问:
$f:S^n\to X$ is nulhomotopic if and only if there is an extension $\tilde{f}:D^{n+1}\to X$ of $f$.

$f:S^n\to X$ is nulhomotopic if and only if there is an extension $\tilde{f}:D^{n+1}\to X$ of $f$. ($X$ is a topological space)

I know how to prove the forward direction, but I don't know how to prove the reverse direction directly. Could you give any hint? (I'm a beginner of Algebraic topology so I don't know the higher concept of Algebraic topology. I just wanted to extend some theorem in Munkres' topology textbook.)

答:
Let $\widetilde{f}\colon D^{n+1}\to X$ be an extension of $f\colon S^n\to X$.

If you already know that $D^{n+1}$ is contractible, you're essentially done since that implies that there is a homotopy $$F\colon D^{n+1}\times[0,1]\to X$$ such that $$F(\cdot,0) =\widetilde{f},\ F(\cdot,1) = \operatorname{const}$$ which means precisely that $\widetilde{f}$ is nullhomotopic. Since $\widetilde{f}|_{S^n} \equiv f$, i.e. $\widetilde{f}$ agrees with $f$ on $S^n \subset D^{n+1}$, $f$ is nullhomotopic.

If you do not want to presume that $D^{n+1}$ is contractible, you can pick an arbitrary point $y \in \operatorname{int}(D^{n+1})$ (the interior) and homotope $\widetilde{f}$ via a homotopy $F$ so that it becomes constant on a neighbourhood $V_y$ of $y$.

Now pick a smaller copy $U$ of the disk that sits inside $V_y$ and remove its interior $\operatorname{int}(U)$. Since the homotopy $F$ has been arranged to be constant on $V_y$, it's certainly constant on the remaining boundary $\partial U$ of the removed disk $U$.

Now observe that $D^{n+1}$ with the removed open disk $U$ becomes a cylinder. So this new cylinder gives a homotopy between $f\colon S^n \to X$ and the constant map $\operatorname{const}$ via $$F\colon S^n\times[0,1] \to X$$ with $$F(\cdot,0) = f,\ F(\cdot,1) = \operatorname{const.}$$


问:
Question about a proof from Spivaks calculus book 4th edition
I am currently reading Spivaks Calculus book 4th edition, and am on the chapter with limits. Theorem one of this chapter says that as $x$ approaches $a$, you can not approach two different limits $l$ and $m$ unless they are equal. The proof of this theorem makes sense except for one thing. He ends up with a contradiction, but I have no idea what he contradicted. Would someone be able to explain it to me? I know I am a little vague, but their are pdf's of his book, so I figured it could be googled.</p>

答: He contradicted the existence of two limits $\ell$ and $m$ with $\ell\ne m$. That is, assuming $\ell\ne m$ results in the nonsensical conclusion that a real number is less than itself.</p>

ContentLicense: CC BY-SA 3.0


问:
Showing $f=\overline{f}$ a.e. with some finite measure hypothesis.
So given the sequence $\{ f_n \}_{n\in \mathbb{N}}$, I found that there exists a subsequence $\{ f_{n_k} \}_{k\in \mathbb{N}}$ converging weakly to an $\overline{f}\in \textit{L}^p(E)$, so by definition

$$\int_Egf_{n_k} \longrightarrow \int_Eg\overline{f} \mbox{ per } k\rightarrow +\infty \mbox{, } \forall g\in \textit{L}^p(E) \mbox{ (for Riesz's rapresentation, with } \frac{1}{p}+\frac{1}{q}=1\mbox{).}$$

I also know that

$$\int_Ff_n \longrightarrow \int_Ff \mbox { per } k\rightarrow +\infty \mbox{, } \forall F \mbox{ measurable s.t. } |F| <+\infty$$

I would like to conclude that $f=\overline{f}$ a.e. Any suggestions? Also I put my (brief) attempt below.

I was about to conclude saying that taking $g=\chi_F$, then
$$\int_Ff_{n_k} \to \int_F\overline{f}=\int_Ff$$
means that $f=\overline{f}$ a.e., but I then realized that I cannot do that becouse I did not suppose they were non-negative.

So could I procede like this?

Taking $g=\chi_{F\cap \{ f-\overline{f}\ge 0 \}}$ and conclude that $f=\overline{f}$ a.e. on $\{ f-\overline{f}\ge 0 \}$, and in the same way $f=\overline{f}$ a.e. on $\{ f-\overline{f}< 0 \}$?

Becouse if I can do that then $f=\overline{f}$ a.e. as I wanted.

答: You are using  the fact that if $\mu (E\cap F)=0$ for every set $F$ of finite measure then $\mu (E)=0$. This requires a proof. There exist sets $F_n$ of finite measures increasing to $\mathbb R^{n}$ (e.g.: $F_n =\{x; \|x\| \leq n\}$). We get $\mu (E)=\lim \mu (E\cap F_n)=0$.

ContentLicense: CC BY-SA 4.0


问:
A certain magical substance that is used to make solid magical spheres
A certain magical substance that is used to make solid magical spheres costs 700\$ per cubic foot. The power of a magical sphere depends on its surface area, and a magical sphere can be sold for 40\$ per square foot of surface area.

If you are manufacturing such a sphere, what size should you make them to maximize your profit per sphere?

答:
Let:

$r$ be the radius of a magical sphere.

$p(r)$ be the profit per magical sphere.

$s(r)$ be the sale price of a magical sphere

$c(r)$ be the cost of making one magical sphere.

$\Rightarrow p(r)=s(r)-c(r)$

Thus, according to the question and the known formulas for the surface area and volume of a sphere:

$s(r) = 40\times4\pi r^2=160\pi r^2$

$c(r) = 700\times\frac{4}{3}\pi r^3=\frac{2800}{3}\pi r^3$

And then:

$p(r)=160\pi r^2-\frac{2800}{3}\pi r^3$

Now, for finding the maximum profit:

$p'(r)=320\pi r-2800\pi r^2=0$

Solving for $r$ yields: $$r=\frac{4}{35}$$


问:
Intuition on group homomorphisms
So I'm studying for finals now, and came across the idea of homomorphisms again. This is not a new idea for me at all, having seen them in groups, rings, fields ect. However, on reevaluating them I realized suddenly that I really don't understand them on the same level as I thought. While isomorphisms seem to be very natural to think about, I can't visualize what's happening in homomorphisms. At this time, I only have a sketchy idea that the existence of a homomorphism between groups (or between groups and what they're acting on) means that they both somehow combine the same way..

Does someone mind sharing their intuition on the concept? As always, any help is appreciated, thanks

答:
This answer will be painfully similar to the one here , but in any case the details might be of help.

Say we have groups $ (G, *) $ and $ (G', \cdot) $ and a map $ \phi : G \rightarrow G' $. We can ask ourselves the following question : Under what kinds of $ \phi $ do we have that $ (\phi(G) , \cdot ) $ is a group, and that " relations in $ G $ give corresponding relations of images in $\phi(G)$ " i.e. that $ ( x*y = z \text{ in group } G ) \implies ( \phi(x) \cdot \phi(y) = \phi(z) \text{ in group } \phi(G) ) $ ?

Proceeding as in the answer linked above, we see maps satisfying $ \phi(x*y) = \phi(x)\cdot\phi(y) $ are precisely the ones we were looking for. Such maps are conventionally called "group homomorphisms".

Title: Intuition on group homomorphisms


问:
how to solve this trigonometric equation involving one parameter?
I was trying to solve this question:

Let $x$ be defined from $0$ to $\pi$, in this equation $\sin x$ must be equal to "$2-k$". It has at least one solution if $k$ is in the range

(1) from 1 to 2 (both included),

(2) from 2 (included) to positive infinity,

(3) from 1 to 3 (both included),

(4) from negative infinity to 1 (included).

Here's my attempt to solve it:

Since x must be in the interval from 0 to pi,

if I plug 3 in the equation I'd get a negative value, so 2 and 3 are both incorrect.

the other options could be both correct,

but if I plug 0, I get "2-0 = 2", 2 it's not in the interval, and if I plug a negative value I got a value that is greater than 2, so it is still not correct.

The first option is correct for each value inside the interval. To prove that, I plug 1.5 (half) and it is acceptable (according to the equation), therefore my solution is (1).

is it correct?
Let me know, and if you know a different method (not trial and error) let me know.

答:
If $x \in [0, \pi],$ then $0 \le \sin x \le 1.$

Hence:

$$0 \le 2-k \le 1 \iff -2 \le -k \le -1 \iff  1 \le k \le 2.$$


问:
$F(x,y)=f(x)g(y)$, $F(r\cos\theta,r\sin\theta)=h(\theta)$, find $F(x,y)$
My progress:
$$F_x(x,y)=f'(x)g(y)$$  $$F_y(x,y)=f(x)g'(y)$$ also, $$F(r\cos\theta,r\sin\theta)=h(\theta)$$
taking partial derivative with respect to $r$ and using chain rule:
$$F_x(x,y)\cos\theta+F_y(x,y)\sin\theta=0$$
$$F_x(x,y)r\cos\theta+F_y(x,y)r\sin\theta=0$$
$$F_x(x,y)x+F_y(x,y)y=0$$
$$f'(x)g(y)x+f(x)g'(y)y=0$$
And......I got stuck. if only the positive sign were negative.....then that would be easy. Any help would be appreciated!

答: I don't know if I  am missing something but this looks trivial. Put $r=0$ to see that $h(\theta)=F(0,0)$. So $h$ is constant and this means $F$ is a constant.


问:
How do i show that this fuction is not injective?
The function is given as:
$$f: \begin{cases}
\mathbb{R} \rightarrow \mathbb{R}^2 \\
t \mapsto (t²-1,t³-t)
\end{cases}$$
i know that this is function class $C^\infty$

答: We have $f(1)=(0,0)=f(-1).$  Conclusion ?


答:
$f(-1)=(0,0), f(1)=(0,0)$
Hence, the function is not injective.

Title: How do i show that this fuction is not injective?

Body:
The function is given as:
$$f: \begin{cases}
\mathbb{R} \rightarrow \mathbb{R}^2 \\
t \mapsto (t²-1,t³-t)
\end{cases}$$
i know that this is function class $C^\infty$


问:
Proving that no term is an initial segment of another term.

The Question:

If $s$ and $t$ are strings, we say that $s$ is an initial segment of $t$ if there is a nonempty string $u$ such that $t = su$, where $su$ is the string $s$ followed by the string $u$. For example, $KUMQ$ is an initial segment of $KUMQUAT$ and $+24$ is an initial segment of $+24u - v$.

The task at hand is to prove, by induction on the complexity of $s$, that if $s$ and $t$ are terms, then $s$ is not an initial segment of $t$.

[Suggestion: The base case, when $s$ is either a variable or a constant symbol, should be easy. Then suppose that $s$ is an initial segment of $t$ and $s := f t_1 t_2 t_3\dots t_n$, where you know that each $t_i$ is not an initial segment of any other term. Look for a contradiction.]

What I have so far:

For bookkeeping, let us first use a definition for terms. A term of $L$ is a nonempty finite string $t$ of symbols from $L$ such that either:

$t$ is a variable

$t$ is a constant symbol, or

$t := f t_1 t_2 t_3\dots t_n$, where $f$ is an n-ary function symbol of $L$ and each of the $t_i$ terms is a term of $L$. This is a recursive definition.

Base Case:

We begin our inductive proof with the base case, as you would expect. What this looks like, is actually, considering the case where $s$ is a variable or a constant symbol.

Suppose that $s$ is a variable. Then, $t := su$, where $u$ is non-empty, can only mean that $t$ is a variable term followed by some other sequence $\in L$. However, since there is no function symbol preceding $s$, this is not possible, since $t$ too is a term. Therefore, if $s$ is a variable, and $t$ is a term, $s$ can not be an initial segment of $t$.

The case for when $s$ is a constant symbol is identical to the case above.

Inductive Hypothesis:

Let the statement of theorem we are trying to prove be that if $s$ and $t$ are terms, then $s$ is not an initial segment of $t$. We define the induction property $P$ such that $P$ holds for term $s$ if $\forall i$, $s$ is not an initial segment of $t_i$ (where $t_i$ is a term) i.e. including a non-empty string $u$, $t_i := su$ is not possible, and conversely, $t_i$ is not an initial segment in $s$.

Inductive Step:

The inductive step of a proof by induction on complexity of a term takes the following form: Assume that $s$ is a term by virtue of clause (3) in the definition of terms above. Also assume that the statement of the theorem is true when applied to the terms $t_1, t_2, \dots, t_n$. Suppose, towards contradiction, that $s$ takes the form $f t_1, t_2, \dots, t_n u$ and is indeed an initial segment for $t$. This implies that $t$ must, by definition of an initial segment, must begin with $f t'_1, t'_2, \dots, t'_n$.

Since we said that:

$$t = su$$

It must be that:

$$f t'_1, t'_2, \dots, t'_n = f t_1, t_2, \dots, t_n u$$

We can then get rid of the $n$-ary function $f$ to achieve the following:

$$t'_1, t'_2, \dots, t'_n = t_1, t_2, \dots, t_n u$$

How to proceed..?

I got some leads from this answer while I was writing this post, but it seems like my answer is missing something and the aforesaid didn't really affirm my understanding or confidence in the correctness of my answer so far. Moreover, I'm looking for a bit more intuitive, and dare I say, simple-worded explanation of how to carry the aforesaid induction forward.

答:
Very very long comment

Yes, KUMQ is an initial segment of KUMQUAT.

Regarding the arithmetical examples, you have to use PREFIX notation, i.e. you have to compare $+24u$ with $−+24uv$. Both + and − are binary: $+(24,u)$ and $−(+(24,u),v)$.

The idea is quite simple... but the formal proof seems complex than it really is.

Consider a simple example regarding the language of arithmetic with two binary function symbols: $+,×$, to be used prefix form.

Having done the base cases with variables and constants, i.e. single-symbol terms, we have to consider the "complex" ones: $fx_1,…,x_n$.

Consider three examples:

$t_1 :=+x,y$ and $t_2 :=×x,y$ and $t_3 :=+x,z$.

What happens when we "compare" $t_1$ and $t_2$? (the case when one of the terms does not have an initial function symbol gives us an immediate answer: the corresponding term is ill-formed).

We start comparing the two function symbols: $+$ vs $×$. They are different, so the answer is immediate: neither $t_1$ is a proper initial segment (an initial substring) of $t_2$ nor $t_2$ is of $t_1$. End of games.

Consider now $t_1$ and $t_3$: the function symbols are the same; thus, we can move on. The first pair of terms: $x$ and $x$, are the same; thus we move on. The last pair is: $y$ and $z$. Now they are different, thus we can conclude that no one is a proper... of the other.

What is left: the hypothetical case: $t_1:=+x,y$ and $t_4:=+x,y,u$ with $u$ nonempty.

We will prove that this case cannot happen, unless we infringe the rules of syntax.
In this case, when we have checked the pair $y$ and $y$ we have verified that $t_1$ actually is a proper initial segment of $t_4$, because up to now the two strings are identical and $t_1$ is shorter than $t_4$.

But we have also the condition that $t_1$ and $t_4$ are terms, and here we need the syntax rules. term $t_4$ starts with $+$, which is binary, and in $t_4$ we have three arguments following $+$.

Thus, either $t_4$ is not a well-formed term, or it is not possible that $u$ is notempty.

But the assumption is that $t_4$ is a term and thus (if $P \lor Q$ and $\lnot P$; therefore $Q$) it is not possible that $u$ is notempty, i.e. $u$ must be empty.


问:
Evaluate $\sum\limits_{n=1}^{+\infty} \frac{\left( \frac{3-\sqrt{5}}{2} \right)^{n}}{n^{3}}$
Evaluate

$$\sum\limits_{n=1}^{+ \infty} \frac{  \left( \frac{3-\sqrt{5}}{2} \right)^{n} }{n^{3}}$$

We can use the Fourier series to calculate this sum, because it converges.

Also, we know that $\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. What is going on about this number $\frac{3-\sqrt{5}}{2}$ ? we know something else ?

Thank you for your answer but we know something else without about trilogarithm function?

答:
Let $a = \frac{3-\sqrt5}{2}$,

There is an interesting bound to this in terms of generalised exponential integral $E_n(x) = \int_{1}^{\infty}\frac{e^{-tx}}{t^n} \,dt$
We do this by applying Abel's partial summation formula. Let the sum of series be $S$. Then
$$
S = lim_{x\to \infty} \frac{A(x)}{x^3} + 3\int_{1}^{\infty} \frac{A(t)}{t^4}\,dt
$$
where $A(x) = \sum_{1\leq n \leq x } a^n = \frac{a}{a-1}(a^{\lfloor x \rfloor}-1)$ and the limit obviously goes to $0$.

Next by substituting the above we get
$$
S = \frac{a}{a-1}\left(1 + 3\int_{1}^{\infty} \frac{a^{\lfloor t \rfloor}}{t^4} \,dt\right)
$$

Next it is sufficient to observe that $a^{t-1} \leq a^{\lfloor t \rfloor } \leq a^t$ and consequently
$$
\frac{a}{a-1}+ 3(a-1)E_4(-ln(a)) \leq S \leq \frac{a}{a-1} + \frac{3a}{a-1}E_4(-ln(a))
$$


问:
Does $\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!}$ converge or diverge
$\sum_{n=1}^\infty(-1)^{n}\frac{n^n}{n!} $

I got that it diverges but I am not sure

答:
There are already four answers on the page, but all of them refer to either Stirling's approximation, or to the famous limit for $e$. Here is a totally elementary argument. Compare two terms with indexes $n$ and $n+1$:
$$
a_n=\frac{n^n}{n!}\quad ?\quad a_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!}
$$
Note that
$$
a_{n+1}=\frac{(n+1)^{n+1}}{(n+1)!}=\frac{(n+1)^n(n+1)}{(n+1)!}>\frac{n^n(n+1)}{(n+1)!}=\frac{n^n}{n!}=a_n,
$$
which proves that the terms of the series do not tend to zero. Hence the series diverge. </p>


问:
Confused regarding the (dual) basis vectors of the cotangent space $T^*_pM$ of a smooth manifold $M$ and components.
Let $M$ be a smooth $n$-manifold with $p\in M$. Consider the cotangent space $$\operatorname{Alt}^{1}T_pM = \operatorname{Hom}(T_pM,\mathbb{R}) = T^*_pM$$

Now, choosing a chart $(U,\varphi)$ around $p$ on $M$ determines coordinates $x^1,...,x^n$.

I know that having $\partial_i = \frac{\partial}{\partial x^i}$ as the standard basis vectors of $T_pM$, the differentials ($1$-forms) $\mathrm  dx^i \in T_p^*M$ are the dual basis vectors of the cotangent space.

Therefore any element $\alpha$ of the contangent space $T^*_pM$ can be expressed with respect to the standard basis vectors $\mathrm dx^i$ as follows

$$\alpha = \sum_{i=1}^n \alpha_i \mathrm dx^i$$

where $a_i$ are the components of $\alpha$ with respect to the basis $\mathrm dx^1,...,\mathrm dx^n$.

Now my question is the following. In one of our solutions to an exercise, the tutor phrased the above expression as

$$\alpha_q = \sum_{i=1}^n \alpha_q\left(\frac{\partial}{\partial x^i}\right) \mathrm dx^i$$

And this is also something that got brushed over in the lecture, but i don't see how this holds since assuming $\alpha_q \in T_p^*M$ to be a $1$-form, it can certainly be evaluated at basis vectors via $ \alpha_q\left(\frac{\partial}{\partial x^i}\right)$ but what it returns is a real number, so (if i am not mistaken)
$$ \alpha_q\left(\frac{\partial}{\partial x^i}\right) \in \mathbb{R}$$

how does that make sense? The components with respect to the basis vectors $\mathrm dx^i$ are supposed to be elements of $T_p^*M$ and not real numbers, aren't they?

答: As pointed out in the comments, the cotangent space $T_p^*M$ is a vector space over $\mathbb{R}$. Thus any linear combination of elements in $T_p^*M$ has real valued coefficients.


问:
Equivalent condition for continuity on a metric (or normed) space.
Let $X$ be a metric (or normed) space and let $f : X → X$. Show that $f$ is continuous at $x \in X$ if and only if, whenever $x_n \to x$ in X, some subsequence of $(f(x_n))$ converges to $f(x).$

The forward direction can be done easily. I am having trouble for the backward direction. I am thinking to approach it using contradiction, assuming that $f(x_n)$ does not converge to $f(x)$ and hence getting some trouble with the continuity.

答:
suppose $f$ is not continuous at $x$, i.e. there is $\varepsilon >0$ s.t. for all $\delta >0$, there $y\in X$ s.t. $\|y-x\|<\delta $ and $\|f(x)-f(y)\|\geq \varepsilon $. In particular, for all $n\in\mathbb N$ there is $x_n\in X$ s.t. $\|x-x_n\|<\frac{1}{n}$ and $\|f(x_n)-f(x)\|\geq \varepsilon $.

In particular, $x_n\to x$ but $f(x_n)\not\to f(x)$ which is a contradiction.


问:
how to solve $\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$?
$$\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$$

I tried this this:
$$\int_{y}^{1} e^{x^2}dx=\sum_{0}^{\infty}\frac{1}{n!}(\frac{1}{2n+1}-\frac{y^{2n+1}}{2n+1})$$
I don't know what to do after or even if this way works...

Body:
OP's way
$$I=\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$$
Then
$$I=\int_{0}^{1}\int_{y}^{1} e^{x^2}dx=\int_{0}^{1}\sum_{0}^{\infty}\frac{1}{n!}\left(\frac{1}{2n+1}-\frac{y^{2n+1}}{2n+1}\right) dy$$
$$\implies I=\sum_{0}^{\infty}\frac{1}{n!}\left(\frac{1}{2n+1}-\frac{1}{(2n+1)(2n+2)}\right).$$
$$I=\sum_{n=0}^{\infty}\frac{1}{n!(2n+2)}=\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{(n+1)!}= \frac{1}{2}\sum_{m=1}^{\infty}\frac{1}{m!}=\frac{e-1}{2}$$


问:
How can I define these functions?
How do I define these functions using absolute value, floor, ceiling, modulo, divides, and arithmetic operators?

$$f(x) =
\begin{cases}
 1&~\text{ if 5 divides x }\\
 0&~ \text{if 5 does not divide x}
\end{cases}
$$

This is what I tried, but it only satisfies '1 if 5 divides x'.

f(x) = (x mod 5) + 1

答: Take $f(x) = 1-(x^4 \bmod 5)$. This follows from Fermat's theorem.

Title: How can I define these functions?


问:
Definite integration technique
What's wrong with this solution of the
$$ \int_0^1 15 \sqrt{9-5x} \, dx \enspace ?$$
$$= \enspace 15\int_0^1 \sqrt{9-5x} \, dx $$
$$ u=9-5x $$
$$ du = -5dx $$
$$ \frac{-15}{5} \int_0^1 \sqrt{u} \, du $$
$$ -3 \int_0^1 \sqrt{u} \, du $$
Which will later lead to the integral of $80$. Other answers say it is $38$. What technique did the use?

答: The bounds on $u$ are wrong. $u=9-5x$ ranges from $9-5\cdot0$ to $9-5\cdot1$ since $x$ ranges from $0$ to $1$.

Title: Definite integration technique


问:
Counting the terms and growth of a nondecreasing sequence
Given a nondecreasing sequence $\{\lambda_n\}_n$, consider the function
$$n(r) = \#\{n\mid \lambda_n \leq r\}.$$

I want to prove that the statements
$\lim_{n\to \infty}\frac{n}{\lambda_n} = L$
and $\lim_{r\to\infty} \frac{n(r)}{r} = L$ are equivalent.

To do so, I've tried the following:

Assuming the first limit $\lim_{n\to \infty}\frac{n}{\lambda_n} = L$, for each $\varepsilon > 0$ there is some $n_0$ such that for all $n\geq n_0$, $|\frac{n}{\lambda_n} - L|<\varepsilon$. Then,
$$n(r) = \#\{n\mid \lambda_n \leq r\} = \#\left\{n\mid \frac{n}{r} \leq \frac{n}{\lambda_n}\right\},$$
so this function should be "roughly equal" to
$$\#\left\{n\mid \frac{n}{r}\leq L\right\} = \#\{n\mid rL\geq n\} = [rL]$$
and so the other limit follows.

I think the reverse implication is immediate if this reasoning is true, by going the other way. My problem is: how can I justify the shaky step properly? Is this the right path?

答:
Here is one way by which you can formalize the "shaky" step:

Fix $\epsilon>0$ and find $n_0$ such that $n>n_0$ implies $n/\lambda_n\in (L-\epsilon, L+\epsilon)$. As you've pointed out,
$$n(r) = \#\left\{n:\lambda_n\leq r\right\} = \#\left\{n:\frac{n}{r}\leq \frac{n}{\lambda_n}\right\}.$$
If we define the set to be $A(r)$ (so that $n(r) = \# A(r))$, then we have the following equation:
$$n(r) = \#(A(r)\cap\{n\leq n_0\})+\#(A(r)\cap\{n>n_0\}).$$
For $r$ sufficiently large, the first term is just some constant depending only on $n_0$. Let's call it $C(n_0)$. The second term, however, can be estimated using the fact $n/\lambda_n\in (L-\epsilon, L+\epsilon)$ for $n>n_0$. We have
$$\left\{n:\frac{n}{r}\leq L-\epsilon \right\}\cap\{n>n_0\}\subset A(r)\cap \{n>n_0\}\subset \left\{n:\frac{n}{r}\leq L+\epsilon\right\}\cap\{n>n_0\}.$$
It is not difficult to see that the cardinality of the left term is $\lfloor r(L-\epsilon)\rfloor - n_0$ while the cardinality of the right term is $\lfloor r(L+\epsilon)\rfloor - n_0$. Putting this altogether, we have
$$\lfloor r(L-\epsilon)\rfloor - n_0\leq n(r)-C(n_0)\leq \lfloor r(L+\epsilon)\rfloor - n_0.$$
Dividing this inequality by $r$ and taking $\limsup$, we conclude that
$$L-\epsilon\leq \limsup_{r\to\infty} \frac{n(r)}{r}\leq L+\epsilon.$$
Since our choice of $\epsilon>0$ at the beginning was arbitrary, we conclude the $\limsup$ is exactly $L$. This same logic works with $\liminf$, so the limit exists and is equal to $L$, the common value of the $\limsup$ and $\liminf$.

This logic doesn't seem immediately reversible for the other direction, but perhaps a similar idea works.


问:
When does restriction of scalars induce flat Modules?
Let $f:Y\rightarrow X$ be a ring homomorphism, and let us consider $X$ to be a $Y$-module via restriction of scalars, i.e. given some $y\in Y$, this element acts on $X$ via $f(y)\cdot x$, for $x\in X$

I'm trying to determine what conditions we must put on $f$ in order for $X$ to be a flat $Y$-module. By definition, this means that given an injective morphism $m:N\rightarrow M$ of $Y$-modules, we have that $1\times m: X\otimes_Y N \rightarrow X\otimes M$ is also injective.

I did a bit of computation that seemed a bit off to me, and I just wanted a bit of a sanity check. To prove what conditions on $f$ would be necessary in order for $1\times m$ to be monic, suppose we had $x_1\otimes m(n_1)= x_2\otimes m(n_2)$. We want to see how we might obtain the equation $x_1\otimes n_1 = x_2\otimes n_2$.

I would like to prove this in a setting more general than vector spaces (really, I'd like to work in the category of sup-lattices, as found in Joyal-Tierney), but for now let's just follow the characterisation provided in the answer here: Necessary and sufficient condition for equality of two tensor products

Then, there are three cases in which $x_1\otimes m(n_1)= x_2\otimes m(n_2)$:

Suppose $m(n_1)\neq 0$, and $\exists y\in Y$ such that $y\cdot m(n_1)=m(n_2)$. Then this forces the relation that $f(y)\cdot x_2 = x_1$. Since $Y$-module morphisms respect the action, we have that $m(n_2)=y\cdot m(n_1)= m(y\cdot n_1)$, which implies that $n_2=y\cdot n_1$ since $m$ is injective. Since $f(y)\cdot x_2=x_1$, this means that $x_2\otimes n_2 = x_2\otimes y\cdot n_1 = f(y)\cdot x_2\otimes n_1 = x_1\otimes n_1$.

Suppose $m(n_1),m(n_2)$ are linearly independent. Then, $x_1=x_2=0$, which implies that $x_1\otimes n_1 = x_2\otimes n_2$.

Finally, suppose $m(n_1)=0$. Then, we have that either $m(n_2)=0$ or $x_2=0$. Since $m$ is monic, we have that $n_1=0$, and thus $x_1\otimes n_1=0$. If $m(n_2)=0$, then $n_2=0$ as well, and thus $x_2\otimes n_2=0$; and if $x_2=0$, then $x_2\otimes n_2=0$ as well.

This little computation seems to imply that restriction of scalars always induces a flat module in this way - is that right? But that seems strange. For instance, take the quotient map $f:\mathbb{Z}\rightarrow\mathbb{Z}/n\mathbb{Z}$. It is well-known that for $n>1$, $\mathbb{Z}/n\mathbb{Z}$ is not flat over $\mathbb{Z}$ since while $n:\mathbb{Z}\rightarrow\mathbb{Z}$, $x\mapsto nx$ is injective, the map $1\times n:\mathbb{Z}/ n\mathbb{Z}\otimes \mathbb{Z}\rightarrow \mathbb{Z}/ n\mathbb{Z}\otimes \mathbb{Z}$ is not injective.

Have I done something silly in the process?

答:
The problem is that linearly dependency is subtler when you work with rings.

You could have $m(n_1)\neq 0$ and $m(n_1),m(n_2)$ linearly dependent, but having $m(n_2)\neq y\cdot m(n_1)$ for all $y\in Y$.

For example, in $\mathbb{Z}$, $2,3$ are linearly dependent since $2.3-3.2=0$, but $2\neq m\cdot 3$ for all $m\in\mathbb{Z}$.

(Also, I hope you know that even if your three cases work (for example in the case of vector spaces), it is not enough to prove injectivity of of $1\times m$, since an element of the tensor product is a sum of elementary tensors.)

What you need to prove is for all $\sum_i x_i\otimes n_i$, $\sum_i x_i\otimes m(n_i)=0$ implies $\sum_i x_i\otimes n_i=0$.

You hae compelte characterization of flat modules, but the easiest case to handle is when your module is finitely presented (ie isomorphic to $F/F'$, where $F$ is free of finite rank and $F'$ is finitely generated). In this case, flat is equivalent to finitely generated and projective (which is equivalent to : isomorphic to a direct factor a free module of finite rank)


问:
How to set up equations to find the volume for solid of revolution?

Let $R$ be the region bounded by the graphs of $y =\sqrt x$, $y=2$, and $x=0$.

Find the volume of the solid generated by revolving $R$ about the $x$-axis;

Find the volume of the solid generated by revolving $R$ about the $y$-axis;

Find the volume of the solid generated by revolving $R$ about the line $x = 4$;

Find the volume of the solid generated by revolving $R$ about the line $y = 2$.

I am trying to set up these integrals, I know how to solve them but I always have a hard time setting them up, any help would be greatly appreciated :)

答:
First off, a (correct-ish) drawing always helps. Always.

Also, practice. There is no way around it: in order to really learn how to do this well enough to be confident you can do it on an exam, you just have to do it a bunch of times. No amount of answers on math forums or videos on youtube is going to change that.

With that said, here is a quick breakdown of the standard method:

Decide whether you want to integrate with respect to $x$ or $y$ (the final form of the integrand depends on this choice in relation to the axis of rotation)

Divide whatever axis you chose to integrate with respect to into little pieces

Slice up the plane region we're after into strips according to the division form point 2

Rotate these slices the same way as the whole region, find the volume of each small corresponding figure, this is your integrand

The dependence mentioned in point 1 determines the basic shape of the figures in point 4, and from here the two names "washer method" and "(cylindrical) shell method" are born. I'll do the first two problems, both times integrating with respect to $x$, to illustrate both methods.

So, without further ado, here is a drawing of the relevant region of the plane:

This you should be able to do on a piece of paper. If not exactly, then at least a close enough sketch to recognize the basic shape, to determine distances, and to determine intersection points. It is also more important than the two other figures in this answer.

The actual volumes are a bit harder to draw, but if you can sort-of imagine them from the above figure, you'll be in pretty good shape. Anyways, here is the result of taking the above figure, and rotate it about the $x$-axis:

The volume we are after is the region that is between the green parabola and the blue plane, and inside the red cylinder.

Since we are integrating with respect to $x$, we divide the $x$-axis into many small pieces of width $\Delta x$ (for simplicity, say they are all equally wide). For each narrow piece, there is a corresponding strip of the planar region. Now rotate such a strip about the $x$ axis. It will look a like a coin with a hole in it (except perhaps the very first and the very last slice). If you've ever been to a hardware store and bought washers, that's exactly what they look like. Hence the name for this method: the washer method.

We want to add together the volume of each little washer-shaped slice. If the slices are thin enough, it won't matter much that the inner hole in the wacher has a slanted edge. We can just pretend that it is a flat cylinder with a cylindrical piece missing. The volume of such a slice is the volume of the large cylinder minus the volume of the small cylinder.

More symbolically, let's take the slice at some particular value $x$ along the $x$-axis. The slice has width $\Delta x$, the outer radius of the cylinder is $2$, and the inner radius is $\sqrt x$. The volume of the slice is therefore
$$
\pi\cdot 2^2\cdot \Delta x - \pi\cdot \sqrt x^2\cdot \Delta x = \pi\left(2^2 - \sqrt x^2\right)\Delta x
$$
This is basically all done now. The range of valid $x$-values (i.e. the $x$-values for which our volume exists) is $0$ to $4$. As we take thinner and thinner slices, the sum of the volume of all of them turns into an integral as $\Delta x$ turns into $dx$, and we get
$$
\int_0^4\pi\left(2^2-\sqrt x^2\right)dx
$$

What about rotation about the $y$ axis? In that case, the region ends up looking something like this:

Which is to say, the volume we're after is the upside-down mountaintop that lies between the green bowl and the red plane. Again, we integrate with respect to $x$, so we divide the $x$-axis into small pieces of width $\Delta x$. Again slice the planar region into vertical strips, but this time rotate them about the $y$-axis. The resulting figure will now be a thin-walled tube / pipe, rather than a flat washer. This is where the name comes from: the cylindrical shell method.

We are after the volume of each shell, and from there want to sum them up to get the volume of the entire figure. So, take an arbitrary $x$-value. What is the volume of the corresponding shell? The shell is a cylinder with a smaller cylinder taken out. The volume is the difference between these two.

More symbolically, they both have height $2-\sqrt x$, the inner cylinder has radius $x$, and the outer cylinder has radius $x + \Delta x$. So its volume is
$$
\pi\left(x + \Delta x\right)^2(2-\sqrt x) - \pi\cdot x^2\cdot(2-\sqrt x) = \pi\left(2x\Delta x + \Delta x^2\right)(2-\sqrt x)
$$
And here we get to another part of integral calculus that might seem hand-wavey: If $\Delta x$ is small enough, then $\Delta x^2$ is vanishingly small compared to $2x\Delta x$ for all but the innermost few shells. So as we pass to the limit of $\Delta x\to 0$, we can just ignore the squared $\Delta x^2$ term. And thus we have our integrand. Again, the bounds for $x$ is $0$ to $4$. So the volume becomes
$$
\int_0^4\pi\cdot 2x\cdot(2-\sqrt x)dx
$$


问:
Factorisation of fourth degree polynomial in mod p
If $p$ is an odd prime with $p \equiv 1 ( \mod 4)$ then $x^4 + 4$ factors into 4 linear parts $\mod p.$

I was breaking $x^4 + 4 = x^4 + (p-1)x^2 + x^2 + 4 (\mod p)$ then using $p \equiv 1 ( \mod 4)$ we have $ x^4 + 4 = x^4 + (4l)x^2 + x^2 + 4 (\mod p)$ after this I am stuck.

Any help.

答:
Hint: If $p \equiv 1 \bmod 4$, then there is $j \in \mathbb Z$ such that $j^2 \equiv -1 \bmod p$.
Also, the complex roots of $x^4+4$ are $\pm 1 \pm i$.


问:
Basic question about $\sup_{x\neq 0}{} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\| = 1}{\|Ax\|} $, $x \in\mathbb{R}^n$
I am having trouble with understanding the following definition while studying some basic things related with matrix norms:

For every matrix $A\in M_n(\mathbb{R})$

$$\sup_{x\neq 0}{} \frac{\|Ax\|}{\|x\|} = \sup_{\|x\| = 1}{\|Ax\|},\; x \in\mathbb{R}^n$$

Why we are taking $\|x\| = 1$?  Is there any proof of above statement?

Added: Why to take sup in the definition of norm?

I need help to understand above mentioned fact.

Thank you very much.

答:
As julien pointed out, what follows is not a proof that

$$\sup_{x\neq0}\frac{||Ax||}{||x||}=\sup_{||y||=1}||Ay||,\quad\quad(1)$$

it is a discussion about the consequences of $(1)$. For a proof of $(1)$ see MichaelNgelo's answer.

$(1)$ implies that if we know that a given vector achieves either of the two supremums in $(1)$, then we can deduce from it a vector that achieves the other supremum.

Specifically, $x^*$ is such that

$$\frac{||Ax^*||}{||x^*||}=\sup_{x\neq0}\frac{||Ax||}{||x||}\quad\quad(*)$$

if and only if $y^*:=\frac{x^*}{||x^*||}$ is such that

$$||Ay^*||=\sup_{||y||=1}||Ay||.\quad\quad(**)$$

The above is useful because it allows to compute one of the supremums by computing the other instead, in particular, we can substitute a supremum over $\mathbb{R}\backslash \{0\}$ with one over the unit sphere.

An easy way to argue the above is by contradiction. For example, one direction follows from:

Suppose that $x^*$ satisfies $(*)$, but $y^*:=\frac{x^*}{||x^*||}$ does not satisfy $(**)$. Then, there exists some $\hat{y}\neq y^*$ such that  $||\hat{y}||=1$ and

$$||A\hat{y}||>||Ay^*||\Rightarrow \frac{||A\hat{y}||}{1}=\frac{||A\hat{y}||}{||\hat{y}||}>||Ay^*||=\left|\left|A\frac{x^*}{||x^*||}\right|\right|=\frac{||Ax^*||}{||x^*||}=\sup_{x\neq0}\frac{||Ax||}{||x||}$$

which gives a contradiction.

EDIT: The $\sup$ is taken in order to make the function $\sigma$, that maps from the vector space of real $n\times n$ matrices, $M_n(\mathbb{R})$, to $[0,\infty)$, and is defined as

$$\sigma(A):=\sup_{x\neq 0}\frac{||Ax||}{||x||},$$

be a norm (that is, satisfy the norm axioms) on $M_n(\mathbb{R})$. If we drop the sup and instead use a fixed $x$, then $\sigma$ doesn't satisfy the first axiom ($\sigma(A)=0\Leftrightarrow A=0)$ any more.


问:
Interesting real integrals from $ I=\oint\limits_{C}\sin(\frac{1}{z})\,\mathrm{dz} = 2\pi i $
I have separated and equated Real and Imaginary Parts of the following equation by plugging $\mathrm{z}=e^{it}$    :

$$I=\oint\limits_{C}\sin\frac{1}{z}\,\mathrm{dz} = 2\pi i,$$

where $C$ is the boundary of unit circle centered at origin.

which gives $$ K=\int_{-\pi}^{+\pi}e^{it}\sin(e^{-it})\mathrm{dt} = 2\pi$$

simplifying gives the following:-

$\int_{0}^{\frac{\pi}{2}}\cos(t)\sin(\cos(t))\cosh(\sin(t))+\sin(t)\cos(\cos(t))\sinh(\sin(t))\mathrm{dt} = \frac{\pi}{2}$

the first and second part of the above integral translates to $I_1$ and $I_2$ respectively under a simple substitution for each.

$$ I_1 = \int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\mathrm{dx}\\ I_2 = \int_{0}^1 \cos(x)\sinh(\sqrt{1-x^2})\mathrm{dx}$$

which implies, $I_1 + I_2 = \frac{\Large\pi}{2}$

Interestingly, I have found that $I_1 $ and $ I_2$ are equal (using a calculator), but stuck proving them.
Any ideas how they are equal or finding another method to calculate $I_1 $ or $ I_2$?

EXTRA: some integrals that show up but eventually cancel out( as they are odd functions, $f(x)=f(-x)$ ) in calculating Imaginary part of equation $K$:-

$$ I_3 = \int_{0}^1 \sin(x)\cosh(\sqrt{1-x^2})\mathrm{dx}\\ I_4 = \int_{0}^1 \sinh(x)\cos(\sqrt{1-x^2})\mathrm{dx}$$

however $I_3 $ and $ I_4$ are not equal and their numerical values are approx. 0.584 and 0.418 , can $I_3 $ and $ I_4$ be calculated in closed form?

Body:
Partial answer:

$$\begin{align}
I_1&=\int_{0}^1 \cosh(x)\sin(\sqrt{1-x^2})\,dx\\
&\stackrel{x\to\sqrt{1-t^2}}{=}
\int_{0}^1 \cosh(\sqrt{1-t^2})\sin(t)\frac{t\,dt}{\sqrt{1-t^2}}\\
&=-\int_{0}^1 \sin(t)\,d(\sinh(\sqrt{1-t^2}))\\
&=\left[-\sin(t)\sinh(\sqrt{1-t^2})\right]_0^1+\int_{0}^1 \sinh(\sqrt{1-t^2})\,d (\sin(t))\\
&=\int_{0}^1 \sinh(\sqrt{1-t^2})\cos(t)\, dt=I_2.
\end{align}$$

It follows $I_1=I_2=\frac\pi4$.

Applying the same method one can show $I_3+I_4=\cosh(1)-\cos(1)$.


问:
Projective representations of 1-loop quiver
I study (finite) representations (over an algebraically closed field $k$) of the 1-loop quiver $Q$ which is defined by having a single vertex and a single edge. So, representations of $Q$ are just endomorphisms of finite dimensional $k$-vectorspaces. I need to determine all indecomposeable representations of $Q$, and by using the structure theorem of finitely generated PID's I'm pretty sure in saying that a representation $f : V \to V$ of $Q$ is indecomposeable exactly if the matrix of $f$ is a Jordan block $J_{a,\dim V}$ in some basis for $V$. My question is now: Which of these indecomposeables are projective? Here a projective representation $A : k^n \to k^n$ means that given any commuting diagram of $k$-linear maps as below with $\pi$ surjective, we should be able to construct a $k$-linear map $g$ (denoted by the dotted line below) such that the resulting diagram commutes.

I have tried to deduce some nessecery conditions for the maps in such diagrams if the representation is to be projective, but I have not been getting any wiser as to decide what I should think. My problem is that it seems (to me!) kind of arbitrary when representations of quivers are projective/not-projective so any ideas or hints as to how I could/should approach this problem would be highly appreciated.

Thank you in advance for your time.

答:
Since your algebra is isomorphic to the polynomial algebra $k[x]$, which is a PID, let's answer a slightly more general question:

Question: Let $R$ be any PID.  What are the finitely generated projective $R$-modules?

Answer: The free $R$-modules.

To see why a torsion module cannot be projective, let $M$ be an indecomposable torsion module.  By the structure theorem for finitely generated modules over PIDs, there exists a non-zero, non-invertible element $a$ of $R$ such that $M$ is isomorphic to $R/(a)$.  Then there is a short exact sequence

$$
 0 \to R/(a) \to R/(a^2) \to R/(a) \to 0,
$$

where the first morphism is multiplication by $a$ and the second is the natural projection.  Clearly, this sequence is not split.  Thus the right-most module, $R/(a)$, cannot be projective.

Going back to representations of the 1-loop quiver: over $k[x]$, the finitely generated torsion modules are preciely the modules which are finite-dimensional over $k$.  Thus there are no projective finite-dimensional representations of the 1-loop quiver.


问:
Can anyone give some pointers about what is wrong with this proof?
Given a function $g$ defined on the real line satisfying these conditions:

$g'(0) = k$

$g(0) \neq 0$

$g(x + y) = g(x)g(y)$ for all $x, y$

Show that $g(0) = 1$ and $g'(x) = kg(x)$ for all x.

My "proof" is as follows:

Define $f(x) = g(x)g(x)$, then $f'(x) = g'(x)g(x) + g(x)g'(x) = 2g'(x)g(x)$

$\implies f'(0) = 2g'(0)g(0) = 2kg(0)$ (from definition 1)

But $f(x) = g(2x)$ (definition 3) so $f'(0)=2g(0)=2k$

$\implies f'(0) = 2kg(0) = 2k \implies g(0) = 1$

For the second part:

$\frac{d}{dx}g(x+y) = g'(x+y) = g'(x)g(y)$ (from definition 3)

$\implies g'(x+y) = g'(x)g(y)$

$\implies g'(0 + y) = g'(0)g(y)$

$\implies g'(y) = kg(y)$ (from definition 1)

Any pointers on where the flaw in the argument is? I'm new to proofs and would appreciate any help!

答: The fact that $g(0) = 1$ can be demonstrated in a less circuitous fashion.  Observe directly that property 3 implies for $x = y = 0$ $$g(0) = g(0 + 0) = g(0)g(0) = g(0)^2.$$  Consequently, $$0 = g(0)^2 - g(0) = g(0) (g(0) - 1),$$ and since property 2 excludes $g(0) = 0$, we are forced to conclude $g(0) = 1$.  Note this result does not rely on differentiability.


问:
Finite Concepts that Require Infinity.
Answers to this question, I hope, would address the objections one might have towards the (many different) foundations of infinity by examples of finite concepts - theorems, definitions, intuitions, perspectives, etc. - that require the infinite, whether that be in a proof, an important source of motivation, a key example (like some group like $S_3$ can be for non-abelian group theory), etc.

The examples-counterexamples tag could preempt criticism that the question is too broad.

What objections?

Well, for example, there're those that posit that infinity has no place in mathematics as, otherwise, we would be - and I quote - "quantifying the unquantifiable".

The Question:

What examples are there of finite concepts where infinity does some essential heavy-lifting?

Thoughts:

Nothing springs to mind.

An example might take the following form:

Theorem: (Only finite concepts hypothesised.) Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. (Finite consequences.)

Proof: Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. (Infinity.) Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum. $\square$

(I used filler text but I hope this captures the sort of thing I'm looking for, without belabouring the point.)

What I want is a bunch of things to point to and say, "here: important finite things can require infinity" - and bring something to the table that is tangible in an area that is largely seen as intangible.

A Potential Example:

Finitely-presented groups can be infinite. For example, the presentation $$\langle a\mid \rangle$$ is finite, whereas the group $(\Bbb Z, +)$ it corresponds to is infinite.

Preliminary Discussion:

Before asking here, I brought it up in chat.

Please help :)

答:
In the sourcebook literature you can find Poincare's criticisms of Russell and Hilbert (Ewald's "From Kant to Hilbert" volume 2). Although his most important criticism is often taken to point to the distinction between predicative and impredicative mathematical reasoning, he also points out the circularity of trying to ground arithmetical induction.

This is now resolved with metatheory. It can be contrasted with the constructive mathematics of the Russian school. What the latter cannot represent will be the arithmetized mathematics that requires infinity in its assumptions. This probably cannot be made precise.

What clearly changes is the approach to quantification. In Hilbert's writing, "axiomatic" is  contrasted with "genetic." This reflects the backlash of nineteenth century mathematics against Kant's association of arithmetic with temporal intuition. But, it also relates to the Aristotelian distinction between "demonstrative" and "dialectical" reasoning. Aristotle views the succession of natural numbers in the manner Hilbert would call "genetic." Smaller numbers have "existential priority" with respect to larger numbers. And, earlier statements in a pedagogical demonstration are prior knowledge with respect to the statements which follow.

This is why the comparison with Russian constructive mathematics ought to be made.  Its feasible reasoning does not involve the usual notion of induction.

Title: Finite Concepts that Require Infinity.


问:
Predicate model that satisfies different formulas
Consider the formulas:

$$F_1:=\ \exists y (P(y)\wedge Q(y))$$
and
$$F_2:=\ \exists y\exists x \neg(\neg P(x)\rightarrow Q(y))$$

Provide a model that satisfies both formulae $F_1$ and $F_2$.

Provide a model that does not satisfy formula $F_1$ and also does not satisfy formula $F_2$.

Provide a model that satisfies formula $F_1$, but does not satsify $F_2$.

Provide a model that does not satisfy formula $F_1$, but does satsify $F_2$.

I don't quite understand how to find model satisfying the implication part in the second formula.

答:
Maybe you can try to rewrite $F_2$ avoiding implication, indeed $F_2$ is equivalent to $\exists x\exists y(\neg P(x) \wedge \neg Q(y))$, in this way the exercise becomes easier.

As $P$ and $Q$ are unary predicates, the interpretation of the formula $\exists y P(y)$ is simply "there exists an element in $P^I$" (where $P^I$ is the subset of the domain that interprets $P$), so it simply means that $P^I$ is not empty. In this way, the first formula says "the intersection between $P^I$ and $Q^I$ is not empty", while the second one says "there is an element outside of $P^I$ and another one outside of $Q^I$": now I think it's easy to give an answer to the problem.


问:
Eigenvalues and Eigenvectors for the following matrix
$$\begin{bmatrix}
1 & 1 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{bmatrix}$$
I'm getting 3 eigenvalues 1, 1, 1 (all same) and
on solving for eigenvector I'm getting
$$\begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$
$x_{2}$=0
how to proceed further?

答: The eigenspace is defined by the single linear equation $x_2$, so it has dimension $2$, and two obvious linearly independent eigenvectors are made up of the  first and last column of the matrix. The second column is a generalised eigenvector.


问:
Differential of the product smooth map from arbitrary manifold
Suppose $f:M\to N\times P$ defined by $f(x)=(f_1(x),f_2(x))$ where $f_1:M\to N$ and $f_2:M\to P$ are smooth maps. Then show that $df=df_1\times df_2$ i.e. $df(X)=(df_1(X), df_2(X))$. ($df$ is the differential of $f$ & $M,N,P$ are smooth manifolds)

I was trying by the formula $df(X)(g)=X(g\circ f)$ where $X$ is a tangent vector and $g$ is a real valued function. But did not able to complete the proof. Please help to do this.

答:
Suppose $f : M \to P \times Q$ is a smooth map defined as $f(x) = (g(x),h(x))$ where $g \in C^{\infty}(M, P)$ and $h \in C^{\infty}(M,Q)$. Let $x \in M$ and $g(x)=p \in P$, $h(x)=q\in Q$. First, we know that $T_{(p,q)}(P \times Q) \cong T_pP \times T_qQ$ by isomorphism
$$
\alpha : T_{(p,q)}(P \times Q) \longrightarrow T_pP \times T_qQ, \quad \alpha(v) = \Big(d(\pi_1)_p(v), d(\pi_2)_q \Big),
$$
where $\pi_1 : P \times Q \to P$ and $\pi_2 : P \times Q \to Q$ are the canonical projections.

Let's define $dg_x \times dh_x : T_xM \to T_pP \times T_qQ $ as the map $(dg_x \times dh_x)(v) = \Big(dg_x(v),dh_x(v)\Big)$. Since the codomain of $df_x : T_xM \to T_{(p,q)}(P \times Q)$ and $dg_x \times dh_x$ are not really the same, what we really need to show is that the composition map
$$
T_xM \xrightarrow{df_x} T_{(p,q)}(P \times Q) \xrightarrow{\alpha} T_pP \times T_qQ
$$
is equal to $dg_x \times dh_x$. But this is just a modest computation using the chain rule :
\begin{align}
\alpha \circ df_x (v) &=  \big( d(\pi_1)_p \circ df_x(v), d(\pi_2)_q \circ df_x(v) \big)\\
&= \big( d(\pi_1 \circ f)_x(v),d(\pi_2 \circ f)_x(v) \big) \\
&= \big( dg_x(v),dh_x(v) \big)\\
&= dg_x \times dh_x(v).
\end{align}
See here for a more general case.


问:
Convolution of $1/(1+x^2)$ and $\exp(-x^2/(4t))$
Is there a closed form formula for the convolution of $1/(1+x^2)$ and $\exp(-x^2/(4t))$, where $t>0$, i.e. the integral
$$\int_{-\infty}^\infty \frac{1}{1+u^2}\exp\left(-\frac{(x-u)^2}{4t}\right)\,du?$$
I doubt it, but someone may be able to use special functions.</p>

答:
An equivalent integral,
$$\int_{-\infty}^{\infty}\frac{e^{-x^2}dx}{(x-a)^2+b^2},$$
for $a,b>0$ can be expressed in terms of error function:
$$\frac{\pi}{b}e^{b^2-a^2}\left\{\cos2 ab-\mathrm{Re}\left[e^{2iab}\mathrm{erf}(b+ia)\right]\right\}.\tag{1}$$

Added derivation: Write
$$\frac{1}{(x-a)^2+b^2}=\frac{1}{2ib}\left(\frac{1}{x-a-ib}-\frac{1}{x-a+ib}\right).
\tag{2}$$
Henc our problem reduces to calculating $\displaystyle\int_{-\infty}^{\infty}\frac{e^{-x^2}dx}{x-\alpha}$ with $\alpha\notin\mathbb{R}$. Now suppose e.g. that $\mathrm{Im}\,\alpha>0$ and then deform the integration contour by translating it upwards. The final contour will consist of 3 parts: two horizontal segments $C_-=(\alpha-\infty,\alpha-\epsilon]$ and $C_+=[\alpha+\epsilon,\alpha+\infty)$ and a demi-circle $C_{0}$ of radius $\epsilon$ centered at $\alpha$.

The integral over $C_0$ in the limit $\epsilon\rightarrow0$ gives simply $i\pi \, e^{-\alpha^2}$.

The integrals over $C_+$ and $C_-$ give
\begin{align}
\int^{\infty}_{\epsilon}\frac{e^{-(\alpha+s)^2}ds}{s}+\int^{-\epsilon}_{-\infty}\frac{e^{-(\alpha+s)^2}ds}{s}&=-2e^{-\alpha^2}\int_{\epsilon}^{\infty}\frac{\sinh2\alpha s}{s}e^{-s^2}ds\rightarrow\\&\rightarrow -2e^{-\alpha^2}\int_{0}^{\infty}\frac{\sinh2\alpha s}{s}e^{-s^2}ds\quad \mathrm{as}\;\epsilon\rightarrow0.
\end{align}

The last integral can be rewritten as
\begin{align}
\int_0^{\infty}\frac{\sinh2\alpha s}{s}e^{-s^2}ds&=2\int_0^{\infty}e^{-s^2}\left(\int_0^{\alpha}\cosh 2\nu s \,d\nu\right)ds=\\ &=\int_0^{\alpha}\left(\int_{0}^{\infty}2\cosh2\nu s\,e^{-s^2}\right)d\nu=\\ &=\sqrt{\pi}
\int_0^{\alpha}e^{\nu^2}d\nu=-\frac{i\pi}{2}\,\mathrm{erf}(i\alpha).
\end{align}

Summarizing all of this, we get
$$ \int_{-\infty}^{\infty}\frac{e^{-x^2}dx}{x-\alpha}=i\pi e^{-\alpha^2}\left[1+\mathrm{erf}(i\alpha)\right].\tag{3}$$
Combining (2) and (3), we then get (1). $\blacksquare$


答:
Edit: Replaced $x$ with $|x|$ in case some program treats negative $x$ differently.

$f(x) := 1-\lceil \frac{|x| \mod 5}{5} \rceil.$

Title: How can I define these functions?

Body:
How do I define these functions using absolute value, floor, ceiling, modulo, divides, and arithmetic operators?

$$f(x) =
\begin{cases}
 1&~\text{ if 5 divides x }\\
 0&~ \text{if 5 does not divide x}
\end{cases}
$$

This is what I tried, but it only satisfies '1 if 5 divides x'.

f(x) = (x mod 5) + 1


问:
Inverse indicator function set theory
I have this True\False question:
if A,B are subsets of a universal set U, then $\chi^{-1}_A( \{1\})\cap\chi^{-1}_B( \{0\})=A\setminus B$

I feel like this is False because it intersects a group with elements that are not in A with elements that are in B, which is $B\setminus A$ .

That being said, I am still unsure about the definition of the indicator set, and how it's being used here.

答:
As to the definition of the indicator set function:
$$\chi_A (x) = \begin{cases} 1 \quad \text{if } x\in A\\
0 \quad \text{if } x\not\in A\\
\end{cases}$$
It practically tells you whether an element is on the set or not.

Therefore, $\chi_A^{-1}(\lbrace 1\rbrace) = \lbrace x\in U: \chi_A(x)=1\rbrace = A$. So the condition translates, directly from definition, into $A\cap B^c = A$\ $B$.

Title: Inverse indicator function set theory


问:
Calculate the angle between two curves $f(x)=x^2$ and $g(x)=\frac{1}{\sqrt{x}}$
I want to Calculate the angle between two curves on their intersect $f(x)=x^2$ and $g(x)=\frac{1}{\sqrt{x}}$, what I did so far is:

$$x^2=\frac{1}{\sqrt{x}} \rightarrow x=1$$then :
$$\tan(a)=\left |\frac{f'(a)-g'(a)}{1+f'(a)*g'(a)}\right|
$$ after set $x=1$ I get zero at the denominator.

Any suggestions? Thanks!

答: Also, just note that the slope of $ f(x) $ is $ 2 $ and the slope of $ g(x) $ is $ -\frac{1}{2} $ at $ x = 1 $. Their slopes are perpendicular so the angle is $ \frac{\pi}{2} $. That is why the denominator of your expression is $ 0 $ - $ \tan\left(\frac{\pi}{2}\right) $ is similarly undefined. </p>


问:
How to find Graph that is not isomorphic?
I need to find Graph G with degree sequence (5,5,4,4,4,4,4,2,2,2,2,2) constructed with Havel Hakimi method, that will not be isomorphic with G. Is it even possible?

答:
Using the Havel-Hakimi Theorem we can build graph $H$. After some thought I came up with graph $G$, such that: $G \ncong H$. Why?

In graph $G$, we have: $v_{i}$ adjacent to two vertices of degree $2$ and a vertex of degree $5$, for $i = 1, 2$. Where as in graph $H$, we can clearly see that vertex $u_{2}$ doesn't have such a configuration.

Furthermore, note that both graphs have degree sequence; $S: 5, 5, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2$.


问:
Is neighborhoods on a subspace is the same as the intersection of neighborhoods and the subspace?
Suppose $(\Omega,\tau)$ be a topological space and $A$ is subset of $\Omega$, the relative topology(subspace) is $(A,\tau_A)$, $x$ any point of $A$. Question is

$$\mathcal{N}_{\tau_A}(x)= \{N\cap A:N\in \mathcal{N}_\tau (x)\}$$

holds or not?

Where $\mathcal{N}_\tau (x)$ denoted all the neighborhoods of $x$ in $\tau$

答:
Yes, that holds.

Let $\,x\in A.\,$ If $\,N\,$ is a neighborhood of $\,x\,$ in $\Omega$, then there exists an open set $\,U\,$ of $\,\Omega\,$ such that $\,x\in U \subseteq N$, and therefore

$$ x\in U\cap A\;\subseteq\; N\cap A, $$

so $\,N\cap A\,$ is a neighborhood of $x$ in $\,A$, being $\,U\cap A\,$ an open set of $\,A\,$ by definition.

Conversely, if $\,N'\,$ is a neighborhood of $\,x\,$ in $\,A$, there exists an open set $\,U'\,$ of $\,A\,$ such that

$$\,x\in U'=U\cap A\subseteq N'$$

for some open set $\,U\,$ of $\,\Omega$. Now, if we put

$$ N:= N'\cup U, $$

we obtain

$$ N\cap A = \big(N'\cup U\big)\cap A = \big(N'\cap A\big) \cup \big(U\cap A\big) = N'\cup U' = N', $$

and we are done, since $\,N\,$ is a neighborhood of $\,x\,$ in $\Omega\,$ because $\,N\supseteq U\ni x\,$ and $\,U\,$ is open in $\,\Omega$.


答:
Hint: why not factor $x^4+4$ in the real domain first?

  $$x^4+4=(x^2+2)^2 - 4x^2 = (x^2+2+2x)(x^2+2-2x) \\= ((x+1)^2+1)((x-1)^2+1)$$ Can you start from here?

More hint: $p\equiv 1 \pmod 4 \implies \exists a \text{ such that } a^2 \equiv -1 \pmod p$.

 $$(x+1)^2+1 \equiv (x+1)^2-a^2 \pmod p\\ (x-1)^2+1 \equiv (x-1)^2-a^2 \pmod p$$ Can you end it now?

Title: Factorisation of fourth degree polynomial in mod p

Body:
If $p$ is an odd prime with $p \equiv 1 ( \mod 4)$ then $x^4 + 4$ factors into 4 linear parts $\mod p.$

I was breaking $x^4 + 4 = x^4 + (p-1)x^2 + x^2 + 4 (\mod p)$ then using $p \equiv 1 ( \mod 4)$ we have $ x^4 + 4 = x^4 + (4l)x^2 + x^2 + 4 (\mod p)$ after this I am stuck.

Any help.


问:
How many ways can the letters of the word $XQRZX$ be arranged?
How many ways can the letters of the word $XQRZX$ be arranged?

This is taken from the book on Combinatorics, by Daniel Marcus. The answer given is $\frac{5!}2=60$. But, am getting different answer.

I know by rote that for repeated characters the no. of arrangements are divided by no. of repetitions of each type, i.e. for a word of $n$ characters, s.t. $n \geq mk$ characters (where, $m$ characters are repeated $k$ times), then arrangements are: $\frac{n!}{m.k!}$. Similarly, for $m$ characters being repeated $k_1,\cdots, k_m$ times; leads to arrangements as: $\frac{n!}{k_1!\cdots k_m!}$.

My logical reasoning is that for each set of repeated characters, the total no. of arrangements are divided by the lost arrangements of the repeated ones.

But, if try to have an alternative to the above logic, for the given word, then falter as below:

Considering the two $X$ distinct, leading to $5!$ chances, and subtract from that the chance of taking the arrangements of two $X$.

Divide into two cases:

Case(a): Take two as a block, with ordering not important in that block. This leads to $4!=24$ cases.

 This leaves us with $5!= 120 - 24= 96$ cases.

Case (b): The left cases (to be subtracted) come
from the other positions that two $X$ can take, apart from being placed next to each other. There are five such ways to place $X$ seperately.

$X\_---=3$ choices $\times 2=6$,
 $\_X\_--=2$ choices $\times 2=4$,
 $-\_X\_-= 1\implies 11$ choices for placing the two $X$ seperately, but in each case the three left spaces can be filled with $3\times 2\times 1= 3!$ ways.
So, get $11\times 3!= 66$ ways.

The last case ($-\_X\_-$) has one choice (instead of two), due to clash in one choice with the first case ( $X\_---$) .

But, seems to match need subtract $36$ cases rather than $66$ in Case (b), i.e. $96-36 =60$.

答:
Just following your logic. Say they are $X$ and $Y$ for argument sake -

Case(a): $XY$ together leads to $2 \times 4! = 48$ cases. If they are not distinct, they cannot be permuted within and so you need to subtract $24$ times.

Case (b):

$X \_  − − - \,$ has $3$ choices for $Y$ and so does $Y_ − − − -$ has $3$ choices for $X$ (total of $6$). If they are same, you overcounted $3 \times 3!$ times.

Similarly in $ \_ X \_ − − \,$  you overcount $2 \times 3!$ times.

$− \_ X \_ − $ has only one place of choice for $Y$ and similarly for $− \_ Y \_ − $, there is one choice for $X$. If they are same, you overcounted $1 \times 3!$ times.

If you add them up, you get $36$ cases you need to subtract from $(b)$.

So it is $ 120 - 36 - 24 = 60$.


问:
How to prove that a form like $0^\infty$ is not an indeterminate form? What makes a form indeterminate?
I just wanted to know why some particular forms (seven of them) are called indeterminate forms. Why are there only seven indeterminate forms? Can anyone please prove why $0^\infty$ is not an indeterminate form?

答:
For these "indeterminate forms" it is important to note that numbers like $0$ and $1$ in these expressions are not necessarily identically $0$ or $1$ but are rather sequences of numbers "near" $0$ and $1$ who approach their respective values.  As for $\infty$ it is in fact sequences of numbers who keep growing larger and larger without bound.

Expressions like "$1^\infty$" being indeterminate is referring to how things like $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n$ is in fact equal to $e$.  Compare to $\lim\limits_{n\to\infty}(1-\frac{1}{n})^n$ which is $\frac{1}{e}$ and $\lim\limits_{n\to\infty}1^n$ which is identically $1$.  These each gave different values despite all being of the form "a number close to $1$ raised to a large number."  Just being told "$1^\infty$" without more context we don't know what the final result will be.  It could be literally anything.  That is to say if we know that $f(n)\to 1$ and $g(n)\to\infty$ as $n\to\infty$ we have no way of knowing what the value of $\lim\limits_{n\to\infty}f(n)^{g(n)}$ is equal to.

In the case of "$0^\infty$", a small number "near" $0$ raised to larger and larger exponents will always result in numbers near or equal to $0$.  In every case, if you have $f(n)\to 0$ and $g(n)\to\infty$ as $n\to\infty$ you have $\lim\limits_{n\to\infty}f(n)^{g(n)}=0$.  Tweaking the exact behavior of $f$ or $g$ does not change the overall result.  In that way it is not "indeterminate"... it is very much determined.  We know how $0^{\infty}$ acts regardless what form exactly "$0$" or "$\infty$" take.

As for a proof of why given that $f(n)\to 0,g(n)\to 0$ as $n\to\infty$ implies that $\lim\limits_{n\to\infty}f(n)^{g(n)}=0$... note that there will necessarily be some $N$ such that for all $n>N$ you have $g(n)>1$ and $|f(n)|<1$.  At that point you have $|f(n)^{g(n)}|\leq |f(n)|$ and the fact that $f(n)\to 0$ implies the rest.


问:
Milnor's exercise: for any manifold $M$, $\mathrm{Hom}(C^\infty(M,\mathbb{R}),\mathbb{R})\cong M$
I am studying some chapters of Kolar-Michor-Slovak, Natural Operations in Differential Geometry, in particular the one on Weil bundles, where they present the famous "Milnor's exercise". They give two similar proofs, one much shorter than the other, supposedly because of omitted details. I would like to understand it better.

So we have to prove the:

Theorem For any manifold $M$, $\mathrm{Hom}(C^\infty(M,\mathbb{R}),\mathbb{R})\cong M$

Where $\mathrm{Hom}$ refers to morphisms of commutative unital algebras and $\cong$ to bijection of sets (I suppose).

Now the short proof goes like this:

Consider an homomorphism of algebras $C^\infty(M,\mathbb{R}\xrightarrow{\phi}\mathbb{R})$, by the usual short exact sequence argument, $\ker(\phi)\leq C^\infty(M,\mathbb{R})$ is an ideal of codimension $1$ (notice that such an homomorphism cannot be the zero map, since $\phi(1_\mathbb{R})=1\in\mathbb{R}$).

Now, consider the following system of subsets of $M$, where $V(f)=f^{-1}(0)$:
$$\{V(f)\mid (M\xrightarrow{f}\mathbb{R})\in \ker(\phi)\}$$
this is clearly a filter of closed sets of $M$, since $V(f^2+g^2)=V(f)\cap V(g)$.

Find a function $M\xrightarrow{f^*}\mathbb{R}\in \ker(\phi)$ which is unbounded on each non-compact closed subset of $M$ (for instance a positive proper function, such as the square of the geodesic distance w.r.t. any Riemannian metric).  Then $V(f^*)$ is a compact set contained in  $\bigcap_{f\in \ker(\phi)} V(f)$, which is then nonempty.

Let $x_0$ be one of its elements.
Then for any $f\in C^\infty(M,\mathbb{R})$ $f-\phi(f)1$ is in $\ker(\phi)$ so $(f-\phi(f)1)(x_0)=0$, and $\phi(f)=f(x_0)$.
So $x_0$ completely determines $\phi$ and the map $\phi\mapsto x_0$ is the desired bijection.

I have fetched the italicised details. What I am having trouble understanding is the bold passage. Moreover I don't see where we use the fact that the given system of sets is a filter and that $\ker(\phi)$ has codimension 1: I suspect these are relevant for the bold passage also.

Any help would be much appreciated.

答:
First we observe that what we really need in the proof is that $\bigcap_{f \in \operatorname{ker}(\phi)} V(f)\ne  \emptyset$. The proof follows from that just as you written it. Now
$$V(f^*) \supset \bigcap_{f \in \operatorname{ker}(\phi)} V(f)=\bigcap_{f \in \operatorname{ker}(\phi)} V(f)\cap V(f^*),$$
so we can use the following characterization of compactness: Any collection of closed subsets of $V(f^*)$ with the finite intersection property has nonempty intersection.

Now we prove that each finite subset $\{V(f_i)\}_{i=1}^n\subset \ker \phi$ has non empty intersection. Otherwise $g:=\sum_i f_i^2>0$ and $g\in \ker \phi$. This can't be true because if $g$ is an invertible element then $1_\mathbb{R} \in \ker \phi$( Contradiction to $\ker \phi$ has codimension $1$).

We can conclude that $\bigcap_{f \in \operatorname{ker}(\phi)} V(f)=\bigcap_{f \in \operatorname{ker}(\phi)} V(f)\cap V(f^*)$ has a non empty intersection by the compactness of $V(f^*)$.

Some observations:

The hard part is to construct $f^*$ as this shows that ${f^*}^{-1}(0)$ is compact.

Another equivalent definition of compactness of a space $X$ is: Every filter on $X$ has a cluster point. You could probably use that for the proof. We have seen in the proof that I  imitate the same reasoning used ot show $\{V(f) \mid(M \stackrel{f}{\rightarrow} \mathbb{R}) \in \operatorname{ker}(\phi)\}$ is a filter. I was not confident enough with the definitions of filters to go in that direction.


答:
$$(1)\;\;x^2=\frac1{\sqrt x}\implies x=1$$

$$(2)\;f'(1)=2\;,\;f(1)=1\implies\;\text {the tangent line to this function at the intersection point is}$$

$$y-1=2(x-1)\implies y=2x-1$$

$$(3)\;g'(1)=-\frac12\,,\,g(1)=1\implies\;\text {the tangent line to this function at the intersection point is} $$

$$y-1=-\frac12(x-1)\implies y=-\frac12x+\frac32$$

Since the product of the tangent lines' slopes is $\,-1\,$ these lines, and thus the curves...etc.

Title: Calculate the angle between two curves $f(x)=x^2$ and $g(x)=\frac{1}{\sqrt{x}}$

Body:
I want to Calculate the angle between two curves on their intersect $f(x)=x^2$ and $g(x)=\frac{1}{\sqrt{x}}$, what I did so far is:

$$x^2=\frac{1}{\sqrt{x}} \rightarrow x=1$$then :
$$\tan(a)=\left |\frac{f'(a)-g'(a)}{1+f'(a)*g'(a)}\right|
$$ after set $x=1$ I get zero at the denominator.

Any suggestions? Thanks!


问:
Continuous functions are closed
I am wondering whether every continuous real valued functions is closed? Apparently it seems that it is really that way. Let B be a closed subset of the domain of f. We show that f(B) is closed. If not then there exists a sequence of values $y_n$  in f(B) that converges to a point b, outside f(B). Then there exists a sequence of values $x_n$ in B such that $f(x_n)=y_n$. Since $y_n$ is convergent and f is continuous, so $x_n$ is also convergent to a point a in B (since B is closed) . Since f is continuous, $f(x_n)=y_n$ converges to $f(a)$ which happens to be a point in B, thus contradicting the fact that $y_n$ converges to a point $b$ outside f(B). Is my thinking correct?

答:
Then there exists a sequence of values $x_n$ in $B$ such that $f(x_n)=y_n$. Since $y_n$ is convergent and $f$ is continuous, so $x_n$ is also convergent to a point a in $B$ (since $B$ is closed).

Not quite. It's true that if $x_n$ converges then $f(x_n) = y_n$ also converges, but the converse does not hold. For example, if $f$ is the constant map $0$, then $y_n = 0$ is the only option. But, we can make literally any sequence $x_n$, and it will satisfy $f(x_n) = y_n$. So, think of a sequence that's as non-convergent as you can make it, use it for $x_n$, and it will be a counterexample to this logic.

That said, constant functions are indeed closed. If you want a counterexample, consider
$$f(x) = \frac{1}{1 + x^2}.$$
Then $\Bbb{R}$ is closed, and $f(\Bbb{R}) = (0, 1]$, which is neither open nor closed. In particular, if we take any $y_n \to 0$, then inevitably we must have $|x_n| \to \infty$, and therefore $x_n$ cannot be convergent.

Title: Continuous functions are closed


问:
Geometric Interpretation of Catalan Number
There seems to be a simple recurrence relation for Catalan Numbers:

$$C_{n+1} = \frac{4 n+2}{n+2} C_n$$

Any geometric interpretation in terms of Dyck Paths?

https://mathworld.wolfram.com/DyckPath.html

答:
There is a beautiful geometric interpretation of the identity $(n+2)C_{n+1}=2(2n+1)C_{n}$ in terms of triangulations, see e.g. this historical account.

The idea can be translated to Dyck paths (where it gets slightly less beautiful in my opinion). In the following, I will write Dyck paths as parenthesis expresseions, where „(„ means taking a step to the right and „)“ means taking a step up.

We first note that we can define Dyck paths recursively as $D=(D_1)D_2$, where $D_1$ and $D_2$ are (potentially empty) subpaths of the Dyck path $D$. Using this, we can show that a Dyck path of length $2n$ has $2n+1$ subpaths, $n+1$ of which are empty (where the entire path is also counted).
Let us look more closely at the empty subpaths. An empty subpath $0$ can be of two forms, namely $\ldots ((0)D_2)\ldots$ or $\ldots ((D_1)0)\ldots$. In particular, we can interpret the empty subpaths as vertices in the Dyck path, where the first form corresponds to a „peak“ and the second form corresponds to a vertex between two up-steps.

We now consider two sets $B_1$ and $B_2$. The set $B_1$ consists of Dyck paths of length $2n+2$ with one empty subpath marked. It follows that $B_1$ has cardinality $(n+2)C_{n+1}$. The set $B_2$ consists of Dyck paths of length $2n$ with one subtree marked and oriented left or right. It follows that the cardinality of $B_2$ is $2(2n+1)C_{n}$. We get a bijection between $B_1$ and $B_2$ as follows: let $0$ be the marked empty subpath in the Dyck path in $B_1$. If it is of the form $\ldots((0)D_2)\ldots$, map to $\ldots (D_2)\ldots$, with $D_2$ oriented left. Similarly, map $\ldots ((D_1)0)\ldots$ to $\ldots (D_1)\ldots$ with $D_1$ oriented left. As this is reversible, we have a bijection and thus $(n+2)C_{n+1}=2(2n+1)C_{n}$.

Geometrically, the bijection is the following: interpret the empty subpath as a vertex $v$ as above. If the vertex $v$ is a peak, just delete the peak, mark the subpath after the peak and orient it to the left. If the vertex $v$ is of the second type, consider the subpath ending at that vertex (that is, walk down-left on the diagonal with slope 1 and stop at the first vertex $w$ you encounter. The subpath is the path between those two vertices $w$ and $v$). Now, delete the right-step after $w$ and the up-step before $v$, mark the remaining subpath and orient it to the right.


问:
Counting the terms and growth of a nondecreasing sequence
Given a nondecreasing sequence $\{\lambda_n\}_n$, consider the function
$$n(r) = \#\{n\mid \lambda_n \leq r\}.$$

I want to prove that the statements
$\lim_{n\to \infty}\frac{n}{\lambda_n} = L$
and $\lim_{r\to\infty} \frac{n(r)}{r} = L$ are equivalent.

To do so, I've tried the following:

Assuming the first limit $\lim_{n\to \infty}\frac{n}{\lambda_n} = L$, for each $\varepsilon > 0$ there is some $n_0$ such that for all $n\geq n_0$, $|\frac{n}{\lambda_n} - L|<\varepsilon$. Then,
$$n(r) = \#\{n\mid \lambda_n \leq r\} = \#\left\{n\mid \frac{n}{r} \leq \frac{n}{\lambda_n}\right\},$$
so this function should be "roughly equal" to
$$\#\left\{n\mid \frac{n}{r}\leq L\right\} = \#\{n\mid rL\geq n\} = [rL]$$
and so the other limit follows.

I think the reverse implication is immediate if this reasoning is true, by going the other way. My problem is: how can I justify the shaky step properly? Is this the right path?

答:
I'll denote the counting function
$$
N(r) = \#\{n\mid \lambda_n \leq r\}.
$$
with a capital $N$ to avoid confusion with $n$ being used as an index. We must also assume that the sequence $(\lambda_n)$ is unbounded (i.e. it diverges to $+\infty$) because otherwise $N(r) = \infty$ for large $r$.

The problem with your approach is to make the “roughly equal” argument strict, i.e. replace it with concrete estimates. You must also consider that $|\frac{n}{\lambda_n} - L|<\varepsilon$ holds only for sufficiently large $n$. Also it is not immediately obvious (at least not to me :) how proving the reverse implication by “by going the other way” would work.

I'll suggest a slightly different approach, where $N(r)$ is estimated from  above and from below.

First assume that $\lim_{r\to\infty} \frac{N(r)}{r} = L$ exists. It follows from the definition of $N(r)$ that
$$
 N(\lambda_n - 1) < n \le N(\lambda_n)
$$
and therefore
$$
 \frac{\lambda_n - 1}{\lambda_n} \cdot \frac{N(\lambda_n-1)}{\lambda_n-1}
< \frac{n}{\lambda_n} \le \frac{N(\lambda_n)}{\lambda_n} \, .
$$
Both the left-hand side and the right-hand side of that inequality chain converge to $L$, therefore $\lim_{n\to \infty} \frac{n}{\lambda_n} = L$.

Now assume that $\lim_{n\to \infty} \frac{n}{\lambda_n} = L$ exists. Let $(r_k)$ be an increasing sequence of positive real numbers converging to $+\infty$. For each $k$ we can find an $n_k$ such that
$$
 \lambda_{n_k} \le r_k <  \lambda_{n_k+1} \, .
$$
Then $N(r_k) = n_k$ and therefore
$$
\frac{n_k+1}{\lambda_{n_k+1}} \cdot \frac{n_k}{n_k+1} < \frac{N(r_k)}{r_k} \le
\frac{n_k}{\lambda_{n_k}} \, .
$$
Since $n_k \to \infty$ for $k\to \infty$, both the left-hand side and the right-hand side of the inequality chain converge to $L$. It follows that $\lim_{k \to \infty} \frac{N(r_k)}{r_k} = L$. This holds for all increasing unbounded sequences $(r_k)$, so that $\lim_{r\to\infty} \frac{N(r)}{r} = L$.

Remark: A slight variation of the above arguments shows that the following relationships hold, even if the limits do not exist:
$$
\liminf_{r\to\infty} \frac{N(r)}{r} = \liminf_{n\to \infty} \frac{n}{\lambda_n}
$$
and
$$
\limsup_{r\to\infty} \frac{N(r)}{r} = \limsup_{n\to \infty} \frac{n}{\lambda_n} \, .
$$


问:
$\min_{\beta_1,...,\beta_3,\beta}\sum_{i=1}^{3}\frac{1}{2}||y_i-X_i\beta_i||_2^2+\lambda||\beta||_1$ in augmented Lagrangian form
Write $\min_{\beta_1,...,\beta_3,\beta}\sum_{i=1}^{3}\frac{1}{2}||y_i-X_i\beta_i||_2^2+\lambda||\beta||_1$ in augmented Lagrangian form. s.t. $\beta_1=\beta_2=\beta_3=\beta$

Augmented Lagrangian is defined as:

$L(x,u;p)=f(x)+u^T(Ax-b)+\frac{p}{2}||Ax-b||_2^2$

$f(x)=\sum_{i=1}^{3}\frac{1}{2}||y_i-X_i\beta_i||_2^2$

$g(x)=\lambda||\beta||_1$

I'm not entirely sure where to go from here. My guess of solution is:

$$\min_{\beta_1,...,\beta_3,\beta}\sum_{i=1}^{3}\frac{1}{2}||y_i-X_i\beta_i||_2^2+\lambda||\beta||_1+\frac{p}{2}||(\beta_1-\beta) + (\beta_2-\beta) + (\beta_3-\beta)||_2^2$$
or perhaps the constraints have to be separated
$$\min_{\beta_1,...,\beta_3,\beta}\sum_{i=1}^{3}\frac{1}{2}||y_i-X_i\beta_i||_2^2+\lambda||\beta||_1+\frac{p}{2}(||(\beta_1-\beta)||_2^2 + ||(\beta_2-\beta)||_2^2 + ||(\beta_3-\beta)||_2^2)$$

答:
For a problem of the form:
$$\tag{1}\min_{\mathbf{x}, \mathbf{z}} f(\mathbf{x}) + g(\mathbf{z}) \\ s.t.: A\mathbf{x}+B\mathbf{z}=\mathbf{c}$$
we construct the augmented Lagrangian:
$$\mathcal{L}_\rho(\mathbf{x},\mathbf{z},\mathbf{y})=f(\mathbf{x})+g(\mathbf{z})+\mathbf{y}^T(A\mathbf{x}+B\mathbf{z}-\mathbf{c})+\frac{\rho}{2}\|A\mathbf{x}+B\mathbf{z}-\mathbf{c}\|^2.$$
This could almost fit your model, by setting \begin{aligned}&A=I,\;\mathbf{x}=(\beta_1,\beta_2,\beta_3)^T\\&B=-I,\mathbf{z}=(\beta,\beta,\beta)^T\\& \mathbf{c}=\mathbf{0}.\end{aligned}
However, your $f(\cdot)$ component has a linear transformation. So in fact, your problem is $$\min_{\mathbf{x}, \mathbf{z}} f(\mathbf{Ax}) + g(\mathbf{z}) \\ s.t.: \mathbf{x}-\mathbf{z}=\mathbf{0}.$$
We can still follow this framework and construct the augmented Lagrangian:
$$\mathcal{L}_\rho(\mathbf{x},\mathbf{z},\mathbf{s})=\frac{1}{2}\|A\mathbf{x}-\mathbf{y}\|^2+\frac{\lambda}{3}\|\mathbf{z}\|_1+\mathbf{s}^T(\mathbf{x}-\mathbf{z})+\frac{\rho}{2}\|\mathbf{x}-\mathbf{z}\|^2.$$
The question is what do you plan to do next. The above formulation is without doubt the augmented Lagrangian. But general ADMM frameworks that are proven to converge to an optimal solution are designed to fit problem $(1)$. Your problem is somewhat different, therefore a generic ADMM algorithm might not work.


问:
When will two cars pass each other
There was a question in my math text book the other day that stated:

$2$ cars each travelling at a constant velocity around a ring ,
complete exactly $4$ and $7$ rounds in one hour. If they start at exactly the same time from the same place but travel in opposite directions around the track, how many times will they
pass each other in the one hour?

I tried various things using $\displaystyle v = \frac{d}{t}$ but couldnt get very far. Any help is appreciated, thanks.

答: You're on the right track. In the time the faster driver completes his first lap, they've obviously crossed paths. In time $t$ hours, one has completed $4t$ laps and the other $7t$. Remembering they're going in opposite directions, when do they cross first?</p>

Title: When will two cars pass each other


问:
Young diagram for standard representation of $S_d$
I'm working through Fulton-Harris and I'm kind of "stuck" at the following question. I'm looking for representations of $S_d$, the symmetric group on $d$ letters via Young Tableaux. The question is:
"Show that for general $d$, the standard representation $V$ of $S_d$ corresponds to the partition $d = (d-1)+1$."

When I look at the hint I see that they give a basis:
$$v_j = \sum_{g(d)=j} e_g - \sum_{h(1)=j} e_h$$
for $j=2,...,d$.
Is there any way I can "see" that this should be a basis corresponding to the partition? Because, right now, they just seem to appear from magic.
The representation should be the image of $c_{(d-1,1)} = a_{(d-1,1)} \cdot b_{(d-1,1)}$ in the group algebra of $S_d$ (here $a_\lambda$ corresponds to the permutations of $S_d$ that preserves the rows of the young tableaux and $b_\lambda$ the columns).

I do see how one might try for smaller cases by simply multiplying each element by $c_\lambda$ ( The young symmetrizer) , but even for $d=4$, this becomes quite impractical. Are there any other ways to see that this "should be" a basis? In general, what can we say about a young diagram and the corresponding basis for its irreducible representation?

答:
Although the other answer is correct, I found it a bit confusing. In this answer I'll provide a step-by-step explanation of the proof written by @user382668. I'll denote $e_g$ by simply $g$ and $(d - 1, 1)$ by $\lambda$.

Here it goes:

Let $w_i = (i \; d)a_\lambda$. First note that $\{w_i\}_{i = 1}^d$ is a basis for $\mathbb{C}[S_d]a_\lambda$ (check the other answer for further details).

Let $v_i = w_i b_\lambda$. Since $\{w_i\}_{i = 1}^d$ is a basis for $\mathbb{C}[S_d]a_\lambda$, $\{v_i\}_{i = 1}^d$ generates $\mathbb{C}[S_d]a_\lambda b_\lambda = V_\lambda$.

Note that $\sum_{i = 1}^d v_i = 0$, i.e. $v_1 = - \sum_{i = 2}^d v_i$ (check the other answer for further details). This means $\{v_i\}_{i = 2}^d$ generates $V_\lambda$.

It follows from the Hook length formula that $\dim V_\lambda = d - 1 = |\{v_i\}_{i = 2}^d|$, so $\{v_i\}_{i = 2}^d$ is a basis for $V_\lambda$.

Finally, note that $v_i = \sum_{g(d) = i} g - \sum_{h(1) = i} h$. In other words, $\{v_i\}_{i = 2}^d$ is precisely the basis in the back of Fulton-Harris. We are done.

Here's a proof of the last item:
$$
\begin{split}
v_i
& = (i\;d) a_\lambda b_\lambda \\
& = (i\;d) \left( \sum_{p \in P_\lambda} p \right) \left( \sum_{q \in Q_\lambda} \operatorname{sgn}(q) q \right) \\
& = (i\;d) \left( \sum_{g(d) = d} g \right) (1 - (1\;d)) \\
& = \sum_{g(d) = d} (i\;d)g - \sum_{h(d) = d}(i\;d)h(1\;d) \\
& = \sum_{g(d) = i} g - \sum_{h(1) = i} h
\end{split}
$$


答: The two cars are essentially travelling towards each other; They will collide when the sum of the distances they each cover is equal to one round. Consider how far each car travels in $t$ hours to get an idea.</p>

Title: When will two cars pass each other

Body:
There was a question in my math text book the other day that stated:

$2$ cars each travelling at a constant velocity around a ring ,
complete exactly $4$ and $7$ rounds in one hour. If they start at exactly the same time from the same place but travel in opposite directions around the track, how many times will they
pass each other in the one hour?

I tried various things using $\displaystyle v = \frac{d}{t}$ but couldnt get very far. Any help is appreciated, thanks.


问:
A function where $f(n)$ is the sum of the remainders of $n$ divided by natural numbers up to $n$.
The question asks you to prove that there exists infinitely many $f(x) = f(x-1)$. I've conjectured that $f(2^k)=f(2^k-1)$ for positive integers $k$ but I cannot find a proof for this. I would like some help with this question.

答:
Nice problem.

Sketch of solution:

To prove your conjecture, start by showing that $$  n\pmod d-(n-1)\pmod d =
\begin{cases}
1,  & \text{if $d\nmid n$ } \\
1-d, & \text{if $d\,|\,n$ }
\end{cases}$$

Where we are writing, e.g., $a\pmod b$, to denote the remainder on dividing $a$ by $b$.

(this is not difficult)

Now, letting $n=2^k$  and summing from 1 to $n-1$ we get $$F(2^k)-F(2^k-1)=\sum_{d\nmid n}1+\sum_{d\,|\,n,\,\,d<n}(1-d)=\sum_{i=1}^{2^{k}-1}1-\sum_{i=0}^{k-1}2^i$$

But these sums are easily computed and we get $$(2^{k}-1)-(2^k-1)=0    $$


问:
2020 Australian Maths Competition Senior Problem 30
When I drive to school every day, I pass eight traffic lights, each either green, yellow,
or red. I find that, because of synchronization, a green light is always followed
immediately by a yellow, and a red light is never immediately followed by a red.
Thus a sequence of lights may start with GYRY, but not RRGG.
How many possible sequences of the eight lights are there?</p>

答:
This can be solved through recursive combinatorics, let's call:

$$G_n=\text{ valid sequences of lights long } n \text{ that terminate with G}$$
$$Y_n=\text{ valid sequences of lights long } n \text{ that terminate with Y}$$
$$R_n=\text{ valid sequences of lights long } n \text{ that terminate with R}$$

You can find out that(if you don't know how tell me and I'll help you):

$$G_{n+1}=Y_n+R_n$$
$$Y_{n+1}=G_n+R_n+Y_n$$
$$R_{n+1}=Y_{n}$$
And clearly $G_1=Y_1=R_1=1$


问:
Number of strings containing each vowel at least once?
Consider the English alphabet {$a,b,c,...,z$}. How many strings of length $n$ can be formed in which each vowel {$a,e,i,o,u$} appears at least once?

答:
The two-set version of Inclusion-Exclusion principle states that $|A\cup B|=|A|+|B|-|A\cap B|$.  This can be extended to more and more terms as needed.  The three-set version being $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$ and for your specific case of five sets:

$$\begin{array}{rl}|A\cup E\cup I\cup O\cup U| = &|A|+|E|+|I|+|O|+|U|\\&-|A\cap E|-|A\cap I|-\dots - |O\cap U|\\&+|A\cap E\cap I|+|A\cap E\cap O|+\dots+|I\cap O\cap U|\\&-|A\cap E\cap I\cap O|-\dots-|E\cap I\cap O\cap U|\\&+|A\cap E\cap I\cap O\cap U|\end{array}$$

That is to say, you add the size of each possible singleton, then subtract the size of each intersection of pairs of sets, then add back the size of the intersection of each triple of sets, subtracting the next size intersections, adding the next size intersections, so on so forth until you've covered all cases.

Now.  Letting $A$ be the event that no $a$'s were used in the string... let us try to compute $|A|$.  Well... the number of strings of length $n$ without any $a$'s are precisely those strings of length $n$ who were taken from the alphabet $\{b,c,d,e,f,\dots,x,y,z\}$.  That is just an alphabet with $25$ letters.  The same logic that led you to $26^n$ being the number of length $n$ strings over an alphabet with $26$ letters should lead you to the number of length $n$ strings over an alphabet with $25$ letters as being $25^n$.

The same logic will let you find $|E|,|I|,|O|,$ and $|U|$ as being the same.

Now... $|A\cap E|$ is the set of strings of length $n$ who had no $a$'s and had no $e$'s.  Apply exactly the same logic as before... noting now that there are only $24$ valid letters to choose from from the alphabet.

Continue in this fashion and note how many pairs, how many triples, etc... there are to reach a final count.

 $$26^n - \binom{5}{1}25^n+\binom{5}{2}24^n-\binom{5}{3}23^n+\binom{5}{4}22^n-21^n$$


问:
Prove the identity $\csc2\theta=\frac{\sec\theta\csc\theta}{2}$

Prove the identity $$\csc2\theta=\frac{\sec\theta\csc\theta}{2}$$

I've started by using a double angle identity, but I'm not sure how to continue from here or if this is right approach.

$$\csc2\theta=\frac{1}{\sin2\theta}=\frac{1}{2\sin\theta\cos\theta}$$

答:
You are already on the right track. Here is another approach.
GO!
$$\begin{aligned}\csc{2\theta}&=\frac{1}{\sin{2\theta}}\\&=\frac{1}{2\sin\theta\cos\theta}\\&=\frac12\csc\theta\sec\theta\\&=\frac{\csc\theta\sec\theta}{2}\end{aligned}$$
I hope this helps.


问:
Which one is the correct set builder notation for prime numbers
I have seen this problem in the IYMC 2018 final.
which of these is the prime set?

A. $ \{n ∈ \Bbb N : 1|n ∧ n|n \} $

B. $ \{n ∈ \Bbb N : 1|n ∧ n|n ∧ n > 1\}  $

C. $ \{n ∈ \Bbb N : |{t ∈ \Bbb N : t|n}| = 2\} $

D. $ \{n ∈ \Bbb N : |{t ∈ \Bbb N : t|n ∧ t > 1}| = 2\} $

Which one is the correct one?

A screenshot of the question here

答:
$A = \mathbb{N}$ because any natural number is divisible by $1$ and itself.

$B = \{2,3,4,5\ldots\}$. It's the same as $A$ except with $1$ removed.

$D$ doesn't work because it contains, for example, $4$. Its divisors are $1,2,$ and $4$, and as we're only looking for those over $1$, we're left with $\{2,4\}$, a set with two elements.

Clearly $C$ is the correct answer.


问:
Minimum to maximum in optimization
I have a simple doubt, we know that $$\min f(x)=-\max(-f(x)),$$ but suppose we have an optimization problem, say a linear programming problem, such that the objective is to minimize $f(x)$. If we want to write it as a maximization problem, we write it as $\max(-f(x))$. My doubt is, why it will not be $-\max(-f(x))$.

This may be a very silly question, my apologies, but I have always wondered why is it so.

答:
Not a silly question. It is not possible to write $- \max (-f(x))$ in advance as the value of the maximum is simply not known.

So what we do is, we first calculate $\max (-f(x))$ and then multiply by $-1$.

Title: Minimum to maximum in optimization


问:
Solvable equivalent to nilpotency of first derived Lie algebra?
The Wikipedia "Solvable Lie Algebra" page lists the following property as a notion equivalent to solvability: $\mathfrak{g}$ is solvable iff the first derived algebra $[\mathfrak{g},\,\mathfrak{g}]$ is nilpotent.

I can't see this at all and it does sound dodgy: solvability and nilpotency different by "only one rung" in the derived series. Is this correct (I suspect I'm missing something trivial- I can't seem to find anything relevant in Knapp "Lie Groups: Beyond an Introduction" (mine is reprint of the first edition))? If not universally correct, is it correct with further assumptions about the field which $\mathfrak{g}$ is a vector space over? (For example, a field of characteristic nought, algebraically closed, ...). Can someone either give a proof, a counterexample or a reference to one?

I'd also like to put a reference on the Wikipedia page quotation, or clearly have it stricken from the page if incorrect.

答:
There is a counterexample given in Jacobsons book, i.e., an example of a finite-dimensional solvable Lie algebra $L$ in characteristic $p>0$ such that $[L,L]$ is not nilpotent.
More concretely, this Lie algebra $L$ has a basis $(x,y,e_1,\ldots ,e_p)$ with Lie brackets
\begin{align*}
[x,y]   & = x, \\
[x,e_1] & = e_p, \\
[x,e_i] & = e_{i-1}, \; i\ge 2,\\
[y,e_i] & = (i-1)e_i, \; 1\le i\le p.
\end{align*}
Since $V=\operatorname{span}(e_1,\ldots ,e_p)$ is solvable (even abelian) and the quotient $L/V$ is solvable, so is $L$. But $[L,L]=kx\oplus V$ is not nilpotent.
We have $[L,L]^1=[L,L]^2=\cdots =V$.


问:
How do you show $f(x) = 1 - a + ax - x^a \geq 0$ if $a \in (0, 1)$ and $x \geq 0$?
How do you show $f(x) = 1 - a + ax - x^a \geq 0$ if $a \in (0, 1)$ and $x \geq 0$?

True for $x = 0$ but for $x \neq 0$ wts. $ax - x^a \geq a - 1$ ?

Any help would be apprecaited thanks.

Body:
hint

For $ x>0$
$$f'(x)=a(1-e^{(a-1)\ln(x)})$$

$$x>1\implies f'(x)>0$$
$$x<1\implies f'(x)<0$$
$$f'(1)=0\; \; f(1)=0$$

$ f $ is continuous at $ x=0$, with $ f(0)=1-a>0$.

$ f$ attains its absolute minimum at $ x=1$.


问:
Parent and childs of a full d-node tree
i have a full d-node tree (by that mean a tree that each node has exactly d nodes as kids).
My question is, if i get a random k node of this tree, in which position do i get his kids and his parent?

For example, if i have a full binary tree, the positions that i can find the parent,left and right kid of the k node are $\dfrac k2, 2k, 2k+1$ respectively.

Thanks in advance.

答:
Start by labeling your $k$-ary tree with the root as $0$, and continue in a breath-first-search pattern.

Given a node $n$, with index $n\text{.i}$, $n$'s children have indices:
$$\{(n\text{.i})k+1,\, (n\text{.i})k+1,\, \ldots,\, (n\text{.i})k + k\}$$

We can also see that given a node $n$ with index $n\text{.i}\gt0$, we can find the index of the parent by the formula:
$$p\text{.i}=\left\lfloor\frac{n\text{.i} - 1}{k}\right\rfloor$$

I haven't proved these, but these formulas appear to be correct based on several different drawings.

Title: Parent and childs of a full d-node tree


问:
Let $z \in \mathbb C$, show using de Moivre's that the product of the roots of $z^n=1$ is $1$ when $n$ is odd.
I am taking a course which introduces complex numbers and some of the theorems relating to them. I wish to prove the following result:

Prove, using de Moivre's Theorem, that the product of the roots of $$z^n=1$$ is $1$ when $n$ is odd for $z \in \mathbb C$.

I do not see how de Moivre is very helpful here. Any ideas, since De Moivre only introduces a sine and cosine, I do not see how this helps met get farther.
$$\cos(n\pi) + \sin(n\pi)i =1. $$

答:
Using Vieta's formula on $$z^n-1=0,$$ We get that:
$$P:=\prod_{r=0}^{n-1}z_r=(-1)^n\cdot\dfrac{-1}1= (-1)^{n+1}.$$
Where $z_r$ denotes the complex roots of unity.

If $n=2k+1$ for $k\in \mathbb Z$ we see using the rule $(x^a)^b= x^{ab}$:
$$ P=(-1)^{2k+2}=  \left((-1)^2 \right)^{k+1}= 1.$$


问:
Showing a solution to this differential equation is never zero
Say I want to find a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for some $k \neq 0$, $f' = kf$. Every time I have seen a book solve this, the expression is rewritten  as $$\frac{1}{f}f' = k$$ and then solved using u-substitution. To be technical though, must one first know that $f$ is never zero? If so, how could I go about showing that $f$ is never zero? I know that $f$ must be infinitely differentiable and if for some $a \in \mathbb{R}$, $f(a) = 0$, then for each $n \in \mathbb{N}_0$, $f^{(n)}(a) = 0$. I think I remember from complex analysis, if $f$ is holomorphic with this property, then $f$ is zero everywhere, but I'm not sure about the usual case of $\mathbb{R} \rightarrow \mathbb{R}$.

答:
In addition to the many useful answers I may add this localization trick. Suppose $f(x_0)\neq 0$ at a particular $x_0$. By continuity of $f$ this is the case for all $x$ in an interval containing $x_0$. So, it is safe to divide by $f$ there. Then you proceed to find the solution. Only then does it turn out that the solution you've found is actually valid on the whole real line and that it is never zero!

BTW: You are completely right in being cautious about $f(a)=f'(a)=f''(a)=\cdots=0$ not implying $f=0$. This is because $f$ need not be equal to its Taylor series around $a$ for any value other than at $a$. If we knew $f$ was analytic this would have been the case.


问:
finding an example of an equivalence relation
Find an example of an equivalence relation on the set $\mathbb{N}$.

What I know so far:

An equivalence relation is reflexive, symmetric and transitive. However I have no idea of an example.

答:
There are lots of examples. Perhaps the least instructive are:

the minimal equivalence relation where $m\sim n$ only if $m=n$;

the maximal equivalence relation where $m\sim n$ for every $m,n$.


问:
Simplify $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$
Please show how to simplify this type of expression without using a calculator. I'm new to trigonometry and I don't know how to simplify this expressions: $\frac{\cos \theta}{\sec \theta} + \frac{\sin \theta}{\csc \theta}$.

答:
First, we need to make a common denominator of $\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}$.

Let's go!
$$\require{cancel}\begin{aligned}\frac{\sin\theta}{1-\cos\theta}-\frac{\sin\theta}{1+\cos\theta}&=\frac{\sin\theta(1+\cos\theta)-\sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}\\&=\frac{\cancel{\sin\theta}+\sin\theta\cos\theta\cancel{-\sin\theta}+\sin\theta\cos\theta}{1-\cos^2\theta}\\&=\frac{2\cancelto{1}{\sin\theta}\cos\theta}{\cancelto{\sin\theta}{\sin^2\theta}}\\&=\frac{2\cos\theta}{\sin\theta}\\&=2\cot\theta\end{aligned}$$
I hope this helps.


问:
Short trick to find the number of Sylow p subgroups in permutation groups.
I know, $|S_4 |=24 =2^3\cdot3$.

So here the order of sylow $2$ subgroup is $8$ and by the third sylow theorem we can say the number of sylow $2$ subgroups is $1$ or $3$. Then by finding the sylow $2$ subgroups explicitly we can conclude that its $3$. But here we also know the number of sylow $3$ subgroups is $1$ or $4$.

My question is can we find the exact number of sylow $2$ and $3$ subgroups without calculating the the sylow subgroups explicitly, just by element counting of the subgroups, because sometimes we need only the exact number in of sylow subgroups ?

答: Suppose $S_4$ has a unique Sylow $2$-subgroup say $K$. By Sylow Second Theorem, $K$ must be normal in $S_4$. But $S_4$ does not have any normal subgroup of order $8$ (Refer here). Therefore the number of Sylow $2$-subgroups in $S_4$ must be three.


问:
Convergence of random variable to a negative constant
Let $X_n$ be the sequence of R.Vs and $X_n\overset{P}{\rightarrow}A$ (or $X_n\rightarrow A$ almost surely) where $A<0$

I want to prove that $Pr[X_n < 0] \rightarrow 1$ (or $X_n  0\right) \lim_{n\rightarrow\infty} Pr\left[ |X_n - A| \leq \varepsilon \right] = 1$$

答:
From your definition of convergence in probability, we must have

$\lim_{n\rightarrow\infty}\Pr[|X_n - A| \le \frac{|A|}{3}] = 1$

But since $A < 0$, we know that $|X_n - A| \le \frac{|A|}{3}$ implies that $X_n < 0$. So, we have:

$\{|X_n-A| \le \frac{|A|}{3}\} \subset \{ X_n < 0 \}$.

Thus, $\Pr[ X_n < 0] \ge \Pr[|X_n-A| \le \frac{|A|}{3}]$ for all $n$. Since the latter probability converges to $1$, we must have:

$\lim_{n\rightarrow\infty}\Pr[X_n < 0] = 1$.


答:
Another example is congruence modulo $a$, where $a$ is a positive number:

$m\sim n\iff a|m-n\iff \exists k\in\mathbb Z$ such that $m-n=ka$.

Can you show this is an equivalence relation?

Title: finding an example of an equivalence relation

Body:
Find an example of an equivalence relation on the set $\mathbb{N}$.

What I know so far:

An equivalence relation is reflexive, symmetric and transitive. However I have no idea of an example.


问:
Global sections of Proj
In constructing the relative Proj for a graded ring $S$, one inevitably has that the distinguished opens, $D_+ (f)$, where $f \in S_+$ gives rise to a scheme that is isomorphic to $\textrm{Spec } S_{(f)}$, the latter denoting the degree zero elements in the localization $S_f$.

In all constructions I have seen of this, including Hartshorne and Liu, they usually leave it at that with no mention of global sections. I was tempted to handle this by saying $f = 1$ then $D_+ (f) = \textrm{Proj } S$ is isomorphic to $\textrm{Spec } S_{(1)} = \textrm{Spec } S_0$ (degree 0 elements of $S$). The Wikipedia article mentions this without proof or justification under the "Twisting Sheaf of Serre" section (https://en.wikipedia.org/wiki/Proj_construction#The_twisting_sheaf_of_Serre) lending credence to this fact.

However, the condition that $f \in S_+$, where $1$ typically does not reside (take the natural grading on the polynomial ring for example) seems to invalidate this proof. What justification should I use instead?

Background: I am interested in justifying that for $X = \textrm{Proj } A[x_0,\dots,x_n]$, that $\mathcal{O}_X (1)$ can be generated by the global sections $x_0, \dots, x_n$.

答:
Let $A$ be a commutative unital ring and let $E:=A\{e_0,..,e_n\}$ be the free $A$-module of rank $n+1$ on the elements $e_i$. Let $E^*:=A\{x_0,..,x_n \}$ be the dual of $E$ and let $S:=Sym_A(E^*)\cong A[x_0,..,x_n]$. Let $\mathbb{P}(\mathcal{E}^*):=Proj(S)$ be the projective space bundle of $\mathcal{E}$ (the sheafification of $E$). There is a canonical exact sequence of graded $S$-modules

$\phi: \oplus_{i=0}^n S x_i \rightarrow S(1) \rightarrow 0$

defined by

$\phi( f_0,f_1,..,f_n):= \sum_{i=0}^n f_i x_i .$

Let $\pi:\mathbb{P}(\mathcal{E}^*) \rightarrow Y$ (here $Y:=Spec(A)$) be the projection map. When sheafifying the map $\phi$ we get the tautological sequence

$\pi^*\mathcal{E}^* \rightarrow^{\phi} \mathcal{O}(1) \rightarrow 0.$

Since $\pi^*\mathcal{E}^*$ is a free $\mathcal{O}$-module it follows this proves the invertible sheaf $\mathcal{O}(1)$ is generated by the global sections $s_i:=\phi(x_i)$.

Lemma: Let $X$ be a scheme. For a surjection $\phi: \mathcal{O}_X \{y_1,..,y_l\} \rightarrow \mathcal{L}$ where $\mathcal{L}$ is an invertible sheaf on $X$ and $\mathcal{O}_X\{y_1,..,y_l\}$ is the free sheaf of rank $l$ on the elements $y_i$, it follows $\phi(y_i)$ are global sections generating $\mathcal{L}$. Conversely given a set of global sections $s_1,..,s_l$ of $\mathcal{L}$ that generated $\mathcal{L}$ it follows there is a canonical surjective map of $\mathcal{O}_X$-modules
$\phi: \mathcal{O}_X\{y_1,..,y_l\}\rightarrow \mathcal{L}$ defined on an open set $U$ by $\phi(\sum_i u_iy_i):=\sum_i u_i(s_i)_U$ where $(s_i)_U$ is the restriction of $s_i$ to $U$. The details may be found in Hartshorne's book  Chapter II.7.

Title: Global sections of Proj


答:
Note that every partition of a set corresponds to an equivalence relation on that set (the equivalence relation being two elements are related iff they belong to the same part in the partition).  Similarly every equivalence relation on a set defines a partition on the set (the parts being the equivalence classes of the relation).

So... a way to reword the question is to ask you for an example of a way to partition $\Bbb N$.

You could do so in a number of ways... partition it into the even numbers and the odd numbers, $\{1,3,5,7,\dots\},\{2,4,6,8,\dots\}$.

You could partition it into the prime numbers versus the composite numbers vs $1$, $\{1\},\{4,6,8,9,10,12,\dots\},\{2,3,5,7,11,13,\dots\}$

You could partition it into groups of numbers who all have the same number of digits: $\{1,2,3,\dots,9\},\{10,11,\dots,99\},\{100,101,\dots,999\},\dots$

You could partition it however you like.  Some of these examples will be more important than others and will occur more often.  Some of these examples will be easier to explain or think about.  Some of these will be obscure or impossible to effectively work with.  They all exist however and that is the main point.


答:
It looks like you're starting numbering at 1 for the root, and numbering "left to right" on each level/depth.

If the root has depth $0$, then there are $d^{t}$ nodes with depth $t$ from the root in a full $d$-dimensional tree. Also, the depth of node $k$ is $\ell_{k}=\lceil\log_{d}(k-1)\rceil$.

The number of nodes at depths below the depth of node $k$, then, is
$$n_{k} = \sum_{i = 0}^{\ell_{k-1}}d^{i} = \frac{d^{\ell_{k}}-1}{d-1}.$$
So indexing on the row containing node $k$ starts at $n_{k}+1$.

Children of node $k$:

The position of $k$ in its row is just $p_{k}=k-n_{k}$. The children of node $k$ have depth one more than that of $k$, and in their respective row, the first child $c$ has position $d(p_{k}-1)+1=d(k-n_{k}-1)+1$, so the $j^{th}$ child of $k$ has position
$$j + \sum_{i = 0}^{\ell_{k}+1}d^{i} + d(k-\sum_{i = 0}^{\ell_{k}}d^{i}-1) = j + \frac{d^{\ell_{k}+1}-1}{d-1} + d(k-\frac{d^{\ell_{k}}-1}{d-1}-1)$$
$$=j +dk-d +1$$

Parent of node $k$:

Since this formula applies for the parent $p$ of $k$, for $1 \leq j \leq d$
$$1+dp-d +1\leq k \leq d+dp - d + 1 = dp+1$$
$$\Rightarrow dp - d \leq k-1 \leq dp$$
$$\Rightarrow p-1 \leq \frac{k-1}{d} \leq p$$
so $p = \lceil\frac{k-1}{d}\rceil$.

Title: Parent and childs of a full d-node tree

Body:
i have a full d-node tree (by that mean a tree that each node has exactly d nodes as kids).
My question is, if i get a random k node of this tree, in which position do i get his kids and his parent?

For example, if i have a full binary tree, the positions that i can find the parent,left and right kid of the k node are $\dfrac k2, 2k, 2k+1$ respectively.

Thanks in advance.