问:
Why does the diagonal morphism of a map between affine schemes correspond to the following morphism of rings?
Suppose I have a morphism of affine schemes $(f, f^\sharp) : \operatorname{Spec} A \to \operatorname{Spec} B$, then my question is - why does the diagonal morphism $(\Delta_f, \Delta_f^\sharp) : \operatorname{Spec} A \to \operatorname{Spec} A \times_{\operatorname{Spec} B} \operatorname{Spec} A = \operatorname{Spec} (A \otimes_B A)$ correspond the following morphism of rings $\phi : A \otimes_B A \to A$ given by $\phi(a \otimes a') = a\cdot a'$?

This was my attempt to try and asnwer that.
I know that we have the result that $$\operatorname{Hom}_{\textsf{Sch}}\left(\operatorname{Spec} A, \operatorname{Spec} (A \otimes_B A)\right) \cong \operatorname{Hom}_{\textsf{Ring}}(A \otimes_B A, A)$$
with the bijection given by $(g, g^\sharp) \mapsto g^\sharp_{\operatorname{Spec} A}$ so one possible way to show that the diagonal morphism does correspond to the claimed morphism of rings would be to show that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp = \phi$ but I was unable to actually verify this.

Another thing I was thinking of was the following. If I let $(p_1, p_1^\sharp) : \operatorname{Spec} (A \otimes_B A) \to \operatorname{Spec} A$ denote the projection map that comes with the fibered product, then we know that $p_1 \circ \Delta_f = 1_{\operatorname{Spec} A}$ the identity map on $\operatorname{Spec} A$. Now I guess, (but I cannot show), that the morphism $(p_1, p_1^\sharp)$ corresponds to the morphism of rings $\iota : A \to A \otimes_B A$ given by $\iota(a) = a \otimes_B 1_A$, because then we see that $\phi \circ \iota = 1_A$ the identity map on $A$.

If it were the case that the morphism $(p_1, p_1^\sharp)$ corresponds to $\iota$, then since taking global sections is a contravariant functor (and really what that bijection above is) I would see that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp\circ \iota = 1_A$ as well, but this isn't enough to prove that ${\Delta_f}_{_{\operatorname{Spec} A}}^\sharp = \phi$.

So I have in addition two further follow up questions in addition to the original one:

How does one prove that the morphism $(p_1, p_1^\sharp)$ corresponds to $\iota$?

Is there a more efficient way to prove that a morphism of schemes corresponds to a morphism of rings?

If it is relevant, you can assume that the construction of the fibered product I am working with is the same one given in Theorem 3.3 in Hartshorne's book.

答:
Note that $\Delta : \operatorname{Spec}(A) \to \operatorname{Spec}(A \otimes_B A)$ corresponds to a map $\Delta^*: A \otimes_B A \to A$ as you say. The map $\Delta$ is defined via the universal property of the fibre product with $(\mathrm{id}_{\operatorname{Spec}(A)}, \mathrm{id}_{\operatorname{Spec}(A)})$, so $\Delta^*$ is given (recall that the correspondence is natural, it is an equivalence of categories) by $(\mathrm{id}_A, \mathrm{id}_A)$ via the universal property of pushouts. In particular, it is the unique $B$-algebra homomorphism such that
$$ a \otimes 1 \mapsto a \quad\text{and} \quad 1 \otimes a \mapsto a$$ for every $a \in A$. In other words, $\Delta^*: a \otimes a' \mapsto aa'$.

The important part here is that the correspondence $$\operatorname{Hom}_{\mathbf{AffSch}}(\operatorname{Spec}(A), \operatorname{Spec}(B)) \cong \operatorname{Hom}_{\mathbf{Ring}}(B,A)$$ is not only a bijection but rather a natural bijection, it comes from an equivalence of categories $\mathbf{AffSch} \simeq \mathbf{Ring}^{\mathrm{op}}$. It really preserves diagrams! That should answer your first additional question. Unfortunately, this is often not stressed enough. In most textbooks I only see a proof that there is a bijection whereas the naturality is extremely important. It was definitely one part that wasn't entirely clear to me until I realized that the functoriality is an essential part of this assertion and is used over and over again.

I do not know of a more efficient way to prove it than to define the mutually inverse maps and checking that they define an equivalence of categories by hand. I haven't thought about it but you can probably also check that there is a fully faithful, essentially surjective functor $\mathbf{Ring}^{\mathrm{op}} \to \mathbf{AffSch}$, but the first approach should still be the preferred approach to understand both maps in the correspondence.


问:
If the complement of any singeton is finite, the set is finite
If for any singleton set, the complement is finite then the set is finite right? This seems very intuitive. If the set was infinite, then removing a single element would still be infinite. But what if the statement only says it is true for a specific singleton set? Would the set still be finite?</p>

答:
I think you are confused because of the quantification. 'for any' expresses the universal property of the quantified object, whereas 'there is' only notifies the existence of such an object. Therefore, we should utilize only the existence of the set in solving your problem.

Let $X$ be a set and suppose there is a subset $A$ of $X$ such that $|A| = 1$ and $|X\setminus A| < \infty$ where $|\cdot|$ represents the cardinality of a set. Since $A$ and $X\setminus A$ are disjoint, we have $|X| = |A| + |X\setminus A|$. Therefore,  $|X|< \infty$ and this conclusion is not dependent on what $A$ is actually is. Hence, we can say that $X$ is finite.

ContentLicense: CC BY-SA 4.0


问:
Easy way to check for a valid solution in this triple equality?
Let's say I have the following equalities

$a_1x_1 + a_2x_2 + a_3x_3 + a_4x_4 = b_1x_1 + b_2x_2 + b_3x_3 + b_4x_4 = c_1x_1 + c_2x_2 + c_3x_3 + c_4x_4$

Where the $a$'s, $b$'s, and $c$'s are known, non-negative integers.

Is there an efficient way to check if a solution exists (the $x$'s) such that they are non-negative real numbers (except for the trivial case of all $x$'s being 0)? I don't need to actually calculate them, just need some way to see if a solution even exists.

答:
After subtracting one of your expressions from the other two, you have two homogeneous linear equations, which you can write as
$ A x = 0$, where $A$ is a $2 \times 4$ matrix.  By a theorem of Gordan (or the duality theory of linear programming) this has a solution with $x \ge 0$ and $x \ne 0$ if and only if
the system $A^T y < 0$ has no solution, i.e. there are no real $y_1, y_2$ with
all four entries of $A^T y < 0$.  By homogeneity it suffices to look at the three cases
$y_1 = 1$, $y_1 = 1$ and $y_1 = -1$.  In each case, the four entries of $A^T y$ give us four inequalities on $y_2$, and it is easy to check if a solution exists.

EDIT: In your example, subtract the third expression from the other two to get
$$ \eqalign{x_1 - x_2 + x_3 &= 0\cr -x_1 \phantom{- x_2} + x_3 &= 0\cr}$$
The system $A^T y < 0$ says
$$ \eqalign{y_1 - y_2 &< 0 \cr
            -y_1 \phantom{-y_2} &< 0\cr
             y_1 + y_2 &< 0\cr}$$
The second row says $y_1 > 0$, so we take $y_1 = 1$, and then we need
$1 - y_2 < 0$, i.e. $y_2 > 1$, and $1 + y_2 < 0$, i.e. $y_2 < -1$.  These
are contradictory, so we conclude that $A^T y < 0$ has no solution, and your
system does have a nonnegative nonzero solution.

ContentLicense: CC BY-SA 3.0


问:
Find $\lim_{n \to \infty} \sqrt[n]{n!}$.
I am playing around with the root/ratio test to practice with series. I just showed that $\sum \frac{1}{n!}$ converges by using the ratio test. I decided to see how things would go with the root test and I got stuck at something that I can't find on google. Right away while running the root test I encountered $\lim_{x \to +\infty}\left(\frac{1}{n!}\right)^\frac{1}{n}$. I have made the claim that $\left(n!\right)^\frac{1}{n}\geq 1 \hspace{3mm}\forall n\in \mathbb{N}.$ I have begun the proof and I think it is on the right track but some verification would be nice.

Proof:
We can see that for n=1, this obviously holds, $\left(1!\right)^1\geq 1.$ Now suppose that this is the case for some $n=k$, then we have $\left((k+1)!\right)^{\frac{1}{k+1}}= \left((k+1)k!\right)^\frac{1}{k+1}$. It is at this point that I start to have a little trouble. Can somebody give me a push in the right direction? Thanks!

答: The question seems to be about $n!^{1/n}\ge 1$. This is fairly obvious, as $n!\ge1$. There is no need for induction.


问:
Combining Sinusoidal waves
Is there a general way to add sinusoidal waves? I'm trying to covert the following expression

$A_1 \cos(x+y)+A_2\sin(x-y)$

into

$A_3\cos(z)$ where $z$ is a function of $x$ and $y$

答:
In general when you add two sinusoidal waves together, the result is not necessarily a sinusoid.

In places where it is defined, the function $z=f(x,y)$ is just setting your two expressions equal together and solving for $z$:
$$
z = \arccos \big (\frac{1}{A_3}(A_1 \cos(x+y) + A_2 \sin(x - y)\big )
$$

If $A_3 \ge (A_1+A_2)$, then the input to $\arccos$ will be $\in [-1,1]$ and $z$ is defined.

However it is misleading to think that $cos(f(x,y))$ is a sinusoid (plane wave) of $x$ and $y$.

Title: Combining Sinusoidal waves


问:
Supremum on the positive roots of polynomials with bounded integer coefficients
Let see this post as a complement of Roots of polynomials with bounded integer coefficients.

Let $k$ be a positive integer. Let $R_k$ be the set of positive roots of (nonzero) polynomials with integer coefficients in $[-k,k]$:

$$R_k:=\left\{r > 0 \ \middle| \ \sum_{i=0}^na_ir^i=0,n\in\mathbb N,a_i\in\mathbb Z,|a_i|\leq k, a_n \neq 0\right\}.$$

Let $s_k$ be the supremum of $R_k$. The post mentioned above states (without proof) that $s_k=k+1$.

Question: What is the (detailed) proof of this statement?

Below is the proof that $s_k \ge k+1$ (so it remains to prove that $s_k \le k+1$):

Let $n$ be a positive integer. Consider the polynomial $$X^n-k(1+X+X^2+\dots+X^{n-1}) = X^n-k\frac{X^n-1}{X-1} = \frac{X^{n+1}-X^n-kX^n+k}{X-1},$$ then $r$ is a root iff $r^{n+1}+k = (k+1)r^n$, iff $r + \frac{k}{r^n} = k+1$, so that if $r_n$ is its largest positive root then $\lim_{n \to \infty}r_n=k+1$. $\square$

Body:
Suppose $a_nx^n+\cdots+a_0=0$. Then $$
|x|^n\le\frac{k}{|a_n|}(|x|^{n-1}+\cdots+1)\le k\frac{|x|^n-1}{|x|-1}\le \frac{k|x|^n}{|x|-1}$$

$$\therefore\ |x|^{n+1}-|x|^n\le k|x|^n$$
$$\therefore\ |x|\le k+1$$


问:
How does this function translate into a sum？
I really don't understand how $f(x)=\arctan2x  $ and $f'(x)$$ $ translates into $\sum_{n=0}^{\infty}(-1)^n2^{2n+1}x^{2n}$.
$$
f'(x)=\frac{2}{1+(2x)^2}=\frac{2}{1+4x^2}= 2\sum_{n=0}^{\infty}(-4x^2)^n=\sum_{n=0}^{\infty}(-1)^n2^{2n+1}x^{2n}
$$

Thanks

答:
Let's just work with $\text{arctan}(x)$ for now, as it makes formulas easier.

As you have noted, its derivative is $\frac{1}{1+x^2}$. We can now use a little trick, namely the formula for a geometric series:
$$\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^nx^{2n}$$
This works as long as $|-x^2| < 1$, i.e. $x\in (-1, 1)$.

For your case, the calculation is similar, with just a few multiplicative factors more.
Integrating this should then give you the function you are searching for; you seem to have a typo, but at the end of the day you should get
$$f(x) = \sum^{\infty}_{n=0}(-1)^n\frac{2^{2n+1}}{2n+1}x^{2n+1}$$


问:
Write the list of all the normal series of $\mathbb{Z}_{105}$
In this problem they ask to do a list of all the normal series of $\mathbb{Z}_{105}$ and their respective factor group. I also have to point out which series are from a composition.

Is there any hint so I can solve it myself?

My Path:

We observe that $\mathbb{Z}_{105} = \mathbb{Z}_3 \times \mathbb{Z}_5 \times \mathbb{Z}_7$ and do the Hasse diagram

then the normal series are:

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{15}⟩ \subseteq ⟨\overline{3}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{15}⟩ \subseteq ⟨\overline{5}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{21}⟩ \subseteq ⟨\overline{3}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{21}⟩ \subseteq ⟨\overline{7}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{35}⟩ \subseteq ⟨\overline{5}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{35}⟩ \subseteq ⟨\overline{7}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{15}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{21}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{35}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{3}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{5}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{7}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

and the only normal Series from a composition is the trivial

$\{e\}=⟨\overline{105}⟩ \subseteq ⟨\overline{1}⟩ = \mathbb{Z}_{105}$

答:
Hint 1:

if $G_1$, ... , $G_n$ are pairwise non-isomorphic simple groups, then all normal subgroups of $\prod_{i=1}^n G_i$ are of the form $\prod_{i \in S} G_i$ for some $S \subset \{1, ... , n\}$ (one for each such $S$)

Hint 2:

 $\mathbb{Z}_{105} = \mathbb{Z}_3 \times \mathbb{Z}_5 \times \mathbb{Z}_7$


问:
Calculate this limit without L'Hôpital's rule.

Calculate $$\lim_{x\to0}\frac{(x+32)^{1/5}-2}{x}$$ without L'Hôpital's rule.

My attempt: I first rationalized the expression to get $$\left(\frac{(x+32)^{1/5}-2}{x}\right)\left(\frac{(x+32)^{1/5}+2}{(x+32)^{1/5}+2}\right)=\frac{x+28}{x((x+32)^{1/5}+2)}$$ How should I get rid of the singular $x$ in the denominator now? Should I factor something here?

Body: Let $y=(x+32)^{1/5}$. You can write the limit as $\lim_{y \to 2} \frac {y-2} {y^{5}-2^{5}}$. It is easy to write down this limit using the formula $y^{5}-2^{5}=(y-2)(y^{4}+2y^{3}+2^{2}y^{2}+2^{3}y+2^{4})$


Title: Calculate this limit without L'Hôpital's rule.

Body:
Calculate $$\lim_{x\to0}\frac{(x+32)^{1/5}-2}{x}$$ without L'Hôpital's rule.

My attempt: I first rationalized the expression to get $$\left(\frac{(x+32)^{1/5}-2}{x}\right)\left(\frac{(x+32)^{1/5}+2}{(x+32)^{1/5}+2}\right)=\frac{x+28}{x((x+32)^{1/5}+2)}$$ How should I get rid of the singular $x$ in the denominator now? Should I factor something here?

Body:
Hint.

Making $x+32=y^5$ we have

$$
\lim_{y\to 2}\frac{y-2}{y^5-32}
$$


问:
System of (trigonometric) equations
I have a system of two equations, involving trigonometric terms of two variables $\theta$ and $\phi$:

$$ d_1 = a_1\sin\theta \cos\phi + b_1\sin\phi + c_1\cos\theta\cos\phi $$
$$ d_2 = a_2\sin\theta \cos\phi + b_2\sin\phi + c_2\cos\theta\cos\phi $$

$d_1$, $d_2$, $a_1$, $a_2$, $b_1$, $b_2$, $c_1$ and $c_2$ being known, real constants.

(if we assume $\theta$ and $\phi$ to be spherical angles, you will probably recognize spherical harmonics of degree 1 at work here).

Other than somewhat brute-force numerical resolution methods, is there a way to more or less elegantly solve this system of equations in $\theta$ and $\phi$ ?

Thanks!

Title: System of (trigonometric) equations

Body:
The solution of problem is:

$\tan{\theta}=\frac{b_{1} c_{3} d_{2}-b_{2} c_{3} d_{1}+b_{3}(c_{2} d_{1}-c_{1} d_{2})}{ a_{1} b_{3} d_{2}-a_{2} b_{3} d_{1}+a_{3}(b_{2} d_{1}-b_{1} d_{2})}$

$\tan{\phi}=\frac{z}{\sqrt{x^{2}+y^{2}}}$

$x=-\frac{b_{1} c_{3} d_{2}-b_{2} c_{3} d_{1}+b_{3}(c_{2} d_{1}-c_{1} d_{2})}{a_{1}(b_{2} c_{3}-b_{3}c_{2})+a_{2}(b_{3}c_{1}-b_{1} c_{3})+a_{3}(b_{1} c_{2}-b_{2}c_{1})}$

$y=-\frac{a_{1} b_{3} d_{2}-a_{2} b_{3} d_{1}+a_{3}(b_{2} d_{1}-b_{1} d_{2})}{a_{1}(b_{2} c_{3}-b_{3}c_{2})+a_{2}(b_{3}c_{1}-b_{1} c_{3})+a_{3}(b_{1} c_{2}-b_{2}c_{1})}$

$z=+\frac{a_{1} c_{3} d_{2}-a_{2} c_{3} d_{1}+a_{3}(c_{2} d_{1}-c_{1} d_{2})}{a_{1}(b_{2} c_{3}-b_{3}c_{2})+a_{2}(b_{3}c_{1}-b_{1} c_{3})+a_{3}(b_{1} c_{2}-b_{2}c_{1})}$

$a_{3}=b_{1} c_{2}-b_{2} c_{1}$

$b_{3}=a_{1} b_{2}-a_{2}b_{1}$

$c_{3}=a_{2}c_{1}-a_{1} c_{2}$


Body:
$$L=lim_{x \to 0} \dfrac{(x+32)^{\frac{1}{5}}-2}{x}$$

$$L=lim_{x \to 0} \dfrac{(x+32)^{\frac{1}{5}}-32^{\frac{1}{5}}}{(x+32) -32}$$

Therefore,

$$L=\frac{1}{5} \cdot 32^{\frac{-4}{5}}=\frac{1}{5 \cdot 2^4}$$


答:
Consider the function $f(x) = (x+32)^{1/5}$. By definition, $f'(0) = \lim_{x \to 0} \dfrac{(x+32)^{1/5} - 2}{x}$.

By the power and chain rules for differentiation, you can show that $f'(x) = \dfrac{1}{5}(x+32)^{-4/5}$ where $x \neq -32$.

Therefore, $\lim_{x \to 0} \dfrac{(x+32)^{1/5} - 2}{x} = f'(0) = \dfrac{1}{80}$.


问:
Upper bound on number of edges in graph with no clique of size k as subgraph
Let $G = (V, E)$ be an undirected graph such that $|V| = n$ and $G$ doesn't have a clique of size k as a subgraph.
I need an upper bound on $|E|$. Any ideas?
I'm hoping for $O(nk)$ but can't prove it.

Thanks!

答:
Since Turán's theorem is the right hint and it is already given in the comments, I am going to give an easy counterexample, which actually is a direct consequence of Turán's theorem, to your "hope" that the upper bound is $O(nk)$. First, let us recapitulate the statement as provided here.

Theorem 1 (Turán): Let $G$ be a graph on $n$ vertices that does not contain a $K_{r+1}$ as a subgraph. Then $|E(G)| \leq \frac{r-1}{r}\cdot \frac{n^2}{2}=(1-\frac{1}{r})\cdot \frac{n^2}{2}$

For the special case $r=2$, a triangle free graph, the maximum number of edges edges for a $n$-vertex graph is $\lfloor \frac{n^2}{4} \rfloor $. To formalize:

Corollary 1:
Let $G_n$ be a bipartite graph on $n$ vertices, then $|E(G_n)| \leq \lfloor \frac{n^2}{4} \rfloor$.

It is easy to see that $\lfloor \frac{n^2}{4} \rfloor$ is not bounded from above by $nk$ for fixed $k \in \mathbb{N}$ (in this case $k=3$) for all $n$. To be precise, it fails for all $n>4$.


问:
Show $f(x,y)=2x-\frac{7}{y}$ is continuous on the set $E=\{\,(x,y) \, : \, y>0 \, \}$
Let $f(x,y)=2x-\frac{7}{y}$

Show that $f$ is continuous on the set $E=\{\,(x,y) \, : \, y>0 \, \}$

I did the following:

$\forall \epsilon>0$, $\exists \delta>0$ s.t. $\forall (a,b)\in \mathbb{R^2}$ with $\sqrt{(x-a)^2+(y-b)^2}<\delta$ implies$ |f(x,y)-f(a,b)|<\epsilon$

\begin{align}
|f(x,y)-f(a,b)|&= \left|2x-2a+\frac{7}{b}-\frac{7}{y}\right| \\
&\leq 2\big|x-a\big|+\left|\frac{7}{b}-\frac{7}{y}\right| \\
&<2 \delta +\big(\frac{7\delta}{b|y|}\big)  \quad \text{(since $b>0$)}
\end{align}

What should be the following steps?

答:
\begin{align}
&|f(x,y)-f(a,b)|\\
&= \left|2x-2a+\frac{7}{b}-\frac{7}{y}\right| \\
&\leq 2|x-a|+\left|\frac{7}{b}-\frac{7}{y}\right| \\
&<2 \delta +\frac{7\delta}{|b|\,|y|}
\end{align}

In the second term $\dfrac{7\delta}{|b|\,|y|}$, we need to keep the infinitesimal $\delta$, and $\dfrac{1}{|b|}$ is just a constant, so we can use $|y-b| < \delta$ to give a lower bound to $|y| > |b| - \delta$.  When $\delta$ is sufficiently small ($\delta < \dfrac{|b|}{2}$), we have $|y| > \dfrac{|b|}{2}$.  Hence we have

$$ |f(x,y) - f(a,b)| < 2 \delta +  14 \delta / |b| ^2 = 2(1 + 7/|b|^2) \delta.$$

If you want $ |f(x,y) - f(a,b)| < \epsilon$, you just need $2(1+7/|b|^2) \delta \le \epsilon$, or $\delta \le \dfrac{\epsilon}{2(1+7/|b|^2)}$.

To guarantee that $\delta < \dfrac{|b|}{2}$, you just need to set $\delta = \min\left\{\dfrac{|b|}{2}, \dfrac{\epsilon}{2(1+7/|b|^2)}\right\}$.


Body:
If you want detailed proof:

$\dfrac{(2^{5}(\frac{x}{32}+1))^{\frac{1}{5}}-2}{x}=\dfrac{2(\frac{x}{32}+1)^\frac{1}{5}-2}{x}$

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2}\cdot x^2.....$

$\lim_{x\to0}=\dfrac{2\bigg(1+\frac{1}{5}\frac{x}{32}+\dfrac{\frac{1}{5}(\frac{1}{5}-1)}{2}\cdot\bigg(\frac{x}{32}\bigg)^2.....\bigg)-2}{x}=\lim_{x\to0}\dfrac{\frac{2x}{5\cdot32}+...}{x}=\dfrac{2}{5\cdot 32}$


问:
number of matrices in $GL_{3}(\mathbb{Z}_{3})$ such that $[1 \ 1 \ 1]^{T}$ is an eigenvector.
Find the number of matrices in $GL_{3}(\mathbb{Z}_{3})$ such that $[1 \ 1 \ 1]^{T}$ is an eigenvector.

This question was asked to me in an interview which I wasn't able to answer.

I approached as follows,

If $A$ is a matrix in $GL_{3}(\mathbb{Z}_{3})$  such that $[1 \ 1 \ 1]^{T}$ is an eigenvector then $0$ can't be an eigenvalue as $A$ is invertible. So $A$ can have $1$ and $2$ as eigenvalues. Let $v=[1 \ 1 \ 1]^{T}$.

If $1$ is an eigenvalue of $A$ then we have $Av=v$ and if $2$ is an eigenvalue of $A$ then $Av=2v$.

So we need to find the number of matrices $A$ in $GL_{3}(\mathbb{Z}_{3})$  such that $Av=v$ and $Av=2v$.

Then I tried to find out such A's but I got stuck at a point.

Can you please let me know how to solve this question ? Thanks.

答:
Hint: instead of thinking of the matrix, think of the corresponding (invertible) linear transformation $\mathbb{F}_3^3 \to \mathbb{F}_3^3$.  Then, use the fact that $\{ (1, 1, 1), (1, 0, 0), (0, 1, 0) \}$ is a basis of $\mathbb{F}_3^3$.  So, the linear transformations are uniquely determined by the images of these basis vectors (and each combination comes from a linear transformation); and the transformation is invertible if and only if the image of the basis is another basis (which is further equivalent to the image of the basis being linearly independent).

So:

How many choices are there for the image of $(1, 1, 1)$?

Given a choice from the previous step, how many choices are there for the image of $(1, 0, 0)$?  (Further hint: the only restriction is that it must not be a scalar multiple of the image of $(1, 1, 1)$.)

Given a choice from the previous two steps, how many choices are there for the image of $(0, 1, 0)$?  (This must not be in the span of the previous two vectors.)


问:
$((x+2)^2 - (x+1)^2) - ((x+1)^2-x^2) = 2$ Is this proof valid for this identity?
Claim:
$$((x+2)^2 - (x+1)^2) - ((x+1)^2-x^2) = 2$$

Proof:

$$x \times x = x^2$$
$$(x+1) \times x = x^2 +x$$
$$(x+1) \times (x+1) = x^2 + x + (x+1)$$
$$(x+1)^2 = x^2 + 2x + 1$$

As we know, $2x+1$, is the definition of an odd number, since an odd number is always one more than a multiple of $2$.

$$(x+1)^2 - x^2 = 2x+1$$

The difference between two consecutive odd numbers will always be $2$, like e.g. $3-1 =2$, $5-3 = 2$, $7-5 = 2$, and so on.

Therefore:

$$((x+2)^2 - (x+1)^2) - ((x+1)^2-x^2) = 2$$

Is this proof valid? Also, have you ever seen this proof before? Has the same identity been proven in a different, better way?

Body:
Let's evaulate
$$((x+2)^2 - (x+1)^2) - ((x+1)^2-x^2) $$
using the difference of $2$ squares factorisation:
$$((x+2)^2 - (x+1)^2) - ((x+1)^2-x^2) =(x+2+x+1)(x+2-x-1)-(x+1+x)(x+1-x)$$
$$=(2x+3)-(2x+1)=3-1=2$$
as required.


问:
How to proof: if $A\subset B \rightarrow f^{-1}(A)\subset f^{-1}(B)$ and show that it can't work conversely?
Hello so I was wondering how you can prove $A\subset B \rightarrow f^{-1}(A)\subset f^{-1}(B)$. I found someone with a similar problem on stack exchange but didn't  found it much of a proof.

I know if you have $f:A\rightarrow B$ is a function with e.g. $K\subset Y$ then the inverse image is $f^{-1}(K)=\{x\in X \mid f(x)=K \}$. Which also means that $x\in f^{-1}(K)\leftrightarrow f(x)\in K$. And that for $A\subset B$ means that $x\in A \rightarrow x\in B$.

But knowing this how can I proof this? Because I can say $x\in f^{-1}(A)\rightarrow x\in f^{-1}(B)$ So therefore $f(x)\in A \rightarrow f(x)\in B$ But now I don't know if I automatically can say that $A\subset B$ (I don't think so).

Secondly I want to show using a counterexample that $ f^{-1}(A)\subset f^{-1}(B) \rightarrow A\subset B $ Doesn't work.

If someone knows the answer please share your wisdom :)

Thanks in advance

答:
You are given $A\subseteq B$ and need to show $f^{-1}(A)\subseteq f^{-1}(B)$. You can show it by proving that every $x\in f^{-1}(A)$ also belongs to $f^{-1}(B)$.

Let $x\in f^{-1}(A)$. Then $f(x)\in A$, so $f(x)\in B$ since $A\subseteq B$. Thus $x\in f^{-1}(B)$.

For a counterexample of $f^{-1}(A)\subseteq f^{-1}(B)\implies A\subseteq B$, consider $f:\Bbb R\to\Bbb R$ with $f(x)=x^2,x\in\Bbb R$. Let $A=[-1,1],B=[0,4]$. Then $f^{-1}(A)=[-1,1],f^{-1}(B)=[-2,2]$ but $A\not\subseteq B$.


问:
Is there a way of connecting manifolds together in a higher dimension?
This question might be very naive since I haven't taken much topology or differential geometry yet; so please make the answer more conceptual than mathematical. We can think of a two dimensional sheet as joining an infinite number of one dimensional lines along a second dimension that doesn't exist for the lines. This also works if the surface is curved, only now, the lines are one dimensional curves.

So, in a sense, spaces can be joined together along a higher dimension if the lower dimensional spaces change slowly enough as we move along the higher dimension - or else we would not have a smooth surface at the end. If what I meant is not clear, the image below might explain it better:

In a sense, what I mean by the curves changing too fast is that after combining them, the resulting surface might not be continuous or differentiable (but I suppose I can only say this if the surface can be embedded in a euclidean space).

So my question is this: given an infinite set of manifolds (reasonably smooth or continuous and not necessarily reimannian) of dimension "n", is there a way to connect them like above in a higher dimension to give us a n+1 dimensional manifold?

If so, what conditions do the manifolds need to satisfy (conceptually)? If this question is too mathematical to answer conceptually, please list a few sources or let me know what this is called so that I can read about this myself.

答:
This is a good question and there are a couple of different ways to make it precise, all of which lead to quite interesting mathematics. One is to ignore everything happening in the "middle" and focus on the "ends"; that is, we will settle for connecting two manifolds, forget about infinitely many for now. Then what you are talking about is known as a cobordism between two manifolds $M, N$, meaning a manifold in one higher dimension whose boundary is $M \sqcup N$. If a cobordism exists between $M, N$ they are said to be cobordant. As a simple example, the pair of pants is a cobordism between a circle $S^1$ and two circles $S^1 \sqcup S^1$:

If you slice up a pair of pants and think of it as a "movie" going forward in time you'll note that in the middle you get a wedge of two circles, which is not a manifold. This sort of thing is investigated in Morse theory, which is a whole fascinating topic in itself and about which I will say nothing more.

As a general question we can ask:

Question: When are two $n$-manifolds $M, N$ cobordant?

You might think that this is a hard and deep problem in manifold topology. Remarkably, for closed $n$-manifolds this question has a more-or-less complete answer due to Thom. Associated to any such $n$-manifold is a bunch of numbers called the Stiefel-Whitney numbers, which take values $\bmod 2$, and Thom showed:

Theorem (Thom): Two closed $n$-manifolds $M, N$ are cobordant iff their Stiefel-Whitney numbers are the same.

The simplest Stiefel-Whitney number to explain is called $w_n$, and it's given by the Euler characteristic $\bmod 2$. It's a nice exercise to show that this is a cobordism invariant, meaning that if $M$ and $N$ are cobordant then $\chi(M) \equiv \chi(N) \bmod 2$.

Corollary: A closed manifold with odd Euler characteristic is not cobordant to a closed manifold with even Euler characteristic.

For example, the real projective plane $\mathbb{RP}^2$ has Euler characteristic $1$ and so is not cobordant to the torus $T^2$ (Euler characteristic $0$) or the $2$-sphere $S^2$ (Euler characteristic $2$). However, the torus and the $2$-sphere are cobordant (exercise), and in fact every pair of closed orientable surfaces is cobordant (exercise). This is a very special case of Thom's theorem (which can be proven much more easily than what Thom did), since it turns out that for such surfaces both of the Stiefel-Whitney numbers (which are called $w_1^2$ and $w_2$) vanish.

There's a whole lot more to say here about cobordism rings and so forth but this is a good place to stop for now, I think.


问:
$A_5$ is generated by all 3 cycles
There are some similiar questions, but it seems that there is some different equal definition to $A_5$.

The definition I'm using for $A_5$ is that this is the kernel of the sign homomorphism- i.e all the permutations with even number of transpositions.

I would like to get help with the equivalence of this definition , to the definition of subgroup of $S_5$ , generated by all cycles of order 3.

EDIT- THERE IS AN ANSWER ON THE LINK IN COMMENTS

答:
A5 is the group [3,3,3]+ or [3,5]+ in Coxeter notation.

You could make it, I suppose, out of (AB)³ = (BC)³ = (CD)³ = I, and (AC)² = (BD)² = (DA)² = I, which is what the first symbol means.

The second one means (AB)³ = (BC)^5 = (AC)² = I, which is what the second definition.

A5 is a relatively primitive group, but you can indeed generate A5 from purely cycles of three.

Title: $A_5$ is generated by all 3 cycles


问:
Sequence of M-Lipschitz functions
Let $\alpha$ , $\beta \in \mathbb R$ with $\alpha < \beta$, $M \geq 0$ and $(f_n)$ a sequence of M-Lipschitz functions of [$\alpha , \beta$] in $\mathbb R$. If $(f_n)$ converges simply (or "pointwise" in English I think) to a function $f$ on [$\alpha , \beta$], show that $f$ is also M-lipschitz.

We have that $(f_n)$ converges simply so $\forall x \in [\alpha , \beta], \forall \epsilon > 0, \exists n_0$ such that $\forall n > n_0, \lvert f_n(x) - f(x) \rvert \leq \epsilon$

And we also have that the $f_n$ are M-Lipschitz so $\forall x,y \in [\alpha , \beta], \lvert f_n(x) - f_n(y) \rvert \leq M\lvert x-y \rvert$

We want to show that $\forall x,y \in [\alpha , \beta], \lvert f(x) - f(y) \rvert \leq M\lvert x-y \rvert$

I tried to write $\lvert f(x) - f(y) \rvert = \lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert$

Then we have $\lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert \leq \epsilon + M\lvert x-y \rvert + \epsilon$ but it is not really the form we wanted and I think I made some mistakes with the quantifiers. Can someone help me ? (The correction solves this exercise just by saying : when going to the limit the inequality stays true. I tried to find something more precise and clear).

答:
If $x \leq M|x-y| + \epsilon$ for every $\epsilon$, then $x \leq M|x-y|$.

Proof:

Suppose $x > M|x-y|$.

Then $x> M|x-y| - \epsilon$ for every $\epsilon > 0$. Hence $M|x-y| - \epsilon  < x\leq M|x-y| + \epsilon \implies$

$-\epsilon <x-M|x-y| \leq \epsilon \implies x-M|x-y|=0$ (which is proven using a basic contradiction).

Hence $x = M|x-y| \implies x > x$, a contradiction.

Your problem uses $2\epsilon$ which is a special case of this.

Title: Sequence of M-Lipschitz functions


答:
$\newcommand{\inv}[1]{f^{-1}[#1]}$
This is my answer.

The inverse image of the mapping $f:X \rightarrow Y$ is defined as, for any $B \subseteq Y$,

$$
f^{-1}[B] = \{x \in X~|~\exists y \in B: f(x) = y\}
$$
Therefore, for all $x\in X$, $x \in f^{-1}[B]$ is equivalent to $f(x) \in B$. We can solve your problem using this property.

Let $A \subseteq B \subseteq Y$. Then, for all $x \in X$, $x \in f^{-1}[A]$ implies $f(x) \in A \subseteq B$, which gives $x\in f^{-1}[B]$. Therefore, $f^{-1}[A] \subseteq f^{-1}[B]$.

Conversely, suppose $f^{-1}[A] \subseteq f^{-1}[B]$. In this case, we don't have much to say. However, if we assume $f$ to be subjective we can conclude $A \subseteq B$. Let $f$ be surjective. Then, for all $y \in A$, there exists $x \in X$ such that $f(x) = y$, hence $x \in \inv{A} \subseteq \inv{B}$, which gives $y = f(x) \in B$.

Title: How to proof: if $A\subset B \rightarrow f^{-1}(A)\subset f^{-1}(B)$ and show that it can't work conversely?

Body:
Hello so I was wondering how you can prove $A\subset B \rightarrow f^{-1}(A)\subset f^{-1}(B)$. I found someone with a similar problem on stack exchange but didn't  found it much of a proof.

I know if you have $f:A\rightarrow B$ is a function with e.g. $K\subset Y$ then the inverse image is $f^{-1}(K)=\{x\in X \mid f(x)=K \}$. Which also means that $x\in f^{-1}(K)\leftrightarrow f(x)\in K$. And that for $A\subset B$ means that $x\in A \rightarrow x\in B$.

But knowing this how can I proof this? Because I can say $x\in f^{-1}(A)\rightarrow x\in f^{-1}(B)$ So therefore $f(x)\in A \rightarrow f(x)\in B$ But now I don't know if I automatically can say that $A\subset B$ (I don't think so).

Secondly I want to show using a counterexample that $ f^{-1}(A)\subset f^{-1}(B) \rightarrow A\subset B $ Doesn't work.

If someone knows the answer please share your wisdom :)

Thanks in advance


问:
Square matrix $\|Ax-Ay\|\le \|x-y\|$
Could you give me an example of a square matrix $A\in \mathcal{M}_{2 \times 2}$ or $\mathcal{M}_{3 \times 3}$ for which we have $\|Ax-Ay\|\le \|x-y\|$, $ \ \ x, y \in \{0, e_1, . . . , e_n\}, \ \ e_1, . . . , e_n$ - canonical basis of $\mathbb{R}^n$, $\|\cdot\|$ is the Euclidean norm, but $\|Az\| > \|z\|$ for a certain vector $z$?

I think $A =  \left( \begin{array}{ccc}
1 & \frac{1}{3}  \\
0 & 1  \end{array} \right) $ will be good for the first case, but I have problem for the case where matrix has degree $3$.

答:
Note: My original answer contained an error; correcting the error led to this simpler solution. Most of the comments below refer to the original answer.

You can project onto the subspace of vectors with constant entries, using a matrix with constant entries. Then $Ae_i$ is the same for all $e_i$, so the left-hand side of the desired inequality is $0$ unless one of $x$, $y$ is $0$, in which case the inequality is fulfilled if you take less than $\sqrt n$ times the orthogonal projection. On the other hand, you can make $\lVert Az\rVert/\lvert z\rVert$ anything from $1$ to $\sqrt n$ for $z$ a vector with constant entries by using the corresponding multiple of the orthogonal projection.

Title: Square matrix $\|Ax-Ay\|\le \|x-y\|$


问:
Factor SYMBOLIC third degree polynomial
I apology in advance for the long post...

Say I have a very general 3rd-degree polynomial

$$x^3 + a_1x^2 + a_2x+a_3$$

I would really like to factor that polynomial as a product of a 1st order and a 2nd order polynomial, i.e.

$$(x-b_1)(x^2+b_2x+b_3)$$

And do all of that with the symbolic equation: finding the values of $b_i$ as a function of the $a_i$.

I have tried 3 things:

My first idea was to use a symbolic computation software (Wolfram Mathematica) and finds the symbolic roots of the first equation, let's call them $\alpha_i$. The roots are awful, but it works ...

Then I tried to keep one root alone to build up $(x-\alpha_i)$ and identify $b_1$ (easy so far). And next compute $(x-\alpha_2)(x-\alpha_3)$ and identify $(x^2 + b_2x +b_3)$ (or equivalently computing $b_2 = -(\alpha_2+\alpha_3)$ and $b_3 = \alpha_2\alpha_3$).

That is where things go wrong... Wolfram Mathematica just does not want to "simplify" $-(\alpha_2+\alpha_3)$ and $\alpha_2\alpha_3$ (it just literally writes that answer).

Alternatively, I tried to keep one of the root, and compute

$$ \frac{x^3+a_1x^2+a_2x+a_3}{x-\alpha_1}$$

In this case, $b_1$ is simply $\alpha_1$ (as before) and the result of the division would be $x^2+b_2x+b_3$. But here again, Mathematica does not explicitly compute or "simplify" the division and leaves me with $ \frac{x^3+a_1x^2+a_2x+a_3}{x-\alpha_1}$ as the answer.

Another way around my problem (too long to explain, just believe me) would be to compute the magnitude and phase of the two complex roots $\alpha_i$ (I know I have a complex conjugate pair and a real root). But again, Whenever I enter Abs[$\alpha_i$], Wolfram Mathematica gives me ... Abs[$\alpha_i$] as an answer and does not "simplify" or "compute" explicitly what the magnitude would be.

Does anyone have an idea of how I could "solve" that?

I realize that I am asking for a very "theoretical" thing and that if the exact answer would exist, it would probably be given somewhere in Wikipedia or in textbooks... Maybe Mathematica does not simplify or compute the product because things start to be too complex even for a computer...

Thanks so much for all of you that could read this post to the end. Any help, advice or hint would be very much appreciated!

Anthony.

Title: Factor SYMBOLIC third degree polynomial

Body:
Consider $$\Delta=(x^3 + a_1x^2 + a_2x+a_3)-(x-b_1)(x^2+b_2x+b_3)$$ Expand and group terms to have
$$\Delta=(a_3+b_1 b_3)+ (a_2+b_1 b_2- b_3)x+ (a_1+b_1-b_2)x^2$$ So, we need to solve $$a_3+b_1 b_3=0 \tag 1$$
$$a_2+b_1 b_2- b_3=0 \tag 2$$
$$a_1+b_1-b_2=0 \tag 3$$

Using $(2)$ and $(3)$, you can have
$$b_2=a_1+b_1 \qquad \text{and} \qquad  b_3=a_2+a_1b_1+b_1^2$$ and, replacing in $(1)$ get
$$a_3+a_2b_1+a_1b_1^2+b_1^3=0$$ which is a cubic in $b_1$ !!!


问:
partial sum of $\sum_{i=1}^k \frac{i}{a^i}$
From "How to find the partial sum of n/2n?" we know that $S_n=2\left(1-\frac1{2^n}-\frac n{2^{n+1}}\right)$ so how can we prove that $\sum_{i=1}^k \frac{i}{a^i} = \frac{(a^{(k+1)}-a(k+1)+k)}{a^k(a-1)^2}$

Body:
$$S_n=~~r+2r^2+3r^3+4r^4+...+nr^n~~~~(1)$$
$$rS_n=~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~r^2+2r^3+3r^4+...+(n-1)r^n+nr^{n+1}~~~~(2)$$Suntrcting (2) from (1), we get
$$(1-r)S_n=r+r^2+r^3+r^4+...r^n-nr^{n+1}$$
$$(1-r)S=r[1+r+r^2+r^3+...+r^{n-1}]-nr^{n+1}$$
$$(1-r)S_n=r\frac{1-r^n}{1-r}-nr^{n+1}$$
$$S_n=\frac{r-(n+1)r^{n+1}+nr^{n+2}}{(1-r)^2}$$
Let $r=1/a$, then
$$S_n=\frac{a^{n+1}-(n+1)a+n}{(1-a)^2a^n}=\sum_{k=1}^{n} \frac{k}{a^k}$$


Body:
Let $\sqrt[5]{x+32}-2=y\implies x+32=(2+y)^5$ and $x\to0\implies y\to0$  to find

$$\lim_{x\to0}\frac{(x+32)^{0.2}-2}{x}$$

$$=\lim_{y\to0}\dfrac y{(2+y)^5-32}$$

$$=\lim_{y\to0}\dfrac y{\binom51y\cdot2^4+\binom52y^2\cdot2^3+\binom53y^3\cdot2^2+\binom51y^4\cdot2+y^5}$$
$$=?$$


问:
Calculating Average shortest distance between random points in a rectangle
My question is quite similar to this one. But what would be the solution in case of a rectangle with width W and length L??

Q1

I would like to find the average shortest distance between randomly distributed n points in a rectangle L  xW. I would like to take each point the Euclidean distance to its nearest neighbor,and calculate the average distance over all points?

I was hoping if I could do the same simulation similar to @Henry's answer but I am not quite familiar to computing languages.

Q2

The second part of my question is if I could assign the coordinates of some points in the rectangle.
For example, I would like to add some conditions that points cannot be distributed in the grey shaded area in this rectangle. Image here Thus the points can be distributed randomly over the area of the rectangle except for the shaded gray area. What would be the average shortest distance between randomly distributed n number of points ?

I would sincerely appreciate some help.

答:
Approximate solution to Q1 for a large amount of points

If you have N, a large amount of points, then the shape you use becomes irrelevant. In that case you can assume a Poisson distribution of points and consult the literature for the nearest neighbor distribution ( starting point:  https://en.wikipedia.org/wiki/Nearest_neighbour_distribution).

In order to decide what "large amount" means, you need to have enough points to ignore the circumference of the shape. This means you need to have a lot more points "inside" the shape than "on the edges". To find out if this holds, lets assume we know $l$,  the typical distance between points (we'll estimate  the value in a moment).
If the total circumference of the shape is $S$, there are approximately $N_{edge} \sim {S\over l}$  points near the edge, so as long as $N>>N_{edge}$ you can use the Poisson approximation.

To estimate $l$ note that the mean area a point "occupies" is $A_{mean} = {W*L\over N}$. Then the typical distance between points is $l=\sqrt{A_{mean}}$. If instead of a rectangle you have some other shape, replace $W*L$ with the area of the shape.


问:
Find the first 5 terms of the expansion in a power series
Find the first 5 terms of the expansion in a power series
$$y′=xe^{x}+2y^{2}$$

I've got a riccati equation $$ x e^{x}+2y^{2},  y(0)=0$$
After solving: $$y=e^{x}(x-1)+\frac{2}{3}y^{3} - 1$$

And I don't know how to go forward. Please help me.

答:
You assume that there exist $a_i$ such that $$y = a_0 +a_1x + a_2x^2 + \cdots.$$

From here it is very easy to calculate power series expansions for $y', xe^x,$ and $2y^2$.  Then you equate like powers of $x$ on both sides of the equality to get a relation between the $a_i$, and get a sequence of fairly simple equations in the $a_i$.  The initial condition $y(0)=0$ tells you immediately that $a_0 = 0$ and from there you can usually get the other coefficients.
Once you have $a_0,\ldots a_4$, you have the answer to the question.

To take a much simpler example that gives an idea of the method, suppose the equation were $$y' = y + 1$$ with the initial condition $y(0) = 0$.

We start with:

$$\begin{array}{rlrrr}
y &= &a_0 &+a_1x &+ a_2x^2 &+ \cdots \\
y+1 &=& 1+ a_0 &+a_1x &+ a_2x^2 &+ \cdots \\
y' &=& a_1 &+2a_2x &+ 3a_3x^2 &+ \cdots
\end{array}$$

Then equating the last two we have $$a_1 + 2a_2x + 3a_3x^2 + \cdots = 1 + a_0 + a_1x + a_2x^2 + \cdots$$

and equating coefficients of like powers gives equations in the $a_i$:
$$
\begin{align}
a_1 & =& 1+a_0\\
2a_2 & = &a_1 \\
3a_3 & = & a_2\\
&\vdots&
\end{align}
$$

Now $y(0) = 0$ tells us immediately that $a_0 = 0$.  And the equations above then give us $a_1 = 1, a_2 = \frac12, a_3 = \frac16,\ldots$.  So the series for $y$ is $$y = x + \frac12x^2 + \frac16x^3 + \cdots$$

which is in fact the power series for $y=e^x-1$.


问:
Derive the inequality $x<\arcsin \left(x\right)<\frac{x}{\sqrt{1-x^2}}$
I found that I could do this with the Mean Value theorem and also found a similar question here in the stackexchange and Socratic but couldn't solve the right hand side of the inequality.

I figured that by specifically choosing the function and the interval (combinations of the given question) we could prove that. But  I am lost tin selecting the appropriate function.

Can you guys please suggest some ideas and also some example sums where I could check these kind of sums.

Thanks

答:
This is pretty standard property of $\arcsin$ which is equivalent to the more famous inequality $\sin x <x<\tan x$ for $x\in(0, \pi/2)$.

A direct proof can be given by noting that if $x\in(0,1)$ and $t\in(0,x)$ then $$1<\frac{1}{\sqrt{1-t^2}}<\frac{1}{\sqrt{1-x^2}}$$ integrating the above with respect to $t$ on interval $[0,x]$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$ for all $x\in(0,1)$.

You can avoid integrals and achieve the same result via mean value theorem. If $x\in(0,1)$ then by mean value theorem we have $$\arcsin x=\arcsin x - \arcsin 0=\frac {x}{\sqrt{1-c^2}}$$ for some $c\in(0,x)$. But then we can obviously see that $$1<\frac{1}{\sqrt{1-c^2}}<\frac{1}{\sqrt{1-x^2}}$$ Multiplying by positive $x$ we get $$x<\arcsin x<\frac{x} {\sqrt{1-x^2}}$$


问:
Can a set have more than one ultrafilter?
https://proofwiki.org/wiki/Definition:Ultrafilter_on_Set

Given the definition that a filter $\mathcal F $ on a set $\mathcal S$ is  an ultrafilter on $\mathcal S$ iff there is no filter on  $\mathcal S$ which is strictly finer than $\mathcal F$, doesn't that just mean that the only ultrafilter on a set $\mathcal S$ is simply $\mathcal {P(S)} \setminus \{\emptyset\}$, the power set of $\mathcal S$ with the empty set removed?

答:
This is not a filter (it's not closed under intersection unless $S$ has one element).

Every set $S$ has at least $|S|$ many ultrafilters, namely, for any $s \in S$ the principal ultrafilter given by the subsets containing $s$. These are the only ultrafilters when $S$ is finite. When $S$ is infinite there are (assuming the axiom of choice) non-principal ultrafilters, which let you do lots of fun stuff.


问:
Corestriction of a continuous function remains continuous
Let $\beta: J \to \mathbb{R^3}$ be a continuous function and $M \subset \mathbb{R^3} $. How can I show that $\beta|^{M}: J \to M$ is still continuous?

答: We should assume $\beta[J] \subseteq M$ where $[\cdot]$ represents the image of a mapping. Let $V$ be an open subset of $M$. Then, there is an open subset $S$ of $ \mathbb{R}^3$ such that $S \cap M = V$. Furthermore, $(\beta|^{M})^{-1}[V]  = \beta^{-1}[S]$. Meanwhile, $\beta^{-1}[S]$ is open in $J$ since $\beta$ is continuous, hence $(\beta|^{M})^{-1}[V]$.


问:
How do you prove that the mean of the co-primes of a number is half the number?
Say $n = 6$, The set of co-primes is $\{1, 5\}$, $\text{mean} = 3$

For $n = 9$, the set of co-primes is $\{1, 2, 4, 5, 7, 8 \}, \text{mean} = 4.5$

Question: Prove that the mean of co-primes of $n$, which are less than $n$ is half the number itself.

I computed all values until $10000$ and it seems to hold good.

答:
I assume $n\neq2$. The case $n=2$ should be treated separately (see Goos' comment).

Since $\gcd(a,n)=1\iff\gcd(n-a,n)=1$ it follows that if $C=\{a_1,a_2,\ldots,a_r\}$ is the set of the co-prime positive integers with $n$ (and smaller of $n$) then we can write $$C=\{a_1,a_2,\ldots,a_s,n\color{brown}{-a_s},\ldots,n\color{brown}{-a_2},n\color{brown}{-a_1}\}$$
for $s=\dfrac{r}{2}$.

In your examples for $n=6$, $C=\{1,5\}=\{1,6\color{brown}{-1}\}$ and

for $n=9$, $C=\{1,2,4,5,7,8\}=\{1,2,4,9\color{brown}{-4},9\color{brown}{-2},9\color{brown}{-1}\}$.

Therefore $\sum C=s\cdot n$ and the mean is $\dfrac{\sum C}{2s}=\dfrac{s\cdot n}{2s}=\dfrac{n}{2}$.


问:
prove that $f^{-1} = f*$ as follow
edit:

Let $f: A \to B$. Let $f*$ be the inverse relation, i.e.
\begin{equation*}
f* = \{(y,x) \in B \times A \mid f(x)=y \}.
\end{equation*}
Show that if $f* : B \to A$ is a function,
then $f^{-1} = f*$.

Attempt:

Let $f*: B \to A$ be a function.
Then, $f$ is bijective and hence $f$ is invertible, i.e. $f$ have an inverse, say $f^{-1}$.
It is clear that $f^{-1} \circ f = i_A$
and $f \circ f* = i_B$ where $i$ is the identity function. Hence,
\begin{equation*}
f^{-1} = f^{-1} \circ i_B = f^{-1} \circ (f \circ f*) = (f^{-1} \circ f) \circ f* = i_A \circ f* = f*.
\end{equation*}
Thus, $f^{-1} = f$.

Is the above correct?

答:
First, suppose that $f*$ is a function. That means that each for each $y \in B, \exists!x \in A$ such that $f^*(y)=x.$ Then $(f^* \circ f)(x) = f^*(f(x)) = f^*(y)=x,$ and similarly, $(f \circ f^*)(y) =f(f^*(y))= f(x)=y$, so $f*$ is the inverse of $f$. Or perhaps I should say $f*$ is an inverse of $f$, since you've also been tasked with proving the uniqueness of the inverse.

To prove that the inverse of f is unique, suppose that f has two inverses, g and h. We need to show that g=h. Since g is the inverse of f, that means $g∘f=id_A$  and $f∘h=id_B$. Then $g=g∘id_B=g∘(f∘h)=(g∘f)∘h=id_A∘h=h$.

Title: prove that $f^{-1} = f*$ as follow


答:
If $f_n$ is $M$-Lipschitz (with $M$ that does not depend on $n$) then for every $x,y$ in your domain you have
$$
|f_n(x)-f_n(y)|\leqslant M|x-y|\quad \forall n\in\mathbb{N}.
$$
But the absolute value is a continuous function, as well as the sum of real numbers. Therefore, if $f_n$ converges pointwise to $f$ you can pass to the limit for $n\to\infty$ in the previous relation to get
$$
|f(x)-f(y)|\leqslant M|x-y|.
$$

Title: Sequence of M-Lipschitz functions

Body:
Let $\alpha$ , $\beta \in \mathbb R$ with $\alpha < \beta$, $M \geq 0$ and $(f_n)$ a sequence of M-Lipschitz functions of [$\alpha , \beta$] in $\mathbb R$. If $(f_n)$ converges simply (or "pointwise" in English I think) to a function $f$ on [$\alpha , \beta$], show that $f$ is also M-lipschitz.

We have that $(f_n)$ converges simply so $\forall x \in [\alpha , \beta], \forall \epsilon > 0, \exists n_0$ such that $\forall n > n_0, \lvert f_n(x) - f(x) \rvert \leq \epsilon$

And we also have that the $f_n$ are M-Lipschitz so $\forall x,y \in [\alpha , \beta], \lvert f_n(x) - f_n(y) \rvert \leq M\lvert x-y \rvert$

We want to show that $\forall x,y \in [\alpha , \beta], \lvert f(x) - f(y) \rvert \leq M\lvert x-y \rvert$

I tried to write $\lvert f(x) - f(y) \rvert = \lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert$

Then we have $\lvert f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y) \rvert \leq \epsilon + M\lvert x-y \rvert + \epsilon$ but it is not really the form we wanted and I think I made some mistakes with the quantifiers. Can someone help me ? (The correction solves this exercise just by saying : when going to the limit the inequality stays true. I tried to find something more precise and clear).


问:
Could somebody elaborate "dimensional space" and "hyperplane"?
I am reading a text related to SVM, and the mathematical language is giving me a little hard time.

Here training vectors xi
  are mapped into a higher (maybe
  infinite) dimensional space by the
  function $\theta$. SVM finds a linear
  separating hyperplane with the maximal
  margin in this higher dimensional
  space.

I do not understand the term "dimensional space" in this case. Drawing on a paper is 2D. We are living in 3D space. In Mathematics, when we say "higher dimensional space", what are we actually implying in Mathematics?

Another term "hyperplane" is also giving me a bit of hard time to understand. Is it simply just a 2D plane? I try to search for its definition, and most of the time, I get a term that leads to many more terms (and more confusing) Frankly, mathematical language is difficult for me.

Could somebody simplify and relate "dimension" and "hyperplane" in the text above in an easier way to understand?

Thank you very much.

答:
Karl, higher dimensions are defined so as to be analogous to 2-D and 3-D. This means analytical definitions of orthogonality, distance, and so on must be preserved.

To "visualise" 4-D and up, you can start by visualisation exercises with 4-cubes or think about arrays in programming. $array[i][j][k][l] is a four-dimensional array for example. If you took all of the possible combinations of 0 and 1 in that array -- for instance [0,1,0,0], [1,1,0,0], [0,1,1,1] and so on -- you would have a data structure that's equivalent to a 4-cube or tesseract. You can generate these in R using the combn() function.

Here's another example: imagine a 4-sequence of 4 stock prices progressing in parallel, second by second. Record all the seconds for a single trading day and you have a day-long 4-D vector. That's four-dimensional data. I could have looked at 500 stock prices progressing in parallel and that would be 500-dimensional data.

The Titanic data set in R is another accessible example of 4-D data (age, sex, class, survival).

As for hyperplanes, think about this:

a line splits a circle

a plane splits a sphere

a hyperplane splits a ... whatever comes next.

Squiggly or blobby higher-dimensional shapes are called "manifolds", by the way.

Below are examples of functions that map data to higher dimensions.

$f(x,y) = (x, x, x, y, y)$. $f$ is like copying and pasting whole columns in a table or array. $f$ maps 2 dimensions to 5.

$g(x,y) = (x, \ x+1, \ x^2, \ x \cdot y, \ y^3 + y^5 - y^4 + 55)$.  $g$ maps 2 dimensions to 5.

$h(a, b, c, d, e) = (a + b + c + d + e, \ a \cdot b \cdot c \cdot d \cdot e)$. $h$ maps 5 dimensions to 2.

Here is another function $j$ that maps 5 dimensions to 2: $j(a, b, c, d, e) = (a, e)$. $j$ is like erasing the middle columns of your data. In math books this may be called projection.

Here is one, $\theta$, that's used in a common SVD example: separating an inner ring from an outer ring. $\theta(x,y) = (x^2, \ \sqrt{2} \cdot x \cdot y, \ y^2)$ (that's mapping how many to how many dimensions?)

Here is one more function just to demonstrate that letters, numbers, or greeks can be used: $\gamma_{\eta}(a, b, c, d, e) = (0, 5)$. This one has a complicated name but it maps any 5-dimensional input to the same 2-D point, which is a "trivial" thing to do.

As you can see you just count the commas inside the parentheses to figure out how many dimensions you've got (plus one).

Hope this helps.

ContentLicense: CC BY-SA 2.5


答:
I'm posting a separate answer because I think Inceptio's is far too complicated for a problem this simple.

Note that $a$ is relatively prime to $n$ if and only if $n - a$ is relatively prime to $n$.  Also, generally $a \ne n - a$ (though there is one exception, which you have to deal with separately.  Can you find the exception?)
Anyway, this means we can make a list of everything relatively prime to $n$ in pairs of the form $(a, n - a)$, i.e. our list is:
$$
a_1, n - a_1, a_2, n - a_2, a_3, n - a_3, \cdots , a_k, n - a_k
$$

We can also guarantee that this list is all-inclusive (why?) and that no number appears twice (how?).
Then simply add up the numbers in the list and divide by the total.  You should get that the average is $\frac{n}{2}$.

Title: How do you prove that the mean of the co-primes of a number is half the number?

Body:
Say $n = 6$, The set of co-primes is $\{1, 5\}$, $\text{mean} = 3$

For $n = 9$, the set of co-primes is $\{1, 2, 4, 5, 7, 8 \}, \text{mean} = 4.5$

Question: Prove that the mean of co-primes of $n$, which are less than $n$ is half the number itself.

I computed all values until $10000$ and it seems to hold good.


问:
number of matrices in $GL_{3}(\mathbb{Z}_{3})$ such that $[1 \ 1 \ 1]^{T}$ is an eigenvector.
Find the number of matrices in $GL_{3}(\mathbb{Z}_{3})$ such that $[1 \ 1 \ 1]^{T}$ is an eigenvector.

This question was asked to me in an interview which I wasn't able to answer.

I approached as follows,

If $A$ is a matrix in $GL_{3}(\mathbb{Z}_{3})$  such that $[1 \ 1 \ 1]^{T}$ is an eigenvector then $0$ can't be an eigenvalue as $A$ is invertible. So $A$ can have $1$ and $2$ as eigenvalues. Let $v=[1 \ 1 \ 1]^{T}$.

If $1$ is an eigenvalue of $A$ then we have $Av=v$ and if $2$ is an eigenvalue of $A$ then $Av=2v$.

So we need to find the number of matrices $A$ in $GL_{3}(\mathbb{Z}_{3})$  such that $Av=v$ and $Av=2v$.

Then I tried to find out such A's but I got stuck at a point.

Can you please let me know how to solve this question ? Thanks.

答:
Daniel Schepler's +1 answer maps probably the simplest route to the destination.

As an alternative I proffer the use of group actions on the projective plane $\Bbb{P}^2(\Bbb{Z}_3)$. There are $3^2+3+1=13$ elements on $\Bbb{P}^2(\Bbb{Z}_3)$ — the lines through the origin in $\Bbb{Z}_3^3$.

It is known (and easy to show) that $G=GL_3(\Bbb{Z}_3)$ acts transitively on that projective plane. The question asks about the stabilizer of the line spanned by $L=(1,1,1)^T$. By the orbit-stabilizer theorem
$$
\operatorname{Stab}_G(L)=\frac{|G|}{13}.
$$
I trust you know $|G|$ already :-)


问:
Prove an equation has a solution in within specific interval (IVT)
Let $b_1,b_2,\dots,b_n$ be positive real numbers s.t $b_1+b_2+\dots+b_n=n$.
How can I prove that the equation: $\sum_{k=1}^n k \cdot \ln(b_k)\cdot x^{k-1}\cdot b_k^{x^k}  = 0$
has a solution over the interval of $[0,1]$?

答: Define $\displaystyle f(x) := \sum_{k = 1}^n b_k^{x^k}$. Note that $\displaystyle f(0) = \sum_{k = 1}^n b_k^0 = n$ and $\displaystyle f(1) = \sum_{k = 1}^n b_k = n$. Thus, using Rolle's Theorem, we can conclude that for some $c \in [0,1]$, $f'(c) = 0$. In other words, $\displaystyle f'(x) = \sum_{k =1}^n k \cdot \ln(b_k) \cdot x^{k-1} \cdot b_k^{x^k} = 0$ has a solution in $[0,1]$.


问:
Does (A and B) entails (A if and only if B) and what is intuition?
Is the next statement correct: (A∧B)⊨(A⇔B) ?

Formal definition of entailment is this: α⊨β if and only if, in every model in which α is true, β is also true.

I used a truth table to show that there is only one model {A=True, B=True} where (A∧B)=True. (A⇔B) is also True in this model. It means (A∧B)⊨(A⇔B) is correct.

There exist model2 = {A=False, B=False}, (A⇔B) is True in model2. But (A∧B) is not True in model2.

Can you please provide intuition on what does it mean? Does it extend our knowledge base (A∧B) if we add a new statement (A⇔B)?

答:
The statement is correct and you have shown it via truth table. Knowing that $A \land B$ implies that both $A$ and $B$ are true, thus $A \iff B$ follows, i.e. it does not extend our knowledge base if $A \iff B$ is added as a new statement, since we already know both $A$ and $B$ are true.

There is a different way to derive $A\iff B$ from $A\land B$, that is, via natural deduction:

\begin{align}
 1. &\quad A\land B&\text{Premise}\\
 2. &\quad A &\text{Simplification}\\
 3. &\quad B &\text{Simplification}\\
 4. &\quad \neg B \lor A &\text{Conjunction}\\
 5. &\quad \neg A \lor B &\text{Conjunction}\\
 6. &\quad B \implies A &\text{Material Implication}\\
 7. &\quad A \implies B &\text{Material Implication}\\
 8. &\quad A \iff B &\text{Material Equivalence}
\end{align}

or, if your definition of biconditional includes the alternate definition $(A \land B)\lor (\neg A \land \neg B)$ (instead of only the standard $(A \implies B) \land (B \implies A)$), then the conclusion can be drawn from one simple use of the Conjunction rule.


问:
Is $\sup\{ \max\{f(x)\}\} = \sup\{f(x)\}$?
In some problem from my differential equations course, I stated that, given $f(x)$ a continuous function defined in a compact set $K$,

$$\sup_{x\in K} \{ \max_{x\in K}\{f(x)\}\} = \sup_{x\in K}\{f(x)\}.$$
Is this true? Is it trivial? If it's false, could you give a counterexample? Any help will be appreciated, thanks in advance.

答: $$\sup_{x\in K}f(x)=\max_{x\in K}f(x)=\sup_{x \in K}\max_{x\in K}f(x)$$ where the first equality is true since $K$ is compact(Weierstrass theorem) and the second is obvious since $\max_{x\in K}f(x)$ is constant


问:
Disproving that $\mathbb N$ is a complete Set
I want to disprove that $\mathbb N$ is a complete Set.

It makes sense to me that there are Cauchy Sequences in $\mathbb Q$ that converge to a number that is irrational and therefore not in $\mathbb Q$ (e.g. to $\sqrt{2}$), which makes the Set $\mathbb Q$ incomplete.

Is there a way to use this property when working with natural numbers?

Also, there is the Definition, that if every nonempty subset $A \subseteq S$, where $S$ is an ordered field has a $\sup(A) \in S$ and an $\inf(A) \in S$, the Set $S$ is complete (sorry if I understand this incorrectly).

Using this I say:

The set $A=\{5-x:x\in \mathbb N\}$ proves that $\mathbb N$ is incomplete, because $\inf(A) = -\infty \notin \mathbb N$

Is this a valid claim?

答: In fact $\mathbb N$ is complete (when endowed with the metric induced by the usual one of $\mathbb R$). A Cauchy sequence taking values in $\mathbb N$ is eventually constant and therefore converges.


答:
The only way to get a cauchy sequence of natural numbers is to eventually have $|x_{m}-x_{n}| = 0 < \epsilon$ for $m \geq n \geq N$.

Since otherwise there exist $m_{1} \geq n_{1} \geq N$ such that $|x_{m_1} - x_{n_1}| \geq 1 > \epsilon = \frac{1}{2}$ and then the sequence is not a cauchy sequence. This implies that every cauchy sequence in $\mathbb{N}$ converges and so $\mathbb{N}$ is a complete set.

Title: Disproving that $\mathbb N$ is a complete Set

Body:
I want to disprove that $\mathbb N$ is a complete Set.

It makes sense to me that there are Cauchy Sequences in $\mathbb Q$ that converge to a number that is irrational and therefore not in $\mathbb Q$ (e.g. to $\sqrt{2}$), which makes the Set $\mathbb Q$ incomplete.

Is there a way to use this property when working with natural numbers?

Also, there is the Definition, that if every nonempty subset $A \subseteq S$, where $S$ is an ordered field has a $\sup(A) \in S$ and an $\inf(A) \in S$, the Set $S$ is complete (sorry if I understand this incorrectly).

Using this I say:

The set $A=\{5-x:x\in \mathbb N\}$ proves that $\mathbb N$ is incomplete, because $\inf(A) = -\infty \notin \mathbb N$

Is this a valid claim?


答:
If you have a complete metric space $X$ (such as $X=\mathbb R$) and a closed subset $Y$ in it (say $Y=\mathbb N$), then $Y$ will be a complete subspace of $X$.

Proof: Let $(x_n)$ be a Cauchy sequence in $Y$. It follows that it is also a Cauchy sequence in $X$. From the completeness of $X$, it follows that $(x_n)$ converges to some element $x=\lim_{n\to\infty}x_n\in X$. However, as all $x_n\in Y$ and $Y$ is closed, then $x=\lim_{n\to\infty}x_n\in Y$ - in other words, $(x_n)$ converges in $Y$.

Thus, $Y$ is complete. $\blacksquare$

In particular, $\mathbb N$ is complete as it is a closed subset of $\mathbb R$.


问:
A property of the midpoint of the hypotenuse in a right triangle

Prove that in a right angled triangle the mid point of the hypotenuse is equidistant from its vertices.

I have asked similar question but with no satisfactory result.

So I solved it by myself. But I am not sure whether its correct or not. So please help me.

Solution:

Let in triangle $\Delta OAB$ right angled at $A$. Where $O$ is origin and $\vec{a}$ is vector along $OA$ and $\vec{b}$ is vector along $OB$.

Let $C$ be the mid point of hypotenuse $OB$.

We have to prove $OC = BC = AC$.

As $C$ is mid point so $OC = BC =  \frac{\vec{b}}{2}$

$AC =  \frac{\vec{b}}{2} - \vec{a}$

I am not getting the value of AC as $\frac{\vec{b}}{2}$

Sorry because I am using my mobile unable to draw and include image.

Anything wrong please tell me with explanation. Thanks.

答:
Let the triangle be ABC with angle B = 90 and AC as hypotenuse. Join B to midpoint of AC. Midpoint of AC is called D.

Let angle C = x. So, angle A = 90 - x.
Let angle DBC = y. So, angle DBA = 90 - y

Now, according to Sine rule of triangle, in ∆DBC:

BD/Sin(C) = DC/Sin(angle DBC)
==> BD/Sin(x) = DC/Sin(y)
==> BD = (DC * Sin(x))/Sin(y) -------------------- (equation 1)

Now, according to Sine rule of triangle, in ∆DBA:

BD/Sin(A) = AD/Sin(angle DBA)
==> BD/Sin(90 - x) = AD/Sin(90 - y)
==> BD = (AD * Cos(x))/Cos(y) -------------------- (equation 2)

We know that AD = DC, So:
BD = (DC * Cos(x))/Cos(y) -------------------- (equation 3)

So, now from equation 1 and equation 3, we get:

(DC * Sin(x))/Sin(y) =  (DC * Cos(x))/Cos(y)
==> Sin(x)/Sin(y) = Cos(x)/Cos(y)
==> Sin(x)Cos(y) = Cos(x)Sin(y)
==> Sin(x)Cos(y) - Cos(x)Sin(y) = 0
==> Sin(x - y) = 0
==> x - y = 0
==> x = y

So, angle C = angle DBC.
So, ∆DBC is isosceles triangle and so,
BD = DC.

So, BD = DC = AD --------- (Proved)

------End of proof--------


问:
The values of the trigonometric functions of $45^\circ+\alpha$
I have a new idea about the question I posted earlier today.

If $\sin\alpha=\dfrac{4}{5}$ and $\alpha\in(45^\circ;90^\circ)$, find the values of the other trig functions of $\alpha$ and $45^\circ+\alpha$.

Since $\alpha\in\left(45^\circ;90^\circ\right)$, the angle $45^\circ+\alpha$ is obtuse. I was thinking about using the unit circle. The angle $\measuredangle KOO_x^+=\alpha$ and $\measuredangle POO_x^+=45^\circ+\alpha$ where $O$ is the origin and $O_x^+$ denotes the positive $x$-axis. Cosine and sine corespond, respectively, to the $x$- and $y$-coordinate. I have already calculated that $\cos\alpha=\dfrac{3}{5}$ so $K$ is defined as $K(\dfrac{3}{5};\dfrac{4}{5}).$ Can we use some geometric methods from here to find the coordinates of $P$?

答:
The slope of $\text{OK}$ $= \tan\theta = 4/3$, which implies that slope of $\text{OP}$ would be $\tan (\theta+45^\circ)$ which would be equal to $$ \frac{1 + 4/3}{1-4/3} = -7$$
Finding intersection of $y=-7x$ and $x^2+y^2=1$, we get coordinates of $\text{P}$ as $\left(\dfrac{-1}{\sqrt{50}}, \dfrac{7}{\sqrt{50}}\right)$.

Another purely geometrical way would be taking centre as $\text{K}$, draw a circle of radius $\pi/4$, i.e.,
$$\left(x-\frac35\right)^2 + \left(y-\frac45\right)^2 = \left(\frac\pi4\right)^2$$ And find intersection of this circle with circle of unit radius with origin as centre.

Finding intersection, $\newcommand{\S}{\text{S}}$ $\S_1 - \S_2$:
$$3x+4y = 5 -\frac{5\pi^2}{32}$$
Then find the intersection of this line with any of the circles, you would get $2$ points of which one would be $+45^\circ$ and other would be for $-45^\circ$.


问:
Changing order of derivatives
I would like to rewrite the following expression

$$\frac{d^i}{dx^i}\left\{f(x)\left[\frac{d^jf(x)}{dx^j}\right]\left[\frac{d^kf(x)}{dx^k}\right]\right\}$$

into the form

$$D f(x)^3,$$

with $D$ being an unknown differential operator (to be determined).

$f(x)$ is a real-valued function.

答:
I'll take your comment that you want $D$ "to contain only derivatives, sums and products of derivatives of various orders" at face value, and I'll throw in real-valued coefficients, too. Then what you want is impossible.

Any such expression can be multiplied out to obtain a linear combination of derivatives of various orders. To have the right behaviour under scaling of $x$, the operator can only contain the derivative of order $i+j+k$. (Another way to think about this is to "check the units/dimensions", with each derivative contributing an inverse unit of length). But clearly you can't get all the expressions you want to represent just from scalar multiples of the derivative of order $i+j+k$.

Title: Changing order of derivatives


问:
proving the limit when x tends to +,- infinity of arctan(x)

Would love if someone could look at my proof.

答:
Looks fine to me, every value you chose is well defined during the entire proof.
I would denote though that when you let $\epsilon\ge\pi/2$ ,

you want to keep the $\ge$ , as concluding that:

$\pi/4<\pi/2<\epsilon$

is false when choosing $\epsilon=\pi/2$


问:
Can every continuous curve (or surface) be parametrised?
I read here (on page 3) that

"Every plane parametric curve (or surface) can be expressed as an implicit curve (or surface). Some, but not all implicit curves (or surfaces) can be expressed as parametric curves (or surfaces)".

It also states that this fact comes from Algebraic Geometry.

I was interested in the parameterisation of continuous curves (or surfaces). Can every continuous curve (or surface) be parameterised? If not, then under what conditions they can't be?

答:
There are obstructions to algebraic curves (and more general varieties) being parametrizable. I will discuss the case of projective curves for simplicity, but all of the main ideas extend to higher dimension. Let us even assume we have a smooth projective curve $C$.

A parameterization for $C$ amounts to a surjective morphism $\phi : \mathbb P^1 \to C$. By the Riemann--Hurwitz formula, this forces the genus of $C$ to be $0$ (and in fact forces $\phi$ to be an isomorphism). In the case of smooth curves over the complex numbers, the geometric, arithmetic, and topological notions of genus all agree, so you can think of this as being the "number of holes" appearing in the compact Riemann surface associated to a smooth projective curve over $\mathbb C$. When the curve is not smooth, more care has to be taken, but I'll ignore those subtleties here.

So a good class of examples of curves that cannot be parametrized are smooth curves of nonzero genus, e.g. elliptic curves. The example given in Jan-Magnus's comment above is an elliptic curve. These are the smooth projective curves of genus $1$. In characteristic $>3$, these are precisely the curves of the form $y^2=x^3+ax+b$ where the RHS does not have a double root. Curves of higher genus will also give examples.


问:
Difference between topological space and topology
I am studying for my exam in metric and topological spaces, but got confused about two definition. I felt like they were contradicting each other. I know that a topology on X is a collection of open subsets of X which satisfy three properties. The first property is that X and the empty set are elements of the topology (from which follows that they are both open right?). But then I read further in the book and found this definition

Definition. Let X be a topological space. Then

X, empty set are closed in X

... (not relevant for my question)

... (not relevant for my question)

This really confused me, because I thought that X and the empty set had to be open sets??
Could someone please clarify this? Thank you (in advance)!

答:
You are wrong when you claim that “a topology on $X$ is a collection of open subsets of $X$ which satisfy three properties”. A topology on $X$ is a colletion $\tau$ of sets which satisfy three properties, and then we say that a subset $A$ of $X$ is open when (and only when) $A\in\tau$.

And we say that $F\subset X$ is closed when $F^\complement$ is open. So, since $\emptyset^\complement=X$ and $X\in\tau$, $\emptyset$ is closed. And, since $X^\complement=\emptyset$ and $\emptyset\in\tau$, $X$ is closed.


答: Since $X=\emptyset^\complement\;$ and $\;\emptyset=X^\complement\;$ so in any topological space they are both closed and open sets(clopens). In a topological space these are the only clopens $\iff$ the space is connected

Title: Difference between topological space and topology

Body:
I am studying for my exam in metric and topological spaces, but got confused about two definition. I felt like they were contradicting each other. I know that a topology on X is a collection of open subsets of X which satisfy three properties. The first property is that X and the empty set are elements of the topology (from which follows that they are both open right?). But then I read further in the book and found this definition

Definition. Let X be a topological space. Then

X, empty set are closed in X

... (not relevant for my question)

... (not relevant for my question)

This really confused me, because I thought that X and the empty set had to be open sets??
Could someone please clarify this? Thank you (in advance)!


答:
A subset of a topological space $X$ can be both open and closed. The properties are not mutually exclusive. $\varnothing$ and $X$ are open subsets of $X$ by definition. The closed sets are those whose complement is open. Therefore, $\varnothing$ and $X$ are also closed.

Comments) You should distinguish $X$ from $\tau$. We call a set $X$ together with its topology $\tau$ a topological space. We often write it as $(X,\tau)$ to clarify this. $\tau$ is a family of every open subset of $X$ whereas $X$ is just a set. But in many books, they just simply call $X$ a topological space assuming $\tau$ is understood in the context.

Refer to Clopen set


问:
Prove the derivative of the delta distribution is not a 'regular' distribution.
In the terminology of Gel'fand, Generalized Functions Volume 1, a regular distribution, $\Lambda_f$, is one that is obtained from a locally integrable function, $f$, by integration of the function times a test function:
$$\Lambda_f(\phi)=\int_{\mathbb{R}^n}\!f(x)\phi(x)\,dx\qquad(\phi\in C^\infty_c(\mathbb{R}^n)).$$
I know that the Dirac delta distribution, defined by
$$\delta_p(\phi)=\phi(p)\qquad(\phi\in C^\infty_c(\mathbb{R}^n),\,p\in\mathbb{R}^n)$$
is not a regular distribution. For the $n=1$ case, there is a proof in Gel'fand on page 4 of Volume 1, there is this post: How to prove that the Dirac delta is not a function?, and it is not hard to prove using the $C^\infty$-Urysohn Lemma, even for general integer $n$. (Find a mollified sequence of $C^\infty_c(\mathbb{R}^n)$ functions converging pointwise to the characteristic function of $\{p\}$, use Dominated Convergence and the fact that the Lebesgue measure of $\{p\}$ is zero, to get a contradiction.)

I believe that it must be true as well for all derivatives of the delta distribution. Again, for the $n=1$ case, there is this website from Professor Tao: https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/ in which one finds:

Exercise 11 (Derivative of the delta function) Let . Show that the functional  for all test functions  is a distribution which does not arise from either a locally integrable function or a Radon measure. (Note how it is important here that  is smooth (and in particular differentiable, and not merely continuous.) The presence of the minus sign will be explained shortly.

I have been trying to prove that $\delta_p'$ is not regular in $\mathbb{R}^1$; that is, that there does not exist $f\in L^1_{\rm loc}(\mathbb{R})$ such that $\delta_p'=\Lambda_f$, hoping that such a proof would be generalizable to $\mathbb{R}^n$. However, I haven't had any luck even with $n=1$. So my simplified first question, setting $p=0$ and $n=1$, is "why can't $\delta'$ be $\Lambda_f$ for some $f\in L^1_{\rm loc}(\mathbb{R})$?" If that can be answered, then the next question is can it be extended to $\mathbb{R}^n$? Is it true that, given any multi-index $\alpha$, there is no $f\in L^1_{\rm loc}(\mathbb{R}^n)$ such that $D^\alpha\delta_p=\Lambda_f$?

In working on the $n=1$, $p=0$ case, I start with the obvious:
\begin{equation}\tag{1}
  -\phi'(0)=-\delta(\phi')=\delta'(\phi)=\Lambda_f(\phi)=\int_{-\infty}^\infty\!f(x)\phi(x)\,dx
\qquad(\phi\in C^\infty_c(\mathbb{R})),
\end{equation}
and look for a sequence of test functions that generate a contradiction.
The trick of letting $\{\phi_k\}$ be a mollified sequence in $C^\infty_c(\mathbb{R})$ such that $\phi_k$
converges pointwise to $\chi_{\{0\}}$ doesn't seem to help because it appears that $\phi'_k(0)$ will
likely have to approach $0$ and therefore not generate a contradiction. Letting $\{\phi_k\}$ just
be a bump that narrows down to $0$ but whose height remains constant, also doesn't seem to work, since again, the derivatives will be $0$ at the origin. I also hoped I might be able to make use of the fact that
if $\phi_1$ is a fixed element of $C^\infty_c(\mathbb{R})$ such that
$$\int_{-\infty}^\infty\!\phi_1(x)\,dx=1,$$
then any $\phi\in C^\infty_c(\mathbb{R})$ can be expressed as
\begin{equation}\tag{2}
  \phi=\phi_0+\Biggl[\int_{-\infty}^\infty\!\phi(x)\,dx\Biggr]\phi_1,
\end{equation}
where $\phi_0\in C^\infty_c(\mathbb{R})$ and is the derivative of a function in
$C^\infty_c(\mathbb{R})$,
as used by Gel'fand on page 40 of Volume 1; however, I couldn't find a way where that would help.

答:
If $\delta'$ was in $L^1_{\mathrm{loc}}$, then it would have a continuous primitive which would have the properties of the Dirac delta, which is a contradiction.

You can iterate this reasoning $n$ times to get the same result for the higher derivative $\delta^{(n)}$.


问:
Pose estimation by aligning non-collinear points
As shown in Fig.1, consider $3$ line segments of  equal length each containing $4$ points. Let $j = 1,\dots,3$ and $i = 1,\dots,4$ be the index for the line segment and the points on the line segment respectively.  $^j\mathbf{T}$ is the (rigid) transformation matrix   that describes the pose (position and orientation) of the $j^\mathrm{th}$ line segment with respect to origin $O$. Similarly $^j\mathbf{x}_i = \,^{\!j}(x_i, y_i)$ is the coordiate of the $i^\mathrm{th}$ point on $j^\mathrm{th}$ line segment with respect to origin.

As illustrated in Fig.2 suppose that after measurement only the pose of the second line segment  deviates from its true value and is therefore  denoted by $^2\mathbf{T'}$.

What I know after the measurement (Fig.2) are as follows:

Coordinates of $^1\mathbf{x}_i$, $^2\mathbf{x'}_i$ and $^3\mathbf{x}_i$ for $i = 1,\dots,4$ with respect to $O$.

Values of $^1\mathbf{T}$, $^2\mathbf{T'}$, $^3\mathbf{T}$

Constraints

The points on the line segment are fixed. That is when I transform the line segment, the points move together with the line segment.

My goal is to displace (translate + rotate) the second line segment with pose $^2\mathbf{T'}$ such that the points become collinear as illustrated in Fig.1. Or in other words displace the second line segment such that $^2\mathbf{T'}$ coincides with $^2\mathbf{T}$.

Approach

This can be treated as an optimization problem where we wish to find $^2\mathbf{T}$ such that the shortest distance between the points $^2\mathbf{x'}_i$ and the straight line joining $^1\mathbf{x}_i$ and $^3\mathbf{x}_i$ is minimized. The problem is if I have over 10000 points and over 30 line segments, the computational time is huge.

Since the points are non-collinear, the grey dotted line  connecting the points form an angle at points that lie on second line segment. Is it possible to estimate $^2\mathbf{T}$ iteratively just by using the angle information formed by this grey dotted line as shown in Fig.2? For instance I can assign a vector to each point on the second line segment that divides the angle in half. Then use this vector to find how or in which direction the line segment should be moved.

Any comments or suggestions are welcome.

答:
Optimization problem

Not sure if the cost function for the optimization problem is formulated properly but here is the basic idea:

If I define the points $\{^{2}\mathbf{x_1}, ^{2}\mathbf{x_2}, \dots ^{2}\mathbf{x_4}\}$ on second line segment as $^{2\!}X$, what I am looking for is to find $\{^{2}\mathbf{x_1'}, ^{2}\mathbf{x_2'}, \dots ^{2}\mathbf{x_4'}\}$ or $^{2\!}X'$ such that
$$^{2\!}X' = [\mathbf{R}\, \vert\,\mathbf{t}]_{3\times 3} \, ^{2\!}X$$
or
$$
\underbrace{
\begin{bmatrix}
^2 x_1' & ^2 x_2' & \dots & ^2 x_4'\\
^2 y_1' & ^2 y_2' & \dots & ^2 y_4'\\
1      & 1   &   \dots    & 1
\end{bmatrix}}_{^{2\!} X'}
=
\underbrace{
\begin{bmatrix}
r_{11} & r_{12}  & t_x\\
r_{21} & r_{22}  & t_y\\
0      &   0    & 1
\end{bmatrix}}_{\mathbf{^2 T'}}
\,
\underbrace{
\begin{bmatrix}
^2 x_1 & ^2 x_2 & \dots & ^2 x_4\\
^2 y_1 & ^2 y_2 & \dots & ^2 y_4\\
1      & 1   &   \dots    & 1
\end{bmatrix}}_{^{2\!} X}
$$
Here the components of Rotation Matrix $\mathbf{R}$ and $\mathbf{t}$ are also unknown.

By using the points on first and the third line segment I obtain the equations of straight lines that passes through these points or in other words, the equations of lines joining the corrosponding points on first and third line segments. By denoting the straight line with $f(x)$, for a line segment with 4 points (as shown in the illustration) we obtain
$$
\begin{bmatrix}
f_1(x)\\
f_2(x)\\
\vdots\\
f_4(x)
\end{bmatrix}
$$
The cost function then takes the form
$$
\begin{pmatrix}
\begin{bmatrix}
f_1(^2 x_1)\\
f_2(^2 x_2)\\
\vdots\\
f_4(^2 x_4)
\end{bmatrix}
-
\begin{bmatrix}
^2 y_1\\
^2 y_2\\
\vdots\\
^2 y_4
\end{bmatrix}
\end{pmatrix}^2
\leftarrow \mathrm{min}
$$
that satisfies the condition
$$^{2\!}X' = [\mathbf{R}\, \vert\,\mathbf{t}]_{3\times 3} \, ^{2\!}X$$
as mentioned above.


问:
Calculating a term using a multilinear map
How to calculate the value of the term

$\Delta u:=u_{xx}+u_{yy}+u_{zz}=\large\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}+\frac{\partial^2u}{\partial z^2}$

for the function $u$ on $\mathbb{R}^3 \backslash \{0\}$ with

$u(x,y,z):=\large\frac{1}{\sqrt{x^2+y^2+z^2}}$  ?

答:
Answer

$\nabla^2{\left(\frac{1}{\sqrt{x^2 + y^2 + z^2}} \right)} = -4\,\pi\,\delta{\left(\mathbf{x}\right)}$

Context

The Laplacian of OP's $u$ has a singular nature. This singular nature is not captured by @user1337's answer.  As explained in [1], we can use a limiting process to integrate around the singularity as follows.

Derivation

By $B_r{(\mathbf{0})} \subset \mathbb{R}^3$, I denote an open ball of radius $r > 0$ centered at a point $\mathbf{0}$. By $S^2{(r)}$, I denote a 2-sphere of radius $r$. Then,
\begin{align}
\int_{B_r{(\mathbf{0})}} \nabla^2{u} \left|d\mathbf{x}^3\right|
&=
\int_{S^2{(r)}} \mathbf{n}\cdot  \boldsymbol{\nabla}{u}  \, dS
&&
\text{divergence theorem}
\\
&=
\int_{S^2{(r)}} \mathbf{n}\cdot\boldsymbol{\nabla}{\frac{1}{\sqrt{x^2 + y^2 + z^2}}}  \, dS
&&
\text{substituting given}
\\
&=
\int_{S^2{(r)}} \mathbf{\widehat{r}}\cdot\boldsymbol{\nabla}{\frac{1}{r}}  \, \left[r^2 \,\sin{\theta}\, d\theta\,d\phi \right]
&&
\text{radial coordinates}
\\
&=
\int_{S^2{(r)}} \mathbf{\widehat{r}}\cdot
\frac{\partial \frac{1}{r}}{\partial r}   \mathbf{\widehat{r}}
\, \left[r^2 \,\sin{\theta}\, d\theta\,d\phi \right]
&&
\text{gradient }
\\
&=
-\int_{S^2{(r)}} \mathbf{\widehat{r}}\cdot
  \frac{1}{r^2}    \mathbf{\widehat{r}}
\, \left[r^2 \,\sin{\theta}\, d\theta\,d\phi \right]
&&
\text{derivative }
\\
&=
-\int_{S^2{(r)}}
   \sin{\theta}\, d\theta\,d\phi
&&
\text{simplifying }
\\
&=
-4\,\pi
\end{align}
From [1], ``It is now established that $\nabla^2 {\left(1/r\right)}$ is 0 for $r\neq 0$, and that its volume is $-4\,\pi$. Consequently, we can write the improper [but mathematically justifiable] equation $\nabla^2{(1/r)} = -4\,\pi\,\delta{\left(\mathbf{x}\right)}$''

Bibliography

[1] Jackson, John D. Classical Electrodynamics


问:
Prove inequality with powers
$4^{79}<2^{100}+3^{100}<4^{80}$

The only thing I can think of is the following:
$\ln(3) = 1.098612288$ and $\ln(4) = 1.38629436112$ so

$\frac {\ln(3)}{\ln(4)} = 0.79248125036$ so

$\frac {79}{100} < \frac {\ln(3)}{\ln(4)} < \frac {80}{100} $

I don't know how to continue (and algebra was never my strong point!!).

Thank you!

答:
Set $\alpha := 200 \frac{\log 3}{\log 4} = 158.496...$. Writing all terms as powers of $2$ your inequalities become
$$
2^{158} < 2^{100} + 2^\alpha < 2^{160}.
$$
The first inequality is clearly satisfied (since $\alpha > 158$).

For the second inequality, observe that
$$
2^{160} - 2^{\alpha} = 2^{\alpha}(2^{160 - \alpha} - 1) > 2^{\alpha} > 2^{100},
$$
where the first inequality follows from the fact that $160-\alpha > 1$, hence
$(2^{160 - \alpha} - 1) > 2 - 1 = 1$.

Title: Prove inequality with powers


问:
Combining Sinusoidal waves
Is there a general way to add sinusoidal waves? I'm trying to covert the following expression

$A_1 \cos(x+y)+A_2\sin(x-y)$

into

$A_3\cos(z)$ where $z$ is a function of $x$ and $y$

答:
Taking a simpler example, $A_1 \cos(x)+A_2\sin(x)$, is it possible to write this as $A_3 \cos(x)$? No, because $\sin$ and $\cos$ are linearly independent functions and you are attempting to say $\cos$ is sufficient to span the space of all their linear combinations.

In special cases, yes, e.g. take $A_1=0$ or $A_2=0$.

Title: Combining Sinusoidal waves


问:
Algebra question from HK Math Preliminary Contest 2020
The contest has already concluded more than 6 months ago. I am trying to figure out a smart way to solve the following question below:
Let $k$ be an integer. If the equation
$$(x-1)|x+1|=x + \frac{k}{2020}$$
has 3 distinct real roots, how many different possible values of $k$ are there?

Here is my initial try:
I have split the LHS using test point to get the piece wise function as below:
for $x\geq1$:
$$
x^2-1 = x + \frac{k}{2020},
$$
for $x<-1$:
$$
x^2 -1 = -\left(x+ \frac{k}{2020}\right).
$$
Plotting the LHS as upward opening parabola shifted down by 1 units and RHS as 2 straight lines with slope $+1$ and $-1$ with $y$-intercept of $\frac{k}{2020}$ and $-\frac{k}{2020}$.
To meet the 3 solutions criteria, the parabola cuts the positively sloped line at 2 points while being tangent to the line at $x=-1/2$ with negative slope.

How should I graphically get to the solution from here?

答:
The function $$f(x)=(x-1)|x+1|-x$$
passes through any horizontal line three times provided that the horizontal line is in the range of $y\in(-\frac54,1)$.  Therefore, it suffices to find the number of integers $k$ such that $-\frac54<\frac k{2020}<1.$

Solving the inequality, we get $$-2525<k<2020$$ and we see that $k$ can take on any one of $\boxed{4544}$ integers.


问:
Is there a pattern/formula to this?
I came across a nice "sequence" on a mathematics Facebook page and it intrigued me somewhat. This is not homework or anything of the sort, just something fun to start the year off I guess.

Observe the following:

\begin{align}
(1)^1 & =
1
\\[5 mm]
(8 + 1)^2 & =
81
\\[5 mm]
(5 + 1 + 2)^3 & =
512
\\[5 mm]
(2 + 4 + 0 + 1)^4 & =
2401
\end{align}

Some questions came to my mind almost immediately.

In general, is there a pattern/formula to this? For example, can we find the next (5-digit) number easily? Does it even exist? In particular, can we find a formula for the n-digit number?

For some of the numbers, the solutions are non-unique. For example, for the first power, obviously any digit would fit. Can we generalise this observation? In particular, do we know if the solution for every n-digit number, if it exists, is always non-unique?

P.S. The many brilliant minds on Math SE have never failed to amaze me, so I thought I should put this one up here to see what you guys can come up with :)

Title: Is there a pattern/formula to this?

Body:
Notice that we always have:

$$10^k = 1\underbrace{00...0}_{k \text { zeroes}}$$

which has $k+1$ digits. Hence for $n^k$ to have exactly $n$ digits, we must have $n < 10$.

For fifth powers, the first $9$ fifth powers are (http://oeis.org/A000584):

$$1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049$$

Now we look at the digit sums of $16807, 32768, 59049$, which are $22, 26, 27$ respectively, and are not equal to $7,8,9$. Hence we cannot have $(1+6+8+0+7)^5 = 16807$, etc.

A further question is, are there any more numbers in this sequence? If $9^k$ has less than $k$ digits, there are no $k$-digit $k$-th power. The number of digits (decimal) of a number $N$ is given by:

$$\lfloor\log_{10}N\rfloor+1$$

so if $9^k$ has less than $k$ digits, we have:

$$\lfloor\log_{10}9^k\rfloor+1 < k$$
$$k\log_{10}9 < k-1$$
$$k > \frac 1{1- \log_{10}9}\approx 21.854$$

hence we just need to verify the result up to $22$nd powers. (Observe that $9^{22}$ has only $21$ digits).

EDIT: the rest of the candidates are:

\begin{align}
7^6 &= 117649\\
8^6 &= 262144\\
8^7 &= 2097152\\
8^8 &= 16777216\\
8^9 &= 134217728\\
8^{10} &= 1073741824\\
9^6 &= 531441\\
9^7 &= 4782969\\
9^8 &= 43046721\\
9^9 &= 387420489\\
9^{10} &= 3486784401\\
9^{11} &= 31381059609\\
9^{12} &= 282429536481\\
9^{13} &= 2541865828329\\
9^{14} &= 22876792454961\\
9^{15} &= 205891132094649\\
9^{16} &= 1853020188851841\\
9^{17} &= 16677181699666569\\
9^{18} &= 150094635296999121\\
9^{19} &= 1350851717672992089\\
9^{20} &= 12157665459056928801\\
9^{21} &= 109418989131512359209
\end{align}

and none of these powers have digit sums equal to their bases.


问:
Determining equivalence classes of a matrix relation.
Problem:

Given the equivalence relation, defined on $\mathbb{R}^{n\times n}$:

$A \sim B$ $\Leftrightarrow$ $\exists S,T \in GL_n(\mathbb{R}): S B T$

Show that there are n+1 equivalence classes of "$\sim$" on $\mathbb{R}^{n\times n}$. Specifically $[0]_\sim$ and $[M_k]_\sim$ for $1 \leq k \leq n$, where $M_k$ is a block matrix of the form:

$$M_k := \begin{pmatrix} I_k & 0 \\ 0 & 0 \end{pmatrix} \in \mathbb{R}^{n \times n} $$

Questions:

Hello ! I currently have the problem of proving that the equivalence classes are exactly as given on the problem itself. I got hints that i should think about using the Gauss-Algorithm as well as elementary row and column operations to figure it out. Yet right now I do not know where to even start. I would require some help understanding this problem in general.

Thanks for the help in advance.

答: Let $(e_1,\cdots,e_n)$ be the canonical basis of ${\mathbb R}^n$. Let $A$ be a matrix and $k = \text{rank}(A)$. Let $(u_{k+1},\cdots,u_n)$ be a basis of $\ker A$. Complete this basis in a basis $(u_1,\cdots,u_n)$ of ${\mathbb R}^n$. Let $T$ be the matrix of the linear map such that $T(u_i) = e_i$. Complete the basis $(v_1, \cdots, v_k) = (Au_1,\cdots, A u_k)$ of $\text{Im}(A)$ in a basis $(v_{1},\cdots, v_n)$. Let $S$ be the matrix of the linear map such that $S(e_i) = v_i$. Then one has $A = S M_k T$, hence $A\sim M_k$.


问:
Proving a function has no extreme points.

Is there any possible way I can make this proof shorter?

Question 5

a.

Consider the function $f:\ R\ ➜\ R,\ f\left(x\right)\ =\ \frac{x}{\left|x\right|+1}$

Note that by the algebra of continuous functions f is continuous.

We show the limits of f as x tends to $∞,-∞$ exist.

$\lim _{x\to \infty }\left(\frac{x}{\left|x\right|+1}\right)=\lim _{x\to \infty }\left(\frac{x}{x+1}\right)=\lim _{x\to \infty }\left(\frac{x}{x\left(1+\frac{1}{x}\right)}\right)=\lim \:_{x\to \infty \:}\left(\frac{1}{1+\frac{1}{x}}\right)=\left(\frac{1}{1+0}\right)=1$
$\lim \:\:_{x\to \:-\infty \:\:}\left(\frac{x}{\left|x\right|+1}\right)=\lim \:_{x\to -\infty \:}\left(\frac{x}{-x+1}\right)=\lim _{x\to -\infty }\left(\frac{x}{x\left(-1-\frac{1}{x}\right)}\right)=\lim \:_{x\to -\infty \:}\left(\frac{1}{-1-\frac{1}{x}}\right)=\left(\frac{1}{-1-0}\right)=-1\:$

Now we show that for every $x ∈ R \ \ $ we have $-1 < f(x) < 1$.

Assume toward contradiction that there exist $x ∈ R$ such that $f(x) ≥ 1$ or $f(x) ≤ -1$

Case 1: $x = 0$. In this case $f$ will equal to $0$ which is obviously in $(-1,1)$, so we have a contradiction.

Case 2: $x > 0$.

$\frac{x}{\left|x\right|+1}\ge1 \ ➜ \ \frac{x}{x+1}\ge1 \ ➜ \ x\ge x+1 \ ➜ \ 0\ge1$ Contradiction!

$\frac{x}{\left|x\right|+1}\le-1 \ ➜ \ \frac{x}{x+1}\le-1 \ ➜ \ x\le-x-1 \ ➜ \ 2x\le-1$ Contradiction! Since $2x > 0$.

Case 3: $x < 0$.

$\frac{x}{\left|x\right|+1}\ge1 \ ➜ \ \frac{x}{-x+1}\ge1 \ ➜ \ x\ge-x+1 \ ➜ \ 2x\ge1$ Contradiction! since $2x < 0$.

$\frac{x}{\left|x\right|+1}\le-1 \ ➜ \ \frac{x}{-x+1}\le-1 \ ➜ \ x\le x-1 \ ➜ \ 0\le-1$ Contradiction!

Therefore -1 < f(x) < 1 for every $x ∈ R$.

Now we show f(x) is strictly increasing.
Assume toward contradiction that f(a) ≥ f(b) for every a,b ∈ R such that a < b.

$\frac{a}{\left|a\right|+1}\ge\frac{b}{\left|b\right|+1}\ \ ➜\ \ a\left(\left|b\right|+1\right)\ge b\left(\left|a\right|+1\right)\ $

Case 1: b = 0, 0 > a.

$a\left(\left|0\right|+1\right)\ge0\left(\left|a\right|+1\right)\ \ ➜\ \ a\ \ge0$ Contradiction!

Case 2: b > 0, a = 0.

$0\left(\left|b\right|+1\right)\ge b\left(\left|0\right|+1\right)\ \ ➜\ \ 0\ \ge b$ Contradiction!

Case 3: b > 0, a > 0

$a\left(b+1\right)\ge b\left(a+1\right)\ \ ➜\ \ ab\ +a\ge ab+b\ \ \ ➜\ \ a\ \ge b\ $ Contradiction!

Case 4: 0 > b, 0 > a.

$a\left(-b+1\right)\ge b\left(-a+1\right)\ \ ➜\ \ -ab\ +a\ge-ab+b\ \ \ ➜\ \ a\ \ge b\ $ Contradiction!

Case 5: b > 0, 0 > a.

$a\left(b+1\right)\ge b\left(-a+1\right)\ \ ➜\ \ ab\ +a\ge-ab+b\ \ \ ➜\ \ 2ab\ \ge b-a$ Contradiction! Since $2ab < 0$ and $b - a > 0$

We've reached a contradiction for every possible case of $a,b$, therefore $f(a) < f(b)$ for every $a,b ∈ R$, such that $a < b$ which means $f$ is strictly increasing.

Now we show $f$ has no extreme points.

Assume toward contradiction that $f$ takes a maximum at a point, we will denote it $a$.

Since $f$ is strictly increasing, there exist a point $b ∈ R$ such that $a < b$ for which $f(a) < f(b)$,

in contradiction to the minimality of $a$.

Assume toward contradiction that $f$ takes a minimum at a point, we will denote it $c$.

Since $f$ is strictly increasing, there exist a point $d ∈ R$ such that $d < c$ for which $f(d) < f(c)$,

in contradiction to the minimality of $c$.

Therefore we've shown that $f$ satisfies all requirements.

答:
You can have much simpler examples... For instance, $f(x)=\arctan x$. Here, $L_{1,2} = \pm \frac{\pi}{2}$ and, since $f$ is diffrenciable in an open set and $f' \ne 0$, there cannot be any max / min.

Regarding your example, you can make your proof way shorter... Your function is
$$
f(x)=\begin{cases}\dfrac{x}{-x+1}, & x \leq 0 \\[1em] \dfrac{x}{x+1}, & x>0 \end{cases}
$$
and it is actually differentiable. Since it has no critical points, there can be no max/min.

If you cannot use derivatives, just take $a - b < 0$ and check that $f(a)-f(b)< 0$

$$
f(a)-f(b) = \frac{a}{|a|+1}-\frac{b}{|b|+1} = \frac{a-b + (a|b|-b|a|)}{(|a|+1)(|b|+1)}
$$

If $a$ and $b$ are both positive or both negative $a|b|-b|a| =0$ and it follows that $f(a)-f(b) < 0$. If $a<0$ and $b>0$,  $a|b|-b|a| = 2a b < 0$ and you still get $f(a)-f(b)<0$.


问:
Strang's proof of SVD and intuition behind matrices $U$ and $V$
In lecture 29 of MIT 18.06, Professor Gilbert Strang "proves" the singular value decomposition (SVD) by assuming that we can write $A = U\Sigma V^T$ and then deriving what $U$, $\Sigma$, and $V$ must be based on the eigendecomposition of
$$ AA^T = U\Sigma ^2 U^T$$
and
$$ A^TA = V\Sigma ^2 V^T$$

My intuition tells me there's something wrong with first assuming that we can write $A$ in this form. As in, we are finding $U$, $\Sigma$ and $V$ for those matrices that have this form, but what if some matrices couldn't be written in this form in the first place?

Also, is there some intuition to be found about $U$ and $V$ by only thinking about them as eigenvectors of $AA^T$ and $A^TA$? In the sense that we should be able to know why $V$ will have its properties (be mapped orthogonally by $A$) by just looking at it as an eigenvector of $A^TA$ (which somehow implicitly "encodes" the basis $U$). Not sure if this intuition makes sense either.

答:
(More of a comment but too long to format. )

It's wrong to prove something by assuming it's true and showing no contradiction.

But if you can reverse every step, then it can be counted as a proof. I.e. assuming some non-contradictory facts, show that the proposition is true.

In that line of thought:

Let $v_{i}$ be the orthonormal eigenvectors of $A^TA$, which always exists.

Let $\sigma_{i}^2$ be the corresponding eigenvalues, which always exists and is positive.

Above definitions imply: $$A^TAv_{i}=\sigma_{i}^2v_{i}$$

Let $$u_{i} = \frac{Av_{i}}{\sigma_{i}} $$ which always exists, and can be proved orthonormal.

Now, the following is always true:

$$AA^Tu_i=AA^T \frac{Av_i}{\sigma_i} = \frac{A}{\sigma_i}A^TAv_i = \frac{A}{\sigma_i}\sigma_{i}^2v_i = \sigma_{i}^2\frac{Av_{i}}{\sigma_{i}} = \sigma_{i}^2u_i$$

Which means $u_i$ is a eigenvector of $AA^T$ and $\sigma_{i}^2$ is the eigenvalue.

As defined, for any $A$ we have, $$Av_i=\sigma_{i}u_i$$ And it can be proved there're $r$ such triplets ($v_i, \sigma_{i}, u_i$). Then it becomes $$A_{m \times n}V_{n \times r}=U_{m \times r} \Sigma_{r \times r}$$
Which is the reduced SVD form.

It can be further shown that $v_i$ span the row space of $A$. ($v_i$ is in the row space of $A$ since $A^TAv_{i}=\sigma_{i}^2v_{i}$, and there're $r$ number of orthogonal $v_i$ due to $r$ non-zero eigenvalues of $A^TA$ due to $rank(A^TA)=rank(A)=r$)
Use the orthogonal complement subspaces to complete the $V, U$ to a square shape then it's the full SVD form.

These steps constructs the SVD for any A and didn't assume SVD exists.


问:
pullback of (r,s)-tensor fields
Let $F:M \to N$ be a smooth map between smooth manifolds and let $w\in T_s^0(N)$ be a $(0,s)$-tensor field on $N$. Then let us define the pullback of $w$ by $F$ as the $(0,s)$-tensor field $F^*w\in T_s^0(M)$ on $M$ defined, for any $p\in M$, by
$$F^*w|_p := dF^*_p(w|_{F(p)})$$
where
$$(\forall v_1,\dots,v_s \in T_pM) \quad dF^*_p(w|_{F(p)})(v_1,\dots,v_s) :=w|_{F(p)}(dF_pv_1,\dots, dF_pv_s).$$

Thus $F^* : T_s^0(N) \to T_s^0(M)$. Is there a way to show that this is actually just a specific case of the following generation of the pullback of $(r,s)$-tensor fields by $F$:
$$F^* :=(F^{-1})_* : T_s^r(N) \to T_s^r(M)?$$

答:
In your last line, you need $F^{-1}$ to be well-defined (at least differentiable), which is not the case in general.

The way you defined the pull-back operations is the most common one, and your last line should be the definition of push-forward when $F$ is a diffeomorphism, for example.

Title: pullback of (r,s)-tensor fields


问:
Image of a union of collection of sets as union of the images
I am having problems with establishing the following basic result. Actually, I found a previous post that is close in nature (it is about inverse image), but I was interested in this specific one, with the following notation, because it is the one I found in the book I am self-studying and I guess that most of my problems are actually related with the notation itself. Morevoer I would like to take it as an opportunity to find out how to write proof with equalities, because I tend in those cases to write them in a cumbersome way, with necessary and sufficient conditions.

[I assume that this is partly due to the influence of "How to prove it: a structured approach", a book that I loved, but that left me the tendency to be quite mechanical in proving any sort of result].

Theorem:

Let $X$ and $Y$ be nonempty sets and $f \in Y^X$. Prove that, for any (nonempty) classes $\mathcal{A} \subseteq 2^X$, we have
$ f(\cup \mathcal{A}) = \cup \{f(A):A \in \mathcal{A} \} $

Here, it's how I approach the problem.

Tentative Proof:

First of all, by definition of (direct) image of a function, we have
$$f(\cup \mathcal{A}) := \{f(x):x\in \cup \mathcal{A} \}. \hspace{1cm} (*)$$
Let $X$, $Y$ be arbitrary (nonempty) sets. Let $f$ be an arbitrary function that maps from $X$ to $Y$ and let $\mathcal{A}$ be a family of sets, subset of the powerset of $X$. By $(*)$, the result we have to prove becomes
$$ \{f(x):x \in \cup \mathcal{A} \} = \cup \{ f(A):A \in \mathcal{A}\} \hspace{1cm} (1)$$
In order to prove it, rephrase $(1)$ as
$$ \forall y ( y \in \{f(x):x \in \cup \mathcal{A} \} \leftrightarrow y \in \cup \{ f(A):A \in \mathcal{A}\}). \hspace{1cm} (2)$$
We start by proving the necessary condition. Let $y$ be an arbitrary element and assume that $y$ is a member of $\{f(x):x \in \cup \mathcal{A} \}$. This means that $\exists x(x \in \cup \mathcal{A} \land y=x)$. At the same time, we have to prove that $\exists A(A \in \mathcal{A} \land y\in f(A))$.

And here I got stuck...

I really don't see how from my premises I can get the desired result. Indeed, what $A$ should be?

I assume there is a problem with the way in which I translate the problem in logical terms.

答:
$$\frac{f(A)\subseteq B}{A\subseteq f^{-1}(B)}$$

Hence direct image is left-adjoint to inverse image, hence it preserves colimits, thus in particular it preserves unions.


问:
Ugly functions are non-measurable
I am trying to understand proof of theorem 5.5 from page 119, everything is clear until ,,This implies that $B_0$ (and hence $A_0$) are non-measurable ... $1=\mu([0,1])=\sum_{n\in Z} \mu(B_0) \leq \mu([-1,2])=3$ which is impossible". Pls could someone explain me why it's impossible? Thanks in advance.

I will wrote the whole proof for people who have problem with link:

Let $f$ be ugly. Assume* w.L.o.g that there exist real numbers $a\neq0$ and $b\neq 0$ with $f(a)=1$ and $f(b)=0$. For $n \in Z$, define $A_n=f^{-1}[n,n+1)$ and choose $q_n \in Q$ with $|n*a-q_n*b|<\frac{1}{2}$. Define $B_0=A_0 \cap[-\frac{1}{2},\frac{3}{2}]$ and $B_n=B_0+n*a-q_n*b=\{x+n*a-q_n*b|x\in B_0\}$ for $n\neq 0$.
Then $x \in (A_n \cap [0,1])$ implies that $y=x-(n*a-q_n*b) \in (A_0 \cap [-\frac{1}{2},\frac{3}{2}])$, i.e., $y\in B_0$, and thus $x=y+(n*a-q_n*b) \in B_n$. Consequently $(A_n \cap [0,1]) \subseteq B_n \subseteq [-1,2]$.
Thus: $[0,1]=[0,1] \cap \bigcup _{n \in Z}A_n=\bigcup _{n \in Z} ([0,1] \cap A_n) \subseteq \bigcup_{n\in Z} B_n\subseteq[-1,2]$.
This implies that $B_0$ (and hence $A_0$) are non-measurable, since otherwise the $B_n$'s, being pairwise disjoint and pairwise congruent, would have the same measure $\mu(B_n)=\mu(B_0)$ and thus** (5) Disasters with Choice $1=\mu([0,1])=\sum_{n\in Z}\mu(B_0)\leq\mu([-1,2])=3$, which is impossible. Consequently $f$ is not measurable.

(*) If necessary, choose real numbers $a\neq 0$ and $b \neq 0$ with $\frac{f(a)}{a} \neq \frac{f(b)}{b}$ and replace $f$ by the function $g(x)=\frac{b*f(x)-f(b)*x}{b*f(a)-a*f(b)}$.

(**) Here we use $\sigma$-additivity of Lebesgue-measure (which requires some choice principle). However, our use of choice principles can be avoided. (some exercise about that)

(5) If $f:R \to R$ is ugly, then its graph is dense in $R^2$.

答: $1 \leq \mu[0,1] \leq \sum \mu(B_n)$. Since $B_n$'s are disjoint and they all have the same measure, $\sum \mu(B_n)=0$ or $\infty$. The first possibility is ruled out because $ 1 \leq \sum \mu(B_n)$ and the second one is ruled out because $\sum \mu(B_n) \leq 3 <\infty$.

Title: Ugly functions are non-measurable