问:
Marginal Distributions obtained by restricting a 2D Gaussian to a circle
Suppose I have a 2D Gaussian
$$
f(x, y) = \frac{1}{2\pi\,\sqrt{\text{det}(\Sigma)}}\exp\left\{-\frac{1}{2}(\boldsymbol{x}- \boldsymbol{\mu})^\top \Sigma^{-1} (\boldsymbol{x}- \boldsymbol{\mu})\right\} \qquad \text{with } \boldsymbol{\mu}=(\mu_x, \mu_y)^\top
$$
and then I restrict it to be on a circle. For simplicity, I assume the circle is centered at the mean of the Gaussian
$$
\mathcal{C}=\left\{(x, y)\in \mathbb{R}^2\, :\, (x - \mu_x)^2 + (y - \mu_y)^2 = 1\right\}
$$
What does the marginal distributions become? Here's a short picture of what I mean:

First, we have a 2D Gaussian, denoted here by light grey contours

Then we restrict the distribution to only take values on the unit circle (blue)

Finally, we marginalize on the x and y axis.

I have seen here that maybe the distribution on the circle is a Bingham distribution but that's as far as I can go. Any hint?

Following the Answer

I think the idea is that if we have a point $(x, y)\in \mathbb{R}^2$ then we can map this to be on the unit circle centred at the origin using the mapping
$$
g:\mathbb{R}^2 \to \mathbb{R}^2 \qquad (x, y) \longmapsto \left(\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}}\right)
$$
The Jacobian of this transformation is given by
$$
J_g(x, y) =
\begin{pmatrix}
     \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}} & - \frac{xy}{(x^2+y^2)^{\frac{3}{2}}} \\
- \frac{xy}{(x^2+y^2)^{\frac{3}{2}}} & \frac{x^2}{(x^2+y^2)^{\frac{3}{2}}}
\end{pmatrix} = \begin{pmatrix} \end{pmatrix}
$$
Its determinant is, unfortunately, zero
$$
\text{det}(J_g(x, y)) = \frac{x^2y^2}{(x^2+y^2)^{3}} - \frac{x^2y^2}{(x^2+y^2)^{3}} = 0
$$
Since what we want is the distribution of
$$
u = \frac{x}{\sqrt{x^2+y^2}}
$$
we now instead consider a different function:
$$
g_2:\mathbb{R}^2\to \mathbb{R}^2 \qquad (x, y)\mapsto \left(\frac{x}{\sqrt{x^2+y^2}}, y\right)
$$

This function does not map $\mathbb{R}^2$ into the unit circle centred at the origin. However, conveniently, its first coordinate is the same as the one we want to find the distribution of. Our goal is then this:

Use the change of variable formula to find the joint distribution of
$$
(u, v) = \left(\frac{x}{\sqrt{x^2+y^2}}, y\right)
$$

Marginalize over $v$ to obtain the distribution for $u$
$$
p(u) = \int p(u, v) dv
$$

The Jacobian of this transformation is
$$
J_{g_2}(x, y) = \begin{pmatrix}
\frac{y^2}{(x^2+y^2)^{\frac{3}{2}}} & - \frac{xy}{(x^2+y^2)^{\frac{3}{2}}} \\
0 & 1
\end{pmatrix}
$$
It's determinant is
$$
\text{det}(J_{g_2}(x, y)) = \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}}
$$
The change of variable formula then tells us
$$
\begin{align}
f_{U\times V}(u, v)
&= f_{X\times Y}(g_2^{-1}(u, v)) \cdot |\text{det}(J_{g_2}(x, y)|^{-1}_{(x, y) = g_2^{-1}(u, v)}
\end{align}
$$
One can easily find $g_2^{-1}(u, v)$ since $y = v$ and
$$
u^2 = \frac{x^2}{x^2+y^2} \quad \implies \quad u^2(x^2 + v^2) = x^2 \quad \implies \quad x^2 = \frac{u^2v^2}{1 - u^2}\quad \implies \quad x = \frac{|uv|}{(1 - u^2)^{\frac{1}{2}}}
$$
Pluggin this into the Jacobian we get
$$
|\text{det}(J_{g_2})(x, y)|_{(x, y) = g_2^{-1}(u,v)}  = \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}} = \frac{(1 - u^2)^{\frac{3}{2}}}{|v|}
$$
Thefore the joint distribution on $(u, v)$ is
$$
f_{U\times V}(u, v) = f_{X\times Y}\left(\frac{|uv|}{(1 - u^2)^{\frac{1}{2}}}, v\right) \frac{|v|}{(1 - u^2)^{\frac{3}{2}}}
$$

Since $(x, y) \mapsto \left(\frac{x}{\sqrt{x^2+y^2}}, y\right)$ we can see that this distribution is defined over $[-1, 1]\times \mathbb{R}$. Now we can perform the second step and marginalize with respect to $v$
$$
\begin{align}
    f_U(u)
    &= \int_{\mathbb{R}} f_{U\times V}(u, v) dv \\
    &= \int_{\mathbb{R}} f_{X\times Y}\left(\frac{|uv|}{(1 - u^2)^{\frac{1}{2}}}, v\right) \frac{|v|}{(1 - u^2)^{\frac{3}{2}}} dv
\end{align}
$$
Consider the bivariate normal distribution
$$
{\displaystyle f_{X\times Y}(x,y)={\frac {1}{2\pi \sigma _{X}\sigma _{Y}{\sqrt {1-\rho ^{2}}}}}\mathrm {e} ^{-{\frac {1}{2(1-\rho ^{2})}}\left[({\frac {x-\mu _{X}}{\sigma _{X}}})^{2}-2\rho ({\frac {x-\mu _{X}}{\sigma _{X}}})({\frac {y-\mu _{Y}}{\sigma _{Y}}})+({\frac {y-\mu _{Y}}{\sigma _{Y}}})^{2}\right]}}
$$
Choosing $\Sigma = I_2$ (i.e. $\sigma_x = \sigma_y = 1$ and $\rho=0$) and centered at $\mu = (0, 0)$ we get
$$
f_{X\times Y}(x,y)=\frac{1}{2\pi}\exp\left\{-\frac{\left(x^2+y^2\right)}{2}\right\}
$$
Now plugging in
$$
x^2 + y^2 = \frac{v^2}{1 - u^2}
$$
we obtain
$$
f_{X\times Y}(g_2^{-1}(u, v)) = \frac{1}{2\pi}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}
$$
Marginalizing we get
$$
\begin{align}
    f_U(u)
&= \int_{\mathbb{R}} \frac{1}{2\pi}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}  \frac{|v|}{(1 - u^2)^{\frac{3}{2}}} dv \\
&= \frac{1}{2\pi \sqrt{1 - u^2}} \int_{\mathbb{R}}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\} \frac{|v|}{(1 - u^2)} dv \\
&= - \frac{1}{2\pi \sqrt{1 - u^2}} \int_{-\infty}^{\infty} - \frac{|v|}{(1 - u^2)}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}  dv \\
&= -\frac{1}{\pi\sqrt{1-u^2}} \int_{0}^{\infty} - \frac{|v|}{(1 - u^2)}\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}  dv && \text{even}\\
&= -\frac{1}{\pi\sqrt{1-u^2}}\left[\exp\left\{-\frac{v^2}{2(1 - u^2)}\right\}\right]_0^{\infty} \\
&= -\frac{1}{\pi\sqrt{1-u^2}}\left[0 - 1\right] \\
&= \frac{1}{\pi\sqrt{1-u^2}}
\end{align}
$$

答:
I assume that $\mu_x=\mu_y=0$ for simplicity. The joint pdf of $U:=X/\sqrt{X^2+Y^2}$ and $V:=Y$ is given by
\begin{align}
f_{U,V}(u,v)&=\frac{f_{X,Y}(x(u,v),y(u,v))}{|\partial (u,v)/\partial(x,y)|_{x=u|v|/(1-u^2)^{1/2},y=v}} \\
&=f_{X,Y}\!\left(u|v|/(1-u^2)^{1/2},v\right)\times\frac{|v|}{(1-u^2)^{3/2}}
\end{align}
over $(-1,1)\times\mathbb{R}$. Then the marginal density of $U$ is
$$
f_U(u)=\int_{-\infty}^{\infty}f_{X,Y}\!\left(u|v|/(1-u^2)^{1/2},v\right)\times\frac{|v|}{(1-u^2)^{3/2}}\, dv.
$$
For example, when $\Sigma=I_2$,
$$
f_{U}(u)=\frac{1}{\pi\sqrt{1-u^2}}.
$$


问:
Consider the linear system $ \begin{Bmatrix} x - z = -2a \\ 2x- y+ 2z= 0\\ 3x+ 4y -19z= -a \\ \end{Bmatrix}$

$$
   \begin{Bmatrix}
    x - z = -2a \\
    2x- y+ 2z= 0\\
   3x+ 4y -19z= -a \\
   \end{Bmatrix}
$$

For which values of a does the linear system have

a) a unique solution

b) infinitely many solutions

c) no solution

答:
By Gaussian elimination,

$$
   \begin{cases}
    x - z = -2a \\
    2x- y+ 2z= 0\\
   3x+ 4y -19z= -a
   \end{cases}
$$

$$
   \begin{cases}
    - y+ 4z= 4a\\
    4y -16z= 5a
   \end{cases}
$$

$$
    0= 21a.
$$

Hence if $a\ne 0$ there are no solutions. Else $x=z,y=4z$ make infinitely many solutions.

ContentLicense: CC BY-SA 4.0


问:
Upper bound on the cover number in statistical learning theory.
I am looking at cover numbers for statistical learning theory. Specifically, I refer to the definitions in this tutorial (Section 5.1, Definition 4).

Sticking to their definition, I read from another source that if $\mathcal{G}$ lies in a $d$-dimensional subspace of $\mathbb{R}^m$ and $\max_{g\in\mathcal{G}}\|g\|=M$, then setting $\epsilon=1/\sqrt{n}$ gives us
\begin{align*}
N(\mathcal{G},\epsilon,n)&\leq(2M\sqrt{dn})^d.
\end{align*}

Why is this the case? I have trouble drawing the connection. Thanks.

答:
Your other source probably uses the Euclidean metric somewhere not the Hamming metric to get $\sqrt{d}$.

Each coordinates $x_i$ is bounded by $M$, so you can cover $[-M,M]$ with $\lceil M/(\epsilon/\sqrt{d})\rceil\leq 2M/(\epsilon/\sqrt{d})$ intervals(*) of length $2\epsilon/\sqrt{d}$ (i.e., ball in $\mathbb{R}$ of radius $\epsilon/\sqrt{d}$).  So you look at the $\lceil M/(\epsilon/\sqrt{d})\rceil^d$ points in $\mathbb{R}^d$ where each coordinate is one of the centre of these intervals, and each cube with these intervals as side will be contained in a Euclidean ball of radius $\leq\sqrt{(\frac\epsilon{\sqrt{d}})^2+\dots+(\frac\epsilon{\sqrt{d}})^2}=\epsilon$.

(*) closed intervals and closed balls here.  If you want the open version you need to insert some $+1$s to make sure internal boundaries are covered.


问:
Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$.
Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the family are drawn on the same picture, one gets (click to enlarge):

What I'm calling "pillars" make their appearance. They are located on precise abscissas $x_{c/d}$ which are algebraic numbers, roots of $X^c-(1-X)^d$ polynomials for some positive rational $c/d$ (in irreducible form). To each pillar is associated a vertical wavelength (clearly visible on the picture) which is equal to $-\frac{\ln(x_{c/d})}{d}=-\frac{\ln(1-x_{c/d})}{c}$. I compared the wavelength of the $(c/d=5)$ pillar (for which $x_5$ is the multiplicative inverse of $\rho$, the plastic number) with the wavelength of the $(c/d=3/2)$ pillar (marked green), and found numerically that they were equal. So, the asked question is to prove that, in any of its forms:

$$ -\frac{\ln(x_{3/2})}{2} = -\frac{\ln(x_{5})}{1} $$

$$ x_{3/2} = x_{5}^2 $$

the real root of $X^3-(1-X)^2$ is the square of the real root of $X^5-(1-X)^1$

the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$

答:
Let $\alpha$ denote the real root of $X^3+X^2-1$. Then $\alpha^3 = 1-\alpha^2$. If we plug in $\alpha^2$ into $X^3-X^2+2X-1$, we obtain
\begin{align*}
   (\alpha^2)^3-(\alpha^2)^2+2(\alpha^2)-1 &= (1-\alpha^2)^2-\alpha^4+2\alpha^2-1\\
   &=(1-\alpha^2)^2-(1-\alpha^2)^2=0,
\end{align*}
so $\alpha^2$ is a root of $X^3-X^2+2X-1$.


问:
What values can $\max(\alpha)$ and $\min(\alpha)$ take?
In $\{0,0.2,0.4,0.6,0.8,1\}$-valued logic, define $\\f(\alpha)=\min\{\bar{v}(\alpha): v \text{ is some truth valuation on the atoms}\}$. $g(\alpha)=\max\{\bar{v}(\alpha): v \text{ is some truth valuation on the atoms}\}$.

Suppose the connectives evaluate as follows: $\bar{v}(\alpha \wedge\beta) = \min\{\bar{v}(\alpha),\bar{v}(\beta)\}, \bar{v}(\alpha \vee\beta) = \max\{\bar{v}(\alpha),\bar{v}(\beta)\}, \bar{v}(\neg \alpha) = 1-\bar{v}(\alpha)$.

Given a wff $\alpha$, assume it only contains connectives $\neg, \vee,\wedge$, what are the possible values that $f(\alpha)$ and $g(\alpha)$?

I know $f$ can take:

0 ($A \wedge \neg A$, where A=0), $0.6$(e.g. the possible truth values of $A\vee \neg A$ is $1, 0.8, 0.6$, so $f(A\vee \neg A)$ is the min of the three, which is $0.6$).

$g$ can take $1$ ($A\vee \neg A$, A=1), $0.4$ ($A\wedge \neg A$, $A=0.4$ or $0.6$).

Somehow the answer is these are all the values in the ranges of $f$ and $g$. Why?

答:
Here's one way to show this. First, your logic has the following property:

(i) If a formula $\alpha$ takes value $\le 0.4$ somewhere, then it also takes value $0$ somewhere.

Why is this true? For any valuation $v$, consider the valuation $v'$ defined via $v'(x)=\lfloor v(x)+0.5\rfloor\in\{0,1\}$. In other words, $v'$ makes $v$ into a boolean valuation by setting the splitting point at $0.5$. Then by induction on formulas, you can show that $v'(\alpha)=\lfloor v(\alpha)+0.5\rfloor$. Hence if $v(\alpha)\le 0.4$, then $v'(\alpha)=\lfloor v(\alpha)+0.5\rfloor\le\lfloor 0.9\rfloor=0$.

From (i) it follows that $f$ cannot take any value in $\{0.2,0.4\}$.

By a similar argument you can show that

(ii) If a formula $\alpha$ takes value $\ge 0.6$ somewhere, than it also takes value $0.6$ somewhere.

(hint: project all truth values $\ge 0.6$ to $0.6$ and all truth values $\le 0.4$ to $0.4$)

From (ii) it follows that $f$ cannot take a value in $\{0.8,1\}$.

Finally, the possible values for $g$ can be inferred from the above observations, as $g(\alpha)=1-f(\lnot\alpha)$. E.g. $g$ cannot take value $0.2$ as otherwise $f$ would have to take value $0.8$ somewhere, which is not possible.


问:
Proving a uniform bound on a sequence of elements in $\ell_1^+$.
Let $A\subset\ell_1^+$ and let $\{f_n\}_{n=1}^\infty \in C(X)^+:=\{f\in C(X): f(x)\geq 0\,\, , \forall x\in X\}$, where X is a compact metric space. Suppose the following hold:
\begin{gathered}
1. \sup_{x\in X} |f_n(x)|\leq M_1.\\
2. \sup_{w\in A, x\in X}|\sum_{n=1}^\infty w_nf_n(x)|\leq M_2.\\
\end{gathered}

Prove that there exists $M_3 $ such that $\sup_{w\in A}|\sum_{n=1}^\infty w_n|\leq M_3$.

I think this should be true primarily because I can't think of a counter example. The uniform bound on $\sum_{n=1}^\infty w_n^kf_n(x)$ as well as the non-negativity are strong conditions.

答:
I assume that implicitly $k$ is a positive integer. We only need to prove that for every $n \in \mathbb{N}$ the number $w_n$ is less than $1$, since we have then $$ \sum_{n \in \mathbb{N}} w_n^k \leq \sum_{n \in \mathbb{N}} w_n$$ and the sequence $(w_n)$ is summable by assumption.

Let us assume that $w_{n_0} > 1$ for some integer $n_0$, and choose $x \in X$. I guess that $\mathcal{C}(X)^+$ denotes the set of continuous functions on $X$ taking strictly positive values (the result is clearly false if we only assume non-negative). Then $$\sum_{n \in \mathbb{N}} f_n(x) w_n^k \geq f_{n_0}(x) w_{n_0}^k$$ tends to $+ \infty$ when $k$ tends to $+ \infty$, which contradicts our assumptions.


问:
The Witch of Agnesi and its cartesian equation.
I want to learn to simply take an ancient geometer's description of an interesting curve and try to find its cartesian equation. I'm struggling to find the equation of the locus of all points described by the 'Witch of Agnesi' The description is as follows:

"Starting with a fixed circle, a point O on the circle is chosen. For any other point A on the circle, the secant line OA is drawn. The point M is diametrically opposite to O. The line OA intersects the tangent of M at the point N. The line parallel to OM through N, and the line perpendicular to OM through A intersect at P. As the point A is varied, the path of P is the Witch of Agnesi."

I'd really appreciate insight and not plug and chug kind of answers if any. I'm looking for your ideas and tips. It would be very helpful if you could explain the answer step by step.

(I know the answer thanks to Wikipedia but I'd like to know how to arrive at the result)

答:
Note that the coordinates are formed as projections of known lengths.

The line through origin cuts the circle and line $y= 2a$ at $ K,N $ respectively.

The curve is found by drawing lines parallel to $x-,y-$ axes through $K,N$ to get cutting point at $P$.

$$x= 2a \tan \theta$$

$$ OK=OM \cos \theta= 2a \cos \theta$$

$$ y= OQ=PL=OK \cos \theta = 2a \cos^2 \theta $$

To get this into Cartesian form, eliminate $\theta$ using trig relation $ \sec^2(...)= 1 + \tan^2(...) $

$$ \dfrac{2a}{y}=1+\dfrac{x^2}{4a^2} \to y=\dfrac{8a^3}{x^2+4a^2}$$

Notice that the Cartesian equation form is obtained only through this trig identity after the construction procedure is over.

ContentLicense: CC BY-SA 3.0


问:
Conditional Binomial Probability
Haven't looked at statistics in a while, and came across the following problem:

It is estimated that approximately 20% of marketing calls result in a sale. Assume that 10 out of 12 marketing calls on a day resulted in a sale. What is the probability that the first call made is one of the 10 that resulted in a sale?

As each call is independent, I am incorrect in thinking that the answer is simply 10/12? I just cant figure out how to prove it mathematically. I assume that it has something to do with proving independence of the choosing of the 1st from the assumption that 10 of the 12 calls were successes, but im just not sure where to start.

Thanks in advance!

答:
the result is independent of the prior probability of $20\%$. In fact any of the 66 combination with 10 success among 12 calls are equiprobable.

Thus the result is only a % of the favourable combinations on the possible combinations

$$\frac{55}{66}=\frac{5}{6}$$

this is the sampling space

Of course it is not necessary to write down all the sample space to derive that the favourable events are 55. It is enough to observe that, if the first element in the succession is "1" there are $\binom{11}{9}=\binom{11}{2}=55$ different ways to combine the remaining 11 elements with 9 successes

Title: Conditional Binomial Probability


问:
NP hard/complete
I have never been very clear on this concept. Please help:

At the end of the day, we should want to identify useful problems for which we don't have polynomial solution so far and only have exponential solutions.We want to keep finding if we can find a polynomial solution to these problems and the use of reductions is only to be able to identify new only exponential solvable problems with help of existing known exponential problems.??

Why is concept of determinism and non determinism brought into this whole concept of exponential and polynomial solvable problems?? How is this notion useful?

What do we call the problems for which we have only exponential solution so far, but we haven't been able to find reduction to a known NP - hard/complete problems..??

Thanks,

答:
The: If $\text{P} \neq \text{NP}$ implies NP-Hard problems require exponential time algorithms is a common myth, which seems quite prevalent.

In fact, we have known NP-Hard problems which have sub-exponential algorithms! For instance see this here: http://hal.inria.fr/docs/00/47/66/86/PDF/cops_journal_2008_10_16.pdf

That said, there is a conjecture, called the Exponential Time Hypothesis which states that 3-SAT (and some of it's variants) cannot be solved in sub-exponential time.

Now you might wonder why does the above sub-exponential algorithm for an NP-Hard problem not falsify the Exponential Time Hypothesis, as we can ultimate form a reduction from 3-SAT?

The reason is that reductions are allowed to blow up the input polynomially. Say $A$ reduces to $B$ and $B$ has a $\theta(2^{\sqrt{n}})$ time algorithm. Now if the reduction makes the size of the input to $B$ as $n^2$ (it was $n$ for $A$), that still gives us a $\theta(2^n)$ time algorithm for $A$. So, even though $A$ never has sub-exponential time algorithms, we could still have it reduce to a problem which has sub-exponential time algorithms.

As to why we have non-determinism, I am guessing this was first started by Rabin and Scott in their papers about deterministic finite automaton and non-deterministic finite automaton.

Once Cook/Levin introduced the concept of NP-Completeness and Karp et al gave a seminal list of 21 natural problems which are NP-Complete, NP became a very important class.

I believe it was Edmonds who first characterized NP as 'problems with polynomial certificates'.

(Note: The above three paragraphs are from memory and I haven't verified them).

In fact, there are characterizations of NP which do not use non-determinism. The current hot topic has been the use of PCP Theorem to prove various NP-Hardness results for approximation versions of NP-Hard problems.

So, non-determinism is still part of the definition of NP and is historically important, but the current status is that it is not mandatory.

Title: NP hard/complete


问:
Primality of the numbers in the form of $2n^2-1$
I have a question about primality of integers in the form of $2n^2-1$.

I can prove that for the certain type of n such integers are always composite. For example, if $n=7k+2$ or $n=7k+5$, the whole expression would be always divisible by $7$.

The same is applicable to the whole (probably infinite) set of numbers in the form of $n=ak+b$, where $a$ is $7$, $17$, $23$ etc. and $b$ usually has two values (like $2$ and $5$ for $a=7$).
($a$ is prime here and $b \le a-1$)

I also suspect that the only composite values of $2n^2-1$ are those whose factors are from that set of a (like $7$, $17$, $23$ etc.).

I am trying to see if there anything else can be said about primality or compositness of those numbers $n$. Is there any other forms of n that can guarantee compositness (or primality)?

I would appreciate any ideas.

Thanks!

答:
Let $n=ak+b$, as you have it.  Suppose prime $p|a$ and $2b^2\equiv 1\pmod{p}$.  Then $2n^2-1=2(ak+b)^2-1=a(2ak^2+4abk)+2b^2-1\equiv 0\pmod{p}$, so $p|2n^2-1$ and hence $2n^2-1$ is composite.

We can test if $2b^2\equiv 1$ using Legendre symbols.  Such a $b$ exists (in fact two exist) if $\left(\frac{1/2}{p}\right)=1$.  We multiply both sides by $\left(\frac{2}{p}\right)$ to get $1=\left(\frac{1}{p}\right)=\left(\frac{2}{p}\right)$.  This has solutions if $p\equiv 1,7\pmod{8}$.

Example: Let $a=51=3\times 17$.  Since $17\equiv 1\pmod{8}$, there are two choices for $b$ such that $2b^2\equiv 1\pmod{17}$, namely $3,14$.  Hence $n=51k+3$ and $n=51k+14$ will always be composite for all $k$.


问:
Range and kernel of a linear transformation are ALWAYS disjoint
Is  it true that the Range and kernel of a linear transformation are ALWAYS disjoint. I think they are not but I remember in my notes that the ker L= Im (L') this was under projections. So I am unsure about the answer</p>

答:
In complex vector space, range and kernel of a linear transformation A are identical
if and only if range(A) + range(A*) form the whole space.
Goetz Trenkler, Dortmund, Germany</p>


问:
Solution in $H^2 \cap H^{1}_{0}$
Consider the following problem:
$$-\Delta \phi + \Delta^2 \phi = 4\pi u^2, \ \Omega$$
$$ \Delta \phi = 0, \partial \Omega$$
$$\phi = 0, \partial\Omega$$
How can I prove that the space of weak solutions of this problem is $H^2\cap H^{1}_0(\Omega)$?

It's clear for me that $\phi \in H^1_0$, but why $\Delta \phi = 0$ leads to the space $H^2\cap H^{1}_0(\Omega)$?

答:
The 'correct' choice of the solution space has to fulfill two basic requirements:

It must be complete (and typically, reflexive) space such that you can get good existence results.

Weak and strong solutions should (up to regularity issues) coincide.

It is clear that the first point is satisfied by $V = H^2(\Omega) \cap H_0^1(\Omega)$.

Now, let $\phi$ be strong / classical solution with regularity $\phi \in C^4(\bar\Omega)$ and it satisfies the equations in a pointwise way. Then, it is clear that $\phi \in V$.
For an arbitrary $v \in V$, we have
$$
\int_\Omega -\Delta \phi \cdot v \, \mathrm{d}x
=
\int_\Omega \nabla\phi \cdot \nabla v \, \mathrm{d}x
-
\int_{\partial\Omega} \frac{\partial\phi}{\partial n} \cdot \underbrace{v}_{=0} \, \mathrm{d}s
$$
and
$$
\int_\Omega \Delta^2 \phi \cdot v \, \mathrm{d}x
=
\int_\Omega \Delta \phi \cdot \Delta v \, \mathrm{d}x
+
\int_{\partial\Omega}
 \frac{\partial\Delta\phi}{\partial n} \cdot \underbrace{v}_{=0}
-
 \frac{\partial v}{\partial n} \cdot \underbrace{\Delta \phi}_{=0}
 \, \mathrm{d}s.
$$
By adding these identities you can plug in the right-hand side of your PDE and arrive at the weak formulation.
(Note that this process is very similar to the incorporation of a (natural) Neumann boundary condition for Poisson's equation)

Along the same lines one can show that a weak solution with enough regularity is also a strong solution.

Title: Solution in $H^2 \cap H^{1}_{0}$


问:
Is the expectation of X finite, infinite or undefined?
The annual maximum one-day rainfall can be modeled by a random variable X with p.d.f.

\begin{equation}
f(x)=
\begin{cases}
\frac{2}{\pi}\frac{1}{x^2+1}\quad\text{if}\;x\geq0\\
0\quad\quad\quad\text{if}\;x<0
\end{cases}
\end{equation}
Is the expectation of X finite, infinite or undefined?

From the p.d.f., I know that this is a continuous random variable. Thus, I can use the following formula to determine the expectation:
$$\mathbb{E}(X)=\int_{-\infty}^\infty x f(x)dx$$
Evaluating, I get
$$$$$$\mathbb{E}(X)=\int_{-\infty}^\infty \frac{2x}{\pi}\frac{1}{x^2+1}dx=\frac{1}{\pi}[\ln(x^2+1)]^\infty_{-\infty}=0$$.

Would this mean that the expectation is finite? What does it mean for an expection to be $0$? I know that expectation is analygous to mean, but to me it doesn't make sense for the average rainfall to be $0$. I think my question is basically how to interpret the concept expectation. Thank you.

答:
First you have to look out for the fact that $f(x) =0$ for $x<0$. That means that
$$\mathbb{E}[X] = \int_{-\infty}^{\infty} xf(x) \ dx = \int_0^{\infty} \frac{2x}{\pi (x^2+1)}$$
Now you figure out that this integral is not finite, so the expectation of the given distribution is not finite.

Actually the integral you calculated is also undefined as you have a $\infty - \infty$ situation that is not defined. The integral
$$\int_{-a}^a \frac{2x}{\pi (1+x^2)}$$
exists for every $a >0$, but that does not mean that the integral from $-\infty$ to $\infty$ also exists. It is just the so-called "Cauchy principal value" that exists in this case, but not the improper integral.


问:
Question about split idempotent in triangulated categories
I've read about split idempotents, and I know they are morphisms $e:X\to X$ such that there exists $Y$ and morphisms $i:Y\to X$, $p:X\to Y$ with $pi=\mathrm{id}_Y$, $ip=e$.

In the setting of triangulated categories (which are additive but not abelian), many sources give this notion (or at least a notion with the same name) saying that $e$ provides a splitting of $X=\mathrm{Im}(e)\oplus\mathrm{Ker}(e)$. That would means $\mathrm{Ker}(e)$ and $\mathrm{Im}(e)$ exist, and give a decomposition of $X$.

I don't know how to prove that a splitting idempotent $e$ has a kernel and an image, and neither that these give such a decomposition of $X$.

My intuition says that $Y$ should be the image of $e$, but I'm stuck trying to compute the kernel and the cokernel. It seems $X$ with morphism $\mathrm{id}_X-e$ works as cocone for both the pullback a the pushout diagrams defining kernel and cokernel of $e$, but is it limiting? My geometric intuition of an idempotent as a projection endomorphism of, say, a vector space, says not, and I don't know how to proceed.

Thanks in advance for any help

答:
Ok, it has to be said slighly better. In a category where any idempotent split, for any idempotent $e:X\to X$ there is a canonincal decomposition $X=W\oplus Y$ where Y is the splitting of $e$ (image of $e$), and $W$ is the splitting of $\mathrm{id}_X-e$ (the kernel of $e$).

Canonical monomorphisms of the two splits provide the inclusion morphisms for the coproduct, and for morphisms $h:Y\to Z$ and $h':W\to Z$ the unique morphism $X\to Z$ proving $X$ to be actually the coproduct is the sum $h'p'+hp$, where $p$ and $p'$ are the epimorphisms of the two splits.


问:
Number of reduced residue classes with given order?
In general, is there a way to compute the number of reduced residue classes $\pmod n$ of a certain order?

For example, say $n=385=5\times7\times11$. Then
$$
(\mathbb Z/385\mathbb Z)^\times\cong(\mathbb Z/5\mathbb Z)^\times\times (\mathbb Z/7\mathbb Z)^\times\times (\mathbb Z/11\mathbb Z)^\times.
$$
In $(\mathbb Z/5\mathbb Z)^\times$ there are $1,1,2$ elements of orders $1,2,4$.

In $(\mathbb Z/7\mathbb Z)^\times$ there are $1,1,2,2$ elements of orders $1,2,3,6$.

In $(\mathbb Z/11\mathbb Z)^\times$ there are $1,1,4,4$ elements of orders $1,2,5,10$.

Then, if I wanted to see how many elements of order $4$ (say) there were in $(\mathbb Z/385\mathbb Z)^\times$, I can 'brute' force and check all the triples $(a\pmod5,b\pmod7,c\pmod{11})$ for which $\text{lcm}(\text{ord}_5a,\text{ord}_7b,\text{ord}_{11}c)=4$ and conclude that there are $8$ elements of order $4$.

Is there a faster method/theorem to compute this? And if there isn't, what about for $\lambda(385)=60$, the maximum order? Is there a way to compute how many elements there are of order $60$?

答:
Actually, it is well known that if $n=\prod_{i=1}^{t}p_i^{\alpha_i}$ ,

then $\phi(n)=n\prod_{i=1}^{t}(1-\frac{1}{p_i})$


问:
Probability of Independent Events (binomial)
If someone could just do a very basic walkthrough on how you would go about answering this question it would be greatly appreciated as I'm practising for an exam!

'''

When a company bids for contracts it estimates the probability of winning each contract is $0.18$, independent on whether other contracts have been won or lost.

(a) If the company bids for 5 contracts, what is the probability it wins:

i) at least one contract

ii) at least two contracts

'''

i)I used the OR rule for this question so if there's a $0.18$ chance then surely it's $0.18$ OR $0.18$, etc which is just $0.18 + 0.18 + 0.18 + 0.18 + 0.18 = 0.90$

ii)For winning at least two contracts I thought it was the AND rule as you have to win at least one contract AND another, so I did $0.18 * 0.18 = 0.0324$

If someone could just correct me if I took my stupid pills this morning I would be very grateful!

答:
i)I used the OR rule for this question so if there's a 0.18 chance then surely it's 0.18 OR 0.18, etc which is just 0.18+0.18+0.18+0.18+0.18=0.90

Using your reasoning, if you had 6 events the resulting probability would have resulted in $0.18\times 6=1.08>1$

It is just a matter to study (or re-study) the binomial probability.

(i) wins AT LEAST one contract, it is simply the complement to have lost all the contacts, that is

$$1-0.82^5$$

if you want to count this probability in a direct way you can do

$$P(X\geq 1)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)=$$

$$=\binom{5}{0}\cdot 0.18^0\cdot 0.82^5+\binom{5}{1}\cdot 0.18^1\cdot 0.82^4+\dots+\binom{5}{5}\cdot 0.18^5\cdot 0.82^0$$

With this hint I think you can proceed by yourself with (ii)


答:
You can't use the "or" rule, as it applies only to disjoint events, not independent events.

Instead, you can use the "and" rule to work out the probability of not winning at least one contract. Each contract independently has probability $0.82$ of not being won, so the probability of winning none is $0.82^5$, and so the probability you want is $1-(0.82^5)$.

For the second question, you need to work out the probability of winning exactly one, and subtract it from the above answer. Now there are five ways to win exactly one (five different contracts you can win). In order to win the first but no others, you have probability $0.18$ of winning the first, $0.82$ of not winning the second, and so on, so the probability of winning only the first is $0.18\times 0.82^4$. The other four options are similar.

Now these five ways are disjoint, so the "or" rule applies.

Title: Probability of Independent Events (binomial)

Body:
If someone could just do a very basic walkthrough on how you would go about answering this question it would be greatly appreciated as I'm practising for an exam!

'''

When a company bids for contracts it estimates the probability of winning each contract is $0.18$, independent on whether other contracts have been won or lost.

(a) If the company bids for 5 contracts, what is the probability it wins:

i) at least one contract

ii) at least two contracts

'''

i)I used the OR rule for this question so if there's a $0.18$ chance then surely it's $0.18$ OR $0.18$, etc which is just $0.18 + 0.18 + 0.18 + 0.18 + 0.18 = 0.90$

ii)For winning at least two contracts I thought it was the AND rule as you have to win at least one contract AND another, so I did $0.18 * 0.18 = 0.0324$

If someone could just correct me if I took my stupid pills this morning I would be very grateful!


问:
Trying to understand an example where unbiased estimators don't exist
I am new to statistics especially in the topic of estimators and sufficient statistic.

I am reading a note which says "unbiasedness is a desirable (but not necessary) property of a good estimator". Then it gives an example where unbiased estimators don't exist. Since there is no further explanation in the example, I am struggling to understand it.

The example says:

If $X\sim B(\varphi(\theta))$ for some function $\varphi$, then $\theta^*\in K_0\iff\mathbb{E}_\theta\theta^*\equiv\theta^*(0)(1-\varphi(\theta))+\theta^*(1)\varphi(\theta)=\theta,\forall\theta\in\Theta)$, (where $K_0$ is the class of all unbiased estimators).

This is just for reference:

An estimator $\theta_0^*=\theta_0^*(X)$ from a class $K$ of estimators of $\theta$ is called efficient in $K$ if, for any $\theta^*\in K$,
$$\mathbb{E_\theta}(\theta_0^*-\theta)^2\le\mathbb{E_\theta}(\theta*-\theta)^2,\forall\theta\in\Theta.$$

For a function $b=b(\theta),\theta\in\Theta$, let
$$K_b=\{\theta^*:\mathbb{E}_\theta\theta^*=\theta+b(\theta),\forall\theta\in\Theta\}$$ be the class of all estimators with the bias $b(\theta)$.

I understand the basic idea of estimators, the property for being biased and unbiased. But I have no clue why unbiased estimators don't exist in the given example above. May anyone kindly explain it to me please?

Many thanks!

答:
We have,

$\theta = (1-\varphi(\theta))\theta^\ast(0) + \varphi(\theta)\theta^\ast(1), \quad\forall \theta\in\Theta,$

if and only if

$\varphi(\theta) = \frac{1}{\theta^\ast(1)-\theta^\ast(0)}\theta - \frac{\theta^\ast(0)}{\theta^\ast(1)-\theta^\ast(0)}, \quad\forall \theta\in\Theta$.

That is, $\varphi$ must be linear in $\theta$. Otherwise an unbiased estimator does not exist.


问:
How to prove that $\frac{\ln 5}{\ln 2}$ is irrational
I have to show that $\frac{\ln 5}{\ln 2}$ is irrational

I have tried the following:

Assume it's rational so

$\frac{\ln 5}{\ln 2} = \frac{p}{q}$

which becomes

$\log _2\left(5\right)=\frac{p}{q}$

Therefore

$5=2^{\frac{p}{q}}$

This is where I'm stuck, I don't know how to advance my proof any further

答: $$\frac{\log 5}{\log 2}=\frac pq\iff 5^q=2^p$$ for naturals $p,q$. By the fundamental theorem of arithmetic and primality of $2$ and $5$, this is not possible.


问:
A doubt about differentiability and increasing function
Suppose $f:(a,b)\rightarrow (a,b)$ is differentiable on $(a,b)$ and for an $x_{0}$ such that $a<x_{0}<b$ , $f'(x_{0}) >0 $ then is $f$ increasing in some neighborhood of $x_{0}$?

I have seen examples on this site on disproving this for the interval $(0,1)$ by taking the function $x+2x^{2}\sin(\frac{1}{x})$ when $x\neq 0$ and $0$ if $x=0$. But I have a doubt whether this would be true for $x_{0}$ being an interior point of the open interval $(a,b)$ . Can someone please clarify . I have tried to prove it using LMVT but since nothing is said about continuity on $[a,b]$ I am unable to proceed.

答:
If we can assume that $f'$ is continuous at $x_0$, then by $f'(x_0)>0$ we know that there exists a closed interval $[c,d]$ satisfying $x_0 \in [c,d]\subset (a,b)$ and s.t. $\forall t \in [c,d], f'(t)>0$. Moreover $f$ is differentiable on $(c,d)$, $f$ is continuous at $[c,d]$. Then we apply LMVT: if $x,y\in[c,d], x<y$, then $f(y)-f(x)=f'(\xi)(y-x)$ where $\xi \in [c,d]$, and thus $f'(\xi)>0, (y-x) >0$, finally for $x,y\in [c,d]$, $f(y)>f(x)$ whenever $y>x$.

As noted, for counterexample just take $(a,b)=(-1,1), x_0=0,f(x)=x+2x^2 \sin(\frac{1}{x})$ when $x \neq 0$, $f(x)=0$ when $x=0$.


问:
Uniform convergence of $f_n(x)=\frac1{1+nx}$ on $(0,\infty)$
Problem: Let $f_n(x)=\displaystyle\frac{1}{1+nx}$ for $x\in(0,\infty)$ and $n\in\mathbb N$.

$\textbf{a)}$ Show that for each $\delta>0$, $\{f_n\}_{n=1}^\infty$ is uniformly convergent on $[\delta,\infty)$, and determine to which function.

This is the easy part. Fix $x\in[\delta,\infty)$. Then $\delta\leq x$ implies that
$$\vert f_n(x)\vert=\left\vert\frac{1}{1+nx}\right\vert\leq\frac{1}{1+n\delta}\to0\quad\text{as }n\to\infty.$$
Since $x\in[\delta,\infty)$ was arbitrary, it follows that $\|f_n-f\|_\infty\to0$ as $n\to\infty$ where $f(x)=0$ for all $x\in[\delta,\infty).$

$\textbf{b)}$ Show that $\{f_n\}_{n=1}^\infty$ is not uniformly convergent on $(0,\infty)$.

Fix $x\in(0,\infty)$ and let $\varepsilon>0$ be given. Choose $N\in\mathbb N$ such that $N>\displaystyle\frac{1}{\varepsilon x}$. Then for all $n\geq N$ we have
$$\vert f_n(x)\vert=\frac{1}{1+nx}\leq\frac{1}{nx}<\varepsilon.$$
Therefore, $f_n\to0$ as $n\to\infty$ pointwise on $(0,\infty).$ So we assume that the convergence is uniform and try to derive a contradiction. Pick $\varepsilon=1/2.$ Then there is some $N(\varepsilon)\in\mathbb N$ such that $\vert f_n(x)\vert<\varepsilon$ for all $n\geq N(\varepsilon)$ and all $x\in(0,\infty)$. Therefore, we may choose $n$ and $x$. Pick $n=N(\varepsilon)$ and $x=1/3N(\varepsilon)$. Then $\vert f_n(x)\vert=\displaystyle\frac{1}{1+1/3}>1/2=\varepsilon$ and we have a contradiction. It follows that $\{f_n\}_{n=1}^\infty$ does not converge uniformly on $(0,\infty).$

Do you agree with my proof above? I am specifically concerned about the last argument in part (b).

Thank you in advance for any help, your time is much appreciated.

答:
Your solutions are correct. But here is another solution for part b:
It's clear that $f_n \to 0$ pointwise on $(0,+\infty)$, so for uniform convergence, we would need the uniform norm to converge to $0$ as well. But:
$$\lVert f_n \rVert = \sup_{x \in (0, +\infty)} |f_n(x)|=\sup_{x \in [0,+\infty)} |f_n(x)|=f_n(0)=1$$


答:
$2n^2-1\leftarrow$ Looks like Mersenne Primes. You can have numbers $n$ of form $2^k$ such that $2k+1$ is not a prime.

$2(2)^{2k}-1=2^{2k+1}-1 \implies 2k+1=mj$, you have $2n^2=(2^m)^j-1=(2^m-1)(2^{n-1}+2^{n-2}+ \dots 1) $, so, you need such $k$, which makes $2k+1$ composite.

When you have prime of the form $2k+1$, you have a Mersenne prime(Not always true).

Title: Primality of the numbers in the form of $2n^2-1$

Body:
I have a question about primality of integers in the form of $2n^2-1$.

I can prove that for the certain type of n such integers are always composite. For example, if $n=7k+2$ or $n=7k+5$, the whole expression would be always divisible by $7$.

The same is applicable to the whole (probably infinite) set of numbers in the form of $n=ak+b$, where $a$ is $7$, $17$, $23$ etc. and $b$ usually has two values (like $2$ and $5$ for $a=7$).
($a$ is prime here and $b \le a-1$)

I also suspect that the only composite values of $2n^2-1$ are those whose factors are from that set of a (like $7$, $17$, $23$ etc.).

I am trying to see if there anything else can be said about primality or compositness of those numbers $n$. Is there any other forms of n that can guarantee compositness (or primality)?

I would appreciate any ideas.

Thanks!


问:
Stochastic integral with respect to local martingale
Hey I need to define stochastic integral with respect to a local martingale. Can anyone recommend me some books in which I could find all needed information?</p>

答: Brownian Motion and Stochastic Calculus by Ioannis Karatzas and Steven Shreve.</p>


问:
Does the order of the Fibonacci sequence's initial values matter?
I am reading about generating functions in this reputable engineering textbook and the author uses the Fibonacci sequence as an example:

The Fibonacci sequence is defined by the initial conditions $a_0=0$,
$a_{-1}=1$, and the recursion relation $a_n=a_{n-1}+a_{n-2}$ for
$n>0$.

Using this definition, I am able to work through the author's example to get the correct generating function:
$$G(x)=\frac{1}{1-x-x^2}$$

However, everywhere I read online states the initial conditions of the Fibonacci sequence as (0,1) and not (1,0). So I tried that instead, but I get a different result:

$$G(x)=\frac{1+x}{1-x-x^2}$$

This generating function seems to produce the Fibonacci sequence, but starting from the second number instead of the first $(1,2,3,5,8,13,...)$. Where did I go wrong? I would have thought that the sequence should be identical in both cases.

Many thanks for any help!

My calculations:

The author starts with a general expression for a generating function for a sequence $\{a_n\}$:
$$G(x)=\sum_{n=0}^{\infty}a_nx^n$$

He then looks at the special case where $\{a_n\}$ is a linear recurrence of range $r$:
$$a_n=\sum_{i=1}^rc_ia_{n-i}$$

and proves that this leads to:
$$G(x)=\frac{g(x)}{f(x)}=\frac{\sum_{i=1}^rc_ix^i\left(a_{-i}x^{-i}+\cdots+a_{-1}x^{-1}\right)}{1-\sum_{i=1}^rc_ix^i}$$

So, for the Fibonacci sequence, I guess we have $r=2$ and $c_1=c_2=1$. So we get:

$$g(x)=a_{-1}+a_{-2}+a_{-1}x$$

Now I'm a bit confused because the subscripts don't seem to match up. In all the author's discussion, the "initial conditions" have strictly negative subscripts ($a_{-i}$ for $i=1,2,\dots,r$). But just for the Fibonacci example, we have initial conditions $a_0=0$ and $a_{-1}=1$. So I assume these are just off by 1. Therefore, substituting $a_{-1}=0$ and $a_{-2}=1$ gives the author's result above and using $a_{-2}=0$ and $a_{-1}=1$ gives my result above.

答:
Although one does not always make this explicit, giving a sequence of numbers really amounts to giving a set of (index,value) pairs; for instance an infinite sequence of real numbers is really a function $\Bbb N\to\Bbb R$, whose graph is an infinite set of such pairs. The generating series for a set $S$ of such pairs $(i,a_i)$ is $\sum_{(i,a_i)\in S}a_iX^i$, so for a series of numbers corresponding to $f:\Bbb N\to\Bbb R$ this is $\sum_{i=0}^\infty f(i)X^i$.

Let me first put aside the easy answer to the question in your title: of course order of initial terms matters, and permuting them in general gives an entirely different sequence. But that is not what your question was really about, which seems to be more about whether a shift in indexing produces a different generating series. Here too the answer is yes: a pure shift in indexing amounts for the generating series in multiplication by a power of $X$. If the shift is accompanied by addition or removal of nonzero values in the sequence, then this amounts to addition of terms to the generating series.

Note that normally a generating series is a formal power series in $X$, with no negative powers of $X$ at all. In this sense there is no generating series at all for a sequence with a nonzero value at some negative index, as your book apparently wants to do by starting at $a_{-1}=1$. If you insist on including that, the generating series would have a term $X^{-1}$, and thereby become a Laurent series rather than a formal power series. But usually those who entertain the idea of Fibonacci numbers at negative indices want them for all negative integer indices, but for such bi-infinite sequences the idea of generating series simply does not work: there is no appropriate algebraic framework. Think of it: if there were such a thing as $S=\sum_{i\in\Bbb Z}a_iX^i$ for a sequence $(a_i)_{i\in\Bbb Z}$ satisfying $a_{i+2}=a_i+a_{i+1}$ everywhere, then one would want to express this by saying $S(1-X-X^2)=0$, but $S=0/(1-X-X^2)$ in no way describes a generating series for the Fibonacci numbers (for one thing that expression gives no information about any "initial terms" that were used, for another it just simplifies to $0$ in any reasonable algebraic setting).


问:
Find mean, variance, skewness and kurtosis relations between related $X$ and $Y$ if $(1+k)Y_i = kY_{i-1} + X_i$
I have a simple recurrence relation that relates a sequence of random variables $X_i$ to another sequence $Y_i$ using a constant $k$:

\begin{equation}
Y_i\left(1+k\right) = kY_{i-1} + X_i \quad \mbox{or} \quad Y_i = \left(\frac{k}{k+1}\right)Y_{i-1} + \left(\frac{1}{k+1}\right)X_i
\end{equation}

Assuming that the $X_i$ are independent and identically distributed, and making minimal assumptions about the distribution of the random variable $X$, then I want to know how to relate $\mu_x$ and $\mu_Y$, $\sigma_Y^2$ and $\sigma_X^2$, $\mathrm{skew}\left(X\right)$ and $\mathrm{skew}\left(Y\right)$ and also $\mathrm{Kurt}\left(X\right)$ and $\mathrm{Kurt}\left(Y\right)$.

I have found a reference that suggests taking simple expectations of the recurrence relationship will allow such relationships to be found, and then proceeds to give the relationship between the means, variances and skewnesses. It is relatively straightforward to show that:

\begin{equation}
\left(1+k\right)\mathbf{E}\left[Y\right] = k\mathbf{E}\left[Y\right] + \mathbf{E}\left[X\right] \quad \mbox{so} \quad \mathbf{E}\left[Y\right] = \mathbf{E}\left[X\right], \quad \mu_Y = \mu_X
\end{equation}

and similarly for the variance:
\begin{equation}
\left(1+k\right)^2\mathbf{Var}\left[Y\right] = k^2\mathbf{Var}\left[Y\right] + \mathbf{Var}\left[X\right] \quad \mbox{so} \left(1+2k\right)\mathbf{Var}\left[Y\right] = \mathbf{Var}\left[X\right], \quad \sigma_Y^2 = \left(\frac{1}{1+2k}\right)\sigma_X^2
\end{equation}

But I am not able to see how to derive the expression for the relationship between $\mathrm{skew}\left(X\right)$ and $\mathrm{skew}\left(Y\right)$.  The reference (a published paper) suggests this is:

\begin{equation}
\mathbf{skew}\left[Y\right] = \left(\frac{\left(1+2k\right)^{3/2}}{3k^2+3k+1}\right)\mathbf{skew}\left[X\right]
\end{equation}

but how is this derived? I want to know this because I also want to be able to derive an expression for the kurtosis.

答:
If $Y_i$ is a sequence, you cannot assume that all $Y_i$ have the same distribution, even if you assume that all $X_i$ have the same distribution. in particular, you cannot just write ${\bf E}[Y_i]={\bf E}[Y]$ as you do.

I assume the requrence starts with some $Y_0$ that is not given by the recurssion (because we don\t have $Y_{-1}$).
Then we have
$$ Y_1 = \frac{k}{k+1} Y_0 + \frac{1}{k+1} X_1  $$
$$ Y_2 = \frac{k}{k+1} Y_1 + \frac{1}{k+1} X_2 = \frac{k^2}{(k+1)^2} Y_0 + \frac{k}{(k+1)^2} X_1 + \frac{1}{k+1} X_2 $$
$$ \dots $$
$$ Y_n = \frac{k^n}{(k+1)^n} Y_0  + \sum_{i=0}^{n-1} \frac{k^{i}}{(k+1)^{i+1}} X_{n-i} $$
Assuming that all $X_i$ have the same distribution, we have then
\begin{align} {\bf E}[Y_n] &= \frac{k^{2n}}{(k+1)^{2n}} {\bf E}[Y_0] +\sum_{i=0}^{n-1} \frac{k^{i}}{(k+1)^{i+1}} {\bf E}[X] =  \\
&= \frac{k^n}{(k+1)^n}  {\bf E}[Y_0] +\sum_{i=0}^{n-1} \left(\frac{k}{k+1}\right)^i \frac{{\bf E}[X]}{k+1} = \\
&= \frac{k^n}{(k+1)^n}  {\bf E}[Y_0] +\frac{1-\left(\frac{k}{k+1}\right)^{n}}{1-\frac{k}{k+1}} \frac{{\bf E}[X]}{k+1} = \\
&= \frac{k^n}{(k+1)^n}  {\bf E}[Y_0] +\left(1-\left(\frac{k}{k+1}\right)^{n}\right) {\bf E}[X]  \end{align}
If you assume $ {\bf E}[Y_0] = {\bf E}[X]$ then you indeed obtain ${\bf E}[Y_n] ={\bf E}[X] $, but otherwise it may vary. We can also notice that
$$ \lim_{n\to\infty} {\bf E}[Y_n] = {\bf E}[X] $$
regardless of the starting condition.

Similarily with the variance; assuming that $X_i$ are independent of each other and of $Y_0$ it is posible to show that $X_I$ and $Y_{i=1}$ are independent and we have
$$ {\bf Var}[Y_n] = \frac{k^2}{(k+1)^2}{\bf Var}[Y_n] + \frac{1}{(k+1)^2}{\bf Var}[X]$$
or, after solving the recurrence
\begin{align} {\bf Var}[Y_n] &= \frac{k^{2n}}{(k+1)^{2n}} {\bf Var}[Y_0]  +\sum_{i=0}^{n-1} \frac{k^{i}}{(k+1)^{i+1}} {\bf Var}[X] = \dots = \\
&= \frac{k^{2n}}{(k+1)^{2n}}  {\bf Var}[Y_0] +\left(1-\left(\frac{k}{k+1}\right)^{2n}\right) \frac{{\bf Var}[X]}{1+2k}  \end{align}
And again, if $ {\bf Var}[Y_0] = {\bf Var}[X] $, or in the limit $n\to\infty$, we get
$$ \sigma_{Y_n}^2 = \frac{\sigma_X^2}{1+2k}$$

The skewness (not yet the curtosis, curtosis would be the next step) is defined as
$$ {\bf skew}[Y] = \frac{{\bf E}[(Y-\mu_Y)^3]}{\sigma_Y^3}$$
Using the fact that
\begin{align} {\bf E}[(Y_n-\mu_{Y_n})^3] &= {\bf E}\left[(\frac{k}{k+1}Y_{n-1}+\frac{1}{k+1}X_n-\frac{k}{k+1}\mu_{Y_{n-1}}-\frac{1}{k+1}\mu_X)^3\right] = \\
&= {\bf E}\left[(\frac{k^3}{(k+1)^3}(Y_{n-1}-\mu_{Y_{n-1}})^3+\frac{3k^2}{(k+1)^3}(Y_{n-1}-\mu_{Y_{n-1}})^2(X_n-\mu_X)+\frac{3k}{(k+1)^3}(Y_{n-1}-\mu_{Y_{n-1}})(X_n-\mu_X)^2+\frac{1}{(k+1)^3}(X_n-\mu_X)^3\right] = \\
&= \frac{k^3}{(k+1)^3}{\bf E}[(Y_{n-1}-\mu_{Y_{n-1}})^3]+\frac{3k^2}{(k+1)^3}{\bf E}[(Y_{n-1}-\mu_{Y_{n-1}})^2]{\bf E}[X_n-\mu_X]+\frac{3k}{(k+1)^3}{\bf E}[Y_{n-1}-\mu_{Y_{n-1}}]{\bf E}[(X_n-\mu_X)^2]+\frac{1}{(k+1)^3}{\bf E}[(X_n-\mu_X)^3] = \\
&= \frac{k^3}{(k+1)^3}{\bf E}[(Y_{n-1}-\mu_{Y_{n-1}})^3]+\frac{1}{(k+1)^3}{\bf E}[(X_n-\mu_X)^3]  \end{align}
and using the same method as before we get
\begin{align} {\bf E}[(Y_n-\mu_{Y_n})^3] &= \frac{k^{3n}}{(k+1)^{3n}}  {\bf E}[(Y_0-\mu_{Y_0})^3] +\left(1-\left(\frac{k}{k+1}\right)^{3n}\right) \frac{{\bf E}[(X-\mu_{X})^3]}{1+3k+3k^2}  \end{align}
If we assume $ {\bf E}[(Y_0-\mu_{Y_0})^3] = {\bf E}[(X-\mu_{X})^3]$, then we get $$ {\bf E}[(Y_n-\mu_{Y_n})^3] = \frac{{\bf E}[(X-\mu_{X})^3]}{1+3k+3k^2} $$
so
$$ {\bf skew}[Y_n] = \frac{1}{\sigma_{Y_n}^3}\frac{{\bf skew}[X]\sigma_X^3}{1+3k+3k^2}$$
Using the previous result on $\sigma_{Y_n}$ we get
$$ {\bf skew}[Y_n] = (1+2k)^{\frac32}\frac{{\bf skew}[X]}{1+3k+3k^2}$$

Note that this holds only with the right starting condition for the recurrence or in the limit $n\to\infty$.


问:
Integrability of increasing function
I am trying to prove the following (from Axler's Measure, Integration and Real Analysis):

"Suppose $f:[a,b]\to\mathbb{R}$ is an increasing function (i.e. $c,d\in [a,b], c<d\Rightarrow f(c)\leq f(d)$).
Prove that $f$ is Riemann integrable on $[a,b]$."

Now, I have tried to use the fact that $f$ is integrable if $\forall\varepsilon>0$ there exists a partition $P$ of $[a,b]$ such that $U(f,P,[a,b])-L(f,P,[a,b])<\varepsilon$.

EDITED ACCORDING TO ANSWER BY Fred BELOW:

if for example I fix $\varepsilon>0$ and use the equally spaced partition of $[a,b], x_j-x_{j-1}=\frac{b-a}{n}<\frac{\varepsilon}{f(b)-f(a)+1}$ (such an $n$ exists since $\frac{const}{n}\overset{n\to\infty}{\to}0$) I get

$$U(f,P,[a,b])-L(f,P,[a,b])=\sum_{j=1}^{n}(x_j-x_{j-1})\sup_{[x_{j-1},x_j]}f-\sum_{j=1}^{n}(x_j-x_{j-1})\inf_{[x_{j-1},x_j]}f=
\sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j) -f(x_{j-1}))=\frac{(b-a)}{n}\sum_{j=1}^{n}(f(x_j)-f(x_{j-1}))=\frac{(b-a)}{n}(f(b)-f(a))<\varepsilon\frac{f(b)-f(a)}{f(b)-f(a)+1}<\varepsilon$$

答:
You wrote $\sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j-f(x_{j-1}))=(b-a)(f(b)-f(a)).$ But this is not correct. Correct is:

$$\sum_{j=1}^{n}(\frac{b-a}{n})(f(x_j-f(x_{j-1}))=\frac{b-a}{n}(f(b)-f(a)).$$

Can you procced ?

Title: Integrability of increasing function


问:
At $x_0 \in \partial \Omega$, $\text{Hess}u(x_0) = \Delta u(x_0) (\nu \otimes \nu)$, $\nu$ outer unit normal
Let $x_0 \in \partial \Omega$ and $\nu(x_{0})$ is the exterior unit normal at $x_0$, with $\nu_1(x_0) > 0$, where $u \in C^{2}(\overline{\Omega_{\epsilon}})$ and $\Omega_{\epsilon}=\Omega \cap \{|x-x_0|<\epsilon\}$, $u>0$ in $\Omega$ and $u=0$ on $\partial \Omega \cap \{|x-x_0|<\epsilon\}$. We know that $u_{x_1}(x_0) = 0$. Why does the following hold?

$$
\Delta u = - f(0) > 0 \implies u_{x_{i},x_{j}}=-f(0)\nu_{i}\nu_{j} \quad \text{ at } x_0
$$

Moreover, in their paper, Gidas, Ni and Nirenberg claim, right before what inspires the question, that $\nabla u(x_0) = 0$. It's a mistake, isn't it? Up to my understanding, we only know that $u_1(x_0) = 0$.

This expression appears specifically in the last part of the demonstration of Lemma 2.1 of this article https://projecteuclid.org/download/pdf_1/euclid.cmp/1103905359 of Gidas-Ni-Nirenberg's and I think that it is related to the second directional derivative and Hessian matrix.

答:
It’s not a mistake. Since $u=0$ on the boundary, the tangential derivative of $u$ is zero on the boundary. If $t_1$, $\ldots$, $t_{n-1}$ are the tangent vectors, you can write $$e_1=\sum_{i=1}^{n-1}c_it_i+c_n\nu$$
Taking the inner product with $\nu$ you get that $c_n=\nu_1>0$. Hence $0=u_1=\nabla u\cdot e_1=c_nu_\nu$ at $x_0$ and so also the normal derivative is zero. Hence the gradient is zero.
For the Laplacian you have to do something similar. All the second derivatives in the tangential directions will be zero.


问:
finitely generated subfield of algebraic closure of the finite field with $p$ elements
Let $\mathbb{F}^{\operatorname{alg}}_p$ be the algebraic closure of the finite field with $p$ elements.

I know that any finitely generated subfield of $\mathbb{F}^{\operatorname{alg}}_p $ is contained in some finite sub-union $\displaystyle{\bigcup_{n=1,2,\dots,N} \mathbb{F}_{p^n}}$.

However, I am lacking intuition for why this is so, and I'm having trouble conceiving what a finitely generated subfield of $\mathbb{F}^{\operatorname{alg}}_p $ would look look like.

Though I will continue reading on the matter, any examples of such a finitely generated, and in turn finite, subfield would be greatly appreciated.

答: For typographical simplicity, let’s write $\mathscr F\!_n$ for $\mathbb F_{p^n}$. Then one sees that $\mathscr F\!_m\subset \mathscr F\!_n$ if and only if $m|n$. It follows that $\mathscr F\!_m \mathscr F\!_n=\mathscr F\!_\ell$, where $\ell$ is the least common multiple of $m$ and $n$. Now, since every simply generated subfield of your algebraically closed field is a finite field $\mathscr F\!_m$ for some $m$, it follows that every finitely generated subfield is also simply generated.</p>


问:
Intersection of a exponential and a sin wave
I need to find the intersection between to functions to alter the domain so that they connect at said intersection.

$$0.01^{x-17.7}+7 = \sin\left(\frac {x}{1.3} + 2.6\right) + 8$$

This is for a year 12 mathematics assessment where I have to design a pool using functions. I sketched the graph to begin with in Desmos using estimated intersections but I need to find the actual intersection using mathematics. I'm not sure if this is possible using normal methods but it would be greatly appreciated if anyone could help me out. (sorry the equation doesn't look nice I'm not sure how to work the formatting)

答:
Instead of thinking about equations, it is often better to think about the zero'ss of a function.

Prefering to work with whole numbers and simplifying what in written in the post, let us consider the function
$$f(x)=10^{\frac{177}{5}-2 x}- \left(1+\sin \left(\frac{10 }{13}x+\frac{13}{5}\right)\right)$$

The quantity in parentheses is bounded by $2$. So, we need to explore around $x$ such that
$$10^{\frac{177}{5}-2 x}=2 \implies x=\frac{177}{10}-\frac{\log (2)}{2 \log (10)}\sim 17.5495$$

$f(x)$ varies so fast and produces so large numbers that it is better to consider instead
$$g(x)=\log \left(10^{\frac{177}{5}-2 x}\right)-\log\left(1+\sin \left(\frac{10 }{13}x+\frac{13}{5}\right)\right)$$ which, more or less, looks like a strainght line.

To have an estimate of the solution, make a series expansion around $x=\frac {177}{10}$. This write
$$g(x)=-\log \left(1+\sin
   \left(\frac{1054}{65}\right)\right)+\left(x-\frac{177}{10}\right) \left(-\log
   (100)-\frac{10 \cos \left(\frac{1054}{65}\right)}{13 \left(1+\sin
   \left(\frac{1054}{65}\right)\right)}\right)+O\left(\left(x-\frac{177}{10}\right)
   ^2\right)$$ Neglecting the higher order terms, the solution of the linear equation is then
$$x=\frac {177}{10}-\frac{13 \left(1+\sin \left(\frac{1054}{65}\right)\right) \log \left(1+\sin
   \left(\frac{1054}{65}\right)\right)}{10 \cos \left(\frac{1054}{65}\right)+26
   \log (10) \left(1+\sin \left(\frac{1054}{65}\right)\right)}$$ which is $x\sim 17.9018$ while the "exact" solution is $17.9103$.

I suppose that this is enough.


问:
Problem about weak* convergence.
Let $1 \leq p \leq \infty.$ Consider the sequence $(f_n)$ in $(\ell_p)'$ defined by $f_n(x_1, x_2, x_3,\dots) = x_n.$

(a) If $p < \infty,$ show that $(f_n)$ is weak* convergence, but it does not converge strongly.

(b) If $p = \infty,$ show that $(f_n)$ is bounded but it has no weak* subsequence convergence.

My efforts:

Item (a): Let $x = (x_n) \in \ell_p,$ i.e, $x_n \to 0.$ Therefore $f_n(x) = x_n \to 0, \forall x.$ This follows that $(f_n)$ is weak* convergence to $f \equiv 0.$ But $(f_n)$ is not convergence strongly, why?

Item (b): I hasn't idea.

Can anyone give me an idea?

答:
Clearly, $\|f_n\| \leq 1$ Since $f_n(e_n)=1$ it follows that $\|f_n\|=1$ for all $n$ Hence, $f_n$ does not tend to $0$ strongly. [If it converges is has to converge to $0$ by weak* convergence].

When $p=\infty$ and $n_1<n_2<...$  let $x_{n_k}=1$ for $k$ even, $x_{n_k}=-1$ for $k$ odd and $x_i=0$ when $i$ is not of the form $n_k$. Then $(f_{n_k} (x))$ does not converge.

Title: Problem about weak* convergence.


问:
Find all x and y that make the equation $y(x^2 + y^2 + 1) + 2(3xy + 2) = 2005$ true
The question says: Find all $x,y \in \mathbb R$ that make $y(x^2 + y^2 + 1) + 2(3xy + 2) = 2005$ true.

At first I've simplified the expression:
$$x^2y + y^3 + y + 6xy = 2001$$
$$y(x^2 + y^2 + y + 6xy + 1) = 2001$$

Then I've made prime factorization of $2001$
$$2001 = 3 \cdot 23 \cdot 29$$
$$2001 = 3 \cdot 667$$
$$2001 = 69 \cdot 29$$
$$2001 = 87 \cdot 23$$

And I've checked every possible variant (6 variants) with $y$. Nothing matches. So is the answer that there are no such $x$ and $y$?

Edit: the question doesn't mention that $x$ and $y$ are integers

p.s. English is not my native language so sorry if there are grammar mistakes.

答:
Since $y=0$ is impossible, the equation is equivalent to the quadratic equation
$$
x^2 + 6x + y^2 + 1 - \frac{2001}{y}=0
$$
This has two real solutions, by the usual formula, if and only if
$$
- y^4 + 8y^2 + 2001y \ge 0.
$$
hence the equation has infinitely many real solutions. For example, take $y=1$ and $x=\pm 2 \sqrt{502} - 3$, and then $y=1+\epsilon$ with infinitely many small $\epsilon \in [0,1]$.


问:
What did I do wrong with this integral problem?

Find the value of the following definite integral
$$\int_0^{\pi/2} \sqrt{1 - \sqrt{3}\sin 2x + 2\cos^2x} \, dx$$

My attempt:

$$\int_0^{\pi/2} \sqrt{1 - \sqrt{3}\sin 2x + 2\cos^2x} \, dx$$
$$= \int_0^{\pi/2} \sqrt{2 - \sqrt{3}\sin 2x + \cos2x} \, dx$$
$$= \int_0^{\pi/2} \sqrt{2 + 2\cos \left(2x + \dfrac{\pi}{3}\right)} \, dx$$
$$= \sqrt{2} \int_0^{\pi/2} \sqrt{1 + \cos \left(2x + \dfrac{\pi}{3}\right)} \, dx$$

Let $u = 1 - \cos \left(2x + \dfrac{\pi}{3}\right) \implies du = 2\sin \left(2x + \dfrac{\pi}{3}\right) \, dx = 2\sqrt{2u - u^2} \, dx \implies dx = \dfrac{1}{2\sqrt{2u - u^2}} \, du$

The integral become:

$$\sqrt{2}\int_{1/2}^{3/2} \dfrac{\sqrt{2-u}}{2\sqrt{2u - u^2}} \, du$$
$$= \sqrt{2}\int_{1/2}^{3/2} \dfrac{1}{2\sqrt{u}} \, du$$
$$= \sqrt{2} \cdot \left(\sqrt{u}\right) \Biggr|_{1/2}^{3/2} = \sqrt{3} - 1$$

However, using my calculator and other online calculators, they're all giving $3 - \sqrt{3}$, which is different from mine.

So what did I do wrong?

A small realization I found out is that:

$$\text{(Calculator answer) = $\sqrt{3} \cdot$ (My answer)}$$

答:
$$\int_0^{\pi/2}\sqrt{1-\sqrt{3}\sin(2x)+2\cos^2(x)}dx$$
then as you did we know that:
$$\cos(2x)=2\cos^2(x)-1\Rightarrow 2\cos^2(x)=\cos(2x)+1$$
so we have:
$$\int_0^{\pi/2}\sqrt{2-\sqrt{3}\sin(2x)+\cos(2x)}dx$$
try substituting $u=2x\Rightarrow dx=du/2$ so:
$$\int_0^\pi\sqrt{2-\sqrt{3}\sin(u)+\cos(u)}\frac{du}2$$
now set up
$$\sqrt{3}\sin(u)+\cos(u)=A\cos(u+\alpha)$$
solving gives $2\cos\left(x-\frac\pi3\right)$
now we have:
$$\frac1{\sqrt{2}}\int_0^\pi\sqrt{1-\sin\left(u-\frac\pi3\right)}du$$
letting $v=u-\frac\pi3\Rightarrow dv=du$ we get:
$$\frac1{\sqrt{2}}\int\limits_{-\pi/3}^{2\pi/3}\sqrt{1-\sin v}\,dv$$


问:
Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$.
Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the family are drawn on the same picture, one gets (click to enlarge):

What I'm calling "pillars" make their appearance. They are located on precise abscissas $x_{c/d}$ which are algebraic numbers, roots of $X^c-(1-X)^d$ polynomials for some positive rational $c/d$ (in irreducible form). To each pillar is associated a vertical wavelength (clearly visible on the picture) which is equal to $-\frac{\ln(x_{c/d})}{d}=-\frac{\ln(1-x_{c/d})}{c}$. I compared the wavelength of the $(c/d=5)$ pillar (for which $x_5$ is the multiplicative inverse of $\rho$, the plastic number) with the wavelength of the $(c/d=3/2)$ pillar (marked green), and found numerically that they were equal. So, the asked question is to prove that, in any of its forms:

$$ -\frac{\ln(x_{3/2})}{2} = -\frac{\ln(x_{5})}{1} $$

$$ x_{3/2} = x_{5}^2 $$

the real root of $X^3-(1-X)^2$ is the square of the real root of $X^5-(1-X)^1$

the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$

答:
We will prove the last statement.

Let $\alpha$ be a root of $p_1:= X^3 + X^2-1$, i.e., $\alpha^3 +\alpha^2-1 = 0$. We will prove that $\beta := \alpha^2$ is a root of $p_2:= X^3-X^2+2X-1$.

We have the following identities:

$$\beta = \alpha^2,$$
$$\beta^2 = \alpha^4 = \alpha\cdot \alpha^3 = \alpha (-\alpha^2 +1) = -\alpha^3 + \alpha,$$
$$\beta^3 = \alpha^6 = (\alpha^3)^2 = (-\alpha^2+1)^2 = \alpha^4 -2\alpha^2 +1 = -\alpha^3 -2\alpha^2 + \alpha +1.$$

Then,

$$\beta^3 - \beta^2 +2\beta -1 = (-\alpha^3 -2\alpha^2 + \alpha +1) - (-\alpha^3 + \alpha) +2(\alpha^2) -1 = 0.$$

Thus $\beta$ is a root of $p_2$. Note that we did not require $\alpha$ to be real, this means that the roots of $p_2$ are precisely the squares of the roots of $p_1$. Particularly, the square of the real root of $p_1$ is a (the) real root of $p_2$.


问:
Solving $\tan(x) + x = \frac{\pi}{2}$
$$\tan(x) + x = \frac{\pi}{2}$$

I want to find the solution where

$$0 < x < 1$$

EDIT: Why there are no closed form solution?
Maybe sin(...) or cos(...) or tan(...)

For example, the equation
$$64x^6 -112x^4 + 56x^2 - 7 = 0$$

have a simple solution

$x = sin\left(\frac{\pi}{7}\right)$

答:
Using basic inspection, you know the the root is close to $\frac \pi 4$.

To have an approximation, use Taylor series around this value to get
$$\tan(x)+x-\frac \pi 2=\left(1-\frac{\pi }{4}\right)+3 \left(x-\frac{\pi
   }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^2\right)$$ and a first estimate is
$$x=\frac{\pi -1}{3}\sim 0.713864$$ which is not so bad (at this point, the expression is $\sim 0.00934$.

Now, use a numerical method or approximation. For the fun of it, I give you one
$$x\sim\frac{-2012160+1477632 \pi +256128 \pi ^2+34752 \pi ^3+3048 \pi ^4+192 \pi
   ^5}{5004288+863232 \pi +119808 \pi ^2+9792 \pi ^3+768 \pi ^4}$$ which is $0.710462753$ (very close to @Gary's accurate result).

Edit

Using an high order series expansion and continuing with series reversion, we end with
$$x \sim \frac \pi 4+\sum_{n=1}^\infty a_n\, t^n \quad \text{where} \qquad t=\frac {\pi -4}{12}$$ and the first $a_n$ are
$$\left\{1,-\frac{2}{3},0,\frac{10}{27},-\frac{56}{405},-\frac{292}{1215},\frac{272}{
   1701},\frac{382}{1701},-\frac{7288}{32805},-\frac{769396}{3444525}\right\}$$ Using it, then
$$x \sim \color{blue}{0.7104627377755}93$$

Title: Solving $\tan(x) + x = \frac{\pi}{2}$


问:
Decide if it is possible, by successively switching the place of rows or columns, to reach to the matrix$ −A$.

Let A be an n × n matrix whose entries are 0,1,−1 and each row and column has exactly one 1 and one −1. Decide if it is possible, by successively switching the place of rows or columns, to reach to the matrix −A.

i don't which method i can chose to use, please give a hint or answer.

答:
The basic move is as follows:

Given
$A = \left(\begin{array}{rr} 1 & 0 \\ 0 & -1\end{array}\right)$.
Switching rows and then columns provides $-A = \left(\begin{array}{rr} -1 & 0 \\ 0 & 1\end{array}\right)$.

Since each row and column has only one entry $\pm 1$, this basic move iterated leads from $A$ to $-A$.


问:
How to use induction to prove this argument?

It is obvious that this grammar will always return an equal number of both a's and b's. But I was wondering how to prove it using induction? I understand induction, but I was finding it hard to apply to this situation.

The ε character denotes an empty.

答:
Let $P(n)$ be the assertion that if $w\in\{S,T,a,b\}^*$ can be derived in $n$ steps, then $|w|_a=|w|_b$, where $|w|_x$ is the number of occurrences of the letter $x$ in $w$.

Initially, when you begin a derivation, you have the string $S$, and $|S|_a=0=|S|_b$, so $P(0)$ is true. Suppose that $P(n)$ holds for some $n\ge 0$, and let $w\in\{S,T,a,b\}^*$ be derivable in $n+1$ steps. Suppose that the last step is $v\Rightarrow w$. Then $v$ is derivable in $n$ steps, so $|v|_a=|v|_b$. Moreover, $v$ must be of the form $xSy$ with $x,y\in\{S,T,a,b\}^*$ or of the form $xTy$ with $x,y\in\{S,T,a,b\}^*$, since a further step is possible. Now see if you can use the induction hypothesis $P(n)$ to finish the induction step by showing that $|w|_a=|w|_b$; I’ve done so below, in the spoiler-protected block.

 If $v=xSy$, then $w=xy$, or $w=xTSy$; in either case $|w|_a=|v|_a=|v|_b=|w|_a$. If $v=xTy$, then $w=xy$ or $w=xaTby$. The first possibility has already been covered, and in the second we have $|w|_a=|v|_a+1=|v|_b+1=|w|_b$. In all cases, then, $|w|_a=|w|_b$, so $P(n+1)$ holds. By induction $P(n)$ holds for all $n\ge 0$. In particular, all terminal $w$ strings produced by this grammar satisfy $|w|_a=|w|_b$, which is the desired result.


答:
Solving your equation is the same as founding the zeros of the function
$$f(x)=0.01^{x-17.7}- \sin\left(\frac {x}{1.3} + 2.6\right) - 1$$
You have already used desmos to estimate a root around $x_0=17$

Thus you can use Newton method setting

$x_0=17$

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)};\;n>0$

where $f'(x)=-4.60517 \cdot0.01^{x-17.7}-0.769231 \cos (0.769231 x+2.6)$

after some steps you get $x^*\approx 17.91$

There is another solution around $x_0=19$

repeat the procedure until you find $x^{**}\approx 18.9875$$

There are infinitely many because $\sin\left(\frac {x}{1.3} + 2.6\right) + 1$ is periodic and oscillates between $0$ and $2$ and exponential tend to zero when $x\to\infty$.

Title: Intersection of a exponential and a sin wave

Body:
I need to find the intersection between to functions to alter the domain so that they connect at said intersection.

$$0.01^{x-17.7}+7 = \sin\left(\frac {x}{1.3} + 2.6\right) + 8$$

This is for a year 12 mathematics assessment where I have to design a pool using functions. I sketched the graph to begin with in Desmos using estimated intersections but I need to find the actual intersection using mathematics. I'm not sure if this is possible using normal methods but it would be greatly appreciated if anyone could help me out. (sorry the equation doesn't look nice I'm not sure how to work the formatting)


问:
Stochastic approximation system with different step sizes
I'm having an issue solving a stochastic approximation scheme. I have a finite set of time-dependent variables $$x^t=(x_1^t, x_2^t,..., x_n^t)$$ and I can express them as a system in the form

$$x_i^{t+1}=x_i^t+\gamma_i^t(f(x^t)+u_i^t)$$

My problem is that the value of $\gamma_i^t$ changes from one variable to the other in a way that can't be scaled (e.g. $\gamma_i^t=\frac{1}{a_i \cdot t + 1}$ with $a_i$ a constant). I'm quite convinced it converges anyway but I can't find any reference on why. In Kushner & Yin 2003, I could not find a section corresponding to what I'm looking for, but it may very well be because I don't have the right keyword. It would be a great help if you have any reference on why a system with different time-steps converge!

EDIT: Let me probably add that contrary to two-time scale environments where one would have $\gamma_i^t/\gamma_j^t \to 0$, my step sizes are of the same order and the ratio converges to a constant.

答: Nevermind the question, theorem 2.3 chapter 5 of Kushner & Yin 2003 provides the answer.</p>


问:
Determine if $ϕ$ has a finite and infinite model or not.

Let $\sigma = \{E\}$ be a signature with a two-digit relation symbol
$E$. Determine for the following $FO-[\sigma]$ formula, whether it has
a finite model and whether it has an infinite model.

$$ϕ := ∀x∃y(E(x, y) ∧ ∀z(z \neq y → ¬E(x, z))) ∧ ∀x∀y∀z(E(x, z) ∧ E(y,
> z) → x = y) ∧ ∃x∀y¬E(y, x)$$

I am quite frankly lost and not sure what to do here. I know that if $ϕ$ has an arbitrarily large number of finite model, then it has an infinite model. So first I need to determine if $ϕ$ has a model - which would imply that $ϕ$ has a finite model. Then I have to determine if $ϕ$ has an arbitrarily large number of finite models - which would imply that $ϕ$ has an infinite model. But that is where I am not sure what to do. How can I show that $ϕ$ has a model and subsequently that it has an arbitrarily large number of models?

答:
First, a correction: Having a model does not imply having a finite model. There are formulas having only infinite models (try describing how $<$ works on $\mathbb{N}$).

Then, while it is true that having finite models of arbitrarily large cardinalities implies having an infinite model, this is very rarely the way to go to prove that a concretely given formula has an infinite model.

The most basic way to prove that a formula has a finite/infinite model is to just exhibit one. So have a look at what your formula says, and try to construct a model. You can assume without limitation of generality that the underlying set is either $\{0,1,\ldots,n\}$ or $\mathbb{N}$. Then you only need to say how $E$ works.

If you get stuck trying to built a finite model, turn the obstacle you keep running into into a proof that any model has to be infinite.

If you get stuck trying to built an infinite model, then it may help finding out the maximal size of a finite model, and to argue that no larger models can exist.


问:
Finding limit of [x]{x} when 'x' tends to 0.
I calculated $\lim_{x\to 0}$ [x]{x} as follows ({x} and [x] mean fractional part function and greatest integer function respectively):-
$$\lim_{x\to 0} [x]\{x\}$$
To find the right hand limit, I replaced 'x' by $0+h$ where $h>0$ and ${h\to 0^+}$. Hence we get,
$$\lim_{h\to 0^+} [0+h]\{0+h\}$$
As ${h\to 0^+}$ and $h>0$,${0+h\to 0}$ and hence $[0+h]=0$.

And again, as ${h\to 0^+}$, $\{0+h\}=h$. So we get the right hand limit as,
$$\lim_{h\to 0^+} (0*h)=0$$

To find the left hand limit I replaced 'x' by $0-h$ where $h>0$ and ${h\to 0^-}$. Hence we get,
$$\lim_{h\to 0^-} [0-h]\{0-h\}$$
Now we will get $[-h]=-1$ because $h>0$ and ${h\to 0}$.

Also, $\{-h\}=1-h$. So we get the left hand limit as,
$$\lim_{h\to 0} (-1\cdot(1-h))=-1$$

This means RHL and LHL are not equal and hence the limit of the function doesn't exist when ${x\to 0}$.But the correct answer is that it does and is equal to $0$. Where did I go wrong?

答:
What you have done is absolutely correct. Another way to see this is that $\{x\} = x - [x]$ and thus, your limit is simply $$\lim_{h \to 0}\left(h[h] - [h]^2\right).$$
The limit of the first term exists since $h \to 0$ and $[h]$ is bounded.

However, the limit of the second term does not exist as $[h]$ is $-1$ on $(-1, 0)$ whereas it is $0$ on $(0, 1)$.

Thus, the overall limit does not exist.

Here's a graph (courtesy of Desmos):


问:
How to calculate R-square from adjusted r-square, n, and p?
Let $\bar{R}^2$ denote the adjusted coefficient of determination.

I have $\bar{R}^2 = 0.9199$ with 15 cases. Now I am trying to find $R^2$ given the results below.

I found the formula for $R^2$ but did not understand it. How do you calculate $R^2$ from $\bar{R}^2$?

$\bar{R}^2 = 1-\dfrac{(n-1)(1- R^2)}{n-p-1}$

答:
Given the equation for $\bar{R}^2$ you have:

$\bar{R}^2 = 1-\dfrac{(n-1)(1- R^2)}{n-p-1}$

You have $3$ regressors and a sample of $15$, thus substituting these and $\bar{R}2$ into the equation yields:

$0.9199 = 1 - \dfrac{(15-1)(1-R^2)}{15-3-1}$

Rearranging this expression and solving for $R^2$ gives:

$R^2 = 0.9371$


问:
How to expand $(1+x+b)^2$ correctly?
I'm trying to expand this trinomial equation and wish for your support on my solution.

I'm trying to prove: given $y = (1+x)^n$, then $\frac{dy}{dx}=n(1+x)^{n-1}$

By starting likeso: $(1+x+\nabla x)^n$ then expanding and cancelling when taking the derivative $\frac{\nabla dy}{\nabla dx}$

Although, to do this I need to know how to do trinomial expansions. Here is my effort with an example:

$(1+x+b)^2=[1+2x^{2-1}+\frac{2(2-1)}{2!}x^2] + [1+2b^{2-1}+\frac{2(2-1)}{2!}b^2]+[x+2b^{2-1}x+b^2]$.

I tried following a similar procedure as if it were determinant expansion (i.e. (1+x), (1+b), (x+b)), although I'm sceptical as to whether this is true.

The purpose was to figure this out, then perform the same expansion with $(1+x+\nabla x)^n$ to derive the derivative, otherwise if this way is not possible an alternative suggestion is helpful.

答:
A simple, straightforward way to expand a trinomial is to think of it as a "nested" binomial:

$$\begin{align}
(1+x+b)^2&=((1+x)+b)^2\\
&=(1+x)^2+2(1+x)b+b^2\\
&=1+2x+x^2+2b+2xb+b^2
\end{align}$$

after which you can regroup terms however you like or need. For the purpose of taking derivatives, it might best to leave things at the $(1+x)^2+2(1+x)b+b^2$ stage though, since that gives

$${(1+x+b)^2-(1+x)^2\over b}={(1+x)^2+2(1+x)b+b^2-(1+x)^2\over b}=2(1+x)+b$$

Can you see from this how to approach the more general problem of differentiating $y=(1+x)^n$?

Additional remark: I'm not sure what's behind your expansion of $(1+x+b)^2$, but the result is wrong. Simplifying the right hand side gives

$$[1+2x+x^2]+[1+2b+b^2]+[x+2bx+b^2]=2+3x+x^2+2b+2bx+2b^2$$

which is not correct. If you don't see what you did wrong in getting the expansion you did, that might be worth a separate question.

Title: How to expand $(1+x+b)^2$ correctly?


问:
Sum of lower semicontinuous functions
I'm a little confused about when the sum of two lower semicontinuous functions is continuous. I couldn't find a neat answer to my question on this site, though there are similar ones.

Say that $f: \mathbb R^n \to [-\infty,\infty]$ is lsc iff $x_n \to x$ implies
$$\liminf_n(f(x_n)) \geq f(x).$$

Now suppose $f:\mathbb R^n \to [0,-\infty]$ and $g: \mathbb R^n \to (-\infty,0]$ are both lsc.

Is $f+g$ lsc as well?

An affirmative answer would follow immediately from
$$\liminf_n(y_n + z_n) \geq \liminf_n(y_n) + \liminf_n(z_n)\tag{1}$$
for any sequence $(y_n)$ in $[0,\infty]$ and sequence $(z_n)$ in $(-\infty,0]$. For any $m \in \mathbb N$, we have
$$y_m + z_m \geq \inf_{n \geq m} y_n + \inf_{n \geq m} z_n,$$
which implies
$$\inf_{m \geq n}(y_m + z_m) \geq \inf_{n \geq m} y_n + \inf_{n \geq m} z_n.$$
Then (1) follows by letting $m \to \infty$.

I don't see what's wrong with that argument, and yet the following counterexample seems to hold as well:
$$y_n = n, \ \ z_n = -n.$$

Can someone please explain what's going on here?

答:
It seems you're saying that $0 \ge \infty+(-\infty)$ is false. Some might call it true. For any $m$, you have $\inf_{n \ge m} (y_n+z_n) = 0$, $\inf_{n \ge m} y_n = m$, and $\inf_{n \ge m} z_n = -\infty$. So indeed $$\inf_{n \geq m}(y_n + z_n) \geq \inf_{n \geq m} y_n + \inf_{n \geq m} z_n$$ $$0 \ge m+(-\infty)$$ is true. If you take the limit as $m \to \infty$, you'll get $0 \ge \infty+(-\infty)$, which we can say is true for our purposes.

So your question comes down to how you define addition on the extended reals.

Title: Sum of lower semicontinuous functions


问:
Number of subgroups of order $14$ in $\Bbb Z_{70} \times\Bbb Z_{98}$
I have to find the number of subgroups of order $14$ in $\Bbb Z_{70} \times \Bbb Z_{98}$.

I can find the number of elements of order $14$. I can then find the number of cyclic subgroups by dividing the number of elements of order $14$ by $6$.

How can I find the number of non-cyclic groups?

答:
The number of non-cyclic such subgroups is $0$: an abelian group of order $14$ is cyclic (by the structure theorem, CRT or some other simple arguments).

An element $(a,b)$ has order $\rm{lcm}(|a|,|b|)$.  So we get the possibilities $(2,7),(7,2),(2,14),(14,2),(7,14),(14,7),(1,14),(14,1)$ and $(14,14)$.  Note that in a cyclic group there are $\varphi(2)=1, \varphi(7)=6$ and $\varphi(14)=6$ elements of order $2,7$ and $14$ respectively.

Thus we get $6+6+6+6+36+36+6+6+36=144$.  But to finish we need to divide by $\varphi(14)=6$, since there are that many elements  of order $14$ in each such subgroup.  Thus we get $24$.


问:
Categorize circle on a plane with "left" and "right" relative to a line.
CONTEXT:

I'm doing a simulation in Object Oriented Programming. I have a bunch of circles with properties:

Location = Vector with x and y coordinate

Angle = Vector détermining "where" the circle is looking at

DetectionRadius = A bigger circle, we want to interact only with others circle located within that circle.

So i have my central circle and secondary circle all around the plane. I have access to all circle properties in an array. My question is:

PROBLEM:

How can i determine wich circle is on the left side of my central circle, and wich are on the right side ?

Here is an image to understand better what i mean: Image1

(Principal circle is in the center, black line is the angle vector (we can extend it to a line if needed), blue circle is the detection radius and other green circle are secondary circles)

I already know wich secondary circle are in the principal circle's detection radius. I just need to know wich are on the left, wich are on the right.

Thank you !

答:
Form the equation of the line. Then the question on which side of the line a point is can be answered by calculating the value of the defining (linear) function of the line. On the line its value will be $0$. The value is positive/negative if the point is on the left/right (when looking to the direction of the vector defining the line).

In this case we get the equation of the line as (notice the coefficients make a 90 deg rotation of the given direction vector, since to be on the line means the inner product with its normal vector be zero, we also need a translation from the point $C$ to the origin in there)

$$l(x,y) = -\sin(\alpha) (x-c_x) + \cos (\alpha) (y-c_y)$$

where $\alpha$ is the angle and $(c_x, c_y)$ is the center of the central circle. Try it out here.


问:
Eigen decomposition of Householder operator
You probably know both "Householder" and "Eigendecomposition", but do you really know how it works geometrically?
I got stuck in  question asks me to provide the eigen decomposition of householder operator H=I-2ww* where ‖w‖=1, geometrically!
I know that the eigen-values are [-1, 1 ...1] but I cannot understand about the eigen-vectors and in particular I cannot depict the solution, geometrically!

Can anybody possibly help me to understand it through drawing a geometrical interpretation?

Thanks

答:
Remark that if $v$ is orthogonal to $w$, then $H \cdot v = v$. So geometrically, a whole hyperplane is left invariant.
Now, also remark that $H \cdot w = -w $. In other words, $w$ gets reflected.

Putting those two together, $H$ is the orthogonal symmetry with respect to the hyperplane orthogonal to $w$.

And indeed, $H$ is an orthogonal symmetry (i.e. $H$ is self adjoint and $H^2 = I$).


问:
expected value for min estimated entropy
We have a random generator that generates independent random bits with probability $P(x=1) = P$ and $P(x=0)=1-P$.

Given $N$ random independent bits, we estimate $P$ by $\hat{P} = N_1/(N_0+N_1)$.
where $N_0$ is the number of $0$'s and $N_1$ is number of $1$'s.
The expected value for $\hat{P}$ can be simply shown to be $P$.

a. What is the expected value for the estimated entropy defined as following:
 $\hat{H}=-[ \hat{P} \log(\hat{P}) + (1-\hat{P}) \log(1-\hat{P}) ]$

b. If we take $M$ independent sets of $N$ random bits as above and each time estimate the entropy using the above equation, What is the expected value for the smallest estimated entropy among those $M$ sets ?

thanks,
MG

P.S. If the solution to the integral for general P is too complicated, a solution to the special case $P=1/2$ would be appreciated as well.

答:
As noted in the comments, there is no general exact formula for $\langle \hat H_N\rangle$ but, for large $N$, one can approach it by a $\chi^2$-type limit.

To wit, $N_1=pN+v\sqrt{N}Z_N$ with $v^2=p(1-p)$, $\langle Z_N\rangle=0$ and $\langle Z_N^2\rangle=1$ for every $N$ and, when $N\to+\infty$, $Z_N$ converging in distribution to a standard normal random variable.

Hence $\hat P_N=p+U_N$ and $1-\hat P_N=q-U_N$ with $q=1-p$ and $U_N=vZ_N/\sqrt{N}$, and
$$
\hat H_N=-(p+U_N)\log(p+U_N)-(q-U_N)\log(q-U_N).
$$
Using the expansions
$$
\log(p+U_N)=\log(p)+U_N/p-U_N^2/(2p^2)+o(U_N^2),
$$
and
$$
\log(q-U_N)=\log(q)-U_N/q-U_N^2/(2q^2)+o(U_N^2),
$$
one gets
$$
\hat H_N=-p\log(p)-q\log(q)+U_N\log(q/p)-U_N^2/(2pq)+o(U_N^2).
$$
Since $\langle U_N\rangle=0$, $\langle U_N^2\rangle=v^2/N$ and $v^2=pq$,
one gets
$$
\langle \hat H_N\rangle=-p\log(p)-q\log(q)-1/(2N)+o(1/N).
$$


问:
faithful irreducible representation of $A_{4} \times Q_{8}$
Construct a faithful irreducible representation of the group $A_4 \times Q_8$

$A_{4}$ is the alternating group
$Q_{8}$ is the quaternions

Body:
Okay, here you go.  The four generators of that group map to the following matrices in $\operatorname{GL}_{12}$.

$$\left(\begin{array}{rrrrrrrrrrrr}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \end{array}\right)$$

$$\left(\begin{array}{rrrrrrrrrrrr}
-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right)$$

$$\left(\begin{array}{rrrrrrrrrrrr}
0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{array}\right)$$

$$\left(\begin{array}{rrrrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
-1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \end{array}\right)$$


问:
Projective modules restricted to smooth curves
I want to prove a coherent sheaf $M$ on $X$ is locally free if and only if this is true for $M|_{X'}$ ,
for all smooth curves $X'$ mapping to $X$. I think the only if direction is obvious. For the if direction, a coherent module is flat if and only if it is projective, for Dedekind domains if and only if torsion free as well. So I am thinking of using Tor$_1$. There is local criterion for flatness, but I am not sure if this will help.

This is used at the beginning of the proof of Proposition 5.13 of Gaitsgory's lecture notes.

答:
In Borel's book "Algebraic D-modules" the following result is proved:

Propositon VI.1.7 Let $X$ be a regular scheme of finite type over an algebraically closed field $k$ of characteristic zero. If $M$ is a $D_X$-module that is coherent as $\mathcal{O}_X$-module, then $M$ is locally free.

It seems you do not need the connection to be flat for the result to hold.
It also seems to hold over any field $k$ of characteristic zero.


问:
What is the formula for the nth number in this series.
$1,3,7,9,11,13,17,19,21\dots$

Basically all the numbers that end in the digits $1,3,7,9$

I am working on a formula for approximating how many factors I have to test to find if a large number is prime. So for example to test the number $229,597$. How many possible factors will I have to check?

So the convention is to take the square root of $229,597$ which is approx $479$. Then I take $(\frac{479}{10}) \times (4) - 1$. I do this because out of every ten numbers there are 4 numbers that end in $1,3,7,9$. I subtract the 1 because prime numbers also have 1 as a factor.

So   $(\frac{479}{10}) \times (4) - 1  = 190$ factors to check to see if $229,597$ is prime.

Then I look at the series $1,3,7,9,11,13,17\dots$ to start checking for possible factors. But how do I find the nth number in this series?

答:
This is another possible formula :: Numbers that are odd but not divisible by 5.

here is small python code can be used to generate nth number. Yu can run and see the output https://www.ideone.com/dm9SPj

```<code>def getNthTerm(n):
    return n*2 + 2 * ((n-3)//4) + 1
def main():
    n=input()
    print getNthTerm(n)
main()
</code>```

Body:
This is another possible formula :: Numbers that are odd but not divisible by 5.

here is small python code can be used to generate nth number. Yu can run and see the output https://www.ideone.com/dm9SPj

def getNthTerm(n):
    return n*2 + 2 * ((n-3)//4) + 1
def main():
    n=input()
    print getNthTerm(n)
main()


问:
Does there exist a topological space Y containing more than one point such that any function $f : X\to Y$ is continuous?
I found this question in one of my past question papers of the college.

Let $X$ be any topological space. Does there exist a topological space $Y$ containing more
than one point such that any function $f : X\to Y$ is continuous? Prove your claim or give
an example of such a topological space.

I feel like this should not be true but I am not able to think of a counterexample. Any help would be appreciated

答: Let $Y$ be any set, and give it the topology in which the only two open subsets of $Y$ are $\emptyset$ and $Y$. For any $f: X \to Y$, we see that $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(Y) = X$. These are both open in $X$, so that $f$ is continuous. This worked for arbitrary functions $f: X \to Y$.


答:
The statement is true, hence it is unspurprising that you could not prove it wrong.

In fact, for every set $Y$ there is a topology $\tau$ on $Y$ such that for every topological space $\mathbf{X}$ all functions $f : \mathbf{X} \to (Y,\tau)$ are continuous.

To prove this, take note that continuity does not care about points at all, but rather about open sets. In particular, the potential obstacles to $f : \mathbf{X} \to (Y,\tau)$ are all coming from the open sets $U \in \tau$.

Title: Does there exist a topological space Y containing more than one point such that any function $f : X\to Y$ is continuous?

Body:
I found this question in one of my past question papers of the college.

Let $X$ be any topological space. Does there exist a topological space $Y$ containing more
than one point such that any function $f : X\to Y$ is continuous? Prove your claim or give
an example of such a topological space.

I feel like this should not be true but I am not able to think of a counterexample. Any help would be appreciated


问:
$K$-theory of union of smooth curves.
Is there a method that one can calculate algebraic $K$ theory of a number of smooth curves with some singularity points corresponding to the intersection points of two different curves, given we know the $K$ groups of each curve? (For example bunch of projective lines with some intersecting points)

答:
Question: "Is there a method that one can calculate algebraic K theory of a number of smooth curves with some singularity points corresponding to the intersection points of two different curves, given we know the K groups of each curve?"

Answer. Let $C:=C_1 \cup C_2$ and let $Z:=C_1 \cap C_2$ with $U:=C-Z$ the open complement. There is a long exact localization sequence

$$ \cdots \rightarrow K_{i+1}(U) \rightarrow K_i(Z) \rightarrow K_i(C) \rightarrow K_i(U) \rightarrow K_{i-1}(Z) \rightarrow \cdots $$

hence if you know the K-theory of the complement $U$ and $Z$ then you can calculate $K_i(C)$. If $Z:=\{p_1,..,p_l\}$ is the $l$ singular points on $C$, let  $C'_1:=C_1-\{p_1,..,p_l\}$ and $C'_2:=C_2-\{p_1,..,p_l\}$.
It follows $U = C'_1 \cup C'_2$ is the disjoint union of the two smooth curves $C_1',C_2'$. Hence $K_i(U) \cong K_i(C_1') \oplus K_i(C'_2)$.
You get the long exact sequence

$$ \cdots \rightarrow K_{i+1}(C_1')\oplus K_{i+1}(C_2') \rightarrow K_i(Z) \rightarrow K_i(C) \rightarrow K_i(C_1')\oplus K_i(C_2') \rightarrow K_{i-1}(Z) \rightarrow \cdots $$

For algebraic K-theory there is the long exact localization sequence and the Mayer-Vietoris sequence. A systematic use of the MV sequence sequences can be used to calculate the algebraic K-theory $K_i(X)$ of $X$ in terms of an open cover $U_i$ of $X$. I believe this holds for singular curves but do not have a precise reference.

Title: $K$-theory of union of smooth curves.


问:
On using Fourier transforms to solve the root of a convolution
In continuation of Lower bounds of laplace transform of characteristic functions.

My question is:

Can anyone point out where i'm going wrong in the derivation below.

It's been a while since I've used transforms and i know I'm missing something as the product of the derivation does not agree with direct calculation.

Thanks in advance.

We wish to solve $K = f(x;a)$ where $K$ is some complex number and
\begin{equation}
f(x;a) = \int_{-\infty}^\infty h(x-\omega;a)g(\omega) \text{d}\omega
\end{equation}
with $g$ some probability distribution, $a,x$ are real numbers and
\begin{equation}
h(x;a) = \frac{1}{a+ i x},\quad a>0
\end{equation}

We let $\bar{\varphi}(t)= \hat{g}(t)$ and calculate the Fourier transform of $h$
\begin{equation}
\hat{h}(t) = 2\pi e^{-at}\Theta(t),
\end{equation}
where $\Theta(t)$ is the Heaviside step function. Thus $\mathcal{F}[K](t) = \mathcal{F}[f](t) $ gives
\begin{equation}
2\pi K\delta(t) =2\pi e^{-at}\Theta(t)\bar{\varphi}(t) \implies \frac{K}{\bar{\varphi}(t)}\delta(t)=e^{-at}\Theta(t)
\end{equation}
Taking the IFT we get
\begin{equation}
\frac{K}{\bar{\varphi}(0)} = \int_0^\infty e^{-(a-ix)t}\text{d}t
\end{equation}
Which gives $\frac{1}{a-ix} = \frac{K}{\bar{\varphi(0)}}$ thus
$$a+ix = \frac{\varphi(0)}{\bar{K}}$$

However this does no agree with direct calculation.
Take the Cauchy distribution with median $0$ $g(\omega;\sigma) = \frac{\sigma}{\pi(\sigma^2+\omega^2)}$ and its analytic continuation $G(\omega;\sigma) =\frac{1}{\pi(\sigma-i \omega)}$
$$f(x;a) =i\int_{-\infty}^\infty \frac{g(z)}{z - i(a +ix)}dz = \pi G(ia-x)$$
by the Schwarz Integral formula.
Which implies $f(x;a) =\frac{1}{x+ia + \sigma}$.
Thus if $f(x;a) = K$, then
$$ x+ia = \frac{1}{K} - \sigma$$
and from the section above, this is not equal to $x+ia = \frac{\varphi(0)}{\bar{K}} = \frac{1}{\bar{K}}$

答:
Then you pass to
$$
2\pi K\delta(t) =2\pi e^{-at}\Theta(t)\bar{\varphi}(t)
$$
it means that you are solving equation
$$
K\equiv \int_{-\infty}^\infty h(x-\omega;a)g(\omega) \text{d}\omega
$$
and it's not that is required. The Fourier transform is not applicable in this way
since the lhs is not given, $K$ being just a value of rhs at some point $x$.</p>


答:
Barry Cipra's answer already gives a great explanation, but I want to provide a slightly different way of expanding brackets which I find to be slightly more elegant, and which will allow you to generalise to $(1+x)^n$ more easily.

In general, suppose you want to expand brackets of the general form $(a_0+a_1+...+a_m)(b_0+b_1+...+b_n)(c_0+c_1+...+c_p)$. The way to get the result is just to multiply together every combination of numbers $a_i \times b_j \times c_k$, and then sum the results. So your result will look something like $a_0 b_0 c_0 + a_0 b_0 c_1 + ... + a_0 b_0 c_p + a_0 b_1 c_0 + ... + a_m b_n c_p$

To understand why, imagine you know the sum of the terms of your first bracket to be $a$, and the sum of terms in your second bracket to be $b$. Then your equation can be simplified to $ab(c_0+c_1+...+c_p)$. From there you probably already remember from algebra that the way to expand out the bracket is to multiply $ab$ with every term of the remaining bracket, so you'll end up with a series of terms $abc_0+abc_1+...+abc_p$. We'll look at $abc_0$ from now on, but the same argument will hold for every term. You can then 'remember' the definition of $b$ again, and you'll see $abc_0 = ac_0(b_0+b_1+...+b_n)$. But from there, you can simply repeat exactly the same process, and you'll get a series of terms $ab_0c_0+ab_1c_0+...+ab_nc_0$. You can also repeat the steps for $a$ to get the full equation.

At the risk of being more confusing than helpful, there are a couple simplifications you can make to make your life easier. The first is a substitution. You can let $u=1+x$, take the derivative of $u$, and undo the substitution on the other side. This turns out to be a much less hairy equation to deal with, though you'll eventually get the same answer out the other side through either route.

The second is to notice that if your $dx$ is meant to be a small number, almost 0, then $dx^2$ or any higher power is going to be tiny. So tiny, in fact, that you can safely ignore those terms and pretend they're equal to 0. That means, instead of needing to expand all of your brackets in the above equation, and then collect like terms, which is just a nightmare, you can instead say "Does this result give me multiple powers of $dx$?" - If the answer's yes, ignore it.

Title: How to expand $(1+x+b)^2$ correctly?

Body:
I'm trying to expand this trinomial equation and wish for your support on my solution.

I'm trying to prove: given $y = (1+x)^n$, then $\frac{dy}{dx}=n(1+x)^{n-1}$

By starting likeso: $(1+x+\nabla x)^n$ then expanding and cancelling when taking the derivative $\frac{\nabla dy}{\nabla dx}$

Although, to do this I need to know how to do trinomial expansions. Here is my effort with an example:

$(1+x+b)^2=[1+2x^{2-1}+\frac{2(2-1)}{2!}x^2] + [1+2b^{2-1}+\frac{2(2-1)}{2!}b^2]+[x+2b^{2-1}x+b^2]$.

I tried following a similar procedure as if it were determinant expansion (i.e. (1+x), (1+b), (x+b)), although I'm sceptical as to whether this is true.

The purpose was to figure this out, then perform the same expansion with $(1+x+\nabla x)^n$ to derive the derivative, otherwise if this way is not possible an alternative suggestion is helpful.


问:
Proving infinite intersection has only one point
Given that:

$$ \begin{cases}a_1 = 1\\ a_{n+1} = \sqrt{a_n + 5} \end{cases} ~~~ \text{and} ~~~\begin{cases}b_1 = 4\\ b_{n+1} = \sqrt{b_n + 5} \end{cases}$$

Then:

$$ \bigcap_{n=1}^{\infty}[a_n,b_n] $$

Has only one point.

The thing is that I am stuck, it looks like the intervals are $$[1,4] \\ [\sqrt{5}, 3] \\ [ \sqrt{\sqrt{5} + 5}, \sqrt{8} ~] \\ \vdots $$

So it does go down (by going down I mean the length of the interval), and finally stops.. but where? How should I get the "final" point, is it just the limits?

$$ \lim a_n = \frac{1+\sqrt{21}}{2} \\ \lim b_n = \lim a_n$$

So the final point is $$ [ \lim a_n, \lim a_n]$$

I am really sure it is wrong, any help would be appreciated, Thanks!

答:
Have a look at the Banach fixed-point theorem; it guarantees existence and uniqueness of a fixed point of contractions.
You just need to show that $f(x) = \sqrt{x+5}$ is a contraction on $[1,4]$ which is not difficult.


问:
Examples of functors which have both adjoints
I'm familiar with a few examples of adjoint functor triples $F\dashv G\dashv H$:

equivalences of categories

left/right Kan extensions

induction/coinduction from a ring homomorphism

the forgetful functor from Top to Set

I'm curious if there are other simple examples I haven't heard about.  Of the four listed, I actually find the last example most interesting, in both simplicity and that the two categories involved are actually very specific, rather than broad classes.

答:
There is a simple example between the category of graphs $\mathbf{Gph}$ and the category of sets $\mathbf{Set}$, see this post. To summarise it: we have the forgetful functor $U: \mathbf{Gph} \to \mathbf{Set}$. It has a left adjoint $F: \mathbf{Set} \to \mathbf{Gph}$, which assigns to a set $X$ the graph $F(X)$ with vertex set $X$ and no edges. Then there is also $C: \mathbf{Gph} \to \mathbf{Set}$ that sends a graph to the set of its connected components. We have $C \dashv F \dashv U$.

Interestingly, we can add a fourth functor to this list. Consider $G: \mathbf{Set} \to \mathbf{Gph}$ that sends a set $X$ to the graph $G(X)$ with vertex set $X$ and edges between any two vertices (and on functions it does nothing). Then $U$ is left adjoint to $G$.

A similar example would be to consider the category of preorders and order-preserving maps $\mathbf{PreOrd}$. Then the forgetful functor $U: \mathbf{PreOrd} \to \mathbf{Set}$ has both a left and right adjoint. Similar to the graphs example the left adjoint adds no order relations to our set, while the right adjoint adds all order relations.

It is worth noting that this last example fails if we would consider the category of partial orders instead. In fact, the forgetful functor from partial orders to the category of sets has no right adjoint, see this post.