答: See the answer is in question itself. The question says 'duplicates' so the two A's are identical. And if we have to choose different letters, we can eliminate the duplicates and find the no. of ways just like you did in the first method.</p>

Title: How many ways are there to choose 4 letters out of the following 8 letters without duplicates?

Body:
There are eight letters: $A BCD EF AG$

How many ways can I choose four letters without duplicates?

(1) Eliminating the duplicate letter $A$, no of ways is ${7 \choose 4} = 35$

(2) No of ways to choose $4$ out of $8$ letters is ${8 \choose 4} = 70$

$\quad $ No of duplicates is ${(8-2) \choose 2} = 15$

$\quad $  No of ways is $70 - 15 = 55$

Both seem right, I don't know why and which one of them is wrong?
Thank you.


问:
Calculate area of the surface $z = kxy/(1-x)$
Calculate area of the surface:
$$z = kxy/(1-x)$$
(https://www.geogebra.org/3d/uca6ehtc)

It's domain is restricted: $0\leq x\leq 1, 0\leq y\leq 1-x$

$k$ is just a parameter controlling its streching. I need a formula or maybe an integral to calculate the area of the surface.

答:
You can use the following formula to calculate the area of a surface:
$$\iint_{S} dS$$
And you can find $dS$ as follow:
$$dS = \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dA$$

So you end up with the following formula:
$$ \iint_{S} dS = \iint_{D} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dA $$

And for your problem it's:
$$Area \space of \space z = \iint_{S} dS = \int_{0}^{1}\int_{0}^{1-x} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2} dydx $$

You can read this also for more explanations:
MIT18_02SC_MNotes_v9.3to4

ContentLicense: CC BY-SA 4.0


问:
Find Taylor polynomial of an integral function
I ran into this problem and I'm absolutely clueless. The problem is:

Let $f$ be: $$f(x)=\int_0^{\ln(1+2x)} \frac{e^t}{1+t^2}\;dt$$ Find the Taylor polynomial of order $1$ around $x=0$.

The only idea I've had is that you could write $f(x)= F(ln(1+2x))-F(0)$, and then I could use that $f'(x)=F'(ln(1+2x))-F'(0)$, but whenever I try that I always just get $0$.

Any help is greatly appriciated.

答:
MacLaurin expansion is the following
$$f(x)=f(0)+f'(0)x+O(x^2)$$
If we have
$$f(x)=\int_0^{\ln(1+2x)} \frac{e^t}{1+t^2}\;dt$$
then $f(0)=0$, and by Leibniz integral rule
$$f'(x)=\frac{e^{\log (2 x+1)}}{\log ^2(2 x+1)+1}\cdot \frac{2}{2 x+1}=\frac{2 }{\log ^2(2 x+1)+1}$$
and $f'(0)=2$.

Thus expansion is $f(x) = 2x +O(x^2)$


问:
Proof about twice differentiable function.
I've been restudying Real Analysis (in one dimension) and got stuck on one problem. I thought that my proof was correct, but, when I checked the solution, the proof that I saw was way more complicated than mine, so I'm assuming mine was incorrect. With that being said, I would like to know why my solution fails to be correct (or if that's actually the case). The problem is the following:

Let $f: [a, + \infty) \rightarrow \mathbb{R}$ be twice differentiable. Prove that if $\lim_{x \rightarrow \infty} f(x)=f(a)$, then there exists $x \in (a, + \infty)$ such that $f''(x)=0$.

Here is my proof:
If $f$ is constant, we must have $f(x)=f(a)$. Hence, $f''$ vanishes everywhere, and we are done.

Let's then suppose that $f$ is not constant. If $f'(a)$ was both the global maximum and minimum of $f'$, then, for every $x \in [a, + \infty)$, we would have
$$f'(a) \leq f'(x) \leq f'(a).$$ That is, $f'$ is constant, and $f(x)=mx+b$ for some $m, b \in \mathbb{R}$ with $m \neq 0$ (since we are assuming $f$ not to be constant). Therefore, $\lim_{x \rightarrow \infty} f(x)=+ \infty$ or $\lim_{x \rightarrow \infty} f(x)=- \infty$, a contradiction. Thus, $f'(a)$ cannot be both the global maximum and minimum of $f'$. In other words, $f'$ attains an extremum at some point $x \in (a, + \infty)$; but since $f'$ is differentiable and $x$ is an interior point of $(a, + \infty)$, we conclude that $f''(x)=0,$ as desired.

答:
Lemma. (Theorem of Darboux) If $f: [a,b]\to\mathbb R$ is differentiable, and $f'(a)<0<f'(b)$, then there exists a $c\in (a,b)$, such that $f'(c)$.

If $f''(x)\ne 0$, for all $x\in(a,\infty)$, then according to the above result, either $f''(x)>0$, for all $x\in(a,\infty)$ or $f''(x)<0$, for all $x\in(a,\infty)$.

Assume that $f''(x)>0$, for all $x\in(a,\infty)$. (The case, $f''(x)<0$, for all $x\in(a,\infty)$ is treated similarly.) Then $f'$ is strictly increasing, and hence there are three possibilities:

I. $f'(x)>0$, for all $x\in(a,\infty)$.

II. $f'(x)<0$, for all $x\in(a,\infty)$.

III. There is a $b>a$, such that $f'(x)<0$, for all $x\in(a,b)$ and $f'(x)>0$, for all $x\in(b,\infty)$.

In the cases I and III, $f$ tends necessarily to infinity, as $x\to\infty$.
Observe that in I, $f'(x)>f'(a+1)>0$, for $x>a+1$, and hence
$f(x)>(x-a-1)f'(a+1)+f(a+1)\to\infty$.
Similarly, in III, $f'(x)>f'(b+1)>0$, for $x>b+1$, and hence $f(x)>(x-b-1)f'(b+1)+f(b+1)\to\infty$.

In case II, $f$ is strictly decreasing, and hence
$$
\lim_{x\to\infty}f(x)\le f(a+1)<f(a).
$$
So, all there cases lead to contradiction, and hence there exists a $x\in(a,\infty)$, where $f''(x)=0$.


问:
Finding the antiderivative of $\frac 1{(1-x^m)^n}$,with $n,m\in\Bbb N$

For $m,n \in \Bbb N$, find the antiderivative of $g:(0,1)\rightarrow\mathbb{R}$ defined by:
  $$g(x)=\frac{1}{(1-x^m)^n}$$

Mathematica gives a result with functions we didn't learn about yet. The question is correct, and  it's over my head. Some ideas might help me find a possible alternative.

答:
The reason Mathematica gives a result with Hypergeometric Functions is that you probably didn't specify $m,n$ had to be natural numbers. That being said, it would appear your integral can be written in the form (when $m$ is even):

$$\sum_{i=1}^{n-1} \frac{\alpha_i x}{(x^m - 1)^i} +\beta \log\left( \frac{x+1}{x-1}\right) + \gamma \tan^{-1} x $$
This can be achieved by expanding the denominator and writing it out as a sum of partial fractions.

ContentLicense: CC BY-SA 3.0


问:
Topologies properties of a closed disc with arcs on the boundary identified
Fix a natural number $n$.

Let $D^2$ denote the unit disc in $\mathbb{R}^2$ and $S^1$ the boundary circle, and consider the quotient space $X=D^2/ \sim$ where we have the quotient on the boundary $$(\cos \theta, \sin \theta) \sim \left(\cos\left(\theta + \frac{2r\pi}{n}\right), \sin\left(\theta + \frac{2r\pi}{n}\right)\right)$$

In other words, we are identifying points on arc of length $\frac{2\pi}{n}$. Denote by $Y$ the image of $S^1$ under the quotient map.

Two questions. First, I'd like to prove that the inclusion $Y \hookrightarrow X \setminus \{(0,0)\}$ is a deformation retract. I'd also like to compute the fundamental group $\pi_{1}(X, x_0)$ for some basepoint $x_0 \in X \setminus Y$.

Some thoughts. I don't think a simple deformation retract such as $((x,y),t) \mapsto (x,y)(1-t) + t\frac{(x,y)}{|(x,y)|}$ works because we have continuity issues where the angle of $(x,y)$ (in polar coordinates) is close to $\frac{2r\pi}{n}$ for some integer $r$. I've tried constructing more complicate retracts, but with little progress.

答:
Let us first understand the identification of points.

We regard $S^1$ as subspace of $\mathbb C$. Then the equivalence relation $\sim $ means $z \sim \zeta^k_n z$ where $\zeta_n = e^{2\pi i/n}$ is the $n$-th root of unity with smallest positive argument. The set of the $\zeta^k_n$ is the set of all $n$-th roots of unity.

In other word, we have $z \sim z'$ if and only if $z^n = (z')^n$. The "only if" part is obvious. For the converse note that $z^n = (z')^n$ means $(z/z')^n = 1$, i.e. $z/z'$ is an $n$-th root of unity.

We conclude that $X = D^2/\sim$ is nothing else than the adjunction space $C^2(n) = D^2 \cup_{\mu_n} S^1$, where $D^2$ is attached to $S^1$ by $\mu_n : S^1 \to S^1,\mu_n(z) = z^n$. In fact, let $D^2 + S^1$ be disjoint union, $p : D^2 + S^1 \to D^2(n)$ be the quotient map and $i : D^2 \to D^2 + S^1, j : S^1 \to D^2 + S^1$ be the canonical embeddings of the two summands. Since $\mu_n$ is surjective, we have $p(D^2) = C^2(n)$. The map $q = p \circ i : D^2 \to C^2(n)$ is a surjection and $j' = p \circ j : S^1 \to C^2(n)$ is an embedding. $C^2(n)$ is obtained from the disjoint union $D^2 + S^1$ by identifying $z \in S^1 \subset D^2$ with $z^n \in S^1$ in the second summand. This means that $z,z' \in D^2$ are sent to the same point in  $C^2(n)$ iff $z, z' \in S^1$ and $z^n = (z')^n$. Thus $q(z) = q(z')$ iff $z \sim z'$. Hence $q$ induces a continuos bijection $ \bar q : X \to C^2(n)$. Since both space are compact Hausdorff, $\bar q$ is a homeomorphism. It also shows that $\bar q$ identifies $Y$ with $j'(S^1) \approx S^1$.

$C^2(n)$ is the mapping cone of $\mu_n$. This can be regarded as the mapping cylinder $Z(\mu_n)$ of $\mu_n$ with the cone $CS^1 = D^2$ attached to the top of $Z(\mu_n)$ (i.e. the points of the base $S^1 \times \{0\}$ of $CS^1$ are identified with the point of the top $S^1\times \{1\}$ of $Z(\mu_n)$). Thus removing $0$ from $D^2/\sim$ is the same as removing the tip of $CS^1$ from $Z(\mu_n) \cup CS^1$. The resulting space deformation retracts to $Z(\mu_n)$ and the latter deformation retracts to the base $S^1$ of $Z(\mu_n)$. This answers your first question. If you consider about the above constructions, you also see that your suggestion of a (strong) deformation retraction works in fact. To see this directly, note that you actually defined a map
$$H : (D^2 \setminus \{0\} )\times I \to D^2 \setminus \{0\} , H(z,t) = (1-t)z + t \frac{z}{\lvert z \rvert}$$
which is clearly a strong deformation retraction onto $S^1$. If $z \sim z'$ with $z \ne z'$, we must have $z, z' \in S^1$, hence $H(z,t) = z \sim z' = H(z',t)$ for all $t$. Going to the quotients, we see that $H$ induces a strong deformation retraction of $X \setminus \{0\} $ onto $Y \approx S^1$.

Your second question can answered by the Seifert-van Kampen theorem. We have $C^2(n) = A \cup B$ with $A  = C^2(n) \setminus \{\text{tip of } CS^1\}, B = C^2(n) \setminus j'(S^1)$, where $A \simeq S^1$, $A \cap B \approx S^1 \times (0,1) \simeq S^1$ and $B \approx D^2 \setminus S^1$. We get
$$\pi_1(C^2(n)) \approx (\pi_1(A) * \pi_1(B)) / N$$
where $N$ is the normal subgroup generated by all words of the form $i_A(g)i_B(g)^{-1}$ with $g \in \pi_1(A \cap B)$ and $i_A: \pi_1(A \cap B) \to \pi_1(A)$ and $i_B: \pi_1(A \cap B) \to \pi_1(B)$ induced by the inclusions $\iota_A: A \cap B \to A, \iota_B: A \cap B \to B$. But $\pi_1(B) = 0$ and thus all  $i_B(g)^{-1} = 0$. Moreover $\pi_1(A) \approx \mathbb Z \approx \pi_1(A \cap B)$. We conclude that
$$\pi_1(C^2(n)) \approx \pi_1(A)/G ,$$
where $G$ is the (normal) subgroup generated by all $i_A(g)$ with $g \in \pi_1(A \cap B)$. Since $\pi_1(A) \approx \mathbb Z$, $G$ is nothing but the image of $i_A$. If we identity $\pi_1(A)$ with $\mathbb Z$, then $G  = k \mathbb Z$ for some $k$. Hence $\pi_1(C^2(n)) \approx \mathbb Z_k$. Let us show that $k = n$.

In fact we have a commutative diagram
$\require{AMScd}$
\begin{CD}
\pi_1(S^1) @>{f_*}>> \pi_1(Z(\mu_n))  @>{}>> \pi_1(A) \\
@A{(\mu_n)_*}AA @A{}AA @A{i_A}AA \\
\pi_1(S^1) @>{g_*}>> \pi_1(S^1 \times \{1\}) @>{}>> \pi_1(A \cap B) \end{CD}
where all unnamed arrows are the inclusion-induced homomorphism, $f: S^1 \to Z(\mu_n)$ embeds of $S^1$ as the base of the mapping cylinder and $g$ is the obvious homeomorphism. Note that all horizontal arrows are isomorphisms.

But is clear that the image of $(\mu_n)_*$ is generated by $(\mu_n)_*([id]) = [\mu_n] = n [id]$.


问:
Finding a limit at negative infinity
$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$

Hello,

I know that the answer to this is $- \infty$, but I'm having trouble showing it. Can anyone please point me in the right direction?

Thanks!

答:
Let $x = -\frac{1}{t}$, now as
$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$ we can say,
as $x$ tends to $-\infty$, $t$ would tend to $0$. Note the negative sign.
Change the problem as:
$$\lim_{t \to 0}{\frac{1}{t^2} - \frac{2}{t^7}}$$
Solve to get:
$$\lim_{t \to 0}{\frac{t^5-2}{t^7}} = -\frac{2}{0} = -\infty$$
Hope this helps...

Title: Finding a limit at negative infinity


问:
How many ways are there to choose 4 letters out of the following 8 letters without duplicates?
There are eight letters: $A BCD EF AG$

How many ways can I choose four letters without duplicates?

(1) Eliminating the duplicate letter $A$, no of ways is ${7 \choose 4} = 35$

(2) No of ways to choose $4$ out of $8$ letters is ${8 \choose 4} = 70$

$\quad $ No of duplicates is ${(8-2) \choose 2} = 15$

$\quad $  No of ways is $70 - 15 = 55$

Both seem right, I don't know why and which one of them is wrong?
Thank you.

答:
Alternative approach.

ABCDEFAG

Imagine each element is a building block.  So normally, there are $\binom{8}{4}$ ways of choosing 4 of the 8 blocks.

When you outlaw having more than 1 A, the physical effect is equivalent to taking the two A blocks and gluing them together, so that you can't use them separately.  Now you only have 7 blocks, so the number of ways is $\binom{7}{4}.$


问:
Did I calculate the size of this group right?
given the group $$\Omega = \{\left((P_1,F_1),...,(P_n,F_n)\right) \ : \ \forall \ j \ , \ P_j \in \{\mathsf{right},\mathsf{left},\mathsf{center}\} \ , \ F_j \in \{\mathsf{eng},\mathsf{sci}\}\}$$

I calculated that $$|\Omega|=  6^n$$

this is because for every pair there is 6 options and than I multiple all the $n$ pairs and getting $$6^n$$ options.

if this is true, why do I multiple the pairs and not summing them up? what is the intuition behind that?

答:
$6^n$ is correct.

Consider $n = 1$. The elements of $\Omega_1$ are:

(right, eng)

(left, eng)

(center, eng)

(right, sci)

(left, sci)

(center, sci)

There are six of them.

Consider $n = 2$. The elements of $\Omega_2$ are:

$((\mbox{right}, \mbox{eng}), \omega)$

((right, eng), (right, eng))

((right, eng), (left, eng))

...

((right, eng), (center, sci))

$((\mbox{left}, \mbox{eng}), \omega)$

((left, eng), (right, eng))

((left, eng), (left, eng))

...

((left, eng), (center, sci))

...

$((\mbox{center}, \mbox{sci}), \omega)$

((center, sci), (right, eng))

((center, sci), (left, eng))

...

((center, sci), (center, sci))

That is to say, $\Omega_2 = \left( \Omega_1 \right)^2$.

We have six choices for the first pair, and for each of these, we have six choices for the second pair.

In general, $\Omega_{n} = \left( \Omega_1 \right)^{n}$.


答: Multiplication occurs when you are trying to find the cardinality of a Cartesian product or a special subset of a Cartesian product. Addition occurs when there is casework, meaning you are working with the partition of a set. In your case, there is enough flexibility that each pair is counting a $3\times 2$ size Cartesian product and the tuple of pairs is counting a $$\underbrace{6\times 6\times \cdots \times 6}_{n \text{ copies}}$$ Cartesian product.

Title: Did I calculate the size of this group right?

Body:
given the group $$\Omega = \{\left((P_1,F_1),...,(P_n,F_n)\right) \ : \ \forall \ j \ , \ P_j \in \{\mathsf{right},\mathsf{left},\mathsf{center}\} \ , \ F_j \in \{\mathsf{eng},\mathsf{sci}\}\}$$

I calculated that $$|\Omega|=  6^n$$

this is because for every pair there is 6 options and than I multiple all the $n$ pairs and getting $$6^n$$ options.

if this is true, why do I multiple the pairs and not summing them up? what is the intuition behind that?


问:
Let X be a set on which the cyclic group with 3 elements acts. If $|X| = 5$, show that it must have at least 2 fixed points.
I do not understand how the cyclic group acts on a set. As I know it's a group generated by a single element, but I do not understand how its group action is defined on a set. If you could help me with the action I can try to approach the problem on my own.</p>

答:
Denote by $G= \{ 1, g, g^2\}$ the group with three elements, which is generated by $g$.

Suppose $x \in X$ is not a fixed point. This means that
$gx \neq x$. Moreover
$$g(g^2 x) = g^3 x = 1x = x$$
This means that $g^2x \neq x$.

In a similar way we can prove that $g^2x \neq gx$. In other words, we have an orbit with three distinct elements.
$$Gx = \{ x, gx, g^2x\}$$ has three elements which are not-fixed.

Since orbits are disjoint each other, this means that the set of non-fixed points has cardinality divisible by $3$ (it's a disjoint union of orbits with three elements).

Now, the set of non fixed points can have 0 elements or 3 elements.
Taking the complement, since $|X|=5$, the set of fixed points can have 5 elements or 2 elements.

In both cases, there are at least two fixed points.


问:
What is wrong with this way of calculating a volume element?
I know that what follows is wrong and I know one needs to use the jacobian to calculate a new volume element $dV$ after a change of variables. However, I don’t understand why the following goes wrong:

Say we have the following change of variables (i.e. we could be doing a multivariable integral):
$$
x_i = x_i(u_0,...,u_n),
$$
where as usual $x_i$ is the $i$th component of some vector $x$ (same for $u$). One learns that the differential of $x_i$ is given by
$$
dx_i= \frac{\partial x_i}{\partial u_j}du_j,
$$
where summation is implied. Then, if $dV$ in $x$ is given by
$$
dV=\Pi_i dx_i
$$
why cant we say
$$
dV = \Pi_i dx_i = \Pi_i (\frac{\partial x_i}{\partial u_j}du_j).
$$

My instinct tells me the problem might lie in the definition of $dV=\Pi_i dx_i$, if incorrect please explain why.

Thanks!

答:
If you are familiar with differential forms the answer is simply that the volume element is a decomposable top form, i.e. the wedge product of $n$ 1-forms. In Euclidean space with the usual coordinates $x^i$ $d V= dx^1\wedge \cdots \wedge dx^n$ and if you plug in the transformation law $dx^i \mapsto \sum _j \partial x^i /\partial u ^j du^j$ you get out a factor $\det (\partial x^i /\partial u^j)$. Duh...

More intuitively and working in 2 dimensions for ease, suppose you have a small surface area of sides $dx$, $dy$ centred at $(x,y)$. Its "volume" (well, area here) is $dx \times dy$, where $\times$ is just good old standard multiplication. Now suppose you work in polar coordinates and the surface element is centred at $(r,\theta)$, and it "extends" by $d\theta$ on the angular part and by $dr$ in the radial one. By drawing a figure you should be able to convince yourself that you have a small portion of an annulus whose "volume" is (approximatively) $rd\theta \times dr$.

Reasoning in terms of volume it should be possible to get the general expression without making use of the language of differential forms, but I am too lazy to try right now. Differential forms are a convenient language as the determinant comes out automatically so you don't have to think hard about the geometry described by the coordinates you are using.

Finally, it may be helpful to point out why Cartesian coordinates behave so nicely: it is because the $dx^i$ are orthonormal stuff - where stuff and hence orthonormal can be made precise but the intuitive meaning should be clear, after all the Euclidean metric is $dx^2+dy^2 +\cdots$. Any other orthonormal stuff (coframe) $e^i$ would lead to a similarly nice volume element $e^1\wedge\cdots\wedge e^n$. Since in 2D $dx^2+dy^2=dr^2+r^2 \sin^2\theta d\theta^2$, an orthonormal coframe is $e^1=dr$, $e^2 = r \sin\theta d\theta$.

If a manifold admits an orthonormal frame which is also coordinate, $e^i=dx^i$, near every point then that manifold is flat. In other words coordinate and orthonormal are both nice properties but you usually cannot have both!


问:
example where $E(E(X \mid \mathcal{F}_1) \mid \mathcal{F}_2) \neq E(E(X \mid \mathcal{F}_2) \mid \mathcal{F}_1)$
Given $\Omega = \{ a,b,c\}$, I want to construct an example where $E(E(X \mid \mathcal{F}_1) \mid \mathcal{F}_2) \neq E(E(X \mid \mathcal{F}_2) \mid \mathcal{F}_1)$. This is from Durrett's probability theory and examples.

Let $\mathcal{F}_1 = \sigma(\{a\}) = \{ \{a\}, \{b,c\}, \{a,b,c\}, \emptyset\}$ and let $\mathcal{F}_2 = \sigma(\{ c\}) = \{\{c\}, \{a,b\}, \{a,b,c\}, \emptyset \}$. Let $X(a) = X(c) = 0$ and $X(b) = 1$, and $P(a) = P(b)= P(c) = 1/3$. Then in the text, we apparently get
$$E(X\mid \mathcal{F}_1)(a) = 0,\quad  E(X\mid \mathcal{F}_1)(b) = 1/2, \quad E(X\mid \mathcal{F}_1)(c) = 1/2 $$
$$E(E(X\mid \mathcal{F}_1)\mid \mathcal{F}_2)(a) = 1/4, \quad E(E(X\mid \mathcal{F}_1)\mid \mathcal{F}_2)(b) = 1/4, \quad E(E(X\mid \mathcal{F}_1)\mid \mathcal{F}_2)(c) = 1/2$$
I think what the book did here is use $E(X \mid \mathcal{F}) = {E(X ; \Omega_i) \over P(\Omega_i)}$ on $\Omega_i$, so for $\{b,c\}$,

$$E(X\mid \mathcal{F}_1) (\{b,c\}) = {E(X ; \{b,c\}) \over P(\{b,c\})} = {1/3 \over 2/3} = {1 \over 2}$$
But how are $E(X\mid \mathcal{F}_1)(b) = 1/2$, $E(X\mid \mathcal{F}_1)(c) = 1/2$ when $\{b\}$ and $\{c\}$ are not even in $\mathcal{F}_1$. Even if we assume $E(X\mid \mathcal{F}_1)(c)$ can be computed, shouldn't $E(X\mid \mathcal{F}_1)(c) = 0$ since $E(X; \{c\}) = 0$?

答:
$E(X|\mathcal{F}_1)$ is by definition $\mathcal{F}_1$- measurable, so we have

$E(X|\mathcal{F}_1)(\omega)=\alpha {1}_{\{a\}}(\omega) + \beta 1_{\{b,c\}}(\omega)$ for all $\omega \in \Omega$ and some $\alpha, \beta\in \mathbb{R}$. Moreover, it easily follows from the definition of the conditional expectation that

$\beta=\frac{E(X \cdot 1_{\{b,c\}} )}{P(\{b,c\})}=\frac{1\cdot \frac{1}{3}+0\cdot \frac{1}{3}}{2/3}=\frac{1}{2}$.

Therefore, $E(X|\mathcal{F}_1)(b)=E(X|\mathcal{F}_1)(c)=\beta=\frac{1}{2}$.


问:
Antisymmetric relation from a real life
Would you know some examples of antisymmetric relation from a real life? That is, relation $$(x,y)\in R\quad
 \text{and}\quad (y,x)\in R \rightarrow  x=y.$$ Thanks for your help.

答:
The three most common elementary ones in my experience are:

Subset relation, like for geometric locus results

Real number inequalities

Divisibility of positive integers

These are all applicable in real life as long as the mathematical concepts model what you are looking at.

Title: Antisymmetric relation from a real life


答:
Let $G$ be a group and $A$ be a non-empty set. Then, a group action is a function $G \times A \to A$ such that

$e \cdot a = a, \forall a \in A$.

$g_1 \cdot(g_2 \cdot a) = (g_1g_2)\cdot a, \forall g_1,g_2 \in G \text{ and } a \in A$.

This means, any function $G\times A \to A$ satisfying these two conditions is a group action of $G$ on a set $A$. So, the question does not ask for a spesific action, but asks for all possible actions.

Also, we say a point $a \in A$ is fixed by $G$ if $g \cdot a = a, \forall g \in G$.

HINT: Orbit Stabilizer Theorem implies that for a finite group $G$ acting on a set $A$, we have $|G_a|\cdot|O_a| = |G|$ for any element $a \in A$, where $O_a$ is the orbit of $a$ and $G_a$ is the stabilizer of $a$. And a point $a$ is fixed if $|G_a| = |G|$ (or equivalently $|O_a| = 1$). Considering orbits are partitions of set $A$, can you prove your statement?

Title: Let X be a set on which the cyclic group with 3 elements acts. If $|X| = 5$, show that it must have at least 2 fixed points.

Body: I do not understand how the cyclic group acts on a set. As I know it's a group generated by a single element, but I do not understand how its group action is defined on a set. If you could help me with the action I can try to approach the problem on my own.</p>


问:
Why is my approach the wrong method?
The equation $\sin^4x-2\cos^2x+a^2=0$ can be solved if :

My approach was to substitute $y$ as $\sin^2x$ and form the quadratic $y^2+2y+a^2-2=0.$

$D\geq0,$ so $-\sqrt3\leq a\leq\sqrt3.$

Which is not the answer $a²\leq2.$

My guess is that $D\geq0$ can only be applied if $y$ has all real values but I am not sure.

And could an alternative method be given? This is the only method I have to approach these kind of problems.

答:
With the quadratic formula you find that
$$\sin^2(x)=-1+\sqrt{3-a^2}\mbox,$$
so first of all $|a|\le \sqrt{3}$ as you found, but you sill have to chech for which of these $a$ you obtain
$$0\le -1+\sqrt{3-a^3}\le 1 \iff 1 \le 3-a^2 \le 4 \iff-1 \le a^2\le 2 \iff a^2\le 2$$

Title: Why is my approach the wrong method?


问:
how to apply hint to question involving the pigeonhole principle
The following question is from cut-the-knot.org's page on the pigeonhole principle

Question

Prove that however one selects 55 integers $1 \le x_1 < x_2 < x_3 < ... < x_{55} \le 100$, there will be some two that differ by 9, some two that differ by 10, a pair that differ by 12, and a pair that differ by 13. Surprisingly, there need not be a pair of numbers that differ by 11.

The Hint

Given a run of $2n$ consecutive integers: $a + 1, a + 2, ..., a + 2n - 1, a + 2n$, there are n pairs of numbers that differ by n: $(a+1, a+n+1), (a + 2, a + n + 2), \dots, (a + n, a + 2n)$. Therefore, by the Pigeonhole Principle, if one selects more than n numbers from the set, two are liable to belong to the same pair that differ by $n$.

I understood the hint but no concrete idea as to how to apply it, but here are my current insights:

My Insights

break the set of 100 possible choices into m number of 2n(where $n \in \{9,12,13\}$ ) consecutive numbers and since 55 numbers are to be chosen , show that even if one choses randomly there will be n+1 in one of the m partitions and if so, by the hint there will exist a pair of two numbers that differ by n.

答:
The easiest example is to do $10$. Then break up your set into $5$ sets, $$\{1,\dots,20\}\\\{21,\dots 40\}\\\vdots\\\{81,\dots,100\}$$

Since you have selected $55$ elements, you must have selected $11$ elements from one of these partitions. Then from the hint, you are done.

The case for $10$ is easy because $20$ divides $100$. The case of $12$ is harder. Partition the numbers as:

$$\{1,\dots,24\}\\\{25,\dots,48\}\\\{49,\dots,72\}\\\{73,\dots, 96\}\\\{97,98,99,100\}$$

Now we know that we've selected at least $51=55-4$ from the first four sets, and hence at least $13$ from one of them, and again we are done by the hint.

The problem for $11$ is that the "remainder" is large.  When we partition $100$ into sets of $22$, we are left with $12$ elements. That's a large enough number of elements left to "break" the logic above - we can pick up to $11$ different elements from that remainder, and we have $44$ to parcel out amongst the first four partitions, which we can do by putting $11$ in each partition. It is not hard to find a counter-example for $11$ precisely this way.

More generally, given $N$ and $k$, let $m=k\left\lfloor \frac{N}{2k}\right\rfloor$. Then you can select from $\{1,\dots,N\}$ the following number of elements:$$m + \min(k,N-2m)$$
and not get a pair that differ by $k$. But it you try to select more than that number, you must get a pair with difference $k$.

In the case $N=100$ and $k=11$, $m=44$ and $\min(k,N-2m)=11$ so we can select $55$ such numbers.


问:
In $\int \frac{1\;dt}{t}=\ln|t|+C$, why do we write $|t|$?
$$\int \frac{1\;dt}{t}=\ln|t|+C$$

Why do we write $|t|$?

答: You want the log function to have, as a domain, all non-zero real numbers.  However, for any real $x, e^x$ is never negative.  Therefore, $\log |x|$ is used instead of $\log x$.  This allows logarithms to be defined for negative numbers.


答: Take the set of all floors from a building. If $f_1$ and $f_2$ are floors, consider$$f_1\mathrel Af_2\quad\text{if}\quad\text{$f_1$ is above than or equal to $f_2$}.$$Then $A$ is an antisymmetric relation.

Title: Antisymmetric relation from a real life

Body:
Would you know some examples of antisymmetric relation from a real life? That is, relation $$(x,y)\in R\quad
 \text{and}\quad (y,x)\in R \rightarrow  x=y.$$ Thanks for your help.


答:
On the set of all people set $(x,y) \in R$ if $x$ is the parent of $y$.

Then for any two people $x$ and $y$ it is not possible that $x$ is the parent of $y$ and that $y$ is the parent of $x$. Therefore the implication
$$(x,y) \in R \quad\text{and}\quad (y,x) \in R \implies x=y$$
is vacuously true so $R$ is antisymmetric.

Title: Antisymmetric relation from a real life


问:
How to find the degree cover of a path-connected space with an infinite simple fundamental group
Let $X$ be a path-connected space with fundamental group $G$, where $G$ is an infinite simple group. Prove that $X$ has no finite degree cover of degree more than $1$.

I know that $G$ has no normal subgroups other that $\{1\}$ and $G$. But how do I go about proving the statement?

答: Hint: if a group has a proper subgroup of finite index then it has a proper normal subgroup of finite index. To see this, let $H$ be a proper subgroup of $G$ with $n = (G:H)$ finite. The action of $G$ on the left cosets of $H$ induces a homorphism of $G$ into a non-trivial subgroup of $S_n$. The kernel of this homomorphism is a proper normal subgroup of finite index.


问:
How and why does one come up with inequalities such as $(n!)^2\leq n^n(n!)<(2n)!\;$?
I just began with combinatorics and so far, I've done nothing more than learn about the definition of the factorial and do a few questions based on it, one of which was this :

Prove that $\forall n\in\Bbb Z^+, (n!)^2\leq n^n(n!)<(2n)!$

I was wondering how and why one comes up with seemingly weird inequalities like these.

Pardon me if my question sounds absurd.

Thank You!

答:
Identities like these are usually discovered by playing with strings of numbers. That's also a good way to discover the intuition behind the identity. For example take $n = 6$. The three expressions are:

$A = \left(6!\right)^2 = \left(1\cdot2\cdot3\cdot4\cdot5\cdot6\right)^2$

$B = 6^6 \left(6!\right) = \left(6\cdot6\cdot6\cdot6\cdot6\cdot6\right)\left(1\cdot2\cdot3\cdot4\cdot5\cdot6\right)$

$C = \left(2\cdot6\right)! = 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10\cdot11\cdot12$

We can go from $A$ to $B$ by replacing one of the $1$s with $6$, and one of the $2$s with $6$, ... and one of the $6$s with $6$. Each of these makes the number bigger, so $A \leq B$.

Similarly, we can go from $B$ to $C$ by replacing one of the $6$s in the first group with $7$, and one of the $6$s in the first group with $8$, ... and the last one of the $6$s in the first group with $12$. Each of these makes the number bigger, so $B \lt C$.

THIS IS NOT A PROOF but it shows WHY the identity is true.


问:
How to prove this inequality $ \Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$
Suppose $a > b > 0$,

how to prove that
$ \Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$

I been working on this for like 2 hours still did not find the trick.

Since when $a = 5$ and $b = 4$, this inequality holds, but what's the trick to actually prove it?

答: Note that$$\left|-a+\sqrt{a^2-b^2}\right|=a-\sqrt{a^2-b^2},$$since $0<b<a$, and that\begin{align}a-\sqrt{a^2-b^2}<b&\iff a-b<\sqrt{a^2-b^2}=\sqrt{(a-b)(a+b)}\\&\iff(a-b)^2<(a-b)(a+b)\\&\iff a-b<a+b,\end{align}which is true, since $b>0$.


答:
One should better write $$\int \frac{dx}{x}=\ln x, x \ne 0~~~(1)$$for the reality of the integral, instead of $$\int \frac{dx}{x}=\ln |x|~~~(2)$$

Note three situations:
$$I_1=\int_{1}^{2} \frac{dx}{x}=\ln 2$$
$$I_2=\int_{-2}^{-1} \frac{dx}{x}=\ln \frac{-1}{-2}=-\ln 2$$
For,
$$I_3=\int_{-1}^{2} \frac{dx}{x}$$
we cannot use (1) as $0\in(-1,2)$, but (2) will give a real value for $I_3=\ln 2$
In fact if one used (1) for $I_3$, one would get $$I_3=\ln 2 +\ln (-1)=\ln 2+i\pi \implies \Re(I_3)=\ln 2,$$ which is more appropriately called principal value..

Title: In $\int \frac{1\;dt}{t}=\ln|t|+C$, why do we write $|t|$?

Body:
$$\int \frac{1\;dt}{t}=\ln|t|+C$$

Why do we write $|t|$?


答: The number in question is the absolute value of the greatest root of $\frac{1}{2}b x^2+ax+\frac{1}{2}b$. Let $\alpha$ denote what is inside the absolute value. Then $\alpha$ must be negative since $a^2-b^2 < a^2$. Meanwhile, by symmetry, if $x_{0}$ is a non-zero root of this polynomial, then so is $1/x_{0}$. Since $\alpha$ is the greatest root and negative, we must have $1/\alpha < - 1 < \alpha <0$, and the result follows.

Title: How to prove this inequality $ \Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$

Body:
Suppose $a > b > 0$,

how to prove that
$ \Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$

I been working on this for like 2 hours still did not find the trick.

Since when $a = 5$ and $b = 4$, this inequality holds, but what's the trick to actually prove it?


答:
Note that $0 < a^2 - b^2 < a^2$. So $\sqrt{a^2-b^2} < a$. Hence

\begin{align}
   \left| \frac{-a + \sqrt{a^2-b^2}}{b} \right|
   &= \frac{a - \sqrt{a^2-b^2}}{b} \\
   &= \dfrac{b}{a+\sqrt{a^2-b^2}} \\
   &< 1
\end{align}


答:
The quadratic equation $bx^2+2ax+b=0$ has two roots:

$r=\frac{-a+\sqrt{a^2-b^2}}{b}$ and $s=-\frac{a+\sqrt{a^2-b^2}}{b}$.

Obviously, $|s|>1$, so $|r|=\frac{1}{|s|}<1$ from Vieta's formulas.


答:
Since $a>b$, put $x=a/b >1$. Then the inequality you are trying to prove becomes
$$
\left|-x + \sqrt{x^2-1}\right| < 1.
$$

Now this is quite straight forward.

Infact, we can see
$$
\left|(-x + \sqrt{x^2-1})(x + \sqrt{x^2-1})\right| = 1,
$$

and therefore

$$
\left|-x + \sqrt{x^2-1}\right| < \frac{1}{(x + \sqrt{x^2-1})}.
$$


问:
Complex integration: pole of order 2 at zero
Fix $a>0$. The following integral seems to be related to a complex integration:
$$
\int_{0}^{a}\frac{e^{-z}+z-1}{z^2}dz.
$$
We see that $f(z):=\frac{e^{-z}+z-1}{z^2}$ has a pole of order 2 at $z=0$. However, the upper end is a fixed positive real number $a$ instead of +∞ in which case usually can be dealt with a Cauchy integral. Is it possible to study the integral by complex analysis?Please give me some help or hints. Thanks a lot.

答:
Since the function is analytic (once the removable singularity is removed), all integrations along curves do not depend on the curve itself, only the endpoints.

Since $a$ is a real number, one curve connecting $0$ and $a$ is the real interval $[0,a]$. I.e., you can treat this as a function on the real line and perform the real integration. Any other curve connecting $0$ and $a$ would give you the same result.


问:
Confusing question on integers and probability
A number $\,d\,$ is chosen at random from the first $10$ positive integers.

What is the probability for the following events?

$d\,$ is an even number.

$d\;$ is an odd number.

$d\;$ is a prime number.

$3 \vert d$

答:
For example for (1): how many even integers are there? Well, now divide this numbers by the total ammount of integers there. (By "there" I mean in the set of the first 10 positive integers).

All the others are done in a similar way.


答:
The natural logarithm function represented by $\ln(x)$ is defined as:
$$f(x)=e^{x}$$
$$\ln(x)=f^{-1}(x)$$

What you notice with the exponential function is that if $x\in\mathbb{R},f:x\in\mathbb{R}$, in other words for any real number $x$ that number exponentiated cannot be negative. When we look at inverse functions in a way we are working backwards from the output to the input of the standard function, and since the above statement is true,it does not make sense for the output of the $\ln$ function to exist in the real domain.

As an example of why we can write it like this lets compare two integrals:
$$\int_{-2}^{-1}\frac 1xdx=\left[\ln|x|\right]_{-2}^{-1}=\ln(1)-\ln(2)=-\ln(2)$$
$$\int_1^2\frac1xdx=\left[\ln|x|\right]_1^2=\ln(2)-\ln(1)=\ln(2)$$
so you can see that the area in the negative domain is the same magnitude but opposite sign to that of the positive domain. This makes sense since:
$$g(t)=\frac1t,g(-t)=-g(t)$$


问:
Calculate norm of a polynomial knowing inner product but not the polynomials themselves
In $\mathbb R[t]_2$ (vector space of all polynomials of up to a second degree), I need to compute the norm $||t^2 - 2t + 1||$, knowing only that:

$$\langle p(t),q(t)\rangle = p(-2)q(-2) + p(-1)q(-1) + p(0)q(0) + p(1)q(1)$$

I have absolutely no clue on how to compute it as I don't have the polynomials $p(t)$ and $q(t)$.
Do I need to assume $p(t)$ and $q(t)$ generic polynomials, such as $p(t) = a + bt + ct^2$ and $q(t) = d + et + ft^2$, where $a,b,c,d,e,f$ would be scalars in $\mathbb R$?

Body:
Just plug into your formula:

\begin{align}
\|t^2-2t+1\|^2 &= \|(t-1)^2\|^2\\
&= \langle (t-1)^2, (t-1)^2\rangle\\
&= (-2-1)^2(-2-1)^2 + (-1-1)^2(-1-1)^2 + (0-1)^2(0-1)^2 + (1-1)^2(1-1)^2\\
&= 81+16+1\\
&= 98
\end{align}
so $\|t^2-2t+1\|=\sqrt{98}$.


问:
How can we define a Metric $d$ ( in space $\mathbb{R}$ of reals) so that sequence $A_n = 1/n, n = 1,2,3,...$ converges to a non-zero real number?
How can we define a Metric $d$,in space $\mathbb{R}$ of real numbers,so that sequence $A_n = \frac 1 n$ , $n = 1,2,3,...$, converges to a non-zero real number ?

答:
Define $f : \Bbb{R} \to \Bbb{R}$ by:
$$
\begin{align*}
f(x) = \left\{
\begin{array}{l@{\quad}l}
1 - x & \mbox{if $x \in \{0, 1\}$}\\
x & \mbox{otherwise}
\end{array}
\right.
\end{align*}
$$
Then define a metric $d$ by:
$$
d(x, y) = |f(x) - f(y)|
$$
Under $d$, $\frac{1}{n} \to 1$ as $n \to \infty$.


问:
Induction proof done, need to simplify $\sqrt[4]{(n+1)^5}\ln(1 + \frac{1}{(n+1)^2}) \leq 4 \sqrt[4]{\frac{n+1}{n}}$
Prove that:

$$\sum_{j=1}^{n} \sqrt[4]{j^5}\ln(1+\frac{1}{j^2}) \leq 4 \sqrt[4]{n}, n \in \mathbb{N} $$

I got everything right and came up with an inequality:

$$\sqrt[4]{(n+1)^5} \ln\left(1 + \frac{1}{(n+1)^2}\right) \leq 4  \sqrt[4]{\frac{n+1}{n}}$$

The problem is I cannot simplify that to get my solution all done. Anybody got an idea?

答:
As regards your final step, just notice that $\ln(1+x)\leq x$.Therefore it suffices to show that
$$\sqrt[4]{(n+1)^5}\cdot \frac{1}{(n+1)^2} \leq 4  \sqrt[4]{\frac{n+1}{n}}$$
that is
$$\sqrt[4]{n}\leq 4 (n+1)^{\frac{1}{4}+2-\frac{5}{4}}=4(n+1)$$
which trivially holds.


答:
I don't really get what you mean by simplify it.

Anyways, in case the last part of the induction part is required, given your inequality and the induction hipothesys you would have:
$$\sum_{j=1}^{n+1}\sqrt[4]{j^5}\ln\biggl(1+\frac{1}{j^2}\biggr)\le 4\sqrt[4]{n}+\sqrt[4]{(n+1)^5}\ln\biggl(1+\frac{1}{(n+1)^2}\biggr)\le$$
$$\le 4\sqrt[4]{n}+4\sqrt[4]{\frac{n+1}{n}}\stackrel{n\ge 1}{\le\le\le}4\sqrt[4]{n}+4\sqrt[4]{n+1}\le 4\sqrt[4]{n+1}\mbox,$$
just as we wanted.

Title: Induction proof done, need to simplify $\sqrt[4]{(n+1)^5}\ln(1 + \frac{1}{(n+1)^2}) \leq 4 \sqrt[4]{\frac{n+1}{n}}$

Body:
Prove that:

$$\sum_{j=1}^{n} \sqrt[4]{j^5}\ln(1+\frac{1}{j^2}) \leq 4 \sqrt[4]{n}, n \in \mathbb{N} $$

I got everything right and came up with an inequality:

$$\sqrt[4]{(n+1)^5} \ln\left(1 + \frac{1}{(n+1)^2}\right) \leq 4  \sqrt[4]{\frac{n+1}{n}}$$

The problem is I cannot simplify that to get my solution all done. Anybody got an idea?


问:
Hilbert space with all subspaces closed
Does there exist an infinite-dimensional Hilbert space with all subspaces closed?</p>

答: If $H$ is an infinite dimensional Hilbert space and $\{x_n:n\geq 1\}$ is a countably infinite set of linearly independent elements, then the abstract span of these elements, which is an infinite dimensional subspace, cannot be closed. If it were closed, it would be a Hilbert space itself, and infinite dimensional Hilbert spaces must have uncountably infinite dimension.</p>

Title: Hilbert space with all subspaces closed


问:
How to express it into set builder method?
$$\left\{\frac{1}{4},\frac{2}{10},\frac{4}{28},\frac{8}{82},\frac{16}{244},\frac{32}{730}\right\}$$

Above is a set  which I have to express into set builder method .

So, I have divided into two parts:

For the numerator, I got the condition $p=2^x$ ,$0\le x \le 5$

For the denominator $q=3^n+1$, $1\le n \le 6$

I can't write it in set builder form. Can anyone help me to write the set in builder form?

答:
You have written correctly. Now, just notice that for each term in the set, $n=x+1$. So, substitute this value and the set becomes
$$\left\{y:\ y=\frac{2^x}{3^{x+1}+1},\ 0\leq x\leq 5,\ x\in\mathbb{Z}\right\}$$

Or simply,
$$\left\{\frac{2^x}{3^{x+1}+1}:\ 0\leq x\leq 5,\ x\in\mathbb{Z}\right\}$$

[If you're unfamiliar $x\in\mathbb{Z}$ means $x$ is an integer.]

Hope this helps :)


答:
How about:
$$\left\{ \frac{2^{n}}{3^{n+1}+1}\mid0\leq n\leq5\right\}$$
or more precize:
$$\left\{ \frac{2^{n}}{3^{n+1}+1}\mid0\leq n\leq5\tag{ },n\in\mathbb{Z}\right\} $$

Title: How to express it into set builder method?

Body:
$$\left\{\frac{1}{4},\frac{2}{10},\frac{4}{28},\frac{8}{82},\frac{16}{244},\frac{32}{730}\right\}$$

Above is a set  which I have to express into set builder method .

So, I have divided into two parts:

For the numerator, I got the condition $p=2^x$ ,$0\le x \le 5$

For the denominator $q=3^n+1$, $1\le n \le 6$

I can't write it in set builder form. Can anyone help me to write the set in builder form?


问:
Exponentials in the category of graphs?
Let $\Gamma$ be the category $e\overset{s}{\underset{t}{\rightrightarrows}}v$ and $\mathbf{Graphs}=\mathbf{Sets}^{\Gamma}$ the category of (directed multi) graphs.

For graphs $G$ and $H$, it's my understanding that the exponential object $H^G$ is the graph whose vertices are arbitrary functions $\varphi:G_v\to H_v$ and whose edges $\theta:\varphi\to\psi$ are functions $\theta:G_e\to H_e$ making these diagrams commute:
$\require{AMScd}$
\begin{CD}
G_v @<s<< G_e @>t>> G_v\\
@V \varphi VV @VV \theta V @VV \psi V\tag{1}\\
H_v @<<s< H_e @>>t> H_v
\end{CD}

The evaluation homomorphism $\epsilon:H^G\times G\to H$ is defined on vertices by $(\varphi,v)\mapsto\varphi(v)$ and on edges by $(\theta,e)\mapsto\theta(e)$. For a homomorphism $f:F\times G\to H$, the transpose homomorphism $\lambda f:F\to H^G$ is defined on vertices by $v\mapsto f(v,-)$ and on edges by $e\mapsto f(e,-)$.

I'm trying to see how this falls out of the general construction of exponentials in categories of sheaves using the Yoneda lemma, but can't quite get things to line up.

Let $y:\Gamma^{\mathrm{op}}\to\mathbf{Graphs}$ be the (contravariant) Yoneda embedding. Then I believe:

$y(v)$ is the "vertex graph" $\bullet$ consisting of a single vertex with no edges

$y(e)$ is the "edge graph" $\bullet\to\bullet$ consisting of two vertices with one edge between them

The graph $\bullet\times G$ is the graph of "vertices of G"

The graph $(\bullet\to\bullet)\times G$ is the graph of "edges of G" consisting of (i) two copies $v_s$ and $v_t$ of each vertex $v$ of $G$ ($v_s$ represents "$v$ as a potential source" and $v_t$ represents "$v$ as a potential target"), and (ii) an edge $s(e)_s\to t(e)_t$ for each edge $e$ of $G$.

Now if $H^G$ is an exponential graph we must have

$$\begin{align*}
(H^G)_v&\cong\mathrm{Hom}_{\mathbf{Graphs}}(y(v),H^G)\\
&\cong\mathrm{Hom}_{\mathbf{Graphs}}(\bullet\ ,H^G)\\
&\cong\mathrm{Hom}_{\mathbf{Graphs}}(\bullet\times G,H)\\
&\cong\mathrm{Hom}_{\mathbf{Sets}}(G_v,H_v)
\end{align*}$$

which checks out, and

$$\begin{align*}
(H^G)_e&\cong\mathrm{Hom}_{\mathbf{Graphs}}(y(e),H^G)\\
&\cong\mathrm{Hom}_{\mathbf{Graphs}}(\bullet\to\bullet,H^G)\\
&\cong\mathrm{Hom}_{\mathbf{Graphs}}((\bullet\to\bullet)\times G,H)
\end{align*}$$

However, in general won't there be more graph homomorphisms from $(\bullet\to\bullet)\times G$ to $H$ than functions $\theta:G_e\to H_e$ satisfying (1), for example if $G$ has isolated vertices?

My understanding is that a graph homomorphism is a mapping of vertices to vertices and edges to edges which respects sources and targets of edges, so for an isolated vertex $v$ in $G$ we could map the isolated vertices $v_s$ and $v_t$ of $(\bullet\to\bullet)\times G$ freely for any given mapping of edges.

What is my error?

答:
You haven’t specified what exactly you think $\bullet\to\bullet \times G$ is, so lets do so: it’s the graph with vertices $G_v\times \{0\}\cup G_v\times \{1\}$ and edges $(x,0)\to (y,1)$ for every edge $x\to y$ in $G$. A map from this to $H$ precisely consists of functions $f_0,f_1:G_v\to H_v$ and $f:G_e\to H_e$ such that $s\circ f=f_0\circ s$ and $t\circ f=f_1\circ t$, as you say.

It is true that $f_0$ and $f_1$ are unconstrained in their action on isolated vertices by $f$, as you also say. Your initial description implies that $f_0$ and $f_1$ should be determined by $f$, and indeed in general they are not. An edge of the exponential graph is the whole triple.

Title: Exponentials in the category of graphs?


问:
If $f(x)=a$ for $x\in A$, and $f(x)=b\neq a$ otherwise, where $A$ and $\Bbb R-A$ are dense in $\Bbb R$, show that there is no limit as $x\to\infty$

Let $A\subseteq \Bbb{R}$ and let $a,b\in \Bbb{R}$ such that $a\neq b$. $A$ and $\Bbb{R}-A$ are dense in $\Bbb{R}$.

Define $f:\Bbb{R}\to\Bbb{R}$:

$$f(x) = \begin{cases} a, &\text{if } x \in A \\
b, & \text{otherwise}\end{cases}$$

Prove that there is no limit when $x\to\infty$.

I tried proving by contradiction, but I couldn't find any claim that contradicts that there's a limit.

I assumed in contradiction that there's a limit and defined an $\epsilon = 0.5$.

Then from the definition of limit:

$|f(x) - L| = | a - L| < \epsilon$

$|f(y) - L| = | b - L| < \epsilon$

$a-0.5 < L < a + 0.5$

$b-0.5 < L < b + 0.5$

Then I got stuck because I don't see any contradiction here.

答: Since $\mathbb{R}\setminus A$ is dense in $\mathbb{R}$ there is a sequence of points $x_n\in\mathbb{R}\setminus A$ such that $x_n\to a$, but $f(x_n)=b\neq f(a)$ for all $n$.


答:
let $y=\ln(x)$

Notice that y is only defined for positive x.

$e^y = x$

By implicit differentiation

$e^y\cdot \frac{dy}{dx} = 1$

$\frac{dy}{dx} = e^{-y}$

$\frac{dy}{dx} = \frac{1}{x}$

So we know that ln(x) is an anti-derivative for $\frac{1}{x}$ when x>0. What about when x<0. How do we get an anti-derivative for $\frac{1}{x}$ when $x<0$. We know the answer can't be $\ln(x)$ because $\ln(x)$ is undefined for negative x. Also plotting $\frac{1}{x}$, it's clear an anti-derivative exists for $\frac{1}{x}$ when $x<0$. So there should be an answer.

We can define a new function

$y = \ln(-x)$

Notice that this function is only defined for $x<0$

We can use our previous result to take the derivative here

$\frac{dy}{dx} = \left(\frac{1}{-x}\right)(-1)$

$\frac{dy}{dx} = \frac{1}{x}$

So we know that when $x<0$, $\ln(-x)$ is an anti-derivative of $\frac{1}{x}$

So if we define a function like this:

$f(x) = \begin{cases} \ln(-x), &x<0 \\ \ln(x), &x>0 \end{cases}$,

then it's an anti-derivative for $\frac{1}{x}$ everywhere that $\frac{1}{x}$ is defined.

We can rewrite this function as

$f(x) = \ln|x|$, where $x \ne 0$


问:
Cardinality of order preserving functions from totally ordered set with dense subset
I know that in the category of continuous functions, if $X$ and $Y$ are Hausdorff topological spaces and $D\subseteq X$ is a dense subset of $X$, then the cardinality of the continuous functions from $X$ to $Y$ is at most $\vert Y^D\vert$.
Can we obtain a similar result for ordered spaces with respect to order preserving functions $A\to B$ where $D\subseteq A$ is dense, in the order sense, and saying there are at most $\vert B^D\vert$ order preserving functions? For example if we assume that $A$ is the completion of $D$ or perhaps something stronger?

答: No.  For instance, consider order-preserving maps $f:\mathbb{R}\to\mathbb{R}\times\mathbb{R}$ where $\mathbb{R}$ has its usual order and $\mathbb{R}\times\mathbb{R}$ has its lexicographic order.  For any function $g:\mathbb{R}\to\mathbb{R}$, the map $f(x)=(x,g(x))$ is order-preserving.  So the number of order-preserving maps $\mathbb{R}\to\mathbb{R}\times\mathbb{R}$ is $(2^{\aleph_0})^{2^{\aleph_0}}=2^{2^{\aleph_0}}$ which is greater than $|(\mathbb{R}\times\mathbb{R})^\mathbb{Q}|=2^{\aleph_0}$ even though $\mathbb{Q}$ is dense in $\mathbb{R}$.


答:
If $H$ is an infinite-dimensional Hilbert space, then $H$ contains $\ell^2(\mathbb{N})$ isometrically. And the property you mention fails for the latter. For instance
$$
F:=\mbox{span} \{e_k=(0,\ldots,0,1,0,\ldots)\;;\; k\in\mathbb{N}\}\subsetneq \overline{F}=\ell^2(\mathbb{N}).
$$

Note: this is the lowtech argument for Hilbert spaces. But Keenan Kidwell's approach, which uses Baire, works for a general infinite dimensional Banach space. And since the converse is clearly true, we get: a Banach space has finite dimension if and only if every subspace is closed.

Title: Hilbert space with all subspaces closed

Body: Does there exist an infinite-dimensional Hilbert space with all subspaces closed?</p>


问:
What is the asymptotic growth of the terms in this expression?
Let's say we have $k$ variables in this expression: $(n_1+n_2+n_3+...+n_k)^2$. When you expand it, you get something like $n_1^2+n_1n_2+n_2^2+...$ , using the multinomial theorem. Now, if we multiply each term of the expansion out like this: $(m_1n_1^2+c_1)(m_2n_1n_2+c_2)(m_3n_2^2+c_3)...$ and expand it, is it true that the number of terms is $O(2^{k^2})$?

答: No.  The expansion of  $A=(\sum_{i=1}^kn_i)^2$ has $k(k+1)/2$ terms in it.  The expansion of $\prod_{i=1}^{k(k+1)/2}(T_i+c_i)$, where $T_i$ is the $i$ term in the expansion of $A$, has $2^{k(k+1)/2}$ terms.


问:
Finding maximum of $x+y+z$
If positive numbers $x, y$ and $z$ satisfy that $xyz=1$, what is the minmum value for $x+y+z$?

From $xyz=1$, we can get $$x = \frac{1}{yz};\space\space\space y = \frac{1}{xz};\space\space\space z = \frac{1}{xy}; $$

Subsitute them into $x+y+z=1$ and I got$$\frac{xy+yz+xz}{xyz} = xy+yz+xz = 1$$

Since we're finding the minimum for $x+y+z$, I thought of using the formula $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+xz)$ due to the fact that we have the value of $xy+yz+xz$.

That's all I've got so far. How can I continue?

答:
By geometry:

The surface of equation $xyz=1$ (don't know its name) is a cubic with an "hyperbolic-like" shape, as any cross section by a plane of one constant coordinate is an hyperbola. It has a symmetry of order $3$ around the axis $x=y=z$, and is open towards infinity.

The sections by the plane $x+y+z=c$ are closed curves, starting from $c=3$ and enlarging monotonously and unboundedly.

The minimum is $c=3$ and there is no maximum.

Title: Finding maximum of $x+y+z$


问:
Writing singleton as countable intersection/union of open sets
I want to write $\{ x \}$ as a finite intersection of open sets. I came up with $\bigcap^{\infty}_{n=1} (x - \frac{1}{n}, x + \frac{1}{n})$, but I am not sure wether this is correct. If incorrect, is there another way to write a singleton as a countable union/intersection of open sets?

答:
That is correct and indeed it expresses $\{x\}$ as a countable intersection of open sets.

On the other hand, there is no way of expressing it as a countable union of open sets, since each of those open sets would have to contain some interval around $x$, and therefore so would their union.


问:
This is a question in number theory. It seemed I can't start the proof.
How can this be solved?

I knew that the concept of an order of an integer is needed, but rather than that I think I'm really lost with the proofs.

Body:
$$a^8-1\equiv 0 \implies (a^4+1)(a^4-1)\equiv 0 \implies a^4 \equiv -1 \pmod p$$

$$c^2=(a-a^3)^2=a^2-2a^4+a^6=a^2(1+a^4)-2a^4\equiv 2 \pmod p$$

$$d^2=(a+a^3)^2=a^2+2a^4+a^6=a^2(1+a^4)+2a^4\equiv -2 \pmod p$$


答:
With these simple probabilities, all you need to do is create a fraction with the total possibilities on the bottom, and then how many possibilites there are of what you want on the top.

For Example:

a) There are 5 even numbers between 1 and 10 - (2,4,6,8,10) - Because there are 5 even numbers and 10 numbers total, take 5/10 and simplify to get 1/2.

Title: Confusing question on integers and probability

Body:
A number $\,d\,$ is chosen at random from the first $10$ positive integers.

What is the probability for the following events?

$d\,$ is an even number.

$d\;$ is an odd number.

$d\;$ is a prime number.

$3 \vert d$


答: Try $\epsilon = 0.5 |b - a|$.  Then use the density of $A$ and $R\setminus A$ to see $L \in (a-\epsilon, a+\epsilon)$ and $L \in (b-\epsilon, b+\epsilon)$, but $\epsilon$ is half of the distance between $a$ and $b$, so these two open intervals won't overlap.  Contradiction.

Title: If $f(x)=a$ for $x\in A$, and $f(x)=b\neq a$ otherwise, where $A$ and $\Bbb R-A$ are dense in $\Bbb R$, show that there is no limit as $x\to\infty$

Body:
Let $A\subseteq \Bbb{R}$ and let $a,b\in \Bbb{R}$ such that $a\neq b$. $A$ and $\Bbb{R}-A$ are dense in $\Bbb{R}$.

Define $f:\Bbb{R}\to\Bbb{R}$:

$$f(x) = \begin{cases} a, &\text{if } x \in A \\
b, & \text{otherwise}\end{cases}$$

Prove that there is no limit when $x\to\infty$.

I tried proving by contradiction, but I couldn't find any claim that contradicts that there's a limit.

I assumed in contradiction that there's a limit and defined an $\epsilon = 0.5$.

Then from the definition of limit:

$|f(x) - L| = | a - L| < \epsilon$

$|f(y) - L| = | b - L| < \epsilon$

$a-0.5 < L < a + 0.5$

$b-0.5 < L < b + 0.5$

Then I got stuck because I don't see any contradiction here.


答:
Use AM-GM inequality,

$$\frac{x+y+z}{3} \ge \sqrt [3]{xyz}$$

$$x+y+z \ge 3$$

The minimum is $3$ and there's no maximum.

Title: Finding maximum of $x+y+z$

Body:
If positive numbers $x, y$ and $z$ satisfy that $xyz=1$, what is the minmum value for $x+y+z$?

From $xyz=1$, we can get $$x = \frac{1}{yz};\space\space\space y = \frac{1}{xz};\space\space\space z = \frac{1}{xy}; $$

Subsitute them into $x+y+z=1$ and I got$$\frac{xy+yz+xz}{xyz} = xy+yz+xz = 1$$

Since we're finding the minimum for $x+y+z$, I thought of using the formula $(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+xz)$ due to the fact that we have the value of $xy+yz+xz$.

That's all I've got so far. How can I continue?


问:
For which $N$ is it possible to arrange all whole numbers from $1$ to $N$ in such a way that every adjacent pair sums up to a Fibonacci number?
Recently I came up with a problem regarding Fibonacci numbers:

For which $N$ is it possible to arrange all whole numbers from $1$ to $N$ in such a way that every adjacent pair sums up to a Fibonacci number?

I have manually tested a bunch of cases and I was also able to prove almost every case. The results that I was able to prove are the following:

If $N$ is a Fibonacci number or exactly one less than a Fibonacci number, I was able to prove that an arrangement exists. I used an argument by induction to achieve this.

If $F_k+2 \leq N \leq F_{k+1} -3$ , it is completely impossible, because the numbers $F_k$, $F_k + 1$ and $F_k + 2$ have only one possible pair and therefore have to be at the end of the arragement. This obviously gives a contradiction, because arrangements only have two ends (the first number and the last number).

The cases I was not able to solve are $N=F_k+1$ and $N=F_k-2$. My theory is that $N=9$ is the only working case of the form $N=F_k+1$, and $N=11$ the only working case of the form $F_k-2$. I expect every other $N$ of these two forms to be impossible.

Does anybody know a full proof to this problem or maybe the name of the official theorem (if this exists)?

答:
Thanks to @Rei Henigman data, I could find OEIS sequence A079734 and from there the "Fibonacci Plays Billiards" paper by Elwyn Berlekamp and Richard Guy, where they state with theorem 1 on page 3, that the only solutions are
$N = 9$, $N = 11$, $N = F_k$, $N = F_k − 1$ for $k \ge 4$. Note that $1$ is excluded from the OEIS sequence.


问:
solve logarithmic inequation using lagrange mean value theorem
I have this inequality: for $ x >0, x \neq 1 : \log(x) < x-1$.
If i understand it right, i should take an interval, like $(0,x)$ and prove it using the formula $ \frac{f(b)-f(a)}{b-a}=f'(c) $. But since $ \frac{\log(x)-\log(0)}{x-0}$ does not exist I don't know how to do it. Maybe I am doing it wrong, but how should i do it?

答: Let $f(x) = \log(x)$ on $(1,x)$, $a = 1$, $b = x$.  Then invoke MVT to find $c \in (1,x)$ such that $1/c = f'(c) = (f(b)-f(a))/(b-a) = (\log x - \log 1)/(x-1)$.  Use $c > 1$ to prove the desired inequality.


问:
characteristic vs minimal polynomial
Let $L$ be a finite field extension of $K$. For every element $\theta$ in $L$ define the characteristic polynomial of $\theta$ as follows
$$\operatorname{char}_{\theta}(X):=\det(X\cdot id_L-r_\theta)$$
where $r_{\theta}$ is the $K$-linear map given by
$$r_{\theta}:L\longrightarrow L\qquad;\qquad x\longmapsto x\theta$$

Let finally $f(X)$ be the minimal polynomial of $\theta$ over $K$.

My question is: $\operatorname{char}_{\theta}=f?$

答: Suppose $d = [L : K(\theta)]$. Then $\chi_{char(\theta)}(x) = \chi_{min(\theta)}(x)^d$ assuming we've normalized both polynomials to be monic. </p>

Title: characteristic vs minimal polynomial


问:
Show that $\bar\int f+g \leq \bar\int f + \bar\int g$
Let $I\subset \Bbb{R^n}$ be a compact interval and $f:I\to \Bbb{R},g:I\to \Bbb{R}$ bounded functions. Show that $\bar\int_I (f+g) \leq \bar\int_I f + \bar\int_I g$.

Let $M_j(f):=\sup f(I_j)$ and $M_j(g):=\sup g(I_j)$. We have that $M_j(f+g)\leq M_j(f)+M_j(g)$. Since $\bar\int_I (f+g):=\inf\{U((f+g),P)\}.$ We could then write  $$\bar\int_I (f+g)\leq U((f+g),P)\leq U(f,P)+U(g,P),$$

which means $\bar\int_I (f+g)$ is a lower bound for $\{U(f,P)+U(g,P)\}$. If we define the infima of this set as $y$, we'd have that $y\geq \bar\int_I (f+g)$. How can I finish this proof?

答:
Since you have only one partition $P$ you couldn't take the infimum. Instead you could try to split the partition as follows. Take two different partitions of $I$, call them $P_1$ and $P_2$ and define $P=P_1 \cup P_2$ which is again a partition of $P$. Then you get the following inequality.
\begin{align*}
\overline{\int} (f+g) \leq U(f+g, P_1\cup P_2)\leq U(f,P_1\cup P_2)+U(g,P_1\cup P_2) \leq U(f,P_1) + U(g,P_2)
\end{align*}
Fixing the partition $P_2$ the left take infimum over $P_1$, we get
\begin{align*}
\overline{\int} (f+g)\leq \overline{\int}f+U(g,P_2)
\end{align*}
Now take infimum over $P_2$, we get
\begin{align*}
\overline{\int}(f+g)\leq \overline{\int}f +\overline{\int} g
\end{align*}
$\textbf{Edit}$:"By fixing partition $P_1$ taking infimum over $P_2$" I meant the following,
\begin{align*}
\inf_{P_2:P_2\text{ a partition of $I$}} \biggl( U(f,P_1)+U(g,P_2)\biggr) = U(f,P_1)+ \inf_{P_2:P_2\text{ a partition of $I$}}U(g,P_2)
\end{align*}
The above is legitimate because $P_2$ is not involved in the term $U(f,P_1)$


问:
Elementary algebraic question
$$\frac{q^{n+1}-1+q^{n+1}(q-1)}{q-1}=\frac{q^{(n+1)+1}-1}{q-1}$$ I don't understand this step above, how can I get the right sight?

答: Right-hand side seems to be $q^{(n+1)+1}-1$.

Title: Elementary algebraic question


问:
How to prove this inequality $ \Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$
Suppose $a > b > 0$,

how to prove that
$ \Big| \frac{-a + \sqrt{a^2-b^2}}{b} \Big| < 1$

I been working on this for like 2 hours still did not find the trick.

Since when $a = 5$ and $b = 4$, this inequality holds, but what's the trick to actually prove it?

Body:
WLOG $b=a\sin2t,$

As $a>b>0,0<\sin2t<1$

WLOG $0<2t<\dfrac\pi2\implies 0<t<\dfrac\pi4 \ \ \ \ (1)$

$$\dfrac{-a+\sqrt{a^2-b^2}}b=\dfrac{-a+a\cos2t}{a\sin2t}=\dfrac{1-2\sin^2t-1}{2\sin t\cos t}=-\tan t$$

By $(1), -1<\tan t<0$


Title: Elementary algebraic question

Body: $$\frac{q^{n+1}-1+q^{n+1}(q-1)}{q-1}=\frac{q^{(n+1)+1}-1}{q-1}$$ I don't understand this step above, how can I get the right sight?

Body: $$\frac{q^{n+1}-1+q^{n+1}(q-1)}{q-1}=\frac{q^{n+1}-1+q^{(n+1)+1}-q^{n+1}}{q-1}$$ Then it equals the RHS, as the term $q^{n+1}$ cancels.


问:
How do I prove the existence of a set of finite measure?
Let $f \in L^{1}(\mathbb{R})$, Prove that then for every $\epsilon >0$,

i) There exists a set of finite measure $B$ such that $\int_{B^c} |f(x)|dx< \epsilon.$

ii) There is a $\delta>0$ such that $\int_{E} |f(x)|dx< \epsilon$, where $m(E)<\delta$.

In this question, I proved that if $f \in L^1$, then the set $\{x: f(x)\neq0\}$ is sigma finite.

Can anyone suggest me me how do I prove the existence of the set $B$?

答:
You know that $\int_\mathbb R |f(x)| dx$ is finite. You also know that the sequence $\int_{[-n,n]} |f(x)| dx$ is monotonic increasing and has limit equal to $\int_\mathbb R |f(x)| dx$. Therefore, for some $n$ it follows that
$$\int_{(-\infty,-n) \cup (n,\infty)} |f(x)| dx = \int_\mathbb R |f(x)| dx - \int_{[-n,n]} |f(x)| dx < \epsilon
$$
Now take $B = [-n,+n]$.


答:
For the first question, according to Monotone Convergence Theorem (MCT), $$\lim_{n\to \infty }\int_{\mathbb R}\boldsymbol 1_{B(0,n)}|f|=\int_{\mathbb R}|f|,$$
where $B(0,n)$ is the ball centered at $0$ and of radius $n$.

For the second one, for all $k>0$,
$$\int_{E}|f|=\int_{E}|f|\boldsymbol 1_{\{|f|\leq k\}}+\int_{E}|f|\boldsymbol 1_{\{|f|>k\}}.$$

According to MCT, There is $k$ big enough s.t. $$\int_{E}|f|\boldsymbol 1_{\{|f|>k\}}<\frac{\varepsilon}{2}. $$

Moreover, $$\int_{E}|f|\boldsymbol 1_{\{|f|\leq k\}}\leq km(E\cap\{|f|\leq k\})\leq km(E).$$

Taking $0<\delta <\frac{\varepsilon }{2k}$ yields $$\int_{E}|f|\boldsymbol 1_{\{|f|\leq k\}}\leq \frac{\varepsilon }{2},$$
whenever $m(E)<\delta $. The claim follow.

Title: How do I prove the existence of a set of finite measure?

Body:
Let $f \in L^{1}(\mathbb{R})$, Prove that then for every $\epsilon >0$,

i) There exists a set of finite measure $B$ such that $\int_{B^c} |f(x)|dx< \epsilon.$

ii) There is a $\delta>0$ such that $\int_{E} |f(x)|dx< \epsilon$, where $m(E)<\delta$.

In this question, I proved that if $f \in L^1$, then the set $\{x: f(x)\neq0\}$ is sigma finite.

Can anyone suggest me me how do I prove the existence of the set $B$?


问:
Commutator subgroup of $S_n$ generated by commutators of transpositions
Is it true that the commutator group $[S_n, S_n]$ is generated by elements of the form $[\tau_1, \tau_2]$ for $\tau_1, \tau_2$ transpositions?

I know that in general, when a group $G$ is generated by a set $S$, then one has to take the normal closure of the group generated by commutators of elements in $S$ to obtain $[G,G]$.

答:
Every element of $S_n$ is a product of transpositions.

I use the definition $[x,y]=x^{-1}y^{-1}xy$; if you use the other convention, the identities below need to be tweaked, but they are essentially the same.

Now notice that
$$\begin{align*}
[xy,z] &= [x,z]^y[y,z] = [x^y,z^y][y,z]\\
[x,yz] &= [x,z][x,y]^z = [x,z][x^z,y^z].
\end{align*}$$
Since the conjugate of transposition is a transposition, this allows you to decompose any commutator into a product of commutators of transpositions.

Note that this does not require you to know that the commutator subgroup is $A_n$, or how $A_n$ can be generated, just that every element of $S_n$ is a product of transpositions, and that the conjugate of a transposition is a transposition.


答:
For (i):

Define
$$B_n=\{ x:|x|\le n \mbox{, } |f(x)|\le n \}  \mbox{, so that } B_1\subseteq B_2\subseteq\mbox{... , and } \cup_{n=1}^{+\infty}B_n=\mathbb{R}.$$
Define then the sequence $f_n=\chi_{B_n}f$, we have that $f_n \rightarrow f$ pointwise and $|f_n|\le |f|\in \textit{L}^1$, so that $f_n \rightarrow f \in \textit{L}^1$ as $n\rightarrow +\infty$ by the DTC. Now it suffices to choose $n_{\epsilon}$ so that $$\int_{B_{n_{\epsilon}}^C}|f|dm=\int_{B_{n_{\epsilon}}^C}|f-f_{n_{\epsilon}}|dm=||f-f_{n_{\epsilon}}||_{\textit{L}^p}<\epsilon$$ since $f_{n_{\epsilon}}=0$ on $B_{n_{\epsilon}}$.
Also $B_{n_{\epsilon}}$ is limited for its definition at the beginning.

For (ii):

Given $\epsilon>0$, let $A$ be a set such that $\int_{A_C}|f|dm <\epsilon /2$, which i can for the part (i), we also know that f is bounded on A for how we defined $A=B_{n_0}:=\{ x:|x|\le n \mbox{, } |f(x)|\le n \}$. Let $M:=\sup_{x\in A}|f(x)|$, and choose $\delta =\epsilon/ (2M)$. Then if F is a measurable set such that $m(F)<\delta $ we obtain
$$\int_F|f|dm=\int_{f\cap A^C}|f|dm + \int_{F\cap A}|f|dm \le \int_{A^C}|f|dm + M\int_{F\cap A}dm \le$$
$$< \frac{\epsilon}{2} + Mm(F)\le \frac{\epsilon}{2}+M\frac{\epsilon}{2M}=\epsilon.$$


问:
Finding the local minima
Consider the function $f(x) = px^3 - q^2x$ where $p$ and $q$ are positive constants. Find all $x$-values where $f$ has local minima (if any).

I graphed this function and there was a local minima at $x = 0,577$ and lots of other values when $p$ and $q$ were changed. How do I find all of them ?

答:
For a general case with $p,q$ being positive constants, you should differentiate the function and find all its critical points.

$$f(x)=px^{3}-q^{2}x \Rightarrow f^{\prime}(x)=3px^{2}-q^{2}$$ Then to find critical points, $$f^{\prime}(x)=0, 3px^{2}-q^{2}=0$$  By rearranging the equation, you get $$x^{2}=\frac{q^{2}}{3p}$$ Since $p$ is positive, $p \neq 0$. Square rooting both sides will get you the values of the critical points of $f$. $x= \pm \frac{q}{\sqrt{3p}}$.

To conclude, the local extremas are $(\frac{q}{\sqrt{3p}},f(\frac{q}{\sqrt{3p}}))$ and $(-\frac{q}{\sqrt{3p}},f(-\frac{q}{\sqrt{3p}}))$. You can do the algebraic manipulations yourself. To test which is the minima and maxima, you can apply the second derivative test to both of these points to determine it.

Title: Finding the local minima


问:
Locally compact subspace is an intersection of an open and closed set
Let X be a locally compact topological space. I need to prove that if $M\subset X$ is a locally compact subspace of X then there exist $U,F\subset X$ such that U is open and F is closed, and $M=U\cap F$.
It can be assumed that for every open $x\in V$ there exist an open subset U such that $x \in U \subset \overline U \subset V$ and $\overline U$ is compact.

My thoughts about the problem are that for every $x\in M$ and for every neighborhood $x\in V_x$ there exist an open neighborhood $U_x$ in M such that $x\in U_x\subset \overline U_x \subset V_x$ and $\overline U_x$ is compact and closed in M. So there exist an open subset $Z_x\subset X$ such that $U_x = Z_x\cap M$. I can take $U = \bigcup _{x\in M}Z_x $, then U is open in X, but i was having trouble finding the closed set F. I thought  to take $F = \overline {\bigcup _{x\in M}U_x}$, that is closure in X. Now $M\subset U\cap F$, but I haven't succeeded in showing that $M\supset U\cap F$, maybe that's not even true.

Thanks!

答:
The obvious choice of closed set here is $\overline M$; it is the smallest closed set containing $M$. With this choice of closed set, it suffices to choose an open set which excludes $\overline M - M$.

I claim each point $m\in M$ has an open neighborhood $U_m$ which excludes $\overline M - M$. For $m \in \text{int}(M)$, this is not hard to see. For $m \in M\cap \partial M$, choose a compact neighborhood $K_m\subseteq M$. Considering $K_m$ as a subspace of $\overline M$, we have that $\text{int}_{\overline M}(K_m)$ is open in $\overline M$, so there is an open set $U_m$ so that $U_m \cap \overline M = K_m\subseteq M$. Hence $U_m\cap (\overline M - M)$ is empty, as desired.

Since each $U_m$ excludes $\overline M - M$, so does their union, which is an open set containing $M$. This is the desired open set.


问:
Why is my approach the wrong method?
The equation $\sin^4x-2\cos^2x+a^2=0$ can be solved if :

My approach was to substitute $y$ as $\sin^2x$ and form the quadratic $y^2+2y+a^2-2=0.$

$D\geq0,$ so $-\sqrt3\leq a\leq\sqrt3.$

Which is not the answer $a²\leq2.$

My guess is that $D\geq0$ can only be applied if $y$ has all real values but I am not sure.

And could an alternative method be given? This is the only method I have to approach these kind of problems.

答:
The flaw in your argument is that setting the discriminant $D$ to be non-negative merely finds the restriction on $a$ such that $$y=\sin^2x\in\mathbb{R};$$ whereas the actual goal is to find the restriction on $a$ such that $$x\in\mathbb{R}.$$ The latter restriction is narrower because $$x\in\mathbb{R}\implies y\in\mathbb{R} \quad\text{ but}\quad y\in\mathbb{R} \kern.6em\not\kern-.6em\implies x\in\mathbb{R}.$$

So to solve the problem, first simplify the given equation by completing the square:
$$\sin^4x-2\cos^2x+a^2=0\\
\iff (\sin^2x)^2-2(1-\sin^2x)+a^2=0\\
\iff (\sin^2x)^2+2\sin^2x+a^2-2=0\\
\iff (\sin^2x+1)^2+a^2-3=0\\
\iff (\sin^2x+1)^2=3-a^2.$$
Then set the given restriction:
$$x\in\mathbb{R}\\
\iff -1\leq \sin x \leq1\\
\iff 1\leq (\sin^2 x+1)^2 \leq4\\
\iff 1\leq 3-a^2 \leq4\\
\iff -1\leq a^2 \leq 2\\
\iff a^2\leq2\\
\iff -\sqrt2\leq a \leq\sqrt2.$$

Title: Why is my approach the wrong method?


问:
Find a group isomorphic to $GL(2, \mathbb{Z}/6\mathbb{Z})$.
The question is phrased exactly as in the title.

I've tried listing the elements of this group, but there are just too many of them and I don't think that it would be reasonable to work with multiplication tables here because of the sheer order of the group.

I'm not interested very much in the actual answer to this question but more so in the way of thinking that one could employ to tackle problems like this. What should I do first? Should I try to think of a suitable $1-1$ homomorphism and then maybe reduce the target space until it becomes onto as well...

I'm really stuck so any and all help would be much appreciated. Thank you in advance!

答:
We are looking at the automorphisms of the group $A:=\mathbb{Z}_6\oplus\mathbb{Z}_6=(\mathbb{Z}_2\oplus\mathbb{Z}_2)\oplus(\mathbb{Z}_3\oplus\mathbb{Z}_3)$.

Clearly, as the orders of elements are preserved, any automorphism of $A$ keeps the $2$- and $3$- parts invariant. So the automorphism group of $A$ is a subgroup of $\text{GL}(2,2)\times\text{GL}(2,3)$; and conversely every element of this product gives an automorphism of $A$.

So the answer to the part you are not very interested in is $\text{GL}(2,2)\times\text{GL}(2,3)$.


问:
Set is closed as for multiplication of matrices
We have the matrices \begin{equation*}s:=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}, \ d:=\frac{1}{2}\begin{pmatrix}-1 & -\sqrt{3} \\ \sqrt{3} & -1\end{pmatrix}\end{equation*} and the points \begin{equation*}p:=\begin{pmatrix}2 \\ 0 \end{pmatrix}, \ q:=\begin{pmatrix}-1 \\ \sqrt{3} \end{pmatrix}, \ r:=\begin{pmatrix}-1 \\ -\sqrt{3} \end{pmatrix}\end{equation*}

I draw the points $p, q, r$ and calculate also the points $sp, sq, sr, dp, dq, dr$ and I noticed that $s$ is a reflection as for the $x$-axis and $d$ is a rotation of $\frac{2\pi}{3}$.

Consider $G:=\{d, d^2, d^3, sd, sd^2, sd^3\}\subseteq \mathbb{R}^{2\times 2}$ and show that $G$ is closed as for multiplication of matrices, i.e. $gh\in G$ for all $g,h\in G$.

Show that the elements of $G$ are invertible and for each $g\in G$ there is $g^{-1}\in G$.

What is the geometric interpretation of $G$ ?

Let $z=\begin{pmatrix}-1 & 0 \\ 0 & -1\end{pmatrix}$ and $H:=G\cup \{zg\mid g\in G\}$. Show that $H\subseteq \mathbb{R}^2$ is closed as for multiplication of matrices.

$$$$

Let's start with question 1. Do we have to consider all possible combinations of the elements of $G$ and show that their product is again in $G$ ?

For example, do we have to do the following?

\begin{align*}d\cdot d^2&=d^3\in G \\ d\cdot d^3&=d^4=4-\text{times rotation about }120^{\circ}=\text{ratotaion about }480^{\circ}\\ & =\text{rotation about }360^{\circ}+120^{\circ}=\text{rotation about }120^{\circ}\\ & =d\in G \\ d\cdot s\cdot d&=\text{rotation about }120^{\circ}\text{ then reflection about x-axis and then rotation about }120^{\circ}\\ & =\text{reflection abour ax-axis}=s\in G\end{align*}

Or is there is shorter way?

At question 2 do we have to show that the matrices are invertible? Or do we have to consider what the matrices represent?

答:
There are several useful relations you can notice.

Since three rotations by $120^\circ$ give rotation by $360^\circ$, we have $d^3 = I.$

Rotating a point by $0^\circ$, $120^\circ$ and $240^\circ$ gives an equilateral triangle around the origin so the sum of those points is $0$. Therefore $d^2+d+I=0$.

Reflecting around $x$-axis twice doesn't change anything so $s^2=I$.

Algebraically, multiplying a matrix by $s$ multiplies a row or a column of a matrix by $-1$ so we get $$sd+ds = \frac12 \begin{bmatrix} -1 & \sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix}+\frac12 \begin{bmatrix} -1 & -\sqrt{3} \\ -\sqrt{3} & 1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = -s$$
and therefore $$ds = -sd-s = -s(d+I)=-s(-d^2) = sd^2.$$

Similarly, $$sd = sd^4 = (sd^2)d^2 = (ds)d^2 = d(sd^2) = d(ds) = d^2s.$$

Therefore, your set $G$ is $$G = \{I,d,d^2,s,sd,sd^2\} = \{I,d,d^2,s,ds,d^2s\}.$$
Denote $S = \{I,s\}$ and $D = \{I,d,d^2\}$ and it is easy to see that $S$ and $D$ are closed under multiplication. Now the crucial observation is that $G$ is the element-wise product of $S$ and $D$ in both orders:
$$G = SD = DS.$$

Take two elements $x,y \in G$. We can write $x = s_1d_1$ and $y = d_2s_2$ with $s_1,s_2 \in S, d_1,d_2 \in G$ so
$$xy = (s_1d_1)(d_2s_2) = s_1(d_1d_2)s_2 = s_2d_3s_2$$
for some $d_3 \in D$ since $D$ is closed under multiplication. Now $d_3 s_2 = s_2'd_3'$ for some $s_2' \in S$ and $d_3' \in D$ so
$$xy = s_2(d_3s_2) = s_2s_2'd_3' = s_3d_3' \in SD = G$$
for some $s_3 \in S$ since $S$ is closed under multiplication. Therefore $xy \in G$ so $G$ is closed under multiplication.

It is also very evident that all matrices in $G$ are invertible. Namely, $s$ and $d$ are invertible with inverses $s^{-1}=s$ and $d^{-1}=d^2$ and hence all matrices in $S$ and $D$ are invertible as products of invertible matrices. Moreover, element-wise we have $S^{-1} = S$ and $D^{-1}=D$. Now for $x \in G$ we have $x = s_1d_1$ for some $s_1\in S$, $d_1 \in D$. We have
$$x^{-1} = (s_1d_1)^{-1}=d_1^{-1}s_1^{-1} \in D^{-1}S^{-1}=DS = G$$
so $x^{-1} \in G$.

The relevant reference is the result that element-wise product $EF$ of two subgroups $E, F$ is itself a subgroup if and only if $EF = FE$.


问:
Let $G$ a connected graph, and $B(G)$ the set of blocks of $G$, $cv(G)$ the set of cut vertex of $G$
Let $G$ a connected graph, and $B(G)$ the set of blocks of $G$, $cv(G)$ the set of cut vertex of $G$, $b(x)$ the number of blocks that contain $x$. Show that $$|B(G)|=1+\sum_{x\in V(G)}(b(x)-1)$$

getting the result.

One way was, consider two cases:

There are no cut vertices. In this case $ b (x) = 1 $, then $b(x)-1=0$ for any $x$.

There are cut vertices. In this case $x$ is in at least $2$ blocks, so $b(x)\ge2$, therefore $b(x)-1\ge1$. Then we can take the sum.

答:
There may be a simpler counting argument, but you can do induction on the number of blocks of $G$. The base case of $1$ block is simple enough.

Note you will need the fact that if $G$ has a cut-vertex, then there is a block $H$ of $G$ containing exactly one cut-vertex, $x$. I believe one can see this fairly quickly by contradiction.

We then remove the vertices of $H - x$ from $G$ to obtain $G'$, a graph with $1$ fewer block than $G$. Then $G'$ satisfies the induction hypothesis. We have
\begin{align}
|B(G)| &= |B(G')| + 1\\
&= \left(1 + \sum_{v \in V(G')} (b(v) - 1)\right) + 1\\
&= 1 + \sum_{v \in V(G)} (b(v) - 1).
\end{align}
The final equality comes from the fact that $b(x)$ in $G$ is one more than it is in $G'$, and for all other vertices $w$ in $H$ (the vertices we removed) we have $b(w) = 1$ in $G$, i.e. $b(w) - 1 = 0$.


问:
Finding asymptotes to general functions
I stumbled upon this problem and didn't quite understand why the solution worked. I have two questions:

Why is it that the taylor series is enough to find the oblique asymptote? Specifically, why do the first two terms suffice? It seems like that may not work for all functions.

From pre-calculus I recall finding asymptotes of rational functions, but how do I find vertical, horizontal, and oblique asymptotes of a more general function (such as an exponential or square root, etc.)?

答:
1) In general it holds for every a that $$\lim\limits_{n->{\infty}} \frac{a}{n} = 0$$. Therefore, all the terms except for the first two disappear.

2) It works in exactly the same way. Just take the limit to infinity and see what result you obtain. However, to take this limit, you sometimes need tricks such as the Taylor polynomials in your example.

Title: Finding asymptotes to general functions


问:
Least squares regression of sine wave
Let's say that a sine-like function of a fixed frequency and zero-mean can only vary in amplitude and offset.

That is, for variables $a \in R^+$ and $z \in [0..2\pi]$

$$ f(t) = a(\sin(kt + z))$$

(where $k$ is a constant proportional to the fixed frequency)

For some set of $n$ samples $(t_1,y_1),(t_2,y_2)...(t_n, y_n)$ we want to fit $a$ and $z$ such that they minimize:

$$ L(a,z) = \sum(f(t_i)-y_i)^2 $$

If f was a linear function we could just use ordinary least squares regression.

Is there some similar technique for the above function?  What approach would you use to minimize $L$ ?

答:
In general because of the presence of periodic functions, this is not the simplest problem in the domain of curve fitting. However, this case is simple because $k$ is not a tunable parameter but a fixed constant.

You have $n$ data points $(t_i,y_i)$ and you want to perform a least square fit based on the model
$$y=a \sin(kt+z)$$
Rewrite is as
$$y=a \cos (z) \sin (k t)+a \sin (z) \cos (k t)$$ and define
$$A=a \cos (z)\qquad B=a \sin (z)\qquad S_i=\sin (k t_i)\qquad C_i=\cos (k t_i)$$ which makes the model to be
$$y=A \,S+B \,C$$ which is just a bilinear regression without intercept.

When solved for $(A,B)$, you have
$$a^2=A^2+B^2\qquad\qquad\text{and}\qquad\qquad z=\tan ^{-1}\left(\frac{B}{A}\right)$$

I tried an example : using $n=21$ and $k=2$, I generated the following table
$$\left(
\begin{array}{cc}
t & y \\
 0.0 & 42 \\
 0.5 & 121 \\
 1.0 & 89 \\
 1.5 & -25 \\
 2.0 & -115 \\
 2.5 & -99 \\
 3.0 & 8 \\
 3.5 & 108 \\
 4.0 & 109 \\
 4.5 & 10 \\
 5.0 & -98 \\
 5.5 & -115 \\
 6.0 & -27 \\
 6.5 & 87 \\
 7.0 & 121 \\
 7.5 & 44 \\
 8.0 & -73 \\
 8.5 & -123 \\
 9.0 & -59 \\
 9.5 & 59 \\
10.0 & 124
\end{array}
\right)$$

This gives
$$A=116.1789094\qquad\qquad\text{and}\qquad\qquad B=41.88715050$$ from which
$$a=123.4992808\qquad\qquad\text{and}\qquad\qquad z=0.3460335933$$

In fact the data  were generated using
$$y_i=\lceil 123.456 \sin (2 t+0.345678)\rceil$$

Title: Least squares regression of sine wave


问:
Calculate norm of a polynomial knowing inner product but not the polynomials themselves
In $\mathbb R[t]_2$ (vector space of all polynomials of up to a second degree), I need to compute the norm $||t^2 - 2t + 1||$, knowing only that:

$$\langle p(t),q(t)\rangle = p(-2)q(-2) + p(-1)q(-1) + p(0)q(0) + p(1)q(1)$$

I have absolutely no clue on how to compute it as I don't have the polynomials $p(t)$ and $q(t)$.
Do I need to assume $p(t)$ and $q(t)$ generic polynomials, such as $p(t) = a + bt + ct^2$ and $q(t) = d + et + ft^2$, where $a,b,c,d,e,f$ would be scalars in $\mathbb R$?

答:
As you have computed once for all the expression

$$\langle p,q \rangle=-2ea+4ad+6af+6eb-8ec-2bd+6cd-8bf+18cf,$$

said otherwise

$$(a \ \ b \ \  c)\underbrace{\left(\begin{array}{rrr}2&-1&3\\-1&3&-4\\3&-4&9\end{array}\right)}_A\begin{pmatrix}d\\e\\f\end{pmatrix},$$

it is not lost time if you are asked to calculate other norms.

Moreover, and this is the most important, due to the fact that the eigenvalues of symmetric matrix $A$ are positive ($0.5815,$$1.4351,$ $11.9834$) it proves that this inner product is associated with a "true" norm, which isn't a priori evident (triangular inequality is lengthy by direct means).


问:
Why is $u\cdot v/(||u||||v||)=\cos\theta$ not giving me the correct result? What am I doing wrong?
This might seem like a stupid question, but
why is theta in the follow picture
$60^\circ$, I understand how it is $60^\circ$ through simple trig, through $2 \cos \theta$, you can find $\theta$ to be $60^\circ$. Yet if you do the equation
$$
\frac{|u\cdot v|}{||u||\cdot ||v||} = \cos\theta
$$
while $v = [1,2]$ and $u = [1,0]$, you get $\theta \approx 63^\circ$.
What am I doing wrong?
P.S. This comes from a question I got where:

Bert can swim at a rate of $2$ miles per hour in still water. The current in a river is ﬂowing at a rate of $1$ mile per hour. If Bert wants to swim across the river to a point directly opposite, at what angle to the bank of the river must he swim?

I know the answer is 60*, yet I am confused why it is 60 but not around 63. Thank you.

答: I see nothing wrong there. Indeed,$$\frac{u.v}{\|u\|.\|v\|}=\frac1{\sqrt5}\quad\text{and}\quad\arccos\left(\frac1{\sqrt5}\right)\simeq63.4^\circ.$$


答:
Your vectors are wrong. In particular, the vector $v$ is not $(1,2)$ but $(1,\sqrt{3})$ so that its magnitude (i.e. the speed of swimming) would be $2$.

With $v=(1,\sqrt{3})$, you have $u\cdot v=1$ and $|u|=1, |v|=2$ so you get $\cos\theta=1/2$ as expected.

Title: Why is $u\cdot v/(||u||||v||)=\cos\theta$ not giving me the correct result? What am I doing wrong?

Body:
This might seem like a stupid question, but
why is theta in the follow picture
$60^\circ$, I understand how it is $60^\circ$ through simple trig, through $2 \cos \theta$, you can find $\theta$ to be $60^\circ$. Yet if you do the equation
$$
\frac{|u\cdot v|}{||u||\cdot ||v||} = \cos\theta
$$
while $v = [1,2]$ and $u = [1,0]$, you get $\theta \approx 63^\circ$.
What am I doing wrong?
P.S. This comes from a question I got where:

Bert can swim at a rate of $2$ miles per hour in still water. The current in a river is ﬂowing at a rate of $1$ mile per hour. If Bert wants to swim across the river to a point directly opposite, at what angle to the bank of the river must he swim?

I know the answer is 60*, yet I am confused why it is 60 but not around 63. Thank you.


问:
Euler's method for second order differential equation
Not really homework but sample exam.

The question is to use Euler's Method to approximate Y:

$Y''(t) = Y'(t) - 2Y(t)$

$Y'(0) = Y(0) = 1$

with $t_0 = 0$ and $h=0.2$

So what I did:

First iteration:

$t_1 = 0.2$

$y(t_1) = y(t_0)+h \cdot y'(t_0) = 1 + 0.2 \times 1 = 1.2$

$y'(t_1) = y'(t_0) + h \cdot (y'(t_0) - 2y(t_0)) = 1 + 0.2 \times (1-2\times1) = 0.8$

Second iteration:

$t_2 = 0.4$

$y(t_2) = y(t_1)+h \cdot y'(t_1) = 1.2 + 0.2 \times 0.8 = 1.36$

$y'(t_2) = y'(t_1) + h \cdot (y'(t_1) - 2y(t_1)) = 0.8 + 0.2 \times (0.8-2\times1.2) = -2.4$

Correct?

答:
There is an arithmetical mistake at the very last step. I get $y'(t_2)=0.48$.

Such problems are easy to code in a spreadsheet.

```<code>A     B       C
0     1       1
0.2   1.2     0.8
0.4   1.36    0.48
0.6   1.456   0.032
</code>```

Here A is for $t$, B is for $y(t)$, and C is for $y'(t)$. The top row are initial values A1=0, B1=1, C1=1. The second row is the Euler step: A2=A1+0.2 , B2=B1+0.2*C1, C2=C1+0.2*(C1-2*B1). Then drag down for as many rows as you wish.

If for some odd reason you can't use spreadsheet software during an exam, at least it gives a way to check your hand computations.

Body:
There is an arithmetical mistake at the very last step. I get $y'(t_2)=0.48$.

Such problems are easy to code in a spreadsheet.

A     B       C
0     1       1
0.2   1.2     0.8
0.4   1.36    0.48
0.6   1.456   0.032

Here A is for $t$, B is for $y(t)$, and C is for $y'(t)$. The top row are initial values A1=0, B1=1, C1=1. The second row is the Euler step: A2=A1+0.2 , B2=B1+0.2*C1, C2=C1+0.2*(C1-2*B1). Then drag down for as many rows as you wish.

If for some odd reason you can't use spreadsheet software during an exam, at least it gives a way to check your hand computations.


问:
Minimum Variance Estimators that Attain the Cramér-Rao lower bound
The probability mass function of a negative binomial random variable $X$ is given by $$P(X=x\mid\theta)= {r+x-1 \choose x}(1-\theta)^x\theta^r$$ and $\Bbb E(X\mid \theta)=\frac{r(1-\theta)}{\theta}$.

Does this model admit a minimum variance unbiased estimator that attains the Cramér-Rao lower bound?

What I have tried:

I have worked out the maximum likelihood estimator $\hat{\theta}$ to be $\frac{r}{r+x}$  and the CRLB to be  $\Bbb V(\hat{\theta}) \geq \frac{1}{I(\theta)}= \frac{\theta^2-\theta^3}{r}$.

I know that if the estimator attains the CRLB then we know that we have found a Minimum Variance Unbiased Estimator. Also if the variance of the estimator is equal to the CRLB then it is an MVUE.  However I am unsure of how to calculate the variance of $\hat{\theta}$.

Thank you for your help!

答:
I know that if the estimator attains the CRLB then we know that we have found a Minimum Variance Unbiased Estimator. Also if the variance of the estimator is equal to the CRLB then it is an MVUE.

An UMVUE (Unformly Minimum Variance Unbiased Estimator) rarely attains the CRLB.

Moreover: if a family of rv's belongs to the exponential family (as it is the case) always exist an estimator that attains CRLB.

To verify that an estimator attains the lower bound there is a IFF condition:

An estimator $T$ attains CRLB iff there is a function $b(n,\theta)$
such that

$$\sum_{i=1}^n\frac{\partial}{\partial\theta}\log f(x_i;\theta)=b(n,\theta)[T-\mathbb{E}[T]]$$

In your case you have

$$\sum_{i=1}^n\frac{\partial}{\partial\theta}\log f(x_i;\theta)=\frac{nr}{\theta}-\frac{\Sigma_i X_i}{1-\theta}=\underbrace{-\frac{n}{(1-\theta)}}_{=b(n,\theta)}\Bigg[\overline{X}_n-\frac{r(1-\theta)}{\theta}\Bigg]$$

Thus $\overline{X}_n$ is UMVUE for $\mathbb{E}[X]$ and it attains CRLB

References taken from Mood Graybill Boes, McGraw Hill


问:
Number of elements of order 4 in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$
I've read that number of elements of order 4 in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ is 8, but I can't figure it out where this number come from. I now that if we're searching number of elements of order 4 in $\mathbb{Z}_{16}$, we use Euler's phi function ($\phi(4)=2$). Do we use something similar here?

答: The order of an element in a direct product like this one is the l.c.m of the orders inside the respective coordinates. So an element of order $4$ in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ must be $1$ or $3$ in the first coordinate. In the second and third coordinate, you could have either $0$ or $1$. In each case you get elements of order $4$. There are $2 \cdot 2 \cdot 2 = 8$ such possibilities.


答:
Suppose that $(a,b,c)$ is an element of order $4$. As $2b=2c=0$, $a$ has to be an element of order $4$. Therefore $a \in \{1,3\}$.

And then $b,c$ can be whatever you want. Therefore $8$ elements of order $4$.

Title: Number of elements of order 4 in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

Body: I've read that number of elements of order 4 in $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ is 8, but I can't figure it out where this number come from. I now that if we're searching number of elements of order 4 in $\mathbb{Z}_{16}$, we use Euler's phi function ($\phi(4)=2$). Do we use something similar here?


问:
Find a certain group $G$ with $H,K\unlhd G$, $H\cong K$ and $(G/H)\ncong(G/K)$
I'm trying to solve a problem from my group theory text book. It says:

Find a certain group $G$ with $H,K\unlhd G$, that verifies $H\cong K$ but
$(G/H)\ncong(G/K)$.

I don't know what group to consider. At first I thought about the quaternion $Q_8$, but I didn't find a solution (in my book, this kind of exercises use to use the $Q_8$). Then I considered the diedric groups $D_n$ and the symmetric $Sn$ but also got nothing. What is a possible solution to this problem? Any help will be appreciated, thanks in advance.

答: So you can just consider $G = \mathbb Z, H= 2 \mathbb Z, K = 3 \mathbb Z$. Then $H \cong K$ but $\lvert G / H \rvert = 2$ and $\lvert G / K \rvert = 3$, hence $G / H \not\cong G / K$.


问:
Composition with inclusion map
Does the composition between the inclusion map and another function $a$ (obviously if I can compose them), work as the function $a$?

答: If $A\subseteq X$ and $\iota$ denotes the inclusion function $A\to X$ and $f$ is a function $X\to Y$ then composition $f\circ\iota$ is a function $A\to Y$ prescribed by: $$a\mapsto f(a)$$

Title: Composition with inclusion map


问:
Example that isometry doesn't preserve extrinsic distance
Is there any example that an isometry doesn't preserve the extrinsic distance? That is, I want some example that for some isometry $f:S\to\tilde{S}$ between connected regular surfaces such that it doesn't satisfy this property: $|f(p)-f(q)| = |p-q|$ for all $p,q\in S$.

答: Take a rectangle in a plane and the corresponding portion of a (right circular) cylinder.</p>


问:
Show that a triangle delimited by parallel lines to the sides of another triangle is similar to said triangle.
I was looking at geometric constructions of similar triangles, and at one point I came across the statement in the question.

A triangle delimited by vertices which are intersections of lines parallel to the sides of some other triangle is similar to said triangle.

I drew an example of the statement as follows:

Where the small green triangle's vertices are given by the intersections of some lines which are parallel to the sides of the blue triangle. The statement seems intuitively obvious, but I can't seem to find a way to justify it. I tried using the intercept theorem which is what I normally see is used to prove statements about similar triangles, but since I'm not sharing a side in both triangles I didn't think I could use it. Does anyone know a simple argument to justify this? Thank you!

答: The angle between two lines is the same as the one between two lines parallel to the original ones. Therefore the 3 angles of the deduced triangle are the same than the ones of the original triangle and the triangles are similar.</p>


问:
What is the difference between $\leftrightarrow$, $\iff$, and $\equiv$, especially when used in the same text?
I think that the difference has something to do with object language and meta language, but I'm not sure.

I've heard that $\leftrightarrow$ is a connective on a proposition level, whereas $\iff$ is a statement about propositions. But where does $\equiv$ come in?

I've seen texts that use either ($\leftrightarrow$ and $\iff$) or ($\iff$ and $\equiv$). But how do those three relate if they are all used at the same time? Would such use imply that there can be a meta language of a meta language? If not, which pairing makes more sense to use?

答:
↔ short left right arrow - has not default meaning, reference If_and_only_if.

⟺ double short left right arrow - has not default meaning, but can be used in books about List of logic symbols.

≡ congruent  - has not default meaning. Look up congruent in the context of modulo division classes. ⟺ double short left right arrow might indicate that too.

For reference purposes look at Operators Without Built-In Meanings. These are symbols available both in LATEX and MathML. In these they have just the names I gave in first place and are entered in a short form. These names are known to google for example, ↔ short left right arrow LATEX or w3.org operator dictionary for MatML as appendix.

Body:
↔ short left right arrow - has not default meaning, reference If_and_only_if.

⟺ double short left right arrow - has not default meaning, but can be used in books about List of logic symbols.

≡ congruent  - has not default meaning. Look up congruent in the context of modulo division classes. ⟺ double short left right arrow might indicate that too.

For reference purposes look at Operators Without Built-In Meanings. These are symbols available both in LATEX and MathML. In these they have just the names I gave in first place and are entered in a short form. These names are known to google for example, <a href="https://www.google.com/search?client=firefox-b-d&sxsrf=ALeKk01iaR3SMjJH-HIWDKXLX9_zIFi0Zg%3A1606579450004&ei=-XTCX_zhPM3ykwXUzKPwAg&q=%E2%86%94%20short%20left%20right%20arrow%20LATEX&oq=%E2%86%94%20short%20left%20right%20arrow%20LATEX&gs_lcp=CgZwc3ktYWIQAzIICCEQFhAdEB4yCAghEBYQHRAeMggIIRAWEB0QHjoRCAAQsAMQigMQtwMQ5QIQiwNQ9xBY-BNg7xZoAXAAeACAAVWIAaIBkgEBMpgBAKABAaABAqoBB2d3cy13aXrIAQm4AQLAAQE&sclient=psy-ab&ved=0ahUKEwi8wq7rzqXtAhVN-aQKHVTmCC4Q4dUDCAw&uact=5" rel="nofollow noreferrer">↔ short left right arrow LATEX or w3.org operator dictionary for MatML as appendix.


问:
Obtention and estimate error of trapezoid rule: Why was necessary to mention that $f''(\xi)$ is constant?
I'm considering the following integral

$$I=\int_a^b[f(a)+\Delta f(a)\alpha+\frac {f''(\xi)}{2}\alpha(\alpha-1)h^2]dx$$

where $a\le x\le b, h=b-a, \Delta f(a)=f(b)-f(a), \xi$ is a number between $a$ and $b$. If $\alpha=\frac{x-a}{h}$ then $dx=hd\alpha$. Thus

$$I=h\int_0^1[f(a)+\Delta f(a)\alpha+\frac {f''(\xi)}{2}\alpha(\alpha-1)h^2]d\alpha$$

If one assumes that for $h$ small, $f''(\xi)$ is approximately constant, then

$$I=h\left[\alpha f(a)+\frac{\alpha^2}{2}\Delta f(a)+\left(\frac{\alpha^3}{6}-\frac{\alpha^2}{4}\right)+f''(\xi)h^2\right]_0^1d\alpha$$

My question is, why is the author making this assumption of "If one assumes that for $h$ small, $f''(\xi)$ is approximately constant"? Wasn't already a constant?

Thank you in advance

答:
In the remainder of a linear interpolation at $a$ and $b$ you get
$$
f(x)=p_1(x)+(x-a)(x-b)r(x).
$$
On the one hand, one can show that for twice differentiable $f$ the function $r$ is continuous in $x$. On the other hand, as you used it, it can be shown that there is a $\xi_x$ in the hull of the points $a,b,x$ so that $r(x)=\frac{f''(\xi_x)}2$.

For the integration you can now take the first version to apply the mean value theorem,
$$
\int_a^b(x-a)(x-b)r(x)dx=r(\eta)\int_a^b(x-a)(x-b)dx.
$$
After that use the connection to the second derivative with $\xi=\xi_\eta$. So there is no need to consider $h$ small enough or $\xi$ nearly constant or such.


问:
Haar measure on Lie Group is unique
My question is

Why is the Haar measure on a Lie Group unique upto scalar multiple?

I know how to show it for $\mathbb R^n$ because there I have countable many open balls that form a base and the measure of the unit ball around 0 gives the constant scalar.

How to show for general Lie Group?

答:
Let $\mu,\nu$ be two left Haar measures on a locally compact group. Then $\mu+\nu$ is a left Haar measure, and $\mu$ has a density $f$ with respect to it. Then $f$ is left-invariant, so is constant. Hence $\mu$ is a scalar multiple of $\mu+\nu$.

I don't really see how to make it simpler, even with additional assumptions?

Title: Haar measure on Lie Group is unique


问:
Simplifying $\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7}$

$$\sqrt {11 - 4 \sqrt 7} - \sqrt {8 - 2 \sqrt 7} = a$$

I'm having trouble solving this equation. I've tried squaring both sides and got this

$$11 - 4\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} + 8 - 2\sqrt{7} = a^2$$

after simplifying

$$19 - 6\sqrt{7} - 2 \sqrt{(11 - 4\sqrt{7})(8 - 2\sqrt{7})} = a^2$$

and that's where I got stuck.

Body:
Hint:

$11=(\sqrt7)^2+2^2$  and $\sqrt7-2>0$

$8=(\sqrt7)^2+1^2$ and  $\sqrt7-1>0$

Finally

$$\sqrt{a^2+b^2-2ab}=|a-b|=a-b\text{  if } a-b\ge0$$


问:
Is there (or can there be) a general algorithm to solve Rubik's cubes of any dimension?
I love solving Rubik's cube (the usual 3D one). But, a lecture by Matt Parker at the Royal Institute (YouTube Link) led me to an app that can simulate a four dimensional rubik's cube. But unfortunately it was so complex, that I soon got bored as I failed to solve it.

The website to the 4D cube : http://superliminal.com/cube/cube.htm

Following which today, I also found an app that can simulate a 5D cube!!

The website to the 5D cube: http://www.gravitation3d.com/magiccube5d/

So, my two questions are :

Is there a general (non brute-force) algorithm that can be used to solve a well-scrambled cube of any dimension (even though it may not be very efficient, but yet is not a simple search over all the available space of move sequences) ? [Point modified after reading @RavenclawPrefect's answer :D ]

Mathematically, what is common in all these cubes, and "hypercubes"?

NOTE: By dimensions I mean physical dimension and not the number of rows of pieces on a face of the cube. So a four-dimension cube is a cube in x,y,z,$\delta$ dimensions where $\hat x,\hat y,\hat z,\hat \delta$ are mutually orthogonal unit vectors in the respective dimensions.

For example, here is how the aforementioned 4D cube moves:
https://miro.medium.com/max/2552/1*ga32DoV_Hc6e8t6PC1hFHw.gif

Addendum:

List of resources that may help finding an answer to this question:

Solving Rubik's cube and other permutation puzzles

答:
Firstly, are you familiar with the commutator-based solution to general permutation puzzles? You certainly can do the same here for each specific dimension and size, and probably can generalize it to arbitrary dimension and size with some effort. It comes down to finding suitable commutators for each of the orbits and showing that parities can be resolved. Here an orbit is a set $S$ of positions such that any piece at a position in $S$ can be moved to every position in $S$ but no position outside $S$. (In group-theoretic terminology, the action of the group of all scrambles of the cube on the cube is transitive.)

While I have not thought about the details, the general idea would be to show the following:

There is some algorithm to make it so that for every orbit $S$ the pieces in $S$ are in some even permutation.

For each orbit $S$, there is some 3-cycle commutator on some positions $P,Q,R$ in $S$ and some algorithm to move pieces at any given positions $A,B,C$ in $S$ to $P,Q,R$ respectively. This lets you perform any even permutation on the pieces in $S$.

For each orbit $S$, I think you can define the orientation of pieces in each orbit such that each face turn does not change their total orientation, and there is some commutator that only changes the orientation of the pieces at some positions $P,Q$ by $+1$ and $-1$ respectively, and some algorithm to move pieces at any given positions $A,B$ in $S$ to $P,Q$ respectively. This would suffice to let you fix the orientation of all pieces in $S$.

Once you establish the above points, you can simply apply them in that order to solve the cube. For example, it is not hard to prove the above points for the 3d cube of arbitrary size, where you use the slice turns to achieve the first point (fixing parity for all orbits), even if it is a little messy. I think it generalizes to arbitrary dimension in a similar fashion.


问:
Alternative proof for $(m+1)^3 \leq2m^3$
This is the proof on the textbook:

Until here, I was able to get to the same conclusion. What I tried to do was to prove by induction that $(m+1)^3 \leq2m^3$. I started from $(m + 2)^3$, but got nowhere.

The author then does the following:

I don't understand the first three lines of the previous picture (before the proof starts) nor how he uses it in the proof.

Is it possible to prove $(m+1)^3 \leq2m^3$ by induction? Or is there any other simpler proof that I'm not seeing?

Could someone please throw some light on the proof used in the textbook? I find it highly unintuitive.

答:
The author made it too complicated.

$$(m+1)^3 \le 2m^3 \iff m+1 \le \sqrt[3]{2}m \iff m \ge \frac{1}{\sqrt[3]{2}-1}$$

Using a calculator you get $\frac{1}{\sqrt[3]{2}-1} \approx 3.8,$ but it's not difficult to prove it's between 3 and 4:
$$ 3 < \frac{1}{\sqrt[3]{2}-1} < 4 \iff \frac 43 > \sqrt[3]{2} > \frac 54 \iff \frac{64}{27} > 2 > \frac{125}{64}.$$

So $(m+1)^3 \le 2m^3$, if $m\ge 4.$