{X}$, but $a$ is supposed to be an element of $A$, not any kind of map on $\tilde{X}$, right? Or, digging deeper, the reference that page seems to be getting that from is this paper, which appears to say around page 7 (again with some translation of the notation) that $\mathcal{A}(U)=\lbrace a \in A_{\tilde{X}}(p^{-1}(U)):\text{ for every } x\in p^{-1}(U) \text{ and }g\in G, a(g \cdot x) = g \cdot a(x) \rbrace$. This seems to once again assume that elements of $A_\tilde{X}$ are some kind of functions from $\tilde{X}$ to somewhere else, but that doesn't make any sense to me at all.

My (admittedly limited) understanding of (pre)sheaves is that they're supposed to be functors from the collection of all the open sets of the space to some other category, in this case to the category of $G$-modules, so how can some element of the image of a functor like that also be a map from an element of the open set the functor came from to a topological space?

答:
$\mathcal{A}$ is a sheaf of abelian groups (or vector spaces), not a sheaf of $G$-modules, and an easy way to answer your questions is that : yes sections of a sheaf can be seen as some kind of functions. Let me expand a bit on this and then take a look at the definitions you linked.

If you have a very general sheaf $\mathcal{F}$ on a space $X$, a section $s$ indeed define a "function" $x\mapsto s_x$. The source is $X$, but the target is not a single set, instead, for any $x\in X, s_x\in\mathcal{F}_x$. Yet these "functions" very much behave as usual functions, and in particular if two "functions" agree then their corresponding section are the same. I am just saying here that if for any $x\in X, s_x=t_x$ then $s=t$.

But here, you have much more : $A_{\tilde{X}}$ is a constant sheaf and the best way to define a constant sheaf is indeed as a sheaf of usual functions : $A_{\tilde{X}}(U)$ is the set of locally constant function $U\to A$. So if $s\in A_{\tilde{X}}(U)$ is any section $s(x)$ simply means the value of $s$ at $x$ and will be an element of $A$.

Note that the two descriptions are not incompatible since $A_{\tilde{X},x}$ is canonically isomorphic to $A$ and $s_x=s(x)$ using this identification.

So now, let us have a look at the two definitions of the local sheaf associated to a $\pi_1(X,x)$-module $A$. Well in fact this is exactly the same definition (but the wikipedia article contains an error : it should be $a\circ g$ instead of $g\circ a$).

As you said, any $g\in G$ can be seen as deck transformation $g:\tilde{X}\to\tilde{X}$, and for any $U\subset X$, it restricts to a transformation $g:p^{-1}(U)\to p^{-1}(U)$. If $a$ is a locally constant function on $p^{-1}(U)$ then $a\circ g$ is also locally constant and thus define a new section $a\circ g\in\mathcal{A}(U)$. You can also use the $G$-structure on $A$ : if $a$ is a locally constant fonction on $p^{-1}(U)$, then for any $x\in p^{-1}(U), a(x)$ is a element of $A$ so it makes sense to write $g.a(x)$. The correspondance $x\mapsto g.a(x)$ is a locally constant function written $g.a$. So finally : $$\begin{align}\mathcal{A}(U)&=\{a: p^{-1}(U)\to A \text{ locally constant}~|~ a\circ g=g.a\}\\
&= \{a: p^{-1}(U)\to A \text{ locally constant}~|~ \forall x\in p^{-1}(U),  a(g.x)=g.a(x)\}\end{align}$$

An example : let $X=S^1=\mathbb{R/Z}$ bet the circle and $p:\mathbb{R}\to S^1$ its universal cover. We have $G=\mathbb{Z}$ and let $A=\mathbb{Z}$. We define a $G$-module structure on $A$ by putting $g.a=(-1)^ga$ (recall that $a,g\in\mathbb{Z}$) (Actually, we don't have much choice for the module structure here, this is the only non trivial one).

Now let $\mathcal{A}$ be the local system on $S^1$ associated to this module structure. We basically have two kinds of open set for which we want to compute $\mathcal{A}(U)$ :

$U=S^1$ is the whole space. Then $p^{-1}(S^1)=\mathbb{R}$. A locally constant function $\mathbb{R}\to A$ is constant and completely determined by its image still denoted $a\in A=\mathbb{Z}$. The composition $a\circ g$ will always yield the same constant function $a$. However $g.a=(-1)^ga$ is not equal to $a$ for odd $g$ and non-zero $a$. It follows that the only $a$ such that for any $g, a\circ g=g.a$ is zero : $\mathcal{A}(X)=\{0\}$.

$U=(x,y)$ is a small interval, then $p^{-1}((x,y))=\coprod_{n\in\mathbb{Z}}(x+n,y+n)$ with no overlapping. Then a locally constant function $a:p^{-1}((x,y))\to A$ will be constant on any $(x+n,y+n)$ and thus corresponds to the data for all $n\in\mathbb{Z}$ of $a_n\in A$ the constant value of $a$ on $(x+n,y+n)$. Now $a\circ g=g.a$ here means that $a_{n+g}=(-1)^ga_n$ (since the deck transformation $g$ corresponds to the translation $+g:\mathbb{R\to R}$). It follows that $a_g=(-1)^ga_0$ and thus $a_n$ is completely determined by $a_0$. Hence $\mathcal{A}((x,y))=A$.

Hence we have $\mathcal{A}((0.1, 0.9))=\mathbb{Z}$ and $\mathcal{A}((0.8, 1.2))=\mathbb{Z}$ and the intervals $(0.1,0.9), (0.8, 1.2)$ cover $S^1$. Consider the two sections corresponding to 1 under the above identifications. Can you see that they do not glue ?


问:
Holomorphic series with its real part positive $f(z)=1+\sum_{n=1}^\infty a_n z^n$
Let $$f(z)=1+\sum_{n=1}^\infty a_n z^n,$$

$f \in H(B(0,1))$, and $\operatorname{Re} f(z)\ge 0$,
$\forall z \in B(0,1) $.

Prove:

(1) $| a_n | \le2$;

(2) $|a_1^2-a_2| \le 2, |2a_1a_2-a_1^3-a_3|\le2$.

答:
Some ideas:

a) Introduce $g(z)=(1-f(z))/(1+f(z))$. Show  that $|g|<1$ in the unit disk.

b) Do some algebra of power series (much like the long division of polynomials) to get
$$g(z)=-\frac{1}{2}a_1z +\frac14(a_1^2-2a_2)z^2-\frac18(a_1^3-4a_1a_2+4a_3)z^3+\dots \tag1$$

c) Conclude that $|a_1|\le 2$ from the Schwarz lemma.

d) Apply (c) to the function $f_n(z)=n^{-1}\sum_{k=0}^{n-1} f(\zeta^k z)$ where $\zeta=\exp(2\pi i /n)$.

The form of expansion (1) suggests it may be relevant for part (2), but you'll have to think more there.


问:
Univariate Kernel in Gaussian Process
I cant seem to get my head around the behaviour of different kernels in an univariate case when using a gaussian process regressor.

For example, this is the RBF kernel: $$K(x, x')= exp( - \frac{ ||x-x'||^2 }{ 2\sigma^2 })$$
In the univariate case $x' = x$, so the kernel becomes $K(x, x)=exp( -\frac{ ||x-x||^2 }{ 2\sigma^2 })$ which resolves to $K(x,x) = e^0$

So from what I know right now, every kernel function yields $1$ for every given value and still, changing the kernel parameters for a kernel changes the outcome of the prediction when using the gaussian process regressor.

I've tried this with scikit-learns GP implementation, using various kernels. Every kernel has the euclidean distance term in its exponent so it should result to 1 in the univariate case. But still, messing around with the various kernel parameters, changes the quality of the learned model.

Please enlighten me! Thanks in advance!

答:
In a single univariate, $x'$ need not be equal to $x$.

For example you can have $0$ and $1$ and $$K(0,1)=\exp\left(-\frac1{2\sigma^2} \right)$$

Changing $\sigma$ would change the kernel value.

Title: Univariate Kernel in Gaussian Process


问:
Integrating the Difference between Sine and its Maclaurin Series
The integral in question:

Let $n$ be a nonnegative integer.

$$
\int_0^\infty
\frac{1}{x^{2n+3}}
\left (
\sin x -
\sum_{k=0}^n \frac{(-1)^k x^{2k+1}}{(2k + 1)!}
\right )
\mathrm{d}x
$$

I was given this problem to try for fun from a professor about two months ago and haven't made a dent in it.
If possible, I'd like someone to show how they get to their answer, but please keep in mind that I haven't learned contour integration or complex analysis, but know of a few tricks like Feynman's.

What I've tried:

Writing out the terms to integrate by segments, using the Cosine Maclaurin series and substituting (pi/2 -x) to get the Sine series in a different way, integration by parts (2n+3) times or such to simplify, and a lot more I'm probably forgetting.

I know this site is full of integral-calculating gods, so if this piques your interest please have an attempt and share below.

答:
This can be done via Integration by Parts.
$$
\begin{align}
&\int_0^\infty\frac1{x^{2n+3}}\left(\sin(x)-\sum_{k=0}^n\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right)\mathrm{d}x\\
&=\frac{-1}{(2n+1)(2n+2)}\int_0^\infty\frac1{x^{2n+1}}\left(\sin(x)-\sum_{k=0}^{n-1}\frac{(-1)^kx^{2k+1}}{(2k+1)!}\right)\mathrm{d}x\tag1\\
&=\frac{(-1)^{n+1}}{(2n+2)!}\int_0^\infty\frac1x\,\sin(x)\,\mathrm{d}x\tag2\\[3pt]
&=\frac{(-1)^{n+1}}{(2n+2)!}\frac\pi2\tag3
\end{align}
$$
Explanation:

$(1)$: integrate by parts twice

$(2)$: repeat $(1)$ $n$ more times

$(3)$: see $(9)$ from this answer


问:
Computing Relative Homology
I'm sort of uncertain about how to compute relative homologies, so I would like some feedback on my work. As I understand it, you entirely use the idea that they form long exact sequences.

Suppose $X = S^1 \times [0, 1]$, the cylinder. And let $A = S^1 \times \{0, 1\}$ the upper and lower boundary circles. We want to compute $H_n(X,A)$ for every $n$.

So we use the idea that the homology groups form a long exact sequence $$...H_n(A) \rightarrow H_n(X) \rightarrow H_n(X, A) \rightarrow H_{n-1}(A) \rightarrow H_{n-1}(X) .....$$

Next since $X$ deformation retracts to a circle, $H_n(X) = \mathbb{Z}$ for $n = 1, 2$ and trivial otherwise. Similarly, as $A$ is a disjoint union of two circles, $H_n(A) = \mathbb{Z} \oplus\mathbb{Z}$ for $n = 1,2$ and trivial otherwise. This means that $H_{n}(X,A) = 0$ if $n \geq 4$. For $n = 3$,

$$.. 0 \rightarrow H_3 (X, A) \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$$

is what we obtain. Which means that $H_3(X, A)$ is a subgroup of $\mathbb{Z} \oplus \mathbb{Z}$. Moreover, the map $i_{*} : \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$ is induced by the inclusion $i : C_2(A) \rightarrow C_2(X)$.

However, I'm not sure which properties of $i_{*}$ are to be used to get what $H_3(X, A)$ is exactly. I'm facing similar problems for the other $H_n(X,A)$ ($n = 0,1,2$).

I think I could get it done after looking at a concrete example but such examples are surprisingly hard to find. Most examples, like the ones in Hatcher, are trivial and quick. I'm sure this one is easy as well and I'm just being an idiot but I would still appreciate the help.

答:
You have already received hints to do the computation using the homology LES. Here is another way to compute the relative homology group: if A is deformation retract of a neighborhood in X, then $H_n(X, A) \approx \widetilde{H}_n(X/A)$. In our case, this is indeed true as $S^1 \times [0, \frac12) \sqcup S^1 \times (\frac12, 1]$ is a neighborhood in $X$ that deformation retracts to $A$.

Observe that $X/A$ can be obtained by collapsing the two boundary circles. This is like a 2-sphere with its north and south poles identified, which is homotopy equivalent to $S^2 \vee S^1$. It should be easy to compute the reduced homology of this.

$H_0(X, A) = \widetilde{H}_0(S^2 \vee S^1) = 0$

$H_1(X, A) = \widetilde{H}_1(S^2 \vee S^1) = \mathbb{Z}$

$H_2(X, A) = \widetilde{H}_2(S^2 \vee S^1) = \mathbb{Z}$

$H_k(X, A) = \widetilde{H}_k(S^2 \vee S^1) = 0$ for $k \ge 3$.

Title: Computing Relative Homology


问:
Show that the ring $\Bbb C[x,y,z]/\langle 2x^2y-5xy^3+z\rangle $ is a UFD
I need to show that the ring
$\Bbb C[x,y,z]/\langle 2x^2y-5xy^3+z\rangle $
is unique factorisation domain.

I think this ring is isomorphic to $\Bbb C[x,y]/\langle 2x^2y-5xy^3\rangle $ so it's enough to show the later ring is a UFD. I am stuck at this part.

答:
Write $\Bbb C[x,y,z]=D[z]$, where $D=\Bbb C[x,y]$. Let $a=2x^2y-5xy^3 \in D$.

Then $\Bbb C[x,y,z]/\langle 2x^2y-5xy^3+z\rangle = D[z]\langle a+z\rangle \cong D$ is a UFD.


问:
Subextensions of Finitely Generated Fields
Let $K$ be an extension of a field $F$, and assume that there exist $\alpha_1,\dots,\alpha_n \in K$ such that $K=F(\alpha_1,\dots,\alpha_n)$. Let $E$ be a subfield of $K$ containing $F$.

(I) Do there exist $\beta_1,\dots,\beta_m \in E$ such that $E=F(\beta_1,\dots,\beta_m)$?

(II) Assume that the previous question has (in general or in some specific example) a positive answer. Do there exist $\gamma_1,\dots,\gamma_k \in E$ such that $E=F(\gamma_1,\dots,\gamma_k)$ with $k \leq n$?

Any help is welcome. Thank you very much for your attention.

Comments. I have only a basic knowledge of field theory, which does not allow me to answer these general questions.
The only thing which is clear to me is that if $[K : F] < \infty$, then (I) has trivially a positive answer, since in this case $[E : F] < \infty$, so if $\beta_1,\dots,\beta_m$ are a basis of the $F$-vector space $E$, then clearly $E=F(\beta_1,\dots,\beta_m)$. As a corollary, if we assume that $K$ is an algebraic extension, then (I) has a positive answer, since being $K=F(\alpha_1,\dots,\alpha_n)$, we have in this case $[K : F] < \infty$. I do not know what can happen when $[K : F] = \infty$. As for (II), I have no idea at all of the answer.

答:
For a finite extension $A/B$ then $A$ is finitely generated (as a field over $F$) iff $B$ is finitely generated. One direction is obvious. For the other one, if $A$ is finitely generated, then consider the extension $C$ generated (over $F$) by the coefficients of the $B$-minimal polynomials of the generators of $A$, then $A/C$ is a finite extension thus so is $B/C$, and hence $A$ finitely generated gives $C$ finitely generated thus $B$ finitely generated.

Going back to the fields in your question, let $L=E(a_{k_1},\ldots,a_{k_s})$ where $a_{k_1},\ldots,a_{k_s}$ is a minimal subset of the $a_i$ such that $K/L$ is algebraic. $K$ is finitely generated and algebraic over $L$, thus $K/L$ is a finite extension, whence $L$ is finitely generated, by finitely many rational functions in $s$ variables (the $a_{k_j}$) and coefficients in $E$. Let $D$ be the extension of $F$ generated by the (finitely many) coefficients appearing in those rational functions. Then $D=E$ (use induction in $s$ to make it rigorous)

For part II, I don't know. In characteristic $0$ we'll have that $E$ is a finite extension of a subfield of $K$ isomorphic to $F(t_1,\ldots,t_m)$ thus (by the primitive element theorem) $E\cong F(t_1,\ldots,t_m)[\alpha]$ is generated by $m+1\le n+1$ elements.

Is there an algorithm to check if a field like $\Bbb{Q}(x^3,y^3,x+y)$ is generated by 2 elements ?


答:
The ideal is the kernel of the (clearly surjective) ring homomorphism
$$\Bbb{C}[x,y,z]\ \longrightarrow\ \Bbb{C}[x,y]:\ f\ \longmapsto\ f(x,y,5xy^3-2x^2y),$$
so the quotient is isomorphic to $\Bbb{C}[x,y]$, which is a UFD.

Title: Show that the ring $\Bbb C[x,y,z]/\langle 2x^2y-5xy^3+z\rangle $ is a UFD

Body:
I need to show that the ring
$\Bbb C[x,y,z]/\langle 2x^2y-5xy^3+z\rangle $
is unique factorisation domain.

I think this ring is isomorphic to $\Bbb C[x,y]/\langle 2x^2y-5xy^3\rangle $ so it's enough to show the later ring is a UFD. I am stuck at this part.


答:
Hint.

For $x>0$ small we have

$$
\left(2\sin(\sqrt{x})-\sqrt{x}\right)^x\le \sigma(x)\le \left(2\sin(\sqrt{x})+\sqrt{x}\right)^x
$$

Title: Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$

Body:
I need to compute a limit:

$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$

I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.

$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$

I've calculated several values of this function, and it seems to have a limit of $1$.


问:
Primes in an extension over Dedekind domain
I am considering Dedekind domain $\mathbb{Z}[a]/(a^3-a-1)$. I consider the splitting of $23\in \mathbb{Z}$. I know from Daniel Marcas, that if $Q$ is a prime dividing $23$ in the extension, them $Q\cap \mathbb{Z}=(23)$. I need to prove $(23,a-3)$ is a prime dividing $23$, in $\mathbb{Z}[a]/(a^3-a-1)$. Clearly, $(23,a-3)\cap \mathbb{Z}=(23)$. So, does it proves that it is a prime over $23$?

答:
You need to prove that $(23,a-3)$ is a proper ideal (ie. not containing $1$). Then $(23,a-3)\cap \Bbb{Z}$ is a proper ideal of $\Bbb{Z}$ containing $23$ so it has to be $(23)$.

The proof will follow from that $\mathbb{Z}[a]/(a^3-a-1)/(23,a-3)\cong \Bbb{F}_{23}[x]/(x^3-x-1,x-3)\cong \Bbb{F}_{23}[x]/(x-3)\cong \Bbb{F}_{23}$. The latter is a field thus $(23,a-3)$ was a maximal ideal.


问:
Topological interior of a standard $n$-simplex
A topological standard $n$-simplex is a subset $\Delta^n = \{ (x_0,x_1,...,x_n) \in \mathbb{R}^{n+1} \mid x_i \geq 0, \sum_{i = 0}^n x_i = 1 \}$ of $\mathbb{R}^{n+1}$ endowed with the subspace topology. Simplices have their own definition of boundaries and interiors, but little is said about the topological interior and the topological boundary of $\Delta^n \subseteq \mathbb{R}^{n+1}$ (the closure is, of course, $\Delta^n$ itself since it is closed in $\mathbb{R}^{n+1}$).

Accepted answers to this question (and many similar proofs) implicitly assumes that it is nonempty since a theorem they invoke requires a convex subset of $\mathbb{R}^{n+1}$ to have a nonempty interior.

This question and the answer to it suggest that the interior of $\Delta^n$ to be the subset of those $(x_0,...,x_n)$ for which $x_i > 0$, though I'm not sure whether they are talking about a topological or a simplicial interior.

I've also heard that it is empty, but then it would render all proofs I've seens of the fact that $\Delta^n$ is homeomorphic to $D^n$ invalid, which I find unlikely since it appears to be a folk theorem, which rely on the following theorem:

Let $K \subseteq \mathbb{R}^n$ be a compact convex subset with nonempty interior. Then, for any interior point $p$ of $K$, there is a relative homeomorphism $(D^n,S^{n-1}) \to (K,\partial K)$ which maps $0$ to $p$.

About a year ago I tried to understand the issue, to no avail: googling provides little to none information on the matter. This time I hope to get to the bottom of this, and to provide this answer as a reference to all in my situation.

答:
I shall attempt to answer your first question on the topological interior of the standard $n$-simplex. More specifically, I will show that every point in
\begin{equation}
x\in\mathrm{Int}(\Delta^{n})=\left\{(x_{0},\dots,x_{n})\in\mathbb{R}^{n+1}:\sum_{j=0}^{n}x_{j}=1,x_{j}>0\right\}
\end{equation}
is contained in some open ball $B(x,\delta)$ for some $\delta>0$, and $B(x,\delta)\subseteq\mathrm{Int}(\Delta^n)$. I am in fact working with the above definition (as in the equation above) of the interior of a simplex.

Let $a=(a_{0},\dots,a_{n})\in\mathrm{Int}(\Delta^n)$. Let
\begin{equation*}
2\delta=\inf\left\{\Vert\alpha-a\Vert_{2}:\alpha\in\mathrm{Int}(\Delta^n),\alpha\text{ has exactly one coordinate }0,\text{ and the rest positive}\right\}
\end{equation*}
where $\Vert\cdot\Vert_{2}$ is the usual Euclidean norm. We need to show that $B(x,\delta)\subseteq\mathrm{Int}(\Delta^n)$. Let $y\in B(x,\delta)\cap\Delta^{n}$. Then we see that from the definition of $\delta$, we must have $y\in\mathrm{Int}(\Delta^n)$. This shows that the open set $B(x,\delta)\cap\Delta^n$ is completely contained in $\mathrm{Int}(\Delta^n)$, where $B(x,\delta)\subseteq\mathbb{R}^{n+1}$ is the open ball centered at $x$ of radius $\delta$.

To answer your second question on the topological boundary, it follows that
\begin{equation*}
\partial\Delta^{n}=\Delta^{n}\setminus\mathrm{Int}(\Delta^n)=\left\{(x_{0},\dots,x_{n})\in\mathbb{R}^{n+1}:\sum_{j=0}^{n}x_{j}=1,x_{j}=0\text{ for some }j\right\}
\end{equation*}

Remark: Note that I am working with the topology on $\mathbb{R}^{n+1}$ induced by the usual Euclidean metric, and hence working with the subspace topology on the $n$-simplex $\Delta^n$ induced by the (Euclidean) metric topology on $\mathbb{R}^{n+1}$.

Any comments are welcome.


问:
Convergence in distribution space and tempered distribution space
I can neither prove nor disprove the following claim.

Let $u_k \in \mathcal{S}'(\mathbb{R})$ and $u_k \to 0$ in $\mathcal{D}'(\mathbb{R})$.
Then, $u_k \to 0$ in $\mathcal{S}'(\mathbb{R})$.

I expect this is false.
Define $u_k := \frac{d^k\delta_k}{dx^k} \in \mathcal{S}'(\mathbb{R})$, i.e. $\langle u_k, \varphi \rangle := (-1)^k\frac{d^k\varphi}{dx^k}(k)$ for all $k \in \mathbb{N}$.
Then, it is clear that $u_k \to 0$ in $\mathcal{D}'(\mathbb{R})$. I expect $u_k \nrightarrow 0$ in $\mathcal{S}'(\mathbb{R})$, however, I cannot prove it.

I would be grateful if anyone could give me some advice.

答:
$\frac1k e^{t^2} 1_{|t|< k^2}$ converges to $0$ in $D'$ but not in $S'$ (pairing with $e^{-x^2}$).

It is true that
$\delta^{(k)}(x-k)$ doesn't converge to $0$ in $S'$, take $\phi \in C^\infty_c$ constant $=1$ on $[-1/2,1/2]$ and $=0$ on $|x|>3/2$, let $\psi(x) = \frac{\phi(x)}{1-x}$ and look at $\varphi(x)=\sum_{k\ge 0}2^{-k} \psi(x-k)$, all the derivatives are rapidly decreasing, but $\varphi^{(k)}(k)=2^{-k}\psi^{(k)}(0) = 2^{-k}k!$


问:
An example of posterior distribution
I am learning statistics myself. The question might be trivial but it's not a homework problem.

Let $n$ be a positive integer, and $k$ be a smaller one. Suppose $X_1, X_2, \cdots, X_n$ are IID Bernoulli random variable $B(\theta)$ with an unknown $\theta$. Let $X$ be $\frac{X_1 + \cdots X_n}{n}$. Suppose $k$ is the observed $X$. Then

$$P(X=\frac{k}{n}\,|\,\theta) = {n \choose k} \theta^k (1-\theta)^{n-k} \, [k \in \{0 \cdots n\}]. $$

Fix $a \in [0,1]$. I'm interested in calculating how likely the underlying model $\theta$ is $a$. I hope to do it by the Bayesian formula:

$$
P(\theta = a| X=k) = \frac{P(X=k|\theta = a) P(\theta=a)}{P(X=k)}.
$$

However, I cannot think of what $P(X = k)$ and $P(\theta = a)$ are. Am I missing anything?

答:
Your model is Bernoulli. Thus the likelihood you observe will be

$$p(\mathbf{x}|\theta)\propto \theta^k(1-\theta)^{n-k}$$

The posterior distribution is the following

$$\pi(\theta|\mathbf{x}) \propto\pi(\theta) p(\mathbf{x}|\theta)$$

To derive your posterior now you have some different choices to set your prior

You have no information about the parameter so it can take any values in $[0;1]$ with the same probability: $\pi(\theta)=1$ and your posterior is

$$\pi(\theta|\mathbf{x}) \propto \theta^k(1-\theta)^{n-k} \sim Beta(k+1;n-k+1)$$

That is

$$\pi(\theta|\mathbf{x})=\frac{\Gamma(n+2)}{\Gamma(k+1)\Gamma(n-k+1)}\theta^k(1-\theta)^{n-k}$$

$\theta \in [0;1]$

You have some information about the values of the parameter...it is more probabily that $\theta \in A \subset [0;1]$ so you can use a conjugate prior and do the same calcualtion as in 1.

You can use a non informative prior (EDIT: not exactly what Jeffrey proposed) that in this case can be in the form

$$\pi(\theta)\propto \frac{1}{\theta(1-\theta)}$$

and your posterior becomes

$$\pi(\theta|\mathbf{x})\propto\theta^{k-1}(1-\theta)^{n-k-1}$$

Which is always a beta with differen paramenters:

$$\pi(\theta|\mathbf{x})\sim Beta(k,n-k)$$

in this case, your posterior is a density only if $0<k<n$ because in the extremes situation of no successes or all successes, the posterior is not integrable

et cetera et cetera

Title: An example of posterior distribution


问:
Critical points of a square function.
What are all the critical points of $ (x-y)^2? $? How can we prove that they are saddle points?

I'm confused at what they will be because the partial derivative is x=y, so is there infinite critical points?

答:
Let $f(x,y)=(x-y)^2$. Then $f \ge 0$ on $ \mathbb R^2$ and $f(x,x)=0$

Conclusion: in each critical point $(x,x)$ the function $f$ has an absolute minimum and $(x,x)$ is not a saddle point.

Title: Critical points of a square function.


问:
Simplify $xyz’w’+xy’zw+xyz’w+xy’z’w’+xy’z’w+zw’xy’$

Simplify $xyz’w’+xy’zw+xyz’w+xy’z’w’+xy’z’w+zw’xy’$

( ` = not)

Using a karnaugh map I found that it's equal to $xy'+xz'$, but I can't find how to get there with simple algebraic simplifications.
Help would be appreciated.

答:
Let
$$f(x,y,z,w) = xyz'w + xyz'w' + xy'zw + xy'zw' + xy'z'w + xy'z'w'.$$
Then,

\begin{align}
f(x,y,z,w)
&= x(yz'w + yz'w' + y'zw + y'zw' + y'z'w + y'z'w')\\
&= x( yz'(w + w') + y'z(w + w') + y'z'(w + w') )\\
&= x( yz' + y'z + y'z')\\
&= x( (y + y')z' + y'(z + z') )\\
&= x(z' + y').
\end{align}
The steps are justified by distributivity (which apparently you are already used to) and the laws:
$$a+a'=1\quad\text{and}\quad a1=a,$$
for each $a$, and the usual idempotence, commutativity and associativity.

I suppose you can fill in these justifications.


Body:
$$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... $$

$$\therefore \ (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$

$$ = \left(2 \left(x^{1/2} - \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} -...\right) + x^{1/2} \sin \frac{1}{x}\right)^x$$

$$ = \left(x^{1/2} \left(2 + \sin \frac{1}{x} - \frac{2x^{3/2}}{3!} + \frac{2x^{5/2}}{5!} -...\right) \right)^x$$

$$ = \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x.$$

Now $ \left(x^{1/2}\right)^x = \left(x^{x}\right)^{1/2} = (e^{x \ln x})^{1/2} = e^{\frac{1}{2} x \ln x},\ $ and $\ \lim\limits_{x\rightarrow 0^{+}}e^{\frac{1}{2}x\ln(x)}=e^{\lim\limits_{x\rightarrow 0^{+}}\frac{1}{2}x\ln(x)}=1. $

Lastly, for $ \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x,\ $  make the substitution $u = \frac{1}{x}$ to get:

$\lim\limits_{u\rightarrow \infty} \left( 2 + \sin u - \frac{2}{3!u} + \frac{2}{5!u^2} - ...\right)^{\frac{1}{u}}$, and the bracket oscillates between $1$ and $3$ for large $u$, so this limit is $1$. Thus:

$$ \lim\limits_{x\rightarrow 0^{+}} \left(x^{1/2}\right)^x \left(2 + \sin \frac{1}{x} - \frac{2x}{3!} + \frac{2x^2}{5!} -...\right)^x = 1 \times 1 = 1.$$


问:
Dominant eigenvalue definition -- is this paper wrong?
I have been reading this paper by Neubert et al. (2004) on calculating reactivities of dynamical models. On page 31, equation 11, they define the two matrices:

\begin{equation}
\begin{pmatrix}
-1 & 0 \\
0 & -10\\
\end{pmatrix}
\end{equation}

\begin{equation}
\begin{pmatrix}
-1 & 15 \\
0 & -10\\
\end{pmatrix}
\end{equation}

and claim that both matrices "have the same dominant eigenvalue" of $\lambda_{1} = -1$.

It was my understanding that the dominant eigenvalue has the be the greatest in absolute value (i.e. magnitude), which would make $\lambda_{1} = -10$.

Can anyone clear this up for me? Thanks

答: You are right. If $\lambda_1,...., \lambda_n$ are the eigenvalues of a matrix with $|\lambda_1| \ge |\lambda_2| \ge ....|\lambda_n|$, then $\lambda_1$ is the dominant eigenvalue.


问:
if $\sin24^\circ = p$ what is $\cos24^\circ$?
Let $p=\sin 24^\circ$

Then what would $\cos (24^\circ)$ be in terms of $p$?

What would $\sin (168^\circ) \cdot  \sin(-78^\circ)$ be in terms of $p$?

I'm not sure how to approach these as we have only started this section.

Thanks.

答:
Part 2

Method 1:

$\sin168^\circ=\sin(180^\circ-168^\circ)$ ( as $\sin(180^\circ-\theta)=\sin\theta$)

So, $\sin168^\circ=\sin12^\circ$

and $\sin(-78^\circ)=-\sin78^\circ$ as $\sin(-\theta)=-\sin\theta$

So, $\sin(-78^\circ)=-\sin(90^\circ-12^\circ)=-\cos12^\circ$

$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\sin12^\circ\cdot(-\cos12^\circ)=-\frac{\sin (2\cdot12^\circ)}2=-\frac{\sin24^\circ}2$$

Method 2:

$\sin168^\circ=\sin(90^\circ+78^\circ)=\cos78^\circ$ as $\sin(90^\circ+\theta)=\cos\theta$

Again, $\cos78^\circ=\cos(-78^\circ)$ as $\cos(-\theta)=\cos\theta$

$$\implies \sin168^\circ\cdot\sin(-78^\circ)=\cos(-78^\circ)\cdot\sin(-78^\circ)=\frac{\sin 2(-78^\circ)}2$$
$$=\frac{\sin(24^\circ-180^\circ)}2=-\frac{\sin24^\circ}2$$ as $\sin(\theta-180^\circ)=-\sin(180^\circ-\theta)=-\sin\theta$


问:
Examples of irreducible holomorphic function in more than one variable.
I'm studying analysis in several complex variables and in particular Weierstrass preparation theorem caught my interest (I'll include the theorem for clarity).

In the examples I came up with I only found functions that could be written as the product of an unit and linear Weierstrass polynomials.

It's never implied that every Weierstrass irreducible polynomial is of degree 1 or that in general irreducible functions are the degree 1 polynomial so I was looking for some examples of holomoprhic functions that are irreducible (at the origin) but are not polynomials of degree 1.

答:
I actaully don't remember about details on why this is but I've been told in several occasions that polynomials like $z^2-xy^2$ are irreducible.
I hope this helps


问:
How does the proof of the Second Isomorphism theorem of rings prove this theorem?
The theorem is :

Let $I$ be an ideal of the ring $R$.
Then there is a one-to-one correspondence between the subrings of $R/I$ and the subrings of $R$ containing $I$, given as follows: to a subring $S$ of $R$ containing $I$ corresponds the subring $ S/I $ of $R/I$. Under this correspondence, ideals of $R/I$ correspond to ideals of $R$ containing $I$; and, if $J$ is an ideal of $R$ containing $I$, then
$(R/I)/(J/I) \cong R/J$

I understood everything in the proof, however they prove it as if it were an equivalence, and not a bijection, what am I missing?
The equivalence would be
$$ (\exists S)(S\leq R\ \land\ I\subset S\ \land\ A=S/I)\quad \iff\quad A\leq R/I.$$
But in the theorem they stated a function like this:
$$ \Phi : \{S:S \leq R \land I \subset S \} \mapsto \{A: A\leq R/I \} $$

Would be a bijection. I've seen this in many points of view, asked my colleagues, teacher, older colleagues and didn't get satisfied with any answer.

I tried to see it, the implication in this direction $\rightarrow$ prove that it's well defined and $\leftarrow$ proves that it's surjective, but where is injectivity?
Another way is, $\rightarrow $ proves that it's a function and $\leftarrow$ proves that it's inverse is also a function, so it's a bijection, but I'm not so conviced with this argument, could someone help me?
Can I prove bijections with equivalences?

答:
On its own, given a ring $R$ and an ideal $I\subset R$, the logical statement
$$S \leq R\quad \land\quad I \subset S\qquad \iff\qquad S/I \leq R/I,$$
doesn't quite make sense; if the left hand side is not satisfied, then the right hand side may be undefined. What is $S/I$ supposed to mean if $I\not\subset S$? Or worse, if $S\not\leq R$? (I take it that $S\leq R$ means that $S$ is a subring of $R$.)

As you suggest in the comments, you might instead prove the two equivalences
\begin{eqnarray*}
&&S \leq R\ \land\ I \subset S\ &\Longrightarrow&\qquad S/I \leq R/I,\\
(\exists S)(&&S\leq R\ \land\ I\subset S\ \land\ A=S/I)\qquad &\Longleftarrow&\qquad \hspace{9pt}A\leq R/I.
\end{eqnarray*}
The first is equivalent to $\Phi$ being well-defined, the second is equivalent to $\Phi$ being surjective. Neither shows that $\Phi$ is injective; this is equivalent to the $S$ in the second implication being unique. So the claim that $\Phi$ is a bijection is equivalent to the implication
$$(\exists!S)(S\leq R\ \land\ I\subset S\ \land\ A=S/I)\qquad \Longleftarrow\qquad A\leq R/I.$$


问:
Puzzlement about a step in WW Rouse Ball's explanation of Hero's proof
I have a 4th edition copy of "A Short Account of the History of Mathematics" by W. W. Rouse Ball.  This work, written in the late 1800's (the 4th edition dates to the early 1900's) is a highly respected history, and contains accessible explanations of many historically important results.

In his explanation of how Hero derived the formula $\sqrt{s(s-a)(s-b)(s-c)}$ for the area of a triangle with sides $a, b, c$ and semi-perimeter $s$, on page 90 in chapter 4, Ball sets up a triangle $ABC$ with inscribed circle centered at $O$, touching side $AB$ at point $F$.  He then defines point K as the meeting of two lines:  The line passing through $C$ perpendicular to side $BC$, and the line passing through $O$ perpendicular to $BO$.  Then the following statement is made:

He then shews (sic) that the angle $OAF = $ angle  $CBK$; hence the triangles $OAF$ and $CBK$ are similar.

(Note that "shews" rhymes with "sews," not with "news.")  The formula then follows from a sequence of ratio manipulations.

I cannot prove or even convince myself that those two angles are equal.  I don't restrict the proof to classical geometry; I have tried the usual vector tricks (which for example easily demonstrate that if side $BC = \vec{u}$ and $BA = \vec{v}$ the angle bisectors meet at a point which differs from $B$ by $\frac{\vec{u}+\vec{v}}{3}$).  I have also tried using analytic geometry; the coordinates of $K$ get pretty messy and I was unable to "shew" that the two discussed angles are equal.

Could it be that this respected author simply erred?  I doubt it, because the subsequent ratio manipulations would then hint that Hero's formula does not work.

So I am looking for any simple demonstration that those two angles are equal.

答:
As @YNK observes, the key is that, because $O$ and $C$ subtend equal angles with $\overline{BK}$, we have $\square OBKC$ is cyclic. Here's an alternative angle chase to the result:

$$\angle CBK \underbrace{=}_{\text{Insc}\angle\text{Thm}} \angle COK \underbrace{\;=\;}_{\triangle OBC} 180^\circ - \left(\tfrac12B+\tfrac12C+90^\circ\right)=\tfrac12(180^\circ-B-C)=\tfrac12 A$$


问:
Proving function for Stirling Numbers of the Second Kind
I need to proof the following formula for Stirling Numbers of the Second Kind:

$\sum\limits_{n \geq 0} S(n,k) x^n = \frac{x^k}{(1-x)(1-2x)\cdots(1-kx)}$

It is used widely around formularies but I neither have found any proof nor was I able to figure it out on my own.

Could you please help me finding a solution?

Thank you in advance!

答:
Hmm, I cannot see the problem which came up in the comments. If we assume simply an error  in the notation and that the actually the Stirling numbers 2'nd kind are meant (as verbally exposed in the question) then the identity holds. This can even be checked simply using negative integer $x$ and computing Eulersums of appropriate order.

Let's write S2 the matrix of that numbers as

$ \qquad \small
\begin{array} {rrrrrrr}
 1 & . & . & . & . & . & . & . \\
 1 & 1 & . & . & . & . & . & . \\
 1 & 3 & 1 & . & . & . & . & . \\
 1 & 7 & 6 & 1 & . & . & . & . \\
 1 & 15 & 25 & 10 & 1 & . & . & . \\
 1 & 31 & 90 & 65 & 15 & 1 & . & . \\
 1 & 63 & 301 & 350 & 140 & 21 & 1 & . \\
 1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1 \\
 \ldots &
 \end{array}
$

where we use zero-based row- and columnindexes.

Then the problem can be restated as summing by building the dot product of one column by one row vector $V(x) = [1,x,x^2,x^3,...]$ with manageable (ideally infinite) dimension.

The numbers along a column can be seen as composed by finite compositions of geometric series.

Column 0 is $[1,1,1,1,1,...]$ and the dot-product with the V(x)-vector is then
$V(x)*S2[,0] = {1 \over 1-x}$

Column 1 is $[1-1,2-1,4-1,8-1,16-1,...]$ and the dot-product with the V(x)-vector is then
$V(x)*S2[,1] = {1 \over 1-2x} - {1 \over 1-x} = { (1-1x) - (1-2x) \over (1-1x)(1-2x) }  = {x \over (1-1x)(1-2x) }$

One needs the simple composition of the other columns (see for instance in wikipedia) to see more examples for that decompositions, and also a general description for that compositions (where the text is:"Another explicit expanding of the recurrence-relation(...)").

I think the idea behind that homework-assignment was, that the student should find the compositions of powers such that the problem is ocnverted to describe the (finite) composition of closed forms of geometric series.


问:
Is $x\in\emptyset$ defined?
We know, $A\cup B=\{x:x\in A\lor x\in B\}$

In this definition, however if A or B is empty or both A and B are empty, we are getting the term $x\in \emptyset$
This poses the problem of defining that term, which I tried by saying that if $x\in \emptyset$ then $x=null$. But, it just doesn't work because then I am just allowing x as a variable to have a 'null state' which means if $x\in \{1\}$, then it means that $x=1$ or $x=null$.

This same problem also happened while defining subsets, but then rather than defining what is a subset I ended up defining what is not and so avoided using $x\in \emptyset$

A is not subset of B if there is a x such that $x\in A$ and $x\notin B$, otherwise A is subset of B.

答:
Let's take the proposition again:

$A\cup B=\{x:x\in A\lor x\in B\}$

It means that $\mathbf A$ union $\mathbf B$ is a set, which only has all the element(s) which belong to $\mathbf A$ or to $\mathbf B$.
This one clearly defines the case when only one of the set is empty. For example, if $A$ is empty and $B=\{1\}$ then no element belongs to $A$, but $1$ belongs to $A$ or to $B$, so $1$ belongs to their union.

Now, if both are empty then we can say that no element belongs to $A$ or to $B$. If we assume that still $A\cup B$ has element(s), then it clearly becomes case of contradiction. That leaves us with only one option, that is $A \cup B$ is empty set. Now, if the union is empty set does it follow the second condition that it only has all the element(s) which belong to $\mathit A$ or to $\mathit B$. This was a little surprise to me since it does follow that condition, we can see it from the article on vacuous truth.

That means that the element of an empty set do have all property including the fact that it belongs to $A$ or $B$. This is because in first place there is no element of an empty set.

If we see the term $x\in \emptyset$, it has always truth value false. So, no it isn't defined, and neither do we need to define it.

$A=\{x \mid ...\}$, here '$A$ is a set which only has all the element(s)  such that the element(s) have ...' is the interpretation I used above. This interpretation needs a person to know about vacuous truth, to allow for $A$ to be empty set. However, there is another interpretation that ' $A$ is a set, and an element belongs to this set if and only if the element has ....'. This makes the interpretation of union as '$\mathbf A$ union $\mathbf B$ is a set, and an element belongs to this set if and only if the element belongs to $\mathbf A$ or to $\mathbf B$' , and we don't need a person to know about vacuous truth.

It's no doubt that the similar problem also occured to me while defining subsets. However what I did to define subsets is kind of interesting, because now I have realised that I can similarly define what is union by simply defining which element it should not contain rather than defining which element it should contain as:

$\mathbf A$ union $\mathbf B$ is a set which don't has an element which don't belong to both $\mathbf A$ and to $\mathbf B$.

Acknowledgement: I want to acknowledge the answer by Eric Towers, it helped me a lot.

Title: Is $x\in\emptyset$ defined?


答:
Let us begin with notation needed below: By $e_i^m$, $i = 0,\ldots,m-1$, we denote the standard basis vectors of $\mathbb R^m$.

The topological interior of $\Delta^n$ in $\mathbb R^{n+1}$ is empty. To see that, take any $x \in \Delta^n$ and any $r > 0$. Let $y = x + \frac{r}{2}e_0$. Then $y \notin \Delta^n$ and $\lVert y - x \rVert =  \frac{r}{2}$. This shows that $B_r(x) = \{ y \in \mathbb R^{n+1} \mid \lVert y - x \rVert < r \} \not\subset \Delta^n$.

The $i$-th face of  $\Delta^n$ is defined as
$$\partial_i \Delta^n = \{ x \in \Delta^n \mid x_i = 0 \} .$$
Let $p_i : \mathbb R^{n+1} \to \mathbb R^n$ denote the projection omitting the $i$-the coordinate from $x = (x_0,\ldots,x_n)$ which is clearly continuous. It is easy to see that $p_i$ maps $\partial_i \Delta^n$ bijectively onto $\Delta^{n-1}$, thus establishes a homeomorphism $\bar p_i : \partial_i \Delta^n \to \Delta^{n-1}$.

The simplicial boundary of $\Delta^n$ is
$$\partial \Delta^n = \bigcup_{i=0}^n \partial_i \Delta^n = \{ x \in \Delta^n \mid x_i = 0 \text{ for some } i \} .$$

You should draw pictures for $n =1, 2$ to see what is going on here. The simplicial interior of $\Delta^n$ is
$$\mathring{\Delta}^n = \Delta^n \setminus \partial \Delta^n = \{ x \in \mathbb R^{n+1} \mid \text{ All } x_i > 0 , \sum_{i=0}^n x_i = 1 \} .$$

You are right, the proofs in your links do not really work.

However, it is easy to see that $\mathring{\Delta}^n$ is the topological interior of $\Delta^n$ in the hyperplane $H \subset \mathbb R^{n+1}$ described by the equation $\sum_{i=0}^n x_i = 1$, and thus $\partial \Delta^n$ is the topological boundary of $\Delta^n$ in $H$. To see this, observe that $O  = \{ x \in \mathbb R^{n+1} \mid \text{ All } x_i > 0 \} = (0,\infty)^{n+1}$ is open in $\mathbb R^{n+1}$ and therefore  $\mathring{\Delta}^n = O \cap H$ is open in $H$. Moreover, no $x \in \partial \Delta^n$ is an interior point of $\Delta^n$ in $H$. In fact, $x \in \partial \Delta^n$ means that some $x_j = 0$. Pick any $k \in \{0,\ldots,n\} \setminus \{j\}$. For any $r > 0$ the point $y = x - \frac{r}{2}e_i^{n+1} + \frac{r}{2}e_k^{n+1}$ is contained in $H$. We have $y_i < 0$, thus $y \notin \Delta^n$. Also $\lVert y - x \rVert = \sqrt{2(\frac{r}{2})^2} = \frac{r}{\sqrt 2} < r$ so that $B_r(x) \cap H \not\subset \Delta^n$.

The map $h : H \to \mathbb R^n, h(x_0,\ldots,x_n) = (x_0,\ldots,x_{n-1})$, is a homeomorphism (its inverse being $(x_0,\ldots,x_{n-1}) \mapsto (x_0,\ldots,x_{n-1},1 - \sum_{i=0}^{n-1} x_i)$). Considering the compact convex set $h(\Delta^n) \subset \mathbb R^n$ which has non-empty interior in $\mathbb R^n$ reduces everything to what has been proved in your links.

You can also proceed as follows. $\Delta^n$ is the convex hull of the $n+1$ standard basis vectors $e_i^{n+1}$ of $\mathbb R^{n+1}$. Thus $h(\Delta^n)$ is the convex hull of the $n+1$ vectors $v_i = h(e_i^{n+1})$. But $v_i = e_i^n$ for $i < n$ and $v_n = 0$. Drawing a picture is helpful. Now have a look at my answer to Homeomorphism between $k$-simplex and a product of $k$ unit intervals. There I denoted $h(\Delta^n)$ by $\Delta^n$, so please be not be confused. Anyway, I constructed a homeomorphism $h(\Delta^n) \to I^n$. It should be well-known that $I^n$ is homeomorphic to $D^n$.

Title: Topological interior of a standard $n$-simplex

Body:
A topological standard $n$-simplex is a subset $\Delta^n = \{ (x_0,x_1,...,x_n) \in \mathbb{R}^{n+1} \mid x_i \geq 0, \sum_{i = 0}^n x_i = 1 \}$ of $\mathbb{R}^{n+1}$ endowed with the subspace topology. Simplices have their own definition of boundaries and interiors, but little is said about the topological interior and the topological boundary of $\Delta^n \subseteq \mathbb{R}^{n+1}$ (the closure is, of course, $\Delta^n$ itself since it is closed in $\mathbb{R}^{n+1}$).

Accepted answers to this question (and many similar proofs) implicitly assumes that it is nonempty since a theorem they invoke requires a convex subset of $\mathbb{R}^{n+1}$ to have a nonempty interior.

This question and the answer to it suggest that the interior of $\Delta^n$ to be the subset of those $(x_0,...,x_n)$ for which $x_i > 0$, though I'm not sure whether they are talking about a topological or a simplicial interior.

I've also heard that it is empty, but then it would render all proofs I've seens of the fact that $\Delta^n$ is homeomorphic to $D^n$ invalid, which I find unlikely since it appears to be a folk theorem, which rely on the following theorem:

Let $K \subseteq \mathbb{R}^n$ be a compact convex subset with nonempty interior. Then, for any interior point $p$ of $K$, there is a relative homeomorphism $(D^n,S^{n-1}) \to (K,\partial K)$ which maps $0$ to $p$.

About a year ago I tried to understand the issue, to no avail: googling provides little to none information on the matter. This time I hope to get to the bottom of this, and to provide this answer as a reference to all in my situation.


问:
The conjugate function of the $e^x$

The conjugate function of the $f(x) = e^x$.

The conjugate function is defined as $f^*(y) = \sup_{x \in \operatorname{dom}(f)}\{x^Ty-f(x)\}$. In this case we should find the supremum of the $xy-e^x$ over the domain of $f(x)$ that is over $\mathbb{R}$. Now $e^x$ grows faster than $yx$ so if $y > 0$ then $xy-e^x$ will go to the $-\infty$ as $x \to \infty$. At the same time if $y < 0$ we will have as $x \to \infty$ that $xy-e^x$ goes to $-\infty$. If $y > 0$ and $x \to -\infty$ we will have $-\infty$ since $e^x \to 0$ in that case. If $y < 0$ and $x \to -\infty$ we will have $\infty$ for the same reason. If $y = 0$ we will have $\infty$.

答:
IA AM ASSUMING that $f^*(y):=\sup\{xy-f(x):\, x\in {\rm Dom}\,(f)\}$.

For $y>0$, take $g_y(x)=xy-e^x$. Then $g_y'(x)=y-e^x=0$ $\iff$ $x=\ln y$ and it is a MAXIMUM ($g_y'(x)>0$ if $x<\ln y$ and $g_y'(x)<0$ is $x>\ln y$).

So,  for $y>0$, one has that
$$f^*(y)=g_y(\ln y)=y\ln y-e^{\ln y}=y(\ln y-1).$$

If $y=0$, it is clear that $f^*(y)=+\infty$ (it does not exist) since $\lim_{x\to+\infty}e^x=+\infty$.

If $y< 0$, $xy-e^x\le0$ and $\lim_{x\to-\infty}(xy-e^x)=+\infty-e^{-\infty}=+\infty$, so $f^*(y)=+\infty$ (so, it does not exists).

Title: The conjugate function of the $e^x$


问:
In this probability question about counting using partitions and the multinomial coefficient in a probability question, are the events disjoint?
I designed a question where there are 12 employees, and they are to be assigned to 3 distinct teams (let's call them team 1, 2 and 3). Suppose the twelve people are $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$, $X$, $X$, and $X$ (for brevity I'm not adding the whole question here but this is its basic design).

I wanted to know how many different teams can be formed such that all the $X$'s are in one team. Then I wanted to find the probability that all $X$'s go into one team.  I started out grouping the $X$'s in one team at a time, and using the concept of partitioning the employees into teams. I'll summarize my approach below:

Let's define the events $E_{X,n}=\text{\{All X's go to team n\}},n=1,2,3$.

$$\text{Total number of different teams of those employees} = \binom{12}{4,4,4} = |\Omega|$$

$$\text{Number of different arrangements of all employees in teams such that all X's go to team 1}=1\times \binom{9}{1,4,4} = |E_{X,1}|$$

$$\text{Probability that all X's go into team 1} = \frac{|E_{X,1}|}{|\Omega|}$$

$$\text{Number of different arrangements of all employees in teams such that all X's go to team 2}=1\times \binom{9}{1,4,4}=|E_{X,2}|$$

$$\text{Probability that all X's go into team 2} = \frac{|E_{X,2}|}{|\Omega|}$$

$$\text{Number of different arrangements of all employees in teams such that all X's go to team 3}=1\times \binom{9}{1,4,4} = |E_{X,3}|$$

$$\text{Probability that all X's go into team 3} = \frac{|E_{X,3}|}{|\Omega|}$$

since there is only one way to combine all the $X$'s into any team, and the rest can join in any order.

I can't figure out of all the $E_{X,n}$ are disjoint or not, because I don't know what the sample space will be like.

答:
As you solved, the probability $P(A) = \displaystyle \frac{3 \times \binom{9}{1,4,4}}{\binom{12}{4,4,4}} = \frac{3}{55}$.

The other way to look at it is that whichever team first $X$ gets assigned to, there are $3$ more places on that team. The probability that remaining two $X$ will get $2$ of those $3$ seats $= \displaystyle \frac{3}{11} \times \frac{2}{10} = \frac{3}{55}$.


问:
How to solve a second order BVP with functions as boundary values?
Problem:
$ u_{tt} = 16 u_{xx}, 0 \leq x \leq 1
\\u(0,t) = u(1,t) = 0, t \geq 0
\\u(x,0) = \sin(3 \pi x)
\\u_{t}(x,0) = x(1-x^2)
$

Attempt:
$u = X(x)T(t)
\\ X(x)T''(x)=16X''(x)T(t)
\\ \frac{X''(x)}{X(x)} = \frac{T''(t)}{16T(t)} = -\lambda $

I first got this general equation for X: $ X(x) = C_{1}\cos(\sqrt{\lambda} x) + C_{2}\sin(\sqrt{\lambda} x) $
Plugging in the initial conditions $u(0,t)$ and $u(1,t)$, I got $ X(x) = C\sin(n\pi x)$

I then got this general equation for T: $ T(t) = C_{1}\cos(4\pi nt) + C_{2}\sin(4\pi nt) $

However, I don't know how to proceed from here. When I plug in $t = 0$ into $u(x,t)$, I can't find a way to solve any variables

$ u(x,0) = X(x)T(0) = C\sin(n\pi x) * C_{1} = \sin(3\pi x)$

I'm lost and I feel like I don't have the right approach

答:
The separation of variables solution starts with
$$
                u(x,t)=X(x)T(t).
$$
Then
$$
                u_{tt}=16u_{xx} \implies X(x)T''(t)=16X''(x)T(t) \\
             \implies \frac{T''}{16T}=\lambda = \frac{X''}{X} \\
             \implies T''=16\lambda T,\;\; X''=\lambda X.
$$
The condition $u(0,t)=u(1,t)=0$ requires $X(0)=X(1)=0$, which gives solutions which are constant multiples of
$$
                    X_0(x)= 0,\\
                   X_n(x)=\sin(n\pi x),\;\;\; n=1,2,3,\cdots
$$
The values of $\lambda$ are determined by $X(0)=X(1)=0$ to be
$$
             \lambda_n = n^2\pi^2,\;\;\; n=1,2,3,\cdots.
$$
The corresponding solutions in $T$ are
$$
                      T_n(t)=A_n\cos(4n\pi t)+B_n\sin(4n\pi t).
$$
So, the general solution is
$$
                 u(x,t)=\sum_{n=1}^{\infty}A_n\sin(n\pi x)\cos(4n\pi t)+B_n\sin(n\pi x)\sin(4n\pi t)
$$
The coefficients $A_n, B_n$ are determined by the initial conditions:
$$
        \sin(3\pi x)=u(x,0) = \sum_{n=1}^{\infty}A_n\sin(n\pi x) \\
        x(1-x^2)=u_t(x,0) = \sum_{n=1}^{\infty}4n\pi B_n\sin(n\pi x)
$$
So $A_{3}=1$ and all other $A_n$'s are $0$. The $B_n$'s are determined by the orthogonality of the $\sin(n\pi x)$ terms:$$
        \int_0^1 x(1-x^2)\sin(n\pi x)dx= 4n\pi B_n \int_0^1 \sin^2(n\pi x)dx, \\
            B_n = \frac{\int_0^1 x(1-x^2)\sin(n\pi x)dx}{4n\pi\int_0^1\sin^2(n\pi x)dx},\;\;\; n=1,2,3,\cdots.
$$
Without computing the coefficients,
$$
       u(x,t)=\sin(3\pi x)\cos(12\pi t)+\\+\sum_{n=1}^{\infty}\frac{\int_0^1 x(1-x^2)\sin(n\pi x)dx}{4n\pi\int_0^1\sin^2(n\pi x)dx}\sin(n\pi x)\sin(4n\pi t).
$$


问:
If $f(x)=x^2+ax+b$ has integer roots, and $f(x+\frac1x)=f(x)+f(\frac1x)$, then find $a^2+b^2$

Let $f(x) = x^2 + ax + b$. If for all non-zero real $x$
$$f\left(x + \frac1x\right) = f(x) + f\left(\frac1x\right)$$
and the roots of $f(x) = 0$ are integers, what is the value of $a^2 + b^2$?

Here is the photo of the question if you are finding difficult to look at it here

EDIT: The answer must be in two-digits

EDIT: What I have done so far is that I added temporary values to the equation
$$f(x) = 0 = x^2 + ax + b$$ and tried to solve it in the form of $a + b$ to which then I squared and solved and got,
$$a^2 + b^2 = x^2 - 2ab$$

I don't really know what I am doing is correct, so help me out please!

答:
This is a completely routine matter of writing out the given ingredients; you have
$$f(x)=x^2+ax+b\qquad\text{ and }\qquad f(x+\tfrac1x)=f(x)+f(\tfrac1x),$$
for all $x$, so simply write out what $f(x+\tfrac1x)$ and $f(\tfrac1x)$ are. We have
\begin{eqnarray*}
f(x+\tfrac1x)&=&(x+\tfrac1x)^2+a(x+\tfrac1x)+b\\
&=&(x^2+2+(\tfrac1x)^2)+ax+a\tfrac1x+b\\
&=&(x^2+ax+b)+((\tfrac1x)^2+a\tfrac1x+b)-b+2\\
&=&f(x)+f(\tfrac1x)-b+2,
\end{eqnarray*}
and so $f(x+\tfrac1x)=f(x)+f(\tfrac1x)$ implies that $b=2$. The roots of $f(x)=0$ are integers, so
$$0=f(x)=x^2+ax+2,$$
has integer roots. These must then be divisors of $2$, i.e. at least one of $1$, $2$, $-1$ and $-2$ is a root of $f(x)=0$. These options correspond to four linear equations in $a$:
\begin{eqnarray*}
0=f(1)&=&1^2+a\cdot1+2=3+a,\\
0=f(2)&=&2^2+a\cdot2+2=6+2a,\\
0=f(-1)&=&(-1)^2+a\cdot(-1)+2=3-a,\\
0=f(-2)&=&(-2)^2+a\cdot(-2)+2=6-2a,\\
\end{eqnarray*}
and so either $a=3$ or $a=-3$, hence $a^2+b^2=13$.


问:
$\lim_{x \to 0}x^2 (a^{1/x} - a^{1/(x-1)}); a > 0$
First post in this forum. This is from my first semester calculus course.... I'm completely lost.

$$\lim_{x \to 0}x^2 (a^{1/x} - a^{1/(x-1)}); a > 0$$

Any hints or steps would be cool

Thanks!

答:
Note the correct question would be when $x \to \infty$, to get a finte limit:
$$L=\lim_{x \to \infty}x^2 (a^{1/x} - a^{1/(x-1)}); a > 0$$
Let $x=1/t$, then
$$L=\lim_{t\to 0} \frac{a^t-a^{t/(1-t)}}{t^2} =\lim_{t\to 0} \frac{a^t-a^{t+t^2}}{t^2}=\lim_{t \to 0} -a^t \frac{a^{t^2}-1}{t^2}\\=\lim_{t \to 0} -a^{t} \frac{1+t^2 \ln a-1}{t^2}=-\ln a, a>0$$
In the above we have used binomial approximation $\frac{t}{1-t}\to t(1+t)$ and $a^z=1+z\ln a+...$ when $|t|$ and $|z|$ are very small.