问:
Sufficent condition that a function from $\mathbb{R}^2$ is continuous
I was wondering if it is sufficient for a function $f:\mathbb{R}^2 \rightarrow{\mathbb{R}}$ to be continuous that
$$\lim_{h \to 0} f(h, 0) = \lim_{k\to 0} f(0, k)$$
As far as I knew this was the case, yet the function $f(x, y) = \frac{2xy^3}{(x^2+y^2)^2}$ suffices this condition since both limits are $0$, but $\lim_{m \to \infty}f(\frac{1}{m}, \frac{1}{m}) = \frac{1}{2} \neq 0$, so the function can't be continuous. Maybe someone can help me figure out where I'm wrong...
Thanks!

答:
You have provided a counter-example yourself. Here is a simpler one:$$f(x,y)=\begin{cases}0&\text{ if $x=0$ or $y=0$}\\1&\text{ otherwise.}\end{cases}$$It is clearly discontinuous at $(0,0)$, right?! However, you always have $f(h,0)=f(0,k)=0$. In particular,$$\lim_{h\to0}f(h,0)=\lim_{k\to0}f(0,k)=0.\tag1$$

The equalities $(1)$ only tell you something about what happens to $f(x,y)$ when $(x,y)$ approaches $(0,0)$ along the $x$-axis or the $y$-axis, but nothing about when $(x,y)$ approaches $(0,0)$ in general.


问:
Finite extension of completions and prime ideal factorization
Let $K$ be a number field (i.e. a finite extension of $\Bbb{Q}$). Let $\mathfrak{p}$ be a prime ideal of the ring of integers $\mathcal{O}_K$. Also, let us denote the completion of $K$ with respect to $\mathfrak{p}$ (i.e. wrt. the absolute value defined by $\mathfrak{p}$) by $K_\mathfrak{p}$.

Let $L/K_\mathfrak{p}$ be a finite extension and suppose
$$
\mathfrak{p} \mathcal{O}_L = \prod_i \mathfrak{p_i}^{m_i}
$$
is the factorization of $\mathfrak{p} \mathcal{O}_L$ into prime ideals where $\mathfrak{p}_i \subseteq \mathcal{O}_L$ are prime ideals of $\mathcal{O}_L$.

Question: Is there a $\mathfrak{p}_i$ such that $L = K_{\mathfrak{p}_i}$?

I was interested in this question because in this comment of my other post, it sounded like this is a general statement. That's why I would like to try the theory behind that better.

答:
$K_{\mathfrak{p}}$ is a local field, hence so are its finite extensions. So if you have a finite extension $L$, then $\mathfrak{p}$ splits as $\mathfrak{P}^{e}$ for a single lifted prime $\mathfrak{P}$ of $L$, with ramification index $e$. To ask whether $L$ arises as a localisation of $K$ by $\mathfrak{P}$ seems slightly senseless because $\mathfrak{P}$ doesn't live in $K$.

A better way to phrase the question is whether $L$ is still the completion of a global field at a prime, which is indeed the case, as you learned a few days ago. :)

ContentLicense: CC BY-SA 4.0


问:
Does every abelian group of order $16$ have an element of order $4$?
Can we solve this by saying that $|G|=16$ is divisible by $o(g)=4$ and therefore there must exist an element of order $4$ in this group?

答: $\mathbb Z_2^4$ has only elements of orders $1$ and $2$.


问:
Linear independence of $(x\sin x)^{\frac{n-1}{2}}$ and $(x\sin x)^{\frac{n+1}{2}}$
Could you tell me why  $(x\sin x)^{\frac{n-1}{2}}$ and $(x\sin x)^{\frac{n+1}{2}}$ are lineraly independent?

I've tried $\alpha(x \sin x)^{\frac{n-1}{2}} + \beta (x\sin x)^{\frac{n+1}{2}} =0$

$(x\sin x)^{\frac{n-1}{2}}(\alpha + \beta (x\sin x))=0$

but I don't know how that implies linear independence.

$x \in [0, \frac{\pi}{2}]$, so $(x\sin x) \ge 0$ on this interval.

答: If $\alpha(x \sin x)^{\frac{n-1}{2}} + \beta (x\sin x)^{\frac{n+1}{2}} =0$ (the zero function), clearly it cannot happen that exactly one of $\alpha,\beta$ is zero. If both $\alpha$ and $\beta$ are nonzero, then by rearranging terms, we have $x\sin x=-\alpha/\beta$ for all $x$ such that $x\sin x\neq0$. I think it is easy enough to rule out this case.

ContentLicense: CC BY-SA 3.0


问:
Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$
I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.

The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$

How am I supposed to solve this problem?

答:
$$\frac{2 \tan x}{1-\tan ^2 x}+\tan x=^*\frac{2}{\sqrt{3}}$$.
$$\tan x=-1.45424, \tan x=0.35178, \tan x =2.25716$$
$L=\tan x \cot 2x$ can be written$^{**}$ as $\frac{1}{2} \left(1-\tan ^2 x\right)$ thus possible values are $$L_1=-0.557406,L_2=0.438125,L_3=-2.04739$$

$$(^*)\quad \tan x=u \to u^3-\frac{2 u^2}{\sqrt{3}}-3 u+\frac{2}{\sqrt{3}}=0$$
see solutions here. (press Approximate forms button)
$$(^{**})\quad\tan x \cot 2x=\frac{\sin x\left(\cos ^2 x-\sin ^2 x\right)}{\cos x(2 \sin x \cos x)}=\frac{\cos ^2 x-\sin ^2 x}{2\cos^2 x}=\frac{1}{2}\left(1-\tan^2 x\right)$$


答:
No, and in fact the statement is false. The group
$\mathbb{Z}_2^4$ has only elements of order $2$ and $1$.

You are trying to apply Cauchy's theorem, but this only works when the divisor is prime. For example, $2$ is a prime dividing $16$, so Cauchy's theorem tells you there is an element of order $2$, but you can't do the same thing for $4$ because it is not prime.

Title: Does every abelian group of order $16$ have an element of order $4$?

Body: Can we solve this by saying that $|G|=16$ is divisible by $o(g)=4$ and therefore there must exist an element of order $4$ in this group?


问:
Polynomials of degree $n$ as a vector space

Let $P$ be the set of polynomials $p: \Bbb{R} \to \Bbb{R}, x \mapsto a_n x^n + \cdots + a_1x + a_0$ equipped with addition and scalar multiplication. Show that $P$ is a vector space.

Since $P$ is equipped with scalar multiplication I have that $$cp = ca_n x^n + \cdots + ca_1x + ca_0$$ so it is closed under multiplication.

Now let $p = a_n x^n + \cdots + a_1x + a_0$ and $q = b_n x^n + \cdots + b_1x + b_0$ from which one gets that $$p+q = (a_n+b_n)x^n+(a_{n-1}+b_{n-1})x^{n-1} + \dots + a_0+b_0$$

so its closed under addition. I'm not really sure how to proceed from here, how can I show that this is a vector space?

答: $P$ is the set of all polynomials of degree at most $n$. It is thus a subset of the set of all polynomials that forms a vector space equipped with addition and scalar multiplication. In order to show that $P$ is a vector space, it suffices to prove it is a subspace of the vector space of all polynomials so you only need to check for closure under addition and scalar multiplication, as you have done.


问:
When is a category the colimit of its slice categories
My question: is it true in general for a small category $\mathcal{C}$ that the canonical functor
\begin{aligned}
\operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x \to \mathcal{C}
\end{aligned}
is an equivalence of categories? If so, is there a canonical way of seeing this without explicitly constructing an inverse?

Let me explain the statement in more detail. For an object $x$ in $\mathcal{C}$, let $C/x$ denote the over-category of $\mathcal{C}$ over $x$, with objects given by morphisms $f: y \to x$ in $\mathcal{C}$ and morphisms from $f: y \to x$ to $f': y'\to x$ given by morphisms $g: y \to y'$ in $\mathcal{C}$ such that $f = f' \circ g$. These over-categories assemble into a functor
\begin{aligned}
\mathcal{C} \to \operatorname{Cat}: c \mapsto \mathcal{C}/x
\end{aligned}
to the (large) category of small categories. Indeed, for a morphism $h: x \to x'$, we get a functor $\mathcal{C}/x \to \mathcal{C}/x': (f: y \to x) \mapsto (h \circ f: y \to x')$.

For every $x$, there is a canonical functor source functor $s: \mathcal{C}/x\to \mathcal{C}$ given by sending $f: y \to x$ to its source $y$. Essentially by definition, these source functors assemble into a cocone over $\mathcal{C}$ in $\operatorname{Cat}$, and thus give rise to a functor
\begin{aligned}
\operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x \to \mathcal{C}.
\end{aligned}
This is the canonical functor referred to in the statement above.

Attempt: one can try to simply write down an inverse functor $\mathcal{C} \to \operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x$ as follows. An object $x$ is sent to the object in the colimit system represented by the object $(\operatorname{id}_x: x \to x)$ in $\mathcal{C}/x$. If $f: x \to x'$ is a morphism in $\mathcal{C}$, notice that this object $(\operatorname{id}_x: x \to x)$ gets identified in the colimit system with the object $(f: x \to x')$ in $\mathcal{C} / x'$. Also notice that $f: x \to x'$ forms a morphism in $\mathcal{C}/x'$ from $(f: x \to x')$ to $\operatorname{id}_{x'}: x' \to x'$. It thus seems that the assignment $x \mapsto [\operatorname{id}_x: x \to x]$ extends to a functor $\mathcal{C} \to \operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x$. I think this should be the desired inverse.

Question: Does the above proof attempt work? If so, is there a more invariant way of proving that the canonical functor is an equivalence? I am interested in a version of the statement for $\infty$-categories, for which one cannot simply write down an inverse functor by giving what it does on objects and morphisms.

答:
At this point, I am quite sure that the statement is true (both in the sense of a strict colimit as well as in the more relevant sense of a pseudo colimit.) I'll give a proof below. Nevertheless, I'd be happy if someone could provide an alternative (shorter/more elegant) proof that has the hope of also providing a proof in the setting of $\infty$-categories.

Proof: We show that $\mathcal{C}$ has the universal property of the pseudo colimit. (To prove it has the universal property of the strict colimit, replace all appearing natural isomorphisms by identity transformations.)

So consider a collection of functors $F_x: \mathcal{C}/x \to \mathcal{D}$ into a small category $\mathcal{D}$ with natural isomorphisms
\begin{aligned}
\alpha_f: F_{y} \circ f_* \overset{\cong}{\implies} F_x
\end{aligned}
for all $f: x \to y$ in $\mathcal{C}$ satisfying

$\alpha_{\operatorname{id}_x} = \operatorname{id}_{F_x}$

$\alpha_{gf} = \alpha_g \circ \alpha_f$ for $f:x\to y$ and $g: y \to z$.

We now have to show that up to natural isomorphism there is a unique functor $F: \mathcal{C} \to \mathcal{D}$ with natural isomorphisms $\beta_x: F_x \overset{\cong}{\implies} F \circ s_x$ such that the two natural equivalences
\begin{aligned}
F_{y} \circ f_* \overset{\alpha_f}{\implies} F_x \overset{\beta_x}{\implies} F \circ s_x
\end{aligned}
and
\begin{aligned}
F_{y} \circ f_* \overset{f_*(\beta_y)}{\implies} F \circ s_y \circ f_* = F \circ s_x
\end{aligned}
are equal.

Existence: On objects, we define $F$ by
\begin{aligned}
F(x) := F_x(\operatorname{id}_x: x \to x).
\end{aligned}
On a morphism $f: x \to y$, we define $F(x)$ as the composite
\begin{aligned}
F(x): F_x(\operatorname{id}_x: x \to x) \xrightarrow[\alpha_f]{\;\;\cong\;\;} F_y(f_*(\operatorname{id}_x: x \to x)) \xrightarrow{F_y(f: f \to \operatorname{id})} F_y(\operatorname{id}_y: y \to y).
\end{aligned}
This forms a functor. Indeed, we see:

We have $F(\operatorname{id}_x) = \operatorname{id}$ as $\alpha_{\operatorname{id}} = \operatorname{id}$ and $F_y(\operatorname{id}) = \operatorname{id}$.

For $f: x \to y$ and $g: y \to z$, we have $F(gf) = F(g) \circ F(f)$. This can be seen by the following commutative diagram:
$\require{AMScd}$
\begin{CD}
F_x(\operatorname{id}: x \to x) @>{=}>> F_x(\operatorname{id}: x \to x) @>{=}>> F_x(\operatorname{id}: x \to x) \\
@V{\alpha_f}VV @V{\alpha_{gf}}VV @VV{\alpha_{gf}}V\\
F_y(f: x \to y) @>{\alpha_g}>> F_z(gf: x \to z) @>{=}>> F_z(gf: x \to z) \\
@V{F_y(f: f \to \operatorname{id})}VV @V{F_z(f: gf \to g)}VV @VV{F_z(gf: gf \to \operatorname{id})}V\\
F_y(\operatorname{id}: x \to y) @>>{\alpha_g}> F_z(g: y \to z) @>>{F_z(g: g \to \operatorname{id})}> F_z(gf: x \to z) \\\end{CD}
Here the left composite is $F(f)$, the bottom one is $F(g)$ and the right one is $F(gf)$.

Furthermore, we have natural isomorphisms $\beta_x: F_x \implies F \circ s_x$ of functors $\mathcal{C}/x \to \mathcal{D}$ by
\begin{aligned}
\beta_x: F_x(f: y \to x) = F_x(f_*(\operatorname{id}_x: x \to x)) \xrightarrow[\cong]{\alpha_f} F_y(\operatorname{id}_y: y \to y) = F(s_x(f)).
\end{aligned}

Since $\alpha$ is associative, it is clear that $\beta_x \circ \alpha_f = f_*(\beta_y)$. This finishes the proof of existence.

Uniqueness: We now assume that $G: \mathcal{C} \to \mathcal{D}$ is another functor with natural equivalences $\gamma_x: F_x \overset{\cong}{\implies} G \circ s_x$ satisfying $\gamma_x \circ \alpha_f = f_*\gamma_y$. We show that there is a natural isomorphism $\delta: F \overset{\cong}{\implies} G$.  On objects, this is given by the composite
\begin{aligned}
F(x) \xrightarrow[\cong]{\beta_x^{-1}} F_x(\operatorname{id}_x: x \to x) \xrightarrow[\cong]{\gamma_x} G(x).
\end{aligned}
Note that the relations $\beta_x \circ \alpha_f = f_*(\beta_y)$ and $\gamma_x \circ \alpha_f = f_*\gamma_y$ imply that for every morphism $f: x \to y$, this composite agrees with the composite
\begin{aligned}
F(x) = F(s_y(f:x \to y)) \xrightarrow[\cong]{\beta_y^{-1}} F_y(f: x \to y) \xrightarrow[\cong]{\gamma_y} G(s_y(f:x \to y)) = G(x).
\end{aligned}
The fact that these morphisms give a natural transformation follows from the following commutative diagram for every $f: x \to y$:
\begin{CD}
F(x) @>F(f)>> F(y) \\
@V{=}VV @VV{=}V \\
F(s_y(f:x \to y)) @>{F(s_y(f: f \to \operatorname{id}_y}))>> F(s_y(\operatorname{id}_y: y \to y)) \\
@V{\beta_y^{-1}}VV @VV{\beta_y^{-1}}V \\
F_y(f: x \to y) @>{F_y(f: f \to \operatorname{id}_y)}>> F_y(\operatorname{id}_y: y \to y) \\
@V{\gamma_y}VV @VV{\gamma_y}V \\
G(s_y(f:x \to y)) @>{G(s_y(f: f \to \operatorname{id}_y}))>> G(s_y(\operatorname{id}_y: y \to y))\\
@V{=}VV @VV{=}V \\
G(x) @>G(f)>> G(y)
\end{CD}
So $\delta: F \implies G$ is indeed a natural equivalence. Basically by definition of $\delta$, the composite $F_x \overset{\beta_x}{\implies} F \circ s_x \overset{\delta \circ s_x} G \circ s_x$ agrees with $F_x \overset{\beta_x}{\implies} G \circ s_x$, proving the uniqueness (up to isomorphism) of the functor $F: \mathcal{C} \to \mathcal{D}$ with its natural isomorphisms $\beta_x: F_x \cong F \circ s_x$.


问:
Solving equation involving finite continued fraction
I'm interested in solving the equation
$$
\color{red}{x}=1+\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\ddots \cfrac{a_n}{b_n+\color{red}{x}}}},
$$
where $a_i,b_i$ are positive real numbers. Is there a formula to simplify this continued fraction?

Any help will be appreciated.

答:
$$b_{n-1}+\frac{a_n}{b_n+x}=\frac{b_{n-1}b_n+a_n+b_{n-1}x}{b_n+x}.$$

From this you can conclude that your equation has the form

$$x=\frac{px+q}{rx+s}$$ where the coefficients can be computed by recurrence. This equation can be rewritten as quadratic.


问:
The set of irreducible representations (over C) determines a finite group
I'm trying to understand group representations, and I think that this statement is correct (where determines means up to isomorphism), although I couldn't find a proof online (without references to more complicated things like "Tannaka duality"), so I tried to prove it, but I may have overlooked something:

Let $G = \{x_1, ..., x_n\}$, $H = \{y_1, ..., y_n\}$ be two finite groups of same cardinality.

Suppose G and H have the same set of irreducible representations in the following sense :
there is a finite set of morphisms $g_i$ (resp. $h_i$), each one from G (resp. H) to
some subgroup of $GL(n_i, \Bbb{C})$ and we can write $g_i(x_j) = P_i h_i(y_j) P_i^{-1}$ for some invertible matrix $P_i$ and all $j$.

Then if we consider the regular representation of G, we can write it as a sum of the $g_i$ representations
(with $n_i$ giving the multiplicity) and do the same for H.

That shows that the two regular representations are isomorphic as we can build a new base using a block-diagonal transition matrix using $(P_i)$.
As they are also faithful (isomorphic to the groups themselves), then G and H are isomorphic.

Is that correct ?

EDIT: I updated the "same set of irreducible representations" following the comments (that also requires the additional assumption of same cardinality)

答:
Your theorem is correct: we can reconstruct a group's operation from a full set of irreducible representations (interpreted as mere functions from $G$ to $GL(V)$s). And your proof is also correct: we can reconstruct the regular representation (as a function) by taking direct sums of the irreps (with multiplicities equal to their dimensions); then $G$ is isomorphic to its image (under the regular rep) so its operation can be recovered by transporting matrix multiplication back to it.

We can probably phrase this in categorical terms, by considering the category whose objects are $(X,\mathcal{F})$ where $X$ is a set of $\mathcal{F}$ is a family of functions on $X$, and whose morphisms are tuples $(\alpha,\beta,\Phi)$ where $\Phi$ is a family of functions $\phi_f$ indexed by $\mathcal{F}$, and $\alpha:X\to Y,\beta:\mathcal{F}\to\mathcal{G}$ satisfy the diagram $f=\phi_f\circ\beta(f)\circ\alpha$.

Given a group $G$, we can construct such an object as $(G,\mathcal{F})$ with $\mathcal{F}$ a complete set of irreducible representations, then any isomorphism $(G_1,\mathcal{F}_1)\to(G_2,\mathcal{F}_2)$ with $\Phi$ consisting of linear operators can be converted into an isomorphism $G_1\to G_2$.

Remiss to not explain why Tannaka-Krein duality (a) is natural and (b) not that complicated for finite groups. First of all, just as every (finite-dimensional) vector space is (not naturally) isomorphic to its dual, every (finitely-generated) abelian group is (not naturally) isomorphic to its dual, aka its character group. However, the dual of the dual is naturally isomorphic in these cases. Indeed, the Pontryagin dual of the Pontryagin dual of a locally compact abelian group is naturally isomorphic to the original group. This is achieved with the Fourier transform on $\mathbb{R}$ (which is self-dual) or between $S^1$ and $\mathbb{Z}$ (which are dual).

The question, then, is if we can generalize this to nonabelian $G$. The characters of an abelian group are its irreps, and pointwise multiplying characters is equivalent to tensoring them. Can we recover a finite group (or more generally, a compact group, if not locally compact) from its irreducible representation and tensor product? The first issue is that irreps are not closed under tensoring, so (being category theorists as well as represenation theorists) we consider the monoidal category of representations of $G$ with the tensor product operation. This turns out to not be enough, but it is if we equip it with the fiber functor $\Phi$ (forgetful functor to the category of vector spaces).

Then $\mathrm{Aut}(\Phi)\cong G$ (tensor-preserving natural functors). Pick a $v\in V$ and consider evaluation-at-$v$ as a morphism $\mathbb{C}[G]\to V$ given by $x\mapsto xv$, to show $\eta\in\mathrm{Aut}(\Phi)$ satisfies $\eta_V(v)=\eta_{\mathbb{C}[G]}(e)v$, then consider the "comultiplication" morphism $\Delta:\mathbb{C}[G]\to\mathbb{C}[G]\otimes\mathbb{C}[G]$ given by $\Delta(g)=g\otimes g$ to show $\eta_{\mathbb{C}[G]}(e)=g$ for some group element $g$. Then $\eta\leftrightarrow g$ is an isomorphism $\mathrm{Aut}(\Phi)\cong G$.

There is still "duality" present here. All intertwiners intertwine with all $\pi_V(g)$ operators, and conversely the natural transformations which intertwine with all intertwiners are precisely of the form $\eta_V=\pi_V(g)$.


问:
Find constants $a$, $b$, $c$, $d$ and $e$ such that $\cos4x=a\sin^4x+b\sin^3x+c\sin^2x+d\sin x+e$ for all angles $x$
Basically, write $\cos4x$ as a polynomial in $\sin x$.

I've tried the double angles theorem and $\cos2x=\cos^2x-\sin^2x$. I'm still having trouble right now though.

Please help! Thanks!

答:
If you prefer we can use complex numbers (yes!).

Let $z=\cos x+i\sin x$. Using de Moivre's theorem:
$$\begin{align} z^4&=\cos 4x+i\sin 4x=(\cos x+i\sin x)^4\\&=\cos^4x+4\cos x(i\sin x)+6\cos^2x(i\sin x)^2+4\cos x(i\sin x)^3+(i\sin x)^4\end{align}$$
Since we're only concerned with the real part of this expression, $\cos 4x$, we see that
$$\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x$$
and now use $\cos^2x=1-\sin^2x$ repeatedly.


问:
Counting Cosets of $\langle\tfrac12\rangle$, in $\Bbb{R}$ and in $\Bbb{R}^{\times}$
Describe the cosets of the subgroups described:

The subgroup $\langle\frac{1}{2}\rangle$ of $\mathbb{R}^{\times}$, where $\mathbb{R}^{\times}$ is the group of non-zero real numbers with multiplication.

The subgroup $\langle\frac{1}{2}\rangle$ of $\mathbb{R}$, where $\mathbb{R}$ is the group of real numbers with addition

答:
A coset of this subgroup is a set of all real numbers who's pairwise ratio is a power of $2$ (really a power of $\frac{1}{2}$, but that's the same thing). For instance, the coset containing $\pi$ also contains $16\pi$ and $\frac{\pi}{128}$, because $16$ and $\frac{1}{128}$ are both powers of $2$.

Same thing, but this time the defining feature is that pairwise differences is a multiple of $\frac{1}{2}$. That means that the coset containing for instance $\log 53$ also contains $\frac{2197}{2}+\log 53$.


问:
Is the union of two circles homeomorphic to an interval?
Let $Y$ be the subspace of $\Bbb R^2$ given by $Y=\{(x,y): x^2+ y^2=1\}\cup \{(x,y): (x−2)^2+ y^2=1\}$. Is $Y$ homeomorphic to an interval?

I have previously already shown that the unit circle is not homeomorphic to any interval and I think this is also true for $Y$. Basically if we remove a point from $Y$ that is not the intersection, and remove a point from any interval that is not an endpoint, the interval becomes disconnected, but the union of two circles are still connected so there is no homeomorphisms. Is that correct?

答: Just for a change: Let $D$ be the open disk in $\Bbb R^2$ centered at $p=(1,0)$ with radius $1$ and let $E=D\cap Y.$ Then $E$ is a connected subspace of $Y$  and $E\setminus \{p\}$ is the union of $4$ pairwise-disjoint non-empty connected open subspaces of $E$. But if $E'$ is any connected subspace of $\Bbb R$ and $p'\in E'$ then $E'\setminus \{p'\}$ is either connected or is the union of $2$ disjoint non-empty connected open subspaces of $E'$.


问:
What is the meaning of vector (A cross B) cross A?
Answered (I think!):

The triple product's purpose is to find a direction to the origin, perpendicular to the baseline, which is super trivial in 2D as there is only two perpendicular orientations, but the "cylinder" distinction is made in 3D because there are infinite perpendicular orientations - hence the triple product. Diagram in one of the answers shows this nicely!

Original Question:

I was looking at a video explaining the G-J-K algorithm for finding the two closest features on two convex shapes - when it came to a bit about, given a line, what is the direction in 3D space, from a point on the line, which points to the origin. Seems easy enough: just invert the position vector to point backwards to the origin - and that's exactly what the code does in an earlier step of the algorithm. Bizarrely, they say that as "the origin could be in a cylinder around the line in 3D space", the direction to the origin is given by the following:

Let X be the vector between points A and B on the line (from A to B), and Y be the inverse position vector from A to the origin. The direction to the origin, apparently, is (X cross Y) cross X! I expanded this out, saw an interesting pattern but could not link it to the problem:

For two vectors A, B; (A cross B) cross A I found to be equal to:

Ax(AyBy + AzBz) - Bx(AyAy + AzAz)

Ay(AxBx + AzBz) - By(AxAx + AzAz)

Az(AxBx + AyBy) - Bz(AxAx + AyAy)

Notated as a column vector, with "Vn" meaning the Nth coordinate of vector V. By "cross" I mean the 3D cross product.

I know this isn't the computer science stack exchange - I would like the mathematical meaning of "(A cross B) cross A", and a reason why the code does this, and not just another inverse position vector as it did previously, to find a direction to the origin. Many thanks for any help, and I will repost this on a different stack if it's too offensive here!

答:
I must be missing something, but taking the question in isolation, namely "Let X be the vector between points A and B. The direction to the origin, apparently, is:"

... Welp, If we know all about the endpoints of the line, namely $\vec{A}$ and $\vec{B}$, then some point on the line connecting the endpoints must be $$\vec{X} = \alpha \vec{A} + \left(1-\alpha\right) \vec{B} \:,$$ where $\alpha$ is a dimensionless parameter between zero and one, inclusive. If that form is agreeable, just take $-\vec{X}$ to reach back to the origin, or calculate $-\hat{X}$ to simply point there.


答:
Double cross product is a very common technique to project a vector onto the surface. Consider the triple product

$$-\vec{n} \times \vec{n} \times \vec{v}$$

where $\vec{n}$ is a normal vector to some surface. $\vec{n}$ is typically normalized, meaning that its length $|\vec{n}| = 1$. Any cross-product with a normal is going to be orthogonal to that normal, so it will land on the surface the normal is orthogonal to. It will also be orthogonal to the original $\vec{v}$. Now the second cross product will rotate the vector counter-clockwise 90 degrees, still within that surface. The result of that will be that it will now be parallel to $\vec{v}$ in some sense. More precisely, we will have obtained the projection of the vector $\vec{v}$ onto the surface, for which $\vec{n}$ is a normal. It turns out that the cross product rotates the vector in the wrong direction, so we also need to add a minus sign in front to correct for that.

Title: What is the meaning of vector (A cross B) cross A?

Body:
Answered (I think!):

The triple product's purpose is to find a direction to the origin, perpendicular to the baseline, which is super trivial in 2D as there is only two perpendicular orientations, but the "cylinder" distinction is made in 3D because there are infinite perpendicular orientations - hence the triple product. Diagram in one of the answers shows this nicely!

Original Question:

I was looking at a video explaining the G-J-K algorithm for finding the two closest features on two convex shapes - when it came to a bit about, given a line, what is the direction in 3D space, from a point on the line, which points to the origin. Seems easy enough: just invert the position vector to point backwards to the origin - and that's exactly what the code does in an earlier step of the algorithm. Bizarrely, they say that as "the origin could be in a cylinder around the line in 3D space", the direction to the origin is given by the following:

Let X be the vector between points A and B on the line (from A to B), and Y be the inverse position vector from A to the origin. The direction to the origin, apparently, is (X cross Y) cross X! I expanded this out, saw an interesting pattern but could not link it to the problem:

For two vectors A, B; (A cross B) cross A I found to be equal to:

Ax(AyBy + AzBz) - Bx(AyAy + AzAz)

Ay(AxBx + AzBz) - By(AxAx + AzAz)

Az(AxBx + AyBy) - Bz(AxAx + AyAy)

Notated as a column vector, with "Vn" meaning the Nth coordinate of vector V. By "cross" I mean the 3D cross product.

I know this isn't the computer science stack exchange - I would like the mathematical meaning of "(A cross B) cross A", and a reason why the code does this, and not just another inverse position vector as it did previously, to find a direction to the origin. Many thanks for any help, and I will repost this on a different stack if it's too offensive here!


问:
Non-unique solutions for ODE with boundary conditions at infinity
Ello,

I am looking for solutions to equations such as

$$ U ^ \prime (y) - y^{\prime \prime} + \beta y^{\textit{IV}} = 0$$
where $\beta >0$ is a constant and $U(x)$ is a function whose Taylor expansion at the origin equals $ \alpha x^2 + \dots$ for $\alpha >0$ and such that $\lim_{x \to \infty} = -\infty$.
A prototypical example would be the quartic double-well, $$ U(x) = x^2 -x^4$$.

There is a trivial solution $y(x) = 0$. On intuitive grounds (please see physical background below), as well as noting that the origin is a hyperbolic point for a dynamical system governed by the related system of equations, another solution is expected as well, with boundary conditions $$ \begin{cases} \lim_{x \pm \infty} y(x) = 0  \\
\lim_{x \pm \infty} y^ \prime (x) = 0 \end{cases}$$

This fact does not contradict the unicity theorem, as far as I understand, as the latter deals with boundary conditions on the real line, I believe.

Now to my problem, what numerical scheme could be used to get the non-trivial solution? Anything I could think of will pick the trivial solution branch only.
Further, I would be grateful for theory references on the possibility of multiple solutions for boundary conditions at infinity.

Things would be different if $\beta$ were equal to $0$.
In that case it is easy to determine two boundary conditions (also clarified in the physical background section below) for $x = 0$, and any numerical method will handle it easily.

Physical Background

The ODE is the Euler-Lagrange equation for the functional

$$ \int _{- \infty} ^{\infty } \frac{1}{2} y^{\prime 2} + U(y) + \frac{\beta}{2} y^{\prime \prime 2} \mathrm{d}x $$
representing the energy of a stretchable beam in the potential $U$. The first term represents the stretching energy, the second the potential energy, the third the bending energy.
Given this physical picture, the trivial solution represents a beam lying in the potential energy "valley" at the origin, un-stretched and straight. The non-trivial solution represents a beam lying on the valley as before, then deforming over the potential energy "bump", kept in the equilibrium by the pull from the descending potential energy on the other side, and going back to the valley at the origin.

If $\beta = 0$ the non-trivial solution is amenable to a simple solution.
The energy functional reduces in such case to

$$ \int _{- \infty} ^{\infty } \frac{1}{2} y^{\prime 2} + U(y)  \mathrm{d}x $$ which could be interpreted as the action of a particle rolling in a potential energy $-U(x)$ (with $x$ now representing time). The turning point is then easily found by solving the equation $ U(x_1) = 0$: that is, the particle starts from $x = 0$, rolls down and up until $x_1$, and rolls back in a homoclinic orbit.
This gives two boundary conditions

$$ \begin{cases}  y(0) = x_1  \\
 y^ \prime (0) = 0 \end{cases}$$

and these allow any numerical scheme to pick the non-trivial branch.

If $\beta \neq 0$, this simple reasoning fails for the full fourth-order ODE (I also thought about converting the fourth-order ODE to a system of equations, to attempt a similar interpretation for a system of four particles, but did not get anywhere yet, apart confirming that a second non-trivial solution exists). I am hence left with the boundary conditions a infinity, and cannot get any solution other than the trivial one.

To sum up, I would like to be pointed towards references for the non-unicity of solutions when boundary conditions at infinity are applied, and to understand how to numerically solve similar problems.

The ODE has a physical interpretation and I was unsure whether the question were more apt for Physics StackExchange, or even SciComp StackExchange. I would certainly re-route if advised, I thought to start here as ultimately the question is primarily on the maths, thanks a lot.

答:
In the far field at $x=+\infty$ you have that everything, $y$, $y'$, ..., is zero, so on the approach you get small values. This allows to ignore the higher-degree terms in $U(y)$ resp. $U'(y)$ and reduce the equation to an approximate linear DE
$$
βy^{(4)}-y''+2αy=0.
$$
This has a characteristic polynomial
$$
0=βλ^4-λ^2+2α\implies (βλ^2+\sqrt{2αβ})^2=β(1+2\sqrt{2αβ})λ^2
$$
etc. Better find the roots numerically.
Looking closely, one can read off that the solutions come in pairs with positive and negative real part. For the part to positive infinity the coefficients to the roots with positive real part have to be zero. This can be translated into a boundary condition at some large finite point.

The same can be done for the far field in direction of negative infinity.

So taking the 4 roots $\lambda_{\pm k}$, $k=1,2$, with $Re(λ_{+k})>0$ etc. then the far field pieces are for $x\ge T$
$$
y_+(x)=A_+e^{λ_{-1}x}+B_+e^{λ_{-2}x}
$$
and for $x<-T$
$$
y_-(x)=A_-e^{λ_{+1}x}+B_-e^{λ_{+2}x}
$$
Leaving the coefficients as free parameters, or as constant components in the ODE system, gives a total of 8 components and thus 8 slots for boundary conditions at $\pm T$.
With these one can establish the general boundary value problem with the first 6 boundary conditions being the trivial
\begin{align}
y(T)&=y_+(T),& y(-T)&=y_-(-T) \\
y'(T)&=y_+'(T),& y'(-T)&=y_-'(-T) \\
y''(T)&=y_+''(T),& y''(-T)&=y_-''(-T) \\
\end{align}
Using the remaining 2 slots to ensure 3rd order derivative continuity would force the solution to be the zero function. Instead one could use one slot to fix the integral over $y^2$ or similar and use the remaining degree of freedom to minimize the jumps in the 3rd derivative. This should give convincing solution approximations.

(added 2022/4/2) Continuing from the partial factorization, completing it gives factors
$$
βλ^2\pm \sqrt{β(1+2\sqrt{2αβ})}λ+\sqrt{2αβ}=0.
$$
Applying Viete rules tells that the version with the positive sign combines the roots with negative real part and the version with negative sign the roots with positive real part. The corresponding second order DE are those that the solution approximately follows going to $\pm\infty$, respectively, if it stays bounded there. Thus one can get sensible boundary conditions without introducing additional parameters as
$$
βy''(T)+ \sqrt{β(1+2\sqrt{2αβ})}y'(T)+\sqrt{2αβ}y(T)=0
\\
βy''(-T)- \sqrt{β(1+2\sqrt{2αβ})}y'(-T)+\sqrt{2αβ}y(-T)=0
$$
and their versions in one derivative higher.
To force a non-zero solution add the energy integral as
$$
I'(x)=Ky(x)^2,~~ I(-T)=-1,~~I(T)=1.
$$
The parameter $K$ is also a free parameter or a constant state component, depending on what the solver permits to use.


答:
Vector trple product is defined as  $$\vec P \times (\vec Q \times \vec R)= (\vec P \cdot \vec R)\vec Q-(\vec P\cdot \vec Q) \vec R.$$ Then
$$\vec V=(\vec A \times \vec B) \times \vec A=-\vec A \times (\vec A \times \vec B)=-(\vec A \cdot \vec B)\vec A+(\vec A \cdot \vec A)\vec B.$$
If $\vec A$ is unit vector then,
$$\vec V=\vec B-(\vec B \cdot \hat A)\hat A.$$
Then $\vec V$ is the component of $\vec B$ perpendicular to $\vec A$.

Otherwise ingeneral $\vec V$ is perpendicular to $\vec A$ and coplanar with $\vec A$ anf $\vec B$.

Edit:
$\vec V$ is also the direction of projection of a line (with direction  as $\vec B$) in the plane $ \vec A. \vec r=d$.


问:
Prove that every natural number is either even or odd using induction
It is a pretty basic question at first sight. It seems to be intuitively correct but I can't figure out a way to prove it. I think we have to use strong induction but even after sitting on this question for a long time, I still haven't got any leads.</p>

答:
Base case: $0$ is even since $0 = 2 \times 0$.

Assumption: Assume that $n \in \mathbb{Z}$ is even or odd. Hence, $n=2k$ or $n=2k+1$ with $k \in \mathbb{Z}$.

Inductive step: If $n$ is even, then $n=2k$ with $k \in \mathbb{Z}$. Adding one to both sides of the equality, we find that $n+1=2k+1$ (i.e. $n$ is odd). Hence, if $n$ is even, then $n+1$ is odd.* If $n$ is odd, then $n=2k+1$ with $k \in \mathbb{Z}$. Adding one to both sides of the equality, we find that $n+1=2k+2=2(k+1)$. Since $n+1$ is the product of $2$ and an integer, $n+1$ is even.** Therefore, regardless of whether $n$ is even or odd, $n+1$ is even or odd.

Hence, if the statement is true for $n$, then it is true for $n+1$. Since the statement is true for $n=0$, it must be true for all $n \in \mathbb{N}$ by mathematical induction.

*We have actually proven the stronger result that if $n$ is even, then $n+1$ is odd, but this is not strictly necessary for the proof.

**This actually assumes that if $k$ is an integer, then $k+1$ is an integer. This fact is often adopted as an axiom.


问:
$W_0^{1,p} \cap W^{1,q} = W_0^{1,q}$?
Let $\Omega \subset \mathbb R^n$ be an open set. We denote by $W^{1,p}(\Omega)$ the usual Sobolev spaces and $W_0^{1,p}(\Omega)$ is the closure of $C_c^\infty(\Omega)$ in $W^{1,p}(\Omega)$.

Let $1 \le p < q \le \infty$. Do we have $W_0^{1,p}(\Omega) \cap W^{1,q}(\Omega) = W_0^{1,q}(\Omega)$?

Of course, "$\supset$" is clear, but "$\subset$" is hard.
I do not see that this inclusion can be tackled by applying the definitions directly.

If the boundary of $\Omega$ possesses some regularity (e.g. Lipschitz boundary), then we can define a trace operator (which is independent of the regularity exponent) and the conclusion follows from
$$W_0^{1,r}(\Omega) = \{ u \in W^{1,r}(\Omega) \;\mid\; u = 0 \text{ on }\partial\Omega\}.$$

Is there some easier argument? In particular, I would like to see what happens in case that $\partial\Omega$ has low regularity (or "no regularity" at all).

答:
The answer is negative for general open sets; to see this let $\Omega = B_1 \setminus \{0\} \subset \Bbb R^n$ be the punctured ball in $\Bbb R^n,$ and take $p < n < q.$ Fix any $\varphi \in C^{\infty}_c(B_1)$ such that $\varphi(0)=1.$

By Sobolev embedding convergence in $W^{1,q}$ implies uniform convergence, so $\varphi \in W^{1,q}(\Omega) \setminus W^{1,q}_0(\Omega).$ On the other hand I claim $\varphi \in W^{1,p}_0(\Omega)$ by considering
$$ v_r(x) = 1 - \frac{\int_{|x|}^1 t^{(1-n)/(p-1)} \,\mathrm{d} t}{\int_r^1 t^{(1-n)/(p-1)} \,\mathrm{d} t} $$
for $0 < r < |x| < 1,$ setting $v_r(x) = 0$ for $|x| \leq 1.$ Then $v_r \in C^{1}_c(\Omega)$ and one can compute that
$$ \int_{B_1} |\nabla v_r(x)|^p \,\mathrm{d} x = \omega_{n-1} \left(\frac{p-1}{n-p}\right)^{1-p} |1 - r^{(p-n)/(p-1)}|^{1-p} \to 0$$
as $r \to 0.$ Hence $v_r \to 1_{\Omega}$ in $W^{1,p}(\Omega),$ and by mollifying $v_r \varphi$ we see that $\varphi \in W^{1,p}_0(\Omega)$ as required.

The above counterexample is furnished using the theory of Sobolev capacity, and I've adapted the calculations from 2.11 of the following reference.

Heinonen, Juha; Kilpeläinen, Tero; Martio, Olli, Nonlinear potential theory of degenerate elliptic equations, Oxford Mathematical Monographs. Oxford: Clarendon Press. v, 363 p. (1993). ZBL0780.31001.

More generally from chapters 2 and 4 of the text we have the following results, which provide some insight about when we may or may not expect results of this type to hold true.

Theorem (2.43): If $\Omega \subset \Bbb R^n$ is open and $E \subset \Omega$ is relatively closed, then we have
$$ W_0^{1,p}(\Omega) = W^{1,p}_0(\Omega \setminus E) $$
if and only if $E$ has $p$-capacity zero.

Theorem (4.5): If $\Omega \subset \Bbb R^n$ is open, then $\varphi \in W^{1,p}(\Omega)$ lies in $W^{1,p}_0(\Omega)$ if and only if there is a $p$-quasicontinuous function $\tilde \varphi$ on $\Bbb R^n$ agreeing with $\varphi$ almost everywhere in $\Omega,$ such that $\tilde\varphi = 0$ $p$-quasieverywhere in $\Bbb R^n \setminus \Omega.$

Using Theorem 2.43 we can generate more counterexamples as follows: suppose $E \Subset \Omega$ has $p$-capacity zero but non-zero $q$-capacity, and choose $\varphi \in C_c^{\infty}(\Omega)$ such that $\varphi \equiv 1$ on $E.$ Then we see that $\varphi \not\in W^{1,q}_0(\Omega \setminus E),$ but $\varphi \in W^{1,p}_0(\Omega \setminus E) = W^{1,p}_0(\Omega).$ The above counterexample does this by noting the point $\{0\}$ has $p$-capacity zero if and only if $p < n,$ reproducing the necessary calculations.

Title: $W_0^{1,p} \cap W^{1,q} = W_0^{1,q}$?


问:
How to minimize $\Big( \frac{2}{a^2}+\frac{1}{b^2}\Big)(a^2+b^2)$
I've tried completing the square but I can't however i think there's something called a cauchy schwarz inequality that could help me solve this. Help?

But how $\frac2{a^2}$ is not a square number for the inequality $$(++)^2\le(^2+^2+^2)(^2+^2+^2)$$ do you split it into $\frac1{a^2}+\frac1{a^2}$?

Body:
AM-GM

$3+2(b^2/a^2)+(a^2/b^2) \ge$

$3+2\sqrt{2(b^2/a^2)(a^2/b^2)}=$

$3+2\sqrt{2};$

Equality:

$2(b^2/a^2)=(a^2/b^2);$

$2b^4=a^4,$ or $a^2=\sqrt{2}b^2.$


问:
What is the Picard group of a variety of type G/P with $G$ a semi simple algebraic group and $P$ a parabolic subgroup?
Let $G$ be a simple algebraic group and let $P$ be a parabolic subgroup of $G$. It follows $X:=G/P$ is a smooth projective variety - the flag variety of $G$ corresponding to $P$.

Is it true that the following holds:

Pic($X$) has rank $1$ iff $P$ is a maximal parabolic subgroup.

Why? Where do i find a reference?

答:
I do not have a precise reference for this. Let $G:=SL(V)$ with $V\cong k\{e_1,..,e_d\}$ with $k$ a field. Given any parabolic subgrop $P$ of $SL(V)$, there is a flag $V_{\bullet} \subseteq V$ such that $P$ is the subgroup of elements stabilizing the flag: Let

$$ 0 \neq V_1 \subsetneq V_2 \subsetneq \cdots \subsetneq V_{l} \subsetneq V $$

be such a flag  $V_{\bullet}$ in $V$ with $dim(V_i):=d_i$. Let $P \subseteq SL(V)$ be the (parabolic) subgroup fixing the flag $V_{\bullet}$. It follows $SL(V)/P$ is the "flag variety" parametrizing flags of type $V_{\bullet}$ in $V$. I believe there is an isomorphism

$$ u:\mathbb{Z}^l \cong \operatorname{Pic}(SL(V)/P) $$

induced by the closed embedding

$$ \phi: SL(V)/P \rightarrow \mathbb{G}(d_1,V) \times \cdots \times \mathbb{G}(d_l,V)$$

defined by

$$\phi([V_{\bullet}]):=([V_1],..., [V_l]).$$

There is an isomorphism

$$\operatorname{Pic}(\mathbb{G}(d_i,V))\cong \mathbb{Z}$$

with generator $\mathcal{O}(1)$. Hence if

$$p_i: \mathbb{G}(d_1,V) \times \cdots \times \mathbb{G}(d_l,V)\rightarrow \mathbb{G}(d_i,V)$$

is the projection, it follows any invertible sheaf $\mathcal{L}\in \operatorname{Pic}(SL(V)/P)$ is on the form

$$\mathcal{L}\cong \phi^*(p_1^*\mathcal{O}(n_1)\otimes \cdots \otimes p_l^*\mathcal{O}(n_l)).$$

Hence the isomorphism $u$ is the following map:

$$u(n_1,..,n_l):= \phi^*(p_1^*\mathcal{O}(n_1)\otimes \cdots \otimes p_l^*\mathcal{O}(n_l)).$$

Question: "$Pic(X)$ has rank $1$ iff $P$ is a maximal parabolic subgroup.
Why? Where do i find a reference?"

Anser: The answer is "no" for the following reason:

The quotient $\operatorname{Pic}(SL(V)/P)\cong \mathbb{Z}$ iff $l=1$ which is iff $SL(V)/P\cong \mathbb{G}(d_1,V)$ is the grassmannian of $d_1$-planes in $V$.

Over the complex numbers I believe you may find this in:

Akhiezer, Dmitri N. "Lie group actions in complex analysis". Aspects of Mathematics. E27. Braunschweig: Vieweg. vii, 201 p. (1995).

In particular page 65 in the above book mentions the following: The quotient $SL(V)/P$ is projective iff $P$ is parabolic. Moreover for any parabolic subgroup $P$, there is a flag $V_{\bullet}$ in $V$ with the property that $P$ is the subgroup stabilizing the flag.

If $G$ is an "affine algebraic group" (of finite type) over a field $k$ you may always realize $G$ as a closed subgroup of $GL_k(V)$, where $V$ is a finite dimensional $k$-vector space, and if $G$ is simple (or semi simple) you may ask for similar results: "Is every parabolic subgroup of $G$ a subgroup stabilizing a flag in $V$?" and "Is the quotient $G/P$ projective iff $P$ is parabolic?". I believe the answers to these questions is "well known" but I do not have a precise reference.


问:
Find $\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})$

Find $\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})$

I know that the solution is $1$, but that’s what I got:

$\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})= \lim_{x \to 0} 1+ \lim_{x \to 0} x \sin {\frac{\pi}{x}}$
$=1+\pi \lim_{x \to 0} \frac{\sin {\frac{\pi}{x}} }{\frac {\pi}{x}}$
$=1+\pi (1)$
$=1+\pi $

Can you tell me where was I mistaken?

EDIT:

$|\sin \frac{\pi}{x}| \leq 1$

$-1\leq \sin \frac{\pi}{x} \leq 1$

$-|x| \leq |x \sin \frac{\pi}{x}| \leq |x|$

So $  \lim_{x \to 0} x \sin {\frac{\pi}{x}} =0$

And therefore $\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})=1$

Thanks

答:
To compute
$$
\lim_{x \to 0}\frac{\sin(\pi/x)}{\pi/x} \, ,
$$
presumably you wrote that $\sin(\pi/x) \approx \pi/x$ and proceeded from there. However, this is not valid. Can you see why?


问:
Sum of Logarithms with different bases
it is my exercise to compute the numerical value of $x$ and $y$, where $$x:=\log_{20}(100)+\log_{100}(20),\\
y:=\log_{\frac{1}{2}}(70)+\log_{\frac{1}{10}}(200)+\log_{\frac{3}{2}}\frac{1}{100}.$$

In order to do so, I tried to simplify the expressions in various ways (by using the usual rules of logarithm), but I can not do it in a satisfactory way, such that the values are obvious. Is there any trick you can tell me which might work?

Let my show you, for example, what I tried to find $x$ explicitly. First I observed that the terms $\log_{20}(100)$ and $\log_{100}(20)$ do not seem to be computable individually. As I said before, I tried then to manipulate it in various ways, but I always get stuck, e.g.:

$$x= \log_{20}(100) + \frac{1}{\log_{20}(100)}=?\\
x= 2 \log_{20}(5)+ 2\log_{20}(2)+ 2\log_{100}(2)+ \log_{100}(5) = ?$$
etc.

I would appreciate your help very much!

Best

Title: Sum of Logarithms with different bases

Body:
There is a formula to change the base of logarithms
$$\log_a b=\frac{\log_c b}{\log_c a}$$
for instance
$$\log_{20} 100+\log_{100} 20=\frac{\log_{100} 100}{\log_{100} 20}+\log_{100} 20=\frac{1}{\log_{100} 20}+\log_{100} 20$$
Note that $$\log_{100}x=\frac{\log_{10}x}{\log_{10}100}=\frac{1}{2}\log_{10}x$$


答:
Hint:

For real $x,$  $$\left|\sin\dfrac{\pi}x\right|\le1$$

$$\implies\lim_{x\to0}x\cdot \sin\dfrac{\pi}x=?$$

Body:
Find $\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})$

I know that the solution is $1$, but that’s what I got:

$\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})= \lim_{x \to 0} 1+ \lim_{x \to 0} x \sin {\frac{\pi}{x}}$
$=1+\pi \lim_{x \to 0} \frac{\sin {\frac{\pi}{x}} }{\frac {\pi}{x}}$
$=1+\pi (1)$
$=1+\pi $

Can you tell me where was I mistaken?

EDIT:

$|\sin \frac{\pi}{x}| \leq 1$

$-1\leq \sin \frac{\pi}{x} \leq 1$

$-|x| \leq |x \sin \frac{\pi}{x}| \leq |x|$

So $  \lim_{x \to 0} x \sin {\frac{\pi}{x}} =0$

And therefore $\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})=1$

Thanks


问:
Calculate surface area of $f(x+y+z)$

Calculate the following surface integral:
$$\iint_{\sigma}(x+y+z)\mathrm{d}S$$
When $\sigma$ is the plane $x+y=1$ (x,y positive) between $z=0$ and $z=1$.

For some reason I don't manage because I can't express $z$ in terms of $x$ and $y$.

答:
While this is a more straightforward case but given your question that you are not able to parametrize as $z =f(x, y)$, you always do not have to express $z$ in terms of $x, y$. We can also express as $x = f(y,z)$ for example. It depends on the surface and in which plane, the projection would make more sense.

For this question, let's take the projection in $YZ$ plane.

Here the given surface is $x+y=1, x, y \geq 0$, so we can express it as $x = f(y) = 1-y$.

So $\sqrt{1+ (\frac{\partial f}{\partial y})^2 + (\frac{\partial f}{\partial z})^2} \ = \sqrt2$

Also integrand $x+y+z = 1-y+y+z = 1+z$

So the surface integral is

$\displaystyle \int_0^1 \int_0^1 \sqrt2 \ (1+z) \ dy \ dz = \frac{3}{\sqrt2}$

Title: Calculate surface area of $f(x+y+z)$


问:
An area of a convex polyhedral surface homeomorphic to a sphere vs an area of its shadow
Consider a convex polyhedral surface $P$ in $\mathbb{R}^3$. If $\pi_v : \mathbb{R}^3\rightarrow
\{ x\in \mathbb{R}^3| x\perp v\}$ where $v\in \mathbb{S}^2$ is an orthogonal projection, then $$
\frac{1}{{\rm area}\ \mathbb{S}^2 }\int_{v\in \mathbb{S}^2} \ {\rm area}\ \pi_v(P)\ d{\rm Vol}_v = \frac{1}{4}\cdot {\rm area}\ (P) $$

This can be proved by using mutivariable calculus. But by using the fact that unit ball in $\mathbb{R}^2$ has an area, which is $\frac{1}{4}$ of area of $\mathbb{S}^2$, how can prove this ?

答:
Assume that $F$ is a face in convex polyhedral surface $P$ homeomorphic to a sphere.
If $v\in \mathbb{S}^2$, and $\pi_v$ is an orthogonal projection of
direction $v$, then

\begin{align*} &
{\rm Average\ area\ of\ the\ shadow\ of}\ P
\\&=   \frac{1}{2} \sum_{F\subset P} \frac{1}{{\rm area} \ \mathbb{S}^2}\int_{v\in \mathbb{S}^2}\ {\rm area}\ \pi_v
(F)\ dv
\\&=  \frac{1}{2} \sum_{F\subset P} \frac{1}{{\rm area} \ \mathbb{S}^2}
\int_{v\in \mathbb{S}^2}\ {\rm area}\ (F) |v\cdot (0,0,1)|\  dv
\\&= \frac{1}{2} {\rm area}\ (P) \frac{1}{{\rm area}\
\mathbb{S}^2 }\int_v\ |v\cdot (0,0,1) |\ dv
\\&= \frac{1}{4} {\rm Area} (P) \end{align*}


答:
The final limit is just 1. Due to Sandwich theorem:
$$-x \le x \sin (\pi/x) \le x \implies \lim_{x \to 0} x \sin(\pi/x) =0$$

Title: Find $\lim_{x \to 0} (1+x \sin {\frac{\pi}{x}})$


问:
Show that if $\forall x \in [0,1], P(x) \ge 0$ therefore $\exists A,B,C,D \in \Bbb R[X], P=A^2+XB^2+(1-X)C^2+X(1-X)D^2$.

Let $P \in \Bbb R[X]$. Show that if $\forall x \in [0,1], P(x) \ge 0$ therefore $\exists A,B,C,D \in \Bbb R[X], P=A^2+XB^2+(1-X)C^2+X(1-X)D^2$.

I showed that this is true for $\deg P=2$ but I don't see how to generalize my proof. I tried to introduce coefficients for the polynomials, also tried Lagrange interpolation without success. Does someone have any hint?

If we consider $P \ge 0$ on $\Bbb R$ a classical problem shows that $P=A^2+XB^2$. Adapting the proof of this result seems difficult.

Edit: I found this reference (a very similar problem with solution): George Pólya and Gábor Szegő, Problems and Theorems in Analysis II (problem $47$, p.$78$).

答:
HINT:

The interval $[0,1]$ is defined by the inequality $x(1-x)\ge 0$.

It is better ( and easier in fact) to show a stronger statement: every
polynomial that is $\ge 0$ on $[0,1]$ is of the form

$$P(x)= A^2(x) + x(1-x)D^2(x)$$

( the terms $B$, $C$ may be made $0$).

It's easier to show by induction since the multiplicative property is immediate. Let's see how we can prove it.

A real polynomial that is positive on $[0,1]$ is a product of polynomial of this form

$x^2 + b x + c$, with non-real roots $b^2 - 4 c< 0$

$(x-\alpha)$ where $\alpha \le 0$

$-(x-\beta)$, where  $\beta \ge 1$.

$Q^2\ $   a square

Now you need to show that every  polynomials of each of the four types has a writing as above.

Obs: Any polynomial of form $x B^2(x) + (1-x) C^2(x)$ is clearly positive on $[0,1]$. So it will have a representation
with $A$, $D$. And conversely!  So we have two possible representations for polynomials positive on $[0,1]$.


问:
is every antiderivarive an integral?
Given a function $f:\mathbb R\to\mathbb R$, can every antiderivative $F$ of $f$ be expressed as $$F(x)=f(x_0)+\int_{x}^{x_0}f(t)\mathrm dt$$ for some $x_0$?

答:
Assuming Riemann integration:

No. There are functions which can have an anti-derivative without being (Riemann) integrable.

In other words, a function may be differentiable without the derivative being integrable. One such example is Volterra's function.

Another example could be $F : [-1, 1] \to \Bbb R$ defined by

$$F(x) := \begin{cases} x^2 \sin\left(\frac{1}{x^2}\right) & x \neq 0 \\ 0 & x = 0 \end{cases}$$

In this case, you can verify that $F$ is differentiable but $F'$ is not bounded. (The earlier example is nicer in that the derivative is bounded and still not integrable.)

Title: is every antiderivarive an integral?


问:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$ subject to $x+y=23$

Solve the system of equations:
$\sqrt{x^2+12y}+\sqrt{y^2+12x}=33$, $x+y=23$

The obvious way to solve it is by substituting for one variable. However I was looking for a more clever solution and went ahead and plotted two graphs.

The first graph looks pretty weird so please help as to how to proceed with this graphically or an easier algebraic method.
Thanks :)

答:
The obvious (simple) solution is not repeated; instead we try to see the full picture.

Whenever there are radicals with polynomials of the form including $\pm$

$$ \pm \sqrt {P(x,y)}\pm \sqrt {Q(x,y)}= 1, $$

a higher degree polynomial should be recognized whose factors/ equations/constituents appear separately. They can  be graphed separately as below with different colors and labels of each part. $x\pm y=0$ lines and an asymptote are drawn included for reference:

EDIT1:

Continuous segments can be shown graphed together (it can be plotted in desmosas well) in Mathematica.

```<code>ContourPlot[{(Sqrt[x^2 + 12 y] +Sqrt[y^2 + 12 x] +33) *
    (Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] +33)*
    (-Sqrt[x^2 + 12 y] +Sqrt[y^2 + 12 x] +33)*
    (+Sqrt[x^2 + 12 y] +Sqrt[
       y^2 + 12 x] -33)*(+Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] -33)
    (+Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] -33) == 0
  }, {x, -p, p}, {y, -p, p}, GridLines -> Automatic,
 ContourStyle -> {Thick, Red}]
</code>```

Six factors and a product polynomial plot with uncut segments are shown.

A neat rational cubic product polynomial equation is cumbersome and unless seen in that form I feel as having not taken to the finish. A symmetric graph with respect to y-axis could be obtained by $45^{\circ}$ CW rotation.

Body:
The obvious (simple) solution is not repeated; instead we try to see the full picture.

Whenever there are radicals with polynomials of the form including $\pm$

$$ \pm \sqrt {P(x,y)}\pm \sqrt {Q(x,y)}= 1, $$

a higher degree polynomial should be recognized whose factors/ equations/constituents appear separately. They can  be graphed separately as below with different colors and labels of each part. $x\pm y=0$ lines and an asymptote are drawn included for reference:

EDIT1:

Continuous segments can be shown graphed together (it can be plotted in desmosas well) in Mathematica.

ContourPlot[{(Sqrt[x^2 + 12 y] +Sqrt[y^2 + 12 x] +33) *
    (Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] +33)*
    (-Sqrt[x^2 + 12 y] +Sqrt[y^2 + 12 x] +33)*
    (+Sqrt[x^2 + 12 y] +Sqrt[
       y^2 + 12 x] -33)*(+Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] -33)
    (+Sqrt[x^2 + 12 y] -Sqrt[y^2 + 12 x] -33) == 0
  }, {x, -p, p}, {y, -p, p}, GridLines -> Automatic,
 ContourStyle -> {Thick, Red}]

Six factors and a product polynomial plot with uncut segments are shown.

A neat rational cubic product polynomial equation is cumbersome and unless seen in that form I feel as having not taken to the finish. A symmetric graph with respect to y-axis could be obtained by $45^{\circ}$ CW rotation.


问:
What's For Multiplication...
As summation symbol ($\sum$) is for summing up terms, is there a similar notation for multiplication (except factorial notation)?

答:
$\prod$ (\prod) is the multiplicative analogue of $\sum$. It multiplies things together, where multiplication is assumed commutative.

But there's more! For some other commutative binary operations, the result of applying that operation to a list of objects may be denoted in a $\sum$-like notation, but with an enlarged operator taking $\sum$'s place. For example, $\bigwedge$ performs a logical AND ($\wedge$) of whatever follows, $\bigvee$ logical OR ($\vee$) and $\bigoplus$ logical XOR ($\oplus$).

Title: What's For Multiplication...


问:
How to find the correct equations for solving the following geometry question?
I am trying to solve the following question.

My attempt is as follows.

\begin{cases}
x+a+b = 30\\
y+z+c+d=30\\
x+y+z+a=30\\
b+c+d=30\\
x+z+b+c=40\\
y+a+d=20\\
x+y+z+a+b+c+d=60
\end{cases}

Now I am confused in how to get $x-y$.

答:
Draw $EM$ parallel to $AB$ we know $EH:BH=EM:BF=EM:AF=2:3$

Therefore $S_{CEH}={2\over5}S_{CEB}={2\over5}\cdot 40=16$

Also since $I$ is centroid we have $CI:IF=2:1$ so $S_{AIF}={1\over3}S_{CAF}={1\over3}\cdot 30=10$

Therefore the area difference is $16-10=6$.


问:
Clear and convenient notation for choice of signs $(\pm 1)$ in an expression
We often encounter an arbitrary choice of signs in solutions of equations and other things. For example, the orthogonal group $O(2)$ can be written as a double cover
$$
O(2)=SO(2)\cup \begin{bmatrix} 1&0\\0&-1 \end{bmatrix}SO(2).
$$
So, when we want to write done the general form of a matrix in $O(2),$ we need to write
$$
\begin{bmatrix} 1&0\\0&-1 \end{bmatrix}^{j}\begin{bmatrix} \cos \theta&-\sin \theta\\\sin \theta&\cos \theta \end{bmatrix},
$$
where $j=0,1.$

It is often very awkward to use the right notation for this choice of sign. For example, if we use a simplier notation
$$
\begin{bmatrix} 1&0\\0&\pm 1 \end{bmatrix}\begin{bmatrix} \cos \theta&-\sin \theta\\\sin \theta&\cos \theta \end{bmatrix},
$$
then we will run into trouble when we are trying to prove, for example, $O(2)$ is closed under multiplication, since if we write done two elements of $O(2)$ of the form
$$
\begin{bmatrix} 1&0\\0&\pm 1 \end{bmatrix}\begin{bmatrix} \cos \theta&-\sin \theta\\\sin \theta&\cos \theta \end{bmatrix},\begin{bmatrix} 1&0\\0&\pm 1 \end{bmatrix}\begin{bmatrix} \cos \theta'&-\sin \theta'\\\sin \theta'&\cos \theta' \end{bmatrix},
$$
we are assuming that  the two elements take the same sign, and therefore they do not represent an arbitrary choice of two elements, which is annoying.

If we instead just do separate calculations for each sign, then we are essentially doing the same thing several times, so this is not a good option as well.

What is the best way to manage signs like this?

I put done the group theory tag because that's where many questions about signs arise.

答:
To give a "general" answer, we can combine both the comments of mine and Cameron to first note that we can keep $(-1)^n$ and $(-1)^m$ for this particular case.

Of course, along the way, we would use the "closure" property like $(-1)^n \cdot (-1)^m = (-1)^{m+n}$ and the fact that $(-1)^2 = 1$.

In general, when working with $k$-th roots of unity, you could assume them as $\omega^{n}$ and $\omega^{m}$, where $\omega = \exp\left(\dfrac{2\pi\iota}{k}\right)$ is a primitive $k$-th root of unity.


问:
Integral of $1/x^2$ without power rule
I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents

\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}

When using integration by parts, you end up with the same integral in the rhs so it seems out of luck

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}

This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}

you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.

The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that

\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}

and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)

答:
$$I=\int \frac{dx}{x^2} $$ Let $x=\sin(t)$ to make
$$I=\int \cot (t) \csc (t) \,dt=-\csc (t)$$

Title: Integral of $1/x^2$ without power rule


问:
Real Analysis Differentiation
Real analysis question about differentiation

Assume that $f$ and $f'$  are differentiable on $\mathbf R$ and that for every x in $\mathbf R$,  $f(x) + f''(x) = 0 $. Show that $g (x)= f^2(x) + (f'(x))^2$ is a constant.

答:
Suppose that $f:\Bbb R\to\Bbb R$ and $f+f''=0$.

Then $g=f^2+(f')^2$ is constant.

Proof: $g'=2ff'+2f''f'=2f'(f+f'')=0$.

It's amusing to note that this proves uniqueness for the solution:

Cor. If $f(0)=f'(0)=0$ then $f=0$.

And hence

Cor. If $f''+f=0$, $f(0)=a$ and $f'(0)=b$ then $f(x)=a\cos(x)+b\sin(x)$.

Proof: The previous corollary shows that $g=0$, if $g(x)=f(x)-(a\cos(x)+b\sin(x))$.

Title: Real Analysis Differentiation


答:
Updated:

I found a new second substitute much faster:

Method $-1$

Let $x=\tan t$, then we have

$$\int \dfrac {1}{x^2} dx=\int \dfrac{1}{\sin ^2(t)} dt=-\cot (t)+C=-\cot (\arctan x)+C=-\dfrac 1 x+C$$

Method $-2$

How about this substitution?

Let $x=\dfrac {1}{\ln t}$, then we have
$$\int \dfrac {1}{x^2} dx=-\int \dfrac{1}{t} dt=-\ln t+C=-\dfrac 1x+C.$$

Title: Integral of $1/x^2$ without power rule

Body:
I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents

\begin{equation*}
  \int \frac{1}{x^2} \; dx
\end{equation*}

When using integration by parts, you end up with the same integral in the rhs so it seems out of luck

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}

This question is inspired by this blackpenredpen's video

Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write

\begin{equation*}
  \int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}

you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.

The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that

\begin{equation*}
    \int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}

and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)


Title: Integral of $1/x^2$ without power rule

Body: You could use $u=x^{-2}.\,v=x$ so $\int x^{-2}dx=x^{-1}+2\int x^{-2}dx\implies\int x^{-2}dx=-x^{-1}+C$.


问:
Question about poincare-Hopf theorem and trajectories of hamiltonian system
In this Wiki Link: https://en.wikipedia.org/wiki/Poincaré–Hopf_theorem. It says in the following picture that

"According to the Poincare-Hopf theorem, closed trajectories can encircle two centres and one saddle or one centre, but never just the saddle. (Here for in case of a Hamiltonian system)"

My question is how can we apply the Poincare-Hopf theorem here. The area enclosed by the outer blue curve is a manifold with a boundary. However, the vector field is not always pointing outward along the boundary or the blue curve. How to apply the theorem in this example?

答:
Poincare-Hopf can be also applied to compact manifolds with boundary $M$ and vector fields $X$ which are tangent to the boundary. Assuming that $X$ has no zeroes on $\partial M$, $\chi(M)= ind(X)$. You can reduce this to the case of closed manifolds by gluing to copies of $M$ together along the boundary: The resulting manifold $DM$ has a symmetry fixing the boundary pointwise. Extend $X$ to $DM$ by symmetry. The result is a continuous vector field $DX$ on $DM$ and
$$
ind(DX)=2ind(X), \chi(DM)= 2\chi(M).
$$
In your case, $M$ is the 2-dimensional disk $B^2$ (bounded by a closed trajectory of $X$) and $\chi(B^2)=1$. Hence, $X$ cannot have a single saddle zero in $B^2$.


答:
Another way is to substitute $x=e^t \implies dx = e^t dt $:

$$\int \frac{1}{e^{2t}} e^t dt = \int e^{-t} dt = -e^{-t} + C = -\frac 1x +C$$

Here, only the fact that $e^x$ is its own derivative is used.

Title: Integral of $1/x^2$ without power rule


问:
Find sum of an unknown sequence

Given $a_1,a_2...,a_n$ to be integers bigger than $0$ with different of $2$ from each other and have prime factor smaller than $5$. Let $S_n=\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}$ then find $S_n$

We are giving 4 answers
\begin{equation}
A.S_n<3\\ B.8<S_n<12\\ C.3\leq S_n\leq 8\\ D.12\leq S_n<20
\end{equation}
I dunno how to approach this but think for this kind of sum of fraction it'd be hard to have the sign $\leq$ and since in the giving it has prime factor smaller than $5$ so I think $B.8<S_n<12$ is the answer. ( This problem might have some issue at " different of $2$ with each other" because it is not in English and moreover I also have some trouble understand that phrase too But I think that's how it is ) please help

Edit as stated in the comment by @Oussema, he/she suggested that " different of $2$ from each other" actually means pairwise distinct, as in each two terms are different.

答:
If the numbers $a_n$ are pairwise distinct and they have $2$ or $3$ as prime factors, then the sum is $$\frac12+\frac13+\frac14+\frac16+\frac18+\frac19+\ldots +\frac{1}{2^i3^j}< 3$$
Indeed $$ \sum _{i=0}^{\infty } \frac{1}{2^i3^j}=3$$

Edit

$$\sum _{j=0}^{\infty }\sum _{i=0}^{\infty } \frac{1}{2^i3^j}=\sum _{j=0}^{\infty }\frac{1}{3^j}\sum _{i=0}^{\infty } \frac{1}{2^i}=\sum _{j=0}^{\infty }\frac{1}{3^j}\cdot \frac{1}{1-\frac12}=2\sum _{j=0}^{\infty }\frac{1}{3^j}=$$
$$=2\cdot\frac{1}{1-\frac13}=2\cdot\frac{3}{2}=3$$

Title: Find sum of an unknown sequence


答:
Normally, I wouldn't give more than a hint, because the OP hasn't really shown work.  However, this problem is weird because I'm unsure how to interpret:

Given $a_1,a_2...,a_n$ to be integers bigger than $0$ with different of $2$ from each other and have prime factor smaller than $5$.

It's been suggested that one of the constraints is that $a_1, a_2, \cdots, a_n$ are all distinct.

However, it is very unclear what the other constraint signifies.  Does this mean that for every element $a_k \in \{a_1, a_2, \cdots, a_n\}$, that there exist non-negative integers $r,s$ such that $a_k = 2^r \times 3^s$?

In an attempt to reverse-engineer an answer, I'm going to assume that the answer to the question posed in the previous paragraph is yes.

Partition the possible values of $a_n = 2^r \times 3^s$ by the exponent $r$.  That is, let $T_0$ denote all integers of the form $(2^0 \times 3^s)$, $T_1$ denote all integers of the form $(2^1 \times 3^s)$ and so forth.

Then, the sum of the reciprocals of the elements taken from $T_k$ must be less than $(1/2^k) \times [1 + (1/3) + (1/9) + \cdots] = (1/2^k) \times (3/2)$.

Since $k$ can range through all of the non-negative integers, we have that the sum of the reciprocals of $T_0 \cup T_1 \cup T_2 \cdots$ must be less than

$$ (3/2) \times [1 + (1/2) + (1/4) + \cdots] = (3/2) \times 2 = 3.$$

Title: Find sum of an unknown sequence

Body:
Given $a_1,a_2...,a_n$ to be integers bigger than $0$ with different of $2$ from each other and have prime factor smaller than $5$. Let $S_n=\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}$ then find $S_n$

We are giving 4 answers
\begin{equation}
A.S_n<3\\ B.8<S_n<12\\ C.3\leq S_n\leq 8\\ D.12\leq S_n<20
\end{equation}
I dunno how to approach this but think for this kind of sum of fraction it'd be hard to have the sign $\leq$ and since in the giving it has prime factor smaller than $5$ so I think $B.8<S_n<12$ is the answer. ( This problem might have some issue at " different of $2$ with each other" because it is not in English and moreover I also have some trouble understand that phrase too But I think that's how it is ) please help

Edit as stated in the comment by @Oussema, he/she suggested that " different of $2$ from each other" actually means pairwise distinct, as in each two terms are different.


问:
If $(X,A)$ has homotopy extension property, then so does $(X\times Y,A\times Y)$?
If $(X,A)$ is a pair of spaces that has the homotopy extension theory, then so does $(X\times Y,A\times Y)$ for all spaces $Y$. How do I prove this?

答: Hatcher, on page 14 of his book, proves that a pair $(X,A)$ has the homotopy extension property if and only if $X \times \{0\} \cup A \times I$ is a retract of $X \times I$. Using the latter characterisation, we can easily form a retraction of $\widetilde{r} \colon X \times Y \times \{0\} \cup A \times Y \times I$ onto $X \times Y \times I$ using whatever retraction $r\colon X \times \{0\} \cup A \times Y \to X \times I$ we have (I guess by just letting $\widetilde{r}$ be constant on $Y$.) This should answer your question positively.


问:
What is larger -- the set of all positive even numbers, or the set of all positive integers?
We will call the set of all positive even numbers E and the set of all positive integers N.

At first glance, it seems obvious that E is smaller than N, because for E is basically N with half of its terms taken out.  The size of E is the size of N divided by two.

You could see this as, for every item in E, two items in N could be matched (the item x and x-1).  This implies that N is twice as large as E

On second glance though, it seems less obvious.  Each item in N could be mapped with one item in E (the item x*2).

Which is larger, then?  Or are they both equal in size?  Why?

(My background in Set theory is quite extremely scant)

答:
They are both the same size, the size being 'countable infinity' or 'aleph-null'. The reasoning behind it is exactly that which you have already identified - you can assign each item in E to a single value in N. This is true for the Natural numbers, the Integers, the Rationals but not the Reals (see the Diagonal Slash argument for details on this result).

-- Added explanation from comment --

The first reasoning is invalid because the cardinality of infinite sets doesn't follow 'normal' multiplication rules. If you multiply a set with cardinality of aleph-0 by 2, you still have aleph-0. The same is true if you divide it, add to it, subtract from it by any finite amount.

ContentLicense: CC BY-SA 2.5


问:
The direct sum $\oplus$ versus the cartesian product $\times$
In the case of abelian groups, I have been treating these two set operations as more or less indistinguishable. In early mathematics courses, one normally defines $A^n := A\times A\times\ldots\times A$; however in, for example, the fundamental theorem of finitely generated abelian groups, we normally write that every such group is isomorphic to one of the form
$$
\mathbb{Z}^n \oplus \mathbb{Z}_{r_1} \oplus \cdots \oplus \mathbb{Z}_{r_t}
$$
where $\mathbb{Z}^n$ now means $\mathbb{Z}\oplus\mathbb{Z}\oplus\cdots\oplus\mathbb{Z}$.

From an intuition perspective, and in the sense of sets, is this more or less the same as $\mathbb{Z}\times\mathbb{Z}\times\cdots\times\mathbb{Z}$? (Bear in mind I am normally using these ideas in relation to homology groups.)

答:
As long as you restrict to finite index sets, the direct sum and the direct product of commutative groups are identical.

For a general index set $I$, the direct product of commutative groups $\{G_i\}$ is the full Cartesian product $\prod_{i \in I} G_i$, whereas the direct sum $\bigoplus_{i \in I} G_i$
is the subgroup of the direct product consisting of all tuples $\{g_i\}$ with $g_i = 0$
except for finitely many $i \in I$.

The coincidence of the direct sum and the direct product in additive categories can also be explained in categorical terms, but I'll let someone else take a crack at that if they so desire.


答:
Let $x=\cos \theta +i\sin \theta$. By De Moivre's Theorem, $$\frac{1}{x^2}=x^{-2}=\cos (-2\theta)+i\sin (-2\theta)=\cos (2\theta)-i\sin(2\theta)$$
Also $$dx=(-\sin \theta+i\cos \theta) d\theta$$
The integral becomes $$\int( \cos 2\theta-i\sin 2\theta)(-\sin \theta+i\cos\theta)d\theta\\
=\int (-\sin \theta\cos 2\theta+i\cos \theta\cos 2\theta+i\sin\theta\sin 2\theta+\cos \theta\sin2\theta)d\theta\\
=\int -\frac{1}{2}(\sin 3\theta-\sin \theta)+i\frac{1}{2}(\cos 3\theta+\cos \theta)+i\frac{1}{2}(\cos\theta-\cos 3\theta)+\frac{1}{2}(\sin 3\theta+\sin\theta)d\theta \\ =\int \sin \theta +i\cos \theta d\theta=-\cos \theta +i\sin\theta +C\\
=-(\cos \theta-i\sin \theta)+C=-(\cos (-\theta)+i\sin (-\theta))+C=-\frac{1}{x}+C$$

You can find $\int x^a\:dx,a\in \mathbb Z$ in a similar manner.

Title: Integral of $1/x^2$ without power rule


问:
The arcsine is "not really a function"
My math teacher said that later on in life, if you pursue math, you would see that the arcsine function isn't actually a function. I do not see how it's not a function.</p>

答:
Not only is $\sin$ a perfectly well-defined function in the set-theoretic sense, we actually have an effective formula to compute the values of this function. Taking a wild guess here on what the teacher might have meant, perhaps it would be that relying on the geometric meaning of the sine function is not a terribly accurate way of doing things. That would be a correct statement. There is a very rigorous way to define the sine function but it relies on a pretty heavy theorem.

Here we go: The sine function is the unique function $f\colon \mathbb R \to \mathbb R$ satisfying the initial condition $f(0)=0,f'(0)=1$ and the differential equation $f''=-f$. In order for this to be a legitimate definition we invoke a theorem from the theory of function that guarantees that this particular differential equation does have a unique solution given the initial condition.

So, the sine function exists as a particular case of a theorem that allows us to construct many other functions. Now, from this information about the sine function one can find its Taylor series, compute its radius of convergence to be $\infty$, and use the remainder form in order to compute $\sin(t)$ for any value of $t\in \mathbb R$ and to any degree of accuracy (given sufficient computational time).

I would not say that the existence of the sine function is trivial. But it certainly is a function. Perhaps it is worth noting here that the common definition of $\sin(t)$ as the $Y$ coordinate of the point on the unit circle at $t$ radians from the positive $X$ axis measuring counterclockwise is correct but it relies on the notion of distance/angle. That is a tricky business. These geometric subtleties are avoided by the theorem mentioned above, essentially converting the whole thing to a more algebraic situation and good dose of analysis.

Title: The arcsine is "not really a function"


Title: Integral of $1/x^2$ without power rule

Body:
$$ \frac{1}{x^2} = e^{\ln \left ( \frac{1}{x^2} \right ) }  = e^{\ln \left ( x^{-2} \right ) } = e^{-2\ln (x  ) } = (e^{-2})^{\ln(x)} $$

$$ \text{when} ~~a \in \mathbb{R^+}: ~~~\int a^{\ln(x)} dx = \int \frac1x x^{\ln(a) + 1} dx \\ u=\ln x \Rightarrow du = \frac1x dx \\ = \int e^{(\ln(a) + 1)u} \\v = ( \ln(a) + 1)u \Rightarrow du = \frac{1}{( \ln(a) + 1)} dv \\ = \frac{1}{ \ln(a) + 1} \int e^v dv = \frac{1}{\ln(a) + 1} e^v+C= \frac{1}{\ln(a) + 1}e^{(\ln(a) + 1)u} + C = \frac{e^{(\ln(a) + 1)u}}{\ln(a) + 1} + C = \frac{x^{\ln( a ) + 1}}{\ln(a) + 1} + C $$

Substituting $a = e^{-2}$ we get:

$$ \frac{x^{-2 + 1}}{-2 + 1} + C = - \frac{1}{x} + C$$


问:
Probability generating function and expectation
Let $X$ be Poisson random variable with parameter $Y$, where $Y$ is Poisson random variable, with parameter $\mu$. Prove that, $G_{X+Y}(s)=\exp\{\mu (s\exp^{s-1}-1)\}$

I know that, Poisson r.v. generating function is $G(s)=\exp\{\lambda(s-1)\}$. Do I need to calculate joint probability distribution first ($P(Z)$), where $Z=X+Y$?

答: Hint: As recalled in the post, $E[s^X\mid Y]=\exp\{Y(s-1)\}$ hence $G_{X+Y}(s)=E[s^Xs^Y]$ is also $G_{X+Y}(s)=E[E[s^X\mid Y]s^Y]=E[\exp\{Y(s-1)\}s^Y]=E[t^Y]$ for $t$= $____$. Furthermore, for every $t$, $E[t^Y]=\exp\{\mu(t-1)\}$ hence $G_{X+Y}(s)=$ $______$.</p>


问:
When is a category the colimit of its slice categories
My question: is it true in general for a small category $\mathcal{C}$ that the canonical functor
\begin{aligned}
\operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x \to \mathcal{C}
\end{aligned}
is an equivalence of categories? If so, is there a canonical way of seeing this without explicitly constructing an inverse?

Let me explain the statement in more detail. For an object $x$ in $\mathcal{C}$, let $C/x$ denote the over-category of $\mathcal{C}$ over $x$, with objects given by morphisms $f: y \to x$ in $\mathcal{C}$ and morphisms from $f: y \to x$ to $f': y'\to x$ given by morphisms $g: y \to y'$ in $\mathcal{C}$ such that $f = f' \circ g$. These over-categories assemble into a functor
\begin{aligned}
\mathcal{C} \to \operatorname{Cat}: c \mapsto \mathcal{C}/x
\end{aligned}
to the (large) category of small categories. Indeed, for a morphism $h: x \to x'$, we get a functor $\mathcal{C}/x \to \mathcal{C}/x': (f: y \to x) \mapsto (h \circ f: y \to x')$.

For every $x$, there is a canonical functor source functor $s: \mathcal{C}/x\to \mathcal{C}$ given by sending $f: y \to x$ to its source $y$. Essentially by definition, these source functors assemble into a cocone over $\mathcal{C}$ in $\operatorname{Cat}$, and thus give rise to a functor
\begin{aligned}
\operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x \to \mathcal{C}.
\end{aligned}
This is the canonical functor referred to in the statement above.

Attempt: one can try to simply write down an inverse functor $\mathcal{C} \to \operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x$ as follows. An object $x$ is sent to the object in the colimit system represented by the object $(\operatorname{id}_x: x \to x)$ in $\mathcal{C}/x$. If $f: x \to x'$ is a morphism in $\mathcal{C}$, notice that this object $(\operatorname{id}_x: x \to x)$ gets identified in the colimit system with the object $(f: x \to x')$ in $\mathcal{C} / x'$. Also notice that $f: x \to x'$ forms a morphism in $\mathcal{C}/x'$ from $(f: x \to x')$ to $\operatorname{id}_{x'}: x' \to x'$. It thus seems that the assignment $x \mapsto [\operatorname{id}_x: x \to x]$ extends to a functor $\mathcal{C} \to \operatorname{colim}_{c\in \mathcal{C}}\mathcal{C}/x$. I think this should be the desired inverse.

Question: Does the above proof attempt work? If so, is there a more invariant way of proving that the canonical functor is an equivalence? I am interested in a version of the statement for $\infty$-categories, for which one cannot simply write down an inverse functor by giving what it does on objects and morphisms.

答:
Apparently I confused myself. The claim is true for both the strict colimit and the pseudo colimit, as you realised.

Consider the codomain projection $\operatorname{codom} : [\mathbf{2}, \mathcal{C}] \to \mathcal{C}$.
This is a Grothendieck opfibration whose fibres are the slice categories $\mathcal{C}_{/ x}$ as $x$ varies, and the opcartesian morphisms are the ones whose domain part is an isomorphism.
There is a canonical splitting, exhibiting this as the Grothendieck construction of the diagram $x \mapsto \mathcal{C}_{/ x}$ you are interested in.
It is a general fact that the Grothendieck construction of a diagram is its oplax colimit, and one obtains the pseudo colimit by inverting the opcartesian morphisms.

Proposition. The domain projection $\operatorname{dom} : [\mathbf{2}, \mathcal{C}] \to \mathcal{C}$ is a localisation functor.

Proof. The domain projection is the right adjoint of the functor $\mathcal{C} \to [\mathbf{2}, \mathcal{C}]$ that sends each object $x$ to $\textrm{id}_x : x \to x$. The latter is fully faithful, so the domain projection is a coreflector. But reflectors are automatically localisation functors, so we are done. ■

Corollary. The pseudo colimit of $x \mapsto \mathcal{C}_{/ x}$ is equivalent to $\mathcal{C}$ via the canonical cocone. ■

The strict colimit can be constructed in a similar way: instead of inverting the opcartesian morphisms, we force them to be identity morphisms.

I am informed that similar general results hold in the setting of quasicategories, so the pseudo colimit version should translate directly.


问:
Is this family of functions equicontinuous in $\mathbb{R}$?
Let $\{ \varphi_{\lambda}(x,y): \lambda \in \mathbb{R} \}$ be the family of functions defined by $$\varphi_{\lambda}(x,y):= \frac{1}{1 + x^4 + \frac{1}{2}\arctan(\lambda)\sin(y^6)}$$ for all $\lambda \in \mathbb{R}$. Is this family of functions equicontinuous in $\mathbb{R}$?

It is easy to see that this family of functions is uniformly bounded by $1$, but I don't know how to see if is equicontinuous. I have thought about using subsequences and maybe Arzela-Ascoli theorem, but I do not get the point... For example, it is not equi-Lipschitz as its partial derivatives are not bounded in a convex open set, so this doesn't prove nothing neither...

Any hint will be appreciated. Thanks in advance

答:
Take $\lambda,\mu$.
Since $\frac1a-\frac1b = \frac{b-a}{ab}$ we have
$$
|\phi_\lambda(x,y)-\phi_\mu(x,y)|\le \frac1{(1-\frac\pi4)^2} \frac12 |\arctan(\lambda)-\arctan(\mu)|.
$$
This implies $\phi_\mu\to \phi_\lambda$ uniformly for $\arctan(\mu) \to \arctan(\lambda)$.

Since the set $\{\arctan \lambda: \ \lambda\in \mathbb R\}$ is bounded, we can find for each  sequence $(\lambda_n)$ a subsequence such that $(\arctan(\lambda_{n_k}))$  converges, and so $(\phi_{\lambda_{n_k}})$ converges uniformly. Then by Arzela-Ascoli, the family is equicontinuous.


答:
A direct calculation/estimate works: The denominator is bounded below by $1 - \pi/4 > 0$, so that
$$
| \varphi_{\lambda}(x_1,y_1) - \varphi_{\lambda}(x_2,y_2)|
\le  \frac{|x_2^4 + \frac{1}{2}\arctan(\lambda)\sin(y_2^6) - x_1^4 - \frac{1}{2}\arctan(\lambda)\sin(y_1^6)|}{(1-\pi/4)^2} \\
\le \frac{|x_2^4 -x_1^4|+ \frac{\pi}{4}|\sin(y_2^6)-\sin(y_1^6)|}{(1-\pi/4)^2} = C |f(x_2)-f(x_1)| + D|g(y_2)-g(y_1)|
$$
with some constants $C, D$ and the continuous functions $f(x) = x^4$ and $g(y) = \sin(y^6)$.
For given $(x_1, y_1) \in \Bbb R^2$ this becomes arbitrary small if $(x_2, y_2)$ is sufficiently close to $(x_1, y_1)$, independently of $\lambda$.

Body:
Let $\{ \varphi_{\lambda}(x,y): \lambda \in \mathbb{R} \}$ be the family of functions defined by $$\varphi_{\lambda}(x,y):= \frac{1}{1 + x^4 + \frac{1}{2}\arctan(\lambda)\sin(y^6)}$$ for all $\lambda \in \mathbb{R}$. Is this family of functions equicontinuous in $\mathbb{R}$?

It is easy to see that this family of functions is uniformly bounded by $1$, but I don't know how to see if is equicontinuous. I have thought about using subsequences and maybe Arzela-Ascoli theorem, but I do not get the point... For example, it is not equi-Lipschitz as its partial derivatives are not bounded in a convex open set, so this doesn't prove nothing neither...

Any hint will be appreciated. Thanks in advance


答:
Hint: Use that
$$
{\rm E}[s^{X+Y}]=\sum_{y=0}^\infty{\rm E}[s^{X+Y}\mid Y=y]\cdot P(Y=y)=\sum_{y=0}^\infty s^y\cdot {\rm E}[s^X\mid Y=y]\cdot P(Y=y)
$$
along with the fact that $X\mid Y=y\sim \mathrm{po}(y)$.

Title: Probability generating function and expectation

Body:
Let $X$ be Poisson random variable with parameter $Y$, where $Y$ is Poisson random variable, with parameter $\mu$. Prove that, $G_{X+Y}(s)=\exp\{\mu (s\exp^{s-1}-1)\}$

I know that, Poisson r.v. generating function is $G(s)=\exp\{\lambda(s-1)\}$. Do I need to calculate joint probability distribution first ($P(Z)$), where $Z=X+Y$?


问:
PDF or CDF of $B^2-B$ for Beta-distributed $B$
I'm looking for a way to compute the CDF (or PDF) of the following random variable:

Let $B\sim Beta(\alpha, \alpha)$ for some $\alpha \geq 1$. Then the random variable in question is $B^2-B$.

Trying to compute the CDF directly from the PDF of the beta distribution results in some nasty integration which I'm not sure how to complete. Ideas would be appreciated!

答:
Let's set

$X\sim Beta(a;a)$ and $Y=X^2-X$

Y is defined on the support $y \in(-0.25;0]$

The transformation function is not monotonic but we can use the Fundamental transformation theorem getting Y density inverting the function as a piecewise one getting

$$f_Y(y)=\frac{\Gamma(2\alpha)}{\Gamma(\alpha)\Gamma(\alpha)}\Bigg[\Bigg(\frac{1}{2}+\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg(\frac{1}{2}-\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}+\Bigg(\frac{1}{2}-\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg(\frac{1}{2}+\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg]\frac{1}{2\sqrt{y+\frac{1}{4}}}$$

$$ \bbox[5px,border:2px solid black]
{
f_Y(y)=\frac{\Gamma(2\alpha)}{\Gamma(\alpha)\Gamma(\alpha)}\frac{\Bigg(\frac{1}{2}+\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}\Bigg(\frac{1}{2}-\sqrt{y+\frac{1}{4}}\Bigg)^{\alpha-1}}{\sqrt{y+\frac{1}{4}}}\cdot\mathbb{1}_{(-0.25;0]}(y)
\
}
$$

Examples:

Fixed $\alpha=1$  you get

$$f_Y(y)=\frac{1}{\sqrt{y+\frac{1}{4}}}\cdot\mathbb{1}_{(-0.25;0]}(y)$$

Fixed $\alpha=2$ you get

$$f_Y(y)=\frac{-6y}{\sqrt{y+\frac{1}{4}}}\cdot\mathbb{1}_{(-0.25;0]}(y)$$

You can easy check that the above expressions are nice densities


问:
Resolving Forces With Vectors. Find the Resultant Force
In this system, a particle on a horizontal surface is acted on by two forces, $F_1$ and $F_2$. Find the $\hat i$ and $\hat j$ components of the resultant force, where $\hat j$ represents due north and $\hat i$ due east. $F_1 = 3\hat i + 2\hat j$, $F_2$ has a magnitude $5$ N and acts at a bearing of 60 degrees. I thought that I had to resolve $F_2$ into horizontal and vertical components $5\cos(60^\circ)$ and $5\sin(60^\circ)$ and then add these to the $\hat i$ and $\hat j$ parts of $F_1$, giving me $5.5\hat i$ and $6.33\hat j$ for the resultant force, but the mark scheme says that the answer is $7.33\hat i + 4.5\hat j$. I don't know how to get to these, and I don't think I've made any stupid mistakes, but if I have, please let me know.

Any advice would be appreciated.

答:
The possible way of getting the answer you have in the key is considering that $F_{2}$ "acts at a bearing of 60 degrees" means that this angle is not formed with the positive direction of x-axis, but with the one of the y-axis. Equivalently, we can say that the direction of the applied force is not north of east relative to the coordinate system, but the east of north. Then, The horizontal component of $F_{2}$ is $5sin(60^\circ) = 5*1.73/2 = 4.33$, and the vertical one is $5cos(60^\circ) = 5*0.5 = 2.5$. Then, adding these components to the corresponding ones of $F_{1}$:
$$R = (3+4.33)\hat{i} + (2+2.5)\hat{j} = 7.33\hat{i} + 4.5\hat{j}$$

Because of this ambiguity, in this kind of problems, besides the angle, it is usually indicated the direction too (in our example, north of east vs. east of north), so the condition is not really accurate.


问:
How to read $\frac{d^2y}{dx^2}$
Is it correct to say 'dee squared y over dee x squared'?

I am just trying to make sure because I have to teach this to someone.

I learnt that $f''(x)$ is pronounced 'f double prime of x' but I couldn't find how to pronounce Leibniz notation. Thanks!

答:
I'm pretty sure you read it either

dee two y by dee x squared

or

dee two y over dee x squared

(either is fine)

Title: How to read $\frac{d^2y}{dx^2}$


问:
If $E(|X+Y|^p)<\infty$, then $E(|X^p|)<\infty$ and $E(|Y^p|)<\infty$.
If $X$ and $Y$ are independent and for some $p>0$: $E(|X+Y|^p)<\infty$, then $E(|X^p|)<\infty$ and $E(|Y^p|)<\infty$.

How can I go from $E(|X+Y|^p)<\infty$ using independence to conclude something about $X$ all alone?

答: Use Fubini-Tonelli's theorem and the fact that if $f$ is integrable, then $|f(x)|$ is finite for almost every $x$.</p>


问:
Overpopulated apartment
$119$ people live in a building with $120$ apartments. An apartment is called overpopulated if there are more than $15$ people living in it. Each day the tenants of some overpopulated apartment have a quarrel and they all move to different apartments in the same building. Is it true that eventually none of the apartments will be overpopulated?

Attempts and observations:

If we organize the building into a $15\times 8$ grid with $8$ floors, after a quarrel at least $2$ of new floors(floors other than the floor on which the quarrel happened) occupy $1$ more tenants each(floors might be same)

If an apartment is called overpopulated if there are more than $14$ people in it, then the answer is no

If we prove that this process terminate, then we are done

答:
To try and prove that it is false would entail beginning from an overpopulated apartment, dispersing the inhabitants to different apartments  and getting another overpopulated apartment, and thus dispersing the inhabitants can cause the population of an apartment to increase by one for each abandoned apartment.

Say, you wanted a sequence of overpopulated apartments with $n_i$ people living in each one,
$$ n_i \geq 16 ,1 \leq i \leq m+1.$$

Beginning from $n_1$, $n_1$ people are dispersed to $n_1$ different apartments, if you want an apartment next with $n_2$ people in it, you would of needed to have an apartment with at least $n_2 -1$ people in it at first go. And then an apartment with at least $n_3 -2$ people in it at first go... and so on and so on.

Thus in general there must be $n_1 + n_2 -1 + n_3 -2 + ... =\sum\limits_{i=1}^{m+1} n_i - \sum\limits_{i=0}^{m}i \geq 16(m+1)-\frac{m(m+1)}{2}= 136> 119$ for $m=16$, since an apartment cannot have negative people in it. Thus if you wanted a sequence other than $16,16,16,...$ you would need even more people.

Title: Overpopulated apartment


问:
problem involving centroid and perpendiculars

in triangle $ABC$, $G$ is the centroid and $L$ is an arbitary line through the centroid.     $K,I,J$ are the foot of the perpendiculars from $A, B, C$

i need to prove that $KA+JC=BI$

how to use the midpoints of the sides and is there any way to use the fact that centroid cut the median in 2:1 ratio

答: In any Cartesian coordinatisation of the plane, the centroid of a point set is simply the arithmetic mean. In particular, if $G$ is the origin and $L$ the $x$-axis, $0$ must be the arithmetic mean of the $y$-coordinates of $A,B,C$, i.e. with signed distances $IB+KA+JC=0$. If $B$ is alone on one side of $L$ then with unsigned distances $IB=KA+JC$.


问:
The degree of an irreducible representation of a finite solvable group
I know that the degree of an irreducible representation of a finite group $G$ over a field of characteristic $0$ divides the order of $G$. I also know that such a result is false when the characteristic $p$ of the field is positive (even if $p$ does not divide the order of $G$). However, such examples seem all to be non-solvable and actually searching online I've found clues of the fact that for a solvable group it is indeed possible that the degree of an irreducible representation (over a field of positive characteristic) divides the order of the group itself. Since I cannot find anything better than clues, I'm asking: is this true? Where is it possible to find a proof of this fact?

答:
I'll convert my comments to an answer though whether it sufficiently answers the question might be a bit subjective. If $k$ is an algebraically closed field of characteristic not dividing $|G|$, then the dimensions of the irreducible representations of $G$ over $k$ should always divide $|G|$. In fact, I don't think it should matter which $k$ you choose: over algebraically closed fields with nonmodular characteristic, the general structure of representation theory is always pretty much the same; the representation categories are semisimple, for instance.

If the field is not algebraically closed, funny things can happen. I mentioned that the cyclic group $C_3$ has an irreducible $2$-dimensional representation over $\mathbb{R}$. In positive characteristic, equally funny things can happen. For instance, if we take the cyclic group $C_n$ of order $n$, and we let $p$ be a prime not dividing $n$, then the irreducible representations of $C_n$ over the finite field of $\mathbb{F}_p$ are pretty fun to study, and their degrees aren't so well-behaved.

I do not know what structural statements one can make about degrees of irreducible representations over non-algebraically closed fields (though I'd be interested to learn about it). In addition, I'm not aware of any connections with solvability of the group considered. (For what it's worth all examples I gave are abelian and hence trivially solvable. This raises the question for me: What does the representation theory of $A_5$ look like over say finite fields of characteristic $p > 5$?)


问:
Prime Numbers...
Are there any functions whose output is a prime number?

For eg. The function $x^2$ spits out squares of $x$; is there a similar function for Prime Numbers as well(which gives output as Prime Numbers)?

答:
A function is, in essence, is a well-defined rule.

Do prime numbers follow a well-defined rule? No.

So, there can't be a function whose output is a prime number , atleast not one which can be written out algebraically ( you could define a function however; for example you can define a function f(n) = n iff Φ(n) = n-1 and so on. Φ(N) here denotes the totient function)

You can have a function which gives primes for atleast some values of n algebraically ( i.e. with a restricted domain )
e.g. p(n) = ${n^2+n+41}$
which gives primes for inputs upto 39, but fails at 40.

Edit 1: By algebraically, I mean not by using simple/elementary operations, i.e. not in the form of a polynomial.

Edit 2: By algebraically, I mean not by using simple/elementary operations, i.e. not in the form of a finite polynomial.

Title: Prime Numbers...


问:
Functional division $\max(f(x+y),f(x-y))\mid \min(xf(y)-yf(x), xy)$
As the title suggests, the problem here is:

Find all functions $f:\mathbb{Z}\to\mathbb{N}$ such that, for every $x,y\in\mathbb{Z}$, we have
  $$\max(f(x+y),f(x-y))\mid \min(xf(y)-yf(x), xy)$$

I have already solved the problem - but, with seven pages full of calculations and case splitting, not quite the elegant proof I wished I'd have. I've put the answer in the yellow box below (I've hidden it for the people that wish to try this themselves first). Hover your mouse over the box to reveal its contents.

 The only solution is $f(x)=1$ for all $x\in\mathbb{Z}$.

I'd love to see a shorter proof for this. I'd be happy with any suggestions!

答:
For any $a, b > 0$, $x=a, y=-b$ gives $\max(f(a+b), f(a-b)) | \min(af(-b)+bf(a), -ab)$. Since $f$ takes positive values, $\max(f(a + b), f(a-b)) | ab$.

For $a=b=1$ this shows that $f(0)=f(2)=1$.

For $a=b$ this shows that $f(2a) | a^2$, and for $b=1$ this shows that $\max(f(a+1), f(a-1))|a$. Combined together these imply that $f(x) = 1$ for all even $x \geq 0$.

Now we claim that either $f(1) = 1$ or $f(3) = 1$. Suppose that neither equalled one. By the above inequality with $a=2$ and $b=1$, $\max(f(1), f(3)) | 2$, so $f(1)=f(3)=2$. By the original inequality with $x=3$ and $y=2$, $\max(f(1), f(5)) = \min(3f(2)-2f(3), 6)$. Since $3f(2)-2f(3) = 3 \cdot 1 - 2 \cdot 2 = -1$, $f(1)=1$, a contradiction. Therefore either $f(1) = 1$ or $f(3) = 1$.

Now for any odd $x \geq 1$ with $f(x) = 1$ let $m = \max(f(x+2), f(x-2))$. By the original inequality $m| \min(xf(2)-2f(x), 2x) =x-2$. If $x=1$ then this shows that $m=1$. Otherwise, if $x > 1$, by the above inequality ($a=x, b=2$) $m | 2x$, so since $\gcd(x-2, 2x) = 1$, $m = 1$. Either way $f(x \pm 2) = 1$, so by induction $f(x) = 1$ for all odd $x > 0$.

Therefore $f(x) = 1$ for all integer $x \geq 0$.

Now consider the case of negative arguments to $f$. For $x > 0$ and any $k > 0$, use the above inequality with $a = kx$ and $b = (k+1)x$. We know that $f((2k+1)x) = 1$, so the above inequality shows that $f(a-b) = f(-x) | k(k+1)x^2$. Since this holds for all such $k$, $f(-x) | x^2$.

In particular this shows that $f(-1) = 1$.

For $x > 0$, the above inequality with $a = 1$ and $b=x$ yields $\max(f(1+x), f(1-x)) = f(1-x) | x$. So for $x > 1$, $f(-x) | x + 1$. We also know $f(-x) | x^2$, so since $\gcd(x^2, x + 1) = 1$, $f(-x) = 1$.

Thus $f(x)=1$ for all $x \in \mathbb{Z}$. QED


问:
The number 3 can be written as $3$, $2+1$, $1+2$ or $1+1+1$ in four ways. In how many ways can the number $n$ be written?
Attempt
Let $x$ be any variable
$X+0=n ; X+Y=n ; X+Y+Z=n ; \dots; X+Y+Z+A+\dots=n$ (sum of n-1 terms); $1+1+1+.......+1=n$ (sum of n terms).
So total number of ways=
$$(n-1) C (1-1)+(n-1) C (2-1)+\dots+(n-1) C (n-1)$$
This equation is correct & gives correct answer if we have a value of $n$ but equation in $n$ needs to be proven equal to $2^{n-1}$.
Answer is $$2^{n-1}$$
What can be further done to solve it ?

答: Write down $n$ marks (e.g. "$x$"). Between any two $x$ you may put a "$+$" or not. Now replace each block of $x$ by their count, and you obtain a valid solution. (E.g., for $n=6$ we start with $xxxxxx$, then obtain prhaops $xx+x+xxx$ and finally $2+1+3$). As there are $n-1$ gaps between two consequtive $x$'s, we have $$2^{n-1}$$ ways to proceed.


答:
A simple function is

${\displaystyle f(n)=\left\lfloor {\frac {n!{\bmod {(}}n+1)}{n}}\right\rfloor (n-1)+2}$

(for positive integer n, where ${\displaystyle \lfloor \ \rfloor }$ is the floor function, which rounds down to the nearest integer.)

Proof-By Wilson's theorem, +1 is prime if and only if ${\displaystyle n!\equiv n{\pmod {n+1}}}$. Thus, when n+1 is prime, the first factor in the product becomes one, and the formula produces the prime number n+1. But when n+1 is not prime, the first factor becomes zero and the formula produces the prime number 2.

Title: Prime Numbers...

Body:
Are there any functions whose output is a prime number?

For eg. The function $x^2$ spits out squares of $x$; is there a similar function for Prime Numbers as well(which gives output as Prime Numbers)?


答:
By independence,
$$
E[|X+Y|^p]=\int\limits_{}E[|X+y|^p]\mathrm dP_Y(y).
$$
Hence, if $E[|X+Y|^p]$ is finite then $E[|X+y|^p]$ is finite for $P_Y$-almost every $y$. In particular, there exists some $y$ such that $E[|X+y|^p]$ is finite. Maybe you can finish from here.</p>

Title: If $E(|X+Y|^p)<\infty$, then $E(|X^p|)<\infty$ and $E(|Y^p|)<\infty$.

Body:
If $X$ and $Y$ are independent and for some $p>0$: $E(|X+Y|^p)<\infty$, then $E(|X^p|)<\infty$ and $E(|Y^p|)<\infty$.

How can I go from $E(|X+Y|^p)<\infty$ using independence to conclude something about $X$ all alone?