问:
Combinations & probability: dividing 9 people into 3 groups
I'm struggling to understand the procedure to solve the following problem:

The group consists of 9 people, 3 men and 6 women. They are divided randomly into 3 groups (3 persons each). What's the probability that..

a) all men are in the same group?
b) all men are in different groups?

I somehow managed to calculate a) as follows:

all combinations: 9!/(3!^4) = 280

combinations of all men in one group: 6!/(3!*3!*2!) = 10

probability: 10/280 = 0,03571..

Please, do let me know if I am doing this wrong. Also, I do not know how to figure out b) all men in different groups. Asking for your advice, thank you!

答:
I think this can be done in a straight-forward way. The order in which groups are formed does not matter, so we can assume that the process begins by randomly drawing people from the pool to add to the first group. This makes it easier because now you simply have to compute {M is drawn} * {M is drawn again} * {M is drawn a third time}. This is

$$\frac39 \cdot \frac28 \cdot \frac17 = \frac1{84}$$

Then the probability that any group consists of only men is 3 times that, so $\frac{1}{28}$. Alternatively, begin after the first man has been assigned, and just multiply $\frac28 \cdot \frac17 = \frac{1}{28}$.

For the second, again we can just fix one order. Assume we first draw people for just group one, then the probability to have exactly one man in there (which is necessary to have one men in each group) is

$$3 \cdot \frac39 \cdot \frac68 \cdot \frac57$$

(Compute the probability for MWW, then multiply it by 3 to account for the permutations WMW and WWM.) Similarly for group two:

$$3 \cdot \frac26 \cdot \frac45 \cdot \frac34$$

and group three:

$$3 \cdot \frac13 \cdot \frac22 \cdot \frac11$$

This one just yields 1, which makes sense because after the first two groups have one man in each, the last one will inevitably also have one.

In total, we get

$$3 \cdot \frac39 \cdot \frac68 \cdot \frac57 \cdot 3 \cdot \frac26 \cdot \frac45 \cdot \frac34 = \frac9{28}$$


答:
First of all one should determine the meaning of "they are divided randomly into 3 groups (3 persons each)."

I would assume that the procedure is following: there are 9 positions separated in the groups of the size 3: $(123)(456)(789)$ and the people are randomly distributed between the positions.

Then

there are altogether $9!$ ways to distribute the people.

there are $3\times 3!\times 6!$ ways to distribute the people in such a way that all men are in the same group.

Explanation: $3$ ways to choose the group, $3!$ ways to permute the men inside the group, $6!$ ways to place women on remaining positions.

there are $3!\times 3^3\times 6!$ ways to distribute the people in such a way that all men are in different groups.

Explanation: $3!$ ways to permute the men between the groups, $3$ ways to choose the position of each men inside his group, $6!$ ways to place women on remaining positions.

ContentLicense: CC BY-SA 4.0

Title: Combinations & probability: dividing 9 people into 3 groups

Body:
I'm struggling to understand the procedure to solve the following problem:

The group consists of 9 people, 3 men and 6 women. They are divided randomly into 3 groups (3 persons each). What's the probability that..

a) all men are in the same group?
b) all men are in different groups?

I somehow managed to calculate a) as follows:

all combinations: 9!/(3!^4) = 280

combinations of all men in one group: 6!/(3!*3!*2!) = 10

probability: 10/280 = 0,03571..

Please, do let me know if I am doing this wrong. Also, I do not know how to figure out b) all men in different groups. Asking for your advice, thank you!


问:
For $x, y, z \in R^n$, show if $x \cdot y = z \cdot y$ , $x = z$
Is it right for me to do the following?

$x \cdot y = z \cdot y$

$x \cdot y - z \cdot y =0$

$(x-z) \cdot y= 0$

Let y = (x-z)

$(x-z) \cdot (x-z) = 0$

x = z

Similarly for another question, For $x, y, z$ in $R^3,$ show if  $ x \times y = z \times y$ , $x = z$, how do I begin proving this question

答:
If the question is:

If $x,z\in\Bbb R^n$ are such that $(\forall y\in\Bbb R^n):x.y=z.y$, then $x=z$.

then your answer is correct.

In the case of the cross-product, it's a little different. Assuming this time that the question is:

If $x,z\in\Bbb R^n$ are such that $(\forall y\in\Bbb R^n):x\times y=z\times y$, then $x=z$.

this can be done as follows: if $x\ne z$ take a vector $y\ne0$ which is orthogonal to $x-y$. Then $\|(x-z)\times y\|=\|x-z\|.\|y\|\ne0$, and therefore $(x-z)\times y\ne0$. It follows that $x\times y\ne z\times y$.


问:
Minimum distance from a point to a subspace
The subspace V in $R^4$ is the span of vectors $u = \frac 1{3} [1, -1, 1, 2]^T$ and $v = \frac 1 {5}[1, 2, 1, 1]^T$. Find the shortest distance from the point $P(1, 1, 1, 1)$ and the subspace.

Don't really know where to start here so would really appreciate some hints and tips. I've looked up similar questions like this: Distance from point $(1,1,1,1)$ to the subspace of $R^4$ but don't really get the idea of it.

答:
A vector of the span has coordinates
$$a (1,-1,1,2)+b (1,2,1,1)$$
the distance from $P(1,1,1,1)$ is minimum when the function
$$d^2(a,b)=2 (a+b-1)^2+(2 a+b-1)^2+(-a+2 b-1)^2\tag{1}$$
has a minimum. Its derivatives wrt $a$ and $b$ must be zero
$$\begin{cases}
14 a+4 b-6=0\\
2 a+7 b-5=0
\end{cases}
$$
$$a= \frac{11}{45},b= \frac{29}{45}$$
substitute in $(1)$ and get the distance $P,V=\frac{1}{3}\sqrt{\frac{2}{5}}$

ContentLicense: CC BY-SA 3.0


问:
Why are these NO counterexamples for Baire's Theorem?
Baire's Theorem states: Let $X$ be a complete metric space and let $U_n \subset X$ be open and dense ($n \in \mathbb{N}$). Then $\bigcap_{n=1}^\infty U_n\subset X$ is dense.

In the lectures, we discussed the well-known counterexample of the rational numbers. We got two more examples that are NO counterexamples:

$X=\mathbb{N}\subset \mathbb{R}$ with subspace topology, $U_n:=\mathbb{N}\setminus \{n\}$

$X=\{\frac{1}{n} \mid n\in \mathbb{N}\} \cup \{0\}, U_0:=X\setminus \{0\}, U_n:= X\setminus \{\frac{1}{n}\}$

My problem is that I don't see any difference between the counterexample of the rationals and these two examples, e.g. $\mathbb{N}$ is complete, $\mathbb{N}\setminus \{n\}$ is open since singletons are closed and the closure of $\mathbb{N}\setminus \{n\}\subset \mathbb{N}$ is $\mathbb{N}$ (imo).

Where's my mistake in thinking?

答: $\mathbb N \smallsetminus\{n\}$ is not dense, since no sequence of points in that set approaches $n$ as a limit.


问:
Why does the $T_1$ Axiom imply that a point is a closed set?
According to my reading the $T_1$ axiom say that given a topological space $X$ and $\forall x, y \in X$ We have:

$ \exists U, V \subset X: [(x \in U \land y \notin U) \land (x \notin V \land y \in V)].$

Here, $U, V$ are open sets, and they're not assumed to be disjoint.

Why does this imply that $\forall x \in X$, the set $\{ x \}$ is closed?

答:
For every $y\ne x$ choose an open subset $V_y$ such that $y\in V_y,x\notin V_y$. Then $$X\setminus\{x\}=\bigcup_{x\ne y\in X}V_y$$
is open as it is the union of open sets, hence $\{x\}$ is closed.


问:
Finding Force from acceleration at a point given constant speed
Prompt: An object of mass $m$ travels along a parabola $y = x^2$ with constant speed $5$ units/sec. What is the force on the object due to acceleration at point $(\sqrt{2}, 2)$? Recall $\vec{F} = m\vec{a}$.

This was an "applied theory" problem I got incorrect on an generated exam I recently took. This question seems to be a similar idea, excepting that I need to work backwards from speed rather than velocity and that speed is constant. Here is what I understand:

Since we're working in 2-D, $|\vec{v}(t)| = \sqrt{v_x^2 + v_y^2} = 5 \implies |\vec{v}(t)|^2 = 25 = v_x^2 + v_y^2$. I don't know if this helps me.

The parabola can be reparameterized as a trajectory/displacement curve: $\vec{r}(t) = \langle t, t^2 \rangle$

From the reparameterization, we can differentiate $\frac{d \vec{r}}{dt} = \vec{v}(t) = \langle 1, 2t \rangle$ and once more $\frac{d \vec{v}}{dt} = \vec{a}(t) = \langle 0, 2 \rangle$.

If I compute the supposed magnitude of $\vec{v}$ at point $(\sqrt{2}, 2) \Longleftrightarrow t = \sqrt{2}$, I find $$
|\vec{v}(t=\sqrt{2})| = \sqrt{1^2 + (2\sqrt{2})^2} = \sqrt{1 + 8} = \sqrt{9} = 3 \neq 5
$$ A contradiction of sorts.

Point ($3$) above would lead one to naively assume that acceleration is $2$ units/sec$^2$ upwards for all times $t$. From there, one could conclude $\vec{F} = m \langle 0, 2 \rangle = \langle 0, 2m \rangle$. Alas the automated grader reports the correct answer is:

$$\displaystyle \vec{F} = \langle-\frac{-100\sqrt{2}}{81}m, \frac{50}{81}m\rangle$$

The connection between 3 (or 9) and 81 is not lost on me ($3^4 = 9^2 = 81$), but I don't see how to bridge the gap. Any help would be greatly appreciated. I've already got my grade and the "answer". What I seek now is understanding.

Many thanks in advance.

答:
The curvature of the parabola $y=x^2$ is:

$$k=\dfrac{2}{(4x^2+1)^{3/2}}$$

So the centripetal force is:

$$F_c=V^2k$$


问:
Confusion about domain of definition for rational functions
Many times I get confused about the domain of function when we deal with limit or continuity or differentiation at particular point. I want to get clear about this confusion.

From baby rudin.

In definition $5.1$, to define derivative of real valued $f$, it is first assured that $f$ is defined on $[a,b]$ and then derivative is defined for $x\in [a,b]$.

Theorem $5.3$ part (c) says, Suppose $f$ and $g$ are defined on $[a,b]$ and are differentiable at point $x\in [a,b]$, then $(\frac{f}{g})'(x)=\frac{g(x)f'(x)-g'(x)f(x)}{g(x)^2}$, provided $g(x)\neq 0$.

Here's my doubt : When we define $\frac{f}{g}$, we alter the domain by removing those points at which $g$ takes value $0$. Then, it is possible that the alterted domain for $\frac{f}{g}$ may not be even interval. Then how to define derivative? I think along with assumption $g(x)\neq 0$, we should put one more assumption that in some neighbourhood of $x$, $g$ doesn't take value $0$.

Such kind of confusion pops up many times in my mind. For simplicity, can I always make the assumption that given function(even if rational) is nicely defined in some neighbourhood of point $p$, (i.e. when domain is real, function makes sense for $(a,b)$ and $p\in (a,b)$), at which we wish to determine continuity/differentiability/limit.

答: The derivative of a function is defined from a limit. You need to use the rules that hold for limits. In particular, you only use the values where the function is defined, i.e you remain in the domain of the fuction. So if there are zero denominators, you can still compute the limit and see if the derivative exists.</p>


答:
Here is an alternate method. We will approximate the parabola as a circle at $(\sqrt{2}, 2)$. The radius of curvature is given by:

$$R = \bigg\vert\frac{(1 + y'^{2})^{\frac{3}{2}}}{y''}\bigg\vert$$

Plugging in $y = x^{2}, y' = 2x, y'' = 2$:

$$R = \frac{(1 + 4x^{2})^{\frac{3}{2}}}{2}$$

At $x = \sqrt{2}$:

$$R = \frac{9^{\frac{3}{2}}}{2} = \frac{27}{2}$$

Now, the centripetal force on the object is given by:

$$\vert F\vert = \frac{mv^{2}}{R} = \frac{50m}{27}$$

The force is directed towards the center of the circle. From $(\sqrt{2}, 2)$, the direction towards the center of the circle is normal to the tangent of the parabola. Thus, the line between the center of the circle and $(\sqrt{2}, 2)$ has slope $-\frac{\sqrt{2}}{4}$. Also, the center of the circle will be "inside" the curve of the parabola. Using this information, we find the force to be:

$$\boxed{\vec{F} = \bigg(-\frac{100m\sqrt{2}}{81}, \frac{50m}{81}\bigg)}$$

Title: Finding Force from acceleration at a point given constant speed

Body:
Prompt: An object of mass $m$ travels along a parabola $y = x^2$ with constant speed $5$ units/sec. What is the force on the object due to acceleration at point $(\sqrt{2}, 2)$? Recall $\vec{F} = m\vec{a}$.

This was an "applied theory" problem I got incorrect on an generated exam I recently took. This question seems to be a similar idea, excepting that I need to work backwards from speed rather than velocity and that speed is constant. Here is what I understand:

Since we're working in 2-D, $|\vec{v}(t)| = \sqrt{v_x^2 + v_y^2} = 5 \implies |\vec{v}(t)|^2 = 25 = v_x^2 + v_y^2$. I don't know if this helps me.

The parabola can be reparameterized as a trajectory/displacement curve: $\vec{r}(t) = \langle t, t^2 \rangle$

From the reparameterization, we can differentiate $\frac{d \vec{r}}{dt} = \vec{v}(t) = \langle 1, 2t \rangle$ and once more $\frac{d \vec{v}}{dt} = \vec{a}(t) = \langle 0, 2 \rangle$.

If I compute the supposed magnitude of $\vec{v}$ at point $(\sqrt{2}, 2) \Longleftrightarrow t = \sqrt{2}$, I find $$
|\vec{v}(t=\sqrt{2})| = \sqrt{1^2 + (2\sqrt{2})^2} = \sqrt{1 + 8} = \sqrt{9} = 3 \neq 5
$$ A contradiction of sorts.

Point ($3$) above would lead one to naively assume that acceleration is $2$ units/sec$^2$ upwards for all times $t$. From there, one could conclude $\vec{F} = m \langle 0, 2 \rangle = \langle 0, 2m \rangle$. Alas the automated grader reports the correct answer is:

$$\displaystyle \vec{F} = \langle-\frac{-100\sqrt{2}}{81}m, \frac{50}{81}m\rangle$$

The connection between 3 (or 9) and 81 is not lost on me ($3^4 = 9^2 = 81$), but I don't see how to bridge the gap. Any help would be greatly appreciated. I've already got my grade and the "answer". What I seek now is understanding.

Many thanks in advance.


问:
A 2nd countable metric space is separable

A 2nd countable metric space is separable

Can anyone give an idea on how to prove that, I could prove the converse quite easily but I am not sure which set to consider in order to show that it is dense.

答:
I am sorry. As you only asked for a hint, I now hide my answer in the following spoiler box:

 Let $(U_n)_{n \in \mathbb{N}}$ be a countable basis of nonempty sets for the 2nd countable metrik space $(X,d)$. Fix a choice function $f: \bigcup_{n \in \mathbb N} U_n \mapsto V$, where $V$ is the universal class. We claim that the set $(f(U_n))_{n \in \mathbb N}$ is dense. In fact, for any open $U \subseteq X$ there is an $n_0 \in \mathbb{N}$ such that $U_{n_0} \subseteq U$ (by definition of a basis) and thus $f(U_{n_0}) \in U$, showing density.


答:
$$a\cdot x=0$$ is true for all $x$ perpendicular to $a$. In 2D, this is a straight line and in 3D a plane.

$$a\times x=0$$ is true for all $x$ parallel to $a$, which is a straight line in 3D.

So $x=0$ is not necessarily true.

Title: For $x, y, z \in R^n$, show if $x \cdot y = z \cdot y$ , $x = z$

Body:
Is it right for me to do the following?

$x \cdot y = z \cdot y$

$x \cdot y - z \cdot y =0$

$(x-z) \cdot y= 0$

Let y = (x-z)

$(x-z) \cdot (x-z) = 0$

x = z

Similarly for another question, For $x, y, z$ in $R^3,$ show if  $ x \times y = z \times y$ , $x = z$, how do I begin proving this question


问:
Having trouble proving that $N \trianglelefteq G$

I am having a bit trouble showing $N$ is normal in $G$.

The group $G$, of order 18, and generated by $x,y$ and $z$, of order 2, 3 and 3, respectively, satisfy the following conditions:
$$yz=zy, \quad x^{-1}yx=y^{-1}, \quad x^{-1}zx=z^{-1}$$
Let $N:=\langle y,z \rangle$.

I know that for $N$ to be normal in $G$, I have to show that $gNg^{-1} \subseteq N$ for all $g\in G$, i.e. showing $xyx^{-1}, yyy^{-1}, zyz^{-1}, xzx^{-1}, yzy^{-1}$ and $zzz^{-1} \subseteq N$. I have already noticed that $yyy^{-1}, zyz^{-1},yzy^{-1}$ and $zzz^{-1} \subseteq N$, because $N$ is generated by $y,z$, and by the given conditions $y$ and $z$ commute. Now I only have to show that $xyx^{-1}, xzx^{-1} \subseteq N$, but I am kinda stuck.

I have noticed by the given conditions, that $yz=zy \Longleftrightarrow y=zyz^{-1}$, and I have tried to substitute $y$ in $xyx^{-1}$, (and same for $z$), but that didn't get me far. I know that I am missing something simple, but I can't figure out what. Maybe somebody can help with a hint?

答: You can do this almost hands-down. If $G=\langle x,y,z \rangle$, then a subgroup $N \leq G$ is normal if and only if $x,y$ and $z$ normalize $N$. Since in this case $N=\langle y,z \rangle$, you only need to verify this for $x$. But this follows from the second and third presentation equations.


问:
Reparametrize the curve by arc length
We have the following curve $\alpha(t) = (e^t\cos(t), e^t\sin(t))$. And I used the following formula to reparametrize the curve by arc length:

$$s(t) =\int_0^t \|\alpha'(\tau)\| \, d\tau.$$ Then I got $t = \ln\left(\dfrac{s+ \sqrt{2}}{2}\right).$

But according to our solutions we replace $t$ with $\ln(\frac{s}{\sqrt{2}})$. Is it possible to have more than one reparametrization?

答: Neither your answer nor the one that was given to you is correct. Note that\begin{align}s(t)&=\int_0^t\sqrt2e^t\,\mathrm dt\\&=\sqrt2\left(e^t-1\right).\end{align}So, you should take$$t=\log\left(1+\frac s{\sqrt2}\right).$$With this choice, $s=0\iff t=0$ as it should be, which fails to happen with any of those two solutions.

Title: Reparametrize the curve by arc length


问:
Finding the curvature & torsion of the derivative of a smooth regular curve in $\mathbb{R}^3$
Let $\beta(s)$ be a smooth regular curve in $\mathbb{R}^3$ parameterized by arclength with nowhere vanishing curvature. Let $\gamma(s) = \beta'(s)$. Find $\kappa_\gamma(s)$ and $\tau_\gamma(s)$ in terms of the curvature and torsion of $\beta(s)$ which we denote by $\kappa(s), \tau(s)$ respectively.

I think I am supposed to use the Frenet Serret equations to simplify the following expressions:

$\Large
\kappa = \frac{|\beta' \times \beta''|}{|\beta'|^3}$

$\Large \tau = \frac{det_3(\beta', \beta'', \beta''')}{|\beta' \times \beta''|}$

but I have not worked with determinants or cross products in a while.

I found a similar question here but I do not understand the computation of the cross product. Also, it does not contain anything about the torsion.

答:
There is also another way without using cross-product.

First of all notice that $||\gamma'(s)||=||\beta''(s)||=k_\beta(s)$ and so we have the relation $\frac{dt}{ds}=k_\beta(s)$ where $t$ is the arclenght variable for $\gamma$,  indeed $||\frac{d}{dt}\gamma(s(t))||=||\beta''(s)\frac{ds}{dt}||=1$. Then to compute the curvature $k_\gamma(t)$ we can use the formula $k_\gamma(t)=||\frac{d^2}{dt^2}\gamma(s(t))||$ (since now $t$ is an arclenght variable for $\gamma$). If we set $\frac{d}{dt}\gamma'(s(t))=:\gamma'$, by a long calculation \begin{equation*}\frac{d\gamma'}{dt}=\frac{d\gamma'}{ds}\frac{ds}{dt}=\frac{\beta'''(s)}{k_\beta(s)^2}-\frac{\beta''(s)k_\beta'(s)}{k_\beta(s)^2}=-T(s)+\frac{k_\beta(s)}{\tau_\beta(s)}B(s)
\end{equation*}
where $T(s)=\beta'(s)$ and $B(s)=\beta'(s)\times N(s)$ and $N(s)=\frac{\beta''(s)}{k_\beta(s)}$. Finally \begin{equation*}k_\gamma(t)=||\frac{d^2}{dt^2}\gamma(s(t))||=\sqrt{1+\frac{k_\beta(s(t))^2}{\tau_\beta(s(t))^2}}\end{equation*}

A similar calculation can be made to compute $\tau_\gamma(t)$ in terms of $k_\beta(s), \tau_\beta(s)$


问:
If $f,g\in C[a,b]$, then is $\max_{a\le t\le b} |f(t) + g(t)| \le \max_{a\le t\le b} |f(t)| + \max_{a\le t\le b} |g(t)|$ true?

If $f,g\in C[a,b]$, then is $$\max_{a\le t\le b} |f(t) + g(t)| \le \max_{a\le t\le b} |f(t)| + \max_{a\le t\le b} |g(t)|$$ true?

Intuitively, this feels right. Also, I think that $$\min_{a\le t\le b} |f(t) + g(t)| \ge \min_{a\le t\le b} |f(t)| + \min_{a\le t\le b} |g(t)|$$ must be true. What if we remove the absolute values, i.e. we talk about
$$\max_{a\le t\le b} f(t) + g(t) \le \max_{a\le t\le b} f(t) + \max_{a\le t\le b} g(t)$$ and
$$\min_{a\le t\le b} f(t) + g(t) \ge \min_{a\le t\le b} f(t) + \min_{a\le t\le b} g(t)$$

Do any of these hold? I'm not able to come up with formal methods to prove/disprove these, and I'd appreciate any help! I ended up using the very first inequality in a proof, so I hope it is right. About the rest, I am only speculating, but I would love to know how to approach inequalities like these. Thanks a lot!

P.S. $C[a,b]$ denotes the set of continuous functions on $[a,b]\subset \mathbb R$.

答:
Yes, all you wrote is correct except for the second inequality (with the $\min$'s and the absolute values). To see e.g. $$\max_{a\leq t\leq b}|f(t)+g(t)|\leq\max_{a\leq t\leq b}|f(t)|+\max_{a\leq t\leq b}|g(t)|$$
note that for all $s\in [a,b]$ we have $$|f(s)+g(s)|\leq |f(s)|+|g(s)|\leq\max_{a\leq t\leq b}|f(t)|+\max_{a\leq t\leq b}|g(t)|$$
Now taking the maximum over all $s\in [a,b]$ we get the inequality above. The other inequalites can be proven in the same way.

Notice that the inequality may be strict since $f$ and $g$ can attain their maxima at different points, consider e.g. $f(t)=t, g(t)=1-t$ on $[0,1]$.

To see why the second inequality fails, take e.g. $f=-g$.


问:
Is $\sum\limits_{k=1}^{n-1}\binom{n}{k}x^{n-k}y^k$ always even?
Is
$$ f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$
always divisible by $2$?</p>

答:
Hint: Recall binomial formula
$$
(x+y)^n=\sum\limits_{k=0}^n{n\choose k} x^{n-k} y^k
$$


答:
Since $G=\langle x,y,z\rangle$ we must have $G=\langle x^{-1},y,z\rangle$ and thus $\langle T_{x^{-1}},T_{y},T_{z}\rangle$ is precisely the $G$ inner automorphism group $\text{Inn}(G)$ where  $$T_{x^{-1}}(g)=x^{-1}gx\;\;\;\;T_y=ygy^{-1}\;\;\;\;T_z(g)=zgz^{-1}$$
Therefore, to see $N\unlhd G$, it suffices to show $$T_{x^{-1}}[N]=T_y[N]=T_z[N]=N$$ but $y,z\in N=\langle y,z\rangle$ and thus $T_y[N]=T_z[N]=N$. Now, since $T_{x^{-1}}(y)=y^{-1}$ and $T_{x^{-1}}(z)=z^{-1}$, we must have $$\therefore\;T_{x^{-1}}[\langle y,z\rangle]=\langle T_{x^{-1}}(y),T_{x^{-1}}(z)\rangle=\langle y^{-1},z^{-1}\rangle=\langle y,z\rangle$$ $$\therefore\;x^{-1}\;\text{lies in the normalizer $\mathcal N(N)$ of $N$}$$
$$\therefore\;x\;\text{lies in the normalizer $\mathcal N(N)$ of $N$}$$
$$\therefore\;T_x[N]=N$$

Title: Having trouble proving that $N \trianglelefteq G$

Body:
I am having a bit trouble showing $N$ is normal in $G$.

The group $G$, of order 18, and generated by $x,y$ and $z$, of order 2, 3 and 3, respectively, satisfy the following conditions:
$$yz=zy, \quad x^{-1}yx=y^{-1}, \quad x^{-1}zx=z^{-1}$$
Let $N:=\langle y,z \rangle$.

I know that for $N$ to be normal in $G$, I have to show that $gNg^{-1} \subseteq N$ for all $g\in G$, i.e. showing $xyx^{-1}, yyy^{-1}, zyz^{-1}, xzx^{-1}, yzy^{-1}$ and $zzz^{-1} \subseteq N$. I have already noticed that $yyy^{-1}, zyz^{-1},yzy^{-1}$ and $zzz^{-1} \subseteq N$, because $N$ is generated by $y,z$, and by the given conditions $y$ and $z$ commute. Now I only have to show that $xyx^{-1}, xzx^{-1} \subseteq N$, but I am kinda stuck.

I have noticed by the given conditions, that $yz=zy \Longleftrightarrow y=zyz^{-1}$, and I have tried to substitute $y$ in $xyx^{-1}$, (and same for $z$), but that didn't get me far. I know that I am missing something simple, but I can't figure out what. Maybe somebody can help with a hint?


问:
Design of Experiment vs Sensitivity Analysis
What is the difference between the Design of Experiments (DOE) and sensitivity analysis? By reading the explanation of both the analyses it seems they are the same thing.</p>

答:
As far as I see it, Design of experiments require some sort of info on the sensitivity in order to give an answer.
By this I mean that in order to optimally select/design an input to an experiment, we need an estimate on how the output reacts to input (and how uncertainty is propagated), i.e. some sensitivity analysis.
Alternatively, if no such info is available, methods of randomization or spreading the experiments out can be said to be "experiment design".

I would however not say that sensitivity analysis is a subdicipline of DoE, since it is a broad subject.
DoE can be done hueristically from knowledge on sensitivities, but also analytically, using optimization techniques and info on your system that goes far deeper than what is "needed" for analyzing sensitivities.

As you indicate, they are connected, though with different objectives. Experiment design is about trying to make good information with little effort, sensitivity analysis seems to rather analyze your responses and/or explain interactions (and interactions is a also sub-part of DOE).


问:
Where did I mistake ? Calculate $\displaystyle\int_{-\infty}^{\infty} \dfrac{x \sin(\pi x)}{(x-3)(x-2)} dx$ using Residue Theorem.
I want to calculate $\displaystyle\int_{-\infty}^{\infty} \dfrac{x \sin(\pi x)}{(x-3)(x-2)} dx.$

Let $f(z)=\dfrac{ze^{i\pi z}}{(z-3)(z-2)}$ .

Integral Contour;

$C$ : $[-R, R]$

$C_R$ : $z=Re^{i\theta}, \theta : 0 \to \pi.$

$\displaystyle\int_{C_R} f(z) \to 0.$

$\text{Res}(f, 3)=-3, \text{Res}(f, 2)=-2$.

From Residue Theorem,
$\displaystyle\int_{-\infty}^{\infty} f(x) dx=2\pi i (-3-2)=-10\pi i.$

Therefore,
$\displaystyle\int_{-\infty}^{\infty} \dfrac{x \sin(\pi x)}{(x-3)(x-2)} dx=-10\pi.$

But according to wolfram alpha, $\displaystyle\int_{-\infty}^{\infty} \dfrac{x \sin(\pi x)}{(x-3)(x-2)} dx=-5\pi.$
https://www.wolframalpha.com/input/?i=%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D+xsin%28%5Cpi+x%29%2F%7B%28x-3%29%28x-2%29%7D+dx&lang=ja

Where did I mistake ?

答: You are trying to apply here the theorem that says that, in order to compute an integral of the sort $\int_{-\infty}^\infty f(x)e^{i\alpha x}\,\mathrm dx$, with $\alpha>0$, you see $f$ as an analytic function from $f\colon\Bbb C\setminus F\longrightarrow\Bbb C$, where $F$ is a finite subset of $\Bbb C\setminus\Bbb R$, then, if $\lim_{z\to\infty}f(z)=0$,$$\int_{-\infty}^\infty f(x)e^{i\alpha x}\,\mathrm dx=2\pi i\sum_{z_0\in F,\ \operatorname{Im} z_0>0}\operatorname{res}_{z=z_0}\left(f(z)e^{i\alpha z}\right).$$But you cannot apply it here, since your $F$ is the set $\{2,3\}$. It should be a set of complex non-real numbers.


问:
Is a recursively given series monotone?
The series is given as $a_{n+1}=\frac{8a_{n}-3}{2a_{n}+1}$ and $a_{1}=1$

I have no idea how to prove if this series is monotone or not.

I have tried rearranging the equation to form $a_{n+1}>a_{n}$ but nothing seems to work.

Please if you know how to solve this it would help a ton!

答:
$$a_{n+1}-3=\frac{2(a_n-3)}{2a_n+1}\tag1$$
$$2a_{n+1} - 1 = \frac{7(2a_n-1)}{2a_n+1} \tag 2$$

Since $a_1=1 > \frac 12>0$, from $(2)$ we know $2a_n-1>0, \forall n$. We also know $a_n < 3, \forall n$ from $(1)$.

Therefore
$$a_{n+1}-a_n = \frac{(3-a_n)(2a_n-1)}{2a_n+1}>0.\blacksquare$$

Remark 1: We derive $(1)$ and $(2)$ based on the solutions of the characteristic equation $\lambda = \frac{8\lambda - 3}{2\lambda + 1}$ which are $3$ and $\frac 12$.

Remark 2: We can derive the closed form solution for $a_n$ easily as follows

Divide $(1)$ by $(2)$ and we obtain

$$\frac{a_{n+1}-3}{2a_{n+1}-1} = \frac 27 \cdot \frac{a_n-3}{2a_n-1} \tag 3$$

Now $(3)$ is a geometric sequence,
$$
\frac{a_n-3}{2a_n-1} = \left( \frac 27 \right)^{n-1} \cdot \frac{8a_1-3}{2a_1+1}
$$
and the rest is straightforward.

Title: Is a recursively given series monotone?


问:
Decomposition of products of monomial symmetric polynomials into sums of them
I'm trying to make sense of the answer given in: this question

I am stuck at the phrase 'where the partitions $\gamma$ result from adding, respectively, from $\alpha$ all distinct partitions obtained by permuting in all possible ways the parts of $\beta$.'

If I permute (in all possible ways) the parts of a partition, then I obtain a set of integers that is not weakly decreasing anymore; adding that part-by-part to a proper partition does again not conserve the property 'weakly decreasing'.

Q: how should I interpret this?

Yet, there is a tempting structure in this product decomposition that I would like to understand.

Example: $m[1,1]^2= m[2,2]+2 m[2,1,1]+6 m[1,1,1,1]$ (* using 4 variables *)

added 19/05/2013 15:42 CET: @MarcVanLeeuwen: thanks Marc, I got it now. The 'Dim' function was new to me, it turns out to be just the multinomial of the contents (of the zero-padded partitions).

答:
It is natural to try to manipulate symmetric polynomials as one does polynomials, representing them as a sum of monomials, and to take advantage of the symmetry by representing a whole orbit of monomials only once, as a minimal symmetric polynomial$~m_\lambda$. Also one can now leave the number of indeterminates unspecified, and thus effectively manipulate symmetric functions. This being said, I should warn that the algorithms resulting form this head-on approach to symmetric functions are neither particularly beautiful nor efficient; representation on other bases with more pleasant theoretical properties seems preferable.

Anyway, the best way to understand the multiplication of $m_\mu$ and $m_\nu$ is to try deriving the rule for the coefficients yourself, since only this gives you the right feeling for what kind of complications are needed; reading a ready-made description of the multiplication risks misinterpreting the subtle notions involved. Here I will try to deduce the rule. The first thing to realise it that if $l(\lambda)$ denotes the number of nonzero parts of$~\lambda$, then in the product of symmetric functions $m_\mu m_\nu$ one get terms $m_\lambda$ with $l(\lambda)$ going up to $l(\mu)+l(\nu)$, so one should consider at least that number of indeterminates to faithfully represent that product as a polynomial. If you are only interested in multipliying symmetric polynomials in a fixed number of indeterminates one can of course use that number, which amounts to ignoring the $m_\lambda$ where $l(\lambda)$ exceeds that number.

If $n$ is the number of indeterminates used, it is useful to consider scalar multiples of the $m_\lambda$ obtained by a sum over the whole group $S_n$, namely $$M_\lambda(n)=c(\lambda,n)m_\lambda=\sum_{\pi\in S_n}X^{\pi(\lambda)},$$ where $\def\N{\mathbf N}\pi(\lambda)\in\N^n$ is the result of permuting the parts of$~\lambda$ (including $n-l(\lambda)$ parts zero) by $\pi$; the number $c(\lambda,n)$ is the size of the stabiliser of $\lambda$ in this action, the product of the factorials of the multiplicities of the distinct parts of $\lambda$, including $(n-l(\lambda))!$ for the parts equal to $0$ involved. Now clearly
$$
  M_\mu(n)M_\nu(n)=\sum_{\pi,\sigma\in S_n}X^{\pi(\mu)+\sigma(\nu)}
 =\sum_\lambda C^\lambda_{\mu,\nu}M_\lambda(n),
$$
and once these coefficients $C^\lambda_{\mu,\nu}$ are determined one deduces that
$$
  m_\mu m_\nu=\sum_\lambda c^\lambda_{\mu,\nu}m_\lambda,\qquad
 \text{where }
  c^\lambda_{\mu,\nu}=\frac{c(\lambda,n)}{c(\mu,n)c(\nu,n)}C^\lambda_{\mu,\nu}.
$$
The defining equation for the coefficients $C^\lambda_{\mu,\nu}$ is
$$
  \sum_\lambda C^\lambda_{\mu,\nu}\sum_{\pi\in S_n}X^{\pi(\lambda)}
 =\sum_{\pi,\sigma\in S_n}X^{\pi(\mu)+\sigma(\nu)}
 =\sum_{\pi,\sigma\in S_n}X^{\pi(\mu+\pi^{-1}(\sigma(\nu)))},
$$
which by symmetry can be simplified to
$$
  \sum_\lambda C^\lambda_{\mu,\nu}X^\lambda
 =\sum_{\rho\in S_n}X^{(\mu+\rho(\nu))^+},
$$
where for a composition $\alpha$ the notation $\alpha^+$ means that partition obtained by sorting its parts decreasingly.
Now we can simplify this further by factoring the stabiliser subgroup of $\nu$, to
$$
  \sum_\lambda \frac{C^\lambda_{\mu,\nu}}{c(\nu,n)}X^\lambda
 =\sum_{\alpha\in S_n(\nu)}X^{(\mu+\alpha)^+},
$$
where $S_n(\nu)$ is the permutation orbit, without repetitions. In the end, the recipe is
$$
  c^\lambda_{\mu,\nu}=\frac{c(\lambda,n)}{c(\mu,n)}c'(\lambda,\mu,\nu)
\quad\text{where}\quad
  \sum_{\alpha\in S_n(\nu)}X^{(\mu+\alpha)^+}=\sum_\lambda c'(\lambda,\mu,\nu)X^\lambda;
$$
the right hand side asks for adding all distinct permutations of $\nu$ (which was padded with zero) to $\mu$, sorting the results decreasingly and grouping by equal outcome$~\lambda$.


问:
How do you handle $f(x)$ when inside an integration?
I'd like to integrate the following, which contains a function. I'm not sure which rule I need to review? I thought maybe I could bring it out of the integral - but not sure what it would become?
$$\int^n_1\frac{f(u)}{u^{1+p}}du \Rightarrow f(?)\cdot\int^n_1\frac{1}{u^{1+p}}du \Rightarrow \bigg[\frac{u^{-1+p+1}}{-1-p+1}\bigg]^n_1\Rightarrow \bigg[-\frac{n^{-p}}{p}+\frac{1}{p}\bigg]$$

答: If it's just an arbitrary function, then you unfortunately cannot separate it outside of the integral. To spell it out, if $f(u) = u$, then clearly $\int f(u) du\neq u \int du = u \cdot (u+c) = u^2 + uc$ for a constant c, since the correct answer would instead be $\frac{u^2}{2} + c$. You can expect this incongruency with most if not all choices of f(u).


问:
How to take the derivative of $ y = xA^T + b$ w.r.t. $A$ and $b$, where $x,b$ are vectors and $A$ is a matrix
How exactly could I take the derivative of the following expression?

$$ y = xA^T + b$$

Let's say that I have $x \in \mathbb{R}^{n}$, $A \in \mathbb{R}^{m,n}$, $y \in \mathbb{R}^{m}$, and $b \in \mathbb{R}^{m}$. And, I wish to take the derivative of $y$ with respect to $A$ and $b$, i.e. $\frac{\partial y}{\partial A}$ and $\frac{\partial y}{\partial b}$. I understand that $\frac{\partial y}{\partial A}$ would a rank-3 tensor containing $\frac{\partial y_i}{\partial A_{jk}}$, although I'm not entirely sure how to get to the solution. I've tried looking through the matrix cookbook but the only other solution (that I can find at least) is for $\frac{\partial x^Ta}{\partial x}$ where $x$ and $a$ in this case are both vectors. So, I'm a little confused!

With regards to the second term, $\frac{\partial y}{\partial b}$, I would assume that this is just the identity matrix ($\mathbb{I} \in \mathbb{R}^{m \times m}$) as the terms is just element-wise addition?

Thank you in advance!

答:
$\def\E{{\cal E}}\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\v#1{\operatorname{vec}(#1)}$Doing
the calculation using index notation (i.e. element-wise) is your best option
$$\eqalign{
y_i &= A_{ij}x_j + b_i \\
\p{y_i}{b_k}
  &= \p{b_i}{b_k}
   = \delta_{ik} \quad\implies \p{b}{b} = I
\\
\p{y_i}{A_{k\ell}}
  &= \p{A_{ij}}{A_{k\ell}}x_j
   = \delta_{ik}\delta_{j\ell}\;x_j
   = \delta_{ik}\,x_\ell
\\
}$$
You could also vectorize the equation using Kronecker products
$$\eqalign{
a &\doteq \v{A} \\
y &= (x\otimes I) a + b \\
\p{y}{a} &= (x\otimes I) \p{a}{a} = (x\otimes I) I = (x\otimes I) \\
}$$
Or you could use indexed matrices (sort of a half-index notation)
$$\eqalign{
y &= Ax + b \\
\p{y}{A_{jk}} &= \left(\p{A}{A_{jk}}\right)x = E_{jk}\,x \\
}$$
where $E_{jk}$ is a matrix containing all zeros except for a single ${\tt1}$ at the $(j,k)$ element.

Or you can go into full-tensor mode by giving a name $(\E)$ to the fourth-order tensor that we encountered using index notation, i.e.
$$\eqalign{
\E_{ijk\ell} &\doteq \delta_{ik}\delta_{j\ell}
\quad\implies\quad
\p{y}{A} &= \E x \\
}$$


问:
Specific values or asymptotic behaviour of ${}_{1}F_{1}$
I'm interested in analysing the behaviour of the hypergeometric function ${}_{1}F_{1}$ for the following cases:
$${}_{1}F_{1}\Big[\frac{1}{2} - \frac{k}{2}, \frac{1}{2}; -z^2\Big]$$
and
$${}_{1}F_{1}\Big[1 - \frac{k}{2}, \frac{3}{2}; -z^2\Big]$$
for integer values, $k \in \mathbf{N}$. So far I've used some properties I found on NIST for small values of $k$ and uncovered error functions.

Can these functions be simplified more generally for this special case of integer or half-integer first arguments -- even getting some asymptotic form as $k \rightarrow \infty$ would be very welcome information!

EDIT:

It seems that I can use the identities
$$\begin{align}
{}_{1}F_{1}\Big[\frac{1}{2} - \frac{k}{2}, \frac{1}{2}; -z^2\Big] &= \textrm{e}^{-z^2}{}_{1}F_{1}\Big[\frac{k}{2}, \frac{1}{2}; z^2\Big]\\
{}_{1}F_{1}\Big[1 - \frac{k}{2}, \frac{3}{2}; -z^2\Big] &= \textrm{e}^{-z^2}{}_{1}F_{1}\Big[\frac{k}{2}+\frac{1}{2}, \frac{3}{2}; z^2\Big]
\end{align}
$$
to change sign of the first argument - does this help uncover any additional properties I can use?
Thanks for your suggestions!

答:
By http://dlmf.nist.gov/13.6.iv, we have
$$
e{}^{ - z^2 }\!{}_1F_1 \!\left( {\frac{k}{2},\frac{1}{2};z^2 } \right) \!= \frac{{2^{\frac{k}{2} - 1} }}{{\sqrt \pi  }}\Gamma\! \left( {\frac{{k + 1}}{2}} \right)\!e^{ - \frac{{z^2 }}{2}} \!\!\left( {U\!\left( {k - \frac{1}{2},\sqrt 2 z} \right) \!+ U\!\left( {k - \frac{1}{2}, - \sqrt 2 z} \right)} \right)
$$
and
$$
e{}^{ - z^2 }\!{}_1F_1 \!\left( {\frac{{k + 1}}{2},\frac{3}{2};z^2 } \right) \!= \frac{{2^{\frac{k}{2} - 2} }}{{z\sqrt {\pi } }}\Gamma\! \left( {\frac{k}{2}} \right)\!e^{ - \frac{{z^2 }}{2}} \!\!\left( {U\!\left( {k - \frac{1}{2}, - \sqrt 2 z} \right) \!- U\!\left( {k - \frac{1}{2},\sqrt 2 z} \right)} \right)\!.
$$
The parabolic cylinder functions then may be expressed in terms of the derivatives of the complementary error function (cf. http://dlmf.nist.gov/12.7.E6). There is probably no simpler "explicit" expression for these functions.


问:
Sequent calculus: What does it mean to say “choose the sequents from which we can derive some other sequent via the right conjunction rule”?
My question is probably quite silly really, but I am puzzled what this exercise is asking:

Excercise. Choose all sequents from which we can derive some other sequent via the right-side conjunction rule.

$\color{green}{\lnot (A\land B), \lnot A \vdash A\land B}$ (correct)

$\lnot (A\land B), A \vdash A\land B$

$\color{green}{\lnot (A\land B), \lnot A \vdash \lnot A \land \lnot B}$ (correct)

$\lnot (A\land B), B \vdash \lnot A \land \lnot B$

$\lnot (A\land B), \lnot B \vdash A\land B$

(Source: homework excerise in since-discontinued course, translated)

The correct answers are marked green.

Here is the referenced

Right-side conjunction rule:
$$\cfrac{\Gamma \vdash \mathcal{F}\quad \Gamma \vdash \mathcal{G}}{\Gamma\vdash \mathcal{F}\land \mathcal{G}}$$
where $\mathcal{F}, \mathcal{G}$ are zeroth order formulae, and $\Gamma$ is a list of zeroth order formulae or is empty.

I believe I do understand the basics of Gentzen's sequent calculus. I am aware of the axioms and rules for allowed derivations, including the right-side conjunction rule; the intuition behind the formalisation is also clear. I have derived dozens upon dozens of formulae in sequent calculus (as trees), and proved, for example, the correctness and completeness of sequent calculus wrt tautologies in propositional calculus.

With this in mind,

the problem is not how to solve the excercise. Rather, my problem is in understanding what is even being asked in the first place. So: What exactly is the excercise after?

Even with the “correct” answers known, I am admittedly confused. Firstly, the multiple choices are all of the form $\Omega \vdash \mathcal{F} \land \mathcal{G}$ (with varying $\Omega, \mathcal{F}, \mathcal{G}$). One usually uses the right-side conjunction rule to derive sequents of the form $\Omega \vdash \mathcal{F} \land \mathcal{G}$, not use them as “sequents from which” to derive other sequents. Secondly, what are the “some other sequents” which I should check? Others from the multiple-choice list? Other sequents in general?

Perhaps the phrasing of the exercise is a bit off from what is common... Is this a common type of exercise, maybe you can recognize the intended idea behind it? To reiterate, what is the exercise asking for?

EDIT

Here is another exercise of the same type. I also know the “correct” answer(s) to the test excercise but I will omit this information at first. That way I have an elementary check if an answer has the intended interpretation.

Test excercise.
Choose all sequents from which we can derive some other sequent via the left-side conjunction rule.

$\vdash (X \lor \lnot X) \land (Y \lor \lnot Y)$

$Z, X \land Y \vdash \lnot X \lor \lnot Y \lor \lnot Z$

$X \land (Y \lor Z), Y \land (X \lor Z) \vdash X \land Y \land Z$

$\lnot Z \land \lnot X \vdash (Y \lor \lnot Z) \land (\lnot Y \lor \lnot X)$

$X\land Y \vdash X\lor Y$

(Source: homework excerise in since-discontinued course, translated)

Left-side conjunction rule:
$$\cfrac{\Gamma, \mathcal{F}, \mathcal{G} \vdash \mathcal{H}}{\Gamma, \mathcal{F} \land \mathcal{G} \vdash \mathcal{H}}$$
where $\mathcal{F}, \mathcal{G}$ are zeroth order formulae, and $\Gamma$ is a list of zeroth order formulae or is empty.

The correct answers were indeed (spoiler)

 2 and 3.

答: Having the answer, I think I can figure out what the question was intended to be. An application of the right-$\land$ rule needs two premises, and those two premises must have the same set $\Gamma$ of formulas to the left of $\vdash$. The two green sequents have the same formulas to the left of $\vdash$, and no two others on your list of five options do that.  So the two green sequents, and only those two, can serve as the two premises in an application of right-$\land$.


答:
(Hint)
An odd number raised to any power is odd, and an even number raised to any power is even.
In particular,
$$
(x + y)^n \equiv (x + y) \pmod 2
$$
Using this along with the binomial formula, you should be able to prove the result.</p>

Title: Is $\sum\limits_{k=1}^{n-1}\binom{n}{k}x^{n-k}y^k$ always even?

Body:
Is
$$ f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$
always divisible by $2$?</p>


问:
Alternative Definition for the Derivative
In one class, we said that a function $f : \mathbb{R} \to \mathbb{R}$ is differentiable at $x_0$ if there exists a map $g : U\subset \mathbb{R} \to L(\mathbb{R}, \mathbb{R})$ continuous at $x_0$ such that
$$
f(x) = f(x_0) + g(x)(x - x_0)
$$
where $U$ is a neighborhood of $x_0$. This is obviously just a restatement of the usual definition of a derivative with $f'(x_0) = g(x_0)$, but it has distinct advantages. For example, the proof of the chain rule becomes much clearer.

Can we play a similar game in multiple variables? That is, given a function $f : \mathbb{R}^n \to \mathbb{R}^m$, can we say that it is differentiable at $x_0$ if there exists a map $g : U \subset \mathbb{R}^n \to L(\mathbb{R}^n , \mathbb{R}^m)$ continuous at $x_0$ such that
$$
f(x) = f(x_0) + g(x)(x - x_0)
$$
I tried to prove the equivalence myself, but it's not as obvious as one the one-variable case (if it's true at all), since in the one variable case we could use the fact that $\mathbb{R}$ is a field and $L(\mathbb{R} , \mathbb{R}) \cong \mathbb{R}$.

答:
One direction is clear: if such $g$ exists, then $g$ is approximately constant $g(x)=g(x_0)+o(1)$ so $f$ is approximately linear $f(x_0)+g(x)(x-x_0)=f(x_0)+g(x_0)(x-x_0)+o(\lVert x-x_0\rVert)$ which is the usual definition of $f$ being differentiable at $x_0$.

For the other direction, for $x\neq x_0$, we only know $g(x)$ on the linear span of $x-x_0$, and the easiest way to extend this to the whole of $\mathbb{R}^n$ is to make $g(x)(v)=f'(x_0)(v)$ for all $v\perp (x-x_0)$.  Then set $g(x_0)=f'(x_0)$ and it clearly works.


问:
Finding Force from acceleration at a point given constant speed
Prompt: An object of mass $m$ travels along a parabola $y = x^2$ with constant speed $5$ units/sec. What is the force on the object due to acceleration at point $(\sqrt{2}, 2)$? Recall $\vec{F} = m\vec{a}$.

This was an "applied theory" problem I got incorrect on an generated exam I recently took. This question seems to be a similar idea, excepting that I need to work backwards from speed rather than velocity and that speed is constant. Here is what I understand:

Since we're working in 2-D, $|\vec{v}(t)| = \sqrt{v_x^2 + v_y^2} = 5 \implies |\vec{v}(t)|^2 = 25 = v_x^2 + v_y^2$. I don't know if this helps me.

The parabola can be reparameterized as a trajectory/displacement curve: $\vec{r}(t) = \langle t, t^2 \rangle$

From the reparameterization, we can differentiate $\frac{d \vec{r}}{dt} = \vec{v}(t) = \langle 1, 2t \rangle$ and once more $\frac{d \vec{v}}{dt} = \vec{a}(t) = \langle 0, 2 \rangle$.

If I compute the supposed magnitude of $\vec{v}$ at point $(\sqrt{2}, 2) \Longleftrightarrow t = \sqrt{2}$, I find $$
|\vec{v}(t=\sqrt{2})| = \sqrt{1^2 + (2\sqrt{2})^2} = \sqrt{1 + 8} = \sqrt{9} = 3 \neq 5
$$ A contradiction of sorts.

Point ($3$) above would lead one to naively assume that acceleration is $2$ units/sec$^2$ upwards for all times $t$. From there, one could conclude $\vec{F} = m \langle 0, 2 \rangle = \langle 0, 2m \rangle$. Alas the automated grader reports the correct answer is:

$$\displaystyle \vec{F} = \langle-\frac{-100\sqrt{2}}{81}m, \frac{50}{81}m\rangle$$

The connection between 3 (or 9) and 81 is not lost on me ($3^4 = 9^2 = 81$), but I don't see how to bridge the gap. Any help would be greatly appreciated. I've already got my grade and the "answer". What I seek now is understanding.

Many thanks in advance.

答:
This is how you can proceed without using the notion of curvature, which is not necessary. I write some hints.

Write $\mathbf x(t)=(\alpha(t),\alpha^2(t))$

In general terms:

$\dot{ \mathbf x}=(\dot \alpha(t),2\alpha(t) \dot \alpha(t)) [1]$

Can you get the analog of [1] for $\ddot{ \mathbf x}$ ? Hint: Just differentiate [1]

Now imposing the constrain on the velocity:

$v^2=\dot{ \mathbf x}^2(t)=\dot \alpha^2(t)+4\alpha^2(t)\dot \alpha^2(t)$

So that:

$\dot \alpha^2(t)=v^2/(1+4\alpha^2(t))$ [2]

Can you get the analog of [2] for $\ddot \alpha(t)$ ? Hint: just differentiate [2].

If you did all the steps you could substitute the analog of [2] into the analog of [1] and obtain your answer :) ?


问:
Evaluate $\int\frac{da}{a\sqrt{a+1}}$

$$\int\dfrac{da}{a\sqrt{a+1}}$$

I don't know how to solve this integral. The fact that $\dfrac1a$ is the derivative of $\ln(a)$ and $\dfrac{1}{\sqrt{a+1}}$ is the derivative of $\cos^{-1}a$ suggested Integration by Parts.

$$\int\dfrac{da}{a\sqrt{a+1}} = \dfrac{\ln(a)}{\sqrt{a+1}}-\int $$
However I got stuck after this.
Any help with the Integral would be greatly appreciated. Many thanks!

答:
Note
$$\int\dfrac{da}{a\sqrt{a+1}}\overset{a=\frac1{t^2}}=-\int \frac2{\sqrt{1+t^2}}dt=-2\sinh^{-1}t+C
$$

Title: Evaluate $\int\frac{da}{a\sqrt{a+1}}$


问:
Difference between parallel transport and derivative of the exponential map
Given a Riemannian manifold $M$, let $c(t) = \exp_p(tX)$ be the geodesic emanating from $p \in M$ with initial value $X$. Let $t_0$ be small enough, then we have to ways to map $T_pM$ to $T_{c(t_0)} M$ isomorphically. One is the parallel transport along $c$, let's call it $P_{c, 0, t_0}$ and the other is given by
$$ d \exp_p|_{tX}: T_{tX}T_pM \cong T_p M \longrightarrow T_{\exp_p(tX)}M = T_{c(t_0)}.$$

My question is: What is the relation between those two?

The parallel transport is a linear isometry (in all dimensions), and the derivative of the exponential map is a radial isometry by the Gauss lemma (also in all dimensions), meaning
$$ \langle d \exp_p|_{tX} \cdot Y, \dot{c}(t) \rangle = \langle Y, X \rangle $$
for all $Y \in T_pM$.

In two dimensions, this means that the two mappings coincide up to scaling, as there is only on orthogonal direction to the radial one. However, in higher dimensions, this is not true, i guess, though I find it hard to actually compute examples...

Are there formulas which relate the two concepts with curvature terms?

\Edit: Computed that on $S^3$, conincides with the parallel transport except that vectors orthogonal to the direction of parallel transport are multiplied by $\frac{\sin r}{r}$.

答:
The relation between the differential of the exponential map $\text{d } \text{exp}$ and parallel transport $P$ (along geodesics) is expounded in the two papers (in the context of optimization on manifolds):

1 ''An accelerated first-order method for non-convex optimization on manifolds'' and

2 ''Curvature-Dependant Global Convergence Rates for Optimization on Manifolds of Bounded Geometry''

Both also provide bounds on $||\text{d } \text{exp}||$, see Proposition A.3 in 1 or Theorem 3.12 in 2.  The following is from 1:

Let $$f_{K_{\text{low}}}(t) =  \left\{ \begin{array}{lII} r^2\left(1 -
 \frac{\sin(t/r)}{t/r}\right) & \quad K_{\text{low}} = 1/r^2 > 0  \\
 \frac{t^2}{6} & \quad K_{\text{low}} = 0  \\
 r^2\left(\frac{\sinh(t/r)}{t/r} - 1\right) & \quad K_{\text{low}} =
 -1/r^2 < 0  \end{array} \right\}.$$

Let $(M, g)$ be a Riemannian manifold with sectional curvatures
bounded below by $-K$ and above by $K$.  Let $x \in M$, $s \in T_x M$,
and $\gamma(t) = \text{exp}_x (t s)$.  If $\gamma$ is defined and has
no interior conjugate point on the interval $[0,1]$, then $$||(\text{d
 } \text{exp})_x (s) - P_s)[\dot{s}]|| \leq K ||\dot{s}_{\perp}||
 f_{-K}(||s||) \quad \forall \dot{s} \in T_x M$$ where $\dot{s}_{\perp}
 = \dot{s} - \frac{\langle s, \dot{s} \rangle}{||s||^2} s$ is the component of $\dot{s}$ perpendicular to $s$, and $P_{t s}$ denotes
parallel transport along $\gamma$ from $\gamma(0)$ to $\gamma(t)$.


答:
In general, you have the Mean value theorems for definite integrals. It says that

if $f : [a, b] → \mathbb{R}$ is continuous and $g$ is an integrable function that does not change sign on $[a, b]$, then there exists $c$ in $(a, b)$ such that
$$
{\displaystyle \int _{a}^{b}f(x)g(x)\,dx=f(c)\int _{a}^{b}g(x)\,dx.}
$$

But this theorem only tells you the existence of $c$; there is no "algorithm" that tells you what $c$ is.

Title: How do you handle $f(x)$ when inside an integration?

Body:
I'd like to integrate the following, which contains a function. I'm not sure which rule I need to review? I thought maybe I could bring it out of the integral - but not sure what it would become?
$$\int^n_1\frac{f(u)}{u^{1+p}}du \Rightarrow f(?)\cdot\int^n_1\frac{1}{u^{1+p}}du \Rightarrow \bigg[\frac{u^{-1+p+1}}{-1-p+1}\bigg]^n_1\Rightarrow \bigg[-\frac{n^{-p}}{p}+\frac{1}{p}\bigg]$$


答:
It has already been pointed out in leoli1's answer that the second inequality is wrong. A correct lower bound for $\min |f(t) + g(t)| $ can be obtained by using the “reverse triangle inequality”:
$$
|f(t) + g(t)| \ge |f(t)| - |g(t)| = |f(t)| + (- |g(t)|)
$$
implies
$$
\min_{a\le t\le b} |f(t) + g(t)| \ge \min_{a\le t\le b} |f(t)| + \min_{a\le t\le b} (-|g(t)|) =  \min_{a\le t\le b} |f(t)| - \max_{a\le t\le b} |g(t)| \, .
$$

Title: If $f,g\in C[a,b]$, then is $\max_{a\le t\le b} |f(t) + g(t)| \le \max_{a\le t\le b} |f(t)| + \max_{a\le t\le b} |g(t)|$ true?

Body:
If $f,g\in C[a,b]$, then is $$\max_{a\le t\le b} |f(t) + g(t)| \le \max_{a\le t\le b} |f(t)| + \max_{a\le t\le b} |g(t)|$$ true?

Intuitively, this feels right. Also, I think that $$\min_{a\le t\le b} |f(t) + g(t)| \ge \min_{a\le t\le b} |f(t)| + \min_{a\le t\le b} |g(t)|$$ must be true. What if we remove the absolute values, i.e. we talk about
$$\max_{a\le t\le b} f(t) + g(t) \le \max_{a\le t\le b} f(t) + \max_{a\le t\le b} g(t)$$ and
$$\min_{a\le t\le b} f(t) + g(t) \ge \min_{a\le t\le b} f(t) + \min_{a\le t\le b} g(t)$$

Do any of these hold? I'm not able to come up with formal methods to prove/disprove these, and I'd appreciate any help! I ended up using the very first inequality in a proof, so I hope it is right. About the rest, I am only speculating, but I would love to know how to approach inequalities like these. Thanks a lot!

P.S. $C[a,b]$ denotes the set of continuous functions on $[a,b]\subset \mathbb R$.


问:
Why does the $T_1$ Axiom imply that a point is a closed set?
According to my reading the $T_1$ axiom say that given a topological space $X$ and $\forall x, y \in X$ We have:

$ \exists U, V \subset X: [(x \in U \land y \notin U) \land (x \notin V \land y \in V)].$

Here, $U, V$ are open sets, and they're not assumed to be disjoint.

Why does this imply that $\forall x \in X$, the set $\{ x \}$ is closed?

答: Let $x \in X$. We'll show that $\{x\}$ is closed: let $y \neq x$ be arbitrary, then there is an open set $O_y$ such that $y \in O_y$ and $x \notin O_y$. This means that $O_y \cap \{x\} = \emptyset$, so that $y$ is not an adherence point (closure point) of $\{x\}$. So no point except $x$ itself lies in the closure of $\{x\}$, and the singleton is closed.


问:
Prove or disprove these two statements.

If $f(x)$ is differentiable for every $x$, then $f(x^2)$ is continuous for every x.

If $f(x)$ is differentiable and not negative, then $\sqrt {f(x)}$ is differentiable.

My Work:

Intuition just drived me to think this is right so I tried to prove. $f(x)$ is differentiable for all $x$ then it's continuous for all $x$, is it just a matter of substituting $x^2$ and that's it? or I'm missing something?

I got this one wrong (it's a false statement), I can't still think of a function that counters it.

It would mean alot to me if you show me your thought process from when you face these type of questions, since they're my weakness and I most of the time just follow intuition and don't have any idea where to start.

Thanks in advance.

答:
For the first you should use that the composition of continuous functions is continuous to conclude the exercise.

For the second note that $x^2$ is a counterexample, as $\sqrt{x^2}=|x|$ which is clearly non differentiable.

Title: Prove or disprove these two statements.


问:
Some characterization of finite complement topology
Let $Y:=\{(u,v)\in \mathbb{R}^2: u^2+v^2\le r\}$ where $r>0$ is finite complement topology.
Now, If I want to characterize all the connected subsets of $Y$, then how should I start for such characterization?
I don not know how to formally write it.
Any help is highly appreciated.

FYI, I am learning it for the first time.

答:
First of all the finite complement topology depends only on the cardinality of the underlying set. Meaning that it doesn't matter if you consider it on $\mathbb{R}$ or your $Y$ or $[0,1]$. The finite complement topology makes them all homeomorphic.

Now assume that $X$ is an infinite set with the finite complement topology. Let $U$, $V$ be two nonempty and open subsets. By the definition $X\backslash U$ and $X\backslash V$ are finite. But since $X$ is infinite then there is $x_0\in X$ such that $x_0\not\in X\backslash U$ and $x_0\not\in X\backslash V$. Or equivalently $x_0\in U$ and $x_0\in V$. And thus any two nonempty open subsets in $X$ have nonempty intersection. In particular $X$ is connected.

On the other hand a finite set $X$ with finite complement topology is discrete. And the only discrete connected space is the singleton $\{*\}$.

Finally, every subspace of a finite complement topological space has finite complement topology. Which leads to:

A subset of a finite complement topological space $X$ is connected if and only if it is a singleton $\{*\}$ or infinite.


问:
Let $G$ be a group and $H = \{ g \in G \mid gx = xg \ \forall x \in G \}$. Show that $H \leqslant G$ and that $H$ is commutative.

Let $G$ be a group and $H = \{ g \in G \mid gx = xg \ \forall x \in G \}$. Show that $H \leqslant G$ and that $H$ is commutative.

How can I show that $H \leqslant G$? This seems to me that I would want to show that $H$ is a subgroup of $G$? That would seem to imply the statement $H \leqslant G$. So I would need to show that $H$ is non empty and closed under products and inverses?

Let $h\in H$, then $hh^{-1} = e_G$ right? So the inverse is there. How can I show the closure under products here?

答:
$H$ is the $\it{center}$ of $G$ and is denoted $\mathcal Z(G)$. Let $g$ be variable over $G$ and note $$xgx^{-1}=g\iff x^{-1}gx=g$$
$$\begin{align}&\therefore\;x,y\in H\implies g=xgx^{-1}=y^{-1}gy\implies (xy^{-1})g(yx^{-1})=x(y^{-1}gy)x^{-1}=g\\&\therefore\;x,y\in H\implies xy^{-1}\in H\\ &\therefore H\leq G\end{align}$$


问:
I need help with solving this differential equation
Solve diffential equation: $\frac{\partial f}{\partial x}-3\frac{\partial f}{\partial y}\:=\:0$

where $u = ax + y$ and $v = x$ (hint: pick a reasonable $a$)

I just started studying calculus 3, I don't know how to solve this problem. I'm looking for help, thank you.

答:
$$f_x - 3 f_y =0$$

$$f_x =f_u u_x + f_v v_x = af_u + f_v$$

$$f_y = f_u u_y + f_v v_y = f_u$$

$$af_u + f_v - 3 f_u = 0.$$

Now, let $a=3$.

$$f_v = 0.$$

Integrate with respect to $v$ so $$f=g(u), \quad \text{a function of } u \text{ only}.$$

$$f=g(3x+y).$$


问:
Sigma/Sum transformation
If you have
$$\sum_{j=2}^{n} c_{j} \times a_{j} + \left(1-\sum_{j=2}^{n}c_j\right)a_{1} $$
How can you transform it into
$$ a_{1}+ \sum_{j=2}^{n} \left(c_j \times (a_{j}-a_{1})\right)$$
I know the basics of sum calculations, but I don't understand, how to transform it. Especially with the 1-Sigma term. Can someone explane it step by step ?

Title: Sigma/Sum transformation

Body:
\begin{align}
\sum_{j=2}^{n} c_{j} \times a_{j} + \left(1-\sum_{j=2}^{n}c_j\right)a_{1} &= \left(\sum_{j=2}^{n} c_{j} \times a_{j} \right) + a_{1} - \sum_{j=2}^{n}c_{j}a_{1}\\
&=a_{1} + \sum_{j=2}^{n}c_{j}a_{j} - \sum_{j=2}^{n}c_{j}a_{1}\\
&=a_{1} + \sum_{j=2}^{n}c_{j}(a_{j} - a_{1})
\end{align}


问:
In a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ is larger than $s_2$
While solving a totally different problem , I made observation that in a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ say to $θ_1$ is larger than angle opposite to $s_2$ say $θ_2$.

I verified by drawing several triangles . Can someone tell if it's right and prove it ?

答: It is a well-known property. It is in fact Euclid's Proposition 18 in his first book of the Elements. See this link for Euclid's argument.</p>


答:
Following formula shows that your claim is correct:

$\frac a{sin (\alpha)}=\frac b{sin (\beta) }=\frac c{sin (\gamma)}=2R$

Where R is radius of circumcircle of triangle ABC with sides a, b and c and opposite angle to these sides $\alpha, \beta, \gamma$.

Title: In a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ is larger than $s_2$

Body:
While solving a totally different problem , I made observation that in a triangle , if side $s_1$ has length greater than $s_2$ , then angle opposite to $s_1$ say to $θ_1$ is larger than angle opposite to $s_2$ say $θ_2$.

I verified by drawing several triangles . Can someone tell if it's right and prove it ?


问:
Can we write each number between $1$ and $n$ in one of the forms $2^j, \ 2^j+1, \ \ldots , \ 2^{j+1}-1$?
I want to find a natural number $n$ such that $$\sum_{k=1}^{n}\frac{1}{k}\geq \frac{15}{1.1}$$

There is a hint, but I don't really know how to use this.

Hint : Write together the terms of the sum with $k=2^j, \  2^j+1, \ \ldots , \ 2^{j+1}-1$, $\ j\in \mathbb{N}_0$ give an upper bound for these.

So can we write each number between $1$ and $n$ in one of the forms $2^j, \  2^j+1, \ \ldots , \ 2^{j+1}-1$ ?

$$$$

EDIT:

Do we find it as follows?

\begin{equation*}\sum_{k=1}^{n}u_k\geq 15 \Rightarrow r\cdot \sum_{k=1}^{n}\frac{1}{k}\geq 15\Rightarrow 1.1\cdot \sum_{k=1}^{n}\frac{1}{k}\geq 15 \Rightarrow  \sum_{k=1}^{n}\frac{1}{k}\geq \frac{15}{1.1} \end{equation*}

It holds that \begin{align*}\sum_{k=1}^{n}\frac{1}{k}&=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\cdots \\ & \geq 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots\\ & =1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots\end{align*}
Für jedes $j\in \mathbb{N}_0$ folgt dann dass \begin{equation*} \sum _{k=1}^{2^{j}}{\frac {1}{k}}\geq 1+{\frac {j}{2}}\end{equation*}
So we find the smallest $j$ such that $1+{\frac {j}{2}}\geq \frac{15}{1.1} \Rightarrow j\geq 2\cdot \left (\frac{15}{1.1}-1\right ) \Rightarrow j\geq 25.27$.

So the minimum result is $2^{26}$.

Body:
I think what that are trying to say is

$\sum_{k=1}^n\frac 1k = \sum_{k=1}^{2^1-1} \frac 1k + \sum_{k=2^1}^{2^2 -1} \frac 1k + \sum_{k=2^2}^{2^3 -1} \frac 1k + .......  + \sum_{k=2^j}^{2^{j+1}-1} \frac 1k + ......$

and to consider what $\sum_{k=2^j}^{2^{j+1}} \frac 1k$.  As $2^j \le k < 2^{j+1}$ then $\frac 1{2^{j+1}} < \frac 1k \le \frac 1{2^j}$ and $2^j\cdot \frac 1{2^{j+1}} < \sum_{k=2^j}^{2^{j+1}} \frac 1k \le 2^j\cdot \frac 1{2^j}$ and we can conclude that $\frac 12 < \sum_{k=2^j}^{2^{j+1}} \frac 1k \le 1$.

Can you continue?

This assumes that if $2^m \le n < 2^{m+1}$ then

$\sum_{k=1}^n\frac 1k = \sum_{k=1}^{2^1-1} \frac 1k + \sum_{k=2^1}^{2^2 -1} \frac 1k + \sum_{k=2^2}^{2^3 -1} \frac 1k +  ......+  \sum_{k=2^{m-1}}^{2^{m}-1} \frac 1k + \sum_{k=2^m}^n \frac 1k$.

And so

$\sum_{k=1}^{2^1-1} \frac 1k + \sum_{k=2^1}^{2^2 -1} \frac 1k + \sum_{k=2^2}^{2^3 -1} \frac 1k +  ......+  \sum_{k=2^{m-1}}^{2^{m}-1} \frac 1k \le \sum_{k=1}^n\frac 1k \le \sum_{k=1}^{2^1-1} \frac 1k + \sum_{k=2^1}^{2^2 -1} \frac 1k + \sum_{k=2^2}^{2^3 -1} \frac 1k +  ......+  \sum_{k=2^{m-1}}^{2^{m}-1} \frac 1k$

$\frac 12 + \frac 12 +....+\frac 12 < \sum_{k=1}^n\frac 1k \le 1+1+ 1+...... + 1$

$\frac m2 < \sum_{k=1}^n\frac 1k \le m+1$

Anyhoo.   If $\frac m2 \ge \frac {15}{1.1}$ then $m\ge \frac {30}{1.1} > 27$.

So if $n = 2^{28}$ then that should be large enough.  After all...

$\frac 11 + $

$\frac 12 +  $

$\frac 13 +\frac 14 +$

$\frac 15 + \frac 16 + \frac 17 + \frac 18 + $

$\frac 19 + .... + \frac 1{16} + $

......

$\frac 1{2^{27}+ 1} + \frac 1{2^{27}+2} + .... + \frac 1{2^{28}} >$

$1 +$

$\frac 12+$

$2\cdot \frac14+$

$4\cdot \frac18+$

$8\cdot \frac 1{16}+$

$...... $

$2^{27}\cdot \frac 1{2^{28}} =$

$1 +\frac 12 + \frac 12 + \frac 12 + \frac 12 + ..... + \frac 12 = 1+ 28\frac 12 = 1 + 14 = 15 > \frac {15}{1.1}$.

.....

But I feel I am being a bit ham fisted.  were we expected to find a tighter range?


问:
Completing the proof of a theorem
My question is about the proof of the Theorem 1.3 which is on the page 9 of the book "Topics in Real Analysis", which is available electronically through the previous link.

I want to demonstrate that theorem in the case $\mu (X)=\infty$. In that book there's the following tip:

To extend our result to the general case observe that the finite case implies $\mu (A\cap X_j)=\tilde \mu (A\cap X_j)$ (just restrict $\mu,\tilde\mu$ to $X_j$). Hence

$$\mu (A)=\lim_{j\to\infty}\mu (A\cap X_j)=\lim_{j\to \infty}\tilde \mu(A\cap X_j)=\tilde \mu (A)$$

Below is my attempt to use that tip to prove that theorem in the case $\mu (X)=\infty$.

Let $j\in\mathbb{N}$ be any element and define $S_j:=\big\{E\cap X_j:E\in S\big\}$. It's trivial to verify that $S_j$ is a $\pi$-system (because $S$ is by hypothesis a $\pi$-system). Besides, since $X_j\in S$, then $S_j\subseteq S$ which implies that $\Sigma (S_j)\subseteq\Sigma (S)=\Sigma $ in which $\Sigma (S_j)$ means the $\sigma$-algebra generated by $S_j$.

Let $\mu _{S_j}$ and $\tilde \mu _{S_j}$, respectively, be the restrictions of $\mu$ and $\tilde \mu$ to $\Sigma(S_j)$. It's easy to see that $(X_j,\Sigma (S_j),\mu_{S_j})$ is a measure space.

The elements of $\{X_j\cap X_n\}_{n\in\mathbb{N}}$ belongs to $S_j$ and satisfy $(X_j\cap X_n)\nearrow X_j$ and $\mu_{X_j}(X_j\cap X_n)<\infty $ for all $n\in\mathbb{N}$.

Because of the above considerations and $\mu_{S_j}(X_j)<\infty$, we can use the first part of the proof of the Theorem 1.3 which is in the book mentioned above to prove that $\mu_{S_j}(E)=\tilde \mu_{S_j}(E)\,\,\color{red}{(1)}$ for all $E\in \Sigma (S_j)$.

Let $A\in\Sigma $ be any element.

If there's a $j\in\mathbb{N}$ such that $A\cap X_j\in \Sigma (S_j)$, then I can use $(1)$ and that tip above to prove that $\mu (A)=\tilde \mu (A)$. However, if $A\cap X_j\,\,{\color{red}{\notin}} \,\,\Sigma (S_j)$ for all $j\in\mathbb{N}$, then I can't use $(1)$ and, therefore, that tip is useless.

Please help me to finish the proof. At least indicate some reference that contains the proof of that theorem.

Thank you for your attention!

答:
Actually, the $S_j$ you've defined are the best they can be, in the sense that the condition you speak of as a problem, cannot happen! Basically , your work so far is correct, and what I'm going to show is that for every $A \in \Sigma$, we have $A \cap X_j \in \Sigma(S_j)$ for every $j$.

Why can't that condition happen? Well, $S$ generates $\Sigma$. $S_j$ is just every element of $S$ intersected with $X_j$.  Now because any intersections, unions etc. behave nicely with respect to just an intersection with some $X_j$, the sigma-algebra generated by $S_j$ creates every possible intersection of an element of $\Sigma$ , and $X_j$. In one line : $S$ approximates $\Sigma$, and intersecting with an $X_j$ (or any set in $\Sigma$) should still retain the approximation.

The precise statement is this :

$$
\Sigma = \{F \in \Sigma : F \cap X_j \in \Sigma(S_j) \ \ \forall j = 1,2,...\}
$$

That is, there's no set $A \in \Sigma$ such that $A \cap X_j \notin \Sigma(S_j)$ for any single (forget about all) $j$. Note that $\Sigma(S_j)$ is a sigma-algebra of subsets of $X_j$ by construction (you are using this because the probability space is set up that way). Therefore $X_j \in \Sigma(S_j)$.

To prove this, note that $S$ is a subset of the RHS by definition of $S_j$. So we only need to prove that the RHS is a sigma-algebra. Let's call the RHS as $T$.

Note that $\emptyset \cap X_j = \emptyset \in \Sigma(S_j)$ for all $j$ and $X \cap X_j = X_j \in \Sigma(S_j)$. So $0,X \in T$.

Let $F \in T$. We know that $F \cap X_j \in \Sigma(S_j)$ for all $j$. Now, remember that $\Sigma(S_j)$ are sigma-algebras of $X_j$, so when I take a complement, $(F \cap X_j)^c = F^c \cap X_j$ for each $j$, because I am taking the complement corresponding to each $X_j$. Therefore, $F^c \cap X_j \in \Sigma(S_j)$ for all $j$ and thus $F^c \in T$.

Let $F_i \in T$ be countably many sets. Then , note that :
$$
\left(\cup_{i=1}^\infty F_i\right) \cap X_j = \cup_{i=1}^\infty(F_i \cap X_j)
$$
now if I fix a $j$, then $F_i \cap X_j \in \Sigma(S_j)$ for all $i$, so from the above we get $(\cup_{i=1}^\infty F_i) \cap X_j \in S_j$ for all $j$ so $\cup_{i=1}^\infty F_i \in T$.

Therefore, $T$ is a sigma-algebra, so $T$ contains $\Sigma(S) = \Sigma$. So $T = \Sigma$.

Now, let $A \in \Sigma$. Then $A \in T$. So $A \cap X_j \in \Sigma(S_j)$ for all $j$. From the result for $\Sigma(S_j)$, we know that $\mu(A \cap X_j) = \tilde{\mu}(A \cap X_j)$ for every $j$. Now $A \cap X_j \uparrow A$, you can conclude.

I actually ran into a very similar issue when I dealt with a similar problem. It turns out that this has to do with a far more general statement concerning pull-backs of sigma-algebras.

I state that as follows :

Let $f : (S,\mathcal S) \to (F , \mathcal F)$ be measurable. Let $\mathcal E \subset \mathcal F$. Then, we have :
$$
f^{-1}(\Sigma(E)) = \Sigma(f^{-1}(E))
$$

That is, pulling back every set in $\Sigma(E)$ to give a sigma-algebra is equivalent to pulling back every set of $E$ and generating a sigma-algebra out of that. We can use this result to prove the above equality of sigma-algebras very easily, using the inclusion map from $X_j \to X$ for each $j$. Use this to prove that $\Sigma(S_j) = \{A \cap X_j : A \in \Sigma\}$ for each $j$, which of course leads to the theorem I wrote down.

Being comfortable with simple lemmas and theorems around sigma-algebras is very important if you want to far into probability, especially when conditioning, martingales etc. come into play.

Title: Completing the proof of a theorem


问:
The total number of subarrays
I want to count the number of subarrays for a vector (not combinations of elements).

Ex.

```<code> A[1,2,3]
</code>```

It has 6 subarrays :

```<code>{1}, {2}, {3}, {1,2}, {2,3}, {1,2,3}
</code>```

I think that for a vector of N elements the total number of subarrays is N*(N+1)/2.

I am not able to prove it, can someone do it?

答:
Trying to explain in layman terms.

Let's say f(0) = 0

```<code>f(1) = 1
f(2) = 3
f(3) = 6
f(4) = 10
f(5) = 15
</code>```

By observation, you can see that each result is just an addition of previous result and current number.

```<code>f(n) = n + f(n-1)
</code>```

So with this formula let's expand f(5).

```<code>f(5)
=> 5 + f(4)
=> 5 + 4 + f(3)
=> 5 + 4 + 3 + f(2)
=> 5 + 4 + 3 + 2 + f(1)
=> 5 + 4 + 3 + 2 + 1 + f(0)
=> 5 + 4 + 3 + 2 + 1 + 0
===> This is equal to sum of n numbers = n(n+1)/2
</code>```

Title: The total number of subarrays

Body:
Trying to explain in layman terms.

Let's say f(0) = 0

f(1) = 1
f(2) = 3
f(3) = 6
f(4) = 10
f(5) = 15

By observation, you can see that each result is just an addition of previous result and current number.

f(n) = n + f(n-1)

So with this formula let's expand f(5).

f(5)
=> 5 + f(4)
=> 5 + 4 + f(3)
=> 5 + 4 + 3 + f(2)
=> 5 + 4 + 3 + 2 + f(1)
=> 5 + 4 + 3 + 2 + 1 + f(0)
=> 5 + 4 + 3 + 2 + 1 + 0
===> This is equal to sum of n numbers = n(n+1)/2


问:
Homology Groups of $S^n$
Is it possible to consider $S^n$ as a $0$-simplex and a singular map of an $n$-simplex so that the $n$-simplex forms the surface of $S^n$ minus a point, and the point is the singular map of the $0$-simplex? Can a (singular map of an) $n$-simplex have zero boundary for $n>1$? I am trying to understand the homology groups of $\mathbb{R}^n$ with finitely many points removed. I imagine I could use the Mayer-Vietoris exact sequence and homotopy invariance, but I want to try to find the groups directly from the definition to confirm that I am right. So, I have started by showing that $X=\mathbb{R}^n\backslash\{x_1,\ldots,x_m\}$ is homotopy equivalent (deformation retracts) to the wedge sum of $m$ copies of $S^{n-1}$ meeting at a common point, as in the diagram for $n=3,m=4$.

答: Certainly you can construct $S^n$ as a cell complex with one 0-cell and one $n$-cell, and compute cellular homology that way. Many algebraic topology books use this as their first example of computing cellular homology.

Title: Homology Groups of $S^n$


问:
Do infinity and zero really exist?
From the first day that I entered college, I was wondering about the relationship between some basic mathematical abstract concepts and nature. I'm going to explain them here and you may find them a little bit opinion-based, but it's not. So please guide me in case you have good scientific-based answers/examples.

Infinity:
Does infinity exist in reality? We all know it is an abstract concept but what I'm curious to know is if there are any physical phenomena out there that can show, stimulate or somehow help us understand the concept of infinity in reality/nature/the physical world.

Zero:
What about Zeno's paradox? In nature (our physical world) there is a "smallest distance". It's about $1.6 \times 10^{-35}$ meters. Another example would be when someone says "there are three apples on the desk, if you take all of them there are $0$ apples on the desk". Obviously, it's an abstract concept, but what I would like to know is that if there is any observable physical event that can anyhow show us the concept of nothingness or absolute zero. E.g., we don't have absolute zero temperature in thermodynamics, absolute zero distance between 2 points in mechanics, or absolute zero gravity in a given space, etc.

Axiom of empty set: This axioms states:

"There is a set such that no element is a member of it."

One can imagine an empty set in nature as an absolutely empty box. But one can see it the other way around: the element "nothing" is there (quoting from Prof. Lawrence Krauss's book "nothing is something"). So, the same way one can say the element "nothing" is a member of a set. Obviously, it's an abstract concept and this may look like a play on words, but it's also an interesting paradox.

$\mathbb R$?! (the set of real numbers): There are millions of mathematical theorems which are based on $\mathbb R$! I wonder if there are any non-countable physical phenomena in the real world/nature?

Again, I know these mathematical concepts are just abstractions and help us to solve real-world problems! However, I'm more interested in the relation between these abstract concepts and our physical world/nature.

答:
Claim: At least one of $0$, $\infty$ exists.

Proof: If $\infty$ exists, we are done.  If not, the number of instances of $\infty$ is $0$.

I am actually somewhat serious.  This is an adaptation of a certain philosopher's (unfortunately I cannot remember who) ontological proof of the existence of the empty set, which is itself an extraordinarily perceptive and witty riff on St. Anselm's ontological argument for the existence of God.  Strangely, while theists and atheists alike tend to find St. Anselm's argument intriguing but not convincing (I seem to recall that Anselm's paper begins by apologizing -- to God -- for making the argument), when you invert it to prove the existence of nothingness, it seems quite convincing!

Title: Do infinity and zero really exist?


问:
Lagrange's Theorem for further elementary consequences
Question: Let $G$ be a finite group, and let $H$ and $K$ be subgroup of $G$. Prove: suppose $H$ and $G$ are not equal, and both have order the same prime number $p$, Then $H\cap K=\{e\}$.

This is my proof steps:

Proof: $G$ is finite, $H, K$ subgroup of $G$. therefore $H\cap K$ subgroup of $G$, also is subgroup of both $H$ and $K$
Then the order of $H\cap K$ is a common divisor of order of $H$ and the order of $K$. since $p$ is prime number. Show that $G$ is cyclic group and that any a belong to $G$ is its generator.  Therefore $H,K, H\cap K$ are cyclic group too, because every subgroup of a cyclic group is cyclic. Since $H$ not equal $G$, Then …..
I just did here, because I feel I miss some part. Can anyone give me some advises! Thanks so much

答:
You can't assume $G$ is cyclic, nor can you prove it, becasue not all finite groups are cyclic.

Your proof is very good up to where you say the order of $H \cap K$ is a common divisor of the orders of both the groups $H$ and $K$.  At this point, note that the only divisors of a prime $p$ are 1 and $p$ itself.  What does this mean about $H \cap K$?  How many elements can it have?


问:
$\operatorname{tr}(BA)$ where $A$ is low rank and has a subset of eigenvalues from $B^{-1}$
Most of the question is in the title.
Suppose that I have $B$ a square $N\times N$ positive definite matrix, and $A$ a low-rank square matrix with its non-zero eigenvalues being a subset of the spectrum of $B^{-1}$. Is there a simple way to compute the trace : $\mathrm{tr}(BA)$?

My intuition is that if $M$ is the rank of $A$, then $B$ and $A$ can share all $N$ eigenvectors and that one can diagonalize $A$ and $B$ and get automatically $\mathrm{tr}(BA) = \sum_{i:\lambda_i>0} \frac{\lambda_i}{\lambda_i} = M$ where $\lambda_i$ are the eigenvalues of $A$. But I am not sure how to obtain the proper ordering etc.

Thanks for your help!

答:
Without further information, there is no way to use the eigenvalues of $A$ in this computation. However, we can make use of the fact that $A$ is low-rank. In particular, if $A = CF$ is a rank factorization (so that $C$ is $N \times r$ and $F$ is $r \times N$), then we have
$$
\operatorname{tr}(BA) =
\operatorname{tr}(BCF) = \operatorname{tr}(FBC).
$$
That is, we may rewrite the expression as the trace of an $r \times r$ matrix.


问:
Number of roots of the equation $f(xf(x)) = \sqrt{9 - x^2f^2(x)}$

Given the following graph of a $5$th degree polynomial $y = f(x)$:

Find the number of solutions of the equation: $$f(xf(x)) = \sqrt{9 - x^2f^2(x)}$$

My attempt was to figure out what $f(x)$ is, by noticing the roots of its derivative based on the graph to know $f'(x)$, then find $\int f'(x)$, but the function I found did not match the original graph. (Even if I founded, I think it would be very messy to plug it in the equation)

One other idea is that let $t = xf(x)$, then the equation become:

$$f(t) = \sqrt{9-t^2}$$

But I can't process from here.

Is my idea correct? If yes, then how do I continue? Or is there a better way to solve this?

答:
$f(x)=a_5x^5+\cdots+a_0\Rightarrow xf(x)=a_5x^6+\cdots+a_0x$ has degree $6$ so $$F(x)=f(xf(x))=a_5(a_5x^6+\cdots+a_0x)^5+\cdots+a_1(a_5x^6+\cdots+a_0x)+a_0$$ has degree $30$. We have $$(F(x))^2=9-(xf(x))^2$$ from which the resulting polynomial has degree $60$.

Thus there are $\color{red}{60}$ roots, not all distinct maybe because of possible multiplicity.


问:
How can one characterise the subsets of $R^d$ with non-decreasing indicator function?
I am looking to characterise the subsets of $R^d$ for which the indicator function is non decreasing.

For example, if $d=1$, then I think that the subsets of $R$ with non-decreasing indicator function are of the form $(r, \infty)$ or $[r, \infty)$ for $r \in R$.

My question is, can this characterisation be generalised to $R^d$ where a function is nondecreasing if it is nondecreasing for all of it's elements.

i.e. $\forall i, \epsilon > 0$ $f(x_1,..., x_i + \epsilon, ... x_d ) \geq f(x_1,..., x_i, ... x_d )$.

答:
The transitive, reflexive closure of your relation defines the following order: for any two points $x=(x_1, \dots, x_d),\; y=(y_1, \dots, y_d) \in \mathbb R^d$,
$$x \leq y \Leftrightarrow \forall 1 \leq i \leq d,\; x_i \leq y_i $$

Hence, it is not too hard to see that for any subset $A \subseteq \mathbb R^d$, the indicator function $\chi_A$ is non-decreasing if and only $A$ is a axis-parallel polytope of the form $(r_1, +\infty) \times \dots \times (r_d, +\infty)$, each of those being potentially closed interval, with $r_i \in \mathbb R \cup \{-\infty\}$

This is because as soon as you have a point $r=(r_1, \dots, r_d) \in A$, your order imposes that all the polytope of which $r$ is the "smallest corner" is included in $A$.

In dimension 2, this would just mean that if $r \in A$, then all the points above and on the right of it are also in $A$. By extension, you get this whole quarter of the plane.


问:
Inequality involving magnetic derivative
I need to understand the following inequality:

\begin{align*}
\|\partial_j \psi\|_2 &\leq \|D_j\psi\|_2 + \|A_j\|_6\|\psi\|_3 \\ &\leq \|D_j\psi\|_2 + C \sum_{k=1}^3\|\partial_k A_j\|_2\sum_{k=1}^3\|\partial_k \psi\|^{1/2}_2\|\psi\|^{1/2}_2
\end{align*}

We are in $\mathbb{R}^3$ and $D_j = \partial_j-iA_j$, $j=1,2,3$. The first inequality follows from triangle and Hölder inequalities but I have a hard time seeing the second one. I tried Sobolev embedding but that gives (for example) $\|A_j\|_6 \leq \|A_j\|_{H^1}$ and not $\|A_j\|_6 \leq \|\nabla A_j\|_2$. For the other terms I tried $$\|\psi\|_3 =\|\psi^{1/2}\psi^{1/2}\|_3 \leq \|\psi^{1/2}\|_{12}\|\psi^{1/2}\|_4 = \|\psi\|^{1/2}_6\|\psi\|_2^{1/2}$$ by Hölder but again I don't know how to establish $\|\psi\|_6\leq \|\nabla\psi\|_2$.

答:
I will just assume that the $A_j$ are smooth enough. Then the Gagliardo-Nirenberg-Sobolev inequality
$$
||u||_{L^{p^*}(\mathbb{R^n})} \leq  C ||Du||_{L^{p}(\mathbb{R^n})}
$$
gives you
$$
||A_j||_{L^{6}(\mathbb{R^n})} \leq C \sum_k ||\partial_k A_j||_{L^{2}(\mathbb{R^n})}
$$
since the exponents match perfectly: Here $p^*$ is the usual Sobolev conjugate
$$
p^*=\frac{np}{n-p}  \iff 6=\frac{2\cdot3}{3-2}
$$
The second part follows from the Hölder interpolation inequality (I found it in my German textbook, shouldn't be too hard to find)
$$
||\psi||_{L^{q}(\mathbb{R^n})} \leq ||\psi||_{L^{p}(\mathbb{R^n})}^{\theta} ||\psi||_{L^{p^*}(\mathbb{R^n})}^{1-\theta}
$$
Plugging in the exponents, we have
$$
\theta=\frac{\frac{1}{q}-\frac{1}{p}}{\frac{1}{p^*}-\frac{1}{p}}=\frac{1/3-1/2}{1/6-1/2}=\frac{1}{2}
$$
And so
$$
||\psi||_{L^{3}(\mathbb{R^n})} \leq ||\psi||_{L^{2}(\mathbb{R^n})}^{\frac{1}{2}} ||\psi||_{L^{6}(\mathbb{R^n})}^{\frac{1}{2}}
$$
Then you acn again use the first inequality to see that
$$
||\psi||_{L^{3}(\mathbb{R^n})} \leq C^{\frac{1}{2}}\sum_k||\psi||_{L^{2}(\mathbb{R^n})}^{\frac{1}{2}} ||\partial_k \psi||_{L^{2}(\mathbb{R^n})}^{\frac{1}{2}}
$$
Combining all the estimates you have your desired estimates.
A note on the inequality:

Gagliardo-Nirenberg-Sobolev inequality only holds for $C_c^{\infty}$; however, you can find a smooth, compactly supported sequence of functions $\psi_i \to \psi$ in $W^{1,p}(\mathbb{R}^n)$. Then you have
$$
||\psi_i -\psi_m||_{L^{p^*}(\mathbb{R^n})} \leq  C ||D\psi_i -D\psi_m||_{L^{p}(\mathbb{R^n})}
$$
and so the sequence also converges in $L^{p^*}(\mathbb{R^n})$, since Cauchy sequences converge. Take limits and conclude.


问:
Surjective endomorphisms of finitely generated modules are isomorphisms

My Problem: Let $M$ be a finitely generated $A$-module and $T$ an endomorphism. I want to show that if $T$ is surjective then it is invertible.

My attempt:
Let $m_1,...,m_n$ be the generators of $M$ over $A$. For every $b = b_1 m_1 + ... + b_n m_n$ with $b_i \in A$ there is $a = a_1 m_1 + ... + a_n m_n$ with $a_i \in A$ such that
$$ T(a)=b $$
or in matrix-vector notation
$$ T \vec{a} = \vec{b} $$
where $\vec{x}$ is the column vector of $x_1,...,x_n$ where $x = x_1 m_1 + ... + x_n m_n$. I multiply by the adjugate matrix to get
$$ \mathrm{adj}(T) \vec{b} = \mathrm{adj}(T) T \vec{a} = \det(T) I_n \vec{a} = \det(T) \vec{a} \ .
$$
Now take $\vec{b}=0$. Then $\vec{0} = \det(T) \vec{a}$ and thus $T$ is injective if and only if
$\det(T)$ is not a zero divisor.

If I prove that $T$ is injective, then I'll get it is invertible. For that, I think the way is to prove that $\det(T)$ is not a zero divisor.

The importance of finitely generated condition:

Let $M = A^{\aleph_0} =\{ ( a_1 , a_2 , ... ) \mid a_i \in A \}$ be a not finitely generated $A$-module. Let $T : M \to M$ defined by
$$ T(a_1, a_2, a_3, ... ) = (a_2, a_3, ... ) \ . $$
Then clearly $T$ is surjective but not injective ($\ker T = \{ ( a , 0 , 0 , ... ) \mid a \in A \}$),
and thus not invertible.

The importance of surjective and not injective condition:

Need to find a counter-example.

答:
Consider the commutative ring $B\subset\mathrm{End}_A(M)$ generated by $T$ and all multiplication by $a\in A$ maps, i.e. the maps $\mathsf{m}_a:M\to M, m\mapsto am$. Let $m_1,\dots,m_n$ generate the $A$-module $M$. We can find a matrix $(b_{ij})\in M_n(A)$ with $$\forall j\in\{1,\dots,n\},\quad Tm_j=\sum_{i=1}^n b_{ij}m_i$$ The usual "multiplication by the comatrix" trick done in the commutative ring $B$ shows that $T$ satisfies a monic polynomial equation $$T^n+\sum_{i=0}^{n-1}b_i T^i=T^n+\cdots+(-1)^n\mathsf{m}_\beta=0$$ with coefficients $b_i$ in $A$ and where we wrote $\beta=\det(b_{ij})$.

Since $T$ is onto, the $Tm_i$ generate $M$ and there is a matrix $(a_{ij})\in M_n(A)$ with $$\forall j\in\{1,\dots,n\},\quad m_j=\sum_{i=1}^n a_{ij}Tm_i$$ so that $$(a_{ij})(b_{ij})\begin{pmatrix}m_1\\\vdots\\m_n\end{pmatrix}=\begin{pmatrix}m_1\\\vdots\\m_n\end{pmatrix}$$ and so (again using the comatrix trick) $\mathsf{m}_\alpha\circ\mathsf{m}_\beta=\mathrm{id}_M$ where $\alpha=\det(a_{ij})$. Therefore, composing the monic polynomial relation from earlier with $\mathsf{m}_\alpha$, we get $$\mathrm{id}_M=ST=TS$$ with $S=(-1)^{n-1}\alpha\big(T^{n-1}+\sum_{i=1}^{n-1}b_i T^{i-1}\big)$ showing that $T$ is invertible with inverse $S$.


问:
CLT for $P\{X_k = \pm 2^k\} = \frac{1}{2}$
I want to show that the following family of random variables either satisfy or don't satisfy the central limit theorem $P\{X_k = \pm 2^k\} = \frac{1}{2}$.

For the CLT I know that I want to use the Lindeberg condition. However, I'm having trouble understanding the definition given in equation 5.6 of Feller's book volume 1.

答:
Let $S_n=\sum_{k=1}^n X_k$. Then
$$\require{cancel}
\frac{S_n}{\sqrt{\text{Var }S_n}}
=\frac{\sum_{k=1}^n X_k}{\sqrt{\sum_{k=1}^n2^{2k}}}
=\frac{2^{n+1}\sum_{k=1}^nX_k2^{-(n+1)}}{\sqrt{(2^{2(n+1)}-1)/3}}
= \cancelto{\to 1}{\color{gray}{\frac{2^{n+1}}{\sqrt{2^{2(n+1)}-1}}}}\sqrt{3}\cdot\sum_{k=1}^n X_{n+1-k}2^{-k}
$$
Now, notice that $X_{n+1-k}2^{-(n+1)}\sim \text{Unif}\{+2^{-k},-2^{-k}\}$. Therefore, $S_n/\sqrt{\text{Var }S_n}$ converges in distribution to
$$
\sqrt{3}\sum_{i=1}^\infty \xi_i\left(\frac12\right)^i, \qquad \xi_i\sim \text{Unif} \{-1,+1\}\;\text{i.i.d}
$$
One can verify this summation is not normally distributed, so CLT does not apply. In fact, it is distributed like $2C-1$, where $C$ has the Cantor distribution.


问:
Showing set inclusion holds via topology
I am trying to show the following set inclusion holds for $X:=\mathbb{R}$,

$\mathscr{A}:=\left\{A\subseteq X:\exists U\in\tau\text{ such that } 0\in U\subseteq A\right\}
\subseteq
\left\{A\subseteq X:X\setminus A\text{ is finite}\right\}=:\mathscr{A}',$

where the basic nhbds at the origin will be the sets: $(-\epsilon, \epsilon) \cup (-\infty, -n) \cup (n, \infty)$, for all possible choices $\epsilon > 0$ and $n \in \mathbb{N}$.

This is something I need to hold true, as part of a larger problem. My attempt is as follows,

Let $A\in \mathscr{A}$.

By definition of the set,

$$\exists U\in\tau \text{ such that } 0\in U\subseteq A.$$

By definition of the basic nhbds at the origin,

$$U:=\left(\bigcup_{\epsilon>0}(-\epsilon, \epsilon)\right) \cup (-\infty, -n) \cup (n, \infty).$$

Therefore, $X\setminus A\subseteq X\setminus U$, and by De Morgan,

\begin{align*}
      X\setminus U&=\left(\bigcup_{\epsilon>0}(-\epsilon, \epsilon)\right)^c \cap (-\infty, -n)^c \cap
      (n,\infty)^c. \\
\end{align*}

This is where I get stuck. I see that if I can show that $X\setminus U$ is finite, then I can conclude that $X\setminus A$ is finite and thus $A\in \mathscr{A}'$ giving the desired inclusion. I have a feeling I may be taking $U$ to be of the wrong form however.

答: The claim is not true: for instance $(-1/2,1/2)\cup(-\infty,-1)\cup(1,\infty)\in\mathscr{A}\setminus\mathscr{A}^\prime$.


问:
Is $\sum\limits_{k=1}^{n-1}\binom{n}{k}x^{n-k}y^k$ always even?
Is
$$ f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$
always divisible by $2$?</p>

答:
In fact, Goos and Norbert have given the answer. (And you should also assume $n\in \mathbb{N}$)

$$
f(n,x,y) = (x+y)^n - x^n -y^n
$$

If

both $x$ and $y$ are even: even - even - even = even;

both $x$ and $y$ are odd: even - odd - odd = even;

$x$ is even and $y$ is odd: odd - even - odd = even;

$y$ is odd and $x$ is even: just like the case above.

So, $f(n,x,y)$ is always even.


问:
Equivalent definitions of convex hull
$\newcommand{\N}{\mathbb{N}}$
The convex hull $H$ of a subset $A$ of $X$ is defined as
$$
H = \{z \in X~|~\exists x,y \in A:\exists t \in [0,1]:z=(1-t)x+ty\}
$$

Then, can we show that any convex combination of elements of $A$ is in $H$? That is,

$$
\forall n \in \N: P(n)
$$
where
$$
P(n) \Leftrightarrow \forall (x_i)_{i \leq n} \subseteq A: \forall (t_i)_{i \leq n} \subseteq [0,1]:
\sum_{i=1}^n{t_i = 1} \Rightarrow \sum_{i=1}^n {t_i x_i} \in H
$$

I tried to show by mathematical induction:

First, it is obvious that $A \subseteq H$ because we can pick an element of $A$ twice; for all $x \in A$, $(1-t)x+tx = x \in H$.

Let $n = 1$. Then, $x_1 \in A$ and $t_1 = 1$ implies that $\sum_{i=1}^n {t_i x_i} = t_1 x_1 =x_1 \in A \subseteq H$. Therefore, $P(1)$ holds.

Now, let $n$ be arbitrary and suppose $P(n)$ holds. Let $(x_i)_{i \leq n+1} \subseteq A$, $(t_i)_{i \leq n+1} \subseteq [0,1]$, and $\sum_{i=1}^{n+1}t_i = 1$.

Case 1) $t_{n+1} = 1$. Then, $t_i = 0$ for all $i \leq n$, which gives $\sum_{i=1}^{n+1}t_ix_i = x_{n+1} \in A \subseteq H $.

Case 2) $t_{n+1} \neq 1$. From $\sum_{i=1}^{n+1}t_i = \sum_{i=1}^{n}t_i + t_{n+1} = 1$, we have $\sum_{i=1}^{n}\frac{t_i}{1-t_{n+1}} = 1$. Thus, by $P(n)$, $\sum_{i=1}^{n}\frac{t_i}{1-t_{n+1}}x_i \in H$. However, we cannot conclude that
$$
\sum_{i=1}^{n+1}t_i x_i = (1-t_{n+1})\sum_{i=1}^{n}\frac{t_i}{1-t_{n+1}}x_i + t_{n+1}x_{n+1} \in H
$$
because it is not guaranteed that
$$
\sum_{i=1}^{n}\frac{t_i}{1-t_{n+1}}x_i \in A
$$
whereas $x_{n+1} \in A$.

答:
There is a major problem with your definition of the convex hull $H$: it is not a convex set!

As an example, let $A$ be 3 points in the plane that form the vertices of a triangle. Now, all of the edge points are included in $H$ because they are a convex combination of two of the vertices. However, no interior point of the triangle can be written as a convex combination of two of the vertices (since a convex combination of two points is collinear with them). This is a problem: now the line connecting the midpoints of two edges is not contained in $H$, breaking convexity.

I realized this issue while thinking about the end of your argument. Because you showed that $y := \sum_{j = 1}^n \frac{t_j}{1 - t_{n + 1}} x_j \in H$, you'd be done if you could conclude
$$(1 - t_{n + 1})y + t_{n + 1}x_{n + 1} \in H$$
using the convexity of $H$. So I tried to verify that $H$ is convex and failed for a while.

Title: Equivalent definitions of convex hull


问:
Maximum value of $f(x,y)=e^{-y} - \ln x$ on $(x,y) \in (0,1) \times (0,1).$
Let us consider a function $~f(x,y)$ defined by $$f(x,y) = e^{-y} - \ln x,~~ (x,y) \in [0,1] \times [0,1].$$ Now calculate the minimum and maximum of $f(x,y)$ on $[0,1] \times [0,1]$.

My Attempt: For minimum, $~\lim_{y \to 1} e^{-y} = \frac{1}{e}$ and $~\lim_{x \to 1} -\ln x = 0$ and this follows that minimum value of $f(x,y)$ is $\frac{1}{e}.$ Now I am not able to find maximum but the answer is given $2$. Please help me to find this.

答: The function does not have (absolute) maximum because it is unbounded above (note that $f(x,y)\rightarrow +\infty$ as $x\rightarrow 0$).


问:
Show that $A = \{0, 1, 1/2, 1/3, 1/4,...,1/n...\}$ is compact
I write down my gull answer for this question, however my main question is on this part:"Therefore for each number 1/n of this sequence we can associate one open set $G_k$ such that 1/k $\in G_k". How can I be sure that for any arbitrary open cover I can pick one open set that has one of these elements?

Fell free to point out if something else is wrong..

Show that $A = \{0, 1, 1/2, 1/3, 1/4,...,1/n...\}$ is compact

Let first review our definitions:

Finite - A set B is finite if it is empty or if there is a bijection with domain B and range in an initial segment ot N.

Compactness - A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover. More explicitly, the requirement is that if $\{G_a\}$ is an open cover of K, then there are finitely many indices $a_1, ..., a_n$ such that K $\subset G_a1\ \cup...\cup G_an$

Open Set - E is open if every point of E is an interior point of E

Interior point - A point p is an interior point of E if there is a neighborhood N of p such that N

proof (Direct Method)

Consider the open cover of the set a to be G = $\{G_a\}$ and let $G_0$ be the open set such such that 0
$\in G_0$. By the definition of open set, we know that there is a B(0, r) with r $>$ 0 around 0, such that 1/$n_0$ $\in$ B(0, r) = (0 - r, 0 + r) $\subset$ $G_0$ for some $n_0 > 1/r$ in the naturals. By the definition of finite, we know that the sequence $\{1/n\}$ for n $>$ $n_0$ have a finite number of elements. Therefore for each number 1/n of this sequence we can associate one open set $G_k$ such that 1/k $\in G_k$ considering that this sequences is finite the union of $G_0$ with all open sets $G_k$ for k $>$ 0 is a finite open cover.

答: Your idea and answer are correct. More generally, if $X$ is a metric (topological) space and $\{x_n\}_{n=1}^\infty$ is a sequence in $X$ with $\lim_n x_n = x$, then $\{x\}\cup \{x_n: n \ge 1\}$ is compact.


问:
Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable.
Show that $X :=$ {$0, 1$}$^{\mathbb R}$ is not first-countable.

According to Munkres Topology:
Definition: A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\beta$.
A space that has a countable basis at each of its points is said to be first countable.

I can also proceed indirectly by showing that there exists a real-valued function on some subspace of $[0,1]^{\mathbb R}$ that is sequentially continuous but not continuous. It is the case of $F(f) = \int_0^1 f(t) dt$ where $f$ is measurable. But I don't know how to prove that there exists such a function. Any help please?

答:
You can also argue indirectly as follows:
$\{0,1\}^\mathbb R$ is compact because it is a product of compacts, by Tychonoff's theorem.  However, it is not sequentially compact (see this question). Since first countable, compact spaces are sequentially compact (see this question), it follows that $\{0,1\}^\mathbb R$ is not first countable.


问:
Why do rectangular matrices have no inverse that can solve Ax = v?
We can reverse a linear transformation using an inverse matrix to find the input vector used in the original linear transformation:

$$A^{-1}A\vec{x} = A^{-1}\vec{v}$$

Let's say we have a 2D vector that we transform into a 1D "vector":
$$\begin{pmatrix}
a & b
\end{pmatrix}*\begin{pmatrix}
x  \\
y
\end{pmatrix} = \begin{pmatrix}
ax + by
\end{pmatrix}       $$

Isn't there some matrix that can then convert that 1D vector back into the original 2D vector? Something like this:

$$\begin{pmatrix}
c  \\
d
\end{pmatrix}*\begin{pmatrix}
ax + by
\end{pmatrix}  = \begin{pmatrix}
x  \\
y
\end{pmatrix}      $$

My TA says that the original transformation does not have an inverse because a single input vector would map to multiple  output vectors. But I'm not seeing those multiple outputs in my algebra.

答: Some rectangular matrices have left inverses which can be used to solve the equation $Ax=v$ where $v$ is in the span of the columns of $A$. Specifically, if $A$ is a rectangular matrix of full rank having more rows than columns, then the "left inverse" of $A$ equals $(A^TA)^{-1}A^T$ so that $$Ax=v \implies x=(A^TA)^{-1}A^Tv$$ This is only true if $v$ belongs to the column space of $A$. On the other hand, if $v\notin \text{Col}(A)$, then $x^*=(A^TA)^{-1}A^Tv$ happens to be the least squares solution to the inconsistent system $Ax=v$ i.e. it's "the best you can do," and $Ax^*$ is the projection of $b$ onto the span of the columns of $A$.


答:
By the definition of open cover, $A\subset\bigcup G_\alpha$. So, if $a\in A$, there is ome $\alpha$ such that $a\in G_\alpha$.

And your proof is fine.

Body:
I write down my gull answer for this question, however my main question is on this part:"Therefore for each number 1/n of this sequence we can associate one open set $G_k$ such that 1/k $\in G_k". How can I be sure that for any arbitrary open cover I can pick one open set that has one of these elements?

Fell free to point out if something else is wrong..

Show that $A = \{0, 1, 1/2, 1/3, 1/4,...,1/n...\}$ is compact

Let first review our definitions:

Finite - A set B is finite if it is empty or if there is a bijection with domain B and range in an initial segment ot N.

Compactness - A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover. More explicitly, the requirement is that if $\{G_a\}$ is an open cover of K, then there are finitely many indices $a_1, ..., a_n$ such that K $\subset G_a1\ \cup...\cup G_an$

Open Set - E is open if every point of E is an interior point of E

Interior point - A point p is an interior point of E if there is a neighborhood N of p such that N

proof (Direct Method)

Consider the open cover of the set a to be G = $\{G_a\}$ and let $G_0$ be the open set such such that 0
$\in G_0$. By the definition of open set, we know that there is a B(0, r) with r $>$ 0 around 0, such that 1/$n_0$ $\in$ B(0, r) = (0 - r, 0 + r) $\subset$ $G_0$ for some $n_0 > 1/r$ in the naturals. By the definition of finite, we know that the sequence $\{1/n\}$ for n $>$ $n_0$ have a finite number of elements. Therefore for each number 1/n of this sequence we can associate one open set $G_k$ such that 1/k $\in G_k$ considering that this sequences is finite the union of $G_0$ with all open sets $G_k$ for k $>$ 0 is a finite open cover.


问:
If $ab+bc+ca\ge1$, prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$
The following problem is from CHKMO 2018 Problem 1:

If $ab+bc+ca\ge1$, prove that $$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{\sqrt{3}}{abc}$$

I tried to use Cauchy–Schwarz inequality, by try multiplying different things, such as $1^2+1^2+1^2$, $(abc)^2+(abc)^2+(abc)^2$. But I still can’t solve it. Can someone help me?

答:
Just as what was done in Post 1, we can substitute;
$$a=\frac{1}{x}$$
$$b=\frac{1}{y}$$
$$c=\frac{1}{z}.$$

Then, we have that;
\begin{align*}
& \text{ } \text{ } \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } ab+bc+ca \geq 1\\
&\implies \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\geq 1\\
&\implies \frac{xyz}{xy}+\frac{xyz}{yz}+\frac{xyz}{zx}\geq xyz\\
&\implies x+y+z\geq xyz.\hspace{7.7cm} \rightarrow 1
\end{align*}

We can now proceed as in this link Linked Question

We know by the QM-AM inequality that;
\begin{align*}
\sqrt{\frac{x^2+y^2+z^2}{3}}
&\geq \left(\frac{x+y+z}{3}\right)^2\\
&\geq \left(\frac{xyz}{3}\right)^2.\hspace{2cm} \left(\text{Note that }x+y+z\geq xyz\right)\\
\end{align*}

Squaring both sides and multiplying by 3 gives;
$$x^2+y^2+z^2\geq \frac{x^2y^2z^2}{3}$$

\begin{align*}
\implies\left(x^2+y^2+z^2\right)^\frac{1}{4}
&\geq \left(\frac{x^2y^2z^2}{3}\right)^\frac{1}{4}\\
&\geq \frac{\sqrt{xyz}}{3^\frac{1}{4}}.\hspace{6cm} \rightarrow 2\\
\end{align*}

By AM-GM, we know that;

$$\frac{x^2+y^2+z^2}{3}\geq \sqrt[3]{x^2y^2z^2}$$
$$\implies x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}$$
$$\implies \left(x^2+y^2+z^2\right)^3\geq 3^3\left(x^2y^2z^2\right)$$
\begin{align*}
&\implies \left(x^2+y^2+z^2\right)^\frac{3}{4}\geq 3^\frac{3}{4}\sqrt[4]{x^2y^2z^2}\\
&\implies \left(x^2+y^2+z^2\right)^\frac{3}{4}\geq 3^\frac{3}{4}\sqrt{xyz}.\hspace{5.05cm} \rightarrow 3
\end{align*}

We can now multiply Inequality 2 and 3 to have;

\begin{align*}
x^2+y^2+z^2
&=\left(x^2+y^2+z^2\right)^\frac{1}{4}\times \left(x^2+y^2+z^2\right)^\frac{3}{4}\\
&\geq \frac{\sqrt{xyz}}{3^\frac{1}{4}}\times 3^\frac{3}{4}\sqrt{xyz}\\
&\geq xyz\sqrt{3}.
\end{align*}

Substituting back the original variables yields the desired result;
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\sqrt{3}\frac{1}{abc}$$
$$\implies\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq\frac{\sqrt{3}}{abc}.$$