问:
Finding the value of $ax^4+by^4$

If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
 ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.

Here is my attempt:

$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.

答:
At the core of contest problems like

Given $x^n+1/x^n=p$, evaluate $x^m+1/x^m$

Given $x^n+y^n=p$, evaluate $x^m+y^m$

for most likely, integers $m,n$, lies the recurrence relation
$$x^{n+1}+y^{n+1}=(x+y)(x^n+y^n)-xy(x^{n-1}+y^{n-1})$$

Our question seems a bit generalized but in the same spirit, the following recurrence relation can be formed :
$$ax^{n+1}+by^{n+1}=(x+y)(ax^{n}+by^{n})-xy(ax^{n-1}+by^{n-1})$$

Thus the problem reduces to finding the linear coefficients $(x,y)$ or $(x+y,xy)$.

To ease the calculation, we do the following :
$$217=79(x+y)-23xy \tag{1}$$
$$691=217(x+y)-79xy \tag{2}$$

Lets do $3\times(1)-(2)$. We get
$$-40=20(x+y)+10xy$$
$$\Rightarrow (x+2)(y+2)=0$$

Thus one of $x,y$ is $-2$ but not both. (Why?) Putting $x=-2$ in $(1)$ gives $y=3$. Our recurrence relation is
$$ax^{n+1}+by^{n+1}=(ax^{n}+by^{n})+6(ax^{n-1}+by^{n-1})$$

which means this ladder goes up in following manner :

$$79+6\cdot 23=271$$
$$271+6\cdot 79=691$$
$$691+6\cdot 217=1993$$

Thus we have observed a systematic method to tackle these type of problems.

Title: Finding the value of $ax^4+by^4$


问:
Show every compact subspace of $m$-manifold has topological dimension at most $m$.
Notation $\bar{U}$ is the closure of $U$.

Definition $U$ is a neighborhood of $x$ means that $U$ is an open set containing $x$.

Definition An $m$-manifold is a Hausdorff space $X$ with a countable basis such that each point $x$ of $X$ has a neighborhood that is homeomorphic with an open subset of $\mathbb{R}^m$.

答:
There is the countable closed sum theorem in dimension theory (see Engelking's books):

Let $X$ be a normal (i.e. $T_4$) space. If $X = \bigcup_{n \in \Bbb N} X_n$ writes $X$ as an at most countable union of closed subspaces with $\dim(X_n) \le k (\in \Bbb N)$ for all $n$, then $\dim(X) \le k$ as well.

Apply this to the countable cover (of the closed subset) of the manifold by closed subsets of $\Bbb R^N$, which exists by having a countable base. By the subspace theorem all of these sets also have dimension $\le N$ etc.

ContentLicense: CC BY-SA 4.0


问:
Directed bipartite graphs with weighted edges with uniform input weight
Here is the set-up:

Let $G = (V(G), E(G))$ be a bipartite graph with $ V_1 \sqcup V_2$ denoting the vertices with respect to 2-colouring of the graph. Consider the induced directed graph $G' = (V(G),A(G))$ with the vertices $V(G)$ and with an arrow of each direction replacing each edge of $G$. Furthermore, we add a weighting $ wt : A(G) \rightarrow \mathbb{R}$ to each arrow such that $wt(u \rightarrow v)*wt(v \rightarrow u) = 1$ for all adjacent $u,v \in V(G)$.

Here is the question:

Does there exist a weighting function such that such that if $u, v \in V_i $ then

$$\sum_{x \in N(u)} wt(x\rightarrow u) = \sum_{x \in N(v)  } wt(x\rightarrow v) $$

where $N(v)$ is the set of adjacent vertices to $v$ in $G$ and $V_i \in \{V_1,V_2\}$.

I'm very ignorant of this general area. It seems like this might be a well-known example of some discrete version of some kind of heat equation or harmonic function. Any references to anything related would be extremely useful to me! Thanks!

答:
The constraint on the weighting is not very easy to deal with... because it's about the product of the weights, right ? Somehow it doesn't seem natural.

Anyway : sometimes it's impossible (take $V_1 = \{a,b,c\}, V_2 = \{d,e\}, E=\{(ad), (bd), (ce)\}$ : the constraints on $V_1$ say that all weights are equal, and the constraints on $V_2$ say that they are not.), sometimes it will be easy (you have $|V|-2+|E|$ constraints and $2|E|$ variables) so you should have enough degrees of freedom in most instances of the problem).

You could transform this into an instance of semidefinite programming (your constraints are quadratic, defined with a dot product) so this can be solved "efficiently", for example with interior point methods.

Maybe there is a clever way to reformulate this into a classical problem, but I can't seem to find an idea that could make this quadratic constraint appear magically.


问:
Find inflection point of parametric cubic
I'm familiar with turning points and inflection points for "normal" graphs (i.e. those that relate y and x) but how would I get an inflection point for a graph where each dimension has a separate equation?

In my case, given a parametric cubic curve $$\vec r(t) = \vec At^3 + \vec Bt^2 + \vec Ct + \vec D$$ , how would I determine the inflection point? Also, what does this point actually represent? Change in concavity? I am struggling to conceptualize this in 3D Vector space.

Thanks!

答:
A parametrized curve has a tangent vector $$\overrightarrow{T} = \frac{\overrightarrow{dr}/dt}{|\overrightarrow{dr}/dt|} $$ divided by its length.
The curvature $\kappa$ (a scalar) is defined as $$\kappa =\frac{|\overrightarrow{dT}/dt|}{|\overrightarrow{dr}/dt|}$$
The unit normal vector is $$\overrightarrow{N} = \frac{\overrightarrow{dT}/dt}{\vert{\overrightarrow{dT}/dt}\vert} $$
An inflection point would be where the curvature is zero.


问:
Is there a duality principle for classical first-order logic? (and/or free logic?)
In order theory, there's a variety of duality principles, like:

If a sentence involving only the meet and join operations is a
  consequence of the lattice axioms, then: the dual sentence, obtained by

replacing all meets in the sentence with joins, and

replacing all joins in the sentence with meets

is also a consequence of the lattice axioms.

Is there something similar for classical first-order logic? (and/or free logic that allows for empty domains?)

答:
In propositional logic you have the notion of a dual sentence.

Let $\varphi$ be a sentence (in propositional logic) written only by using $\land,\lor,\lnot$, we define the dual sentence $\varphi^*$ by replacing all the $\lor$ with $\land$, and vice versa.

Given an assignment, $\sigma$, we define the dual assignment to be $\sigma^*=\mathbf t_\lnot\circ\sigma$, that is $\sigma^*(p)=\sf T$ if and only if $\sigma(p)=\sf F$. (Where $\mathbf t_\lnot$ is the truth function of the $\lnot$ symbol.)

Now we have a nice duality theorem:

$$\operatorname{val}(\varphi,\sigma)^*=\operatorname{val}(\varphi^*,\sigma^*)$$

That is, $\operatorname{val}(\varphi^*,\sigma^*)=\mathbf t_\lnot\circ\operatorname{val}(\varphi,\sigma)$.

ContentLicense: CC BY-SA 3.0


问:
Problem on deriving $E(x)$ of a geometric distribution
I want to derive the result that for a geometric distribution
$$X \sim Geo(p)$$
The expected value is equal to
$$E(x)=\frac{1}{p}$$

I have started off by using the definition of $E(x)$ being the mean

$$E(x)=\sum xP_x$$

And we can write $\sum xP_x$ as

$$\sum xP_x= p + 2qp + 3q^2p + 4q^3p......+ nq^{n-1}p $$

Multiplying $\sum xP_x$ by $q$ and then taking the difference of the two expressions

$$\left[\sum xP_x= p + 2qp + 3q^2p + 4q^3p......+ nq^{n-1}p \right] $$

$$-\left[ q\sum xP_x= qp + 2q^2p + 3q^3p + 4q^4p......+ (n-1)q^{n-1}p + nq^np \right]$$
$$\text{_________________________________________________________________}$$
$$\sum xP_x \left(1-q \right)=p+qp+q^2p.....+q^{n-1}p +nq^np$$
$$\text{_________________________________________________________________}$$

From here we can see that the right hand side excluding the last term forms the sum of a geometric progression hence we can use the formula for the sum of first $n$ terms of a geometric progression which is in this case $$\frac{p(1-q^n)}{1-q}$$
$$\because p=1-q$$ $$\therefore \frac{p(1-q^n)}{1-q} \iff 1-q^n$$

Now the expression simplifies to
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^np$$
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^n(1-q)$$
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^n - nq^{n+1}$$

At this point it is apparent I have made a mistake somewhere as even by simplifying it further

$$\sum xP_x \left(1-q \right)= 1-q^n(1 + n - nq)$$
$$\sum xP_x \left(1-q \right)= 1-q^n(1 + n(1 - q))$$
$$\sum xP_x = \frac{1-q^n(1 + np)}{\left(1-q \right)}$$

And I have no clue on how to proceed, where have I messed up or have I taken a wrong method entirely?

答:
there is a very simple way to do that:

$$\mathbb{E}[X]=\sum_{x=1}^{\infty}xpq^{x-1}=p\sum_{x=1}^{\infty}xq^{x-1}=$$

$$=p\sum_{x=1}^{\infty}\frac{d}{dq}q^x=p\frac{d}{dq}\sum_{x=1}^{\infty}q^x=$$

$$=p\frac{d}{dq}\frac{q}{1-q}=p\frac{1}{(1-q)^2}=\frac{1}{p}$$

Doing in your manner you get

$$\mathbb{E}[X]=p+2pq+3pq^2+4pq^3+\dots=p[1+2q+3q^2+4q^3+\dots]=pS$$

Now let's focus on the sum S

$$S=1+2q+3q^2+4q^3+\dots$$

$$Sq=q+2q^2+3q^3+4q^4+\dots$$

Subtracting member to member you get

$$S(1-q)=1+q+q^2+q^3+\dots$$

remembering the sum of a geometric series you immediately get

$$S(1-q)=\frac{1}{1-q}$$

thus

$$\mathbb{E}[X]=pS=\frac{p}{(1-q)^2}=\frac{1}{p}$$


问:
The equation of a sphere tangent to a plane at a point whose center belongs to a plane
Determine the equation of a sphere that is tangent to the plane $x+y+z-9=0$ at the point $M(2,3,4)$ and whose center belongs to the plane:
$$ (P) : 7x-4y+5z-14=0 $$
My solution: it is clear that the point $A(2,5,4)$ belongs to the plane $(P)$. Then, let $d$ be the line
passing through $A$, $d$ orthogonal to $(P)$ (so that the director vector is equal with the normal vector of the plane, $v(7,-4,5)$ ). From the parametric equations of this line I found that the center of the sphere is of the form $C(2+7t, 5-4t, 4+5t)$. At this point, if I calculate the distance from the center of the sphere to the first plane I get ${|8t+11|\over\sqrt{3}}=r$, where $r$ is the radius of the sphere. And I'm stuck at this point because I don't really see how I can use the point $M$ to solve this problem. Any ideas? Thanks!

答: The line orthogonal to the plane $x+y+z-9=0$ and passing through $M$ is the line$$\{(2,3,4)+\lambda(1,1,1)\mid\lambda\in\Bbb R\}.\tag1$$The center of the sphere must belong to $(1)$ and also to the plane $P$. So, let us compute the point at which they intersect. In order to do that, one solves the equation$$7(2+\lambda)-4(3+\lambda)+5(4+\lambda)-14=0;$$its only solution is $\lambda=-1$. So, the sphere is centered at $(1,2,3)$ and its radius is the distance from $(1,2,3)$ to $(2,3,4)$, which is $\sqrt{3}$. So, it's the sphere described by$$(x-1)^2+(y-2)^2+(z-3)^2=3.$$


问:
Show that a particular symmetric sum is $0$ or $1$
Given $x_1, \ldots, x_n$ some numbers, and $0 \leq d \leq n-1$, how can I show that \begin{align*}\sum_{i=1}^n \frac{x_i^d}{\prod\limits_{\substack{j=1\\j\neq i}}^n(x_i-x_j)} = \begin{cases} 0 &\text{if } 0 \leq d < n-1\\1 &\text{if }d=n-1\end{cases}\end{align*} (or is this even true - as far as I can tell it holds for any choices of $x_1,\ldots, x_n$ distinct)

I see the case $d=0$ has been answered here. Thanks!

答:
Similar to as in the linked post, let:
\begin{align*}
q(T) = \sum_{i=1}^n \left(\prod_{j\neq i} \left(\frac{T-x_i}{x_j-x_i} - 1 \right)\right)
\end{align*}
$q(T)$ has the following properties:

The degree of $q$ (as a polynomial in $T$) must be $\leq n-1$, as each term in the sum is a product of $n-1$ monic terms in $T$.

At $T=x_k$ say, in each term in the sum where $i \neq k$, $j=k$ occurs in the inner product. So that product contains a factor of $\left(\frac{x_k-x_i}{x_k-x_i} -1\right) = (1-1) = 0$, and hence that term vanishes, leaving only the term $i=k$. Hence:
\begin{align*}
q(x_k) = \prod_{j\neq k}\left(\frac{x_k-x_k}{x_j-x_k}-1\right)=\prod_{j\neq k}(-1) = (-1)^{n-1}
\end{align*}

So $q(T) - (-1)^{n-1}$ is a polynomial of degree $\leq n-1$ with $n$ distinct zeros (at each of the $x_k$), so is constant zero, and so $q(T) \equiv (-1)^{n-1}$.

Now consider the coefficients of $T^{(n-1)-d}$.

When $d <n-1$, the coefficient of $T^{(n-1)-d}$ is zero. On the other hand, expanding the product, the contributions to the $T^{(n-1)-d}$ coefficient from the $i$th term in the sum is $\frac{\binom{n-1}{d}(-1)^{d}x_i^d}{\prod_{j\neq i}(x_j-x_i)}$, and so
\begin{align*}
\binom{n-1}{d}(-1)^d\sum_{i=1}^n \frac{x_i^d}{\prod_{j\neq i}(x_j-x_i)} &= 0\\
\implies \sum_{i=1}^n \frac{x_i^d}{\prod_{j\neq i}(x_j-x_i)} &= 0
\end{align*}

When $d = n-1$, the coefficient of $T^{(n-1)-d}$ is $(-1)^{n-1}$. Expanding as above:
\begin{align*}
\binom{n-1}{n-1}(-1)^{n-1}\sum_{i=1}^n \frac{x_i^{n-1}}{\prod_{j\neq i}(x_j-x_i)} &= (-1)^{n-1}\\
\implies \sum_{i=1}^n \frac{x_i^{n-1}}{\prod_{j\neq i}(x_j-x_i)} &= 1
\end{align*}


问:
Problem with lowercase versus uppercase.
I often attack problems on two levels, a verbal level and a visual level. At a visual level, I try to imagine how everything is related - I arrange the ingredients of the problem into a two (or sometimes three) dimensional configuration - sometimes even a movie.

I then try to reason about the configuration verbally. Like: 'Since $A$ is a proper subset of $B$, I can find $b \in B$ such that $b \notin A$. Fix any such $b$.'

However, the verbal approach suffers from the following problem: I can't distinguish $b$ from $B$, because they both sound the same, 'bee.' So I need to actually visualize those glyphs - I need to visualize $b$ and $B$. But this visualization of glyphs then interferes with the previous visualization whereby the ingredients of the problem were organized into a spatial configuration. I can't keep both images in mind at the same time.

My general question is: What can be done about this problem?

More specifically: Has anyone bothered to come up with letter names that distinguish between lowercase and uppercase? I mean, we could say 'little bee' for $b$, but that's just too many syllables.

答:
Since I was effectively answering the question in the comments, I decided to move it and expand a bit.

What's wrong with saying "small $b$", "capital $B$", "bold $\Bbb B$", "curly/scripted $\mathscr B$", "gothic (small) $\mathfrak b$"?

That's short enough for me (and probably shorter than describing what these letters stand for anyway, and you'll have to do that too).

Should this not match your preferences regarding brevity then -- since I get the impression you are only "visualizing" things for yourself by talking -- you are of course entirely free to introduce some nice abbreviations (e.g. "cap $B$" for "capital $B$"), and/or some conventions (unless I say "cap", the letter is implicitly small).

I hope this is of some help.


问:
reflexive, symmetric and not transitive relation
If a relation on a set $R$ is reflexive, symmetric but not transitive, does it imply that all the classes $[x] = \{ y\mid xRy\}$ are mutually exclusive, in the sense that for no $[x]$ and $[y]$, they both share an element, and so their intersection is always empty?

答:
I am not pretty sure whether I understood your question correctly, but on the set $\Omega = \{1,2,3\}$ the relation
$$R = \{ (1,1), (1,2), (2,1), (2,2), (2,3), (3,2), (3,3)\}$$
might be a counterexample. The classes
$$[1], [2], [3]$$
are pairwise different, but $2$ is in all classes.


问:
Minimize the $l_1$ norm
The task is to minimize $|x1 - b|_{l1}$ in $l_1$ norm. Here $1 \in R^{n}$ is a vector of ones, $x \in R$ is a scalar, $b \in R^{n}$.

First of all, I wrote it in a general form without specifying the norm: minimize $|x1-b|$.

Then by the introduction of the equality constraint, it can be written in an equivalent form as:  $$\text{minimize }|y|$$
$$\text{subject to: } y=x1-b $$.

The dual of the problem is $L(\lambda) = |y|+\lambda^T(y-x1+b)$. Taking the infimum we get $g(\lambda) = \inf_{x,y} L(\lambda) = \inf_{x,y}|y|+\lambda^T(y-x1+b) = \inf_{x}\lambda^T(b-x1) \text{ subject to }|\lambda|_{*} \le 1$.

So, the dual problem is $$\text{maximize }\inf_{x}\lambda^T(b-x1)$$
$$\text{subject to: } |\lambda|_{*} \le 1 $$.

Where $|.|_{*}$ is the dual norm.

Is this right so far? What should I do next?

答:
The little issue with $y$ in the Lagrangian aside, the steps so far are correct. The next step is to focus on $\inf_x \lambda^T(x1) = \inf_x \lambda^T1 x$. When is the infimum finite?

Update based on the discussion:
$$\inf_x \lambda^T1 x = \begin{cases} 0 & \text{if } \lambda^T1 = 0 \\ -\infty & \text{otherwise.}\end{cases}$$

Title: Minimize the $l_1$ norm


问:
High School Math, Unit Circle, why is this negative?
I'm having trouble understanding the solution

I don't understand how x^2 = 0.5 changes into x = - 1 / 2^0.5

答:
Because "in the quadrant III", $x<0$.

For $x^2=\frac{1}{2}$, there are two solutions:
$$
x=\frac{1}{\sqrt{2}},\quad x=-\frac{1}{\sqrt{2}}
$$
The second one is negative.


问:
Uniform convergance for $f_n(x)=x^n-x^{2n}$
the function $f_n(x)=x^n-x^{2n}$ converge to $f(x)=0$ in $(-1,1]$. Intuativly the function does not converge uniformally in (-1,1]. How can I prove it?
I tried using the definition $\lim \limits_{n\to\infty}\sup \limits_{ x\in (-1,1]}|f_n(x)-f(x)|$ function is continial fractional on $[-1,1]$ and $x=0,(\frac 1 2 )^{\frac 1 n}$ are the roots of the derivative. I found that the second derivative is negative in the second point. then $\sup=1/4$ and the function does not converge uniformally?</p>

答:
An idea: say for $\,n>2\,$

$$f_n(x)=x^n-x^{2n}\implies f'_n(x)=nx^{n-1}-2nx^{2n-1}=nx^{n-1}\left(1-2x^{n}\right)=0\iff$$

$$x=0\,,\,\frac1{\sqrt[n]2}$$

$$f_n''(x)=n(n-1)x^{n-2}-2n(2n-1)x^{2n-2}\implies\begin{cases}f''(0)=0\\{}\\f''\left(\frac1{\sqrt[n]2}\right)=\frac{n\sqrt[n]4}4\left(-2n\right)<0\end{cases}\;\implies$$

$$\implies\text{at}\;\;\left(x=\frac1{\sqrt[n]2}\;,\;y=f_n\left(\frac1{\sqrt[n]2}\right)=-\frac14\right)\;\text{we have a maximum for}\;\;f_n(x)\;,\;\forall\,n>2 .$$

And from here it follows at once that since $\,\displaystyle{f(x):=0=\lim_{n\to\infty}f_n(x)}\;$ , then

$$ \lim_{n\to\infty}\sup_{n}|f_n(x)-f(x)|=\lim_{n\to\infty}\frac14=\frac14$$

and thus the convergence isn't uniform.


问:
Trying to prove this inequality involving exponential function
Given :

A continuous non-negative function $f(x)$ on $\mathbb R^{+} \cup \{ 0\}$

$\alpha > 0$

$h(x) = \displaystyle \int \limits_0^x f(u) du + \alpha$

$\forall x$, $f \leq h$

To prove : $$f(x) \leq \alpha e^x$$

Attempt :

$$\dfrac {d}{dx} [\ln h(x)] = \dfrac {h'(x)}{h(x)}$$

By the fundamental theorem of integral calculus, $h'(x) = f(x)$ and $\therefore $

$$\dfrac {d}{dx} [\ln h(x)] = \dfrac {f(x)}{h(x)}$$

From the 4th given point, $f \leq h$,

$$\dfrac {d}{dx} [\ln h(x)] \leq 1$$

i.e. the function $\ln [h(x)]$ never grows faster than the linear function, with slope 1.

From this step, I wrote $$\ln [h(x)] \leq x + C, \text{ for some } C$$

Is the above step correct?

$$\therefore h(x) \leq e^{x+C} = e^C e^x = B e^x$$

$$\therefore f(x) \leq h(x) \leq B e^x$$

Also, am unable to get in $\alpha$ in this inequality.

答:
This is just a consequence of Grönwall's inequality, see for instance: https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality.

In your case, you can take $u(x) = h(x)$, $b(x) = f(x)/h(x) > 0$. Your condition then becomes:
\begin{equation}
u(x) = \alpha +  \int_{0}^{x} b(s)u(s)ds.
\end{equation}
By your first three assumptions,  $\alpha$ is positive and non-decreasing, $b$ is continuous and non-negative. Therefore, Grönwall's inequality yields:
\begin{equation}
u(x) \leq \alpha \exp \left( \int_{0}^{x} b(s)ds \right).
\end{equation}
By your final assumption, $b(s)\leq 1$,  which yields the inequality you wanted.


问:
Correspondance between section of covering space and points in the preimage
Let $p:\tilde{X}\to X$ be a covering map (where $X$ is path conncetd and locally path connected). We call a continuous map $s:X\to\tilde{X}$ a section if $p\circ s=id_X$. Fix a point $x_0\in X$. Prove that there exists a bijective correspondance between the set of points $y\in p^{-1}(x_0)$ such that every loop $f$ with base point $x_0$ lifts to a loop $\tilde{f}$ with basepoint $y$ and the sections of $p$.

Attempt: I defined a map $s_y:X\to\tilde{X}$ with $s_y(x)=\tilde{f}(1)$ where $f$ is a path with $f(0)=x_0,f(1)=x$ and $\tilde{f}(0)=y$. Can we prove this correspondance is onto?

答:
Hints:

To show that the correspondence is onto, start with an arbitrary section $s'$. Your goal is to find a point in $p^{-1}(x_0)$ with correct lift-looping behavior that maps to $s$ under your correspondence $y \mapsto s_y$.

Since $s'(x_0)$ is in the fiber over $x_0$, and $s'(x_0)$ does have correct lift looping (why?), it remains to show that this point maps to $s'$ under the correspondence. That is, you want to show $s_{s(x_0)} = s'$. How can you use properties of covering spaces to show this?


问:
Is the punctured plane a normal space? Is a normal space without a point normal?
At first I thought is was obvious that $\mathbb R^2 \setminus \{ P \}$ is normal as a topological subspace of $\mathbb R^2$. But what about if we have two closed sets in $\mathbb R^2$, $F$ and $E$, such that $F \cap E = \{ P \}$. How can we show in general that we can separate $F \setminus \{ P \}$ from $E \setminus \{ P \}$ with open sets in $\mathbb R^2 \setminus \{ P \}$? Is this even true?

To generalise, is it true in general that the subspace obtained from a normal space by removing a point stays normal?

答:
No, in general remvoing a point from a normal space can make the resulting subspace non-normal. The classic case is the Tychonoff plank, the subspace

$(\omega+1) \times (\omega_1+1) \setminus \{\omega, \omega_1)\}$
of the compact Hausdorff (hence normal) space $(\omega+1) \times (\omega_1+1)$. (Each factor in the order topology).

But for metric and ordered spaces normality is hereditary. So for the plane it's no issue.


问:
How many bits of randomness needed to sample from $\operatorname{Bernoulli}(1/3)$
The title says it all. How can you take a sequence of random coin flops $x_1,x_2,\ldots,x_n\sim\operatorname{Bernoulli}(1/2)$ and generate a sample from a coin $X \sim \operatorname{Bernoulli}(1/3)$ with $n$ as small as possible?

If this isn't possible exactly, how about a sample of $\tilde X \sim \operatorname{Bernoulli}(1/3 \pm\epsilon)$ for some small $\epsilon>0$?

I know of algorithms to sample $X$ using uniform random numbers but I can't think of a procedure using finite randomness.

答:
Although any method to simulate a Bernoulli(1/3) variable using Bernoulli(1/2) "coin flips" will require an unbounded number of flips in the worst case, an optimal algorithm's expected number of flips is exactly 2.

In fact, given a fair coin and a greedy strategy, the expected number of fair coin flips to simulate a known probability p is exactly 2, unless p is a dyadic rational number (Kozen 2014).

REFERENCES:

Kozen, D., "Optimal Coin Flipping", 2014.


问:
A generalization of the conditional expectation to kernels
Let $\left(\Omega_1,\mathcal{A}_1, P\right)$ be a probability space, let $\left(\Omega_2,\mathcal{A}_2\right)$ be a measurable space and let $\kappa:\Omega_1\times\mathcal{A}_2\rightarrow\left[0,1\right]$ be a Markov kernel from $\mathcal{A}_1$ tp $\mathcal{A}_2$. Suppose $\mathcal{A}_3$ is a sub-$\sigma$-algebra of $\mathcal{A}_1$. Is there a Merkov kernel $\kappa'$ from $\mathcal{A}_3$ to $\mathcal{A}_2$ such that for all $B\in\mathcal{A}_2$,

$$
\kappa\left(\omega,B\right)=\kappa'\left(\omega,B\right)\hspace{10mm}\left[\mathcal{A}_1,P\right]
$$

Possible plan of attack: We could view $\kappa$ as a family of $\mathcal{A}_1$-measurable real functions indexed by $\mathcal{A}_2$ (viz. for every $B\in\mathcal{A}_2$, $\kappa_B$ is the function $\omega\mapsto\kappa\left(\omega,B\right)$) and define $\kappa'\left(\omega,B\right):=E\left(\kappa_B\mid\mathcal{A}_3\right)$. That would meet half the requirements for a Markov kernel, except $\kappa'$ is not guaranteed to be a probability in the second argument. In fact, for some combinations of $\omega\in\Omega$, $B\in\mathcal{A}_2$, $\kappa'\left(\omega,B\right)$ may not even be defined.

答:
The answer is yes, provided $\mathcal{A}_2$ is a Borel space.

This is an application of definition 5.1 and theorem 5.1 from Bahadur, R. R. Sufficiency and Statistical Decision Functions. Ann. Math. Statist. Volume 25, Number 3 (1954), 423-462 (p. 434).

Definition 5.1. A field $\mathbf{S}_0\subseteq\mathbf{S}$ is sufficient for the measures $P$ on $\mathbf{S}$ (briefly, $\mathbf{S}_0$ is sufficient for $P$) if corresponding to each $\mathbf{S}$-measurable set $A$, there exists and $\mathbf{S}_0$-measurable function $\varphi_A(x)$ such that
$$
\left(5.1\right)\hspace{10mm}\varphi_A(x)=E_p\left(\chi_A(x)\mid\mathbf{S}_0\right)\hspace{5mm}\left[\mathbf{S}_0,p\right]
$$
for each $p$ in $P$.

Theorem 5.1. When $R$ is the real line, and $\mathbf{R}$ the class of Borel sets of $R$, the following statements are mutually equivalent.

(i) $\mathbf{S}_0$ is sufficient for $P$.

(ii) (...) [Omitted. --E.A.]

(iii) Corresponding to each function $\mu\left(B,x\right)$, defined for $B\in\mathbf{R}$ and $x\in X$ such that $\mu$ is $\mathbf{S}$-$P$ integrable for each $B$ and a measure on $\mathbf{R}$ for each $x$, there exists a function $\nu\left(B,x\right)$ such that $\nu$ is $\mathbf{S}_0$-$P$-integrable for each $B$ and a measure on $\mathbf{R}$ for each $x$, and such that
$$
\left(5.3\right)\hspace{10mm}\nu\left(B,x\right)=E_p\left(\mu\left(B,x\right)\mid\mathbf{S}_0\right)\hspace{5mm}\left[\mathbf{S}_0,p\right]
$$
for each $B\in\mathbf{R}$ and $p\in P$.

Note that the order of parameters of a kernel in Bahadur's paper is non standard.

The proof of theorem 5.1 is rather involved, but apparently standard: Bahadur notes (p. 435) that it is a generalization of an argument used by Doob for the existence of a conditional probability measure on the real line.

Before applying these results to the problem at hand, it should be noted that theorem 5.1's proof, as given in the paper, can be readily modified to allow for the following enhancements:

$\left(R,\mathbf{R}\right)$ may be taken to be any Borel space (i.e. a measurable space isomorphic to a Borel subset of $\left[0,1\right]$; see Kallenberg's Fondations of Modern Probability (2002), p. 7).

If $\mu\left(B,x\right)$ is a probability measure in $B$ for any fixed $x$, $\nu$ can be chosen likewise.

Now all that is left is to notice that when $P$ consists of a single measure, every field $\mathbf{S}_0\subseteq\mathbf{S}$ is trivially sufficient.

As a corollary, since any $\sigma$-algebra admits of some probability measure, we get the following result.

If $\kappa$ is a Markov kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$, with $\mathcal{A}_2$ being a Borel space, and if $\mathcal{A}_3\subseteq\mathcal{A}_1$ is a sub-$\sigma$-algebra, there exists some Markov kernel from $\mathcal{A}_3$ to $\mathcal{A}_2$.


问:
find $f$ such $f(f(x)+y)=2x+f(f(f(y))-x)$
Find all function $f:R\to R$,such for any real number $x,y$ have
$$f(f(x)+y)=2x+f(f(f(y))-x)$$

I have prove this one result:

$$f(f(0))=f(0)$$
proof:let $x=y=0$,then we have
$$f(f(0))=f(f(f(0)))$$
and take $x=f(f(0)),y=0$,we have
$$f(f(f(f(0))))=2f(f(0))+f(0)$$
so we have
$$f(f(0))=-f(0)$$

use this links Ivan Loh methods to prove two reslut:

$f(x)$ is also injective

proof: For convenience, let $P(x, y)$ represent the equation $f(f(x)+y)=2x+f(f(f(y))-x)$.

$P(x, -f(x))$ gives $f(f(-f(f(x)))-x)=f(0)-2x$. Thus $f(x)$ is surjective.

Suppose $f(a)=f(b)$ for some $a, b$. Then $P(x, a)$ and $P(x, b)$ give
$$f(f(x)+a)=2x+f(f(f(a))-x)=2x+f(f(f(b))-x)=f(f(x)+b)$$
Since $f(x)$ is surjective, $f(x+a)=f(x+b) \, \forall x \in \mathbb{R}$.

Now using the above equation and $P(a, y), P(b, y)$ give

\begin{align}
2a=f(f(a)+y)-f(f(f(y))-a) & =f(f(a)+y)-f(f(f(y))-(a+b)+b) \\
&=f(f(a)+y)-f(f(f(y))-(a+b)+a) \\
& =f(f(b)+y)-f(f(f(y))-b) \\
& =2b
\end{align}

Thus $f(x)$ is also injective.

Finally, $P(0, y)$ gives $f(y+f(0))=f(f(f(y)))$, so since $f(x)$ is injective, $$f(f(y))=y+f(0) \, \forall y \in \mathbb{R}$$.

then I can't  continue ,and I think this answer is $f(x)=x$.so How to prove it.and this problem is  from   my teacher to give me exercise .Thanks for you help!
Now when post this post,This site automatically displays a prompt for a similar question to me(But it's different problem) :links

答:
let
$$p(x,y)=f(f(x)+y)-2x=f(f(f(y))-x)$$
we have from $p(x,-f(x))$ that $f$ is surgective. assume $f(a)=f(b)$ then from
$$p(x,a)-p(x,b)$$
We have
$$f(f(x)+a)=f(f(x)+b)$$
so we have if $a-b=t$ then
$$f(x+t)=f(x)$$
because $f$ is surgective.now we have :
$$p(x+t,y)-p(x,y)=0=2t$$
so $t=0$ now we have from
$$p(0,y)=f(f(0)+y)=f(f(f(y)))$$
so we have
$$T(y)=y+f(0)=f(f(y))$$
we have from $T(0)$ that
$$f(0)=f(f(0))$$
so $f(0)=0$ and we rewrite $T(y)$ as
$$y=f(f(y))$$
now rewrite $p(x,y)$ as
$$p(x,y)=f(y+f(x))=2x+f(y-x)$$
let
$$q(x,y)=p(x,y+x)=f(y+f(x)+x)=2x+f(y)$$
from $q(x,0)$ we have
$$f(f(x)+x)=2x$$
so $f(x)+x$ is injective and surgective aswell. so we can just write $q(x,y)$ as
$$q(x,y)=f(x+f(x)+y)=f(x+f(X))+f(y)$$
as $x+f(x)$ is injective and surgective. we can rewrite it as
$$W(x,y)=f(x+y)=f(x)+f(y)$$
so we have that $f$ is additive. remember we had $f(f(x)+x)=2x$ so we have
$$f(f(x))+f(x)=x+f(x)=2x$$
which gives $f(X)=x$ for all real $x$ and we are done


答:
There are four issues that I can see:

$\sum xP_x \left(1-q \right)=p+qp+q^2p.....+q^{n-1}p +nq^np$ should have the last term as $-nq^p$.

Your simplifications with $\frac{p(1-q^n)}{1-q}$ to $1-q^n$ are not valid because you can't multiply and divide terms by arbitrary things. I think you're trying to rearrange as if it were $\frac{p(1-q^n)}{1-q}=0$.

What is $q$? If it is $1-p$ (as the symbol would generally be used to mean in this context) then you can simplify all cases of $1-q$ to $p$ and all $1-p$ to $q$.

The sum shouldn't be up to $n$ in the first place, but infinite (from $k=1$ to $\infty$), because a geometric random variable can have arbitrarily large value rather than a maximum value of $n$.

Does this give you enough information to make another attempt?

Title: Problem on deriving $E(x)$ of a geometric distribution

Body:
I want to derive the result that for a geometric distribution
$$X \sim Geo(p)$$
The expected value is equal to
$$E(x)=\frac{1}{p}$$

I have started off by using the definition of $E(x)$ being the mean

$$E(x)=\sum xP_x$$

And we can write $\sum xP_x$ as

$$\sum xP_x= p + 2qp + 3q^2p + 4q^3p......+ nq^{n-1}p $$

Multiplying $\sum xP_x$ by $q$ and then taking the difference of the two expressions

$$\left[\sum xP_x= p + 2qp + 3q^2p + 4q^3p......+ nq^{n-1}p \right] $$

$$-\left[ q\sum xP_x= qp + 2q^2p + 3q^3p + 4q^4p......+ (n-1)q^{n-1}p + nq^np \right]$$
$$\text{_________________________________________________________________}$$
$$\sum xP_x \left(1-q \right)=p+qp+q^2p.....+q^{n-1}p +nq^np$$
$$\text{_________________________________________________________________}$$

From here we can see that the right hand side excluding the last term forms the sum of a geometric progression hence we can use the formula for the sum of first $n$ terms of a geometric progression which is in this case $$\frac{p(1-q^n)}{1-q}$$
$$\because p=1-q$$ $$\therefore \frac{p(1-q^n)}{1-q} \iff 1-q^n$$

Now the expression simplifies to
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^np$$
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^n(1-q)$$
$$\sum xP_x \left(1-q \right)= 1-q^n + nq^n - nq^{n+1}$$

At this point it is apparent I have made a mistake somewhere as even by simplifying it further

$$\sum xP_x \left(1-q \right)= 1-q^n(1 + n - nq)$$
$$\sum xP_x \left(1-q \right)= 1-q^n(1 + n(1 - q))$$
$$\sum xP_x = \frac{1-q^n(1 + np)}{\left(1-q \right)}$$

And I have no clue on how to proceed, where have I messed up or have I taken a wrong method entirely?


问:
Find the volume of a sphere with triple integral
Consider this figure:

As shown in the figure. The center of the sphere is at $(0,0,-2)$ and the radius of the sphere is $4$. My question is how to calculate the volume of the sphere when $z\geq 0$ (upper half space). How can i apply triple integral on this problem?

My idea is shifting the sphere to above up to $2$. First i need to find the angle between the cone and $xy$ plane. That is:
$$\arcsin\left(\frac 12\right) = \frac{\pi}{6}$$
It means the complement of the angle is $\frac{\pi}{3}$ and calculating this integral:

$$\int_0^{\frac{\pi}{3}} \int_0^{2\pi} \int_0^{\frac{\pi}{6}} \rho^2 \sin(\phi) \,\Bbb d\rho\,\Bbb d\theta\, \Bbb d\phi$$

Finally, the volume of the sphere when $z\geq 0$ is that integral minus the volume of the cone. Anyway, do you have the shortest way to solve this problem using triple integral that requires one step only? (Mine is 2 steps, calculate the integral and the cone)

答:
Can you integrate an interior of a circle with a double integral so that you calculate its area $S(r)$ for any given $r$...?

If so, you can present the problem by integrating 'horizontal' slices of a sphere by $dz$ above the $XY$ plane.
By a Pythagoras' theorem, the radius of a $z=\text{const.}$ section of a sphere at a chosen $z$ is $$r(z)=\sqrt{R^2-(z-z_A)^2}$$ where $R=8$ is a radius of a sphere and $z_A=-2$ is a height of the sphere's center.

Then the volume of the upper spherical cap is $$V_{z>0}=\int\limits_{z=0}^{R+z_A}S(r(z))\,dz$$
Just insert a double integral over the $XY$ circle with radius $r$ for $S(r)$ and you'll have a triple integral.

Which doesn't mean you must integrate all three at once. You can resolve them one by one from the innermost outwards.


问:
Prove or disprove that $a_{n}=\int_{2019}^{n+2019}\frac{\sin{x}}{x\ln{x}}dx$ has a limit
Let $n$ be postive integer. If we take
$$a_{n}=\int_{2019}^{n+2019}\dfrac{\sin{x}}{x\ln{x}}dx$$
how can I prove or disprove that the sequence
$\{a_{n}\}_{n\ge 1}$ has a limit?

It seem I must use this well known: $|\sin{x}|\le 1$ but I can't go on.

答:
By an integration by parts, you have
$$a_n = \left[- \frac{\cos(x)}{x\ln(x)} \right]_{2019}^{2019+n} - \int_{2019}^{2019+n} \cos(x)\frac{\ln(x)+1}{x^2 \ln^2(x)} \,dx$$

The first term has a finite limit, and the second one also, since for every $x \geq 2019$,
$$\left| \cos(x)\frac{\ln(x)+1}{x^2 \ln^2(x)}\right| \leq \frac{1}{x^2}$$

which is integrable on $[2019, +\infty)$.

So $(a_n)$ has a finite limit.


问:
First order logic: Formulating sentences for graph properties

Let $\sigma = \{ E \}$ be a signature with a binary relational symbol $E$. Give for each of the following statements a sentence $\varphi \in FO[\sigma]$, such that for a $\sigma$-strucutre $\mathcal{A}$, $\mathcal{A} \models \varphi$ iff the statement is true for $\mathcal{A}$.

$i$) $E^\mathcal{A}$ is an equivalence relation and  $E^\mathcal{A}$ has at most three equivalence classes.

$ii$) $\mathcal{A}$ is a $\sigma$-structure  $\mathcal{G}$, which is un undirected graph and there exists for its universe $U$ a bipartition $A \cup B = U$ in two disjoint, non-empty sets, such that $(A \times B) \cup (B \times A) = E^{\mathcal{G}}$.

For $i$), I divided  $\varphi$ in two parts: $\varphi = \varphi' \land \varphi''$ where
$\varphi'$ describes the equivalence relation: $$ \varphi' = \forall x E(x, x) \land \forall x \forall y \forall z (E(x, y) \land E(y, z) \rightarrow E(x, z))   \land (\forall x \forall y E(x, y) \rightarrow E(y, x))$$

$\varphi''$ should describe the property of having at most three equivalence classes, but I'm not sure how I can write it. What I understand:

If the graph only has one equivalence class, then we must have a complete graph (since the relation is reflexive, transitive and symmetric) that is connected. If we have two equivalence classes, then we must have two complete and disconnected subgraphs within the original graph. And for three equivalence classes, we must have three complete and disconnected complete subgraphs. Is my understanding correct? How can I formulate this property mathematically?

For $ii$), my understanding is that $\mathcal{A}$  should be a bipartit graph with non-empty color classes. Again, I'm not sure how I can formulate this. Any help would be much appreciated.

答:
First, let's focus on $(1)$.

The trick to writing $\phi''$ out is to shift from second-order to first-order thinking, via representatives.

Specifically, "The structure is the disjoint union of three complete graphs" is really a second-order statement as phrased, since it essentially quantifies over sets ("There exist three sets which partition the universe and such that ..."). However, I don't really need to think about the whole "pieces" of the universe: every equivalence class is determined by any of its members.

For example, "There are at least five equivalence classes" can be expressed as "There are at least five elements whose equivalence classes are different," and - granting that we've already written down $\phi'$, so we know that $E$ is an equivalence relation in the first place - this is really just a first-order claim:

 $\exists x_1,x_2,x_3,x_4,x_5(\bigwedge_{1\le i<j\le 5}\neg x_iEx_j)$.

Now do you see how to write "There are exactly three equivalence classes" in a first-order way (again, taking $\phi'$ for granted)?

 The sentence $$\exists x_1,x_2,x_3[(\bigwedge_{1\le i<j\le 3}\neg x_iEx_j)\wedge\forall y(\bigvee_{1\le k\le 3}yEx_k)]$$ will do the job. I've chosen to use "big-connective" notation here, namely $\bigwedge$ and $\bigvee$, since this emphasizes how the solution generalizes if we replace $3$ with any larger number.

OK, now what about $(2)$?

Well, here the challenge is the same: the statement as written involves quantification over sets, so we need a way to think about representatives somehow.

The obvious notion of "representative" is "member of each 'side,'" i.e., we're looking for a sentence beginning "$\exists x_1,x_2$" where intuitively $x_1$ and $x_2$ lie in the two parts of our bipartite graph. Here the trouble is that we need to figure out a way to identify each "side." So:

Suppose $\mathcal{G}$ is a bipartite graph with domain $U$ and $A,B$ are nonempty sets partitioning $U$ such that $E^\mathcal{G}=(A\times B)\cup(B\times A)$. How can I tell whether two elements of $U$ lie in the same "piece" (= either both in $A$ or both in $B$)?

Now do you see how to use this to solve your problem?


问:
An application of Portmontau's theorem
Let $\phi$ be a continuous bounded function. Show that if $(X_n)_{n \geq 1}$ converges in probability to $c \in \mathbb{R}$ then $\mathbb{E}[\phi(X_n)]$ converges to $\phi(c)$.

It seems to me that it is slightly modified version of Portmanteau's theorem, but I can't manage to see how the boundedness is used in the proof.

答:
A sequence of random variable $(X_n)_{n \geq 1}$ which converges in probability to some random variable $X$, also converges in distribution to $X$. In particular, if $X_n$ convergence in probability to a constant  $c$, then $X_n$ converges to $c$ in distribution (in this case, the converse is also true). Now, since $\phi$ is a bounded continuous function, we may apply Portmontau's theorem to get

$$
\lim_{n \to \infty} \mathbb{E}(\phi(X_n))  = \mathbb{E}(\phi(c)) = \phi(c).
$$

Title: An application of Portmontau's theorem


问:
Maximizing angle between three unit vectors

Find the maximum angle between three equally inclined unit vectors in the 3D space.

My attempt

Define unit vectors $\hat{a},\hat{b},\hat{c}$ and their components $a_i,b_i,c_i, i\in \{1,2,3\}$

Note that $a_1b_1+a_2b_2+a_3b_3=c_1b_1+c_2b_2+c_3b_3=a_1c_1+a_2c_2+a_3c_3$

So maximize

$$\arccos(a_1b_1+a_2b_2+a_3b_3) \text{ or } \arccos(c_1b_1+c_2b_2+c_3b_3) \text{ or } \arccos(a_1c_1+a_2c_2+a_3c_3)$$

under the constraint

$$\sqrt{a_1^2+a_2^2+a_3^2}=\sqrt{b_1^2+b_2^2+b_3^2}=\sqrt{c_1^2+c_2^2+c_3^2}=1$$

I'm not sure what to do next. Any help is appreciated!

Update

WLOG it can be assumed that any one of the vectors is $\hat{i}$ or $\hat{j}$ or $\hat{k}$

This would reduce the number of variables to 6

答:
Lets not complicate it...

Let $p,q,r$ be  unit the vectors indeed $$|p+q+r|^2\ge 0$$

 $$\implies 3+2(p\cdot q+q\cdot r+r\cdot p)\ge 0$$ $$3+6\cos \theta\ge 0$$ Can you end it now?


问:
Quasicompact over affine scheme
Let $X$ be a scheme and $f : X \rightarrow \mathrm{Spec}\, A$ a quasicompact morphism. Are there any easy conditions on $A$ under which we can say that $X$ is quasicompact?

Quasicompact morphism means only that there is an affine cover $\cup_{i \in I} \mathrm{Spec}\, A_i$ where $f^{-1}(\mathrm{Spec}\, A_i)$ is quasicompact. It doesn't seem to be enough.

答: $X$ is always quasi-compact. This is a standard result which can be found in any good and complete introduction to algebraic geometry. But you can also prove it yourself. In the end, it is just an exercise in general topology. Hint: Affines schemes are quasi-compact. So choose a finite subcover of $\{\mathrm{Spec} A_i \to \mathrm{Spec} A\}$ and observe that $X$ is a finite union of quasi-compact open subspaces.</p>

Title: Quasicompact over affine scheme


问:
By choosing a suitable binomial expansion of the form $(1±kx)^n$ and a suitable value of x. Find $(1.01)^{20}$ correct to 3 s.f.
I have tried replacing kx by 0.1, square rooting 0.1 and squaring 0.1, but none of these methods worked.
I don’t know where to start from. I would appreciate some assistance on this question. Thanks</p>

答:
If you write out the binomial expansion
$$1.01^{20}=1^{20}+{20 \choose 1}1^{19}\cdot 0.01+{20 \choose 2}1^{18}\cdot 0.01^2 \ldots$$
You can note that the binomial increases by at most a factor $20$ while the power of $0.01$ divides by $100$.  You can bound the tail by a geometric series of ratio $\frac 15$, so the tail is less than $\frac 14$ of the last term you include.  The first two terms give $1.2$, so once a term is less than $0.01$ you can stop.


问:
A general question about inequalities
I am studying Sobolev spaces, and always appear inequalities like the following
$$
||u||_{\mathcal{X}}\leq C ||u||_{\mathcal{Y}},
$$
where $\mathcal{X}=L^\infty $ and  $\mathcal{Y}=W^{1,p}$  for example.

Why if I have this kind of inequalities is possible conclude that $\mathcal{Y}\subset\mathcal{X}$ with continuous injections?

答:
We have the following well-known result:

Let $L:X \to Y$ be a linear map between normed vector spaces $X, Y$. Then, $L$ is continuous if and only if $L$ is Lipschitz continuous, i.e.
$$\exists C > 0,\forall x \in X, \quad \|L(x)\|_Y \le C \|x\|_X.$$

Let $X, Y$ be normed vector spaces. We note $\mathcal L(X, Y)$ the set of all continuous map from $X$ to $Y$ with the usual norm, for $L \in \mathcal L(X, Y)$,
$$\|L\|_\mathcal L = \sup_{\|x\|_X \le 1} \|L(x)\|_Y$$
which is well-defined since continuous map are Lipschitz continuous.
Now to answer your question, if we have a canonical linear injection $\iota: X \to Y$, show that the injection is continuous is equivalent to show that there is $C > 0$ with
$$\|\iota(x)\|_Y = \|x\|_Y \le C \|x\|_X.$$

Title: A general question about inequalities


答:
Because if such an inequality holds for all $u\in\mathcal{Y}$ then the identity map
$$I:(\mathcal{Y},\|\cdot\|_\mathcal{Y})\to(\mathcal{X},\|\cdot\|_{\mathcal{X}})$$
is an injective, bounded linear operator. For linear maps, bounded and continuous are notions that coincide.

$^*$: Just to make sure that no one gets confused with the word bounded: an operator $T:E\to F$ is called bounded when there exists a constant $c>0$ so that $\|Tx\|_F\leq c\|x\|_E$ for all $x\in E$

Title: A general question about inequalities

Body:
I am studying Sobolev spaces, and always appear inequalities like the following
$$
||u||_{\mathcal{X}}\leq C ||u||_{\mathcal{Y}},
$$
where $\mathcal{X}=L^\infty $ and  $\mathcal{Y}=W^{1,p}$  for example.

Why if I have this kind of inequalities is possible conclude that $\mathcal{Y}\subset\mathcal{X}$ with continuous injections?


问:
How to calculate this log
I'm supposed to calculate $$16^{\log_{0.5}{2.5}}$$

now the problem is it is not .25 rather 2.5. How can I solve this?

Title: How to calculate this log

Body: $16^{\log_{0.5}2.5}=16^{\ln2.5/\ln0.5}=16^{-\ln2.5/\ln2}=e^{-4\ln2.5}=2.5^{-4}$


问:
Equation of ellipse formed by the projection of a tilted circle onto a plane
So I'm assuming that the x-y plane is represented by the normal vector (0,0,1), and I have a circle with radius $\omega$, given by ($\omega$ cos[$\phi$], $\omega$ sin[$\phi$], 0).

I apply rotations about the x and y axis of $\alpha$ and $\beta$ respectively to the circle and project it onto the x-y plane. Which gives me the parametric form of an ellipse:

($\omega$ (cos[$\beta$] cos[$\phi$] +sin$^2$[$\alpha$] sin[$\phi$]), $\omega$ cos[$\alpha$] sin[$\phi$],
0)

All is well so far, I plotted this with a couple of tilt angles and things seem to behave as expected. However, I would like to fit this to 5 data pairs, for which I need a non-parametric version. Is anyone able to massage this into the equation of a rotated ellipse? I need the ellipse to be described by the tilt angles $\alpha$ and $\beta$.

答:
You can rewrite your parametric equations as follows:
$$
x-y{\sin^2\alpha\over\cos\alpha}=\omega\cos\beta\cos\phi,
\quad
y=\omega \cos\alpha \sin\phi.
$$
Dividing these, respectively, by $\cos\beta$ and $\cos\alpha$, then squaring and adding together, we finally get:
$$
x^2\cos^2\alpha+y^2(\sin^4\alpha+\cos^2\beta)
-2xy\sin^2\alpha\cos\alpha=\omega^2\cos^2\alpha\cos^2\beta.
$$


问:
Different definitions of abelian categories
Í'm writing some notes about homological algebra and I started to ponder about two different definitions of abelian categories.

A category is abelian if it's preabelian, every monomorphism is a kernel and every epimorphism is a cokernel (somewhat the standard definition nowadays);

A category is abelian if it's preabelian and, for every morphism $f$, the natural map $\operatorname{coim}f\to\operatorname{im}f$ is an isomorphism. (Grothendieck's definition.)

I know how to prove that they are equivalent but I would like to understand better the pedagogical advantages of each one.

I see two main points: it is (way) easier to prove that $(2)\implies (1)$ than it is to prove that $(1)\implies (2)$ but using approach (1) we can prove early-on that $\ker(\operatorname{coker})$ satisfies the universal property of images, which motivates better the notion of image than simply defining as being $\ker(\operatorname{coker})$. (Or is it obvious that this should be the definition of the image?)

I would like to know what the community here thinks about whether it is better to start with either definition and why.

答:
I don't know what's more standard, but I personnally usually see 2 being used, and I like it better because the intuition behind it is clearer (essentially you're imposing some form of "first isomorphism theorem")

The fact that $\ker(\mathrm{coker})$ is the image is intuitively obvious if you view $\mathrm{coker}$ as being, well $B/\mathrm{im}(f)$ : the kernel of the projection is (or at least should be) $\mathrm{im}(f)$.

With a bit of thought, you can also see why (1) makes intuitive sense, but 2 just seems more intuitive, at least to me.

But that's just a personal point of view. In fact, I think for abelian categories, you soon forget the definition (or at least I do), and mostly you work with them with "everything you can"


问:
Construct a set of local rules in plane such that all permitted tilings are non periodic.(there exists at least one tiling)
This is the problem given to me on a practice.

I imagine the problem is understandable but if there is anything unexplained, please tell me.

I have tried various things to solve this problem but all of them lead to a dead-end, and so far I don't think I have made any good progress to share.

Here are some definitions I think would be helpful to write.

"local rules: a set of $n.n$ squares which are forbidden in the tiling ($n$ is not a constant and can be any positive integer greater than $1$)",

"non periodic : means there is not a non_zero vector such that after adding the tiling to that vector we reach the same tiling",and the plane we are tiling the plane using $1.1$ black or white squares

答:
Allow only "pixelated sections" of Penrose tilings.

To elaborate: Let us first assign integer coordinates to all parallelogram vertices in the above image:

Assign $(0,0)$ to the centre (where five "thick" parallelograms meet)

Assign $(-3,0)$ to its left neighbour and from there assign $(-1,-2)$, $(2,-1)$, $(2,1)$, $(-1,2)$ to its other four neighbours counter-clockwise.

From this, you can extend assigning integer coordinates to parallelogram vertices by using the parallelogram rule, i.e., that we demand
$ x_2-x_1=x_3-x_4$ and $y_2-y_1=y_3-y_4$ for a parallelogram with vertices $(x_1,y_1),\ldots,(x_4,y_4)$.

Now turn this into a tiling of the square latice by black/white tiles, namely make the tile at $(x,y)$ black iff $(x,y)$ is one of the infinitely many vertices labelled above.
This is an aperiodic tiling of the square lattice.

Now forbid all $6\times6$ patterns that do not appera in this aperiodic tiling.
(Or put differently: Allow only such $6\times 6$ patterns that do occur).
It requires checking, but it seems quite clear that all lattice tilings that use only allowed patterns must correspond to a Penrose tiling and therefore are aperiodic.


问:
Proof of separating hyperplane theorem using minimax theorem for zero sum games
Let $x_1,\ldots,x_M\in\mathbb{R}^N$. Let $P \equiv conv \{x_1,\ldots,x_M\}$ denote the convex hull of these points. Using the Minimax Theorem prove that for all $y\in\mathbb{R}^N \setminus P$ there exists a vector $v\in[-1,1]^N$ such that $v\cdot(x-y)>0$ for all $x\in{}P$.

(Hint: Find a way of identifying $v$ and $x$ with mixed
strategies in a zero-sum game.)

My attempt: I have been able to prove the other direction (that too with help from online lecture notes) - i.e. how to use the separating hyperplane theorem to prove the minmax theorem. But I'm unable to find a way to proceed with the direction required in the problem. Any help is appreciated.

答:
Hint: $v^T(x-y)> 0$ for all $x \in P$ is the same as $\min_{x \in P} v^T(x-y) > 0$, and 'there exists a $v$' can turn this into "$\min_x \max_v$" or "$\max_v \min_x$" (one of the two, try to figure out which one). Then you can apply the minimax theorem and use that $x \neq y$.

prove that for all $y\in\mathbb{R}^N \setminus P$ there exists a vector $v\in[-1,1]^N$ such that $v\cdot(x-y)>0$ for all $x\in{}P$.

For any $x \in P$ we have $\max_{v \in [-1,1]^N} v^T(x-y) = \sum_{i=1}^N |x-y|_i >0$ since $x\neq y$. Therefore $ \min_{x \in P} \max_{v \in [-1,1]^N} v^T(x-y) > 0$. By the minimax theorem:
$$\min_{x \in P} \max_{v \in [-1,1]^N} v^T(x-y) = \max_{v \in [-1,1]^N} \min_{x \in P} v^T(x-y)$$
so $\max_{v \in [-1,1]^N} \min_{x \in P} v^T(x-y)>0$, which means there exists a $v\in[-1,1]^N$ such that $\min_{x \in P} v^T(x-y)>0$, which means $v^T(x-y)>0$ for all $x \in P$.


问:
Why I should show that no element of $R$ divides both $2$ and $1 + \sqrt{-5}$?
Here is the question I want to answer:

Let $R = \mathbb Z[\sqrt{-5}]$ and $ I = (2, 1 + \sqrt{-5})$ be an ideal of $R$.

Show that $I$ is not a free $R-$module.

But here in this question How to show that an $\mathbb{Z}[\sqrt{-5}]$-module is not free? in its answer the author said that: to prove that it is free of rank 1, we should try to prove that no element of $R$ divides both $2$ and $1 + \sqrt{-5}.$ could anyone explain to me why this will prove that it is not free of rank $1$?

Here is the answer I am speaking about:

It is generated by two elements so if it were free, it would be free of rank $1$ or $2$.

If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.

If it were free of rank $2$, then $2$ and $1+\sqrt{-5}$ would have to be free generators (can you see why ?). You can then try to prove that this is not the case, that is, that $2$ and $1+\sqrt{-5}$ are not linearly independent.

EDIT:

I know that we are using the contrapositive of this statement for the proof:

$I$ is free $R$ module $\implies$ I is a principal ideal.

And I know that we are trying to show that $I$ is not principal.

答:
If $I$ were free of rank $1$ there would be an element $b \in I$ such that the $R$-linear map $\varphi: R \to I, x \mapsto xb$ would be an isomorphism. This would mean that there exist elements $r_1,r_2 \in R$ s.t. $r_1b=2$ and $r_2b=1+\sqrt{-5}$. If you are able to show that no $b$ in $R$ divides both elements this is a contradiction.

Edit. Let $b=\alpha+\beta \sqrt{-5}$, $\alpha, \beta \in \mathbb{Z}$. It holds that $N(2)=4, N(b)=\alpha^2+5\beta^2$. Since the norm is multiplicative $N(b)$ has to divide $N(2)=4$ in $\mathbb{Z}$ and thus $N(b) \in \{1,2,4\}$. Note that in all of these cases $\beta$ has to be zero and that the norm can never be two. Assume that the norm is one, then $\alpha=1$ or $\alpha=-1$ and $\beta=0$. On the other hand $N(b)$ has to divide (in $\mathbb{Z}$) $N(1+\sqrt{-5})=6$. This means that $b=1$ or $b=-1$ are in fact the only options. This however implies that $I=R$ which is a contradiction.


问:
Rescale linear interval [a, b] to logarithmic [0, 1] scale in stead of linear [0, 1] (normalization)
I can't get my head around this simple problem. I would very much appreciate a good explanation in the answers.

I have a data array in which elements are between [a, b] and I normalize this array between [0, 1] using:

```<code>min_scale = np.min(data)
max_scale = np.max(data)
normalized_data = (data - min_scale) / (max_scale - min_scale)
</code>```

Instead of normalizing to a linear [0, 1] scale, I would like to normalize to a logarithmic scale between [0, 1] in order to increase the importance of values closer to the max and decrease the importance of values closer to the minimum.

I am still looking for a convincing formula.. Does it make sense to first linearly normalize [a, b] to [1, e] and then take the log of the resulting vector?

答:
Your suggested transformation seems fine. You could add a degree of freedom by first linearly transforming $[a,b] \to [1,\theta]$ followed by $\log_\theta$. This gives the transformation

$$x \mapsto \frac{\ln  \left(\frac{a \theta -b}{a-b}+\frac{(1-\theta ) x}{a-b}\right)}{ \ln (\theta) }. $$


问:
Generating lookup tables for Thistlethwaite's algorithm
I'm trying to make a Rubik's cube solver using Thistlethwaite's algorithm. I've already made a working program that solves the cube in 3~ minutes without the lookup tables.

I'll summarize the algorithm to save you time:

There are 4 groups, G0 is any scrambled cube and G4 is a solved cube.
Each group has a goal that is reachable only using the legal moves of said group.
To get from Gi to Gi+1 you can only use the moves from Gi.
This way the cube is solved in groups which makes it easier to find a solution, and each group "contains" the group before it (in G2, G1 is already solved and so on). by limiting the moves of each group, the cube is assured to never un-solve any of the goals that it already reached.

the lookup tables are used as a way to prune the tree out of nodes that lead to "dead ends". if a node is further from a solution, it doesn't need to be checked.

My initial thought was to use a solved cube (G4) and run a BFS that indexes all the possible permutations based on a specific group and their rank (how close they are to a solution). for example G0->G1 can use all the moves, and only needs to worry about edge orientation.

The problem is, I can't just use a solved cube as a starting point, since there are many permutations were G1 would be satisfied besides the solved state.

I then thought about using the permutation of the edges together with the orientation of them to generate an index, but it still leads to the same problem that there isn't a single starting point.

How can I generate a table that tells me based on a position how close or far I am from a solution? (not from a solved cube state, but from a solved group state)

thanks in advance, I hope I was clear.

答:
A BFS starting from the solved cube is the way to go. However, you don't use the complete cube but instead ignore all the aspects that are not affected by the current stage of the algorithm.

In the first stage, where the edge orientations get solved, the only aspect of the cube that you look at are those edge orientations. Ignore the corners, and think of all the edges as being identical pieces so that their permutation is irrelevant. Of course the edges still move around when a move is done, but you cannot distinguish one edge piece from another. Only their orientation matters, so you can still see whether an edge is flipped or not.

Do a BFS with such a simplified cube to build your pruning table. The state of the cube can be uniquely encoded into an 11-bit integer, where each bit is the orientation of an edge at some location on the cube. That integer is the index into your pruning table. The table will store the distance from solved for each state of this simplified cube. Each time the BFS encounters a cube state it has not seen before (the associated table entry has not yet been filled), it then stores the current distance from start into the table. It is basically generating a God's Algorithm table for this simplified cube.

You will need to define what the edge orientation is on your cube. It depends on which pair of opposite faces gets restricted in $G_1$. Let's say it is F and B. On the simplified edge-orientation-only cube any moves in G1= do not flip any pieces (though they are still moved around). The moves F, B, F', and B' however will each not only move four edges around, they also flip those four moving edges. With this definition you can build your pruning table through a BFS search.

When you want to solve a normal cube, and use this pruning table to solve the first stage in the algorithm, you will need to be able to look up the normal cube in your pruning table. So you need to extract the edge-orientation-index from it. For any edge, imagine bringing it to its home location using only moves from $G_1$. Once it is in its home location you can easily tell if it is flipped or not by whether its colours match the face centres. Since the $G_1$ moves did not affect its orientation, you now know whether the edge is flipped even if it was not at its home location.

When you have determined the orientation of all the edges of your cube, you can set the simplified edge-orientation cube to that state, use the pruning table to find a solution for this stage, and then apply this solution to the original cube.

The other stages of the algorithm can be done in essentially the same way, generating one or more pruning tables using a simplified version of the cube that only encodes the aspects being solved in that stage. The most difficult aspect to encode into a pruning table index will be what happens to the corner permutation in the third stage.

For each stage you can either generate a single pruning table, so that solving that stage for a real cube is essentially a God's Algorithm set of table lookups, or you can use several pruning tables, in which case solving that stage will involve a search using those tables (i.e. IDA*, iterative deepening A* search).

I presume you have found the page on my website about the Thistlethwaite algorithm. Another page that might be helpful is the page about computer cubing which has brief descriptions of the techniques involved.

Body:
A BFS starting from the solved cube is the way to go. However, you don't use the complete cube but instead ignore all the aspects that are not affected by the current stage of the algorithm.

In the first stage, where the edge orientations get solved, the only aspect of the cube that you look at are those edge orientations. Ignore the corners, and think of all the edges as being identical pieces so that their permutation is irrelevant. Of course the edges still move around when a move is done, but you cannot distinguish one edge piece from another. Only their orientation matters, so you can still see whether an edge is flipped or not.

Do a BFS with such a simplified cube to build your pruning table. The state of the cube can be uniquely encoded into an 11-bit integer, where each bit is the orientation of an edge at some location on the cube. That integer is the index into your pruning table. The table will store the distance from solved for each state of this simplified cube. Each time the BFS encounters a cube state it has not seen before (the associated table entry has not yet been filled), it then stores the current distance from start into the table. It is basically generating a God's Algorithm table for this simplified cube.

You will need to define what the edge orientation is on your cube. It depends on which pair of opposite faces gets restricted in $G_1$. Let's say it is F and B. On the simplified edge-orientation-only cube any moves in G1=<U,D,R,L,F2,B2> do not flip any pieces (though they are still moved around). The moves F, B, F', and B' however will each not only move four edges around, they also flip those four moving edges. With this definition you can build your pruning table through a BFS search.

When you want to solve a normal cube, and use this pruning table to solve the first stage in the algorithm, you will need to be able to look up the normal cube in your pruning table. So you need to extract the edge-orientation-index from it. For any edge, imagine bringing it to its home location using only moves from $G_1$. Once it is in its home location you can easily tell if it is flipped or not by whether its colours match the face centres. Since the $G_1$ moves did not affect its orientation, you now know whether the edge is flipped even if it was not at its home location.

When you have determined the orientation of all the edges of your cube, you can set the simplified edge-orientation cube to that state, use the pruning table to find a solution for this stage, and then apply this solution to the original cube.

The other stages of the algorithm can be done in essentially the same way, generating one or more pruning tables using a simplified version of the cube that only encodes the aspects being solved in that stage. The most difficult aspect to encode into a pruning table index will be what happens to the corner permutation in the third stage.

For each stage you can either generate a single pruning table, so that solving that stage for a real cube is essentially a God's Algorithm set of table lookups, or you can use several pruning tables, in which case solving that stage will involve a search using those tables (i.e. IDA*, iterative deepening A* search).

I presume you have found the page on my website about the Thistlethwaite algorithm. Another page that might be helpful is the page about computer cubing which has brief descriptions of the techniques involved.


Title: How to calculate this log

Body:
I'm supposed to calculate $$16^{\log_{0.5}{2.5}}$$

now the problem is it is not .25 rather 2.5. How can I solve this?

Body:
Working to base $2$ we have

$$16^{-\log 2.5}=(2^{\log 2.5})^{-4}=2.5^{-4}.$$


Title: How to calculate this log

Body:
First note $$\Large \log_{0.5} 2.5 = \log_{1/2}2.5=-\log_{2}2.5$$

We used $$\Large \log_{a^m}b=(1/m)\log_a b$$

So
$$\Large 16^{-\log_{2}2.5}=2^{-4\log_{2}2.5}=(2.5)^{-4}$$


问:
Let $U$ be an arbitrary subspace of a vector space $V$ over a field $\Bbb F$ as follows.
Proposition:

Let $V$ be a vector space over a field $\Bbb F$. A non-empty set $U \subseteq V$ called subspace of $V$ iff

for all $u,w \in U$, then $u+w \in U$, and

for all $k \in \Bbb F$ and $u \in U$, then $ku \in U$.

Below is the consequent of the proposition above, which I want to prove.

Let $U$ be an arbitrary subspace of a vector space $V$ over a field $\Bbb F$.
Then,

$U$ is an additive subgroup of $V$, and

$U$ contains $0*$, where $0*$ is a zero vector of $V$.

Attempt:
For $1$, let $U$ be an arbitrary subspace of a vector space $V$ over $\Bbb F$. Let $u,w \in U$. Denote $-w$ is the additive inverse of $w$ for all $w \in U$. Want to show: for all $u,w \in U, u - w \in U$.
This is follow by definition and by the proposition. Since $U$ is a subspace of $V$, then $U$ itself is a vector space over $\Bbb F$. So, for all $w \in U, -w \in U$ and hence $u+(-w) = u-w \in U$. Hence proved. $\Box$.

For $2$, by the proposition, there exists $0 \in \Bbb F$ such that for all $u \in U$, we obtain $0\cdot u = 0* \in U$. Thus, $U$ contains $0*$, the zero vector of $V$. Hence proved. $\Box$.

I know this is a very basic concept of vector space chapter. But, I only want to know that whether my answer above is correct. Thanks in advanced.

答:
Remember, to prove (i), we are simply proving that $U$ is in fact a subgroup of $V$ with respect to addition. This is equivalent to showing $U$ is closed with respect to addition and (additive) inverses, and that $U$ is non-empty. But by assumption $U$ is already non-empty, and $U$ is closed w.r.t. to addition. So we just have to show it is closed w.r.t. to inverses. But by the definition of what it means to be a subspace, $U$ is itself a vector space. Hence for any $\bar{u} \in U$, there must exist some additive inverse which we denote $-\bar{u}$. This simply follows from the axioms of what it means to be a vector space. We don't need to use the proposition at all (we just need to know that $U$ is a vector space, and you assume this anyway).

Addressing your actual proof of (i): your proof is correct, but it can be shortened. It's true that proving $\bar{u} - \bar{w} \in U$ shows it's a subgroup, but you just need to show it's closed wrt to negatives since by definition $U$ is non-empty and from the proposition it's obviously closed wrt to addition. I.e. just the final piece of your proof is what you need to actually prove the result.

For the proof of (ii), your proof is correct! Another way to see it is to  note that since $U$ is an additive subgroup of $V$. This means that $U$ must contain the identity element of $V$, since we have already proved that $U$ is closed w.r.t. inverses and addition. Therefore for any $\bar{u} \in U$, we have that $\bar{u} + (-\bar{u}) = 0* \in U$.


问:
How does one show that two functors are *not* isomorphic?
Let $C$ be the category of finite-dimensional vector spaces over some field. It is easy to construct pairs of endofunctors $F, G$ of $C$, of the same variance, such that $F(V)$ and $G(V)$ have the same dimension for every $V \in C$, yet are not naturally isomorphic.

For example, we could take $G(V) := (V^*)^{\otimes 2}$, and $F(V) := (V^{\otimes 2})^*$, where $*$ denotes the dual. Or, I'm sure many of us have, at some time, tried showing that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, before realizing that it is false.  A convincing reason why they are not canonically isomorphic is that there is no obvious reason, in terms of universal properties, why they should be. In other words: if you try proving that $(\Lambda V)^* \simeq \Lambda(V^*)$ canonically, you're going to have a bad time.

However, how does one explicitly show that $F$ and $G$ are not isomorphic? Is there a slick, systematic way of doing this?

Edit: Well, my examples couldn't be worse, as Mariano points out. But the gist of my question still stands (perhaps just get rid of the finite-dimensionality assumption ;))

答:
I would like to address the question posed in the title in its full generality, rather than discussing single cases.

Two parallel functors are simply objects in a functor category.
Citing Awodey:

One way to prove that two objects are not isomorphic is to use
  “invariants”: attributes that are preserved by isomorphisms. If two
  objects differ by an invariant they cannot be isomorphic. Generalized
  elements provide an easy way to define invariants....

So if you are convinced that 2 functors are not isomorphic, try to use that feeling to provide invariants or generalized elements in the functor category that back up your intuition.

If the functor category is locally small, the above procedure amounts to using the Yoneda embedding: 2 objects in a locally small category are isomorphic if and only if their Yoneda embeddings are isomorphic.

This is certainly a slick and systematic procedure, but it may not be that easy.


问:
Python library for large scale convex optimization
I'm trying to find a library for convex optimization in python which can handle a large amount of variables and constraints.

My problem is in the form

$$
\min p^T \cdot q + |c| \\
\text{lower bound} \leq q \leq \text{upper bound} \\
\text{lower bound} \leq c \leq \text{upper bound} \\
Aq + Bc = d \\
\text{lower bound} \leq Rq \leq \text{upper bound}
$$

Where $p, d$ are vectors and $A, B$  are sparse matrices and $q, c$ are variables. $R$ is a special matrix which might be the cause of my issue.

$$
Rq = \\
\begin{pmatrix}
q_1 \\
q_2 \\
q_3 \\
q_4  \\
q_1 + q_5 \\
q_2 + q_6 \\
q_3 + q_7\\
q_4 + q_8 \\
q_1 + q_5 +q_9 \\
q_2 + q_6 + q_{10} \\
q_3 + q_7 + q_{11} \\
q_4 + q_8 + q_{12} \\
\vdots \\
\sum_{k=0}^n q_{4k+1} \\
\sum_{k=0}^n q_{4k+2} \\
\sum_{k=0}^n q_{4k+3} \\
\sum_{k=0}^n q_{4k+4}
\end{pmatrix} \\
=\\
\begin{pmatrix}
1 & 0 & \dots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & \dots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & \dots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & \dots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 0 & \dots & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 1 & 0 & \dots & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & \dots & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & \dots & 0 & 0 & 0 & 0 & 0\\
1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & \dots & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & \dots & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & \dots & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & \dots & 0\\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots &
\end{pmatrix}q
$$

code

```<code>from cvxopt.modeling import variable, op, sum
q = variable(numb_prod, 'q')
c = variable(numb_capacity, 'c')
prod_price = price_purchase_power.trans() * q

problem = op(sum(prod_price) + 1e-6*sum(abs(c)), [constraints])
problem.solve(format='sparse', solver='glpk', options=
{'glpk'{'msg_lev':'GLP_MSG_OFF'}})
</code>```

error

Traceback (most recent call last): File "C:\Users\OsRas\AppData\Local\Continuum\anaconda3\lib\site-packages\IPython\core\interactiveshell.py", line 3325, in run_code exec(code_obj, self.user_global_ns, self.user_ns) File "", line 489, in q, c, total_cost_x, water_values, list_numb_of_capacity_variables = solution(demand, price_purchase_power, supply) File "", line 391, in solution test = problem.solve(format='sparse', solver='glpk', options={'glpk':{'msg_lev':'GLP_MSG_OFF'}}) File "C:\Users\OsRas\AppData\Local\Continuum\anaconda3\lib\site-packages\cvxopt\modeling.py", line 2600, in solve t = self._inmatrixform(format) File "C:\Users\OsRas\AppData\Local\Continuum\anaconda3\lib\site-packages\cvxopt\modeling.py", line 2479, in _inmatrixform for v in variables: vmap[v] = x[vslc[v]] File "C:\Users\OsRas\AppData\Local\Continuum\anaconda3\lib\site-packages\cvxopt\modeling.py", line 240, in getitem return (+self).getitem(key) File "C:\Users\OsRas\AppData\Local\Continuum\anaconda3\lib\site-packages\cvxopt\modeling.py", line 931, in getitem f._linear = self._linear[l] File "C:\Users\OsRas\AppData\Local\Continuum\anaconda3\lib\site-packages\cvxopt\modeling.py", line 1385, in getitem for v,c in iter(self._coeff.items()): OverflowError: Python int too large to convert to C long

Currently I'm having 108575 variables which is too large for cvxopt. I was hopping running an optimization which can handle the amount of variables in the millions...

Is there any library that can handle that? Or does one need to look into another language?

I can sacrifice some accuracy and make this problem LP by removing $|c|$ if that opens up other possibilities.

I've asked this question on stackoverflow as well. But perhaps there is a better auidience here.

Edit
As the problem is formulated now we have

$\dim(A) = 26058 \times 73831 $ with average three $1$:s per row, rest zero, upper triangle.

$\dim(B) =26058 \times 34744$ with average of three elements per row, $1$:s and $-1$:s, rest zero, elements around diagonal and some one the diagonal.

$\dim(R) = 17280 \times 73831$

答:
On Stackoverflow you shared that $|c|$ means $||c||_1$, so the problem can be made linear. I can think of two solution methods.

The first one is to use an interior point algorithm. Due to the size of the problem, the location of the $1$s is important. If either $AA^T + BB^T$ or
$$\begin{pmatrix}A^TA & O \\ O & B^TB\end{pmatrix}$$
has a sparse Cholesky factorization (for example, it can be row/column permuted to an arrow matrix), this approach can be memory efficient and fast. You could try CPLEX or Gurobi. They should tell you the number of nonzeros in the cholesky factor.

The second option is to use ADMM. You could solve the subproblems in cvxpy or use a specialized package. I have never had much luck with ADMM, but your mileage may vary.


问:
reflexive, symmetric and not transitive relation
If a relation on a set $R$ is reflexive, symmetric but not transitive, does it imply that all the classes $[x] = \{ y\mid xRy\}$ are mutually exclusive, in the sense that for no $[x]$ and $[y]$, they both share an element, and so their intersection is always empty?

答:
Take any non-trivial equivalence relation $R$ on a set $X$ (where equivalence relation means a relation that is reflexive, symmetric and transitive, and where, by non-trivial, I mean that there is at least one pair $x, y \in X$ with $\lnot x \mathrel{R} y$). (E.g., define $R$ by $x \mathrel R y \Leftrightarrow x \cong y \;(\mbox{mod $2$})$ on $X = \Bbb{Z}$.) Let's find a way to extend $R$ to a relation on a larger set $X'$ that is not transitive.

Let $X' = X \cup \{*\}$ where $*$ is some object that does not belong to $X$ and define $x \mathrel{R'} y$ to hold iff $x, y \in X$ and $x \mathrel{R} y$ or at least one of $x$ and $y$ is *. So we've added something to $X$ that is related to everything while keeping the original relation on pairs of elements of $X$. $R'$ cannot be transitive, because there are $x, y \in X$ with $\lnot x \mathrel{R'} y$, but $x \mathrel{R'} *$ and $* \mathrel{R'} y$. However, $x \in [*] \cap [x]$ for every $x$ in $x$. So the classes $[z]$ for $z \in X'$ are certainly not disjoint.

I see another answer has appeared while I have been typing this. The example in that answer is a specific instance of the above construction.


答:
Just express the base $16$ in terms of $\frac 12$ and use laws of exponents:

$$16^{\log_{\frac 12}\frac 52}= 2^{4\log_{\frac 12}\frac 52}= \left( \left(\frac 12\right)^{\log_{\frac 12}\frac 52}\right)^{-4} $$$$=  \left(\frac 52\right)^{-4}= \frac{16}{625}$$

Title: How to calculate this log


问:
Showing the Polar form of the Cauchy-Riemann Equations imply the Cartesian Form
Often times, when the Cauchy-Riemann equations $$u_x = v_y; \ u_y = -v_x$$ are introduced, it's an exercise to show that one can arrive at the "polar form" of those equations. That is, the cartesian Cauchy Riemann equations imply the equations $$r \cdot \frac{\partial{u}}{\partial{r}} = \frac{\partial{v}}{\partial{\theta}}; \ \frac{\partial{u}}{\partial{\theta}} = -r \frac{\partial{v}}{\partial{r}}$$

However, if I am to call these second equations the Polar form of the Cauchy-Riemann equations, I'd like there to be an equivalence of the two sets of equations. That is, I would like the polar equations to imply the Cartesian ones as well. However, a proof of this converse is often completely missing from many texts, and the converse proof doesn't seem to be as immediately obvious (at least to me).

Naively, one could try and replicate the same style of proof often given for the forward direction. Namely, we could use chain rule to deduce $$u_x = u_r \cdot r_x + u_{\theta} \cdot \theta_x$$ and do the same for $v_y$, but explictly computing $r_x, \theta_x, r_y, \theta_y$ is admittedly quite a bit trickier and involves dividing by various trig functions or taking inverse trig functions of things and then computing derivatives (at least with the methods I've tried). In either case, we now have to be much more careful making sure every expression is well-defined for values of $\theta$.  So far, I've not been able to attain the Cartesian Cauchy-Riemann equations, so perhaps a little help in the right direction would be greatly appreciated.

答:
Let's try it this way:

$$\begin{aligned}
u_r &= u_x x_r + u_y y_r = u_x \cos \theta +u_y \sin \theta\\
v_r &= v_x x_r + v_y y_r = v_x \cos \theta +v_y \sin \theta
\end{aligned}$$

$$\begin{aligned}
u_\theta &= u_x x_\theta + u_y y_\theta =   u_y r\cos \theta -u_x r \sin \theta\\
v_\theta &= v_x x_\theta + v_y y_\theta =   v_y r\cos \theta -v_x r \sin \theta
\end{aligned}
$$

$$\begin{aligned}
r \, u_r &= v_\theta \\
u_x \cos \theta +u_y \sin \theta &= v_y \cos \theta -v_x  \sin \theta\\
(u_x - v_y) \cos \theta +(v_x+u_y) \sin \theta &= 0
\end{aligned}
$$

$$\begin{aligned}
-r \, v_r &= u_\theta \\
-v_x \cos \theta -v_y \sin \theta &= u_y \cos \theta -u_x  \sin \theta\\
(u_x-v_y) \sin \theta -(v_x + u_y) \cos \theta &= 0
\end{aligned}
$$

Combining the last equation in the third and fourth "sets" of equations:

$$ \pmatrix{(u_x-v_y)\cos \theta +(v_x+u_y) \sin \theta \\ -(v_x+u_y)\cos \theta+ (u_x-v_y) \sin \theta}=\pmatrix{0\\0}$$
or
$$ \pmatrix{\cos \theta & \sin \theta\\ -\sin\theta & \cos \theta}\pmatrix{u_x-v_y\\v_x+u_y}=\pmatrix{0\\0}.$$

Since the matrix on the left is a rotation matrix, it must be that the vector $\pmatrix{u_x-v_y\\v_x+u_y}$ is identically zero.

$$\bbox[5px, border: 1pt solid blue]{\begin{aligned}u_x&=\phantom{-}v_y,\\u_y&=-v_x.\end{aligned}}$$

Notice, in the above, that we've assumed $r\ne 0,$  otherwise, there is not much information in the polar form of the C-R equations.  There (at the origin), both $u_\theta$ and $v_\theta$ are zero.  Actually, at $z=0$, the parameter $\theta$ is anyway not well defined.


问:
Solutions to $A^N+B^N=C^N \pm 1$
Is there a solution to $A^N+B^N=C^N \pm 1$ where $A,B,C,N\in\Bbb{N}$, such that $A,B,C > 1,N \geq 4$ and $gcd(A,B)=gcd(B,C)=gcd(A,C)=1$?

This question was inspired by this one: $A^X+B^Y=C^Z\pm 1$ Beal's conjecture "almost" solutions

A related question was asked here before.

Noam Elkies has published a paper on this subject. The table on page 15 of the paper and his extended table here have some interesting results. If I understood his work correctly, there are no solutions for $4 \leq N \leq 20$ and $C < 2^{23}=8,388,608$.

I also found a MathWorld article on 4th order diophantine equations, but it doesn't cover equations of this form. I also checked that the OEIS has no sequences for such Fermat near misses.

答:
This exact question was a summer research project of mine when I was a second-year undergraduate in 2013. It was inspired by infinitely many solutions parametrized by Ramanujan in the $n=3$ case. I didn't make much progress, but I can show you a conditional result that I found, and some slight progress in the even case (with help from Noam Elkies, if I recall correctly). I haven't look at this in 7 years, so please verify independently if the propositions are true.

Abstract. Srinivasa Ramanujan wrote down in his Lost Notebook infinitely many non-trivial examples of integers $x,y,z$ such that $x^3+y^3=z^3\pm 1$, for each sign. Aside from his miraculous construction, the result itself is remarkable as it shows that the smallest perturbation of Fermat's famed equation of degree $3$, which has been known since the time of Euler to have no solutions, has infinitely many solutions. We study the natural generalization $a^n+b^n=c^n\pm 1$, which we call the Fermat-Ramanujan equation.

Definition 1. Let $x\in\mathbb{Z}$ be non-zero. Then there exists a unique prime factorization $x=\pm\prod_{i=1}^{s}{p_i^{e_i}}$, where the $p_i$ are distinct ascending primes and the $e_i$ are positive integers. Then define, for each $k\in\mathbb{Z}^+$, the $k$-radical of $x$ to be $$\text{rad}_k (x)=\prod_{i=1}^{s}{p_i^{\min(e_i,k)}}.$$

Note that for $k=1$, this is the classic radical function
$$\text{rad}_1 (x)=\prod_{i=1}^{s}{p_i^{\min(e_i,1)}=\prod_{i=1}^{s}{p_i}}=\text{rad}(x).$$

Hu-Yang Conjecture. Let $\{a_i\}_{i=0}^{m}$ be integers with $m\ge 2$, such that $\begin{cases} \gcd(a_0,a_1,\ldots, a_m) =1\\ a_1+\cdots+a_m=a_0\end{cases}$ and no subset of $\{a_1,\ldots,a_m\}$ sums to $0$. Then, independent of $\{a_i\}_{i=0}^m$,
$$\forall \epsilon>0,\exists \kappa >0 : \max_{0\le i\le m}(|a_i|)\le \kappa\cdot \left(\prod_{i=0}^m{\text{rad}_{(m-1)}(a_i)}\right)^{1+\epsilon}. \hspace{1cm} [\textbf{1}]$$

Corollary. The above conjecture would imply that for every $\epsilon>0$, for all but finitely many such $\{a_i\}_{i=0}^m$,
$$\max_{0\le i\le m}(|a_i|)\le \left(\prod_{i=0}^m{\text{rad}_{(m-1)}(a_i)}\right)^{1+\epsilon}.$$

Lemma 1. We have that
$$\forall x\in\mathbb{Z}, \forall n,k\in\mathbb{Z}^+: n\ge k\implies \text{rad}_k (x^n)=\text{rad}(x)^k.$$

Proof. Suppose $x=\pm\prod_{i=1}^{s}{p_i^{e_i}}$ as in the definition, and that $n,k\in\mathbb{Z}^+: n\ge k$. Each $e_i\ge 1$ so $ne_i\ge k$ and
$$\text{rad}_k (x^n)=\prod_{i=1}^{s}{p_i^{\min(ne_i,k)}}=\prod_{i=1}^{s}{p_i^k}=\left(\prod_{i=1}^{s}{p_i}\right)^k=\text{rad}(x)^k,$$
as desired.

Proposition 1. Assuming the conjecture:

For all sufficiently large $n$, there exist no triples of positive
integers $a\le b< c$ such that $a^n+b^n=c^n+1$.

If there are infinitely many triples of positive integers $a\le b< c$ such that
$a^n+b^n=c^n+1$, then $n\le 6$.

Proof. Suppose that $n,a,b,c\in\mathbb{Z}^+$ such that $a^n+b^n-c^n=1$ with $a\le b< c$.

Then $\gcd(a^n,b^n,-c^n,1)=1$ and no subset of $\{a^n,b^n,-c^n\}$ sums to $0$. By the corollary and the lemma, for all but finitely many such triples $(a,b,c)$,
$$c^n=\max(|a^n|,|b^n|,|-c^n|)\le \left(\text{rad}_2 (a^n)\cdot\text{rad}_2 (b^n)\cdot\text{rad}_2 (c^n)\right)^{1+\epsilon}= (a^2 b^2 c^2)^{1+\epsilon}<c^{6(1+\epsilon)}.$$

Fix any $\epsilon>0$. Then it is clear that for sufficiently large $n$, $c$ does not even exist and so neither does a triple.

If the existence of infinitely many solutions is assumed, a triple with $c^n<c^{6(1+\epsilon)}$ exists as there are only finitely many exceptions. Then $n<6(1+\epsilon)$, and as this holds for every $\epsilon>0$, we get $n\le 6$.

Corollary. The exact same proposition and proof holds for the equation $a^n+b^n=c^n -1$.

In order to study the equations for even $n$, we first solve $a^2+b^2=c^2\pm 1$, for each sign.

Lemma 2. The binary quadratic form $ap^2+bpq+cq^2$ is equivalent to

$p^2+q^2$ if and only if $b^2-4ac=-4$.

$p^2-q^2$ if and only if $b^2-4ac=4$ and $a,c$ are not both even.

Proposition 2. The integers $x,y,z$ satisfy $x^2+y^2=z^2\pm 1$, for a fixed sign the opposite of which is $\mp$, if and only if there exist integers $a_1,a_2,b_1,b_2$ such that
$$\begin{cases}
1=a_1 b_2-a_2b_1\\
2x=(a_1^2-a_2^2)\mp (b_1^2-b_2^2)\\
y=a_1 a_2\mp b_1 b_2\\
2z=(a_1^2+a_2^2)\mp (b_1^2+b_2^2)
\end{cases}.$$

This is under the assumption that if not both of $x,y$ are even, then $x$ is odd, which does not compromise generality as $x,y$ are symmetric in the equation.

Proof. Suppose $x,y,z\in\mathbb{Z}$ such that $x^2+y^2=z^2\pm 1$, for a fixed sign. Let $b=y$ and define $a,c$ as
$$\begin{cases}a=z+x\\ c=z-x\end{cases}\iff \begin{cases}2x=a-c\\ 2z=a+c\end{cases}.$$

Substituting yields $$\left(\frac{a-c}{2}\right)^2+b^2=\left(\frac{a+c}{2}\right)^2\pm 1\iff b^2-ac=\pm 1.$$

Then the discriminant of the binary quadratic form $$\phi(p,q)=ap^2+2bpq+cq^2$$ is $D(\phi)=(2b)^2-4ac=4(b^2-ac)=\pm 4$. If $D(\phi)=-4$, then Lemma 2.1 can be immediately applied. If $D(\phi)=4$, then Lemma 2.2 applies. This is because at least one of $a,c$ being odd is equivalent to at least one of $z,x$ being odd. If $z$ is odd, we are done. Otherwise, one of $x,y$ must be odd and in the proposition we assumed that it is $x$ in this case.

So $ap^2+2bpq+cq^2$ is equivalent to $x^2\mp y^2$. In other words, there exist $a_1,a_2,b_1,b_2\in\mathbb{Z}$ such that
\begin{align*}
 ap^2+2bpq+cq^2 &= (a_1 p+a_2 q)^2\mp (b_1 p+ b_2 q)^2\\
 &= (a_1^2\mp b_1^2)p^2+2(a_1 a_2 \mp b_1 b_2)pq+(a_2^2\mp b_2^2)q^2.
\end{align*}

Moreover, since we have integers and each form is improperly equivalent to itself, it can be forced that
$$\begin{pmatrix}a_1&a_2\\ b_1& b_2\end{pmatrix}\in SL_2(\mathbb{Z})\implies\det\begin{pmatrix}a_1&a_2\\ b_1& b_2\end{pmatrix}=a_1 b_2 - a_2 b_1=1,$$ One direction of the proposition follows from comparing coefficients. The other is straightforward by substituting and expanding.

So all solutions can be generated by $SL_2(\mathbb{Z})$.

Lemma 3. We have that $a,b,c,d\in\mathbb{Z}$ satisfy

$a^2+b^2=c^2+d^2$ if and only if there exist $p,q,r,s\in\mathbb{Z}$ such that $$(a,b,c,d)=(pq+rs, pr-qs, pq-rs, pr+qs).$$ This is assuming, without loss of generality, that $a,c$ have the same parity and so do $b,d$.

$a^2+b^2+c^2=d^2$ if and only if there exist $p,q,r,s\in\mathbb{Z}$ such that $$(a,b,c,d)=(ps+qr, p^2-q^2+r^2-s^2, pq-rs, p^2+q^2+r^2+s^2).$$

Proof.

Suppose $a,b,c,d\in\mathbb{Z}$ satisfy $a^2+b^2=c^2+d^2$. Assume, without loss of generality, that $a,c$ have the same parity and so do $b,d$. Define the integers $$(2a_0, 2b_0, 2c_0, 2d_0)=(a+c, d+b, a-c, d-b).$$ so that the equation can be manipulated to $a_0 c_0= b_0 d_0$.

Since $\mathbb{Z}$ is a unique factorization domain, there exist $p,q,r,s\in\mathbb{Z}$ such that $(a_0,b_0,c_0,d_0)=(pq,pr,rs,qs)$. Then $$(a,b,c,d)=(pq+rs,pr-qs,pq-rs,pr+qs).$$

The converse holds by Fibonacci's identity $$(pq+rs)^2+(pr-qs)^2=(pq-rs)^2+(pr+qs)^2.$$

Suppose $a,b,c,d\in\mathbb{Z}$ satisfy $a^2+b^2+c^2=d^2$. Rewrite the equation as $(c+ai)(c-ai)=(d+b)(d-b)$ in Gaussian integers. Since $\mathbb{Z}[i]$ is a unique factorization domain, there exist $x=p+ri$ and $y=q+si$ in $\mathbb{Z}[i]$ such that $$(c+ai,c-ai, d+b, d-b)=(xy, \bar{x}\bar{y},x\bar{x}, y\bar{y}),$$ where the bar denotes complex conjugation. Then $\begin{cases} c+ai=(pq-rs) +(ps+qr)i\\ d+b=p^2+r^2\\ d-b=q^2+s^2\end{cases}$ so
$$(a,2b,c,2d)=(ps+qr, p^2-q^2+r^2-s^2, pq-rs, p^2+q^2+r^2+s^2).$$

The converse holds by Lebesgue's identity $$(2(ps+qr))^2+(p^2-q^2+r^2-s^2)^2+(2(pq-rs))^2=(p^2+q^2+r^2+s^2)^2.$$

Title: Solutions to $A^N+B^N=C^N \pm 1$


问:
Lagrange inversion theorem for the normal cumulative distribution
The Lagrangian inversion theorem gives the Taylor series expansion for the inverse of a function. I was wondering how to apply it to the normal cdf:
$$
f(x)=\int_{-\infty}^x g(t)dt,
$$
where $g(t)=(2\pi)^{-1/2}e^{-t^2/2}$. Following the notation in https://en.wikipedia.org/wiki/Lagrange_inversion_theorem, how that applies to $f(x)$, what are $a$ and $f(a)$ in this case?

cheers.

答:
You have apparently chosen to confuse yourself by reusing $g$ in a second way.  Therefore, we do not use the function $g$ from the wiki page you cite.

Since
$$  f'(x) = g(x) > 0  $$
and $g(x)$ is analytic everywhere, you are free to pick any $a \in \Bbb{R}$.

For instances,

randomly choosing $a = 0$, $f(0) = 1/2$;

randomly choosing $a = -1$, $f(-1) = \frac{1}{2} (1 - \mathrm{erf}(1/\sqrt{2}))$;

randomly choosing $a = \pi$, $f(\pi) = \frac{1}{2} (1 + \mathrm{erf}(\pi/\sqrt{2}))$; and

for any choice of $a$, $f(a) = \frac{1}{2}(1 +\mathrm{erf}(a/\sqrt{2}))$,

Where "erf" is the error function.

Let's apply this with $a = 0$, so $f(a) = 1/2$ and
$$  f^{-1}(z) = 0 + \sum_{n-1}^\infty g_n \frac{(z - 1/2)^n}{n!}  \text{.}  $$
Here we find we are getting a power series centered at $1/2$.  Since $f(x) \in (0,1)$, you may center your power series anywhere in that interval.  (Since you want an inverse function, you cannot center it anywhere else -- the center is a point of convergence of the series and you want a series that converges on the range of $f$, so the series must be centered somewhere in the range of $f$.  I picked $a = 0$ so the center would be at $1/2$, so there was some chance the series would converge over the entire range.  The inverse function has singularities at $0$ and $1$, so both endpoints place constraints on the radius of convergence of the series.)

Then
$$  g_n = \lim_{w \rightarrow 0} \frac{\mathrm{d}^{n-1}}{\mathrm{d}w^{n-1}}\left( \left( \frac{w - 0}{f(w) - 1/2} \right)^n \right)  \text{.}  $$
The first few terms of the resulting series are
$$  f^{-1}(x) = \sqrt{2\pi}(x - 1/2) + \frac{\sqrt{2}\pi^{3/2}}{3}(x-1/2)^3 + \frac{7 \pi^{5/2}}{15\sqrt{2}}(x-1/2)^5 + \frac{127\pi^{7/2}}{315\sqrt{2}}(x-1/2)^7 + \frac{4369 \pi^{9/2}}{11340 \sqrt{2}}(x-1/2)^{9}+ \cdots  $$

Everything in the coefficients except for the sequence $1, 1, 7, 127, 4369, 34807, \dots$ is straightforward to extract into a general term.  That sequence of coefficients in the numerator is OEIS A092676.  The description of that sequence gives a procedure to produce this sequence of coefficients that are part of the $g_n$ coefficients (in the "%p c:=proc(n) ..." item).


问:
Counting nested integer partitions
One partition of 8 is 5 + 3, but if we then partition each of the 5 and 3 we could get (3+2) + (2+1), and then partition again to get ((2+1)+(1+1)) + ((1+1)+1) and finally (((1+1)+1)+(1+1)) + ((1+1)+1). 5+3 could also be expanded as (4+1)+(2+1), then ((2+2)+1)+((1+1)+1), then (((1+1)+(1+1))+1)+((1+1)+1).

This question is about viewing "+" as a binary operation , so 1+1+1+1 would have to written as either (1+1)+(1+1) or ((1+1)+1)+1.

Every partition can be written as a such nested partition of 1s. It is still order independent, but associative dependent.

For a given number $n$, how many associative dependent binary-operation nested partitions of 1s are there of $n$ ?

What is known about this function ?

答:
Marc's certainly correct that the Catalan numbers count full binary trees.  But while every nested partition defines a full binary tree, not every full binary tree corresponds to a nested partition.

Smallest example: 1+(1+1) would come from 1+2, not a partition of 3 following the standard convention of listing parts in non-increasing order.  As in Roy's example, the only two binary-operation nested partitions of four 1's are (1+1)+(1+1) from 2+2 or ((1+1)+1)+1 from 3+1.

Working out more terms, the number of such sums (from 1 to 10) is 1, 1, 1, 2, 3, 6, 11, 24, 47, 103.  This is http://oeis.org/A000992, and Callan's description of restricted binary trees there matches this context.  The sequence has a convolution recurrence similar to Catalan numbers, only going "half-way."  (If Roy's question were about compositions [partitions where "order matters"], allowing 1+2 and then 1+(1+1), the recurrence formula would go the whole way and the answer would be the Catalan numbers.)

ContentLicense: CC BY-SA 2.5

Title: Counting nested integer partitions


问:
Find the value of x which would make triangles ABC and DEF congruent
"For what value of x is △ABC ~ △DEF?"

Apologies for asking a highschool geometry question.

My first thoughts in solving this were to do x^2 -5x = 84, if you were to square root it  -5 and -84 would just end up as decimals. Same for x^2 -8x = 48.

答:
You have reduced it to an algebra problem.

$$x^2-5x=84$$
$$x^2-8x=48$$

Taking difference,
$$3x=36$$
$$\therefore x=12$$


问:
Is product space of lower limit topology separable?
$(\mathbb{R}_l, \tau)$ : lower limit topology.

I wonder whether $\mathbb{R}_l \times \mathbb{R}_l$ is separable.
In my opinion, it is true since $\mathbb{Q} \times \mathbb{Q}$ is dense in $\mathbb{R}_l \times \mathbb{R}_l$:

(sketch of proof)
let $W$ be a open set containing $(q_1,q_2) \in \mathbb{Q} \times \mathbb{Q}$. Then there are basic (open) sets $U=[a, a+r_a)$ and $V=[b, b+r_b)$ such that $q_1 \in U$, $q_2 \in V$. This means $W$ contains another rational pair, namely $(q_3,q_4)$. $\Box$

Please let me know if there are any errors.

答: Yes, your argument in essence proves the correct statement that the product of  two separable spaces is again separable (this even extends to continuum many (!) separable spaces by the Hewitt-Marczewski-Pondiczery theorem,  BTW, but that’s overkill here).</p>


问:
A problem on matrices: Find the value of $k$

If $
        \begin{bmatrix}
        \cos \frac{2 \pi}{7} & -\sin \frac{2 \pi}{7} \\
        \sin \frac{2 \pi}{7} & \cos \frac{2 \pi}{7}  \\
        \end{bmatrix}^k
 =
        \begin{bmatrix}
        1 & 0 \\
        0 & 1 \\
        \end{bmatrix}
$, then the least positive integral value of $k$ is?

Actually, I got no idea how to solve this.

 I did trial and error method and got 7 as the answer.

how do i solve this?
 Can you please offer your assistance? Thank you

答:
I founded out that $$A^k=
        \begin{bmatrix}
        \cos \frac{k.2 \pi}{7} & -\sin \frac{k.2 \pi}{7} \\
        \sin \frac{k.2 \pi}{7} & \cos \frac{k.2 \pi}{7}  \\
        \end{bmatrix}
$$

using appropriate trigonometric formulae. Now for $A=I$, $k$ should be equal to $7$


问:
Why does the average of independent random variables converge to Normal distribution not other distribution?
Why does the average of independent random variables converge to Normal distribution not other distribution? Or in other words, I am looking for an intuitive explanation of the central limit theorem. Why can't the average convert to some other distribution?</p>

答:
Given that the proof of CLT is not so hard and you can find it anyway... I can give you a very intuitive explanation...

Suppose you have to measure the lenght of a table, and you can choose the instrument you prefer...you can use you hands for a very basic estimation of the lenght, you can use a ruler for a more precise measurement or you can use a 10 tons instrument taken from a Physic Research Lab. In any case, the true value of the lenght of your table in unknown, and it can change according to several factors: external temperature, error in the instrument, errors in the man who takes the measure and so on....

But some fact are surely true:

The probability to do the same error underestimating or overestimating the actual measure is the same

The probabilty do to a big error is lower w.r.t do a little error. (if a table has an actual lenght of 1 meter, the probability do get a measure of 2 meter is very close to zero, as it is zero the probability to get a measure of 1 cm)

The probability to estimate a lenght in the interval of [1;1.01] m is the same as the one to estimate it as [0.99;1] m

Doing a lot of measurement and plotting them on a chart, you will realize that the resulting drawing is a bell function, symmetric around the actual value...it's a Gaussian density....


问:
pullback of an Euclidian inner product on $\mathbb{R^{2n+1}}$
If $M$ is a differentiable manifold of dimension n and there is a diffeomorphism $g:M \rightarrow \mathbb{R^{2n+1}}$, and we have $\langle\cdot,\cdot\rangle$ is the Euclidian inner product on $\mathbb{R^{2n+1}}$. Then the pullback of $g$ defines a metric on $M$ by
$t:=g*(\langle\cdot,\cdot\rangle)$, my question is:how is $t$ defined by? I mean can you write it as a mapping and it maps what to what...

答:
Suppose that you have two spaces $X$ and $Y,$ a metric $d$ on $Y$, and a function $f : X \to Y.$ The pullback metric is the following metric on $X$:

$$(f^*d)(x^{(1)}, x^{(2)}) = d(f(x^{(1)}), f(x^{(2)})); \quad x^{(1)}, x^{(2)} \in X$$

Thus, we define a metric on $X$ by mapping points over to $Y$ and taking the distance there.

One example is given by considering different coordinate systems. Let $M = \mathbb R^n$ be the plane with ordinary Euclidean distance, then

$$d_M((x_1, y_1), (x_2, y_2), \ldots (x^n, y^n)) = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2+\ldots +(x_{n-1}-x_n)^2}.$$