问:
When is a relation a function in the sets $\mathbb{N},\:\mathbb{Z},\:\mathbb{R}$?
I have recently started learning about relations and functions in a more formal way than I was used to and I am stuck on the following question:

Let $M = \{\mathbb{N},\:\mathbb{Z},\:\mathbb{R}\}$, $R_A = \{(x, 2x)\:|\:x \in A\}$ and $S_A = \{(2x, x)\:|\: x \in A\}$, where $R_A$ and $S_A$ are relations. For which $A \in M$ are $R_A$ and $S_A$ functions and for which $A$ is $R_A$ a bijective function?

Now, it is obvious that $R_A$ is a function for all $A \in M$, since for any $x$ in the natural, whole and real numbers $2x$ is always well-defined. However, when it comes to checking whether $R_A$ is also a bijective function for any $A \in M$ I am a bit confused. My first thought was that its only bijective for the real numbers, since $\frac{x}{2}$ won't always be a whole number for any $x \in \mathbb{N},\:\mathbb{Z}$, but after thinking about it I am not so sure anymore. Take $A = \mathbb{N}$ for example. Then $R_A$ would define a function $f: \mathbb{N} \to 2\mathbb{N},\:x \to 2x$. If you look at it like that, than of course this would also be bijective. Same for $A = \mathbb{Z}$.

With $S_A$ I'm also a bit confused for the same reason. Take $A = \mathbb{N}$ again, then $S_A$ defines the function $f: 2\mathbb{N} \to \mathbb{N},\:2x \to x$. This is again bijective if you define the $f$ like this, because half of every even number is again a whole number and every whole number will be an even number, when doubled. Am I thinking about these relations in the wrong way?

Another part of this exercise was the following:

For two relations $U \subseteq X \times Y$ and $V \subseteq Y \times Z$ their relational product $V \circ U \subseteq X \times Z$ is defined as $V \circ U = \{(x, z)\:|\:\exists y \in Y: (x, y) \in U\:\land\:(y, z) \in V\}$.

What does this now mean, if I look at $R_A \circ S_A$ and $S_A \circ R_A$ for all $A \in M$?

I know this is a lot, but I am stuck on this and I can't really make sense of what is right and where my thought process is wrong.

答:
The question is ambiguous. For each $A\in M$ it is true that both $R_A$ and $S_A$ are functions, in the sense that if $T$ is any of these relations, and $\langle x,y\rangle,\langle x,z\rangle\in T$, then $y=z$. Thus, each of these relations is a function whose domain is the domain of the relation. However, while $R_A$ is a function on $A$ for each $A\in M$, $S_A$ is a function on $A$ only for $A=\Bbb R$, as you realized.

In much the same vein, each of these relations is a bijection between its domain and its range, but $R_{\Bbb R}$ and $S_{\Bbb R}$ are the only ones of the six that are bijections from $A$ to $A$.

I don’t know how you’re supposed to interpret the question. If I had to guess, your first reactions were probably what was intended: $R_A$ is a function (with domain $A$) for each $A\in M$, but $S_A$ is a function (with domain $A$) only for $A=\Bbb R$, and $R_{\Bbb R}$ and $S_{\Bbb R}$ are the only bijections (from $A$ to $A$). But it I had to answer it without being able to discuss this difficulty with an instructor, I’d probably write something like what wrote above, together with my justifications for my various assertions.

By definition

$$S_A\circ R_A=\{\langle a,b\rangle\in A\times A:\exists c\in A(\langle a,c\rangle\in R_A)\text{ and }\langle c,b\rangle\in S_A\}\,,$$

so $\langle a,b\rangle\in S_A\circ R_A$ if and only if there is a $c\in A$ such that $c=2a$ and $c=2b$, i.e., if and only if $a=b$ and $2a\in A$. For each $A\in M$ and each $a\in A$ it is true that $2a\in A$, so

$$S_A\circ R_A=\{\langle a,a\rangle:a\in A\}\,,$$

the identity relation on $A$.

$R_A\circ S_A$ is a little trickier, but for now I’ll leave it to you to take a stab it it; start just as I did with $S_A\circ R_A$ by writing down the definition and working out what it really says in terms of elements of $a$. This time it will matter which $A\in M$ you’re considering.


问:
Show that a random variable $S\leadsto Geom(s)$
Let $(X_n)_{n\geq1}$ a sequence of independent and identically distributed random variables such that $X_n  \leadsto$ $\mathcal Geom(p)$ and $T\leadsto$ $\mathcal Geom(r)$  another random variable $(X_n \perp T$ for all $n)$. Let $S=\sum_{k=1}^TX_k$.

Show that the charactrestic function of $T$ is given by $$\varphi_T(t)=\frac{re^{it}}{1-(1-r)e^{it}}$$

Find the characteristic function of $S$.

Deduce that $S\leadsto Geom(s)$ with $s$ to define.

\begin{align*} \varphi_T(t)&=E(e^{itT}) \\&=\sum_{n=1}^{\infty} p(T=n)(e^{it})^n \\&= \sum_{n=1}^{\infty}r(1-r)^{n-1}(e^{it})^n\\&=re^{it}\sum_{n=1}^{\infty}(1-r)^n(e^{it})^n\\&=\frac{re^{it}}{1-(1-r)e^{it}}\end{align*}

$S$ is a sum of iid random variables so, \begin{align*} \varphi_S(t)&=E(e^{itS})\\&=\prod_{k=1}^T \varphi_{X_k}(t) \\&=\left( \frac{re^{it}}{1-(1-r)e^{it}} \right)^T\end{align*}
I got stuck here ! So any help is appreciated.

答:
Two hints: First, the characteristic function of $X_k$ should involve $p$, not $r$.

Second, $S$ is a sum of a random number of $X$'s, so you cannot write
$$E(e^{itS})=\prod_{k=1}^T\phi_{X_k}(t),$$ which equates the non-random LHS to a RHS that still depends on the random variable $T$.

Instead, condition on the value of $T$:
$$E(e^{itS})=\sum_{n=1}^\infty P(T=n)E(e^{itS}\mid T=n)
=\sum_{n=1}^\infty P(T=n)E\left(e^{it(X_1+\cdots+X_n)}\right).$$ The final equality exploits the independence between $T$ and $(X_1,\ldots X_n)$.

ContentLicense: CC BY-SA 4.0


问:
A blinear form $H$ is non-degenerate if and only if $H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$
Question: Given a bilinear form $H: V \times V \rightarrow F$ on a finite dimensional V, show that $H$ is nondegenrate if and only if the only $w \in V$ s.t. $H(v,w) = 0 \quad \forall v \in V$ is $w = 0$

(i.e. H is nondegenrate if and only if $H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$)

I know that $H$ is non-degenerate if and only if the following hold:

$H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$

$H(v,w) = 0 \ \forall w \in V \Leftrightarrow v = 0$

What this questions seems to be asking me to prove is that one of the above conditions implies the other (specifically, the first condition implies the second).

Since it suffices to prove that 1) implies 2):

Assume $H(v,w) = 0 \ \forall v \in V \Leftrightarrow w = 0$. Prove that $H(v,w) = 0 \ \forall w \in V \Leftrightarrow v = 0$

I am able to prove one direction for the second if and only if, but unable to get the other direction:

The direction I can do:

If $v = 0$, then $H(v,w) = H(0,w) = H(a,w) - H(a,w) = 0 \quad$ (for some $a \in V$)

$\Rightarrow$  if $v=0$ then $H(v,w) = 0$

The direction I can't do:

If $H(v,w) = 0 \quad \forall w \in W$:

$H(0,w) = H(v,w) - H(v,w) = 0 - 0 = 0$

$H(v,v) = 0$

I feel like one of these above statements should somehow imply that $v = 0$, but I'm not sure how to get there...

Any help would be appreciated, thank you!

EDIT: To add, I would like to know if there is a way to prove this without using a dual space/dual basis argument. Intuitively, I feel like there should be a way to prove it without the dual. (i.e. just proving by going forward then backward direction)

答:
Let's denote by $V^\vee$ to the dual space of $V$, and define linear maps $H_l$ and $H_r$ from $V$ to $V^\vee$ by $H_l(v) = H(v,-)$ and $H_r(w) = H(-,w)$ for all $v,w \in V$.

Note then that the condition $$H(v,w)=0 \textrm{ for all } v \in V \!\iff\! w=0$$ can be written as $H(-,w)=0 \!\iff\! w=0$, which is to say that $H_r$ is injective, and then (because $V$ is finite-dimensional) it is an isomorphism.

Thus, the induced map $H_r^\vee : V^{\vee\vee} \to V^\vee$ of $H_r$ is also an isomorphism, and if we denote by $\operatorname{ev} : V \to V^{\vee\vee}$ to the canonical isomorphism, from $H_r^\vee \circ \operatorname{ev} = H_l$ we also have that $H_l$ is an isomorphism.

In particular, $H_l$ is injective, which is to say that $H(v,-)=0 \!\iff\! v=0$, in other words, $$H(v,w)=0 \textrm{ for all } w \in V \!\iff\! v=0.$$


问:
Iterative Numerical Integration
I am aware that this question has been asked before, (e.g. Iterative integration algorthm), however the answers are a bit dissatisfying. When using an iterative scheme, typically a few desiderata are required, so my question is the following:

Q: What are examples of iterative integration schemes that satisfy the following desiderata?

Constant memory - the memory required by the scheme should not increase with the number of iterations done, or at most increase linearly

Constant compute - each iteration should take approximately the same amount of time to carry out, or at most increase linearly (and not exponentially like with the divide&conquer approach)

Efficacy & Interpretability - the iteration scheme should be constructed in such a way, that each iterate tries to optimally solve a subproblem. (example: in GMRES we have $x_m = \operatorname{argmin}\limits_{y\in \mathcal{K}_m} \| b-Ay \|_2.$)

Stopping Criteria: There should be simple stopping criteria that allow us to make quantitative claims, e.g. something along the lines: "if $f$ is Lipschitz cont. with constant $L$, and $|I_{k+1}(f, a, b]) - \int_a^b f(x) dx|\le \epsilon$, then $|I_k(f, a, b))-I^*(f, a, b)|\le g(\epsilon, L, b-a)$"

答:
A "global" adaptive quadrature scheme is an example of what you want.

Fix: A quadrature scheme $Q(f;I)$ with error estimator $\varepsilon(f;I)$ for the value of $\int_I f$ (e.g., a Newton-Cotes formula), a method $m(I)$ of coming up with a point inside an interval $I$.

Initial data: Start with $[a,b]$ with no subdivisions and calculate $Q(f;[a,b]),\varepsilon(f;[a,b])$, $Q_0=Q(f;[a,b])$, $\epsilon_0=\varepsilon(f;[a,b])$.

Iteration step: Locate $(I,Q,\varepsilon)$ with largest $\varepsilon$, say $I=[s,t]$, and create a point $m\in(s,t)$.  Replace $([s,t],Q(f;[s,t]),\varepsilon(f;[s,t]))$ by $([s,m],Q(f;[s,m]),\varepsilon(f;[s,m]))$ and $([m,t],Q(f;[m,t]),\varepsilon(f;[m,t]))$ in our list.  Update $Q_{new}=Q_{old}-Q(f;[s,t])+Q(f;[s,m])+Q(f;[m,t])$ and total error estimate $\epsilon$ similarly.

Because the scheme $Q$ is fixed, it takes an additional $O(1)$ memory and computation (ignore limitation of fixed-precision arithmetics), except for the sorting of intervals according to error estimate which takes $O(\log n)$ for binary insertion into a sorted list.  If your quadrature scheme $Q(f;I)$ is reasonable, then it is exact on some space of integrable functions on $I$, so each iterate solves the problem on piecewise blah where your intervals define the pieces (and hopefully we have uniform convergence of piecewise blah approximations of $f$ to $f$ itself).  Finally, you have a bound of the error $\epsilon$ and can make claims about, e.g., how many iterations it will take (deterministic or probabilistic) based on $Q,f,m$.

Title: Iterative Numerical Integration


ContentLicense: CC BY-SA 3.0

Title: Show that $\frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} = \frac{16}{15}\frac{1}{(1-\frac{1}{4}z^{-1})}-\frac{16}{15}\frac{1}{(1-4z^{-1})}$

Body:
Can anyone help me clarify how this rewriting is done?

$$\frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} = \frac{16}{15}\frac{1}{(1-\frac{1}{4}z^{-1})}-\frac{16}{15}\frac{1}{(1-4z^{-1})}$$

Body:
Let $x=z^{-1}$

Then

\begin{align}
\frac{-4z^{-1}}{(1-\frac{1}{4}z^{-1})(1-4z^{-1})} =\frac{-4x}{(1-\frac{1}{4}x)(1-4x)}
\end{align}

Suppose

\begin{align}
&\frac{-4x}{(1-\frac{1}{4}x)(1-4x)}=\frac{A}{(1-\frac{1}{4}x)}+\frac{B}{(1-4x)}\tag{1}\\&\frac{-4x}{(1-\frac{1}{4}x)(1-4x)}=\frac{A(1-4x)+B(1-\frac{1}{4}x)}{(1-\frac{1}{4}x)(1-4x)}\\&-4x=A(1-4x)+B(1-\frac{1}{4}x)&\\&-4x=x\left(-4A-\frac{B}{4}\right)+A+B\end{align}

Comparing coefficients, you'll have $2$ equations involving $A$ and $B$.

$$-4A-\frac{B}{4}=-4 \tag{2}$$
$$A+B=0 \tag{3}$$

Solve $(2)$ and $(3)$ to get $A=\frac{16}{15}$ and $B=-\frac{16}{15}$ then replace them in $(1)$ and also revert $x$ back into $z^{-1}$ to get your final result.


问:
Conditional Probability using an urn.
An urn contains $5$ white balls and $6$ black balls. $3$ balls are drawn sequentially without replacement. What are the probabilities...

a. $P(\text{2nd is black} | \text{1st is white})$?

For this one I did the

$$ P(\text{1st is white})=\frac{5}{11} , $$

and

$$ P(\text{second is black}) = \frac{3}{11}*\frac{3}{11} $$

(that is the probability of WB and BB added together). I multiplied the probability of

$$ \frac{P(\text{2nd black} \cap \text{1st white})}{P(\text{1st is white})} , $$

for this I got

$$ \frac{\frac{5}{11}*\frac{3}{11}}{\frac{5}{11}} = \frac{3}{11} . $$

However since this is without replacement I did not think that the answer could be independent, some feedback would help. Alright so I have this figured out that it should be $\frac{6}{10}$.

b. If I am looking now for $P(\text{3rd is black} | \text{1st is white and 2nd is black})$, I ended up with

$$ \frac{\frac{5}{9}*\frac{3}{11}}{\frac{3}{11}} , $$

this is again an independent answer is this correct?

答:
a. P(second is black|first is white)=P(black ball out of 4 white 6 black)=6/10

b. P(third is black|1st is white and 2nd is black)=P(black ball out of 4 white and 5 black)=5/9

Title: Conditional Probability using an urn.


问:
If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.
If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.

I've been trying to solve this problem for a few days, but I haven't been able to see how can I do it. First, $\mathbb{R}_{l}$ refers to $\mathbb{R}$ with the lower limit topology (generated by intervals of the form $[a,b)$ ) and $S_{\Omega}$ refers to the uncountable well-ordered set with the order topology (generated by (a,b) and rays). Until now, I've tried to use the fact that both of those topologic spaces are Hausdorff and suppose that f is injective to get a contradiction, but I don't see where can I get it.

答: If $f$ is injective, then you  have an imbedding of $[a,b]$ onto  $f([a,b])$.  But it is easy to see that an uncountable set with the order topology cannot have the same topology as the lower limit topology.  See here.


问:
Pure states on noncommutative $\mathrm{C}^*$-algebra can "be disturbed"
In a finite dimensional commutative $\mathrm{C}^*$-algebra $A\subset B(\mathbb{C}^N)$, if a vector state $\varphi_x(f)=\langle x,f(x)\rangle$ given by $x\in P(\mathbb{C}^N)$ is pure, then there is no projection $p\in A$ that can disturb the state: that is if $p$ is a projection, then $\varphi_x(p)=\langle x,p(x)\rangle$ is zero or one, so measurement (in the sense of Hilbert space quantum mechanics) of the state $x$ with $p$ gives a value of 1/0 with probability one, if one then $x\in \text{ran }p$ so the wave function collapse is $x\mapsto p(x)=x$... if zero then $x\perp \text{ran }p$ and similarly the wave function collapse is trivial in the same way.

I want to show that a noncommutative $\mathrm{C}^*$-algebra $A$ (separable, infinite dimensional, generated by projections), there are pure states $\varphi$ that do not have this property (hopefully without passing the the enveloping vNa and using Borel functional calculus), that there are projections $p\in A$ such that it is not the case that:

$$\varphi(p)=1\text{ or }0.$$

It seems so obviously true... what I think I would like to do is consider the GNS representation $\pi$... as $A$ is noncommutative at least one of the summands give a Hilbert space $\mathsf{H}_\rho$ with dimension greater than one. Let $\xi$ be the associated vector/cyclic vector... now take $y\in (\mathbb{C}\xi)^\perp$, and form the rank one projection $q$ onto the subspace defined by $(\xi+y)/\sqrt{2}$.

Then $$\displaystyle \langle \xi,q(\xi)\rangle=\frac{1}{2}$$... and so we would be done... but there is nothing to say that there is a projection $q_0\in A$ such that $\pi(q_0)=q$...

答:
If $A$ is non-commutative and it is generated by projections then we may choose  two projections  $p$ and $q$ in $A$ which do
not commute.

Let $A_0$ be the closed *-subalgebra of $A$ generated by $p$ and $q$, and notice that, since $pq-qp\neq 0$, there exists an
irreducible representation $\pi $ of $A_0$ such that $\pi (pq-qp)\neq 0$.

Considering the universal C*-algebra $B$ generated by two projections $e$ and $f$,  there exists a *-homomorphism
$\rho :B\to A_0$ sending $e$ to $p$,  and $f$ to $q$.   The composition $\pi \circ \rho $ is then an irreducible representation of $B$,
whose spectrum is fortunately well known!

Observing that $\pi \circ \rho $ cannot be one-dimensional,  since $\pi (\rho (e))$ and $\pi (\rho (f))$ do not commute, it necessarily has to be
two-dimensional and, up to a unitary conjugation, we must have
$$
  \pi (\rho (e)) = \left[\matrix{1 & 0\cr 0 & 0}\right],\quad\text{and}\quad
  \pi (\rho (f)) = \left[\matrix{t & \sqrt{t-t^2}\cr \sqrt{t-t^2} & 1-t}\right],
  $$
where $t\in (0,1)$.

The vector state $\varphi $ on $A_0$ associated to the first canonical basis vector $e_1$ of $\mathbb C^2$, is therefore pure
and satisfies
$$
  \varphi (q) = \langle \pi (q)e_1,e_1\rangle  = \langle \pi (\rho (f))e_1,e_1\rangle  = t,
  $$
so the desired state may be obtained by  extending $\varphi $ to any  pure state of $A$.


问:
Implications of Asymptotic Lower Bound
Suppose we have the following statement
$$ \lim\inf_{n\to\infty} \frac{a_n}{n^s} >0 $$
for any $0<s<\alpha$ for some given constant $\alpha>0$. I guess one way to write this statement is to use the Big-Omega symbol and write $a_n = \Omega(n^{\alpha-\epsilon})$ for any $0<\epsilon<\alpha$.

I'm wondering if such statement has any implication of the form $a_n = \Omega(n^{\alpha}\delta(n))$ for some function $\delta(n)$ satisfying some properties? For example, a more precise statement than $\delta(n)$ of order less than fractional-polynomial?

答:
It shouldn't, no. If we restrict to the natural case where $\delta(n)$ is monotone:

If $\delta(n) \to 0$ as $n\to\infty$ but not as fast as a fractional power of $n$, then we can choose the counterexample $a_n = n^\alpha \delta(n)^2$;

Otherwise we can choose say $a_n = n^\alpha/\log n$.

I guess the combined example $a_n = n^\alpha \min\{\delta(n)^2,1/\log n\}$ works for a general $\delta(n)$. (still assuming $\delta(n)$ is not $O(n^{-\varepsilon})$ for any $\varepsilon>0$)

Title: Implications of Asymptotic Lower Bound


答: HINT: Let $y=\min S_\Omega$; then $\{y\}$ is open in $S_\Omega$, so if $f:\Bbb R_\ell\to S_\Omega$ is continuous, then $f^{-1}[\{y\}]$ is open in $\Bbb R_\ell$. Does $\Bbb R_\ell$ have any isolated points?

Body:
If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.

I've been trying to solve this problem for a few days, but I haven't been able to see how can I do it. First, $\mathbb{R}_{l}$ refers to $\mathbb{R}$ with the lower limit topology (generated by intervals of the form $[a,b)$ ) and $S_{\Omega}$ refers to the uncountable well-ordered set with the order topology (generated by (a,b) and rays). Until now, I've tried to use the fact that both of those topologic spaces are Hausdorff and suppose that f is injective to get a contradiction, but I don't see where can I get it.


问:
If $A^a + B^b = C$, can we have $A^x + B^y = C$ as well? Where '$x$' not equal to '$a$' and '$y$' not equal to '$b$', $C$ not equal to $1.$
If $A^a + B^b = C$, can we have $A^x + B^y = C$ as well?

Where '$x$' not equal to '$a$' and '$y$' not equal to '$b$', $C$ not equal to $1.$

I know it is true when $C=1.$

$ 3^1 - 2^1 = 1\;;$

$ 2^2 - 3^1 = 4-3 = 1\;;$

$3^2 - 2^3 = 9 - 8 = 1\;.$

Is there any theorem to prove or disprove this?

答: You might look at articles like this one, which shows, that there seems something to be open still. Here a screenshot to see how the status has been at 2004. The screenshot might give you as well many keywords with which you can search for more general and modern (and, perhaps, older and still interesting) results:


问:
How to compute this integral? Dominated convergence theorem doesn't apply
I've met a physics problem and I need to evaluate:

$$\int_0^{\infty}y \sum_{n=1}^{\infty}\frac{(-1)^n}{(n^2r^2 + y^2)^{3/2}} dy$$

If we pretend we can interchange the sum and the integral we get the alternate harmonic series. But I'm not sure there is a theorem that garantees we can interchange them in this case.

And I'm not sure if Taylor expansion would help either, both the integral and the sum of the absolute values are not finite.

Can one show that it diverges?

答:
Since terms are alternating and decreasing in absolute value with respect to $n$, we have

$$\left|\sum_{n=1}^N \frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\right| \leqslant  \frac{y}{(r^2 + y^2)^{\frac32}} < \frac{1}{y^2},$$

and since $y^{-2}$ is integrable on $[1,\infty)$, by the dominated convergence theorem,

$$\int_1^{\infty} \sum_{n=1}^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy =  \sum_{n=1}^{\infty}\int_1^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy $$

Note also that

$$\frac{y}{(n^2r^2 + y^2)^{\frac32}} =\frac{1}{2nr}\frac{2nry}{n^2r^2 + y^2} \frac{1}{\sqrt{n^2r^2 + y^2}} \leqslant \frac{1}{2nr} \cdot 1 \cdot \frac{1}{nr} = \frac{1}{2n^2r^2},$$

and $\frac{y}{(n^2r^2 + y^2)^{\frac32}}\to 0 $ uniformly and monotonically with respect to $n$. By the Dirichlet test, the series is uniformly convergent for $y \in [0,1]$ and we can switch the sum and the integral over the finite interval $[0,1]$, as

$$\int_0^{1} \sum_{n=1}^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy =  \sum_{n=1}^{\infty}\int_0^{1}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy,$$

Thus,

$$\int_0^{\infty} \sum_{n=1}^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy =  \sum_{n=1}^{\infty}\int_0^{1}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy + \sum_{n=1}^{\infty}\int_1^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy \\ = \sum_{n=1}^{\infty}\int_0^{\infty}\frac{(-1)^ny}{(n^2r^2 + y^2)^{\frac32}}\, dy$$


问:
Showing that for every $e > 0$, there exist quasi-equilateral triangles with error $e$ whose vertices have integer coordinates in the plane
I'm currently working on the problem stated below, but i'm still kind of a beginner with proofs, so I would appreciate some tips (not a solution) on how to approach this proof.

A triangle is quasi-equilateral with error $e > 0$, if, for every angle $A$, $B$, $C$ in the triangle, we have that $\vert{A-60^{\circ}}\vert < e$, $\vert{B-60^{\circ}}\vert < e$ and $\vert{C-60^{\circ}}\vert < e $. Show that for every $e > 0$, there exist quasi-equilateral triangles with error $e$ that have vertices with integer coordinates in the plane.

I know how to prove that there are no equilateral triangles with integer coordinates, however, this looks much more difficult and I can't find a way to begin.

答:
First hint:

It suffices to show this for rational coordinates (then we can just scale up by the largest common denominator).

Second hint:

Since we're rescaling anyway, we might as well take two of our points to be exactly one unit apart: fix them and call them $x$ and $y$.

Third hint:

Consider the set $Z$ of all points $z$ that make an $e$-quasiequilateral triangle.

Fourth hint:

$Z$ is an open subset of the plane.

Fifth hint (if you're struggling with proving the fourth hint):

Take such a point $z$. Then let the angles of the triangle so formed be $A(z), B(z), C(z)$ (for future convenience, I'll make those the angles at $x,y,z$, in that order), and define $f(z) := \max(|A - 60^\circ|,|B - 60^\circ|, |C-60^\circ|)$. Show that $f$ is continuous, and you're done.

Sixth hint (if you're struggling with proving the fifth hint above):

It suffices to show that each of $A,B,C$ are continuous. To do this, take some $z'$ near $z$, and consider the triangle $x,z,z'$.

The end of the proof above:

The angle at $x$ of this triangle is $|A(z) - A(z')|$ by definition, and by the sine rule (with $\theta,\varphi$ the angles at $z,z'$ of this triangle), $|A(z) - A(z')| = |zz'|\frac{\theta}{|xz'|} = |z - z'|\frac{\varphi}{|xz|}$. But neither $\varphi$ nor $\theta$ can be more than $180^\circ$, so this is bounded above by $180^\circ|zz'|\left(\frac{1}{|xz|} + \frac{1}{|xz'|}\right)$. Note that (for $e$ smaller than $30^\circ$ - the final result for larger $e$ is easy):  $z$ cannot lie within a distance of $\frac{1}{2}$ of $x$, as all points within that disk give angles at $y$ of no more than $30^\circ$, and similarly for $z'$. By the triangle rule, we have $|xz| \geq |xz'| - |zz'| \geq \frac{1}{2} - |zz'|$, so for $|zz'| < \frac{1}{4}$, we have $|xz| \geq \frac{1}{4}$, and so $\frac{1}{|xz|} \leq 4$. Thus, $|A(z) - A(z')| \leq 180^\circ|zz'|\left(4 + 2\right) = 1080^\circ|zz'| \to 0$ as $|zz'|\to 0$, so $A$ is continuous.

 The proof for $B$ is the same as for $A$ with $y$ replacing $x$ throughout. For $C$, apply the cosine rule to the triangles $xyz$ and $xyz'$ to obtain $2|xz||yz|C(z) = |xz|^2 + |yz|^2 - 1$ and similarly for $z'$, and either combine these to obtain $$|C(z) - C(z')| =\left|\dfrac{|xz|^2 + |yz|^2 - 1}{2|xz||yz|} - \dfrac{|xz'|^2 + |yz'|^2 - 1}{2|xz'||yz'|}\right|$$ and proceed roughly as above with a moderately ridiculous sequence of approximations, or just notice that the functions $w \mapsto |xw|$ and $w \mapsto |yw|$ are continuous and neither is zero anywhere near either $z$ or $z'$, and hence $C$ is continuous by repeated applications of the product, sum, and quotient rules for continuity.

Fifth hint (if you're struggling with finishing after proving the fourth hint):

Since the rationals are dense in the reals, $\mathbb{Q}^2$ is dense in $\mathbb{R^2}$ (think of squares inside circles), and so there's a rational point in every open subset of $\mathbb{R^2}$, and in particular there's one in the above, so scaling everything by the lowest common multiple of the denominator gives an integer solution.


问:
Can two polynomials $p(x)$ and $q(x)\in \mathbb{F}[x]$ have just one factor in common?
Let $p(x),q(x)\in \mathbb{F}[x]$. EDIT: $\mathbb{F}$ is a field of $0$ characteristic.

Let us suppose there is an element $b\notin \mathbb{F}$ such that $p(b)=q(b)=0$. Then, $p(x)$ and $q(x)$ both have the factor $(x-b)$ in $\mathbb{F}[b,x]$.

My question is can $p(x)$ and $q(x)$, both of which belong to $\mathbb{F}[x]$, have only one common factor $(x-b)$ in $\mathbb{F}[b,x]$ and no other?

Motivation behind the question: Let me take you step by step.

If $f(x)\in \mathbb{F}[x]$, and $f(x)=g(x).h(x)$, where if $g(x)\in \mathbb{F}[x]$, then $h(x)$ also has to $\in \mathbb{F}[x]$. This is because if $h(x)$ contains coefficients from outside of $\mathbb{F}$, then, considering the coefficients of $g(x)$ are from $\mathbb{F}$, the coefficients of $g(x).h(x)$ wouldn't be from $\mathbb{F}$.

Let $f(x)\in \mathbb{F}[x]$ such that it has a root $b\notin\mathbb{F}$. Then, $f(x)=g(x).h(x)$ such that $g(x),h(x)\in \mathbb{F}[x]$, and $g(b)=0$. Obviously, $deg(g(x))\geq 2$, as if it is $1$, then it will be the linear factor $(x-b)$, which is not there in $\mathbb{F}[x]$.

Let us factorise $g(x)$ from point $(2)$ in $\mathbb{F}[b,x]$. $(x-b)$ is one factor. Let the other factor be $q(x)$. Then $q(x)\in \mathbb{F}[b,x]$, but $q(x)\notin \mathbb{F}[x]$ (from point $(2)$).

Now let us come back to the original question. If $p(x),q(x)\in \mathbb{F}[x]$ have a common root $b\notin F$, then $p(x)$ will factorise in the form $(x-b)t(x)f(x)$, where $t(x)\in \mathbb{F}[b,x]$ such that $(x-b).t(x)\in\mathbb{F}[x]$, and $q(x)$ will factorise in the form $(x-b)s(x)g(x)$ such that $s(x)\in\mathbb{F}[b,x]$ such that $(x-b)s(x)\in\mathbb{F}[x]$. It is clear that $t(x), s(x)\notin \mathbb{F}[x]$. Obviously$f(x),g(x)\in\mathbb{F}[x]$. $t(x)\neq s(x)$, as otherwise $p(x)$ and $q(x)$ won't have just $(x-b)$ in common. Also $f(x)\neq g(x)$.

Clearly, carrying forward from point $(4)$, $(x-b)t(x)$ and $(x-b)s(x)$ will form distinct irreducible polynomials in $\mathbb{F}[x]$, with no common factors. Hence, as $f(x)\neq g(x)$, $p(x)$ and $q(x)$ will not have any common factors in $\mathbb{F}[x]$, but will have the common factor $(x-b)$ in $\mathbb{F}[b,x]$. Isn't this a contradiction of the theorem that if two polynomials have a common factor in a field extension $\mathbb{K}$, then they have to have a common factor in the original field $F$, where  $\mathbb{F}\subset\mathbb{K}$?

The only way out of this contradiction that I can see is if $(x-b)t(x)$ can be factored into $r(x)e(x)$, and $(x-b)s(x)$ into $r(x)u(x)$, where $r(x), e(x)$ and $u(x)\in \mathbb{F}[x]$. $p(x)$ and $q(x)$ will then have the common factor $r(x)$ in $\mathbb{F}[x]$. I want to know if this is always possible.

答: The minimal polynomial of $b$ divides both $p(x)$ and $q(x)$. And that one can't be linear, as $b \not\in \mathbb{F}$.</p>


问:
Triangles in a cyclic quadrilateral
Consider a cyclic quadrilateral $ABCD$ where

$|DA|\lt |AB|=|BC|\lt |CD|$, $E:|BE|\bot |AC|, E\in |CD|$  , $F:|EF|\text{ }||\text{ }|BC|,F\in |AB|$.

The task is to prove that |FB|=|FD|, that is, that $\Delta FDB$ is isosceles.

I have been able to prove some statements about the quadrilateral, like tha fact that the triangles formed by the diagonals opposite each other have to be similar, but I am struggling to prove anything of use regarding $\Delta ABC$ and $\Delta FDB$.

Thank you for any help.

答:
Let $\mathcal{C}$ be the circle through $A, B, C, D$ and denote by $G$ the point of intersections of the diagonals $AC$ and $BD$. Furthermore, call $\mathcal{X}$ and $\mathcal{Y}$ the circles through $AGD$ and $DGC$, respectively.

Because we have the equality of angles $\angle BAC = \angle BCA = \angle BDA$, where we have used that the triangle $ABC$ is isosceles and carried an angle through the circle, we deduce that $\triangle BGA \sim \triangle BAD$, which implies $BA^2 = BG\cdot BD$. Hence, $\mathcal{X}$ is tangent to $BA$ at $A$. In analogous fashion we deduce that $\mathcal{Y}$ is tangent to $BC$ at $C$.

If we call $M$ the other point of intersection of $BE$ with circle $\mathcal{C}$ we know, since $BE\perp AC$ and $ABC$ is isosceles, that that $BM$ is a diameter of $\mathcal{C}$ and hence $MC\perp BC$ and $MA\perp BA$. Because of the tangencies proves above the center of $\mathcal{X}$ has to lie on $AM$ and that of $\mathcal{Y}$ on $MC$.

Call $X$ the intersection of $AM$ and the parallel line through $E$ to $MC$. We have then $$\angle DAM = \angle DCM = \angle DEX.$$ Hence we deduce $DAEX$ is a cyclic quadrilateral.

Now call $L$ the intersection of $DX$ with $MC$. We have that

$$\begin{align*}
 \angle LDC
&= \angle XDE\\
&= \angle XAE   &\;\mbox{(by previous cyclic quadrilateral)}\\
&= \angle XAC - \angle EAC\\
&= \angle MAC - \angle EAC\\
&= \angle MCA - \angle ECA &\mbox{(since BM is perp. bisector of AC)}\\
&= \angle MCE
\end{align*}$$

Hence, $LD = LC$ and so $L$ lies in the perpendicular bisector of $DC$. But the center of $\mathcal{Y}$, which also lies in that perpendicular bisector, since $DC$ is a chord, has to be in $MC$ since this is orthogonal to the tangent $BC$ at the tangency point $C$. We conclude $L$ is the center of $\mathcal{Y}$.

Now, $\angle DAF = \angle DAB = 180 - \angle BCD = 180 - \angle FED$. Hence, $DAFE$ is a cyclic quadrilateral, but $DAEX$ was also a cyclic quadrilateral, and since these circles share the points $D, A, E$, we conclude $DAFEX$ is a cyclic pentagon.

Now, $\angle FDL = \angle FDX = 180 - \angle XEF$, using the above cyclic pentagon and, furthermore, by contruction $XE || MC$ and $FE || BC$ so that $\angle XEF = \angle MCB = 90$, since $MC$ lies on a diameter of $\mathcal{Y}$ and $BC$ a tangent. We deduce $\angle FDL = 90$.

We have proven above that $L$ is the center of $\mathcal{Y}$, and since $D$ is on $\mathcal{Y}$ then $DF$ is a tangent to this circle since we just proved $\angle FDL = 90$. Hence, transporting angles, first on $\mathcal{Y}$, using this tangent, and then on $\mathcal{C}$ we conclude
$$\angle FDG = \angle DCG = \angle DCA = \angle DBA = \angle DBF,$$ that is, $\triangle FDB$ is isosceles, as desired.

Here is a drawing:

Title: Triangles in a cyclic quadrilateral


问:
Finding $Pr(Y < X < Z)$ where $X, Y, Z$ are random variables with known CDF's
I'm interested in the problem of computing $Pr(Y < X < Z)$ where $X, Y, Z$ are random variables (either discrete or continuous), not necessarily defined over the same support. I found one other thread which concluded that, for the continuous case, we have

$Pr(Y < X < Z) = \int_{-\infty}^\infty Pr\left( x \in [Y, Z)\right)f_X(x)dx = \int_{-\infty}^\infty \left( \int \int_{(x, \infty) \times (-\infty, x)}f_{YZ}(y, z)dy dz\right)$

However, I do not understand the bounds for the RHS. Is $y$ integrated out in $(x, \infty)$ or $(-\infty, x)$?

答: $Y<X<Z$ corresponds to limits of integration of $Y<X$, $Z>X$, and $-\infty<X<\infty$ in the order $\int$dydzdx. So you are right, the limits of integration should be $(-\infty,x)\times(x,\infty)$dydz.


问:
Proving that this Function is Surjective
Let's say we have a function $g: \mathbb{R} \rightarrow [0,1)$ and $g$ is defined by $g(x) = x - \lfloor x \rfloor$.

So far I thought about having $g(a) = b$ meaning we find an $a$ that makes $g(a) = b$.

Thus, $$\begin{align*}g(a) = b \text{ where $b\in [0,1)$} \\ g(a) = a- \lfloor a\rfloor = b \\ \end{align*}$$

Here is where I am stuck. How can I choose an $a$ to show that this function is surjective?

答:
To prove that $g$ is surjective, we must find, for any $b\in[0,1)$, $a\in\mathbb R$ such that $g(a)=b$.

Well, take $a=b$.  Then, $a\in[0,1)\subset\mathbb R$, and, since $a\in[0,1)$,

$g(a)=a-\lfloor a\rfloor=a-0=a=b$.

So we have found $a$ such that $g(a)=b$ for any $b\in[0,1)$, so $g$ is surjective onto $[0,1)$.


问:
Integrating $ a\,f(x) +b\,y(x)=\frac{dy}{dx}$
Can somebody put me on the right track for integrating the following equation? How do I separate the variables?

$$ a\,f(x) +b\,y(x)=\frac{dy}{dx}$$

答: That is a linear differential equation of the first order, non homogeneous. It's solution is standard ODE fare...</p>


问:
Proving the difference between two matrices is positive definite
I'm working on the convex relaxation of a problem, and I came across the following question.

Suppose I have a vector $x \in \mathbb{R}^n$ where $-1 \leq x_i \leq 1$ and a matrix $X$ whose diagonal elements are $1$. How can I prove the following inequality:

$$X \succeq x x^T$$

I've checked it's true for $n=1$, but even for $n=2$ I'm not sure how to show it. Thank you.

答: This seems like it might not be true. Consider $X = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ and $x = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$.  Doesn't $X - x x^T$ have a negative eigenvalue? What is the matrix notion of $\geq$ here?


问:
If $f$ is convex, show that $f(x)/x$ is non-decreasing in $x$
A proof for deducing Lypaunov's inequality seems to be centered on showing that if $f$ is convex on $0< a \le x \le b$, then $f(x)/x$ must be non-decreasing on the interval. I am not aware that $f$ needs to be differentiable.

Starting from a definition of convexity, I can write
$$ f(x) \le f(a) + \frac{(x-a)[f(b) - f(a)]}{b-a} \implies \frac{f(x)}{x} \le \frac{f(a)}{a} +  \frac{(x-a)[f(b) - f(a)]}{x(b-a)}. $$
We can define $g(x) = f(x)/x$ to obtain
$$ \frac{g(x) - g(a)}{x-a} \le \frac{f(b)-f(a)}{x(b-a)}.  $$
None of this seems to be heading in the right direction, what am I doing wrong? Is there another requirement on the function $f$?

答: Without extra assumption, it is false. Let $f:(1,3)\rightarrow \mathbb{R}$ be defined as $f(x) = |x-2|$. Clearly, $f$ is convex. However, $x\mapsto f(x)/x$ is not non-decreasing. Denote $g(x) = f(x)/x$. Obviously $g(1) = 1$ but $g(2) =0$.


问:
Finding $\det (AC + C)$ given the matrix $A$ and $\det(C)$
Given a matrix:
$$ A=\left[
\begin{array}{cc}
  5&3\\
  -4&2
\end{array}
\right] $$
and a matrix $C$ with the assumption that $\det(C)=1/5$.

The question is asking me to calculate the determinant of $AC + C$.

How can I solve this problem?

答:
$$\begin{split}\det(AC+C)&=\det(AC+IC)\\&=\det\left((A+I)C\right)\\
&=\det(A+I)\det C \text { determinants are multiplicative}\\
&=\det\begin{pmatrix}6&3\\-4&3\end{pmatrix}\det C\\
&=(18--12)\det C \text{ use rule for determinant of a 2x2 matrix}\\
&=30\det C\end{split}$$


问:
This series is known? $\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$
This series is known?
I was solving a differential equation by power series and my solution involves that series

$$\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$$

答:
My Wolfram Mathematica code

```<code>Sum[2^(n-1) x^(2 n-1)/Product[2i-1, {i,1,n}],
   {n,1,Infinity}] // InputForm
</code>```

returns the result

```<code>(E^x^2*Sqrt[Pi]*Erf[x])/2
</code>```

which is

$$ f(x) := \frac12 \sqrt{\pi } e^{x^2} \text{erf}(x) $$

and where $\,\text{erf}(x)\,$ is the
Error function. The function $\,f(x)\,$ satisfies the ordinary differential equation

$$ f'(x) = 1 + 2\,xf(x). $$

By the way,  this equation is the basis for an iterative recursion
for the truncated power series of the function using

$$ f_0(x) := O(x), \quad f_{n+1}(x) = \int_0^x 1 + 2\,tf_n(t)\,dt. $$

The function also has another property. Define the differential operator

$$ T[ f(x) ] := e^{x^2} \frac{d}{dx} \left(e^{-x^2} f(x)\right)
= f'(x) - 2\,x f(x). $$
Then your function satisfies $\, T[ f(x) ] = 1.$

Body:
My Wolfram Mathematica code

Sum[2^(n-1) x^(2 n-1)/Product[2i-1, {i,1,n}],
   {n,1,Infinity}] // InputForm

returns the result

(E^x^2*Sqrt[Pi]*Erf[x])/2

which is

$$ f(x) := \frac12 \sqrt{\pi } e^{x^2} \text{erf}(x) $$

and where $\,\text{erf}(x)\,$ is the
Error function. The function $\,f(x)\,$ satisfies the ordinary differential equation

$$ f'(x) = 1 + 2\,xf(x). $$

By the way,  this equation is the basis for an iterative recursion
for the truncated power series of the function using

$$ f_0(x) := O(x), \quad f_{n+1}(x) = \int_0^x 1 + 2\,tf_n(t)\,dt. $$

The function also has another property. Define the differential operator

$$ T[ f(x) ] := e^{x^2} \frac{d}{dx} \left(e^{-x^2} f(x)\right)
= f'(x) - 2\,x f(x). $$
Then your function satisfies $\, T[ f(x) ] = 1.$


问:
dominating antichain
Let $P$ be a finite set with partial orderings $\leq_1$ and $\leq_2$. Then, there is a so-called dominating antichain $A \subseteq P$ where no two distinct elements of $A$ are comparabale in $\leq_1$ or $\leq_2$ and for every $p \in P$ there is an element $a \in A$ such that $p \leq_1 a$ or $p \leq_2 a$.

My approach:

Define $M \subseteq P$ as set of all $\leq_1$-maximal elements of $P$. Let $A \subseteq M$ be the set of all $\leq_2$-maximal elements of $M$.

Then, it is evident that no two distinct elements of $A$ are comparabale in $\leq_1$ or $\leq_2$.

Let $p \in P$.

Case 1: If $p \in A$, then $p \leq_1 p$.

Case 2: If $p \in M\setminus A$, then there is an element $a \in A$ with $p \leq_2 a$.

Case 3: If $p \in P\setminus M$, then - I do not know.

Do you have any advice for me? Thank you for your time and your insights!

答:
Your construction of the dominating antichain $A$ is flawed. Let $P = \{a,b,c,d\}$, and let $b \leq_2 a$, $c \leq_1 a$, $d \leq_1 b$. Then, $M = \{a,b\}$ and $A = \{a\}$. But there is no element $x$ in $A$ such that $d \leq_1 x$ or $d \leq_2 x$.

Starting with the $\leq_1$-antichain $M$ is a good idea. Whenever you find two distinct elements $x,y$ in $M$ such that $y \leq_2 x$, you may modify $M$ as follows:

$$ M \gets M - \{y\} + \{z \in P \mid\ z \lessdot_1 y\ \land\ \lnot \exists m \in M: z \leq_1 m \}$$

Can you prove that this process terminates? This may lead to an inductive proof of the statement.

Title: dominating antichain


问:
Conditional Probability using an urn.
An urn contains $5$ white balls and $6$ black balls. $3$ balls are drawn sequentially without replacement. What are the probabilities...

a. $P(\text{2nd is black} | \text{1st is white})$?

For this one I did the

$$ P(\text{1st is white})=\frac{5}{11} , $$

and

$$ P(\text{second is black}) = \frac{3}{11}*\frac{3}{11} $$

(that is the probability of WB and BB added together). I multiplied the probability of

$$ \frac{P(\text{2nd black} \cap \text{1st white})}{P(\text{1st is white})} , $$

for this I got

$$ \frac{\frac{5}{11}*\frac{3}{11}}{\frac{5}{11}} = \frac{3}{11} . $$

However since this is without replacement I did not think that the answer could be independent, some feedback would help. Alright so I have this figured out that it should be $\frac{6}{10}$.

b. If I am looking now for $P(\text{3rd is black} | \text{1st is white and 2nd is black})$, I ended up with

$$ \frac{\frac{5}{9}*\frac{3}{11}}{\frac{3}{11}} , $$

this is again an independent answer is this correct?

答:
Using the notation $B_n$ to represent the events "the $n$-th ball is black", and $W_n$ likewise for white balls.

$$\mathsf P(W_1)=\dfrac{5}{11}\\\mathsf P(B_2)=\dfrac 6{11}$$

Reason: each individual ball has the same chance to be the second one drawn there are six black balls among the eleven.

(that is the probability of WB and BB added together).

That is true, but for some reason you multiplied.  You should add.
$$\begin{align}\mathsf P(B_2)&=\mathsf P(W_1\cap B_2)+\mathsf P(B_1\cap B_2)\\[1ex] &=\mathsf P(W_1)\cdot\mathsf P(B_2\mid W_1)+\mathsf P(B_1)\cdot\mathsf P(B_2\mid B_1)\\[1ex]&=\dfrac{5}{11}\dfrac{6}{10}+\dfrac{6}{11}\dfrac{5}{10}\\[1ex]&=\dfrac 6{11}\end{align}$$

Here, $\mathsf P(B_2\mid W_1)$ is evaluated as the probability for drawing one from the $6$ black balls when selecting one from the $10$ balls remaining after one white ball has been removed.

I multiplied the probability of

$$ \frac{P(\text{2nd black} \cap \text{1st white})}{P(\text{1st is white})} = \frac{\frac{5}{11}*\frac{3}{11}}{\frac{5}{11}} \\= \frac{3}{11} .
 $$

You had already found  $\mathsf P(\text{1st white$\cap$ 2nd black})$ to be just $3/11$.  So you wanted:$$\mathsf P(B_2\mid W_1)=\dfrac{\mathsf P(W_1\cap B_2)}{\mathsf P(W_1)}=\dfrac{3/11}{5/11}=\dfrac{3}{5}$$

Which, again, you had already evaluated without recognising it.

Title: Conditional Probability using an urn.


问:
If $X$ is such that any étale $X$-scheme is locally factorial, is any smooth $X$-scheme locally factorial?
We say that a local ring is geometrically factorial if its strict henselization is a UFD, and that a scheme is geometrically factorial if all of its local rings are. If $X$ is a normal, locally Noetherian and geometrically factorial scheme, is any smooth $X$-scheme geometrically factorial?

$\textbf{Edit:}$ As Minsheon Shin pointed out in his answer, this question is equivalent to the one in the title by Proposition 1 of Danilov, $\textit{On a conjecture of Samuel}$. In the same article, Danilov conjectures that if $R$ is local, Noetherian and geometrically factorial, then $R[[T]]$ is geometrically factorial; and his main theorem is a special case of this conjecture. If $R$ is excellent, a positive answer to my question would imply the conjecture by https://stacks.math.columbia.edu/tag/07GC.

$\textbf{Other equivalent formulations:}$ By the local structure of smooth morphisms+induction, the question is equivalent to:

"If $X$ is a normal, Noetherian, geometrically factorial scheme, then $\mathbb A^1_X$ is geometrically factorial."

Working locally, this, in turn, reduces to:

"If $R$ is a strictly henselian local UFD, then the strict localizations of $R[T]$ at $(\mathfrak m_R,T)$ and at $\mathfrak m_R$ are UFDs."

The strict localization at $(\mathfrak m_R,T)$ has completion $R[[T]]$ so it is a UFD if Danilov's conjecture holds.

答:
For (1), see Danilov, On a conjecture of Samuel (link), Proposition 1, which I reproduce here: Let $A \to B$ be a faithfully flat ring homomorphism. Then (a) if $B$ is a Krull ring, then $A$ is a Krull ring, and (b) if $\operatorname{Pic}(A) = 0$, then $\operatorname{Cl}(A) \to \operatorname{Cl}(B)$ is injective. The argument for (b) is: if an ideal $\mathfrak{a}$ of $A$ is such that $\mathfrak{a} \otimes_{A} B$ is a principal ideal, then $\mathfrak{a}$ is a finitely generated projective $A$-module of rank 1 (i.e. a line bundle) by faithfully flat descent; then $\operatorname{Pic}(A) = 0$ implies that $\mathfrak{a}$ is principal.

Then as Alex notes, you can conclude using that (i) if $A$ is a UFD then any localization is a UFD, and (ii) if $A$ is a UFD then the polynomial algebra $A[t]$ is a UFD. I was surprised to learn (in the same paper mentioned above) that the formal power series ring $A[[t]]$ need not be a UFD even if $A$ is.

(The property that $R^{sh}$ is a UFD is also called "geometrically factorial".)


问:
Finding probability distribution from given moments
Some quantum mechanics derivation gave me the expectation values of some operator. I got some clean analytical result which makes me think that there is an analytical answer for the probability distribution.The odd moments are all $0$ and the even moments are given by:

$$\langle p^{2n}\rangle=2^{-2n}\binom{2n}{n}(k)^{2n}$$

$ghk$ is just some parameter. The most significant result for me will happen if this will result in some probability distribution based on Bessel $J$ functions but anything is good.

答:
Let $f(x)$ be the (unknown) density function. Then the characteristic function is $h(t)=\int_R e^{itx}f(x)dx=\sum_{n=0}^\infty \frac{(it)^nM_n}{n!}$  Then the density function is given by $f(x)=\frac{1}{2\pi}\int_Re^{-itx}h(t)dt$.

Knowing the moments, $h(t)$ can be computed and then $f(x)$ using the inverse Fourier transform.


问:
How parabolic subgroups are determined by roots?
I am interested in understanding the general construction of important subgroups in reductive groups, and how they are parametrized (Borel, Levi, parabolic, etc.). But for simplicity I take the example of $SL(3)$ over a field $k$.

There are six roots, three positive roots and thus three "root" hyperplanes (do they have a name?). I saw many times the statement that parabolic subgroups correspond to subsets of positive roots, and I would like to understand how this correspondence work. Give a parabolic or a subset of roots, how can I construct the corresponding element on the other side?

More precisely, let $M$ be a standard minimal Levi subgroup. I would like to understand why

the set of parabolic having Levi component $M$ is in bijection with the (6) open Weyl chambers (i.e. the connected sectors between hyperplanes)

the set of Levi subgroups containing M is in bijection with the (5) "subspaces" in the diagram: the plane, the three line, and the origin

the set of parabolics containing $M$ is in bijection with the (13) "half-subspaces": the half-lines, the chambers and the origin.

I am really troubled by these statements but I would like to develop a geometric version of this reductive structure. What is a good reference for it (with lots of examples)?

答:
I'll amend what Andreas Cap describes as the rather simple description of parabolic subalgebras in the Lie algebra $\mathfrak{sl}_3(k)$. Actually, I would have thought that most of this translates over to the group level, but I'm not an expert in that. I will use both terminologies in what follows and hope to be corrected by experts if wrong.

If your minimal Levi is the (maximal torus / Cartan subalgebra) $M = \pmatrix{*&0&0\\0&*&0\\0&0&*}$, and we call $\alpha_1$ the root whose (one-parameter unipotent group / root space) is $\pmatrix{(1/0)&*&0\\0&(1/0)&0\\0&0&(1/0)}$, $\alpha_2$ the one with $\pmatrix{(1/0)&0&0\\0&(1/0)&*\\0&0&(1/0)}$, then the entire root system is $R= \{\pm \alpha_1, \pm \alpha_2, \pm(\alpha_1+\alpha_2) \}$, and we have:

The parabolics with Levi component $M$ (i.e. minimal parabolics = Borels containing $M$) are

$$ \pmatrix{*&*&*\\0&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&0\\*&0&*}, \pmatrix{*&0&0\\*&*&0\\*&*&*}, \pmatrix{*&*&0\\0&*&0\\*&*&*},  \pmatrix{*&*&*\\0&*&0\\0&*&*} $$
which correspond to the six possible root bases
$$\{\alpha_1, \alpha_2\}, \{\alpha_1 +\alpha_2, -\alpha_1 \}, \{\alpha_2, -\alpha_1-\alpha_2\}, \{ -\alpha_1, -\alpha_2\}, \{-\alpha_1-\alpha_2, -\alpha_2 \}, \{ -\alpha_2, \alpha_1+\alpha_2\},$$ respectively. Such root bases, a.k.a. sets of simple roots, are in bijection with Weyl chambers, as every serious source on root systems will tell.

The possible Levis which contain $M$ are
$$\pmatrix{*&0&0\\0&*&0\\0&0&*} = M, L_1 := \pmatrix{*&*&0\\*&*&0\\0&0&*}, L_2:= \pmatrix{*&0&0\\0&*&*\\0&*&*}, L_3:= \pmatrix{*&0&*\\0&*&0\\*&0&*}, \pmatrix{*&*&*\\*&*&*\\*&*&*} = G $$

They generally correspond to subsets $S_L$ of the root system which are saturated (see Andreas Cap's answer) and symmetric ($S_L =-S_L$). Here, the above possibilities obviously correspond to $S_L=$

$$ \emptyset, \{\pm \alpha_1\}, \{\pm \alpha_2\}, \{\pm(\alpha_1+\alpha_2)\}, R,$$

respectively, and you can mattch that with those "subspaces".

The set of all parabolics containing $M$ consists of:

the Borels $$ \pmatrix{*&*&*\\0&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&*\\0&0&*}, \pmatrix{*&0&*\\*&*&0\\*&0&*}, \pmatrix{*&0&0\\*&*&0\\*&*&*}, \pmatrix{*&*&0\\0&*&0\\*&*&*},  \pmatrix{*&*&*\\0&*&0\\0&*&*} $$
from the first question, corresponding to the Weyl chambers;

$$ \pmatrix{*&*&*\\*&*&*\\0&0&*}, \pmatrix{*&*&0\\*&*&0\\*&*&*}, \pmatrix{*&*&*\\0&*&*\\0&*&*}, \pmatrix{*&0&0\\*&*&*\\*&*&*}, \pmatrix{*&*&*\\0&*&0\\*&*&*}, \pmatrix{*&0&*\\*&*&*\\*&0&*},$$

of which the first two contain $L_1$, the second and third contain $L_2$, the last two contain $L_3$ as Levi factor; and finally, of course, the full group / algebra

$$\pmatrix{*&*&*\\*&*&*\\*&*&*}. $$

In general, these correspond e.g. to subsets $S_P \subseteq R$ which are saturated and contain a system of simple roots. This contains of course, first, the six root bases / Weyl chambers /Borels listed in the first part; then, in our case (and I am not sure how that would generalise), for each of the next six, we could make it correspond to a half-plane, which one can make correspond to a single vector, namely e.g. for the first one, $S_P = \{\pm \alpha_1, \alpha_2, \alpha_1+\alpha_2 \}$ which is $\{ \alpha \in R: (\alpha, 2\alpha_2+\alpha_1) \ge 0\}$; ...; for the sixth one, $S_P = \{\pm (\alpha_1 +\alpha_2), \alpha_2, -\alpha_1 \}$ which is $\{ \alpha \in R: (\alpha, -\alpha_1+\alpha_2) \ge 0\}$. I do not see how one would naturally parametrise these via one of the roots. (Of course one can via some convention, after all there's six such parabolics and six roots; but the "natural" choice of parameter here seems to be, rather, a vector which is perpendicular to a certain root.)

If one does that, -- i.e. classify the Borels by Weyl chambers $C$ as belonging to those $S \subset R$ where $S = \{\alpha \in R: (\alpha, c) \ge 0 \text{ for all } x \in C\}$, and the six next bigger parabolics via those $S$ where $S = \{\alpha \in R: (\alpha, v) \ge 0\}$ for certain vectors $v$ perpendicular to one root -- then at least the last one falls into place as well, consisting of the entire root system, which we might phrase as those roots such that $(\alpha, 0) \ge 0$.


问:
Showing an inequality involving a sequence
$(a_n)$ is a positive strictly increasing sequence. I have to show that $(a_{2n-1}a_{2n+1})^{0.5} < a_{2n}$. I can use the AM GM inequality to show that $(a_{2n-1}a_{2n+1})^{0.5}$ $\leq ((a_{2n-1} + a_{2n+1}))/2 < ((a_{2n} + a_{2n+1}))/2$.

答:
It is obviously false. Consider the following counter-example:
1, 2, 10000000000, (the remaining terms are not important)

Do you think $\sqrt{1\times 10000000000} < 2$?


问:
period of a function $\lvert\sin x\rvert+\lvert\cos x\rvert$
I have read that $$y=\lvert\sin x\rvert+ \lvert\cos x\rvert $$ is periodic with fundamental period $\frac{\pi}{2}$.

But Wolfram says it is periodic with period $\pi$.

Please tell what is correct.

答: square it and simplify - gives period</p>

Body: square it and simplify - gives period


问:
Find PDF of $Z = X + Y$ where $X$ and $Y$ are jointly continuous random variables.
I want to check my solution.

Let $X$ and $Y$ be jointly continuous random variables and $Z = X + Y$.

For some $z$, $X = z - Y$.

Thus

$F_Z(z) = Pr(Z < z) = Pr(X < z - Y)= \int_{-\infty}^\infty\int_{-\infty}^{z - y} f_{X,Y}(s, t)dsdt
= \int_{-\infty}^\infty\int_{-\infty}^{z - y} f_{X}(s)f_Y(t)dsdt
= \int_{-\infty}^\infty F_X(z-y)f_Y(t)dt
= E[F_X(z-y)]$

Now, to get the PDF of $Z$,

$ f_Z(z) = {\partial F_Z(z) \over \partial z} = {\partial \over \partial z} \int_{-\infty}^\infty F_X(z-y)f_Y(t)dsdt$

Since $F_X(z-y)f_Y(t)$ is bounded, $ {\partial \over \partial z} \int_{-\infty}^\infty F_X(z-y)f_Y(t)dt = \int_{-\infty}^\infty {\partial \over \partial z}F_X(z-y)f_Y(t)dt $

Hence $f_Z(z) = \int_{-\infty}^\infty f_Z(z-y)f_Y(t)dt = E[f_Z(z-y)]$ , since  ${\partial \over \partial z}F_X(z-y) = f_Z(z-y)$

I just have calculated as I can, but I am not sure about the precision of each steps.

答:
$$\begin{align}
f_{\small Z}(z) & =\dfrac{\mathrm d ~~}{\mathrm d z}F_{\small Z}(z)\\&=\dfrac{\mathsf d~~}{\mathrm d z}\iint_{s+t<z} f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t\\&=\dfrac{\mathsf d~~}{\mathrm d z}\int_\Bbb R\int_{-\infty}^{z-t} f_{\small X,Y}(s,t)\,\mathrm d s\,\mathrm d t\\&=\int_\Bbb R \left\lvert\dfrac{\partial (z-t)}{\partial z}\right\rvert f_{\small X,Y}(z-t, t)\,\mathrm d t\\&=\int_\Bbb R f_{\small X,Y}(z-t, t)\,\mathrm d t
\end{align}$$


问:
Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$
Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$
I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.

答:
$\textbf{Solution.}$ (Khang Nguyen) We have $$(3-a)(3-b)(3-c)\ge 0 \rightarrow abc+9(a+b+c)\le 36$$
Assume that $$(b-1)(c-1)\ge 0 \rightarrow 36\ge 8a+(9+a)(b+c)\Rightarrow b+c\le \dfrac{(36-8a)}{9+a}$$
Therefore
\begin{align*}
\text{LHS}&=\sqrt{2a+b+c+2\sqrt{(a+b)(a+c)}}+\sqrt{b+c}\\&
\le \sqrt{2a+\dfrac{36-8a}{9+a}+2\sqrt{a^2+3}}+\sqrt{\dfrac{36-8a}{9+a}}\\&\le 3+\sqrt{3}.
\end{align*}
Let $\dfrac{36-8a}{9+a}=x^2\, \left(1\le x\le 2\right),$ after simplify it becomes:
$$ \left( {x}^{2}+8 \right)  \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{2
 }+8}\sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10
 {x}^{2}+72}$$
Or $$\sqrt{x^2+8} \left( \sqrt {3}+3-x \right) \ge \sqrt {{x}^{4}+4\sqrt {3}\sqrt {7{x}^{4}-50{x}^{2}+124}-10
 {x}^{2}+72}$$
Since $x\le 2\rightarrow \sqrt{3}+3-x\ge \sqrt{3}+3-2>0\rightarrow \text{VT}>0.$ By squaring both sides, we need to prove
$$-2 \left( \sqrt {3}+3 \right)  \left( \sqrt {3}{x}^{2}+{x}^{3}-6{x
}^{2}-10\,\sqrt {3}+8\,x+6 \right) \ge 4\sqrt {3}\sqrt {7\,{x}^{4}-50\,
{x}^{2}+124}$$
It's easy to prove $\text{LHS}>0$ and from here the inequality is equivalent to
$$24\left( 2+\sqrt {3} \right)  \left( {x}^{2}+8 \right)  \left( 2
\sqrt {3}x+{x}^{2}+8\sqrt {3}-9x-10 \right)  \left( x-2 \right)
\left( x-1 \right) \ge 0,$$
One more time, we can check that this inequality is true for all $1 \le x \le 2$ and done!

Equality holds when $(a,b,c)=(3,1,0)$ and permutations.

See also here.


问:
Showing that two sets have the same Cardinality
Let's say $|[\frac{1}{3},\frac{1}{2}] \cap \mathbb{Q}| = |\mathbb{Q}^+$|.

To show that this is true I want to use Schröder-Bernstein Theorem. First I have to show that there is a injective or surjective function from $[\frac{1}{3},\frac{1}{2}]\rightarrow \mathbb{Q}^+$ and vise versa.

I am struggling to find a formula that is injective for $[\frac{1}{3},\frac{1}{2}]\rightarrow \mathbb{Q}^+$. Looking for a push in the right direction.

答: You will never find an injection from $[1/3 , 1/2]$ to $\mathbb{Q}^+$ because the former is not countable while the later is. However, $[1/3, 1/2] \cap \mathbb{Q}$ is a subset of $\mathbb{Q}^+$ and so the function $\iota : [1/3 , 1/2] \cap \mathbb{Q} \rightarrow \mathbb{Q}^+$ defined by inclusion is injective.


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答: Picard’s Great Theorem: In every neighborhood of an essential singularity of an analytic function, the function takes on every value, with at most one exception.</p>

LastEditorDisplayName: anon


问:
How to compute $\int^{\infty}_{0} t^{(\frac1n-1)}\cos t \,\mathrm{d}t$?
How to calculate the below integral?
$$
\int^{\infty}_{0} \frac{\cos t}{t^{1-\frac{1}{n}}} \textrm{d}t = \frac{\pi}{2\sin(\frac{\pi}{2n})\Gamma(1-\frac{1}{n})}
$$
where $n\in \mathbb{N}$.</p>

答:
It is well-known (and relatively easy to prove) that

Lemma. If $f$ is improperly integrable on $(0, \infty)$, then $F(s) = \int_{0}^{\infty} f(x) e^{-sx} \, dx$ defines an analytic function for $\Re s > 0$ and
  $$ \lim_{s\to0^{+}} \int_{0}^{\infty} f(x) e^{-sx} \, dx = \int_{0}^{\infty} f(x) \, dx. $$

Then for $0 < \alpha < 1$,

\begin{align*}
\int_{0}^{\infty} \frac{\cos x}{x^{1-\alpha}} \, dx
&= \lim_{s\to0^{+}} \int_{0}^{\infty} \frac{\cos x}{x^{1-\alpha}} e^{-sx} \, dx \\
&= \lim_{s\to0^{+}} \frac{1}{\Gamma(1-\alpha)} \int_{0}^{\infty} \frac{\Gamma(1-\alpha)}{x^{1-\alpha}} e^{-sx} \cos x \, dx \\
&= \lim_{s\to0^{+}} \frac{1}{\Gamma(1-\alpha)} \int_{0}^{\infty} \left( \int_{0}^{\infty} u^{-\alpha} e^{-xu} \, du \right) e^{-sx} \cos x \, dx \\
&= \lim_{s\to0^{+}} \frac{1}{\Gamma(1-\alpha)} \int_{0}^{\infty} \left( \int_{0}^{\infty} e^{-(u+s)x} \cos x \, du \right) u^{-\alpha} \, du \\
&= \lim_{s\to0^{+}} \frac{1}{\Gamma(1-\alpha)} \int_{0}^{\infty} \frac{(u+s) u^{-\alpha}}{(u+s)^{2} + 1} \, du \\
&= \frac{1}{\Gamma(1-\alpha)} \int_{0}^{\infty} \frac{u^{1-\alpha}}{u^{2} + 1} \, du.
\end{align*}

Here, the interchange of two integrals is justified by the Fubini's theorem. Then by plugging $u = \tan \theta$, the beta function identity gives us

\begin{align*}
\int_{0}^{\infty} \frac{\cos x}{x^{1-\alpha}} \, dx
&= \frac{1}{\Gamma(1-\alpha)} \int_{0}^{\frac{\pi}{2}} \tan^{1-\alpha}\theta \, d\theta \\
&= \frac{1}{2\Gamma(1-\alpha)} \beta\left( 1-\frac{\alpha}{2}, \frac{\alpha}{2} \right) \\
&= \frac{1}{2\Gamma(1-\alpha)} \Gamma \left( 1-\frac{\alpha}{2} \right) \Gamma \left( \frac{\alpha}{2} \right) \\
&= \frac{\pi}{2\Gamma(1-\alpha)} \csc \left( \frac{\pi \alpha}{2} \right)
\end{align*}

as desired.


问:
Integration by inverse trigonometric substitution
I'm trying to integrate by inverse trigonometric substitution, and I have an answer (though I'm pretty sure it is wrong), but it is not the answer that the solution provides. I need to check where I went wrong and what I should do differently.
This is the question:$$
\int\frac{1}{(a^2-x^2)^2}dx$$
And this is what I got:$$
\frac{1}{a^4}\frac{1}{2}(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|)+C=\frac{x}{2a^4(a^2-x^2)}+\frac{\ln|\frac{a+x}{2a\sqrt{a^2-x^2}}|}{2a^3}+C$$
Whereas the solution should be:$$
\frac{x}{2a^2(a^2-x^2)}+\frac{1}{4a^3}\ln|\frac{x+a}{x-a}|+C$$
What I attempted was to do a by-parts once I got this: $$
\frac{1}{a^4}\int{\sec^3\theta d\theta} = \int\sec^2\theta\sec\theta d\theta
$$
Thank you very much for your time and help.

答:
Substitution $x=a\sin\theta$ to proceed as follows
\begin{align}
\int\frac1{(a^2-x^2)^2}dx
=&\frac1{a^3} \int \sec^3\theta d\theta
= \frac1{2a^3} \int \frac{\sec\theta }{\tan\theta}d(\tan^2\theta)\\
=& \frac1{2a^3}\sec\theta\tan\theta + \frac1{2a^3} \int {\sec\theta }d\theta \\
 = &\frac1{2a^3}\sec\theta\tan\theta + \frac1{2a^3} \ln|\sec\theta+\tan\theta|\\
=  &\frac1{2a^3}\frac{\sin\theta }{\cos^2\theta}+ \frac1{2a^3} \ln|\frac{1+\sin\theta}{\cos\theta}|\\
 =&  \frac{x}{2a^2(a^2-x^2)}
+\frac{1}{2a^3}\ln|\frac{x+a}{\sqrt{a^2-x^2}}|\\
 =&  \frac{x}{2a^2(a^2-x^2)}
+\frac{1}{4a^3}\ln|\frac{(a+x)^2}{a^2-x^2}|\\
=&  \frac{x}{2a^2(a^2-x^2)}
+\frac{1}{4a^3}\ln|\frac{x+a}{x-a}|
\end{align}


问:
What is the best place to start learning Clifford and Lie algebras?
I'm looking for good references for a self-study on Lie and Clifford algebras. I think (but I don't know for sure) that these algebras have some connections (maybe these are homomorphic algebras?) and, if this is the case, it would be nice if the reference discusses such connections.

Note: I don't have a very strong background in algebra, but I have in analysis. As far as I know, these algebraic objects usually can be studied/introduced using either a more algebraic approach (say, using modules, rings and so on) or a more analytic one (using linear algebra, say). Thus, a more analytic approach fits best in my needs.

Thanks in advance!

答:
It so happens that I have been looking into that very same question myself and have encountered resources that you will likely find useful.

Geometirc (a.k.a. Clifford) Algebra Video Series by Prof. Alan Macdonald. Note: He also publishes a book (and has a support website for those taking his course). It starts with linear algebra and works up from there, including all the way up to advanced Calculus based upon Geometric (Clifford) Algebra.

A Swift Introduction to Geometric Algebra [44:22 minute YouTube video].

I hope these references you find helpful and set you on the right course. Good luck!


问:
Velleman's Proof Designer problem 44
This is an exercise I didn't solve from the group of exercises present on Velleman's site that should be solved using Proof Designer (the little program that comes with the book).

Theorem:

If $\forall \mathcal{F} (\cup \mathcal{F} = A \rightarrow A \in \mathcal{F})$, then $\exists x (A=\{x\})$.

Before addressing my line of reasoning and the tentative proof, just few words on this specific problem. I find it interesting because it shows me that I still have problems figuring (and writing) properly existence results. About this, I would say that no matter what book on proof writing you take, this point is usually quickly considered and the examples are trivial, so not particularly useful for real mathematics. Probably this is related to the fact that existence theorems are in general the "hard ones" to be proven and to be thaught (I guess that they are the most difficult to be grasp by students), but I really don't know and I would like to know the opinions of experts.

Line of Reasoning:

I think the idea is that $x=\emptyset$ and $A=\{\emptyset\})$. In this way, what we have to assume should stand. Indeed, the antecedent should "bind" the nature of the family of sets and the consequence would follow.

Tentative Proof:

Let x be the empty set $\emptyset$. Thus, $\forall \mathcal{F} (\cup \mathcal{F} = \{\emptyset\} \rightarrow \{\emptyset\} \in \mathcal{F})$. Take $\mathcal{F}=\{\emptyset,\{\emptyset\}\}$. Then, we have $\cup \{\emptyset,\{\emptyset\}\} = \{\emptyset\} \rightarrow \{\emptyset\} \in \{\emptyset,\{\emptyset\}\}$, and the result is established.

PS: As I pointed out in another post, in general I try to be more consistent grammatically speaking when I write down proofs, and I avoid the use of logical symbols. I tend to do it when I have doubts about the content of the problem and here... I have no clue at all, so forgive it. Any feedback is more than welcome.

答: Yes, Lord_Farin's elegant proof can be done in Proof Designer.  Start by applying the "Universal Instantiation" command to the hypothesis $\forall \mathcal{F}(\bigcup\mathcal{F} = A \to A \in \mathcal{F})$.  This allows you to plug in whatever you want for $\mathcal{F}$.  You can go ahead and plug in exactly the set suggested by Lord_Farin:  $\{\{x\} \mid x \in A\}$.</p>

Title: Velleman's Proof Designer problem 44


问:
How to solve this coupled linear ODE
I've tried using mathematica and for some reason it's not working
I think it's a pretty standard problem but for some reason I'm having a hard time
This is essentially a variant of the Rabbi problem I'll really appreciate the help :)
I will appreciate both explanation on how to solve this using mathematica or a manual solution
\begin{align}
i\hbar \frac{d\alpha}{dt}&=\frac{\hbar\omega_0}{2}\alpha+\frac{dE}{2}e^{i\omega t}\beta\\
i\hbar \frac{d\beta}{dt}&=-\frac{\hbar\omega_0}{2}\beta+\frac{dE}{2}e^{-i\omega t}\alpha
\end{align}

答: I think you are trying to solve simultaneous linear differential equations of first order of the form $$\frac{dx}{dt}=ax+by \\ \frac{dy}{dt}=cx+dy.$$ I would highly recommend to go through the book "An Introduction to Dynamical system and chaos" (Springer) by G. C. Layek.

Title: How to solve this coupled linear ODE


问:
Solving for equation with vector terms?
Is it possible to solve for $v$ in terms of $a$, $b$, $M$, and $e$?

Variables:

$a$ and $b$ are constants

$M$ is an invertible $n × n$ matrix in $R^{nxn}$

$v$ is a $n × 1$ vector in $R^n$

$e$ is a $n × 1$ vector of ones, $(1,1,...,1)^T$

$n$ > 5

Equations:

$a = v^TMe$

$b = v^TMv$

I'm not too sure where to begin to isolate $v$, because I know $v$ and $e$ are vectors and therefore are not invertible.

答:
Let $j=\min\{i\in [n]:e_i^TMe \neq 0\}$. We know such a $j$ exists since $M$ is invertible. You can check by direct substitution that the vector $v$ defined by $v=\frac{ae_j}{e_j^TM^Te}$ is a solution to the equation $v^TMe=a$.

Consider the equation $v^TMv=b$ when $b>0$. Assume we can identify $v_{(+)}\in \mathbb{R}^{n\times 1}$ such that $v_{(+)}^TMv_{(+)}>0$. Put $v=v_{(+)}\sqrt{\frac{b}{v_{(+)}^TMv_{(+)}}}$. Then we have $$v^TMv=\Bigg(\sqrt{\frac{b}{v_{(+)}^TMv_{(+)}}}\Bigg)^2v_{(+)}^TMv_{(+)}=b$$ which means $v$ is in fact a solution to $v^TMv=b$. Now assume $b<0$. Assume there exists a vector $v_{(-)}\in \mathbb{R}^n$ such that $v_{(-)}^TMv_{(-)}<0$. Set $v=v_{(-)}\sqrt{\frac{b}{v_{(-)}^TMv_{(-)}}}$ Then $$v^TMv=\Bigg(\sqrt{\frac{b}{v_{(-)}^TMv_{(+-}}}\Bigg)^2v_{(-)}^TMv_{(-)}=b$$ Hence $v$ is a solution to $v^TMv=b$. Taking $v=0$ provides us with a solution for the case when $b=0$.

Remark: The vectors $v_{(-)},v_{(+)}$ need not exist. To see this, consider when $$M=A\oplus A \oplus A\in \mathbb{R}^{6 \times 6}$$ where $A$ is the matrix of the linear transformation which rotates vectors in the $xy-$plane by $\pi/2$ radians.

Title: Solving for equation with vector terms?


答:
Might be easier to prove there are injective functions from $f:[\frac 13, \frac 12]\cap \mathbb Q \to \mathbb Q$ and from $g:\mathbb Q\to [\frac 13,\frac 12]\cap Q$.

Obviously $f(x) = x$ is an injective function from $[\frac 13, \frac 12]\cap \mathbb Q\to \mathbb Q$. So we are, in theory, half done.

We can squeeze $(1, \infty)\to (0,1)$ via $x \to \frac 1x$ and that maps all rationals in $(1,\infty)$ to all rationals in $(0,1)$. And we can map it to any $(q,r);q,r\in \mathbb Q$ but multiplying by $r-q$ and adding $q$.

We can do the some thing for $(-\infty, -1)\to (-1,0)$ via $x\to \frac 1x$ and the shift/squeeze that to $(s,t)$ by multiplying by $-(t-s)$ and adding $s$.

And we can map $(-1,1)\to (a,b)$ by adding $1$, dividing by $2$, multiply $(b-a)$ and adding $a$.

.......

So consider $g:\mathbb Q \to [\frac 13, \frac 12]\cap \mathbb Q$ via.

If $x < -1$ let $g(x) = \frac 1x(-\frac 1{27}) + \frac 13$.  That will map $(-\infty,-1)\to (\frac 13, \frac 13 + \frac 1{27}=\frac {10}{27})$.

Let $g(-1)=\frac {10}{27}$.

If $-1< x < 1$ let $g(x)=x\cdot \frac 1{27}+ \frac{11}{27}$.  This maps $(-1,1)\to (\frac {10}{27}, \frac {12}{27}$.

Let $g(1)= \frac {12}{27}$.

If $g > 1$ let $g(x)=\frac 1x\cdot \frac 1{27} + \frac {12}{27}$.  This maps $(1,\infty)\to (\frac {12}{27}, \frac {13}{27})$.  And it maps

$\mathbb Q \to (\frac 13, \frac {13}{27})\cap \mathbb Q\subset [\frac 13, \frac 12]\cap \mathbb Q$.  And it's injective.

SO that's it.  $f(x) = x$ is incjective so $|[\frac 13, \frac 12]\cap \mathbb Q| \le \mathbb Q$ and $g(x)$ is injective so $|\mathbb Q| \le |[\frac 13, \frac 12]\cap \mathbb Q|$

So $|\mathbb Q| =|[\frac 13, \frac 12]\cap \mathbb Q|$

....

Not elegant but works.

Title: Showing that two sets have the same Cardinality

Body:
Let's say $|[\frac{1}{3},\frac{1}{2}] \cap \mathbb{Q}| = |\mathbb{Q}^+$|.

To show that this is true I want to use Schröder-Bernstein Theorem. First I have to show that there is a injective or surjective function from $[\frac{1}{3},\frac{1}{2}]\rightarrow \mathbb{Q}^+$ and vise versa.

I am struggling to find a formula that is injective for $[\frac{1}{3},\frac{1}{2}]\rightarrow \mathbb{Q}^+$. Looking for a push in the right direction.


问:
Prove that $\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$
For $a,b,c>0;abc=1.$ Prove that $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$$
I will post my solution in the answer. Now I'm looking forward to another solution.

答:
My solution. Assume $(b-1)(c-1)\ge 0 \rightarrow 1=abc\ge (a+b-1)c\rightarrow a+b \le \dfrac{1}{c}+1.$

\begin{align*} \text{LHS}&=\sqrt{\left(a^2+b^2\right)-(a+b)+2+2\sqrt{\left(a^2-a+1\right)\left(b^2-b+1\right)}}+\sqrt{c^2-c+1}\\&\ge \sqrt{t^2-2t+4}+\sqrt{c^2-c+1}=\text{P}\,(\text{where}\,t=a+b)\end{align*}
Let $f(t)=P-c-t$ then prove $f(t)\ge 0.$

Since $t-1<\sqrt{(t-1)^2+3}=\sqrt{t^2-2t+4},$
we have $$f'(t)=\dfrac{t-1}{\sqrt{t^2-2t+4}}-1<1-1=0\rightarrow f(t)\ge f\left(\dfrac{1}{c}+1\right)\ge 0.$$
By some simple calculations, we need to prove $3c^2(c-1)^2\ge 0,$ which is true!

Done. Equality holds when $a=b=c=1.$


问:
Hard fraction problem
The question is below:

Rebecca has a collection of 45 books. The collection consists of novels and textbooks written in either Chinese or English. $\frac{4}{5}$ of the novels are in English and $\frac{3}{4}$ of the textbooks are in English. The total number of books written in English is 35. How many of her books are textbooks written in English?

ANSWER: 15 books

And here are my initial thoughts:

Initially I thought that since $\frac{4}{5}$ of novels are in English and $\frac{3}{4}$ of textbooks are in English, and there is a total of 35 books written in English. So I thought that if I add  $\frac{4}{5}$ and $\frac{3}{4}$  which is equal to $\frac{31}{20}$.This will represent 35 English books, but $\frac{31}{20}$ is improper fraction so it cannot represent 35 English books. Am I interpreting the question wrong? Help would be greatly appreciated.

Also this question should not require any algebra since it's only Year 8 mathematics.

答:
Let $T$ be the number of textbooks and $N$ the number of novels.

Then we have $T+N=45$ and $\frac34T+\frac45N=35$.

Do you know how to solve this system for $T$?

Then the answer to the question is $\frac34T$.

Title: Hard fraction problem


答:
Pure logic, no algebra:

Since fractions of a book doesn't make sense, the number of novels must be a multiple of 5 and the number of textbooks a multiple of 4.  Also the total number of books 45 is a multiple of 5, so we see the number of textbooks must be multiple of 5 too giving it a multiple of 20.  So that leaves only 0, 20 or 40 textbooks and you can check individually which is correct, or use a similar argument for the number of English books to narrow that down.

Title: Hard fraction problem

Body:
The question is below:

Rebecca has a collection of 45 books. The collection consists of novels and textbooks written in either Chinese or English. $\frac{4}{5}$ of the novels are in English and $\frac{3}{4}$ of the textbooks are in English. The total number of books written in English is 35. How many of her books are textbooks written in English?

ANSWER: 15 books

And here are my initial thoughts:

Initially I thought that since $\frac{4}{5}$ of novels are in English and $\frac{3}{4}$ of textbooks are in English, and there is a total of 35 books written in English. So I thought that if I add  $\frac{4}{5}$ and $\frac{3}{4}$  which is equal to $\frac{31}{20}$.This will represent 35 English books, but $\frac{31}{20}$ is improper fraction so it cannot represent 35 English books. Am I interpreting the question wrong? Help would be greatly appreciated.

Also this question should not require any algebra since it's only Year 8 mathematics.


问:
How to simply this radical expression $\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$
$$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$

I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it?

Thanks!

答:
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
Since you want to clear the denominator, try multiplying both numerator and denominator by $(2^{\frac{2}{3}} - 2^{\frac{1}{3}} + 1)$
That will give you 2-1 for the denominator.


问:
Is $a(a+1)=b(b-1) \implies b = a + 1$ correct? ($a, b > 0$)
I've seen a website wrote: $$a(a+1)=b(b-1) \implies b = a + 1$$
I don't know where the second part come from. How to prove it?

答:
Let $n = b - a$.  Then your equation becomes
$$a(a + 1) = (a + n)(a + n - 1).$$
After expanding this you can cancel an $a^2$ term from each side, then move everything to one side and it factors giving
$$0 = (2a + n)(n - 1).$$
Thus either $n = 1$ or $n = -2a$, so either $b = a + 1$ or $b = -a$.


答:
You can't add the fractions together like that because the fractions do not represent a number of books of each type, but rather a proportion of books with a certain characteristic (in your case, written in English), among the total number of books of each type (novels and textbooks).

To illustrate with an example, suppose I have a bowl of candies, $50$ of which are red and $16$ of which are yellow.  If $4/5$ of the red candies are sour flavored and $3/4$ of the yellow candies are sour flavored, then would it make sense to add $4/5 + 3/4$ to get the total number of sour candies in the bowl?  Of course not, because when we say $4/5$ of the $50$ red candies are sour, that means $(4/5)(50) = 40$ of the red candies are sour; similarly, $(3/4)(16) = 12$ of the yellow are sour, and now we can add $40 + 12 = 52$ because now we are adding whole numbers of candies that share the same characteristic together.

Therefore, the same principle applies here with the books.  However, we don't know how many of the books are novels and how many are textbooks, so this is why the question is a bit more challenging.  Indeed, it may be a bit much to ask to avoid using algebra entirely.  One thing we could try is to observe that the number of novels is a multiple of $5$, and the number of textbooks is a multiple of $4$; otherwise, we can't take $4/5$ and $3/4$ of these to get an integer number of English-language books of each type.

Since there are $45$ books in total, we can compare the multiples of $5$ and $4$ less than $45$ and see if there is a combination can add up to $45$:

$$\{0, 5, 10, 15, 20, 25, 30, 35, 40, 45\} \\ \{0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44\}.$$

Notice that there is no way to pick a number from the multiple of $4$s that does not end in a $0$, because there is no corresponding number in the multiple of $5$s that will make a sum of $45$.  So the only possible choices in the second list are $\{0, 20, 40\}$.  This means there are only three possible choices:  either there are:

$45$ novels and $0$ textbooks,

$25$ novels and $20$ textbooks, or

$5$ novels and $40$ textbooks.

In the first case, $4/5$ of $45$ novels is $36$ English novels, which is too many, so this won't work.

In the second case, $4/5$ of $25$ is $20$ English novels, and $3/4$ of $20$ is $15$ English textbooks, resulting in $20 + 15 = 35$ English language books.  This gives $15$ English language textbooks as the solution.

In the third case, $4/5$ of $5$ is $4$ English novels, and $3/4$ of $40$ is $30$ English textbooks, and this adds up to $4 + 30 = 34$ English language books, which is one short, so this is not a solution.

No algebra used!

Title: Hard fraction problem


问:
Minimal prime ideals of $\hat R$ contract to minimal primes of $R$?

In what conditions from $P\in\operatorname{Min}\widehat R$ can we deduce that $P\cap R\in \operatorname{Min} R$ ?

Here $(R,\mathfrak m)$ is a commutative Noetherian local ring, $\widehat R$ its $\mathfrak m$-adic completion and  $\operatorname{Min} R$ the minimal primes of $R$.

答: As I've already told you, the going-down theorem holds for flat $R$-algebras (see, for instance, Matsumura, Commutative Ring Theory, Theorem 9.5). Since $R\to\hat{R}$ is (faithfully) flat and $P$ a minimal prime in $\hat R$, then $\mathfrak p=P\cap R$ is also minimal. Otherwise let $\mathfrak q$ be a prime ideal of $R$ with $\mathfrak q\subset \mathfrak p$. Going-down theorem says that there exists a prime $Q$ of $\hat R$ with $Q\subset P$ such that $Q\cap R=\mathfrak q$ (obviously $Q\neq P$) and this contradicts the minimality of $P$.


问:
Cancellation in the linear regression hat matrix
While studying linear regression, I've been introduced to the so-called hat matrix: $H = X(X^TX)^{-1}X^T$.

Why doesn't it reduce to $I$?

$H = X(X^TX)^{-1}X^T = X(X^{-1}(X^T)^{-1})X^T = XX^{-1}X^{-T}X^T = II = I$

Am I incorrectly distributing the inverse of $(X^TX)^{-1}$?

答: $(AB)^{-1} = B^{-1} A^{-1}$ only if both $A$ and $B$ are invertible matrices. Since $X \in \mathbb{R}^{n \times p}$, where $n$ is the number of observations, and $p$ the dimension of each observation, is it possible for $X$ be invertible?


问:
Find minimum of $f(\mathbf{v})= \mathbf{v}^\top A \mathbf{v} - \sum_{i=1}^N \log((\mathbf{v}^\top \mathbf{u}_i)^2) $
Consider the function:
$$ f(\mathbf{v})= \mathbf{v}^\top A \mathbf{v} - \log((\mathbf{v}^\top \mathbf{u})^2), $$
where $\mathbf{v}, \mathbf{u} \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ is symmetric and positive definite.

I would like to find the solution(s) in $\mathbf{v}$ forming the minimum of this function.

Differentiating, we have
$$f’(\mathbf{v}) = 2A \mathbf{v} - \frac{2 \mathbf{u}}{\mathbf{v}^\top \mathbf{u}}.$$

Setting this to zero and multiplying the LHS by $\mathbf{v}^\top$ gives the solution set as:
$$\mathbf{v}^\top A \mathbf{v} = 1.$$

However, I’m unsure how to interpret the above and find the vectors that constitute as the minimum of the original function.

Now, can this then be extended to find the minimum of:
$$f(\mathbf{v})= \mathbf{v}^\top A \mathbf{v} - \sum_{i=1}^N \log((\mathbf{v}^\top \mathbf{u}_i)^2),$$
with the known and independent $\mathbf{u}_i \in \mathbb{R}^n$?

答:
Rearranging a bit and renaming $\lambda=1/v^T u$ we obtain

$$Av=\lambda u\iff v=\lambda A^{-1}u$$

Dotting with $u$ yields the equation that determines $\lambda$:

$$\lambda^2=\frac{1}{u^T A^{-1}u}$$

which if $A$ is positive definite yields two solutions:

$$v=\pm\frac{A^{-1}u}{\sqrt{u^TA^{-1}u}}$$

both of which are minima of the function.

EDIT: To reflect the update of the question I will dive into the solution of the problem when there are multiple vectors $u_i$ that form an independent set.

We find that the critical points of the function are given by

$$\nabla_\mathbf{v}f=2\left[Av-\frac{\sum_{i=1}^N(v^T u_i)u_i}{\sum_{i=1}^N(v^Tu_i)^2}\right]=0$$

We note that this equation yields the constraint $v^T Av=1$, which is going to be important for the determination of the minimum. We also note that $N\leq n$ or we cannot have an independent set of vectors. For notational convenience we set $v^Tu_i=\lambda_i~,~ r^2=\sum_i\lambda^2_i$. These numbers have to be determined. Solving the equation for the critical points yields

$$v=\frac{\sum_i\lambda_i A^{-1}u_i}{r^2}$$

Dotting on the left with $u_j~,~j=1,...,N$ we obtain the set of equations
$$r^2\lambda_j=\sum_{i}\lambda_i( u_j^TA^{-1}u_i)$$

We note that when we define the matrix $Q_{ij}:=u_i^TA^{-1}u_j$ this can be written as an eigenvalue problem

$$Q\lambda= r^2 \lambda$$

which basically says that the projections on the set of vectors $u$ have to be eigenvectors of the matrix $Q$- the representation of $A^{-1}$ in the subspace generated by the $u$'s. Since $A$ is invertible, $Q$ must be invertible as well (proof?) in which case it contains no zero eigenvalues. Furthermore, $A^{-1}$ is symmetric and all it's eigenvalues are real.

We are left with considering the space of solutions of the eigensystem above. Note that upon a first look, this is quite the eccentric eigenvalue problem; the eigenvalues are constrained to be the norm of the eigenvectors themselves! Not to worry however, for suppose the orthonormal solution to the eigenproblem of $Q$ is given by the set $\{\gamma_i\in\mathbb{R}, e_i\in \mathbb{R}^N\}$ (one should be always able to find such a solution for the matrices in question). Since the eigensystem is homogeneous we can divide it by a constant, so we will solve the equivalent problem

$$Q\frac{\lambda}{||\lambda||}=r^2\frac{\lambda}{||\lambda||}$$

We use the orthonormal solution above to construct the solution. Indeed we find that the solution is given by the set

$$\{r_i=\sqrt{\gamma_i},\lambda_i=\pm\sqrt{\gamma_i}e_i,~~ \gamma_i>0\}$$

There are no solutions allowed corresponding to negative eigenvalues. Since the value of the function on the critical points is

$$f(v)=1-\log(r_i^2)$$

The minimum is given by the vector that corresponds to the maximum positive eigenvalue of the matrix $Q$:

$$r=\max_i(\sqrt{\gamma_i})~,~ \lambda=\pm \sqrt{\gamma_{\max}}e_{\gamma_{\max}}~,~ f=1-\log\gamma_{\max}$$

Note that the minimum is always at least doubly degenerate, however it can have higher degeneracy if the maximum eigenvalue has multiplicity $>1$.


Title: How to simply this radical expression $\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$

Body:
$$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$

I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it?

Thanks!

Body:
Cubing both sides, we wish to prove:

$$\sqrt[3]{2}-1 = \bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3}$$

We have:

$$\bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3} = \frac{3}{1 + 3\sqrt[3]{2} + 3\sqrt[3]{2}^{2} + 2}=\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{1}{1+1\cdot\sqrt[3]{2} + \sqrt[3]{2}^{2}}\cdot\frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}$$

Because $(a - b)(a^{2} + ab + b^{2}) = a^{3} - b^{3}$, this simplifies to:

$$=\frac{\sqrt[3]{2}-1}{\sqrt[3]{2}^{3} - 1} = \frac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1$$

Thus, this identity is true. $\blacksquare$


问:
Convergence of Fourier series at discontinuities
For the Fourier series in this question:
Math Question

How do I find out where the Fourier series converges, I initially thought I just apply this equation:  $ f\left(x\right)=\frac{1}{2}\left\{f\left(x+\right)+f\left(x-\right)\right\}$ at points of discontinuity

From the graph I thought the only point of discontinuity is at $x=2$ and the Fourier series converges to 3/2 at $x=2$.

But the answer says it also converges to -1/2 at $x=4$. I'm not sure why it does this at $x=4$

答:
The function f(x) can be evaluated as defined in formula (1) below.

(1) $\quad f(x)=
 \left\{
\begin{array}{cc}
 x & 0\leq x<2 \\
 3-x & 2\leq x<4 \\
 f((x \bmod 4)) & \text{True} \\
\end{array}\right.$

The following figure illustrates the function $f(x)$ defined in formula (1) above has discontinuities at $(x \bmod 4)=0$ as well as $(x \bmod 4)=2$. The jump at $(x \bmod 4)=0$ is from $-1$ to $0$ so the half-step at $(x \bmod 4)=0$ evaluates to $-\frac{1}{2}$ which is illustrated by the red dots in Figure (1) below. The jump at $(x \bmod 4)=2$ is from $2$ to $1$ so the half-step at $(x \bmod 4)=2$ evaluates to $\frac{3}{2}$ which is illustrated by the green dots in Figure (1) below.

Figure (1): Illustration of $f(x)$ defined in formula (1)


答:
Your work suggests the problem is equivalent to maximizing $|v^\top u|$ subject to the constraint $v^\top A v = 1$.

Making the change of variables $z=A^{1/2} v$ and $y=A^{-1/2} u$, this is equivalent to maximizing $|z^\top y|$ subject to $z^\top z = 1$. By Cauchy-Schwarz, this is attained when $z = \pm y/\|y\|$. Working backwards, this leads to $v = A^{-1/2} z = \pm A^{-1/2} y/\|y\| = \pm \frac{A^{-1} u}{\sqrt{u^\top A^{-1} u}}$.

Title: Find minimum of $f(\mathbf{v})= \mathbf{v}^\top A \mathbf{v} - \sum_{i=1}^N \log((\mathbf{v}^\top \mathbf{u}_i)^2) $

Body:
Consider the function:
$$ f(\mathbf{v})= \mathbf{v}^\top A \mathbf{v} - \log((\mathbf{v}^\top \mathbf{u})^2), $$
where $\mathbf{v}, \mathbf{u} \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times n}$ is symmetric and positive definite.

I would like to find the solution(s) in $\mathbf{v}$ forming the minimum of this function.

Differentiating, we have
$$f’(\mathbf{v}) = 2A \mathbf{v} - \frac{2 \mathbf{u}}{\mathbf{v}^\top \mathbf{u}}.$$

Setting this to zero and multiplying the LHS by $\mathbf{v}^\top$ gives the solution set as:
$$\mathbf{v}^\top A \mathbf{v} = 1.$$

However, I’m unsure how to interpret the above and find the vectors that constitute as the minimum of the original function.

Now, can this then be extended to find the minimum of:
$$f(\mathbf{v})= \mathbf{v}^\top A \mathbf{v} - \sum_{i=1}^N \log((\mathbf{v}^\top \mathbf{u}_i)^2),$$
with the known and independent $\mathbf{u}_i \in \mathbb{R}^n$?


Body:
Note $(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)=(\sqrt[3]{2})^3-1=1$ and

\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}
=& \sqrt[3]{\frac1{\sqrt[3]{4}+\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{(\sqrt[3]{2}+1)^3}}=
 \frac{\sqrt[3]{3}}{\sqrt[3]{2}+1}
\end{align}


问:
Applied example of the weak law of large numbers
Consider a sequence of random variables $X_{n}$ such as $P({X_{n}} = a) = 0$ for $ \forall a, a \ne 0, a\ne 1 $.
Is it possible to define a common distribution but have $$X_{n} \ne X_{p}, \forall (n,p), n \ne p$$

My understanding is that if two random variables are equal they are not independent so to use the law of large numbers you must have them different from one each other.

If this is the case could we apply the law of weak numbers to this sequence ?

答:
Define $U \sim Unif([0,1])$ and define Bernoulli random variables for each $n \in \{1, 2, 3, ...\}$ by
\begin{align}
X_n &= \left\{\begin{array}{cc}
1 & \mbox{ if $U \in \left[\frac{1}{2n}, \frac{1}{2n}+1/3\right]$} \\
0 & \mbox{ else}
\end{array}\right.
\end{align}
The random variables $\{X_n\}_{n=1}^{\infty}$ are distinct but they have the same distribution because $P[X_n =1] =1/3$ for all $n \in \{1, 2, 3, ...\}$.

And $\frac{1}{n}\sum_{i=1}^n X_i$ does not converge to $1/3$ in probability.

Aside: If by $X_n \neq X_p$ you mean that $X_n(\omega) \neq X_p(\omega)$ for all $n\neq p$ and all $\omega \in S$, then note that if $P[X_n=0]=\theta>0$ for all $n \in \{1, 2, 3, ...\}$, then the events $$\{X_1=0\}, \{X_2=0\}, ..., \{X_k=0\}$$ cannot be disjoint if $k$ satisfies $k>1/\theta$.


问:
Which is the algorithm for knn density estimator?
I am reading Pattern Recognition and Machine Learning by Christopher Bishop. In chapter two he talk about using knn to density estimation. I want to replicate a plot using python/R/matlab.  He is doing it with synthetic data, but I do not know how to update the value V (volume of Region containing X from p(x)) in the following formula $$P(x)=\frac{K}{NV}$$. I could not find any implementation of this algorithm for density estimation.
This is the plot:

答:
If your data lives in $p$-dimensional space, then $V = V_p(x)$ is the volume of a $p$-dimensional ball with radius equal to the distance of $x$ from its $k$-th nearest neighbour. So assume $x_k$ is the $k$-th nearest neighbour of $x$, then

\begin{align*}
p_k(x)  =\frac{k}{n} \frac{1}{ \frac{\pi^{p/2}}{\Gamma(p/2+1)}  \|x-x_k \|}.
\end{align*}

Here is an example taken from these notes: Assume $p=1$, and we have data $\mathcal{X} = \{1,2,6,11,13,14,20,33 \}$, and we wish to find the knn density estimator at $x=5$ and $k=2$. The distance from $x=5$ to each data point is
$$
\{4,3,1,6,8,9,15,28 \}
$$
So its nearest neighbour is 6, and it's second nearest neighbour is 2, which is a distance of 3 away. Then we have
$$
p_{k}(x) = \frac{2}{8} \frac{1}{ \frac{\pi^{1/2}}{\Gamma(3/2)} \times 3} = \frac{1}{24}.
$$

Here is a (naive) Python implementation where I sample randomly from a Gaussian and then build the knn density estimator on top of that sample for varying k, producing the following plot:

The code used is:

```<code>import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

gaussian = norm(loc=0.5, scale=0.2)
X = gaussian.rvs(500)
grid = np.linspace(-0.1, 1.1, 1000)

Ks = [4,10,60]
fig, axes = plt.subplots(3,1, figsize=(10,10))

for i, ax in enumerate(axes.flat):

    # choose K value
    K = Ks[i]

    # run knn density estimation with chosen K
    p = np.zeros_like(grid)
    n = X.shape[0]
    for i, x in enumerate(grid):
        dists = np.abs(X-x)
        neighbours = dists.argsort()
        neighbour_K = neighbours[K]
        p[i] = (K/n) * 1/(2 * dists[neighbour_K])

    # True dist
    ax.plot(grid, gaussian.pdf(grid))

    # plot density estimate
    ax.plot(grid, p)

    ax.set_title(f'<span class="math-container">$k={K}$</span>')
plt.savefig("knn-density_est.png", dpi=300)
plt.show()
</code>```

Body:
If your data lives in $p$-dimensional space, then $V = V_p(x)$ is the volume of a $p$-dimensional ball with radius equal to the distance of $x$ from its $k$-th nearest neighbour. So assume $x_k$ is the $k$-th nearest neighbour of $x$, then

\begin{align*}
p_k(x)  =\frac{k}{n} \frac{1}{ \frac{\pi^{p/2}}{\Gamma(p/2+1)}  \|x-x_k \|}.
\end{align*}

Here is an example taken from these notes: Assume $p=1$, and we have data $\mathcal{X} = \{1,2,6,11,13,14,20,33 \}$, and we wish to find the knn density estimator at $x=5$ and $k=2$. The distance from $x=5$ to each data point is
$$
\{4,3,1,6,8,9,15,28 \}
$$
So its nearest neighbour is 6, and it's second nearest neighbour is 2, which is a distance of 3 away. Then we have
$$
p_{k}(x) = \frac{2}{8} \frac{1}{ \frac{\pi^{1/2}}{\Gamma(3/2)} \times 3} = \frac{1}{24}.
$$

Here is a (naive) Python implementation where I sample randomly from a Gaussian and then build the knn density estimator on top of that sample for varying k, producing the following plot:

The code used is:

import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm

gaussian = norm(loc=0.5, scale=0.2)
X = gaussian.rvs(500)
grid = np.linspace(-0.1, 1.1, 1000)

Ks = [4,10,60]
fig, axes = plt.subplots(3,1, figsize=(10,10))

for i, ax in enumerate(axes.flat):

    # choose K value
    K = Ks[i]

    # run knn density estimation with chosen K
    p = np.zeros_like(grid)
    n = X.shape[0]
    for i, x in enumerate(grid):
        dists = np.abs(X-x)
        neighbours = dists.argsort()
        neighbour_K = neighbours[K]
        p[i] = (K/n) * 1/(2 * dists[neighbour_K])

    # True dist
    ax.plot(grid, gaussian.pdf(grid))

    # plot density estimate
    ax.plot(grid, p)

    ax.set_title(f'$k={K}$')
plt.savefig("knn-density_est.png", dpi=300)
plt.show()


问:
Relating hyperspherical and Hopf coordinates for the unit three sphere
I thought that my question would be very elementary and I could find the answer online somewhere, but I have been searching for it and still I have not found anything.

What is the relation between the hyperspherical and the Hopf coordinates for the unit three sphere?

Let's be more precise.

Starting from the four dimensional Euclidean flat space

\begin{equation}
ds^2 = \sum^3_{i=0} dx^2_i
\end{equation}

$\textbf{Hyperspherical coordinates}$

We can write a unit three-sphere simply by embedding it as
\begin{equation}
\begin{aligned}
x_0  &= \cos \chi_1  \, ,\\
x_1  &= \sin \chi_1 \cos \chi_2  \, ,\\
x_2  &= \sin \chi_1 \sin \chi_2 \cos \chi_3  \, ,\\
x_3  &= \sin \chi_1 \sin \chi_2 \sin \chi_3  \, ,\\
\end{aligned}
\end{equation}
and substituing the above yields

\begin{equation}
ds^2 = d \chi^2_1 + \sin^2 \chi_1 (d \chi^2_2 + \sin^2 \chi_2 d \chi^2_3)
\end{equation}
The coordinates take values as $0 \leq {\chi_1, \chi_2} \leq \pi$ and $0 \leq \chi_3 \leq 2 \pi$.

$\textbf{Hopf coordinates}$

These are the cooridnates for the embedding of $S^3$ in the complex plane $\mathbb{C}^2$. However, the coordinate change is easily implemented as an embedding in four dimensional Euclidean space as
\begin{equation}
\begin{aligned}
x_0  &= \cos \xi_1 \sin \vartheta \, , \\
x_1  &= \sin \xi_1 \sin \vartheta \, , \\
x_2  &= \cos \xi_2 \cos \vartheta \, , \\
x_3  &= \sin \xi_2 \cos \vartheta \, , \\
\end{aligned}
\end{equation}
and substituing the above once more in the invariant line element gives us

\begin{equation}
ds^2 = d \vartheta^2 + \sin^2 \vartheta d \xi^2_1 + \cos^2 \vartheta d \xi^2_2
\end{equation}

The above is the Hopf bundle
\begin{equation}
S^1 \rightarrow S^3 \rightarrow S^2
\end{equation}

The coordinates range as follows: $0 \leq \vartheta \leq \pi/2$ and $0 \leq {\xi_1, \xi_2} \leq 2 \pi$

How are these two coordinate sets related to one another?

答:
Notice the last two coordinates imply $\chi_3=\xi_2$. Indeed, the hyperspherical coordinate system is just the usual 3D spherical coordinate system (with the $x_0$-axis the polar axis) but revolved using $x_2x_3$-plane rotations, and similarly the Hopf coordinate system is also a 3D spherical coordinate system (with $x_2$-axis the polar axis) also revolved using $x_2x_3$-plane rotations. Thus, it suffices to examine a 3D cross section:

$$ \begin{array}{rrrrr}
x & = & \cos\chi_1 & = & \cos\xi\sin\vartheta \\
y & = & \sin\chi_1\cos\chi_2 & = & \sin\xi\sin\vartheta \\
z & = & \sin\chi_1\sin\chi_2 & = & \cos\vartheta
\end{array} $$

Again, these are both just spherical coordinates, but with perpendicular polar axes. They become different coordinate systems in 4D since the first coordinate system is rotated between a fourth dimension and an axis perpendicular to the polar one (the $z$-axis), while the second coordnate system is rotated between a fourth dimension and the polar axis (again the $z$-axis).

Besides $\chi_3=\xi_2$, we can solve for one coordinates in terms of the other pretty easily by solving for parameters in the correct order. For instance, to get $\chi_1,\chi_2$ solve for $\chi_1$ first then $\chi_2$, and to get $\xi,\vartheta$ solve for $\vartheta$ first and then $\xi$ second.

If you stereographically project the two spherical coordinate systems (using the $x$-axis as the polar axis, say), you get polar coordinates for the first system and bipolar coordinates for the second.


问:
show that $(0.5)^x+(0.5)^{1/x}\le 1.$ if $x>0$
let $x>0$ show that
$$(0.5)^x+(0.5)^{1/x}\le 1.x>0,\tag{1}$$

Here is what I tried:let $f(x)=(0.5)^x+(0.5)^{1/x}$ so $f(x)=f(\frac{1}{x})$.si we only prove when $x\ge 1$.have $f(x)\le 1=f(1)$.it is sufficient to prove that
$$f'(x)=(0.5)^x\ln{0.5}-\dfrac{1}{x^2}\cdot(0.5)^{\frac{1}{x}}\ln{0.5}\le 0,x\ge 1$$
since $\ln{0.5}<0$,if we can show that
$$(0.5)^{\frac{1}{x}-x}\le x^2,x\ge 1$$
if we let $g(x)=x^2-(0.5)^{\frac{1}{x}-x}$,then $$g’(x)=2x+\ln{0.5}(0.5)^{\frac{1}{x}-x}\left(\dfrac{1}{x^2}+1\right)$$
it seem not easy to prove $g'(x)\ge 0,x\ge 1$ because $\ln{0.5}<0$.other words,this methods following more ugly!,so  How to prove this inequality $(1)$.Thanks.

答:
We can make the OP's approach work. Let $f(x) = (0.5)^x+(0.5)^{1/x}$, so that $f(x) = f(\frac1x)$, and it suffices to prove that $f(x)<1$ for $0<x<1$. We will do so by showing that $f(x)$ is decreasing to a local minimum and then increasing on this interval.

Since $f'(x) = (2^{-1/x}x^{-2}-2^{-x}) \ln 2 = (2^{x-1/x}-x^2)\frac{\ln 2}{x^22^x}$, we see that $f(x)$ is decreasing when $x^2 > 2^{x-1/x}$ and increasing otherwise. Equivalently (and this is, perhaps, the crucial step), $f(x)$ is decreasing when $2\ln x > (x-\frac1x)\ln 2$ and increasing otherwise. If we set $h(x) = 2\ln x - (x-\frac1x)\ln 2$, we therefore want to show that $h(x)$ starts off negative, has a unique zero, and is then positive over the interval $(0,1)$.

But now $h'(x) = \frac2x - (1+\frac1{x^2})\ln 2 = -(\ln2)(x^2-\frac2{\ln2}x+1)/x^2$; this makes it easy to see that $h'(x)>0$ precisely when
$$
0.40281 \approx \frac{1-\sqrt{1-(\ln2)^2}}{\ln2} < x < \frac{1+\sqrt{1-(\ln2)^2}}{\ln2} \approx 2.48258,
$$
which concludes the proof.


问:
Linear Independence of Angles Corresponding to Primes $p \equiv 1 \pmod 4$
Any prime $p \equiv 1 \pmod 4$ can be written uniquely as the sum of two squares $p = a^2 + b^2$. In the complex plane, this corresponds to the 8 Gaussian integers with norm $p$. We can write these points as
$$
\sqrt{p}e^{2\pi i (\pm\Phi + t/4)}
$$
where we define $\Phi$ as the argument of the point in the first octant ($\Phi = \arg(a+bi)$ where $a^2 + b^2 = p$ and $a \geq b$) and $t \in \{0,1,2,3\}$ corresponds to multiplication by a unit.

I want to show that any finite set of the type $\left\{ 1, p_1, \ldots , p_N \right\}$ where $p_j \equiv 1 \mod 4$, $j = 1, \cdots, N$ is linearly independent over $\mathbb{Q}$. In other words, we want to show
$$
\sum_{j=1}^{N} \alpha_j \Phi_j + \alpha_0 = 0
\iff
\alpha_0 = \alpha_1 = \cdots =\alpha_N = 0
$$
where the $\Phi_j$ is defined for each $p_j$ as described above.

I encountered this problem while self-studying some number theory and have been stuck for a bit. I'd appreciate some hints! The approach I tried is to go by contradiction and try to show that something will contradict the definition of a prime number, but I think I've gotten lost!

答:
There was a previous answer here from @TheoBendit (I suppose he deleted it) but that was enough when combined with a hint from Ramin Takloo-Bighash to help me solve this.

Since we can multiply through by a common denominator, we can assume all the $\alpha_j$ are integers.

Let $a_j + ib_j = \sqrt{p_j}e^{2\pi i \Phi_j}$ (These coefficients come from $p_j = a_j^2 + b_j^2$) Suppose the angles are linearly dependent, then we can calculate
\begin{align*}
\prod_{j=1}^{n} (a_j + ib_j)^{2\alpha_j}
&=
\prod_{j=1}^{n} \left(\sqrt{p_j} e^{2\pi i \Phi_j} \right)^{2\alpha_j} \\
&=
\prod_{j=1}^{n} p_j^{\alpha_j} e^{2\pi i (-2\alpha_0)} \\
&=
\prod_{j=1}^{n} p_j^{\alpha_j} \\
&=
\prod_{j=1}^{n} (a_j + ib_j)^{\alpha_j} (a_j - ib_j)^{\alpha_j}
\end{align*}
Cancelling terms from both sides, we have
\begin{align*}
\prod_{j=1}^{n} (a_j + ib_j)^{\alpha_j}
=
\prod_{j=1}^{n} (a_j - ib_j)^{\alpha_j}
\end{align*}
This is a contradiction since $\mathbb{Z}[i]$ is a UFD.

Credit to Theo for motivating me to consider a product of this type, and credit to Ramin for the hint of raising everything to the $2\alpha_j$.