问:
Simple but challenging probability identity
I am reading E.T. Jayne's Probability Theory book, and I am stuck on a "trivial" fact that he mentions at the end of the derivation of the sum rule in Chapter 2 (page 33):

He states that it is a trivial fact that given

$$
\begin{align}
w(AB|C) = w(A|BC)w(B|C) = w(B|AC)w(A|C) \\
w(A\bar{B}|C) = w(A|C)w(\bar{B}|AC)
\end{align}
$$

one can show that

$$
w^{m}(A|C) - w^m(A\bar{B}|C) = w^m(B|C) - w^m(B\bar{A}|C)
$$

where w is a probability measure.

Can someone demonstrate how this result follows?

Note, I think it's clear in the case that $m = 1$, both sides of the final equation are equal to $w(AB|C)$. I just don't see how you can derive this from the given equations for all powers of $m$.

答:
Suppose $x,y,z,w$ are complex numbers and $x^m - y^m = z^m-w^m$ for both $m=1$ and $m=2$.
Then I claim either $x=y$ and $z=w$, or $x=z$ and $y=w$.

Proof: $$x^2 - y^2 - z^2 + w^2 = 2(w-y)(w-z) + (x-y-z+w)(x+y+z-w)$$
so $(w-y)(w-z) = 0$, i.e. $w=y$ or $w=z$.  If $w=y$ we must have $x=z$ and if $w=z$ then $x=y$.

Thus in your case, it is impossible to have your equation true for both $m=1$ and $m=2$, let alone all positive integers $m$, unless either $w(A\mid C) = w(B \mid C)$ and $w(A \overline{B}\mid C) = w(B\overline{A} \mid C)$, or $w(A \mid C) = w(
A \overline{B} \mid C)$ and $w(B \mid C) = w(B \overline{A} \mid C)$.


问:
Verify the markov property for a sum of processes
I'm given such a problem and I'm seeking for help.

Let {$X_n:n>1$} be an i.i.d. process with Poisson marginal PMF $p_x(k)$ = $e^{-\lambda}\frac{\lambda^k}{k!}$, $k \geq 0$, and let $N_l = \sum_{k=1}^{l}X_k$ and where $N_0 = 0$. Is $\{N_l\}_{l\geq 0}$ a Markov process? Explain in detail.

I'm wondering what the random process with poisson marginal pmf means, does it mean that X is a poisson process? Then if this is true, then I can try to prove the sum of poisson processes is also poisson, then N is a homogeneous poisson process, next homogeneous poisson process is markov process. I'm not sure whether this is the right way.

Can someone give me some help on this one? Thanks.

答:
You can use the following fact:

If $(X_n)_n$ are independent random variables and the stochastic process $(Y_n)_n$ respects an identity of the form $Y_{n+1} = f_n(Y_n,X_{n+1})$, then $(Y_n)_n$ is a markov process.

Proof:

$\mathbb{E}[\phi(Y_{n+1})|Y_1,\ldots,Y_n] = \mathbb{E}[\phi(f_n(Y_n,X_{n+1}))|Y_1,\ldots,Y_n] = \mathbb{E}[\phi(f_n(y,X_{n+1}))]_{y=Y_{n}}$, which is $\sigma(Y_n)$-measurable.

Can you finish now?

ContentLicense: CC BY-SA 4.0


问:
Deducing the approximation using double greater-than
I've been struggling against the below deduction of the approximation.

$0<a\in \mathbb{R}$

$d:=\sqrt{x^2+y^2} \hspace{0.3cm}$ where $ \hspace{0.1cm} 0\leq x,y\in \mathbb{R}$

$a\ll d$

$$\frac{1}{\sqrt{1-\frac{2ax}{d^2}}}+\frac{1}{\sqrt{1+\frac{2ax}{d^2}}}\simeq \left(\left( 1+\frac{ax}{d^2}\right)-\left(1-\frac{ax}{d^2}\right)\right)$$

Can anyone tell me some hint(s) so that I can deduce it myself?

答:
Hint:

Can you show $\frac{2ax}{d^2} \ll 1$?

If so, then since $(1+\epsilon)^{-1/2} = 1 -\frac{\epsilon}{2} + O(\epsilon^3)$, you can compute

$$
\frac{1}{\sqrt{1+\frac{2ax}{d^2}}} =
\left ( 1 + \frac{2ax}{d^2} \right )^{-1/2} =
1 - \frac{ax}{d^2} + O \left ( \left ( \frac{2ax}{d^2} \right )^3 \right )
$$

And similarly for $\left ( 1 - \frac{2ax}{d^2} \right )^{-1/2}$.

Do you see how to finish the problem from here?

I hope this helps ^_^


问:
Isomorphisms and Partitions of Permutation
Let $X$ be a finite set, and define $G$ to be the symmetric group $S_X$.

Now, for any $A \subset X$, we define $B$ as the complement of $A$ in $X$, i.e. $B = X \setminus A$. Define $H = \{g \in G \mid g(A) = A \}$, i.e. the subgroup of permutations of $G$ under which $A$ is invariant.

Show that $H \cong S_A \times S_B$.

Intuitively, I can see this being true, in that $A, B$ partition $X$, and so for any $h \in H$, it's clear that the "part of" $h$ permuting $A$ is isomorphic to $S_A$ and the "rest of" $h$ is isomorphic to $S_B$. This is quite a hand-wavy argument, and I'm wondering if there's a way to formalize it.

答:
Hint:  It suffices to show $S_AS_B=H, S_A\cap S_B=\{e\}$ and $S_AS_B=S_BS_A$.

Note: there's a natural way to embed $S_A$ and $S_B$ into $H$.


问:
number of positive integer solutions of $x+y+z+w=24$
Determine the number of positive interger solutions of $x+y+z+w=24$ such that $x\leq 6, y \leq 7, z\leq 8, w\leq 9$
My try:
I used generating polynomial as
$$f(x)=(x+x^2+\cdots+x^6)(x+x^2+\cdots+x^7)(x+x^2+\cdots+x^8)(x+x^2+\cdots+x^9$$
$\implies$
$$f(x)=x^4(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}$$
We need to collect the coefficient of $x^{20}$ in $(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}$
So we can write $$(1-x^6)(1-x^7)(1-x^8)(1-x^9)(1-x)^{-4}=(1-x)^{-4}(1-(x^6+x^7+x^8+x^9)+(x^{13}+x^{14}+2x^{15}+x^{16}+x^{17})-..)$$
Also using $$(1-x)^{-4}=\sum_{k=0}^{\infty}\binom{k+3}{k}x^k$$
We get required coefficient of $x^{20}$ as
$$\binom{23}{3}-\left(\binom{14}{11}+\binom{15}{12}+\binom{16}{13}+\binom{17}{14}\right)+\left(\binom{6}{3}+\binom{7}{3}+2\binom{8}{3}+\binom{9}{3}+\binom{10}{3}\right)=83$$
But the answer is not matching.

答:
Your answer is right here is an easier way to go about this which may bring less calculation errors

set $x=6-u.y=7-v,z=8-t,w=9-p$

then with $0\le u\le 5,0\le v\le 6,0\le t\le 7,0\le p\le 8 $ we have to find the number of solutions of $$u+v+t+p=6$$ but $u\neq 6$ hence number of solutions is $\binom{6+4-1}{4-1}-1=83$


问:
Need help proving a function is absolutely integrable.
Prove that $f(x)=\frac{sinx}{x^2}$ is absolutely integrable on $[\pi,\infty)$.

答: $\left \vert \frac{\sin x}{x^2} \right \vert \leq \frac {1}{x^2}$ and $\frac {1}{x^2}$ is integrable over that domain.


问:
Prove that if $c$ is any element in a field $\mathbb{F}$ and $f(x) \in \mathbb{F}$, then $f(x) \equiv f(c)$ mod $(x-c)$
I believe I'm supposed to produce something of the from $f(x)-f(c) = (x-c)p(x)$ where $p(x) \in \mathbb{F}[x]$, however I'm not sure where to start.

答:
Start by using the division algorithm to show that there exists polynomials $g(x), h(x)\in \mathbb{F}[x]$ such that $f(x) = (x-c)g(x) + h(x)$ and $\deg{h(x)} < \deg{(x-c)} = 1$.

Thus, $h(x)$ must be a constant, lets call it $a$.

Then, plug in the value $c$ for $x$ to get, $f(c) = (c-c)g(x) + a = 0\cdot g(x) + a = a$

$\Rightarrow f(c) = a$

$\Rightarrow f(x) = (x-c)g(x) + h(x) = (x-c)g(x) + a = (x-c)g(x) + f(c)$

Thus, you have proved your result.


问:
$\forall \phi\in C_o^{\infty}(\Omega)$, $\lvert\tilde{f}(\phi)\rvert\le \lVert \phi\rVert_{L^{p'}}\implies f\in L^p$
Let $\Omega\subset\mathbb{R}^d$ open and $\tilde{f} : L^{p'}(\Omega)\to \mathbb{R}$ defined by $\tilde{f}(\phi) :=\int f\phi$ with the property that $\lvert\tilde{f}(\phi)\rvert\le \lVert \phi\rVert_{p'}$ for all $\phi\in C_0^{\infty}(\Omega)$. Now I want to show that $\tilde{f}\in (L^{p'})^*$ so I think I have to show that $\tilde{f}$ is bounded. I almost have this because $\sup\limits_{\phi_\in C_0^{\infty}(\Omega)}\frac{\lvert\tilde{f}(\phi)\rvert}{\lVert \phi \rVert}\le 1$. Also I know that $C_0^{\infty}(\Omega)$ is dense inside $L^{p'}(\Omega)$. But then I am not sure how to show that $\tilde{f}$ is bounded, i.e. $\sup\limits_{g\in L^{p'}(\Omega)}\frac{\lvert\tilde{f}(g)\rvert}{\lVert g \rVert}\le 1$.

答:
I suppose that $f$ is assumed to be a function in $L^1_{\text{loc}}(\Omega)$, that is, $f$ is measurable, and integrable on any compact subset of $\Omega$.

Incidentally this is one of the biggest spaces of functions of interest in Analysis since it contains every $L^p(\Omega)$.   But still, if $f$ lies in $L^1_{\text{loc}}(\Omega)$, then $f\phi$ is integrable for every $\phi$ in $C^\infty_c(\Omega)$ (smooth functions with compact support).

Ok, so we are given that the functional
$$\tilde f: \phi\in  C^\infty_c(\Omega) \mapsto \int_\Omega f\phi\in \mathbb R$$
is continuous relative to the $p'$-norm on $C^\infty_c(\Omega)$, and we need to prove that $f$ lies in $L^p(\Omega)$.

Since $\tilde f$ is continuous, and $C^\infty_c(\Omega)$ is dense in $L^{p'}(\Omega)$, then $\tilde f$  extends uniquely to a
continuous linear functional $F$ on $L^{p'}(\Omega)$.  Since the dual of $L^{p'}(\Omega)$ is isomorphic to
$L^{p}(\Omega)$, there is some $f_1$ in $L^{p}(\Omega)$, such that
$$
  F(g) = \int_\Omega f_1g, \quad \forall g\in  L^{p'}(\Omega).
  $$
In particular,  for every $\phi$ in $C^\infty_c(\Omega)$  we have that
$$
  \int_\Omega f\phi =
  \tilde f(\phi) =
  F(\phi) =
  \int_\Omega f_1\phi,
  $$
so
$$
  \int_\Omega (f-f_1)\phi =   0,
  $$
and we conclude that $f=f_1\in L^{p}(\Omega)$.


问:
Can we say that $\frac{0}{0}$ is every number?
Suppose we have an equation $ab=0$. This equation is true when statements $a=0$ or $b=0$ are true.

If $a=0$, then $b=\frac{0}{0}$. That means $b$ could be any number for $ab=0$ to be true. If the set which groups all the numbers is the complex set, then $b$ will every number within $\mathbb{C}$, so $\forall z\in\mathbb{C}:b=\frac{0}{0}=z$. Therefore, $\frac{0}{0}$ is every number.

I know it really is not defined as number but conceptually it is every number, right?

Is this right, or am I missing something?

答:
There's actually nothing too wrong with taking $\frac{0}{0}$ to have every real number as a value. That is, as you point out, the most logical thing to do when you consider the defining equation for division:

$$b \left(\frac{a}{b}\right) = a$$

which gives, when $a = b = 0$,

$$0 \left(\frac{0}{0}\right) = 0$$

and any substitution for the putative quantity $\frac{0}{0}$ will work, and make this equation hold good true (something that does not happen when $a \ne 0$ and $b = 0$!). Moreover, in many geometric limiting contexts, this makes a lot of sense, too: if you do this, the graph of the equation

$$y = \frac{0}{x}$$

is a cross, which is the exact limiting case of the hyperbolas

$$y = \frac{a}{x}$$

as $a \rightarrow 0$.

So why isn't it done? Simple: We like to have division be a function, and not just a one-to-many (many many!) relation. Having it take all real values at even one point would destroy that. So the standard option is the alternative: limit the domain so that $(0, 0)$ is absent therefrom. This leaves $\frac{0}{0}$ undefined.

But there is nothing inherently wrong with considering such non-function operations: the inverse trigonometric functions, the square root, etc. are all most properly thought of as one-to-many relations, and they pop up in many other areas as well.

ContentLicense: CC BY-SA 3.0


问:
Chance-Constrained Optimization
Let $x \in \mathbb{R}^n$ and  $\xi \in \Xi \subset \mathbb{R}^n$ be a random vector with discrete probability distribution $p(\cdot)$, assuming a finite number of values $\{\xi_i\}_{i = 1}^S$. Let $A \in\mathbb{R}^{n\times n}$ be a Positive Semi-definite matrix.

Formulate the following chance constrained optimization:

$$\inf_{x}x^TAx\\ \text{ s.t. } \mathbb{P}[\xi^\top x>0]>1-\delta $$

with $\delta\in[0,1]$.

What is the correct way to solve this problem? Does a closed-form solution exist?

What I tried :

Note that the constraint set can be rewritten as

$$
\sum_{i=1}^S\mathbb{1}\{\xi_i^\top x>0\}p(\xi_i) > 1-\delta.
$$

The constraint set can be decomposed in $2^S$ disjoint sets. Based on the $p(\xi_i)$ only some of these sets will satisfy the above inequality. Thus I tried to solve each of the cases and took the minimum of these solutions.

Is it correct? I suspect that this is not the best way to proceed and there should be a way to transform the problem into a Mixed Linear programming problem, but I don't know how to do that exactly.

Any help would be greatly appreciated. Thanks!

答:
You can introduce a binary variable $y_i$ for each scenario $S$ indicating if $\zeta_i^Tx > 0$. Let $\varepsilon$ be the smallest positive value for which $P(\zeta_i^Tx > 0) = P(\zeta_i^Tx \geq \varepsilon)$, and let $y_i \in \{0,1\}$ be a binary variable, then the constraint:
$$\zeta_i^Tx \geq \varepsilon - (1-y_i)M $$
enforces that $y_i$ can take the value $1$ only if $\zeta_i^Tx \geq \varepsilon$. So if you combine that constraint with $\sum_i p_i y_i \geq 1-\delta$ (or with a small number added to the right hand side to get a strict inequality), you can model your constraint.

Title: Chance-Constrained Optimization


问:
On partial ordering of integers
In the book by A.N. Kolmogorov and S.V. Fomin: "Real Analysis" (Dover Edition), Example 4 on p. 20 in the Section on Ordered Sets and Ordinal Numbers says that

"The set of all the integers greater than 1 is partially ordered if $a \leq b$ means that "b is divisible by a" i.e. $a|b$."

The text adds that:

"An element $a$ of a partially ordered set is said to be ${maximal}$ if $a \leq b$ implies $b=a$ and ${minimal}$ if $b \leq a$ implies $b=a$. Thus in Example 4 every prime number (greater than 1) is a minimal element."

I would like to verify if my interpretation of this Example 4 is correct:

Since for example, 2 divides 4, 3 divides 6, 4 divides 8, 5 divides 10, ... we have $ 2 \leq 4, 3 \leq 6, \ldots $. This could be expanded as $\pm 2$ divides $\pm 4$, $\pm 3$ divides $\pm 6$, $\pm 4$ divides $\pm 8$, $\pm 5$ divides $\pm 10$, ... we have $ \pm 2 \leq \pm 4, \pm 3 \leq \pm 6, \ldots $.  We conclude the integers are partially ordered. Is this a correct (perhaps too detailed) interpretation?

Definitions of ${maximal}$ and ${minimal}$ elements of a partially ordered set: Taking the integers as a partially ordered set, the text says that "every prime number (greater than 1) is a minimal element." So, the prime numbers are elements of a partially ordered set the integers. Let $a = 3$ (could be any other prime number) so $a = 3 \leq b= 3$ means that $a= 3$ divides $b=3$ according to the meaning of $a \leq b$ in Example 4 and it does $3/3 = 1$ but we also have that $b=3 \leq a=3$ so $b=3$ divides $a =3$ so we have $b=a=3$. Therefore $a= 3$ is a minimal element of the integers with respect to the meaning of the relation $\leq$ in that Example 4.

Now, can we define a maximal element $a$ in the integers that is if $a \leq b$ then $b=a$? I believe not but I may be mistaken or do not interpret the definition of maximal element properly. Any comment on that would be much appreciated.

答: $3$ being minimal means for all $x$ greater than $1$  if $x$ divides $3$ then $x=3.$ Your version doesn't mention the "for all $x$" part at all so looks wrong (to me).

Title: On partial ordering of integers


问:
Find a metric $D$ on $X=\mathbb{R}^2-\{(0,0)\}$ so $(X,D)$ is complete and topologically equivalent to $(X,d)$
I'm trying to find a metric $D$ on the set of $\mathbb{R}^2$ without the origin, so that it forms a complete metric space. There is also the restriction that this metric space has the same open sets of $(X,d)$ where $d$ is the Euclidean metric on $\mathbb{R}^2$.

In other words, I wish to find a metric $D$ on $X=\mathbb{R}^2-\{(0,0)\}$ so $(X,D)$ is complete and topologically equivalent to $(X,d)$

My attempts so far have revolved around the function $$f(x,y) = \frac{1}{x^2+y^2},$$
so that the Cauchy sequences in $\mathbb{R}^2$ converging to the origin under the Euclidean metric are no longer Cauchy in $X$ under whatever metric $D$ I eventually find.

I let $u,v\in\mathbb{R}^2$. Then, I tried naively defining $D(u,v)=|f(u)-f(v)|$, but it doesn't work because $D(u,-u)=0$. Then, I tried $D=\frac{1}{(d(u,v))^2}$, but it's not defined when $u=v$.

I'm not sure what direction to take. Any help would be greatly appreciated!

答:
To help you make @yoyo's suggestion in the comments more concrete, consider the map $ F $ which sends $ (x,y) \in X $ to the point $ (r, \theta) \in \tilde{X} = (0,\infty) \times[0,2\pi) $ given by the cylindrical coordinates.  We can now take the distance between two points $ \mathbf{x}_1 = (x_1, y_1), \mathbf{x}_2 = (x_2, y_2) $ to be the distance
$$ D(\mathbf{x}_1, \mathbf{x}_2) :=  \tilde{D}(F(\mathbf{x}_1), F(\mathbf{x}_2))
$$ between their images on the cylinder.  I will leave it to you to devise the explicit formulas, if those are necessary for your problem.

Let me know if you want a hint for proving topological equivalence.


问:
Set of left cosets is a group
Let $N$ be a normal subgroup of $G$. Let $G/N = \{gN : g \in G \}$. Show that the $(G/N, \circ, 1_{G/N})$ is a group, where:

$\circ : G/N \times G/N \rightarrow G/N$ is defined by $(gN \circ hN) = ghN$
and $1_{G/N} = 1*N$.

Thanks for your help.

答:
First verify that the group operation is well-defined, that is, show that the correspondence from $G/N\to G/N$ is actually a function. Assume for $a,a',b,b'\in G$ we have $aN=a'N,bN=b'N$. We must verify that $aNbN=a'Nb'N$, or $abN=a'b'N$. This step is necessary because when we are dealing with cosets, we must ensure that our function definition depends only on the cosets, and not the coset representatives.

For this part, note that if $aN=a'N$, $bN=b'N$, then $a=a'n_1$ and $b=b'n_2$ for some $n_1,n_2\in N$. Therefore $a'b'N=an_1bn_2N=an_1bN=an_1Nb=aNb=abN$, using associativity and the fact that $N$ is normal in $G$.

From here it should be easy to verify that the three group properties hold:

Identity: $N\quad ((gN)(N)=gN)$

Inverses: $(gN)^{-1}=g^{-1}N\quad (gNg^{-1}N=gg^{-1}N=N)
$

Associativity: $gN(hNkN)=ghkN=(gNhN)kN$

Title: Set of left cosets is a group


问:
Wiener Process with conditional probability
I am currently learning Random Processes and have encountered one such exercise about Wiener process that I can't figure out why the answer is like this. The question is as follows:

Suppose W(t) is a Wiener Process such that $R_{w}(t_1,t_2) = \alpha\, min(t_1, t_2)$. $R_w(t_1,t_2)$ should stand for auto-correlation function of W(t).
Let the event B be $W(4) = 1$. Give the conditional PDF $f_{W(t)}(w|B)$ for $t > 4$.

And the answer for it is $f_{W(t)}(w|B) = \frac{1}{\sqrt{2\pi\alpha(t-4)}}e^{\frac{-(w-1)^2}{2\alpha(t-4)}}$

Can someone tell me why it is $N(1, \alpha(t-4))$? Thanks a lot.

答: I have asked my professor about this and got the answer. Basically the method is to obtain the derivative of cdf for W(t)|B to get the pdf. Then one can easily obtain that the cdf is of the form $P(Z \leq w-1)$, where Z ~ $N(0,\alpha(t-4))$.


答:
To show that the integers greater than 1 are partially ordered (with the order relation you specified), you need to check the definition of a partial order. The formal definition can be found here. Note that there are three axioms that need to be verified. In your analysis, you just showed how some of the integers are ordered, but did not prove that $\le$ is a partial order.

To show that every prime number (greater than 1) is a minimal element, note that if $a$ is prime and $b\le a$ (which is to say that $a$ is divisible by $b$), then $b$ must equal $a$. This is because the only two divisors of a prime number $a$ are $1$ and $a$. Note that $b\ne 1$ since the set for which the order is defined does not contain $1$. Thus, by the definition you gave, $a$ is a minimal element.

Title: On partial ordering of integers

Body:
In the book by A.N. Kolmogorov and S.V. Fomin: "Real Analysis" (Dover Edition), Example 4 on p. 20 in the Section on Ordered Sets and Ordinal Numbers says that

"The set of all the integers greater than 1 is partially ordered if $a \leq b$ means that "b is divisible by a" i.e. $a|b$."

The text adds that:

"An element $a$ of a partially ordered set is said to be ${maximal}$ if $a \leq b$ implies $b=a$ and ${minimal}$ if $b \leq a$ implies $b=a$. Thus in Example 4 every prime number (greater than 1) is a minimal element."

I would like to verify if my interpretation of this Example 4 is correct:

Since for example, 2 divides 4, 3 divides 6, 4 divides 8, 5 divides 10, ... we have $ 2 \leq 4, 3 \leq 6, \ldots $. This could be expanded as $\pm 2$ divides $\pm 4$, $\pm 3$ divides $\pm 6$, $\pm 4$ divides $\pm 8$, $\pm 5$ divides $\pm 10$, ... we have $ \pm 2 \leq \pm 4, \pm 3 \leq \pm 6, \ldots $.  We conclude the integers are partially ordered. Is this a correct (perhaps too detailed) interpretation?

Definitions of ${maximal}$ and ${minimal}$ elements of a partially ordered set: Taking the integers as a partially ordered set, the text says that "every prime number (greater than 1) is a minimal element." So, the prime numbers are elements of a partially ordered set the integers. Let $a = 3$ (could be any other prime number) so $a = 3 \leq b= 3$ means that $a= 3$ divides $b=3$ according to the meaning of $a \leq b$ in Example 4 and it does $3/3 = 1$ but we also have that $b=3 \leq a=3$ so $b=3$ divides $a =3$ so we have $b=a=3$. Therefore $a= 3$ is a minimal element of the integers with respect to the meaning of the relation $\leq$ in that Example 4.

Now, can we define a maximal element $a$ in the integers that is if $a \leq b$ then $b=a$? I believe not but I may be mistaken or do not interpret the definition of maximal element properly. Any comment on that would be much appreciated.


问:
Proof that if $d,d'$ are gcd($f,g$) where $f,g$ are polynomials, then $d = cd'$ for some polynomial $c$
Here is my current thought; Let $f = df', g = dg'$ for some polynomials $f',g'$

by division theorem

$f = gq + r$ for some polynomial $r,q$. We have $r = f - gq = df' - dg'q = d(f'-dg')$; hence, $d$ divides $r$; similarly, $d'$ divides $r$.

So theree are polynomial $a,b$ such that $r = da = d'b$.

My plan is to show that $r=gcd(f,g)$ as well, so I have degree($r$) = degree($da$) = degree($d'b$) = degree($d$) = degree($d'$) implies that $a,b$ just constant.  I  can write $d = \frac{b}{a}d'$

But I'm stuck at how to prove $r=gcd(f,g)$ Any Ideas ?

答: This follows directly from the definition of a greatest common divisor, which is that it is a common divisor, and every common divisor divides it. Thus as $d'(x)$ is a common divisor, and $d(x)$ is a greatest common divisor, $d'(x)|d(x)$, which is equivalent to saying that there exists a $c(x)$ such that $d'(x)c(x) = d(x)$.


问:
Method of least squares question with trigonometric function

We measure a projected signal $y\in\mathbb R$ with an analogue antenna at three different angles $x=0,\ y=1$; $x=\pi/4,\ y=0$; $x=\pi/2,\ y=1$; find $a$ and $b$ such that the function
$$p(x) = a \cos(x) + b \sin(x)$$
best approximates the data points $(x, y)$, using the method of least squares.

I don't understand how to go about this question. i tried the matrix transformation a bunch of different ways but just kept getting stuck.

答:
You have $n$ data points $(x_i,y_i)$ and you want to model
$$y_i = a \cos(x_i) + b \sin(x_i)$$ If this could help you, define $u_i=\cos(x_i)$ and $v_i \sin(x_i)$ to make the model to be
$$y_i=a u_i +b v_i$$ that is to say a bilinear regression with no intercept.

Now, set the matrix.


问:
For an uncountable subset $E$ in $[0,1]\times[0,1]$, show there is a condensation point of $E$.
I'm not sure how to approach this question:

Given $E\subseteq[0,1]\times[0,1]$ and $E$ is uncountable, show that there is some $p\in [0,1]\times[0,1]$ such that $\forall \epsilon>0, E\cap B(p,\epsilon)$ is uncountable.

I tried using the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$. I defined a set $A$ to be the collection of all open balls centered at rational points with some $\epsilon$ such that the unit square is completely covered. Surely at least one of the balls (centered at say $x$) contains uncountably many points in $E$. But, I don't know how to make this work (or even if it could work). I'm concerned about the overlap of the balls and how to find some $x$ that doesn't move around as $\epsilon$ changes. This is my introduction to analysis, so I don't have a good intuition for much of these concepts yet. Any help would be greatly appreciated!

答:
Let $B_Q=\{B(p,q):p\in \Bbb Q^2\land q\in \Bbb Q^+\}.$ Then $B_Q$ is a countable base (basis) for the topology on $\Bbb R^2.$

Let $C$ be the set of all $b\in B_Q$ such that $E\cap b$ is countable. Then $C$ is countable (... because $C\subseteq B_Q$...). So $\cup_{b\in C}(E\cap b)=E\cap(\bigcup C)$ is countable. So $E\setminus (\bigcup C)$ is uncountable.

Every  $p'\in E\setminus (\bigcup C)$ is an accumulation point of $E.$

Proof: Suppose not. Take $p'\in E\setminus (\bigcup C)$ and an  open set $U\subset \Bbb R^2$ such that $p'\in U$ and such that $E\cap U$ is countable. Since $B_Q$ is a base, there exists $b\in B_Q$ such that $p'\in b\subseteq U.$ But then $E\cap b$ is countable (... because $E\cap b\subseteq E\cap U$ ...) so $b\in C$ so $$p'\in E\cap b\subseteq E\cap (\bigcup C)$$ contrary to $p'\in E\setminus (\bigcup C).$

Remarks: This works verbatim for any uncountable $E\subset \Bbb R^2.$ Note that the set of members of $E$ that are $not$ accumulation points of $E$, that is, $E\cap (\bigcup C),$ is only countable. And $\bigcup C$ is the $\subseteq $-largest open set to have countable intersection with $E.$


问:
Finding the perpendicular distance from a chord to the circumference

Given a circle of circumference 60 and a chord of length 12, what is the perpendicular distance from the chord to the circumference?

What I don't understand about this the distance from the chord to the circumference means geometrically. Wouldn't that vary based on where on the chord you are?

(This question is apparently from BM 85194 (A Babylonian tablet).)

答:
Let $O$ be the center of the circle,
$C$ be the center of the chord,
and $A$ be one endpoint of the chord.

Then $OCA$ is a right angle
so triangle $OCA$ is a right triangle.

Therefore
$OC^2+CA^2 = OA^2$,
so
$OC = \sqrt{OA^2-CA^2}$.

In this case,
$OA = 60/(2\pi)
=30/\pi$
and
$AC = 12/2 = 6$
so
$OC = \sqrt{(30/\pi)^2-6^2}$.

(added later)

So the distance from
the center of the chord
to the circle
is the radius minus the
distance from the
center of the circle
to the chord or,
in this case,
$30/\pi-\sqrt{(30/\pi)^2-6^2}$.


问:
Understanding an answer for a function which is in $L^p$ for only one $p$
I am looking at Robert's answer here. The question is to find a function $f$ which is in $L^p$ for only one value of $p$. The answer given by Robert is:

$$f(x)=\frac{1}{x^{1/p}((\log x)^2+1)} \text{ on } (0,\infty)$$

I am assuming this $f$ is supposed to be in $L^1$ and not in $L^p$ for $p>1$? Assuming the previous line is correct, I can see that $f\in L^1$ and don't understand why $f\notin L^p$ for $p>1$?

Because:

$$\int_{(0,\infty)}\left( \frac{1}{x^{1/p}((\log x)^2+1)}\right)^pdx = \int_{(-\infty,\infty)}\frac{1}{(y^2+1)^p}dy\leq \int_{(-\infty,\infty)}\frac{1}{(y^2+1)}dy<\infty$$

May I know what's wrong in my thinking? Thanks.

答:
No, this function, with a $p$ in its definition (and therefore fixed, since letting one change the function for each $p$ would make the question trivial), is in $L^p$ but not $L^q$ for $q\neq p$. The integral under consideration should be $$\int_0^\infty\frac1{x^{q/p}((\log x)^2+1)^q} dx.$$

Your calculation shows that it belongs to $L^q$ if $q=p$. For the other cases, you just replace the $\log$ term with an arbitrarily small power of $x$: for $q > p$, the problem is at $0$,there is some $C>0$ and $r\in(1,q/p)$ such that $x^{q/p}((\log x)^2+1)^q<Cx^r$ for $x\in[0,1]$. It follows that $\int_0^1\frac1{x^{q/p}((\log x)^2+1)^q} dx=\infty.$ For $q<p$, the problem is at infinity.


问:
Infinite shortest paths in graphs
From Wikipedia:

"If there is no path connecting two
  vertices, i.e., if they belong to
  different connected components, then
  conventionally the distance is defined
  as infinite."

This seems to negate the possibility that there are graphs with vertices connected by an infinite shortest path (as opposed to being not connected).

Why is it that for every (even infinite) path between two vertices there is a finite one?

Note that infinite paths between vertices do exist - e.g. in the infinite complete graph -, but they are not the shortest.

答:
The point I made in the comments is that standard graph-theoretic terminology does not allow you to make sense of the notion of two points being connected by an infinite path. However, for what it's worth, there is a formalism that allows you to make sense of this related to end compactification: see Diestel's survey Locally finite graphs with ends: a topological approach for details. Briefly, we have to add extra vertices "at infinity," then introduce a nontrivial topology, after which we can speak of two vertices connected by a continuous function from $[0, 1]$ that passes through infinitely many vertices.

In any case, the excerpt from the Wikipedia article is about the standard graph-theoretic context. In the standard graph-theoretic context, we sometimes want to think of a graph as an extended metric space, and a natural way to do that is to define the distance to be infinite if two vertices aren't connected by an edge; certainly there's no sensible finite value, and if you choose infinity then the triangle inequality holds.

A very general context in which to understand this assignment is to think of the distance $d(u, v)$ between two vertices as the infimum over the lengths of all paths between $u$ and $v$. If $u, v$ are not connected by a path, this infimum is empty, and the empty infimum in a poset is always the greatest element, if it exists. This is a special case of a general fact in category theory that the empty limit is the terminal object. (Here the category is the poset $0 < 1 < 2 < ... < \infty$, regarded as a category where there is a single arrow $a \to b$ if $a \le b$ and no arrows otherwise.)

Title: Infinite shortest paths in graphs


问:
What is the function given by $\sum_{n=0}^\infty \binom{b+2n}{b+n} x^n$, where $b\ge 0$, $|x| <1$
For a nonnegative integer $b$, and $|x|<1$, what is the function given by the power series
$$
\sum_{n=0}^\infty \binom{b+2n}{b+n} x^n.
$$

For $b=0$, this post shows
$$
\sum_{n=0}^\infty \binom{2n}{n}x^n = (1-4x)^{-1/2}.
$$

How do we proceed for an integer $b>0$?

I tried to start from the power series of
$$
(1-x)^{-(b+2n)},
$$
but it doesn't work.

答:
My suggestion : avoid binomial coefficient. Instead, write your series in its hypergeometric form :
$$
\begin{eqnarray}
\sum_{n=0}^\infty \binom{b+2n}{b+n} x^n &=& {}_2F_1\left( \left. \begin{array}{c} \frac 12 + \frac b2, 1 + \frac b2\\1+b\end{array}\right| 4x\right)\\
&=& \frac {2^b \left(\sqrt {1-4x} + 1\right)^{-b}}{\sqrt {1-4x}}
\end{eqnarray}
$$

This identity is well-known. You can find it on the Wolfram web site.

If you want to deduce your identity, you will have to apply twice the geometric sum $${}_1F_0\left( \left. \begin{array}{c} a\\-\end{array}\right| z\right)
= \sum_{n=0}^{\infty} \frac{\left(a\right)_n z^n}{n!} = \left(1-z\right)^{-a}$$

where $(a)_n$ is the Pochhammer symbol.

Hope this helps.


问:
If $f$ is a function differentiable at $a$ find: $\underset{h\rightarrow 0}{\lim} \frac{f(a+7h)-f(a-9h^2)}{h}$
If $f$ is a function differentiable at $a$ find:
$\underset{h\rightarrow 0}{\lim} \frac{f(a+7h)-f(a-9h^2)}{h}$

I am struggling to understand what to do. I tried brute forcing this question and I get $\frac{f(a)-f(a)}{0}$ which makes no sense to me. Any clarification would be appreciated.

Body:
Why not to try with Taylor series ?

$$f(a+7h)=f(a)+7 h f'(a)+\frac{49}{2} h^2 f''(a)+O\left(h^3\right)$$
$$f(a-9h^2)=f(a)-9 h^2 f'(a)+O\left(h^3\right)$$
$$f(a+7h)-f(a-9h^2)=7 h f'(a)+h^2 \left(\frac{49 f''(a)}{2}+9 f'(a)\right)+O\left(h^3\right)$$
$$\frac{f(a+7h)-f(a-9h^2)}h=7  f'(a)+h \left(\frac{49 f''(a)}{2}+9 f'(a)\right)+O\left(h^3\right)$$ and $h \to 0$.


问:
Explicit Formula for the Integral of the Legendre Polynomial
Question: Expand the sign function between $-1$ and $1$ in a series of Legendre polynomials. Obtain an explicit formula for the expansion coefficients.

Attempt: I found that

$$a_l=(2l+1)\int_0^1 P_l(x)$$

where $a_l$ is the coefficient of degree l. However, I am still confused how to find an explicit formula for this value. Can someone please help me?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\on{P}_{\ell}:\ Order\mbox{-}\ell\ Polynomial}$.

\begin{align}
a_{\ell} & \equiv \bbox[5px,#ffd]{\left.\pars{2\ell + 1}
\int_{0}^{1}\on{P}_{\ell}\pars{x}\,\dd x
\,\right\vert_{\,\ell\ \in\ \mathbb{N}_{\,\geq\ 0}}}
\\[5mm] & =
\pars{2\ell + 1}
\int_{0}^{1}\braces{\bracks{h^{\ell}}{1 \over
\root{1 - 2xh + h^{2}}}}\,\dd x
\\[5mm] & =
\pars{2\ell + 1}\bracks{h^{\ell}}
\int_{0}^{1}{\dd x \over\root{1 - 2xh + h^{2}}}
\\[5mm] = &\
\pars{2\ell + 1}
\bracks{h^{\ell}}{\root{1 + h^{2}} + h - 1 \over h}
\\[5mm] & =
\pars{2\ell + 1}\bracks{\ell\ odd}
\bracks{h^{\ell + 1}}\root{1 + h^{2}} + \delta_{\ell 0}
\\[5mm] & =
\delta_{\ell 0} +
\pars{2\ell + 1}\bracks{\ell\ odd}
{1/2 \choose \bracks{\ell + 1}/2}
\end{align}


问:
Metric and the triangle inequality
Let $A,B$ be finite subsets of the natural numbers. If we let $d(A,B)=\sum_{x\in A\mathbin\Delta B} 2^{-x}$, where $A\mathbin\Delta B=(A\cup B)\setminus (A\cap B)$ is the symmetric difference between $A$ and $B$, does $d$ satisfy the triangle inequality? That is, if $A,B,C$ are finite, open subsets of the natural numbers, is $d(A,B)\leq d(A,C)+d(C,B)$? Or do we have to assume that the subsets are disjoint for the triangle inequality to be satisfied?</p>

答:
Martin provided one answer in the comments. Here’s another way to show that $d$ is a metric. For each $A\subseteq\Bbb N$ let $\chi_A$ be the indicator (or characteristic) function of $A$. Then

$$d(A,B)=\sum_{n\in\Bbb N}2^{-n}|\chi_A(n)-\chi_B(n)|\;$$

for any $A,B\subseteq\Bbb N$. (There’s no need to restrict oneself to finite subsets.) Hence

$$\begin{align*}
d(A,B)&=\sum_{n\in\Bbb N}2^{-n}|\chi_A(n)-\chi_B(n)|\\
&\le\sum_{n\in\Bbb N}2^{-n}\Big(|\chi_A(n)-\chi_C(n)|+|\chi_C(n)-\chi_B(n)|\Big)\\
&=\sum_{n\in\Bbb N}2^{-n}|\chi_A(n)-\chi_C(n)|+\sum_{n\in\Bbb N}2^{-n}|\chi_C(n)-\chi_B(n)|\\
&=d(A,C)+d(C,B)\;.
\end{align*}$$

This is actually just a special case of the proof that if $\{\langle X_n,d_n\rangle:n\in\Bbb N\}$ is a family of metric spaces such that each $d_n$ is bounded by $1$, and $X=\prod_{n\in\Bbb N}X_n$, then the function

$$d:X\times X\to\Bbb R:\langle x,y\rangle\mapsto\sum_{n\in\Bbb N}2^{-n}d_n(x_n,y_n)$$

satisfies the triangle inequality. (It is of course a metric on $X$.) Here each of the spaces $X_n$ is a copy of $\{0,1\}$ with the metric inherited from $\Bbb R$, and the map $A\mapsto\chi_A$ is an isometry.

Title: Metric and the triangle inequality


问:
Prove that the function $f(x)=x \ln x$ is not a polynomial
Prove that the function $f(x)=x \ln x$ is not a polynomial

My thought process is that I know polynomials have specific properties when I take the derivative. Namely, for $f(x)=x^n$ the derivative is $f^{'}(x)=nx^{n-1}$. That or $f(x)=0$ and that derivative is equal to $0$ (itself).
I know that $f^{'}(x)=\ln(x) + 1$ which is not a degree less that $f(x)$ and which doesn't equal f(x). Therefore, I have that $f(x)=x\ln x$ is not a polynomial.
How accurate is this proof? I feel like the proof is clunky, but can't get a finger on what the issue is.

答:
First off, it's not a polynomial because it's not defined for $x \leq 0$. If you mean for $x > 0$, then you can go as follows:

Assume $x\log x$ is a polynomial.

$$x\log x = a_n x^n + a_{n-1}x^{n-1} + ... + a_1x+a_0$$

$$\Rightarrow \log x = a_nx^{n-1} + a_{n-1}x^{n-2}+...+a_1+\frac{a_0}{x}$$

Use the fact that $g(x) = \log x$ satisfies $f(x^2) = 2f(x)$ by log laws:

$$\Rightarrow a_nx^{2n-2} + a_{n-1}x^{2n-4} + ... +a_1x^2+\frac{a_0}{x^2} = 2a_nx^{n-1}+2a_{n-1}x^{n-2}+...+2a_1 + \frac{2a_0}{x}$$

Compare coefficients. This means $\log x = 0$ for all $x$, which is nonsense.  There's many ways to do it, but here's strictly from the definition of a polynomial.


问:
Minimum Value of Reciprocal of Roots
If $f(x) = x^6+a_5x^5+...a_0$ be such that $f(0)=f(1)$ and each root of $f(x)=0$ is real and is between 0 and 1.Then find the minimum value of reciprocal of product of roots.

I tried to write $f(x)=\prod_{i=1}^{6}(x-x_{i})$ . Then with $f(0)=f(1)$ gives $\prod_{i=1}^{6}x_{i}=\prod_{i=1}^{6}(1-x_{i})$.I am not sure how to proceed from here.

答:
Instead of minimizing product of reciprocal of roots, we will try to maximize $\prod\limits_{k=1}^6 x_k$.

We can safely assume all $x_k \in (0,1)$. This is because if any $x_k = 0$, then $\prod\limits_{k=1}^6 x_k = 0$ which is clearly not the maximum. Similiarly, if any $x_k = 1$, then $f(1) = 0 \implies \prod\limits_{k=1}^6 x_k = f(0) = 0$ which is again not we want.

For other configuration of $x_k$, we have

$$\prod_{k=1}^6 x_k = \prod_{k=1}^6 \sqrt{\frac{x_k}{1-x_k}}\sqrt{x_k(1-x_k)}$$
Notice for all $\lambda \in (0,1)$, we have $\sqrt{\lambda(1-\lambda)} \le \frac12$. This leads to

$$\prod_{k=1}^6 x_k \le \frac1{2^6}\prod_{k=1}^6\sqrt{\frac{x_k}{1-x_k}}
= \frac1{64}\sqrt{\frac{f(0)}{f(1)}} = \frac1{64}$$

This implies $a_0 = \prod\limits_{k=1}^6 x_k$ is bounded from above by $\frac{1}{64}$. Since this upper bounded is attained when all $x_k = \frac12$, we can
deduce the maximum value of $a_0$ is $\frac{1}{64}$. As a result,
the minimum value of product of reciprocals of roots is $64$.

Title: Minimum Value of Reciprocal of Roots


问:
Simplifying $\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$
My teacher gave me the following expression to change into a single expression:

$$\log_x 25 - \left(\frac{\log_5 x}{4} - 2\log_5 x\right)$$

This is what I got after working it out:

$$\log_x 25 - \log_5 \left(\frac{x}{4x^2}\right)$$

Body:
$$\begin{align}
& \log_{x}25 - \left(\dfrac{\log_{5}x}{4} - 2\log_{5}x\right) \\
= & \log_{x}5^2 + \dfrac{7}{4}\log_{5}x \\
= & 2\log_{x}5 + \dfrac{7}{4}\log_{5}x \\
= & \dfrac{2}{\log_{5}x} + \dfrac{7}{4}\log_{5}x \\
= & \dfrac{7\log_{5}x + 8}{4\log_{5}x} \\
= & \dfrac{\log_{5}x^7 + \log_{5}5^8}{\log_{5}x^4} \\
= & \dfrac{\log_{5}(5^8 \cdot x^7)}{\log_{5}x^4} \\
= & \log_{x^4}(5^8 \cdot x^7) \\
\end{align}$$


问:
Eigenvalues of circulant using spectral mapping
Is there a way to utilize the spectral mapping theorem to find the eigenvalues of a circulant matrix if eigenvectors of P are known?

From this wiki page, I know that $C=c_0I+c_1P+c_2P^2+...+c_{n-1}P^{n-1}$ (P is a shift matrix, it is shown in the link)

I think with the given equality, I can use the spectral mapping theorem to find the eigenvalues of $C$ but not quite sure how to get started.

答:
$P$ is unitary and thus normal, and hence admits an orthonormal basis of eigenvectors. However, from the fact that $C$ is a polynomial of $P$, it would follow that each of these eigenvectors is also an eigenvector of $C$. Furthermore, we can easily calculate the coefficients $c_0, c_1, \cdots, c_{n - 1}$ by just inspecting the matrix $C$. (Hint: they are exactly the elements of the first column, from top to bottom. You can prove this iteratively by showing that the only contribution to the elements on the diagonal are $c_0 I$, the contribution to the elements right under the diagonal and in the top right corner are from the term $c_1 P$, and so on.)

It is helpful to note that the eigenvalues of the shift matrix are the $n$ roots of unity, where $n$ is the order of the matrix. So for example, if $\vec{v}$ is the vector such that $P \vec{v} = e^{2\pi i/ n} \vec{v} = \omega \vec{v}$, then
$$C \vec{v} = (c_0 + c_1 \omega + c_2 \omega^2 + \cdots + c_{n - 1} \omega^{n - 1}) \vec{v}$$
Can you sort out the general case from here? (Another remark: if you want to efficiently calculate all eigenvalues, you can use the Fast Fourier transform to do this in $\Theta(n \log n)$ operations.)


问:
Why is $1,x,x^2,x^3,\cdots$ a bad basis?

For functions of $x$, the first basis I would think of contains the powers $1, x, x^2 , x^3 , ...$. Unfortunately this is a terrible basis. Those functions $x^n$ are just barely independent. $x^{10}$ is almost a combination of other basis vectors $1, x, ... , x^9$. It is virtually impossible to compute with this poor "ill-conditioned" basis.

This is said in Page 426, Chapter 8-Linear Transformations, Introduction to Linear Algebra, Gilbert Strang, 5th Edition.

Why do we say that $1,x,x^2,x^3,\cdots$ is a bad basis ?

Taking the Wronkian
$$
\begin{vmatrix}x&x^2\\1&2x\end{vmatrix}=2x^2-x^2=x^2\\
\begin{vmatrix}x^2&x^3\\2x&3x^2\end{vmatrix}=3x^4-2x^4=x^4\\
\begin{vmatrix}x^3&x^5\\3x^2&5x^4\end{vmatrix}=5x^7-3x^7=2x^7
$$

Looks like linear independence is almost fine. So what does it mean to say it an ill-conditioned basis ?

And how does one show that "those functions $x^n$ are just barely independent" ?

答:
The vectors $(1,0,0), (1,0.01, 0), (1,0.01, 0.01)$ are linearly independent in $\mathbb{R}^3$, but are very close to each other.

Strang may be saying that the monomials $1,x,x^2,\ldots$ are not close to "orthogonal." Other bases like the Chebychev polynomials are orthogonal with respect to the $L^2$ inner product.


问:
Asymptotically $n(1-\frac{a\log n}{n})^{n/2-1}(1-\frac{b\log n}{n})^{n/2}\sim n^{1-\frac{a+b}{2}}$
Suppose $a,b>0$ and $\frac{a+b}{2}<1$

Why is the following relation true as $n\rightarrow\infty $ $$n(1-\frac{a\log n}{n})^{n/2-1}(1-\frac{b\log n}{n})^{n/2}\sim n^{1-\frac{a+b}{2}}$$

The $1$ in the exponent on the right hand side value is obviously clear... So Why does $(1-\frac{a\log n}{n})^{n/2}(1-\frac{b\log n}{n})^{n/2}$ reduce to $n^{\frac{a+b}{2}}$?

答:
Consider
$$y=\left(1-a\frac{ \log (n)}{n}\right)^{\frac{n}{2}-1}$$
$$\log(y)=\left(\frac{n}{2}-1\right)\log\left(1-a\frac{ \log (n)}{n}\right)\sim -\left(\frac{n}{2}-1\right)a\frac{ \log (n)}{n}$$
So, taking the logarithm of the rhs of you expression $A$ we have
$$\log(A)\sim \log(n)-\left(\frac{n}{2}-1\right)a\frac{ \log (n)}{n}-\frac n 2 b \frac{ \log (n)}{n}$$ Expanding and grouping
$$\log(A) \sim \left(1-\frac{a+b}{2} \right) \log (n)+a\frac{ \log (n)}{n}\sim \left(1-\frac{a+b}{2} \right) \log (n)$$
$$A=e^{\log(A)} \sim n^{1-\frac{a+b}{2}}$$


问:
Conditional expectation of Poisson Process given two counts
Let ${N(t),t\ge 0}$ be a Poisson Process with rate $\lambda=2.$ How do I go about calculating $E[N(6)\mid N(4)=4,N(10)=12]$?

答:
$$N(6) - N(4) \sim\operatorname{Poisson}(\lambda(6-4)) = \operatorname{Poisson}(2\lambda).$$

\begin{align}
& \operatorname E[N(6)\mid N(4)=4,N(10)=12] \\[10pt]
= {} & 4 + \operatorname E(N(6)-N(4)\mid N(10) - N(4) = 8)
\end{align}

So a question is: What is the conditional distribution of $N(6)-N(4)$ given that $N(10)-N(4)=8\text{?}$

\begin{align}
& \Pr(N(6)-N(4) = n \mid N(10)-N(4) = 8) \\[10pt]
= {} & \frac{\Pr(N(6) - N(4) = n\ \&\ N(10)-N(4) = 8)}{\Pr(N(10)- N(4) =8) } \tag 1 \\[8pt]
= {} & \frac{\Pr(N(6) - N(4) = n\ \&\ N(10)-N(6) = 8-n)}{\Pr(N(10)- N(4) =8)} \tag 2
\end{align}
The reason the probabilities in the numerators in lines $(1)$ and $(2)$ are equal is that the actual events whose probabilities are taken are equal. On line $(2),$ you can exploit the fact that the two events with "and" between them are independent.

After that, when you simplify the fraction, everything depending on $\lambda$ will cancel out and you will get a binomial distribution with parameters $8$ and $1/3,$ the latter number coming from the fact that the length of the interval $[4,6]$ is $1/3$ the length of the interval $[4,10].$


问:
Show that if $f$ is a differentiable function whose third derivative is continuous, then there exists $c\in (0,1)$ such that $f'''(c)\geq 24$

Let $f:[0,1] \to \mathbb{R}$ be a function whose third derivative $f'''(x)$ is continuous on $[0,1]$. Suppose that
$$ f(0)=f'(0)=f''(0)=f'(1)=f''(1)=0 \text{ and } f(1)=1$$
Then there exist $c\in(0,1)$ such that $f'''(c)\geq 24$

I am trying to prove the above statement via Taylor's theorem, and see if I can make use of the Darboux's theorem, but I don't know how to come up with a 24 in the last sentence of the statement. Any kind of hints are much appreciated, thank you.

答:
I was just playing around with Taylor's theorem, and I came up with something that seems right to me. Using Taylor's theorem, we can say that there exists $c_1\in (0,\frac{1}{2})$ such that
$$f\left(\frac{1}{2}\right) = f(0) + f'(0)\left(\frac{1}{2}-0\right)+ \frac{f''(0)}{2!}\left(\frac{1}{2}-0\right)^2 + \frac{f'''(c_1)}{3!}\left(\frac{1}{2} - 0\right)^3$$
which is just
$$f\left(\frac{1}{2}\right) = \frac{f'''(c_1)}{48}$$
Another application of Taylor's theorem guarantees the existence of $c_2 \in(\frac{1}{2},1)$ such that
$$f\left(\frac{1}{2}\right) = f(1) + f'(1)\left(\frac{1}{2}-1\right) + \frac{f''(1)}{2!}\left(\frac{1}{2}-1\right)^2 + \frac{f'''(c_2)}{3!}\left(\frac{1}{2}-1\right)^3$$
which is just
$$f\left(\frac{1}{2}\right) = 1 - \frac{f'''(c_2)}{48}$$
So we have $$f'''(c_1) + f'''(c_2) = 48 \text{ where }c_1\in \left(0,\frac{1}{2}\right) \text{ and }c_2\in \left(\frac{1}{2},1\right)$$

Now, we claim that one of $f'''(c_1)$ and $f'''(c_2)$ must exceed or equal $24$. We give a proof by contradiction.

Suppose, if possible, that $f'''(c_1) < 24$ and $f'''(c_2)<24$. This would mean that $f'''(c_1)+f'''(c_2) < 48$ which is a contradiction. Hence, we have at least one of $f'''(c_1)\geq 24$ and $f'''(c_2)\geq 24$.

This shows the existence of $c\in (0,1)$ such that $f'''(c)\geq 24$.


问:
How can I evaluate the following sum: $\sum_{k=0}^{n-1}\binom{2n-1}{k}\left[\text{sgn}(2n-2k-2)+\text{sgn}(2n-2k)\right]$??
I was trying to prove the equality below:
$$
\int_0^\infty\frac{1-\cos^{2n}(x)}{x^2}\mathrm dx=\frac{n\pi}{4^n}\binom{2n}{n}.
$$
First, I proceeded using IBP making $u=1-\cos^{2n}(x)$ and $\mathrm dv=\frac{1}{x ^ 2}$, so i got:
$$\int_0^\infty\frac{1-\cos^{2n}(x)}{x^2}\mathrm dx=2n\int_0^\infty\frac{\cos^{2n-1}(x)\sin(x)}{x}\mathrm dx. $$
Here, I used the identity for the term $\cos^{2n-1}(x)$:
$$\cos^{2n-1}(x)=\frac{2}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n-1}{k}\cos\left[(2n-k-1)x\right]$$
And, in short, using the property $2\cos(\theta)\sin(\varphi)=\sin(\theta+\varphi)+\sin(\theta-\varphi)$ and Dirichlet Integral, i arrived at:
$$\int_0^\infty\frac{1-\cos^{2n}(x)}{x^2}\mathrm dx=\frac{2n\pi}{4^n}\sum_{k=0}^{n-1}\binom{2n-1}{k}\left[\text{sgn}(2n-2k-2)+\text{sgn}(2n-2k)\right]$$

Here is my difficulty, I know the definitions of the Sign Function $\text{sgn}(z)$, but I do not know how to approach using the cases of the definition of this function. By hypothesis, we would have:
$$\sum_{k=0}^{n-1}\binom{2n-1}{k}\left[\text{sgn}(2n-2k-2)+\text{sgn}(2n-2k)\right]=\frac{1}{2}\binom{2n}{n}$$

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{\sum_{k = 0}^{n - 1}{2n - 1 \choose k}
\bracks{\on{sgn}\pars{2n - 2k - 2} + \on{sgn}\pars{2n - 2k}}}:
\ {\Large ?}}$.

\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{n - 1}{2n - 1 \choose k}
\bracks{\on{sgn}\pars{2n - 2k - 2} + \on{sgn}\pars{2n - 2k}}}
\label{1}\tag{1}
\\[5mm] = &\
-\sum_{k = 0}^{n - 1}{2n - 1 \choose k}
\bracks{\on{sgn}\pars{k - \bracks{n - 1}} + \on{sgn}\pars{k - n}}
\\[5mm] = &\
2\sum_{k = 0}^{n - 2}{2n - 1 \choose k} + {2n - 1 \choose n - 1} =
-{2n - 1 \choose n - 1} + 2\
\underbrace{\sum_{k = 0}^{n - 1}{2n - 1 \choose k}}
_{\ds{\equiv {\tt J}}}\label{2}\tag{2}
\\[5mm] = &\
-{2n - 1 \choose n - 1} +
2\sum_{k = 0}^{2n - 1}{2n - 1 \choose k} -
2\sum_{k = n}^{2n - 1}{2n - 1 \choose k}
\\[5mm] = &\
-{2n - 1 \choose n - 1} +
2\pars{2^{2n - 1}} -
2\sum_{k = 0}^{n - 1}{2n - 1 \choose k + n}
\\[5mm] = &\
-{2n - 1 \choose n - 1} + 2^{2n} -
2\sum_{k = 0}^{n - 1}{2n - 1 \choose
\bracks{n - 1 - k} + n}
\\[5mm] = &\
-{2n - 1 \choose n - 1} + 2^{2n} -
2\
\underbrace{\sum_{k = 0}^{n - 1}{2n - 1 \choose k}}
_{\ds{=\ {\tt J}}}
\label{3}\tag{3}
\end{align}
With (\ref{2}) and (\ref{3}):
\begin{equation}
{\tt J} \equiv \sum_{k = 0}^{n - 1}{2n - 1 \choose k} =
2^{2n - 2}
\label{4}\tag{4}
\end{equation}
With (\ref{1}), (\ref{2}) and (\ref{4}):
\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{n - 1}{2n - 1 \choose k}
\bracks{\on{sgn}\pars{2n - 2k - 2} + \on{sgn}\pars{2n - 2k}}}
\\[5mm] = &\
\bbx{2^{2n - 1} - {2n - 1 \choose n - 1}} \\ &
\end{align}


问:
What happens when you take the limit with a big-oh error?
What happens if you have some error in your expression and you wish to take the limit? Does the big-oh go away or are you unable to take a limit at all?

This may seem very obvious as it is likely I am misunderstanding the definition of big-oh?

答:
What you understand under $\lim_\limits{n \to \infty} O(g(n))$? Set of limits of all members?

If exists $\lim_\limits{n \to \infty} g(n)$, then  $O(\lim_\limits{n \to \infty} g(n))=O(1)$, so here are bounded functions. On another hand even if we assume that all members of $O(g(n))$ have limits, then we obtain set of numbers, not bounded functions.

$$\lim_\limits{n \to \infty} O(g(n)) \ne O(\lim_\limits{n \to \infty} g(n))$$


问:
Diagonal matrices that belong to conjugates of T
Let $G = SU(3)$, and let T be the subgroup of G of diagonal matrices.

How can we prove that  $\lbrace T \cap gTg^{-1}, g \in  G \rbrace = \lbrace T, T_e, T_1 , T_2, ,T_3 \rbrace $?

Where $T_e$ is the set of matrices of the form  $\lambda I$, $\lambda \in \mathbb{C}, I $ is the identity matrix.

$T_1$ are diagonal matrices in T s.t the first and second components are equals .

$T_2$ are diagonal matrices in
T s.t the second and third components are equals .

$T_3 $ are diagonal matrices in T s.t the first and third components are equals .

答:
$\newcommand{\inv}{^{-1}}$Case 1. Suppose first that $T\cap gTg\inv$ contains a matrix $a$ with three unequal diagonal entries.

Observing that  $g\inv a g$ is also diagonal, and that it has the same spectrum as  $a$, we see that the diagonal entries of
$g\inv a g$ must be a permutaion of the entries of $a$.  Therefore there exists a permutation matrix $\sigma $ (i.e. the
matrix representing a linear transformation permuting the canonical basis) such that
$$
  g\inv a g = \sigma \inv a \sigma .
  \tag1
  $$
It follows that $g\sigma\inv$ commutes with $a$, but it is well known that only diagonal matrices have this property.
Therefore $g\sigma\inv=d$, for some diagonal matrix  $d $, hence $g = d\sigma $, and one may now easily prove that $gTg\inv=T$, so
$T\cap gTg\inv=T$.

Case 2.  Suppose now that $T\cap gTg\inv$ contains only matrices $a$ with three identical  diagonal entries.

If $a\in  T\cap gTg\inv$, then we may write $a=\lambda I$, and it is clear that $\lambda ^3=\det(a) =1$, so
$$
  \lambda \in \Lambda := \{1,  e^{2\pi i/3},  e^{4\pi i/3}\}.
  $$
This shows that
$$
  T\cap gTg\inv \subseteq  \Lambda I,
  $$
and the reverse inclusion is obviously true (regardless of $g$).

Case 3.  Suppose now that $T\cap gTg\inv$ does not contain any matrix  with three unequal diagonal entries, but that it
contains a matrix $a$ with two unequal diagonal entries.  Without loss of generality we may assume that
$$
  a=\pmatrix{\lambda  & 0 & 0\cr 0 & \lambda  & 0\cr 0 & 0 & \mu }
  $$
where $\lambda \neq \mu $
(otherwise we may conjugate everything by a cyclic permutation matrix).

Just like case (1), we may find a permutation matrix $\sigma $  satisfying (1), so that $g\sigma\inv$ commutes with $a$.
The commutator of $a$ in $SU(3)$ may be shown to coincide with  the set of matrices of the form
$$
  v=\pmatrix{u_{11} & u_{12} & 0\cr u_{21} & u_{22} & 0\cr 0 & 0 & \tau },
  $$
where
$$
  u:=\pmatrix{u_{11} & u_{12} \cr u_{21} & u_{22}}\in  U(2),
  $$
and $\tau =\det(u)\inv$.
So we deduce that there is some $v$, as above, such that $g\sigma\inv=v$, whence $g=v\sigma $.

With a little more effort one may now show that
$T\cap gTg\inv$ coincides with $T_1$ (as defined in the original post).

The other cases correspond to alternatives to case (3), depending on which of the two diagonal entries of $a$ agree, and they
lead to $T\cap gTg\inv$ coinciding with $T_2$ or $T_3$.

This shows that
$$
  \lbrace T \cap gTg^{-1}: g \in  G \rbrace \subseteq  \lbrace T, T_e, T_1 , T_2, ,T_3 \rbrace,
  $$
and the reverse inclusion may be easily shown by choosing  $g$ of the apropriate form, as suggested by the above cases.


问:
How do you express a single equation as a system of equations?
I have the equation $(2x-3)(ax+b)+(x^2-2)(cx+d) = 1$, and somehow I have to arrive at the following system:

$c = 0$

$2a+d = 0$

$2b-3a-2c = 0$

$-3b - 2d = 1$

I'm completely blanking on the math necessary.

答: Multiply everything out and collect terms to $Ax^3 + Bx^2 + Cx + D = 1$, and then compare coefficients to find A, B, C, and D.


问:
Matrix representation wrt bases
Consider the function $T:P_{2}\to P_{3}$ given by $T(p(x)) = xp(x)$.
Let $\mathcal{B} = \{1, x, x^{2}\}$ be the ordered standard basis of $P_{2}$. Let $\mathcal{C} = \{2,x/2,3x^{2}, x^{3}-2\}$ be an ordered basis of $P_{3}$. Find the matrix representation of $T$ with respect to the bases $\mathcal{B}$ and $\mathcal{C}$.
enter image description here

答: Hint:  The columns will be $T$ applied to the elements of $\mathcal B$ (in order) expressed in terms of $\mathcal C$.

Title: Matrix representation wrt bases


问:
I have a function $f(x,y,z)$. Can this be rewritten as $f(y,z)$ if $x$ is not present in the function?
Let's say that I have $f(x,y,z)$ which is derived from another function g. And this is true for all cases.

For a certain case, $f(x,y,z) = yz$

Is this the same as writing/ defining the function as $f(y,z) = yz$  ?

Or is it imperative that I write it as the $f(x,y,z)=yz$   as its definition?

答:
$
\newcommand{\R}{\mathbb{R}}
\newcommand{\ff}{\widetilde{f}}
\newcommand{\dom}{\operatorname{dom}}
\newcommand{\cod}{\operatorname{cod}}
$
You can't, because it goes back to the definition of the function. The fact that you have $f(x,y,z)$ rather than $f(y,z)$ means $f$ is a function with domain $\R^3$.

(Or, in general, the domain of $f$ is $A \times B \times C$ where $A$ is the set in which $x$ comes from, $B$ is that of $y$, and $C$ is that of $z$, but, for simplicity, let's work in vector spaces over $\R$.)

On the other hand, if you have a function $\ff(y,z)$, this means that the function has domain $\R^2$.

This matters because we say two functions $g,h$ are equal if and only if...

They have the same domain (i.e. $\dom(g) = \dom(h)$)

They have the same codomain (i.e. $\cod(g) = \cod(h)$}

For every element in the domain, each function (i.e. $g(x) = h(x)$ for all $x$ in $\dom(g)$ or, same difference, $\dom(h)$)

In this case, $f$ in this discussion has $\dom(f) = \R^3$ and $\ff$ has $\dom(\ff) = \R^2$ -- even if the former does absolutely nothing with the $x$ component of its vector, it's still sending ordered triples or vectors $(x,y,z)$ as a whole somewhere.

So it doesn't make sense, really, to say you can write $f(x,y,z)$ or $f(y,z)$ interchangeably when $f$ does not depend on $x$. One represents a function with domain $\R^3$ and the other one with $\R^2$. They're not really the same thing.

To take it to the extreme, I could define a function $F : X^n \to Y \ne \varnothing$ by

$$F \big( (x_1,x_2,\cdots,x_n \big) = y$$

for some fixed $y \in Y$. Sure, this is a constant map and does not depend on any of the $x_i$. However, to say you could just "ignore" the $x_i$ and write $F(\varnothing) = y$, just begins to look silly. You're still sending stuff somewhere, it just happens to be a place that doesn't depend on the $n$-tuple.

Luckily, this is easily handled, as mentioned in the comments. So you have the function

$$f(x,y,z) = yz$$

and want to define a function purely in terms of $y,z$ for whatever reason. Simply define a new function! That being, you could say something like this:

Define the function $\ff$ by $\ff(y,z) = f(x,y,z)$, i.e. $\ff(y,z) = yz$.


问:
Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$
I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:

$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?

Thank you very much!

答:
Using algebra

Rewriting
$$a_{n+1}=\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)$$ as
$$a_{n+1}=a_n-\left(a_n-\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)\right)=a_n-\frac{a_n^2-3}{2 a_n}$$ we can recognize the itegrative scheme for solving $a^2=3$; so the limit is $\sqrt 3$.

Considering the function $f(a)=a^2-3$, starting with $a_0=1$, we have
$f(1)=-2$ and $f''(1)=2$, $f(1)\times f''(1)=-4<0$, then, by Darboux theorem, we shall have one (and only one overshoot of the solution). This corresponds to $a_2$ and, later, we shall converge quadratically to the root.
$$\left(
\begin{array}{cc}
n & a_n \\
 1 & 1.0000000000000000000 \\
 2 & 2.0000000000000000000 \\
 3 & 1.7500000000000000000 \\
 4 & 1.7321428571428571429 \\
 5 & 1.7320508100147275405 \\
 6 & 1.7320508075688772953 \\
 7 & 1.7320508075688772935
\end{array}
\right)$$


问:
Database of unsolved problems in mathematics
Is there a good database of unsolved problems in mathematics?</p>

答:
The site called scilag (https://www.scilag.net/) is another more recent one too.

"SciLag is a free web-based platform for facilitating dynamic organisation of scientific problems at a research level. It represents an online service for scientists for sharing their knowledge about the forefront of research. The current version is set up for the Mathematical community only to test and refine the concept in action."


问:
Determine the value of $p\ge0$ for which the series $\sum_{n=1}^\infty\frac{n(-1)^n}{(1+n^2)^p}$ is divergent, conditionally convergent.

Determine the value of $p\ge0$ for which the series $$\sum_{n=1}^\infty\frac{n(-1)^n}{(1+n^2)^p}$$ is divergent, conditionally convergent or absolutely convergent.

I have done the absolute convergent part. $$\left|\frac{n(-1)^n}{(1+n^2)^p}\right|\le \left|\frac{n}{(1+n^2)^p} \right|\le\left| \frac{1}{(1+n^2)^p}\right|\le\left|\frac{1}{n^{2p}} \right|$$ Hence, by comparison test, the series converges absolutely when $p>\frac{1}{2}$.

Is this series diverges when $p\le\frac{1}{2}$? I have no idea how to show that for what $p$ it will converges conditionally. Thanks for any comments.

答:
Absolute convergence comes from
$$\left|\frac{n(-1)^n}{(1+n^2)^p}\right| \sim \frac{1}{n^{2p-1}}$$
so you need $2p-1>1$.

Conditional convergence we will have when $2p-1>0$

Addition.

For conditional convergence we need $\frac{n}{(1+n^2)^p} \to 0$ monotonously. Possibly most easy way to prove is consider  $f(x)=\frac{x}{(1+x^2)^p}$ and  calculate $f'(x)=\frac{1+x^2(1-2p)}{(1+x^2)^{p+1}}$, from where, for $2p-1>0$ we obtain desired.

On another hand if there is some reason to want solve inequality $\frac{n+1}{(1+(n+1)^2)^p} \leqslant \frac{n}{(1+n^2)^p}$ straight ahead, then we can write $(n+1)(1+n^2)^p \leqslant n(1+(n+1)^2)^p$ and notice, that while maximal power of $n$ on both sides is $n^{2p+1}$ with same coefficient $1$, then member $n^{2p}$ on left side have coefficient $1$ and on right side coefficient $2p$. So, again, we obtain decreasing when $2p-1>0$.


问:
A problem on probability of a biased coin on winning if HH is observed
A biased coin is tossed repeatedly. The probability of landing a head is p. The game ends the first time when two consecutive heads (HH) or two consecutive tails (TT) are observed. A player
wins if (HH) is observed and loses if (TT) is observed. Find the probability that he wins.
(For example if the outcome is THTHTT, he loses. On the other hand, if the outcome is s THTHHH,he wins)

In this question of probability ,I am a bit confused with the language of the question. Can someone help me out. What does the problem Want and how to frame that probability?I mean how to proceed off with the probkem

Can anyone kindly help me out

答:
Calculate the probability that the game ends with HH. If the game begins with H, the possible outcomes are: HH, HTHH, HTHTHH, HTHTHTHH, ... But if the game begins with T, the possible outcomes are: THH, THTHH, THTHTHH, ... So, the probability of two heads from the first outcome is $$P(HH)(1+P(HT)+P(HTHT)+P(HTHTHT)+\cdots )$$ While the probability of winning after starting with T is $$P(THH)(1+P(TH)+P(THTH)+\cdots)$$

This yields:

$$(P(HH)+P(THH))\cdot \sum_{k=0}^\infty (P(HT))^k = \dfrac{P(HH)+P(THH)}{1-P(HT)}$$


问:
Are all non-associative (not necessarily associative) finite division rings finite fields?
According to the Artin–Zorn theorem, any finite alternative division ring is a finite field, but I'm interested in the general non-associative (i.e. not necessarily associative) case. Are there any non-associative finite division rings different from finite fields, or are all non-associative finite division rings finite fields?

Edit: We define a not necessarily associative division ring to be a set $S$ equipped with two binary operations $+$ and $\cdot$ such that

$S$ with $+$ is an abelian group

$S/\{0\}$ with $\cdot$ is a loop (unital quasigroup)

$\cdot$ distributes over $+$.

This comes from the following defintion of a division ring: a set $S$ equipped with two binary operations $+$ and $\cdot$ such that

$S$ with $+$ is an abelian group

$S/\{0\}$ with $\cdot$ is a group (associative loop)

$\cdot$ distributes over $+$.

答:
Nonassociative division rings with unity are known as semifields. They come up in the coordinatization of projective planes. Their study in the finite case started with

Donald Knuth, Finite semifields and projective planes. J. Algebra 2 (1965), 182-217.

This published version was based on Knuth's 1963 PhD dissertation.

There is a considerable literature on semifields, but note that the term "semifield" is also used in a conflicting sense in other parts of mathematics as an associative semiring with unity in which every nonzero element has a multiplicative inverse. Papers on semifields in the nonassociative sense are generally tagged with the MSC classification 17A35 (Nonassociative division algebras).


问:
Prove that the function $f(x)=x \ln x$ is not a polynomial
Prove that the function $f(x)=x \ln x$ is not a polynomial

My thought process is that I know polynomials have specific properties when I take the derivative. Namely, for $f(x)=x^n$ the derivative is $f^{'}(x)=nx^{n-1}$. That or $f(x)=0$ and that derivative is equal to $0$ (itself).
I know that $f^{'}(x)=\ln(x) + 1$ which is not a degree less that $f(x)$ and which doesn't equal f(x). Therefore, I have that $f(x)=x\ln x$ is not a polynomial.
How accurate is this proof? I feel like the proof is clunky, but can't get a finger on what the issue is.

答:
As I mentioned in the comment section, one way to prove it is that if the order of the polynomial is $n$ then $f^{(n+1)}(x) \equiv 0$ which we can  easily verify is not the case.

Another way, based on the solution of Luna145, is as follows:

Suppose $f(x)$ is an order-$n$ polynomial, then

$$\ln x^2 = 2 \ln x \implies f(x^2)=x^2\ln x^2 = 2x \cdot x\ln x = 2x f(x).$$

Compare the order of both sides we have $$2n=1+n \implies n=1 \implies x \ln x = ax+b$$
which is absurd.


问:
Why the graph of $\sin(\sin^{-1}(x))$ is the identity?
I just made the following plots on Mathematica:

This may be really silly but I don't understand why the plot of $\sin(\sin^{-1}(x))$ is the identity. Isn't $\sin^{-1}(x)$ is defined only on the interval $[-1,1]$? How come it has values outside of that interval? Here on wikipedia, it says:

The trigonometric functions are periodic, and hence not injective, so strictly speaking, they do not have an inverse function. However, on each interval on which a trigonometric function is monotonic, one can define an inverse function, and this defines inverse trigonometric functions as multivalued functions. To define a true inverse function, one must restrict the domain to an interval where the function is monotonic, and is thus bijective from this interval to its image by the function. The common choice for this interval, called the set of principal values, is given in the following table.

So is Mathematica using more than one set of principal values? I'm confused.

答:
The sine function maps the real axis onto the interval $[-1, 1]$, but it maps the complex plane onto itself. From
$$
 \sin(z) = \frac{e^{iz} - e^{-iz}}{2i}
$$
it is not difficult to see that the equation $\sin(z) = w$ has (infinitely many) solutions for every $w \in \Bbb C$.

ArcSin[z] is defined for arbitrary complex values $z$ (the precise formula is given in the reference) and returns some (complex) number such that Sin[ArcSin[z]] == z.


问:
Gre counting question
Two men and two women are to be selected from $5$ sets of twins, where each set consists of $1$ man and $1$ woman. If only $1$ person from each set of twins may be selected for the group, what's the total number of distinct group that can be formed?

It seems a lot if I count by hand, the answer is $20$, but I keep counting either excess or lower to it, anyone see the pattern could help me to count quickly and accurately?

答:
Select the 2 men from the 5 sets in $\binom 52=10$ ways.

Now, you can't select a woman from those 2 sets already selected, so you select 2 women from the remaining 3 sets in $\binom 32=3$ ways.

By the rule of product, you can form the committee in $10\times 3=30$ ways.

Another approach would be to first select 4 sets in $\binom 54=5$ ways. Now, select two of these 4 sets for the men in $\binom 42=6$ ways and the women are automatically determined from the remaining 2 sets, so by the rule of product, form the committee in $5\times 6=30$ ways.

Title: Gre counting question


问:
simply connected open as a union of convex open subsets
Let $U$ be a simply connected and connected open subset of $\mathbb R^2$. Is it always possible to write
$U$ as a finite (non-disjoint) union of convex open subsets of $\mathbb R^2$?

答:
Here are a couple of bounded examples.

Let $U=\{(x,y):0<x<1, 0<y<x^2\}.$

Let $P=\{(x,y):1<x^2+y^2<2, 0<y\}.$

The proofs are similar, here is a proof for $P$.
Let $C=\{(x,y):1=x^2+y^2, 0\le y\}.$ If $V$ is
any convex open subset of $P$ then its closure
$K$ is a compact convex subset which intersects $C$
in at most one point. If $V_1,...,V_n$ are
finitely many convex open subset of $P$, then
their closures $K_1,...,K_n$ intersect $C$ in at most finitely many points, and hence there is some $(p,q)\in C\setminus \cup_{i=1}^n K_i.$
The set $K=\cup_{i=1}^n K_i$ is compact, hence
the distance from $(p,q)$ to $K$ is positive,
hence there are points $(x,y)$ near $(p,q)$ in
$P\setminus K$, so such $(x,y)$ are not covered
by any of the $V_1,...,V_n.$


问:
Find a metric $D$ on $X=\mathbb{R}^2-\{(0,0)\}$ so $(X,D)$ is complete and topologically equivalent to $(X,d)$
I'm trying to find a metric $D$ on the set of $\mathbb{R}^2$ without the origin, so that it forms a complete metric space. There is also the restriction that this metric space has the same open sets of $(X,d)$ where $d$ is the Euclidean metric on $\mathbb{R}^2$.

In other words, I wish to find a metric $D$ on $X=\mathbb{R}^2-\{(0,0)\}$ so $(X,D)$ is complete and topologically equivalent to $(X,d)$

My attempts so far have revolved around the function $$f(x,y) = \frac{1}{x^2+y^2},$$
so that the Cauchy sequences in $\mathbb{R}^2$ converging to the origin under the Euclidean metric are no longer Cauchy in $X$ under whatever metric $D$ I eventually find.

I let $u,v\in\mathbb{R}^2$. Then, I tried naively defining $D(u,v)=|f(u)-f(v)|$, but it doesn't work because $D(u,-u)=0$. Then, I tried $D=\frac{1}{(d(u,v))^2}$, but it's not defined when $u=v$.

I'm not sure what direction to take. Any help would be greatly appreciated!

答:
$D(x,y)=\|x-y\|+|\frac 1 {\|x\|} -\frac 1 {\|y\|}|$ is such a metric.

If $U$ is any open set in  complete metric space  $(X,d)$ then $D(x,y)=d(x,y)+|\frac  1{d(x,U^{c})}-\frac  1{d(y,U^{c})}|$ defines a new metric which has same open sets as $(X,d)$ and $(U,D)$ is complete.


问:
$I$ a maximal ideal in ring $R$ if and only if for all $a \in R\setminus I, \exists r\in R, b \in I$ such that $b+ar=1$.
Let $R$ be a commutative Ring with $1\neq 0$. Let $I$ be a proper ideal of $R$. Show that $I$ is maximal $\iff$ for all $a \in R\setminus I$ there are $r\in R$ and $ b \in I$ such that $b+ar=1$.

I'm not sure how to approach proving this.

答:
It is helpful to view this as an ideal analog of $\,B\,$ prime, $\,B\nmid A\Rightarrow \gcd(A,B) = 1.\,$ Indeed, in PIDs  we have $\, B\supseteq A\iff B\mid A\,$ and $\,A+B = \gcd(A,B)\,$ (and similarly in Dedekind domains). Keep this analogy in mind when reading the generalization below.

Theorem $\ $ The following are equivalent for an ideal $\,B\,$ in a commutative ring $\,R$.

$(1)\ \ A\not\subseteq B \Rightarrow \ \, A\,+\,B = (1)$

$(2)\ \ \ a\not\in B \Rightarrow\ (a)+B=(1)$

$(3)\ \ \ a\not\in B \Rightarrow\ \  ar\,+\,b\, =\, 1,\,$ for some $\,b\in B.\ r\in R$

$(4)\ \ \,B\,$ is a maximal ideal

Proof $\ \  (1\Rightarrow 2\Rightarrow 3)\  $ clear.

$(3\Rightarrow 4)\ \,$ If $\,A\supsetneq B\,$ then there is $\,a\in A,\,a\not\in B\,$ so $\,1 = ar+b\in A\,$ so $\,A = (1)$

$(4\Rightarrow 1)\ \ $ The ideal $\,A+B\supsetneq A\,$ so $\,A+B = (1)$ by maximality of $B$.


问:
Computing Beta Function for Polya's Urn Scheme
Suppose we have an urn that contains $r$ red and $g$ green balls. At each time we draw a ball, we look at the color, replace it, and add $c$ more balls of that color.

Let $X_n$ be the ratio of green balls to the total number of balls after the $n^{th}$ draw.

It is not hard to see that $X_n$ is a martingale, and since $X_n \geq 0$, it will converge absolutely to some $X_\infty$.

It is also not hard to see that the probability of getting a green ball on the first $m$ draws and a red ball on the next $l = n-m$ draws is
$$\frac{g}{g+r}\cdot \frac{g+c}{g+r+c} \cdots \frac{g+(m-1)c}{g+r+(m-1)c} \cdot \frac{r}{g+r+mc}\cdot \frac{r+c}{g+r+(m+1)c}\cdots\frac{r+(l-1)c}{g+r+(n-1)c}.$$

Also, any other outcome of $n$ total draws with drawing $m$ green and $l$ red balls will have the same probability.

Can anyone help me see why the above probability should be
$$\frac{\beta(\frac{g}{g+r+c},\frac{r}{c+n-m})}{\beta({\frac{g}{c}, \frac{r}{c}})},$$
where $\beta(a,b) = \int_0^1 x^{a-1}(1-x)^{b-1} dx$?

I know that $\gamma(a+1)=a\gamma(a)$, but I can't figure out a nice way to compute the betas.

答:
I think the required equality is wrong. For instance, when $g=r=c=m=l$ then the product of the fractions equals $\frac 16$, whereas
$\frac {B\left(\tfrac 13,\tfrac 12\right)} {B\left(1,1\right)}= B\left(\tfrac 13,\tfrac 12\right)=4.2\dots$.

On the other hand, I obtained a different expression with $B$-function for the product.

Below we shall use a well-known equality that $B(x,y)=\tfrac{\Gamma(x) \Gamma(y)}{ \Gamma(x+y)}$ for any $x,y>0$. Thus

$$\frac{g}{g+r}\cdot \frac{g+c}{g+r+c} \cdots \frac{g+(m-1)c}{g+r+(m-1)c} \cdot \frac{r}{g+r+mc}\cdot \frac{r+c}{g+r+(m+1)c}\cdots\frac{r+(l-1)c}{g+r+(n-1)c}=$$
$$\frac{g/c}{g/c+r/c}\cdot \frac{g/c+1}{g/c+r/c+1} \cdots\frac{g/c+(m-1)}{g/c+r/c+(m-1)/c}\times$$ $$\frac{r/c}{g/c+r/c+m}\cdot \frac{r/c+1}{g/c+r/c+(m+1)}\cdots\frac{r/c+(l-1)}{g/c+r/c+(n-1)}=$$
$$\frac{\tfrac{\Gamma(g/c+m)}{\Gamma(g/c)}\cdot \tfrac{\Gamma(r/c+l)}{\Gamma(r/c)}}
{ \Gamma(g/c+r/c+n)/\Gamma(g/c+r/c)}=\frac{B(g/c+m,r/c+n-m)}{B(g/c,r/c)}.$$


问:
What is the dependent formula of the vectors
Are vectors $(1,0,3,4)$, $(1,0,1,1)$, $(4,1,2,3)$ and $(6, 1,8,11)$ linearly
dependent? If yes, what is  the dependence formula.

I found out the determinant is $0$ isn't is supposed to be independent and what is dependance formula.
What is the dependance formula in this case?

答:
You can either calculate the determinant, or you can put you vectors as rows in matrix $A$ and try to find the row reduced echelon form.

If the row reduced echelon form leaves one row with only zeros, what does this mean?

What you are doing when you find the row reduced echelon form, is actually to try to eliminate entries in your matrix by writing the rows as linear combination of each other. And a set of vectors are linearly independent if you cant write one vector of them as a linear combination of the other vectors, of more formal:

let $v_1, v_2 \cdots v_n$ be a set of vectors. Then they are linearly independent if
$$ \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n = 0 $$ implies
$$ \alpha_1 = \alpha_2 = \cdots \alpha_n = 0. $$

Remember that if the determinant of the row reduced echelon form is zero then also the determinant of the original matrix is zero.


问:
Help to show that $\bigoplus_{i\in\mathbb{N}}\mathbb{Z}\cong\operatorname{Hom}(\prod_{i\in\mathbb{N}}\mathbb{Z},\mathbb{Z})$.
I am trying to show that $\bigoplus_{i\in\mathbb{N}}\mathbb{Z}\cong\operatorname{Hom}(\prod_{i\in\mathbb{N}}\mathbb{Z},\mathbb{Z})$, as Abelian groups. I know that there is the map $f:\bigoplus_{i\in\mathbb{N}}\mathbb{Z}\rightarrow\operatorname{Hom}(\prod_{i\in\mathbb{N}}\mathbb{Z},\mathbb{Z})$ that takes an element $a=(a_1,a_2,...)$ to the function $f_a:\prod_{i\in\mathbb{N}}\mathbb{Z}\to\mathbb{Z}$, where $f_a(x_1,x_2,...)=\sum_{i=1}^\infty a_ix_i$. By definition, every element of $\bigoplus_{i\in\mathbb{N}}\mathbb{Z}$ has only finitely many non-zero entries. This means that the sum in $f_a$ will be of only finitely many non-zero terms and hence always converge, so $f_a$ is well-defined. I am able to show that each $f_a$ is a group homomorphism for every $a$ in $\bigoplus_{i\in\mathbb{N}}\mathbb{Z}$, and that $f$ defines an injective group homomorphism, but I can't at all see why $f$ is surjective...

答:
This is a lovely theorem, due – if I'm not mistaken – to Reinhold Baer. For convenience, we will denote $\prod_{i\in\mathbb{N}}\mathbb{Z}$ by $\mathbb{Z}^\mathbb{N}$ and $\bigoplus_{i\in\mathbb{N}}\mathbb{Z}$ by $\mathbb{Z}^{\oplus\mathbb{N}}$. Let $e_0=(1,0,0,\dots)$, $e_1=(0,1,0,\dots)$ denote the canonical basis for the submodule $\mathbb{Z}^{\oplus\mathbb{N}}<\mathbb{Z}^\mathbb{N}$. We wish to define an inverse to your map $f$; the naive way of doing this is to try to define $g:\operatorname{Hom}(\mathbb{Z}^\mathbb{N},\mathbb{Z})\rightarrow\mathbb{Z}^{\oplus\mathbb{N}}$ by $\phi\mapsto (\phi(e_1),\phi(e_2),\dots)$. Unfortunately, a priori, there is no guarantee that this is well-defined; we need to show that only finitely many of the $\phi(e_i)$ will be non-zero.

Claim 1: for any morphism $\phi:\mathbb{Z}^\mathbb{N}\rightarrow\mathbb{Z}$, $\phi(e_i)$ is non-zero for only finitely many $i$.

Proof: Suppose for contradiction we have some $\phi$ for which this is not the case. Then, by considering $-\phi$ if necessary, we may assume that $\phi(e_i)$ is strictly positive for infinitely many $i$. Say $n_1<n_2<\dots\in\mathbb{N}$ are such that $\phi(e_{n_i})>0$. Then, by considering the homomorphism $\mathbb{Z}^\mathbb{N}\rightarrow\mathbb{Z}^\mathbb{N}$ taking the $i$-th copy of $\mathbb{Z}$ in the first direct product to the $n_i$-th copy of $\mathbb{Z}$ in the second direct product, we may assume that $\phi(e_i)>0$ for all $i$.

Now, choose $(n_i)_{i\in\mathbb{N}}$ a sequence of integers such that $2^{n_{i+1}}>2^{n_i}\phi(e_i)$ for each $i$, and let $\bar{x}=(2^{n_0},2^{n_1},2^{n_2},\dots)\in\mathbb{Z}^\mathbb{N}$. Note that $\bar{x}-\sum_{i=0}^{k-1}2^{n_i}e_i$ is divisible by $2^{n_k}$ in $\mathbb{Z}^\mathbb{N}$, for every $k\in\mathbb{N}$. This means that $$\phi(\bar{x}-\sum_{i=0}^{k-1}2^{n_i}e_i)=\phi(\bar{x})-\sum_{i=0}^{k-1}2^{n_i}\phi(e_i)$$ is divisible by $2^{n_k}$ in $\mathbb{Z}$ for every $k$. Along with the construction of the $n_i$, this means that, in the binary expansion of $\phi(\bar{x})$, the entries between the $2^{n_i}$-th place and the $2^{n_{i+1}}$-th place coincide with the binary expansion of $\phi(e_i)$. In particular, since every $\phi(e_i)$ is strictly positive, $\phi(\bar{x})$ has strictly positive entries arbitrarily high in its binary expansion, a contradiction. (Every integer has all but finitely many entries in its binary expansion equal to zero.)

Thus the map $g$ is indeed well-defined. It is now easy to verify that $g$ is a $\mathbb{Z}$-module morphism, and also that $g\circ f=\operatorname{id}_{\mathbb{Z}^{\oplus\mathbb{N}}}$; I'll let you check this. Thus we need only show that $f\circ g=\operatorname{id}_{\mathbb{Z}^{\mathbb{N}}}$. For this, we claim it suffices to show that $\operatorname{Hom}(\mathbb{Z}^\mathbb{N}\big/\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})\cong\{0\}$. Indeed, if this is true, then any two maps $\mathbb{Z}^\mathbb{N}\rightarrow\mathbb{Z}$ that coincide on $\mathbb{Z}^{\oplus\mathbb{N}}$ are in fact the same (their difference will map $\mathbb{Z}^{\oplus\mathbb{N}}$ to $0$, and hence by the desired result be identically zero). Since the $e_i$ are a basis for $\mathbb{Z}^{\oplus\mathbb{N}}$, showing that two maps coincide on $\mathbb{Z}^{\oplus\mathbb{N}}$ amounts to showing that they coincide on the $e_i$, and it is easy to compute that $(f\circ g)(e_i)=\operatorname{id}_{\mathbb{Z}^\mathbb{N}}(e_i)$, as needed. Thus, after the claim below, we will be done.

Claim 2: $\operatorname{Hom}(\mathbb{Z}^\mathbb{N}\big/\mathbb{Z}^{\oplus\mathbb{N}},\mathbb{Z})\cong\{0\}$.

Proof: For an arbitrary $\mathbb{Z}$-module $M$, we say that $m\in M$ is infinitely divisible if, for infinitely many $k\in\mathbb{Z}$, there is some $n_k\in M$ such that $m=kn_k$. The only infinitely divisible element of $\mathbb{Z}$ is $0$, so any $\mathbb{Z}$-module morphism from $M$ to $\mathbb{Z}$ must send all infinitely divisible elements of $M$ to $0$. Thus, to show our claim, it will suffice to show that $M:=\mathbb{Z}^\mathbb{N}\big/\mathbb{Z}^{\oplus\mathbb{N}}$ is generated by infinitely divisible elements. (Indeed, if a module morphism maps some collection of generators of a module to $0$, then it maps every element of the module to $0$.)

$M$ has an ample supply of infinitely divisible elements. For instance, given a fixed $k\in\mathbb{N}\setminus\{0\}$, and any $\bar{x}=(x_0,x_1,\dots)\in\mathbb{Z}^\mathbb{N}$, denote the coset of $(x_0,kx_1,k^2x_2,\dots,k^ix_i,\dots)$ in $M$ by $\bar{x}^{(k)}$. We claim that $\bar{x}^{(k)}$ is divisible by every power of $k$, and thus is infinitely divisible. To see this, fix $i\in\mathbb{N}$. We have that $$(x_0,kx_1,k^2x_2,\dots)=k^i(0,\dots,0,x_i,kx_{i+1},k^2x_{i+2},\dots)+(x_0,kx_1,\dots,k^{i-1}x_{i-1},0,0,\dots).$$ The first term in the sum on the right hand side is divisible by $k^i$, while the second term is an element of $\mathbb{Z}^{\oplus\mathbb{N}}$. Thus $(x_0,kx_1,k^2x_2,\dots)$ differs from a $k^i$-divisible element of $\mathbb{Z}^\mathbb{N}$ by an element of $\mathbb{Z}^{\oplus\mathbb{N}}$, and so $\bar{x}^{(k)}$ (its image in $M$) is $k^i$-divisible in $M$.

We claim now that $M$ is generated by the collection $\{\bar{x}^{(2)},\bar{x}^{(3)}:\bar{x}\in\mathbb{Z}^\mathbb{N}\}$. To see this, recall that $2^i$ and $3^i$ are coprime for any $i$. Thus, given an arbitrary element $\bar{y}\in\mathbb{Z}^\mathbb{N}$, for each $i$ we can find some $a_i,b_i\in\mathbb{Z}$ such that $y_i=a_i2^i+b_i3^i$. This means that $\bar{y}+\mathbb{Z}^{\oplus\mathbb{N}}=\bar{a}^{(2)}+\bar{b}^{(3)}$ as elements of $M$, as desired. Thus $M$ is generated by a collection of infinitely divisible elements, so any map $M\rightarrow\mathbb{Z}$ must be trivial, and so we are done.


问:
Set of surjective bounded linear operator between two Banach spaces.
I've to prove that the set of all subjective bounded linear operator is a closed subset of B(X, Y).

I tried with a sequence $T_n$ of surjective bounded maps, suppose it converges to some T $\in$ B(X, Y).

I've to show now T is surjective and bounded.

Let y $\in$ Y, for each $T_n$ there exists $x_n \in$ X such that $T_n(x_n)=y$.

Can someone help me how to proceed from here?

答: This is false. Take $Y$ to be the scalar field. Let $f$ be any non-zero continuous linear functional on $X$ And $f_n=\frac  f n$. Then $f_n \to 0$ in operator norm and each $f_n$ is surjective.


问:
Find a metric on $X = \mathbb{R}^2 - \{(0,0), (0,1)\}$ so $(X,D)$ is complete and topologically equivalent to $(X,d)$.
I originally asked this question with only one "pole" at the origin, however I wanted to understand the problem in the case of multiple "poles." The advice I got in the other post was to convert to cylindrical coordinates. While I'm currently trying to understand their advice, I don't think this would generalize readily to an $X$ with multiple holes.

I'm trying to find a metric $D$ on the set of $\mathbb{R}^2$ without the origin, so that it forms a complete metric space. There is also the restriction that this metric space has the same open sets of $(X,d)$ where $d$ is the Euclidean metric on $\mathbb{R}^2$.

In other words, I wish to find a metric $D$ on $X=\mathbb{R}^2−\{(0,0)\}$ so $(X,D)$ is complete and topologically equivalent to $(X,d)$

My attempts so far have revolved around the function
$$f(x,y)=\frac{1}{x^2+y^2},$$
so that the Cauchy sequences in $\mathbb{R}^2$ converging to the origin under the Euclidean metric are no longer Cauchy in $X$ under whatever metric $D$ I eventually find.

I let $u,v\in\mathbb{R}^2$. Then, I tried naively defining $D(u,v)=|f(u)−f(v)|$, but it doesn't work because $D(u,−u)=0$. Then, I tried $D=\frac{1}{(d(u,v))^2}$, but it's not defined when u=v.

I'm not sure what direction to take. Any help would be greatly appreciated!

So in this case, where $X = \mathbb{R}^2 - \{(0,0), (0,1)\}$ so $(X,D)$, I would look at the function
$$g(x,y)=\frac{1}{x^2+y^2}+\frac{1}{(x-1)^2+y^2},$$
at which point I'm stuck. I want to find an explicit formula, but all my attempts so far failed to even satisfy the conditions for being a metric, let alone making $(X,D)$ complete and whatnot. I would really appreciate any help in the right direction.

答:
HINT: Let $p=\langle 0,0\rangle$ and $q=\langle 0,1\rangle$, for $x\in X$ let

$$f(x)=\frac1{\min\{d(x,p),d(x,q)\}}\,,$$

and for $x,y\in X$ let

$$D(x,y)=d(x,y)+|f(x)-f(y)|\,.$$

Show that $D$ generates the same topology on $X$ as $d$ and that $\langle X,D\rangle$ is complete.

More generally, if $G$ is open in a complete metric space $\langle X,d\rangle$, let

$$f(x)=\frac1{d(x,X\setminus G)}$$

and

$$D(x,y)=d(x,y)+|f(x)-f(y)|\,;$$

then $D$ generates the same topology on $X$ as $d$, and $\langle X,D\rangle$ is complete.


问:
Maximum/Minimum of a quadratic form on the unit sphere
Prove that, the maximum/minimum of the quadratic form
$$
f(x, y, z)=A x^{2}+B y^{2}+C z^{2}+2 D y z+2 E z x+2 F x y
$$
on the unit sphere surface
$$
x^{2}+y^{2}+z^{2}=1
$$
is exactly the corresponding max/min eigenvalue of the matrix
$$
\Phi=\left(\begin{array}{lll}
A & F & E \\
F & B & D \\
E & D & C
\end{array}\right)
$$

I don't even know where to start. Can anyone help?

答:
The key fact is that every real symmetric matrix is orthogonally diagonalisable, i.e. we can express $\Phi$ as $Q\Lambda Q^T$, where $Q$ is an orthogonal matrix and $\Lambda=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$ is a diagonal matrix whose diagonal entries the eigenvalues of $\Phi$. The proof of this fact can be found in most introductory texts on linear algebra.

Since $Q$ is orthogonal, $(x,y,z)\mapsto(u,v,w):=(x,y,z)Q$ is a bijective mapping on the unit sphere. Hence
$$
\max_{x^2+y^2+z^2=1}f(x,y,z)
=\max_{x^2+y^2+z^2=1}\pmatrix{x&y&z}\Phi\pmatrix{x\\ y\\ z}
=\max_{u^2+v^2+w^2=1}\pmatrix{u&v&w}\Lambda\pmatrix{u\\ v\\ w}
$$
and the rest should be straightforward.


问:
Evaluate this power series
Evaluate the sum

$$x+\frac{2}{3}x^3+\frac{2}{3}\cdot\frac{4}{5}x^5+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}x^7+\dots$$

Totally no idea. I think this series may related to the $\sin x$ series because of those missing even powers. Another way of writing this series:

$$\sum_{k=0}^{\infty}\frac{(2k)!!}{(2k+1)!!}x^{2k+1}.$$

Title: Evaluate this power series

Body:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
$\ds{\pars{2k}!! = 2^{k}\,\Gamma\pars{k + 1}}$ and
$\ds{\pars{2k + 1}!! = 2^{k + 1}\,{\Gamma\pars{k + 3/2} \over \Gamma\pars{1/2}}}$.
\begin{align}
&\bbox[5px,#ffd]{\sum_{k = 0}^{\infty}{\pars{2k}!! \over
\pars{2k + 1}!!}\,x^{2k + 1}}
\\[5mm] = &\
{1 \over 2}\sum_{k = 0}^{\infty}{\Gamma\pars{k + 1}
\Gamma\pars{1/2} \over \Gamma\pars{k + 3/2}}\,x^{2k + 1}
\\[5mm] = &
{1 \over 2}\sum_{k = 0}^{\infty}\bracks{%
\int_{0}^{1}t^{k}\pars{1 - t}^{-1/2}\,\,\dd t}\,x^{2k + 1}
\\[5mm] = &\
{1 \over 2}\,x\int_{0}^{1}{1 \over \root{1 - t}}
\sum_{k = 0}^{\infty}\pars{x^{2}\,t}^{k}\,\dd t
\\[5mm] = &\
{1 \over 2}\,x\int_{0}^{1}{1 \over \root{1 - t}}
\sum_{k = 0}^{\infty}\pars{x^{2}\,t}^{k}\,\dd t
\\[5mm] = &\
{1 \over 2}\,x\int_{0}^{1}{\dd t \over
\root{1 - t}\pars{1 - x^{2}t}}
\\[5mm] \stackrel{t\ =\ 1 - \xi^{2}}{=}\,\,\,\,\,&
x\int_{0}^{1}{\dd\xi \over
1 - x^{2}\pars{1 - \xi^{2}}}
\\[5mm] = &\
x\,{1 \over 1 - x^{2}}\,{\root{1 - x^{2}} \over x}\int_{0}^{1}{\pars{x/\root{1 - x^{2}}}\dd\xi \over
\pars{x\xi/\root{1 - x^{2}}}^{2} + 1}
\\[5mm] = &
{1 \over \root{1 - x^{2}}}\arctan\pars{x \over \root{1 - x^2}}
\\[5mm] = &\
\left.{\arctan\pars{\tan\pars{\theta}} \over \root{1 - x^{2}}}
\,\right\vert_{\,x\ =\ \sin\pars{\theta}} =
\bbx{\arcsin\pars{x} \over \root{1 - x^{2}}} \\ &
\end{align}


问:
Are these 2 graphs isomorphic?

They meet the requirements of both having an $=$ number of vertices ($7$).

They both have the same number of edges ($9$).

They both have $3$ vertices of degree $2$ and $4$ of degree $3$.

However, graph two has $2$ simple circuits of length $3$ whereas graph one has only $1$ of length $3$.

Is this not a valid method for checking isomorphism?

My guide says that these two figures are isomorphic.

答:
If two graphs are isomorphic, then if we represent one of them as an matrix, we can find an adjacency matrix for the other which is identical, except for the names of the nodes and edges.

An adjacency matrix for the left graph is:

```<code>   A B C D E F G
A:   1 1 1
B: 1     1 1
C: 1         1
D: 1       1   1
E:   1   1     1
F:     1       1
G:       1 1 1
</code>```

If we identify the vertices as (A B C D E F G) = (7 4 3 6 5 2 1), and write down the adjacency matrix for the second graph in that order, we get the same matrix:

```<code>   7 4 3 6 5 2 1
7:   1 1 1
4: 1     1 1
3: 1         1
6: 1       1   1
5:   1   1     1
2:     1       1
1:       1 1 1
</code>```

Of course, the objects are not isomorphic if the names of the vertices are considered significant in the representation.  Or if there are constraints, such as that A must be identified with node 1 (because, say, the graph are part of some larger object, and how they connect to it is not negotiable).

Two entities that are considered isomorphic are never identical unless they are the same entity.

An isomorphism is always based on caring about some possible differences, while declaring that others do not matter.

Title: Are these 2 graphs isomorphic?

Body:
If two graphs are isomorphic, then if we represent one of them as an matrix, we can find an adjacency matrix for the other which is identical, except for the names of the nodes and edges.

An adjacency matrix for the left graph is:

   A B C D E F G
A:   1 1 1
B: 1     1 1
C: 1         1
D: 1       1   1
E:   1   1     1
F:     1       1
G:       1 1 1

If we identify the vertices as (A B C D E F G) = (7 4 3 6 5 2 1), and write down the adjacency matrix for the second graph in that order, we get the same matrix:

   7 4 3 6 5 2 1
7:   1 1 1
4: 1     1 1
3: 1         1
6: 1       1   1
5:   1   1     1
2:     1       1
1:       1 1 1

Of course, the objects are not isomorphic if the names of the vertices are considered significant in the representation.  Or if there are constraints, such as that A must be identified with node 1 (because, say, the graph are part of some larger object, and how they connect to it is not negotiable).

Two entities that are considered isomorphic are never identical unless they are the same entity.

An isomorphism is always based on caring about some possible differences, while declaring that others do not matter.


问:
Inverse function of a polynomial $x(x-1)(x+1) = 0 $
How to get the interval to get the inverse function of the polynomial $$x(x-1)(x+1) = f(x) $$, to show that the function is surjective.

The interval where the function is surjective is $x > 1$ and $x < -1$, how to get inverse on this interval ?

答:
Consider $$x(x-1)(x+1) =x^3-x= y$$ and then th cubic equation
$$x^3-x-y=0$$ for which the discriminant $\Delta=4-27y^2$ must be negative in order to have only one real root. This defines
$$-\frac{2}{3 \sqrt{3}} <y <\frac{2}{3 \sqrt{3}}$$

Using the hyperboic method for only one real root, we then have
$$x=\frac{2}{\sqrt{3} }\frac{ |y|}y   \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{3 \sqrt{3}
   }{2}|y|\right)\right)$$


问:
Prove that if the partition $\textbf{P}'$ is finer than $\textbf{P}$, then $f$ keeps piecewise constant
Let $I$ be a bounded interval, let $\textbf{P}$ be a partition of $I$, and let $f:I\to\textbf{R}$ be a function which is piecewise constant with respect to $\textbf{P}$. Let $\textbf{P}'$ be a partition of $I$ which is finer than $\textbf{P}$. Then $f$ is also piecewise constant with respect to $\textbf{P}'$.

MY ATTEMPT

Let $\textbf{P} = \{J_{1},J_{2},\ldots,J_{m}\}$ be a partition of $I$. Since $\textbf{P}'$ is finer than $\textbf{P}$, each bounded interval $J_{i}$ can be written as the disjoint union of bounded intervals
\begin{align*}
J_{i} = \bigcup_{j=1}^{n_{i}}K_{ij}
\end{align*}
where $\textbf{P}' = \{K_{11},\ldots,K_{1n_{1}},\ldots,K_{m1},\ldots,K_{mn_{m}}\}$. Thus, if $x\in K_{ij}$, then $x\in J_{i}$. Consequently, $f|_{K_{ij}} = f|_{J_{i}} = c_{i}$. That's to say, $f$ is also piecewise constant with respect to $\textbf{P}'$, and we are done.

Is there any theoretical flaw in my reasoning? Please let me know if it is the case.

答:
How about proving it by using just definitions of piecewise constant function and refinement partitions?

Let P and $P'$ be partitions of I such that P' is finer than P.
Suppose $f : I \to R$ is a piecewise constant function with respect to P. So, $\forall$ K $\in$ $P'$ $\exists$ a $J\in P$ such that K$\subseteq$J.

Since f is a constant function on J and K$\subseteq$J, f is also constant on K $\forall$  $K\in P'$. Therefore, f is also piecewise constant with respect to $P'$.


问:
Scale Invariance of the Distribution of First Digits
Why is the distribution of first digits scale invariant? That is, if we multiply all our numbers by an arbitrary constant then why does the distribution of first digit frequencies remain unchanged. This is related to Benford's Law. I was reading through https://plus.maths.org/content/looking-out-number-one#:~:text=Using%20this%20reasoning%2C%20Pinkham%20went,Benford's%20Law%20really%20does%20exist. and was confused when going through the first two paragraphs in the section "Pinning down Scale Invariance"</p>

答: You need to look at what is assumed and what is derived.  I believe that we assume or discover empirically that the distribution is scale invariant.  We then derive Benford's law from this.  The simple idea is that if the distribution is scale invariant, we can multiply everything by a scale factor.  If we choose that factor to be $5$, we take the points that had a leading digit $1$, meaning they were in the interval $[1,2)$  and spread them among the interval $[5,10)$ with leading digits $5$ through $9$.  Looking at this in detail, we find that the fraction of points between $a$ and $b$ is $\log b - \log a$.  Specializing that to leading digits gives Benford's law.


问:
Limit of $0^0$ (Indeterminate form)
Well this format of a limit $0^0$ is an indeterminate form.

I claim that whatever this limit is (which depends on the exact question) should always be in between $[0,1]$.

Is my claim correct?

I have no mathematical proof for it but just a basic idea, that any number base $0$ should try to pull towards $0$, whereas any number power $0$ should pull towards $1$.

答:
That's not true in general. Take $$y(t)=e^{-1/t}$$ and $$x(t)=-5t$$ say.
Then $$\lim_{t \rightarrow 0^+} y(t)  = \lim_{t \rightarrow 0^+} x(t)=0,$$ but $$\lim_{t \rightarrow 0^+} (e^{-1/t})^{-5t}= e^5 >1.$$

For an even more dramatic example take $y(t)=e^{-1/t^2}$ and $x(t)=-t$.

HOWEVER, if $y(t)$ and $x(t)$ as above are both positive for then infact the limit will indeed fall between 0 and 1, as any positive number less than 1 raised to a positive power is in [0,1].

Title: Limit of $0^0$ (Indeterminate form)


答:
Your claim is wrong. You can choose any nonnegative number as the limit:
$$\lim_{x\to0^+}\left(e^{-1/x}\right)^{ax} = e^{-a}.$$

This example is from the Wikipedia article on $0^0$, as @PrasunBiswas suggested. There are other examples there showing a limit of infinity, etc.

Title: Limit of $0^0$ (Indeterminate form)

Body:
Well this format of a limit $0^0$ is an indeterminate form.

I claim that whatever this limit is (which depends on the exact question) should always be in between $[0,1]$.

Is my claim correct?

I have no mathematical proof for it but just a basic idea, that any number base $0$ should try to pull towards $0$, whereas any number power $0$ should pull towards $1$.


问:
Decreasing from the horizontal asymptote
The function $f(x) = x^2/(x^2 - x -2)$ has the following graph. It has a horizontal asymptote $y=1$. For $x$ less than $-4$, the function is decreasing and its graph is under the asymptote. How is this possible when $\lim_{n \to -\infty} f(x) = 1$? Can a function decrease away from its vertical asymptote?</p>

答:
The definition of increasing is
$$
x_1<x_2\implies f(x_1)\ge f(x_1).
$$
As $x$ increases, $f(x)$ decreases. But as $x$ decreases to $-\infty$, $f(x)$ increases to $1$.

Maybe the following example will make things clear. $x^2$ decreases on $(-\infty,0)$, but $\lim_{x\to-\infty}x^2=+\infty$.


问:
How to solve this mathematical programming problem?
Suppose each point in the interval $[0,1]$ has some of a set of $N$ properties.  The total length of the intervals with property $i$ is $x_i$ ($0\le x_i \le 1$).

Questions:

What is the maximum and minimum of the length of intervals of points with exactly $m$ properties?

Is there some none-zero-length interval with $m$ properties?  This second question may be seen as part of the first one.

Not sure about what is the proper name of such kind of problems.

An example:

In a school, 80% of the students play basketball, 60% play football, 60% play tennis, 60% play baseball.

Question: Is it true that, there must be some students play all four of the sports?  What is the maximum and minimum fraction of students play three of the four sports?

答:
Introduce $2^4=16$ nonnegative decision variables, one for each subset of the four sports.  They must sum to $1$, and the given percentages imply four additional linear constraints.

To determine whether some students must play all four sports, minimize the corresponding decision variable by using linear programming.  The minimum is positive if and only if some students must play all four sports.

The fraction that play three sports is a sum of four decision variables and hence a linear function.  You can again use linear programming to minimize or maximize.


问:
Application problem of natural logarithms

A student has 3 hours to cram for an examination and during this time wishes to memorize a set of 60 facts. According to psychologists, the rate at which a person can memorize a set of facts is proportional to the number of facts remaining to be memorized. Thus, if a student memorizes y facts in t minutes, then

$$ \frac{dy}{dt} = k (60 - y) $$
where k is a positive constant and  and y < 60 for all values t $ \ge $ 0

Assume that zero facts are memorized initially and the student memorizes 15 facts in the
first 20 minutes .
a.  Express y as a function of t.
b.  How many facts will the student memorize in 1 hour?
c.  How many facts will the student memorize in 3 hours?
d.  How many facts will the student memorize in the long run?
e.  How long will the student memorize 40 facts?

My main problem is a.

I used the differential equation $ y(t) = k (y(t) - A) $ and I got $ y(t) = 60 -60e^{-kt} $.

However, it took few trial and error methods to do this. But is there more detailed solution, especially by integrating $ \frac{dy}{dt} = k (60 - y) $?

答:
You are provided that,

at $t=0$, $y=0$ and

at $t=20$, $y=15$

Also, $\frac{dy}{dt} = k(60-y)$ Integrating using the limits as $y(0)= 0$ and taking $y(t) = y$ at some $t$,

$ \begin{align} &\int_0^y\frac{dy}{60-y} = \int_0^t kdt \\&\Rightarrow -\ln(\frac{60-y}{60}) = kt \\&\Rightarrow \ln(1- \frac{y}{60}) = -kt\\ &\Rightarrow \boxed{y\equiv y(t)= 60 - 60^{-kt}}\end{align}$

Now using the fact $y(20) = 15$, you can get $k$