问:
Understanding fundamental theorem of calculus and its generalization
I am reading the Wikipedia page on the fundamental theorem of calculus and have a hard time understanding the logic of the theorem and its generalization (section 9). I browsed over the other questions on the site,  Understanding the Fundamental theorem of Calculus in plain english, but it seems irrelevant to mine.

First, I am going to phrase the theorem in my own way so others may spot the problems in my understanding:

Let $f$ be a function defined on $[a,b]$, define  $$F(x):=\int^x_{a}f(x)dx$$
the fundamental theorem of caluclus states that:

$f$ is continuous on $[a,b] \Rightarrow F(x)$ is uniformly continuous on $(a,b)$ and $F'(x)=f(x)$ for all $x\in(a,b)$.

$f$ is Riemann integrable $\Rightarrow \int^b_af(x)dx=F(b)-F(a)$.

(Purpose of Generalization) Since the fundamental theorem of calculus primarily serves to link integration and differentiation, and continuity of $f$ on $[a,b]$ is a very strong condition, it makes sense to relax this condition a bit and still make $F'(x)=f(x)$ and $\int^b_af(x)dx=F(b)-F(a)$ work.

Here comes to my understanding of the Generalizations part of Wikipedia:

"Part 1 of the theorem then says... is differentiable for $x=x_0$ with $F'(x_0)=f(x_0)$"

This part means we can relax the assumption of $f$ is continuous on $[a,b]$ to $f$ is Lebesgue integrable on $[a,b]$. To my understanding, Lebesgue integrable function only requires the function to be measurable, and satisfies either $\int_{X} f^{+} d \mu<\infty $ or $ \int_{X} f^{-} d \mu<\infty$. (Reference). So it seems the condition is indeed weaker. However, I am not sure any $x_0\in[a,b]$ satisfies $F'(x_0)=f(x_0)$ make any different to $F'(x)=f(x)$ for all $x\in[a,b]$.

"we can relax the conditions on $f$ still further... the function $F$ is differentiable a.e and $F'(x)=f(x)$"

This part means $f$ need not be Lebesgue integrable on the whole $[a,b]$ as long as it can be decomposed into small Lebesgue integrable sections. ($F'(x)=f(x)$ will still hold a.e).

I hope so far I am not far off the correct understanding, I have bigger problems understanding the generalization of the second part of the theorem.

"Part II of the theorem is true for any Lebesgue integrable function f...., then $F(b)-F(a)=\int_{a}^{b} f(t) d t$".

Does it mean the statement will hold true when we relax $f$ being Riemann integrable to Lebesgue integrable? However, I am really confused when trying to understand the next paragraph, where states:

This result may fail for continuous functions $F$ that admit a derivative $f(x)$ at almost every point $x$, as the example
of the Cantor function shows. However, if $F$ is absolutely continuous, it admits a derivative $F^{\prime}(x)$ at almost every
point $x,$ and moreover $F^{\prime}$ is integrable, with $F(b)-F(a)$ equal to the integral of $F^{\prime}$ on $[a, b] .$ Conversely, if $f$ is any
integrable function, then $F$ as given in the first formula will be absolutely continuous with $F^{\prime}=f$ a.e.

This paragraph seems to suggest there exists function $f$ with continuous antiderivative $F(x)$ but will fail the equation $F(b)-F(a)=\int_{a}^{b} f(t) d t$. What exactly is $f$ is not explicitly stated. And it seems also suggest as longas $f$ is Lebesgue integrable, $F(x)$ will be absolutely continuous, hence $F(b)-F(a)=\int_{a}^{b} f(t) d t$ will still hold.

Is my logic correct?

答:
In general $F$ is only differentiable almost everywhere if $f$ is Lebesgue integrable. Its derivative is given by the "local average" of $f$, if this local average is well defined independent of the path used to shrink the domain of integration. The precise statement is called the Lebesgue differentiation theorem.

Part II, if $F$ is defined by the integral anyway, is rather trivial, it just follows from the additivity property of the integral. What is more interesting and useful is that if you have $F'=f$ everywhere, then $\int_a^b f(x) dx = F(b)-F(a)$. It doesn't matter where $F$ came from. In this regard I would say your original statement of FTC is a bit deceptive.

But this property doesn't work if you only assume $F'=f$ almost everywhere, because there are non-absolutely continuous functions. In particular, if $F$ is the Cantor function then $F'$ exists almost everywhere and is $0$ where it exists, but $F(x) \neq \int_0^x 0 \, dt$.


问:
Alternative characterisation of weakly complete Banach spaces
Let $V$ be a topological vector space over some locally compact field $\Bbb K$. Let $V'$ denote all continuous functionals on $V$, the weak topology is defined to be the coarsest topology on $V$ making the elements of $V'$ continuous.

Definition (weakly-complete-1)

$V$ is called weakly complete if it is complete in the uniform structure given by the weak topology.

I have seen another definition:

Definition (weakly-complete-2)

A topological vector-space $V$ is called weakly complete if it is isomorphic to $\Bbb K^J$ for some set $J$ (here take the product topology).

I think the mumbo-jumbo you do for $\Bbb K \in\{\Bbb R, \Bbb C\}$ also works for the other cases to show that $(\Bbb K^J, \bigoplus_J \Bbb K)$ are a dual pair which you can use to show that if $V$ is weakly-complete-1 if it is weakly-complete-2. Is there a way to turn this around?

Question

If $V$ is a Banach space that is weakly-complete-1, does it follow that $V$ equipped with its weak topology is weakly-complete-2? Meaning $(V,\tau_{\text{weak}})\cong \Bbb K^J$ for some $J$?

I think a reflexive Banach-space is weakly-complete-1, so every reflexive Banach-space must be (weakly) isomorphic to $\Bbb K^J$ for some $J$ if this were true. For separable Banach-spaces you may only have $J=\Bbb N$ for cardinality reasons, so this would imply that all separable reflexive Banach-spaces are weakly isomorphic, that seems unlikely!

If the answer is false, is there some other connection between the two notions?

答:
If $E$ and $F$ are Banach spaces and $\phi:E\to F$ is weakly continuous, then it is norm-continuous by the uniform boundedness principle.

Therefore the reasoning given by the OP implies that all separable reflexive Banach spaces are in fact norm-isomorphic, which is patently false.  The conclusion is that weakly-complete-2 is not equivalent to weakly-complete-1.

ContentLicense: CC BY-SA 4.0


问:
why $n^{\log{\log{n}}}=\log{n}^{\log{n}}$?
I was reading the solution of one exercise on my book on algorithms.
What I need to do in short is order some function from fast to slow.
This is the link of the solution: site.

At some point I get the following property:
$$n^{\log{\log{n}}}=\log{n}^{\log{n}}$$
but how do you prove that?

I've tried to use the properties of logarithm, but I wasn't able to get that result.
Can you give me some hints?

答:
$$n^{\ln x}=x^x, x=\ln n$$

$$\implies \ln n^{\ln x}=\ln x^x$$

$$\implies \ln x \ln n=x \ln x$$

$$\implies x=\ln n$$

Reverse the steps and generate the formula.


问:
For what s does the series converge $\sum_{1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{(-1)^nn^s}$
$$\sum_{1}^{\infty}\frac{\sqrt{n+1}-\sqrt{n}}{(-1)^nn^s}$$

For what $s>0$ does the series converge

My try:

The  $\lim_{n\rightarrow \infty}\frac{\sqrt{n+1}-\sqrt{n}}{n^s}=0$ if $s\geq\frac{1}{2}$ and thus the series is also decreasing series and it converges.

答:
For each $n\in\Bbb N$,$$\frac{\sqrt{n+1}-\sqrt n}{(-1)^nn^s}=\frac{(-1)^n}{\left(\sqrt{n+1}+\sqrt n\right)n^s}.$$If $s>-\frac12$, then the sequence $\left(\frac1{\left(\sqrt{n+1}+\sqrt n\right)n^s}\right)_{n\in\Bbb N}$ is decreasing and it converges to $0$; therefore, your series converges.

Otherwise, you don't have $\lim_{n\to\infty}\frac{(-1)^n}{\left(\sqrt{n+1}+\sqrt n\right)n^s}=0$, and therefore your series diverges.


问:
Can I simplify the following vector?
Let $\boldsymbol{a}$ and   $\boldsymbol{b}$ be $n \times 1$ vectors and $M_1, M_2, ..., M_n$ be $n\times n$ matrices.

Is there a way to write the following vector:
\begin{equation}
v = \begin{bmatrix}
a^T M_1 b
\\
a^T M_2 b
\\
\vdots
\\
a^T M_n b
\end{bmatrix}
\end{equation}

in such a way that reads $v = a^T M_{new} b$, where $M_{new}$ is just a manipulation of the $M_i$ matrices?

I know I can do:
\begin{equation}
v = \begin{bmatrix}
a^T M_1
\\
a^T M_2
\\
\vdots
\\
a^T M_n
\end{bmatrix}
b
\end{equation}

But then I cannot get the vector $a$ "out of the brackets".

答:
You could write
$$
v = \pmatrix{a^T \\& \ddots & \\&&a^T} \pmatrix{M_1\\ \vdots \\ M_n}b.
$$
If you like, you could write the matrix on the left compactly using the fact that
$$
\pmatrix{a^T \\& \ddots & \\&&a^T} = a^T \otimes I,
$$
where $I$ denotes a size $n$ identity matrix and $\otimes$ denotes the Kronecker product.

Title: Can I simplify the following vector?


问:
Negative equal positive paradox
I was just bored and started practicing even more the exponentiations, and as I was working, I went ahead and did this:

$(-1)^{2} = ((-1)^{\sqrt{2}})^{\sqrt{2}}$

So, I entered in my calculator of what is negative one to the power of the square root of two and it gave me negative one, so it goes like:

$(-1)^{2} = ((-1)^{\sqrt{2}})^{\sqrt{2}} = (-1)^{\sqrt{2}} = (-1)$

But in other hand, we have:

$(-1)^2 = (-1)(-1) = 1$

And that implies that $-1 = +1$... Is there something wrong in it? If not, could someone explain me how can that happen?

Thanks for the answers.

答:
Here you have $2$,$3$ mistakes:

$(-1)^{2} = ((-1)^{\sqrt{2}})^{\sqrt{2}} = (-1)^{\sqrt{2}} = (-1)$

The first equality is true, since $\sqrt 2\cdot \sqrt 2=2$. The expression $(-1)^{\sqrt{2}}$ is not real and certainly not $-1$.

Title: Negative equal positive paradox


问:
Show that [3,4] x [5,6] is closed?
[3,4] is closed in R <-- R-[3,4] is open

[5,6] is closed in R <-- R-[5,6] is open

Show that [3,4] x [5,6] is closed in R x R by writing it as the complement of the intersection of two open sets in R x R.

(R - [3,4]) x (R - [5,6]) not equal R x R - [3,4] x [5,6]

答:
$$\Bbb R^2 - ([3,4] \times [5,6]) = ((\Bbb R - [3,4]) \times \Bbb R) \cup (\Bbb R \times(\Bbb R - [5,6]))$$ so that $\Bbb R^2 - ([3,4] \times [5,6])$ is a union of basic open sets hence open, and so $[3,4] \times [5,6]$ is closed.

You can also directly write $$[3,4] \times [5,6] = \pi_1^{-1}[[3,4]] \cap \pi_2^{-1}[[5,6]]$$

which is the intersection of closed sets (as the projections are continuous) and hence closed.

Title: Show that [3,4] x [5,6] is closed?


问:
How to verify if this is a autocovariance function?
Is this$$γ(h) = 1(h = 0) − 0.5 · 1(|h| = 2) − 0.25 · 1(|h| = 3)$$an autocovariance function?

How to check this? Is there a method one can use to check if a given function is an autocovariance funtion? I have found nothing in my "time series analysis" notes.

答:
There is a another way to solve this.

We know that $\gamma$ is a valid autocovariance function iff, for every $n > 0$, the matrix $\Gamma_n$ defined as $\Gamma_n(i,j) = \gamma(|i-j|)$ is positive semidefinite.

In particular we must have that, for every $n > 0$, $\sum_{i,j} \Gamma_n(i,j) \geq 0$

Now let's assume $\gamma(h) = 0$ for all $h$ such that $|h|>3$ and $\gamma(0)=1$ (this includes your case).

Then for $n \geq 3$,
$$
\sum_{i,j} \Gamma_n(i,j) = n (1 + 2 \gamma(1) + 2 \gamma(2) + 2 \gamma(3)) + C
$$
where $C$ does not depend on $n$.

Since this has to be non-negative for every $n$, so must it be its limit for $n \to \infty$.

This implies that
$$
\gamma(1) + \gamma(2) + \gamma(3) \geq - \frac 12
$$
which is not satisfied in your case
(which is $\gamma(1) = 0, \gamma(2) = -0.5, \gamma(3) = -0.25$)

ContentLicense: CC BY-SA 3.0


问:
What is the difference between the ideal $I=(x-1,x^2-y)$ and the ideal $J=((x-1)(x^2-y))$
In the ring, $\mathbb{R}[x,y]$, what is the difference between the ideal $I=(x-1,x^2-y)$ and the ideal $J=((x-1)(x^2-y))$?

This is probably a silly or obvious question but I am not yet familiar with multivariable polynomials. Can someone help me acknowledge the difference between $I$ and $J$. I am trying to find the common roots of the polynomials than belong to each of them.

答:
The ideal $I$ is the smallest ideal containing both the elements $a = x - 1$ and $b = x^2 - y$.

On the other hand, the ideal $J$ is the smallest ideal containing the product $ab$.

(The above is the difference in definition.)

More explicitly, we have $I = \{ar + bs : r, s \in \Bbb R[x, y]\}$ and $J = \{rab : r \in \Bbb R[x, y]\}$.

(To see why, argue that $I$ and $J$ are indeed ideals. Moreover, they must be contained in any ideal containing $\{a, b\}$ and $\{ab\}$, respectively.)

To see the difference as sets, note that $a \in I$ (by definition) but $a \notin J$. Indeed, suppose that $a \in J$. Then, $a = rab$ for some $r \in \Bbb R[x, y]$.

Analyse the degree of $y$ to get a contradiction.


问:
Proving the pointwise limit of a monotonically increasing sequence of Riemann integrable functions is Riemann integrable
I am trying to prove the following statement

Let $A \subset \mathbb{R}^n$ be a bounded set that has volume. Let $f : A \mapsto \mathbb{R}, f \ge 0$ be an unbounded function. Suppose $C_i$ is a sequence of compact sets with volume, $C_i \subset A$ with $C_i$ increasing to $A$. Then $f$ is Riemann integrable iff $f$ is Riemann integrable on each $C_i$ and $\lim_{i \to \infty} \int_{C_i} f$ exists.

My attempt. Let $M > 0$ be given.
Define
$$
f_M(x) = \min \{ f(x), M\}
$$
$$
f_i = f \chi_{C_i}
$$
Then $\{f_i\}$ is a sequence of non-negative functions such that $f_i \le f_{i+1}$ and $f_i \to f$ pointwise. Now, if $f$ is integrable on each $C_i$ and $\lim_{i\to \infty} \int_{C_i} f < \infty$, then the set of discontinuities of $f_M$ on $C_i$ has measure zero for all $i$. Let $x_0 \in \{ \text{discontinuities of} \ f_M \ \text{on $A$} \}$ be given. Since $C_i$ increases to $A$, there exists $N$ such that $f_M$ is discontinuous at $x_0$ on $C_N$. So $x_0 \in \bigcup_{i=1}^\infty \{ \text{discontinuities of} \ f_M \ \text{on $C_i$} \ \}$. Since $\bigcup_{i=1}^\infty \{ \text{discontinuities of} \ f_M \ \text{on $C_i$} \ \}$ has measure zero, the set of discontinuities of $f_M$ on $A$ has measure zero. Thus, $f_M$ is integrable. Since this is true for all $M > 0$, $f$ is integrable. Then by Lebesgue's monotone convergence theorem,
$$
\lim_{i\to \infty} \int_{C_i} f = \lim_{i\to \infty} \int_A f_i = \int_A f
$$
Conversely, suppose $f$ is integrable. Since each $C_i$ is a compact subset of $A$ with volume, then $\{\text{discontinuities of} \ f_i \} \subset \partial C_i \cup \{ \text{discontinuities of} \ f \}$ has measure zero, so $f_i$ is integrable, i.e. $f$ is integrable on each $C_i$. Then by Lebesgue's monotone convergence theorem,
$$
\int_A f=\lim_{i\to\infty} \int_A f_i = \lim_{i\to\infty} \int_{C_i} f
$$.

Is my proof mathematically correct? Is there a more direct way of proving the statement? Any comment would be appriciated.

答:
For the reverse implication, suppose $D \subset A$ is any compact set with volume (i.e., rectifiable).  Since $D \subset \cup_{i=1}^\infty C_i=A$, there exists by compactness a set $C_m$ such that $D \subset C_m$.

Since $f$ is Riemann integrable on $C_m$ , it is Riemann integrable on $D$. (Here you would make the argument that the discontiuities of $f\chi_D$ must be a set of measure zero.)  Thus, since $f \geqslant 0$,

$$\int_D f \leqslant \int_{C_m} f \leqslant \lim_{i\to \infty}\int_{C_i} f < \infty,$$

and

$$\int_A f := \sup_{D \subset \mathcal{D}} \int _D f\leqslant  \lim_{i\to \infty}\int_{C_i} f,$$

where the supremum is taken over all compact rectifiable subsets of $A$.  This proves the existence of the improper integral of $f$ over $A$ -- as it is usually defined.


问:
Proof of Lemma 6.3 in Carothers' Real Analysis

Lemma 6.3 (Chapter 6 - Connectedness): Let $E$ be a subset of a metric space $(M,d)$. If $U$ and $V$ are disjoint, open
sets in $E$, then there are disjoint open sets $A$ and $B$ in $M$ such that $U= A \cap E$
and $V = B \cap E$.

The proof roughly goes as follows (I haven't typed it out exactly, but I have captured all the important details) -

Since $U$ is open in $E$, for each $x\in U$, $\exists\epsilon_x > 0$ such that $E\cap B(x,\epsilon_x) \subset U$. Since $V$ is open in $E$, for each $y\in V$, $\exists\delta_y > 0$ such that $E\cap B(y,\delta_y) \subset V$. Since $U\cap V = \varnothing$, we also have $E\cap B(x,\epsilon_x) \cap B(y,\delta_y) = \varnothing$.

Now, the author claims that:

For every $x\in U$ and $y\in V$, $$B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right) = \varnothing$$

Why is this true? I'm trying to show this in the following way. Suppose $z \in B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right)$. So, $d(x,z) < \epsilon_x/2$ and $d(y,z) = \delta_y/2$. To show that the intersection is non-empty would amount to showing $d(x,z) < \epsilon_x/2$ and $d(y,z) = \delta_y/2$ cannot hold together. How do I do this?

Thus, $$A = \bigcup_{x\in U} B\left(x,\frac{\epsilon_x}{2}\right) \text{ and } B = \bigcup_{y\in V} B\left(y,\frac{\delta_y}{2}\right)$$ work.

Why does this choice of $A, B$ work? All I need to show for this, is that $U = A\cap E$ and $V= B\cap E$ right? Is anything else needed to be done?

Thank you.

Picture for Reference:

Attempting a Proof by Contradiction:

As mentioned in an answer, $B(x,r/2)\subset B(x,r)$. Suppose $\exists x \in U, \exists y\in V$ for which $B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right) \ne \varnothing$. Also, $$E\cap B(x,\epsilon_x) \subset U \implies E\cap B(x,\epsilon_x/2) \subset U$$ and
$$E\cap B(y,\delta_y) \subset V \implies E\cap B(y,\delta_y/2) \subset V$$
What's next? I'm not able to find a contradiction. $U\cap V = \varnothing \implies E\cap B(x,\epsilon_x) \cap B(y,\delta_y) = \varnothing$, and I believe this is what we want to contradict.

答:
Question 1

Remember that for any $r \gt 0$, $B(x, r/2) \subseteq B(x,r)$. Therefore any point that would belong to $$B\left(x,\frac{\epsilon_x}{2}\right) \cap B\left(y,\frac{\delta_y}{2}\right)$$ would also belong to
$$B\left(x,\epsilon_x\right) \cap B\left(y,\delta_y\right)$$ in contradiction with the initial hypothesis.

Question 2

Is a consequence of the previous point.


问:
Figuring out Mechanics of How Chain Rule was Applied
How do I go from system $(1)$ to system $(2)$ below?

\begin{gather}
\begin{aligned}
S &= K e^{x}\\[7pt]
V(S,t) &= K v(x, \tau)\\
\tau &= (T-t) \frac{\sigma^{2}}{2}
\end{aligned}\tag{1}
\end{gather}

\begin{gather}
\begin{aligned}
\frac{d}{dt}V &= -K \frac{\sigma^{2}}{2} \frac{d}{d \tau} v\\
\frac{d}{dS}V &= \frac{K}{S} \frac{d}{dx} v
\end{aligned}\tag{2}
\end{gather}

答:
Let's start with the first one:
$$V(S,t) = K v(x, \tau).\tag{1}$$
Differentiating each side, using the chain rule:
$$
\frac{d}{dt}V(S,t) = \frac{d}{dt}Kv(x,\tau)= K\frac{dv}{d\tau}(x,\tau)\cdot\frac{d\tau}{dt}(t).\tag{2}\\
$$

Note that $x$ does not depend on $t$, which is why we only have to consider $\frac{dv}{d\tau}$ above.  Now, using $$\tau= (T-t) \frac{\sigma^{2}}{2}$$ we can obtain $$\frac{d\tau}{dt} = -\frac{\sigma^{2}}{2},$$ and so $(2)$ becomes $$\boxed{\frac{d}{dt}V(S,t) = -K\frac{\sigma^{2}}{2}\frac{dv}{d\tau}(v,\tau).}$$

For the second, we differentiate $(1)$ with respect to $S$, again using the chain rule.  This time we only have to consider $\frac{dv}{dS}$ because $\tau$ does not depend on $S$.
$$\frac{d}{dS}V(S,t) = \frac{d}{dS}Kv(x,\tau) = K\frac{dv}{dx}(x,\tau)\cdot \frac{dx}{dS}(S).\tag{3}$$  Now, to compute $\frac{dx}{dS}$ we differentiate implicitly:
\begin{align}
S &= Ke^{x}\\
1 &= Ke^{x}\frac{dx}{dS}\\
\end{align}
Solving for $\frac{dx}{dS}$, making using of $S = Ke^{x}$, we get $$\frac{dx}{dS} = \frac{1}{Ke^{x}}=\frac{1}{S}.\tag{4}$$  Plugging $(4)$ into $(3)$ gives us $$\boxed{\frac{d}{dS}V(S,t) = \frac{K}{S}\frac{dv}{dx}(x,\tau).}$$


问:
What does it say about a multivariate polynomial to be zero on a linear subspace?
If I univariate polynomial $f(x)$ that vanishes at a point $x_0$, we conclude that $x - x_0$ divides $f(x)$, and in particular that $f$ is reducible if $\deg f > 1$.  Can anything of significance be said if a multivariate polynomial $g : \mathbb{R}^d \to \mathbb{R}$ vanishes along an entire linear subspace $V < \mathbb{R}^d$?

The case $\dim V = 1$ can occur quite easily: if $g$ is homogenous, any zero of $g$ produces such a $V$.  Thus, I am particularly interested in $\dim V = 2$ or higher, or if there is a taxonomy of different ways the $\dim V = 1$ case can occur.

This question is related to an earlier question on the Schwartz-Zippel lemma linked below.  The motivation for both is that I have an algorithm that chooses a random linear subspace and succeeds whenever a particular multivariate polynomial doesn't vanish entirely on that subspace.

Analogue of the Schwartz–Zippel lemma for subspaces

答:
If $V$ has codimension $1$ and is cut out by a single linear equation $\sum_{i=1}^d a_i x_i = 0$, then you can conclude that $\sum a_i x_i$ divides $f$. The proof is fairly short: by applying a suitable change of coordinates you can assume WLOG that the linear equation is $x_1 = 0$, and then it's clear. </p>

<p>In higher codimension you can't conclude much. For example, when $d = 2$ and $V$ has codimension $2$ then $V$ is a point, and vanishing on a point doesn't tell you too much in this case. </p>

Body:
If $V$ has codimension $1$ and is cut out by a single linear equation $\sum_{i=1}^d a_i x_i = 0$, then you can conclude that $\sum a_i x_i$ divides $f$. The proof is fairly short: by applying a suitable change of coordinates you can assume WLOG that the linear equation is $x_1 = 0$, and then it's clear.

In higher codimension you can't conclude much. For example, when $d = 2$ and $V$ has codimension $2$ then $V$ is a point, and vanishing on a point doesn't tell you too much in this case.


问:
Find the limit of $\lim_{x\to 0}\left(\frac{\cos x}{\cos 2x}\right)^\frac{1}{x^2}$
Find limit of $$\lim_{x\to 0}\left(\frac{\cos x}{\cos2x}\right)^{x^{-2}}.$$
After some algebra I get $$\lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right).$$ Maybe I am missing something but I can't continue from this.

答:
Hint:

Let $y = \left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}$.
We need to find $\lim_{x\to 0} y$, however, this might get super ugly.

So instead, try to compute $\lim_{x\to 0} \ln(y) = \lim_{x\to 0} \frac{1}{x^2} \ln\left(\frac{\cos(x)}{\cos(2x)}\right) = \lim_{x\to 0} \frac{\ln\left(\frac{\cos(x)}{\cos(2x)}\right) }{x^2}$.

You will have to use L-Hospital twice, and it gets messy algebraically, but very much doable. Then say your limit is $\alpha$, now since $\ln$ is continuous, we know that $\lim_{x \to 0} y = e^{\alpha}$


答:
Rearrange the latter form of the limit that you are getting so that the limit becomes
\begin{equation}
\lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right)=\exp\left(\lim_{x\to 0}\dfrac{1-\cos x}{x^2}\times\lim_{x\to0}\dfrac{1+2\cos x}{\cos2x}\right)=\boxed{e^{\frac{3}{2}}}
\end{equation}

Title: Find the limit of $\lim_{x\to 0}\left(\frac{\cos x}{\cos 2x}\right)^\frac{1}{x^2}$

Body:
Find limit of $$\lim_{x\to 0}\left(\frac{\cos x}{\cos2x}\right)^{x^{-2}}.$$
After some algebra I get $$\lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right).$$ Maybe I am missing something but I can't continue from this.


答:
First of all $$\lim _{x\to 0}\left(\left(\frac{cosx\:}{cos2x\:}\right)^{\frac{1}{x^2}}\right)=\lim _{x\to 0}\left(e^{\frac{ln\left(\frac{cosx}{cos2x}\right)}{x^2}}\right)=e^{\lim _{x\to 0}\left(ln\left(\frac{cosx}{cos2x}\right)\right)}$$

Now, we will calculate:
$$\lim _{x\to 0}\left(\frac{ln\left(\frac{cosx}{cos2x}\right)}{x^2}\right)=\lim \:_{x\to \:0}\left(\frac{ln\left(cosx\right)-ln\left(cos2x\right)}{x^2}\right)=\lim \:\:_{x\to \:\:0}\left(\frac{-tanx+2tan2x}{2x}\right)=\lim \:\:_{x\to \:\:0}\left(\frac{4sec^2x-sec^{2x}}{2}\right)=\frac{3}{2}$$

So the final answer is $e^{\frac{3}{2}}$


问:
Minimum over a dense subset
Let $X$ denote a vector space and let $S \subset X$ denote a dense subset of $X$.
Does the following result hold?
\begin{equation*}
\min_{x \in X} F(x) = \min_{y \in S} F(y).
\end{equation*}
Do we need some assumptions on the function or the vector space?

Any help is appreciated.

答:
Provided $F$ is continuous, then we can conclude
$$\inf_{x \in X} F(x) = \inf_{x \in S} F(x). \tag{$\star$}$$
Clearly
$$\inf_{x \in X} F(x) \le \inf_{x \in S} F(x),$$
as the right hand side is the infimum of a subset of the left hand side. Further, for all $\varepsilon > 0$, there exists some $y \in X$ such that
$$\inf_{x \in X} F(x) \le F(y) < \inf_{x \in X} F(x) + \frac{\varepsilon}{2}.$$
Since $F$ is continuous at $y$, there exists some $\delta > 0$ such that
\begin{align*}
\|z - y\| < \delta &\implies |F(z) - F(y)| < \frac{\varepsilon}{2} \\
&\implies F(z) < F(y) + \frac{\varepsilon}{2} \\
&\implies F(z) < \inf_{x \in X} F(x) + \varepsilon.
\end{align*}
Since $S$ is dense in $X$, there exists some $z \in S$ such that $\|z - y\| < \delta$, and hence
$$\inf_{x \in S} F(x) \le F(z) < \inf_{x \in X} F(x) + \varepsilon.$$
This holds for all $\varepsilon$, so $(\star)$ holds.

Now, if we assume that both the infima are achieved, then we get
$$\min_{x \in X} F(x) = \min_{x \in S} F(x).$$
However, be warned that $\min_{x \in X} F(x)$ existing doesn't necessarily imply $\min_{x \in S} F(x)$. For example, $f : \Bbb{R} \to \Bbb{R} : x \mapsto x^2$ achieves a minimum at $0$, but does not achieve a minimum on the dense set $\Bbb{R} \setminus \{0\}$.

Title: Minimum over a dense subset


问:
Deformation Retract of $I^2 - \{c_0\}$
Consider $I^2 \setminus \{c_0\}$, unit square with the center point removed. Call this space $X$.
Consider $I^2$ with a small open disk removed about $c_0$ Call this space $Y$.
Instead of taking open disk in $Y$, take closed disk instead. Call this space $Z$.

Now, I want to construct a deformation retract from X to Y and deformation retract from X to Z. Let us say that interior of the disk is radially retracted (retraction is well-defined since the disk has $c_0$ removed). Now I have a little issue.

Intuitively, deformation retract from $X$ to $Z$ does not make sense; for any $p$ in the interior of the disk with $c_0$ removed, $p$ needs to land somewhere on $S^1$. However, $S^1$, which is the boundary of the disk, is removed and $p$ is “lost” where to go under the retraction. In this sense, deformation retract of $X$ to $Z$ is not well-defined.

Question 1: Does this rule out the possibility of retraction/deformation retract from $X$ to $Z$?

Now, let us investigate what happens if we embed $X, Y, Z$ in $I^2$. $X$ is open, $Y$ is closed, and $Z$ is open.

Question 2: Under deformation retraction $r: X \rightarrow Y$, an open space is retracted to a closed space. Is thist he right thing?

Thank you very much in advance.

答:
Q1 : There does not exist a retraction from $X$ to $Z$.

Let $i : Z \to X$ denote inclusion. Assume there exists a retraction $r : X \to Z$. The map $R = ir : X \to X$ has the property $R(x) = x$ for $x \in Z$. Since $Z$ is Hausdorff, the set $\{ x \in X \mid R(x) = x \}$ is closed in $X$, thus it must contain $\overline Z = Y$. Thus for all $x \in Y \setminus Z$ we must have $x = R(x) = r(x) \in Z$, a contradiction.

Q2: As Tyrone comments, it does not make sense to call $X$ an open space and $Y$ a closed space.

Yes, $X$ is an open subspace of $I^2$ and $Y$ a closed  subspace of $I^2$, but that is completely irrelevant in this context. Points outside of $X$ and $Y$ do not play any role for maps $X \to Y$. The space $X$ is non-compact and $Y$ is compact, but that does not play a role either. The punctured closed disk is radially retracted to its boundary - this gives you the desired (strong) deformation retraction $r : X \to Y$.

Title: Deformation Retract of $I^2 - \{c_0\}$


问:
Coordinate Functions and Coordinate Bases
Let $M$ be a smooth manifold of n dimension. And let $(U,\varphi)$ and $(V,\psi)$ be two smooth charts on $M$.

Define the coordinate functions of $\varphi$ and $\psi$ as $(x^i)$ and $(\tilde{x}^i)$ respectively, i.e. $\forall p \in U\cap V$, $\varphi(p)=(x^1(p), ..., x^n(p)\,)$ and $\psi(p)=(\tilde{x}^1(p), ..., \tilde{x}^n(p)\,)$.

And then the author says that any tangent vector at $p \in U\cap V$ can be represented with respect to either basis $(\partial\,/ \, \partial x^i \vert _p)$ or $(\partial\,/ \, \partial x^i \vert _p)$. Later he also uses notation like $\frac{\partial}{\partial \widetilde{x}^j}$.

So I guess the author here identifies the coordinate bases with the coordinate functions. But how is such identification valid? I.e. Is there a one-to-one correspondence between the set of coordniate bases and the set of coordinate functions?

I'm really confused. Thanks for help.

Edit: What I really wanted to ask was that, if $f\in C^\infty[\varphi(U \cap V)]$, what does $\partial _{x^i} \vert _{\varphi(p)}f$ mean?

答:
This answer is for my own future reference as well as someone who gets stuck here too.

What I really wanted to ask was that, if $f\in C^\infty[\varphi(U \cap V)]$, what does $\partial _{x^i} \vert _{\varphi(p)}f$ mean?

It turns out that we can regard $\partial _{x^i} \vert _{\varphi(p)} f$ as a partial derivative w.r.t the coordinate axis determined by $x^i $(recall that $x^i$ is a function but not an element of a basis for each $i$) at $\varphi(p)$.

Lemma: Suppose we have two bases $\{a_1,a_2, ..., a_n\}$ and $\{b_1,b_2, ..., b_n\}$ for $\mathbb R^n$ and for each $\alpha$ in some small open ball centered at the origin, the coordinates of $\alpha$ w.r.t  these two bases are identical. Then these two bases are the same.

Proof of lemma: Let $i\in \{ 1,2, ..., n\}$. For each $a_i$, there exists some $k_i >0$ s.t. $k_i a_i$ lies in the open ball. Its coordinate w.r.t the basis $\{a_1,a_2, ..., a_n\}$ is clearly $(0,...,0,k_i,0,...,0)$, where $k_i$ is at the $i$-th slot. By our assumption, its coordinate w.r.t the basis $\{b_1,b_2, ..., b_n\}$ should also be $(0,...,0,k_i,0,...,0)$, where $k_i$ is at the $i$-th slot. Hence, $k_i a_i = k_i b_i$, which implies $a_i=b_i$. Since $i$ is arbitrary, we are done.

By subtracting the constant vector $\varphi (p)$ if necessarily, we may assume that the coordinate chart $(U,\varphi)$ is centered at $p$, which means $\varphi(p)=0$. A chart actually gives a way to assign coordinates to each point in a small ball centered at $\varphi(p)$, which is the origin of $\mathbb{R}^n$. Since we can write $\varphi (p)$ and every point in some small open neighborhood of $\varphi (p)$ as n-tuples, it implicitly implies that there exists at least one local basis. By the lemma, we see a chart does produce a unique (local) basis.

An n-tuple has $n$ slots, each of which corresponds to an axis. Recall that we define $x^i$ by $\varphi(p)=(x^1(p), ..., x^n(p)\,)$.

So for each $i$, $x^i$ can represents an axis. And so we can regard $\partial _{x^i} \vert _{\varphi(p)} f$ as a partial derivative w.r.t the coordinate axis determined by $x^i$ at $\varphi(p)$.


问:
How many field homomorphisms?
Let $F$ and $F′$ be two finite fields with nine and four elements respectively.
How many field homomorphisms are there from $F$ to $F′$?</p>

答:
Hint $1$: A homomorphism of fields is injective.  Can you see why?

Hint $2$:  Hint $1$ answers your question.  Can you see why?

Title: How many field homomorphisms?


问:
Are continuous convex functions subharmonic?
We say that a continuous function $u:\mathbb{R}^d\to \mathbb{R}$ is subharmonic if it satisfies the mean value property
$$u(x)\leq \frac{1}{|\partial
 B_r(x)|}\int_{\partial B_r(x)}u(y)\,\mathrm{d}y \qquad (\star)$$  for
any ball $B_r(x)\subset \mathbb{R}^d$.

Let $u:\mathbb{R}^d\to \mathbb{R}$ be a convex function (hence,
continuous). Is $u$ subharmonic?

If $u\in C^2(\mathbb{R}^d)$, this is  true. Using a second-order Taylor expansion we have
\begin{align*}\int_{\partial B_r(x)}(u(y)-u(x))\,\mathrm{d}y&=\int_{\partial B_r(x)}\left(\nabla u(x)\cdot(y-x)+\frac{1}{2}(y-x)D^2u(\xi)(y-x)^t\right)\,\mathrm{d}y.\end{align*}
The first term in the above integral vanishes by symmetry, the second is non-negative because $D^2u(\xi)$ is a positive semi-definite matrix. Therefore, ($\star$) is proven.

If $d=1$, the statement is true when $u$ is continuous, in general. Indeed since balls reduce to intervals, ($\star$) is easily shown to be equivalent to $u$ being midpoint-convex.

I'm not sure how to attack the problem in higher dimensions. Of course $(\star)$ is true for affine functions in any dimension, and I'd like to use the fact that the graph of a convex function lies below that of an affine function, loosely speaking. However, to close the estimate I would need $u$ to be equal to the affine function at the boundary of the ball, and this is not necessarily possible.

答:
Using that $u$ is midpoint-convex works in higher dimensions as well.

$y \mapsto x - (y-x) = 2x-y$ maps the sphere $\partial B_r(x)$ bijectively onto itself (each point is mapped to the “opposite” point on the sphere). It follows that
$$
\int_{\partial B_r(x)} u(y) \, dy = \int_{\partial B_r(x)} u(2x-y) \, dy
$$
and therefore
$$
\int_{\partial B_r(x)} u(y) \, dy = \int_{\partial B_r(x)} \frac 12\bigl(u(y) + u(2x-y)\bigr) \, dy \\
\ge \int_{\partial B_r(x)} u\left(\frac{y + (2x-y)}{2}\right) \, dy = \int_{\partial B_r(x)} u(x) \, dy = |\partial
 B_r(x)| \cdot u(x) \, .
$$


问:
continued fraction formula . pls help
I'm self studying this book "Methods of Solving Number theory Problems by Elina" since many days but currently stuck on this formula of continued fractions.

For example $a=87/
55
= [1,1,1,2,1,1,4]$ in continued fraction. but when I put the values according to the formula I don't get the answer to being   $a=87/55$. what am I missing, if someone can explain this in simple words, it will be very helpful. Thanks

Title: continued fraction formula . pls help

Body:
We have

$$\frac{P_0}{Q_0}=\frac{1}{1},\frac{P_1}{Q_1}=\frac21, \frac{P_2}{Q_2}=\frac32, \frac{P_3}{Q_3}=\frac85, \frac{P_4}{Q_4}=\frac{11}{7}$$

and

$$\alpha_0 = \frac{87}{55}, \alpha_1 = \frac{55}{32}, \alpha_2 = \frac{32}{23}, \alpha_3 = \frac{23}{9}, \alpha_4 = \frac95, \alpha_5 = \frac54.$$

In particular, we have $\alpha_5 = \frac{5}{4}, P_4 =11, P_3=8, ,Q_4=7, Q_3 = 5 $.

We can verify that $$\frac{P_4\cdot \alpha_5+P_3}{Q_4\cdot \alpha_5 + Q_3}=\frac{11 \cdot \frac54 + 8}{7 \cdot \frac54 + 5}=\frac{55+32}{35+20}=\frac{87}{55}=\alpha$$

Another example, we have $\alpha_4=\frac95, P_3=8, P_2=3, Q_3=5, Q_2=2.$

We can verify that

$$\frac{P_3\cdot \alpha_4+P_2}{Q_3\cdot \alpha_4 + Q_2}=\frac{8 \cdot \frac95 + 3}{5 \cdot \frac95 + 2}=\frac{72+15}{45+10}=\frac{87}{55}=\alpha$$


问:
Binomial coefficients that are powers of 2
I would like a proof that
$$ {{n}\choose{k}} = \frac{n!}{k!(n-k)!} = 2^m $$
for $n,k,m\in \mathbb{N}$, only if $k=1$ or $k=n-1$.

It seems to me that this must be true since for other values of $k$ the numerator contains more factors that are not powers of 2 than the denominator. Furthermore, the numerator also contains larger factors than the denominator and thus can't all be cancelled. Nevertheless, I have been unable to form an elegant, watertight proof.

答:
I have an elementary proof. I don't know how to do math notation typesetting in this window [REWRITE UNDERWAY BELOW USING MATHJAX], but I shall endeavor to communicate my proof, that there are no powers of 2 in the interior of Pascal's Triangle.

Let B(a, b) be the binomial coefficient 'a above b',
known to be a!/(b! (a - b)!) when 0 <= b <= a (integers).

Let B*(a, b) be the exponent of 2 in the prime factorization of B(a, b),
let n# be the exponent of 2 in the prime factorization of n!,
and let $(n) be the exponent of 2 in the prime factorization of n itself.

Then B*(a, b) = a# - b# - (a - b)#
and $(n) = n# - (n - 1)#.

The function # satisfies (2n + 1)# = (2n)# = n + n#.
For there are n even numbers from 1 to 2n, and if we factor out one 2 from each of them, their remaining parts together are n!.

The strategy of the proof that an interior binomial coefficient
(that is B(a, b) with 2 <= b <= a - 2)
is not a power of 2 is to work with the four parity combinations of a and b,
B(even, even), B(odd, even), B(odd, odd), B(even, odd),
that is B(2a + r, 2b + s) with (r, s) an ordered pair from {0, 1},
use the property (2n + 1)# = (2n)# = n + n# of # to relate B*(2a + r, 2b + s) to B*(a, b), obtain an inequality from that relation if B(2a + r, 2b + s) is assumed to be a power of 2, bounding B(2a + r, 2b + s) above by a multiple of B(a, b), and contradict that bound by
analyzing how much larger B is in the neighborhood of (2a, 2b) than in the neighborhood of (a, b).

Let's start with the simplest case to analyze, the B(even, even) case.

B*(2a, 2b) = (2a)# - (2b)# - (2a - 2b)# = a + a# - b - b# - (a - b) - (a - b)# = B*(a, b).

So if B(2a, 2b) is a power of 2, say 2^n, then B(a, b) = (2^n)D for some odd number d, but with 2^b being B(2a, 2b), that implies B(2a, 2b) <= B(a, b).

But B(2a, 2b) / B(a, b) works out to, um, typesetting of fractions would be very nice here, but I'll describe it and trust you can do it.

B(2a, 2b) / B(a, b) is the ratio of f(a) to f(b) f(a - b) where f(n) = (2n)!/(n!).

That's a factors in the numerator, and a factors in the denominator, and when you write out it is easy to see that the factors in the numerator can be paired with the factors in the denominator to make it all a product of a fractions that each exceed 1, so the whole product exceeds 1, so B(2a, 2b) > B(a, b) and therefore B(2a, 2b) can't be a power of 2.

Next let's look at B*(2a + 1, 2b).

B*(2a + 1, 2b) = (2a + 1)# - (2b)# - (2a - 2b + 1)#

= (2a)# - (2b)# - (2a - 2b)# = B*(2a, 2b)

and we already know B*(2a, 2b) = Ba, b), so B(2a + 1, 2b) = B*(a, b).

Now the same argument as with B(even, even): if B(2a + 1, 2b) = 2^n, then B(a, b) = (2^n)D for some odd D, and therefore B(2a + 1, 2b) <= B(a, b). But B(2a + 1, b) / B(a, b) works out to (2a + 1)/(2a + 1 - 2b) times B(2a, 2b) / B(a, b), and both of those exceed 1, so B(2a + 1, 2b) > B(a, b), incontrovertibly contradicting the consequence of the assumption that B(2a + 1, 2b) is a power of 2. Therefore B(odd, even) is never a power of 2 in the interior of Pascal's Triangle.

For B(2a + 1, 2b + 1), again the B* is equal to B*(a, b), implying that if B(2a + 1, 2b + 1) is a power of 2 then B(2a + 1, 2b + 1) <= B(a, b). This time the fraction B(2a + 1, 2b + 1) / B(a, b) works out to (2a + 1)/ (2b + 1) times B(2a, 2b) / B(a, b), again giving B(2a + 1, 2b + 1) > B(a, b) and hence B(odd, odd) is not a power of 2.

The only case left is B(even, odd). This one plays out different.

B*(2a, 2b + 1) = (2a)# - (2b + 1)# - (2a - 2b - 1)#
= (2a)# - (2b)# - (2a - 2b - 2 + 1)#
= (2a)# - (2b)# - (2(a - b - 1))#
= a + b - (a - b - 1) + a# - b# - (a - b - 1)#
= 1 + a# - b# - (a - b)# + (a - b)# - (a - b - 1)#
= 1 + B*(a, b) + $(a - b)

which means that if B(2a, 2b + 1) = 2^n and B(a, b) = (2^k)D with D odd, and a - b = (2^t)d with d odd, then n = k + t + 1 so B(2a, 2b + 1) divides 2^(1+t) B(a, b).

Now 2^t divides a - b and a - b <= a - 2, and we may have a 'feel' that B(2a, 2b + 1) is so much larger than B(a, b) that we can replace 2^(1 + t) by the larger but simpler 2(a - 2) and still prove B(2a, 2b + 1) > 2(a - 2) B(a, b). But how to prove it? The factorials analysis is trickier than in the other three cases. I tried it and then tried a different way: use the Pascal Triangle Property B(m, n) = B(m - 1, n) + B(m - 1, n - 1) on (m, n) = (2a + r, 2b + s) and keep repeating the use of the Pascal Triangle Property on the summands resulting from using the Pascal Triangle Property, until the inputs to B get down to a and b. The coefficients of the summands in the process are themselves binomial coefficients, and the result after k rounds of use is the Vandermonde convolution,

B(m, n) = Sum{t=0..k} B(k, t) B(m - k, n - t).

With (m, n) = (2a + r, 2b + s), to get m - k down to a,
take k = a + r.

B(2a + r, 2b + s) = Sum{t=0..a+r} B(a + r, t) B(a, 2b + s - t).

Now separate out the term indexed by t for which 2b + s - t = b, that is t = b + s. We then have

B(2a + r, 2b + s) = B(a + r, b + s) B(a, b) +

Sum{t=0..b+s-1,b+s+1..a+r} B(a + r, t) B(a, 2b + s - t)

The indexed sum still has terms if 2 <=  2b + s <= 2a + r, and they are positive terms, so

B(2a + r, 2b + s) > B(a + r, b + s) B(a, b),

which in case (r, s) = (0, 1) is B(2a, 2b + 1) > B(a, b + 1) B(a, b),

so if B(a, b + 1) > 2(a - 2) then B(2a, 2b + 1) is not a power of 2.

B(2a, 2b + 1) is interior iff 2 <= 2b + 1 <= 2a - 2, which works out to 1 <= b <= a - 2. In that case, B(a, b + 1) >= a(a - 1)/2; so if a(a - 1) > 4(a - 2) then B(2a, 2b + 1) is not a power of 2.

a^2 - 5a + 8 > 0 for all real a,
since that quadratic has negative discriminant and positive leading coefficient. And with that, the proof is complete, that interior binomial coefficients are never powers of 2.

REWRITE UNDERWAY USING MATHJAX:

Let $E(n)$ be the exponent of $2$ in the prime factorization of $n$, and let $n^*$ be the composite $E(n!)$.

When input to the function $E$ written in the form of a binomial coefficient  $\binom{a}{b}$, let the output notation $E\left(\binom{a}{b}\right)$ be acceptable in the form $E\binom{a}{b}$.

The functions $E$ and $^*$ satisfy $$E\binom{a}{b}=a^*-b^*-(a-b)^*\tag{1}\label{1};$$ $$(2n+1)^*=(2n)^*=n+n^*\tag{2a,b}\label{2a,b};$$ $$E(n)=n^*-(n-1)^*\tag{3}\label{3},$$
when $2<=b<=a-2$ (integers). That is all we'll need about $E$ and $^*$.

$(1), (2a),$ and $(3)$ seem to me obvious enough to you so we don't have to take up space here proving them. For $(2b),$ that is $(2n)^*=n+n^*,$ I'll just say that there are $n$ even numbers from $1$ to $2n,$ and if we factor out one $2$ from each of them, their remaining parts together are $n!.$

The proof I am giving that no interior binomial coefficients are powers of 2 can be done more efficiently than I'm about to do it (and more generally, proving that no interior binomial coefficient is any prime power), but I think the ideas will come through best by sticking close to how I went about it in the first place (and then, after the proof, I also want to say why I wanted to prove it, what in my math life long ago prepared me to have the key idea in this proof now, how else I tried to prove it before proving it this way, and wonder how these two approaches to proving this theorem might relate to the theorems of Kummer or Sylvester quoted above).

Or should I not wait until after the proof to say why I wanted a proof? I'm new to writing or reading in this forum and have been reading our in-house guidelines about posting but am still unsure of protocols. Why don't I write it how I feel it wants me to write it, and leave it to readers to tell me if I wrote it wrong for readers here.

How I came to the problem of proving that interior binomial coefficients are never powers of 2 was that I am studying a function I'm calling $G(h)$, defined by $$G(h):= the\ least\ k\ for\ which\ \binom{h+k}{h}>2^h$$ and I simply wanted to know that I could rely on the starkness of two $<$ in $$\binom{h+G(h)-1}{h}<2^h<\binom{h+G(h)}{h}$$ and not have to muddle along with $$\binom{h+G(h)-1}{h}<=2^h<\binom{h+G(h)}{h}.$$ I just felt more comfortable about G with the two <. That was my reason for wanting to know that there are no powers of 2 in the interior of Pascal's Triangle. Whether it is a good reason or not, it is a stated reason and thereby takes some heat off the question of why or whether to care if there are any powers of 2 in the interior of Pascal's Triangle or not. After I get this proof written well enough, I want to resume my study of G and a companion function with it, $$L(h):=the\ least\ k\ for\ which\ \binom{h}{h-k}>2^{h-k},$$ and post some questions about the respective sequences of lengths of runs of constancy of G and L.

I noticed a long time ago, like 40 years ago, that property $(2b)$ of the function * easily implies $E\binom{2a}{2b}=E\binom{a}{b}$, but I did not make anything of it, nor work out anything about the three other parity combinations $\binom{even}{odd}, \binom{odd}{even},\binom{odd}{odd}$, until the urge to try to prove the lower < in $$\binom{h+G(h)-1}{h}<2^h<\binom{h+G(h)}{h}$$ came on me these 40 years later.

I knew $(2b)$ back then from having the formula $\frac{(2n)!}{n! 2^n}$ for the number of 2-uniform partitions of 2n elements, working out the first few values (by hand, not having anything like Maple in 1980), and proving that those numbers were what they looked like: the product of the first n odd numbers, satisfying the same recursion and having the same initial value as the number of 2-uniform partitions of a 2n-element set.

And why I was thinking about 2-uniform partitions of sets 40 years was partly from reading the concept of an involution in a group, making me think about involutory permutations of a set, and partly from a breakup with a girlfriend, making me think about singles and couples.

That was a vivid learn in my math life, but after 40 years it still took me a few days now to think of using $E\binom{2a}{2b}=E\binom{a}{b}$ to prove $\binom{2a}{2b}\not=2^n$ once I set out to prove $\binom{m}{k}\not=2^n$ for all parities of $m$ and $k$ with $2<=k<=m-2$.

If hypothetically $\binom{2a}{2b}=2^n$ then $\binom{2a}{2b}$ divides $\binom{a}{b}$ since $$E\binom{2a}{2b}=(2a)^*-(2b)^*-(2a-2b)^*$$$$=\cdots=(a+a^*)-(b+b^*)-((a-b)+((a-b)^*)$$$$=a-b-(a-b)+a^*-b^*-(a-b)^*$$$$=a^*-b^*-(a-b)^*=E\binom{a}{b},$$so $\binom{2a}{2b}<=\binom{a}{b}$. But there are many ways to prove, non-hypothetically, that $\binom{2a}{2b}>\binom{a}{b}$. The non-hypothetically actual takes priority over the hypothetical, forcing $\binom{even}{even}\not=2^n$.

One of the ways to prove $\binom{2a}{2b}>\binom{a}{b}$ is $$\frac{\binom{2a}{2b}}{\binom{a}{b}}=\frac{\frac{(2a)!}{(2b)!(2a-2b)!}}{\frac{a!}{b!(a-b)!}}=\frac{(2a)!b!(a-b)!}{(2b)!(2a-2b)!a!}$$$$=\frac{(2a)!}{a!}\frac{b!}{(2b)!}\frac{(2(a-b))!}{a-b)}$$$$=\frac{(a+1)(a+2)\cdots(a+a)}{(b+1)(b+2)\cdots(b+b)\cdot(a-b+1)(a-b+2)\cdots(a-b+(a-b))}$$$$=\left(\prod_{n=1}^b\frac{a+n}{b+n}\right)\cdot\left(\prod_{n=1}^{a-b}\frac{a+b+n}{a-b+n}\right)>1,$$ each of those factors $\frac{a+n}{b+n}$ and $\frac{a+b+n}{a-b+n}$ being more than $1$ when $a>b>1$.

And so it is proved that $\binom{2a}{2b}$ is not a power of $2$ when $2<=2b<=2a-2$.

The general case $E\binom{2a+r}{2b+s}$, with $(r,s)\in\{0,1\}\times\{0,1\}$ goes $$E\binom{2a+r}{2b+s}=(2a+r)^*-(2b+s)^*-(2a-2b+r-s)^*$$$$=(2a)^*-(2b)^*-(2(a-b)+r-s)^*.$$ If $r-s>=0$ then $(2(a-b)+r-s)^*=(2(a-b))^*$ and $$E\binom{2a+r}{2b+s}=(2a)^*-(2s)^*-(2(a-b))^*=E\binom{a}{b};$$in that case, if $E\binom{2a+r}{2b+s}$ is a power of $2$, then $\binom{2a+r}{2b+s}<=\binom{a}{b}$, but that can't happen, since $$\binom{2a+r}{2b+s}=\frac{(2a+r)!}{(2b+s)!((2a-2b+r-s)!}$$$$=\binom{2a}{2b}\times\begin{cases}\frac{2a+1}{2b+1}&\text{if r=s=1}\\1&\text{if r=s=0}\\\frac{2a+1}{2a-2b+1}&\text{if (r,s)=(1,0)}\end{cases}$$$$>\binom{a}{b}\times\begin{cases}\frac{2a+1}{2b+1}&\text{if r=s=1}\\1&\text{if r=s=0}\\\frac{2a+1}{2a-2b+1}&\text{if (r,s)=(1,0)}\end{cases}$$$$>\binom{a}{b}.$$

That leaves us just the $\binom{even}{odd}$ case, $\binom{2a}{2b+1}$, to prove can't be a power of 2.

... Now I'm two days later writing this sentence than the last sentence before this one, and I understand more about this math now, but am socially more unsure how to proceed writing this math. I'm worried about the length of this, if I recast the math at hand as the case p = 2 of there being no prime powers in the interior of Pascal's Triangle. I just now read our Code of Conduct and it was more reassuring than not. I don't still have that window open but two things it said that are standing out to me in beginning this writing session are that it can be hard to learn how to fit in this community, and that some kinds of jokes are acceptable. I'll try a joke here now, one that I thought yesterday without trying to think of a joke. It popped into my head yesterday to resume this writing today something like the following, spoofing myself as an old man reminiscing:

"With very little extra effort we can do this for all prime powers, not just powers of 2. (All I wanted was $\binom{h+G(h)-1}{h} \not= 2^h$, and it's looking like I'm getting the whole $\binom{a}{b} \not= p^h$. Reminds me of the time Threepio and Artoo shut down all the garbage mashers on the detention level just to shut one of them down.)"

I feel a social obligation to finish the proof for $2$ before gallivanting with $p$, and the $2$ can be done without the $p$, but going with $p$ can give us a better mathematical look at what's going on with $2$ in the $\binom{even}{odd}$ case, and setting things up for $p$ is not far out of the way from simply continuing with $2$.

For every prime number $p$ and positive integer $n$, let $E_p(n)$ be the exponent of $p$ in $n$, and $F_p(n)$ the exponent of $p$ in the prime factorization of $n!$. So $F_p$ is the partial sums of $E_p$, and $E_p(n)=F_p(n)-F_p(n-1)$. If $0<=r<p$ then $F_p(ap+r)=F_p(ap)=a+F_p(a)$; readers who agree with that, readers who accept it for the sake of argument, and readers who ignore $p>2$ and keep reading $2$ for $p$, $n^*$ for $F_p(n)$, and $E$ for $E_p$, this is for all of you: if $ap+r\geq{bp+s}\geq0$ with $0\leq{r}<p$ and $0\leq{s}<p$ and $a\geq0$ and $b\geq0$ ($a, b, r, s$ integers) then $a\geq{b}$ and $$E_p\binom{ap+r}{bp+s}=F_p(ap+r)-F_p(bp+s)-F_p\left((ap+r)-(bp+s)\right)$$$$=\cdots=F_p(ap)-F_p(bp)-F_p\left((a-b)p+(r-s)\right).$$

In case $r-s\geq0$ then $p>r-s\geq0$ since $p>r$ and $s\geq0$,
so $F_p\left((a-b)p+(r-s)\right)=F_p\left((a-b)p\right)$ and hence $$E_p\binom{ap+r}{bp+s}=F_p(ap)-F_p(bp)-F_p\left((a-b)p\right)$$$$=\cdots=\left(a+F_p(a)\right)-\left(b+F_p(b)\right)-\left((a-b)\right)$$$$=\cdots=\left(a-b-(a-b)\right)+\left(F_p)(a)-F_p(b)-F_p(a-b)\right)$$$$=\cdots=E_p\binom{a}{b}.$$

But if $r-s<0$ then$$E_p\binom{ap+r}{bp+s}=F_p(ap)-F_p(bp)-F_p\left((a-b)p-p+p+(r-s)\right)$$$$=F_p(ap)-F_p(bp)-F_p\left((a-b-1)p + p+r-s\right)$$$$=F_p(ap)-F_p(bp)-F_p\left((a-b-1)p\right)$$$$=\left(a+F_p(a)\right)-\left(b+F_p(b)\right)-\left((a-b-1)+F_p(a-b-1)\right)$$$$=\left(a-b-(a-b-1)\right)+\left(F_p(a)-F_p(b)-F_p(a-b-1)\right)$$$$=1+\left(F_p(a)-F_p(b)-F_p(a-b)+F_p(a-b)-F_p(a-b-1)\right)$$$$=1+E_p\binom{a}{b}+E_p(a-b)$$

That is,$$E_p\binom{ap+r}{bp+s}=\begin{cases}E_p\binom{a}{b}&\text{if ${r-s}\geq{0}$}\\E_p\binom{a}{b}+1+E_p(a-b)&\text{if $r-s<0$}\end{cases},$$so if $\binom{ap+r}{bp+s}$ is a power of $p$, then $$\binom{ap+r}{bp+s}\text{divides}\binom{a}{b}p^{1+E_p(a-b)},$$and $p^{E_p(a-b)}$ divides $a-b$, so $$\binom{ap+r}{bp+s}\text{divides}\binom{a}{b}\cdot{p}\cdot(a-b).$$

It is difficult to overstate how contradictory that looks, but that it is indeed contradictory still has to be proved. It suffices to prove $$\binom{ap+r}{bp+s}>\binom{a}{b}\cdot{p}\cdot(a-b).$$

The two sides are so far apart that to compare them is hard like trying to focus sight on a near object and a far object in the same look. To me it's like having to find intermediates between them that that are close enough to each other so we can carry out a comparison. Someone could work to contradict that divisibility relation from $\binom{ap+r}{bp+s}$, but I am going to work to contradict the magnitude relation.

If I want to prove Left Side > Right Side and I can decrease Left to New Left and increase Right to New Right and prove New Left > New Right, then that would suffice to prove Old Left > Old Right. I'll start by increasing a - b to a, so it suffices to prove $$\binom{ap+r}{bp+s} > \binom{a}{b}\cdot p \cdot a.$$

Now use $\binom{x+y}{u+v}>\binom{x}{u}\cdot\binom{y}{v}$ which holds when $x\geq{u}$ and $y\geq{v}$ and $uv\not=0$.$$\binom{ap+r}{bp+s}>\binom{a(p-1)+r}{b(p-1)+s}\cdot\binom{a}{b},$$ so it suffices to prove $$\binom{a(p-1)+r}{b(p-1)+s}>p\cdot a.$$ Also, we know $$\binom{a(p-1)+r}{b(p-1)+s}>{\binom{a}{b}}^{k} \cdot \binom{a(p-k)+r}{b(p-k)+s}$$ while $b(p-k)+s$ stays between 0 and $a(p-k)+r$, which is when $k\leq{\left(p-\frac{r-s}{a-b}\right)}$. (If $r-s<0$, then k could be more than p.)

In the case where $p = 2$, $r-s$ could be at most $1$, so if $a - b > 1$ then $k$ could be $2$, and it would suffice to prove $$\binom{a}{b}^2 \cdot\binom{r}{s}>2a$$,with r and s being only 0 or 1, so it would suffice to prove$$\binom{a}{b}^2 >2a,$$which could be done by the team of $\binom{a}{b}>2$ and $\binom{a}{b}>a$, if these two inequalities hold where $\binom{2a+r}{2b+s}$ is interior, that is where $$2 \leq 2b+s \leq 2a+r-2.$$

Friends, enemies if this has made any, and even that girl 40+ years ago if she reads this (and whichever she is, friend or enemy) I'll leave the proof here needing some finishing touches but answered the question posed at the top of this reading window. The answer is that the only way it can happen that $\binom{n}{k}=2^m$ is for $n$ itself to $2^m$, and $k$ to be $1$ or $2^k-1$.

It seems I have to post this to save it. I wanted to save it just now, so I posted it, accepting that I have to share it to save it. But I wasn't done writing it to post. Even though leaving off the finish of the proof that answered the question, there is still the matter of the other proof I tried, and there is the matter of sooner or later removing or condensing the first draft of the answer in this posting, so I am hoping that I will not be blocked for continuing to edit a posting for maybe quite a while of time.

But I will rest this here for now awhile, to watch for comments and think on these math and social matters.


问:
How is the area of a country calculated?
As countries' or states' borders are not straight lines but they are irregular in nature, I wonder how anyone can calculate the area of a country or state.

When do you think the area of a country or state was first calculated?  Was it before satellites provided us accurate picture of the earth?

Note: I am not asking about surface area of a country. They are assumed as flat while calculating the area.

答:
I guess you could ask Google or WolframAlpha.  Interestingly, these answers differ substantially.  Perhaps, that's just a matter of how territories are interpreted but it illustrates the point that, there's really no easy answer.  The question is at once terribly elementary and, on the other hand, fabulously interesting.  Mandelbrot asked the question "How long is the coast of Britain?"  Turns out that it depends strongly on how carefully you measure it.

So, the short answer is - it's super simple, in that you do it just like any other spherical polygon.  Dealing with data at this level, as well as territorial disputes, is a bit more complicated.  To illustrate more concretely, consider the image below.

You can see that, in a very simple sense, each "country" can be triangulated.  These triangles can then be further sub-divided to get a good approximation and the areas of the triangles can be added up.  Thus, again, on the most basic level it's no harder than adding up the areas of some triangles.  To fully implement this, though, requires a fair understanding of computational geometry as well as access to solid data, which is likely disputable anyway.

The real question is - can you get good upper and lower bounds?

Addendum

Per request, here's a look at the other side of the planet:

And here's a look on the map of just India and its neighboring countries.


问:
Confusion in one of the combinatorics/probability problem, with my approach for its solution.

The only contents of a container are 10 disks that are each numbered
with a different positive integer from 1 through 10, inclusive. If 4
disks are to be selected one after the other, with each disk selected
at random and without replacement, what is the probability that the
range of the numbers on the disks selected is 7?

My approach: (Using the concept of conditional probability and combinatorics)

Possible ranges: 1-8, 2-9, 7-10 (Total 3)

Now, taking one range at a time, (say 1-8):
Hence, among the 4 disks to be selected, the probability of selecting the first disk will 1/10 (for 1).
Similarly, 1 is selected, hence second place could be filled by 8. Prob(second place is 1/9).
Now, since out of range 1-8 1 and 8 are selected, therefore we have only 6 choices (out of the remaining 8) for third place. Henceforth, only 5 choices (out of 7 choices) for fourth place.
Combining the probability of all 4 places, (1/10)(1/9)(6/8)(5/7).

Now, the number in four places can be arranged in 4! ways. Also, apart from the (1-8) range, there is also other two range of numbers.
Therefore, combining all, the required probability should be (3) x (1/10)(1/9)(6/8)(5/7) x (4!).
But my answer is wrong. My answer is double of actual answer.

答:
I'm not exactly sure where you're double counting.  The method you've elected to use is rather confusing.

You've got all the pieces, but there's a simpler way to put them together.  There are, as you said, $3$ possible ranges.  There are $\binom62$ ways to choose $2$ numbers in the range other than the endpoints, so there are $3\binom62$ possible choices.  There are $\binom{10}4$ equiprobable ways to choose $4$ numbers, so the probability is $$\frac{3\binom62}{\binom{10}4}=\frac3{14}$$

EDIT

I'm having difficulty pinpointing your error.  As I said, the method is confusing.  I think the problem arises when you compute the probability of choosing $2$ numbers from the middle of the range.  You say the probability is $\frac{6\cdot5}{8\cdot7}$ which is true, but this incldues both the possibility of choosing $5$ and then $6$ and the possibility of choosing $6$ and then $5$.  When you then multiply by $4!$, I think you're doubling up.  I'm not $100\%$ certain though.


问:
Is it "technically consistent" to interpret a Gentzen-style proof caclulus as a Hilbert system with $\vdash$ as a variadic logical connective?
Is it internally consistent to interpret $\vdash$ in Gentzen-style systems as a variadic-but-otherwise-ordinary connective? If it is consistent, is there a way to show that interpreting $\vdash$ this way leads to really bad predictions or is substantially less parsimonious than the correct interpretation?

Alternatively, what's the most concrete way to show that the interpretation of $\vdash$ as a variadic logical connective instead of a metalogical symbol is "wrong" in some sense, assuming that such a way exists?

I think misunderstanding that $\vdash$ and the comma are metalogical symbols and instead interpreting them as part of the syntax is common among folks who are new to logic or studying it on their own. I, at least, thought this way for years before learning the correct interpretation.

What follows is a more complete explanation of the question with an example.

Here's a Gentzen-style system for propositional calculus for classical logic with the connectives $\to$ for implication and $\nrightarrow$ for nonimplication.

This is a variant of system LK as presented in the Wikipedia article, but with fewer connectives and hence fewer rules. The structural rules are omitted entirely for brevity.

$$ \frac{}{A \vdash A} \;\; \text{is the identity rule} $$

$$ \frac{\Gamma \vdash \Delta, A \;\; \text{and} \;\; A, \Sigma \vdash \Pi}{\Gamma, \Sigma \vdash \Delta, \Pi} \;\;\text{is cut elimination} $$

$$ \frac{\Gamma \vdash A, \Delta \;\; \text{and} \;\; \Sigma, B \vdash \Pi}{\Gamma, \Sigma, A \to B \vdash \Delta, \Pi} \;\; \text{is left implication introduction} $$

$$ \frac{\Gamma, A \vdash B, \Delta}{\Gamma \vdash A \to B, \Delta} \;\; \text{is right implication introduction} $$

$$ \frac{\Gamma, A \vdash B, \Delta}{\Gamma, A \nrightarrow B \vdash \Delta} \;\; \text{is left nonimplication introduction} $$

$$ \frac{\Gamma \vdash A, \Delta \;\; \text{and} \;\; \Sigma, B \vdash \Pi}{\Gamma, \Sigma \vdash A \nrightarrow B, \Delta, \Pi} \;\; \text{is right nonimplication introduction} $$

In this notation, $A$ is a well-formed formula, $\Gamma, \Delta, \Sigma, \Pi$ are finite sets of well-formed formulas, $\to$ and $\nrightarrow$ are logical connectives, and $\vdash$, the comma, the line of inference and the $\text{and}$ that separates premises are metalogical symbols.

However, if we squint, $( \text{formula, $\cdots,$ formula} \vdash \text{formula, $\cdots,$ formula})$ looks like a connective similar to $\to$ or $\nrightarrow$, just one that takes an arbitrary number of left arguments and an arbitrary number of right arguments.

I'm now wondering whether interpreting $\vdash$ as a variadic connective with an unlimited number of left arguments and an unlimited number of right arguments is internally consistent.

By the faux-Hilbert-system interpretation, the premise of the right implication introduction rule $\Gamma, A \vdash B, \Delta$ would be a well-formed formula with a toplevel connective $\vdash$ and the interpretation of the right implication introduction rule would just be a rule that takes this well-formed formula and outputs the related well-formed formula $\Gamma \vdash A \to B, \Delta$. The rule would be interpreted the same way more familiar rules like modus ponens are interpreted. In modus ponens, shown below, $A$ and $B$ are syntactic metavariables that bind to particular subformulas in an application of modus ponens.

$$ \frac{A \;\;\text{and}\;\; A \to B}{B} \;\; \text{is Hilbert-style modus ponens} $$

In faux-Hilbert-style modus ponens $A$ and $B$ are syntactic metavariables, and $\Gamma$ and $\Delta$ are a special kind of syntactic metavariable that stands for a collection of arguments to a variadic function, which are themselves well-formed formulas.

答:
The sequent calculus comes in two layers: A structural layer, which describes how formulas are shifted around in sequents, and a logical layer, which describes how formulas can be decomposed depending on their outermost logical connective. Your suggestion amounts to removing the structural layer by treating $\vdash$ as a logical connective. In this way, you obtain a Hilbert-style proof system which is based on formulas instead of sequents.

While this is certainly possible in some way or another, it does not seem to be very beneficial.

First, there are some technical obstacles. As you mention yourself, $\vdash$ seen as a binary connective does not have a fixed arity. A more substantial issue is that you do not have nestings of $\vdash$, since $\vdash$ appears only as the outermost symbol in sequents. Consequently the formulas of your logic would not be closed under substitution, as it is usually the case. E.g., $p\vdash A$ and $q\vdash B$ are formulas but $(q\vdash B)\vdash A$ is not. So even if you take $\vdash$ to be a logical connective, its treatment will have to be quite different from that of the standard logical connectives.

More importantly though, there are conceptional reasons why the two-layered presentation of the sequent calculus is adequate.

It is a nice property of the logical rules of sequent calculus that they depend only on the outermost logical connective of formulas; we never have to look deep inside a formula to see which rules are applicable. This property will be obfuscated in the "faux-Hilbert"-approach, where the outermost connective is $\vdash$.

The vital subformula property will be obfuscated as well. The subformula property (established via cut elimination) states that every sequent has a proof in which all occuring formulas are subformulas of the endsequent. Note that the "structural layer" is swept under the carpet here: We only care about whether a formula appears in a sequent or not, but not how often or in which order. Consider for example the proof:
$$
\frac{A\vdash A}{\vdash \lnot A,A}
$$
This proof satisfies the subformula property because every formula appearing in the upper sequent (i.e., $A$) is a subformula of one of the formulas in the lower sequent (i.e., $A$ and $\lnot A$). On the other hand, $A\vdash A$ seen as a single formula with main connective $\vdash$ is not a subformula of $\vdash\lnot A,A$, and so at least the naive subformula property fails in the "faux-Hilbert" approach. You sometimes read this point being expressed as analyticity holds only modulo the structure of the proof system $-$ that is, if we ignore $\vdash$ and $,$ (comma) and concentrate on the formulas only. This is another good reason for keeping the the structural layer of the sequent calculus separate from the logical one.


问:
Onto and one-to-one
Let $T$ be a linear operator on a finite dimensional inner product space $V$. If $T$ has an eigenvector, then so does $T^*$.

Proof. Suppose that $v$ is an eigenvector of $T$ with corresponding eigenvalue $\lambda$. Then for any $x \in V$,

$$
0 = \langle0,x\rangle = \langle(T-\lambda I)v,x\rangle
= \langle v, (T-\lambda I)^*x\rangle = \langle v,(T^*-\bar{\lambda} I)x\rangle
$$
This means that $(T^*-\bar{\lambda} I)$ is not onto. WHY?

(Of course the proof is not completed in here)

答: If you want to know why $T^{*}$ is not onto if $T$ is not onto, just observe that the matrix of $T^{*}$ is just the conjugate transpose of the matrix of $T$ and we know that a matrix and its conjugate transpose have the same rank (which is equal to the dimension of the range space).</p>

Title: Onto and one-to-one


问:
$\Bbb R[x]/(x^2-1)\cong \Bbb R[x]/(x-1)\times \Bbb R[x]/(x+1)\cong \Bbb R^2\,$ via Chinese Remainder Theorem
I'm currently following my 2nd algebra course at uni where we are discussing rings.

The assignment goes like this
$$\text{Show that} \ \mathbb{R}[x]/(x^2-1) \ \text{is isomorphic to} \ \mathbb{R}\times\mathbb{R}$$

$\mathbb{R}[x]$ is the polynomial ring and $\mathbb{R}[x]/(x^2-1)$ is then a quotient ring. A hint to solve it is to use the Chinese Remainder Theorem.

I am stumped. The hint implies to me that $(x^2-1)$ must be the kernel of some ring homomorphism. It also to me implies that there is an ideal in $\mathbb{R}[x]$, let's call it K, so that $\mathbb{R}[x]/K=\mathbb{R}$ and I can't make either of those things to make sense. Is it also right to assume that $\mathbb{R}[x]/(x^2-1)=\{f(x)+x^2-1\vert f(x)\in\mathbb{R}[x]\}=\mathbb{R}[x]$, since all equivalence classes will contain just one element, since for two elements to be in the same equivalence class, the polynomials must be the same? Like if,

$$q(x)+x^2-1=r(x)+x^2-1\Leftrightarrow q(x)=r(x)$$
for $q(x),r(x)\in\mathbb{R}[x]$.

答:
Hint:

As you know, $X^2-1=(X-1)(X+1)$ and the factors are coprime, hence by the Chinese remainder theorem,
$$\mathbf R[X]/(X^2-1)\simeq \mathbf R[X]/(X-1)\times\mathbf R[X]/(X+1).$$


问:
Questions about the Koszul differential in this case
The following statement is from the article by Kumar and Vergne on equivariant cohomology page 133:

Let us now consider the lie group G with lie algebra $\mathfrak{g}$. Recall the definition of te Koszul differential $c_L$ on the space $T^*= L \otimes \wedge^*\mathfrak{g}^*$ calculating the cohomology of a $\mathfrak{g}$-module L: for $ \alpha \in L \otimes \wedge^p\mathfrak{g}^*$, $c_L \alpha \in L \otimes \wedge^{p+1}\mathfrak{g}^*$ is defined by
$$ (c_L \alpha)(X_1,...,X_{p+1}) = \sum_i (-1)^{i+1}X_i.\alpha(X_1,..., \hat{X_i},...,X_{p+1}) + \sum_{i<j}(-1)^{i+1}\alpha([X_i,X_j]X_1,..., \hat{X_i},...\hat{X_j},...,X_{p+1})$$ where $X_1,...,X_{p+1}$ are elements of $\mathfrak{g}$

What I don't understand in this paragraph is why are the entries of $c_L \alpha$ elements of $\mathfrak{g}$. I mean $c_L \alpha$ is defined on the tensor product of L and $\wedge^* \mathfrak{g}^*$, why isn't  kind of this  $ c_L \alpha(l_1 \otimes X_1 ,....,l_k \otimes X_k)$ , where $l_i \otimes X_i \in L \otimes \wedge^*  \mathfrak{g}^*$ ?

I understand from that statement that the Koszul differential on the space $ L \otimes \wedge^*\mathfrak{g}^*$ helps to  calculate the cohomology of a  L, am I right ? Please could you explain to me why it is useful to work with Koszul differential in this setting .

Thanks!

答:
If $\mathfrak{g}$ is a Lie algebra, its graded differential algebra is $\Lambda^* \mathfrak{g}^*$, and its elements are alternate -or differential- forms: they take value in the field of scalar $\mathbb{K}$ of $\mathfrak{g}$.

But sometimes, you may want to consider the same sort of structures but with value in another space than the field $\mathbb{K}$. For example, if $V$ is a $\mathbb{K}$ vector space, you may want to study "alternate forms" with values in $V$. A way to do that is considering the tensorial product $\left(\Lambda^*\mathfrak{g}^*\right)\otimes_{\mathbb{K}}V$: its elements are of the form $\sum \omega^i \otimes v_i$ where $\omega^i$ is a differential form on $\mathfrak{g}$ and $v_i$ is an element of $V$. For instance, in the simplest case, if $\omega$ is a linear form on $\mathfrak{g}$ (i.e $\omega \in \mathfrak{g}^*$) and if $v \in V$, then one can think of $\omega\otimes v$ as a linear function on $\mathfrak{g}$ with value in $V$: indeed, if $X \in \mathfrak{g}$, then $\left(\omega\otimes v \right)(X) = \omega(X)\cdot v \in V$.

One can extend this construction to many algebraic objects. In the case you are talking about, consider $L$ a $\mathfrak{g}$-module. Then we can think of $\mathfrak{g}^*\otimes_{\mathfrak{g}}L$ as the set of linear functions from $\mathfrak{g}$ to $L$, and more generally, if $\omega \in \Lambda^p \mathfrak{g}^*$, and $l\in L$, $\omega\otimes l$ can be thought as a $p$-alternate form with value in $L$, etc. I just wrote things on simple tensors $\omega \otimes l$, but of course, the linearity is the key to define everything.

There is a canonical isomorphism between $A \otimes B$ and $B\otimes A$, thus the order does not really matter. I would rather consider the alternate form on the left to keep thinking about it as a coefficient, but it seems the authors of the cited article prefer writing it to the right.

It is not surprising that this can help computing some cohomology classes: it is a generalisation of the de Rham complex and its exterior differential.


问:
Show that [3,4] x [5,6] is closed?
[3,4] is closed in R <-- R-[3,4] is open

[5,6] is closed in R <-- R-[5,6] is open

Show that [3,4] x [5,6] is closed in R x R by writing it as the complement of the intersection of two open sets in R x R.

(R - [3,4]) x (R - [5,6]) not equal R x R - [3,4] x [5,6]

答:
Okay, I think the exercise was badly written and incorrect.  But we can do it.

Let $A = \{(x,y)| x < 3\}; B= \{(x,y)|y< 5\}; C= \{(x,y)|x>4\}; D=\{(x,y)|y>6\}$ then $U = A\cup B \cup C \cup D$ is open.

And $\mathbb R \times \mathbb R$ is also open.

So $U\cap (\mathbb R \times \mathbb R) = U$ and $U^c = \{(x,y)| 3\le x\le 4$ and $5 \le y \le 6\}= [3,4]\times [5,6]$ is closed.

....

which is really stupid and convoluted.

....

To be an iota less contrived we could add $E= \{(x,y)| y > x-2\}$ and $F=\{(x,y)|y < x-2\}$ and if $U= A\cup B \cup C \cup D \cup E$ and $W = A\cup B \cup C \cup D \cup F$ then $U \cap W= ([3,4]\times [5,6])^c$.

Title: Show that [3,4] x [5,6] is closed?


问:
Sigma algebra generated by $X+Y$
Let $X,Y$ be two random variables with values in $\{0,1\}$.

I want to show that $XY$ is measurable regarding $\sigma(X+Y)$ but I am not sure how $\sigma(X+Y)$ looks like.

I tried $\sigma(X+Y)=\sigma(\emptyset,0,1,2,\Omega)$.

Now, to show $XY$ is measurable I need to check if $(XY)^{-1}(\mathcal{P}(\{0,1\}))  \in \sigma(X+Y):$
 $\Rightarrow$ $(XY)^{-1}(1)=2 \quad \in \sigma(X+Y) \quad$
 (since $2=\{X=1\}+\{Y=1\} \Rightarrow \{X=1\}\cdot\{Y=1\} =1) $
$(XY)^{-1}(0)= 0 \cup 1 \quad \in \sigma(X+Y)$.

I am honestly not quite sure, if my idea is right; may someone comment on this?

答:
Your idea is in principle correct, you are just being a bit sloppy in notation.

When you write "$(XY)^{-1}(1) = 2 \in \sigma(X+Y)$", what you actually mean is the set $\{X+Y=2\}$ (or more verbosely $\{\omega\in\Omega\mid (X+Y)(\omega) = 2\}$) and that lies indeed in $\sigma(X+Y)$.

So formally you can write
$$(XY)^{-1}(\{1\}) = \{XY=1\} = \{X=1,Y=1\}=\{X+Y=2\}=(X+Y)^{-1}(\{2\})\in\sigma(X+Y)$$
and similar for $(XY)^{-1}(\{0\})$.

Here is another way to phrase it:
We have
$$XY=\begin{cases}1, & X+Y=2 \\
0, & X+Y = 0 \text{ or } X+Y=1
\end{cases}$$
Hence for the function $f\colon\mathbb{R}\to\mathbb{R}$ given by
$$f(x)=\begin{cases}1, & x=2 \\
0, & \text{otherwise}
\end{cases}$$
we can write $XY=f(X+Y)$.

Generally speaking, for any real-valued random variables $S$, $T$ the following statements are equivalent:

$S=f\circ T$ for some measurable function $f\colon \mathbb{R}\to\mathbb{R}$

$S$ is $\sigma(T)$-measurable (i.e. $\sigma(S)\subseteq\sigma(T)$)

It is not hard to see that (1) implies (2) which is what we use here, but in fact the reverse implication also holds. This equivalence is sometimes known as Doob-Dynkin lemma.

Title: Sigma algebra generated by $X+Y$


问:
Linear equation with prescribed precision of result
Let $x$ be an unknown real vector of size n.

Suppose we can find n vectors $v_i$, and are given the values $x^Tv_i$.
Then we can simply solve for $x$ by Gauss or some other method and determine $x$ uniquely and precisely( iff $v_i$ are lin indep).

However, suppose now that we are only given the values $x^Tv_i$ up to some finite precision, then we might still find $x$, but also only up to a certain finite precision.

In the ideal infinite precision case, we would not gain anything by working with more than n vectors $v_i$, as we can already compute x uniquely.
However, it seems that increasing n might help to determine $x$ with higher precision or maybe arbitrary precision even though the values $x^Tv_i$ have a fixed bounded precision.
Of course this would be easy if we could freely choose the  $v_i$, be just choosing very large values.

So suppose this is not possible. ie suppose that $v_i$ are given randomly and the norm of  $v_i$ is bounded. What numerical methods are available, to determine x.

Assume that computation can be done with infinite precision

答:
Using more vectors (say, $m$) than the dimension of the space (say, $n$) means using a frame (as opposed to a basis). Your question is about the recovery from noisy frame coefficients.

Let $x_1,\dots,x_m$ be the frame vectors, and let $e_1,\dots e_n$ be a basis of $\mathbb R^n$. Then for each $j=1,\dots,n$ we have an $m$-dimensional vector vectors $(x_1^Te_j,\dots , x_m^Te_j)$. These vectors span the $n$-dimensional subspace $A$ in $\mathbb R^m$.  The most basic idea (which seems to work well sometimes) is to orthogonally project your vector of noisy coefficients  onto $A$, and reconstruct $x$ from this projection.

The book Frames and bases: an introductory course by Ole Christensen discusses this topic and gives references to the literature. A relevant subject is quantized frame expansions, in which the coefficients are subject to rounding. E.g., this paper.


问:
Lim sup of independent identically distributed Poisson random variables
I want to prove that if $X_n$ is sequence of collectively independent Poisson random variables and $\mathbb{E}(X_n) = 1$ than $$\mathbb{P}\left(\lim{\sup{}_{n\to\infty}\frac{X_n\ln{\ln{n}}}{\ln{n}}} = 1\right)=1$$I don't know where to start. I would be very grateful for help!

答:
In general, to show that $P(\limsup_n Y_n=M)=1$ for random variables $Y_n$, you need to show two things:

For all $\epsilon>0$, $P(Y_n>M+\epsilon \text{ infinitely often})=0.$

For all $\epsilon>0$, $P(Y_n>M-\epsilon \text{ infinitely often})=1.$

To prove both of these, we will use the Borel-Cantelli lemmas, which involve looking the convergence of $\sum_{n\ge 1}P(X_n \ln\ln n/\ln n>1\pm \epsilon)$. Therefore, you will need a good estimate on $P(X_n\ge k)$, and apply this to $k=(\ln n/\ln \ln n)(1\pm \epsilon)$.

Assume $k$ is an integer for now. On the one hand, $P(X\ge k)\ge P(X=k)=\frac{e^{-1}}{k!}$. On the other hand, you can upper bound $P(X\ge k$ by a geometric series:
$$
P(X\ge k)
=\frac{e^{-1}}{k!}\sum_{i\ge k}{1 \over (k+1)(k+2)\cdots (i-i)\cdot i}
\le \frac{e^{-1}}{k!}\sum_{i\ge k}\frac1{k^{i-k}}
=\frac{e^{-1}}{k!}\cdot \frac{1}{1-1/k}
$$
This shows that $\lim_{k\to\infty}\frac{P(X\ge k)}{P(X=k)}=1$, so you can use the simple estimate $P(X\ge k)\approx e^{-1}/k!$ to determine the summability of $P(X_n\ge \ln n/(\ln \ln n)(1\pm \epsilon))$.

Therefore, you just need to show that the sum
$$
\sum_{n\ge 1}P\left(X_1\ge (1\pm \epsilon)\frac{\ln n}{\ln \ln n}\right)
$$
is finite when the sign is $+$, and infinite when the sign is $-$. This  is equivalent to showing the same about the sums
$$
e^{-1}\sum_{n\ge 1}\frac1{\Big((1\pm \epsilon)\ln n/\ln \ln n\Big)!}
$$
To prove this sum is infinite/finite, use Stirlings approximation, which says $\lim_{k\to\infty}\frac{k!}{k^ke^{-k}\sqrt{2\pi k}}=1$, so you can replace $k!$ with $k^ke^{-k}\sqrt{2\pi k}$ without affecting convergence. The math will get a bit messy, but everything works out nicely in the end. There is also the small issue that the argument to the factorial is not an integer, so you really need to round down, but this does not affect things in the long run.


问:
Find the limit of $\lim_{x\to 0}\left(\frac{\cos x}{\cos 2x}\right)^\frac{1}{x^2}$
Find limit of $$\lim_{x\to 0}\left(\frac{\cos x}{\cos2x}\right)^{x^{-2}}.$$
After some algebra I get $$\lim_{x\to 0}\exp\left(\left(\frac{\cos x-2(\cos x)^2+1}{\cos2x}\right)\cdot\frac{1}{x^2}\right).$$ Maybe I am missing something but I can't continue from this.

答:
I thought it might be instructive to present an approach that relies on straightforward arithmetic and two standard limits and circumvents use of logarithms, L'Hospital's Rule, and Taylor's Theorem.  To that end we now proceed.

We begin by writing the term of interest as

$$\begin{align}
\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}&=\left(\frac{2\cos^2(x)-1}{\cos(x)}\right)^{-1/x^2}\\\\
&=\left(1-\underbrace{\left(\frac{1-\cos(x)}{x^2}\right)\left(1+\frac{1+\cos(x)}{\cos(x)}\right)}_{\text{denote this as}\,f(x)}x^2\right)^{-1/x^2}\\\\
&=\left(\left(1-f(x)x^2\right)^{-1/f(x)x^2}\right)^{f(x)}
\end{align}$$

Noting that $\lim_{x\to 0}f(x)=3/2$ so that $\lim_{x\to 0}f(x)x^2=0$, and using $\lim_{t\to 0}\left(1-t\right)^{-1/t}=e$, we conclude that

$$\lim_{x\to 0}\left(\frac{\cos(x)}{\cos(2x)}\right)=e^{3/2}$$

NOTE:  We tacitly used the "standard limit" $\displaystyle \lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\frac12$


问:
Butcher's Tableau to Diagonally Implicit Runge Kutta method
I'm trying to implement a function that would input a butcher's tableau for a Runge-Kutta method of order $s$, an ode and its jacobian, as well as the initial values and return the next step.
If we have a butcher tableau,
$$ \begin{array}{c|cc}
\alpha & \beta \\
\hline
& \gamma \\
\end{array} $$
we'd like to convert that into a diagonally implicit butcher tableau
$$
y_{n+1} = y_n + \sum_{i=1}^s \gamma_i \kappa_i
$$
with
\begin{align}
\kappa_1 &= f\left(t_0 + \alpha_1 h, y_0 + \beta_{1j}\kappa_j\right) \\
\kappa_2 &= f\left(t_0 + \alpha_2 h, y_0 + h\sum_{j=1}^2 \beta_{2j}\kappa_j\right)\\
&\vdots \\
\kappa_s &= f\left(t_0 + \alpha_s h, y_0 + h\sum_{j=1}^s \beta_{sj}\kappa_j\right).
\end{align}
Keep in mind for the i-th step we need to solve for $\kappa_i$, where we use Newton Method to solve for a vector $\delta_i$ s.t. $F(\delta_i) = 0$,
$$
F(\delta_i) = \delta_i - f\left(t_0 + \alpha_i h, y_0 + \beta_{ii}\delta_i + h\sum_{j=1}^{i-1} \beta_{ij}\kappa_j\right)
$$
where DF is our Jacobian matrix.

```<code>def dirk(f, Df, t0, y0, h, alpha, beta, gamma):
    #---# Parameters #-------------------------------------------#
    #f(t,y) : lambda function of the ODE y'(t) = f(t, y)
    #Df(t,y): lambda function for the Jacobian of f
    #t0     : initial time
    #y0     : initial coordinate
    #alpha  : Butcher tableau vector of size s (weights of y step)
    #beta   : Butcher tableau matrix of size s*s (weights of steps)
    #gamma  : Butcher tableau vector of size s (weights of k)
    #------------------------------------------------------------#

    n = len(beta)
    m = len(f(t0, y0))
    K = np.zeros(len(alpha))
    if beta[0][0] != 0:
        K0,_ = newton(lambda delta: delta - f(t0 + alpha[0]*h, y0+h*beta[0][0]*delta), lambda delta: np.eye(len(y0))-Df(t0+alpha[0]*h,y0+h*beta[0][0]*delta), f(t0, y0), 1e-8,1000)
    else:
        K0 = y0 + h*f(t0, y0)
    K = np.insert(K, 0, K0)
    for i in range(1, n):
        sum = 0
        for j in range(0, i):
            sum += beta[i][j]*K[i-1]
        if beta[i][i] != 0:
            F =lambda delta: delta - f(t0+alpha[i]*h, y0+h*(delta*beta[i][i]+sum))
            DF = lambda delta: np.eye(m)-Df(t0+alpha[i]*h,y0+h*(beta[i][i]*delta+sum))
            sol,_ = newton(F, DF, f(t0, y0), 1e-8,1000)
            print(sol)
        else:
            sol = f(t0 + alpha[i]*h, y0+h*sum)

        K = np.insert(K, i, sol)
    phi = 0
    for i in range(n):
        phi += gamma[i]*K[i]
    return y0 + h*phi
</code>```
```<code>def newton(F,DF,x0,eps,K):
    #---# Parameters #-------------------------------------------#
    #F   : lambda function F(delta)
    #DF  : our lambda function of the Jacobian of F
    #x0  : our initial point f(t0, y0)
    #eps : our tolerance for 0
    #K   : maximum number of steps
    #------------------------------------------------------------#

    k  = 0
    x  = x0.copy().astype(float)
    delta = np.zeros([len(x0)])
    Fx = F(x)
    while Fx.dot(Fx) > eps*eps and k<K:
        delta[:] = np.linalg.solve(DF(x), Fx)
        x[:] -= delta[:]
        Fx = F(x)
        k += 1
    return [x,k]
</code>```

We can also use this function evolve to find the value for a given $T>t_0$ over $N$ steps so $h=(T-t_0)/N$.

```<code>def evolve(phi, f,Df, t0,y0, T,N):
    #---# Parameters #-------------------------------------------#
    #phi: our method
    #f(t, y): lambda function
    #Df(t, y): Jacobian of our function
    #t0: starting time
    #y0: initial coordinates
    #T: final time
    #N: number of steps
    #------------------------------------------------------------#

    h = T/N
    y = np.zeros( [N+1,len(y0)] )
    y[0] = y0
    t = 0
    for i in range(N):
        y[i+1] = phi(f,Df, t,y[i], h)
        t = t+h
    return y
</code>```

I'm testing on the code

```<code>stepper = lambda f,Df,t0,y0,h: dirk(f,Df,t0,y0,h,aCN,bCN,gCN)
y = evolve(stepper,
           lambda t,y: array([ -y[0],
                                y[0]-exp(-y[2]),
                                1 ]),
           lambda t,y: array([[-1,0,0],[1,0,exp(-y[2])],[0,0,0]]),
           0,array([1.,1.,0.]), 1,10)
print("%d " % len(y), end="")
print("%1.3e  %1.3e  %1.3e  " % tuple(y[0]),   end="")
print("%1.3e  %1.3e  %1.3e  " % tuple(y[-1]),   end="")
</code>```

I should be getting the vector (0.3667, 1.001, 1.000) but instead I get (0.9955,  0.9955,  -0.004536). I assume my code is breaking down inside the dirk function, particularly the input for Newton's Method.

答:
Your big $F$ function and its derivative are slightly wrong. $F$ is only wrong in the formula, it should be
$$
F(\delta_i) = \delta_i - f\left(t_0 + \alpha_i h, y_0 + h\beta_{ii}\delta_i + h\sum_{j=1}^{i-1} \beta_{ij}\kappa_j\right)
$$
This is correct in the code. However, the derivative should be
$$
F'(\delta_i) = I - h\beta_{ii}\partial_yf\left(t_0 + \alpha_i h, y_0 + h\beta_{ii}\delta_i + h\sum_{j=1}^{i-1} \beta_{ij}\kappa_j\right)
$$
The factor $h\beta_{ii}$ is missing in the code. In both places.

Also, there is something strange happening with the K array. The insert operation is not doing what you think it is doing. Why not just declare K = np.zeros([n,m], dtype=float) and then assign K[0]=K0, K[i]=sol.

With these changes I get for y[-1] the values

```<code>3.67339302e-01  1.00054511e+00  1.00000000e+00
</code>```

with the 3rd order DIRK table from https://www.math.auckland.ac.nz/~butcher/ODE-book-2008/Tutorials/IRK.pdf

((Note that the format specifier %1.3e has 1 as the total length, thus is ignored. Typical formats leading to mostly straight columns are 8.5 or 12.8.))

Body:
Your big $F$ function and its derivative are slightly wrong. $F$ is only wrong in the formula, it should be
$$
F(\delta_i) = \delta_i - f\left(t_0 + \alpha_i h, y_0 + h\beta_{ii}\delta_i + h\sum_{j=1}^{i-1} \beta_{ij}\kappa_j\right)
$$
This is correct in the code. However, the derivative should be
$$
F'(\delta_i) = I - h\beta_{ii}\partial_yf\left(t_0 + \alpha_i h, y_0 + h\beta_{ii}\delta_i + h\sum_{j=1}^{i-1} \beta_{ij}\kappa_j\right)
$$
The factor $h\beta_{ii}$ is missing in the code. In both places.

Also, there is something strange happening with the K array. The insert operation is not doing what you think it is doing. Why not just declare K = np.zeros([n,m], dtype=float) and then assign K[0]=K0, K[i]=sol.

With these changes I get for y[-1] the values

3.67339302e-01  1.00054511e+00  1.00000000e+00

with the 3rd order DIRK table from https://www.math.auckland.ac.nz/~butcher/ODE-book-2008/Tutorials/IRK.pdf

((Note that the format specifier %1.3e has 1 as the total length, thus is ignored. Typical formats leading to mostly straight columns are 8.5 or 12.8.))


问:
Evaluating $\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$
What options do I have for this series? No idea how to do it.

$$\sum_{k=1}^{\infty} \frac{\cos^2(k)}{4k^2-1}$$

Body:
Avoiding contour integration:
$$
\begin{align}
\sum_{k=1}^\infty\frac{\cos^2(kx)}{4k^2-1}
&=\sum_{k=1}^\infty\frac{\cos(2kx)+1}{2(4k^2-1)}\\
&=\frac12\mathrm{Re}\left(\sum_{k=1}^\infty\frac{e^{2ikx}+1}{4k^2-1}\right)\\
&=\frac14\mathrm{Re}\left(\sum_{k=1}^\infty\frac{e^{2ikx}+1}{2k-1}-\frac{e^{2ikx}+1}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(e^{2ix}+1+\sum_{k=1}^\infty\frac{e^{2i(k+1)x}-e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{2ix}-1)+(e^{2ix}-1)\sum_{k=1}^\infty\frac{e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{2ix}-1)\sum_{k=0}^\infty\frac{e^{2ikx}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+(e^{ix}-e^{-ix})\sum_{k=0}^\infty\frac{e^{i(2k+1)x}}{2k+1}\right)\\
&=\frac14\mathrm{Re}\left(2+i\sin(x)\log\left(\frac{1+e^{ix}}{1-e^{ix}}\right)\right)\\
&=\frac14\mathrm{Re}\left(2+i\sin(x)\log\left(i\tan(x/2)\right)\right)\\
&=\frac14\mathrm{Re}\left(2-\frac\pi2\sin(x)+i\sin(x)\log(\tan(x/2))\right)\\
&=\frac12-\frac\pi8\sin(x)
\end{align}
$$
Plugging in $x=1$ yields
$$
\sum_{k=1}^\infty\frac{\cos^2(k)}{4k^2-1}=\frac12-\frac\pi8\sin(1)
$$


问:
Let $F$ be a probability distribution function. Prove that $\int\limits_{-\infty}^{\infty} F(x+a)-F(x)\ dx= a\ \forall a\in\Bbb{R}$
A real-valued map $F$ is called a probability distribution if

$F$ is monotonically increasing

$F$ is right continuous

$\lim\limits_{x\to\infty} F(x)=1$ and $\lim\limits_{x\to-\infty} F(x)=0$

Again I know that for such $F$, there is a probability measure $P$ on borel $\sigma$-field of $\Bbb{R}$. such that $P((a,b])=F(b)-F(a)\ \forall -\infty\le a\le b\le \infty$.

So we have to prove that $\int\limits_{-\infty}^{\infty}P((x,x+a])\ dx=a$

Here $F$ may not be absolutely continuous, so we cannot work with the probability density function.

I'm not getting any idea how to prove this result. Can anyone help me in this regard? Thanks for your help in advance.

答:
Let $\mu$ be the probability measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$
induced by $F$ (i.e, $\mu(a,b]=F(b)-F(a))$. Let $m$ be the usual
Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Consider
the product measure space $(\mathbb{R}\times\mathbb{R},\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R}),m\times \mu)$.
Fix $a>0$ (the case that $a\leq0$ can be treated similarly). Let
$A=\{(x,y)\in\mathbb{R}^{2}\mid x<y\leq a+x\}$ which is a Borel subset
of $\mathbb{R}^{2}$ (see remark below). By Fubini-Tonelli Theorem,
$$
\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s).
$$
Observe that for each $s$, $\int1_{A}(s,t)d\mu(t)=\mu\left((s,s+a]\right)=F(s+a)-F(s)$.
Therefore $\int\left(\int_{A}(s,t)d\mu(t)\right)dm(s)=\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds.$
On the other hand, for each $t$, $1_{A}(s,t)=1$ iff $s<t\leq s+a$
iff $s\in[t-a,t)$. Therefore, $\int1_{A}(s,t)dm(s)=m\left([t-a,t)\right)=a$.
It follows that $\int\left(\int1_{A}(s,t)dm(s)\right)d\mu(t)=\int ad\mu(t)=a$.

That is, $\int_{-\infty}^{\infty}\left[F(s+a)-F(s)\right]ds=a$.

Remark: We go to show that $A$ is a Borel set.

Let $A_{1}=\{(x,y)\in\mathbb{R}^{2}\mid x<y\}$ and $A_{2}=\{(x,y)\in\mathbb{R}^{2}\mid y>a+x\}$.
Note that
\begin{eqnarray*}
A_{1} & = & \cup_{r\in\mathbb{Q}}\{(x,y)\mid x<r<y\}\\
 & = & \cup_{r\in\mathbb{Q}}\left((-\infty,r)\times\mathbb{R}\right)\cap\left(\mathbb{\mathbb{R}}\times(r,\infty)\right)
\end{eqnarray*}
which is a countable union of measurable rectangles, so it is measurable.
Similarly, $A_{2}$ is measurable. Finally, $A=A_{1}\setminus A_{2}$,
so $A$ is measurable too.


Body:
Hint

Let
$$1+y=1+\dfrac{\cos x-\cos2x}{\cos2x}=1+\dfrac{2\sin\dfrac x2\sin\dfrac{3x}2}{\cos2x}$$

$$(1+y)^{1/x^2}=((1+y)^{1/y})^{y/x^2}$$

Now use $\lim_{y\to0}(1+y)^{1/y}=e$

and $\lim_{h\to0}\dfrac{\sin h}h=1$


问:
Examples of the difference between Topological Spaces and Condensed Sets
There is apparently cutting-edge research by Dustin Clausen & Peter Scholze (and probably others) under the name Condensed Mathematics, which is meant to show that the notion of Topological Space is not so well-chosen, and that Condensed Sets lead to better behaved structures.

What is a simple low-tech example to see the difference?

I am looking for some explicit construction with quite simple topological spaces where some bad behaviour occur, and how their condensed analog fix that.

I am aware of the nlab entry and of an introductory text by F. Deglise on this page but it goes quite far too quickly and I am missing knowledge to grasp it.

答:
Condensed sets not being "simple low-tech" (to put it mildly) in the first place, I'm not sure to what extent this question can be answered satisfactorily. However, it is easier to address the question of why we might be unhappy with the usual category of topological spaces.

As Tyrone commented above, the introduction to Scholze's notes on condensed mathematics mentions three major issues with the category ${\bf Top}$ of toplogical spaces, from the perspective of developing a theory of "topological algebras" analogous to the rich theory we already have for purely algebraic strutures (his Question $1.1$). The second and third are rather technical (developing theories of derived categories and quasicoherent sheaves), but the first is relatively snappy: namely, the failure of the abelian category axioms. For example, the category of abelian group objects in ${\bf Sets}$ (= the category of abelian groups) is an abelian category, but the category of abelian group objects in ${\bf Top}$ is not. This means that the whole machinery of abelian categories can't be applied as we would hope.

(My original version of this answer said that the point is that ${\bf Top}$ itself is not abelian. Of course that's silly, as Connor Malin pointed out below: neither is ${\bf Sets}$!)

One important point here is that the issues here do not result from pathological objects or morphisms; we're not going to improve things by restricting to a subcategory of "nice" spaces and continuous functions. Instead, in order to get a better-behaved category we need to shift attention to a larger category, whose objects may be wilder but whose overall structure is better. This is exactly what we get in shifting from topological spaces to condensed sets, and from topological groups/rings/etc. to condensed groups/rings/et cetera (see Example $1.5$ and Proposition $1.7$ to check that this is in fact a case of passing to a larger context).


问:
A-Level Mechanics: Finding the Weight of a Non-Uniform Rod
A non-uniform rod has length 2m. The rod is placed on a support at its midpoint. It is in equilibrium when a particle of weight 50N is placed at one end and a particle of weight 30N on the other end. The particles are then removed and the rod is then suspended by two light inextensible strings attached to its ends. When the rod is horizontal, the tension in one of the strings is 120Nn. Find the two possible values for the weight of the rod. Trying to use moments hasn't been useful to me, because it introduces another unknown, namely, the distance between the centre of mass and the midpoint. I don't think this is what I need to do, but I may be wrong.

I have also tried to use simultaneous equations to find one value. M = mass of rod. R = reaction force

9.8M = 170 + R

9.8M = R - 80

This gives me an answer of 125N, but according to the mark scheme, 220N and 260N are the answers, and I have no clue how to get these values.
If you could offer any help, thank you very much.

答:
```<code>50     |  x    30
      2m
</code>```

Let x represent the center of mass from distance from the midpoint and W weight of the rod. By the equation of torque, we get xW+30-50=0, xW=20. There are two scenarios. Scenario 1 is the 120N string is attached to the left end of the rod.

```<code>120               T1
________|____x______
</code>```

xW+120-T1=0

T1=140

So the weight is, by using a not-torque equation, W-T1-120=0, W=260 N.

Scenario 2 is with 120 on the opposite side.

```<code>T2                120
________|____x_______
</code>```

Torque equation gives xW+T2-120=0, T2=100. So by the standard force diagram, W-120-100=0, W=220 N.

Body:
50     |  x    30
      2m

Let x represent the center of mass from distance from the midpoint and W weight of the rod. By the equation of torque, we get xW+30-50=0, xW=20. There are two scenarios. Scenario 1 is the 120N string is attached to the left end of the rod.

120               T1
________|____x______

xW+120-T1=0

T1=140

So the weight is, by using a not-torque equation, W-T1-120=0, W=260 N.

Scenario 2 is with 120 on the opposite side.

T2                120
________|____x_______

Torque equation gives xW+T2-120=0, T2=100. So by the standard force diagram, W-120-100=0, W=220 N.


问:
Why is the endomorphism ring of $\mathbb{Z}\times\mathbb{Z}$ noncommutative?
So I hear that the endomorphism ring of an abelian group is not always commutative. In particular, I'm looking at the abelian group $A=\mathbb{Z}\times\mathbb{Z}$, and considering $\text{End } A$. I can't find a counterexample to show that $\text{End } A$ is not commutative. Does anyone know of two such endomorphisms that don't commute?</p>

答: Every ring acts by left multiplication as endomorphisms of its underlying abelian group ("Cayley's theorem for rings"), so if you believe that there exist non-commutative rings, then you already believe that there exist non-commutative endomorphism rings. </p>


答: Hint: Consider the map $\phi:\mathbb R[x] \to \mathbb{R}\times\mathbb{R}$ given by $\phi(f)=(f(1),f(-1))$. Prove that $f$ is a surjective ring homomorphism and compute its kernel.

Title: $\Bbb R[x]/(x^2-1)\cong \Bbb R[x]/(x-1)\times \Bbb R[x]/(x+1)\cong \Bbb R^2\,$ via Chinese Remainder Theorem

Body:
I'm currently following my 2nd algebra course at uni where we are discussing rings.

The assignment goes like this
$$\text{Show that} \ \mathbb{R}[x]/(x^2-1) \ \text{is isomorphic to} \ \mathbb{R}\times\mathbb{R}$$

$\mathbb{R}[x]$ is the polynomial ring and $\mathbb{R}[x]/(x^2-1)$ is then a quotient ring. A hint to solve it is to use the Chinese Remainder Theorem.

I am stumped. The hint implies to me that $(x^2-1)$ must be the kernel of some ring homomorphism. It also to me implies that there is an ideal in $\mathbb{R}[x]$, let's call it K, so that $\mathbb{R}[x]/K=\mathbb{R}$ and I can't make either of those things to make sense. Is it also right to assume that $\mathbb{R}[x]/(x^2-1)=\{f(x)+x^2-1\vert f(x)\in\mathbb{R}[x]\}=\mathbb{R}[x]$, since all equivalence classes will contain just one element, since for two elements to be in the same equivalence class, the polynomials must be the same? Like if,

$$q(x)+x^2-1=r(x)+x^2-1\Leftrightarrow q(x)=r(x)$$
for $q(x),r(x)\in\mathbb{R}[x]$.


问:
Laurent expansion of $\frac{16}{z^2(z-3)}$

What is Laurent series of $\frac{16}{z^2(z-3)}$ at $z=3$ in the inner annulus?

$\frac{16}{z^2(z-3)}=\frac{-16}{3z^2}-\frac{16}{9z}+\frac{16}{9(z-3)} = \frac{16}{9}(z-3)^{-1}+\frac{16}{27}\cdot\frac{1}{1-\frac{(z-3)}{3}}-\frac{16}{3z^2}$

What can I do to turn $\frac{16}{3z^2}$ to a geometric series?

答:
$${1 \over z^2} = \left( {1 \over 3 + (z - 3)} \right)^2.$$
Now,
$$
{1 \over 3 + (z - 3)}  = {1 \over 3} \left({1 \over 1 + \left({1\over 3}\right)(z-3)}\right)
= {1 \over 3} \left({1 \over 1 - \left(-{1\over 3}\right)(z-3)}\right) =
{1 \over 3} \sum_{k \geq 0} \left[ \left(-{1\over 3}\right)(z-3) \right]^{k}.
$$
Finally, you will need to square the geometric series, and this can be done by using the Cauchy product.

I heartily recommend Cartan's textbook for this material.


问:
How to find the radius of convergence of different powers $\sum_{n=0}^{\infty} b_{n} z^{2 n+2}$
If I have a power series on the form $$\sum_{n=0}^{\infty} a_{n} z^{n}$$ A theorem tells me it has a radius of convergence given by $$R=\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{a_{n+1}}\right|$$

How can I use this formula to find the radius of convergence of a series with different powers? Consider fx. $$\sum_{n=0}^{\infty} b_{n} z^{2 n+2}$$

Is it possible to determine the radius of convergence of this series using the above formula? I don't have the ratio test at my disposal. If the power series just had even powers like $z^{2n}$ it could easily be shown that the radius of convergence would be $R=\sqrt{\lim _{n \rightarrow \infty}\left|\frac{b_{n}}{b_{n+1}}\right|}$. But I don't know what to do with shifted powers.

答: You can simply do a relabeling of your coefficients to shift everything down 2.  For instance, let $c_0=c_1=0$ and $c_n=b_{n-2}$ for $n\geq 2$.   Then your series is the exact same as the series with terms $c_nz^{2n}$


答: Rewriting $z^{2n+2}=z^2(z^2)^n=z^2w^n$, if you get common factor $z^2$, you get as a result $z^2 \sum b_nw^n$

Title: How to find the radius of convergence of different powers $\sum_{n=0}^{\infty} b_{n} z^{2 n+2}$

Body:
If I have a power series on the form $$\sum_{n=0}^{\infty} a_{n} z^{n}$$ A theorem tells me it has a radius of convergence given by $$R=\lim _{n \rightarrow \infty}\left|\frac{a_{n}}{a_{n+1}}\right|$$

How can I use this formula to find the radius of convergence of a series with different powers? Consider fx. $$\sum_{n=0}^{\infty} b_{n} z^{2 n+2}$$

Is it possible to determine the radius of convergence of this series using the above formula? I don't have the ratio test at my disposal. If the power series just had even powers like $z^{2n}$ it could easily be shown that the radius of convergence would be $R=\sqrt{\lim _{n \rightarrow \infty}\left|\frac{b_{n}}{b_{n+1}}\right|}$. But I don't know what to do with shifted powers.