问:
Conditional expectation with $\sigma$-field
I'm reading Bass' book about Stochastic Processes. Since I haven't yet acquired a firm understanding of some basics, I'm currently looking into Conditional Expectation in the Appendix.

The proof of the following proposition (A.21) gives me trouble:

If X and XY are integrable and Y is measureable with respect to $\mathcal F$, then
$$
\mathbb E [XY | \mathcal F] = Y \mathbb E[X|\mathcal F].
$$

The proof:

If $A\in\mathcal F$, then for any $B \in \mathcal F$,
$$
\mathbb E[\ 1_A \mathbb E [X|\mathcal F];B\ ]=\mathbb E[\ \mathbb E [X|\mathcal F];A\cap B\ ] =  \mathbb E [X; A\cap B] = \mathbb E [1_A X; B].
$$
(and more arguments $\dots$)

Three questions:

Does $\mathbb E[X; A]$ mean $\mathbb E[X 1_A]$ in terms of notation?

How does the second "$=$" work? What is the argument here?

After the part of the proof I wrote above, the author then writes that this argument is enough two show the statement for indicator functions (i.e. $Y=1_A$) due to linearity. The original statement does not have the expectation there though, so why is this argument enough? What does linearity have to do with it?

Any help is highly appreciated!

答:
For your first question, yes $\mathbb E[X;A]:\mathbb E[X\boldsymbol 1_A]$. So,
$$\mathbb E[\boldsymbol 1_A\mathbb E[X\mid \mathcal F];B]=\mathbb E[\boldsymbol 1_{A}\mathbb E[X\mid \mathcal F]\boldsymbol 1_B]=\mathbb E[\mathbb E[X\mid \mathcal F]\boldsymbol 1_{A\cap B}]=\mathbb E[\mathbb E[X\mid \mathcal F],A\cap B].$$

Why is this enough ? What you've done before allows you to conclude that the formula holds for simple function, i.e. for function of the form $$Y:=\sum_{i=1}^n a_i\boldsymbol 1_{A_i},$$ where $A_i$ are $\mathcal F-$measurable.

Therefore, if $Y\geq 0$, there is a a sequence $(Y_n)$ of simple function s.t. $Y_n\nearrow Y$, and thus, for $Y\geq 0$, the formula follow by monotone convergence theorem. Finally if $Y$ is only $\mathcal F-$measurable, then you can write it as $Y^+-Y^-$ where $Y^+=\max\{Y,0\}\geq 0$ and $Y^-=-(\min\{Y,0\})\geq 0$, and thus apply apply the formula to $Y^+$ and $Y^-$.


问:
Euler product over a finite subset of composite numbers
The product $$\prod_{p\text{ prime}}\frac{p^n}{p^n-1}=\zeta(n)$$ is well-known. This works over primes, and only over the entire set of them.

Is any similar expression known for composites? For a finite subset of them?

e.g. $$\prod_{c\text{ composite, c}=c_{i}}^{c=c_{j}}\frac{c^n}{c^n-1}=f(m)$$ for some $f$.

This is generic. If such an expression is not known, is anything known for the case $i=1$ and $n=1$? If not, then for $i=n=1$ and $j=\infty$? If that is also not known, is anything known on $\lim \inf f(m)$ or $\lim \sup f(m)$?

答:
This Answer addresses the coefficients of the Dirichlet-Series of $f(s) = \prod_{c \text{ comp}} \frac 1{1-c^{-s}}$.

We introduce
$$g(s) = \prod_{n=2}^\infty (1 + n^{-s} + n^{-2s} + \dots ),$$ which is a holomorpic function on $\Re s > 1$, as $\left | 1 - \frac 1{1-n^{-s}}\right | \ll n^{-s}$.
This can be re-written as a dirichlet-series
$$g(s) = \sum_{n=1}^\infty c(n) n^{-s},$$
where $c(n) = \#\{(e_2, e_3, \dots) \in \mathbb N_{\geq 0}^* | 2^{e_2} 3^{e_3} 4^{e_4} \dots = n \}$ is the amount of multiplicative partitions of $n$. With $\mu$ (the möbius-function) we obtain
$$\prod_{c \text{ comp}} \frac 1{1-c^{-s}} = \zeta(s)^{-1} g(s) = \sum_{n=1}^\infty a(n) n^{-s},$$
where $a(n) = (c \star \mu)(n) = \sum_{d \mid n} \mu(d) c(\frac nd)$.

ContentLicense: CC BY-SA 4.0


问:
What lies beyond the Sedenions
In the construction of types of numbers, we have the following sequence:

$$\mathbb{R} \subset \mathbb{C} \subset \mathbb{H} \subset \mathbb{O} \subset \mathbb{S}$$

or:

$$2^0 \mathrm{-ions} \subset 2^1 \mathrm{-ions} \subset 2^2 \mathrm{-ions} \subset 2^3 \mathrm{-ions} \subset 2^4 \mathrm{-ions} $$

or:

"Reals" $\subset$ "Complex" $\subset$ "Quaternions" $\subset$ "Octonions" $\subset$ "Sedenions"

With the following "properties":

From $\mathbb{R}$ to $\mathbb{C}$ you gain "algebraic-closure"-ness (but you throw away ordering).

From $\mathbb{C}$ to $\mathbb{H}$ we throw away commutativity.

From $\mathbb{H}$ to $\mathbb{O}$ we throw away associativity.

From $\mathbb{O}$ to $\mathbb{S}$ we throw away multiplicative normedness.

The question is, what lies on the right side of $\mathbb{S}$, and what do you lose when you go from $\mathbb{S}$ to one of these objects ?

答:
You can build tessarines of any dimension of the for $2^n$. They still will be commutative ans associative. This is used in digital signal processing.

Cayley-Dickson's construction, used in quaternions, on the other hand, loses most of its useful properties with the growth of the number of dimensions.

ContentLicense: CC BY-SA 3.0

Title: What lies beyond the Sedenions


问:
How to solve integrals of the form $\sqrt{ax^4+bx^2+c}$?
How do you integrate $\sqrt{ax^4+bx^2+c}$? I tried substituting $x^2=t$, but it didn't work. Thanks in advance.

答:
Some of these are elementary
$$
\int\sqrt{x^4+2 x^2+1}\;dx = \frac{x^3}{3}+1+C
$$
But in general these are elliptic integrals .  For example,
$$
\int\sqrt{-x^4+1}\;dx = \frac{x\sqrt{1-x^4}}{3} +
\frac{2}{3}\;F(x,1) + C
$$
where $F$ is the incomplete elliptic integral of the first kind


问:
Is a line, that's contained in a plane, parallel to it?
Have got into a pretty heated debate with a friend, and looking online there's lacking proof
Line contained in a plane

Is a line that is contained within a plane, considered parallel to it? By my understanding it is parallel , if at all points the line has equal distance to the plane, which in this case is always 0, but they are  debating that it intersects the plane at infinitely many points, so it cannot be considered parallel, is there some mathematical proof that would prove or disprove either statement?

I see there is heated debate online about whether a line that matches another line is parallel to it and I'm wondering if there's a clean answer about this.

答: In Euclidean space for two lines to be parallel they must never intersect. Since a line in a plane intersects the plane it cannot be parallel to it. Note that the lack of intersection is necessary but not sufficient in higher dimensions.</p>


问:
Monotone likelihood ratio without densities
I would like to find a generalization of the monotone likelihood ratio ordering that does not require that the probability distributions admit a density and have the same support, which would allow me to deal with degenerate probability distributions, and with the case where the supports are disjoint.

For instance, if $T_0$ is a degenerate probability distribution localized at $0$ and $T_1$ is a probability distribution with support included in $[1,+\infty)$, I would like the definition to guarantee that $T_1$ dominates $T_0$ according to this ordering. This seems to be a natural extension of the standard definition.

I would also like to the definition to boil down to the standard criterion if the distributions admit a density and have a common support.

Are there any existing generalizations of the monotone likelihood ratio order that would be appropriate for this purpose? Thanks!

答:
If we take in full generality two probability measures $\mu , \nu$ on $\Bbb R$ we can define the likelihood ratio order as in Def. 1.C.1 in Stochastic Orders by Shaked and Shanthikumar (2007):

For two measurable sets $A,B \subset \Bbb R$ we write $A\leq B$, if $(x,y) \in A\times B$ implies $x\leq y$ for any such pair. We say $\mu$ is smaller than $\nu$ in the likelihood ratio order, if
$$A \leq B \Rightarrow \mu (A) \nu (B) \geq \mu (B) \nu (A).$$

If $\mu , \nu$ have Lebesgue densities $f, g$, this definition is equivalent to
$$f(u)g(v) \geq f(v) g(u) \qquad \text{for almost all } u\leq v.$$


问:
Notation on definition of spherical symmetry of spacetime
In reading something about relativity and in particular looking at first to the definition of spacetime as "spherically symmetric" I have encountered the following notation referred to an element of the orthogonal special group of $\mathbb{R}^3$ ($SO(3)$):
$$R\in SO(3)\iff \begin{cases} {}^{t}_{}R I R=I\\ det(R)=1 \end{cases} $$
What is the meaning of the superscript $t$? Which operation of matrices correspond?

答: That superscript $^t$ is (bad, in my opinion) notation for the matrix transpose. That is, if $$R = \pmatrix{a & b & c \\ d & e & f \\ g & h & i}$$, then $^tR$ (urgh, the kerning) is simply the matrix $$\pmatrix{a & d & g\\b & e & h\\c & f & i}.$$


问:
Other proof that $h(x)=x^{p}\chi_{[0,\infty)}(x)$ is convex.
Let $p\geq 1$. The function $h:\mathbb{R}\rightarrow\mathbb{R}$ given by
\begin{align*}
    h(x) =
    \begin{cases}
        x^{p}   & \text{if } x\geq0\\
        0       & \text{otherwise}
    \end{cases}
\end{align*}
is convex. Is there a proof for this that does not consider cases then points are to the left and/or to the right of $0$? E.g. via the epigraph or operations that preserve convexity.

答:
For $p>1$ a continuous derivative $h'$ exists and $h'$ is increasing This implies that $h$ is convex. The case $p=1$ is trivial.

[To prove that $h(tx+(1-t)y) \leq th(x)+(1-t)h(y)$ apply MVT to $h(y)-h(tx+(1-t)y)$ and $h(tx+(1-t)y)-h(x)$. You will see immediately  that $t(h(tx+(1-t)y)-h(x)) \leq (1-t) h(y)-h(tx+(1-t)y)$$ which is what we want to prove].


问:
Quaternion Matrix Multiply
My question is about applying rules of quaternions to quaternion matrices.

I know that for some rotation quaternion q = [w, x, y, z], I can find the rotation of point p1 by p2 = q*p1*conjugate(q). I also know that I can do quaternion multiplication to get the combined rotation of 2 quaternions q1 and q2 by using quaternion multiplication q = q1*q2 as expressed here.

Now suppose I construct a quaternion matrix as described here. Furthermore suppose I do that for three coordinate axes to construct three quaternion matrices Qx Qy and Qz. Could I then get a combined quaternion matrix to achieve the same result by doing Q = Qz*Qy*Qx and then applying that to p1 (i.e. p2 = Q*p1)? Obviously the ordering here matters.

This is somewhat similar to this question, but I am specifically interested in knowing if the method I have shown produces the correct result. This question is also not about whether or not this approach is the preferred way to do it.

This question has some nice references, but I didn't find the answer to what I'm asking at any of the links.

答:
First of all, let's note that what you linked to is a quaternion derived rotation matrix of an axis-angle representation. That is, given a vector $\vec{a}$ and an angle $\theta$, you get a rotation matrix $A$ such that $Ap_1=p_2$. The only way quaternions came into play is during the translation between the axis-angle and the rotation matrix.

Therefore the question appearing after the first paragraph is about regular rotation matrices, and the quaternion stuff mentioned in the first paragraph doesn't really come into play.

Your question seems to be:

"Suppose I have a rotation of $\theta$ around axis $\vec{a}$ and I make a rotation matrix $R$ like in the link that produces the rotation for $\vec{a}$ and $\theta$. Further suppose that I have figured out three rotations around the axes which compose to the same rotation, and I find their matrices individually and multiply them together to get another rotation matrix. Is this product matrix the same as $R$?"

The answer, provided you have used a single basis the entire time, is "yes." This is because the $n\times n$ special orthogonal matrices faithfully represent the group of rotations in $\Bbb R^3$ for a fixed basis. That is, each matrix corresponds to exactly one rotation.

As an aside, this can be contrasted with the quaternion case, since for every rotation there are two quaternions that represent it, so the unit quaternions don't faithfully represent the rotations (although they come pretty close.) It turns out that if $q$ represents a certain rotation, then so does $-q$. So, if you come up with two unit quaternions $x$ and $y$ via two different methods, and you know they produce the same rotation, then you can only conclude that $x=\pm y$

Title: Quaternion Matrix Multiply


问:
Evaluating $\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \cdots}}}}$

$$x =\sqrt{1+\sqrt{2 - \sqrt{3 + \sqrt{4 - \sqrt{5+ \sqrt{6 - \cdots}}}}}}$$

Find $x$.

I am not sure how to proceed.

Is this a sort of Arithmetico-Geometric Progression? Will this converge at a point?

Any help would be sincerely appreciated.

答:
The computations will involve complex numbers. Starting at stage $4$. So
$$
\sqrt{1+\sqrt{2-\sqrt{3+\sqrt{4}}}} \approx 1.03 + i\;0.24
$$
Always choosing the "principal" square root, the limit seems to be  $1.005667607+ i\; 0.1066177058$ .

added

In fact, $\sqrt{3+\dots}$ converges to a real value, but $\sqrt{2-\dots}$  and $\sqrt{1+\dots}$ to non-real values.


答: This is entirely a question of convention. I'd say yes, because I like "parallelness" to be an equivalence relation, but it frankly does not matter in the slightest, providing you're clear about how you are using the word.</p>

Title: Is a line, that's contained in a plane, parallel to it?

Body:
Have got into a pretty heated debate with a friend, and looking online there's lacking proof
Line contained in a plane

Is a line that is contained within a plane, considered parallel to it? By my understanding it is parallel , if at all points the line has equal distance to the plane, which in this case is always 0, but they are  debating that it intersects the plane at infinitely many points, so it cannot be considered parallel, is there some mathematical proof that would prove or disprove either statement?

I see there is heated debate online about whether a line that matches another line is parallel to it and I'm wondering if there's a clean answer about this.


问:
How to determine the leading-order asymptotic behaviour of this integral?
$\int^{1}_{0} \cos(xt^{3})\tan(t)\, dt$

as x → ∞

I am stuck on how to apply the stationary phase method when $\tan(t)$ vanishes at the stationary point. Should I expand tan in its Taylor series? Is the stationary point $0\,$? Thanks in advance

答:
Just to show something different from @joriki's answer to [this question], working first the antiderivative
$$I=\int\cos(xt^{3})\tan(t)\, dt$$ let $x t^3=u^3$ to make
$$I=\frac 1{x^{1/3}}\int\cos \left(u^3\right) \tan \left(\frac{u}{x^{1/3}}\right)\,du$$ Since $x\to \infty$, using $\tan(\epsilon)\sim \epsilon$,we have
$$I\sim \frac 1{x^{2/3}}\int\ u\, cos \left(u^3\right)\,du$$ The integral can be computed using the gamma function or the exponential integral function to make
$$\int\ u\, cos \left(u^3\right)\,du=-\frac{1}{6} u^2 \left(E_{\frac{1}{3}}\left(-i
   u^3\right)+E_{\frac{1}{3}}\left(i u^3\right)\right)$$ Back to $t$
$$I \sim -\frac{1}{6} t^2 \left(E_{\frac{1}{3}}\left(-i t^3
   x\right)+E_{\frac{1}{3}}\left(i t^3 x\right)\right)$$ Using the bounds
$$J=\int_0^1\cos(xt^{3})\tan(t)\, dt\sim\frac{1}{6} \left(\frac{\Gamma
   \left(\frac{2}{3}\right)}{x^{2/3}}-E_{\frac{1}{3}}(-i x)-E_{\frac{1}{3}}(i x)\right)$$

Limited to the first order
$$E_{\frac{1}{3}}(-i x)+E_{\frac{1}{3}}(i x)\sim -\frac {e^{-i x }} x$$

So the leading order of $J$ is
$$\frac{\Gamma \left(\frac{2}{3}\right)}{6\, x^{2/3}}$$


答:
This is a matter of opinion and convention, and one definition may be more useful than another definition for some purposes. This is common
in mathematics.

One definition of parallel lines in
a Euclidean plane is that they are equidistant from each other. This
breaks down in the context of the elliptic or hyperbolic plane and
it depends on a distance metric. Perhaps a better definition is that
they do not intersect. Again, this breaks down in the context of the
elliptic or hyperbolic plane. However, in the context of the projective
plane, this must be modified to state that parallel lines intersect
only at one point at infinity.

Moving up to three dimensional space, there is the concept of skew lines which don't intersect each other, and are not parallel since they are
not in the same plane. Introducing a plane at infinity, the definition
of parallel lines is the same as in a fixed plane. That is, they only intersect at one point at infinity. Any plane intersects the plane at
infinity in a line at infinity. Two parallel planes only intersect at
the same line at infinity.

It now makes sense that each line in a fixed plane intersects the same
line at infinity as the fixed plane does and exactly at one point.
This also holds for each line in a plane parallel to the fixed plane.
Thus, there are good reasons to consider every line in a plane to be
parallel to that plane.


问:
Relationship between projection of $y$ onto $x_1, x_2$ individually vs. projection on both?
This is essentially similar to the question I just asked on cross validated, but here I am going to pose it in a linear algebra way.

Consider $y \in \mathbb{R}^n$ and $x_1, x_2, 1_n \in \mathbb{R}^{n}$. Suppose you orthogonally project $y$ onto $x_1, 1_n$ and find the projection of $y$ onto the subspace spanned by $x_1, 1_n$ can be written as $\hat{y}_1 = \hat{\beta}_1 x_1 + b_1$, i.e., a linear combination of $x_1$ plus some offset. Now do the same for orthogonal projection of $y$ onto $x_2, 1_n$ and find $\hat{y}_2 = \hat{\beta}_2 x_2 + b_2$.

Now consider projecting $y$ onto the subspace spanned by both $x_1, x_2, 1_n$ and find $\hat{y}_{12} = \hat{\gamma}_1 x_1 + \hat{\gamma}_2x_2 + b_{12}$.

If $x_1 \perp x_2$, then I know $\hat{\beta}_i = \hat{\gamma}_i$. But what if they're not orthogonal?

What can I say about the relationship between $\hat{\beta}$ and $\hat{\gamma}$ in this case?

Some specific questions that I am also interested in is if $\hat{\beta} >0 $, does this imply $\hat{\gamma} > 0$? If $x_1, x_2$ are linearly dependent, then I don't think this won't be true for one of the coefficients.

答:
I cannot say I completely understood what those constants $b_1$, $b_2$ or $b_{12}$ are for. But I understood the gist of your question and I'll try my best.

Say the orthogonal projection of $y$ onto the subspace spanned by $x_1$ can be written as $\hat{y}_1 = \hat{\beta}_1 x_1$, i.e., a linear combination of $x_1$. Now we do the same for orthogonal projection of $y$ onto $x_2$ and find $\hat{y}_2 = \hat{\beta}_2 x_2$.

Also we have the projection of $y$ onto the subspace spanned by both $x_1, x_2$ and find $\hat{y}_{12} = \hat{\gamma}_1 x_1 + \hat{\gamma}_2x_2$.

Without loss of generality we can say the vectors $x_1$ and $x_2$ are unit vectors and represent them by $\hat{x_1}$ and $\hat{x_2}$. If you don't want to do this, then rewrite all the vectors in terms in terms of $\hat{x_1}$ and $\hat{x_2}$. So for example, $\hat{\beta_1}$ will become $\hat{\beta_1} ||x_1||$

Now, consider this statement. The orthogonal projection of $\hat{y_{12}}$ onto $x_1$ would be the same as $\hat{y_1}$ and the orthogonal projection of $\hat{y_{12}}$ onto $x_2$ would be the same as $\hat{y_2}$.

So, by the definition of projection,

$$ \hat{y_{12}}.\hat{x_1} = ||\hat{y_1}|| $$

$$ \implies (\hat{\gamma}_1 \hat{x_1}  + \hat{\gamma}_2 \hat{x_2}).\hat{x_1} = ||\hat{y_1}||$$

$$ \implies \hat{\gamma}_1 \hat{x_1}.\hat{x_1}  + \hat{\gamma}_2 \hat{x_2}.\hat{x_1} = \hat{\beta}_1$$

$$ \implies \hat{\gamma}_1  + \hat{\gamma}_2 \hat{x_2}.\hat{x_1} = \hat{\beta}_1 \tag{1}$$

Similarly we can solve $ \hat{y_{12}}.\hat{x_1} = ||\hat{y_2}|| $ to get

$$ \implies \hat{\gamma}_1 \hat{x_1}.\hat{x_2} + \hat{\gamma}_2  = \hat{\beta}_2 \tag{2}$$

There you go. We have 2 equations and 2 unknowns.

Obviously we should know the value of $\hat{x_1}.\hat{x_2}$, in other words the cosine of the angle between them, to get the required relations. In the case where
$\hat{x_1}$ and $\hat{x_2}$ are orthogonal, $cos \frac{\pi}{2}=0$ and hence the result you gave $\hat{\beta}_i = \hat{\gamma}_i$.


问:
If a finite set tiles the integers, must it be an arithmetic progression?
Let $P$ be a finite subset of ${\mathbb Z}$ containing at least three elements. I say that $P$ tiles $\mathbb Z$ if $\mathbb Z$ can be written as a disjoint union of translates of $P$.

Trivially if $P$ is an arithmetic progression of the form $\lbrace t+sk \ | \ 1\leq k \leq M \rbrace$ where $s$ and $M$ are coprime, then $P$ tiles $\mathbb Z$ (because of Euclidean division and the Chinese remainder theorem).

Is the converse true ?

My thoughts : It is easy to see that $P=\lbrace 1,2,4\rbrace$ can never tile $\mathbb Z$. For suppose it did, and suppose $\bigcup_{t\in T}t+P$ is such a tiling. We call the $t+P$ (for $t\in T$) distinguished translates of $P$. Choose any $t_1\in T$. Then $t_1+3$ must be in some distinguished  translate of $P$, which can only be $(t_1+2)+P$ or $(t_1+1)+P$ or $(t_1-1)+P$. But all those intersect $t_1+P$, contradiction.

答: $$\{1;2;5;6 \}$$ tiles $\Bbb Z$ and is not an arithmetic progression.


问:
Why should one use the quotient rule instead of the power rule to differentiate a quotient?
There's a lot of emphasis on the difference quotient, as it's on the AP test and all that, but honestly using the power rule plus the product rule is soo much easier and gives you the same answer and slope. What am I missing here?

For example, if you wanted to differentiate
$$f(x)=\frac{x^2+4x-2}{x-1},$$
you could either do
$$f'(x)=\frac{d}{dx}[(x^2+4x-2)(x-1)^{-1}]=(2x+4)(x-1)^{-1}-(x^2+4x-2)(x-1)^{-2}, $$
or you could do
$$
f'(x)=\frac{(x-1)(2x+4)-(x^2+4x-2)(1)}{(x-1)^2}.
$$

答:
Maybe if you go up to the AP test, and then never do any math after that, it will not matter.  But is that what you plan?

For example, when you work on differential equations, it is often helpful to recognize an expression of the form $u v' - u' v$ as related to the derivative of a quotient.


问:
Different answers after different methods in solving a limit

Evaluate $$L=\lim_{x \to 0} \frac{e^{\sin(x)}-(1+\sin(x))}{(\arctan(\sin(x)))^2}$$

Method $1$: $$\frac{h^2\left(\frac{e^{h}-1}{h^2}-\frac1{h}\right)}{(\arctan(h))^2}=1^2\left(\frac{1*1}{h}-\frac1{h}\right)=\frac1{h}-\frac1{h}=0$$ Therefore $L=0$.

The identities I have used here to simplify the expression are $$\lim_{x \to 0} \frac{\arctan\left(x\right)}{x}=1$$ $$\lim_{x \to 0} \frac{a^{x}-1}{a}=\ln\left(a\right) \implies \lim_{x \to 0} \frac{e^{x}-1}{x}=\ln\left(e\right)=1$$

Method $2$: $$L=\frac{e^{h}-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(1+h+\frac{h^{2}}{2!}\right)-\left(1+h\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{\left(\frac{h^{2}}{2!}\right)}{\left(\arctan\left(h\right)\right)^{2}}=\frac{1}{2}$$ Therefore $L=\frac12$.

I am not at all familiar with $O(n)$ notation BTW.

答:
My (personal) favored approach is composition of series one piece at the time.
$$y=\frac{e^{\sin(x)}-(1+\sin(x))}{\big[\arctan(\sin(x))\big]^2}$$ Since the denominator is $\sim x^2$, let us use expansions to $O(x^4)$. Working one piece at the time
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$
$$e^{\sin(x)}=1+x+\frac{x^2}{2}+O\left(x^4\right)$$
$$\arctan(\sin(x))=x-\frac{x^3}{2}+O\left(x^4\right)$$
$$\big[\arctan(\sin(x))\big]^2=x^2-x^4+O\left(x^5\right)$$ Putting all together
$$y=\frac { \frac{x^2}{2}+\frac{x^3}{6}+O\left(x^4\right)} {x^2-x^4+O\left(x^5\right) }=\frac{1}{2}+\frac{x}{6}+O\left(x^2\right)$$ which shows not only the limit but also how it is approached.


问:
Cutting sticks puzzle
This was asked on sci.math ages ago, and never got a satisfactory answer.

Given a number of sticks of integral length  $ \ge n$ whose lengths
  add to $n(n+1)/2$. Can these always be broken (by cuts) into sticks of
  lengths $1,2,3, \ldots ,n$?

You are not allowed to glue sticks back together. Assume you have an accurate measuring device.

More formally, is the following conjecture true? (Taken from iwriteiam link below).

Cutting Sticks Conjecture: For all natural numbers $n$, and any given sequence $a_1, .., a_k$ of
  natural numbers greater or equal $n$ of which the sum equals
  $n(n+1)/2$, there exists a partitioning $(P_1, .., P_k)$ of $\{1, .., n\}$
  such that sum of the numbers in $P_i$ equals $a_i$, for all $1 \leq i \leq k$.

Some links which discuss this problem:

http://www.iwriteiam.nl/cutsticks.html

答:
There are variations on the problem where the division is always possible and a proof using complete induction.

The first variation is: [..original problem ...] where at least one of the sticks is >=2n.

The second variation is: [..original problem ...] where it is allowed to glue two of the sticks together and break once at an arbitrary position.

Proof for the variations:

The case n=1 is trivial - one stick of length 1. Then we suppose the assumption holds for n and consider the case n+1. I.e. given a number of sticks of integral length >= n+1 whose lengths add to (n+1)(n+2)/2 - can these be divided into sticks of lengths 1,2,3,…,n+1?

Break off from one of the sticks a length n+1. In the first variation we use the stick with lenght >= 2n+2. This part will be the required stick of length n+1 in the solution. Because we broke off a length n+1 the remaining total length is (n+1)(n+2)/2 - (n+1) = n(n+1)/2. In the first variation the inducution step is valid, in the second variation we may not yet use the induction step because the other part of the broken stick may now be smaller then n+1. In that case the second variation allows us to glue together this part with one of the other sticks. Now the induction step is valid and we break the rest of the collection of sticks in {1,..n} and have succeeded in the division {1,.. n+1}.

Title: Cutting sticks puzzle


问:
Calculate interior, closure and boundary
As part of an exercise, I want to calculate the interior, closure and boundary of the following sets in $\mathbb{R}^2$ (with the standard topology).

1. $\mathbb{Z} \times \mathbb{Z}$

2. $\mathbb{Q} \times (\mathbb{Q}\cap]0,+\infty[)$

I found the following solution and would like to verify whether those are correct.

1. Interior: $\emptyset$; Closure: $\mathbb{Z} \times \mathbb{Z}$; Boundary: $\mathbb{Z} \times \mathbb{Z}$

2. Interior: $\emptyset$; Closure: $\mathbb{R} \times (\mathbb{R}\cap[0,+\infty[)$; Boundary: $\mathbb{R} \times (\mathbb{R}\cap[0,+\infty[)$

答: You are correct. You're basically done after interior and closure: the boundary follows easily as the difference between the  closure and the interior. So empty interior implies boundary = closure, as we have here.</p>


答:
I'll throw my hat into the ring.

I certainly agree that it is a matter of convention.

However, it is also reasonable to presume that 3 dimensional Geometry is (in effect) an extension of 2 dimensional Geometry.

In 2 dimensional Geometry, I have never heard of the specification that a line is parallel to itself.  This (inconclusively) suggests that in 3 dimensional Geometry, a line can not be considered parallel to the plane that contains the line.

This is where the fun starts.  Normally, definitions, like the concept of parallel lines in 2 dimensional Geometry, are created in order to facilitate solving problems.  So, a hidden issue is the question of why the concept, in 3 dimensional Geometry, of a line being parallel to a plane, was created in the first place.

That is, what math problems is the definition designed to help solve.  Having identified these problems, you would then ask which convention facilitates solving these problems, whatever they are.


问:
Finding the (condensed/reduced) SVD of a given matrix by inspection
$
\newcommand{\tp}{^\mathsf{T}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\RR}[2]{\mathbb{R}^{#1 \times #2}}
\newcommand{\diag}{\mathrm{diag}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\m}[1]{\begin{pmatrix} #1 \end{pmatrix}}
$The Problem: We are given the matrix
$$
A := \begin{pmatrix}
3 & 6 & 9 \\
4 & 8 & 12
\end{pmatrix}
$$
From this, we want to find - via inspection, and not any detailed calculations - the singular value decomposition of $A$. (Specifically, the reduced or condensed form, in the sense later elaborated upon.)

Context: This comes up as problem $4.1.14$ in Watkins' text Fundamentals of Matrix Computations. In the text at this point, we've not come up with any of the tools typically associated with SVDs. For instance, it has not yet been developed at this point that the singular values of $A$ are tied to the eigenvalues of $A\tp A$, for instance (which has made searching for help quite difficult). We are just given a few basic theorems and definitions...

Relevant Theorems & Definitions:

Definition (Orthogonal): We say a matrix $A \in \RR n n$ is orthogonal if $A\tp A = I$, or equivalently $A\tp = A^{-1}$.

Definition (Isometry): We say a matrix $A \in \RR n m$ is an isometry if $A \tp A = I$.

Theorem $4.1.1$: (the usual SVD theorem)

Let $A \in \RR n m$ be nonzero with rank $r$. Then we may write
$$
A = U \Sigma V \tp
$$
for $U \in \RR n n, V \in \RR m m$ orthogonal, and where $\Sigma = \diag(\sigma_1,\cdots,\sigma_r,0,\cdots,0)$ with $\sigma_1 \ge \cdots \ge \sigma_r > 0$.

Theorem $4.1.3$: (a geometric interpretation)

Let $A \in \RR n m$ be nonzero of rank $r$. Then $\R^m$ has an orthonormal basis $\set{v_i}_{i=1}^m$ and $\R^n$ has orthogonal basis $\set{u_i}_{i=1}^n$, and there are $\sigma_1 \ge \cdots \ge \sigma_r > 0$ such that
$$
Av_i = \begin{cases}
\sigma_1 u_i & i = 1,\cdots,r \\
0 & \text{otherwise}
\end{cases} \qquad
A\tp u_i = \begin{cases}
\sigma_1 v_i & i = 1,\cdots,r \\
0 & \text{otherwise}
\end{cases}
$$

Theorem $4.1.10$: (the condensed SVD theorem)

Let $A \in \RR n m$ be nonzero and of rank $r$. Then $\exists U \in \RR n r, V \in \RR m r$ which are isometrics, and $\Sigma \in \RR r r$ where $\Sigma = \diag(\sigma_1, \cdots, \sigma_r)$ with $\sigma_1 \ge \cdots \ge \sigma_r > 0$, all such that
$$
A = U \Sigma V \tp
$$

Theorem $4.1.12$: (a possibly relevant alternate SVD form?)

Let $A \in \RR n m$ be nonzero of rank $r$, with $\set{\sigma_i}_{i=1}^r$ the singular values of $A$ (ordered in the usual way). Each $\sigma_i$ is associated with right singular vectors $\set{v_i}_{i=1}^r$ and left singular vectors $\set{u_i}_{i=1}^r$. Then we may write
$$
A = \sum_{\ell = 1}^r \sigma_\ell u_\ell v_\ell \tp
$$

Questions:

How in the world should the SVD of $A$ (as given at the outset) have an SVD that is obvious on inspection? (Again, keep in mind the limited tools I'm working with; essentially only the above.) How should I see it or look for it? Clearly, $\mathrm{rank}(A) = 1$, so we only have one singular value. I can set up the decomposition (though not determine the values): it takes the form
$$
A = U \Sigma V \tp = \m{u_1 \\ u_2 } \m{\sigma_1} \m{v_1 & v_2 & v_3}
$$
but what can I glean beyond this?

答:
$\newcommand{\tp}{^{\mathsf{T}}}$
Looking at Watkins' text, the question you're trying to answer is to find the condensed SVD of A.

Here we have $A = uσv\tp$ and you want to find $u$ and $v$ subject to the isometry conditions:

$u_1^2 + u_2^2 = 1$

$v_1^2 + v_2^2 + v_3^2 = 1$

The way I'd approach this is by looking at the columns of $uσv\tp$ individually
i.e. for the first column: $σu_1 v_1  = 3$ & $σu_2 v_1 = 4$.

Solving for $u_1$ & $u_2$ and using $(1)$ we get: $σ^2 v_1^2 = 25$.

Similarly, we can use the other columns to get $σ^2 v_2^2 = 100$ & $σ^2 v_3^2 = 225$.

Adding these together gives $σ^2(v_1^2 + v_2^2 + v_3^2) = 350$ and using $(2)$ we get: $σ^2 = 350$.

Finally we can solve for our $v_1,v_2,v_3$ followed by our $u_1,u_2$ and you're done - this is how you get the condensed SVD by inspection. What's left to do is to compare to Theorem $4.1.12.$


问:
Dualizing complex of a connected smooth Artin Stack of Dimension Zero
In an article I am reading I found a claim that I do not know how to prove, and reading a solution to it would probably teach me a lot. The context is the following:

$f:\mathcal{X}\to Spec(k)$ is a connected smooth $0$-dimensional Artin stack over an algebraically closed field $k$. We work on a (suitable) derived category with $\ell$-torsion coefficients $D_{et,c}=D_{et,c}(\mathcal{X},\mathbb{F}_\ell)$. And we are trying to compute the dualizing complex $K_\mathcal{X}$.

The claim is that $K_\mathcal{X}\cong \mathbb{F}_\ell[0]$.

The article uses the following definition of dualizing complex for an Artin stack:

A dualizing complex in $D_{et,c}$ is an object $K$ equipped with isomorphisms $f^!K\cong K_X$ for any locally finitely presented morphism $f:X\to \mathcal{X}$ where $X$ is itself locally of finite presentation over $Spec(k)$. The isomorphisms are required to be functorial.

The wording of the article makes it seem like this is a general fact about connected smooth 0-dimensional Artin stacks, but it might end up being only a fact of the specific stack they work with.

答: If you're working in a context that satisfies the six functor formalism (I think there's a six functor formalism in the context you're in, have a look at the papers of Liu and Zheng), then for any smooth map $f:X\to Y$ of dimension $d$ we have $f^!\simeq f^*[2d](d)$. In particular, if $X$ is smooth, i.e. $\pi_X:X\to\text{pt}$ is smooth, then the dualising sheaf $\pi_X^!k=k[2d_X](d_X)$ is  a shift of $k$, where $d_X$ is the dimension of $X$.


问:
Total numbers formation by permutation method

How many numbers greater than $3400$ can be formed using first five natural numbers?

Without and with repetition.

答:
Let's talk without repetition first.

All 5 digit numbers would give $^5P\ _5=5!$

And 3, 2, and 1 digit numbers are out of the question.

And in 4 digit numbers, keeping the first digit as 4 or 5 would straight-up give an acceptable number. Hence, we have $2\times^4P\ _3$(fixing the first digit and arranging the remaining 3 digits using 4 available digits).

If the first digit is 3 then the second digit must be 4 or 5. So there goes another $2\times^3P\ _2$.

The first digit cannot be 1 or 2.

So we have
$$ 5P\ _5 + 2\times ^4P\ _3 + 2\times^3P\ _2 $$
$$ =120 + 2\times 24 + 2\times 6 $$
$$ =120 + 48 + 12 $$
$$ =180 $$

So 180 such numbers are possible without repetition.

As for with repetition, infinite numbers are possible. To give you an idea, the following are some such numbers. 11111,111111,1111111,11111111,...


问:
Choosing a bound when it can be plus or minus? I.e. $\sqrt{4}$
My textbook glossed over how to choose integral bounds when using substitution and the value is sign-agnostic. Or I missed it!

Consider the definite integral: $$ \int_1^4\! \frac{6^{-\sqrt{x}}}{\sqrt x}  dx $$

Let $ u = -\sqrt{x} $ such that $$ du = - \frac{1}{2\sqrt{x}} dx $$

Now, if one wishes to alter the bounds of the integral so as to avoid substituting $ - \sqrt{x} $ back in for $ u $, how is the sign of the integral's bounds determined?

Because: $ u(1) = -\sqrt 1 = -(\pm 1) = \pm 1 $ and $ u(4) = -\sqrt{4} = -(\pm2) = \pm2 $

How does one determine the correct bound? My textbook selected $ -1 $ and $-2 $ without explaining the choices.

答:
When $x$ varies between $1$ and $4$ (as in this integral), $\sqrt{x}$ varies between $1$ and $2$, and $-\sqrt{x}$ varies between $-1$ and $-2$.

$\sqrt{x}$ is not a multi-valued function on the reals.  Its input is a nonnegative number, and its output is a nonnegative number.  This is different from solving $x^2=y$, which DOES typically have two solutions in the reals.


问:
If a function is holomorphic in a certain domain, does it mean it necessarily has an antiderivative
Say I have a holomrphic function $f(z)$ in domain $D$, say $|z|>4$ and my function is not holomorphic for $|z|\le4$ - it has some poles there. Does it mean there is necessarily an antiderivative for $f(z)$?

I know I can use Morera's theorem and calculate all the residues inside $|z|\le4$ and check if their sum is zero. However what seemed more intuitive to me is to prove that since $\int_{\gamma}f(z)dz$ for every $\mathbb{\gamma}$ contained in $D$, depends only on the start and end points, there exists an antiderivative for $f(z)$.

Is my prof correct? Can I conclude that in general, a holomorphic function in a domain $D$ not only has infinitely many derivatives in $D$, but it has also infinitely many antiderivatives there?

答:
No. If $f(z)=\frac1z$ (on $\{z\in\Bbb C\mid |z|>4\}$), then $f$ has no antiderivative.

But if $f$ has the additional property that $\int_\gamma f(z)\,\mathrm dz$ dpends only upon the initional and the end point of $\gamma$, then, yes, $f$ has an antiderivative. It does not follow, however, that such an antiderivative will have an antiderivative as well. Take $\frac1{z^2}$, for instance.


问:
If a random variable $Z$ is measurable w.r.t. $\sigma{Y})$ for some random vector $Y:\Omega \to \mathbb{R}$, then it is of the form $g\circ Y$
I've encountered this claim at the beginning of a course on stochastic processes but no reference or proof is given and I can't see why this is true.</p>

答:
Step 1: Suppose $Z=\sum_{i=1}^na_i\boldsymbol 1_{A_i}$ is simple. Let $B_i\in \mathcal B(\mathbb R)$ s.t. $A_i=Y^{-1}(B_i)$. Then, $$g(x):=\sum_{i=1}^na_i\boldsymbol 1_{B_i}$$ is a measurable s.t. $Z=g(Y)$.

Step 2: Suppose $Z\geq 0$ being $\sigma (Y)-$measurable. There is a sequence $(Z_n)_{n\in\mathbb N}$ of $\sigma (Y)-$measurable functions s.t. $Z_n\nearrow Z$. By the first step, there is $g_n:\mathbb R\to \mathbb R$ measurable s.t. $Z_n=g_n(Y)$. Then $$g(x):=\limsup_{n\to \infty }g_n(x)$$
is s.t. $Z=g(Y)$.

Step 3: If $Z$ is only $\sigma (Y)$ measurable, then you can write it as $Z^+-Z^-$ where $Z^+:=Z\vee 0$ and $Z^-:=-(Z\wedge 0)$ and apply the Step 2.


问:
Is every axiom in the definition of a vector space necessary?

Definition: A vector space over a field $K$ consists of a set $V$ and two binary operations $+: V \times V \to V$ and $\cdot: K \times V \to V$ satisfying the following axioms:

Commutativity of $+$.

Associativity of $+$.

Existence of an identity element $\mathbf{0}$ for $+$.

Existence of inverses for $+$.

Compatibility of $\cdot$ with multiplication in $K$.

Distributivity of $\cdot$ over $+$.

Distributivity of $\cdot$ over addition in $K$.

$1_K$ is a left identity of $\cdot$.

Question: Are all seven of the previous axioms necessary (in the sense that weakening any one of them permits a structure which is not a vector space)? If not, which can be weakened (or removed)?

EDIT: user7530 has quite cleverly shown that the commutativity of $+$ can be derived from axioms 2-8. Supposing we throw this out, can the remaining axioms all be proven necessary?

EDIT 2: It was pointed out that axiom 3 cannot simply be thrown out, as the definition of an inverse in axiom 4 depends on the existence of $\mathbf{0}$. What if we tweak the statement of axiom 4 to axiom 4': "For every $x \in V$, there exists $y \in V$ such that $(x+y)+x = x$ and $(y+x)+y = y$"? Is this weakened version equivalent to the original, and if so, does it allow the removal of axiom 3?

答:
The answer of user7530 about commutativity axiom removal may not cover all the cases yet.

For, $K$ is any field, so it may also have characteristic  $char(K)=2$. In that case we have $1+1=0$, and $(1+1)(\vec u+ \vec v)$ is simply equal to $0(\vec u+ \vec v) = \vec 0$.

In his original work [1] Bryant considered the spaces over the real field $\mathbb R$ only, so that the mentioned issue with the field characteristic did not appear.

[1] Bryant, Victor, Reducing Classical Axioms, The Mathematical Gazette, Vol. 55, No. 391, 38-40


答:
It is convention that $\sqrt{x} = + \sqrt{x}$. Thus, you set $u(1) = -\sqrt{1}=-1$ and $u(4) = -\sqrt{4}=-2$.

The only situation where you introduce the $\pm$ signs is when you are finding the root of a quadratic such as $y^2=x$ in which case both $y=+\sqrt{x}$ and $y=-\sqrt{x}$ satisfy the original equation.

Title: Choosing a bound when it can be plus or minus? I.e. $\sqrt{4}$

Body:
My textbook glossed over how to choose integral bounds when using substitution and the value is sign-agnostic. Or I missed it!

Consider the definite integral: $$ \int_1^4\! \frac{6^{-\sqrt{x}}}{\sqrt x}  dx $$

Let $ u = -\sqrt{x} $ such that $$ du = - \frac{1}{2\sqrt{x}} dx $$

Now, if one wishes to alter the bounds of the integral so as to avoid substituting $ - \sqrt{x} $ back in for $ u $, how is the sign of the integral's bounds determined?

Because: $ u(1) = -\sqrt 1 = -(\pm 1) = \pm 1 $ and $ u(4) = -\sqrt{4} = -(\pm2) = \pm2 $

How does one determine the correct bound? My textbook selected $ -1 $ and $-2 $ without explaining the choices.


问:
Prove: $T: X \rightarrow Y$ with $Tx=x$ is bounded
I want to show that $T: X \rightarrow Y$ with $Tx=x$ is bounded and surjective.

where we have $||x||_X = \sum_{n=1}^{\infty}|x_n|$ and $||x||_Y = sup_{n \in \mathbb{N}} |x_n|$

$X:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$

$Y:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$

For boundedness I know that $T \in B(X,Y) \leftrightarrow^{def} \exists M s.t. ||Tx||_Y \leq M||x||_X
  \forall x \in X$.

That is, if $T$ is bounded, the number $||T|| = sup_{x \neq 0} \frac{||Tx||_Y}{||x||_X} \leq M < \infty$ is finite.

Now I am stuck on how to proceed. Help would be grateful. Thanks in advance.

答:
Take $x \in X$ with $$\Vert x \Vert_X = \sum_{n=1}^{\infty}\vert x_n \vert$$ finite.

For all $n \in \mathbb N$ we have

$$\vert x_n \vert \le \Vert x \Vert_X = \sum_{n=1}^{\infty}\vert x_n \vert$$ hence
$$\Vert x \Vert_Y \le \Vert x \Vert_X$$ proving that $M=1$ satisfies the required inequality and that $T$ is bounded.

Now the identity is clearly surjective as $X=Y$ in your question. So for any $y \in Y$, $y$ itself is an inverse image of $y$ under $T$.


问:
What is the min number of edges in a strongly connected directed graph of N nodes?
Given a strongly connected component of a directed graph which contains N nodes:

What is the min needed number of edges as a relation of number of nodes ?

How to prove it ?

答: A directed cycle is strongly connected, so at most $n$ edges are needed – and this is sharp, since if you have fewer edges you either have a tree or can't connect all nodes.


问:
Why is 8 so special?
I have been reading about multi-dimensional numbers, and found out that it's been proven that the Octonions are the composition algebra of the largest dimension. I was wondering why, despite having infinitely many different dimensions of numbers, the only composition algebras are of 1, 2, 4, and 8 dimensions. What's so special about 8?</p>

答: Your information is wrong: tessarines of any $2^n$ dimension (for instance, 16-dimensional tessarines) is a commutative, associative, composition algebra.

Title: Why is 8 so special?


问:
Orthonormal Basis - Angle of Rotation with respect to Standard Orthonormal Basis
I have an orthonormal basis ${\bf{b}}_1$ and ${\bf{b}}_2$ in $\mathbb{R}^2$. I want to find out the angle of rotation. I added a little picture here. I essentially want to find $\theta$

I know that I can compute the angle between two vectors but then there are $4$ combinations here

$\theta_{11} = \arccos({\bf{b}}_1^\top{\bf{e}}_1)$

$\theta_{12} = \arccos({\bf{b}}_1^\top{\bf{e}}_2)$

$\theta_{21} = \arccos({\bf{b}}_2^\top{\bf{e}}_1)$

$\theta_{22} = \arccos({\bf{b}}_2^\top{\bf{e}}_2)$

How would one know which angle is correct? Importantly, here I used the labels $b_1$, $b_2$ in the same order as $e_1$ and $e_2$ but that's not necessarily the same order geometically!

答:
You may want to consider using the complex plane in this situation. Each
point in the $\,x\,y\,$ plane is associated with a complex number. Thus,
$\,\mathbf{e}_1 \equiv 1\,$ and $\,\mathbf{e}_2 \equiv i.\,$ Similarly, for any two orthogonal unit basis vectors $\,\mathbf{b}_1 \equiv z_1\,$ and
$\,\mathbf{b}_2 \equiv z_2\,$ where $\,z_2 = z_1 i.\,$ More precisely, if
$\,\mathbf{b}_1 := (x_1,y_1),\,$ then $\,\mathbf{b}_2 := (x_2,y_2) = (-y_1,x_1).$
You may need to reverse the roles of $\,\mathbf{b}_1\,$ and $\,\mathbf{b}_2\,$ to
ensure this.

Because $\,\mathbf{b}_1\,$ is a unit vector, we get $\,z_1 = e^{i\theta}\,$
for some angle $\,\theta.$ Compute this angle using
$\,\theta = \text{atan2}(y,x)\,$ where
atan2 is the "2-argument arctangent". If $\,\text{atan2}\,$ is not available, use the
identity
$$\text{atan2}(y,x) =
2\arctan\left(\frac{y}{x+\sqrt{x^2+y^2}}\right).$$


问:
Find the integral of $P(X<Y)$
Let X,Y be r.v following a normal distribution ($\mu_X,\mu_Y$ and $\sigma_X^2,\sigma_Y^2 >0$) that are independent. Let $z:=X-Y$. What integral shows $P(X<Y)$. Do not compute it

I can not find the integral but I am fairly sure on how to compute it. What is the integral? Is it just the integral of the normal distribution from $\pm \infty$? I know the linear property applies for $Z$.

答:
I think it is useful for you a numerical example:

Let $X\sim N(1;1)$ and $Y\sim N(2;1)$ and let's $X,Y$ independent.

We want to calculate $P(X<Y)=P(X-Y<0)$

To do that we first derive $f_Z(z)$.

As per what I told you in my previous post, $Z\sim N(-1;2)$

Thus

$$P(Z<0)=\Phi\left(\frac{0-(-1)}{\sqrt{2}}\right)=\Phi\left(\frac{1}{\sqrt{2}}\right)=76.025\%$$

Title: Find the integral of $P(X<Y)$


问:
How to prove that $(A \cup B) \oplus (B \cup C) = (A \oplus C) - B$?

Prove

$(A \cup B) \oplus (B \cup C) = (A \oplus C) - B$

It's a problem I met when studying discrete mathematics. I'm a starter so I'm still unfamiliar with some operations.
Here is how I tried and failed:

$$\begin{align}
&(A \cup B - B \cup C)\cup(B \cup C - A \cup B) \\
&=B\cup(A-C)\cup B \cup(C-A)\\
&=B\cup(A-C)\cup(C-A)\\
&=B\cup(A \oplus C)\\
\end{align}$$

答:
We want $(A \cup B) \oplus (B \cup C) = (A \oplus C) - B$

case 1: If $x \in B$, then $x \in (A \cup B) \oplus (B \cup C)$ cannot hold, as $x$ is in both $A \cup$ and $B \cup C$. Also $x \notin (A \oplus C) - B$ by definition.

So the only interesting case is when $x \notin B$. Then $x$ is in the left hand side iff $x \in A \oplus C$ because e.g. $x \in B \cup C$ implies $x \in C$ in this case and ditto for $A \cup B$ and $A$.
And so it's clearly in the right hand side and vice versa too. QED.


问:
About Riemann rearrangement theorem!
How to show that $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{\sqrt n}$ converges to $+\infty$, by changing the order of its terms.

I think that the idea is changing the order of terms such as: taking enough positive terms that there sum is bigger than 1/2 ,then the first negative term in the series, and so on:
$(1+1/\sqrt 3 -1/\sqrt2)+(1/\sqrt 5+1/\sqrt7+\cdots+1/\sqrt 21-1/\sqrt 4)+...$.
Now, if our series is $\sum_{k=1}^{\infty} b_k$.
Then, for every $k\in N$, $b_k \geq 1/2- 1/\sqrt(2k)$.

For $k\geq 8 , b_k \geq 1/2-1/4=1/4$.
So, $S_K=\sum_{k=1}^{K} b_k$ does not converge to 0.
Which lead to that the series converges to $\infty$.

答: Hint: already a grouping like two positive terms followed by one negative term would lead to a divergent series, in this case (for the alternating harmonic series, it wouldn't).</p>

Title: About Riemann rearrangement theorem!


问:
Does the law of large numbers hold for covering numbers?
I am self-studying empirical process theory.
I have encountered the covering number $N(\delta,\mathcal{G},P)$, as well as the empirical version $N(\delta,\mathcal{G},P_n)$.
It seems intuitive to expect some kind of convergence:
$$
N(\delta,\mathcal{G},P_n)\rightarrow N(\delta,\mathcal{G},P)
$$
Yet, I have no idea how to prove this. Can such a result be shown? Or are there counterexamples?

Definitions

Covering number:
Let $P$ be a probability measure on the Borel-$\sigma$-algebra over $\mathbb{R}$.
For $p\in[1,\infty)$ let $L^p(P)$ be the set of Borel-measurable mappings $\mathbb{R}\rightarrow\mathbb{R}$, for which $\int_\mathbb{R} |f|^p dP<\infty$.
Let $\mathcal{G}$ be a totally bounded subset of $L^p(P)$.
For some $\delta>0$, we can define the covering number of $\mathcal{G}$ as the smallest $N\in\mathbb{N}$, such that there exists a finite subset $G\subset \mathcal{G}$ with the following property:
For any $g\in\mathcal{G}$, there exists a $h\in G$, such that $||g-h||_p<\delta$.
This number is denoted by $N(\delta,\mathcal{G},P)$.

Empirical measure:
Let $P$ be as above. Let $\{X_n\}_{n\in\mathbb{N}}$ be a sequence of independent $P$-distributed random variables. If $\delta_{X_i}$ denotes the dirac-measure, the empirical measure $P_n$ is defined as:
$$
P_n:\mathcal{B}(\mathbb{R})\rightarrow[0,1],\quad E\mapsto \frac{1}{n}\sum_{i=1}^n\delta_{X_i}(E)
$$

答:
The answer is no, due to problems with equivalence classes.
We will look at two cases:
First, the elements of $L^p(P)$ are understood as equivalence classes ($f=g$ if $P(f(x)=g(x))=1$), then we take them as individual functions.

Case 1: Interpreting $L^p(P)$ as a set of equivalence classes.

In this case, calculating the empirical covering number makes no sense, unless $P$ is discrete.

To demonstrate that this is nonsense, let $\mathcal{B}$ be the set of all Borel-measurable mappings $\mathbb{R}\rightarrow\mathbb{R}$.
take $P$ as the Lebesgue measure on $[0,1]$ and $\mathcal{G}=\{[0]\}$. Here, $[0]$ is the set of all Borel-measurable mappings $\mathbb{R}\rightarrow\mathbb{R}$, which are $P$-almost surely equal to zero.
Take a finite subset $\{x_1,\dots,x_n\}\subset \mathbb{R}$.
For any $g\in\mathcal{B}$,
$$
x\mapsto \sum_{i=1}^n g(x)1\{x=x_i\}\in[0]
$$
Therefore, for any $\delta>0$ and any $n\in\mathbb{N}$, the element $[0]$ can cover all of $\mathcal{B}$ in the empirical $L^p$-norm:
$$
N(\delta,\mathcal{B},P_n)=1
$$
Trivially,
$$
N(\delta,\mathcal{B},P)=+\infty
$$
In fact $\mathcal{B}$ is so large that we cannot even define a countable covering for it.

Case 2: Interpreting $L^p(P)$ as a set of functions.

In this case, there is also a counterexample based on equivalence classes.
But now it goes the other way.

Let $\mathcal{G}$ be the set functions parametrized by $\alpha\in [0,1]$, $\beta>0$, which map:
$$
g_{\alpha,\beta}:[0,1]\rightarrow\mathbb{R},\quad x\mapsto \beta 1\{x=\alpha\}
$$
Let $P$ be the Lebesgue measure on $[0,1]$.
For any $g,h\in\mathcal{G}$, it holds that $g=h$, $P$-almost surely. So, for any $\delta>0$,
$$
N(\delta,\mathcal{G},P)=1
$$
At the same time, suppose we have $n=1$ and we observe $x_1$.
The empirical distance between elements of $\mathcal{G}$ with $\alpha=x_1$ is unbounded: $$
||g_{x_1,\beta_1}-g_{x_1,\beta_2}||_{P_1}=|\beta_11\{x=x_1\}(x_1)-\beta_21\{x=x_1\}(x_1)|=|\beta_1-\beta_2|
$$
So,
$$
N(\delta,\mathcal{G},P_1)=+\infty,\quad\text{$P$-almost surely}
$$
It is easy to see that the same holds for any $n$:
$$
N(\delta,\mathcal{G},P_n)=+\infty,\quad\text{$P$-almost surely}
$$


答:
The simplest way is to construct orthogonal matrix with column vectors $B=[b_1 \ \ b_2 \ \ b_1 \times b_2]$.

(I assume here that $b_1$ and $b_2$ are normalized to unit length)
and to use trace of such matrix $\text{tr}(B)$.

Then we have immediately  $\text{tr}(B)=1+2\cos(\theta)$.

See also https://en.wikipedia.org/wiki/Rotation_matrix#Determining_the_angle

Title: Orthonormal Basis - Angle of Rotation with respect to Standard Orthonormal Basis

Body:
I have an orthonormal basis ${\bf{b}}_1$ and ${\bf{b}}_2$ in $\mathbb{R}^2$. I want to find out the angle of rotation. I added a little picture here. I essentially want to find $\theta$

I know that I can compute the angle between two vectors but then there are $4$ combinations here

$\theta_{11} = \arccos({\bf{b}}_1^\top{\bf{e}}_1)$

$\theta_{12} = \arccos({\bf{b}}_1^\top{\bf{e}}_2)$

$\theta_{21} = \arccos({\bf{b}}_2^\top{\bf{e}}_1)$

$\theta_{22} = \arccos({\bf{b}}_2^\top{\bf{e}}_2)$

How would one know which angle is correct? Importantly, here I used the labels $b_1$, $b_2$ in the same order as $e_1$ and $e_2$ but that's not necessarily the same order geometically!


问:
Why is the maximum functional $f \mapsto \max_{t \in [0,1]} f(t)$ continuous on $C[0,1]$?
Define the maximum functional on the normed space $(C[0,1], \|\cdot\|_\infty)$ as follows:
$$
m: C[0,1] \to \mathbb R, \quad f \mapsto m(f) := \max_{t \in [0,1]} f(t).
$$
Why then is $m$ continuous?

答:
Take $f, g \in X= (\mathcal C[0,1],\Vert \cdot \Vert_\infty)$ and $t \in [0,1]$. You have

$$ f(t) - \Vert f-g \Vert_\infty \le g(t) \le f(t) + \Vert f-g \Vert_\infty.$$ Hence

$$\sup\limits_{t \in [a,b]} f(t) - \Vert f-g \Vert_\infty  \le \sup\limits_{t \in [a,b]} g(t)\le \sup\limits_{t \in [a,b]} f(t) + \Vert f-g \Vert_\infty.$$

From there the required continuity follows immediately.


答:
One can write $m$ as composition of norms and other (trivially?) continuous functions:
$$
m(f) = \Big\|f + \|f\|_\infty \Big\|_\infty - \|f\|_\infty,
$$
where I used that $f+ \|f\|_\infty\ge0$.

Title: Why is the maximum functional $f \mapsto \max_{t \in [0,1]} f(t)$ continuous on $C[0,1]$?

Body:
Define the maximum functional on the normed space $(C[0,1], \|\cdot\|_\infty)$ as follows:
$$
m: C[0,1] \to \mathbb R, \quad f \mapsto m(f) := \max_{t \in [0,1]} f(t).
$$
Why then is $m$ continuous?


问:
Double recurrence of first order
The sequences $(a_n)_{n\geq 0},(b_n)_{n\geq 0}$ are given by $a_0>0,b_0>0$ and

$$a_{n+1}=3a_nb_n(a_n+b_n),~~~b_{n+1}=a_n^3+b_n^3,~~~ \forall n\geq 0.$$
The problem is to prove that the following limits exist and to compute them:
$$\lim_{n\to\infty}\frac{a_n}{a_n+b_n};~~~\lim_{n\to\infty}\frac{a_0^3+a_1^3+\dots +a_{n-1}^3}{a_n}.$$
I have noticed that $a_{n+1}+b_{n+1}=(a_n+b_n)^3$ and thus
$$a_n+b_n=(a_0+b_0)^{3^n},~~~\forall n\geq 0.$$
I found no way to decide the monotony or boundedness of the sequences and to compute the limits.

答:
Here's how to do the first one. Note that

$$ \frac{b_{n+1}}{a_{n+1}} = \frac{a_n^2-a_nb_n +b_n^2}{3a_n b_n} = \frac{1}{3}\bigg(\frac{a_n}{b_n} - 1 + \frac{b_n}{a_n}\bigg).$$
The obvious substitution $w_n = \frac{a_n}{b_n}$ now yields
$$ \frac{1}{w_{n+1}} = \frac{1}{3}\bigg(w_n- 1 + \frac{1}{w_n}\bigg).$$
The limit $w = \lim_{n\to\infty}w_n$, if it exists, should then satisfy
$$ 2w^{-1} = w-1; \qquad w = 2. $$
Maybe you can prove by induction that $w_n$ is monotonous and bounded. This allows you to compute the first limit.

For the second one: using the above and the relation for $a_{n+1}$, you can see that $$\lim_{n\to\infty} \frac{a_{n-1}^3}{a_{n}} = 4/9. $$
But then the second-to-last term in the sum is $$\lim_{n\to\infty} \frac{a_{n-2}^3}{a_{n}} = \lim_{n\to\infty} \frac{a_{n-2}^3}{a_{n-1}^3}\frac{a_{n-1}^3}{a_{n}}$$
Using $a_n(1+\frac{b_n}{a_n}) = (a_0 + b_0)^{3^n}$ we get
$$\lim_{n\to\infty}\frac{a_{n-2}^3}{a_{n-1}^3} = \lim_{n \to \infty} (a_0 + b_0)^{(3^{n-1}-3^{n})} =0.$$
Similarly, all other terms in the sum also approach zero. So the second limit would be $4/9$.

Title: Double recurrence of first order


问:
Is this proof by mathematical induction correct?
For any natural numbers $x,y,p$

$x<y$ if and only if $x+p<y+p$. It is asked to prove by induction.

My attempt:
Base case: Let $p=1$. Then $1+x < y+1$.

Inductive step: suppose $x<y$

Then, $S(x)<S(y)$

Adding $p$ to both side of inequality,

$S(x)+p < S(y)+p$

Thanks in advance

答:
As a proof? No, it's is not correct.

The form of induction is this:

First you prove the base case. Here you have stated it but not proved it.

Then you assume that the statement is true for some $k$ and using that assumption (the "inductive hypothesis") prove that it is also true for $k+1$.You haven't done this either.

It seems likely that the intention was for you to run induction on values of $p$. It's not clear what foundational truths you are relying on here but your use of $S()$ suggests that you can rely on $S(x) = x+1$ and $S(x)>x$.

Then the base case for example might need two cases: $y=S(x)$ and $y>S(x)$.

The inductive step could use associativity, $x+(p+1) = (x+p)+1$.

Hope this is enough to get you on track.


答:
Ok: Now that I know your background it is easier for me to help you...

Observe that your exercise explicitly states

What integral shows P(X<Y)? Do not compute it

thus  what they want from you is this

$$\mathbb{P}[X<Y]=\int_{-\infty}^{0}\frac{1}{\sqrt{2\pi(\sigma_X^2+\sigma_Y^2)}}e^{-[z-(\mu_X-\mu_Y)]^2/[2(\sigma_X^2+\sigma_Y^2)]} dz$$

...and nothing else.

Title: Find the integral of $P(X<Y)$

Body:
Let X,Y be r.v following a normal distribution ($\mu_X,\mu_Y$ and $\sigma_X^2,\sigma_Y^2 >0$) that are independent. Let $z:=X-Y$. What integral shows $P(X<Y)$. Do not compute it

I can not find the integral but I am fairly sure on how to compute it. What is the integral? Is it just the integral of the normal distribution from $\pm \infty$? I know the linear property applies for $Z$.


答:
No, it is not correct, because you use the theorem inside the proof (doing $S(x)<S(y)\implies S(x)+p<S(y)+p$).

The inductive proof requires $x<y\implies S(x)<S(y)$, as well as $S(a+b)=a+S(b)$. If this is taken for granted, the inductive step is

$$x+p<y+p\implies S(x+p)<S(y+p)\implies x+S(p)<y+S(p).$$

Then as the claim is true for $p=0$, it is true for all $p$.

Title: Is this proof by mathematical induction correct?

Body:
For any natural numbers $x,y,p$

$x<y$ if and only if $x+p<y+p$. It is asked to prove by induction.

My attempt:
Base case: Let $p=1$. Then $1+x < y+1$.

Inductive step: suppose $x<y$

Then, $S(x)<S(y)$

Adding $p$ to both side of inequality,

$S(x)+p < S(y)+p$

Thanks in advance


问:
Derivation of term conversion rule
I've started to read Egbert Rijke's HoTT lecture notes which can be found here.

In the first lecture, some inference rules and structural rules are given, and in Exercise 1.1 it is asked to derive following term conversion rule using the rules given before:

If $\Gamma \vdash A\equiv A'$ type and $\Gamma \vdash a:A$, then $\Gamma \vdash a:A'$.

I'm not sure how can prove it formally. Here is my attempt:

Since $A\equiv A'$ in the context $\Gamma$, we have $\Gamma \vdash A'$. Using the variable rule, we gave $\Gamma, x:A' \vdash x:A'$ (1). Using the symmetry of judgmental equality of types, we have $\Gamma \vdash A'\equiv A$ (2). Using the conversion rule with generic judgment A/A' on (2) and (1), we obtain $\Gamma, x:A \vdash x:A'$ (3). Now, from (3) and $\Gamma \vdash a:A$, using substitution rule, we conclude $\Gamma \vdash a:A'$.

I'm not so familiar with the notation, so it can be written better. I hope, nevertheless, it makes sense to a reader. If someone gives a feedback or correction, it would be helpful.

答:
That looks right.

However, I have to say this exercise looks bizarre to me.  Note that Rijke's "variable conversion rule" can also be derived from the term conversion rule together with substitution, so we might as well have taken the term conversion rule as the primitive one instead of the variable conversion rule.  This is what's is done in nearly all presentations of dependent type theory that I've seen; I don't know why Rijke did it a different way.

Title: Derivation of term conversion rule


答:
You have drawn the angle $\theta$ as an arc from ${\bf e}_1$ to ${\bf b}_1,$
so apparently you already know intuitively that this is the angle you want,
implying that you want you use the angle
$\theta_{11} = \arccos( {\bf b}_1^\top {\bf e}_1).$
But this is only half the story.

You are rotating both basis vectors in the same direction in the same plane, so you will always find that

$$
\cos(\theta_{11}) = {\bf b}_1^\top {\bf e}_1 = {\bf b}_2^\top {\bf e}_2 = \cos(\theta_{22}).
$$

Choose either one, it's the same either way.

Another way to see this is to observe from the figure that the pair of vectors $({\bf e}_2, {\bf b}_2)$ is just the
pair of vectors $({\bf e}_1, {\bf b}_1)$
rotated by a right angle counterclockwise.
The angle between each pair of vectors is the same.

Now for the other half of the story:

When you take the arc cosine, you get an angle $\theta$ in the range
$0 \leq \theta \leq \pi,$
indicating that you can rotate ${\bf e}_1$ and ${\bf e}_2$
to ${\bf b}_1$ and ${\bf b}_2$ by rotating them through $\theta$ radians
in either the counterclockwise or clockwise direction.
But this result does not tell you whether the rotation is clockwise or counterclockwise.

By convention, we usually want $\theta$ to be positive for a counterclockwise rotation, negative for a clockwise rotation.

Assuming the vectors ${\bf e}_1$ and ${\bf e}_2$ are arranged in the usual counterclockwise orientation as shown in your figure,
a counterclockwise rotation more than $0$ but less than $\pi$ radians will put
${\bf b}_1$ on the same side of ${\bf e}_1$ as ${\bf e}_2$,
so ${\bf b}_2^\top {\bf e}_1$ will be positive.
With a clockwise rotation, ${\bf b}_2^\top {\bf e}_1$ will be negative.
So a rule to decide the direction of rotation is:

Counterclockwise if ${\bf b}_2^\top {\bf e}_1 > 0$,
so let $\theta = \arccos({\bf b}_1^\top {\bf e}_1).$

Clockwise if ${\bf b}_2^\top {\bf e}_1 < 0$,
so let $\theta = -\arccos({\bf b}_1^\top {\bf e}_1).$

If ${\bf b}_2^\top {\bf e}_1 = 0$ then either direction of rotation has the same result, so you can choose arbitrarily.

Note that in general you will find that
${\bf b}_2^\top {\bf e}_1 = - {\bf b}_1^\top {\bf e}_2.$


问:
Matrix equation involving a Pauli matrix
I should solve the following problem:
find the matrix $A$ that satisfies the following equation:
$$\sin(\pi A)+\cos(\pi A)^2= \left( \begin{array}{ccc}
0 & 1 \\
1 & 0 \end{array} \right)$$
How can I solve the problem?
Thanks in advance.</p>

答:
The properties of Pauli matrices can be handled abstractly.  You need not, at any time, write them out explicitly.

You want $A$ such that

$$\sin \pi A + \cos^2 \pi A = \sigma_1$$

Edit: I was mistaken earlier; we can't assume that $A$ is a linear combination of $\sigma_1, \sigma_2, \sigma_3$.  But the problem does hinge on the behavior of the trig functions on various kinds of inputs.  Unit "vectors"--linear combinations of $\sigma_1, \sigma_2, \sigma_3$--will pull out of the trig function, for instance.  Bivectors--linear combinations of $\sigma_1 \sigma_2, \sigma_2 \sigma_3, \sigma_3 \sigma_1$.  The logic extends to products of three Pauli matrices.

But that's not all.  There are idempotents of the form $I_\pm = (1 \pm \sigma_1)/2$ (you can understand $1$ as the identity).  These also pull out of trig functions.  There may be other such objects that pull out that I'm not aware of, but I think this will get us in the right direction.

For various reasons, I will take as an ansatz that $B = \pi A = c I_+ + d I_-$.  If $\pi A$ is a "vector" in this terminology, this can be accomplished with $c = -d$; if $\pi A$ is a "scalar" (that is a multiple of the identity), then $c = d$.  I ignore the possibility that $\pi A$ has any other terms, as these are unlikely to cancel out through the power series.

The reason for this decomposition is as follows.  $B^n = c^n I_+ + d^n I_-$.  Thus, the trig functions become

$$\sin B = I_+ \sin c + I_- \sin d, \quad \cos B = I_+ \cos c + I_- \cos d - 1$$

We plug this into the original equation to get the following:

$$I_+ \sin c + I_- \sin d + I_+ \cos^2 c + I_- \cos^2 d - 2(I_+ \cos c + I_- \cos d) + 1 = \sigma_1$$

Breaking down in terms of $1, \sigma_1$, we get

$$\frac{1}{2} (\sin c + \sin d + \cos^2 c + \cos^2 d - 2 \cos c - 2 \cos d) + 1 = 0$$

and

$$\frac{1}{2}(\sin c - \sin d + \cos^2 c - \cos^2 d - 2 \cos c + 2 \cos d) \sigma_1 = \sigma_1$$

Cancel the $\sigma_1$ and add/subtract the two equations to cancel out everything involving $c$ or $d$ to get

$$\sin c + \cos^2 c - 2 \cos c = 0, \quad \sin d + \cos^2 d - 2 \cos d = -2$$

Well, I must concede that this problem was not as simple as I anticipated, so I think julien's approach, which makes fewer assumptions, is probably the best course.


问:
Poisson process. Compute $E[N_z-N_t | N_s=3]$
Let $(N_t)_{t  \in[0,\infty)}$ be a Poisson process. Let $t<s<z$. Compute

$E[N_z-N_t | N_s=3]$

My idea:
Using the fact that Poisson process has independent increments:

$E[N_z-N_t| N_s=3]=E[N_z-N_s+N_s-N_t | N_s=3]= E[N_z-N_s]+ E[N_s-N_t | N_s=3]$

答: As Henry pointed out $E[N_z-N_s] = E[N_{z-s}] = \lambda (z-s)$ (assuming $\lambda$ is the intensity). Also $E[N_s-N_t | N_s=3] = \frac{3(s-t)}{s}$ (since given $N_s=3$ the arrival times are uniformly distributed in $[0,s]$).


问:
Is $D = \{(x,y) \in\Bbb R^2\mid x^2y=1\}$ a closed set or open set?

Is $D = \{(x,y) \in\Bbb R^2\mid x^2y=1\}$ an open or closed set?

Correct me if I'm wrong, an open set is a set that doesn't contain all its limits. From graphing $y = 1/x^2$ I can see that the curves converge towards the $x$- and $y$-axis but never touch the axes.

Therefore, the set $D$ is an open set.

Thats my intuition but I don't know how to formally prove it.

答:
In the first place, it's a closed set. If $((x_n,y_n))_{n\in\Bbb N}$ is a sequence of points of $D$ which converges to some $(x,y)\in\Bbb R^2$, then\begin{align}x^2y&=\lim_{n\to\infty}x_n^{\,2}y_n\\&=\lim_{n\to\infty}1\\&=1,\end{align}and therefore $(x,y)\in D$.

In the second place, even if it was not closed, you would not be able to deduce from that that $D$ is open. Sets are not doors: they can be both and open and they can also be not closed nor open.


答:
Yes that is the idea.

$\begin{align}\mathsf E[N_z-N_t\mid N_s=3]&=\mathsf E[N_z-N_s+N_s-N_t \mid N_s=3]\\[1ex]&=\mathsf  E[N_z-N_s\mid N_s=3]+\mathsf  E[N_s-N_t \mid N_s=3]\\[1ex]&=\mathsf  E[N_z-N_s]+\mathsf  E[N_s-N_t \mid N_s=3]\\[1ex]&=\mathsf  E[N_{z-s}]+\mathsf  E[N_s-N_t \mid N_s=3]\end{align}$

Now you need to determine what is the conditional distributions of $N_s-N_t$ given $N_s=3$.

 Binomial(3,(s-t)/s)

Title: Poisson process. Compute $E[N_z-N_t | N_s=3]$

Body:
Let $(N_t)_{t  \in[0,\infty)}$ be a Poisson process. Let $t<s<z$. Compute

$E[N_z-N_t | N_s=3]$

My idea:
Using the fact that Poisson process has independent increments:

$E[N_z-N_t| N_s=3]=E[N_z-N_s+N_s-N_t | N_s=3]= E[N_z-N_s]+ E[N_s-N_t | N_s=3]$


问:
Definition of associated filtered algebra
Consider the following fragment from wikipedia:

Can someone justify the choice of multiplication on this algebra? For me, it seems more natural define
$$(x+ F_{n-1)})(y+ F_{m-1}):= xy + F_{n+m-2}$$

Why do they use $F_{n+m-1}$ instead of $F_{n+m+2}$?

答:
Because the product of an element of $G_n$ and an element of $G_m$ must be in $G_{n+m}$, that is something of the form $z+F_{n+m-1}$ for $z\in F_{n+m}$.

Moreover, taking $xy+F_{n+m-2}$ would not even be well-defined: adding an element $u\in F_{n-1}$ to $x$ adds $uy\in F_{n+m-1}$ to the product, so $xy$ is really well defined up to something in $F_{n+m-1}$, not $F_{n+m-2}$.


问:
Why when we approximate an analytic function has poles outside the interval of the integration give poor accuracy?
Why when an analytic function has poles which are outside the domain of integration but lie very close to it, its numerical integration by means of a quadrature rule can give very poor? Why the function should be also defined outside the interval of the integration?</p>

答:
The question asked about poor performance of numerical integration of
analytic functions with poles near the interval of integration.
The Wikipedia article Runge's phenomenon gives a good example of this where a
rational function with poles is evaluated at equidistant points and
approximated with a polynomial. Integrating the polynomial is easy
but the polynomial to be integrated is a poor fit for the function
and hence yields a poor integration result. The article suggest some
alternative ways to approximate the function to give better fit.

The question also asked

Why the function should be also defined outside the interval of the integration?

is not clear to me. In general we are dealing with
Meromorphic functions. The Wikipedia article states

In the mathematical field of complex analysis, a meromorphic function on an open subset $D$ of the complex plane is a function that is holomorphic on all of $D$ except for a set of isolated points, which are poles of the function.

Notice the use of "open subset $D$ of the complex plane".
The behavior of the function is very closely related to
the topology of the complex plane. Its natural domain is
open subsets of the complex plane. Thus, while the interval
of integration is of primary importance, a real interval is
not an open domain in the complex plane. The behavior
of the function on open domains containing the interval has
an impact on the behavior in the interval itself and vice
versa. That is why poles outside the real interval can have
a substantial effect on fitting by polynomials. A much better
fit could be found if rational functions are allowed to better
account for poles.


答: $D$ is the inverse image under the continuous map $f : (x,y) \mapsto x^2y$ of the closed singleton $\{1\}$. Hence $D$ is closed.

Title: Is $D = \{(x,y) \in\Bbb R^2\mid x^2y=1\}$ a closed set or open set?

Body:
Is $D = \{(x,y) \in\Bbb R^2\mid x^2y=1\}$ an open or closed set?

Correct me if I'm wrong, an open set is a set that doesn't contain all its limits. From graphing $y = 1/x^2$ I can see that the curves converge towards the $x$- and $y$-axis but never touch the axes.

Therefore, the set $D$ is an open set.

Thats my intuition but I don't know how to formally prove it.


问:
The Cantor ternary set is totally disconnected
A set $S$ in a metric space $X$ is called totally disconnected if for any distinct $x,y\in S$, there exists separated sets $A$ and $B$ with $x\in A$, $y\in B$ and $S=A \cup B$.

Let $C=\bigcap_{n=1}^\infty C_n$ be the Cantor ternary set.

Given $x,y \in C$ with $x\lt y$, set $\epsilon=y-x$. For each $n\in$ N, $C_n$ consists of a finite union of closed intervals. Explain why there must exist an N large enough so that it is impossible for $x$ and $y$ both to belong to the same closed interval in $C_N$.

I know that the Cantor set is constructed by removing the middle open thirds for each n. And each $C_n$ has $2^n$ closed intervals. As you go on, the closed sets get significantly small, so it's safe to assume that for some N, $x$ and $y$ will be "separated" into two different closed intervals. I'm not sure how to show this formally though.

答:
If $\underline D$ is Cantor Discontinuum, then we have that $D = \bigcap D_m, m \in N$, where: $$D_m=\{ \sum_{n=1}^{\infty} \frac{x_n}{3^n} \ | \ (x_n) -\text{ sequence in } \{0,1,2\} \text{ with } \{x_1,x_2,...,x_m\} \subset \{0,2\}   \}$$

Since $\underline D$ is equipped with a subspace topology of $\underline ℝ$, then any connected set $A$ in $\underline D$ must be an interval.

If $card(A) \gt 1 \Rightarrow \exists a, b \in A, \text{(wlog) } a < b$. So, it must then hold that $ [a,b] \subset A \subset D$.

Since $[a,b] \subset D = \bigcap D_m \Rightarrow$ it must hold that $\forall m \in N: [a,b] \subset D_m$.

However, one can always find such $k \in N : |b-a| \gt \frac{1}{3^k}$ meaning  $[a,b] \not \subset D_k$.

Latter implying a contradiction: $A \not \subset D$. Therefore, $card(A) \le 1$. $\blacksquare$


问:
What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$?
What is the number of solution of the equation $\ x^3 - \lfloor x\rfloor = 3$ ? (where $\lfloor x \rfloor\ $ is the greatest integer $\le x$)

I tried plotting the graphs of these equations on Desmos graph calculator and that they intersect each other in between x = 1 and 2 but I couldn't figure out a way to get to this conclusion on my own.

Is there any way by which I can determine where these functions intersect?

答:
Here is another answer I came up with:

Notice that $\lfloor x\rfloor$ and $3$ are integers, so $x^3$ is an integer, say $x=\sqrt[3]{y^3+z}$, where $0\leq z<(y+1)^3-y^3$, and y and z are integers. Now we have $y^3+z-y=3$, or $(y-1)y(y+1)=3-z.$ Since $z\geq0$, $(y-1)y(y+1)\leq 3$ and $y\leq1$. However, notice that if $y<-1$, the inequality $z<(y+1)^3-y^3$ fails, so $y\geq-1$, $(y-1)y(y+1)=0$, $z=3$, and $y=1$ to make the inequality true, so  $x=\sqrt[3]4$.


问:
Derivative of function containing parametric integral
I have the following function:

$$p_{D}(t_{D}) = \frac{4}{\pi^2}\int\limits_0^\infty\frac{1-\exp\bigl(-u^2t_{D}\bigr)}{u^3\bigl(J_{1}^2(u) + Y_{1}^2(u)\bigr)}\,du $$

for which I need the derivative $p_{D}'(t_{D})$. I am stumped so would appreciate any help anyone could give.  Thanks!

答:
After some internet searching I came across this link

It seems that Liebnitz got here some time ago, so following the info in the link, the analytical derivative of my function is

.

I checked the resulting derivative by comparing it to the derivative of the spline of the pD function and the two match pretty closely:


问:
Questions on the dual of the composition $R;S$ of binary relations.
I recently started reading about relation algebras. This is a very new and unfamiliar, but interesting field to me. I'm not taking a course in it, it's self learning on my spare time, so please bear with me. My level is undergraduate-master with no real knowledge of relation algebras or formal order theory, but with some minor knowledge of category theory.

Let $R, S$ be binary relations (subsets of $X\times X$) on a finite set $X=\{x_1,\dots,x_n\}$. We can then define the composition of relations as

$\begin{equation*}
R;S:= \{\,(x,y)\in X\times X\mid (x,z)\in R \operatorname{and}\, (z,y)\in S \text{ for some } z\in X\}.
\end{equation*}$

This corresponds nicely to multiplication of logical matrices with elements in the two element boolean algebra.

Example: The following directed graph represents a relation $P$ on $X=\{A,B,C,D\}$.

This can be represented by the logical matrix

$P=\begin{pmatrix} 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 &
 0 & 0 & 0 \end{pmatrix}$.

Say that $X$ represents a family. Then, we could possibly express the relation of being siblings as
$P^\top;P$ where $P^\top$ means the converse of the relation $P$, and
corresponds naturally to a transpose. However, we don't say that you
are your own sibling, so we would set the diagonal to 0, perhaps like
\begin{equation} (P^\top;P)\wedge \operatorname{di}_X \end{equation}
where $\operatorname{di}_X=\{(x,y)\mid x\not=y\}$, here represented by
$\operatorname{di}_X=\begin{pmatrix} 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1\\ 1
 & 1 & 0 & 1\\ 1 & 1 & 1 & 0 \end{pmatrix}$, and $\wedge$ denotes
elementwise 'and'.

Then being someones aunt/uncle could possibly be represented by
\begin{equation} ((P^\top;P)\wedge\operatorname{di}_X);P=\\
 \Big(\Big(\begin{pmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 1 & 0 & 0 &
 0\\ 0 & 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 1 & 0\\ 0 & 0
 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{pmatrix}\Big)\wedge
 \begin{pmatrix} 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1\\ 1 & 1 & 0 & 1\\ 1 & 1
 & 1 & 0 \end{pmatrix}\Big) \begin{pmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 &
 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}=\\ \begin{pmatrix} 0 &
 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0
 \end{pmatrix} \begin{pmatrix} 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 &
 0 & 1\\ 0 & 0 & 0 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 & 0 & 0\\ 0
 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix}
 \end{equation} which does make sense, since $B$ is the aunt/uncle of
$D$ and it's the only such relation.

One can however define a dual notion of composition by exchanging and/or and exist/for all, let's call it $R\oplus S$ for now, as

$\begin{equation*}
R\oplus S:= \{\,(x,y)\in X\times X\mid (x,z)\in R \operatorname{or}\, (z,y)\in S \text{ for all } z\in X\}.
\end{equation*}$

As I understand it, this is the dual of $R;S$ in the same sense as $\vee$ and $\cup$ are duals of $\wedge$ and $\cap$ in order theory/logic and set theory, since we have that

\begin{equation}
R\oplus S = \neg(\neg R;\neg S).
\end{equation}

Just as $R;S$ can be represented by "ordinary" matrix multiplication, this dual can be represented by defining composition of matrices dually by exchanging operations in the usual definition, i.e.
\begin{equation}
(M\oplus N)(i,j):=\prod_{k=1}^{n}(M(i,k)+N(k,j))
\end{equation}
where multiplication of elements means 'and' and addition 'or'. It then turns out that

\begin{equation}
M\oplus N=\neg(\neg M\cdot \neg N)
\end{equation}

if we let $\cdot$ denote the usual multiplication of logical matrices.

Questions:

What is $R\oplus S$ as defined above called in general (cocomposition)? I've found very few references to it, for example in Relation Algebras by Steven Givant, where it's called relational sum. However, I cannot even seem to find the definition anywhere else, yet it seems like a very general thing. I looked at different types of joins of relations but I can't seem to find this one. Perhaps it's not so widely used, or I am lousy at looking. I've also looked a bit in Relational Topology by Winter and Schmidt, but that book was way above my level at the moment. Perhaps 'sum' is fine, but it sounds more like ordinary union to me.

Is this dual multiplication of matrices used in other areas, i.e. over other structures, especially with ordinary real matrices, and if so, what is it called?

Would anyone try and give (if possible) some intuition for the relational sum in a similar sense as one could for the composition, e.g. in the example of uncle above?

Less important: Is my way of "setting the diagonal to zero", i.e. taking away the reflexive property of a relation $P$ by doing $P\wedge \operatorname{di}_X$ correct and standard, or would one express it differently?

答: The "relational sum" is often called "relative sum" and denoted $R \dagger S$ in (for example) Maddux Relation Algebras (2006). Maddux explains a lot about the history of the operation, which is also called "relative addition", and quotes Pierce in a useful extract. As Maddux says (p8) "It is fairly easy to express relative products in everyday language ... but relative sums present a greater challenge" and the examples provided do not make the operation sound conceptually obvious when modelling examples like uncles and aunts etc.


问:
Value of $\sum_{k=1}^\infty \frac{1}{2^k k (k+1)}$.
I need to compute this value, but I have no idea how to do it.</p>

答: Hint: $\int_{0}^{0.5}\int_{0}^{x}y^{k-1}dy =\frac{{0.5}^{k+1}}{k(k+1)}= $. Also, you can interchange the integration and summation.


问:
Notation to say "this set is closed for all defined operations"
Is there a notation to express "this set is closed for all it's operations", in the sense of, given a set with it's defined operations, after a change in the set, the set retains the "closedness" of the operations it previously had, either by just using the same definitions on the new set, or by creating new definitions for the operations, yet to be said.

For use in the examples, let's say we have the operator $\overset*\setminus$, similar to \setminus, but capable of removing all the instances of something. For example:

$\mathbb C\overset*\setminus(\mathbb Ri) =: \mathbb R$, $\mathbb C$ without all the imaginary numbers becomes $\mathbb R$ (but $\mathbb R$ is not closed under the exponentiation, diferenttly than $\mathbb C$).

$\mathbb C\overset*\setminus\mathbb R$ becomes something where the square of the unity (the imaginary unity) is not present on the set, so it's not closed.

Supposing if $\left[\mathbb E\right]_*^*$ is the standard notation I'm looking for, we would have:

$\left[\mathbb C\overset*\setminus\mathbb R\right]_*^*$ is closed, with something like $i^2 = 0$ (with the $0$ from $\mathbb Ri$).

$\left[\mathbb H\overset*\setminus\mathbb R\right]_*^*$ mighty be defined as something like $\mathbb V^3$, the set of the 3d vectors, with $i^2 = j^2 = k^2 = ijk = 0$ (instead of $- 1$).

Is there a standard notation for this managing of sets?

答:
Minus operator

For your examples of $\overset{*}{\setminus}$, it seems like you are positing the existence of an operation that is not well-defined. Treating $\mathbb{R}$ and $\mathbb{C}$ as real vector spaces, it is essentially true that $\mathbb{C}=\mathbb{R}\oplus(\mathbb{R}i)$, where $\oplus$ is the direct sum. This is how I understand your intent for $\mathbb{C}\overset{*}{\setminus}\left(\mathbb{R}i\right)$ to be $\mathbb{R}$ and $\mathbb{C}\overset{*}{\setminus}\mathbb{R}$ to be $\mathbb{R}i$.

However, it is also true that, say, $\mathbb{C}=\left(\mathbb{R}(3+4i)\right)\oplus(\mathbb{R}i)$. Given $a+bi\in\mathbb{C}$, we have $a+bi=\dfrac{a}{3}(3+4i)+\dfrac{3b-4a}{3}(i)$. The notation $\mathbb{C}\overset{*}{\setminus}\left(\mathbb{R}i\right)$ doesn't seem to have any way to know that the answer you wanted is $\mathbb{R}$ instead of $\mathbb{R}(3+4i)$.

There may be a careful way to define this sort of inverse of direct sum (in some context) so that it has a unique result, but it's not obvious to me what that would be, and I'm skeptical that there would be such a definition at all.

Brackets notation

You essentially wrote "$\left[\mathbb{R}i\right]_{*}^{*}$ is closed, with something like $i^{2}=0$". If you redefine multiplication so that $i^{2}=0$, then you have the $\varepsilon$-axis of the "dual numbers" , which is indeed closed under addition and multiplication.

But if you redefine $i^{2}=0$, then the original meaning of $i$ is lost completely. So it seems at first glance that you want $\left[X\right]_{*}^{*}$ to mean something like "make whatever algebraic redefinitions of the operations are necessary to turn $X$ into something closed under addition and multiplication". There may be a careful way to define that (in some context) so that it always exists and has a unique result, but it's not obvious to me what that would be either.

Conclusion

In any case, since even the existence of these ideas as well-defined concepts is not obvious to me, I'm pessimistic that there's a common notation for it that I'm unaware of, but I'd be happy to be proven wrong by another answer.


问:
Create Gaussian distribution from random number generator output
I have a random number generator with output between -1 and +1 having a uniform distribution.

Is there a simple mathematical formula that would convert this into an estimated Gaussian distribution (around 0)?

答:
Usually a random sample from $U(0;1)$ is used. If you have a random sample from $U(-1;1)$ you can consider the absolute values of your output as from a $U(0;1)$

Then use the integral transform theorem


问:
Combinatorics problem - please check my answer and reasoning
Just want to check if my answer and reasoning is correct for the following problem (Not a homework problem - it is a sample question for a test I'm preparing for)

In a survey, viewers were given a list of 20 TV Shows and are asked to label 3 favourites not in any order. Then they must tick the ones that they have heard of before, if any. How many ways can the form be filled, assuming everyone has 3 favourites?

My reasoning:

1) Choose 3 shows out of 20: $c(20,3)$

2) Choosing 0-17 shows from 17 choices:
$c(17,0) + c(17,1) + c(17,2) + ... + c(17,16) + c(17,17)$

and add 1) and 2) together for the final answer.

Would this be correct? Is there a better way of doing the second part that doesn't involve so many calculations?

答:
Yes, your reasoning is correct up to the last part (implicit in your second part is that they must have heard of their favorites, so that they only need to check or not check the remaining 17). However, the final part is incorrect: you don't add (1) and (2), you multiply them: each person must select both their three favorites and those other programs that they have heard of. Multiplication, not addition, is appropriate here.

As for fewer calculations: what you need to do is pick a subset of the remaining 17 shows to represent the shows you have heard of. There are $2^{17}$ possible subsets, so that's what you want.  Alternatively, for each of the remaining 17 programs, you can either have heard of it before or not; so you have one of two choices for each of the remaining 17 programs. That means making a choice from 2 possibilities, 17 times, or $2^{17}$ possibilities.

And alternatively,
$$C(17,0) + C(17,1) + \cdots + C(17,17) = (1+1)^{17} = 2^{17}$$
by the binomial theorem, so that's another way to see that the big sum you have is simply $2^{17}$.

So the correct answer is $2^{17}\times\binom{20}{3}$.


答:
Yes, you are right. Although we tend to avoid using decimals and fractions with modular arithmetic, and thus you just have to keep your integers as integers. This would've been a cleaner way of writing what you just did (the math is basically the same):
$$
77 = 64 + 13 = 4\cdot 16 + 13 \equiv 13 \not\equiv24 \mod{16}
$$</p>

Title: Modulo question about equality

Body:
True or false?

$$24 \equiv 77 \mod 16 $$

$1.$ $77/16 = 4.8125 $

$2.$ $4.8125 - 4 = .8125$

$3.$ $0.8125 \times 16 = 13$

$4.$ $24 != 13$

So the answer is false? Am I right?


问:
Open sets separating points in a metric space?
I was solving my Real Analysis assignment on Basic Topology, and I had the following statement to prove

Let (X, d) be metric space. Show that any two distinct points in X are separated by disjoint open sets.

At first, naively, I thought that we'd obviously have some positive distance (by definition of metric), say $d$ between the points, and then we can construct open balls around each of the points of radius, say $d/4$. And then we are essentially done if we can show that the balls are disjoint using the triangle inequality, which all metrics satisfy.

But, I realized that this argument might not work if we have X consisting of discrete isolated points. Then we don't have any open ball surrounding a point, and the above proof will fail. I am afraid that this will serve as a counterexample to the theorem I am required to prove, and I am skeptical that I am missing something obvious.

So, could someone help me out with this?

Thank you.

答: Your original argument is completely correct. If $x \neq y$ then $d=d(x,y)>0$ and the triangle inequality shows that $B(x, \frac{d}{2})$ and $B(y, \frac{d}{2})$ are disjoint and open balls are by definition open in the metric topology, so we're done. We always have a ball if the radius is $>0$, even if it reduces to $\{x\}$ , as it can in the discrete metric. It doesn't change the fact that that ball is then open. In the discrete metric all subsets are open and point separation becomes trivial. But the proof works in any metric space.


问:
Similarity Transformation when the Linear Operator is not written in the standard basis doesn’t lead to a Diagonal Matrix. Why?
Consider a diagonalizable linear operator $T$, and let $\{\alpha\}$ be the basis of eigenvectors of $T$. Let $\{\beta\}$ be the standard basis for $\mathbb R^n$ and let $\{\gamma\}$ be another arbitrary basis. Now the similarity transformation when the $T$ is written in standard orders basis  proceeds as follows -

\begin{equation}
S^{-1}[T]_\beta S =  [S]_\beta^\alpha [T]_\beta^\beta [S]_\alpha^\beta = [STS]_\alpha^\alpha........................(1)
\end{equation}

Equation $(1)$ leads to a diagonal representation of $T$.

Now consider the similarity transformation when $T$ is written in the basis $\{\gamma\}$. This proceeds as follows :

\begin{equation}
S’^{-1}[T]_\gamma S’=  [S’]_\gamma^\alpha [T]_\gamma^\gamma [S’]_\alpha^\gamma = [S’TS’]_\alpha^\alpha........................(2)
\end{equation}

Equation $(2)$ doesn’t lead to a diagonal matrix.

The question is simple - why ?

For e.g.:

Let $T(x,y) = (2x+y,11x+12y)$. In the standard basis this is $T = \begin{pmatrix}2 &1 \\ 11 & 12\end{pmatrix}$. The basis of eigenvectors is $[1,11],[-1,1]$ in which the matrix is $\begin{pmatrix} 13 &0 \\ 0 & 1 \end{pmatrix}$. Let $B$ be a new basis, $\{[1,1],[1,0]\}$. Then in this basis one sees the matrix for $T$ is $M=\begin{pmatrix} 23 & 11 \\ -20 & -9 \end{pmatrix}$. Now, the change of basis from $B$ to the eigenbasis is the matrix $S = \begin{pmatrix} 1 & -1 \\ 11 & 1\end{pmatrix}^{(-1)} \begin{pmatrix}1 & 1\\ 1 & 0\end{pmatrix} = \frac 1{12}  \begin{pmatrix} 2 & 1\\ -10 &-11\end{pmatrix}$. Now, $S^{-1} MS$ should be diagonal but is not : both the off-diagonal entries is non-zero.

Why does the Similarity Transformation not lead to a diagonal matrix when the the linear operator is written in a basis other than the standard ordered basis?

Why is the standard ordered basis special in getting a diagonal matrix using a similarity transformation?

答:
The symbols you use don't make sense and they are very confusing.

When $\alpha$ and $\beta$ are bases of two finite-dimensional vector spaces $V$ and $W$ respectively and $T:V\to W$ is a linear map, the matrix representation of $T$ with respect to the bases $\alpha$ and $\beta$ is denoted by $[T]_\alpha^\beta$. Note that $T$ is a linear map, not a matrix, and $[T]_\alpha^\beta$ is not a linear map, but a matrix representing a linear map.

When one writes expressions like $[S]_\alpha^\beta,[STS]_\alpha^\alpha$ or $S^{-1}[T]_\beta S$, unless one overloads the same symbol to mean different things, $S$ should mean either a linear map or a matrix. However, if $S$ is a matrix, it doesn't make any sense to write $[S]_\alpha^\beta$ or $[STS]_\alpha^\alpha$. Similarly, if $S$ is a linear map, it doesn't make sense to write $S^{-1}[T]_\beta S$. The symbol $S$ in your equality $(1)$ thus have contradictory meanings.

A correct and clearer way to deal with matrix representations under changes of bases is to stick with the convention that the $T$ in $[T]_\alpha^\beta$ is a linear map. When $\alpha$ and $\beta$ are bases of the same vector space $V$, the change of basis matrix from $\alpha$ to $\beta$ is $[\operatorname{id}]_\alpha^\beta$ --- since we only change the basis of $V$ without mapping any vector in $V$ to a different vector, the linear map inside the pair of square brackets is the identity map $\operatorname{id}$, not some mysterious $S$. Given the matrix representation $[T]_\alpha^\alpha$ of a linear operator $T$ with respect to the basis $\alpha$, the matrix of $T$ with respect to another basis $\beta$ is obtained by
$$
[T]_\beta^\beta
=[\operatorname{id}\circ T\circ \operatorname{id}]_\beta^\beta
=[\operatorname{id}]_\alpha^\beta [T]_\alpha^\alpha [\operatorname{id}]_\beta^\alpha
=\left([\operatorname{id}]_\beta^\alpha\right)^{-1} [T]_\alpha^\alpha [\operatorname{id}]_\beta^\alpha.
$$

When $\beta$ is an eigenbasis for $T$, the matrix $[T]_\beta^\beta$ is diagonal. Hence $[T]_\alpha^\alpha=[\operatorname{id}]_\beta^\alpha[T]_\beta^\beta \left([\operatorname{id}]_\beta^\alpha\right)^{-1}$ is a diagonalisation of $[T]_\alpha^\alpha$. This is true regardless of whether $\alpha$ is the standard basis or not.

In your example, let $\alpha$ be the standard basis, $\beta$ be the eigenbasis from $T$ and $\gamma$ be the new basis. Then
$$
[\operatorname{id}]_\beta^\alpha=\pmatrix{1&-1\\ 11&1},
[\operatorname{id}]_\gamma^\alpha=\pmatrix{1&1\\ 1&0}.
$$
Hence the matrix representation of $T$ with respect to $\gamma$ is
$$
[T]_\gamma^\gamma
=\left([\operatorname{id}]_\gamma^\alpha\right)^{-1} [T]_\alpha^\alpha [\operatorname{id}]_\gamma^\alpha
=\pmatrix{1&1\\ 1&0}^{-1}\pmatrix{2&1\\ 11&12}\pmatrix{1&1\\ 1&0}
=\pmatrix{23&11\\ -20&-9}.
$$
The change-of-basis matrices from $\gamma$ to $\beta$ and from $\beta$ to $\gamma$ are given by
\begin{align}
[\operatorname{id}]_\gamma^\beta
&=[\operatorname{id}]_\alpha^\beta[\operatorname{id}]_\gamma^\alpha
=\left([\operatorname{id}]_\beta^\alpha\right)^{-1}[\operatorname{id}]_\gamma^\alpha
=\pmatrix{1&-1\\ 11&1}^{-1}\pmatrix{1&1\\ 1&0}
=\frac{1}{12}\pmatrix{2&1\\ -10&-11},\\
[\operatorname{id}]_\beta^\gamma
&=\left([\operatorname{id}]_\gamma^\beta\right)^{-1}
=\pmatrix{11&1\\ -10&-2}.
\end{align}
One can verify that the similarity transform
$$
[\operatorname{id}]_\gamma^\beta
[T]_\gamma^\gamma
[\operatorname{id}]_\beta^\gamma
=\frac{1}{12}\pmatrix{2&1\\ -10&-11}\pmatrix{23&11\\ -10&-9}\pmatrix{11&1\\ -10&-2}=\pmatrix{13&0\\ 0&1}
$$
indeed a diagonalises $[T]_\gamma^\gamma$.


问:
Q: Find the relation betwen $P(X=1)P(Y=1)$ and $P(X=1,Y=1)$
Q: An experiment has two optional result, $1$ and $0$. The result is $1$ with the probabily of $Z$, where $Z$ is a continuous r.v. with the following density:

$f_Z(z) =
\begin{cases}
  cz^a(1-z^b)& ,0 \le \ z \le 1 \\
  1 & ,o.w.
\end{cases}$
, when $a,b,c \ge0$

The experiment is done twice, let $X,Y$ be the the results of the first and the second time respectively.

Find out for which values of $a,b,c$ the relations $\ge,=,<$ are holding between $P(X=1)P(Y=1)$  and $P(x=1,y=1)$.

My attempts:

$P(X=1)=\int\limits_{0}^{1}P(X=1\mid Z=z)f_Z(z)dz=\int\limits_{0}^{1}cz^{a+1}(1-z^b)=\frac{bc}{(a+2)(a+b+2)}$ .

For symmetry reasons $P(X=1)=P(Y=1)$, so $P(X=1,Y=1)=\Bigl(\frac{bc}{(a+2)(a+b+2)}\Bigr)^2$.

What I don't understand is how to find $P(X=1,Y=1)$, since $f_{X,Y}$ is not given.

Please try to keep your answers simple enough for an undergraduate taking his first course in probability.

Thanks

答:
As you correctly reduced (in spite of the repaired typo):

$P(Y=1)=P(X=1)=\int P(X=1\mid Z=z)f_Z(z)dz=\int zf_Z(z)dz=\mathbb EZ$.

Similarly we find:

$P(X=1,Y=1)=\int P(X=1,Y=1\mid Z=z)f_Z(z)dz=\int z^2f_Z(z)dz=\mathbb EZ^2$.

Then:$$P(X=1,Y=1)-P(X=1)P(Y=1)=\mathbb EZ^2-(\mathbb EZ)^2=\mathsf{Var}Z>0$$


问:
Braided Monoidal Category Axioms
I'm having trouble understanding the following proof from Joyal and Street's paper Braided Monoidal Categories.

B1 and B2 are the hexagon axioms:

To get a triangular diagram starting from a hexagon diagram, it seems like the associators are getting removed somehow (by using the identity triangle axioms of monoidal categories? not sure). I'm new to this and I'm having trouble figuring out exactly how the proof works. I'd appreciate it if someone could provide the actual details.

答: To solve it, you need to show that given the Hexagonal equation $B1$, you can somehow build $B3$ and $B4$ insides $B1$. And since the outer diagram $B1$ commutes and some of the inside diagrams because of the monoidal structure of category commutes, then $B3$ and $B4$ commute. Please find below a proof sketch.

Title: Braided Monoidal Category Axioms


答:
A simpler approach: $a\equiv b\pmod{n}$ if and only if $n$ divides $(b-a)$.

In this problem, $77-24=53$, which is not a multiple of $16$ ($53/16=3.3125$), so $77\not \equiv 24\pmod{16}$.

Title: Modulo question about equality


问:
Is the phrase "Weakly/strongly Non-linear" qualitative or quantitative?
When it comes to differential equations, I notice that "Weakly Nonlinear" seems to be code for 'Relatively well behaved, easy to qualitatively predict, even if the solution requires an obscure special function or doesn't exist in closed form at all', whereas "Strongly Nonlinear" seems to mean 'Complicated, prone to chaos, don't even try finding a closed-form expression'.

Approaching this naively, if a differential equation has a nonlinear term in it, it is nonlinear. This is all I know from my limited maths education.

Do the terms 'Strong' and 'Weak' have a quantitative basis? Is it possible to inspect the differential equation and evaluate how weakly or strongly nonlinear the system is?

Or is 'Strong' and "Weak' nonlinearity just a qualitative descriptor of its behaviour and relative stability?

答:
I have only a remark, but the standard comment above does not allow such a long text.

I have heard about strong or weak nonlinearity only in the qualitative (and often very subjective) way. However, what comes to my mind is for example the Kunkel path-following method (similar to the more popular arc-length method), which has adaptive step adjustment and for this purpose nonlinearity of the problem at the current step is measured. It is called $nonlinearity~index$ (Nichtlinearitätsindex) and it is defined in the step $t$ by
\begin{align}
^tL=\frac{||^tu-^t\!\!u^{\text{1.it.}}||}{||^tu^{\text{1.it.}}-^{t-1}\!u||}
\end{align}
where $u^{t-1}$ is a value from the previous step, $u^t$ is the current iterate and $u^{t,\text{1.it.}}$ is the first iteration. Taken from the disseration of Eckart Resch, Zur numerischen Simulation von Holzverbindungen, 2008, pp.129.


问:
Confusion about subsets of Hausdorff spaces being closed or open

Every finite subset of a Hausdorff space is closed.

If $X$ is Hausdorff then in order to prove that every finite subset is closed I can show that the set $\{x\}$ is closed for every $x \in X$ since finite union of closed sets is closed.

Let $y\ne x \in X$ thus since $X$ is Hausdorff there exists $U_x, U_y$ such that $U_x \cap U_y = \emptyset$. Now this implies that $y$ is not in the closure of $\{x\}$ which further implies that the closure of $\{x\}$ is $\{x\}$ hence it's closed.

This seems a bit weird since every subsets of $X$ are open in $X$. Every singleton in a Hausdorff space has it's own open neighbourhood hence subsets of $X$ are just an union of open sets thus open. What does this mean? Every Hausdorff space is both open and closed?

答: It means that in a finite Hausdorff space all subsets are open and closed, indeed. It's a discrete space. All $\{x\}$ are closed; this easily follows from Hausdorffness. So all finite subsets are closed because closed sets are closed under finite unions (not arbitrary unions). And all subsets of a finite set are finite etc.


答: No, it's not true that all subsets of a Hausdorff space are open. For example, take $\mathbb{R}$ with the usual topology, nonempty finite sets are not open there. To be honest, I don't really understand your argument to why do you think all sets must be open.

Title: Confusion about subsets of Hausdorff spaces being closed or open

Body:
Every finite subset of a Hausdorff space is closed.

If $X$ is Hausdorff then in order to prove that every finite subset is closed I can show that the set $\{x\}$ is closed for every $x \in X$ since finite union of closed sets is closed.

Let $y\ne x \in X$ thus since $X$ is Hausdorff there exists $U_x, U_y$ such that $U_x \cap U_y = \emptyset$. Now this implies that $y$ is not in the closure of $\{x\}$ which further implies that the closure of $\{x\}$ is $\{x\}$ hence it's closed.

This seems a bit weird since every subsets of $X$ are open in $X$. Every singleton in a Hausdorff space has it's own open neighbourhood hence subsets of $X$ are just an union of open sets thus open. What does this mean? Every Hausdorff space is both open and closed?


问:
Geometrical shape of level curve
Let $A, B$ be two distinct points in the Euclidean plane $\mathbb{R}^2$. Describe geometrically the level sets of the mapping $f(M)=\frac{AM}{BM}$.

My reasoning so far:

Except in a degenerate case, $ABM$ is a triangle. So $\frac{AM}{BM}$ is the ratio between two sides of a triangle. But I don't know how to describe the level sets geometrically.

答:
Let$ f(M)  = \tau $ a constant for a level curve.

If angle at M is bisected internally the constant ratio locus is called the Apollonian Circle.

To locate this point bipolar coordinates are convenient.

Const $\tau$ curves have this ratio maintained constant.

We can make a simple construction and  look to the intersection of Apollonian circle with x-axis at M1. The second point of intersection is for the the external angle bisector at M2.

The Apollonian Circle has its diameter as $M1-M2$ as shown.

Note that constant  $\tau $ locus of level curve  permits M to be anywhere on the two green bisectors shown due to the property

$$\dfrac{AM_{int}}{M_{int}B}=\dfrac{AM_{ext}}{M_{ext}B}=\tau.$$

Title: Geometrical shape of level curve


答:
Please, stop using fractions when evaluation of modulus...

$$24 \equiv 8 \mod 16$$

$$77 \equiv 13 \mod 16$$

thus your equation is not valid.

Title: Modulo question about equality