问:
Are directional derivatives symmetric around the gradient?
I have read that the definition of the directional derivative of a function $f$ on a vector space at a point $x$ along a vector $\vec{v}$ is given by:

$$D_\vec{v}f(x)=\nabla f(x) \cdot\vec{v}=|\nabla f(x)||\vec{v}|\cos(\theta)$$

where $D_\vec{v}f$ is the directional derivative of $f$ in the direction $\vec{v}$, $\nabla f$ is the gradient of $f$ and $\theta$ is the angle between $\vec{v}$ and $\nabla f(x)$. The last equality follows by the definition of the dot product.

I don't get this... does this mean that for any vector $\vec{a}$ with equal length as $\vec{v}$ such that the angle between $\nabla f(x)$ and $\vec{a}$ is $\theta$, $D_\vec{v}f(x)=D_\vec{a}f(x)$?!

This seems crazy! The directional derivative is kind of symmetric about the gradient?! Surely this can't be true - if I'm defining a function can't I define it to be anything I like with no symmetry like this?

This makes no sense to me... have I misunderstood?

答:
I think what you are missing is that for the purpose of the derivative you can replace the function with a (hyper-)plane. In other words, the first-order Taylor approximation
$$f(x) \approx f(x_0) + \nabla f(x_0)(x-x_0)$$ has the same value and derivatives as $f(x)$ at $x = x_0$. The plane is entirely determined by the gradient and you cannot change the shape of the plane to be some arbitrary shape.

Consider the simplest case, which is a function of a single variable. The plane in this case is the tangent line, and the directional derivatives are in the left and right directions. And they are the same because the tangent line is the same line in both directions. Changing the shape of the function would not change this.


问:
Trouble visualizing a subset of $\mathbb{N}$
I have the following problem:

Fix $ C>0 $.

I want to figure out as much as possible about the following set
\begin{equation}\label{key}
S= \left  \{ n \in \mathbb{N} : \exists m \in \mathbb{N}:   \left| \dfrac{n}{Cm} -1  \right| < \dfrac{1}{2m}  \right \}
\end{equation}

I am looking for any particular way to visualise this set, in particular as $m \to \infty$.

I want to say that it is the set of integers that are divisible by $C$, but I am not quite sure. More specifically, I suspect that the relative density of $S$ to $\mathbb{N}$ is $1/C$, but I am not sure if this is true or how to prove it.

答:
The inequality
$$\left| \frac{n}{Cm} - 1 \right| < \frac{1}{2m}$$ can be multiplied by $m$ to get
$$\left| \frac{n}{C} - m \right| < \frac{1}{2}$$
Which defines the nearest integer from $n/C$. Such an integer $m$ exists if and only if $n/C$ is not a half-integer (i.e. an integer $+1/2$).

If $C$ is irrational, then $n/C$ is irrational and in particular it is not an half-integer. In such a case $S= \Bbb N$.

If $C$ is rational, things get more complicated. In any case you can write
$$S= \left\{ n \in \Bbb N : \frac{n}{C} + \frac{1}{2} \notin \Bbb N \right\}$$

ContentLicense: CC BY-SA 4.0


问:
Lower bounds on the modified Bessel function of the first kind
It is known that the modified Bessel function $I_{n}(x)$ satisfies the lower bound

\begin{eqnarray*}
I_{n}(x) > \frac{1}{\Gamma(n+1)} \left( \frac{x}{2} \right)^n
\end{eqnarray*}

for $x > 0$, $n > -\frac{1}{2}$. This lower bound is pretty good when $x$ is small, but it loses tightness when $x$ is about the same size as $n$; the plot below shows this discrepancy in a logarithmic scale for $I_{20}(x)$ (the dashed curve is the lower bound):

A better lower bound for $I_n(x)$ likely has to incorporate the term $e^{x}$ that is found in its asymptotic expansion:

\begin{eqnarray*}
I_{n}(x) \sim \frac{e^{x}}{\sqrt{2 \pi x}}\left( 1 - \frac{4n^2 -1}{8x} + \cdots \right)
\end{eqnarray*}

I have not been able to find such a lower bound thus far. To paraphrase, the bound I am seeking about is good for $x$ in the order of $n$, and I hypothesize that such a bound must include the term $e^x$.

答:
You can get this for $n=0$ as follows (I needed it in my research just now). The result is $I_0(x) \ge \frac{2\sqrt{2}}{3\pi}\frac{e^x}{\sqrt{x}}$, for all $x\ge 1/2$.
Take the integral rep. https://dlmf.nist.gov/10.32#E2
whose integrand is the ratio of two positive functions. The dominant contribution is from a region of size $1/x$ approaching $t=1$, and we can ignore the rest of the domain, and use simple tangent line bounds for the two functions.
Use a lower bound $e^{xt} \ge e^x(1-(1-t)x))$, and the upper bound $1-t^2 \le 2(1-t)$. Change variable to $u=1-t$ and shrink the domain to $u\in(0,1/x)$, which must lie in $(0,2)$, giving the restriction on $x$. The integrals become simple algebraic ones, and you should get the constant I did, unless I made a mistake. The constant is about 0.3001 which is pretty close to the asymptotic $1/\sqrt{2\pi} \approx 0.3989$.

For $n>0$ the same integral rep should work, but I suggest you leave $e^{xt}$ exact, giving something like a Gamma function bound. Let me know if it works out, or if you improve/fix the above...


问:
Deriving Polya’s Random Walk Constants
It is a well known theorem of Pólya that a random walk in 1 or 2 dimensions has a probability of 1 of returning to the origin. However, the probability in the 3-dimensional case is given by a strange triple integral.

How is this integral derived?

答:
The integral for Polya's constant $P$ is derived as follows.  First you need a formula for $p(n)$, the probability that a random walk of length $n$ starting at the origin ends at the origin in $\mathbb{Z}^3$.  Then the probability $P$ of returning to the origin at least once satisfies $1 + P + P^2 + \cdots = \sum_{n=0}^\infty p(n) = u(3) = \frac{1}{1-P}$, so $P = 1-\frac{1}{u(3)}$.

The main thing is to give a formula for $p(n)$.  Namely it is the average value
of $f(x,y,z)^n$, where $f(x,y,z) = \frac{\cos(x) + \cos(y) + \cos(z)}{3}$ and $x,y,z$ each range in $[0,2\pi]$.  Once you know this, the sum of $p(n)$ is the average value of $1+f+f^2 + \cdots = \frac{1}{1-f}$, which gives a single integral formula for $u(3)$.

Now where does the formula for $f(x,y,z)$ come from?  Consider first the simpler case of a random walk on $\mathbb{Z}$.  Then the same formula works with $f(x) = \cos(x)$, that is, the probability of return to the origin after $n$ steps in the average value of $\cos(x)^n$.  To see this write $\cos(x) = \frac{z+z^{-1}}{2}$, where $z = e^{ix}$.  Then $\cos(x)^n$ is a sum of powers $z^i$, each times the probability that a random walk of length $n$ ends in position $i$.  The average of $z^i$ is zero unless $i=0$.  The same reasoning applies in dimension 3 using $f(x,y,z)$.

Title: Deriving Polya’s Random Walk Constants


问:
Does series diverge
So, a series of $\sum_{n=1}^{\infty}a_n$, but $a_n=2n^{-1}$ if n is even, but $a_n=2^{-n}$ if $n$ is odd. So is $\sum_{n=1}^{\infty}$ diverge. I think it diverges, because it's the same as $\sum_{n=1}^{\infty}a_n=(-1)^n$.

答: Since all $a_n$ are positive, the series converges iff all subseries do, but the odd-$n$ subseries doesn't, as can be seen by comparing it with the harmonic series viz. $2/(2k-1)>1/k$.

Title: Does series diverge


问:
What is an elliptic curve, and how are they used in cryptography?
I hear a lot about Elliptic Curve Cryptography these days, but I'm still not quite sure what they are or how they relate to crypto...</p>

答: You can see this article as well: www.cs.umd.edu/Honors/reports/ECCpaper.pdf</p>

ContentLicense: CC BY-SA 2.5

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 3.0


问:
What is the symmetry between the definitions of the bounded universal/existential quantifiers?
What is the symmetry between the definitions of the bounded universal/existential quantifiers?

$\forall x \in A, B(x)$ means $\forall x (x \in A \rightarrow B(x))$

$\exists x \in A, B(x)$ means $\exists x (x \in A \land B(x))$

These make intuitive sense, but I would expect there to be some kind of symmetry between how the definitions of the bounded quantifiers work, and I can't see one. $A \rightarrow B$ means $\lnot A \lor B$ which doesn't seem to have a direct relationship with $A \land B$. What am I missing?

答:
You might think of universal as a mega-intersection, and existential as a mega-union. E.g., if $A=\lbrace x_1,x_2,\dots\rbrace$ then your first formula is $B(x_1)$ and $B(x_2)$ and ..., while the second is $B(x_1)$ or $B(x_2)$ or ....

There is also some sort of symmetry in noting that "not for all" is the same as "there exists ... not," and "not there exists" is the same as "for all ... not."


问:
Let $f$ differentiable on $(a, \infty)$ such that $\lim_{x\to\infty}\frac{f(x)}{x}=0$. Show that $\lim\inf_{x\to\infty}|f'(x)|=0$

Let $f$ differentiable on $(a, \infty)$ such that $\lim_{x\to\infty}\frac{f(x)}{x}=0$. Show that $\lim\inf_{x\to\infty}|f'(x)|=0$. I would like to know if my proof holds, please and have a feedback.

As $\lim_{x\to\infty}\frac{f(x)}{x}=0$, we have by definition that: $\forall \epsilon>0 \ \exists x_0 \ \forall x\ge x_0$: $|\frac{f(x)}{x}|<\epsilon$.

Let $k \in (a,\infty)$. We have then $\forall \epsilon>0, \ \forall x\ge x_0$:

$|\inf|f'(x)||=\inf|f'(x)|\le|f'(x)|=\Big|\lim_{x\to k}\frac{f(x)-f(k)}{x-k}\Big|\le \lim_{x\to k}\Big|\frac{f(x)-f(k)}{x-k}\Big|< \lim_{x\to k}\epsilon=\epsilon$

So $\lim\inf_{x\to\infty}|f'(x)|=0$ as $k$ is arbitrary.

I'm not sure at all of this proof. The last inequaility seems to me false, but I didn't find a counter example. Another approach was to distinguish cases: $f$ converges or $f$ diverges (then apply L'Hôpital).

答:
Ok, Your computations/intuition are good except that you are looking at the wrong limit by sending $x \to k$. You should send $x \to \infty$ because that is where you have some assumptions available. So, let me spell that out.

Given $\epsilon >0 $ I will show that for any $y$ there is a $c$ such that $y < c$ and $|f'(c)| < \epsilon$. I leave it to you to confirm that this implies $\liminf_{x \to \infty} |f'(x)| = 0. $

Given $\epsilon>0$ and $y>a$, we know that
$$
\lim_{x\to \infty} \frac{f(x)-f(y)}{x} = \lim_{x \to \infty} \frac{f(x)}{x} - \lim_{x \to \infty} \frac{f(y)}{x} = 0 \, .
$$
On the other hand,
$$
\lim_{x\to \infty} \frac{x}{x-y} = 1 \, .
$$
So, together they give
$$
\lim_{x \to \infty} \frac{f(x)-f(y)}{x-y} = \lim_{x \to \infty} \frac{f(x)-f(y)}{x} \cdot \frac{x}{x-y} = 0 \, .
$$
So, there exists an $x>y$ such that
$$
\left \vert \frac{f(x)-f(y)}{x-y} \right \vert < \epsilon \, .
$$
However, by mean value theorem we do have a $c \in (y,x)$ -- of course ultimately $x>y$ -- such that
$$
f'(c) = \frac{f(x)-f(y)}{x-y} \, .
$$
Therefore,
$$
|f'(c)| < \epsilon \, .
$$
So, to wrap up, for any $\epsilon >0$ and any $y>a$ I found a $c$ such that $ c>y$ and $|f'(c)| < \epsilon.$ Informally, this means that there will be arbitrarily small values for $|f'(x)|$ arbitrarily late at the tail.


问:
Solving $x^5 - 2x^3y^2 + xy^4 = y^2$

Find all integer solutions to the equation $$x^5 - 2x^3y^2 + xy^4 = y^2.$$

I wasn't quite sure how to start on this problem, as factoring out $y^2$ would likely give me a headache and mean that I would have to deal with nasty fractions. I also tried subtracting, but that didn't get me anywhere either.

答:
I use Ben's hint. Clearly, $x=0$ iff $y=0.$  Henceforth, x and y are nonzero. Since $x$ is a square, we write $x=t^2$ and get
$$t{(t^4-y^2)}=y.$$
Thus, $t| y$ and, therefore, $t^2| y.$ Writing $y=t^2u$ and cancelling $t^2,$ we get $$u=t^3(1-u^2),$$ which is impossible because of at least two contradictions (using that $u$ is nonzero):

(1) $|u| +1$ divides $|u|$;

(2) $t^3\,|\, u \,|\, t^3$ because $u$ and $u^2-1$ are coprime. ($1-u^2$ would have to be $\pm 1.$)

Title: Solving $x^5 - 2x^3y^2 + xy^4 = y^2$


问:
$\operatorname{Aut}(G)$ construction is not functorial
I'm trying to prove what the title says: There is no functor $F$ such that  $G \mapsto \operatorname{Aut}(G)$. I know that there is an equivalent question: Taking the automorphism group of a group is not functorial. but I really don't understand the proof, and the author's answer was last seen several years ago. Do you have another proof, or could you give me a little explanation of that?

答:
If there is such a functor $F$, then for every sequence of group homomorphisms
$$G_1\ \stackrel{f}{\longrightarrow}\ G_2\ \stackrel{g}{\longrightarrow}\ G_3,$$
we get another sequence of group homomorphisms
$$\operatorname{Aut}G_1\ \stackrel{F(f)}{\longrightarrow}\ \operatorname{Aut}G_2\ \stackrel{F(g)}{\longrightarrow}\ \operatorname{Aut}G_3,$$
where $F(g)\circ F(f)=F(g\circ f)$ because $F$ is a functor. Also $F(\operatorname{id}_G)=\operatorname{id}_{\operatorname{Aut}G}$ for every group $G$ because $F$ is a functor.

In particular, it follows that if $G_3=G_1$ and $g\circ f=\operatorname{id}_{G_1}$, then also $F(g\circ f)=\operatorname{id}_{\operatorname{Aut}G_1}$. So we could prove that no such functor exists if we can find explicit groups $G_1$ and $G_2$ with explicit group homomorphisms
$$G_1\ \stackrel{f}{\longrightarrow}\ G_2\ \stackrel{g}{\longrightarrow}\ G_1,$$
such that $g\circ f=\operatorname{id}_{G_1}$, but such that there exist no group homomorphisms
$$\operatorname{Aut}G_1\ \stackrel{F(f)}{\longrightarrow}\ \operatorname{Aut}G_2\ \stackrel{F(g)}{\longrightarrow}\ \operatorname{Aut}G_1,$$
such that $F(g\circ f)=\operatorname{id}_{\operatorname{Aut}G_1}$. The answer to the question you link gives precisely such groups and homomorphisms.


问:
Claiming that $\int_c^\infty I(y)dy = \int_a^\infty J(x)dx$

Claim: Let a real-valued function $f(x,y)$ be definied on $D = [a,+\infty)\times [c,+\infty)$.

If $I(y) = \int_a^\infty f(x,y)dx$ and $J(x) = \int_c^\infty f(x,y)dy$ are continuous and at least one of the following
$$\int_c^\infty I(y)dy = \int_c^\infty dy \int_a^\infty f(x,y)dx \\ \int_a^\infty J(x)dx = \int_a^\infty dx \int_c^\infty f(x,y)dy$$ improper integrals conveverges, then so does the other and
$$\int_c^\infty I(y)dy = \int_a^\infty J(x)dx$$

I haven't found anything very related to this theorem in my textbooks. I was thinking of taking two arbitrary sequences $(a_n)$ and $(c_n)$ such that $a_n \to +\infty$ and $c_n \to +\infty$ and defining
$$I_n(y) = \int_a^{a_n} f(x,y) dy$$ and $$J_n(x) = \int_c^{c_n} f(x,y)dy$$ Then, obviously $I_n(y)_\rightarrow^\rightarrow I(y)$ and $J_n(x)_\rightarrow^\rightarrow J(x)$. However, it seems I cannot go any further. I wonder if after the definition of the functional sequence above, one could use the ready results from that section.

答:
This is false under the given hypotheses.  Take $D = [1,\infty)\times [1,\infty)$ and $f(x,y) = \frac{x-y}{(x+y)^3}$.

Noting that

$$\frac{x-y}{(x+y)^3} = \frac{\partial}{\partial x}\frac{-x}{(x+y)^2} = \frac{\partial}{\partial y}\frac{y}{(x+y)^2},$$

we have

$$I(y) = \int_1^\infty f(x,y) \, dx = \int_x^\infty\frac{\partial}{\partial x}\frac{-x}{(x+y)^2}\, dx = \frac{1}{(1+y)^2},\\  J(x) = \int_1^\infty f(x,y) \, dy = \int_x^\infty\frac{\partial}{\partial y}\frac{y}{(x+y)^2}\, dy = \frac{-1}{(1+x)^2}$$

Here, $I$ and $J$ are continuous and both iterated integrals exist, with

$$\int_1^\infty I(y) \, dy = \int_1^\infty \frac{dy}{(1+y)^2} = \frac{1}{2}, \\\int_1^\infty J(x) \, dx = -\int_1^\infty \frac{dx}{(1+x)^2} = -\frac{1}{2} $$

However, $\int_1^\infty I(y) \, dy \neq \int_1^\infty J(x) \, dx$.

The theorem is true if we add the hypothesis that $f$ is nonnegative.  Here is an elementary  proof for improper Riemann integrals without reference to Lebesgue integration and Tonelli's theoem.

Since $f$ is nonnegative, we have  $J_n(x) = \int_c^{c+n}f(x,y) \, dy \nearrow \int_c^\infty f(x,y) \, dy =J(x)$. Since $J$ is continuous, it is Riemann integrable and by the monotone convergence theorem it follows that for all $A > a$,

$$\int_a^A\left(\int_c^\infty f(x,y) \, dy \right) \, dx =\int_c^\infty\left(\int_a^A f(x,y) \, dx \right) \, dy$$

Suppose that $\int_c^\infty I(y) \, dy$ exists. Then we have

$$\int_a^A\left(\int_c^\infty f(x,y) \, dy \right) \, dx= \int_c^\infty\left(\int_a^A f(x,y) \, dx \right) \, dy \leqslant \int_c^\infty\left(\int_a^\infty f(x,y) \, dx \right) \, dy = \int_c^\infty I(y) \, dy$$

Thus,

$$\int_a^\infty J(x) \, dx = \lim_{A \to \infty}\int_a^A\left(\int_c^\infty f(x,y) \, dy \right) \, dx \leqslant \int_c^\infty I(y) \, dy$$

The argument can be repeated when the other iterated integral exists to obtain the reverse inequality

$$\int_c^\infty I(y) \, dy \leqslant \int_a^\infty J(x) \, dx, $$

and, thus,

$$\int_c^\infty I(y) \, dy = \int_a^\infty J(x) \, dx$$


问:
show function does not converge as h tends to 0
I need to show that $3h\sin(1/h) - \cos(1/h)$ does not have a limit as $h \to 0$. I'm fairly new to the epsilon delta proofs and am not sure how to do it for this question, would someone be able to walk me through it please?

答:
We have to prove $\lim_{x \to 0} (3hsin(\frac{1}{h}) - cos(\frac{1}{h}))$= Limit does not exist

So,

1).Lets prove $\lim_{x \to 0} 3hsin(\frac{1}{h})=0$

$$-1\leq sin\frac{1}{h}\leq 1$$

$$-3h\leq 3hsin\frac{1}{h}\leq 3h$$

$$0=\lim_{h \to 0}(-3h)\leq 3h\lim_{h \to 0}(sin\frac{1}{h})\leq \lim_{h \to 0}(3h)=0$$

By sandwich theorem we get, $$3h\lim_{h \to 0}sin\frac{1}{h}=0$$

2).Lets prove $\lim_{x \to 0} cos(\frac{1}{h})=Does  \: not \: exist$

$$\forall \varepsilon >0 \;  \exists \delta \; s.t \;  0<\left | x \right |  < \delta \mapsto |cos\frac{1}{h}-L|<\varepsilon$$

Let $\varepsilon=\frac{1}{2}$,

$$0<\left | x \right |  < \delta \mapsto |cos\frac{1}{h}-L|<\frac{1}{2}$$

So lets assume that, $$\lim_{x \to 0} cos(\frac{1}{h})=L \; \in \mathbb{R}$$

Let $$x_{1}=\frac{1}{2n\pi} \; \;  (n\in Z^{+})\; \; and \; \; x_{2}=\frac{1}{2n\pi+\pi } \; \;  (n\in Z^{+})$$
So we have to get range of n,
$$0<|\frac{1}{2n\pi +\pi}| < |\frac{1}{2n\pi }|<\delta$$
$$0<\frac{1}{|2n\pi| }<\delta$$
So we get,
$$0<\frac{1}{|2\pi\delta| }<n\; \;  (n\in Z^{+})$$

Now,
$$|cos(2n\pi)-L|=|cos(x_{1})-L|=|1-L|<\frac{1}{2}\Rightarrow \mathbf{A}$$
$$|cos(2n\pi+\pi)-L|=|cos(x_{2})-L|=|-1-L|=|1+L|<\frac{1}{2}\Rightarrow \mathbf{B}$$
By A + B we get,
$$|1-L|+|1+L|< 1$$
$$|(1-L)+(1+L)|\leq |1-L|+|1+L|< 1$$
$$2\leq |1-L|+|1+L|< 1$$
$$2<1 \; \;\; \; (\therefore This \;is \;a \;contradiction)$$

So our assumption is false, Therefore $\lim_{x \to 0} cos(\frac{1}{h})$ has no limit.

So $\lim_{x \to 0} (3hsin(\frac{1}{h}) - cos(\frac{1}{h}))=\lim_{x \to 0} 3hsin(\frac{1}{h})-\lim_{x \to 0} cos(\frac{1}{h})=$Limit does not exist


问:
Spliting Field over $\mathbb{F}_3$
How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?</p>

答:
first observe that $f(x) = x^3 - x + 1$ is irreducible over $F_3$ since the derivative of $f(x)$ is equal to 2, which is nonzero in $F_3$. Therefore, the extension of $F_3$ given by $L = F_3[x]/(f(x))$ must have order equal to $3^{\deg(f)} = 27$ and hence is isomorphic to $F_{3^3}$. So for your isomorphism question, remark that any two finite fields of the same size are isomorphic.

To actually "see" the splitting field, the theorem of the primitive element says that $$L = F_3[y],$$ for some $y\in F_3^{sep}$. One way to try and find such a $y$ is to ask, first of all, if $x\in L$ with $x^3 - x + 1 = 0$ is a generator. To do so, see if you can write down all powers of $x$ as $$x^n = \alpha_n.x+\beta_n,$$ for some $\alpha_n, \beta_n\in F_3$ and $\alpha_n\neq 0$ for $n<\deg(f)$. We would need to check for $1< n < \deg(f)= 27$...It's instructive to work out what $x^2$ is...(this check for such $\alpha,\beta$ is from linear algebra, i.e. we are trying to see if $\langle x\rangle_{F_3}$ spans $L$ as an $F_3$-vector space; we know that if $y^n = \gamma$ for some $0<n<\deg(f)$ and $\gamma\in F_3$, then $y\in L$ is not a generator of $L/F_3$).

Title: Spliting Field over $\mathbb{F}_3$


问:
Is the curve $\left[ \cos \left( t^2 \right), \sin t \right]$ a square?
I found this marvellous thing while playing around in Desmos!

It seems to me that the curve $\vec{r}(t) = \left[ \cos ( t^2 ), \sin t \right] \left( t \in [0, +\infty) \right)$ is filling a square. But, as @BarryCipra kindly pointed out, it crosses the X axis only when $t$ is an integer multiple of $\pi$, so it crosses the X axis countably many times. And we know that $[-1, 1]$ is uncountable, so it seems the curve doesn't actually fill the square. A question that would be fair to ask: Does the curve get arbitrarly close to every point in the square? Or: Is $[-1, 1]^2$ the infinite closure of the curve?

答:
Intuitively, as $t$ gets larger and larger, we see that $\cos(t^2)$ oscillates faster and faster relative to $\sin(t)$, so we can choose very large $t$ that gives the right $y$-value and then just perturb it a bit to get the right $x$-value while only changing $y$ by a very small amount.

Let's make this more precise. Let $(x_0, y_0) \in [-1, 1]^2$ and $\varepsilon > 0$ be arbitrary. Let $t_0 \geq \frac{\pi}{\varepsilon}$ such that $\sin(t_0) = y_0$. We have $\cos(t_0^2 + c) = x_0$ for some $c \in [0, 2\pi]$. Let $t = \sqrt{t_0^2 + c}$. Then $\cos(t^2) = x_0$. Moreover,
$$\lvert \sin(t) - y_0 \rvert = \lvert \sin(t) - \sin(t_0) \rvert \leq \lvert t - t_0 \rvert = \sqrt{t_0^2 + c} - t_0$$
$$ \mathrel{\leq} \sqrt{t_0^2 + 2\pi} - t_0 = \frac{2\pi}{\sqrt{t_0^2 + 2\pi} + t_0} < \frac{\pi}{t_0} \leq \varepsilon.$$
Thus, the distance between $(\cos(t^2), \sin(t))$ and $(x_0, y_0)$ is less than $\varepsilon$.

In particular, the curve gets within distance $\varepsilon$ of every point in $[-1, 1]^2$ as $t$ ranges over the interval $[0, \frac{\pi}{\varepsilon} + 2\pi]$. (I'm not claiming this bound is necessarily optimal, to be clear, but I'd guess it's at least pretty close.)


问:
Prove if $X_n \stackrel{\mathbb P}\to X$ then $\mathbb E|X_n - X|^c \to 0$
Let $c > 0$ and $\mathbb E|X_n|^c$ exists. Prove if $X_n \stackrel{\mathbb P}\to X$ then $\mathbb E|X_n - X|^c \to 0$.

And the opposite seems to be true too.

For $c = 1$ this seems clear, somewhere I even saw the answer to this question, but overall it is not clear.

答: This is not true, even for $c=1$.  Let $\Omega = [0,1]$ with Lebesgue measure, and set $X_n(\omega) := n1_{\omega \le \frac{1}{n}}$.  Then $(X_n) \stackrel{\mathbb{P}}{\rightarrow} 0$ so $X(\omega)=0$ and $\mathbb{E}[|X_n|] = 1$, but $\mathbb{E}[|X_n - X|]=1 \not\rightarrow 0$.


问:
Evaluating $\sec(\tan^{-1}\sqrt3)$ exactly

Find the exact value of the expression
$$\sec(\tan^{-1}\sqrt3)$$

I'm confused about how to start the problem and what quadrant to use.

Body: By definition, $\tan^{-1}\sqrt(3)$ is in quad  $1$ ( since $\sqrt(3) > 0$), so $\sec(\tan^{-1}\sqrt(3))=\frac{2}{1}~(>0)$.


问:
I need to prove $x^{2}+ y^{2}= 1$ is a closed set in $R^{2}$ using the complementary set proof with sphere $B(a, r)$ with interior points
$x^{2}+ y^{2}= 1$ is a closed set with complimentary set open sphere proof in $R^{2}.$

答: Let $(x,y)$ be a point not in the set. Let $r=x^2+y^2 \neq 1$. Try to show that the ball around $(x,y)$ of radius $r-1$ (if $r>1$) or $1-r$ (if $r < 1$) shows the point to be an interior point of the complement.


答:
HINT

\begin{align*}
\sec^{2}(x) = \tan^{2}(x) + 1
\end{align*}

Title: Evaluating $\sec(\tan^{-1}\sqrt3)$ exactly

Body:
Find the exact value of the expression
$$\sec(\tan^{-1}\sqrt3)$$

I'm confused about how to start the problem and what quadrant to use.


问:
How to prove continuity with respect to Mackey topology on dual pairing?
In my attempt to properly learn about topological vector spaces, I tried to generalize a well-known concept from Banach spaces to dual pairings:

Let $(X,Y,b)$ be a dual pairing of vector spaces (i.e. $b \colon X \times Y \to \mathbb{R}$ is a bilinear form with respect to which $X$ and $Y$ mutually separate points). Furthermore, let $R\colon X \to Y$ be a positive-definite symmetric mapping, in the sense that $b(x,Ry)=b(y,Rx)$ and $b(x,Rx) \geq 0$ for all $x,y \in X$, with $b(x,Rx) = 0$ if and only if $x=0$. It follows that $R$ is linear and, if I'm not mistaken, continous from $X$ to $Y$ when both are equipped by any polar topology between the weak and Mackey topologies (the same type of topology on both spaces).

Now complete $X$ with respect to the inner product $(x,y) = b(x,Ry)$ to obtain a Hilbert space $H$, so that in particular $X$ is densely contained in $H$.

My question is: Does it follow without any further assumptions on $X$ and $Y$ that the natural embedding $i \colon X \to H$ is continuous, where $X$ is equipped with the Mackey topology $\tau(X,Y)$?

Using heavy machinery, I think this is true when $(X,\tau(X,Y))$ is metrizable as a well-known consequence of Banach-Steinhaus (on separate continuity vs joint continuity of bilinear forms). Does the conclusion hold in full generality?

Thanks a lot in advance!

答:
The answer to the question about continuity of the embedding is negative in its most general form, and in fact  it is not hard to  characterize precisely when
this is so.

But before that,  it is interesting to rephrase the hypotheses in a different way.

As indicated in the question, $X$ is a pre-Hilbert space, so there is no harm in stating this as hypothesis number 1.

Next,  notice that each $y$ in $Y$ defines a linear functional
$$
  \varphi _y: x\in  X\mapsto b(x,y)\in {\bf R}.
  $$
In addition,  the correspondence
$y\mapsto \varphi _y$ is injective, thanks to the fact that the duality $b$ mutually separates points.

So, instead of emphasizing the bilinear form $b$, we make the hypothesis that we are given a linear subspace $Y$ of the algebraic dual $X^\dagger$.

For each $y$ in $X$, the linear functional $R(y)$ is given by
$$
  R(y)|_x = \langle x,y\rangle , \quad\forall x\in  X,
  $$
so that $R(x)$ is precisely the canonical image of $x$ within $X^\dagger$ given by the inner-product.

We may therefore add to our new set of hypothesis that  the given subspace $Y\subseteq X^\dagger$ contains the canonical
image of $X$ in $X^\dagger$.

En passant, the above picks up the part of the original set of hypothesis where $b$ is required to separate points in
its first variable.

There is nothing to guarantee that the functionals in $Y$ are continuous on $X$
(relative to its pre-Hilbert space structure), so we shall denote by $Y_c$ the subset of $Y$ formed by the continuous
linear functionals.  Notice that
$$
  X\subseteq Y_c = X^*\cap Y \subseteq Y,
  $$
where the $X$ in the LHS is to be interpreted as the canonical image of $X$ in $X^\dagger$, and $X^*$ refers to the
topological dual (which happens to be naturally isomorphic to $H^*$).

We are therefore ready to state the main result:

Theorem The inclusion $\iota :X\to H$ is continuous with the Mackey topology on $X$, and the norm topology on $H$, iff
$X^*\subseteq Y$ or,  equivalently,  $Y_c=X^*$.

Proof.  Assuming continuity, pick any $\varphi $ in $X^*$,  and let $\tilde \varphi $ be the unique continuous linear functional on
$H$ extending $\varphi $.
The composition
$\require{AMScd}$  \begin{CD}  X @>{\iota}>> H@>{\tilde\phi}>> \mathbb R\\  \end{CD}
which is nothing but $\varphi $,
is therefore continuous wrt the Mackey topology $\tau (X, Y)$.  It is well known that the dual of $X$ relative to $\tau (X, Y)$
coincides with $Y$ (actually $\tau (X, Y)$ is the finest locally convex topology on $X$ whose dual is $Y$),  so we conclude that $\varphi $ lies in $Y$.

Conversely, assuming that $X^*\subseteq Y$, we immediately see that $\iota $ is weakly continuos (i.e. continuos relative to
$\sigma (X,Y)$ and $\sigma (H,H^*)$), and it is not hard to see that this implies continuity relative to
$\tau (X,Y)$ and $\tau (H,H^*)$, the latter topology being precisely the norm topology on $H$.  QED

In order to present a concrete counter-example,
let $X$ be the space of all sequences $x=(x_n)_{n\in {\bf N}}$ of real numbers with finitely many nonzero coordinates, and let
$Y=X$.  Define a duality between $X$ and $Y$ by
$$
  b(x,y)=\sum_{n\in {\bf N}}x_ny_n,
  $$
and let $R$ be the identity map, so that the inner-product
$
  \langle x,y\rangle =b(x,Ry)
  $
is the usual one,  and hence $H=\ell ^2$.

With the conventions above, we have that $Y$ is a proper subspace of $X^*$, so  $\iota $ is not continuos relative to $\tau
(X,Y)$ and the norm topology on $H$.


问:
Why excluding the end points in Short Calculus?
I’m reading a book about calculus Short Calculus by serge lang.

I’ve realized that the author always excludes the end points while talking about the existence of maxima and minima or the mean value theorem .

Is this something related to these theorem or it is just the choice of the author?

答: It is very useful not to require differentiability at the endpoints. Consider Rolle's theorem (a special case of the mean value theorem) applied to the function $f : [-1, 1] \to [0, 1]$ defined by $f(x) = \sqrt{1 - x^2}$ (so the graph is a semicircle). $f$ is continuous on $[-1, 1]$ and differentiable on $(-1, 1)$, so we can apply Rolle's theorem, even though $f$ is not differentiable at $0$ or $1$, to conclude that $f'(x) = 0$ for some $x \in (-1, 1)$.


答:
The best way to start is to recall the definition:
$$
\sec x=\frac1{\cos x}.
$$
Since the range of $\tan^{-1}(x)$ is $[-\frac\pi2,\frac\pi2]$ we conclude that the value of cosine is positive. Hence:
$$
\frac1{\cos x}=\sqrt{1+\tan^2x}=2.
$$


问:
How Many Ways to Choose Dance Partners if Choosing Both Men and Women?
I was working through supplementary exercises in Grimaldi, and I came across a question about dance partners--it asked how to choose a clean-up team of 8 from a group of 12 men, and then asked how many different ways to pair up 8 women from those 12 men.  The answer was simple enough: find the number of permutations of size 8 from a set of 12 men (i.e., $\frac{12!}{4!})$.  But that got me wondering, if we want to choose both the men and the women from a set of 12 each, how does the answer change?  Since I'm completely new to self-studying math, I thought I'd pose the question to see if my approach to answering the question is valid (I thought it might still have something to do with permutations, but I wasn't sure).

So let's say there are 12 men and 12 women at a dance.  How many ways are there to pair off 8 men and 8 women as dance partners?

答:
Here's my guess.  First we choose 8 men from the 12 total men, which is $\binom{12}{8}$.  Then, once we have the 8 men chosen, we find the permutations of 8 women from the set of 12, which is $\frac{12!}{(12-8)!} = \frac{12!}{4!}$.  The total number of ways should therefore be:

$$
\binom{12}{8} \times \frac{12!}{4!}
$$


问:
Roots of a polynomial with integer coefficients
Let $f(x)$ be a polynomial in the ring $\mathbb{Z}[x]$. Suppose that $f(i)=0$, where $i^2=-1$.
Can I conclude that $x^2+1$ is a factor of $f(x)$? If so, how can be proven?</p>

答:
Hint: Apply the remainder factor theorem.

Hint: Apply the conjugate root theorem.


问:
Induction Proof $4^k < 5^k$
So I need to use induction to prove that $4^n<5^n$ for all natural numbers $n > 0$. I have proved the base case to be true. For the induction hypothesis, we assume that when $n=k$, then $4^k < 5^k$. However, I am stuck in the inductive step when I try to show that it needs to be true for $n+1$. Therefore, we need to show that $4^{k+1} < 5^{k+1}$. However I get stuck on the first step trying to prove it as in

$4^{k+1} = 4 \cdot 4^k$, then by substituting $5^k$ instead of $4^k$ from the inductive hypothesis and I get stuck right here and I do not know how to proceed.

Any help would be appreciated.

答: Since $4^k < 5^k$ by the inductive hypothesis, we have $4^{k + 1} = 4 \cdot 4^k < 4 \cdot 5^k < 5 \cdot 5^k = 5^{k + 1}$. This completes the induction.

Title: Induction Proof $4^k < 5^k$


问:
Dimension and basis of set of linear codes of even weight
I am working on questions about coding theory. The set C is the set of all words in binary code that have an even weight and are of length n.

I have already proven that this is a linear code. Now I have to determine the dimension and basis of C. However, I cannot seem to wrap my head around this problem. Can anyone help me with this?

答:
This is the set of all even paritys.
So you can creat any message as long as the last bit is the sum of previous bits.
$H = \begin{bmatrix} 1_{n\times1}\end{bmatrix}$.
And if you solve $GH^T = 0$,
the basis will be
$G=\begin{bmatrix} I_{n-1} & 1_{(n-1)\times1}\end{bmatrix}$. So the dimension as mentioned is $n-1$.


答:
When $n = 1$, clearly $4 < 5$. Suppose inductively that $4^k < 5^k$. Then
\begin{align*}
4^{k+1} = 4^k \cdot 4 < 5^k \cdot 4 < 5^k \cdot 5 = 5^{k+1}.
\end{align*}

Title: Induction Proof $4^k < 5^k$

Body:
So I need to use induction to prove that $4^n<5^n$ for all natural numbers $n > 0$. I have proved the base case to be true. For the induction hypothesis, we assume that when $n=k$, then $4^k < 5^k$. However, I am stuck in the inductive step when I try to show that it needs to be true for $n+1$. Therefore, we need to show that $4^{k+1} < 5^{k+1}$. However I get stuck on the first step trying to prove it as in

$4^{k+1} = 4 \cdot 4^k$, then by substituting $5^k$ instead of $4^k$ from the inductive hypothesis and I get stuck right here and I do not know how to proceed.

Any help would be appreciated.


问:
Show limit exists if and only if $m+n>2$.
Let $m,n \in \mathbb{N}$.

Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.

Attempt:

$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\sqrt{x^2+y^2})^{n-2}(\sqrt{x^2+y^2})^m=(\sqrt{x^2+y^2})^{m+n-2}$.

So for $\epsilon>0$, choose $\delta=\epsilon^{\frac{1}{m+n-2}}$, then $0<\sqrt{x^2+y^2}<\delta \implies |\frac{x^ny^m}{x^2+y^2}|<\epsilon$.

$(\Rightarrow)$.If $n+m\leq 2$. $f(\frac{1}{k},\frac{1}{k})=(\frac{1}{2})(\frac{1}{k})^{n+m-2}$ which does not approach zero as $k \rightarrow 0$, while $f(\frac{1}{k},0)=0$ approaches $0$ as $k \rightarrow 0$. Since these limits are different, the limit does not exist in this case.

答:
Your proof seems to be correct! Another more intuitive way would be to use polar coordinates in this situation (which always is a good thing to consider when you have a $x^2+y^2$ term somewhere): $$\lim \limits_{(x,y) \to 0} \frac{x^ny^m}{x^2+y^2}=\lim \limits_{r\to 0}\frac{(r\cos(\theta))^n(r\sin(\theta))^m}{(r\cos(\theta))^2+(r\sin(\theta))^2}=\lim \limits_{r\to 0} r^{n+m-2}\cos(\theta)^n\sin(\theta)^m \rightarrow 0 \quad \text{when} \; n+m>2$$

(since $cos(\theta)^n\sin(\theta)^m$ is bounded)


问:
Example of $\mathbb{E}[X_t|\mathcal{F}_0]=X_0$ but $X_t$ is not a martingale?
Edit: Some great examples in the comments and in the answer below by @NullUser: it would be great to see even more examples, if anyone can come up with more please. Happy to award a small bounty on this if incentive needed. Trying to collect as many different examples as possible.

I am looking for examples of processes that satisfy the condition that:

$$\mathbb{E}\left[X_t|\mathcal{F}_{t_0}\right]=X_{t_0}$$

But where $X_t$ is not a martingale, i.e. where for some $0<s\leq t$:

$$\mathbb{E}\left[X_t|\mathcal{F}_s\right]\neq X_s$$

I can't think of any such process at the moment, although I distinctly remember seeing examples of such processes in the past. As many interesting examples as possible would be very much appreciated here,

Thank you so much.

答: If we let $X_0 = \mu$ and $(X_t)$ be i.i.d. random variables with mean $\mu$ for $t > 0$ (it may make sense to do this in discrete time so we don't have to worry about questions like whether or not we can find uncountably many i.i.d. random variables on the same probability space), then we have $\mathbb{E}[X_t|\mathcal F_0]=\mu=X_0$, but also $\mathbb{E}[X_t|\mathcal F_s] = \mu \ne X_s$ unless we chose a trivial distribution.


问:
Faithful representations and the adjoint representation of a Lie algebra
Let $L$ be a Lie algebra over a field $k$ with char$(k)\neq 2$. The adjoint representation is a map
$$
\mathrm{ad}:L \to {\frak gl}(V)
$$
whose kernel is the center of $L$. This means that
$$
L/Z(L) \simeq \mathrm{ad}(L)
$$
as Lie algebras. From this we can conclude that
$$
L \simeq Z(L) \oplus \mathrm{ad}(L),
$$
meaning that every Lie algebra is the direct sum of a trivial Lie algebra and a Lie algebra admitting a faithful representation.

The trouble is that this is not true . . .  as far as I can understand from the literature . . .where is my mistake? I guess I have assumed that exact sequences of Lie algebras split . . . which is probably not true!

答: Indeed, in Lie algebras exact sequences do not necessarily split. Take, for instance, the Heisenberg Lie algebra:$$\mathfrak h=\left\{\begin{bmatrix}0&a&c\\0&0&b\\0&0&0\end{bmatrix}\,\middle|\,a,b,c\in\Bbb C\right\}.$$Then,$$Z(\mathfrak h)=\left\{\begin{bmatrix}0&0&c\\0&0&0\\0&0&0\end{bmatrix}\,\middle|\,c\in\Bbb C\right\}.$$Furthermore, $[\mathfrak h,\mathfrak h]\subset Z(\mathfrak h)$. So, you can't have $\mathfrak h=Z(\mathfrak h)\oplus\operatorname{ad}(\mathfrak h)$, since $Z(\mathfrak h)$ is in the center of $Z(\mathfrak h)\oplus\operatorname{ad}(\mathfrak h)$, and $\bigl[Z(\mathfrak h)\oplus\operatorname{ad}(\mathfrak h),Z(\mathfrak h)\oplus\operatorname{ad}(\mathfrak h)\bigr]\not\subset Z(\mathfrak h)$.


问:
Proving no homomorphism exists
I am trying to understand the below problem.

Take $A = B = \mathbb{Z}$ and $\phi$ given by $n \longmapsto 2 \cdot n$. Show that there there does not exist a group homomorphism $\psi: B \to A$ such that $\psi \circ \phi = \mathrm{id}_A$. Hint: suppose that $\psi$ existed. Consider the element $\psi(1)$ and arrive at a contradiction.

I've written up the question verbatim, but there are two ambiguities that I cannot solve (and don't have the intuition to discern). First is the domain/codomain of $\phi$, though I believe they are $A$ and $B$, respectively. Second, $A$ and $B$ are clearly groups, I assume under multiplication (or, at the very least, $B$ is, since $\phi$ maps to a product of two integers.

Injectivity isn't hard to prove. If $\phi(a) = \phi(b)$, then $2a = 2b$, so $2(a - b) = 0$, so $a - b = 0$, hence $a = b$.

Proving that no such $\psi$ existed is harder. That the problem asks me to consider $\psi(1)$ suggests to me that $1$ is the identity in $A$, in which case $A$ is a multiplicative group of integers. Homomorphisms have to carry identities to identity, so assuming $B$ is also multiplicative, we have $\psi(1) = 1$. I don't know how this yields a contradiction, however.

Any hints?

答:
Let me rephrase the problem: $\DeclareMathOperator{id}{id}$

Take $G = \mathbb{Z}$, the group of integers under the $+$ operation. Define the group homomorphism $\phi : G \to G$ by $\phi(n) = 2 \cdot n$.

Show that there is no homomorphism $\psi : G \to G$ such that $\psi \circ \phi = \id_G$. Hint: consider the value of $\psi(1)$ and derive a contradiction.

If you want an even bigger clue, consider $\psi(1) + \psi(1)$.

Title: Proving no homomorphism exists


问:
Proof that $S_3 \simeq{\rm Aut}(S_3)$
After consulting another post on this website, e.g., here Proving that ${\rm Aut}(S_3)$ is isomorphic to $S_3$, I've come up with a proof that $S_3 \simeq {\rm Aut}(S_3)$. Could someone look it over to tell me if there are any holes? I don't know if my proof of surjectivity is the best way to go about it. It almost feels like "cheating."

I claim that $\phi: S_3 \to \mathrm{Aut}(S_3)$ sending $g \longmapsto c_g (x) = gxg^{-1}$ is an isomorphism. In general, $\phi: G \to \mathrm{Aut}(G)$ sending $g \longmapsto c_g$ is a homomorphism, so it suffices to show that $\phi$ is a bijection. We have:
\begin{align*}
g \in \mathrm{ker}(\phi) & \iff \phi(g) = c_g (x) = x, \; \forall x \in G \\
& \iff gxg^{-1} = x, \; \forall x \in G \\
& \iff gx = xg, \; \forall x \in G \\
& \iff g \in Z(G) = \{e\},
\end{align*}
so the kernel of $\phi$ is trivial, so $\phi$ is injective. Since $S_3$ has exactly $6$ elements which must be sent to distinct elements of the ${\rm Aut}(S_3)$, this implies that $|{\rm Aut}(S_3)| \leq 6$. I claim that $\phi$ is also surjective. We have $S_3 = \langle a = (12), b = (13), c = (23) \rangle$, each of which has order $2$. Since an automorphism $f \in{\rm Aut}(S_3)$ must preserve the order of elements, $f$ must send transpositions to transpositions. Furthermore, upon fixing where $f$ send these transpositions, the rest of the map is determined. Since there are $3!$ possibilities for where to send the permutations and these are the only possible automorphisms, we have $|{\rm Aut}(S_3)| \leq 6$. Hence, $|{\rm Aut}(S_3)| = 6$. But $\phi(S_3) \leq \text{Aut}(S_3)$, so that they have the same order immediately implies that $\phi(S_3) = \text{Aut}(S_3)$, so $\phi$ is surjective, hence bijective.

答:
There are a few problems, but all are fixable.

It is not clear that every element $g$ can be written as $a^ib^jc^k$ with $i,j,k\in\{0,1,2\}$. While every element is a product of factors, each of which is either $a$, $b$, or $c$, you would need to show that you can rewrite one in which they occur in the "wrong" order so that they show up in the right order. For example, $ba$ happens to be equal to $ac$ so you are fine, but that is not immediate and needs to be checked.

The expressions for $g$ are clearly not unique, since you describe $8$ elements but $S_3$ only has six. So there are some repeats in there. You would need to show your maps are well-defined. You do not do so.

It is not necessarily true that you are totally free to decide where $a$, $b$, and $c$ go. After all, $bab=c$, so if you know what happens to $a$ and $b$, then what happens to $c$ is forced. You need more than just deciding that you will map the transpositions arbitrarily: you are claiming all such maps are automorphisms, and you would need to check that. You do not do so. You do not even check that they are homomorphisms.

Now: you actually never use the fact that elements can be written in the form $a^ib^jc^k$, so why are you mentioning it? No need. In general, if $X$ generates $G$, then any group homomorphism is completely determined by what happens on $X$. This would take care of both 1 and 2: it doesn't matter whether you can write them in that specific way, and it doesn't matter that the expressions are not unique.

As for 3, you are asserting too much again. You already know that $\mathrm{Aut}(S_3)$ has at least six elements. Now, as you note, under any automorphism $\phi$, the three transpositions must be permuted (because $\phi$ sends elements of order $2$ to elements of order $2$, and because two different transpositions cannot be mapped to the same transpositions, so the restriction of $\phi$ to the set $\{(12),(23),(13)\}$ gives a permutation of that set). That means that there are at most $6$ homomorphisms, since, as noted above, what $\phi$ does to the transpositions completely determines $\phi$, and there are only six ways you can permute the elements of this set.

Thus, you know there are at least $6$ automorphisms, and at most $6$ automorphisms, so $|\mathrm{Aut}(S_3)|=6$, and so the map you have $S_6\to\mathrm{Aut}(S_6)$, which is an injective group homomorphism between finite groups of the same size must be an isomorphism.

Title: Proof that $S_3 \simeq{\rm Aut}(S_3)$


答: Suppose $\psi : \mathbb{Z} \rightarrow \mathbb{Z}$ existed. Let $\psi(1) = a \in \mathbb{Z}$. Then $\psi(\phi(1)) = \psi(2) = \psi(1 + 1) = \psi(1) + \psi(1) = 2a = 1 = \text{Id}(1)$. This forces $a = 1/2$, contradiction.

Title: Proving no homomorphism exists

Body:
I am trying to understand the below problem.

Take $A = B = \mathbb{Z}$ and $\phi$ given by $n \longmapsto 2 \cdot n$. Show that there there does not exist a group homomorphism $\psi: B \to A$ such that $\psi \circ \phi = \mathrm{id}_A$. Hint: suppose that $\psi$ existed. Consider the element $\psi(1)$ and arrive at a contradiction.

I've written up the question verbatim, but there are two ambiguities that I cannot solve (and don't have the intuition to discern). First is the domain/codomain of $\phi$, though I believe they are $A$ and $B$, respectively. Second, $A$ and $B$ are clearly groups, I assume under multiplication (or, at the very least, $B$ is, since $\phi$ maps to a product of two integers.

Injectivity isn't hard to prove. If $\phi(a) = \phi(b)$, then $2a = 2b$, so $2(a - b) = 0$, so $a - b = 0$, hence $a = b$.

Proving that no such $\psi$ existed is harder. That the problem asks me to consider $\psi(1)$ suggests to me that $1$ is the identity in $A$, in which case $A$ is a multiplicative group of integers. Homomorphisms have to carry identities to identity, so assuming $B$ is also multiplicative, we have $\psi(1) = 1$. I don't know how this yields a contradiction, however.

Any hints?


问:
Is the product of uniquely decipherable codes uniquely decipherable?
A code $C$ is prefix-free if no codeword in $C$ is a prefix of another codeword in $C$. For example, $C=\{0,10,110\}$ is prefix-free but $D=\{0,10,100\}$ is not.

A code $C$ is uniquely decipherable if given any combination of codewords in $C$ there is a unique way to decipher it. For example, $C=\{0,011,101 \}$ is uniquely decipherable but $D=\{010,10,101\}$ is not.

The product of codes $C$ and $D$ is the set $C \times D = \{ cd : c \in  C\ and\ d \in D \}$

It is also known that prefix-free implies uniquely decipherable, but the converse is false.
I've proved that the product of prefix-free code is prefix-free, so it would imply that the product of prefix-free code is uniquely decipherable.

I'm pretty sure the answer to the  question is no, but I still cannot find any counter-examples.

答:
Your hunch is right. Consider this case
$A = {0, 10, 11}$ and $B = {1,00, 10}$ (B is the reverse of a prefix free)
Now
$A \times B = {01,000,010,101,111,1000,1010,1100,1110}$.

There should be easier ones as well but this is what I got,

01 010 101 01

010 1010 101


问:
Topological spaces similar to but not actual manifolds
I'm looking for topological spaces that are similar to but not actual manifolds.
Is there a way to construct topo. spaces that are locally homeomorphic to, say, $S^n$ (rather than $\mathbb R^n$)? By this I mean that every point in my space $X$ has an open neighborhood which is homeomorphic to (in this example) $S^n$.

For example, I can think of gluing circles into a torus where the open sets are the circles that get glued together.
In this case, every point has an open neighborhood (which is $S^1$ by construction), which is not homeomorphic to $\mathbb R$. Note that in this topology, small "patches" are not open sets, only combinations of "cylindrical slices" of the torus. Therefore, there is no open set in my $T^2$ that is homeomorphic to $\mathbb R^n$.

However, this space is not Hausdorff (not even $T_0$), which is a property that I would like to preserve.

Is it possible to construct such Hausdorff spaces?
How similar would they be to typical manifolds?

答:
The relation "locally homeomorphic" is a transitive relation. Note that $\mathbb{R}^n$ is locally homeomorphic to $\mathbb{S}^n$ and vice versa. Therefore, for any topological space $Q$, $Q$ is locally homeomorphic to $\mathbb{R}^n$ iff $Q$ is locally homeomorphic to $\mathbb{S}^n$.

In fact, for any $n$-manifold $M$ with at least one point, $M$ is locally homeomorphic to $\mathbb{R}^n$ and vice versa. Therefore, $Q$ is locally homeomorphic to $M$ iff $Q$ is locally homeomorphic to $\mathbb{R}^n$.

So the project of studying topological spaces which are locally homeomorphic to a manifold $M$ is in fact the project of studing topological spaces which are locally homeomorphic to $\mathbb{R}^n$.

Edit: this answer is no longer applicable since it does not use the OP's definition of "locally homeomorphic".


问:
Are directional derivatives symmetric around the gradient?
I have read that the definition of the directional derivative of a function $f$ on a vector space at a point $x$ along a vector $\vec{v}$ is given by:

$$D_\vec{v}f(x)=\nabla f(x) \cdot\vec{v}=|\nabla f(x)||\vec{v}|\cos(\theta)$$

where $D_\vec{v}f$ is the directional derivative of $f$ in the direction $\vec{v}$, $\nabla f$ is the gradient of $f$ and $\theta$ is the angle between $\vec{v}$ and $\nabla f(x)$. The last equality follows by the definition of the dot product.

I don't get this... does this mean that for any vector $\vec{a}$ with equal length as $\vec{v}$ such that the angle between $\nabla f(x)$ and $\vec{a}$ is $\theta$, $D_\vec{v}f(x)=D_\vec{a}f(x)$?!

This seems crazy! The directional derivative is kind of symmetric about the gradient?! Surely this can't be true - if I'm defining a function can't I define it to be anything I like with no symmetry like this?

This makes no sense to me... have I misunderstood?

答: You can define a function with arbitrary directional derivatives that don't satisfy the given equation, but such a function is not considered differentiable at $x$ (even if $D_{\vec v}f(x)$ exists for every nonzero $\vec v$). The symmetry you noticed results from the linear relationship between $D_{\vec v}f(x)$ and $\vec v$, which is essentially the definition of differentiability of $f$ at $x$.


答:
You can certainly define structures that are locally homeomorphic to something different than $\mathbb{R}^n$, for example you may have complex manifolds where the local charts are homeomorphic to $\mathbb{C}^n$, or you can have Banach or Hilbert manifolds where the local charts are homeomorphic to a Banach or a Hilbert space respectively.

However, in general, it makes little sense from a differentiable point of view to require the local homeomorphism to be with respect to a (topological) space that does not have a linear structure, since the linear structure is the one that allows you to have difference quotients and, hence, to have differential calculus (Absolute calculus on manifolds started with this aim with the Italian school of Ricci Curbastro, Bianchi, Levi Civita and so on...).

Take away: yes, in general you can, but it makes little sense if the goal is to look for a notion of derivative outside $\mathbb{R}^n$.

Title: Topological spaces similar to but not actual manifolds

Body:
I'm looking for topological spaces that are similar to but not actual manifolds.
Is there a way to construct topo. spaces that are locally homeomorphic to, say, $S^n$ (rather than $\mathbb R^n$)? By this I mean that every point in my space $X$ has an open neighborhood which is homeomorphic to (in this example) $S^n$.

For example, I can think of gluing circles into a torus where the open sets are the circles that get glued together.
In this case, every point has an open neighborhood (which is $S^1$ by construction), which is not homeomorphic to $\mathbb R$. Note that in this topology, small "patches" are not open sets, only combinations of "cylindrical slices" of the torus. Therefore, there is no open set in my $T^2$ that is homeomorphic to $\mathbb R^n$.

However, this space is not Hausdorff (not even $T_0$), which is a property that I would like to preserve.

Is it possible to construct such Hausdorff spaces?
How similar would they be to typical manifolds?


答: Problem like this is fun to solve for sure. Let's do the backward direction: If $m+n > 2$, then either $m \ge 2$ or $n \ge 2$. Let's say $m \ge 2$, then rewrite it as $\dfrac{x^2y}{x^2+y^2}\cdot x^{m-2}y^{n-1}$. Observe that $x^{m-2}y^{n-1} \to 0$ or $1$, and $0  \le \left|\dfrac{x^2y}{x^2+y^2}\right| \le |y|$, thus it goes to $0$, and the whole thing goes to $0$. Conversely, if the limit exists, say $L < \infty$ and if $m+n \le 2 \implies m = n = 1\implies \dfrac{x^my^n}{x^2+y^2} = \dfrac{xy}{x^2+y^2}$ does not have a limit as you can choose $2$ paths to $(0,0)$, contrary to the existence of $L$, thus it is the case that $m+n > 2$.

Title: Show limit exists if and only if $m+n>2$.

Body:
Let $m,n \in \mathbb{N}$.

Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.

Attempt:

$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\sqrt{x^2+y^2})^{n-2}(\sqrt{x^2+y^2})^m=(\sqrt{x^2+y^2})^{m+n-2}$.

So for $\epsilon>0$, choose $\delta=\epsilon^{\frac{1}{m+n-2}}$, then $0<\sqrt{x^2+y^2}<\delta \implies |\frac{x^ny^m}{x^2+y^2}|<\epsilon$.

$(\Rightarrow)$.If $n+m\leq 2$. $f(\frac{1}{k},\frac{1}{k})=(\frac{1}{2})(\frac{1}{k})^{n+m-2}$ which does not approach zero as $k \rightarrow 0$, while $f(\frac{1}{k},0)=0$ approaches $0$ as $k \rightarrow 0$. Since these limits are different, the limit does not exist in this case.


问:
Limit of $\sin(1/x)$ - why there is no limit?
$$ \lim_{x\to 0+} \sin\left(\frac{1}{x}\right)$$
I know that there is no limit.

but, why there is no limit?
I tried $x=0.4$, $x=0.3$, $x=0.1$, it looks like the limit is $0$.

And how can I show that there is no limit? I tried to calculate it like all the other functions, and I got wrong result and I don't know why:

$$\lim_{x \to 0+} \sin\left(\frac{1}{x}\right) = \sin\left(\frac{1}{0^+}\right) =  \sin\left(\frac{1}{\infty}\right) = \sin(0) = 0.$$

答:
Lets look at $sin(\frac{1}{x}) $ function,

Now you can get a idea how it behaves near to 0. That is oscillating near 0. THat's why we can't get limit when x goes to 0 of $sin(\frac{1}{x}) $ function. So let's move into it's proof,

Have to prove $\lim_{x \to 0+} sin(\frac{1}{x})=Does  \: not \: exist$

$$\forall \varepsilon >0 \;  \exists \delta \; s.t \;  0< x  < \delta \mapsto |sin\frac{1}{x}-L|<\varepsilon$$

Let $\varepsilon=\frac{1}{2}$,

$$0< x   < \delta \mapsto |sin\frac{1}{x}-L|<\frac{1}{2}$$

Assume that, $$\lim_{x \to 0+} sin(\frac{1}{x})=L \; \in \mathbb{R}$$

Let $$x_{1}=\frac{1}{2n\pi+\frac{\pi}{2}} \; \;  (n\in Z^{+})\; \; and \; \; x_{2}=\frac{1}{2n\pi+\frac{3\pi}{2} } \; \;  (n\in Z^{+})$$
So we have to get range of n,
$$0<\frac{1}{2n\pi +\frac{3\pi}{2}} < \frac{1}{2n\pi+\frac{\pi}{2} }<\delta$$
$$\frac{1}{2n\pi+\frac{\pi}{2} }<\delta$$
So we get,
$$\frac{1}{2\pi\delta }-\frac{1}{4}<n\; \;  (n\in Z^{+})$$

Now,
$$|sin(x_{1})-L|=|1-L|<\frac{1}{2}\Rightarrow \mathbf{C}$$
$$|sin(x_{2})-L|=|-1-L|=|1+L|<\frac{1}{2}\Rightarrow \mathbf{D}$$
By C + D we get,
$$|1-L|+|1+L|< 1$$
$$|(1-L)+(1+L)|\leq |1-L|+|1+L|< 1$$
$$2\leq |1-L|+|1+L|< 1$$
$$2<1 \; \;\; \; (\therefore contradiction)$$

Assumption is not true. $\: \: \therefore\:\lim_{x \to 0+} sin(\frac{1}{x})$ has no limit.


问:
given a sequence converging to 0, show the series converge and that its value is a0.
I'm having trouble proving the following problem, and also I'm not exactly sure how to choose/arrange an's in order to show the given series, as suggested in the problem..

答:
Let's begin with

$\displaystyle \{a_{n}\}^{\infty }_{n=0}\rightarrow 0$

We now choose $\displaystyle x_{n} =a_{n} -a_{n+1}$

Let's look at the sum $\displaystyle \sum ^{\infty }_{i=0} x_{n}$

This is

$\displaystyle \sum ^{\infty }_{i=0} x_{n} =\sum ^{\infty }_{i=0} a_{n} -a_{n+1} =( a_{0} -a_{1}) +( a_{1} -a_{2}) +...+( a_{n} -a_{n+1}) +...$

We see that $\displaystyle a_{1} ,a_{2} ,...,a_{n} ,...$ cancel so we are left with

$\displaystyle \sum ^{\infty }_{i=0} x_{n} =a_{0}$

Now for the sequence, I would suggest you try to think of how it got in that form.

One way to get $\displaystyle n( n+1)$ in the denominator is when we for example

substract two fractions together $\displaystyle \frac{1}{n} -\frac{1}{n+1} =\frac{n+1-n}{n( n+1)} =\frac{1}{n( n+1)}$.

Because we get this alternating sum, you can see that the first term

of the sequence $\displaystyle a_{n} =\frac{1}{n}$ is $\displaystyle a_{1} =\frac{1}{1} =1$ and the other terms cancel out

like in the first example.


问:
Proof of non-prime Mersenne numbers
I have to prove that $2^n-1$ is not a prime number if $n$ wasn't a prime number.
I have done it.

If $n = 1$, then $2^n-1 = 1$. And if I write $2^{n+1}$, so $n+1$ will be the next non-prime number and that will be $4$. $2^4-1 = 15$. So far so good.

But when I am doing the induction I am wondering why this would be possible:

$2^{n+1} - 1 = 2^n * 2 - 1$

$16-1 = 2^1 * 2 -1 $

$15 = 3$

It's false for sure, but why?

答:
Your proof doesn't work as given. Actual proofs of this tend to be algebraic rather than inductive but you can potentially squeeze induction in there in various ways.

Let's prove a small well-known equality first:

Claim: $2^n-1 = \sum_{k=0}^{n-1} 2^k$

Proof: By induction.

Base case $n=1$, with $2^1-1 = 1$ and $\sum_{k=0}^{0}2^k = 2^0 = 1$ as required.

Assume $2^n-1 = \sum_{k=0}^{n-1} 2^k$, then $2^{n+1}-1 = 2^n+2^n-1 = 2^n+\sum_{k=0}^{n-1} 2^k = \sum_{k=0}^{n} 2^k$ as required.

Now observe that $\sum_{k=0}^{n-1} 2^k$ has $n$ terms. Assume that $n$ is composite and we have $n=ab$ with $a,b>1$.Then we can split this sum in $b$ parts of $a$ length:

$$\begin{align}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
2^n-1 = \sum_{k=0}^{n-1} 2^k &= \sum_{\ell=0}^{b-1}  \sum_{m=0}^{a-1} 2^{\large a\ell+m} \tag{groups of $a$}\\
  & = \sum_{\ell=0}^{b-1}  \sum_{m=0}^{a-1} \left(2^{\large a\ell}2^{\large m} \right)  \tag{separate exponent}\\
  & = \sum_{\ell=0}^{b-1} \left(2^{\large a\ell} \sum_{m=0}^{a-1} 2^{\large m} \right)  \tag{constant term}\\
  & = \left(\sum_{\ell=0}^{b-1} 2^{\large a\ell}\right) \left(\sum_{m=0}^{a-1} 2^{\large m} \right) \tag{independent sums}\\
\end{align}$$

Now since $a,b>1$, both of these factors are also greater than one and $2^n-1$ is composite when $n$ is composite.

The other non-prime case, $n=1$, gives $2^n-1=1$ and that is also not prime.

Title: Proof of non-prime Mersenne numbers


问:
How to calculate this PV complex integral?
$$P.V. \int_{-\infty}^{\infty}\frac{xe^{ix}}{x^2-\pi^2}.dx$$
I find this a challenging integral to solve. I tried solving by extending to the complex plane and integrating over a semicircle contour (real axis, but infinitesimally small hoops (radii $\rho$) around points $x=-\pi$ and $x=\pi$,  plus the arc $Re^{i\phi}$, $\phi \in [0,\pi]$), but this leads me nowhere as i find that the integral on all of the three arcs vanishes, in the limit $R->\infty$, $\rho ->0$ .

答:
You may define a contour that looks like $\mathcal{C}=[-R,-\varepsilon]\cup\varepsilon e^{i[-\pi-\varepsilon,-\pi+\varepsilon]}\cup \varepsilon e^{i[-\varepsilon,\varepsilon]}\cup\varepsilon e^{i[\pi-\varepsilon,\pi+\varepsilon]}\cup [\varepsilon,R]\cup Re^{i[-R,R]}$.
$$\oint_\mathcal{C}=\int_{-R}^{\varepsilon}+\int_{-\varepsilon}^{\varepsilon}+\int_{-\varepsilon}^{R}+\int_{-R}^{R}+\int_{-\pi-\varepsilon}^{-\pi+\varepsilon}+\int_{\pi-\varepsilon}^{\pi+\varepsilon}.$$
Note that the sum of the first and the third path intergals correspond your integral, wheras he second and the fourth one go to zero, if you let $R\to\infty$ and $\varepsilon\to0$. This can be easily proven by the ML-inequality and Jordans Lemma. Now you just have to find the other two semicircles that will give you $\pm i\pi$ if you parametrize them correctly:
with
$$x=\pm\pi+\varepsilon e^{i\varphi}.$$

Also not that the integral over $\mathcal{C}$ equals zero.


问:
Calculate the distance $h(t)=1+e^{-t}\cos(t)$ that drops
A weight is hanging from a spring, and oscillates with heights modeled by the function $h(t)=1+e^{-t}\cos(t)$ at time $t\geq 0.$ Does it travel an infinite distance (in other words, does it converge or diverge) as $t\to\infty$ and also, what is the total distance that it drops ?

Note: the distance dropped indicates only the downward motion of the function, not when the function increases.

I have tried to plot this and it seems from the graph that it only drops to $\frac{3\pi}{4}$ and increases from that point on. I got a final answer of $\frac{1}{1-e^{-\pi}}.$ For the other question of whether the weight travels infinitely, the answer is a simple no.

答:
We have $\cos(t)=\frac{1}{2}\cdot(e^{it}+e^{-it}),$ so rewrite to
$$1+e^{-t}\cdot\frac{1}{2}\cdot(e^{it}+e^{-it})=1+\frac{1}{2}\cdot(e^{t(i-1)}+e^{t(-i-1)})$$
Now integrate over t to get $t+1/2(\frac{e^{t(i-1)}}{i-1}-\frac{e^{t(-i-1)}}{i+1})$ from $0$ to infinity to get infinity and thus it travels an infinite distance.


问:
Probabilities of related events
I can't figure out which rules of probability to apply to solve for my probability of the second event.
My information for the question is as follows:

$P(A) = 0.3,
P(A \cap B) = 0.1$,
$P(A\cup B) \setminus (A \cap B) = 0.75$

I need to find the probability of B. Any guidance is appreciated

答:
You have been given $\mathsf P(A), \mathsf P(A\cap B),$ and $\mathsf P((A\cup B)\smallsetminus(A\cap B))$, and wish to evaluate $\mathsf P(B)$ .

Well you can express some of these events as unions of the disjoint events: $(A\cap B), (A\smallsetminus B), (B\smallsetminus A)$.
$$\begin{align}A&=(A\cap B)\cup(A\smallsetminus B)\\B &= (A\cap B)\cup(B\smallsetminus A)\\(A\cup B)\smallsetminus(A\cap B)&=(A\smallsetminus B)\cup(B\smallsetminus A)\end{align}$$

So applying the additive rule :

$$\begin{align}\mathsf P(A)&=\mathsf P(A\cap B)+\mathsf P(A\smallsetminus B)\\\mathsf P(B)&=\mathsf P(A\cap B)+\mathsf P(B\smallsetminus A)\\\mathsf P((A\cup B)\smallsetminus(A\cap B))&=\mathsf P(A\smallsetminus B)+\mathsf P(B\smallsetminus A)\end{align}$$

Title: Probabilities of related events


问:
Is the following optimization problem convex?
I'm not an mathematician so sorry for the possibly trivial question.

I have written the following integer programming model:

\begin{align}
\max z &= \sum_{i=1}^M \left(\sum_{j=1}^N b_j x_{ij}- c_i y_{i}\right)\\
\text{s.t.} & \sum_{j=1}^N a_{ij}x_{ij} \le d_i, & i=1,\dots,M,\\
& \sum_{i\in\mathcal{F}} x_{ij} = 1, & j=1,\ldots,N,\\
& \sum_{j\in\mathcal{N}} x_{ij} \le N y_{i}, & i=1,\ldots,M,\\
& x_{ij} \in \{0,1\}, & i=1,\ldots,M, j=1,\ldots,N,\\
& y_{i} \in \{0,1\}, & i=1,\ldots,M.
\end{align}
where $a_{ij}$, $b_j$, $c_i$ and $d_i$ are suitable real parameters.

I suspect this problem is not convex for the third constraint, do you?

A possible alternative formulation is to replace in the objective function the $y_i$ decision variables with the following:
\begin{equation}
\operatorname{sgn}\biggl(\sum_{j=1}^N x_{ij}\biggr)
\end{equation}
and thus I can remove the third constraint.
The rationale for using the $\operatorname{sgn}$ function is that whenever $y_i=1$ in the original problem, there must be some $x_{ij}=1$ (for some $j$); while, if $y_i=0$ in the original problem, then $x_{ij}=0$ for all $j$.

But, again I suspect that this version is not convex too, due to the presence of the $\operatorname{sgn}$ function, do you?

Thank you very much for the help

Best Regards

答:
No, it is not convex, but not because of the third constraint. The third constraint is, in fact, convex in $x$ and $y$. It is not a convex optimization problem because the variables are binary. A convex program must have a convex feasible set, which cannot be the case if the variables are not continuous.

Its continuous relaxation, however---that is, the model obtained by replacing each constraint of the form $z\in\{0,1\}$ with $z\in[0,1]$---is convex. In fact, the relaxation  is a linear program. This is important in practice, because there is software available to globally solve binary linear programs. Of course, these solvers cannot guarantee a bound on execution time---and since the number of binary variables here is $MN+M$, it wouldn't take long for this to grow to unwieldy proportions.

That said, the structure of this problem resembles certain linear assignment problems so there may be some specialized methods out there that could be adapted to it.


问:
Uncountability in smallest sigma algebra
There are a bunch of questions and answers about the "smallest $\sigma$-algebra" argument. My question is focused on uncountability.

Claim: Given a set $\Omega$ and a collection $A$ of subsets of $\Omega$, then there is a smallest $\sigma$-field containing $A$.

I think the argument is something like:

take the set $S$ of subsets $S_i$ of $\Omega$ with $A\subseteq S_i$.

$S$ has a $\sigma$-algebra, so you can make another $\sigma$-algebra $\bigcap_i S_i$ by intersecting its components

that new $\sigma$-algebra is big enough: $A \subseteq \bigcap_i S_i$

that new $\sigma$-algebra is small enough: all others are bigger

My questions are

In (4), $\sigma$-algebras are only preserved by countable intersections, not uncountable ones. How does this work then?

In (3), how do we know that?

答:
You are not doing this quite right.

Let $A$ be a collection of subsets of $\Omega$. So $A\subseteq \mathbf{P}(\Omega)$, the power set of $\Omega$.

Now, given a $S$ subset of $\mathbf{P}(\Omega)$ we can ask two questions:

Is $S$ a $\sigma$-algebra?

Does $S$ contain $A$ as a subset?

For any subset $S$, the answer to each of those questions is either "yes" or "no". So now we can form a collection of subsets of $\mathbf{P}(\Omega)$, called $\mathscr{S}$, as follows:
$$\mathscr{S} = \{S\subseteq\mathbf{P}(\Omega)\mid S\text{ is a }\sigma\text{-algebra and }A\subseteq S\}.$$
That is, the subsets of $\mathbf{P}(\Omega)$ for which the answer to both question is "yes".

Now, say you have two elements $S_1$ and $S_2$ of $\mathscr{S}$; these are $\sigma$-algebras that contain $A$. Question: is $S_1\cap S_2$ (that is, all subsets of $\Omega$ that are both in $S_1$ and in $S_2$) a $\sigma$-algebra that contains $A$? Well, clearly, since $A\subseteq S_1$ and $A\subseteq S_2$, then $A\subseteq S_1\cap S_2$. Proving that the intersection is a $\sigma$-algebra is more involved, but doable. In fact:

Lemma. Let $\{S_i\}_{i\in I}$ be a nonempty family of subsets of $\mathbf{P}(\Omega)$ such that $S_i$ is a $\sigma$-algebra on $\Omega$ for each $i\in I$. Then
$$T=\bigcap_{i\in I}S_i = \{X\subseteq \Omega\mid X\in S_i\text{ for all }i\in I\}$$
is a $\sigma$-algebra on $\Omega$, regardless of the cardinality of $I$.

To prove this you should verify that $T$ is indeed a collection of subsets of $\Omega$; that $\Omega\in T$ (or equivalently, that $\Omega\in S_i$ for all $i$); that if $X\in T$ then $\Omega-X\in T$; and that if $\{X_j\}_{j=1}^{\infty}$ is a countable collection of subsets of $\Omega$, $X_j\subseteq \Omega$ for all $j$, and $X_j\in T$ for all $j$, then $\cap X_j\in T$. (This is the countable intersection issue).

This is regardless of the cardinality of $I$. Note that the intersection that creates $T$ is not an intersection of sets in a $\sigma$-algebra, it is an intersection of $\sigma$-algebras. That's why the fact that $\sigma$-algebras are closed under countable intersection only is irrelevant: you aren't doing an intersection of sets in a given $\sigma$-algebra.

You say:

"$\sigma$-algebras are only preserved by countable intersections"

and the problem, in part, is that this shorthand is confusing you. The correct statement is that $S$ is a single, particular $\sigma$-algebra, then an countable intersection of elements of $S$ must be in $S$, but an arbitrary, not-countable intersection of elements of $S$ need not be in $S$. That is, the property if being a set in $S$ is preserved under countable intersections, but not necessarily under intersections with uncountably many factors. Note that it is not $S$ that is "preserved", it's the property of being an element of $S$.

And, again, this doesn't apply here because we are not talking about intersecting elements of a particular $\sigma$-algebra, we are talking about intersecting different $\sigma$-algebras to get a new collection of sets.

Once you prove the Lemma, the way we proceed to construct the $\sigma$-algebra generated by $A$ is the following:

Consider the collection $\mathscr{S}$.

Note that $\mathscr{S}$ is not empty, because $\mathbf{P}(\Omega)$ itself is a $\sigma$-algebra on $\Omega$ that contains $A$.

Apply the Lemma to construct $T$.

Show that $A\subseteq T$.

Note that if $S$ is any $\sigma$-algebra that contains $A$, then by construction $S\in\mathscr{S}$, so $T\subseteq S$.

Conclude $T$ is the "smallest" $\sigma$-algebra that contains $A$.

See also this previous question for general remarks about this kind of procedure, which shows up often in mathematics.


问:
"Extremely low probability events happen all the time" - Does this allow us to draw any conclusions?
"Probability of getting any specific hand in poker is the same as getting Royal Flush" - Statements of these kinds seem fairly intuitive to interepret. But recently I thought of the following statement - "There are 10 towns each with its own distinct specialty, like - Food, Landscape, Culture, etc. Now Bob visits exactly one of these towns for his winter vacation."

Now not knowing the fact the bob has no preference whatsoever, I conclude that since he visited town X he must like the town's culture, food, etc. Since the probability of him choosing that specific culture he very less. But this does not make sense as I can say that about any town bob chooses.

Could we or should we draw conclusions like these in a scenario like this? What kind of conclusions we can draw from such an event? What conditions should an event like this satisfy so that meaningful conclusions can be derived?

Thanks in Advance.

答:
I recommend looking at some philosophy of probability.

There's a basic distinction that many authors make between various kinds of subjective/epistemic/Bayesian probability on one hand, and various sorts of objective probability on the other.  (I'm leaving out some other categories that some authors would add because they are less widely thought to matter.)  This distinction is closely related to the distinction in statistics between frequentist and Bayesian methods.

When we talk about probabilities of card hands, that is usually thought of as a kind of objective probability: there is something about cards and shuffling that makes certain hands come up more or less often.  On the other hand, subjective probability, which might be conceived of as a degree of belief that satisfied axioms of probability, could be applicable, too.  (This would be especially so if you didn't know whether someone might have removed cards from the deck.)

However, your question seems to concern subjective, or epistemic, or Bayesian probability.

I agree with some comments that the issue you raise doesn't seem to have to do with small probabilities.  I also think that when you write "since he visited town X he must like the town's culture, food, etc.", this is a place where one would use Bayes theorem in the context of Bayesian probability.

There is a large, multidimensional universe of math, statistics, and philosophy that is relevant to your question.  You are peeking into a tiny corner of it. I say that not to discourage, but to encourage you to look into it as much as you want.

Here is one starting point for the philosophical side:
https://plato.stanford.edu/entries/probability-interpret/
Introductory textbooks on philosophy of probability include Ian Hacking's An Introduction to Probability and Inductive Logic and Brian Skyrm's Choice and Chance.  Other introductory books include Rowbottom's Probability: Key concepts in philosophy (but there are places where I think the presentation is not great), and Childers' Philosophy and Probability (which is a little more difficult).  An old but useful mathematical survey can be found in Terrence Fine's Theories of Probability.  There are many other books that survey these ideas or focus on particular areas of it.

As my remark above suggests, you might also be interested in literature on the difference between frequentist and Bayesian statistics.  Even the statisticians get philosophical at times, in this area of discussion.  If you're interested in the philosophical dimension, here's a starting point:
https://plato.stanford.edu/entries/statistics/


问:
Solving the system $x^2+2y^2-y-2z=0$, $x^2-8y^2+10z-1=0$, $x^2-7yz=0$
How to solve this system using elimination:

$$\begin{align}
x^2+2y^2-y-2z&=0 \\
x^2-8y^2+10z-1&=0 \\
x^2-7yz&=0
\end{align}$$

I feel like it should be easy but I have tried many different ways and cannot seem to solve for any of the variables.

If you take equation 3 and subtract equation 2 you end up getting
$8y^2-7yz-10z+1=0$ and using the quadratic I could solve for y with respect to z but it didn't get me any closer to solving the system.

答:
I actually think eliminating $x^2$ is the easiest first step, since $x$ only appears as $x^2$ and is not multiplied by any other variable.  So from the third equation, $x^2 = 7yz$, hence the first two equations become
$$\begin{align}
2y^2 - y + 7yz - 2z &= 0, \\
-8y^2 + 7yz + 10z - 1 &= 0. \end{align}$$  Then the next step is to observe that these equations are linear in $z$ but quadratic in $y$, so we collect like terms in $z$ and get
$$\begin{align}
(7y-2)z &= y - 2y^2, \\
(7y+10)z &= 8y^2 + 1. \end{align}$$
Therefore, $$(7y+10)(y - 2y^2) = (7y-2)(8y^2+1).$$  Now we have a cubic in $y$ that we can solve:
$$70y^3 - 3y^2 - 3y - 2 = 0.$$  Unfortunately, this cubic is not "nice," so the roots are very complicated to write out.  The unique real root is
$$y = \frac{1}{210} \left(3+\sqrt[3]{135162-945 \sqrt{20165}}+3 \sqrt[3]{5006+35
   \sqrt{20165}}\right) \approx 0.368933448565128\ldots,$$ which can then be substituted back into the earlier equations to get
$$z \approx 0.166015453844704, \\ x \approx \pm 0.654782847449611.$$  Substitution into the system verifies this result.  There are other solutions but they are complex-valued.

A visualization of the three equations in $\mathbb R^3$ is shown below.  The solution set is depicted as purple spheres (they're a bit hard to see but you can follow the surfaces and locate where all three mutually intersect).


问:
Roots of a polynomial with integer coefficients
Let $f(x)$ be a polynomial in the ring $\mathbb{Z}[x]$. Suppose that $f(i)=0$, where $i^2=-1$.
Can I conclude that $x^2+1$ is a factor of $f(x)$? If so, how can be proven?</p>

答: Hint: if not, then $i$ is a root of the linear $\,f\ {\rm mod}\ (x^2\!+\!1) = f - (x^2\!+\!1)g \ne 0,\ $ so $\, i\in \Bbb Q.$</p>


问:
Showing $Kf$ is a compact operator on $L^p(\mu)$

Let $(X, \mathcal{S}, \mu)$ be a finite measure space, $1<p<\infty$, and $1/p+1/q=1$. If $k: X\times X\to\mathbb{F}$ is an $\mathcal{S}\times\mathcal{S}$ measurable function such that
\begin{align*}
    \sup\bigg\{\int_X|k(x, y)|^q\,d\mu(y):\,x\in X\bigg\}<\infty,
\end{align*}
then
\begin{align*}
    (Kf)(x)=\int_X k(x, y)f(y)\,d\mu(y)
\end{align*}
defines a compact operator on $L^p(\mu)$.

I have seen the proof of this result but my hypothesis is a little different and my background in topology is pretty weak. I would like to show that $Kf$ is completely continuous so that I can use the result that any completely continuous operator on a reflexive space is compact. I am wondering if someone wouldn't mind taking a "heres why $Kf$ is completely continuous for dummies" approach. Thanks in advance.

答:
Let $$c= \sup\bigg\{\int_X|k(x, y)|^q\,d\mu(y):\,x\in X\bigg\}.$$
Suppose that $f_n\to0$ weakly. This means that for every $g\in L^q$,
$$
\int_X f_n\overline g\,d\mu\to0.
$$
In particular, for each $x$
$$
\int_Xk(x,y)\,f_n(y)\,d\mu(y)\to0
$$
Since a weakly convergent sequence is bounded, there exists $b>0$ with $\|f_n\|_p<b$ for all $n$. Then, by Hölder,
$$
\bigg|\int_Xk(x,y)\,f_n(y)\,d\mu(y)\bigg|\leq\Bigg(\int_X|k(x,y)|^q\,d\mu(y)\bigg)^{1/q}\|f_n\|_p<bc^{1/q}.
$$
Because we are in a finite-measure space, a bounded function is integrable. Then, using Dominated Covergence,
\begin{align}
\|Kf_n\|_p^p&=\int_X\bigg|\int_X k(x,y)\,f_n(y)\,d\mu(y)\bigg|^p\,d\mu(x)\to0.
\end{align}
so $K$ is completely continuous, and as $L^p$ is reflexive, $K$ is compact.


问:
Does this equation also holds for complex parameters?
I know that the following integral equation holds

$$\int_{-\infty}^{\infty}\frac{\log(a^2+x^2)}{b^2+x^2}dx=\frac{2\pi}{b}\log(a+b)$$

for $a,b>0\in\mathbb{R}$. Does it also hold for $a,b\in\mathbb{C}$ ?

答:
To support the idea of @Conrad about analytic continuation.

If $\Re \,{a,b}>0 \quad{a}=|a|e^{i\phi_a}$ and ${b}=|b|e^{i\phi_b}$, where $\phi_a, \phi_b\in (-\frac{\pi}{2},\frac{\pi}{2})$

$$I=\int_{-\infty}^{\infty}\frac{\log(a^2+x^2)}{b^2+x^2}dx=\int_{-\infty}^{\infty}\frac{\log\bigl((x-ae^{\frac{\pi{i}}{2}})(x-ae^{\frac{-\pi{i}}{2}})\bigr)}{(x-be^{\frac{\pi{i}}{2}})(x-be^{\frac{-\pi{i}}{2}})}dx=$$
$$=\int_{-\infty}^{\infty}\frac{\log(x-ae^{\frac{\pi{i}}{2}})}{(x-be^{\frac{\pi{i}}{2}})(x-be^{\frac{-\pi{i}}{2}})}dx+\int_{-\infty}^{\infty}\frac{\log(x-ae^{\frac{-\pi{i}}{2}})}{(x-be^{\frac{\pi{i}}{2}})(x-be^{\frac{-\pi{i}}{2}})}dx$$

Logarithm in the first integral has the branch point in the upper half of complex plane, so we close the contour in the lower half-plane (to keep the integrand a single-valued function). For the second integral we close the contour in the upper half-plane (integrating counter-clockwise along a half-circle of a big radius $R$ ). Integrals along half-circles $\to0$ at $R\to\infty$, and we get
$$I=-2\pi{i}\,\frac{\log(be^{\frac{-\pi{i}}{2}}-ae^{\frac{\pi{i}}{2}})}{(be^{\frac{-\pi{i}}{2}}-be^{\frac{\pi{i}}{2}})}+2\pi{i}\,\frac{\log(be^{\frac{\pi{i}}{2}}-ae^{\frac{-\pi{i}}{2}})}{(be^{\frac{\pi{i}}{2}}-be^{\frac{-\pi{i}}{2}})}=$$

$$=-2\pi{i}\,\frac{\log(e^{\frac{-\pi{i}}{2}}(b+a))}{-2bi}+2\pi{i}\,\frac{\log(e^{\frac{\pi{i}}{2}}(b+a))}{2bi}=\frac{2\pi}{b}\log(a+b)$$

where $\Re \,{a,b}>0$


问:
What is the group $G=\langle a,b\mid a^2=b^2=1\rangle$?
I am trying to describe all the elements of $G=\langle a,b\mid a^2=b^2=1\rangle$, and I am really struggling with where to start.

I feel like the group could be infinite, with each element in the group, aside from the identity, having order $2$ (I have since learnt that these are called Boolean, or Abelian-$2$, Groups - again, not sure if this is the right track to go down). I would like to go on to build two subgroups, $S_1$ and $S_2$ that are isomorphic to $\mathbb{Z} /2\mathbb{Z}$ and $\mathbb{Z}$, but obviously to do that I need to understand what $G$ actually looks like!

Thank you for any help.

答: Hint: $ab$ will not have order $2$ in $G$. The standard construction of the free group on $a$ and $b$ should suggest to you a normal form for elements of $G$ as words in $a$, $b$, $a^{-1}$ and $b^{-1}$ that admit no cancellation subject to the rules that $a^2 = 1 $ and $b^2 = 1$.


答:
To solve this problem, we find the time intervals where the height increases and when it decreases. The height increases on the intervals (where $k\in \mathbb{N}\cup \{0\}$)
$$[3\pi/4+2k\pi,7\pi/4+2k\pi],$$
and decreases on
$$[0,3\pi/4],\quad [7\pi/4+2k\pi,11\pi/4+2k\pi].$$
This can be seen by noting that $h'(t)=-e^{-t}(\sin(t)+\cos(t))=-\sqrt{2}e^{-t}\sin(t+\pi/4)$. The distance travelled going downwards is therefore given by the sum
$$\scriptsize (2-(1+e^{-3\pi/4}\cos(3\pi/4)))+\sum_{k=0}^{\infty} (1+e^{-(7/4+2k)\pi}\cos((7/4+2k)\pi))-(1+e^{-(11/4+2k)\pi}\cos((11/4+2k)\pi)),$$
which simplifies to
$$1+\frac{e^{-3\pi/4}}{\sqrt{2}}+\frac{1}{\sqrt{2}}\sum_{k=0}^{\infty} (e^{-(7/4+2k)\pi}+e^{-(11/4+2k)\pi})$$
which by the geometric series (which converges since $|e^{-2\pi}|<1$) is given by
$$1+\frac{e^{-3\pi/4}}{\sqrt{2}}+\frac{e^{-3\pi/4}}{\sqrt{2}(e^{\pi}-1)}=1+\frac{e^{\pi/4}}{\sqrt{2}(e^{\pi}-1)}\approx 1.0700467.$$
You can similarly find the distance going upwards by computing the sum
$$\sum_{k=0}^{\infty} (1+e^{-(7/4+2k)\pi}\cos((7/4+2k)\pi))-(1+e^{-(3/4+2k)\pi}\cos((3/4+2k)\pi)),$$
which by a similar method is given by
$$\frac{e^{\pi/4}}{\sqrt{2}(e^{\pi}-1)}\approx 0.0700467.$$
Alternatively, you can use continuity of $h$ and that the difference between the starting height and the height as $t\to \infty$ is $1$ to conclude the same result.

Therefore the total distance travelled by the weight is finite.

Title: Calculate the distance $h(t)=1+e^{-t}\cos(t)$ that drops

Body:
A weight is hanging from a spring, and oscillates with heights modeled by the function $h(t)=1+e^{-t}\cos(t)$ at time $t\geq 0.$ Does it travel an infinite distance (in other words, does it converge or diverge) as $t\to\infty$ and also, what is the total distance that it drops ?

Note: the distance dropped indicates only the downward motion of the function, not when the function increases.

I have tried to plot this and it seems from the graph that it only drops to $\frac{3\pi}{4}$ and increases from that point on. I got a final answer of $\frac{1}{1-e^{-\pi}}.$ For the other question of whether the weight travels infinitely, the answer is a simple no.


问:
Contraction of certain ideals
Let $A \subseteq B$ be two integral domains over a field $k$ of characteristic zero.
Let $p$ be a prime ideal of $A$.

A comment to this question says:
"Extensions of prime ideals pretty much never remain prime. If $A \subseteq B$ are rings, and $p$ is a prime of $A$, then $pB$ is prime in $B$ if and only if $B/pB = B \otimes_A A/p$ is an integral domain, which it seldom is, even if $A$ and $B$ are integral domains."

Further assume that $p$ is a maximal ideal of $A$. Then $B/pB = B \otimes_A k$; What can be said about that tensor product?
Probably not much? When is it an integral domain?

What if we further assume that $p=A \cap P$, for some prime ideal $P$ of $B$; should this help somehow?

Thank you very much.

答:
The situation where $A = \mathbb Z$, and $B$ is the integral closure of $A$ in a finite field extension $K$ of $\mathbb Q$ is particularly enlightening on this matter.

For example, if $K = \mathbb Q(i) = \{ a+bi : a, b \in \mathbb Q\}$, then $B = \mathbb Z[i] = \{a+bi : a,b \in \mathbb Z\}$.

For general $K$, the following results may be interesting to you:

Every nonzero prime ideal of $B$ is a maximal ideal.

Every prime ideal $\mathfrak p$ of $A$ is equal to $P \cap A$ for a (not necessarily unique) prime ideal $P$ of $B$). In fact, $P$ is unique if and only if $\mathfrak p B$ is a power of a prime ideal of $B$ (in which case, we have $P^e = \mathfrak p B$ for some $e \geq 1$).

Let us specialize to the case $B = \mathbb Z[i]$.  If $\mathfrak p$ is a prime ideal of $\mathbb Z$, generated by a prime number $p$, then sometimes $\mathfrak p$ remains prime in $B$, and sometimes not.  Specifically:

If $p = 2$, then $\mathfrak p B$ is not a prime ideal of $\mathbb Z[i]$, but it is equal to $P^2$, where $P$ is the ideal in $\mathbb Z[i]$ generated by $1+i$.

If $p \equiv 1 \pmod{4}$, then $\mathfrak pB$ is not a prime ideal of $B$, but is equal to the intersection of two distinct prime ideals of $B$.

If $p \equiv 3 \pmod{4}$, then $\mathfrak pB$ is a prime ideal of $B$.

This exhausts all the possibilities.

For the general case of a finite field extension $K$ of $\mathbb Q$, the problem of determining for which prime ideals $\mathfrak p$ of $A$ have the property that $\mathfrak p B$ remains a prime ideal in $B$ is unsolved.  When $K$ is a Galois extension of $\mathbb Q$ such that $\operatorname{Gal}(K/\mathbb Q)$ is an abelian group, the problem is essentially solved by the Kronecker-Weber theorem.

Title: Contraction of certain ideals


问:
Sketching a polar curve
Continued off the question I asked earlier, I also have to sketch the curve.

$r^2=−4\sin(2\theta)$

So I have to set up a table of values I'm assuming. How do I know what values to choose for $\theta$. Then I'm just lost all together trying to plot the theta and $r$ values to get the curve.

Oh and one quick other question. If it also asks me to find all the points of intersections and to set up the integral to evaluate the area inside after sketching the curve, what does that mean? Do I have to pick a top and bottom or something? Confused about that.

答: As a rule, it helps to know how often your function repeats itself (if ever). It is usually found by looking at the co-efficient of the θ (in this case 1). You then smoothly connect where it is at 0 to 2π in 4*co-efficient number of even increments. Remember that if r is negative, its reflected over the x axis. Generally also a good idea to have the zeros of the function, but not always necessary.</p>

Title: Sketching a polar curve


问:
Regarding a Paper by Paul A. Clement on Tridiagonal Matrices
I've asked this question at MathOverflow and was told it'd be better suited for here.

In Paul A. Clement's (1959) paper,

A Class of Triple-Diagonal Matrices for Test Purposes, SIAM Review, Vol. 1, No. 1 (Jan., 1959), pp. 50-52

He makes the claim that the eigenvalues of:

$ \begin{pmatrix}
0 & y_{1} & 0 & ... & 0 \\\
x_{1} & 0 & y_{2} &  & ... \\\
0 & x_{2} & 0 & ... & 0 \\\
... &  & ... &  & y_{n} \\\
0 & ... & 0 & x_{n} & 0
\end{pmatrix} $

are $\pm (n), \pm (n-2), ..., (\pm1 \; or \; 0)$ for $x_{k} = k$ and $y_{k} = n-k+1$.

Specifically, and I quote, "then a theorem of Sylvester establishes that the eigenvalues of this An+, are the
numbers".

I can't for the life of me figure out what theorem and/or how it follows from them. I am familiar with Sylvester's formula for matrices in terms of their eigenvalues, but to get Frobenius covariants of a matrix A one needs to know the eigenvalues to start with.

Am I overlooking something trivial here?

答: A better discussion of the (Clement-)Kac(-Sylvester) tridiagonal matrices would be in this paper and this paper; there are a few slick (I could not have come up with them, I'll admit) proofs that the eigenvalues of these matrices are those particular integers.</p>


问:
Chebyshev's inequality for complex random variable
Let $Z = X + Y i$ be a complex random variable, where $X, Y$ are real-valued r.v.'s

Denote by $\mathbb{E}[Z]$ and $\sigma^2 = Var[Z]$ the expectation and variance of $Z$.

By Chebyshev's inequality,

$$P(|Z-\mathbb{E}[Z]| \geq \sigma k) \leq \frac{1}{k^2}$$

If $\mathbb{E}[|Z|] \geq 2 \sigma k$, is it true that

$$P(|Z| \leq \sigma k) \leq \frac{1}{k^2} ? $$

Why am I asking this?

If $Z$ was a real-valued r.v. and $E[Z] \geq 2 \sigma k $ then
$P(Z \leq \sigma k) = P(Z \leq 2\sigma k - \sigma k) \leq P(Z \leq \mathbb{E}[Z] - \sigma k) \leq \underbrace{P(|Z - \mathbb{E}[Z]| \geq   \sigma k) \leq \frac{1}{k^2}}_{Chebyshev}$.

For my question above, we cannot do exactly the same, as $| \ \cdot \ | $ doesn't have the same meaning in the complex numbers...

I think the answer to my question should be yes! but I'm missing something... Anyone can help?

答:
First, a detour to recap: the proof of Chebyshev's inequality is just using Markov's inequality on $X^2$, for

$X := |Z-\mathbb{E}[Z]|^2$.

$X$ is a non-negative, real-valued random variable, so
$$
\mathbb{P}\{ |Z-\mathbb{E}[Z]| \geq a \}
= \mathbb{P}\{ X \geq a^2 \} \leq \frac{\mathbb{E}[X]}{a^2}
= \mathbb{P}\{ X \geq a^2 \} \leq \frac{\mathbb{E}[|Z-\mathbb{E}[Z]|^2]}{a^2}
= \frac{\mathrm{Var}[Z]}{a^2} \tag{1}
$$
still holds.

So we still have Chebyshev's inequality, as you had stated. (Just for the sake of it, it's worth restating.)

Now, for the second (main) part: the modulus does satisfy the triangle inequality, which is what we need:
$$
|\mathbb{E}[Z]| - |Z|
\leq ||\mathbb{E}[Z]| - |Z||
\leq |\mathbb{E}[Z]- Z| \tag{2}
$$

If $|\mathbb{E}[Z]| \geq 2a$, then this implies
$$\begin{align}
\mathbb{P}\{ |Z| \leq 2a - a \} &\leq
\mathbb{P}\{ |Z| \leq |\mathbb{E}[Z]| - a \} =
\mathbb{P}\{ |\mathbb{E}[Z]| - |Z| \geq a \}\\
&\leq \mathbb{P}\{ |Z-\mathbb{E}[Z]| \geq a \}\\
&\leq \frac{\mathrm{Var}[Z]}{a^2} \tag{3}
\end{align}$$


问:
How to solve the integral $\int \frac{y}{\cos^2(y)}~dy$.
I would like to compute the integral of

$$\int \frac{y}{\cos^2(y)}~dy.$$

Here is what I tried:

$$u=\cos(y),\quad du=-\sin(y)~dy,$$

$$dv=y~dy,\quad v=y^2/2,$$

$$uv-\int v~du=\frac{y^2}{2}\cos(y)+\int \frac{y^2}{2}\sin(y)~dy.$$

The expected answer is
$$y(\tan y)+\ln(\cos(y)).$$

How can I get to this? Thanks guys.

答: Since you are taking "u= cos(y) dv= ydy"m=,you appear to be using integration by parts to integrate $\int cos(y) ydy$ rather than $\int \frac{y}{cos^2(y)} dy$.


问:
I don't get how the answer to this problem works, its a practice mathcon problem.
The answer to this problem is c. I don't get how that is the answer.
This is the problem

答: $AB$ is the diameter of the base, so the area of the base is $9\pi$ cm$^2$.  Whatever angle the can is tilted at the volume to raise the level $2$ cm in slant height is the same.  You are raising the level $2$ cm parallel to the axis over the area of the bottom.  This means the volume of interest is $18 \pi$ cm$^3$


问:
Is it true that the transpose of a matrix is the dual map in any basis?
For self adjoint operators, we know that the Hermitian adjoint of the matrix is the adjoint of the operator but this is only true if we choose orthonormal basis for the matrix.  So I was wondering whether we need a specific basis in order for the statement "transpose of a matrix is the dual map" to be true?</p>

答:
Let $V$ be a (finite dimensional) vector space with inner product $\langle\cdot,\cdot \rangle$.  Let $\varphi:V \rightarrow V$ be the linear map.  Let $x_1,\dots, x_n$ be any basis.  Let $A=(a_{ij})$ be the matrix representation of $\varphi$ with respect to $x_1,\dots, x_n$.  Let $\varphi^t$ denote the adjoint of $\varphi$ with respect to $\langle\cdot,\cdot \rangle$. Let $B=(b_{ij})$ be the matrix representation of $\varphi^t$ with respect to the same basis. Let $G$ be the $n\times n$ matrix defined by
$$
G_{ij}=\langle x_i,x_j\rangle
$$
Note $G$ is symmetric because the inner product is symmetric. Let $e_i\in \mathbb{R}^n$ for $i=1,\dots, n$ denote the standard basis in $\mathbb{R}^n$ expressed as column vectors.  Then
$$
\langle \varphi x_i,x_j\rangle = (Ae_i)^TGe_j=e_i^TA^TGe_j = (A^TG)_{ij}
$$
and
$$
\langle x_i,\varphi^t x_j\rangle= e_i^T G(Be_j)=e_i^TGBe_j=(GB)_{ij}
$$
Since
$$
\langle \varphi x_i,x_j\rangle=\langle x_i,\varphi^t x_j\rangle,
$$
we have
$$
(A^TG)_{ij}=(GB)_{ij}.
$$
Hence, $A^TG=GB$ which means
$$
B=G^{-1}A^TG.
$$
On the other hand, if we started with an orthonormal basis, then $G=I_n$ is the $n\times n$ identity matrix and the above equation simplifies to $B=A^T$, that is, the matrix representation of $\varphi^t$ is the transpose of that of $\varphi$.


问:
Markov Chain Reversibility and Stationary Distribution
A librarian has three books, labelled $i$ =1,2,3. Everyday, the librarian picks a book i with probability $p_i$ (regardless of where it is on the shelf). By the end of the day, they return it to the shelf one position to the left of where it was previously (unless it was already the leftmost book, in which case it is put back where it was).

(a)Make Xn be the order of the books at the beginning of day n (before a book is selected that day). Show that (Xn,n≥0) is a reversible Markov chain with equilibrium distribution $\pi_{ijk}$ proportional to $p^2_i p_j$.

(b)Suppose that $p_i$ = i/6 and initially the books are in order 123.  Find the expected amount of days until the books are again in this order

答:
To check the reversibility condition $\pi(x)P(x,y)=\pi(y)P(y,x)$ there are just two types of transitions to consider: Swapping the first two books, and swapping the last two:
$$\pi_{ijk} \, P(ijk \,,\, jik) =(Cp_i^2 p_j)\, p_j=(Cp_j^2 p_i)\, p_i=\pi_{jik} \,P(jik \,,\, ijk) $$
$$\pi_{ijk} \,   P(ijk \,,\, ikj)=(Cp_i^2 p_j)\, p_k=(Cp_i^2 p_k)\, p_j=\pi_{ikj} \, P(ikj \,,\, ijk)$$


问:
Half-line - open or closed set
Half-line is given by

$$A=\{x\in R^n \mid x=x_0 + t(a-x_0), \forall t\geq 0\}$$

Is it open or closed set?

答:
Clearly the complement of $A$ is open (since the point $x_0$ is not in the complement of $A$). This $A$ is closed.

On the other hand, the set $A$ is not open, so it must be closed (not clopen).

Title: Half-line - open or closed set


问:
Show that $A=\{0^m 1^n |\text{ }m\text{ is even or }m > n\}$ has infinitely many equivalence classes to prove non-regularity.
Show that $A=\{0^m 1^n |\text{ }m\text{ is even or }m > n\}$ has infinitely many equivalence classes.

I'm trying to prove that that $A$ is a non-regular language. Alternatively, I could use closure properties - but have been struggling.

答:
I claim that for any $a\neq b$, $0^a\not\equiv 0^b$ via the Myhill-Nerode Equivalence relation.

To see why, observe that if $a$ is even and $b$ is not or vice versa, then the string $z=01^{a+b+1}$ will ensure that either $0^az$ is accepted and $0^bz$ is not, or v.v.

So assume that both $a,b$ are even or odd. Without loss of generality, we can assume that $a,b$ are odd, since we can add a $0$ at the end to both.

Without loss of generality, $a<b$. Then, let $z=1^{a+1}$. Thus, $0^az$ will be accepted, while $0^bz$ will not.


问:
Solutions of $|z+5|=|z-1-i|$
I am trying to prove what the following set
$$|z+5|=|z-1-i|$$
is in the complex plane. Intuitively you can see that it is a straight line, since they are the points whose distance from points a and b are equal.

But what is the equation of a line on the complex plane ? How is your shape? I'm having trouble because of the $i.$

答: Let $a+bi$ satisfy the given equation, then we have $|a+5+bi|=|a-1+(-b-1)i|$ which comes down to $(a+5)^2+b^2=(a-1)^2+(-b-1)^2$ which you can simplify into a relation between $a$ and $b$ yielding a linear equation

Title: Solutions of $|z+5|=|z-1-i|$


问:
Combinatorics: How many non-negative integer solutions are there to the following equation: x1 * x2 * x3 * x4 = 1000000

Combinatorics: How many non-negative integer solutions are there to the following equation: x1 * x2 * x3 * x4 = 1000000

I thought to look at the equation as x1 * x2 * x3 * x4 = $2^6$ * $5^6$.

and then there are 4C2 * 4 * 4 solutions.

what do you think?

答:
You’re missing quite a bit, I’m afraid. We can break it into two problems. If $x_1x_2x_3x_4=10^6$, then the numbers $x_k$ must be of the form $x_k=2^{a_k}5^{b_k}$ for $k=1,2,3,4$. Then we’ll have a solution in non-negative integers to the original equation if and only if

$$a_1+a_2+a_3+a_4=6\tag{1}$$

and

$$b_1+b_2+b_3+b_4=6\,,\tag{2}$$

where the $a_k$ and $b_k$ are non-negative integers. Counting the solution in non-negative integers to $(1)$ is a basic stars and bars problem; the linked article has a pretty good explanation of the reasoning involved. The same goes for $(2)$, of course; in fact, the number of solutions will obviously be the same.

Once you have that number, you’ve done the hardest part. Now, taking into account the fact that the powers of $2$ can be distributed amongst $x_1,x_2,x_3$, and $x_4$ completely independently of the powers of $5$, what do you do with the numbers of solutions to $(1)$ and $(2)$ to finish off the problem?


问:
Logarithm algebra question
The question says: Given $\log_n(9)=x$ and $\log_n(4)=y$, find $\log_n(12)$. I'm just really stuck on how to work this out.

答:
So we have given that $\log_{n} 9=x=x$ $\log_{n} 4=y$.

$$\log_{n} 9=\log_{n} 3^{2}=2\log_{n} 3=x$$
$$\log_{n} 3=\frac{x}{2}$$
$$\log_{n}12=\log_{n}3*4=\log_{n}3+\log_{n}4=\frac{x}{2}+y$$

So,

$$\log_{n}12=\frac{x}{2}+y$$

In this the answer doesen't change because of value of n. Just use normal logarithm laws.
I just used,

$\log_{n} (x^y) = y\log_{n} (x)$ = Log of a power

$\log_n (xy) = \log_{n} (x)+\log_{n} (y)$ = Product rule.

Title: Logarithm algebra question


答: So you know that $z$ is the locus of complex numbers of the form $x + yi$ that are equidistant from $-5$ and $1+i$.  This forms a line that passes through the midpoint $-2 + i/2$, and is perpendicular to the slope of the segment joining the two points, i.e., the slope is the negative reciprocal of $\frac{1 - 0}{1+5} = \frac{1}{6}$, which is $-6$.  So we want a line passing through $(-2, 1/2)$, and slope $-6$, or $$(y - 1/2) = -6(x + 2),$$ or $$y = -6x - \frac{23}{2}.$$  This gives us the parametrization $$z = x -(6x + \tfrac{23}{2})i,$$ for all real $x$.

Title: Solutions of $|z+5|=|z-1-i|$

Body:
I am trying to prove what the following set
$$|z+5|=|z-1-i|$$
is in the complex plane. Intuitively you can see that it is a straight line, since they are the points whose distance from points a and b are equal.

But what is the equation of a line on the complex plane ? How is your shape? I'm having trouble because of the $i.$


答:
Hint

From Geometry:

If $w_1$ and $w_2$ are complex numbers, then the locus of solutions to

$$|z - w_1| = |z - w_2|$$

will precisely be all complex numbers $z$ such that $z$ is on the

perpendicular bisector between $w_1,$ and $w_2$.

Therefore:

form the line segment connecting $w_1$ and $w_2$.

find the midpoint $z_0$ of the line segment.

construct the line that passes through $z_0$ that is perpendicular to the line
segment connecting $w_1$ and $w_2$.  This means that the slope of the perpendicular line
will be the negative reciprocal of the slope of the line segment connecting $w_1$ and $w_2$.

Title: Solutions of $|z+5|=|z-1-i|$


问:
If $f'(t)$ exists then $\lim_{h,k\to0}\frac{f(t+h)-f(t-k)}{h+k}=f'(t)$
Consider $f'$ continuous...

I'm tried to prove it...with Taylor's expansion:

$f(t+h)=f(t)+hf'(t)+h\phi(t,h)$

$f(t-k)=f(t)-kf'(t)-k\phi(t,-k)$

Then:

$\frac{f(t+h)-f(t-k)}{h+k}=f'(t)+\frac{h}{h+k}\phi(t,h)+\frac{k}{h+k}\phi(t,-k)$

I don't know how I can prove $\lim_{h,k\to0}\frac{h}{h+k}\phi(t,h)=0$

答:
If $f'$ is continuous, by the MVT, we have $\frac{f(t + h) - f(t - k)}{(t + h) - (t - k)} = f'(\xi)$ for some
$\xi$ between $t+h$ and $t - k$. As $h, k \to 0$, we have $\xi \to t$.

What if $f'$ is not continuous?

A counterexample

Let $f: (a, b) \to \mathbb{R}$ be differentiable. Let $t \in (a, b)$.
Can we say $\lim_{h, k\to 0} \frac{f(t + h) - f(t - k)}{h + k} = f'(t)$?

No. $\lim_{h, k\to 0} \frac{f(t + h) - f(t - k)}{h + k}$ may not exist.
Let
$$f(x) = \left\{\begin{array}{cc}
                  x^2\sin\frac{1}{x} & x\ne 0 \\
                  0 & x = 0.
                \end{array}
\right.$$
Let $t = 0$. $f$ is differentiable on $\mathbb{R}$ and its derivative is given by
$$f'(x) = \left\{\begin{array}{cc}
                  2x\sin\frac{1}{x} - \cos \frac{1}{x} & x\ne 0 \\[5pt]
                  0 & x = 0.
                \end{array}
\right.$$

Let us prove that
$\lim_{h, k\to 0} \frac{f(h) - f( - k)}{h + k}$ does not exist.
Consider $h_n = \frac{1}{n}$ and $k_n = -\frac{1}{n} + \frac{1}{n^2}$.
Then as $n \to \infty$, we have $(h_n, k_n) \to (0, 0)$.
However, we have
\begin{align}
&\frac{f(h_n) - f( - k_n)}{h_n + k_n} \\
=\ & \sin n - \sin\left(n + 1 + \frac{1}{n-1}\right) + \frac{2n - 1}{n^2}\sin \frac{n^2}{n-1}\\
=\ & -2\cos\left(n + \frac{1}{2} + \frac{1}{2n - 2}\right)
\sin\left(\frac{1}{2} + \frac{1}{2n - 2}\right) + \frac{2n - 1}{n^2}\sin \frac{n^2}{n-1}.
\end{align}
Clearly, it does not converge to zero as $n\to \infty$.

Discussion

However, the following is true:

Suppose that $f$ is differentiable at $x$. Then
$$f'(x) = \lim_{h\to 0^{+},\ k\to 0^{+}} \frac{f(x+h) - f(x-k)}{h+k}.$$
Calculus by Spivak, 3rd Ed., page 164, question 22(b).

See: Prove that $f'(x) = \lim_{h\to 0^+ \\k\to 0^+} \frac{f(x+h) - f(x-k)}{h+k}$


问:
Does the series converge? and what it is calculate to
So, a series of $\sum_{n=1}^{\infty}a_n$, but $a_n=2n^{-1}$ if n is even, but $a_n=2^{-n}$ if $n$ is odd. Does $\sum_{n=1}^{\infty}a_{2n-1}$ converge, and what it's to. So it's only odd numbers because $2n-1.$ And so $a_{2n-1}=2^{-n}$. With a divergence test, I can not see anything, because the limit is 0. So I don't know if it's converge or not. But $2^{-n}$ is a geometric series, so yes it converge. I calculate this to be $a_n$ converges to 1. This is correct?

答:
Actually, $a_{2n-1}=2^{-(2n-1)}$ and not $2^{-n}$. Pay attention to the subscript of $a$, it should agree with what comes up in the exponent of $2$. So you really have
$$ \sum_{n=1}^\infty a_{2n-1}=\sum_{n=1}^\infty \frac1{2^{2n-1}}=\frac12+\frac18+\frac1{32}+\frac1{128}+\cdots $$
which you can recognise as a geometric series. In particular the series is equal to
$$\frac12\left(1+\frac14+\left(\frac14\right)^2+\left(\frac14\right)^3+\cdots\right)=\frac12\frac{1}{1-1/4}=\frac23.$$
Whenever you are confused by the summation $\Sigma$ notation, just write it out and see what it represents. This will be very useful in preventing confusion.