问:
Sums of powers being powers of the sum
I'm looking for literature on solving problems of the form
$$
n_1^\alpha+\cdots+n_k^\alpha=(n_1+\cdots+n_k)^\beta
$$
for positive integers $n_1,\ldots,n_k$ and fixed parameters $k$ and $\alpha\ne\beta.$ Any ideas?

This is perhaps similar to multigrade equations or "equal sums of like powers" but not quite similar enough to use those results (as far as I can see). But surely this is too simple not to have been studied?

Title: Sums of powers being powers of the sum

Body:
... Continuation:

A list of solutions for certain $(\alpha,\beta)$, $k$:

\begin{array}{|c|l|c|}
\hline
(\alpha, \beta), \; k & (n_1, \ldots, n_k) & n_1+\ldots+n_k \\
\hline \color{#003388}{(\alpha,\beta)=(4,3), \; k=1}
& \color{#CC3300}{1} & \color{#CC3300}{1} \\
\hline \color{#003388}{(\alpha,\beta)=(4,3), \; k=2}
& \color{#CC3300}{(4,4)} & \color{#CC3300}{8} \\
\hline \color{#003388}{(\alpha,\beta)=(4,3), \; k=3}
& \color{#CC3300}{(9,9,9)} & \color{#CC3300}{27} \\
\hline \color{#003388}{(\alpha,\beta)=(4,3), \; k=4}
& (7,14,14,14) & 49 \\
& \color{#CC3300}{(16,16,16,16)} & \color{#CC3300}{64} \\
\hline \color{#003388}{(\alpha,\beta)=(4,3), \; k=5}
& (5,15,15,20,20) & 75 \\
& (6,7,12,15,19) &  59 \\
& \color{#CC3300}{(25,25,25,25,25)} & \color{#CC3300}{125} \\
\hline \color{#003388}{(\alpha,\beta)=(4,3), \; k=6}
& (2,2,2,8,8,14) & 36 \\
& (2,9,14,21,22,23) & 91 \\
& (3,3,7,13,16,19) & 61 \\
& (4,5,6,11,13,20) & 59 \\
& (28,32,32,32,36,40) & 200 \\
& \color{#CC3300}{(36,36,36,36,36,36)} & \color{#CC3300}{216} \\
\hline \color{#003388}{(\alpha,\beta)=(4,3), \; k=7}
& (1,1,2,4,7,14,14) & 43 \\
& (1,2,11,14,17,20,26) & 91 \\
& (1,5,6,6,11,11,21) & 61 \\
& (1,8,12,22,23,28,29) & 123 \\
& (1,22,23,26,29,33,39) & 173 \\
& (2,2,4,12,12,14,22) & 68 \\
& (2,14,14,14,23,23,33) & 123 \\
& (3,11,18,21,26,27,35) & 141 \\
& (3,18,21,27,30,36,36) & 171 \\
& (4,12,14,14,14,28,30) & 116 \\
& (5,7,15,23,23,31,31) & 135 \\
& (5,9,16,20,26,30,33) & 139 \\
& (5,10,11,19,23,24,33) & 125 \\
& (6,7,18,21,25,31,33) & 141 \\
& (6,12,21,32,32,33,35) & 171 \\
& (6,16,16,18,24,34,34) & 148 \\
& (7,7,9,9,12,22,27) & 93 \\
& (7,8,8,8,9,10,25) & 75 \\
& (7,9,9,9,17,23,29) & 103 \\
& (8,11,24,25,32,34,37) & 171 \\
& (9,9,16,17,19,20,35) & 125 \\
& (9,10,22,24,26,32,38) & 161 \\
& (9,12,24,24,30,33,39) & 171 \\
& (10,19,30,33,33,37,43) & 205 \\
& (10,21,21,22,24,31,42) & 171 \\
& (10,21,33,36,39,40,40) & 219 \\
& (13,16,17,26,27,33,41) & 173 \\
& (13,16,19,22,27,37,39) & 173 \\
& (13,17,20,22,24,34,41) & 171 \\
& (13,21,31,35,38,39,44) & 221 \\
& (15,20,24,25,26,32,45) & 187 \\
& (17,17,25,30,33,39,44) & 205 \\
& (17,18,22,31,33,38,44) & 203 \\
& (18,19,24,24,26,38,44) & 193 \\
& (18,26,37,40,41,42,47) & 251 \\
& (19,20,29,31,36,39,47) & 221 \\
& (21,21,25,28,41,41,44) & 221 \\
& (21,28,36,39,39,39,51) & 253 \\
& (22,27,32,36,40,47,47) & 251 \\
& (22,30,30,44,44,44,46) & 260 \\
& (25,25,29,30,35,44,49) & 237 \\
& (25,33,35,40,41,42,53) & 269 \\
& (27,36,37,39,46,46,52) & 283 \\
& \color{#CC3300}{(49,49,49,49,49,49,49)} & \color{#CC3300}{343} \\
\hline
\hline \color{#003388}{(\alpha,\beta)=(5,4), \; k=1}
& \color{#CC3300}{1} & \color{#CC3300}{1} \\
\hline \color{#003388}{(\alpha,\beta)=(5,4), \; k=2}
& \color{#CC3300}{(8,8)} & \color{#CC3300}{8} \\
\hline \color{#003388}{(\alpha,\beta)=(5,4), \; k=3}
& \color{#CC3300}{(27,27,27)} & \color{#CC3300}{27} \\
\hline \color{#003388}{(\alpha,\beta)=(5,4), \; k=4}
& (22,22,33,44) & 121 \\
& (50,50,60,65) & 225 \\
& \color{#CC3300}{(64,64,64,64)} & \color{#CC3300}{256} \\
\hline \color{#003388}{(\alpha,\beta)=(5,4), \; k=5}
& (36,36,36,36,72) & 216 \\
& \color{#CC3300}{(125,125,125,125,125)} & \color{#CC3300}{625} \\
\hline \color{#003388}{(\alpha,\beta)=(5,4), \; k=6}
& (1,3,14,21,22,39) & 100 \\
& (3,57,81,96,99,114) & 450 \\
& (4,45,57,90,91,98) & 385 \\
& (5,5,40,45,57,68) & 220 \\
& (7,60,80,85,85,118) & 435 \\
& (10,25,41,55,55,84) & 270 \\
& (14,31,36,44,44,81) & 250 \\
& (16,56,64,64,88,112) & 400 \\
& (18,36,45,72,81,99) & 351 \\
& (21,38,45,75,92,99) & 370 \\
& (25,60,85,115,115,125) & 525 \\
& (30,35,52,77,88,108) & 390 \\
& (30,81,113,115,138,139) & 616 \\
& (34,68,80,110,112,136) & 540 \\
& (35,45,55,75,75,115) & 400 \\
& (36,40,68,82,112,113) & 451 \\
& \color{#CC3300}{(216,216,216,216,216,216)} & \color{#CC3300}{1296} \\
\hline
\end{array}


问:
Applying the Normal Equations to solve the Linear Regression Problems.
I am new to machine learning and I am currently studying the gradient descent method and its application for linear regression. An iterative method known as gradient descent is finding the linear function:
$$
J(\theta)=\underset{\theta}{\operatorname{argmin}}\frac{1}{2}\sum_{i=1}^{n}\left(h_{\theta}(x^{(i)})-y^{(i)}\right)
\tag1$$
However, I came to notice of an explicit non-iterative scheme in $\text{Andrew Ng's}$ lecture notes right here: https://see.stanford.edu/materials/aimlcs229/cs229-notes1.pdf for which he concluded in page $11$ that the gradient descent is now equivalent to finding the vector $\theta$ where:
$$
\theta=(X^{T}X)^{-1}X^{T}y
\tag2$$
Terminology:

$\theta$ are the parameters.

$n$ are number of training examples.

$x$ are input features and $y$ are output variables.

$(x^{(i)},y^{(i)})$ is the $i^{th}$ training example.

$X\in\mathbb{R}^{m\times n}$ is the design matrix where $\mathbf{X}=\begin{bmatrix}
--(x^{(1)})^{T}-- \\
--(x^{(2)})^{T}-- \\
\vdots\\
--(x^{(n)})^{T}--
\end{bmatrix} $

I have two questions.

$(1)$: Say I have $(1,2)$, $(2,1.5)$, and $(3,2.5)$ I wish someone can demonstrate this procedure to find $\theta$ by solving $\theta=(X^{T}X)^{-1}X^{T}y$

$(2)$: I would really hope if a Python,MATLAB, or C$++$ algorithm for this procedure can be provided.

答:
The way to use normal equations is just explicitly solving the equation as it has only one solution. The only reason why its not widely used instead of gradient descent is because it get to be very inefficient for large datasets. Look at the equations you wrote:

$$
\theta=(X^{T}X)^{-1}X^{T}y
\tag2
$$

You have three matrix multiplications and one matrix inversion. Usually, the data matrix can be large, because its composed by the observed samples. Imagine $X$ matrix can be huge, and matrix multiplication has complexity $O(n^3)$.

However, if one wants to solve the linear system of equations by using the normal equation, one just have to do exactly what the equation says. If you are using numpy, for example, you will do something like:

For the $T_1 = (X^{T}X)^{-1}$ matrix:

```<code>x_transpose = np.transpose(x)   #calculating transpose
x_transpose_dot_x = x_transpose.dot(x)  # calculating dot product
T_1 = np.linalg.inv(x_transpose_dot_x) #calculating inverse
</code>```

For the $T_2 = X^{T}y$ matrix:

```<code>T_2 = x_transpose.dot(y)
</code>```

Now just multiply them together:

```<code>theta = T_1.dot(T_2)
</code>```

Normal equation can be easily derived understanding the fact that a linear transformation $A$ respects:

$Ker(A^T) \perp Im(A)$,

which is, image of $A$ is the orthogonal complement of kernel of $A^T$. This means that if the linear system is not solvable (which is, $y$ vector is not spanned by image of $A$) then the best we can do is to find a vector perpendicular to image space of $A$ (since the vector that minimizes a distance between a hyperplane and a point is perpendicular to that hyperplane).

Edit: I just realize now that I think you want some example using some static dumb data. For applying the given example to a sample data, you only have to know how to create a matrix using numpy, which can easily be found on internet. It would look something like this:

```<code>x = np.array([[1,2],
    [2, 1.5],
    [3, 2.5]])
</code>```

ContentLicense: CC BY-SA 4.0

Body:
The way to use normal equations is just explicitly solving the equation as it has only one solution. The only reason why its not widely used instead of gradient descent is because it get to be very inefficient for large datasets. Look at the equations you wrote:

$$
\theta=(X^{T}X)^{-1}X^{T}y
\tag2
$$

You have three matrix multiplications and one matrix inversion. Usually, the data matrix can be large, because its composed by the observed samples. Imagine $X$ matrix can be huge, and matrix multiplication has complexity $O(n^3)$.

However, if one wants to solve the linear system of equations by using the normal equation, one just have to do exactly what the equation says. If you are using numpy, for example, you will do something like:

For the $T_1 = (X^{T}X)^{-1}$ matrix:

x_transpose = np.transpose(x)   #calculating transpose
x_transpose_dot_x = x_transpose.dot(x)  # calculating dot product
T_1 = np.linalg.inv(x_transpose_dot_x) #calculating inverse

For the $T_2 = X^{T}y$ matrix:

T_2 = x_transpose.dot(y)

Now just multiply them together:

theta = T_1.dot(T_2)

Normal equation can be easily derived understanding the fact that a linear transformation $A$ respects:

$Ker(A^T) \perp Im(A)$,

which is, image of $A$ is the orthogonal complement of kernel of $A^T$. This means that if the linear system is not solvable (which is, $y$ vector is not spanned by image of $A$) then the best we can do is to find a vector perpendicular to image space of $A$ (since the vector that minimizes a distance between a hyperplane and a point is perpendicular to that hyperplane).

Edit: I just realize now that I think you want some example using some static dumb data. For applying the given example to a sample data, you only have to know how to create a matrix using numpy, which can easily be found on internet. It would look something like this:

x = np.array([[1,2],
    [2, 1.5],
    [3, 2.5]])


问:
Interpretation Regression Coefficient
I estimated the following specification:
$$ Y = a + bX + e , $$

where $Y$ is a dummy variable and has a mean of $.15$. $X$ is log-transformed with a mean of $3.54$ and SD of $.87$; before transformation $X$ had mean $46.87$ and SD $36.82$. The model estimates $b=-.025$. I am trying to interpret correctly coefficient $b$.

Is it correct to say that: the model estimates that a one standard deviation increase in $X$ (36.82) reduces $Y$ by $(36.82*-.025) = -.9205$ percentage points? Then, if correct, how can I make sense of this result given the baseline probability of $.15$?
Thank you!

答:
Given a regression model of the form $y = a + b\ln x + \varepsilon$, let $\hat{y} = \hat{a} + \hat{b} \ln x $ be the estimated model. Taking the derivative with respect to $x$, we have

$$ d{\hat{y}} = \hat{b} \frac{dx}{x} . $$

This implies that a 1% increase in $x$ is associated with a $\frac{\hat{b}}{100}$ increase in $\hat{y}$.

So, in the context of this problem, given some starting value of $x$, if you increase that value by $1\%$, then the predicted value of $y$ decreases by $0.00025$.

Generally, with log transformations in regressions, this type of percent increase interpretation works so long as the percent changes are near 0 - the interpretation depends on the approximation $\ln(1+x)\approx x$. For bigger changes in $x$, the percent change interpretation becomes less accurate.

Title: Interpretation Regression Coefficient


问:
Explain why the following statements are logically equivalent (2 examples)
I have the following 2 separate statements.

(pvq)-> q = ~(p^~q)

r->(~pvq)=(pvr)->q

I have proved they are both equivalent with a truth table, but I am asked to briefly explain (in words not with a theorem chart) why they are equivalent, and I have no idea why they are. I have looked at the Laws of Algebra of Propositions and they don't really address more 'complex' statements. Could I have some help and maybe someone knows of a more advanced chart (or I will learn to extrapolate better)

Thanks

答:
For the first: You are given (p or q) -> q.

Suppose that p and not q. Then p is true, because both are true, so p or q is true, because at least one of them is. But by the given implication, this implies q. But we assumed not q. Thus we must have not "p and not q."

For the second: You are given r -> ((not p) or q)

Suppose that p and r. Since r, we have not p and/or q. But we don't have not p, we have p! Thus we have q, because we must have at least one of not p or q, and we don't have not p. Which is what we wanted to implicate, q.

Hope this suffices


问:
Proving existence of square roots
In Rudin's proof of the existence of square roots, he assumes for a contradiction that $y^n < x$ where $y = \sup\{t \in \mathbb{R} \mid t > 0, \; t^n < x\}$. He then asserts that there exists $h \in (0,1)$ with $h < \frac{x-y^n}{n(y+1)^{n-1}}$ so that $(y+h)^n < x$. I'm trying to understand exactly where he uses continuity of the function $f(u) = u^n$ and specifically which value of $\epsilon$ he choose. It's clear that $h$ is his $\delta$. From what I can tell, $x - y^n$ seems to be his $\epsilon$. Is this correct?

答:
He didn't really need the concept of continuity to do what he did.

He has a number $y$ such that $y = \sup\{t\mid t^n<x\}$ and $y^n<x$. All he wants to do is to construct a number slightly bigger than $y$ but still satisfy the fact that the $n$th power of that number is still less than $x$.

Here is how he construct this slightly bigger number. He considers $y+h$ which is strictly greater than $y$ provided $h>0$. Clearly, $h$ can't be arbitrarily large if he wants $(y+h)^n<x$. So, he needs to come up with a condition for $h$ in terms of $y$ and $n$. Note that we only need to find one $h$ to complete the contradiction. Hence it suffices to restrict our attention to the values of $h$ that are less than one.

Note that in general, we have the equality

\begin{align}
a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+\ldots + ab^{n-2}+b^{n-1})
\end{align}
this means that
\begin{align}
(y+h)^n-y^n = h((y+h)^{n-1}+(y+h)^{n-2}y+\ldots + y^{n-1})<h((y+1)^{n-1}+(y+1)^{n-1}+\ldots +(y+1)^{n-1}) = hn(y+1)^{n-1}.
\end{align}

Since we want to find $h$ such that $(y+h)^n<x$, then this is the same as wanting to find $h$ such that
\begin{align}
x-y^n>(y+h)^n-y^n.
\end{align}
In fact, it suffices to choose $h$ so that
\begin{align}
x-y^n>hn(y+1)^{n-1}
\end{align}
since
\begin{align}
x-y^n>hn(y+1)^{n-1}>(y+h)^n-y^n.
\end{align}
Then it is clear that the above inequality holds provided
\begin{align}
h<\frac{x-y^n}{n(y+1)^{n-1}}.
\end{align}

Title: Proving existence of square roots


问:
Preparation for Putnam?
Currently, my math training includes Calc 1-3, linear algebra, and some introduction to set theory/discrete math.
What would you recommend that I study over summer in preparation for the Putnam? Real analysis, topology, abstract algebra (all of the above)? What would be the most pertinent? Thanks!</p>

答:
There are lots of good books.

William Lowell Putnam Mathematical Competition: Problems & Solutions: 1938-1964

William Lowell Putnam Mathematical Competition: Problems & Solutions: 1965-1984

William Lowell Putnam Mathematical Competition: Problems & Solutions: 1985-2000

Putnam and Beyond

Problem-Solving Strategies

The first 3 are all the contests from 1938 through 2000, and includes solutions to all of them.  The last 2 have lots of problems that are arranged by topic.  So, those are some topics you can study.  The last 2 are probably better for this reason because it teaches you many strategies/problem solving techniques and gives you problems that you can try involving those techniques.  Then, you can try other problems from the first 3 books where they aren't organized by topic.  If you're in college now, your library may have some or all of these books, or similar books.

ContentLicense: CC BY-SA 3.0

Title: Preparation for Putnam?


问:
Example of series of the form $\sum \frac{1}{n^{1+a_n}}$
How one can find a sequence $(a_n)$ of positive numbers such that $\lim a_n=0$ and

$$\sum^{\infty}_{n=i} \frac{1}{n^{1+a_n}}$$

converges.

Thank you!

答:
If the $a_n$ decrease slow enough to make $\frac1{n^{1+a_n}}$ non-increasing, the the series converges by the Cauchy condensatoin test, provided
$$ \sum_n 2^n\cdot\frac 1{2^{n(1+a_{2^n})}}=\sum_n \frac 1{2^{na_{2^n}}} $$
converges. If we try to make
$$ \frac 1{2^{na_{2^n}}}=\frac1{n^2},$$
we are led to $a_{2^n}=\frac2n\log_2n$ and generalize this to $a_n=\frac{2}{\log_2n}\log_2\log_2n$. Remains to show that this makes $\frac1{n^{1+a_n}}$ non-decreasing ...</p>


问:
Giving an example to the following statement regarding set theory.
Write down in roster notation a set  of cardinality 3, of which all elements are sets, and which
satisfies the following property:
∀,  ∈  ( ⊆  ⇒  ⊆ ).
No justification is needed. Do not use ellipses (“…”) in your answer.

What I understands from this question is that all the 3 sets have to be the same to fulfill the statement.Example of an answer would be { {1,2,3},{1,2,3},{1,2,3}} am i right?

答:
An example could be $U=\{\{1\}, \{2,3\}, \{4,5,6\}\}$.

It has three elements, all three are sets. And there is no inclusion relation between any of them. So it’s voidly true that if one is a subset of the other, the reverse also holds, as your property states. Otherwise put, there is no counterexample among them so the statement holds.

The element sets could have been any size.


问:
Finding $\alpha$-stable distributions by (probably) Levy's continuity theorem
Let $X_1,...,X_n$ be iid random variables uniformly distributed on $[-1,1]$. Now let's put $$Y_n = \frac{sgn X_n}{|X_n|^{1/\alpha}}, \, n= 1,2,...,$$ with a set value $\alpha \in (0,2)$.

The goal is to:

$\textbf{(a)}$ Show that $Z_n := \frac{Y_1 + ... + Y_n}{n^{1/\alpha}} $ converges by distribution.

$\textbf{(b)}$ Find the characteristic function of the limit.

And that would mean that the limiting distribution is stable as it has its own domain of attraction consisting of $Z_n$.

My best guess is to do this by showing that the characteristic function of $Z_n$ converges to a certain function that's continuous at $t=0$ and then using the Levy's continuity theorem (https://en.wikipedia.org/wiki/L%C3%A9vy's_continuity_theorem), but I was sadly unable to do so.

Do You perhaps have other ideas that would answer this problem or maybe it's indeed solvable by Levy's theorem? If so, how?

Edit:

I've calculated that the characteristic function of $Z_n$ should be: $$\phi_{Z_n}(u) = \left[ \frac{1}{2} \int_0^1 exp(iu \frac{-1}{x^{1/\alpha}} n^{-\frac{1}{\alpha}}) dx + \frac{1}{2} \int_0^1 exp(iu \frac{1}{x^{1/\alpha}} n^{-\frac{1}{\alpha}}) dx \right]^n,$$ which is $$\left[ \int_0^1 cos \left( \frac{ u}{(xn)^{1/\alpha}} \right) dx \right]^n$$

Edit 2:

I've found an answered question that shows what the limit should be, but still doesn't have what I need. Symmetric alpha stable distributions with $X_1+X_2+\cdots+X_n \stackrel{d}{=} n^{1/\alpha}X$ as definition

答:
Problem
Let $ X_1,X_2,\cdots $ be a sequence of iid random variables uniformly distributed on $ [-1,1] $.
Now let's put
\begin{equation*} Y_n=\frac{\mathrm{sgn}X_n}{|X_n|^{1/\alpha}}, \qquad n\ge 1, \end{equation*}
with a set value $ \alpha\in(0,2) $.

(a) Show that $ Z_n=\frac{Y_1+\cdots+Y_n}{n^{1/\alpha}} $ converges by distribution.

(b) Find the characteristic function of the limit.

And that would mean that the limiting distribution is stable as it has its own domain of attribution consisting of $ Z_n $

Answer  Let
$$ \phi_n(t)=\mathsf{E}\Big[\exp\Big(\frac{itY_1}{n^{1/\alpha}}\Big)\Big]
=\int_{0}^{1}\cos\Big(\frac{t}{(xn)^{1/\alpha}}\Big)dx. $$
Then
$$ \phi_{Z_n}(t)=\mathsf{E}[\exp(itZ_n)]=(\phi_n(t))^n. $$
Meanwhile,
\begin{align*}
 n[1-\phi_n(t)]&=n\int_0^1\Big[1-\cos\Big(\frac{t}{(nx)^{1/\alpha}}\Big)\Big]\,dx\\
 &=\alpha |t|^\alpha\int_{|t|/n^{1/\alpha}}^{\infty}\frac{1-\cos(z)}{z^{\alpha +1}}\,dz\\
 &\to  C(\alpha) |t|^\alpha, \qquad \text{as}\quad n\to\infty.
\end{align*}
where(cf. Sato, Lévy Processes and Infinitively Divisible Distributions, Cambridge University Press, 1999, Lemma 14.1 p.84,
or Y. S. Chow &  H. Teicher, Probability Theory, 3rd ed., Springer Verlag, 1997, p.469--.)
\begin{align*}
 C(\alpha)&=\alpha\int_{0}^{\infty}\frac{1-\cos(z)}{z^{\alpha +1}}\,dz\\
 &=\begin{cases}
  \cos\Big(\dfrac{\alpha\pi}{2}\Big)\Gamma(1-\alpha),& \alpha \in (0,1)\cup(1,2)\\
  \quad\dfrac{\pi}{2}, &\alpha=1.
 \end{cases}
\end{align*}
Then
$$ \lim_{n\to\infty}\phi_{Z_n}(t)=\lim_{n\to\infty}[1-(1-\phi_n(t))]^n=\exp[-C(\alpha) |t|^\alpha]. $$


问:
Factoring $a^4+b^4+(a-b)^4$
I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.

I rewrite it as:

$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.

Title: Factoring $a^4+b^4+(a-b)^4$

Body:
$$(a-b)^4=(a^2+b^2-2ab)^2=(a^2+b^2)^2+4a^2b^2-4ab(a^2+b^2)$$

$$a^4+b^4=(a^2+b^2)^2-2a^2b^2$$

On addition, $$2((a^2+b^2)^2+a^2b^2-2ab(a^2+b^2))=2(a^2+b^2-ab)^2$$


问:
Combining geometric means from different datasets
In statistics, one sometimes uses the geometric mean which for a dataset $\{x_i\}_{i=1}^N$ is defined as  $$(\prod\limits_{i=1}^N x_i)^{(1/N)}.$$
This is particularly useful when the data of the experiment is distributed across many orders of magnitude, so that it would make more sense to plot histograms on a log scale than a linear scale.

Now suppose I am doing an experiment to determine some experimental variable $X$, which theoretically is  the geometric mean of the data set I measure. Suppose I have done this experiment repeatedly to generate $r$ data sets with geometric means $\mu_1, \mu_2, \ldots, \mu_r$ and geometric standard deviations $\sigma_1,\sigma_2, \ldots, \sigma_r$. How do I combine the data from these different experiments to obtain one geometric mean that is the best estimate for the variable?

Can one simply take the arithmetic mean and standard error of this collection of means? If so, why? Or should we consider a different 'geometric error of the mean'?

答:
Just to be aware that the last two equations in @heropup's answer contain a small error (the difference in the (square) bracket should be a sum):
$$\sum_{i=1}^{N_k} \left(\ln \frac{x_i}{\mu} \right)^2 = N_k \left[ \left(\ln \sigma_k\right)^2 \color{blue}{+} \left(\ln \frac{\mu}{\mu_k}\right)^2\right],$$
leading to
$$N (\ln \sigma)^2 = \sum_{k=1}^r \sum_{i=1}^{N_k} \left( \ln \frac{x_i}{\mu} \right)^2 = \sum_{k=1}^r N_k \left[ (\ln \sigma_k)^2 \color{blue}{+} \left(\ln \frac{\mu}{\mu_k}\right)^2\right].$$
Thus,
$$\sigma = \exp\left[\sqrt{\frac{1}{N} \sum_{k=1}^r N_k \left[ (\ln \sigma_k)^2 \color{blue}{+} \left(\ln \frac{\mu}{\mu_k}\right)^2\right]} \right].$$

It might be worth pointing out that the last term in the square bracket could be either $\left(\ln \frac{\mu}{\mu_k}\right)^2$ or $\left(\ln \frac{\mu_k}{\mu}\right)^2=\left(\ln \frac{\mu_k}{\mu}\right)\left(\ln \frac{\mu_k}{\mu}\right)=\left(-\ln \frac{\mu}{\mu_k}\right)\left(-\ln \frac{\mu}{\mu_k}\right)=\left(\ln \frac{\mu}{\mu_k}\right)^2$, as they are evidently the same.


问:
interpretation of tangent space of level curve manifold
Given a function $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ a smooth function, $c \in \mathbb{R}^m$ a regular value, define the manifold $M_c = \{x \in \mathbb{R}^n| f(x) =c\}$. Then we get that the tangent space at every point $x \in M$ has tangent space of dimension $n-m$. So fix a point $x$ and look at $T_x M_c$, the tangent space at $x$, since $f|M_c$ is a constant function in $C^\infty(M_c)$, we get that for any $v \in T_x(M_c)$, $v(f)=0$, which leads to the conclusion that $D_xf(v) = 0$, so $T_x M_c = Ker(D_xf)$.

Now I have trouble with the step that $v(f)=0$ implies $D_xf(v)=0$. If I think of the tangent vectors as directional derivatives at $x$, then it makes sense that every directional derivative of $f$ at $x$ equals $0$ implies that the differential itself is $0$. However, I cannot interpret this result if I think of $f$ as smooth map between manifolds, and $D_xf$ takes tangent space of $x$ to tangent space of $f(x)$.

答:
As you said so yourself, the tangent vectors can be interpreted as directional derivatives of $f$.

So in particular, if $\forall v \in T_xM: v(f) = 0$, then the directional derivatives of $f$ in any direction $v$ will be $0$, so $D_x f(v) = 0$.

Alternatively, since you're in Euclidean speace, you can work in local coordinates. $v(f)$ can be expressed as $\partial_v f = (D_x f)v$ (if you're not clear on why this is so let me know I'll explain more)


答:
We begin with the special case that $a=x$ and $b=1$ to obtain $f(x)=x^4 +(x-1)^4 =2g(x),$ where $$g(x) =x^4-2x^3+3x^2-2x+1.$$ Since $g(x)$ has no real roots, it equals the product of two irreducible quadratic polynomials. Let us guess that these two  quadratic polynomials are equal, so that $g(x)=(x^2+cx+1)^2.$ Happily, $c=-1$ works. Therefore,
$$f(x)=2(x^2-x+1)^2.$$
In the above factorization replace $x$ by $a\over b$ and multiply both sides by $b^4$ to get the factorization of the original expression.

Title: Factoring $a^4+b^4+(a-b)^4$

Body:
I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.

I rewrite it as:

$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.


问:
Taylor's series for multivariable function: not understanding some arguments
I am following G. B. Folland's Advanced Calculus; the discussion below comes on p.90 of it.

Let $f$ be function from an open convex set $S\subset \mathbb{R}^n$ to $\mathbb{R}$ which is $k$-th derivative continuous on $S$.

Fix $\mathbf{a}\in S$.

Consider $g(t)=f(\mathbf{a}+t\mathbf{h})$.

By chain rule, $g'(t)=\mathbf{h}\cdot \nabla f(\mathbf{a}+t\mathbf{h})$.

And hence $g^{(j)}(t)=(\mathbf{h}\cdot \nabla)^j f(\mathbf{a}+t\mathbf{h})$.

Q. I didn't understand the precise meaning of last equation.
To be precise in my question, I didn't get what is meaning of $(\mathbf{h}\cdot \nabla)^j$? I tried to understand it for $j=2$, but I couldn't! Can one explain what it says for $j=2,3$ in detail?

(I am trying to understand whether $\mathbb{h}\cdot \nabla$ is a function or operator? On what objects it is applied? What do we mean by taking its power? )

答:
It is an technically operator that takes a function to its $j$th derivative time and multiplies this function with $h^j$. Practically it is just an convenient way to write down, what happens if you derive $g$ for $j$ times:

This is just an application of the chain rule, where the inner function is $x(t) = a + th$, so $\frac{dx}{dt}=h$. Every time you derive, you are left with the derivative of the outer function and a factor of $h$.

Edit: As a general rule: If we write $K^j$ for an operator $K$ that means we  apply the operator $j$ times.


问:
Find the $f^{(n)}(0)$ of $\arctan(x)$.
I wish to solve this by using the method as prescribed here:

https://diendantoanhoc.net/topic/192848-lim-sum-frac2n-12n/?p=742824

$$\arctan(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nx^{2n+1}}{2n+1}$$

The general formula of Maclaurin series is

$$\displaystyle\sum_{n=0}^{+\infty} \dfrac{f^{(n)}(0)}{n!}x^n$$

Since both series are equal, their coefficients ought to equal,

So I have

$$\dfrac{f^{(n)}(0)}{n!}=\dfrac{(-1)^nx^{2n+1}}{2n+1}$$

$$f^{(n)}(0)=\left[\dfrac{(-1)^nx^{2n+1}}{2n+1}\right]n!$$

Something is wrong, the n! shouldn't be there.

Reference from:

https://socratic.org/questions/how-do-you-find-f-6-0-where-f-x-arctanx-x

Title: Find the $f^{(n)}(0)$ of $\arctan(x)$.

Body:
$f(x)=\tan^{-1}x \implies f'(x)=\frac{1}{1+x^2}$
So $$f^{n}(x)=D^n(\tan^{-1} x)=\frac{i}{2}D^{n-1}\left(\frac{1}{x+i}-\frac{1}{x-i}\right)$$
Use $$D^{m}(x+a)^{-1}=\frac{(-1)^m m!}{(x+a)^{m+1}}$$
Then $$D^n(\tan^{-1} x)=(-1)^{n-1}(n-1)!\frac{i}{2}\left(\frac{1}{(x+i)^n}-\frac{1}{(x-i)^n}\right)$$
$$\implies f^{n}(0)=(-1)^{n-1}(n-1)! \sin (n\pi/2)$$


问:
Proving a sum of three series is equal to $e^z$
Let

\begin{align*}
    u(z)=&\ 1+\frac{z^3}{3!}+\frac{z^6}{6!}+\frac{z^9}{9!}+\ldots\\
    v(z)=&\ z+\frac{z^4}{4!}+\frac{z^7}{7!}+\frac{z^{10}}{10!}+\ldots\\
    w(z)=&\ \frac{z^2}{2!}+\frac{z^5}{5!}+\frac{z^8}{8!}+\frac{z^{11}}{11!}+\ldots
\end{align*}

show that $e^z= u(z)+ v(z) +w(z) $

We can rewrite the three series as

$$u(z)=\sum^{\infty}_{n=0}\frac{z^{3n}}{(3n)!}$$
$$v(z)=\sum^{\infty}_{n=0}\frac{z^{3n+1}}{(3n+1)!}$$
$$w(z)=\sum^{\infty}_{n=0}\frac{z^{3n+2}}{(3n+2)!}$$

Thus
$$ u(z)+ v(z) +w(z) = \sum^{\infty}_{n=0} z^{3n} \left(\frac{1}{(3n)!}+\frac{z}{(3n+1)!}+\frac{z^2}{(3n+2)!}\right)$$

but I cant really see how to combine the factorial terms together. Any ideas where I should go from here? I suppose I could just claim that the sum of all the terms appears to form the sequence $\frac{z^n}{n!}$, but that doesn't seem rigorous enough to me.

答: Any non-negative integer $n$ is of the form $3k+i$ for some non-negative integer $k$ and some $i \in \{0,1,2\}$. Check that the coefficient of $z^{n}=z^{3k+i}$ in $u(z)+v(z)+w(z)$ is exactly $\frac 1 {(3k+i)!}=\frac 1 {n!}$ in each of the cases $i=0, i=1$ and $i=2$.


问:
For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more
I set out to solve the following question using the pigeonhole principle

Regardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more.

My outline

[1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\cdot 55=165$.

[2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more.

My query

Could someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle.

答:
We will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more.

Let the integers be $\{a_i\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider

$$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$

The sum of these 3 numbers is $2+3 +\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\lfloor \frac{54}{3} \rfloor = 18 $.

I leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do.


答:
I suppose I could just claim that the sum of all the terms appears to form the sequence $\frac{z^n}{n!}$, but that doesn't seem rigorous enough to me.

Well, you could make it more rigorous by appealing to the known calculus result which is:

If $\sum_{i=0}^\infty a_nx^n$ and $\sum_{i=0}^\infty b_nx^n$ are both convergent series, then $\sum_{i=0}^\infty (a_n + b_n)x^n$ is also a convergent sequence and $\sum_{i=0}^\infty a_nx^n+\sum_{i=0}^\infty b_nx^n=\sum_{i=0}^\infty (a_n+b_n)x^n$

Title: Proving a sum of three series is equal to $e^z$

Body:
Let

\begin{align*}
    u(z)=&\ 1+\frac{z^3}{3!}+\frac{z^6}{6!}+\frac{z^9}{9!}+\ldots\\
    v(z)=&\ z+\frac{z^4}{4!}+\frac{z^7}{7!}+\frac{z^{10}}{10!}+\ldots\\
    w(z)=&\ \frac{z^2}{2!}+\frac{z^5}{5!}+\frac{z^8}{8!}+\frac{z^{11}}{11!}+\ldots
\end{align*}

show that $e^z= u(z)+ v(z) +w(z) $

We can rewrite the three series as

$$u(z)=\sum^{\infty}_{n=0}\frac{z^{3n}}{(3n)!}$$
$$v(z)=\sum^{\infty}_{n=0}\frac{z^{3n+1}}{(3n+1)!}$$
$$w(z)=\sum^{\infty}_{n=0}\frac{z^{3n+2}}{(3n+2)!}$$

Thus
$$ u(z)+ v(z) +w(z) = \sum^{\infty}_{n=0} z^{3n} \left(\frac{1}{(3n)!}+\frac{z}{(3n+1)!}+\frac{z^2}{(3n+2)!}\right)$$

but I cant really see how to combine the factorial terms together. Any ideas where I should go from here? I suppose I could just claim that the sum of all the terms appears to form the sequence $\frac{z^n}{n!}$, but that doesn't seem rigorous enough to me.


问:
Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition
I am given the sequence
$$a_{n+1} = \sqrt{2 a_n}, \quad a_1 = \sqrt{2}.$$
That is, the sequence
$$ \sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$$

I am aware of using the recursive method to find the limit: setting $x = \sqrt{2x}$ and getting $x = 2$, rejecting $x = 0$ due to the sequence being monotone and positive.

I now want to show that the limit of this sequence is $2$ using the epsilon-neighbourhood definition for a convergent series:

Let $(a_n)$ be a sequence that converges to a real number $a$. Then, for every number $\epsilon > 0$, there exists a number $N \in \mathbb{N}$ such that for all $n \geq N$, $|a_n - a| < \epsilon$.

I take the logarithm on both sides and arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \ln 2 - \ln 2 \right|  < \epsilon. $$

Using triangle inequality, $|x - y| \geq |x| - |y|$, I arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| - |\ln 2| < \epsilon. $$

Add $\ln 2$ to both sides
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| < \epsilon + \ln 2. $$

The left hand side can be rewritten as a sum of $n$ terms in a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$,
$$ \frac{\frac{1}{2} \left( 1 - \left( \frac{1}{2} \right)^n \right)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} < \epsilon + \ln 2. $$
I am unsure of how to proceed beyond this point.

答:
Let's make an observation: for each natural number $n$, your sequence can be given explicitly by $$a_n = 2^{\sum_{i=1}^n (\frac{1}{2})^i}$$

For example, $a_2 = \sqrt{2\sqrt{2}} = (2(2^{1/2}))^{1/2} = 2^{1/2} \cdot 2^{1/4} = 2^{1/2 + 1/4}$

Furthermore, we know that $${\sum_{i=1}^n \left(\frac{1}{2}\right)^i} \to 1 \ \ \text{as}
 \ \ n \to \infty$$

We also know that the function $f(x) = 2^x$ is a continuous function. In other words, if we pick $\epsilon > 0$, there is $\delta > 0$ such that $|x-1| < \delta$ implies $|f(x) - f(1)|< \epsilon$.

Thus, because $$\sum_{i=1}^n \left(\frac{1}{2}\right)^i \to 1$$there is $M > 0$ so that $n > M$ implies $$\left|{\sum_{i=1}^n \left(\frac{1}{2}\right)^i -1}\right| < \delta$$

Hence, letting $x = \sum_{i=1}^n \left(\frac{1}{2}\right)^i$ in the paragraph where I mentioned continuity, we obtain $$\left|f\left(\sum_{i=1}^n \left(\frac{1}{2}\right)^i\right) - 2\right| < \epsilon$$

This is precisely $|a_n -2 | < \epsilon$


问:
is there a mathematical way to go about this coding question?
i have the following python code(assume x is already defined as an integer):

```<code>numList=[] #creates a list that is empty
for i in range(x,-1,-1): #loops through from i=x to i=0, decreasing one each iteration
    if i%5==1: #if i mod 5 is 1...
        numList.pop(1) #remove the second item from the list
    elif i%5==4: # if i mod 5 is 4...
        numList.pop(-1) #remove the last item from the list
    else: #if i mod 5 is not 1 or 4...
        numList.append(i) #add i to the list
print(numList) #prints out the list

</code>```

i am given that the output is [23, 8, 7, 3, 2, 0](hopefully its clear what this means). i am tasked to find the possible values for x. is there some logical/mathematical way i can go about this without resorting to guess and check? i tried going "backwards" from the end result but it got too convoluted very quickly.

i have also tried setting x to 5a+2, 5a+3, and 5a (we dont consider 5a+1 and 5a+4 because that will terminate the code prematurely) because they give different values mod 5, but i cant seem to get anywhere with that as well.

答:
At least roughly, every consecutive five rounds pop two and append three, so net append 1. For six entries you will need 30-ish rounds (and of course certainly you appended 23 at some point.

Also note that the first term will not change any more after a few rounds.

After clarification of undefined behaviour when poping an illegal position:

Write $x=5m+r$ where $m,n\in\Bbb N_0$ asn $r<5$. As we never insert a number $>x$ into the list, we must have $x\ge 23$ and so $m\ge4$.

If $r=0$, the first few rounds produce
$$[]\to [5m]\to [] \to [5m-2]\to [5m-2,5m-3]\to [5m-2]\to [5m-2,5m-5] $$
and after this, the length will always be $\ge2$ when a pop is made. It follows that the first term $5m-2$ will stay there until the end. As thsi must be $=23$, we conclude $m=5$ and $x=25$.

If $r=1$, already the first round fails as pop(1) from an empty list fails.

If $r=2$, we fail in the second round:
$$ []\to[5m+2]\stackrel{\text{pop}(1)}\longrightarrow ??$$

If $r=3$,
$$ []\to [5m+3]\to[5m+3,5m+2]\to[5m+3]\to\ldots$$
which is in fact the sequence belonging to $x+2$ (see the case $r=0$ above) with the first two rounds dropped. It follows that $x=23$ is also a solution if $x=25$ is.

If $r=4$, we fail immeditely with an illegal pop(-1).

IN summary, there are two possibilities

Both $x=25$ and $x=23$ are solutions

There is no solution.

So let's play through $x=23$:
$$[]\to[23]\to[23,22]\to[23]\to [23,20]\to[23]\to\\
\to[23,18]\to[23,18,17]\to[23,17]\to [23,17,15]\to[23,17]\to\\
\to[23,17,13]\to[23,17,13,12]\to[23,13,12]\to[23,13,12,10]\to[23,13,12]\to\\
\to[23,13,12,8]\to[23,13,12,8,7]\to[23,12,8,7]\to[23,12,8,7,5]\to\\\to[23,12,8,7]\to[23,12,8,7,3]\to[23,12,8,7,3,2]\to[23,8,7,3,2]\to\\\to[23,8,7,3,2,0]$$
Yay!


答:
It suffices to show that for each complex $z \in \mathbb{C}$ the exponential series can be brought to the same form that you obtained in the right-hand side of your final relation. The most elementary method to achieve this is to realise that:
$$\mathrm{e}^z=\sum_{n=0}^{\infty}\frac{z^n}{n!}=\lim_{n \to \infty}\sum_{k=0}^n \frac{z^k}{k!}=\lim_{n \to \infty}\sum_{k=0}^{3n+2}\frac{z^k}{k!},$$
bearing in mind that in general if a complex sequence $u \in \mathbb{C}^{\mathbb{N}}$ has limit $a \in \mathbb{C}$ then the subsequence $(u_{kn+l})_{n \in \mathbb{N}}$ will have the same limit $a$ for any $k \in \mathbb{N}^*=\mathbb{N}\setminus\{0\}$ and $l \in \mathbb{N}$ (in our case $k=3$ and $l=2$).

Finally, since the map:
$$\begin{align*}
\left(\mathbb{N} \cap [0, n]\right) \times \{0, 1, 2\} &\to \mathbb{N} \cap [0, 3n+2] \\
(h, l) &\mapsto 3h+l
\end{align*}$$
is a bijection
it is clear that we can obtain the arrangement:
$$\sum_{k=0}^{3n+2}\frac{z^k}{k!}=\sum_{h=0}^n\left(\frac{z^{3h}}{(3h)!}+\frac{z^{3h+1}}{(3h+1)!}+\frac{z^{3h+2}}{(3h+2)!}\right),$$
which is precisely the general term of the sequence whose limit you are considering in the aforementioned right-hand side.


问:
Understanding proof of infinite countable product of separable space is separable
I am trying to understand the proof showing why infinite product of separable space is separable.

The link is:
https://math.stackexchange.com/questions/244427/proving-separability-of-the-countable-product-of-separable-spaces-using-density#:~:text=Proving%20separability%20of%20the%20countable%20product%20of%20separable%20spaces%20using%20density.,-real%2Danalysis%20general&text=The%20goal%20is%20to%20prove,has%20a%20countably%20dense%20subset.

I am reading the check marked answer. In the proof, the author does product the dense subset of each factor finitely up to $1<n<m$ and then for each factor with index $n>m$, the author fixes a point from each factor. So why is this done? Why can't we use the same method we did with the proof of finite product and simply product all the dense subset of each factor?

答: In that proof if you take all sequence $(x_n)$ where $x_n \in D_n$ for all $n$ then you do get an dense set in the product but this set is uncountable. Even an infinite product of two point sets is uncountable.


答:
Use a brute force method.

```<code>z=numeric(0)
for (x in 0:1000) {
y=numeric(0)
for (i in x:0) {
  if (i%%5==1) {
    y=y[-2]
  } else if (i%%5==4) {
    y=y[-length(y)]
  } else {
    y=append(y, i)
  }
}
if (identical(c(23, 8, 7, 3, 2, 0), y)) {
  z=append(z, x)
}
}
</code>```

The answers are

```<code>> print(z)
[1] 23 24 25 26
</code>```

Title: is there a mathematical way to go about this coding question?

Body:
i have the following python code(assume x is already defined as an integer):

```<code>numList=[] #creates a list that is empty
for i in range(x,-1,-1): #loops through from i=x to i=0, decreasing one each iteration
    if i%5==1: #if i mod 5 is 1...
        numList.pop(1) #remove the second item from the list
    elif i%5==4: # if i mod 5 is 4...
        numList.pop(-1) #remove the last item from the list
    else: #if i mod 5 is not 1 or 4...
        numList.append(i) #add i to the list
print(numList) #prints out the list

</code>```

i am given that the output is [23, 8, 7, 3, 2, 0](hopefully its clear what this means). i am tasked to find the possible values for x. is there some logical/mathematical way i can go about this without resorting to guess and check? i tried going "backwards" from the end result but it got too convoluted very quickly.

i have also tried setting x to 5a+2, 5a+3, and 5a (we dont consider 5a+1 and 5a+4 because that will terminate the code prematurely) because they give different values mod 5, but i cant seem to get anywhere with that as well.

Body:
Use a brute force method.

z=numeric(0)
for (x in 0:1000) {
y=numeric(0)
for (i in x:0) {
  if (i%%5==1) {
    y=y[-2]
  } else if (i%%5==4) {
    y=y[-length(y)]
  } else {
    y=append(y, i)
  }
}
if (identical(c(23, 8, 7, 3, 2, 0), y)) {
  z=append(z, x)
}
}

The answers are

> print(z)
[1] 23 24 25 26


答:
We cannot just pick a dense set $D_n$ in every $X_n$:  yes, the set $D=\prod_n D_n$ is dense, but $D$ is uncountable when we have infinitely many $D_n$ that are not singletons. A set $\{0,1\}^{\Bbb N}$ (Cantor cube) is also uncountable, e.g. because it's in bijection with the power set of $\Bbb N$ (Cantor's theorem). And we need a countable dense set to show separability. For a finite power, a finite product of countable sets is countable, so there the naive idea works. You have to be more clever for infinite products. We do essentially need that all basic open sets only depend on finitely many coordinates.

So we take (as in that proof) all points (i.e sequences ) in $D$ that are constant after a finite index. Of these there are only countably many. It's the difference between all subsets of $\Bbb N$ vs all finite subsets of $\Bbb N$.

Title: Understanding proof of infinite countable product of separable space is separable

Body:
I am trying to understand the proof showing why infinite product of separable space is separable.

The link is:
https://math.stackexchange.com/questions/244427/proving-separability-of-the-countable-product-of-separable-spaces-using-density#:~:text=Proving%20separability%20of%20the%20countable%20product%20of%20separable%20spaces%20using%20density.,-real%2Danalysis%20general&text=The%20goal%20is%20to%20prove,has%20a%20countably%20dense%20subset.

I am reading the check marked answer. In the proof, the author does product the dense subset of each factor finitely up to $1<n<m$ and then for each factor with index $n>m$, the author fixes a point from each factor. So why is this done? Why can't we use the same method we did with the proof of finite product and simply product all the dense subset of each factor?


答:
Let $U_n(z),V_n(z),W_n(z)$ be the partial sums of the $n$ first terms. As all three series are convergent,

$$U_\infty(z)+V_\infty(z)+W_\infty(z)=\lim_{n\to\infty}U_n(z)+\lim_{n\to\infty}V_n(z)+\lim_{n\to\infty}W_n(z)
\\=\lim_{n\to\infty}(U_n(z)+V_n(z)+W_n(z))=\lim_{n\to\infty}E_{3n}(z)=E_\infty(z)=e^z$$

where $E_m(z)$ denotes the sum of the $m$ first terms of the exponential development.

Technical note:

$E_{3n}(z)$ indeed converges to $e^z$ as it is a subsquence of the converging $E_m(z)$.


问:
Stopping times and martingales expectation
Suppose $M_n$ is a martingale and $T$ a stopping time. It is well known that $E[M_0]=E[M_n]$ for all $n\geq 0$ but under what assumptions on $T$ does it hold that $E[M_0]=E[M_{T\wedge n}]$? I don't think it holds for a $T$ that can be infinite on a set of positive probability, but I can't prove it, nor tell if it's actually true.

答:
For any martingale $\{M\}_{n\in\mathbb{N}}$ and any stopping time $T$ it is true that $\mathbb{E}[M_0]=\mathbb{E}[M_{T\wedge n}]$. Therefore, we also have $\mathbb{E}[M_0]=\lim_{n\rightarrow \infty}\mathbb{E}[M_{T\wedge n}]$. On the other hand, it is NOT always the case that $\lim_{n\rightarrow\infty}\mathbb{E}[M_{T\wedge n}]=\mathbb{E}[M_T]$. That requires justifying interchanging the order of limits and expectations.

Conditions under which this is true are established by optional sampling theorem. For example, we have $\lim_{n\rightarrow\infty}\mathbb{E}[M_{T\wedge n}]=\mathbb{E}[M_T]$ if any of the following holds:

$T$ is bounded, or

$\{M\}_{n\in\mathbb{N}}$ is bounded (i.e., there exists $C\in\mathbb{R_+}$ such that for all $n\in\mathbb{N}$ we have $|M_n|\leq C$), or

$\{M\}_{n\in\mathbb{N}}$ is uniformly integrable.


问:
From $\textsf{Prop}$ to $\textsf{Type}$
Let $P,\top:\textsf{Prop}$ (where $\textsf{Prop}$ is the universe of mere propositions), $A=\{P,\neg P\}:\textsf{Type}$ and $B$ is an identity function such that $B(x)=x$. Given the above definitions, $(\star)$ trivially holds.
$$P\rightarrow(\top\rightarrow\Sigma x:A.B(x))\tag{$\star$}$$
By the distribution axiom of $\rightarrow$, we obtain the following:
$$(P\rightarrow\top)\rightarrow(P\rightarrow\Sigma x:A.B(x)).\tag{$\star\star$}$$
But this seems odd to me because $P\rightarrow\top$ is a mere proposition by definition but $P\rightarrow\Sigma x:A.B(x)$ is not a mere proposition but just a type. Thus, it shows that from  a proof of a mere proposition (that is proof-irrelevant), we obtain a proof of a type (that is proof-relevant). This appears odd but it is natural in $(\star\star)$.

My question is: what amounts to a general condition for a proof-irrelevant type (of $\textsf{Prop}$) to imply a proof-relevant type such as in $(\star\star)$? Thanks!

答:
$\newcommand{\inj}{\mathsf{inj}}$
$\newcommand{\refl}{\mathsf{refl}}$
$\newcommand{\isProp}{\mathsf{isProp}}$
$\newcommand{\isContr}{\mathsf{isContr}}$

I'm not sure exactly of the source of confusion, but here are some things to think about.

$⊤ → A$ is equivalent to $A$, so your $(\star)$ is already equivalent to $P → Σx:A.B(x)$

Your $(*)$ already has two propositions implying $Σx:A.B(x)$, so it doesn't seem unreasonable that one proposition would imply it (especially since $⊤$ is rather vacuous).

Some non-propositions are easy to exhibit from any proposition. For instance, $λ x. 0$ has type $⊤ → ℕ$

$P$ actually implies that $T = Σx:A.B(x)$ is contractible (and thus a proposition) by the following argument:

I presumed $A$ means something equivalent to $ΣX:\mathcal U. \Vert (P=X)+(¬P=X) \Vert$ and so $B$ is $π_1$

$T$ is inhabited by $(P, \vert ι_1 \refl \vert, p)$, where $p$ is the premise. Let $(X, t, x)$ be some other witness

$P = X$ is a propsition because $P$ is, using $(B → \isProp(B)) → \isProp(B)$ with $B := (P = X)$

This allows us to deduce that $P = X$ using the truncated witness, because $p$ and $x$ together refute $¬P = X$

$p$ is related to $x$ by the path from $P$ to $X$ because $P$ and $X$ are propositions

The two truncated witnesses are related because the truncated types are propositions

Thus, $(X, t, x)$ is equal to our center of contraction

$¬P$ implies that $T$ is contractible by a very similar argument

$T → \isContr(T)$, because $T$ gives us a value $x : X$, $\isContr(T)$ is a proposition, and by cases on $(P=X)+(¬P=X)$, we can get either a value of $P$ or a value of $¬P$ to use 4 or 5

Thus $T$ is a proposition, because $(T → \isContr(T)) → \isProp(T)$. Here is a cubical Agda formalization.

However, the fact that $Σx:A.B(x)$ is actually a proposition is incidental. If the source of your question is an idea that propositions in HoTT are somehow restricted so that they may be 'erased', as in various other type theories, you should abandon that. A key feature of HoTT is that 'proof irrelevant' does not mean 'computationally irrelevant.' Mere propositions are proof irrelevant in the sense that for any two proofs $p, q$, any construction on $p$ can be systematically translated to an equivalent construction on $q$. However, this does not mean that $p$ and $q$ have no computational content.

An example of this is finite sets. One can define the type of finite sets as:

$$ΣX:\mathcal U. \Vert Σn:ℕ. \mathsf{Fin}(n) \simeq X \Vert$$

I.E. it is the collection of types that are merely equivalent to some set of a particular finite size. However, it can be shown that for a given such $X$, $n$ is unique, and may be extracted from the truncation. There is no mechanism for deducing the size $n$ from $X : \mathcal U$, so the conclusion is that the truncation was carrying it (at least, if you are expecting a computational explanation). There are just obligations for accessing it.

I don't think there is necessarily a general description of 'when a proposition can imply a non-proposition,' though. Propositions aren't arbitrarily separated/distinguished in HoTT. They are just types with a certain defined property, and there are multiple conceptual reasons a type might have that property.

Title: From $\textsf{Prop}$ to $\textsf{Type}$


问:
Lyapunov Stability of Non-autonomous Nonlinear Dynamical Systems
Let $\mathbf{F}:X\times\mathbb{R}^{+}\to X$ be a non-autonomous dynamical system, which is governed by $\dot{\mathbf{x}} = \mathbf{F}(\mathbf{x}, t, u)$, viz,
\begin{equation}
\begin{split}
\dot{x}_1 &= x_2 - 314.2 \\
\dot{x}_2 &=  122\sin{x_1}\cos{x_1} - 154.8x_3\sin{x_1} - 0.5x_2 + 201.2  \\
\dot{x}_3 &= 0.64\cos{x_1} - 0.8x_3 +  0.2u
\end{split}
\end{equation}
where $\mathbf{x} = \mathbf{x}(t) = [x_1(t), x_2(t), x_3(t)]^T \in X \subset\mathbb{R}^3$ are the state variables; $t \in \mathbb{R}^{+}$ denotes time and $u = u(\mathbf{x},t) \in U \subset \mathbb{R}$ is the driving variable dependent both on $\mathbf{x}$ and $t$, which is unknown. We mention here that $X=[0.4, 2.1] \times [298.5, 329.9] \times [-2, 2]$, and $U = [-6, 6]$.

We need to solve the Lyapunov stability problem, namely, how to find a bounded control law $u(\mathbf{x},t)\in U$ and the corresponding invariance set $\mathcal{X}_o\subset X$, such that $\forall \mathbf{x}_0 \in \mathcal{X}_o, \forall t>0$, $\mathbf{F}(\mathbf{x}_0, t, u) \in \mathcal{X}_o$?

Alternatively, given a positive-definite Lyapunov function such as
$$
V(\mathbf{x}) = 0.5(x_2 - 314.2)^2 + 0.5(1.6\sin{x_1}\cos{x_1} - 2x_3\sin{x_1} + 0.6)^2
$$
find $u(\mathbf{x},t) \in U$ and the invariant set $\mathcal{X}_o \subset X$, $\forall \mathbf{x}_0 \in \mathcal{X}_o$, $\forall t>0$, such that
$$
\frac{\mathrm{d}}{\mathrm{d}t} V(\mathbf{x}, t) \leq -\gamma V(\mathbf{x}, t)
$$
where $\gamma>0$  is the decay rate corresponding to the convergence speed to the equilibrium.  For example, it could be in the range $[0.01,1]$.

The analytical solution might not be available. We may discretize the above problem. Hence, a time series of the control law is preferable.

Any hints or comments are appreciated.

答:
Why is the analytic solution not available?

Have you done the time derivative of $V(x)$?

Hint: Once you compute $\frac{d}{dt}V(x)$ (note that it does not depends explicitly on $t$), you can upper bound it with expressions like $\sin x \leq 1$. And for the worst case condition, find such $u$ in order to make the derivative always negative for some compact set $\mathcal{C}=\{x: V(x) \leq c\}$. Actually, $\mathcal{C}$ of interest is given by the bounds of your state vector in the statement of your problem.


问:
selecting k items from X+Y items such that count of X is odd.
Given x and y denoting count of two different items, also given a value K.

We need to select a total of K items from those X+Y items such that the count of chosen X items is odd.

For example, X=4,Y=3,K=3.

then number of ways= (4C1)(3C2)+(4C3)(3C0)

here nCr denotes n!/(r!*(n-r)!)

Is there any formula to calculate this?

答:
Assuming you have a little confusion over the '$!$' sign:

$${n \choose r}=\frac{n!}{r!(n-r)!}$$

Where:
$$n! = n\times(n-1)\times\dotsb\times1$$
$$r! = r\times(r-1)\times\dotsb\times1$$
And:
$$(n-r)! = (n-r)\times((n-r)-1)\times\dotsb\times1$$

For more info, read up about the factorial and combination.


问:
Rudin appendix on Dedekind cuts
In the appendix on Dedekind cuts in Rudin's book, he defines for each $r \in \mathbb{Q}$ the Dedekind cut $r^* = \{q \in \mathbb{Q} \mid q < r\}$. He then asserts that
$$r^* s^* = (rs)^*.$$
Rudin omits the proof and states that it is similar for the proof that $(r+s)^* = r^* + s^*$. I do not understand how this is the case, in part because I need to consider three cases: both positive, positive negative, and one positive and one negative.

Suppose both are positive, either $r^*, s^* > 0*$. Then $r^* s^* = \{ab \mid a \in r^*, b \in s^*, ab > 0\}$. To show inclusion in the right-hand side, I need to show that given an $ab \in r^* s^*$, I have $ab < rs$. Since $a \in r^*$, $a < r$. Since $b \in s^*$, $b < s$. So $ab < rs$, giving the first inclusion.

I'm having trouble with the second inclusion. If $q \in (rs)^*$, then $q < rs$. Attempting to follow the proof for addition, I can then say that $rs - q > 0$ and define $rs - q = t$ .Then I want to construct, with $t$, two positive rational numbers whose product is $q$, the first term of which is in $r^*$, the second of which is in $s^*$. I can't figure out how to do this. Subtracting $t$, as in the addition proof, doesn't work, and multiplying it gives an extraneous $t^2$ factor.

I haven't had much success with the remaining cases, but I think I could make more progress if I could get this first one. Any help would be appreciated.

答:
Here is a simple strategy to find rational numbers $a, b$ with $0<a<r,0<b<s$ such that $ab=q$ where $q$ is a given number with $0<q<rs$. Just choose any rational $a$ which lies between $q/s$ and $r$ and then but $b=q/a$ and you are done.

Dedekind's construction of real numbers is the simplest approach possible. It is really a remarkable feat of Rudin to turn it into something seemingly complex.

Title: Rudin appendix on Dedekind cuts


问:
Let $A$ be a subset of $\mathbb{R}$ with the following property: $a \leq 1$ for $a$ in $A$
Let $A$ be a subset of $\mathbb{R}$ with the following property: $a \leq 1$ for $a$ in $A$

Give an example that sup$A = 1$

Give an example that sup$A < 1$

What does it mean to give examples of such conditions? Does it mean to pick different sets of $A$ that satisfy those conditions?

If so, how does a set have the supremum that $ \leq 1$? Don't all sets have one supremum if exists?

答:
If we take $A = (0, 1)$ this is an example for 1)

If we take $A = (0, 0.5)$ this is an example for 2)

And yes, every subset of $\mathbb{R}$ has a unique supremum (if it exists).


问:
Quotient of abelian groups of rank $2$
Let $A, B$ be abelian groups, $B$ is contained in $A$, both $A$ and $B$ are assumed to have rank $2$. Is there a standard way to show that the quotient group $A/B$ is finite? I think there exists some general theorem about modules finitely generated over a PID, but i can't find the precise theorem i need to apply, so even a simple reference would be very appreciated. Thank you.</p>

答: I think the theorem you want is that there is a generating set $x,y$ for $A$ and there are non-negative integers $e$ and $f$ such that $ex,fy$ is a generating set for $B.$ Then $[A:B] = |ef|$. This should be proved (in somewhat greater generality) in any algebra text that treats modules over a PID.</p>

Title: Quotient of abelian groups of rank $2$


问:
Implicitly Differentiating a System of Matrix Equations
Setup

Let $\mathbf a$ be an arbitrary $m\times 1$ vector, $\mathbf B$ be an arbitrary $n\times m$ matrix, with $m>n$, and $\mathbf C$ be a symmetric $m\times m$ matrix. The scalar $\lambda$ is also given.

The $n\times 1$ vector $\mathbf h$, the $m\times m$ matrix $\mathbf \Gamma$, and the scalars $w$, and $s$ satisfy the following system of equations:
\begin{align*}
\mathbf h =  & \;\left(\mathbf B \,{\mathbf \Gamma}\,\mathbf B'\right)^{-1}\,\mathbf B\,{\mathbf\Gamma}\,\mathbf{a}  ,\\[2ex]
\mathbf {\Gamma} =&\;\mathbf I_m   +w\,\mathbf C,\\[2ex]
w=&\;\frac{1}{1-\lambda s},\\[2ex]
s=&\ \left(\mathbf h'\,\mathbf B-\mathbf a' \right)\mathbf {C}\left(\mathbf B^{\prime}\,\mathbf h -\mathbf a \right).
\end{align*}

Question

I want to find

$$\frac{\partial w}{\partial \lambda}=\mathbf?$$

The problem is that changing $\lambda$ changes $w$ which changes $\mathbf \Gamma$, and so on. So, I have to somehow use the Implicit Function Theorem. For systems of scalar variables, I am used to totally differentiating the system, but I am not sure how to deal with the matrix equations in that case. Any help would be very welcome.

答:
This is a pretty straightforward application of the chain rule. Given

$$\begin{aligned}
0 &= f(\lambda, w(\lambda), s(\lambda)) = (1-\lambda s) w - 1
\\[2ex]
\implies 0 &= \frac{{\rm d\,} f(\lambda, w(\lambda), s(\lambda))}{{\rm d\,}  \lambda}
= \frac{\partial f}{\partial \lambda}
+ \frac{\partial f}{\partial w}\frac{\partial w}{\partial \lambda}
+ \frac{\partial f}{\partial s}\frac{\partial s}{\partial {\mathbf h}}\frac{\partial {\mathbf h}}{\partial w}\frac{\partial w}{\partial \lambda}
\\[2ex]
&\iff \Big(\frac{\partial f}{\partial w}
+ \frac{\partial f}{\partial s}\frac{\partial s}{\partial {\mathbf h}}\frac{\partial {\mathbf h}}{\partial w}\Big)\frac{\partial w}{\partial \lambda} = -\frac{\partial f}{\partial \lambda}
\end{aligned}$$

So, your derivative can be found by solving the linear system with $-\frac{\partial f}{\partial \lambda} = sw$, $\frac{\partial f}{\partial w} = 1-\lambda s$, $\frac{\partial f}{\partial s}=\lambda w$ and $\frac{\partial s}{\partial {\mathbf h}} = 2{\mathbf B}{\mathbf C}({\mathbf B}'{\mathbf h}-{\mathbf a})$. With regard to $\frac{\partial {\mathbf h}}{\partial w}$, we could do it directly, using $\partial X^{-1}= -X^{-1} (\partial X)X^{-1}$, but we also just apply the implicit function theorem a second time:

$$\begin{aligned}
0 &=  g(w, {\mathbf h}(w)) = \left(\mathbf B (\mathbb I + w \mathbf C) \mathbf B'\right) \mathbf h -  \mathbf B(\mathbb I + w \mathbf C)\mathbf{a}
\\[2ex]
\implies 0 &= \frac{{\rm d\,} g(w, {\mathbf h}(w))}{{\rm d\,}  w}
= \frac{\partial g}{\partial w}
+ \frac{\partial g}{\partial {\mathbf h}}\frac{\partial {\mathbf h}}{\partial w}
\\[2ex]
&\iff \frac{\partial g}{\partial {\mathbf h}}\frac{\partial {\mathbf h}}{\partial w} = -\frac{\partial g}{\partial w}
\end{aligned}$$

where $\frac{\partial g}{\partial w} = {\mathbf B}{\mathbf C}{\mathbf B}'h - {\mathbf B}{\mathbf C}a$ and $\frac{\partial g}{\partial {\mathbf h}} = {\mathbf B}{\mathbf \Gamma} {\mathbf B}'$. So in total your derivative is given by solution of the nested linear system

$$\begin{aligned}
(1)&&&{\mathbf B}{\mathbf \Gamma} {\mathbf B}' \tfrac{\partial \mathbf h}{\partial w} = -{\mathbf B}{\mathbf C}{\mathbf B}'\mathbf h + {\mathbf B}{\mathbf C}\mathbf a
\\
(2)&&&\big(1-\lambda s + 2\lambda w (\mathbf a'- {\mathbf h}'{\mathbf B}){\mathbf C}{\mathbf B}'\tfrac{\partial \mathbf h}{\partial w}\big)\tfrac{\partial w}{\partial \lambda} = sw
\end{aligned}$$

If you want to, you can plug everything back in and try to simplify it. As a final remark, you may wonder why $\frac{\partial s}{\partial {\mathbf h}}$ appears transposed in the formula, this is due to how the tensor-contraction between $\frac{\partial s}{\partial {\mathbf h}}$ and $\frac{\partial {\mathbf h}}{\partial w}$ operates. You can either write it out or realize that the result needs to be a scalar again.


答:
In the way you asked the question I think the aim is to give an example of set $A \subseteq \mathbb{R}$ such all its elements are $\le 1$ and such that $\sup A = 1$ or such that $\sup A < 1$

To answer the first question take for example $A :=(0,1) \subseteq \mathbb{R}$. It clearly has $\sup A = 1$

For the second just take $A:= (-1,0)$.

Body:
Let $A$ be a subset of $\mathbb{R}$ with the following property: $a \leq 1$ for $a$ in $A$

Give an example that sup$A = 1$

Give an example that sup$A < 1$

What does it mean to give examples of such conditions? Does it mean to pick different sets of $A$ that satisfy those conditions?

If so, how does a set have the supremum that $ \leq 1$? Don't all sets have one supremum if exists?


问:
What does it mean to have the transpose symbol on the solution set of a linear system?
I know that the superscript $T$ on a matrix indicates that the matrix is to be "transposed", that is its columns turned into rows; however for some reason I am seeing it on some solution sets to linear systems, and I am unsure as to why. For example, this is an example of something I have come across:

\begin{equation}
\mathrm{x} = (1 - \alpha, 1, \alpha)^{T}, \alpha \in \mathbb{R}
\end{equation}

Since this is a solution set and not a matrix, what does this "Transpose" symbol indicate/mean?

答:
You can also transpose a vector. This will turn a row vector into a column vector and vice versa:
$$
(1-\alpha, \alpha)^T = \begin{pmatrix}1-\alpha \\ \alpha\end{pmatrix}
$$

We often transpose vectors like this, so that we can properly multiply them with a matrix:
$$
\begin{pmatrix}a & b \\ c & d\end{pmatrix}(1-\alpha, \alpha)^T
$$
is well defined, while
$$
\begin{pmatrix}a & b \\ c & d\end{pmatrix}(1-\alpha, \alpha)
$$
is not since the dimensions don't match.

edit: For fixed $\alpha\in\Bbb R$
$$
(1-\alpha, \alpha)^T
$$
is not a set, but a fixed vector! The solution set is then something like
$$
\{(1-\alpha, \alpha)^T : \alpha\in\Bbb R\} = \left\{\begin{pmatrix}1-\alpha \\ \alpha\end{pmatrix} : \alpha\in\Bbb R\right\}.
$$


问:
Finding the Root of a Polynomial over $\mathbb F_{16}$
Suppose we have the polynomial $p(x) = x^4 + x + 1$ over $\mathbb F_{16}.$  How does one find a root $\alpha$ of $p(x)$?

I have done this for $\mathbb F_2$ by listing and trying every possible element. That approach doesn't seem practical here.

答:
$p(x)=x^4+x+1$ is irreducible over $GF(2)$ and $GF(16) = GF(2)[x]/\langle p(x)\rangle$, where $GF(16)$ is generated by a zero $\alpha$ of $p(x)$, i.e.,  $\alpha^4+\alpha+1=0$ in
$GF(16)$.

$p(x)=x^4+x+1$ factors in $GF(16)$ into linear factors,

$p(x) = (x-\alpha)(x-\alpha^2)(x-\alpha^4)(x-\alpha^8)$

by the Frobenius automorphism $a\mapsto a^2$.


Body:
Working in $S_7$ solve the equations $(1234)x=(45)(17)$ and $x^2=(45)(17).$

For the first one I got that $(1234)^{-1}(1234)x=(1234)^{-1}(45)(17) \Longrightarrow x=(4321)(45)(17) =(17432).$ Trying to verify the result I ended up with $$(1234)(17432)=(17)(2)(3)(4)=(17)$$ what might I be doing wrong here?

Body: $$(4321)(45)(17) =(174532)\neq (17432)$$


问:
The space [0,1] relativesed with usual topology is Hausdorff
Source: On Minimal Separation Axioms in Topological Spaces (2012)

Definition(Minimal Closed Set): A proper nonempty closed subset $F$ of a topological space $X$ is said to be minimal closed if any
closed set which is contained in $F$ is $\emptyset$ or $F$.

Definition(Minimal Completely Regular Space): For every point $x\in X$ and each minimal closed set $F\subset X$ such that $x\not\in
   F$, there exist a continuous map $f:X\rightarrow [0,1]$ such that
$f(x)=0$ and $f(F)=\{1\}$.

Definition(Minimal c-regular Space): For every point $x\in X$ and each minimal closed set $F\subset X$ such that $x\not\in F$, there
exists open sets $U,V\subset X$ such that $x\in U$, $F\subset V$ and
$U\cap V=\emptyset$.

Theorem: Every min-completely regular space is min c-regular space.

Proof: Let $X$ be a min-completely regular space. To prove $X$ is min c-regular space. Let $x\in X$ and $F$ be any minimal closed set in $X$ such that $x\not\in F$. Since $X$ is min-completely regular space, there exists a continuous map $f:X\rightarrow [0,1]$ such that $f(x)=0$ and $f(F)=\{1\}$. Also, it is easy to see that the space $[0,1]$ with relativesed usual topology is Hausdorff space. Hence, there exist open subsets $G,H\subset [0,1]$ such that $0\in G$, $1\in H$ and $G\cap H=\emptyset$.

Since $f$ is continuous, $f^{-1}(G),f^{-1}(H)\subset X$ are open such that

$f^{-1}(G)\cap f^{-1}(H)=f^{-1}(G\cap H)=f^{-1}(\emptyset)=\emptyset$,

$f(x)=0\in G \implies f(x)\in G \implies x\in f^{-1}(G)$,

$f(F)=\{1\}\subset H \implies f(F)\subset H \implies F\subset f^{-1}(H)$.

Thus, for every $x\in X$ and each minimal closed set $F\subset X$ such that $x\not\in F$, there exists open sets $f^{-1}(G),f^{-1}(H)\subset X$ such that $x\in f^{-1}(G)$, $F\subset f^{-1}(H)$ and $f^{-1}(G)\cap f^{-1}(H)=\emptyset$. Hence, $X$ is min c-regular space.

Now, I want to clarify the sentence "the space $[0,1]$ with relativised usual topology is Hausdorff."

Firstly, I need to know that what the usual topology on $[0,1]$ is.
I have tried to find the base of the usual topology on $[0,1]$ but...

$X=[0,1]$, $\tau=$ Standard Topology

If I consider that the base is $\beta=\{(a,b)\in X : 0<b<a<1\}$, then I cannot generate $[0,1]$.

Thank you for your helps..

答: This means that $[0,1]$ in the inherited topology from $\Bbb R$ (in the usual topology too) is Hausdorff. This is quite clear as that topology is metric (so Hausdorff), ordered (another reason it's Hausdorff) and hence so are all its subspaces as well as Hausdorff-ness is hereditary. No more than that.


问:
Is there a general way to convert a continued fraction into a recursive formula?
I was looking into this continued fraction:

$$e = 3 - \frac{1}{4 - \frac{2}{5 - \frac{3}{6 -...}}}$$

I was able to duplicate the approximations of the continued fraction with this recursive formula:

$$f(n+1) = \frac{f(n)*(n+1)*(n+2)^2*(n+1)!-1}{(n+1)*(n+2)*(n+2)!}\tag{1}$$
Where $f(0)=3$

The first few terms are:

$f(1)=11/4=2.75$

$f(2)=49/18=2.72222$

$f(3)=261/96=2.71875$

$f(4)=1631/600=2.71833$

...

Is there a general way to convert continued fractions into recursive formulas?

Or do these need to be dealt with on a case by case basis?

This is a reply to a comment by WA Don

Thank you for your thoughtful reply.

Which of my terms is off?

In the meantime, I'll provide a more detailed derivation.

The denominator, as a function of n, can be represented as:
$$g(n)=(n+1)*(n+1)!$$
The numerator, as a recursive function of n, can be represented as:
$$h(n)=(h(n-1)*(n+1)^2-1)/n$$
So we now can write the recursive approximation of e as:
$$f(n)=h(n)/g(n)$$
$$=\frac{(h(n-1)*(n+1)^2-1)/n}{(n+1)*(n+1)!}$$
$$=\frac{h(n-1)*(n+1)^2-1}{n*(n+1)*(n+1)!}\tag{2}$$
Now substituting $h(n-1)=f(n-1)*g(n-1)$ into (2), we get
$$f(n)=\frac{f(n-1)*g(n-1)*(n+1)^2-1}{n*(n+1)*(n+1)!}$$
$$=\frac{f(n-1)*n*n!*(n+1)^2-1}{n*(n+1)*(n+1)!}\tag{3}$$
Substituting n+1 for n in (3), we get

$$f(n+1)=\frac{f(n)*(n+1)*(n+1)!*(n+2)^2-1}{(n+1)*(n+2)*(n+2)!}\tag{4}$$
Which is just the original formula (1)

I'm still seeing an exact match between the continued fraction approximation and the recursive function in (1) or (4).

So, if there is an error in the forgoing, I'd be grateful to have it pointed out. And the specific term that deviates from the continued fraction approximation.
The first few terms without removing common factors are:

$f(1)=11/4$

$f(2)=98/36$

$f(3)=783/288$

$f(4)=6524/2400$

$f(5)=58715/21600$

$f(6)=575406/211680$

...

答:
A general approach to evaluating a continued fraction is the following.  Consider the fraction,
\begin{align}
f = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{p_3}{\ddots}}}
\end{align}
If we define functions of $w \in \mathbb C$ by,
\begin{align}
t_0(w) = q_0 + w, \quad t_n(w) = \frac{p_n}{q_n +w}
\end{align}
then when we combine $t_0$ through $t_n$ we obtain,
\begin{align}
c_n(w) = t_0\circ t_1 \circ \cdots \circ t_n(w) = q_0 + \dfrac{p_1}{q_1+\dfrac{p_2}{q_2+\dfrac{\ddots}{q_{n-1}+\dfrac{p_n}{q_n+w}}}}
\end{align}
and we interpret $f$ to be the limit of this as $n \to \infty$ when $w=0$ (if the limit exists).  The number $c_n(0)$ is called the $n$th convergent.

A useful recurrence relation exists for $c_n(w)$, $n \geqslant 0$,
\begin{align}
c_n(w) = \frac{P_n+P_{n-1}w}{Q_n +Q_{n-1} w} \tag{1}\label{eq1}
\end{align}
where we use initial conditions, $P_0 = q_0, Q_0 = 1, P_{-1} = 1, Q_{-1}=0$ and the $P_n, Q_n$ do not depend on $w$ and can be obtained from the recurrence,
\begin{align}
\begin{array}{l}
P_{n+1} = q_{n+1}P_n + p_{n+1} P_{n-1} \\
Q_{n+1} = q_{n+1}Q_n + p_{n+1} Q_{n-1} .
\end{array}\tag{2}\label{eq2}
\end{align}
This is proved by induction.  We observe that the given initial conditions imply equation \eqref{eq1} holds when $n=0$.  Then, if \eqref{eq1} holds for some $n \geqslant 0$, from the definition,
\begin{align}
c_{n+1}(w) &= c_n(p_{n+1}/(q_{n+1}+w)) \\
&= \frac{P_n+P_{n-1}(p_{n+1}/(q_{n+1}+w))}{Q_n+Q_{n-1}(p_{n+1}/(q_{n+1}+w))} \\
&=\frac{P_{n}q_{n+1}+P_n w + P_{n-1}p_{n+1}}{Q_{n}q_{n+1}+Q_n w + Q_{n-1}p_{n+1}} \\
&=\frac{P_{n+1}+P_n w}{Q_{n+1}+Q_n w}.
\end{align}
Thus \eqref{eq1} also holds for $n+1$ with the new coefficients $P_{n+1}, Q_{n+1}$ given by \eqref{eq2}.

Postscipt:  After a false start (apologies) the methodology described in this answer delivers the same recurrence as in the OP question,  $f_{n+1}=\frac{f_n(n+1)(n+2)^2(n+1)!-1}{(n+1)(n+2)(n+2)!}$,
which generates the same fractions, $\frac{3}{1}, \frac{11}{4}, \frac{49}{18}, \frac{261}{96}, \cdots$.

The general formula presented here can be used to evaluate any continued fraction.  But it consists of a ratio of two second order recurrence formulae.
I doubt it can always be simplified to a single first order recurrence, as it can in this example.


问:
trigonometric identity related to $ \sum_{n=1}^{\infty}\frac{\sin(n)\sin(n^{2})}{\sqrt{n}} $
As homework I was given the following series to check for convergence:

$ \displaystyle \sum_{n=1}^{\infty}\dfrac{\sin(n)\sin(n^{2})}{\sqrt{n}}  $

and the tip was "use the appropriate identity".

I'm trying to use Dirichlet's test and show that it's the product of a null monotonic sequence and a bounded series, but I can't figure out which trig. identity is needed.

Can anyone point me towards the right direction?

Many thanks.

答:
Hint: You can show that
$$
\sum\limits_{n=1}^N\sin(n)\sin(n^2)=\frac{1}{2}(1-\cos(N^2+N))
$$
To do this use identity
$$
\sin(\alpha)\sin(\beta)=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta))
$$


问:
Why does the function of a random variable X, have a different probability than the inverse of the function in the PDF of X?
There is probably a very easy answer to this but I'm just not getting it.   Suppose X is a random variable with a PDF = $f_X(x)$.

If $Y$ is a function of $X$ such that:

$$Y=g(X)$$ and $$g^{-1}(Y) = X$$
where $g(x)$ is differentiable and monotonic. The PDF of $Y$, then $f_Y(x)$ is

$$f_Y(y)= f_X(x) / g'(g^{-1}(y))$$

While this formula makes sense, can anyone provide some intuition as to why $$f_Y(y)\neq f_X(g^{-1}(Y))$$

Since $Y=y$ iff $X = g^{-1}(y)$, why would the probability of $X=g^{-1}(y)$ differ from the probability of $Y=y$.

For example if $g(x) = X^2$ and $$f_X(x)=\frac{e^{-x^2/2}}{\sqrt{2\pi}},\forall x\in\Bbb R$$then $$f_Y(y)=\frac{e^{-y/2}}{\sqrt{2y\pi}}$$

Why is it that $f_Y(Y=4) \neq f_X(X=2)+f_X(X=-2)$?

First time poster so any comments on question format and content would also be appreciated. Thanks.

答:
I suppose the pithy answer to your question is that this is a consequence of change-of-variables in integral calculus:
$$\int_\color{gray}{a}^\color{gray}{b} f(g(t))g^\prime(t)dt = \int_{\color{gray}{g(a)}}^{\color{gray}{g(b})} f(u) du,$$
for a monotonically increasing function $g$, and continuous $f$ over $[a,b]$.

If you would like an appreciation for the geometry of what this represents though...

In short, the factor you refer to is needed to preserve total probability density (by rescaling the original density function).

[Using $g^\prime(g^{-1}(y))$ here is perhaps a typo...I think you're trying to express the derivative of the inverse transformation?]

The density function of $Y$ is better presented as
$$f_Y(y)=f_X(g^{-1}(y)) \lvert\frac{d}{dy}g^{-1}(y)\rvert,$$
where $g$ is a monotonic function (and hence so is $g^{-1}$) over the support of $X$. Students love to ignore the $\frac{d}{dy}g^{-1}(y)$ factor, but it is where all the action happens for continuous $X$, as I hope to suggest to you!

(Side note: be careful with your notation: you have written $f_Y(\color{blue}{y})= f_X(\color{red}{x}) / g'(g^{-1}(\color{blue}{y}))$, which is a function of $y$ on the left hand side but still has an $x$ on the right hand side, which may contribute to confusion - the $x$ has been (and should be) replaced by $g^{-1}(y)$, a function of $y$.)

As Shubham Johri pointed out, when $X$ is discrete then (effectively) $\frac{d}{dy}g^{-1}(y)$ "falls away". However, "falling away" doesn't give you much of a feel for what is really going on (geometrically).

A good starting point is to accept that probability masses (associated with discrete variables) and densities (continuous variables) are "incompressible". When you perform a statistical transformation ($Y=g(X)$) you are shifting probability mass or density around, but preserving the total mass or density ($\sum_{\text{all y}} f_Y(y)=1$ or $\int_\mathbb{R} f_Y(y)dy=1$).

For a physical analogy for a continuous random variable, think of taking an inflated balloon: you can reshape the balloon, without popping it, in many different ways (by squashing it, for example), and the total volume in the balloon remains the same (yes, real gasses do compress/expand but that is not the point here - fill the balloon with water in your mind if it helps you get past the limitations of my physical analogy).

Now, for a monotonic transformation of a continuous random variable, consider how the supports map (from $X$ to $Y$). To make it concrete, let's suppose that the support of $X$ is $(0,1)$ and we are going to transform this variable to $Y=e^X-1=g(X)$ (notice that this "$g$" is monotonic). The transformed support of $Y$ is then $(e^0-1,e^1-1)=(0,e-1)$ (we can do this because $e^x-1$ is monotonic in $x$; if this were not the case, we would need to be far more careful in determining the new support). The point is that geometrically the support (not the density directly!) has been "stretched" by this transformation, from a range of $1$ for $X$ to a range of $e-1$ for $Y$. The density simply follows this stretching.

In order to preserve the total probability density (which is incompressible), the density of $Y$ will need to be rescaled. How much to rescale by? That depends on how much local "stretching" has occurred. (Think of the graph of the transformation, $y=e^x-1$, over the support of $X$ (i.e. $(0,1)$.) The amount of local "stretching" that must be compensated for is given precisely by the factor $\frac{d}{dy}g^{-1}(y)$! In this example, $g^{-1}(y)=\ln (y+1)$, and $\frac{d}{dy}g^{-1}(y)=\frac{1}{y+1}$. You will notice that (for this transformation) for values of $y>0$, this factor is less than $1$ i.e. the straight substitution you enquired about, $f_X(g^{-1}(y))$, is decreased by this factor. And it needs to be decreased more the larger $y$ is. Why is that (geometrically)? The transformation we are using ($e^X-1$) stretches (the $x$-axis) more, the larger $x$ is. The height of the density above this needs to be decreased then, in order to preserve the total density in a local region when it goes through the transformation. In non-rigorous terms, the density of $X$ is "smeared" out by this particular transformation.

Since point masses can't be "smeared", there is no need for this scaling factor in the density transformation.

[If the non-linear transformation above is confusing at first, try the same argument using a linear transformation, $Y=g(X)=a+b X$. In this case the scaling factor, $\lvert1/b\rvert$, is constant over the support. And, in the really trivial case where $b=1$, no scaling is required at all...because the transformation is only a shift, $Y=a+X$, of the distribution of $X$.]


问:
Is the countable union of second countable open subspaces second countable?
I need help checking if the statement below is true. I have written a proof to it. If the statement is incorrect I'd greatly appreciate it you could point out where I went wrong in my proof. Thank you.

Let $X$ be a second countable topological space and $(X_{n})_{n\in\mathbb{N}}$ a sequences of open subspaces of $X$ with the subspace topology. If $X=\cup_{n\in\mathbb{N}}X_{n}$ and each $X_{n}$ is second countable then $X$ is second countable.

My proof:

For each $n\in\mathbb{N}$ let $B_{n}$ be a countable basis for $X_{n}$. Let $B=\cup_{n\in\mathbb{N}}B_{n}$. Notice that B is a countable union of countable sets so countable. Now, given $V$ open in $X$, for each $n \in \mathbb{N}$ we have that $V \cap X_{n}$ open in $X_{n}$. So $\exists C_{n} \subset B_{n}$ such that $V \cap X_{n} = \cup_{Y\in C_{n}}Y$. Taking $C=\cup_{n\in\mathbb{N}}C_{n}$, we have that $V  = \cup_{Y\in C}Y$.

$\textbf{QED}$

答: Your collection $B$ need not be countable. Even the  collectoion of all $U$ such that $U\cap X_1 \in B_1$ need not be countable.


问:
Examples of liminf sequences
I am looking to find an example of $(a_n)$ and $(b_n)$ such that

liminf $a_n$ + liminf $b_n$ < liminf $(a_n+b_n)$

Ive found the proof for this thing but I am having the worse trouble coming up with an example for it. I would appreciate any help, thanks.

答:
What about
$$a_n = (-1)^n \quad \text{and} \quad b_n = (-1)^{n+1} \quad ?$$

Title: Examples of liminf sequences


答:
Your union does not consist of open subsets of $X$, so is not a base. You'd have to assume all $X_n$ are open in $X$ for this to work.

Moreover the whole thing is false: there are countable spaces that are not second countable, so you could never proof such a "theorem".

You do have that a union of second countable subspaces has a countable network by essentially this proof. (Look at network weight here, under the topology section.)

Title: Is the countable union of second countable open subspaces second countable?

Body:
I need help checking if the statement below is true. I have written a proof to it. If the statement is incorrect I'd greatly appreciate it you could point out where I went wrong in my proof. Thank you.

Let $X$ be a second countable topological space and $(X_{n})_{n\in\mathbb{N}}$ a sequences of open subspaces of $X$ with the subspace topology. If $X=\cup_{n\in\mathbb{N}}X_{n}$ and each $X_{n}$ is second countable then $X$ is second countable.

My proof:

For each $n\in\mathbb{N}$ let $B_{n}$ be a countable basis for $X_{n}$. Let $B=\cup_{n\in\mathbb{N}}B_{n}$. Notice that B is a countable union of countable sets so countable. Now, given $V$ open in $X$, for each $n \in \mathbb{N}$ we have that $V \cap X_{n}$ open in $X_{n}$. So $\exists C_{n} \subset B_{n}$ such that $V \cap X_{n} = \cup_{Y\in C_{n}}Y$. Taking $C=\cup_{n\in\mathbb{N}}C_{n}$, we have that $V  = \cup_{Y\in C}Y$.

$\textbf{QED}$


问:
In $\mathsf{ZF}$, for each set $A$ exists an ordinal $\alpha$ such that no surjection $A\to \alpha$ exists
Some observations.

The statement can easily be proven in $\mathsf{ZFC}$, but we do not
assume that $\mathsf{AC}$ holds.

If one proves that $S=\{\beta \mid \text{there is a surjection $A\to
   \beta$}\}$ is a set, the statement follows by taking $\alpha=\sup S$.

An obvious candidate would be $\alpha=h(A)$, the Hartogs number of $A$.

All my attempts to prove the statement required some form of $\mathsf{AC}$. I'm starting to wonder if the statement can be proven in $\mathsf{ZF}$.

答:
You can really just repeat the proof of Hartogs theorem, but use surjections instead. An easy proof is simply consider the fact that if there exists a surjection $f\colon X\to Y$, then there is an injection from $Y$ into $\mathcal P(X)$. So the Hartogs number of $\mathcal P(X)$ is an upper bound of this ordinal you're looking for.

Interestingly enough, it is not necessarily the Hartogs number, and it may not be equal to the Hartogs of the power set either.

(And one last piece of trivia, the name of the least non-zero ordinal that $X$ does not map onto is the Lindenbaum number.)


问:
Uniserial modules are determined up to isomorphism by their composition series?
I have a question on uniserial modules. Here, every modules are finitely generated modules over a finite dimensional algebra $A$ over a field $K$.

In the book Elements of the Representation Theory of Associative Algebras: Volume 1 , there is a remark as below. (p.164)

We also notice that uniserial modules are determined up to isomorphism
by their composition series, that is, if M and N are uniserial modules and
have the same composition factors in the same order, then they are isomorphic.
An isomorphism is constructed by an obvious induction on the
common composition length of M and N.

But I cannot figure out how to make an induction step. Obviously, we have an isomorphism $M/radM \cong N/radN$ and by induction hypothesis, have an isomorphism $radM\cong radN$, but I don't know how to make an isomorphism between $M$ and $N$ from these informations. How can I show that those two are isomorphic?

答:
This is not true in general. For example, the Klein four group has infinitely many $2$-dimensional uniserial modules over a field of characteristic $2$. I'm not quite sure what the authors mean in this section. It doesn't look like they are considering specific classes of algebras, where this might well be true.

An explicit example for $G=\langle x,y\rangle$ Klein four is to map
$$ x\mapsto \begin{pmatrix}1&1\\0&1\end{pmatrix}$$
and
$$ y\mapsto \begin{pmatrix}1&\lambda\\0&1\end{pmatrix}$$
for any $\lambda$ in the field. It is an easy exercise to see that there is no invertible matrix that centralizes $x$ and maps $y$ from one such matrix to any other, hence these representations are all inequivalent.


问:
Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition
I am given the sequence
$$a_{n+1} = \sqrt{2 a_n}, \quad a_1 = \sqrt{2}.$$
That is, the sequence
$$ \sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$$

I am aware of using the recursive method to find the limit: setting $x = \sqrt{2x}$ and getting $x = 2$, rejecting $x = 0$ due to the sequence being monotone and positive.

I now want to show that the limit of this sequence is $2$ using the epsilon-neighbourhood definition for a convergent series:

Let $(a_n)$ be a sequence that converges to a real number $a$. Then, for every number $\epsilon > 0$, there exists a number $N \in \mathbb{N}$ such that for all $n \geq N$, $|a_n - a| < \epsilon$.

I take the logarithm on both sides and arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \ln 2 - \ln 2 \right|  < \epsilon. $$

Using triangle inequality, $|x - y| \geq |x| - |y|$, I arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| - |\ln 2| < \epsilon. $$

Add $\ln 2$ to both sides
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| < \epsilon + \ln 2. $$

The left hand side can be rewritten as a sum of $n$ terms in a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$,
$$ \frac{\frac{1}{2} \left( 1 - \left( \frac{1}{2} \right)^n \right)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} < \epsilon + \ln 2. $$
I am unsure of how to proceed beyond this point.

答:
Hint:

For $n>1$, we can write

$$\frac{a_n}2=2^{-2^{-n}}>2^{-\frac1n}>1-\frac1n$$ because $2^n>n$ and because by the binomial theorem

$$\left(1-\frac1n\right)^{-n}>\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\cdots+1>2.$$

Now it suffices to prove convergence of $1-\dfrac1n$ to $1$, which is easy, and will allow to write

$$n>N\implies\left|2-\frac2n\right|<\epsilon\implies\left|2^{1-2^{-n}}-2\right|<\epsilon.$$

The figure illustrates the first inequalities.


问:
Prove that $X$ is connected

Let: $$S=\{(p,q,0)\in \mathbb{R}^3: p^2 +q^2<1, p\in \mathbb{Q}, q\in \mathbb{Q} \}$$ $$I(a,b) \text{ - open line segment between points } a, b$$
Now let's define set $X$: $$X=\{(x,y,0) \in \mathbb{R}^3 : x^2 +y^2 \leq 1\} \cup \Biggl( \bigcup_{(p,q,0)\in S} I\big((p,q,0),(p,q,p)\big) \Biggr) \subset \mathbb{R}^3$$ Prove that $X$ is connected.

I know from intuition that $X$ is linearly connected so also connected. However I don't know how I can prove it in formal way.

答:
First of all, any path connected space is also connected, so it suffices to prove $X$ is path connected.

Now, the subset $
\{(x,y,0) \in \mathbb{R}^3 : x^2 +y^2 \leq 1\}$ is definitely path connected. Furthermore, any point with nonzero height ($z$-coordinate) can be connected to a point in $
\{(x,y,0) \in \mathbb{R}^3 : x^2 +y^2 \leq 1\}$ by a linear parametrization of the corresponding vertical interval. Finally, to connect two points of nonzero height, first connect each of them to (possibly distinct) points in $
\{(x,y,0) \in \mathbb{R}^3 : x^2 +y^2 \leq 1\}$ which can then be connected using another linear path.

This gives a way to connect any two points in $X$ with a piecewise linear path, and such paths are continuous.

Title: Prove that $X$ is connected


问:
Prove that an uncountable subset of a second countable space contains at least one limit point
This is a 2 part problem about second countable space.

i)Prove that the set of all isolated points of a second countable space is countable

Definition: $x$ is an isolated point if ${x}$ is open in the topology.

Since the topology is second countable, the set of basis is countable and open sets are defined to be unions of sets in the basis. Thus if x is an isolated point, then x is part of the basis set. Now if there are an uncountable number of isolated points, they are all basis elements, so we would have an uncountable number of basis elements.

(ii) Hence, show that any uncountable subset A of a second countable space contains at least one point which is a limit point of A.

Suppose A does not contain any point that is a limit point. Since the space is second countable, we have a countable basis set $B_i$, $i \in N$ such that their union is the entire space X. Since A is a subset of X, each point of A must be contained in some $B_i$ and each $B_i$ is open. Now I'm not sure if this is correct, but since A is uncountable and we only have countable number of $B_i$, is it the case that one of the $B_i$ must contain an uncountable number of points of A. I think that part is accurate. But how do I go from here to showing that every basis elements that contains points of A all contain another point of A that they all have in common, which would be the limit point of A?

答: If no point of $A$ was a limit point of $A$, then every point of $A$ would be an isolated point, and therefore $A$ would be countable.


问:
The global section of direct sum of twisted sheaves
I am working with the Exercise 5.14 in Hartshorne's GTM52. And its problem (a) is stated roughly as follows.

Let $X$ be a closed subscheme of $\mathbb{P}^r_k$, a projective $r$-space over an algebraically closed field $k$, let $S$ be the homogeneous coordinate ring of $X$, i.e. $S=k[x_0,\dots,x_r]/I$, where $I=\Gamma_*(\mathscr{I}_X)$, and let $S'=\bigoplus_{n\ge0}\Gamma(X,\mathscr{O}_X(n))$.

Show that $S$ is a domain and that $S'$ is its integral closure.

Here my questions don't involve the $S$. I just want to ask something about the $S'$.

The hint of the problem (a) tells that one can regard the ring $S'$ as the global sections of such a sheaf of rings $\mathscr{S}=\bigoplus_{n\ge0}\mathscr{O}_X(n)$ on $X$, a direct sum of infinitely many sheaves. That is, there is an equation about them:
$$\Gamma(X,\mathscr{S})=\Gamma(X,\bigoplus_{n\ge0}\mathscr{O}_X(n))=\bigoplus_{n\ge0}\Gamma(X,\mathscr{O}_X(n)).$$
Does it true? I know that the arbitrary direct sum of sheaves $\{\mathscr{F}_i\}$ is defined to be the associated sheaf of the presheaf $U\mapsto\bigoplus_{i}\mathscr{F}_i(U)$. In the finite case, such a presheaf is just a sheaf. And else, it may be not, because we can give a counter-example (see here: Direct and inverse limits of sheaves). So I want to query if the global sections of $\mathscr{S}$ is just $\bigoplus_{n\ge0}\Gamma(X,\mathscr{O}_X(n))$, and if the process of sheafification doesn't change the global sections (up to an isomorphism).

Thanks in advance.

答:
If $\{F_i\}_{i\in I}$ is a set of quasi coherent sheaves on a scheme $X$ of finite type over a field, it follows the direct sum $F:=\oplus_{i \in I} F_i$ is a quasi coherent sheaf, and there is an isomorphism of global sections

$$F1.\text{   }\Gamma(X, F) \cong \oplus_{i \in I} \Gamma(X, F_i):$$

I believe taking global sections commutes with arbitrary direct sums of qc sheaves.

Question: "So I want to query if the global sections of $\mathcal{S}$ is just $\oplus_{n \geq 0} \Gamma(X,\mathcal{O}_X(n))$, and if the process of sheafification doesn't change the global sections (up to an isomorphism)?"

In your example $X$ is a scheme of finite type over $k$ and $F:=\oplus_{n \in \mathbb{Z}} \mathcal{O}(n)$ is a direct sum of locally trivial sheaves of rank one, in particular $\mathcal{O}(n)$ is quasi coherent. Hence it seems to me formula F1 should hold in your case. You should try to write down an explicit isomorphism

$$\Gamma(X, \oplus_{n} \mathcal{O}(n)) \cong \oplus_{n} \Gamma(X, \mathcal{O}(n)).$$


答: (ii) is an entierly trivial consquence of (i).  If there is no limit point then every point is isolated and there are only countable number of them. But then $A$ is countable, a contradiction.

Title: Prove that an uncountable subset of a second countable space contains at least one limit point

Body:
This is a 2 part problem about second countable space.

i)Prove that the set of all isolated points of a second countable space is countable

Definition: $x$ is an isolated point if ${x}$ is open in the topology.

Since the topology is second countable, the set of basis is countable and open sets are defined to be unions of sets in the basis. Thus if x is an isolated point, then x is part of the basis set. Now if there are an uncountable number of isolated points, they are all basis elements, so we would have an uncountable number of basis elements.

(ii) Hence, show that any uncountable subset A of a second countable space contains at least one point which is a limit point of A.

Suppose A does not contain any point that is a limit point. Since the space is second countable, we have a countable basis set $B_i$, $i \in N$ such that their union is the entire space X. Since A is a subset of X, each point of A must be contained in some $B_i$ and each $B_i$ is open. Now I'm not sure if this is correct, but since A is uncountable and we only have countable number of $B_i$, is it the case that one of the $B_i$ must contain an uncountable number of points of A. I think that part is accurate. But how do I go from here to showing that every basis elements that contains points of A all contain another point of A that they all have in common, which would be the limit point of A?