答:
The slope should be the value of the derivative y(x)' at point (2,y(2)), where y(2)=-22.
Read How to get derivatives of polynomials for the details , you will learn  ($x^n$)'=n($x^{n-1}$) .

Given  y=$x^4$-5$x^3$+2, $y(x)'$ = $4x^3$-15$x^2$,   y(2)'=-28

From point-slope formula,  the line equation is  y+22=(x-2)(-28),  or   y=-28x+34.

Body:
Find the slope of the tangent line to the graph of the given function at the given value of x. Find the equation of the tangent line. : $y=x^4-5x^3+2; x=2$

I understand that the slope of the line follows the equation of

$$\lim_{h\to 0} \frac{f(a+h)-f(a)}h.$$

But I don't know how to go about the rest


问:
Simplification of sum of exponentials?
Is there a way to simplify the expression $\sum _{k=0}^n 2^k$? That is, is there a way to write it without a $\sum$ or $\prod$ operator?

答:
$$1 + 2 + 4 + \cdots 2^n = 11 \cdots 1_\text{binary}=2^{n+1}-1$$

In general, $\sum_{k=0}^n  a^n$ is a geometric series.

ContentLicense: CC BY-SA 4.0

Title: Simplification of sum of exponentials?


问:
Question on M/M/s queue
customers arrive to a service station according to a poisson prossees and on average 2 during an hour. The service times and independent of the arrivals and internally independent with mean 45 minutes. At must 3 can be in system (and while one is being served the other wait)put xt=number of customer in the system at time
a) Determine the birth and death frequencies in the birth and death chain (Xt)t≥0.
b) Determine lim→∞Pi(Xt=0) for i=0,1,2,3.
can you explain for me solve problem like this  with markov chain </p>

答:
Your model is an M/M/c queue with arrival rate $\lambda=2$ jobs per hour and service rate of each server $\mu=\frac{1}{0.75}=\frac{4}{3}$ jobs per hour. You have $c=3$ servers. Transition rates are given by the $Q$ matrix in the linked article.

"Births" in this model are customer arrivals, so the birth rate is a constant $\lambda$ in all states. "Deaths" are when jobs are serviced. When $X_t \geq 3$ all the servers are busy so the death rate is $3\mu$, when $X_t=2$ the death rate is $2\mu$ and when $X_t=1$ the death rate is $\mu$. There are no deaths when $X_t=0$ because there are no individuals to die.

The probabilities you seek to compute are all the same because for large values of $t$ the starting state doesn't matter,
$$\lim_{t\rightarrow \infty}\mathbb P_i(X_t=0) = \pi_0$$
for $i=0,1,2,3$. Perhaps the probabilities you are looking for are
$$\lim_{t\rightarrow \infty}\mathbb P(X_t = i) = \pi_i.$$
Formulas for the $\pi_i$ are given in the linked article.

Title: Question on M/M/s queue

ContentLicense: CC BY-SA 3.0


问:
Convex function of independent vs. correlated random variables
Let $f\colon\mathbb R \to \mathbb R$ be convex and $X_1,\ldots,X_n$ be i.i.d. random variables. Is it true that $\mathbb E f(n\cdot X_1) \ge \mathbb E f(X_1 + \cdots + X_n)$?

It seems like one approach is to show that $X$ is a mean-preserving spread of the average of $n$ independent copies of $X$. Is there a name for this sort of result? (Also, is there a name for this sort of inequality?)

答:
Applying the  Jensen's inequality for finite form with $a_i = 1$ and $x_i = nX_i$

$$f\left(\sum_{i=1}^nX_i\right) = f\left(\frac{\sum_{i=1}^n(nX_i)}{n}\right) \le \frac{\sum_{i=1}^nf(nX_i)}{n}  $$

Then
$$\mathbb{E}\left[f\left(\sum_{i=1}^nX_i\right) \right] \le \mathbb{E}\left[ \frac{\sum_{i=1}^nf(nX_i)}{n} \right] =\mathbb{E}[f(nX_1)]$$

PS: I don't think there is a specific name for this inequality


问:
Does nesscery for every $V\subseteq A$ open relative to $A$ $f(V)$ is open relative to $B$

Let $f:A\to B$ continuous function and prove or disprove the following statments.

a) Does nesscery for every $V\subseteq A$ open relative to $A$ $f(V)$ is open relative to $B$

b) Does nesscery for every $V\subseteq A$ close relative to $A$ $f(V)$ is close relative to $B$

For b) I think I can choose $A=\mathbb{R}^n$ and setting $f(x)=arctan(x)$ and the image is an open set.

for a) I think it's not true but I don't have an example any hint?

答:
a). Let $A=B=\Bbb R$ and $f(x)=x^2$ and $V=(-1,1).$ Then $V$ is open in $A$ but $f[V]=[0,1)$ is not open in $B.$

b). Let $A=\Bbb R^2$ and $B=\Bbb R.$ For $(x,y)\in A$ let $f(\,(x,y)\,)=x.$ Then $V=\{(x,1/x):0\ne x\in \Bbb R\}$ is closed in $A$ but $f[V]=\Bbb R \setminus \{0\}$ is not closed in B.

A function that maps open (closed) sets to open (closed) sets is called an open (closed) function. Continuous functions need not be open nor closed functions.

If $f:A\to B$ is continuous and $V$ is a compact subset of $A$ then $f[V]$ is a compact subset of $B$. And a compact subset of a Hausdorff space must be closed. So in the example for b) above, we choose a closed non-compact $V$.


问:
Are the stalks of $U \mapsto \prod_{z \in U} I_z$ exactly $I_x$?
I am currently studying a proof of the fact that every sheaf of abelian groups can be injected in an injective sheaf, and it seems like I'm missing something. In the proof, the idea is pretty straight-forward: We take a sheaf $E$ and for each $x \in X$ (X the base topological space), we choose an injective abelian group $I_{x}$ and a monomorphism $\psi_{x}: E_x \to I_x$ and then, with some verifications involved, we construct the preasheaf $I: U\subset X \mapsto I(U) := \prod_{z \in U} I_z$ and then it turns out to be an injective sheaf. My question is: Is the stalk of $I$ at a point $x \in X$ the abelian group $I_x$?

Thank you in advance for all the help :)

答:
Not necessarily. For example, let $X = \mathbb{R}$ and $I_x = \mathbb{Q}$ for all $x \in \mathbb{R}$. Note that the global sections of $\prod_{x \in \mathbb{R}} I_x$ are just functions $\mathbb{R} \to \mathbb{Q}$.

Suppose for contradiction that there exists an isomorphism $\mathbb{Q} \to (\prod_{x \in \mathbb{R}} I_x)_0$. Composing with the map $(\prod_{x \in \mathbb{R}} I_x)_0 \to \mathbb{Q}$ induced by the projection $\prod_{x \in \mathbb{R}} I_x \to I_0$, we get a surjective homomorphism $\mathbb{Q} \to \mathbb{Q}$, which must therefore be an isomorphism. This means that the germ of a local section in a neighborhood of $0$ is determined by its component in $I_0$, which is false. For example, take the two global sections $f, g$ given by the functions

$$f(x) = 0$$
$$g(x) = \begin{cases} 0 &: x = 0 \\ 1 &: x \neq 0 \end{cases}$$


问:
Confusion on phase diagrams of ODES in the complex plane
I am working through some simple complex differential equations to try to get a better understanding of them. The textbook I'm using has given me the following diff-eq:
$$
z' = z^2 - 1
$$
I have solved it using the techniques I'm familiar with, and I got:
$$
z = \frac{1-Ce^{2t}}{Ce^{2t}+1}
$$
Where $C \in \mathbb{C}$.
How would I draw the phase diagram for this? It seems that I would need some assumptions on the sign of $Re(C) \text{ and }  Im(C)$? Or am I mistaken?

答:
As $z= x+i y$ we have

$$
\dot x+ i\dot y = (x+i y)^2-1\Rightarrow \cases{\dot x = x^2-y^2-1\\ \dot y = 2 x y}
$$

In red the orbit associated to the initial conditions $z(0) = 1+0.2 i$


问:
Find the slope of the tangent line to the graph of the given function at the given value of x.
Find the slope of the tangent line to the graph of the given function at the given value of x. Find the equation of the tangent line. : $y=x^4-5x^3+2; x=2$

I understand that the slope of the line follows the equation of

$$\lim_{h\to 0} \frac{f(a+h)-f(a)}h.$$

But I don't know how to go about the rest

答:
If you use the h-method you start with

$$\lim_{h \rightarrow 0} \frac{y(a+h)-y(a)}{h}=\lim_{h \rightarrow 0} \frac{(a+h)^4-5\cdot (a+h)^3+2-(a^4-5\cdot a^3+2)}{h}$$

To resolve  $(a+h)^4$ you can use the binomial theorem: $(x+y)^n=\sum\limits_{k=0}^n\binom{n}x\cdot x^k\cdot y^{n-k}$

$$=\lim_{h \rightarrow 0}\frac{a^4+4a^3h+6a^2h^2+4ah^3+h^4-5a^3-15a^2h-15ah^2-5h^3+2-a^4+5\cdot a^3-2}{h}$$

Next we eliminate the terms with opposite signs, e.g. $a^4$ and $-a^4$

$$=\lim_{h \rightarrow 0}\frac{4a^3h+6a^2h^2+4ah^3+h^4-15a^2h-15ah^2-5h^3}{h}$$

Factoring out $h$ at the numerator.

$$=\lim_{h \rightarrow 0}\frac{h\cdot \left(4a^3+6a^2h+4ah^2+h^3-15a^2-15ah-5h^2\right)}{h}$$

Cancelling $h$

$$=\lim_{h \rightarrow 0}4a^3+6a^2h+4ah^2+h^3-15a^2-15ah-5h^2$$

$$=4a^3-15a^2=y'(a)$$

This is the derivative of $y(x)$ at value $x=a$ and therefore the slope. Finally you can insert $a=2$ to obtain $y'(2)$.


问:
REVISITED$^1$: How is the determinant of a matrix $A\in M_{2\times 2}(\mathbb{R})$ considered a bilinear form?
I'm trying to prove that $B(X,Y)=\det (X+Y) - \det (X) - \det (Y)$ is a blinear form on the vector space $A$ is from, and also trying to determine if it is an inner product space. I think if I know how $\det$ is a bilinear form, then this will become easier.

EDIT$^1$:

I was thinking, if I let $\det : \mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}$ be a bilinear form defined by
$$ \det\pmatrix{\pmatrix{a_{11}\\a_{21}}, \pmatrix{a_{12}\\a_{22}}} =a_{11}a_{22}-a_{21}a_{12},$$
then if I can show that $\det$ is a bilinear form, then this will suffice to show that $B$ is a bilinear form as the sum of bilinear forms is yet again a bilinear form, as is remarked in Friedberg's Linear Algebra (page 423).

答:
I think what your textbook means is something like this:

If $B(X,Y)$ is a finite sum of bilinear forms in $X$ and $Y$, then $B$ is also a bilinear form in $X$ and $Y$.

If this is the case, there is no hope to apply this statement to your question directly, because neither $\det(X+Y),\,\det(X)$ nor $\det(Y)$ is a bilinear form in $X$ and $Y$.

Adding to Will Jagy's comment, if you can show that some $q(x)$ is a quadratic form in $x$, then $b(x,y):=\frac{q(x+y)-q(x)-q(y)}2$ is a (symmetric) bilinear form in $x$ and $y$ with $q(x)\equiv b(x,x)$. In your case, if you can show that $q(X):=2\det(X)$ is a quadratic form in $X$, then you are done. $q(X)$ is obviously a polynomial in the entries of $X$. Can you show that $q(kX)=k^2q(X)$ for all scalar $k$?

Alternatively, let $X=\pmatrix{a&b\\ c&d}$. Can you find a symmetric matrix $M$ such that $2\det(X) = (a,b,c,d)M(a,b,c,d)^T$? If you can, then for $Y=\pmatrix{x&y\\ z&w}$, we can express the (symmetric) bilinear form $B(X,Y)$ as $(a,b,c,d)M(x,y,z,w)^T$.

You may also work from the first principle and evaluate $B(X,Y)$ in terms of the entries of $X$ and $Y$:
\begin{align*}
&\det\pmatrix{a+x&b+y\\ c+z&d+w} - \det\pmatrix{a&b\\ c&d} - \det\pmatrix{x&y\\ z&w}\\
=&(a+x)(d+w) - (b+y)(c+z) - (ad-bc) - (xw-yz).\tag{1}
\end{align*}
Can you simplify $(1)$ and show that it is a bilinear form?

Finally, if $B(X,X)$ defines an inner product, $B(X,X)$ should be positive for all $X\neq0$. Is that true?


问:
In $\mathbb{F}[x]$, $(f(x)) \subseteq I$ if any only if $I=(g(x))$ for some monic $g$ that divides $f$
Let $f(x)$ be a monic polynomial in $\mathbb{F} [x]$. I need to show that $(f(x))\subseteq I$ if and only if $I=(g(x))$ for some monic polynomial $g$ that divides $f$. The reverse direction is clear. I am getting stuck on the forward direction. It certainly seems intuitively true. I know that $r\in (f(x))$ means that $r=p(x) f(x)$ for some $p(x) \in \mathbb{F} [x]$ and I must have that $r\in J$. Where do I go from here?

答:
The leading coefficient of $f$, call it $l(f)$, is $1$. If $f\in (g)$ then $f=gh$ so obviously $g$ divides $f$, and $l(f)=l(g)l(h)$ so $l(g),l(h)$ are units, ie. without loss of generality $l(g)=1$.

Conclude from that $\Bbb{F}[x]$ is a principal ideal domain: this is because $(u,v)=(\gcd(u,v))$, whence your ideal $I$ is principal $=(g)$ for some $g$.


问:
How to compute this convolution without Graphing it
Consider the following :
$$
x(t)=u(t)-2u(t-2)+u(t-5)\qquad\text{and}\qquad h(t)=e^{2t}u(1-t)
$$
I wish to compute $(x*h)(t)$ without graphing it.

So as a first step, I would rewrite $x(t)$ and $h(t)$ as :
$$
x(t)=
\begin{cases}
0&\text{if $t\geq 5$}\\
-1&\text{if $2\leq t<5$}\\
1&\text{if $0\leq t <2$}\\
0&\text{if otherwise}
\end{cases}
\qquad\text{and}\qquad
h(t)=\begin{cases}
e^{2t}&\text{if $t\leq1$}\\
0&\text{if otherwise}
\end{cases}
$$
Now, I must $x(t)\mapsto x(\tau)$ and $h(t)\mapsto h(t-\tau)$. So we get :
$$
x(\tau)=
\begin{cases}
0&\text{if $\tau\geq 5$}\\
-1&\text{if $2\leq \tau<5$}\\
1&\text{if $0\leq \tau <2$}\\
0&\text{if otherwise}
\end{cases}
\qquad\text{and}\qquad
h(t)=\begin{cases}
e^{2(t-\tau)}&\text{if $\tau\geq t-1$}\\
0&\text{if otherwise}
\end{cases}
$$
Now from what I know I must restrict the domain of :
$$
\tau\geq 5
$$
$$
2\leq \tau<5
$$
$$
0\leq \tau <2
$$
$$
\tau>t-1
$$
Therefore, I have $\max(0,2,5,t-1)<\tau<\min(...)$ I don't know how to find the upper bound which is minimum of upper bounds of the inequalities listed above. I would hope someone can assist me in computing the convolution.

答: Use linearity.  (You may have seen linearity as two properties: distributivity of convolution over addition and associativity with scalar multiplication.  Notice that the latter makes convolution distribute over addition and subtraction.)  Treat $x(t)$ as a linear combination of $u(t)$, $u(t-2)$, and $u(t-5)$.  Compute the convolution of $h$ with each of these.  Then your answer is the fist convolution, minus twice the second one, plus the third one.


问:
What is an indefinite integral?
First of all sorry for asking such a dumb question

What is an indefinite integral?

Is it the difference between y=f(a) from f(b)=0? I mean, if f(b)=0 and we are to determine the integral at x=a of f(x) does that mean indefinite integral for this condition is int of f(x) from b to a?

答: The indefinite integral of $f(x)$ is the integral without upper and lower bounds, and is defined to be the antiderivative of $f(x)$ so is equal to the general function whose derivative is $f(x)$ . We must put a constant of integration because then all functions with this form of derivative are accounted for.

Title: What is an indefinite integral?


问:
Is the following set a semiring?
Is the set:

$\mathcal{M}=\{(-\infty, b], b \in \mathbb{R}\}$

a semi-Ring?

In our definition, a semi-Ring is considered an algebraic structure with:

$(1) \, \emptyset \in \mathcal{M} \\\\
(2) \, A, B \in \mathcal{M} \Rightarrow A \, \cap B \in \mathcal{M} \\\\
(3) \, A,B \in \mathcal{M}, A \subset B \Rightarrow \exists n\in \mathbb{N}, \exists C_1,\dots,C_n \in \mathcal{M}, C_i \cap C_j = \emptyset \, (i \neq j) \,\, B \setminus A = \sum_{i=1}^n C_i$

Can anyone help me out? I personally suspect that the second condition cannot be fulfilled.

答:
Neither (1) or (3) holds.

There's no real $b$ such that $(-\infty,b]$  is empty (it contains $b$).

Also, the difference of two such sets is
$$(-\infty, b]\setminus(-\infty,a]=(a,b]$$
if $a<b$ and empty otherwise, can't be expressed as a union of (finitely many) such sets, because each element of $\mathcal M$ is unbounded from below, and so must be their union, whereas the intervals $(a,b]$ and the empty set are bounded.

Title: Is the following set a semiring?


问:
A question about Square integrable martingale
Let $\left\{X_{n}, \mathcal{F}_{n}\right\}_{n \geq 0}-$  square integrable martingale, and
$$
\sum_{n \geq 1} \mathrm{E}\left(X_{n}-X_{n-1}\right)^{2}<\infty.
$$

If exists Random variable $X_\infty$.

Want to prove
$$
\mathrm{E} X_{\infty}^{2}<\infty, \quad X_{n} \rightarrow X_{\infty} \text { a. s. } ,\quad \mathrm{E}\left(X_{n}-X_{\infty}\right)^{2} \rightarrow 0
$$

答:
Here is the full answer note that we need to use martingale property

So as mentioned in my comment above under hypothesis lets show that $X_n$ is Cauchy in $L^2$ which is a Banch space and os is complete :
$$\sum_{n \geq 1} \mathrm{E}\left(X_{n}-X_{n-1}\right)^{2}<\infty$$

So at $\epsilon >0$ fixed there exist $N_0$ s.t. for $n,p > N_0$:

$$\mathrm{E}\left(X_{n}-X_{p}\right)^{2}=\mathrm{E}\left(X_{n}-X_{n-1}+X_{n-1}- ... +X_{p+1}-X_{p}\right)^{2}=\sum_{i=p+1}^n \mathrm{E}\left(X_{i}-X_{i-1}\right)^{2}$$

the reason for that is the following :
if $n=p+1$ then :
$$\mathrm{E}\left(X_{n}-X_{p}\right)^{2}=\mathrm{E}\left((X_{n}-X_{n-1})+(X_{n-1}- X_{n-2})\right)^{2}=\mathrm{E}\left(X_{n}-X_{n-1}\right)^{2}+$$
$$\mathrm{E}\left(X_{n-1}-X_{n-2}\right)^{2}+ 2.\mathrm{E}\left(X_{n}-X_{n-1}\right).\left(X_{n-1}-X_{n-2}\right) $$
But $\mathrm{E}\left(X_{n}-X_{n-1}\right).\left(X_{n-1}-X_{n-2}\right)=0$ as from martingale property :
$\mathrm{E}(X_{n}-X_{n-1}).(X_{n-1}-X_{n-2})=\mathrm{E}(X_{n}X_{n-1}) - \mathrm{E}(X_{n-1}^2)+\mathrm{E}(X_{n-1}X_{n-2})-\mathrm{E}(X_{n}X_{n-2})$

$$=\mathrm{E}(\mathrm{E}(X_{n}|\mathcal{F}_{n-1})X_{n-1}) - \mathrm{E}(X_{n-1}^2)$$
$$+\mathrm{E}(\mathrm{E}(X_{n-1}|\mathcal{F}_{n-2})X_{n-2}) -\mathrm{E}(\mathrm{E}(X_{n}|\mathcal{F}_{n-2})X_{n-2})$$
$=\mathrm{E}(X_{n-1}.X_{n-1}) - \mathrm{E}(X_{n-1}^2)$ (using (1) below)

$+\mathrm{E}(X_{n-2}.X_{n-2}) -\mathrm{E}(X_{n-2}.X_{n-2})$ (using (1) below)
$$=0+0$$

So by recurrence and as the series is convergent we are assured that at fixed $\epsilon>0$ there exists some $N_0$ such that for every $n,p> N_0$ we have :

$$\mathrm{E}\left(X_{n}-X_{p}\right)^{2}=\mathrm{E}\left(X_{n}-X_{n-1}+X_{n-1}- ... +X_{p+1}-X_{p}\right)^{2}=\sum_{i=p+1}^n \mathrm{E}\left(X_{i}-X_{i-1}\right)^{2}<\epsilon $$

The $X_n$ is Cauchy in a complete space $L^2$, which entails existence of a limit $X_\infty$ up to almost sure equivalence class and everything in your conclusion holds true without further ado.

(*) This is true as for $n>k$ and $X \in \mathcal{F}_{n}$, the conditional expectation of $X$ knowing $\mathcal{F}_{k}$ is the (only) r.v. $Z\in\mathcal{F}_{k}$ s.t. for every $Y$ in $\mathcal{F}_{k}$ we have :

$$\mathrm{E}(Z.Y)=\mathrm{E}(X.Y)$$ (1)

Then such a $Z$ is denoted $\mathrm{E}(X|\mathcal{F}_{k})$ and it's unique almost surely.

Regards


问:
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$?
Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?

I thought it would be 12 this as per pemdas rule:

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$

Wanted to confirm the right answer from you guys. Thanks for your help.

答:
Multiplication takes priority over addition, so this

$1+1+1+1+1+1+1+1+1+1+1+1\times0+1$

becomes:

$1+1+1+1+1+1+1+1+1+1+1+0+1$

Now add everything which gives you 12.


问:
Unions in the subobject poset
In any category $\mathcal{C}$ with pullbacks, we can define the subobject fibration $cod:{\sf Sub}(\mathcal{C})\to\mathcal{C}$ and view the fibers ${\sf Sub}(\mathcal{C})_X$ as power-categories of $X$ -- they are poset categories consisting of one member from each equivalence class of monomorphisms into $X$, with the containment relation given by arrows between monos in a slice. Taking $\mathcal{C}={\bf Set}$ yields ${\sf Sub}({\bf Set})_X\cong\mathcal{P}(X)$ as posets, justifying the terminology and suggesting we look for interpretations of various operations on subsets -- for example, intersections correspond to pullbacks.

What are unions in the subobject lattice of an object in a category with pullbacks?

I would think that the union of subobjects $u:U\to X$, $v:V\to X$ would be a coequalizer of a coproduct by a pullback $$U\cup V=(U\coprod V)/(U\times_X V)$$ as suggested here, but it isn't clear to me how to carry out a construction like this if the ambient category only has pullbacks a-priori. If the fiber categories are posets with supremums then we can trivially identify supremums with coequalizers of coproducts and unions, but it isn't clear to me that the subobject lattices have supremums without additional ambient structure on the category. If we're working in a topos as suggested here the extra structure makes this construction easy to carry out; if extra structure on the category is required, do we need the full expressive power of a topos? Any assistance is appreciated.

答:
Consider the special case of a partially ordered set with finite meets (a meet semi-lattice). The poset thus has all finite limits, including pullbacks. If we had the ability to take "unions" of subobjects of 1, we would have the ability to compute joins. But not all meet-semilattices have joins.

Looking back through my notes on topos theory, it looks like relevant categorical property in the proof that all maps have images in a topos is the fact that every monic is an equaliser and that toposes have colimits. However, I'm sure others know more about this than I do.

Title: Unions in the subobject poset


答:
Let me just quote the first sentence from here:

A coherent category A coherent category is a regular category in which the subobject posets all have finite unions which are preserved by the base change functors.

This is precisely the structure you want. Regularity and pullback-stability of the finite unions is not strictly necessary, but it's what makes these things logically reasonable: in particular, the subobject posets become distributive lattices for which pullback functors are lattice homomorphisms. Since images are nice in regular categories you can also "construct" the unions as the image of the coproduct of two subobjects, assuming those coproducts are there at all, in which case the category is extensive.

Distributive lattices are also, themselves, coherent categories so it makes sense to think about coherent categories as categorified distributive lattices in roughly the sense that toposes are categorified Heyting algebras. The nLab has some general examples of coherent categories; non-additive nice categories tend to be extensive even if they're not very close to toposes, such as categories of spaces or of categories.

Title: Unions in the subobject poset

Body:
In any category $\mathcal{C}$ with pullbacks, we can define the subobject fibration $cod:{\sf Sub}(\mathcal{C})\to\mathcal{C}$ and view the fibers ${\sf Sub}(\mathcal{C})_X$ as power-categories of $X$ -- they are poset categories consisting of one member from each equivalence class of monomorphisms into $X$, with the containment relation given by arrows between monos in a slice. Taking $\mathcal{C}={\bf Set}$ yields ${\sf Sub}({\bf Set})_X\cong\mathcal{P}(X)$ as posets, justifying the terminology and suggesting we look for interpretations of various operations on subsets -- for example, intersections correspond to pullbacks.

What are unions in the subobject lattice of an object in a category with pullbacks?

I would think that the union of subobjects $u:U\to X$, $v:V\to X$ would be a coequalizer of a coproduct by a pullback $$U\cup V=(U\coprod V)/(U\times_X V)$$ as suggested here, but it isn't clear to me how to carry out a construction like this if the ambient category only has pullbacks a-priori. If the fiber categories are posets with supremums then we can trivially identify supremums with coequalizers of coproducts and unions, but it isn't clear to me that the subobject lattices have supremums without additional ambient structure on the category. If we're working in a topos as suggested here the extra structure makes this construction easy to carry out; if extra structure on the category is required, do we need the full expressive power of a topos? Any assistance is appreciated.


问:
Calculate $4||x+y||2+||x−y||2 $
Let's take $x, y \in X$ where $X$ is vector space with norm inducted by inner product.

Additionally $||3x+y||= 2$, $||x+3y|| = 2$.

I want to calculate $4||x+y||^2+||x-y||^2$ $\\$

My work so far

Because $X$ has a norm inducted by inner product then for $x,y \in X$ we have $$||x+y||^2 + ||x-y||^2 = 2(||x||^2+||y||^2)$$

So:

$4||x+y||^2 + ||x-y||^2 = 3||x+y||^2 + (||x+y||^2 + ||x-y||^2)  = 3||x+y||^2 + 2(||x||^2 + ||y||^2)$

And I'm not sure why to do next... I'm not sure how can I was trying to rewrite this expression to somehow include information that $||3x+y|| = ||x+3y|| = 2$ but I wasn't able to. Could you please give me a hint how it can be calculated ?

答:
You are on the right track with the parallelogram law. Apply it to $u=3x+y$ and $v=x+3y$:
$$
 16 = 2 (\Vert 3x+y\Vert^2 + \Vert x+3y\Vert^2 ) = \Vert 4x+ 4y \Vert^2 + \Vert 2x - 2y \Vert^2 \\
= 16 \Vert x+y \Vert^2 + 4 \Vert x-y \Vert^2 = 4 (4\Vert x+y \Vert^2 +  \Vert x-y \Vert^2)
$$

Title: Calculate $4||x+y||2+||x−y||2 $


问:
Find roots of $x^4 + 4x^3 - 25x^2 - 16x + 84$
I'm not quite sure how to approach this, how are you supposed to quickly factor a 4th degree polynomial? For $x^4 + 4x^3 - 25x^2 - 16x + 84$ it's factored into $(x-3)(x-2)(x+2)(x+7)$ which like I said, I have no idea how to approach. Would really appreciate some help, thanks in advance!

Body:
You can use synthetic division in $$p(x)=x^{4}+4x^{3}-25x^{2}-16x+84.$$

To do this, notice that the divisors of:

$a=1$: $\{\pm1\}$.

$b=84$: $\{\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 7, \pm 12, \pm 14, \pm 28, \pm 42, \pm 84\}$.

Then the synthetic division gives us that $$p(x)=x^{4}+4x^{3}-25x^{2}-16x+84=(x-3)(x-2)(x+2)(x+7).$$


问:
Determinate perimeter and area of an n-angle in c++
Okay, so this is what the textbook says:
"The user will input n points in $R^2$, the program needs to determine wether or not this is a n-angle polygon and if so calculate the area and perimeter of such a figure".

To determine wether or not this is a n-angle polygon is rather easy, I made a point and a vector class, the program takes the first point and then forms n-1 vectors, if any of this vectors is parallel or antiparallel with any other, well it's not an n-angle polygon, otherwise it is, so far so good.

Now for the real problem, you need to determine which points form amongst them a side, to determine the perimiter and area, but this program is suppoused to work with any type of polygon, wether regular or not.

Is there any efficient way that actually works to know how two points form a side and not a diagonal? Having this I could use Gauss' formula or even Heron's.

Thank you in advance.

答:
As I understand, the vertices are to be assumed to be entered in order otherwise there are multiple $n$-gons possible of different shapes, perimeters and areas with given set of data points.

Under this assumption, the sides will be uniquely determined by consecutive vertices.

The perimeter can be found by using Pythagoras theorem for $n$ side lengths.

The area can be found by using Shoelace formula.


答: The question is not "whether", but "where". Yes, given the extended real numbers. No, given the real numbers.</p>

Title: Is the following set a semiring?

Body:
Is the set:

$\mathcal{M}=\{(-\infty, b], b \in \mathbb{R}\}$

a semi-Ring?

In our definition, a semi-Ring is considered an algebraic structure with:

$(1) \, \emptyset \in \mathcal{M} \\\\
(2) \, A, B \in \mathcal{M} \Rightarrow A \, \cap B \in \mathcal{M} \\\\
(3) \, A,B \in \mathcal{M}, A \subset B \Rightarrow \exists n\in \mathbb{N}, \exists C_1,\dots,C_n \in \mathcal{M}, C_i \cap C_j = \emptyset \, (i \neq j) \,\, B \setminus A = \sum_{i=1}^n C_i$

Can anyone help me out? I personally suspect that the second condition cannot be fulfilled.


答:
Use the fact that
$$\|x \pm y\|^2 = \|x\|^2 + \|y\|^2 \pm 2\Re\langle x, y \rangle.$$

Title: Calculate $4||x+y||2+||x−y||2 $

Body:
Let's take $x, y \in X$ where $X$ is vector space with norm inducted by inner product.

Additionally $||3x+y||= 2$, $||x+3y|| = 2$.

I want to calculate $4||x+y||^2+||x-y||^2$ $\\$

My work so far

Because $X$ has a norm inducted by inner product then for $x,y \in X$ we have $$||x+y||^2 + ||x-y||^2 = 2(||x||^2+||y||^2)$$

So:

$4||x+y||^2 + ||x-y||^2 = 3||x+y||^2 + (||x+y||^2 + ||x-y||^2)  = 3||x+y||^2 + 2(||x||^2 + ||y||^2)$

And I'm not sure why to do next... I'm not sure how can I was trying to rewrite this expression to somehow include information that $||3x+y|| = ||x+3y|| = 2$ but I wasn't able to. Could you please give me a hint how it can be calculated ?


Title: $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$?

Body:
Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?

I thought it would be 12 this as per pemdas rule:

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$

Wanted to confirm the right answer from you guys. Thanks for your help.

Body:
\begin{align*}
& 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + (1\times0) + 1\\
= &1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 0 + 1\\
= &12
\end{align*}


问:
Binomial identity involving central binomial coefficient
I came across this nice binomial identity

$${\sum}_{k=0}^{2n} \frac{(-1)^k {2n \choose k} {2k \choose k}}{n+k \choose k} = 1$$

This is equivalent to the hypergeometric function ${}_2 F_1(-2n, 1/2; n+1; 4)$. I am having a hard time trying to prove this. Can someone help me with a hint to prove this?

答:
Let $F(n,k)=(-1)^k\binom{2n}k\binom{2k}{k}\big/\binom{n+k}{k}$ be the summand, and let $S_n=\sum_{k=0}^{2n}F(n,k)$ be the sum to be computed. Let
$$
G(n,k)=(-1)^k\binom{2n}{k-2}\binom{2k}{k}\Big/\binom{n+k}{k}.
$$
You can show, through tedious algebraic manipulations alone, that for all $n,k\ge 0$,
$$
3(F(n,k)-F(n+1,k))=G(n,k+1)-G(n,k).\tag{*}
$$
Specifically, write both sides in terms of factorials, then cancel common factorial terms until all that remains is a rational equation in $n,k$, which can be proven by clearing denominators and polynomial manipulation.

Summing both sides of $(*)$ from $k=0$ to $2n+2$, you get that
$$
3S_{n}-3S_{n+1}=G(n,2n+3)-G(n,0)=0-0=0.
$$
This proves that $S_{n}$ is independent of $n$, so you need only verify that $S_0=1$ to prove $S_n=1$ for all $n$.

If you are wondering where $G(n,k)$ came from, it is of the form $G(n,k)=F(n,k)\cdot R(n,k),$ where $$R(n,k)=k(k-1)/(2n-k+1)(2n-k+2),$$ after canceling some common factors, necessary to prevent division by zero. This function $R(n,k)$ which causes $(*)$ to be satisfied can be found using Zeilberger's algorithm. For example, in Maxima, you can use the commands

```<code>load(zeilberger)$
Zeilberger((-1)^k * binomial(2*n, k) * binomial(2*k, k) / binomial(n + k, k), k, n);
</code>```

to get both the coefficients $[3,-3]$ on the LHS of $(*)$, and $R(n,k)$. You can see for yourself by copy pasting those commands into this online Maxima compiler.

My purpose in including this explanation is to help spread the word that a proof of "any" summation identity  with binomial coefficients can be found automatically with the right computer program. For anyone who encounters complicated binomial summations like this a lot, this power is very desirable! In fact, if you only have the summation, you can determine if a closed form exists using the same algorithm. The exact requirements on the "any" part  are described in the book $A = B$, available for free online: A = B on Herbert Wilf's website. This is a must-read for anyone who wants this power.

Body:
Let $F(n,k)=(-1)^k\binom{2n}k\binom{2k}{k}\big/\binom{n+k}{k}$ be the summand, and let $S_n=\sum_{k=0}^{2n}F(n,k)$ be the sum to be computed. Let
$$
G(n,k)=(-1)^k\binom{2n}{k-2}\binom{2k}{k}\Big/\binom{n+k}{k}.
$$
You can show, through tedious algebraic manipulations alone, that for all $n,k\ge 0$,
$$
3(F(n,k)-F(n+1,k))=G(n,k+1)-G(n,k).\tag{*}
$$
Specifically, write both sides in terms of factorials, then cancel common factorial terms until all that remains is a rational equation in $n,k$, which can be proven by clearing denominators and polynomial manipulation.

Summing both sides of $(*)$ from $k=0$ to $2n+2$, you get that
$$
3S_{n}-3S_{n+1}=G(n,2n+3)-G(n,0)=0-0=0.
$$
This proves that $S_{n}$ is independent of $n$, so you need only verify that $S_0=1$ to prove $S_n=1$ for all $n$.

If you are wondering where $G(n,k)$ came from, it is of the form $G(n,k)=F(n,k)\cdot R(n,k),$ where $$R(n,k)=k(k-1)/(2n-k+1)(2n-k+2),$$ after canceling some common factors, necessary to prevent division by zero. This function $R(n,k)$ which causes $(*)$ to be satisfied can be found using Zeilberger's algorithm. For example, in Maxima, you can use the commands

load(zeilberger)$
Zeilberger((-1)^k * binomial(2*n, k) * binomial(2*k, k) / binomial(n + k, k), k, n);

to get both the coefficients $[3,-3]$ on the LHS of $(*)$, and $R(n,k)$. You can see for yourself by copy pasting those commands into this online Maxima compiler.

My purpose in including this explanation is to help spread the word that a proof of "any" summation identity  with binomial coefficients can be found automatically with the right computer program. For anyone who encounters complicated binomial summations like this a lot, this power is very desirable! In fact, if you only have the summation, you can determine if a closed form exists using the same algorithm. The exact requirements on the "any" part  are described in the book $A = B$, available for free online: A = B on Herbert Wilf's website. This is a must-read for anyone who wants this power.


问:
A question on parallelism of affine planes
Disclaimer: I am not sure how much of this is common terminology/ notation, so please bear with me if anything is unclear, I will supply the needed information.

By affine plane we mean an affine subspace of dimension $2$.

So, I learnt that if we have an affine space $\mathcal{A}$ and two affine subspaces $\mathcal{A}_1$ and $\mathcal{A}_2$, then we say that $\mathcal{A}_1$ and $\mathcal{A}_2$ are parallel if $\operatorname{dir}(\mathcal{A}_1)\subset \operatorname{dir}(\mathcal{A}_2)$ or $\operatorname{dir}(\mathcal{A}_2)\subset \operatorname{dir}(\mathcal{A}_1)$, where by $\operatorname{dir}(\mathcal{A}_i)$ I denoted the direction of $\mathcal{A}_i$.

Let's consider this problem now: Let $\pi \subset \mathbb{C}^3$, $\pi:\sqrt{2} x +(1-\sqrt 3)y -z -1 =0$ and $P:=(1,0,2)$. Find the equation of the plane $\alpha$ such that $P\in \alpha$ and $\alpha \parallel \pi$.

My instructor said that $\alpha \parallel \pi \iff \operatorname{dir}(\mathcal{A}_1)=\operatorname{dir}(\mathcal{A}_2)$. This isn't consistent with the definition from above. I think that what he meant was that this equivalence holds because we require $P\in \pi$ and by the fifth postulate I think that we get that equivalence. Am I right?

Furthermore, he went on to say that $\operatorname{dir}(\alpha):\sqrt 2 x+(1-\sqrt 3)y-z=0$. Why? Does this hold because $\operatorname{dir}(\alpha)$ is a vector subspace of $\mathbb{C}^3$?

答:
Then direction of an affine plane in $K^3$ with equation $ax+by+cz+d=0$, is the vector  subspace with equation $ax+by+bz=0$, and the affine plane with the same direction that passes though the point $P_0(x_0,y_0z_0)$ is defined by the equation
$$ax+by+bz=ax_0+by_0+cz_0.$$

$\DeclareMathOperator{\dir}{dir}$
Also, $\alpha\parallel\pi$ indeed means $\dir\mathcal A_1\subseteq\dir\mathcal A_2$, but as the spaces have the same dimension, it is equivalent to $\dir\mathcal A_1=\dir\mathcal A_2$.


问:
Why is Euler product destroyed?
Consider the lower branch of the hyperbola $f(x)=\frac{1}{x}.$ We can add up the images on the curve corresponding to pre-images equal to $-{n}$ for $n=1,2,3,\cdot\cdot\cdot.$

Upon doing this we get, $h(x)=-\sum_{n\ge1} n^x.$ It can be shown with some algebra that $h(x)$ can be expressed as a product of primes.

Changing the coordinate system of $\Bbb R^2$ via diffeomorphism, $g:\Bbb R^2\to \Bbb R^2$ with $g(x,y)=(e^x,e^y),$ we can define an analogous hyperbola in the new coordinate system, defined by $v(x)=e^{\frac{1}{\log(x)}}.$

If we decide to add up the images (like before) on $v(x)$ corresponding to pre-images equal to $e^{-\frac{1}{n}}$ for $n=1,2,3,\cdot\cdot\cdot$ then we get $\rho(x)=\sum_{n\ge1} e^{-n^x}.$

Why is the Euler product destroyed for $\rho(x)?$ Is $\rho(x)$ connected to primes?

答: The Euler product "works" because it establishes a 1-1 correspondence between the terms $1+\frac{1}{2^x}+\frac{1}{3^x}+\cdots$ with terms from expanding $\prod_p (1-1/p^x)^{-1}$. When you write $e^{-n^x}$, that correspondence is seemingly lost because exponents add when multiplied. For example replacing it with $\prod_p(1-1/e^{p^x})^{-1}$ doesn't work because you get terms like $\exp(p_1^{x}+p_2^{x}+\cdots)$.

Title: Why is Euler product destroyed?


问:
Is it more common to use $n \times m$ or $m \times n$ for matrices?
I've seen both versions for representing the number of rows by the number of columns of a matrix. I've always used $n \times m$, but I feel like I more often see $m \times n$. Is there some well known standard that people typically follow?

答:
When one needs to represent a (single) natural number, "$n$" seems to be a pretty common choice.  If another integer variable is needed, lots of people would choose "$m$". Since "$n$" came first, it might be sensible to use $n\times m$.

The same dilemma  takes place when dealing with parameters for curves, as "$t$" (for time) is a natural first choice, "$s$" coming in second place.


问:
When can we get $v_p(x+y) > \operatorname{min}\{v_p(x),v_p(y)\}$ in $p$-adic valuation?
Suppose $x,y\in\mathbb{Q}$. We can write $x=p^n(c/d)$, $y=p^m(a/b)$ with $p$ not dividing $a,b,c$ or $d$. When can we get $v_p(x+y) > \operatorname{min}\{v_p(x),v_p(y)\}$ with the inequality strict?
We have, if we assume wlog that $n>m$, that  $v(x+y)= v(p^m(p^{n-m}+1)=v(y)$, so I'm not sure when we'd get a strict inequality here. What am I missing?

答:
You cannot assume wlog that $n > m$ because $n$ might be equal to $m$.  (Indeed this phenomenon does not occur when $n \neq m$ and you have given a correct proof of that.)

For a counterexample, take $p = 7$ and then $14 + 35 = 49$.


答:
First, apply Descartes' Rule of signs.  Now we know this polynomial has either two or zero positive real roots and either two or zero negative roots.  (How does this help?  If we happen to have a long list of candidate roots and, while checking through this list, we find two positive roots, we can immediately discard all remaining positive candidates.)

Then we apply the rational root theorem.  This tells us that the only rational candidate roots are a divisor of $84$ over a divisor of $1$:
$$  \{-84, -42, -26, -21, -14, -12, -7, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 7, 12, 14, 21, 26, 42, 84\}  \text{.}  $$
We see the coefficients $4$, $-25$, and $-16$, which are sums of products of roots.  These are "small", so we should probably start in the middle of the list and work our way out.

\begin{align*}
x&  &&|&  &x^4 + 4x^3 - 25x^2 - 16x + 84  \\ \hline
1&  &&|&  &1+4-25-16+84 > 0  \\
2&  &&|&  &16+32-100-32+84 = 0 &&\checkmark \\
3&  &&|&  &81 + 108 - 225 - 144 + 84 = 0  &&\checkmark
\end{align*}
Having found two real roots, any further real roots are negative.  Continuing the table, starting from $-1$, you will find $-2$ and then $-7$.

But we can go a little faster.  By synthetic division (or long division), once we have the root $x = 2$, we can reduce the polynomial.
$$  x^4 + 4x^3 - 25x^2 - 16x + 84 = (x-2)(x^3 + 6x^2 - 13x - 42)  $$
So any root of $x^3 + 6x^2 - 13x - 42$ is also a root of the original polynomial.  By the rule of signs, this has one positive real root and two or zero negative real roots.  Any rational root is a divisor of $42$ divided by a divisor of $1$.  This reduces the previous list of candidate roots to (excluding $1$, the one candidate we definitely verified is not a root)
$$  \{ -42, -21, -14, -7, -6, -3, -2, -1, 2, 3, 6, 7, 14, 21, 42\}  \text{.}  $$
Testing $2$ in this reduced polynomial, we find $2$ is not a repeated root.  And then we find $3$ is a root:
$$  3^3 + 6\cdot 3^2 - 13\cdot 3 - 42 = 0  \text{.}  $$
That's the last possible positive root and we are left with
$$ \frac{x^3 + 6x^2 - 13x - 42}{x-3} = x^2 + 9x+14  \text{.}  $$

Of course, this is quadratic, so there is no reason to avoid directly applying the quadratic formula...  Alternatively, we compute the discriminant : $9^2 - 4 \cdot 14 = 25$, so the two roots differ by $\sqrt{25} = 5$.  (Since the discriminant is a square, we know the remaining two roots are rational, hence are on our list.)  Looking at the negatives on our list, the only pairs that differ by $5$ are $\{-1, -6\}$ and $\{-2, -7\}$.  Checking $-1$, we find that $-1$ is not a root, so $-2$ and $-7$ are the final two roots.

Title: Find roots of $x^4 + 4x^3 - 25x^2 - 16x + 84$

Body: I'm not quite sure how to approach this, how are you supposed to quickly factor a 4th degree polynomial? For $x^4 + 4x^3 - 25x^2 - 16x + 84$ it's factored into $(x-3)(x-2)(x+2)(x+7)$ which like I said, I have no idea how to approach. Would really appreciate some help, thanks in advance!


问:
Identify the set of points which satisfy $ |z|=\text{Re}z+1 $
Let $ z=x+iy $

I want to Identify the set of points that satisfy $ |z|=\text{Re}z+1 $.

Here's what I have found:

$ \sqrt{x^{2}+y^{2}}=x+1\thinspace\thinspace\thinspace\iff $

$ x^{2}+y^{2}=\left(x+1\right)^{2}\thinspace\thinspace\thinspace\iff $

$ y^{2}-2x-1=0\thinspace $

Now how can I identify the set of points based on the equation? I tried to use desmos, it looks like hyperbola, but its not. So im not sure...

Thanks in advance

答: You're correct that the equation is $y^2 = 2x + 1$. This gives us $x = \frac{y^2 - 1}{2}$. This is a parabola because $x$ is a quadratic function of $y$.


问:
Why is this polynomial a function of $X^3$?
In studying that recent question, I noticed that curious (or perhaps
not so curious) property : if $x,y$ are rational numbers and $a$ is the real
part of a cubic root of $x+iy$, then $Q(a^3)=0$ where $Q$ is a polynomial
of degree three with rational coefficients.

This is a purely algebraic property, and I obtained it by calculating some
resultants with my computer.

Now, can we show  a priori that $a$ is annihilated by a
polynomial of this form, without computer-made computations ? Note that
this property does not hold for the imaginary part of a cubic root.

答:
That's because the solutions to the system $(a+b)^3 = x+y, (a-b)^3 = x-y$, are the $9$ couples that you obtain by choosing a complex cube root $u$ of $x+y$, a complex cube root $v$ of $x-y$, and set $a=(u+v)/2, b = (u-v)/2$ (this is clear after a change of variables)

Since you can multiply both $u$ and $v$ by the same cube root of $1$ to obtain new solutions,
if $(a,b)$ is a solution, so are $(\zeta_3a,\zeta_3b)$ and $(\zeta_3^2 a, \zeta_3^2 b)$, hence the degree $9$ polynomial for $a$ is in fact a polynomial in $a^3$ (and of course the same goes with the one for $b$)


问:
prove limit of ratio of log of power series to x is 1
the proof for $\lim_{{x}\rightarrow 0}\frac{\ln\left(1+x\right)}{x}$ = 1 can be derived immediately from $\lim\limits_{x\to 0} (1+x)^\frac1x = e$

I was asked to whether the following limit, extending the series in ln above from 2 to n+1 terms, is true:

$\lim_{{x}\rightarrow 0}\frac{\ln\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^n\right)}{x}$ = 1

at the first look, it apparently is.
The mathematical induction method does not seem to be applicable here.
Does someone know a mehtod to prove it?

答:
For small x, use the approximation ln (1+x) = x

Note that x dominates higher powers of x for small x


Title: prove limit of ratio of log of power series to x is 1

Body:
the proof for $\lim_{{x}\rightarrow 0}\frac{\ln\left(1+x\right)}{x}$ = 1 can be derived immediately from $\lim\limits_{x\to 0} (1+x)^\frac1x = e$

I was asked to whether the following limit, extending the series in ln above from 2 to n+1 terms, is true:

$\lim_{{x}\rightarrow 0}\frac{\ln\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^n\right)}{x}$ = 1

at the first look, it apparently is.
The mathematical induction method does not seem to be applicable here.
Does someone know a mehtod to prove it?

Body: Hint: $\lim_{{x}\rightarrow 0}\frac{\ln\left(1+x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^n\right)}{x}    = \lim_{{x}\rightarrow 0}\frac{\left(x+x^{2}+x^{3}+x^{4}+x^{5}+...+x^n\right)}{x}$


问:
Side of a square attached to two triangles
Solve for $l$

I tried to solve this for $l$ with trigonometry, but I stopped at $l^2+4-2l\cdot \cos(x)=16, l^2 + 4 +2l\cdot \sin(x)=25$, where $x$ is the angle of the left triangle between 4 and $l$.
So I've tried with Heron's, and got stuck with some square roots...

According to WolframAlpha, the answer must be something around 3.71374

Any ideas?

------SOLVED----------

Okay, so, thanks to Narasimham I've seen that I took cosine formula wrong, and thanks to cosmo5 I've been able to solve it, here's the solution:

Start with cosine formula
\begin{matrix}
5^2 & = & l^2+4-2\cdot2\cdot l\cdot \cos(270^\circ - \alpha))\\
4^2 & = & l^2+4-2\cdot2\cdot l \cdot \cos(\alpha)
\end{matrix}

Rearrange teh two equation
\begin{matrix}
441-42l^2+l^4 & = &16l^2\cdot\sin^2(\alpha))\\
144-24l^2+l^4 & = &16l^2 \cdot \cos^2(\alpha)
\end{matrix}

find this quadratic formula
\begin{matrix}
585-66l^2+2l^4=16l^2
\end{matrix}

and then found this solutions
\begin{matrix}
+\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
+\sqrt{\dfrac{41-\sqrt{511}}{2}}\\
-\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
-\sqrt{\dfrac{41-\sqrt{511}}{2}}
\end{matrix}

Then, discart the negative ones
\begin{matrix}
\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
\sqrt{\dfrac{41-\sqrt{511}}{2}}
\end{matrix}

答:
Rearrange the two equations as
$$l^2-12=2l\cos x \quad , \quad l^2-21=-2l\sin x$$

On squaring both and adding, $x$ is eliminated,
$$(l^2-12)^2+(l^2-21)^2=(2l)^2$$
$$\Rightarrow 2(l^2)^2-70l^2+585=0$$

and a quadratic in $l^2$ is obtained. Its value can be computed using the quadratic formula.

To compute $l$, remember to take the square root of $l^2$. WA confirms $l=3.71374$.


问:
Show that group $G$ is isomorphic to $Z_3$.
I have this question:

Let $w = e^{\frac{2 \pi i}{3}}$, that is $w^3 = 1$ and $G = {w^0, w^1, w^2}$ or $G = {1,\frac{\frac{-1}{2} + i \sqrt{3}}{2},\frac{\frac{-1}{2} - i \sqrt{3}}{2}}$. Show that $G$ is $\mathbb{Z}_3$ isomorphic.

I set up Cayley's table for the two groups, but I can't see or justify why the two are isomorphic.

I saw that the two groups maintain the same cyclicality, but is that enough to say that they are isomorphic?

Thanks.

答:
Hint: The elements $w^0, w^1, w^2$ correspond to the vertices of a triangle in the complex plane. Multiplication by $w^1$ permutes these vertices cyclically.

Addition of $[1]_3$ in $\Bbb Z_3$ permutes the elements of $\Bbb Z_3$ cyclically too.


问:
Is the variant direct image mathematically significant?
Preimages have the property that for an arbitrary function $f : X \rightarrow Y$ and all $B \subseteq Y$ it holds that $$f^{-1}(B^c)=[f^{-1}(B)]^c.$$

However, the analogous statement for direct images is false. Thus, we have two definitions of direct image; the usual one, and a variant. $$f_*(A) = \{b \mid \exists a \in X : a \in A \,\wedge\, f(a)=b\}$$
$$f_\diamond(A) = \{b \mid \forall a \in X : a \in A \,\vee\, f(a) \neq b\}$$

Now I think (though I have not proved it) that, for all functions $f$, it holds that $f_* = f_\diamond$ iff $f$ is an injection bijection. Note also that, under these definitions, we have $$f_*(A^c) = [f_\diamond(A)]^c, \quad f_\diamond(A^c)=[f_*(A)]^c.$$

Here's another observation. If $X$ and $Y$ are topological spaces, then we have that $f_*$ preserves openness iff $f_\diamond$ preserves closedness, and vice versa.

Okay, but does $f_\diamond$ have any real mathematical significance? Like, does it show up naturally in any interesting theorems etc.? If so, what's the standard terminology/notation, and where can I learn more?

Addendum. Two related maps are $f_\cap(A) := f_*(A) \cap f_\diamond(A)$ and $f_\cup(A) := f_*(A) \cup f_\diamond(A).$ If anyone knows where I can learn more, please leave a comment! Note that all four concepts coincide in the case of bijections.

答:
The variant direct image is universal quantification. So if you think universal quantification is significant...

To explain, let $X, Y$ be sets and let $f : X \times Y \to Y$ be the projection onto the second coordinate. If $\varphi(x, y)$ is any proposition involving variables $x \in X, y \in Y$, then it cuts out a subset of $X \times Y$ (namely the subset for which $\varphi(x, y)$ is true). The corresponding direct and variant direct image subsets in $Y$ are precisely the subsets of $Y$ cut out by the existentially and universally quantified statements $\exists x : \varphi(x, y)$ and $\forall x : \varphi(x, y)$ respectively.

Much more generally, let $F : C \to C'$ be a functor between two categories and let $D$ be another category. Then $F$ induces a functor $D^{C'} \to D^C$ between functor categories. A left adjoint to this functor, if it exists, is called left Kan extension along $F$, and a right adjoint is called right Kan extension.

Now let $C, C'$ be sets regarded as discrete categories (categories with only identity morphisms), and let $D = \{ 0 \to 1 \}$ be the category with two objects $0, 1$ and a single non-identity morphism $0 \to 1$. $D$ is a poset, and so are the functor categories $D^C, D^{C'}$, which are naturally identified with the poset of subsets of $C, C'$ under inclusion. If $F : C \to C'$ is a functor (so just a map of sets), the induced functor $D^{C'} \to D^C$ is the usual preimage functor, the usual direct image is the left Kan extension along $F$, and the variant direct image functor is the right Kan extension along $F$.

Left and right Kan extension are ubiquitous in mathematics; Mac Lane famously said that they subsume all the other fundamental concepts in category theory. So it is a little hard to give examples because there are so many. This example is the simplest example I know.


问:
Counting/combinatorics problem: How many people do not eat either tacos, salmon, or a burger?
I'm working on this counting/combinatorics problem. Can figure if it's a trick question or what kind of technique I should use. Please help.

There are these people who eat either Tacos, Salmon or Burger:

11 eats tacos

11 eats salmon

11 eats burger

3 eats taco and salmon

4 eats taco and burger

7 eat salmon and burger

1 eats taco, salmon and burger

If there are 25 people in a set, how many do not eat either tacos, salmon or burger?

答:
By Inclusion-Exclusion principle, the number of people who eat atleast one of the items is

$$|T|+|S|+|B|-|T \cap S|-|S \cap B|-|T \cap B|+|T \cap S \cap B|$$
$$=11+11+11-3-7-4\color{red}{+1}=20$$

To get the number of people who do not eat any, subtract this number from total number of people. $25-20=5$.


问:
Let $q_n$ be the smallest prime that is strictly great than $P_n = p_1 p_2 \dots p_n + 1$, prove $q_n - p_1 p_2 \dots p_n$ is always prime.

Let $q_n$ be the smallest prime that is strictly great than $P_n = p_1 p_2 \dots p_n + 1$, prove $q_n - p_1 p_2 \dots p_n$ is always a prime.

It's from Elementary number theory 7th by David M. Burton and is a conjecture. What is the name of this conjecture?

答:
The conjecture is called Fortune's conjecture, with the smallest integer $m \gt 1$ where $p_n\# + m$ (where $p_n\#$ is the primorial) is a prime number, being called Fortunate numbers.

Note that, in addition, there are also currently at least $6$ related questions on this site, e.g., as shown in this search.


问:
Prove that $\frac{\sqrt{N}}{(\frac{N(N - 1)}{2} + 1)^2} \le 1$ for $N \ge 2$.
I am trying to prove that for natural numbers $N \ge 2$, the following expression would always be true:

$$
\frac{\sqrt{N}}{\Big(\frac{N(N - 1)}{2} + 1\Big)^2} \le 1.
$$

I can clearly see that when I calculate. For example: when $N = 2$,
$$
\frac{\sqrt{N}}{\Big(\frac{N(N - 1)}{2} + 1\Big)^2}  = \frac{\sqrt{2}}{\Big(\frac{2(2 - 1)}{2} + 1\Big)^2}  = .3536.
$$
Or when $N = 10$,
$$
\frac{\sqrt{N}}{\Big(\frac{N(N - 1)}{2} + 1\Big)^2}  = \frac{\sqrt{10}}{\Big(\frac{10(10 - 1)}{2} + 1\Big)^2}  =.0015.
$$
But how do I prove it for the general case? TIA.

答:
By AM-GM inequality,

$$\frac{1}{2} \bigg( \frac{N(N-1)}{2} + 1 \bigg) \geq \sqrt{\frac{N(N-1)}{2}}$$.

All quantities being positive, inequality is preserved when you take the square. Hence,

$$\bigg( \frac{N(N-1)}{2} + 1 \bigg)^2 \geq 2N(N-1)$$

It is now obvious that $2N(N-1) \geq \sqrt{N}$ since $N \geq 2$. Hence you have your result.


问:
Proof of differentiability
I would appreciate if somebody could help me with the following problem:

Q: $f(x)$ continuous in $\mathbb{R}$

Show that  if

$$e^{x+h}f(x+h)-e^xf(x)<h^2 \implies |e^{x+h}f(x+h)-e^xf(x)|<h^2$$

答: This is false if $f$ is the constant function $f(x)=-1$, even if $h$ is positive.</p>

Title: Proof of differentiability


问:
Proof of the well-ordering principle
I tried to prove Well-Ordering Principle by myself, and I finally did it. However, I'm not sure if this proof is correct. Can anyone evaluate my proof?

Proof:
Since the set of natural numbers, $\mathbb{N}$, is bounded below (every element of $\mathbb{N}$ is equal or greater than zero), an arbitrary subset (let's name this set $X$) of $\mathbb{N}$ is also bounded below. So, by the greatest lower bound principle, there exists an infimum of $X$. Let $a:=\inf (X)$. Then, every element of $X$ is equal or greater than $a$. Also, for every real number $y$ which is greater than $a$, there exists an element $k$ (which is equal or greater than $a$) of $X$ which is less than $y$. Therefore, $k=a$, and since $a$ is an infimum of $X$, it becomes a minimum element of $X$. (Q.E.D.)

答:
Actually, the you can prove the well-ordering principle from the least-upper-bound property of $\mathbb{R}$. See the 2nd bullet point from the Wikipedia article here. Steps:

Dedekind Cuts method to construct $\mathbb{R}$ and then show that $\mathbb{Q}$ and thus $\mathbb{N}$ are subfields of $\mathbb{R}$ (standard "baby Rudin" textbook stuff).

Next, since $\mathbb{R}$ has the least-upper bound property, any subset of $\mathbb{N}$
, say $X$, has an infimum, and thus the arguments provided by the author are correct up to $a=inf(X)$

Lastly, we need to show $a$ (which is automatically an element of $\mathbb{R}$) is actually an element of $X$ itself. The author's original argument here needs just a little strengthening or re-wording. Why not use a contradiction proof to show that there cannot be another real number $a^*$ ($\notin X$) which is "in-between" $a$ and any member $n\in X$? For if there was such a gap between $a$ and $n$, then by the "Q is dense in R" property (Rudin theorem 1.20), we could find a rational number $a^*$ such that $a<a^*<n$. This would contradict the fact that $a$ is the infimum of X.

Thus $a$ must be an element of $X$, and so any subset of $\mathbb{N}$ must have a least element.

Perhaps a logician can comment, but the Axiom of Choice is lurking around somewhere in the 3rd bullet point above. As Jerry Bona says: "who can tell...?"

Title: Proof of the well-ordering principle


问:
Probability of an event occuring in a subinterval, given that some number of events ocurred on the interval
I've just started delving into probability, and I was presented this exercise:

Patients get to a hospital in a mean rate of $5$ patients per hour, following a Poisson process. Given that $13$ patients arrived between $12$PM and $3$PM, what is the probability that $10$ patients arrived between $12$PM and $2$PM.

How can I go about solving this?

答:
Hint

If $X\sim Pois(\lambda _1)$ and $Y\sim Pois(\lambda _2)$ are independents, one can prove that $X+Y\sim Pois(\lambda _1+\lambda _2)$.


问:
Proving $\sum_{n=1}^m\sin^2\left(n\frac\pi{m}\right)=\frac{m}2$ for $m\geq 2$ by induction

$$\sum_{n=1}^m \sin^2\left(n\frac{\pi}{m}\right) = \frac{m}{2}$$ for all $m\ge2$ positive integers

I'm doing a maths problem that involves proving by mathematical induction the sum of a sigma. I find it hard because the highest term is included in the sigma so it changes the actual sigma.

If anyone can help me, please respond it would be very useful!! Even just the smallest help will go a long way : )

Here is the Desmos link to show you this is true.

But I want to prove it using induction.

Thanks.

答:
HINTS

Write $$\sin^2(n \pi /m)= \frac{1 - \cos(2 n \pi/m)}{2}.$$
Then by the linearity of finite sums, we have

$$\sum_{n=1}^m \sin^2(n \pi /m) = \sum_{n=1}^m \frac{1 - \cos(2 n \pi/m)}{2}=\frac{m}{2}-\sum_{n=1}^m \frac{\cos(2n \pi/m)}{2},$$ which is almost the answer.

You can now show that $\sum_{n=1}^m \frac{\cos(2n \pi/m)}{2}=0$ for $m \ge 2$ by an induction argument or by noticing it is the real part of the finite geometric series $$\frac{1}{2}\sum_{n=1}^m e^{2 \pi i n/m}.$$


问:
Evaluating $\lim_{x\to-\infty} xe^{-x}$
Not sure how to think about this one.

$$\lim_{x\to-\infty}  xe^{-x}$$

If I place "$-\infty$" in place for $x$ I get
$$(-\infty)e^{\infty}$$
the limit of which would seem to be $0$ as $x$ approaches $-\infty$.

But in fact the limit is $-\infty$ (if looking at the drawn graph) as $x$ approaches $-\infty$.

答:
Using the product law for limits,
\begin{align}
\lim_{x \to -\infty}xe^{-x} &= \lim_{x \to -\infty}x \cdot \lim_{x \to -\infty}e^{-x} \\
&= -\infty \cdot \infty \\
&= -\infty \, .
\end{align}
Keep in mind though that in limiting analysis, '$\infty$' and '$-\infty$' are both just symbols for unbounded growth. When we write
$$
\lim_{x \to -\infty}e^{-x} = \infty \, ,
$$
all we mean is that $e^{-x}$ gets arbitrarily large as $x$ gets arbitrarily large and negative.

Title: Evaluating $\lim_{x\to-\infty} xe^{-x}$


问:
Very confused about the definition of subbasis for a topology in Munkres's book
In Munkres's Topology textbook, it says,

Definition

A subbasis S for a topology on X is a collection of subsets of X whose union equals X.

The topology generated by the subbasis S is defined to be the collection T of all unions of finite intersections of elements of S.

I am having a very hard time understanding the first sentence. Let's say $X = \{1, 2, 3\}$. Then let $A = \{\{1\}, \{2\}, \{3\}\}$, or $A = \{\{1, 2\}, \{3\}\}$. Either way, $A$ is a collection of subsets of $X$, and the union of the elements in $A$ is $X$, but $A$ does not seem like a subbasis here...

What did I misunderstand here? What should be the correct interpretation of "a collection of subsets of X whose union equals X"?

Thank you in advance!

Edit:

So turned out $A$ is a subbasis in both cases, thanks @Surb for your comment.

But here is what I don't understand (which made me mistakenly think A may not even be subbasis in the first place). If $A = \{\{1\}, \{2\}, \{3\}\}$, then wouldn't "the collection T of all unions of finite intersections of elements of $A$" be {$\varnothing$}, since any intersection of any two elements in $A$ is empty? But since T is a topology, T must also contain $X$, but, in this case, it does not?

答:
What you’re missing is that a finite subcollection of $A$ can be a collection with only one element; in your example, for instance, $\big\{\{1\}\big\}$ is a finite subcollection of $A$.

If $\mathscr{C}$ is any family of subsets of a set $X$, by definition

$$\bigcap\mathscr{C}=\{x\in X:\exists C\in\mathscr{C}(x\in C)\}\,;$$

$\bigcap\mathscr{C}$ is the set of all things that belong to at least one member of $\mathscr{C}$. In your example $\big\{\{1\}\big\}$ is a finite subset of $A$, and and its intersection is the set of all members of $\{1,2,3\}$ that belong to some member of the collection $\big\{\{1\}\big\}$. The only member of that collection is $\{1\}$, and the only thing that belongs to it is $1$, so $\bigcap\big\{\{1\}\big\}=\{1\}$.

More generally, if $S$ is any set, $\bigcap\{S\}=S$: the only member of the collection $\{S\}$ is $S$, and the set of its members is just $S$ itself.


答:
$v_p(x+y) > \min v_p(x),v_p(y)$ iff $x=p^n a/b,y=p^n c/d, p\nmid abcd, p |ad+bc$.

Proof: $x+y= p^n \frac{ad+bc}{bd}$, $v_p(x+y)=n+v_p(ad+bc)$.

Title: When can we get $v_p(x+y) > \operatorname{min}\{v_p(x),v_p(y)\}$ in $p$-adic valuation?

Body:
Suppose $x,y\in\mathbb{Q}$. We can write $x=p^n(c/d)$, $y=p^m(a/b)$ with $p$ not dividing $a,b,c$ or $d$. When can we get $v_p(x+y) > \operatorname{min}\{v_p(x),v_p(y)\}$ with the inequality strict?
We have, if we assume wlog that $n>m$, that  $v(x+y)= v(p^m(p^{n-m}+1)=v(y)$, so I'm not sure when we'd get a strict inequality here. What am I missing?


问:
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$?
Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ?

I thought it would be 12 this as per pemdas rule:

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$

Wanted to confirm the right answer from you guys. Thanks for your help.

答:
Your answer of $12$ is numerically correct, but you’ve applied the precedence rules incorrectly. In the absence of parentheses multiplication is done before addition, so

$$1+1+1+1+1+1+1+1+1+1+1+1\cdot 0+1$$

is to be evaluated as

$$1+1+1+1+1+1+1+1+1+1+1+(1\cdot 0)+1\;,$$

not as

$$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot(0+1)\;.$$

Since $1\cdot0=0$, this simplifies to

$$1+1+1+1+1+1+1+1+1+1+1+0+1=12\;.$$


问:
Verifying algebraic construction of regular 4-polytopes
One nice way to construct the 24-cell and 120-cell is to take a tetrahedron or icosahedron, look its rotational symmetries (of order 12 or 60, respectively) in $\textrm{SO}(3)$, and pull them back along the double cover $\Phi : \textrm{SU}(2) \to \textrm{SO}(3)$, identifying $\textrm{SU}(2)$ with the unit quaternions: in this way we get 24 or 120 points lying on $S^3$.

Is there a nice way to verify that these points form the vertices of a regular 4-polytope? It seems plausible to me, and I've seen it stated, but I've never seen a proof. I believe attempting this with the symmetries of the cube give you 48 points which do not form a regular 4-polytope, so ideally an explanation would explain why that fails without being too computational.

If necessary, I'd be ok assuming as given that the 24-cell and 600-cell exist, if that could help in verifying this construction produces them.

答:
The double cover and identification with unit quaternions boils down to associating a rotation in 3-space, around an axis with unit vector $(x, y, z)$ by angle $\theta$, with the quaternion $q = c + xs\mathbf{i} + ys\mathbf{j} + zs\mathbf{k}$, where $c = \cos(\theta/2)$ and $s = \sin(\theta/2)$, and with its negation $-q$. The quaternion $q$ in turn is identified with the point $(c, xs, ys, zs)$ in $\mathbb{R}^4$.

Since we are dealing with the rotation axes of a regular polytope, they'll be symmetrically distributed (and in a "full-dimensional" way, i.e., not like a prism, whose symmetries are essentially lower-dimensional.)

Thus the corresponding points in $\mathbb{R}^4$ will also be "evenly" distributed with a high degree of symmetry. Since the points arise from a group, which acts transitively on its own elements, and since the group action is multiplication by unit quaternions, which are isometries of 4-space, the points will be acted on transitively by isometries of $\mathbb{R}^4$.

So I don't think there's really a fundamental reason that the convex hull of these points has to be a regular 4-polytope, but it does have to be a highly symmetric vertex-transitive one, which suggests that it's pretty likely to be regular.

Tetrahedral symmetry

Let's look at a tetrahedron, which has threefold axes of rotation through each of its vertices, and twofold axes of rotation through the midpoints of each opposite pair of edges. We'll need to pick some coordinates for a regular tetrahedron. Let me take alternate vertices of the cube $[-1,1]\times[-1,1]\times[-1,1]$, say the ones with evenly many minus signs, and scale them to unit length: that gives me vertices
$$\Bigl(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Bigr),
\Bigl(\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Bigr),
\Bigl(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\Bigr),
\Bigl(\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{-1}{\sqrt{3}}\Bigr).$$

Our rotations are:

The identity. We can consider this to have any axis of rotation we like; it doesn't matter since $\sin(0) = 0$. The first coordinate is $\cos(0) = 1$,
so we have quaternions $(1,0,0,0)$ and $(-1,0,0,0)$.

Rotation by $2\pi/3$ around each vertex. The first coordinate is $\cos(\pi/3) = 1/2$, and the vertex coordinates are multiplied by $\sin(\pi/3) = \sqrt{3}/2$. The corresponding quaternions are
$$\Bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\Bigr),\quad
\Bigl(\frac{-1}{2},\frac{-1}{2},\frac{-1}{2},\frac{-1}{2}\Bigr),\\
\Bigl(\frac{1}{2},\frac{-1}{2},\frac{-1}{2},\frac{1}{2}\Bigr),\quad
\Bigl(\frac{-1}{2},\frac{1}{2},\frac{1}{2},\frac{-1}{2}\Bigr),\\
\Bigl(\frac{1}{2},\frac{-1}{2},\frac{1}{2},\frac{-1}{2}\Bigr),\quad
\Bigl(\frac{-1}{2},\frac{1}{2},\frac{-1}{2},\frac{1}{2}\Bigr),\\
\Bigl(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\Bigr),\quad
\Bigl(\frac{-1}{2},\frac{-1}{2},\frac{1}{2},\frac{1}{2}\Bigr)$$

Rotation by $4\pi/3$ around each vertex. The first coordinate is $\cos(2\pi/3) = -1/2$, and $\sin(2\pi/3)$ is still $\sqrt{3}/2$, so we have
$$\Bigl(\frac{-1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\Bigr),\quad
\Bigl(\frac{1}{2},\frac{-1}{2},\frac{-1}{2},\frac{-1}{2}\Bigr),\\
\Bigl(\frac{-1}{2},\frac{-1}{2},\frac{-1}{2},\frac{1}{2}\Bigr),\quad
\Bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{-1}{2}\Bigr),\\
\Bigl(\frac{-1}{2},\frac{-1}{2},\frac{1}{2},\frac{-1}{2}\Bigr),\quad
\Bigl(\frac{1}{2},\frac{1}{2},\frac{-1}{2},\frac{1}{2}\Bigr),\\
\Bigl(\frac{-1}{2},\frac{1}{2},\frac{-1}{2},\frac{-1}{2}\Bigr),\quad
\Bigl(\frac{1}{2},\frac{-1}{2},\frac{1}{2},\frac{1}{2}\Bigr)$$

Rotation by $\pi$ around the edge midpoints, which scaled to unit vectors are $(0,0,1)$, $(0,1,0)$, and $(1,0,0)$. These give us
$$(0,0,0,1),\quad (0,0,0,-1),\quad
(0,0,1,0),\quad (0,0,-1,0),\quad
(0,1,0,0),\quad (0,-1,0,0).$$

The 16 vertices $\bigl(\frac{\pm 1}{2},\frac{\pm 1}{2},\frac{\pm 1}{2},\frac{\pm 1}{2}\bigr)$ taken by themselves form a tesseract. The remaining eight vertices create a pyramid raised above each facet of the tesseract. Each 2-face (square) $F$ is in two facets of the tesseract (cubes); the pyramids raised over those two facets each contain a square pyramid over $F$. It just so happens that these two square pyramids are actually "coplanar", that is, contained in the same 3-dimensional subspace, so that they join together in an octahedron cell, and $F$ is absorbed in its interior.

It might help to visualize an analogous process in 3 dimensions. Add a vertex above each face of a cube, forming pyramids above the faces. This will give you 24 triangular faces, but if the new vertices are at just the right height, the two triangles over each of the original edges will be coplanar, forming a rhombus, and the original edge will be absorbed in its interior. This gives you the rhombic dodecahedron:

That picture (from MathWorld) shows the skeleton of the rhombic dodecahedron balanced on the edges of the cube.

Since there's one of these octahedra for each 2-face of the tesseract, we get 24 octahedral cells. This is actually enough to know that it's a regular 4-polytope, given Theorem 4C4 of Peter McMullen's thesis:

4C4. Theorem. A $d$-polytope $P$ ($d\geq 3$) is regular if and only if its vertices lie on a sphere, and its facets are regular and combinatorially equivalent.

Octahedral symmetry

If we consider a 3-cube with the unit-norm vertices $\bigl(\frac{\pm 1}{\sqrt{3}},\frac{\pm 1}{\sqrt{3}},\frac{\pm 1}{\sqrt{3}}\bigr)$, it has the same twelve rotations that we just discussed as symmetries, along with some more.
There's a fourfold axis of rotation through the midpoints of each pair of opposite faces. These are the same axes $(0,0,1)$, $(0,1,0)$, $(1,0,0)$ that we saw for the tetrahedron, but now we can include rotations of $\pi/2$ and $3\pi/2$ as well as the rotation by $\pi$ that we had. This gives us six additional rotations, so twelve new points in $\mathbb{R}^4$:

$$
\Bigl(\frac{1}{\sqrt{2}},0,0,\frac{1}{\sqrt{2}}\Bigr),\quad
\Bigl(\frac{-1}{\sqrt{2}},0,0,\frac{-1}{\sqrt{2}}\Bigr),\quad
\Bigl(\frac{-1}{\sqrt{2}},0,0,\frac{1}{\sqrt{2}}\Bigr),\quad
\Bigl(\frac{1}{\sqrt{2}},0,0,\frac{-1}{\sqrt{2}}\Bigr),\\
\Bigl(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},0\Bigr),\quad
\Bigl(\frac{-1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}},0\Bigr),\quad
\Bigl(\frac{-1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},0\Bigr),\quad
\Bigl(\frac{1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}},0\Bigr),\\
\Bigl(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0\Bigr),\quad
\Bigl(\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0,0\Bigr),\quad
\Bigl(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0\Bigr),\quad
\Bigl(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0,0\Bigr).
$$

There are also twofold axes of rotation through the midpoints of each pair of opposite edges; there are six such pairs, giving us twelve more points in $\mathbb{R}^4$. The unit vectors are
$\bigl(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\bigr)$,
$\bigl(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\bigr)$,
$\bigl(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\bigr)$,
$\bigl(\frac{-1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\bigr)$,
$\bigl(0, \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\bigr)$, and
$\bigl(0, \frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\bigr)$.
The quaternions are
$$
\Bigl(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Bigr),\quad
\Bigl(0,\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0\Bigr),\quad
\Bigl(0,\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\Bigr),\quad
\Bigl(0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}},0\Bigr),\\
\Bigl(0,\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\Bigr),\quad
\Bigl(0,\frac{-1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}}\Bigr),\quad
\Bigl(0,\frac{-1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\Bigr),\quad
\Bigl(0,\frac{1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}}\Bigr),\\
\Bigl(0,0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Bigr),\quad
\Bigl(0,0,\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}}\Bigr),\quad
\Bigl(0,0,\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Bigr),\quad
\Bigl(0,0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}\Bigr).
$$

This gives us 48 points total.

If we consider only the 24 new points, we can identify the facet-defining hyperplanes: the sixteen variations of $\pm x_1 \pm x_2 \pm x_3 \pm x_4 = \sqrt{2}$, and the eight versions of $x_i = \frac{\pm1}{\sqrt{2}}$.

Each of these hyperplanes contains 6 of the new points.
For instance, $x_1 + x_2 + x_3 + x_4 = \sqrt{2}$ matches
$\bigl(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0,0\bigr)$,
$\bigl(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},0\bigr)$,
$\bigl(\frac{1}{\sqrt{2}},0,0,\frac{1}{\sqrt{2}}\bigr)$,
$\bigl(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\bigr)$,
$\bigl(0,\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\bigr)$, and
$\bigl(0,0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\bigr)$.
These six points form an octahedron.

Thus the 24 new points form another 24-cell, rotated with respect to the first one; in fact each vertex of the first one is above the midpoint of a facet of the new one, e.g. $\bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\bigr)$ is above the octahedron we just considered.

It's not as easy to visualize the structure of the convex hull of the union of these two 24-cells. We can find a facet-defining hyperplane, though, by verifying that all 48 points satisfy $x_1 + x_2 + \sqrt{2}x_3 \leq 1 + 1/\sqrt{2}$.
The four points meeting the equality condition are
$\bigl(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{\pm 1}{2}\bigr)$,
$\bigl(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}},0\bigr)$, and
$\bigl(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0\bigr)$.

These form a disphenoid tetrahedron; two of the edges have length 1 and the others have length $\sqrt{2-\sqrt{2}}$.

So we know this 4-polytope is not regular.
(In fact, it is the Disphenoidal 288-cell.)

Icosahedral symmetry

Considering an icosahedron, its twelve vertices are in six opposite pairs, with an axis of five-fold symmetry through each pair: this gives us 24 rotations.
Its twenty faces are in ten opposite pairs, with an axis of three-fold symmetry through each pair: this gives us 20 more rotations.
Its thirty edges are in fifteen opposite pairs, with an axis of two-fold symmetry through each pair: this gives us 15 more rotations.
Together with the identity, we have 24 + 20 + 15 + 1 = 60 rotations,
and so 120 corresponding points in $\mathbb{R}^4$.

The identity rotation gives us the two points $(1,0,0,0)$ and $(-1,0,0,0)$ in $\mathbb{R}^4$.

We can take unit vectors for the six axes of five-fold symmetry to be
$\frac{1}{\sqrt{10}}(0, \pm \sqrt{5-\sqrt{5}}, \sqrt{5+\sqrt{5}})$,
and the rotations of these coordinates.
Let $\gamma$ be $\frac{1}{4}(\sqrt{5} + 1)$ (half the golden ratio)
and $\delta$ be $\frac{1}{4}(\sqrt{5}-1)$ (half the golden ratio's inverse).
Using the facts that
$\cos(\pi/5) = \gamma$,
$\sin(\pi/5) = \frac{1}{4}\sqrt{10-2\sqrt{5}}$,
$\cos(2\pi/5) = \delta$, and
$\sin(2\pi/5) = \frac{1}{4}\sqrt{10+2\sqrt{5}}$, we get the points

$$
\Bigl(\gamma, 0, \delta, \frac{1}{2}\Bigr),
\Bigl(\delta, 0, \frac{1}{2}, \gamma\Bigr),
\Bigl(-\delta, 0, \frac{1}{2}, \gamma\Bigr),
\Bigl(-\gamma, 0, \delta, \frac{1}{2}\Bigr),\\
\Bigl(\gamma, 0, -\delta, \frac{1}{2}\Bigr),
\Bigl(\delta, 0, \frac{-1}{2}, \gamma\Bigr),
\Bigl(-\delta, 0, \frac{-1}{2}, \gamma\Bigr),
\Bigl(-\gamma, 0, -\delta, \frac{1}{2}\Bigr),\\
\Bigl(\gamma, \delta, \frac{1}{2}, 0\Bigr),
\Bigl(\delta, \frac{1}{2}, \gamma, 0\Bigr),
\Bigl(-\delta, \frac{1}{2}, \gamma, 0\Bigr),
\Bigl(-\gamma, \delta, \frac{1}{2}, 0\Bigr),\\
\Bigl(\gamma, -\delta, \frac{1}{2}, 0\Bigr),
\Bigl(\delta, \frac{-1}{2}, \gamma, 0\Bigr),
\Bigl(-\delta, \frac{-1}{2}, \gamma, 0\Bigr),
\Bigl(-\gamma, -\delta, \frac{1}{2}, 0\Bigr),\\
\Bigl(\gamma, \frac{1}{2}, 0, \delta\Bigr),
\Bigl(\delta, \gamma, 0, \frac{1}{2}\Bigr),
\Bigl(-\delta, \gamma, 0, \frac{1}{2}\Bigr),
\Bigl(-\gamma, \frac{1}{2}, 0, \delta\Bigr),\\
\Bigl(\gamma, \frac{1}{2}, 0, -\delta\Bigr),
\Bigl(\delta, \gamma, 0, \frac{-1}{2}\Bigr),
\Bigl(-\delta, \gamma, 0, \frac{-1}{2}\Bigr),
\Bigl(-\gamma, \frac{1}{2}, 0, -\delta\Bigr)
$$
and all their negations.

We can take unit vectors for the ten axes of three-fold symmetry to be
$\frac{1}{\sqrt{3}}(1,1,1)$, $\frac{1}{\sqrt{3}}(-1,1,1)$ and its permutations,
and $(0, \pm \frac{\sqrt{5}+1}{2\sqrt{3}}, \frac{\sqrt{5}-1}{2\sqrt{3}})$ and the rotations of its coordinates. Rotating about these through $2\pi/3$ or $4\pi/3$ corresponds to the points

$$
\Bigl(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \Bigr),
\Bigl(\frac{-1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \Bigr),\\
\Bigl(\frac{1}{2}, \frac{-1}{2}, \frac{1}{2}, \frac{1}{2} \Bigr),
\Bigl(\frac{-1}{2}, \frac{-1}{2}, \frac{1}{2}, \frac{1}{2} \Bigr),\\
\Bigl(\frac{1}{2}, \frac{1}{2}, \frac{-1}{2}, \frac{1}{2} \Bigr),
\Bigl(\frac{-1}{2}, \frac{1}{2}, \frac{-1}{2}, \frac{1}{2} \Bigr),\\
\Bigl(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{-1}{2} \Bigr),
\Bigl(\frac{-1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{-1}{2} \Bigr),\\
\Bigl(\frac{1}{2}, 0, \gamma, \delta\Bigr),
\Bigl(\frac{-1}{2}, 0, \gamma, \delta \Bigr),\\
\Bigl(\frac{1}{2}, 0, -\gamma, \delta\Bigr),
\Bigl(\frac{-1}{2}, 0, -\gamma, \delta \Bigr),\\
\Bigl(\frac{1}{2}, \gamma, \delta, 0 \Bigr),
\Bigl(\frac{-1}{2}, \gamma, \delta, 0 \Bigr),\\
\Bigl(\frac{1}{2}, -\gamma, \delta, 0 \Bigr),
\Bigl(\frac{-1}{2}, -\gamma, \delta, 0 \Bigr),\\
\Bigl(\frac{1}{2}, \delta, 0, \gamma \Bigr),
\Bigl(\frac{-1}{2}, \delta, 0, \gamma \Bigr),\\
\Bigl(\frac{1}{2}, \delta, 0, -\gamma \Bigr),
\Bigl(\frac{-1}{2}, \delta, 0, -\gamma \Bigr)
$$
and all their negations.

With the coordinates I chose for the icosahedron's vertices (seen above as the axes of five-fold rotation), the edge midpoints end up being
$\frac{1}{\sqrt{10}}(0,0,\sqrt{5+\sqrt{5}})$,
$\frac{1}{2\sqrt{10}}(\sqrt{5-\sqrt{5}},\sqrt{10+4\sqrt{5}},\sqrt{5+\sqrt{5}})$,
and the rotations of these coordinates (with all assignments of +/- signs.)
After scaling to unit vectors, and choosing one from each opposite pair, we get $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ and all the rotations of $(\delta, \gamma, \frac{1}{2})$ with at most one minus sign. Rotating by an angle of $\pi$ about these vectors gives us the quaternions
$$
(0,1,0,0), \quad (0,0,1,0), \quad (0, 0, 0, 1), \\
(0, \delta, \gamma, \frac{1}{2}), \quad
(0, -\delta, \gamma, \frac{1}{2}), \quad
(0, \delta, -\gamma, \frac{1}{2}), \quad
(0, \delta, \gamma, -\frac{1}{2}), \\
(0, \gamma, \frac{1}{2}, \delta), \quad
(0, -\gamma, \frac{1}{2}, \delta), \quad
(0, \gamma, -\frac{1}{2}, \delta), \quad
(0, \gamma, \frac{1}{2}, -\delta), \\
(0, \frac{1}{2}, \delta, \gamma), \quad
(0, -\frac{1}{2}, \delta, \gamma), \quad
(0, \frac{1}{2}, -\delta, \gamma), \quad
(0, \frac{1}{2}, \delta, -\gamma)
$$
and all their negations.

So after going through all the rotations of the icosahedron,
we can summarize all the corresponding points in $\mathbb{R}^4$
the same way that Wikipedia puts it:
the eight permutations of $(\pm 1, 0, 0, 0)$,
the sixteen points $\bigl(\frac{\pm 1}{2},\frac{\pm 1}{2},\frac{\pm 1}{2},\frac{\pm 1}{2}\bigr)$,
and the 96 even permutations of $\bigl(0,\frac{\pm 1}{2},\pm \delta, \pm \gamma\bigr)$.

So what is the convex hull of these points?

We can find a facet-defining hyperplane by verifying that all the points satisfy $x_1 + x_2 \leq \gamma + \frac{1}{2}$, and exactly four of the points are in that hyperplane:
$$
\Bigl(\frac{1}{2}, \gamma, \delta, 0\Bigr),
\Bigl(\frac{1}{2}, \gamma, -\delta, 0\Bigr),
\Bigl(\gamma, \frac{1}{2}, 0, \delta\Bigr),
\Bigl(\gamma, \frac{1}{2}, 0, -\delta\Bigr).
$$

Since these points are all at a distance of $2\delta$ from eachother, they form a regular tetrahedron. If we trust that all the facets are regular tetrahedra (the vertex-transitivity will help a lot if you want to rigorously check this), then McMullen's Theorem 4C4 tells us this is a regular 4-polytope.


问:
Prove: $T \in L(Y,\ell^\infty(\mathbb K))$ is bounded $\implies$ $\forall n \in \mathbb N:g_n \in L(Y,\mathbb K)$ is bounded
Assume $Y$ is a Banach space over $\mathbb{K}$ and $T : Y → \ell^{\infty}$ is a linear operator, such that $Ty = (g_1(y), g_2(y), g_3(y), . . .)$, where  $g_n \in L(Y, \mathbb{K})$.

I am trying to prove/disprove the following statements:

($T$ is bounded) $\rightarrow$ ($\forall n \in \mathbb{N}:g_n$ is bounded )

($\forall n \in \mathbb{N}:g_n$ is bounded ) $\rightarrow$  ($T$ is bounded)

If it is bounded then $\exists$M $\geq 0$ such that $\|Ty\|_{\infty} \leq$M$\|y\|_{\infty}$ $\forall$y$\in Y$.

Then. $\|Ty\| = \|g_n(y)\|$$ = \sup_{n\in\mathbb{N}}|g_n(y)| \forall n \geq1$, so we can easily take $M=1$

Hence, $\|Ty\|_{\infty} \leq$$\|y\|_{\infty}$ and so $\|T\| \leq 1$.

I am not convinced that I am on the right track. Any help would be grateful for both "arrows" in the question. Thanks in advance.

答:
(1) is true. Let $y\in Y$. Let $n\in\mathbb{N}$ be arbitrary, then
$|g_{n}(y)|\leq||Ty||_{\infty}\leq||T||\cdot||y||$. This shows that
$g_{n}$ is bounded and $||g_{n}||\leq||T||$.

(2) is true. For each $n\in\mathbb{N}$, define $T_{n}:Y\rightarrow l^{\infty}$
by $T_{n}y=(g_{1}(y),g_{2}(y),\ldots g_{n}(y),0,0,\ldots)$. Clearly
$T_{n}$ is linear. Let $y\in Y$, then
\begin{eqnarray*}
||T_{n}y|| & = & \max_{1\leq k\leq n}|g_{k}(y)|\\
 & \leq & \max_{1\leq k\leq n}||g_{k}||\cdot||y||\\
 & = & M||y||
\end{eqnarray*}
where $M=\max_{1\leq k\leq n}||g_{k}||<\infty$. Therefore, $T_{n}$
is a bounded linear map. Consider the family of bounded linear maps
$\{T_{n}\mid n\in\mathbb{N}\}$. We assert that for each $y\in Y$,
$\{T_{n}y\mid n\in\mathbb{N}\}$ is a bounded subset of $l^{\infty}$.
Let $y\in Y$. Since $T:Y\rightarrow l^{\infty}$ is a well-defined
linear map, $Ty\in l^{\infty}$. That is, $M_{1}:=\sup_{k\in\mathbb{N}}|g_{k}(y)|<\infty$.
For each $n$, we have that $||T_{n}y||_{\infty}=\sup_{k\leq n}|g_{k}(y)|\leq M_{1}$.
This shows that $\{T_{n}y\mid n\in\mathbb{N}\}$ is a bounded subset
of $l^{\infty}$. By Uniform Boundedness Principle, $M_{2}:=\sup_{n}||T_{n}||<\infty$.
Let $y\in Y$ be arbitrary. For each $n\in\mathbb{N}$, we have that
$\sup_{k\leq n}|g_{k}(y)|=||T_{n}y||_{\infty}\leq M_{2}||y||$. It
follows that $||Ty||_{\infty}=\sup_{k}|g_{k}(y)|\leq M_{2}||y||$.
Hence, $T$ is bounded.


问:
Approximating Large Nonlinear Integer Optimization
Let $Q \in \mathbb{R}^{n \times n}$ be a positiv semi-definite matrix, let $K \in \mathbb{N}$ and let  $u \in \mathbb{R^{n}}$.
I need to solve the integer programming problem:

$$\max_{x \in \mathbb{N}^n} x^Tu - x^TQx \\ \text{s.t.} \sum_{i=1}^{n} x_{i} \le K $$

Here $n$ is very large, too large to allow for any of the standard algorithms to converge in reasonable time.

An approximate solution would already be good enough, but I am unsure how to get it - unfortunately, discrete optimization is really not my field of expertise.

I was thinking about some kind of dimensionality reduction of $Q$ (maybe SVD?) since $Q$ has mostly entries that are close to zero. Another idea I had was to repeatedly fix random entries of $x$ to be zero and optimize over the remaining ones to see which ones are reasonable for a gobal optimum (after all, only at most $K$ can be different from $0$ due to the linear constraint).

Finally, permutation-based "Pseudo-Gradient-Descent" comes to mind (we can easily check how the objective function behaves under small changes of $x$).

Is there a better way to tackle this? Of course, it is clear to me that we have no guarantee to truly find the maximum unless we do standard mixed integer quadratic programming.

Are there more sophisticated approaches?

答:
You could try the following greedy heuristic with at most $K$ steps.  Initialize $x_i=0$ for all $i$.  In each step, loop over $i\in\{1,\dots,n\}$ to find the $i$ whose incremental improvement to the objective value is largest, and update this $x_i$ to $x_i+1$ if the improvement is positive.  This yields a feasible solution.

From any feasible starting solution, you can optionally perform a local search on (random?) pairs $(x_i,x_j)$ to see if perturbing to $(x_i+1,x_j-1)$ or $(x_i-1,x_j+1)$ improves the solution.


问:
Proving that the Laguerre polynomials do indeed solve the differential equation
I am trying to show that the Laguerre differential equation, given in my homework problem as

$xL''_n(x)+(1−x)L'_n(x)+ nL_n(x) = 0$,

is indeed solved by the Laguerre polynomials in their closed sum form:

$L_n(x)=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k$

I've tried substituting the sum into the equation but I haven't been able to resolve it:

$xL''_n(x)+(1−x)L'_n(x)+ nL_n(x)$

$ = x \sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}k(k-1) x^{k-2} + (1-x)\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!}k x^{k-1} +n\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k$

$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [xk(k-1)x^{k-2}+(1-x)kx^{k-1}+nx^k]$

$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [(k^2-k)x^{k-1}+kx^{k-1}-kx^k+nx^k]$

$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [k^2 x^{k-1}+(n-k)x^k]$

$=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} [x^{k-1}(k^2+(n-k)x)]$

What am I doing wrong? How would I show that this equals zero? Or is this the completely wrong approach? I've seen various similar questions here but they always either work with a different version of the differential equation, or with recurrence relations - which I am hesitant to use, since my homework problem has not introduced the recurrence properties of the Laguerre polynomials so I'd have to derive and prove them before being allowed to use it in my proof.

Beyond the closed sum form I am also given the Rodrigues form of the polynomials:

${\displaystyle L_{n}(x)={\frac {\mathrm {e} ^{x}}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} x^{n}}}{\bigg (}x^{n}\mathrm {e} ^{-x}{\bigg )}}$

But substituting those into the differential equation only leads me to terms with $L_{n+1}$ and $L_{n+2}$.

Can anybody tell me how to approach this? Is there any way to prove that the polynomials in either of these two forms satisfy the above differential equation? Or is there no way around the recurrence relations and shall I prove these first, even though they were mentioned neither in my lecture nor the homework problem?

答:
I shouldn't do your homework for you, but I'll  verify it the fast way, which your teacher appears to hint at: the Sheffer sequence. See wikipedia.

The point is it's much easier to use operator (Heaviside) calculus, and the tasteful Rodrigues representation
$$  L_{n}(x)={\frac { e  ^{x}}{n!}} \partial^n   (  e  ^{-x} x^n )=
{\frac { (\partial-1)^n } {n!}}~~  x^n.
$$
The equality holds by virtue of (NB operators acting on everything to their right),
$$
 \partial ~~  e^{-x} f(x) = e^{-x} (\partial-1) f(x) ~,
$$
hence,

$$
e^x \partial ~  e^{-x} = \partial-1, ~~~\leadsto ~~~ e^x \partial^n ~  e^{-x} = (\partial-1)^n~,
$$
by intercalating $e^{-x} e^x=1$s among powers of the gradient.

Now, likewise, easily show the identity (recall the commutator of quantum mechanics!)
$$
\bbox[yellow]{x(\partial-1)^k = (\partial-1)^k x - k(\partial-1)^{k-1}}.
$$

We then, for $n>1$, plug this representation into Laguerre's second order linear equation, multiplied by $(n-1)!$,
$$
{1\over n} x (\partial -1)^n (\partial^2 x^n -\partial x^n) + {1\over n} (\partial -1)^n  (\partial x^n + n x^n )\\
= x (\partial -1)^{n+1}  x^{n-1}    +  (\partial -1)^n  (x^{n-1}+ x^n) \\ = (\partial -1)^{n+1}(x   x^{n-1}) -(n+1) (\partial -1)^{n} x^{n-1} +  (\partial -1)^n  (x^{n-1}+ x^n) \\ =
(\partial -1)^{n} \Bigl((\partial -1) x^n -(n+1) x^{n-1} +x^{n-1}+ x^n\Bigr )=0.
$$
In the penultimate line, we used the above identity for $k=n+1$.

So, the Rodrigues expression satisfies the Laguerre equation, alright. Heaviside calculus is something effective you should be able to use, anyway.


问:
Side of a square attached to two triangles
Solve for $l$

I tried to solve this for $l$ with trigonometry, but I stopped at $l^2+4-2l\cdot \cos(x)=16, l^2 + 4 +2l\cdot \sin(x)=25$, where $x$ is the angle of the left triangle between 4 and $l$.
So I've tried with Heron's, and got stuck with some square roots...

According to WolframAlpha, the answer must be something around 3.71374

Any ideas?

------SOLVED----------

Okay, so, thanks to Narasimham I've seen that I took cosine formula wrong, and thanks to cosmo5 I've been able to solve it, here's the solution:

Start with cosine formula
\begin{matrix}
5^2 & = & l^2+4-2\cdot2\cdot l\cdot \cos(270^\circ - \alpha))\\
4^2 & = & l^2+4-2\cdot2\cdot l \cdot \cos(\alpha)
\end{matrix}

Rearrange teh two equation
\begin{matrix}
441-42l^2+l^4 & = &16l^2\cdot\sin^2(\alpha))\\
144-24l^2+l^4 & = &16l^2 \cdot \cos^2(\alpha)
\end{matrix}

find this quadratic formula
\begin{matrix}
585-66l^2+2l^4=16l^2
\end{matrix}

and then found this solutions
\begin{matrix}
+\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
+\sqrt{\dfrac{41-\sqrt{511}}{2}}\\
-\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
-\sqrt{\dfrac{41-\sqrt{511}}{2}}
\end{matrix}

Then, discart the negative ones
\begin{matrix}
\sqrt{\dfrac{41+\sqrt{511}}{2}}\\
\sqrt{\dfrac{41-\sqrt{511}}{2}}
\end{matrix}

答:
On Mathematica; Using Cosine formula of a triangle, three unknowns and three variables. Numerical solution.

```<code>  eq={20.-16 Cos[al]==L^2, 29-20 Cos[bt]==L^2,41.-40 Cos[al+bt]== 2 L^2};   NSolve[eq,{al,bt,L}]
</code>```

al->0.829607,bt->0.140595,L->3.03271},{al->2.4004,bt->1.71139,L->5.63938}}`

I have checked by approximate Geogebra construction. Two cases arise when the node is  inside and the other when it is outside the square. Rough sketch attached including all trigonometric parameters.

To me it is convincing enough.

Body:
On Mathematica; Using Cosine formula of a triangle, three unknowns and three variables. Numerical solution.

  eq={20.-16 Cos[al]==L^2, 29-20 Cos[bt]==L^2,41.-40 Cos[al+bt]== 2 L^2};   NSolve[eq,{al,bt,L}]

al->0.829607,bt->0.140595,L->3.03271},{al->2.4004,bt->1.71139,L->5.63938}}`

I have checked by approximate Geogebra construction. Two cases arise when the node is  inside and the other when it is outside the square. Rough sketch attached including all trigonometric parameters.

To me it is convincing enough.


答:
The first "subbase" is {{1}, {2}, {3}}.  The set of all possible unions are {{1}, {2}, {3}, {1,2}, {1, 3}, {2, 3}, {1, 2, 3}}.
The set of all possible intersections is {{1}, {2}, {3}, {}}.

The set of all possible unions and intersections is {{}, {1}, {2}, {3}, {1,2}, {1, 3}, {2, 3}, {1, 2, 3}}.  That is a topology, in fact it contains all subsets, the "discrete topology" for {1, 2, 3}.

The second subbase is {{1, 2}, {3}}.

The set of all possible unions is {{1, 2}, {3}, {1, 2, 3}}.

The set of all possible intersections is {{1, 2}, {3}, {}}.

The set of all unions and intersections is {{}, {3}, {1, 2}, {1, 2, 3}}. Again, that is a topology for {1, 2, 3}.

If you could not see that, are you clear what a "topology" for a set is?  A topology for a set, X, is a collection of subsets or X that has four properties:

It contains X itself.

It contains the empty set, {}.

It contains all unions of its sets.

It contains all finite intersections of its sets.

Title: Very confused about the definition of subbasis for a topology in Munkres's book

Body:
In Munkres's Topology textbook, it says,

Definition

A subbasis S for a topology on X is a collection of subsets of X whose union equals X.

The topology generated by the subbasis S is defined to be the collection T of all unions of finite intersections of elements of S.

I am having a very hard time understanding the first sentence. Let's say $X = \{1, 2, 3\}$. Then let $A = \{\{1\}, \{2\}, \{3\}\}$, or $A = \{\{1, 2\}, \{3\}\}$. Either way, $A$ is a collection of subsets of $X$, and the union of the elements in $A$ is $X$, but $A$ does not seem like a subbasis here...

What did I misunderstand here? What should be the correct interpretation of "a collection of subsets of X whose union equals X"?

Thank you in advance!

Edit:

So turned out $A$ is a subbasis in both cases, thanks @Surb for your comment.

But here is what I don't understand (which made me mistakenly think A may not even be subbasis in the first place). If $A = \{\{1\}, \{2\}, \{3\}\}$, then wouldn't "the collection T of all unions of finite intersections of elements of $A$" be {$\varnothing$}, since any intersection of any two elements in $A$ is empty? But since T is a topology, T must also contain $X$, but, in this case, it does not?


问:
Why doesn't the formula for the PMF on geometric distributions make use of ${\mathbb N}$?
If you look online you'll find that the formula for the PMF on geometric distributions looks like this:

${\displaystyle \Pr(X=k)=(1-p)^{k-1}p}$
for $k = 1, 2, 3... $

Why don't people just write
$k, {k \in \mathbb N}$

答:
One reason people write $k=1,2,3,\dots$ instead of $k\in \mathbb N$ is because there is no consensus about whether $\mathbb N=\{1,2,\dots\}$ or $\mathbb N=\{0,1,2,\dots\}$. The potential for confusion is especially high with the geometric distribution, because some people define it like you did, while others define the geometric distribution by
$$
P(Y=k)=(1-p)p^k,\qquad k=0,1,2,\dots
$$
Both conventions appear on the Wikipedia page side by side. Just looking at the $k\in \mathbb N$ part, you would have to know which convention for $\mathbb N$ the author was using to know which geometric distribution condition they were using.

(I personally prefer $0\in \mathbb N$. Anyone who works with power series a lot may agree, since powers of $x$ can be any nonnegative integer. Furthermore, if the natural numbers are for counting, and you want to count the number of elements of a finite set, then zero is a possible answer!)


问:
Probability that $n$ random walks (1D) intersect a single point
I am struggling finding a relation to determine the probability $n$ random walks (1D) intersect in a single point at step $s$. In the method below my attempts. My method is somewhat intuitive based. I am looking for more rigorous proof.

note:
This question arises from someone who claims that a matching cumulative digit sum of: $\pi$, $e$ and $\varphi$ (golden ratio) is unique and "cosmological" [1]. I tend to disprove it. This digit sum can be seen as a random walk if the constants are normal (every digit occurs with same frequency).

Method:

For every step $s$ on the random walk we can determine the probability density function if we know the standard deviation on every step $s$.

The standard deviation of a single step can be calculated it's a discrete uniform distribution "equally likely outcomes", where $q$ is the number of outcomes e.g. the number of digits $[0,1,2,3,4,5,6,7,8,9]$, $q=10$:

$$\sigma=\sqrt{\frac{q^{2}-1}{12}} $$

All the (1D) random walks start in the origin for this example. The standard deviation will grow with every step $s$, the variance is proportional to the number of steps [2].

$$Var(s)=s \cdot \sigma^{2}$$

$$\sigma(s)=\sqrt{s} \cdot \sigma$$

While the bins grow rapid I assume a normal approximation of the Binomial distribution.

$$f(x)=\int_{-\infty}^{\infty} {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}}dx $$

The probability that $n$ random walks intercept in a single point is (not sure):

$$p(s)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^n dx $$

With help of Wolfram Alpha [3] the solution is found for $n=3$ meaning the probability of $3$ point intersecting random walks.

$$p(s)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \frac{1}{s}$$

The total probability $p(n)$ is proportional to the reciprocal sum of $s$. So the total probability is proportional to the harmonic series:

$$p(n)=\frac{1}{2 \sqrt{3} \ \pi \ \sigma^{2}} \cdot \sum_{s=1}^{\infty}\frac{1}{s}$$

This series diverges, meaning there are infinate point intersections of $n=3$ random walks. So a matching cumulative digit sum of $\pi$, $e$ and $\varphi$ is not unique, probability $\sim 8 \%$ for the first 1200 digits (see graph).

Question

Does anyone know the general formula for the probability $p(n)$ that $n$ random walks (1D) intersect a single point?

```<code>import numpy as np

#Elements of digits [0,1,2,3,4,5,6,7,8,9] rescaled to fit random walk
array=[-9,-7,-5,-3,-1,1,3,5,7,9]

#steps, in single random walk, x walks to intercept, number of trial to find intercept
steps=2500
xwalks=3
trials=1500

#Set output array to zero
count=np.full([steps],0)

for n in range(trials):

    #Identify initial array, set total array to zero
    w0=np.random.choice(array,steps)
    w0=np.cumsum(w0)
    total=np.full([steps],0)

    #Select x random walk check for intercept
    for m in range(xwalks-1):

        #Next current random walk
        w=np.random.choice(array,steps)
        w=np.cumsum(w)

        #Compare previous and current random walk
        eq=np.equal(w0,w)
        eq=eq.astype(int)

        #Count intercepts
        total=total+eq

        #Set current walk to previous
        w0=w

    #Sum all interceptions for all trials
    count=count+np.where(total==(xwalks-1),1,0)

#Print output
print(count)
print(np.sum(count))
print(np.sum(count)/trials)
</code>```

答:
After studying I think I answered my own question. The method has similarities corresponding with: SE.

In my earlier answer I demonstrated how to determine the probability of $n$ intersecting random walks with $2$ or more likely outcomes per step. The method used a normal approximation making an error in the probability for small numbers of steps.

The probability of intersection for small number of steps can be calculated with an discrete method. By using generating functions, a solution is found and is based upon: SE.

The probabilities of a single random walk at step $s$ is determined by the coefficients of the generating function below. Where $d$ is the number of equal likely possibilities per step e.g. $d=2$ for random walk: $[-1,1]$:

$$\left( \frac{1}{d}x^{0}+\frac{1}{d}x^{1}+\frac{1}{d}x^{2}...+\frac{1}{d}x^{d-1} \right)^{s}=\frac{1}{ d^{s}} \left( \frac{1-x^{d}}{1-x} \right)^{s}$$

The coefficients can be calculated by applying the generalized binomial theorem. Where $n$ is the number of intersecting random walks. $p(s)$ is the total probability for an intersection occurring on step $s$:

Discrete method, $s<20$:
$$\boxed{p(s)=\sum_{i=0} \left[\frac{1}{d^s}  \sum_{r=0} (-1)^{r} \binom{s}{r} \binom{s+i-dr-1}{i-dr} \right]^{n}}$$

The discrete method is realistic for $s<20$ steps. For more than $s\geq20$ the normal approximation is more efficient.

$$p(s)=\int_{-\infty}^{\infty} \left[ {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x}{\sigma }}\right)^{2}} \right]^n dx$$

An empirical formula was found see; my question and first own answer (confirmation here: SE). The standard deviation is determined from a discrete uniform distribution, growing with $\sqrt{s}$ per step while a random walk SE:

Continuous method, $s\geq20$:
$$\sigma=\sqrt{\frac{(d^{2}-1)}{12}}$$
$$\boxed{p(s)= \frac{1}{\sqrt { n} \cdot (\sigma \sqrt{2 \pi})^{(n-1)}  } \cdot s^{-\frac{1}{2}(n-1)}} $$

Both functions: discrete and continuous have been verified in a simulation. The intersections of $n=4$ random walks where counted. Random walks were used with $d=4$ equal likely outcomes per step $[-3,-1,1,3]$. The graphs show the discrete method is practical and accurate for $s<20$ the normal approximation is useful for $s\geq20$.

Finally, I want to summarize that the total probability: the sum over all steps $p(n)$ can be calculated by using the $p-$series (or Riemann Zeta function):

$$p(n)=\frac{1}{\sqrt { n} \cdot (\sigma \sqrt{2 \pi})^{(n-1)}   } \cdot \sum\limits_{s=1}^{\infty}s^{- \frac{1}{2}(n-1)}$$

$$p(n)=\frac{1}{\sqrt { n} \cdot (\sigma \sqrt{2 \pi})^{(n-1)}  } \cdot\zeta(\small{\frac{1}{2}(n-1)} \normalsize) $$

For $n\leq3$ point intersecting random walks the series diverges resulting in infinite intersections. For $n\geq4$ the series converges resulting in finite number of intersections. This recurrence and transience is also seen with random walks returning to the origin (one more degree of freedom) and related to Markov processes.

Interesting is that $n=2$ intersecting random walks coincides with the critical line on the Riemann Zeta function.


问:
Is a linear map of norm $1$ always an isometry?
Let $(E,\|.\|_{E})$ and $(F,\|.\|_{F})$ be two normed spaces and let $f:E\longrightarrow F$ a linear map such that $\|f\|=1$. I don't know if this means that $f$ is an isometry $(\|f(x)\|_{F}=\|x\|_{E}, \,\, \forall x\in E)$. Thanks </p>

答:
There are plenty of endomorphisms of norm $1$ which are not injective. I give an example.

Let $(E,\|\cdot\|_E)=(F,\|\cdot\|_F)=(\ell_\infty, \|\cdot\|_\infty)$ and consider left shift operator $T\colon \ell_\infty\to\ell_\infty$ defined by $T(x_1,x_2,\ldots)=(x_2,x_3,\ldots)$. Clearly, $\|T\|=1$ but there is a whole subspace which goes to $0$ under $T,$ namely $\{(c,0,0,\ldots)\colon c\in \mathbb{K}\}$. Hence not only you have a vector ${\bf v}\in\ell_\infty$ s.t. $$\|T{\bf v} \|_\infty\neq\|{\bf v}\|_\infty$$ but you can even pick it so its norm equals your favourite number, still positive though.