问:
How to show $\lim _{n \rightarrow \infty} \int_{1-\frac{1}{n}}^{1}f(x) d x=0$
If $f: [0,1]\to R$ is a continuous function, then prove that:
$$\lim _{n \rightarrow \infty} \int_{1-\frac{1}{n}}^{1}f(x) d x=0$$
My try:
Since $f$ is continuous at $x=1$

we have for a given $\epsilon >0$, $\exists, \delta >0$ such that
$$|f(x)-f(1)|<\epsilon$$ when $1-\delta <x <1$

Now $\forall n \geq 1$
$$\left|\int_{1-\frac{1}{n}}^{1} f(x) d x-0\right| \leq \int_{1-\frac{1}{n}}^{1}|f(x)| d x\leqslant \int_{1-\frac{1}{n}}^{1}|f(x)-f(1)| d x+\int_{1-\frac{1}{n}}^{1}|f(1)| d x$$
$\implies$
$$\left|\int_{1-\frac{1}{n}}^{1} f(x) d x-0\right| \leq \int_{1-\frac{1}{n}}^{1} \varepsilon d x+\frac{f(1)}{n}=\frac{\varepsilon+f(1)}{n}$$
Thus we have:
$$\lim _{n \rightarrow \infty} \int_{1-\frac{1}{n}}^{1}f(x) d x=0$$

Is this proof correct?

答:
There is a minor issue with your proof. The equality

$$\int_{1-\frac{1}{n}}^{1} \varepsilon d x+\frac{f(1)}{n}=\frac{\varepsilon+f(1)}{n}$$ is not correct. However you could conclude by writing

$$\int_{1-\frac{1}{n}}^{1} \varepsilon d x+\frac{f(1)}{n}\le\varepsilon+\frac{\vert f(1) \vert}{n}$$
taking $N$ such that for $n \ge N$ you have $\frac{\vert f(1) \vert}{n} \le \varepsilon$ and say that as the absolute value of the integral is less than $2 \varepsilon$ for any $\varepsilon \gt 0$, it has to be equal to zero.

An easier way would be to notice that a continuous map on the closed segment $[0,1]$ is bounded. Therefore it exists $M \gt 0$ such that

$$\vert f(x) \vert \le M$$ for all $x \in [0,1]$. From there you have

$$\left\vert \int_{1-\frac{1}{n}}^{1}f(x) dx \right\vert \le \int_{1-\frac{1}{n}}^{1}\vert f(x) \vert dx \le \int_{1-\frac{1}{n}}^{1}M dx \le \frac{M}{n}$$ and the right end side of those inequalities converges to zero as $n \to \infty$.


问:
subgroup of $\Bbb Q_p^×$ of index p contains $(\Bbb Q_p^×)^p$ as subgroup
Let $H$ be a subgroup of $\Bbb Q_p^×$ of index p.
Then, I would like to prove $H$ contains $(\Bbb Q_p^×)^p$ as subgroup.

I know $(\Bbb Q_p^×)^p$ has index $p^2$ because $ \Bbb Q_p^×/(\Bbb Q_p^×)^p$ has order $p^2$.

I encountered this question when I was trying to count $\Bbb Q_p^×$'s abelian extension of order $p$.

Thank you in advance.

答:
If $G$ is any group (written multiplicatively) and $H$ is any normal subgroup of $G$ of index $n$, then $H$ must contain $G^n$.

This is because the quotient group $G/H$ has order $n$ and hence the $n$-th power of any element of $G/H$ is the identity element.

Translating back to $G$, we see that $g^n \in H$ for any $g \in G$.

ContentLicense: CC BY-SA 4.0


问:
Do these images both represent disk models of Euclidean geometry?

Q: Do these images represent the same thing? The first one is a disk model of Euclidean geometry, and the second one is the warping of a regular grid through a glass ball. If they are the same, why? It seems like the second image is in the realm of spherical geometry. Why are the geodesics circular arcs if it's supposed to represent Euclidean geometry?

This is a disk model of Euclidean geometry (where geodesics are circular arcs).

It looks exactly like the grid on the glass ball in this image.

答:
I'm not sure what ℝ²→ℝ² map your "disc model of Euclidian geometry" is using.

The "warped by glass ball" map is called refraction, and is explained by Ray Tracing math.

To be honest, I am not sure whether an ordinary refractive glass sphere would show the whole plane, but this can be determined using ray tracing.

How do you know the first one is all circular arcs?


问:
Prove that there exists N ∈ (N) such that $a_n$ > 0 for all n ≥ N.
Let $(a_n)$ be a convergent sequence. Suppose that $\lim_{n\to\infty}$ $a_n$ > 0. Use the definition of a
limit to prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.

Use definition of a limit ($|a_n-L|$<$\epsilon$) and definition of convergence of sequences.

I have no idea how to even start this proof, or what I should be doing with it. I know I'm trying to solve that $a_n$ > 0, but I don't know how to get there. Not looking for a full proof, just some help of what I should be doing, what I should choose N to be, etc.

答: Let $L:=\lim_{n\to\infty} a_n >0$ and set $\varepsilon =\frac{L}{2}$: then by definition of limit we know there exists $N$ such that $|a_n-L|<\varepsilon$ for all $n>N$ - conclude from here that $a_n>0$ for all $n>N$-


答: Taking $\epsilon = a = \lim_{n \to \infty} a_n$, we know there is $N$ such that for $n \geq N$ we have $|a_n - a| < a$ hence $-a < a_n - a < a$ and $0 < a_n < 2a$. In particular, $a_n > 0$ for $n$ large enough.

Title: Prove that there exists N ∈ (N) such that $a_n$ > 0 for all n ≥ N.

Body:
Let $(a_n)$ be a convergent sequence. Suppose that $\lim_{n\to\infty}$ $a_n$ > 0. Use the definition of a
limit to prove that there exists N ∈ (N) (naturals) such that $a_n$ > 0 for all n ≥ N.

Use definition of a limit ($|a_n-L|$<$\epsilon$) and definition of convergence of sequences.

I have no idea how to even start this proof, or what I should be doing with it. I know I'm trying to solve that $a_n$ > 0, but I don't know how to get there. Not looking for a full proof, just some help of what I should be doing, what I should choose N to be, etc.


问:
Sequences in the cofinite topology
Let $X$ be an infinite set with the cofinite topology. I must prove that if $(x_n)$ is a sequence and $x_n \ne x_m$ for all $n \ne m$, then any $x$ is the limit of $x_n$.

Can anyone give me some explained examples on this? I don't get how a sequence works in this topology.

答: Let $x\in X$ and $U\in \mathcal{T_{X}}$ an open subset containing $x$.. The definition of convergence of sequences states that $(x_n)$ converges to x if for every open set U containing $x$ there $\exists \delta \in \mathbb{N}$ such that $\forall n\gt \delta x_n\in U$ ..it means that after a finite number of terms of $(x_n)$ all the infinite  remaining terms enter U.. Since the topology here is the cofinite topology then the compliment of every open set U is finite $\implies$ certainly an infinite number of terms of $(x_n)$ enters U. Thus $x$ is a limite of $(x_n)$.

Title: Sequences in the cofinite topology


问:
What's wrong in my approach in evaluating $\cos^{-1}(\cos\ 10)$
The Question

$\large\cos^{-1}(\cos 10)$

My Approach

$\large\cos^{-1}(\cos\ (10-2\pi))$ As $2\pi$ is the period of $\cos x.$ Now our angle is-

Now we know that for $\large\cos^{-1}(\cos x) = x$, $\large x\ \epsilon\ [0,\pi]$ but currently our $\large x$ is in the 3rd quadrant, we need to get it into 1st or 2nd quadrant. Therefore let us subtract $\large \pi$ from our current angle i.e $\large 10-2\pi$

$\large\cos^{-1}(\cos\ -((10-2\pi)-\pi))$ Now you must be wondering why did I put this minus sign, w.k.t $\large \cos x$ is -ve in third quadrant, by subtracting $\large \pi$ we brought it into 1st quadrant and $\large \cos x$ is +ve in 1st quad, so to compensate that I have put a minus sign.

Now its within range so we can use $\large\cos^{-1}(\cos x) = x$.

So Final answer $\large -((10-2\pi)-\pi)$ = $\large 3\pi - 10$

But this answer is wrong and the correct answer is $\large 4\pi - 10$.

They have used the property $\large \cos\  (\pi - \theta) = \cos\  (\pi + \theta)$ But why my approach is wrong?

Kindly help.

答:
I suppose that $10^c$ means 10 radians.

$\pi^c=180^o\implies 1^c \sim57^o\implies 10^c\sim 570^o$

$\cos 10 =\cos (10-2\pi)=-\cos(10-3\pi)=\cos (\pi-(10-3\pi))$
$\arccos(\cos 10)=4\pi-10$

Your answer turns out to be wrong as $\arccos(\cos x)=x$ when $x\in [0,\pi]$. Note that $3\pi-10\notin [0,\pi]$.


问:
A die is throw until 5 followed by 3 appears
A die is thrown until 5 followed by 3 appears. Like 156234753

What is the expected number of throws?

I've been trying to solve this problem but can't find anything similar.

Any help?

My solution:

Let $\displaystyle X$ be the random variable of the qustion (number of throws until 53).

Let the event $\displaystyle A=$after the first $\displaystyle 5$, you get a $\displaystyle 3$.

From the total law of expectation:

$\displaystyle E[ X] =E[ X|A] \cdotp P( A) +E[ W|\overline{A}] \cdotp P(\overline{A}) =E[ X|A] \cdotp \frac{1}{6} +E[ W|\overline{A}] \cdotp \frac{5}{6}$.

$\displaystyle E[ X|A] :$

We have gotten a $\displaystyle 5$ and then a $\displaystyle 3$. So we stop here. The expected number is the expected until $\displaystyle 5$ plus $\displaystyle 1$ for the $\displaystyle 3$: $\displaystyle 6+1=7$.

$\displaystyle E[ X|\overline{A}]$:

We have gotten a $\displaystyle 5$, but now not a $\displaystyle 3$ (assume we got some $\displaystyle x$).

We don't care about all the outputs before $\displaystyle x$, since they're not relevant to question anymore.

We can "pretend" then that we have just started with the output of $\displaystyle x$, and act as if that was our first result. I will count $\displaystyle 6$ for the results until the $\displaystyle 5$, and continue with same expectation.

$\displaystyle E[ X] =7\cdotp \frac{1}{6} +( 6+E[ X]) \cdotp \frac{5}{6}$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
E[ X] =\frac{7}{6} +5+\frac{5}{6} E[ X]\\
\frac{1}{6} E[ X] =\frac{37}{6}
\end{array}$

$\displaystyle E[ X] =37$.

I do see something wrong with my answer, the thing is the $\displaystyle x$ could be a $\displaystyle 5$ itself. This means I don't think it's very correct to say $\displaystyle E[ X|\overline{A}] =6+E[ X]$.

Because if the $\displaystyle x$ was a $\displaystyle 5$, then the possibility of ending the game next round is $\displaystyle \frac{1}{6}$ (with a $\displaystyle 3$), which of course doens't happen in $\displaystyle E[ X]$.

I believe that's my mistake, agreed?

Is the fact that my result is close to the actual result a pure casuality? Or does this mean my answer is close? I have given this answer on a test, trying to understand if it's completely wrong, or instead slighly wrong.

答:
It seems as though this question has been answered well by those above me, however I decided to still post an answer as there is a really cool way of doing this problem for anyone who comes to visit this question in the future.

Imagine we have a Casino and each day the Casino rolls one dice. The question you ask is what is the expected value of days until 5,3 is rolled.

Let us now say we have a sequence of gamblers who bet on the outcome of the dice. The first gambler comes in and bets \$1 that the Dice will be a 5, the second gambler comes in on the second day and bets \$1 that the dice will be a 6 and so on.

Assume the casino offers fair bets, that is the payout if correct for the \$1 bet is $6.

Each gambler works as follows: If they loose their bet that's it, they go home never to return. However if they win their bet then they come back tomorrow and stake the $6 they now have on the next days dice being a 3.

Example: Casino rolls: $4,5,5,1,2,3,5,3$ then Gambler 1 looses his money on day 1 , Gambler 2 wins \$6 on day 2 and then looses it on day 3, Gambler 3 wins \$6 on day 3 but looses it on day 4. Gamblers 4,5,6, all loose instantly but gambler 7 leaves with $36

Let $ \{M_{n} \}_{n \geq 1}$ be the Martingale that is the casinos wealth. It is clear that this is a Martingale as they only offer fair ($0$ expected value) bets. Let us assume they start with wealth $0$, that is $M_0 = 0 $

Let $\tau$ be the stopping time of the event that the sequence $5,3$ is tossed for the first time. We are after $\mathbb{E}[\tau]$

Via the Optimal Stopping Theorem as $\tau$ is almost surely finite and the increments of $M_n$ are bounded we have that:

$\mathbb{E}[M_{\tau}] = \mathbb{E}[M_0] = 0 $

However at time $\tau$ we have that the first $\tau -2$ gamblers have all lost there money so the casino has made $\$+1$ of each of them and also that the (lucky) $(\tau-1)^{th}$ gambler has made $35$ dollars of them, the $36$ in return minus the initial stake of \$$1$ And that the $\tau^{th}$ gambler bet on a $5$ but saw a 3 so he also lost.

Hence $M_{\tau} = 35 - (\tau-2) -1  = 36 - \tau$

Hence $0 = \mathbb{E}[M_{\tau}] = \mathbb{E}[36-\tau] \implies \mathbb{E}[\tau] = 36$

This method can be expanded very nicely to any sequence of tosses. But particular care must be made for the winnings of the final few gamblers. For example if the sequence was $535353$ then the $(\tau -5 )^{th}$ gambler wins his $6^6$ the $(\tau -3 )^{th}$ gambler wins his $6^4$ as he correctly predicted $5353$ and the $(\tau - 2)^{th}$ gambler wins $6^2$ as he correctly predicted the last $53$. In this case $\mathbb{E}[\tau] = 6^6 +6^4 +6^2$


问:
Paradox: Roots of a polynomial require less information to express than coefficients?
A somewhat information theoretical paradox occurred to me, and I was wondering if anyone could resolve it.

Let $p(x) = x^n + c_{n-1} x^{n-1} + \cdots + c_0 = (x - r_0) \cdots (x - r_{n-1})$ be a degree $n$ polynomial with leading coefficient $1$. Clearly, the polynomial can be specified exactly by its $n$ coefficients $c=\{c_{n-1}, \ldots, c_0\}$ OR by its $n$ roots $r=\{r_{n-1}, \ldots, r_0\}$.

So the roots and the coefficients contain the same information. However, it takes less information to specify the roots, because their order doesn't matter. (i.e. the roots of the polynomial require $\lg(n!)$ bits less information to specify than the coefficients).

Isn't this a paradox? Or is my logic off somewhere?

Edit: To clarify, all values belong to any algebraically closed field (such as the complex numbers). And note that the leading coefficient is specified to be 1, meaning that there is absolutely a one-to-one correspondence between the $n$ remaining coefficients $c$ and the $n$ roots $r$.

答:
The core of your paradox seems to be that you claim that in some sense there are "more" (ordered) tuples $(c_{n-1},\dots,c_0)$ than (unordered) sets $\{r_{n-1},\dots,r_0\}$. This is not true for infinite sets like $\mathbb C$, only for finite sets, and even then it would not be significant for your problem as will be seen below.

In the sequel we only consider polynomials having leading coefficient $1$ to avoid a "trival non-uniqueness".

It is in general not true that the set $R(p)$ of roots of a polynomial $p = p(x)$ allows to reconstruct $p$. Note that $R(p)$ is nothing else than a set and does not contain information about the multiplicity of the roots. The number $m(p)$ of elements of $R(p)$ may be anything between $1$ and $n = \deg(p)$, i.e. we have $1 \le m(p) \le n$. [Note that for $\deg(p) = 0$ this says that the set $R(p)$ is empty.] The notation $R(p) = \{r_{n-1},\dots,r_0\}$ is misleading since it suggests that the set $R(p)$ has $n$-elements. Instead we have $R(p) = \{ \rho_1, \dots, \rho_{m(p)}\}$ with $\rho_i \ne \rho_j$ for $i \ne j$.

Therefore the problem is this: To each ordered tuple $\tau = (c_{n-1},\dots,c_0)$ of finite length we can associate a unique polynomial $p_\tau$ having the $c_i$ as coefficients, but  to a finite set $R$ we can associate infinitely many polynomials having $R$ as a root set. There is exactly one such polynomial with minimal degree (in which each $\rho \in R$ occurs with multiplicity $1$, i.e.the degree equals the number of elements of $R$), but if we associate it to $R$, we do not get all polynomials having $R(p) = R$. The only exception is $R =\emptyset$; in that case $p(x) = 1$.

What we really need to consider is the set of pairs $\{ (\rho_1,k_1) \dots, (\rho_m,k_m) \}$ where $k_i \ge 1$ is the multiplicity of the root $\rho_i$. Then we get a unique polynomial $p$ of degree $n = \sum k_i$ such that $R(p) =  \{ \rho_1, \dots, \rho_{m(p)}\}$.

Alternatively we can consider $\mathbb C^n/\mathfrak S_n$, the quotient of $\mathbb C^n$ by the operation of the symmetric group of $n$ elements. That is, tupels are identified if they are related by permuting their components. One might be tempted to think of the elements of $\mathbb C^n/\mathfrak S_n$ as simple (unordered) sets, but it is not true. In fact, $\mathbb C^n/\mathfrak S_n$ can be identified with the collection of pairs $\{ (\rho_1,k_1) \dots, (\rho_m,k_m) \}$ with $k_i \ge 1$ and $n = \sum k_i$.

See also How to show symmetric product of complex space is homeomorphic to the complex space .


问:
Proof of Fundamental Theorem of Finite Abelian Groups?
The only proofs I've seen of this tend to involve a few intermediate results and a couple of induction proofs with some clever constructions in them. They aren't hard to follow and they're pretty short, but they do still seem to remove some of the "why it should be true" aspect of it. The result seems blindingly obvious to me from a generator perspective: If a finite abelian group $G$ has generators $x,y,z...$ then any element can be represented by the form

$$g=x^ay^bz^c...$$

And there's an immediate and clear isomorphism from this to

$$\mathbb{Z}_{|x|}\times\mathbb{Z}_{|y|}\times\mathbb{Z}_{|z|}\times\cdots$$

Done. It can't help but be true. As far as I can see this is immediately extensible to finitely generated abelian groups as well. But the proofs I've seen, while short and simple enough, have never made its proof seem quite so clear.

Is there some major complication I'm missing here? I don't see anything fishy in assuming the existence of the generators, or proving the form of $g$, or actually constructing the isomorphism. But I assume I'm missing something because I've never seen the result proved like this, even though it's an obvious construction. Is the longer proof for pedagogical reasons (i.e. the lemmas and techniques are useful later), or am I missing a big hole in this proof?

答:
Your representation of $g$ as a product of the generators might not be unique, so the map you want from $G$ to the product of cyclic groups isn't well defined.

That is, we may have $g=x^ay^bz^c\ldots=x^{a'}y^{b'}z^{c'}\ldots$, and there is no immediate way to map $g$ to an element of

$$\mathbb{Z}_{|x|}\times\mathbb{Z}_{|y|}\times\mathbb{Z}_{|z|}\times\ldots$$

ContentLicense: CC BY-SA 3.0


问:
Counterexample to finiteness of integral over $R^{n-k}$ for $L^2(R^{n})$-function.
I have been told that it is false that a function $f\in L^2(R^n)$ will have the property that
$$
g(y)=\int_{R^{n-d}} f(x,y)^2 \mathrm{d}x<\infty,
$$
where $x \in R^{n-d}$ and $ y \in R^{d}$. But that this is false feels intuitively very strange. If this is not the case, how can we  marginalize probability distributions? Or is it not enough to assume square integrability? I have tried coming up with a counterexample, but can't do it...

答:
Consider the function $f:
\mathbb{R}^2 \to \mathbb{R}$ which has value $1$ on the line $y = 0$ but vanishes elsewhere. Note that $f^2 = f$.

Then the integeral $\int_{\mathbb{R}^2} f^2 = 0$ since the function vanishes almost everywhere on $\mathbb{R}^2$.

But at $y = 0$, $$g(y) = \int_{x \in \mathbb{R}} f(x, 0)^2 \mathrm{d}x = \int_{x \in \mathbb{R}} 1 \ \mathrm{d}x = +\infty$$ since the real line has infinite measure.

The intuition (at least for me) is that at some values of $y$, the function can go crazy, but the function in general is still square integrable (e.g. the function only goes crazy on a set of measure zero...)


答: Hint: Think of $n=2,d=1.$ The line $\{(x,1):x\in \mathbb R\}$ is a set of measure $0$ in $\mathbb R^2.$ You can take any $f\in L^2(\mathbb R^2)$ and redefine it to be $1$ on this line. Then $g(1)=\infty.$ It's still in $L^2.$

Title: Counterexample to finiteness of integral over $R^{n-k}$ for $L^2(R^{n})$-function.

Body:
I have been told that it is false that a function $f\in L^2(R^n)$ will have the property that
$$
g(y)=\int_{R^{n-d}} f(x,y)^2 \mathrm{d}x<\infty,
$$
where $x \in R^{n-d}$ and $ y \in R^{d}$. But that this is false feels intuitively very strange. If this is not the case, how can we  marginalize probability distributions? Or is it not enough to assume square integrability? I have tried coming up with a counterexample, but can't do it...


问:
Probability question with before and after scenario
Two shops have cupcakes. Shop $A$ have $2$ blue cupcakes and $3$ chocolate cupcakes. Shop $B$ has $2$ blue cupcakes and $5$ chocolate cupcakes. Suppose we randomly choose a cupcake from $A$ and transfer it to $B$. What is the probability that the cupcake we choose is blue from $B$?

My attempt at this was to say that there are two cases:

If a chocolate cupcake was transferred then we would have $8$ total cupcakes at shop $B$, so choosing a chocolate cupcake is $\frac{5}{8}$.

If a blue cupcake was transferred then we still have $8$ total cupcakes at shop $B$, so choosing a blue cupcake becomes $\frac{2}{8}$.

All together we have: $\frac{3}{8}\cdot \frac{2}{7} + \frac{2}{8} \cdot \frac{5}{7} = \frac{2}{7}$.

Is this the correct way to think about this problem?

答:
If  we first randomly choose a cupcake from A and transfer it to B, we can say we transfer
$\frac{2}{5}$blue + $\frac{3}{5}$ chocolate cupcakes.

If we then randomly choose a cupcake from B, then  the probability that the cupcake being blue  is $\frac{2.4}{8}$ = 30%.


问:
Splitting $ L^2(\Omega) $
Suppose that $ \Omega $ is some domain, and is the disjoint union of two sub-domains  $ \Omega_1 $ and $ \Omega_2 $.

I have a few questions about this.

Firstly, is it true that
\begin{equation}\label{key}
 L^2(\Omega) = L^2 (\Omega_1) \oplus L^2(  \Omega_2)?
\end{equation}

Secondly, if the above equation is true, is it an orthogonal decomposition?

Thirdly, suppose I have a self-adjoint operator $ H $ on $ L^2(\Omega) $.  Denote the restriction of $ H $ to $  L^2 (\Omega_j) $  by $ H_j $. Furthermore, suppose that the spectrum of $ H_1 $ and $ H_2 $ are both simple, and we construct an orthonormal eigenbasis for $ L^2 (\Omega_j) $. My question is: will the union of the spectrum of $ H_1 $ and $ H_2 $ coincide with the spectrum of $ H $, and would the spectrum of $ H $ gaurenteed to also be simple?

I am fairly confident that my first two questions are correct, because $ L^2(\Omega_1) $ is a closed subspace of the Hilbert space  $ L^2(\Omega) $, but I'm not completelysure if that is true.

答:
Questions 1 and 2: The decomposition is an orthogonal one.

Question 3: This is not true unless $L^2(\Omega_1)$ and $L^2(\Omega_2)$ is an invariant subspace of $H$. If not here is a counterexample for $\Omega=\{1,2\}$, $\Omega_1=\{1\}$, $\Omega_2=\{2\}$:
$$
H=\pmatrix{2&1\\1&2}.
$$
The restriction of $H$ to both subspaces has spectrum $\{2\}$, but $H$ has a different spectrum.

Title: Splitting $ L^2(\Omega) $


问:
How to multiply $2 \sin(x) \cdot \sin^{2} (x)$
I am just confused on the way you would multiply this out, if you could help that would be fantastic. The pre calc method for doing these is the solution I am looking for</p>

答: Hint: The formula is: $y^m\cdot y^n = y^{m+n}$, use this for $y = \sin x, m = 1, n = 2$ in your problem.


答:
In mathematics, it is very common to write $\sin^2(x)$ to represent $[\sin(x)]^2$.

Therefore, rewriting your question, you want to know the value of

$$2\sin(x)\sin^2(x)=2\sin(x)[\sin(x)]^2=2[\sin(x)]^3=2\sin^3(x).$$

Title: How to multiply $2 \sin(x) \cdot \sin^{2} (x)$

Body: I am just confused on the way you would multiply this out, if you could help that would be fantastic. The pre calc method for doing these is the solution I am looking for</p>


问:
General derivative of a polynomial
Is the following correct?

$$\frac{d^n}{dx^n} \sum_{k=0}^m a_k x^k = \sum_{k=0}^m \frac{\Gamma(k+1)}{\Gamma(k-n+1)} a_k x^{k-n} \quad n \in \mathbb{N}$$

Obviously this encounters $\frac{1}{\Gamma(1-n)}$ which should be zero for the formula to work and indeed

$$\lim \limits_{x \to n} \frac{1}{\Gamma (1-n)} = 0$$

but is this enough to make the formula rigorous or are there other flaws?

答:
Your formula is correct. If you only want to use the classic factorial function, you may write the formula is
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n} \sum_{k=0}^m a_k x^k =
\sum_{k=0}^m \Theta(k-n) a_k \frac{k!}{(k-n)!}x^{k-n}, \quad n \in \mathbb{N},$$
where $\Theta$ is the Heaviside function. It is an (easy) function that returns $0$ or $1$. I modified the input in the formula, so that it gives us the values that we need.

Title: General derivative of a polynomial


问:
Volume bounded by $x^2 + y^2 + z^2 = 4$ and $x^2 + (y-1)^2 = 1$
I tried to solve it by spliting the volume in two, by the $xy$-axis:

$$1/2 V = \iiint r\,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta$$

where the integral is bounded by:

$0 < z < \sqrt{4-r^2}$

$0 < r < 2\sin(\theta)$

$0 < \theta < \pi$

This is equal to $1/2 V = 8\pi/3$. So the whole wolume is:

$$V = 16\pi/3.$$

Is this correct? Or did I do make some wrong presumptions about the boundary of the integral?

One more question:
When I used online calculators I either get $16\pi/3$ or $8/9(3\pi - 4)$, why is that?

答:
The limits are fine but the result is not.

As you may realize that the total volume of the sphere is $\displaystyle \frac{32\pi}{3}$ and volume of half sphere for $y \geq 0$ is $\displaystyle \frac{16\pi}{3}$. The cylinder only intersects the sphere in the region $y \geq 0$ and the intersected volume is not clearly the whole volume of the half sphere. So logically $\displaystyle \frac{16\pi}{3}$ does not seem the right answer.

As there is symmetry with respect to $YZ$ plane, my suggestion is to use the integral with bounds $0 \leq \theta \leq \frac{\pi}{2}$.

So, $V = \displaystyle 2 \int_0^{\pi/2} \int_0^{2 \sin\theta} \int_{-\sqrt{4-r^2}}^{\sqrt{4-r^2}} r \ dz \ dr \ d\theta = \frac{16 \pi}{3} - \frac{64}{9}$.


问:
Given the equation $x^2-14x+m=0$, where m is a positive integer. If $p_1,p_2$ are roots, then find the value of $K=(p_1+p_2)^2+2p_1p_2$
Given the equation $x^2-14x+m=0$, where $m$ is a positive integer. If roots $p_1, p_2$ are distinct, positive, prime integers, then find the value of $K=(p_1+p_2)^2+2p_1p_2$

I tried to solve this question using the Vieta formulas and said $p_1+p_2=14$ and $p_1p_2=m$. But I got stuck there, I assume that you have to work out what the value of $m$ is, but I can't manage to work it out. Could you please explain to me how to solve this question?

答:
Well, if p1 and p2 are distinct, then there is only a single pair of positive primes that satisfies the condition $p_1+p_2=14$, and these are $3$ and $11$.

Else, if $p_1$ and $p_2$ are not distinct, the question could have two possible answers, because both $7$ and $7$ as well as $3$ and $11$ satisfy the equation.

Can you go ahead now?


问:
Finding point in 3D Space on line with fixed distance
I tried to look for similar questions but couldn't find something relating so here goes:

I have 3 points in 3D Space: Point A, Point B and Point O.
Point O is just the origin at [0, 0, 0].

I need to define a point C, which is between A and B, and exactly at distance d from O.

How I tried to approach this at first was to construct a direction from A to B, which can be used to define the line between A and B, then using the formula f = A + t * (A - B) to define the line. But then I got stuck and used the very ugly brute force method of looping over possible values of t until the distance between the point at said t was d from O, but I want to know if it can be done in one single simple mathematical equation to calculate C.

Kind regards and thanks in advance.

答:
The points A and B determine a line AB.

The distance from O to the line AB is a real positive number.

The distance from O to any point on the line is greater than the perpendicular distance between O and the line.

If the given distance d is less than the perpendicular distance between O and the line determined by A and B, the problem has no solution.

If $d=min(OA, OB)\lor d=max(OA,OB)$ the problem has one solution.

If $min(OA,OB)\lt d\lt max(OA,OB)$ the problem has one or two distinct solutions where the resulting point falls between A and B.

Otherwise, the problem still has two distinct solutions where the point resides on the line outside of the segment AB.

Hope this helps.

Now the almost full solution:

You have three equations to write.
First, if the point is on the line then there is a zero cross product:
$D\epsilon AB\Rightarrow \vec{AD}\times \vec{AB}=\vec{0}\Leftrightarrow (\vec{OD}-\vec{OA})\times \vec{AB}=\vec{0}$

You further find out that $sin(\theta)=\frac{\vec{OA}\times \vec{AB}}{d\cdot ||\vec{AB}||}$

The second equation is this:

$\vec{OD}\cdot \vec{AB}=d\cdot||\vec{AB}|| \cdot cos(\theta)=||\vec{AD}||$

The third and last that gives you the coordinates of point D is:
$\frac{x_D-x_A}{x_B-x_A}=\frac{AD}{AB}$. Use a similar equation to find $y_D, z_D$.

There is one extra observation: solving for $\theta$ should lead to two possible solutions. I have not mentioned the constraints for that since I have presented the case analysis at the beginning.


问:
Convex homogeneous function
Prove (or disprove) that any CONVEX function  $f$, with the property that $\forall \alpha\ge 0, f(\alpha x) \le \alpha f(x)$, is positively homogeneous; i.e. $\forall \alpha\ge 0, f(\alpha x) = \alpha f(x)$.</p>

答:
Maybe I'm missing something, but it seems to me that you don't even need convexity. Given the property you stated, we have that, for $\alpha>0$,
$$f(x)=f(\alpha^{-1}\alpha x)\leq \alpha^{-1}f(\alpha x)$$
so that $\alpha f(x)\leq f(\alpha x)$ as well. Therefore, we have that $\alpha f(x)=f(\alpha x)$ for every $\alpha>0$. The equality for $\alpha=0$ follows by continuity of $f$ at zero (which is implied by convexity).</p>

Title: Convex homogeneous function


问:
Is there a Graph theory way to solve this iteration problem
This thought came when i was trying to find the number of functions $f(x)$ from $\{1,2,3,4,5\}$ to $\{1,2,3,4,5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1,2,3,4,5\}$.  What i did was took cases when only one , two...etc  values of x such that all those are identity functions from that i got the answer as 756 which is correct.  But i was thinking if there was some kind of GT use can be done here  which can solve it more easily than case bash?

答: From @Mike Earnest and @Milten approaches, now I am considering a graph on $\{1,2,3,4,5\}$ made by placing an arrow from $i$ to $j$ if $f(i)=j$, and removing all self-loops. Absolutely we can have no cycles, so this graph is in fact a forest. Adding a vertex $r$ to the graph and having all roots of trees in this graph (that is, $i$ with $f(i)=i$) point to $r$. Then what we want to count is trees on $\{1,2,3,4,5,r\}$ with every vertex a distance of at most $3$ from $r$ (this is equivalent by directing all edges towards $r$). Ignoring the distance condition, Cayley's formula gives that there are $6^4=1296$ trees. We now must subtract the trees with at least one vertex of distance $4$ from $r$. Such trees will have a "backbone" of $5$ consecutive vertices ending at $r$ (say $v_4,v_3,v_2,v_1,r$) and then the 6th vertex will have one of these as its parent. If the parent is $r,v_1,v_2,v_4$ there are $5!$ trees each, and if the parent is $v_3$ there will be $\frac{5!}{2}$ because the 6th vertex and $v_4$ can be swapped. Hence required answer is $1296-120\cdot 4-60 = 756$.


问:
Pi Product finding maximum help
Trying to find the critical number for this ln of a pi product. I tried takin the partial derivative of the inner function it turn out a little messy and I'm unsure what how to get all the way to the critical number.

Let $\ell(\vec{x}:\mu, \sigma) = \ln\left(\prod_{k=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x_k-\mu)^2}{2\sigma^2}\right)\right)$.  Treating the values of $\vec{x} = [x_1, x_2, ..., x_n]$ as constants, at what value of $\mu$, $\sigma$ does $\ell(\vec{x}:\mu, \sigma)$ have a maximum?

Here is my what I have as the derivative.

$\frac{\partial \:}{\partial \:s}\left(\frac{1}{\sqrt{2\pi s^2}}e^{-\frac{\left(x_k-u\left(s,\:x\right)\right)^2}{2s^2}}\right)=\frac{e^{-\frac{\left(x_k-u\left(s,\:x\right)\right)^2}{2s^2}}\left(-s^2+\left(x_k-u\left(s,\:x\right)\right)\left(s\frac{\partial \:}{\partial \:s}\left(u\left(s,\:x\right)\right)+x_k-u\left(s,\:x\right)\right)\right)}{2^{\frac{1}{2}}\pi ^{\frac{1}{2}}s\left(s^2\right)^{\frac{3}{2}}}$

How to do get the maximum of $\: \ell(\vec{x}:\mu, \sigma)$

答:
First, using the properties of log, write the expression as \begin{align}\ell(\vec{x}:\mu, \sigma) &= \sum_{k=1}^n \log\left(\frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(x_k-\mu)^2}{2\sigma^2}\right)\right) \\ &= \sum_{k=1}^n \log\left((2\pi\sigma^2)^{-\frac12}\right)+\sum_{k=1}^n \log\left(\exp\left(-\frac{(x_k-\mu)^2}{2\sigma^2}\right)\right) \\ &=-\frac{n}{2}\log(2\pi)-\frac{n}{2}\log(\sigma^2)-\frac{1}{2\sigma^2}\sum_{k=1}^n (x_k-\mu)^2\end{align}

Differentiating w.r.t. $\mu$ gives $$\frac{\partial \ell}{\partial \mu}=\frac{1}{\sigma^2}\sum_{k=1}^n(x_k-\mu)$$ and setting equal to zero yields $\mu=\frac{1}{n}\sum_{k=1}^n x_k=\bar{x}$ at the maximum. Differentiating w.r.t. $\sigma^2$ gives $$\frac{\partial \ell}{\partial \sigma^2}=-\frac{n}{2\sigma^2}+\frac{1}{2\sigma^4}\sum_{k=1}^n (x_k-\mu)^2.$$ Setting equal to zero shows that $\sigma^2=\frac1n \sum_{k=1}^n (x_k-\mu)^2$ at the maximum.

So $\ell$ has attains its maximum at $\mu=\bar{x}$ and $\sigma^2=\frac1n \sum_{k=1}^n (x_k-\bar{x})^2=S_{XX}/n$.

Title: Pi Product finding maximum help


问:
$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}$, what does it mean?
I would like to know what these fractional exponents means in a derivative $\frac{d}{dx}$ operator. Like, I've seen $\frac{d^2}{dx^2}$ but I don't know what $\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}$ means.</p>

答:
What you have is what is usually referred to as a semiderivative. I would say that this notation is a bit incomplete, since one also has to account for a lower limit, in the case of differintegrals whose orders are not nonnegative integers.

In particular, letting ${}_a D_x^{1/2}f(x)$ be the semiderivative, one possible definition is the Riemann-Liouville form,

$${}_a D_x^{1/2}f(x)=\frac{f(a)}{\sqrt{\pi(x-a)}}+\frac1{\sqrt\pi}\int_a^x \frac{f^\prime(t)}{\sqrt{x-t}}\mathrm dt$$

The semiderivative operator can be thought of this way: this is the operator that, when applied to a function twice, gives the ordinary (first order) derivative.

If you want to learn more about this, I would suggest picking up Oldham/Spanier's The Fractional Calculus.


问:
Area inside square
Let ABCD be a square of side length 3. P is a point interior of this square such that each the the angles APB, BPC, CPD and DPA are at least 60 degrees. Find the area of the region where P could lie?

I have tried doing construction whereby creating 4 equilateral triangles on the sides of the square and finding the common intersection area inside didn’t give me the desired answer. Appreciate any smart ideas or solution at hand.

答:
Hint:

If I understood correctly you mean the least measure of angle is $60^o$,that means it can be more; in this case , as can be seen in figure the area inside the boundary QRST is what you are looking for.The circle concentric with square is where four centers of circles locate. These circles pass two vertices of square and all points on curves opposite the related side of square is $60^o$.In figure:

measure of side is 100 (3 in your question). The measure of circles and also the distance from S to BC is given, Also the measure of arch RS is $60^o$. These values are enough for finding the area inside the boundary.

Title: Area inside square


问:
Proving a sheaf is not locally free
Let $X$ be an affine variety defined by $xy=z^2$ in $\mathbb{A}^3$ and $D,E$ be Weil divisors of $X$ defined by $x,z=0$ and $y,z=0$ respectively.

I want to prove that the sheaf $\mathcal{O}_X(D)$ is not locally free of rank 1.

Let $U=X-E$. If $\mathcal{O}_X(D)$ is locally free of rank 1, then there is a generator $h$ of $\Gamma(U, \mathcal{O}_X(D))$. Since $\frac{1}{x}, \frac{1}{z} \in \Gamma(U, \mathcal{O}_X(D))$, I tried to show that $h$ cannot generate both $\frac{1}{x}, \frac{1}{z}$ but I've got stuck here. Can anyone help me how to proceed?

答:
Consider $D = \{x = z = 0\}$. It is not hard to find a locally free resolution
$$
\mathcal{O}_X^{\oplus 2}
\stackrel{\left(\begin{smallmatrix}z & -y\\-x & z\end{smallmatrix}\right)}\to
\mathcal{O}_X^{\oplus 2} \stackrel{(x,z)}\to
\mathcal{O}_X(-D) \to 0.
$$
Tensoring it by $\mathcal{O}_P$, where $P$ is the vertex of $X$, one obtains
$$
\mathcal{O}_P^{\oplus 2}
\stackrel{\left(\begin{smallmatrix}0 & 0\\0 & 0\end{smallmatrix}\right)}\to
\mathcal{O}_P^{\oplus 2} \to
\mathcal{O}_X(-D) \otimes \mathcal{O}_P \to 0,
$$
which implies that
$$
\mathcal{O}_X(-D) \otimes \mathcal{O}_P \cong \mathcal{O}_P^{\oplus 2},
$$
hence $\mathcal{O}_X(-D)$ cannot be locally free of rank 1. On the other hand, $2D$ is a Cartier divisor, so if $\mathcal{O}_X(D)$ would be locally free, the same would be true for $\mathcal{O}_X(-D)$.

Title: Proving a sheaf is not locally free


问:
Does $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converge?
Does $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converge? I used the integral test and the function $f(x)=\frac{sin (\frac  {\pi}{2}x)}{x}$ to determine that the series converges. I wanted to know if this function $f(x)$ is a valid choice to use the integral test on.

答:
The integral test cannot be used here, here is another way of proving the convergence.

Let $ n\geq 1 $, we have the following : \begin{aligned}\sum_{k=1}^{n}{\frac{\left(-1\right)^{k-1}}{k}}&=\int_{0}^{1}{\sum_{k=1}^{n}{\left(-x\right)^{k-1}}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\frac{1-\left(-1\right)^{n}x^{n}}{1+x}\,\mathrm{d}x}\\ \sum_{k=1}^{n}{\frac{\left(-1\right)^{k-1}}{k}}&=\ln{2}-\left(-1\right)^{n}\int_{0}^{1}{\frac{x^{n}}{1+x}\,\mathrm{d}x}\end{aligned}

Since : $ \left\vert\int_{0}^{1}{\frac{x^{n}}{1+x}\,\mathrm{d}x}\right\vert\leq\int_{0}^{1}{x^{n}\,\mathrm{d}x}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0$, then the sequence $ \left(\sum\limits_{k=1}^{n}{\frac{\left(-1\right)^{k-1}}{k}}\right)_{n\geq 1} $ converges, which means the series $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n-1}}{n}} $ converges. Also we have : $$ \sum_{n=1}^{+\infty}{\frac{\left(-1\right)^{n-1}}{n}}=\ln{2} $$


答:
Yes, it is an alternating series, and by the alternating series test, it does indeed converge!

You can verify it yourself, by substituitng $x=1$ into the Maclaurin series for $\ln( x+1)$, and that it converges to $\ln(2)$.

Body: Does $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converge? I used the integral test and the function $f(x)=\frac{sin (\frac  {\pi}{2}x)}{x}$ to determine that the series converges. I wanted to know if this function $f(x)$ is a valid choice to use the integral test on.


问:
$f(x)=x^2-[x^2]$ is continuous?
$f(x)=x^2-[x^2]$ is continuous? If not than say it has  first order or second order breakpoints.

Can we say because $x^2$ is continuous $[x^2]$ is not than $f(x)$ is not continuous?

$[x]$ is $floor$ function.

答:
$f(x)=x^2-\lfloor x^2 \rfloor$ of course is not continuous - consider, for example $x=\pm 1$. In general,  $\pm \sqrt{n}$ are "jump" breakpoints.

When $f$ is continuous and $g$ not, then $\phi=f\pm g$ cannot be continuous, because if it is, then also become continuous $g=\phi -f$, which contradicts assumption.

Title: $f(x)=x^2-[x^2]$ is continuous?


问:
Are there irreducible polynomials of any degree over "most" fields?
Given a field $K$ that is neither algebraically closed nor real closed, must there necessarily be an irreducible polynomial of degree $n$ over $K$ for all positive integers $n$?

By the Artin-Schreier theorem, we know that irreducible polynomials of arbitrary large degrees must exist, but this does not necessarily mean that such polynomials exist for all degrees.

答:
No. Let $K=\bigcup_n{\mathbb{F}_{p^{3^n}}}$. Then any polynomial in $K$ of degree $3^d >1$ is reducible, but the algebraic closure of $K$ is of infinite degree over $K$.

The same could be said by replacing the sequence $(3^n)_n$ with any increasing sequence $a_n$ such that $a_n|a_{n+1}$ and such that there exists a prime $q$ such that $v_q(a_n)$ is bounded: then there will be irreducible polynomials of degree a power of $q$ over $K$, but no irreducible polynomials of degree $r$ if $v_r(a_n)$ diverges.


问:
Prove that the sequence $(a_n)$ converges, where $a_n$ = $\frac{1 · 3 · 5 · · ·(2n − 1)}{2 · 4 · 6 · · ·(2n)}$.
Prove that the sequence $(a_n)$ converges, where $a_n$ = $\frac{1 · 3 · 5 · · ·(2n − 1)}{2 · 4 · 6 · · ·(2n)}$. Use Monotonic Convergence Theorem.

Here is my proof:

Since $a_n = \frac{2n-1}{2n}$, then $a_{n+1} = \frac{2n-1+1}{2n+1} = \frac{2n}{2n+1}$.
Therefore, $a_n$ = $\frac{2n-1}{2n}\le\frac{2n}{2n+1}$ = $a_{n+1}$, and so the sequence, $a_n$, is increasing for all $n\in\mathbb{N}$, and is monotonic. The term $a_n$ has an upper bound (ie. 1) and a lower bound (ie. 0) for all $n$ and thus is bounded. Therefore, $a_n$ converges by the Monotonic Convergence Theorem.

Please check it over and make sure it sounds all right. If there is anything wrong, please let me know what and how I can fix it.

答:
You can start by noting that the sequence is decreasing. In fact,
$$
\dfrac{a_{n+1}}{a_n} = \dfrac{2n+1}{2n+2} < 1 \Rightarrow a_{n+1}< a_n.
$$

Also, the sequence is clearly bounded. Since it is decreasing and positive, we have that $0 < a_n \leq a_1$.

Finally, being monotonic and bounded, the sequence is convergent.


答: Without looking at the dup solution, you can find a recursive relation for $a_{n+1}$ and $a_n$. One has: $a_{n+1} = \dfrac{1\cdot 3\cdot 5\cdots (2n+1)}{2\cdot 4\cdot 6\cdots (2n+2)}= \dfrac{2n+1}{2n+2}\cdot a_n< a_n$ since $2n+1 < 2n+2$. And $a_n > 0$ for all $n$, hence it converges as a monotonically decreasing and bounded below sequence.

Body:
Prove that the sequence $(a_n)$ converges, where $a_n$ = $\frac{1 · 3 · 5 · · ·(2n − 1)}{2 · 4 · 6 · · ·(2n)}$. Use Monotonic Convergence Theorem.

Here is my proof:

Since $a_n = \frac{2n-1}{2n}$, then $a_{n+1} = \frac{2n-1+1}{2n+1} = \frac{2n}{2n+1}$.
Therefore, $a_n$ = $\frac{2n-1}{2n}\le\frac{2n}{2n+1}$ = $a_{n+1}$, and so the sequence, $a_n$, is increasing for all $n\in\mathbb{N}$, and is monotonic. The term $a_n$ has an upper bound (ie. 1) and a lower bound (ie. 0) for all $n$ and thus is bounded. Therefore, $a_n$ converges by the Monotonic Convergence Theorem.

Please check it over and make sure it sounds all right. If there is anything wrong, please let me know what and how I can fix it.


答:
Well the roots of $x^2 - 14x + m = (x^2 -14x +49) + (m-49) = (x-7)^2 -(49-m)$ are $7 \pm \sqrt{49-m}$.

So $p_1+ p_2 = 14$ and $p_1p_2 = 49 - (49-m) = m$.

So $m$ is the product of two distinct prime so that $49 - m$ is perfect square.  Well, quit worrying and start scurrying.

The perfect squares less than $49$ are $1,4,9, 16, 25, 36$ so $m = 48,45, 40, 33,24,13$ and of those the only one that is product of two distinct primes are $33=3\times 11$

So $(p_1 + p_1)^2 + 2p_1p_2 = 14^2 + 2\cdot 33$.

=====

Argh a dope slap to me for getting $49-m$ must be a perfect square but not getting that $14$ is the sum of distinct primes.  It is still trial and error, but $1+13,2+12,3+11,4+10, 5+9, 8+6, 7+7$ so $3,11$ are the only sum of two primes is an easier trial by error.

Title: Given the equation $x^2-14x+m=0$, where m is a positive integer. If $p_1,p_2$ are roots, then find the value of $K=(p_1+p_2)^2+2p_1p_2$

Body:
Given the equation $x^2-14x+m=0$, where $m$ is a positive integer. If roots $p_1, p_2$ are distinct, positive, prime integers, then find the value of $K=(p_1+p_2)^2+2p_1p_2$

I tried to solve this question using the Vieta formulas and said $p_1+p_2=14$ and $p_1p_2=m$. But I got stuck there, I assume that you have to work out what the value of $m$ is, but I can't manage to work it out. Could you please explain to me how to solve this question?


问:
Prove that $3^{n + 1} \ge n^4 + n^2 + 1$ for any $n \ge 5$
Prove that $3^{n + 1} \ge n^4 + n^2 + 1$ for any $n \ge 5$, using mathematical induction.

I've thought about writing the equation like that: $3^{n + 1} \ge (n^2 + 1)^2 - n^2$, but all for nothing.

Body:
You can check the base case of $n=5$ - add that into your question.

For the inductive step, let's re-cast the inequality as $3^{n+1} - n^4-n^2-1 \geq 0$.

$$\begin{align}
3^{n+2} - (n+1)^4-(n+1)^2-1 &= \color{#0B2}{3^{n+1}} + 2\cdot3^{n+1} \color{#0B2}{- n^4} - 4n^3 - 6n^2 - 4n -1\\
&\quad\quad\quad \color{#0B2}{-n^2}-2n-1 \color{#0B2}{-1}\\
&\geq 2\cdot3^{n+1}  - 4n^3 - 6n^2 - 6n - 2\\
&= 2(3^{n+1} \color{#03B}{ - 2n^3 - 3n^2 - 3n - 1})\\
\end{align}$$

Now note that since $n>5$, $n^4+n^2+1 > 5n^3+5n+1 > \color{#03B}{2n^3+3n^2+3n+1}$ & finish


问:
Mazur-Ulam-like theorem for complex Hilbert spaces
The Mazur-Ulam theorem doesn't hold for complex Hilbert spaces because antiunitary operators are origin preserving surjective isometries, but they aren't linear. Is it true, that every $f:H\to H$ surjective isometry of a complex Hilbert space $H$ has the form $$f(x)=f(0)+U(x)$$ where $U$ is unitary (hence linear) or antiunitary (hence antilinear)?

答: The map $(z, w)\in \mathbf C^2 \mapsto (z,\overline{w})\in \mathbf C^2$ is norm-preserving, but not linear nor antilinear.


答:
$f(x)=x^2-[x^2]=\{x^2\}$, where $\{\}$ is fractional part function.

$f$ is not continuous at any points of the type $\pm\sqrt n$, where $n\in  \mathbb N$. Although $f$ is continuous at $x=0$.

At points $x=\pm \sqrt n, g(x)=[x^2]$ is discontinuous but $h(x)=x^2$ is continuous at $x=\pm \sqrt n$ so if $f(x)$ is continuous then so is $g(x)=h(x)-f(x)$, which is contradiction.

Title: $f(x)=x^2-[x^2]$ is continuous?

Body:
$f(x)=x^2-[x^2]$ is continuous? If not than say it has  first order or second order breakpoints.

Can we say because $x^2$ is continuous $[x^2]$ is not than $f(x)$ is not continuous?

$[x]$ is $floor$ function.


问:
Tribonacci generating matrix
I know that $T(0) = 0$ and $T(1) = T(2) = 1$. For $n \ge 3$,

$$T(n)=T(n-1)+T(n-2)+T(n-3)$$

Now I find that

$$\begin{bmatrix}T(n+1)\\T(n)\\T(n-1)\end{bmatrix} = \begin{bmatrix}1&1&1\\1&0&0\\0&1&0\end{bmatrix}\cdot\begin{bmatrix}T(n)\\T(n-1)\\T(n-2)\end{bmatrix} = \cdots=\begin{bmatrix}1&1&1\\1&0&0\\0&1&0\end{bmatrix}^n\cdot\begin{bmatrix}1\\1\\0\end{bmatrix}$$

How do I find now that
$$\begin{bmatrix}1&1&1\\1&0&0\\0&1&0\end{bmatrix}^n = \begin{bmatrix}
T(n+1)&T(n)+T(n-1)&T(n)\\T(n)&T(n-1) + T(n-2)&T(n-1)\\T(n-1)&T(n-2)+T(n-3)&T(n-2)\end{bmatrix}$$

Do I need to guess? Or there is a way to do it?

答:
One option is surely to get your transfermatrix, say $\mathbb M$, (that one with zeros and ones) as solution of one step of iteration.

So for instance $[T_0, T_1, T_2] \cdot \mathbb M = [T_1,T_2,T_3]  \Rightarrow \text{ find } \mathbb M$ . Now this simple ansatz does not yet help, because we cannot apply matrix-inversion on the one-row vector on the lhs.

But instead to have only a one-row vector on the lhs, we can construct square matrices $\mathbb T_0$,$\mathbb T_1$  to attempt $ \mathbb T_0 \cdot \mathbb M = \mathbb T_1  \Rightarrow \mathbb M = \mathbb T_0^{-1} \cdot \mathbb T_1$ :
$$ \begin{array}{ccc}
   &  & \mathbb M  \\
   & * & ========\\
 \left[\begin{array}{}  T_0&T_1&T_2 \\T_1&T_2&T_3 \\T_2&T_3&T_4 \end{array}\right]
   & = &
\left[\begin{array}{}  T_1&T_2&T_3 \\T_2&T_3&T_4 \\T_3&T_4&T_5 \end{array}\right] \\
\end{array}
$$
Then $ \mathbb T_0 \cdot \mathbb M^n = \mathbb T_n$ by construction.

Note: It would have been better to use small letters for scalar values and capital letters for the matrices, but well: I just used the "mathbb"-attribute for distinction

Title: Tribonacci generating matrix


问:
Estimate involving the Markov inequality and the union bound
The context: I'm reading a paper about Brownian Motion and $\epsilon$-hitting times on a Torus $\mathbb T_2$. However, I'm a little bit rusty with the basics of probabilty theory so I am not sure how the "$\exists$" is translated into the "$\sum$" in the following statement.

I'm trying to understand the following estimate that is made in some proof in the paper:
$$
\mathbb P(\exists x\in L_\epsilon \textrm{ s.t. } \mathcal T(x) > n_\epsilon(\delta) ) \leq \sum_{x\in L_\epsilon} \mathbb P ( \mathcal T(x) > n_\epsilon(\delta) )
$$
The author states the he used the union bound and the Markov inequality here.

A little bit of context (if additional necessary, please ask):

$L_\epsilon$ is a square lattice.

$n_\epsilon(\delta)$ is a fixed value that depends on $\epsilon$ and $\delta$.

$\mathcal T(x)$ is some sort of excursion count of the Brownian Motion.

答:
It uses the inequality: for any collection of countably many events $A_n$ (on some probability space)

$$
\mathbb{P}(\cup_n A_n)\leq \sum_{n} A_n.
$$

In your case
$$
\{\exists x\in L_\epsilon \textrm{ s.t. } \mathcal T(x) > n_\epsilon(\delta)\} = \cup_{x\in L_\epsilon}\{\mathcal T(x) > n_\epsilon(\delta)\}.
$$
It is a countable union as it is the square lattice. To see why the equality holds note that, if the event on the left side happens then the event on the right side must happen. Also if the event in the right hand happens then it must happen for at least one $x$, and therefore left side must happen.


问:
What is the period and phase shift of $f(x)=3\sin 4\left(x+\frac{\pi}{4}\right)-1$?
For the function what is the period and the phase shift respectively are:?

$$f(x) = 3\sin 4\left(x +\frac{\pi}{4}\right) - 1$$

My attempt: It is wrong though.

$y = a\sin b(x\pm h) \pm k$

$a$ : amplitude $\implies$ take absolute value

$b$ : angular frequency

$h$ : horizontal shift (phase shift). ($+$ left, $-$ right)

$k$ : mid. line or vertical shift ($+$ up, $-$ down)

period = $\dfrac{2\pi}{b}$ ( for sines, cosines; if tangent $p = \dfrac{\pi}{b}$)

$f(x) = 3\sin 4\left(x + \dfrac{π}{4}\right) - 1$

$b = 4$

period = $\dfrac{2\pi}{4} = \dfrac{\pi}{2}$

phase shift: left $\dfrac{\pi}{4}$

答:
I'm not sure why you say that your attempt is wrong. That's how I learned it.

Edit: As you've mentioned in the comments below, the options provided to you are

(A) period $4$ and phase shift $-\cfrac\pi4$

(B) period $\cfrac\pi2$ and phase shift $-\cfrac\pi4$

(C) period $-\cfrac\pi2$ and phase shift $-\cfrac\pi4$

(D) period $-\cfrac\pi2$ and phase shift $\cfrac\pi4$

Using the method that you were given to find the period, the only possible option is (B). Remember that a phase shift of $-\cfrac\pi4$ is the same thing as a phase shift of $\cfrac\pi4$ to the left, so this is the same as your answer.

Edit: Let me give you some alternate (hopefully clearer) rules. Suppose you are given an equation $$y=a\sin\bigl(b(x-h)\bigr)+k$$ or $$y=a\cos\bigl(b(x-h)\bigr)+k$$ for some real $a,b,h,k$ where $a\ne0$ and $b>0$. (If $a=0$ or $b=0,$ there's really nothing interesting to say. Do you know why we may assume that $b$ isn't negative?) Then:

The amplitude is $|a|$. Graphically, this will be the vertical distance from the max (or min) value to the midline, or half the vertical distance from the max value to the min value.

The phase shift is $h$. Graphically, this is a shift by $|h|$ to the right if $h$ is positive; by $|h|$ to the left if $h$ is negative.

The midline shift is $k$. Graphically, this is a shift by $|k|$ upward if $k$ is positive; by $|k|$ downward if $k$ is negative.


问:
Sequences in the cofinite topology
Let $X$ be an infinite set with the cofinite topology. I must prove that if $(x_n)$ is a sequence and $x_n \ne x_m$ for all $n \ne m$, then any $x$ is the limit of $x_n$.

Can anyone give me some explained examples on this? I don't get how a sequence works in this topology.

答:
If $(x_n)_n$ is a sequence such that all $x_n$ are different, we know that $A:=\{x_n \mid n \in \Bbb N\}$ is an infinite set.

If $O$ is an open set of $X$ that contains $x$ (so is non-empty) we know that $F:=X\setminus O$ is finite and so $A \cap F$ is finite too and there is some $N$ such that $x_N$ is the point with maximal index in $A \cap F$. It follows that for all $n > N$, $x_n \notin F$ so $x_n \in O$ and so for any open neighbourhood $O$ of any $x \in X$, a tail of the sequence is in that neighbourhood, and this means by definition of convergence that $x_n \to x$.

Title: Sequences in the cofinite topology


问:
A weighted L1 minimization linear programming with varying operating regions
I want to minimize the difference between a variable and a constant in my system using LP. $$ \min |x_1 - b|$$  I found that this can be formulated as an LP using: $$ \min t \\s.t. \hspace{0,2cm} t \geq x_1-b\\  \hspace{1.6cm}t \geq -(x_1-b)$$

As I want to penalize one part more or sometimes a certain deviation at one side is no problem I want to broaden my problem too:
example of what I want

The formulation of the splitting (left figure) has been done as follow:
$$\min \hspace{0.5cm}a_1*\max(0,x_1 - b)+ a_2*\max(0, b - x_1)$$
which can be formulated as follow:
$$ \min a_1*t_1 +   a_2*t_2 \\
\\s.t. \hspace{1cm}
t_1 \geq ( x_1 - b) \\
t_1 \geq 0 \\
\hspace{1.8cm} t_2 \geq( -x_1 + b) \\
t_2 \geq 0$$

But when I try to go to the right side of the figure I get infeasibility issues.
My question is: " how do I formulate this problem in order to get a feasible solution?" I tried this:
$$ \min a_1*t_1 +   a_2*t_2 \\
\\s.t. \hspace{1cm}
t_1 \geq ( x_1 - b) - slack_1 \\
t_1 \geq 0 \\
\hspace{2.3cm}t_2 \geq( -x_1 + b) - slack_2\\
t_2 \geq 0$$

答:
Your formulation without slacks looks correct.  For any value of $x_1$, taking $t_1=\max(x_1-b,0)$ and $t_2=\max(b-x_1,0)$ is feasible.

An alternative formulation is to minimize $a_1 t_1 + a_2 t_2$ subject to
\begin{align}
x_1 - b &= t_1 - t_2 \\
t_1 &\ge 0 \\
t_2 &\ge 0
\end{align}


问:
Given the functions $f_0,...,f_n$ with $f_0(x)=\frac{1}{x-2}$ and $f_{k+1}(x)=\frac{1}{1-f_k(x)}$ for every $k=0,1,...,n-1$. Evaluate $f_{2000}(2001)$
Given the functions $f_0,f_1,...,f_n$ with $f_0(x)=\frac{1}{x-2}$ and $f_{k+1}(x)=\frac{1}{1-f_k(x)}$ for every $k=0,1,...,n-1$. Evaluate $f_{2000}(2001)$.

I'm trying to work out this question by attempting to find a pattern, but I can't seem to find one. Based on a pattern which I see, I believe that the number is negative, but I can't work out the question. Could you please explain to me how to solve it?

Body:
Hint:

$$f_1(x)=\frac{1}{1-\frac{1}{x-2}}=\frac{x-2}{x-3}$$
$$f_2(x)=\frac{1}{1-\frac{x-2}{x-3}}=3-x$$
$$f_3(x)=\frac{1}{1-(3-x)}=\frac{1}{x-2}$$


答:
Any function $F(x)$ defined on $[0,2\pi]$ expands as the Fourier series
$$ F(x) = \dfrac{A_0}2 + \sum^\infty_{n=1}A_n\cos(nx) + \sum^\infty_{n=1} B_n \sin(nx),$$
$$\text{where} \qquad A_0 = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)dx, \quad
   A_n = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)\cos(nx)dx, \quad
   B_n = \frac1\pi\!\int_0^{2\pi}\!\!\!\!\!F(x)\sin(nx)dx. $$
Thus, for $f(x) = \pi-x$, these coefficients are computed as follows:
$$\begin{split}
a_0 &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)dx = 0, \\
a_n &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)\cos(nx)dx
     = \frac{1-\cos(2n\pi)-n\pi\sin(2n\pi)}{n^2\pi}, \\
b_n &= \frac1\pi\!\int_0^{2\pi}\!\!\!(\pi-x)\sin(nx)dx
     = \frac{2\cos(n\pi)\big[n\pi\cos(n\pi)-\sin(n\pi)\big]}{n^2\pi}.
\end{split}$$
Therefore, its Fourier series is given by
$$\sum_{n=1}^\infty \frac1{n^2\pi} \Big\{ \big[ 1 -\cos(2n\pi) -n\pi\sin(2n\pi) \big] \cos(nx) + 2\cos(n\pi) \big[ n\pi\cos(n\pi) -\sin(n\pi) \big] \sin(nx) \Big\}.$$

Title: Fourier series of $f(x)=\pi-x$

Body:
$$f(x)=\pi-x \qquad x \in [0,2 \pi[$$

$$a_0=\frac{1}{\pi} \ \int_0^{2 \pi}f(x) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi} (\pi-x) \ dx=0$$

$$a_n=\frac{1}{\pi} \ \int_0^{2 \pi} \cos(nx) \ dx=\frac{1}{\pi} \ \int_0^{2 \pi}(\pi \ \cos(nx)-x \ \cos(nx)) \ dx=$$
$$\frac{1}{\pi} \ \Big( \ \Big[\frac{\pi}{n} \ \sin(nx) \Big]_0^{2 \pi}-\Big[\frac{x}{n} \ \sin(nx)+\frac{1}{n^2} \ \cos(nx) \Big]_0^{2 \pi} \Big)=0 $$

$$b_n=\frac{1}{\pi} \ \Big( \ \Big[\ -\frac{\pi}{n} \ \cos(nx) \Big]_0^{2 \pi}-\Big[-\frac{x}{n} \ \cos(nx)+\frac{1}{n^2} \ \sin(nx) \Big]_0^{2 \pi} \Big)=\frac{2}{n}$$

$$f(x)=2 \sum_{n=1}^{+\infty} \frac{1}{n} \ \sin(nx)$$

Is it correct?


答:
Well, play around when $x$ is just under and just over an integer.

$x^2 -[x^2]\approx 0$ if $x =n + \delta$ for a very small $\delta > 0$.

That is to says if $x = n + \delta$ then $x^2 - [x^2] = (n+\delta)^2 - [(n+\delta)^2] =n^2 + 2\delta n + \delta^2 -[n^2 + 2\delta n + \delta^2] = (n^2+2\delta n + \delta^2) - n^2 = 2\delta n + \delta^2\approx 0$

But if $x = n -\delta$ for a very small $\delta > 0$ then

$x^2 = n^2 - 2n\delta +\delta^2 < n^2$ (assuming $\delta$ is small enough and $x \ne 0$ so $[x^2] = [n^2 - 2n\delta + \delta^2] = n^2 -1$.

So $x^2 -[x^2] = -2\delta + \delta^2 + 1 \approx 0$.

Or more formally $\lim_{x\to n+\in \mathbb Z; n\ne 0} x^2 -[x^2] = 0$ but $\lim_{x\to n-\in \mathbb Z; n\ne 0} x^2 -[x^2] = 1$.

We can see we will also have the same if $x$ is near a square root of an integer (other than $0$).

Title: $f(x)=x^2-[x^2]$ is continuous?


Title: $f(x)=x^2-[x^2]$ is continuous?

Body:
$$f(x)=x^2-[x^2] \implies f(n)=0$$ $$\implies f(n^{+})=\lim_{h \to 0} f(n+h)=\lim_{h \to 0}(n+h)^2-[(n+h)^2]$$
$$=\lim_{h \to 0} (n^2+h^2+2nh-[n^2+2nh+h^2])=\lim_{h \to 0}(n^2+h^2+2nh-n^2)=0$$
Next, $$f(n^-)=\lim_{h \to 0} (n-h)^2-[(n-h)^2]=\lim_{h\to 0} n^2-2nh+h^2-[n^2-2nh+h^2]=\lim_{h \to 0} n^2-2nh+h^2-n^2+1=1$$
As $f(n^+) \ne f(n^-)$, $f(x)$ is discontinuous at every integer.


问:
Why $\alpha_p$ copies?
I was reading this question here: What are the semisimple $\mathbb{Z}$-modules? and I understood everything except why we need  $\alpha_p$ copies here $$
\bigoplus_{\text{$p$ prime}}(\mathbb{Z}/p\mathbb{Z})^{(\alpha_p)}
$$
And not just $$
\bigoplus_{\text{$p$ prime}}\mathbb{Z}/p\mathbb{Z}
$$

Could anyone explain this to me please?

答:
Because you can have more than one copy of a particular simple module in a given decomposition.

For example, $S\times S$ is a semisimple $\mathbb Z$ module where $S=\mathbb Z/(2)$.

In that case $\alpha_2=2$ and $\alpha_i=0$ for every other index.

Title: Why $\alpha_p$ copies?


问:
Notable difference in a multivariable linear approximation
I have to use the linear approximation of $f(x, y, z) = $($5x^2$ + $y^2$)$/(z + 1)$ at $(-2, 1, 1)$ to estimate $f(-1.98, 0.97, 1.03)$

I'm getting $10.1125$, however it is a bit too low (I guess) if i compare this result with the normal substitution:  ~$10,119$

Is it really supposed to be $10.1125$ or I'm doing something wrong?

答:
You have it right.  Linearization doesn't give the exact value, it gives an approximation.

For example, if you take the parabola $y = x^2$ at $x = 2$ the slope of the tangent line is $4$, and the linear estimate of $2.03^2$ is therefore $4 + 4*.03 = 4.12$, while the exact answer is 4.1209.


问:
How does one just find and come up with homeomorphisms?
For example how would I find a homeomorphism from $(0, 1)$ to the reals and one from $(0, 1)$ to $(0, ∞)$?

I know the definition of a homeomorphism is if a function is both one-to-one, onto, continuous, and has a continuous inverse. However, I do not know how to craft my own to fit these standards from the maps above.

答:
There are some basic real homeomorphisms which we can use as building blocks to construct more with various desired properties. The most useful are:

Linear functions $f(x) = ax+b$ map interval endpoints to different interval endpoints.

$\exp : \mathbb{R} \to \mathbb{R}^{+}$ and its inverse $\ln$ go between the full set and one-ended open intervals.

$\tan : (-\frac{\pi}{2}, \frac{\pi}{2}) \to \mathbb{R}$ and its inverse $\arctan$ go between bounded and unbounded open intervals. (For a one-ended open interval, note $\tan : (0, \frac{\pi}{2}) \to \mathbb{R}^{+}$.)

Composing any two homeomorphisms gives another homeomorphism. (If you haven't seen the proof, it's straightforward from definitions.)

For the first example, to map from $(0,1)$ to the reals, $\tan$ has similar properties, but the wrong interval endpoints. So let's compose it with a linear map $L(x) = ax+b$, and see if we can get $f = \tan \mathrel{\circ} L$ with the desired domain and range.

$$ \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \tan L(x) = \lim_{x \to 0^{+}} \tan(ax+b) $$

so we would like $a(0) + b = -\frac{\pi}{2}$. Similarly, for the limit as $x \to 1^{-}$, we would like $a(1) + b = \frac{\pi}{2}$. This gives $b= -\frac{\pi}{2}$, $a = \pi$, $L(x) = \pi(x - \frac{1}{2})$, and finally

$$ f(x) = \tan \left[ \pi \left(x - \frac{1}{2}\right) \right] $$

(Optionally, then notice $f(x) = -\cot(\pi x)$. Possibly drop the negative sign from this if you don't mind it decreasing rather than increasing. These usually won't matter much for further use.)

Can you write a $(0,1) \to (0,\infty)$ homeomorphism now?


问:
Are the functions $g$ and $h$ uniformly continuous on $\mathbb{R}$
I'm working on a question that states:

Let $f: \mathbb{R}^+ \to \mathbb{R}$ be a continuous functions.

(a) Is the function $g:\mathbb{R} \to \mathbb{R}: x \mapsto f(\frac{1}{1+x^2})$ uniformly continuous?

(b) Is the function $h:\mathbb{R} \to \mathbb{R}: x \mapsto g(x)^2$ uniformly continuous (here $g$ is the same function as in part (a))?

I am aware and can prove that the function $\frac{1}{1+x^2}$ is uniformly continuous and my I think that $g$ is uniformly continuous, but $h$ is not. I know that the composition of two uniformly continuous functions is uniformly continuous, but my issue is that $f$ is only continuous.

Any help on this problem is appreciated.

答:
For point a

The map $l : x \mapsto \frac{1}{1+x^2}$ is $2$-lipschitz on $\mathbb R$ and maps $\mathbb R$ onto $(0,1]$. As $f$ is supposed to be continuous on $\mathbb R_+$, it is uniformly continuous on the closed segment $[0,1]$. Therefore for any $\epsilon \gt 0$, it exists $\delta \gt 0$ such that $\vert f(x) - f(y) \vert \lt \epsilon$ for $\vert x - y \vert \lt \delta$ and $x,y \in [0,1]$. Now for $x,y \in \mathbb R$ and $\vert x - y \vert \lt \delta /2 $ we have $\vert l(x) - l(y) \vert \lt \delta$ (as $l$ is $2$-lipschitz) and $l(x) , l(y) \in (0,1] \subseteq [0,1]$ and finally $\vert (f \circ l)(x) - (f \circ l)(y) \vert \lt \epsilon$. This enables to conclude that $f \circ l$ is uniformly continuous as desired.

Point b

Just apply point (a) to $f^2$, the square of $f$.


问:
Adding powers of $i$
I've been struggling with figuring out how to add powers of $i$.

For example, the result of $i^3 + i^4 + i^5$ is $1$. But how do I get the result of $i^3 + i^4 + ... + i^{50}$? Writing it all down would be pretty mundane...

It has to do something with division by 4, since the "power cycle" of $i$ repeats every fourth power.

Thank you for any clues.

答:
From $i^2=-1$ you get $i^{4n}=1$, $\quad i^{4n+1}=i$, $\quad i^{4n+2}=-1$ and $i^{4n+3}=-i$.

Then you just count your positive and negative multiples of $1$ and $i$.

In particular, $i^3+i^4+\cdots+i^{50}=0$.

Title: Adding powers of $i$

LastEditorDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


问:
Motivation for the Mapping Class Group
Question: What is the motivation for studying the mapping class group?  In particular, what types of questions does it attempt to answer and what kind of invariant is it?

Motivation for this Question: Recently I've seen a number of references to things which assume knowledge of the mapping class group.  I've attempted to page through Dan Margalit and Benson Farbs' book on the subject as well as check out the wikipedia page for it, but both sources give the motivation as just "what it is" as opposed to, bluntly, "why should I care about this."

答:
With the risk of being redundant, I'll attempt to give my answer, if for no other reason than to help myself.  Then again, the best people to answer this question are probably people who use but don't exclusively study mapping class groups.

The mapping class group is a group one associates to a surface, and it's true that it distinguishes two surfaces that are "visually different."  However, I don't think this is where the true usefulness of it lies.  For one thing, it is a group of homeomorphisms up to equivalence (isotopy), so studying the mapping class allows us to say that the homeomorphisms of two surfaces are different (or at least one of them has "more" or different relationships between them).  This is still not the "best" use for the mapping class group though.

As mentioned in Matt E's answer, algebraic geometry is frequently interested in moduli space, and the mapping class group is the fundamental group of moduli space.  Therefore it bridges between algebraic geometry and the study of surfaces.

But what it really boils down to is the mapping class group is a group of isotopy classes of homeomorphisms and will show up any time you want to discuss homeomorphisms of a surface and often when you want to discuss homeomorphisms of higher dimension and it's nice to have properties of such a group any time you want to talk about homeomorphisms.  One thing I'm interested in, for example, is relating homeomorphisms of a base space and total space of a covering space and the mapping class group gives a language for this.

I think the Farb and Margalit book does a great job of motivating and in fact, is what first interested me in the topic.

Title: Motivation for the Mapping Class Group


问:
For the sinusoidal graph below, write the equation in form $y = a\cos\left(\frac{2\pi}{p}x\right)+b$
For the sinusoidal graph below, write the equation

$$y = a\cos \left(\frac{2\pi}{p}x\right) + b.$$

The answer I solved it to be looking at the graph is

$$y = 25\cos\left(\frac{2\pi}{12}x\right) + 30$$

Thanks any input or help would be appreciated :D

答: I don't think there's anything to add; your answer is correct.</p>


答:
Your second option would exclude something like $\mathbb Z/2 \oplus \mathbb Z/2$, which is a direct sum of simple $\mathbb Z$ modules and is therefore semisimple. The OP of that answer indexed the direct sum by the set of primes, so to allow a multiplicity like the example I just gave needs an exponent, which they denote $\alpha_p$. You could alternatively write this as $\bigoplus_{i \in I} \mathbb Z/n_i$ where each $n_i$ is prime. Then, for instance, the multiplicity issue can be handled by taking $n_i = n_j = 2$, for some $i \neq j$ in $I$, so there is no need to explicitly mention the exponent.

EDIT: Additionally, your second option doesn't exclude any primes either! It equals precisely $\mathbb Z/2 \oplus \mathbb Z/3 \oplus \mathbb Z/5 \oplus \cdots$, but this is only one example of a module. There is more than just one semisimple $\mathbb Z$ module, for instance $\mathbb Z/2$, which is not isomorphic to the above.

Title: Why $\alpha_p$ copies?

Body:
I was reading this question here: What are the semisimple $\mathbb{Z}$-modules? and I understood everything except why we need  $\alpha_p$ copies here $$
\bigoplus_{\text{$p$ prime}}(\mathbb{Z}/p\mathbb{Z})^{(\alpha_p)}
$$
And not just $$
\bigoplus_{\text{$p$ prime}}\mathbb{Z}/p\mathbb{Z}
$$

Could anyone explain this to me please?


问:
How to evaluate integrals of type $\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$
I think it is by substitution but I don't  have any clue on how to go further. Should I substitute $$\sin^2x=t~?$$

$$\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx$$

答:
In general, for the integral of an arbitrary rational function of trigonometric functions, the trick is to use the Weierstrass substitution $t=\tan(x/2).$ Then
$$\sin x=\dfrac{2t}{1+t^2},\qquad \cos x=\dfrac{1-t^2}{1+t^2},\qquad \tan x=\dfrac{2t}{1-t^2},\qquad dx=\dfrac{2}{1+t^2}dt$$ transform the original integral to an integral of a rational function, which can solve via partial fractions. The problem that you have posted here has even simpler solution. Note that $$\sin^2x=\dfrac{\tan^2x}{1+\tan^2x},\qquad\text{and}\qquad \cos^2x=\dfrac{1}{1+\tan^2x}.$$ Hence $T=\tan x$ would make a perfectly good substation. Then $$\int\frac{\cos^2x}{\cos^2x+ 4\sin^2x}\,dx=\int\dfrac{1}{(1+4T^2)(1+T^2)}\,dT=\dfrac{1}{3}\int\left(\dfrac{4}{1+4T^2}-\dfrac{1}{1+T^2}\right)\,dT$$

Added: This second substitution is secretly again the Weierstrass substitution. This is a consequence of the fact that $$\sin^2x=\dfrac{1}{2}(1-\cos (2x)),\qquad \cos^2x=\dfrac{1}{2}(1+\cos (2x)).$$


问:
Conditional Probability In Markov Chains
Suppose we have a Matrix with transition probabilities:

\begin{bmatrix}
p_{00} & p_{01} & p_{02}\\
p_{10} & p_{11} & p_{12}\\
p_{20} & p_{21} & p_{22}
\end{bmatrix}

I need to calculate: $a)P[X(2) = 0|X(1) = 2, X(0) = 1]$

My thoughts:

$ P[X(2) = 0|X(1) = 2, X(0) = 1] = P[X(2) = 0|X(1) = 2]$ using Markov property.

$P[X(2) = 0|X(1) = 2]  = \frac{P[X(2) = 0 \cap X(1) = 2]}{P[X(1) = 2]}$

Here I am stuck and don't know how to proceed. I suppose, that probability of $P[X(2) = 0 \cap X(1) = 2]$ is just $p_{20}$, because we move from state 2 to state 1 and we don't consider what happened in time 0 because if Markov property. But how do we calculate $P[X(1) = 2]$? Should we take into account all states or just the information from $ P[X(2) = 0|X(1) = 2, X(0) = 1]$(starting point is state 1)?

答:
We have $$P([X(2)=0|X(1)=2)=P([X(1)=0|X(0)=2)=p_{20}$$

Note that $p_{20}$ is a conditional probability.


问:
Cauchy sequence that doesn't converge
I'm trying to find an example of a metric space in which a Cauchy sequence doesn't converge, i.e., an incomplete space. I can explain in words that I can take a sequence of rational approximations to $\pi$ or to $\sqrt{2}$ which don't converge in $\mathbb{Q}$, but I don't know how to prove that explicitly, because I'm essentially using that convergence (in a larger metric space, $\mathbb{R}$, in which $\mathbb{Q}$ emebeds) implies Cauchy, and limits are unique.

The first question I have is: is such an approach sufficiently rigorous? I may need some form of argument about how limits in $\mathbb{Q}$ "carry over" to limits in $\mathbb{R}$.

The other approach I can use is taking a metric space $\mathbb{R}^{>0}$ and the sequence $\left\{\frac {1}{n}\right\}$. Secretely, we know that this sequence converges to $0$, but it definitely doesn't converge in $\mathbb{R}^{>0}$. I managed to prove that it is Cauchy by taking $N = \frac{2}{\epsilon}$ and using the definition. I want to show, however, that it cannot converge to any element in the space. So given a positive number $M$, I want to find an $\epsilon > 0$ so that for any $N$, for specific $m,n > N$, we have $|a_m - a_n| \geq \epsilon$.

I don't have any intuition for how to do this, however. Does anyone have a hint?

答:
Your approach can be made as rigorous as you like, yes: the magic going on under the surface is that the inclusions $\mathbb{Q}\hookrightarrow\mathbb{R}$ and $\mathbb{R}^{>0}\hookrightarrow\mathbb{R}$ (and such incisions in general) are embedding: that is, they preserve the topology, and hence the limits.

Your approach as written can't work: you're trying to show that the sequence isn't Cauchy. Assuming that the $a_n$ is a typo and should be $M$, it's quite doable. For a hint: once the sequence has gotten smaller than something, it will stay below it, so can't converge to anything above it. There's a full solution under the line below, so stop reading at the line if you don't want spoilers.

For an alternative approach, I'll construct an incomplete space (and a Cauchy sequence evidencing that it is incomplete) that is not (naturally) a subset of a "more obvious" complete space.

We'll do this by redefining "distance" on $\mathbb{R}$. We'll define $d(x,y) = |e^{-x} - e^{-y}|$ (using a different name to avoid confusion with $|-|$. You can show that this behaves nicely as a notion of "distance" if you like.

For our sequence to show this, we'll choose the natural numbers $a_n = n$.

This is Cauchy: for any $\varepsilon > 0$, there is some $N$ such that $e^{-N} < \frac{\varepsilon}{2}$. For any $m,n > N$, then, $d(a_n, a_m) = d(n,m) = |e^{-n} - e^{-m}| \leq e^{-n} + e^{-m} < 2e^{-N} < \varepsilon$.

However, if it converged to some $x\in\mathbb{R}$, then for any $\varepsilon > 0$, there would be some $M \in \mathbb{N}$ such that for all $n > M$, we would have $|e^{-n} - e^{-x}| < \frac{\varepsilon}{2}$. But then, for any $n > \max(N,M)$, we would have $e^{-x} < e^{-n} + \frac{\varepsilon}{2} < \frac{\varepsilon}{2} +  \frac{\varepsilon}{2} = \varepsilon$.

But then $e^{-x}$ has to be smaller than any positive $\varepsilon$, so can be at most $0$. There is no such $x$, though, and so this sequence does not converge.

Spoiler line here.

Given $M$, take $\varepsilon = \frac{M}{2}$. Then for any $n > \frac{2}{M}$, we have $a_n < \frac{M}{2}$, so $|a_n - M| = M - a_n > M - \frac{M}{2} = \frac{M}{2} = \varepsilon$.

Title: Cauchy sequence that doesn't converge


问:
Proof of Fundamental Theorem of Finite Abelian Groups?
The only proofs I've seen of this tend to involve a few intermediate results and a couple of induction proofs with some clever constructions in them. They aren't hard to follow and they're pretty short, but they do still seem to remove some of the "why it should be true" aspect of it. The result seems blindingly obvious to me from a generator perspective: If a finite abelian group $G$ has generators $x,y,z...$ then any element can be represented by the form

$$g=x^ay^bz^c...$$

And there's an immediate and clear isomorphism from this to

$$\mathbb{Z}_{|x|}\times\mathbb{Z}_{|y|}\times\mathbb{Z}_{|z|}\times\cdots$$

Done. It can't help but be true. As far as I can see this is immediately extensible to finitely generated abelian groups as well. But the proofs I've seen, while short and simple enough, have never made its proof seem quite so clear.

Is there some major complication I'm missing here? I don't see anything fishy in assuming the existence of the generators, or proving the form of $g$, or actually constructing the isomorphism. But I assume I'm missing something because I've never seen the result proved like this, even though it's an obvious construction. Is the longer proof for pedagogical reasons (i.e. the lemmas and techniques are useful later), or am I missing a big hole in this proof?

答: Consider the group $G=\mathbb Z/2\times\mathbb Z/3\times\mathbb Z/5$ and the elements $x=(1,1,0)$ and $y=(0,1,1)$, whose orders are $6$ and $15$. You can easily check that $x$ and $y$ generate $G$, but there surely is no isomorphism $G\cong\mathbb Z/6\times\mathbb Z/15$, for this last group has the wrong number of elements.</p>


问:
Value of integral greater than $\frac{\pi}{2}$
Is there an elementary way to prove that
$$\int\limits_{-1}^1 \frac{\arccos x}{\exp(x)+1}>\frac{\pi}{2} ~~?$$
The approximate value of the integral is 1.75.

I have tried to restrict to a smaller interval and to take minimum values of $\arccos$ on that interval.

答:
By even/odd symmetry we have that

$$I \equiv \int_{-1}^1 \frac{\arccos x}{e^x+1}dx = \int_{-1}^1 \frac{\pi - \arccos x}{e^{-x}+1}dx$$

$$\implies I = \frac{\pi}{2}\int_{-1}^1 \frac{e^x}{e^x+1}dx + \frac{1}{2}\int_{-1}^1 \arccos x\cdot\frac{e^{-x}-e^x}{2+e^{-x}+e^x}dx$$

$$ = \frac{\pi}{2}+\frac{1}{2}\int_{-1}^1 \left(\frac{\pi}{2}-\arcsin x\right)\frac{e^{-x}-e^x}{2+e^{-x}+e^x}dx $$

$$ = \frac{\pi}{2}+\int_{0}^1 \arcsin x\cdot\frac{e^x-e^{-x}}{2+e^{-x}+e^x}dx$$

The integrand on the right is strictly positive, thus we have $I > \frac{\pi}{2}$


问:
Vectors: Can someone check my answer?

Find an equation of the plane that contains the point $(2, 0, 3)$ and the line $x = −1 + t, y = t, z = −4 + 2t$.
I plugged in $t=0$ to get $(-1,0,-4)$ and $t=1$ to get $(0,1,-2)$
$n1$ (vector from $(0,1,-2)$ to $(-1,0,-4)): (-1,-1,-2)
 n2$ (vector from $(0,1,-2)$ to $(2,0,3)): (2,-1,5)
 n1*n2=(-7,1,3)$
General equation of plane: $(-7,1,3)*(x-2,y-0,z-3)
                              =-7x+y+3z=-5$

Find a plane through the points $P1(−2, 1, 4)$ and $P2(1, 0, 3)$, and perpendicular to the plane $4x−y+3z =2$
I found $v$ to equal $(3,-1,-1)$
and $n$ to equal $v*n2=(-4,-13,1)$
Then, $-4(x+2)-13(y-1)+1(z-4)=0
-4x-13y+z+1=0$

Find the distance between the given parallel planes:
$−2x + y + z = 0 and 6x − 3y − 3z − 5 = 0$
I set $-2x+y+z=0, 6x-3y-3z-5=0
 y=z=0$, so $-2x=0, x=0$ (passes through origin)
Distance between $6x-3y-3z-5=0$ and origin $(0,0,0)$: abs$(6(0)-3(0)-3(0)-5)/\sqrt{y^2+(-3)^2+(-3)^2}
 =5/sqrt(54)
 =5/3sqrt(6) $

答:
Let $A=(2,0,3)$ and the point on the line is $B(-1,0,-4)$ so $\vec {AB}=3 \vec i+7 \vec k$. the direction vector of the line is $\vec L=\vec i+ \vec j+ 2\vec k$. So the equation of the plane is
$$u(x-2)+v(y-0)+w(z-3)=0~~~~(1)$$ where $\vec n=u \vec i+v \vec j+ w \vec k= \vec{AB} \times \vec L=-7 \vec i+ \vec j+ 3\vec k$. From (1), the equation of plane is $$-7(x-2)+1(y-0)+3(z-3)=0 \implies -7x+14+y+3z-9=0 \implies 7x -y-3z-5=0$$

Title: Vectors: Can someone check my answer?


问:
Convergence of eigenvalues and spaces of sequence of compact, szmmetric and positive-semi-definite operators in Hilbert spaces.
I have a compact operator, which is uniformly approximated by (finite-dimensional) compact operators and I am concerned with the question, how the eigenvalues and eigenspaces of the former are approximated by the ones of the latter. Specifically, let $K_m,K:X\to X$ be compact symmetric and positive semi-definite with $X$ a separable Hilbert-space and $||K_m-K||\to 0$ as $m\to\infty$.
Let $\lambda_1\ge\lambda_2\ge...$ and $\lambda_1^m\ge \lambda_2^m\ge ...\ge \lambda_m^m$ be the eigenvalues of $K$ and $K_m$ and $v_1,v_2,...$ and $v_1^m,...,v_m^m$ corresponding Eigenvectors, such that $(v_j)$ form a basis of $\mathcal{N}(K)^\perp$ (the orthogonal komplement of the kernel of $K$) and $(v_j^m)$ forms a (finite) basis of $\mathcal{N}(K_m)^\perp$.
Define iteratively $I_1:=[k~:\lambda_k=\lambda_1]$ and $I_j:= [ k~:~\lambda_k=\lambda_{\max(I_{j-1})+1}]$ for $j\ge 2$. Define the subspaces $E_i:=span(v_k~:~k\in I_i)$ and $E_i^m:=span(v_k^m~:~k\in I_i)$ (thus $E_i$ is the space spanned by the eigenvectors of the i-th largest eigenvector of K) and $E_i^m$ is only defined for $m$ sufficiently large. Denote by $P_{E_i}$ and $P_{E_i^m}$ the orthogonal projections onto the subspaces $E_i$ and $E_i^m$. I cannot find reference for the following statements, which I belive should be true:

1: For all $i\in\mathbb{N}$ there holds $\lim_{m\to\infty}\lambda_i^m = \lambda_i$.

2: For all $i\in\mathbb{N}$ there holds $\lim_{m\to\infty}||P_{E_i}-P_{E_i}^m||=0$.

May someone provide a proof or a counter example for the statements?

Does someone have a proof at least for the finite-dimensional case?

答:
Your intuition is correct and the claim follows from the following statement: Let $A\in L(X)$ be a bounded linear operator on a Banach space $X$ and $\lambda\in {\Bbb C}$ an isolated eigenvalue of finite algebraic multiplicity $d$. Let $P_0$ be the spectral projection associated.

Then there is $r>0$ and $\delta>0$ so that for any $\Delta\in L(X)$ with $\|\Delta\|<\delta$ the operator $A+\Delta$ has precisely $d$ eigenvalues (repeated according to multiplicity) $\lambda_1(\Delta),...,\lambda_d(\Delta)$ in $B=B(\lambda,r)$ which depends continuously upon $\Delta$. The associated projection operator $P_\Delta$ has rank $d$ and depends continuously (in fact, analytically) upon $\Delta$. In finite dimensions this is shown e.g. in II.1.4, T. Kato, Perturbation theory of linear operators.

The proofs rely on the study of the resolvent $R(z,A)=(z-A)^{-1}$. Under the hypothesis there is $r>0$ such that $R_z=R(z,A)$ is analytic in ${B}^*=B(\lambda,r)^*=B(\lambda,r)\setminus\{\lambda\}$ and with
$M=\sup_{z\in \partial B(\lambda,r)} \|R(z,A)\|<+\infty$. Then
a Neumann series expansion
$$ R(z,A+\Delta) = R_z + R_z\Delta R_z + ...$$
converges uniformly for $\|\Delta\|< \delta=1/M$ and
$$P_\Delta = \oint_{\partial B} R(z,A+\Delta) \frac{dz}{2\pi i}$$
gives the desired projection onto the eigenvalues in $B$. This operator depends smoothly upon $\Delta$ (when $\|\Delta\|< \delta$). Much more can be said, but useful only if you are acquainted with the above type of integral. The book of T. Kato gives you one source of information, Dunford and Schwartz (Linear operators) another.


问:
What's wrong in my approach in evaluating $\cos^{-1}(\cos\ 10)$
The Question

$\large\cos^{-1}(\cos 10)$

My Approach

$\large\cos^{-1}(\cos\ (10-2\pi))$ As $2\pi$ is the period of $\cos x.$ Now our angle is-

Now we know that for $\large\cos^{-1}(\cos x) = x$, $\large x\ \epsilon\ [0,\pi]$ but currently our $\large x$ is in the 3rd quadrant, we need to get it into 1st or 2nd quadrant. Therefore let us subtract $\large \pi$ from our current angle i.e $\large 10-2\pi$

$\large\cos^{-1}(\cos\ -((10-2\pi)-\pi))$ Now you must be wondering why did I put this minus sign, w.k.t $\large \cos x$ is -ve in third quadrant, by subtracting $\large \pi$ we brought it into 1st quadrant and $\large \cos x$ is +ve in 1st quad, so to compensate that I have put a minus sign.

Now its within range so we can use $\large\cos^{-1}(\cos x) = x$.

So Final answer $\large -((10-2\pi)-\pi)$ = $\large 3\pi - 10$

But this answer is wrong and the correct answer is $\large 4\pi - 10$.

They have used the property $\large \cos\  (\pi - \theta) = \cos\  (\pi + \theta)$ But why my approach is wrong?

Kindly help.

答:
Note
$$
\cos^{-1}(\cos(x)) =\begin{cases}
2\pi n +x\in (0,\frac\pi2)\>\>\>x \text{ in 1st quadrant}\\
2\pi n +x\in (\frac\pi2,\pi) \>\>\>x \text{ in 2nd quadrant}\\
2\pi n -x\in (\frac\pi2,\pi) \>\>\>x \text{ in 3rd quadrant}\\
  2\pi n -x\in (0,\frac\pi2)\>\>\>x \text{ in 4th quadrant}\\
\end{cases}$$

In particular, if  $x$ is in the 3rd quadrant

$$\cos^{-1}(\cos(x)) =
\cos^{-1}(-|\cos(x)|)
=  \pi -\cos^{-1}(|\cos(x)|)\\
= \pi -\cos^{-1}(\cos(\pi+ x-2\pi n))
=2\pi n -x\in (\frac\pi2,\pi)$$

Thus, for $x=10$ in the 3rd quadrant, let $n=2$ to obtain

$$\cos^{-1} (\cos(10))= 4\pi -10\in (\frac\pi2,\pi)
$$


问:
Proof involving a simple inequality
I'm struggling to prove the following inequality (for $0 \leq a \leq c$ and $0 \leq b \leq c$)
$$|a-b| \leq c.$$
I see that it is true by drawing a number line and putting the numbers $a, b$ and $c$ there, but I have no idea how to prove it formally. It seems so trivial that it's tempting to write "obviously the following holds next to it." I tried playing around with triangle inequality but I got only this:
$$|a-b| \leq |a-c| + |b-c|=2c-a-b,$$
which is not helpful at all.

答:
$$-c \leq -b \leq a-b \leq a \leq c$$
hence $|a-b|\leq c$.

Title: Proof involving a simple inequality


答: $-c=0-c\leq a-b\leq c-0=c$, when you use inequalities that are given for $a$ and $b$.

Title: Proof involving a simple inequality

Body:
I'm struggling to prove the following inequality (for $0 \leq a \leq c$ and $0 \leq b \leq c$)
$$|a-b| \leq c.$$
I see that it is true by drawing a number line and putting the numbers $a, b$ and $c$ there, but I have no idea how to prove it formally. It seems so trivial that it's tempting to write "obviously the following holds next to it." I tried playing around with triangle inequality but I got only this:
$$|a-b| \leq |a-c| + |b-c|=2c-a-b,$$
which is not helpful at all.


问:
The sides of the three squares $ABCD,AEZH$ and $AUIK$ have integer lengths and $(AEZH)-(ABCD)=31$. If $BE=UE$, then calculate $(AUIK)$
The sides of the three squares $ABCD,AEZH$ and $AUIK$ have integer lengths and $(AEZH)-(ABCD)=31$. If $BE=UE$, then calculate $(AUIK)$. Where $(A_1A_2A_3A_4)$ is the area of the quadrilateral $A_1A_2A_3A_4$

I tried solving it as follows:

I state that $BE=a$ and $x=AU$, then $(x-a)^2-(x-2a)^2=31$

$2ax-3a^2=31$

And I got stuck here. I believe the answer is 289, but I can't prove it. Could you please explain to me how to solve this question? This question is a multiple choice, the possible answers are 225, 256, 289, 300, 961

答:
Knowing that the squares have integer lengths, $a,x$ are both positive integers.

Now $31 = 2ax - 3a^2 = a(2x - 3a)$. Hence both $a$ and $2x-3a$ are factors of $31$.

Fortunately $31$ is prime, so $a$ is either $1$ or $31$. This gives $2x - 3a = 31$ or $1$ respectively. What is the corresponding value(s) of $x$?


问:
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$
What is the sum of number of digits of the numbers $2^{2001}$ and $5^{2001}$? (Singapore 1970)

I attempted to solve this question by working out what each digit must be, and maybe find some pattern, but I couldn't find any, apart from the fact that $2^{2001}\mod{10}\equiv 4$ and $2^{2001}\pmod{10}\equiv 5$. Could you please explain to me how to solve this question? This question is multiple choice with options $1999, 2003, 4002, 6003, 2002$

答:
It's $2002$.  It's asking for the sum of the number of digits of $2^{2001}$ and $5^{2001}$ in base $10$, so just take the log base $10$ of each, take the ceiling function and hey presto:

$603+1399=2002$


问:
Show that there exists a sequence with subsequences converging to every rational number.

The total question is:

(a) Give an example of a sequence with subsequences converging to $1, 2$, and $3$.

(b) Give an example of a sequence with subsequences converging to every integer.

(c) Show that there exists a sequence with subsequences converging to every rational
number.

My attempt

For (a) I got $(1,2,3,1,2,3,1,2,3,\dots)$

For (b), I got $(0,1,-1,2,-2,3,-3,4,-4,\dots)$

Both of these I think are right, but I have no idea how to start on (c), much less an example for (c). If someone could explain what I should be doing, I'd be so grateful. Thank you.

答:
The sequence that:

Move up from $-1$ to $1$ by increments of $1/2$

Then down to $-2$ by increments of $1/3$

Then up to $3$ by increments of $1/4$

Then down to $-4$ by increments of $1/5$

...

is such that for any real $a$ it has a subsequence converging to $a$.


问:
How to go about figuring out if a function is one to one or on to?
I'm having trouble grasping this concept when I think I get it, another problem shows up and it feels like the same logic doesn't work. But that shouldn't be the case no? So I feel like there's something I'm not completely understanding.
Though I feel like the changing domains or co-domains is what's throwing me off.

For example a problem such as:

Given $[0,∞) \to [8,∞)$

$f(x) = 5x^2+8$

Figure out if its a bijection.(which means both one to one and onto right?)

How do I go about figuring this out? What should I be looking at? I though I could just put $1$ & $-1$ into $x1$ and $x2$ and go from there to see if it's one to one, no? Math isn't my strongest subject so bear with me.

答:
Asserting that $f$ is injective is to assert that, if $x,y\in[0,\infty)$ and if $f(x)=f(y)$, then $\require{cancel}x=y$. You have\begin{align}f(x)=f(y)&\iff5x^2+\cancel8=5y^2+\cancel8\\&\iff\cancel5x^2=\cancel5y^2\\&\iff x^2=y^2\\&\iff x=y\text{ or }x=-y.\end{align}But, since $x,y\in[0,\infty)$, the only case in which $x=-y$ is when $x=y=0$; in particular, $x=y$. And, if $x,y>0$, then it follows from the previous computations that, again, $x=y$.

Concerning surjectivity, consider the equation $f(x)=y$ (with $y\in[8,\infty)$), that is, $5x^2+8=y$. Since\begin{align}5x^2+8=8&\iff5x^2=y-8\\&\iff x^2=\frac{y-8}5\\&\iff x=\sqrt{\frac{y-8}5},\end{align}the equation always has a solution. That is, $f$ is surjective.


问:
Probability of forming an odd integer from a 5-permutation of integers in [0,9]
My intuition is telling me that the probability of forming an odd integer from a 5-element partial permutation of integers in [0,9] is one-half, because it should be the same probability as the integer being even, which makes those probabilities one-half.

However, I'm having trouble proving it.

I split the problem into five disjoint cases: the number consists of 1, 2, 3, 4, and 5 odd integers.

P(the number is odd and it uses only 1 odd integer) = $\frac{5*4*3*2*\binom{5}{1}}{10*9*8*7*6} = \frac{600}{30240}$, because the last digit must be an odd integer, and there are five candidates, therefore 5 choose 1, and the rest are even integers where the first even integer has 5 candidates, the second, 4 candidates and so on.

P(the number is odd and it uses only 2 odd integer) = $\frac{5*4*3*\binom{4}{1}\binom{5}{1}}{10*9*8*7*6} = \frac{1200}{30240}$, because the last digit must be an odd integer, and there are five candidates, therefore 5 choose 1, and there is another odd integer, which has 4 possible candidates; the rest are even integers where the first even integer has 5 candidates, the second, 4 candidates and so on.

I see a problem with this approach is that I can continue to the case of 3, 4, 5, but the numerator would not add up to half of 30240. In fact, the numerator only adds up to 3720.

I feel it might be the placement of the odd integers other than the last, so I tried:

P(the number is odd and it uses only 2 odd integer) = $\frac{5*4*3*4*\binom{4}{1}\binom{5}{1}}{10*9*8*7*6} = \frac{4800}{30240}$, because the last digit must be an odd integer, and there are five candidates, therefore 5 choose 1, and there is another odd integer, which has 4 possible candidates but it can be in any of the 4 places, therefore multiplied by 4; the rest are even integers where the first even integer has 5 candidates, the second, 4 candidates and so on.

But this time the numbers are too large, and it's way over half of 30240.

What am I doing wrong here?

答:
While it is clear from the unit digit but if you want to still find all favorable permutations and divide by total permutations, why not do

Favorable permutations $ = \displaystyle 5 \cdot {9 \choose 4} \cdot 4!$ and then divide by $30240$.

[Explanation: choose the unit digit from $5$ odd digits and then choose $4$ from remaining $9$ digits and permute them]

Even if we go by your method, it should be

Count of odd integers with only one odd digit (unit place)  = $\displaystyle 5 \cdot {5 \choose 4} \cdot 4!$

Count of odd integers with two odd digits  = $\displaystyle 5 \cdot 4 \cdot {5 \choose 3} \cdot 4!$

Count of odd integers with three odd digits  = $\displaystyle 5 \cdot {4 \choose 2} \cdot {5 \choose 2} \cdot 4!$

Count of odd integers with four odd digits  = $\displaystyle 5 \cdot {4 \choose 3} \cdot {5 \choose 1} \cdot 4!$

Count of odd integers with all digits odd  = $5!$


问:
Proving $\log(x)$ not uniformly continuous
I am trying to solve problem and I keep running into a different proof.

I am trying to prove $\log(x)$ not uniformly continuous (0,1) yet I keep running into the proof for uniform continuity with $\ln(x)$ on (0,1).

I am challenged and would appreciate some help. please and thank you.

答:
Hint

Considering the sequence $\{u_n\}$ where $u_n = 10^{-n}$. The difference between two terms goes to zero, but not the difference $\ln u_{n+1} - \ln u_n$.


问:
Why subsequence $ a_{n_{p}( k (\epsilon) )} $ doesn't converge to $ b $ if its indexes depend on $ \epsilon $?
Lemma used for proof of a theorem:

$ L \in \mathbb{R} $ is a partial limit of sequence $ ( a_n ) $ iff  for all $ \epsilon > 0 $ the set $ \{ n : | a_n - L | < \epsilon \} $ is infinite.

Notation:  $ \mathscr{P}(a_n) $ is the set of all real partial limits of $ ( a_n) $

Theorem: Let $ ( a_n )  $ be a bounded sequence and let $ ( b_m) \subseteq \mathscr{P}(a_n)$ be a sequence of partial limits of $ ( a_n) $ s.t. $ \lim\limits_{m \to \infty} b_m = b$ .

Show that $ b \in \mathscr{P}(a_n)  $.

Proof( from class notes ) :  We have $ \lim\limits_{m \to \infty} b_m = b$, therefore for every $ \epsilon > 0 $ there exists $ b_{k(\epsilon)} $ s.t. $ | b_{k(\epsilon)} - b | < \epsilon/2 $ ( since  $
\forall \epsilon > 0$ there exists an index $k(\epsilon) $ s.t. $ \forall n>k(\epsilon) . $   $| b_n - b | < \epsilon $ , and I just chose one such index ), on the other hand $ b_{k(\epsilon)} \in \mathscr{P}(a_n)  $ ( meaning that $ b_{k(\epsilon)} $ is a partial limit ) so therefore there exists a subsequence $ a_{n_{p}( k (\epsilon) )} $ s.t. $ a_{n_{p}( k (\epsilon) )}  \xrightarrow{ p \to \infty }  b_{k(\epsilon)} $ , therefore there exists $ p_{0(k ( \epsilon) )} \in \mathbb{N} $ s.t.  for all $ p > p_{0(k ( \epsilon) )}  $ we have $  | a_{n_{p}( k (\epsilon) )}  - b_{k(\epsilon)} | < \epsilon/2  $.

So for all $ p > p_{0(k ( \epsilon) )}  $ we have $ | a_{n_{p}( k (\epsilon) )}  - b | \leq | a_{n_{p}( k (\epsilon) )}  - b_{k(\epsilon)} | + | b_{k(\epsilon)} - b | < \epsilon/2 + \epsilon/2 = \epsilon  $, meaning the set $ \{ n : | a_n - L | < \epsilon \} $ is infinite and therefore $ b \in \mathscr{P}(a_n)  $ . Q.E.D.

My question:

at the last steps of the proof, we got " $ p > p_{0(k ( \epsilon) )}. | a_{n_{p}( k (\epsilon) )}  - b | < \epsilon  $  " and my professor said that it is false to think that we got here a subsequence $ a_{n_{p}( k (\epsilon) )} $ that converges to $ b$ and that is because the indexes in a subsequence that converges to $ b $ should not depend on $ \epsilon $ ( And in $ a_{n_{p}( k (\epsilon) )} $ the indexes $ n_{p}( k (\epsilon) ) $ depend on $ \epsilon $  ).

So I didn't really understand this ( I asked him for further clarification but the answer was overall the same and I still didn't understand. ), why the subsequence  $ a_{n_{p}( k (\epsilon) )} $ does not converge to $ b $? so what if the indexes depend on $ \epsilon $ ?

答:
Consider the sequence $\{a_n\}_{n=1}^\infty$ where
\begin{aligned}
a_n=\begin{cases}1 & \text{if }n \neq p^k \text{ for some prime } p,\, k \in \mathbb{N}\\
p^{-1}-k^{-\frac{p+1}{p}} & \text{if }n = p^k \text{ for some prime } p,\, k \in \mathbb{N}\end{cases}
\end{aligned}

So $\lim_{k \to \infty} a_{p^k}=\lim_{k \to \infty} p^{-1}-k^{-\frac{p+1}{p}}=p^{-1}$ and $\lim_{m \to \infty} p_m^{-1}=0$ where $p_m$ denotes the $m$th prime number. Think about what issue might arise in applying your argument to the above example.

Edit:
I am annoyed by my first counterexample, so I want to also consider the sequence $\{b_n\}_{n=1}^\infty$ where
\begin{aligned}
b_n=\begin{cases}1 & \text{if }n \neq p^k \text{ for some prime } p,\, k \in \mathbb{N}\\
p^{-1}-pk^{-\frac{p+1}{p}} & \text{if }n = p^k \text{ for some prime } p,\, k \in \mathbb{N}\end{cases}
\end{aligned}
(I would like to think that there are in some sense better counterexamples to illustrate the problem with the proposed argument but I am a one mediocre idea kind of guy)


答:
Just to give a different approach, note that

$$|a-b|=\max(a,b)-\min(a,b)\le\max(a,b,c)-\min(0,a,b)$$

holds for any three real numbers $a$, $b$, and $c$. (That is, adding extra options can only make max's bigger and min's smaller.) The assumptions $0\le a,b\le c$, so that $\max(a,b,c)=c$ and $\min(0,a,b)=0$, now do the rest.

Title: Proof involving a simple inequality


问:
Can the remainder of difference be used as pigeon-holes?
Given a problem as follows.

Prove that among 8 integers, there are always two whose difference is divisible by 7.

Method 1

The set of all possible remainders when dividing integers by 7 is $\{0,1,2,3,4,5,6\}$. Because there are 8 integers (pigeons) and 7 possible remainders (pigeon-holes), then at least two integers have the same remainder when divided by 7. The difference of this particular pair has zero remainder when divided by 7.

Method 2

The difference of two integers is an integer. Any integer has one of 7 possible remainders (pigeon-holes) when divided by 7. There are ${8\choose 2}=28$ ways to make differencing pairs (pigeons).

How to make sure there is at least one pair (among 28 possible pairs) with zero difference?

答:
This Method 2 doesn't really get us anywhere.

At least $4$ of the $28$ difference values have the same remainder mod $7$. So $(a-b)-(c-d)$ is a multiple of $7$. But then what?

This argument doesn't rule out the case where the paired differences have remainders $1$ through $6$, some remainders used at least $5$ times, but remainder zero used $0$ times.

The Pigeonhole Principle can be used to show some group of "pigeons" have a common unknown "hole" value, like Method 1 shows there are elements with the same unknown remainder. It can't be used to show that any pigeons at all are in a specific hole, like Method 2 is trying to see if any differences are in the specific "remainder zero" hole.


问:
Intuition / Interpretation of Balance Equations (Markov Chains)
I'm currently studying a module on Stochastic Processes and I'm quite confused as to the interpretation of Global/Local Balanced Equations for Discrete Markov Chains.

From what I've learnt,
Global balanced equations satisfy
$$ \pi = \pi\mathbf{P} $$
where $\pi$ is the distribution and $ \mathbf{P} $ is the transition probability matrix.
Element-wise, it is
$$ \pi(j) = \sum_{k\in S} \pi(k)P_{kj} $$
or
$$ P(X_n = j) = \sum_{k \in S}P(X_n = k)P(X_n+1 = j | X_n = k) = P(X_{n+1} = j) \tag{*}$$

I understand the above equation from the context of regular Markov chains and the limiting distribution, but there's also an explanation that the LHS of the equation is the probability of  leaving node $ j$ to other nodes $ i \neq j $, and the RHS is the probability that nodes $ i \neq j $ changes to node $j$. However, I cannot link the equation $(*)$ to this intuition.

Also, for local balanced equations, where $$  \pi(i)P_{ij} = \pi(j)P_{ji} $$
What is the interpretation of the above?
Is it that the probability from state $j$ to $ i  = $ probability from state $ i $ to $j$? How do I interpret it with the equation?

答:
Equation $(*)$ is just a special case of the law of total probability: whenever $B_1, B_2, \dots, B_k$ are a partition of the sample space (that is, whenever exactly one of them always holds) and $A$ is any other event,
$$
   \Pr[A] = \sum_{i=1}^k \Pr[A \mid B_i] \Pr[B_i].
$$
Here, $B_i$ is the event "$X_n = i$" and $A$ is the event "$X_{n+1}=j$".

The local balance equations (or detailed balance equations, or reversibility equations) are not guaranteed to hold for all Markov chains. In general, they hold for those Markov chains which can be interpreted as random walks on weighted, undirected graphs.

If the equation $\pi(i) P_{ij} = \pi(j) P_{ji}$ holds for all $j$, then we can sum over all $j \in S$ to get
$$
    \sum_{j \in S} \pi(i) P_{ij} = \sum_{j \in S} \pi(j) P_{ji} \implies \pi(i) = \sum_{j \in S} \pi(j) P_{ji}
$$
which is just the global condition.

In practice, the local balance equations are much easier to solve, when a solution to them exists. So we often try solving them, and if it works, we're happy - without thinking about why a particular Markov chain does or does not have a reason to satisfy them.

One nice special case where we do have an intuitive reason to expect reversibility is when we are doing a random walk on the number line. Here, $S$ is a finite or infinite interval of integers, and $P_{ij} = 0$ if $|i-j| > 1$: from each state, we can only go to the next integer, or the previous integer, or stay put.

Suppose that we take $N$ steps in this random walk, where $N$ is very large. We expect to see state $i$ about $N \pi(i)$ times, and so we expect to take a step from $i$ to $i+1$ about $N \pi(i) P_{i,i+1}$ times. Similarly, we expect to take a step from $i+1$ to $i$ about $N \pi(i+1) P_{i+1,i}$ times. However, such steps must always alternate: once we go from $i$ to $i+1$, we must go from $i+1$ back to $i$ before we go from $i$ to $i+1$ again! So $N \pi(i) P_{i,i+1}$ and $N\pi(i+1) P_{i+1,i}$ should be off by at most $1$, which motivates the condition $\pi(i) P_{i,i+1} = \pi(i+1) P_{i+1,i}$.


问:
Evaluating triple integral $\int_1^{10}\int_0^z\int_0^{z + y}\frac{1}{(x + y + z)^3}\,dx\,dy\,dz$

Evaluate the triple integral $$\int_1^{10}\int_0^z\int_0^{z + y}\frac{1}{(x + y + z)^3}\,dx\,dy\,dz$$

I am not looking for the entire solution, but hints on how to perform change of variables/any other trick. My first idea was to use $u = x + y + z$, but I don't know how to properly calculate the new limits: if $u = x + y + z$, then is the new lower bound of $dx$ integral $u(0, y, z) = y + z$? In any case I think that there is a snappier trick for substitution other than the $u$ I proposed.

答: Hint: the innermost integral is just the integral of $\frac1{(x+a)^3},$ where $a=y+z.$