问:
Interpretation of regression formula returned by computer software
I have a dataset consisting of 744 records. Data exploring software generated an equation I don't know how to interpret in simple words. I really appreciate if you could help me about this matter. This dataset consists of two variables x and y values ranging from 0 to 1. Software gives me two equations:

```<code>y = 0.929307*x+normal(0.0305467, 0.0435136)
x = normal(0.4767,0.254105)
</code>```

Additional information about variables:

```<code>mean of y = 0.480593
min of y = -0.351811
max of y = 1.30236
stddev of y = 0.237672
mean of x = 0.483732
min of x = -0.397034
max of x = 1.34522
stddev of x = 0.25316
</code>```

It seems ununderstandable that minimum and maximum of x and y are less than 0 and more than 1. This dataset does not include values lower than 0 or greater than 1. How to explain that?

答:
It looks like your software is (implicitly?) assuming that your data are normal distributed and is then trying to estimate parameters for a pair of normal distributed random variables that would looks like your dataset.

Since a normal distribution doesn't have a true maximum and minimum, the software doesn't know that 0 and 1 are supposed to be special values. Instead it's computing "maximum" and "minimum" artificially by adding/subtracting 3.45 standard deviations from the mean. This produces an interval such that the probablity of a value from the normal distribution being outside the interval is $\frac{1}{1,000,000}$.

What you should be noticing here is probably that the standard deviation of the difference between $X$ and $Y$ is quite small at $0.0435$. That means that your data points fit the line
$$ Y = 0.929307\cdot X + 0.0305467$$
quite well (except that those coefficients have too many significant digits to be really meaningful -- you should probably just call it $Y=0.93\,X +0.03$).

A standard deviation of $0.25$ with mean close to $\frac12$ for the $X$ variable might lead one to suspect that $X$ is not really normal distributed but closer to being uniformly distributed in $[0,1]$. (A true uniform distribution would have standard deviation $0.289$). Whether that makes any sense would depend on details about your dataset that you're not showing here, however.


问:
Finding probability using conditional probability

Two cards are drawn from a deck of four red cards labeled $1$, $2$, $3$, $4$ and four white cards labeled $1$, $2$, $3$, $4$. Find probability that two cards are white given that at least one card is labeled $2$.

My try: Let $A : \text{two cards are white} , B: \text{At least one card is labeled 2}$ . So the answer is $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{7+6} = \frac{3}{13}$. Is my answer correct? What are the alternative solutions to this question?

答:
Let $\Omega_0=\{1,2,3,4;1',2',3',4'\}$ be the (set of) cards. (The cards with the prime are red.)

Let $\Omega$ be the set of (ordered) tuples $(a,b)$ with $a,b\in \Omega_0$, $a\ne b$. Let $\Bbb P$ be the uniform probability. (Each subset of $\Omega$ is an event / is measurable.)

Let $W\subset \Omega$ be the set of tuples $(a,b)$ with "white" $a,b\in\{1,2,3,4\}$. Let $T$ be the event of all tuples $(a,b)$ in $\Omega$, so that either $a\in\{2,2'\}$ or $b\in\{2,2'\}$. We can list the elements of $T$, then count them. Or just count them:

for each $a\in\{2,2'\}$ there are $8-1$ chances for $b$ with $(a,b)\in T$, this leads to a count of $2\cdot 7$ tuples.

for each $b\in\{2,2'\}$ there are $8-1$ chances for $a$ with $(a,b)\in T$, this leads to a count of $2\cdot 7$ tuples.

the two tuples $(2,2')$ and $(2',2)$ were counted twice.

So $|T| = 14+14-2 = 26$.

Now let us count the elements in $W\cap T$. Let $(a,b)$ be an element in this intersection. Then either $a=2$ and $b\in \{1,3,4\}$, or $b=2$ and $a\in \{1,3,4\}$. And conversely. (No case is counted twice.) We obtain $3+3=6$ cases.

The wanted conditional probability is:
$$
\Bbb P(W|T)=\frac{|W\cap T|}{|T|}=\frac{6}{26}=\frac 3{13}\ .
$$

This is more or less the same solution. (Tuples $(a,b)$ were used to model the situation, instead of sets $\{a,b\}$ as in the OP. Just some more sentences...)

ContentLicense: CC BY-SA 4.0


问:
Finding change of Basis Polynomials
For P2(R)is the vector space of polynomials of degree at most 2.

a standard basis, B is given by $(1,x,x^2)$

C is also a basis given by $(x^2-x, x, -x^2+x+1)$

How do i find the change of matrix [C]B and then find [B]C ?
It would really mean a lot if I can get a solution to this.

答:
To find the matrix $[C]_B$ , Take each polynomial from C and make this equation :
for example : (definition of c is scalars)

$x^2-x=c_1*1+c_2*x+c_3*x^2$ (after you solve this , take the scalars and put them in a matrix a the first column)

$x=c_1*1+c_2*x+c_3*x^2$

$-x^2+x+1=c_1*1+c_2*x+c_3*x^2$

after you solve the three equations and put them in a matrix , this is the transformation matrix.

Do the same with the other basis but now (or find the inverse of the matrix) $\Rightarrow$ The left side of the equation will be a vector from the basis $B$ and the right side of the equation will be scalars multiplied by the vectors of the basis $C$.

Title: Finding change of Basis Polynomials


问:
Decomposing a normal bundle
I'm reading a paper on symplectic geometry. I reached to a point where the author uses normal bundles ( which I know the definition but I've never work with). He actually decomposes a given normal bundle into a sum of two other normal bundles. I can't give my precise question and the normal bundles I'm talking about, because this will require to introduce a lot of stuff. But i think he uses the following argument:

Let M be a smooth manifold, and let C and P be two submanifold of M such that: $ C \subset P \subset M.$

Let N be the normal bundle of C in M,

$N_1:=$ the normal bundle of C in P.

$N_2:=$ the normal bundle of P in M,  restricted to C.

Then, we get $N = N_1 \oplus N_2.$

My questions are: first, what is the meaning of restricting a normal bundle to a submanifold (as is used to define the normal bundle $N_2)$, and the second one is, is the above argument true, namely is  $N = N_1 \oplus N_2.$ ?

答:
In general, given a vector bundle $\pi:B\to M$ and an embedded submanifold $S\subseteq M$, one can define the restricted bundle as $R=\pi^{-1}(S)$, with the projection given by $\pi|_R:R\to S$. This is also a vector bundle. This restriction can also be defined as the pullback bundle $R=\iota^*B$ where $\iota:S\to M$ is the inclusion map.

To your second qustion, the answer is yes; this can be seen by noting that, for $p\in C$, we have the decompositions $T_pM=T_pP\oplus N_pP$ and $T_pP=T_pC\oplus N_pC$ where $N_pC$ denotes the normal space w.r.t. $P$.

Title: Decomposing a normal bundle


问:
Prove that $\sqrt{a}+\sqrt{b}+\sqrt{c} \ge ab+bc+ac$

If $a, b, c$ are positive real numbers such that $a+b+c=3$, prove that $\sqrt{a} +\sqrt{b} +\sqrt{c} \ge ab+bc+ac$

My try: First I tried dividing both sides by $\sqrt{abc}$. Then I squared both sides. Afterwards...... no afterwards. I haven't managed to do much from here.

Body:
$2(ab+bc+ac)=(a+b+c)^2-a^2-b^2-c^2$

$\sum_{cyc}(a^2+2\sqrt{a})\ge  3\sum_{cyc}a\ge 9$


问:
why $\mu(A_n)$ is written before $\int_{A_n} f d \mu ?$
I have some confusion in RCA  book

My doubt is given below and marked  in red box

My attempt : i think  it should  be like  this  $$\frac{1}{n} \mu(A_n) \le \mu(A_n) \int_{A_n} f d\mu   \le \mu(A_n) \int_{E} f d\mu   $$

My doubt : why $\mu(A_n)$  is written  before  $\int_{A_n} f d \mu ?$

答: Since $f(x)>\frac{1}{n}$ on $A_n$, by monotonicity of the Lebesgue integral we have $\displaystyle\int_{A_n}f\,\text{d}\mu\ge\int_{A_n}\frac{1}{n}\,\text{d}\mu=\frac{1}{n}\mu(A_n)$.


答:
They skipped a step.  Let $1_E$ be the function on $E$ defined by $1_E(x) = 1$ for all $x \in E$.  Then

$$\mu(A_n) = \int\limits_{A_n} 1_E \space d\mu.$$
Since $ \frac{1}{n} 1_E(x) < f(x)$ for all $x \in A_n$, we have

$$\frac{1}{n} \mu(A_n) = \frac{1}{n} \int\limits_{A_n} 1_E \space d\mu \int\limits_{A_n} \frac{1}{n} 1_E \space d\mu \leq \int\limits_{A_n} f \space d\mu.$$

Body:
I have some confusion in RCA  book

My doubt is given below and marked  in red box

My attempt : i think  it should  be like  this  $$\frac{1}{n} \mu(A_n) \le \mu(A_n) \int_{A_n} f d\mu   \le \mu(A_n) \int_{E} f d\mu   $$

My doubt : why $\mu(A_n)$  is written  before  $\int_{A_n} f d \mu ?$


问:
Using Fermat's little theorem to prove that $\sum_{i=0}^{n-3} (-1)^{i}a^{n-2-i}b^{i+1} = 1 + $ multiple of $n$

Using Fermat's little theorem, prove that

$$ a^{n-2}b - a^{n-3}b^2 + a^{n-4}b^4 - \ldots + ab^{n-2} = 1 + M(n) $$

if $n$ is prime and doesn't divide $a$, $b$ or $a+b$ and $M(n)$ means a multiple of n.

I don't know what to do after getting the closed form of the sum.

Hints please!

答:
Hint:

$$
(a+b)\left(a^{n-2}b - a^{n-3}b^2 + a^{n-4}b^4 - ... + ab^{n-2}\right) = a^{n-1}b+ab^{n-1}
$$
Now evaluate mod $n$ using Fermat's Little Theorem.

ContentLicense: CC BY-SA 3.0


问:
Compute $\int_{-\infty}^{\infty} \frac{1}{(x-5)^2}dx$
I want to compute $$I = \int_{-\infty}^{\infty} \frac{1}{(x-5)^2}dx.$$

We can use the following methods

Compute directly $$I = -(x-5)^{-1}\bigg |_{-\infty}^{\infty}=-\bigg(\frac{1}{x-5}\bigg)_{-\infty}^{\infty}=-\bigg(0-0\bigg)=0$$

Compute by residue $$I = 2\pi \text{ Res}(f(z);5)=0, \ \ \ \ f(z) = \frac{1}{(z-5)^2}$$ We know that it has a pole of order $2$. So its residue is $0$.

However, when checking by computer, I get "integral not converge". Not sure where I made mistakes. Thanks!

答:
You cannot use the formula $\int_a^b f(x) dx = F(b)-F(a)$ if $f$ is not regular enough on $(a,b)$. The integral, in fact, does not exist. To realize this, divide the integral for instance as
$$
\int_{-\infty}^4 f(x) dx + \int_4^6 f(x) dx + \int_6^{+\infty} f(x) dx
$$

The first and third integrals are convergent, but not the second.


问:
Is the inclusion of the wedge sum into the reduced cylinder a relative cell complex?
For a CW-complex $X$ the map
$$X\coprod X \hookrightarrow X \times I$$
is a relative cell complex. What I want to know is if this still holds in the pointed case. That is, if $X$ is a pointed CW-complex with non-degenerated base point and $I_+ = [0,1] \coprod *$,  then is it true that
$$X\vee X \hookrightarrow X \wedge I_+$$
is a relative cell complex?

答: Yes, it is a CW pair since $X \wedge I_+$ is the quotient of $X \times I $ by $\{*\} \times I$. It is easy to show that any point in a CW complex can be a 0-cell of a CW structure on it, so this quotient has a CW structure, and the image of $X \sqcup X$ under the quotient map is the subcomplex $X \vee X$.


答:
$$I = \int_{-\infty}^{\infty} \frac{1}{(x-5)^2}dx=\int_{-\infty}^{5} \frac{1}{(x-5)^2}dx + \int_{5}^{\infty} \frac{1}{(x-5)^2}dx$$
Both integrals on the right are $+\infty$, so $I=\infty$

Title: Compute $\int_{-\infty}^{\infty} \frac{1}{(x-5)^2}dx$

Body:
I want to compute $$I = \int_{-\infty}^{\infty} \frac{1}{(x-5)^2}dx.$$

We can use the following methods

Compute directly $$I = -(x-5)^{-1}\bigg |_{-\infty}^{\infty}=-\bigg(\frac{1}{x-5}\bigg)_{-\infty}^{\infty}=-\bigg(0-0\bigg)=0$$

Compute by residue $$I = 2\pi \text{ Res}(f(z);5)=0, \ \ \ \ f(z) = \frac{1}{(z-5)^2}$$ We know that it has a pole of order $2$. So its residue is $0$.

However, when checking by computer, I get "integral not converge". Not sure where I made mistakes. Thanks!


问:
A decreasing sequence of function which converges to 0
The folloing question was asked in my analysis quiz and was hard nut to crack for me even after exam.

Suppose $\{f_n\}$ is a sequence of continuous real valued functions on $[0,1]$ satisfying :

(i) for each $x$ in reals $\{f_n(x)\}$ is decreasing sequence.

(ii) the sequence ${f_n}$ converges uniformly to 0.

Then let $g_n(x)=\sum_{k=1}^n (-1)^k f_k(x)$ for all $x\in \mathbb{R}$. Now, which of the following is correct:

(i) $\{g_n\}$ is Cauchy wrt sup norm.

(ii) $\{g_n\}$ is uniformly convergent.

(iii) $\{g_n\}$ need not converge pointwise.

(iv) $\{g_n\}$ is uniformly bounded.

I tried proving i and ii option but couldn't do using required definitions and gave a lot of time.

So, please tell how should I approach the question. I am stuck.

答:
First, for each $x \in [0,1]$, the series defining $g_n$ is an alternating series, so the special criteria for alternating series tells that the series converges. So $(g_n)$ converges pointwise, so the option $(iii)$ is not true. Let $g$ be the limit pointwise of the sequence $(g_n)$.

For every $x \in [0,1]$, you have
$$|g(x)-g_n(x)| = \left|\sum_{k=n+1}^{+\infty} (-1)^k f_k(x) \right| \leq f_{n+1}(x)$$

again by the criteria for alternating series (criteria on the remainders). So
$$||g(x)-g_n(x)||_{\infty} \leq ||f_{n+1}||_{\infty}$$

so because $(f_n)$ converges uniformly to $0$, you deduce that $||g(x)-g_n(x)||_{\infty}$ converges to $0$, i.e. $(g_n)$ converges uniformly to $g$. In particular, you deduce that $(g_n)$ is Cauchy for $||.||_{\infty}$, and is bounded for $||.||_{\infty}$. So options $(i)$, $(ii)$ and $(iv)$ are correct.


问:
Normal matrices with orthogonal basis
we have a theorem that says that each REAL normal matrix can be written in terms of an orthonormal basis, so that it has its eigenvalues down the diagonal and 2x2 matrices of the form $\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$. now my question is, how do i calculate this basis and the entries of the 2x2 matrices? </p>

答:
You could calculate an orthonormal basis of eigenvectors as a complex matrix. You can leave the real eigenvectors corresponding to real eigenvalues as they are.
   The non-real eiegnvalues occur in complex conjugate pairs. If $\lambda$ and $\overline{\lambda}$ are one such pair, and $\lambda = re^{i\theta},$ then you can replace the diagonal submatrix $\left( \begin{array}{clcr} \lambda & 0\\0 & \overline{\lambda} \end{array}\right)$ by the submatrix
 $\left( \begin{array}{clcr} r \cos \theta & r \sin \theta \\ -r \sin \theta & r \cos \theta\end{array}\right).$
 Let $A$ be the real matrix you started with. Let $u,v$ be non-zero real vectors such that $A(u + iv) = \lambda (u + iv).$ Then comparing real and imaginary components shows that
$Au = r \cos \theta u - r\sin \theta v , etc$. So, to obtain the corresponding real basis, you replace the pair of complex eigenvectors $u + iv, u-iv$ by the pair of real vectors $u,v$ in each case.</p>

Title: Normal matrices with orthogonal basis


问:
Finding convergence of series using Ratio Test
I have this series but Im not too sure whether my calculations are correct
$$ \sum_{n=1}^\infty \left( \frac{(3n)!}{(n!)^3}\right) $$
$$  \left( \frac{(3(n+1)!}{((n+1)!)^3}\right)\cdot\left( \frac{(n!)^3}{(3n)!}\right) $$
$$  \left( \frac{(3(n+1)(3n)!}{((n+1)^3(n!)^3}\right)\cdot\left( \frac{(n!)^3}{(3n)!}\right) $$
and by simplifying I got to
$$  \left( \frac{(3(n+1)}{(n+1)^3}\right) $$
$$  \left( \frac{(3n+3)}{(n^3+3n^2+3n+1)}\right) = 0 $$
so the series is convergent by Ratio test.

答:
There is a mistake in your first step, indeed by ratio test we obtain

$$ \frac{(3n+3)!}{((n+1)!)^3} \frac{(n!)^3}{(3n)!}=\frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} $$

since $(3(n+1))!=(3n+3)!$.

To avoid ratio test we can use that

$(3n)! =3n\cdot(3n-1)\cdot \ldots \cdot (n+1) \cdot n!\ge (n+1)^{2n} \cdot n! \ge n^{2n} \cdot n!$

and then

$$ \frac{(3n)!}{(n!)^3} \ge \left(\frac{n^n}{n!}\right)^2 \to \infty$$


问:
Does an extraneous root of a polynomial in trigonometric functions have meaning? (Eg, the root $\cos\theta=-2$ of $\cos^2\theta+3\cos\theta+2=0$.)

Does an extraneous root of a polynomial in trigonometric functions have meaning?

E.g.,
$$\cos^2{\theta} + 3\cos{\theta} + 2 = 0$$ has two roots, $\cos{\theta} = -1$ and $\cos{\theta} = -2$. The second root is usually discarded as it is beyond the range of the cosine function. But does it have any geometric, analytical, etc, meaning?

答:
You may write $\cos t= -2$ as $$\frac{e^{it}+e^{-it}}{2}=-2 \implies 4^{it}+e^{-it}+4=0 \implies e^{2it}+4e^{it}+1=0 \implies e^{it}= \frac{-4\pm\sqrt{12}}{2}$$ $=-2\pm\sqrt{3}\implies t=-i \ln (-2\pm \sqrt{3}).$ Using $(-1)=e^{2n+1)i\pi}$, we can write two branches of solutions as:
$$t_1=(2n+1)\pi+i \ln(2-\sqrt{3}). ~~t_2=(2n+1)\pi+i\ln (2+\sqrt{3}), n=0,1,2...$$
Or alternatively, two complex conjugate roots as the original equation is real.
$$t=(2n+1)\pi\pm i\ln (2+\sqrt{3}), n=0,1,2,...$$
^n=0$ guves the principal branch.


问:
Applications of Hausdorff distance and 'hyper'-topologies
Lately I've been reading about Hausdorff distance and the Fell topology, which are both essentially topologies on subsets of $2^X\setminus \{ \emptyset \}$.  The use I saw of this notion is of discussing the limit of spectra of operators, but I'm assuming there are plenty of uses which I don't suspect to look for.

Can anyone please give some examples of intersting Applications\uses of Hausdorff distance and such 'hyper'-topologies? Even with relation to simplifying other known mathametical problems.

答:
One of practical applications of Hausdorff distance is image comparison. For instance, the first ten entries for a respective Google query provided me the following references.

Riadh Bouslimi, Jalel Akaichi, Using Hausdorff Distance for New Medical Image Annotation,
International Journal of Database Management Systems 4:1 (Feb. 2012).

Normand Grégoire, Mikael Bouillot, Hausdorff distance between convex polygons, Web project presented to Mr. Godfried Toussaint for the course CS 507 Computational Geometry McGill University, fall 1998.

Daniel P.  Huttenlocher, Gregory A. Klanderman, William J. Rucklidge, Comparing Images Using the Hausdorff Distance, IEEE transactions of pattern analysis and machine intelligence 15:98  (Sept. 1993) 850–863.

Oliver Jesorsky, Klaus J. Kirchberg, Robert W. Frischholz, Robust Face Detection Using the Hausdorff Distance, in: Proc. Third International Conference on Audio- and Video-based Biometric Person Authentication, Halmstad, Sweden, 6–8 June 2001, Springer, Lecture Notes in Computer Science, LNCS-2091, 90–95.

K. Senthil Kumar, T. Manigandan, D. Chitra,  L. Murali, Object recognition using Hausdorff distance for multimedia applications, Multimedia Tools and Applications 79 (2020) 4099–4114.

Xiao Hong Li, Yi Zhen Jia, Feng Wang, Yuan Chen  Image Matching Algorithm Based on an Improved Hausdorff Distance, 2nd International Symposium on Computer, Communication, Control and Automation (3CA 2013), 244–247.

Barnabás Takács, Comparing face images using the modified Hausdorff distance, Pattern Recognition 31:12 (Dec. 1998) 1873–1881.

Chyuan-Huei Thomas Yang, Shang-Hong Lai, Long-Wen Chang, Hybrid image matching combining Hausdorff distance with normalized gradient matching,  Pattern Recognition 40 (2007) 1173–1181.


答:
There is no need to resort to complex analysis, the problem here is that $$\int_{-\infty}^{5-\varepsilon}\frac{1}{(x-5)^2}=\left[\frac{-1}{x-5}\right]_{-\infty}^{5-\varepsilon}=\frac{1}{\varepsilon}\to +\infty$$ as $\varepsilon\to 0$. The same happens when you tend to $5$ from above.

In fact, the integral $$\int_0^{+\infty}\frac{1}{x^\alpha}$$ never converges for any $\alpha\in\mathbb{R}$, since it diverges obviously for $\alpha\leq0$ and diverges at $0$ or $+\infty$ for $\alpha>1$ and $0<\alpha<1$ respectively (and at both ends for $\alpha=1$).


问:
Valid LDPC parity check matrices
I have found many different methods for generating a parity check matrix for LDPC codes and I have implemented some of them (e.g. the original Gallager ensemble, this method and some more).

I have found, that all of them share one problem: The resulting matrix is not always a valid parity check matrix.

As an example, this is a matrix I got from the Gallager ensemble (just a small example, chose bad values for the algorithm):
$$\begin{pmatrix}1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0\\
1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 0 & 1\\
1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0\\
0 & 0 & 1 & 0 & 1 & 1 & 1 & 0 & 0 & 1\end{pmatrix}$$
This matrix got generated through the Gallager ensemble, obviously. But it also is not a valid parity check matrix, because it can't be turned into an equivalent systematic LDPC code matrix, i.e. the determinant of its "right part"
$$\begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 1 & 1 & 1\\
1 & 0 & 0 & 0 & 1 & 0\\
0 & 1 & 1 & 1 & 0 & 1\\
0 & 0 & 0 & 1 & 1 & 0\\
1 & 1 & 1 & 0 & 0 & 1\end{pmatrix}$$
is equal to $0$.

My problem is, that actually most matrices I generate using these algorithms turn out to be invalid.

So, my question is, if there is a modification to all these generator algorithms to always return a valid parity check matrix or if I really have to do a while(invalid) {//generate another one} to always get a valid matrix.

答:
I'm not an expert on this, so please take everything I say with an adequate amount of salt.
I do have a little of practical experience implementing LDPC-codes (from a very limited family), so I will share my impressions and an example from my lecture notes introducing these ideas at the end of a course (the course is concentrating on convolutional codes, and this is just an addition for a bit of extras).

A practical $(n,k)$ LDPC-code has largish parameters. The length $n$ and the rank $k$ are typically in thousands, as is the number $r=n-k$ of check bits. This means that it may not be prudent to store that huge check matrix anywhere, let alone multiply input vectors with it. This is the case in particular, when the elementary row operations required to bring it into a row echelon form (possibly with the echelon at the right end) will destroy the low density-property. You really don't want to use check equations (or rows of a generator matrix) with the expected number $\approx k/2$ bits defining a single check bit instead of, say $12$, bits as would be the case in the low density version.

Instead, you can record procedures for generating the check bits using a low density parity check equation. Consider the following check matrix
$$
H=\left(\begin{array}{cccccccccc}
1&1&1&1&0&0&0&0&0&0\\
1&0&0&0&1&1&1&0&0&0\\
0&1&0&0&1&0&0&1&1&0\\
0&0&1&0&0&1&0&1&0&1\\
0&0&0&1&0&0&1&0&1&1
\end{array}\right).
$$
This is highly symmetric in that all the columns have weight two, and all the ten possibilities appear as columns. Observe that the rank of this matrix is four, so it fits into your other concern of having a check matrix that is not full rank. More precisely, the last row is the sum of the other four. By design, as all the columns have weight two!

Anyway, only the top four rows are relevant for the purposes of encoding (the last check equation can still be used in Belief Propagation decoding, but that is a separate decision). You see that they have the useful property that the last $1$s on the four top rows fall on distinct columns. This is all we need! We can declare that the positions of those last bits of checks, here the positions $4$, $7$, $9$ and $10$ are the check bits, but the other six positions are payload. Of course, it is trivial to shuffle the columns in such a way that the check positions move to the right end.

The encoding procedure is then trivial. Simply flood the remaining positions (here $1,2,3,5,6,8$) with the information bits. Then run a single iteration of belief propagation to fill in the check positions: the first check calculates position $4$, the second will fill position $7$ et cetera. Observe that only original parity checks are needed to do this as long as we arranged them to have distinct "last participating symbols". This toy example hides some aspects:

$4$ out of $10$ is not truly "low density", but hopefully you see the idea

Above matrix was exceptional in that the other bits participating in the top four equations where ALL payload bits, and a single iteration of belief propagation was sufficient. It is possible (read very likely) that with large parameters $(n,k)$ THIS IS NOT THE CASE. You absolutely don't want to run belief propagation here in those cases. But, if each row of the check matrix has a fixed small number, say $w$, $1$s, then you simply calculate the check bit as a sum of $w-1$ earlier bits, some of which are likely to be check bits themselves.

To make this clear: You need to XOR $w-1$ bits for each of the check bits for a total of $(w-1)r$ operations. Observe that by design $w\ll k$. If you manipulated the check matrix to have that identity block at the end, then you would get check equations with a "random number" of bits participating, and consequently using the check matrix (or the generator matrix) in the usual way would require a random number of operations  ($\approx (k/2)$) per check bit instead of $w-1$.

Still, in a practical system the number of information bits is given to you as a system constant. To get full rank matrices from a given ensemble some extra structure is often imposed.

the check matrices defining the LDPC-codes in 2nd generation standard for DVB impose a staircase structure in the check bit area, as well as a cyclically repeating structure elsewhere

IIRC the LDPC-codes in the MediaFlo standard use a structure with circulant blocks with a single $1$ per row within a block.

In both cases the extra properties imposed on the check matrices make it possible to control the full rank property as well as the related Tanner graph. That last property is important in practical systems. Your hardware engineers will be pulling their hair, if you want them to implement a totally random Tanner graph on hardware. The routing of all those messages in the Belief Propagation becomes then very complicated. Cyclically repeating structures simplify that, and enable parallelism in the hardware (many messages running in parallel without disturbing each other).

Title: Valid LDPC parity check matrices


问:
Every positive real number has a unique positive $k$-th root
Here is my proof so far:
Want to show there is a unique positive number $b$ such that $b^k = a$. Define the set $A = \{x>0: x^k>a\}$ of numbers too big to be $k$-th root.
$A$ is nonempty as $x = (1+a)$ is in $A$ as is min $(1+a)$ which $>a$
By definition of set $A$ is bounded below by $0$ so greatest lower bound characterisation of completeness implies $b= \inf A$ must exist. For each natural number $n$, there exists an belonging to A such that $b \le a < b+1/n$.
Past here I need help with writing out the working.
I then have to show $(an)^k$ tends to $b^k$ to conclude $b^k\geq  a$ and I am particularly unsure of this step.
I then have to show $b^k\geq  a$ by contradiction. I assume $b^k > a$then $0$ $<a/b^k< 1$ so may choose $d <b/k(1-a/b^k)$
I then have to achieve a contradiction by choosing $b-d$ belonging to $A$. This will show $b^k = a $and then I have to prove there are no other positive kth roots

答:
Imagine there exists $b>0$ and $b'>0$ with $b' \neq b$ but with $b^k=a$ and $b'^k=a$, then:

$$
b' \neq b
$$

Taking both sides to the power of $k$:
$$
b'^k \neq b^k
$$

And from the hypothesis $b^k=a$ and $b'^k=a$ we get:
$$
a \neq a
$$

Which is a contradiction. Hence, if the positive root $b$ exists, it is unique.

To prove it exists, we can create a sequence that converges to the root. Let $t_0=0$ and $w_0 = a$, then do the following:

$$u_n = (t_{n-1} + w_{n-1})/2 $$

If $u_n^k >a$, then  $w_n = u_n$ and $t_n = t_{n-1}$ otherwise $t_n = u_n$ and $w_n = w_{n-1}$. This is the bisection method. The sequence $t_n$ is non-decreasing and limited, and $w_n$ is non-increasing limited, hence they both converge to the same value as $u_n$. Say $u_n \to u$, then $u^k = a$. Hence the value exists and is positive.


问:
If f and g are maps such that $g \circ f$ and f are continuous, also $f$ is surjective. Then show that $g$ is continuous.
This question came into my mind while I was seeing the proof of Jordan separation theorem and came across through the following,

Let $f:[0,1] \rightarrow U$ be a loop in $U$ with $f(0)=f(1)=x_{0} ,$ where U is a topological space. Let $p:[0,1] \rightarrow S^{1}$ be the map given by
$$
p(x)=(\cos (2 \pi x), \sin (2 \pi x))
$$
Note that $p$ is continuous and is a bijection except at $0$ and $1$.

$p$ induces a map $h: S^{1} \rightarrow U$ such that
$$
h(x)=\left\{\begin{array}{ll}
f(0) & \text { if } x=(1,0) \\
f\left(p^{-1}(x)\right) & \text { if } x \neq(1,0)
\end{array}\right.
$$

Here $f=h \circ p$. How h is continuous?

答: The claim of the title is not true. Consider $f : \mathbb{R} \rightarrow \lbrace 0 \rbrace $ defined by $f(x)=0$ for all $x \in \mathbb{R}$, and $g : \mathbb{R}\rightarrow \mathbb{R}$ defined by $g(x)=1$ if $x \in \mathbb{Q}$ and $g(x)=0$ otherwise. Then $f$ is continuous and surjective, $g \circ f : \mathbb{R} \rightarrow \mathbb{R}$ is the constant (so continuous) map equal to $1$, but $g$ is not at all continuous.


问:
proof about partial order
Let R and S be two partial orders on a set X, and T is a relation on X such that aTb(i.e. a,b ∊ X) if and only if both aRb and aSb hold. Is T also a partial order on X?
how to prove it?</p>

答:
Check each of the required properties for a relation to be a partial order: reflexivity, antisymmetry, and transitivity.

Here, I go through the proof that $T$ is transitive.

Suppose $aTb$ and $bTc$.  We ask whether or not this implies that $aTc$ is also true.

Well... by $aTb$ we know in particular that $aRb$.  Further by $bTc$ we know in particular that $bRc$.  Since $R$ is a partial order, we know that from $aRb$ and $bRc$ and that $R$ is transitive that $aRc$ is also true.

Similarly, we can show that $aTb$ and $bTc$ imply that $aSb$ and $bSc$ from which it follows that $aSc$.

Since both $aRc$ and $aSc$ are true, it follows that $aTc$ is also true.

Thus, the relation $T$ is indeed transitive.

The remaining parts of the problem are just checking the other properties and they are all performed similarly.  Use what you know about how $T$ is defined in terms of $R$ and $S$ and use the fact that $R$ and $S$ are both partial orders themselves and have the desired properties.

Title: proof about partial order


问:
How does one show Caratheodory's criterion for e.g. counting measure?
How does one show Caratheodory's criterion for e.g. counting measure?

I think it's intuitive since every $E \subset X$ can be written for $A \subset X$ like:

$E=(A \cap E) \cup (E \setminus A)$

And then the counting measure on this intuitively must satisfy the criterion.

But is this sufficient for proof?

答: If $E$ is infinite, then one of $A\cap E, E\setminus A$ must be infinite and we get something like $\infty=\infty+..$. On the other hand, if $E$ is finite, then it's just a matter of counting because $A\cap E, E\setminus A$ are both finite and disjoint and the cardinality of $E$ is the sum of cardinalities of $A\cap E, E\setminus A$.


答:
Personally, without going too crazy with details, I'd say that Carathéodory establishes that a set $E$ is $\mu^*$-measurable if and only if $\mu^*(A)\ge\mu^*(A\cap E)+\mu^*(A\setminus E)$ for all $A$ such that $\mu^*(A)<\infty$.

Since by all means $\mu^*(S)=\begin{cases}\lvert S\rvert&\text{if }\lvert S\rvert<\aleph_0\\ \infty&\text{if }\lvert S\rvert\ge\aleph_0\end{cases}$ and certainly $\lvert A\rvert+\lvert B\rvert=\lvert A\rvert+\lvert A\cap B\rvert$ for finite sets $A$, $B$, we end up having, for all $E$ and for all $A$ of finite outer measure, $$\mu^*(A\cap E)+\mu^*(A\setminus E)=\mu^*(A)$$

Body:
How does one show Caratheodory's criterion for e.g. counting measure?

I think it's intuitive since every $E \subset X$ can be written for $A \subset X$ like:

$E=(A \cap E) \cup (E \setminus A)$

And then the counting measure on this intuitively must satisfy the criterion.

But is this sufficient for proof?


问:
How to compute $\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$?
I want to compute $$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$$

My approach is $$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx=\int_{-\infty}^{\infty} \frac{1}{(x -\pi)^2}dx+\int_{-\infty}^{\infty} \frac{\cos(x)}{(x -\pi)^2}dx.$$

However, the first term on the RHS does not exist since there is a singularity $\pi$.

How to overcome this problem?

Now follow the hint, I get $$\int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$

BTW, if I want to use residue calculus to do this problem, how to do it?  I am confused the pole at origin. Thanks!

答:
HINT: With the substitution $x\to x+\pi$, we get

$$\int_{-\infty}^\infty \frac{1-\cos x}{x^2}dx$$

then integrate by parts with $u=1-\cos x$ and $dv=dx/x^2$. You should end up with a familiar integral...


问:
How can I build this function?
I want to build a function $f(t)$. The function is simple but has some constraints:

$t$ goes from $0$ to $1$

$f(0) = 0$, $f(0.5) = 1$ and $f(1) = 0$

$f$ is linear from $0$ to $0.5$ and from $0.5$ to $1$

How can I build such function?

What I thought: I started with $f(t) = -|t|$, then I added $1$, hence $f(t) = 1 - |t|$.

But I don't know how to drag this function.

答:
The green line is the function you ended up with: $f(t) = 1-|t|$.

To transform it to the red function (so it is shrunk in half with respect to the y-axis), we set $g(t) = f(2t) = 1-|2t|$.

Finally to transform this to the blue function, we translate the graph to the right by $0.5$ units.

This is done by setting $h(t) = g(t-0.5) = 1-|2(t-0.5)| = 1-|2t-1|$.

Hence your function would be $h(t)$ defined on the interval $[0,1]$.

Title: How can I build this function?


问:
Should every group be a monoid, or should no group be a monoid?
Question: What is more convenient/useful? Writing mathematics as if every group is a monoid, or as if these two classes are disjoint?

Additional discussion. Define a monoid as follows.

Defn 1. A monoid is a triple $(X,*,e)$ such that $*$ is an associative binary operation on $X$, and $e \in X$, and $e$ has the property that for all $x \in X$ it holds that $x * e = e * x = x$.

From here, there's at least two ways of defining a group.

Defn 2. A group is a monoid $(X,*,e)$ such that for all $x \in X$ there exists $y \in X$ such that $x*y=e$.

Defn 2'. A group is a quadruple $(X,*,e,i)$ such that $(X,*,e)$ is a monoid, and $i$ is a function $X \rightarrow X$, and for all $x \in X$ it holds that $x * i(x) = e$.

Now I understand that minimalism favors Defn 2, while practitioners of universal algebra favor Defn 3. However, this is not my question.

My question is not: which is preferable, Defn 2 or Defn 2'?

Rather, my question is: which is preferable, definitions like Defn 2 such that every group is a monoid, or definitions like Defn 2' such that no group is monoid? So just to clarify, I want to know: what is more convenient/useful? Writing mathematics as if every group is a monoid, or as if these two classes are disjoint?

答:
Previous answers are very complete. Let me just add a model theoretic point of view. (I did myself ask that kind of questions recently.)

For the model theorist, your question can be stated as : considering $\mathcal L_1$ the language $\{\ast, e\}$ and $\mathcal L_2$ the language $\{\ast, e, i\}$, should a group be defined as a $\mathcal L_1$-structure (satisfying some $\mathcal L_1$-theory $T_1$) or as a $\mathcal L_2$-structure (satisfying some $\mathcal L_2$-theory $T_2$) ?

If your goal is to talk about group, you should prefer $\mathcal L_2$. For example, considering a $\mathcal L_2$-model $\mathfrak M \models T_2$ of base set $M$, the substructure generated by some $A \subseteq M$ is the subgroup $\langle A \rangle$ of $M$. But taking a $\mathcal L_1$-model $\mathfrak N \models T_1$ makes the substructure generated by $A \subseteq N$ only the monoid generated by $A$ in the implicit monoid $N$ and hence the substructure generated by $A$ is not necessarily model of $T_1$. It shows that the language $\mathcal L_1$ is more suitable to talk about monoids than groups.

We can even go a little further. Consider the language $\mathcal L_0 = \{\ast\}$ . You have enough to state a theory $T_0$ of groups :
$$ \begin{aligned}
\forall x \forall y \forall z, &\ (x \ast y) \ast z = x \ast (y \ast z) \\
\exists e \forall x, &\ x \ast e = x = e \ast x \\
\forall x \exists y, &\ x \ast y = e = y \ast x .
\end{aligned}
$$
But because of the second statement, the theory is not $\forall\exists$ (i.e. not every sentence of the theory $T_0$ is equivalent to a sentence $\forall x_1 \dots \forall x_n \exists y_1 \dots \exists y_m, \varphi_0(x_1,\dots,x_n,y_1,\dots,y_m)$ with $\varphi_0$ without quantifier). It has for result that a increasing union of groups isn't necessarily a group (here group in the sense a $\mathcal L_0$-model of $T_0$). That is, groups in that sense does not have the simplest induction limit as a group. Actually, what could go wrong is that the union could have no neutral. As you can imagine, the language $\mathcal L_0$ is then more suitable for associative magmas than for monoids or groups.


问:
Modularity of Ramanujan-Sato series
The Ramanujan-Sato series
$$j^*(\tau)=432\frac{\sqrt{ j(\tau)}+\sqrt{j(\tau)-1728}}{\sqrt{ j(\tau)}-\sqrt{j(\tau)-1728}}=432\frac{E_4(\tau)^{\frac32}+E_6(\tau)}{E_4(\tau)^{\frac32}-E_6(\tau)} \\
= \frac{1}{q}-120+10260 q-901120 q^2+91676610 q^3+\mathcal O\left(q^{4}\right)$$
of level 1 generalises Ramanujan's formula for $\frac1\pi$, where $q=e^{2\pi i\tau}$. Here, $j$ is the Klein $j$-invariant and $E_k$ are the Eisenstein series. Due to the square roots, it does not immediately seem to be a modular function for a congruence subgroup of $SL(2,\mathbb Z)$. However, as it relates to the Klein $j$-invariant by
$$j=\frac{(j^*+432)^2}{j^*},$$
it seems to satisfy a modular polynomial equation in $j$ of degree $2$, suggesting that $j^*$ is modular for an index 2 subgroup of $SL(2,\mathbb Z)$. Is this true? And what would be the invariance group?

Any recommendation on the literature would be most helpful, many thanks!

答:
It is $$j^*(\tau)= 432\frac{(\sqrt{ j(\tau)}+\sqrt{j(\tau)-1728})^2}{1728}=432\frac{2 j(\tau)-1728+2\sqrt{j(\tau)}\sqrt{j(\tau)-1728}}{1728}$$
So we are looking at the modularity of $$\sqrt{j(\tau)}\sqrt{j(\tau)-1728}$$

$j(\tau)-1728$ is holomorphic non-zero at $e^{2i\pi /3}$ while $j(\tau)$ has a zero of order $3$ (not $6$!!!) at $e^{2i\pi /3}$, thus $\sqrt{j(\tau)}\sqrt{j(\tau)-1728}$ and hence $j^*(\tau)$ have a branch point at $e^{2i\pi/3}$, they are not modular.

They are automorphic on a double cover of the modular curve, though.

Note that $f(\tau)=\sqrt{j(\tau)-1728}$ is holomorphic because the only zeros of $j(\tau)-1728$ are double at $SL_2(\Bbb{Z})i$. It is $2$-periodic and $f(\gamma(\tau))=\chi(\gamma)f(\tau)$ where $\chi$ is the character $SL_2(\Bbb{Z})\to \pm 1$ defined by $\chi(\pmatrix{1&1\\0&1})=\chi(\pmatrix{0&1\\-1&0})=-1$ which factors through $SL_2(\Bbb{Z/2Z})$, in fact $\ker(\chi)$ are the matrices whose reduction $\bmod 2$ is $\pmatrix{1&0\\0&1},\pmatrix{1&1\\1&0},\pmatrix{0&1\\1&1}$.

Title: Modularity of Ramanujan-Sato series


问:
A question about Characteristic polynomial of $A^2$ , Diagonalizability, Minimal polynomial
Consider this question asked in my masters entrance for which I am preparing.

Let A be a real matrix with characterstic polynomial $(X-1)^3$ . Choose correct statements:

(1) A is always diagonazable.

(2) If minimal polynomial of A is $(X-1)^3$ then A is diagonizable,

(3) Characterstic polynomial of $A^2$ is $(X-1)^3$ .

(4) If A has exactly 2 jordan blocks , then $(A-1)^2$ is diagonizable.

I have proved a false , c true . I don't know how to approach (b),(d)( which concept should be used?).

Kindly give a detailed answer for (D) as i am not good in Jordan Forms

If useful ,I followed Hoffman Kunze

答:
(2) is false. In fact, a matrix is diagonalizable iff its minimal polynomial has no repeated factors.

(4) is true. The key is that for a Jordan block $J$ associated with $\lambda$, the size of $J$ is equal to the smallest $k$ for which $(J- \lambda)^k = 0$. If there are exactly two Jordan blocks for a size $3$ matrix, then each block has size $2$ or smaller.


问:
How many multiples of X lie in the arbitrary range [Y,Z]?
Is there a way to determine this without brute force?

For example

$X=3$

range $= [17,24]$

The multiples of $X$ in this range are $18$, $21$, and $24$, so $3$ multiples total.

答:
find the smallest $Y1$ such that $Y1 \% X == 0$ and $Y1 >= Y$. similarly, find the greatest $Z1$ such that $Z1 \% X == 0$ and $Z1 <= Z$.

Once you have $Y1$ and $Z1$ with you, it's easy:

The number of divisors of $X$ in range $[Y, Z]$ is same as in range $[Y1, Z1]$ which is $((Z1 - Y1) / X) + 1$.

Running this logic on your example:
$X = 3, Y = 17, Z = 24 => Y1 = 18, Z1 = 24 => (Z1 - Y1) = 6$

hence, the number of divisors of $3$ in range $[17, 24]$ are $(6 / 3) + 1 = 3$


问:
Why can't I use FTLI when $\vec{F}=\nabla f$?
Why can't I use Fundamental Theorem of Line Integral (FTLI) when $\vec{F}=\nabla f$?

And what's an easy way to check if an equation is smooth(differentiable)?

答: This is a completely vague question. Of course, the FTLI applies precisely to vector fields that are gradients: $$\int_C \nabla f\cdot d\vec r = f(B)-f(A)\,,$$ where $C$ is a path from $A$ to $B$.</p>


问:
Computing Joint Moment Generating Function
Good morning, Stack Exchange. My problem is that I have a joint distrubtion of discrete random variables with the following PMF and support:

$f(x,y) = \frac{2^{x-y}  e^{-3}}{x!(y-x)!}$ when $x = 0, 1, 2, ... y$ and $y = 0, 1, 2, ... \infty $ and $f(x,y) = 0$ otherwise

So, my problem here is that I know that to get the MGF,  $M(t_1, t_2)$, we must evaluate the sum:

$$\sum_{y=0}^{\infty} \sum_{x=0}^{y} \frac{e^{x_1t_1 + yt_2} *  e^{-3} * 2^{y-x}}{x!{y-x!}}$$

My strategy for evaluating the inner sum was to move all the terms that contain a $y$ to the outermost sum  and to move the constant $e^{-3}$ out of the sum entirely, hence solve the inner sum as $$\sum_{x=0}^{y} \frac{e^{xt_1} * 2^{-x}}{x!{y-x!}}$$

The hint from the textbook tells me that this sum is easy to evaluate if the term $2^{-x}$ weren't there, but I am not sure how to deal with the sum once this term is present. Am I on the right track here, and how do I solve this innermost sum and help simplifying the sum in general would be much appreciated. Thanks for reading.

Body:
You have PMF:
$$
f(x,y) =
\begin{cases}
 \frac{2^{x-y}  e^{-\frac{3}{2}}}{x!(y-x)!},& x = 0, 1, 2, \ldots y\text{ and } y = 0, 1, 2,\ldots\\
0& \text{otherwise}
\end{cases}
$$

And the MGF is:
\begin{align}
M_{X,Y}(t_1,t_2)
&=
E(e^{t_1x+t_2y})\\
&=
\sum\limits_{y=0}^\infty\sum\limits_{x=0}^y
e^{t_1x+t_2y}\frac{2^{x-y}  e^{-\frac{3}{2}}}{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty 2^{-y}e^{t_2y}\sum\limits_{x=0}^y
e^{t_1x}\frac{2^{x}  }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty 2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\frac{\left(2e^{t_1}\right)^{x} }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\frac{y! \left(2e^{t_1}\right)^{x} }{x!(y-x)!}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\begin{pmatrix}y\\x\end{pmatrix}\left(2e^{t_1}\right)^{x} \\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}\sum\limits_{x=0}^y
\begin{pmatrix}y\\x\end{pmatrix}\left(2e^{t_1}\right)^{x} 1^{y-x}\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}2^{-y}e^{t_2y}(2e^{t_1}+1)^y\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}\left(2^{-1}e^{t_2}(2e^{t_1}+1)\right)^y\\
&=
e^{-\frac{3}{2}}\sum\limits_{y=0}^\infty \dfrac{1}{y!}\left(e^{t_1+t_2}+\dfrac{1}{2}e^{t_2}\right)^y\\
&=
e^{-\frac{3}{2}} e^{e^{t_1+t_2}+\frac{1}{2}e^{t_2}}\\
&=
e^{e^{t_1+t_2}+\frac{1}{2}e^{t_2}-\frac{3}{2}}.
\end{align}

Note

I use the fact:

(1) Binomial theorem
$$\sum\limits_{x=0}^n \begin{pmatrix}n\\x\end{pmatrix}a^x b^{n-x} = (a+b)^n$$

(2) The Taylor series about $x=0$ of $e^x$,
$$\sum\limits_{n=0}^\infty \dfrac{x^n}{n!}=e^x.$$


问:
How to show that the set of all polynomial is of the first category?
Just like the title.

How to show the set of all polynomials on $C[a,b]$ (denoted by $P$) is of the first category?

I try to represent $P$ by a union of $P_n$, with $P_n$ is the set of polynomials on $C[a,b]$ which has a highest exponent $n$. How to show that each $P_n$ is nowhere dense?

答:
For each $P_n$, consider $P_n^c$ as its complement.

For $\forall f\in C[a,b]$, if $f\notin P_n$, then $f\in P_n^c$. Otherwise, $f\in P_n$, consider a sequence of polynomials in $P_{n+1}$, having coefficients of items with exponent no greater than $n$ the same with those in $f$, and coefficients of $x^{n+1}$ converge to $0$. Then $f$ is a limit point of $P_n^c$ since $P_{n+1}\in P_n^c$.

Hence $P_n^c$ is dense and $P_n$ is nowhere dense.


问:
$\prod_{i = 1}^{100} \sin ix = 1$
I have encountered this problem and I do not know how to solve it.
$$\prod_{i = 1}^{100} \sin ix = 1$$I have tried multiplying it with various quantities, such as $\cos x,\ \sin x$. I have also tried organizing the product in pairs of $(\sin x, \sin  100x), \ (\sin  2x, \ \sin  99x)$ and so on, but all these efforts proved to be in vain. I have also multiplied it with the product of cosines, and got to the point of $\prod_{i = 1}^{100} \cos ix$ is also 1. I would like to hear your opinion on it.

答:
All the factors are $\leq 1$ in absolute value. The product can only be 1 if all factors are 1 (in absolute value).

$$
\prod_{i=1}^{n} \sin (ix) = 1 \Rightarrow |\sin (ix)| = 1, \quad i = 1, \cdots, n
$$

As @Andrei pointed out in the comments to this post, there is no such $x$.In fact, one can check that $f_n(x) = \prod_{i=1}^n \sin(i x)$ is, at least, point wise convergent to zero.

Title: $\prod_{i = 1}^{100} \sin ix = 1$


问:
Trigonometry: Least square of a tangent function
I have two vectors $x_n$ and $y_n$, when sorted forms a tangent function of $y_n = a\cdot \tan(b\cdot x_n + c) + d$, and I wondered if it was possible to use the least square method, or another method, to make a approximate function and if so how?

答:
The usual way is to use a nonlinear regression method of fitting. They are several softwares. The calculus is iterative starting from "guessed" values of the parameters. Sometimes the numerical calculus fails due to bad initial guessed values and/or not convergent iteration.

The principle of a non conventional method (no iteration, no initial guess) is explained in this paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales , partially translated in : https://scikit-guess.readthedocs.io/en/latest/appendices/references.html. An application to the case of tangent function is shown below.

Numerical example :

If a well defined criteria for fitting is specified ( least mean absolute deviation or least relative deviation or others) the above method is not sufficient. An iterative process is required with guessed values to start. The above calculus can be used to get a set of good initial values.

CAUTION :

In the above method numerical integrations are carried out for the computation of $S_k$ and $T_k$. Such a simplified process cannot be used if the function tends to infinity around a point on the range of the data $(x_k\, ,\,y_k)$. For example if the points are distributed as on the next figure the above method will not be convenient.

FOR INFORMATION :

An integral equation to which $\quad y(x)=a\tan(bx+c)+d\quad$ is solution is :
$$y(x)=\frac{b}{a}\int (y(x))^2dx-2\frac{bd}{a}\int y(x)\,dx+\left(\frac{b\,d^2}{a}+ab\right)x+\text{constant}$$
This allows the linearisation of the regression and the construction of the first 4x4 matrix.


问:
positive Markov Matrix
For a positive Markov matrix which satisfy 1) every element is positive 2) each column sums to 1. It's easy to prove that 1 is a eigenvalue and every $ | \lambda | \leq 1  $. However, is there any way to prove that $ \lambda \neq -1  $ ?

答:
Here is a quick proof of 1:

Because the entries of $A$ are positive, there exists an $\alpha > 0$ for which  $A - \alpha I$ has positive entries. Define $B = \frac 1{1 - \alpha}(A - \alpha I)$. We note that $B$ is a positive Markov matrix, which means that every eigenvalue $\mu$ of $B$ satisfies $|\mu| \leq 1$. On the other hand, we have
$$
A = \alpha I + (1 - \alpha) B,
$$
which means that each eigenvalue $\lambda$ of $A$ satisfies $\lambda = \alpha + (1 - \alpha )\mu$ for some eigenvalue $\mu$ of $B$.  Now, argue that if $|\mu| \leq 1$ and $\mu \neq 1$, then $|\alpha + (1 - \alpha) \mu| < 1$.

Fact 2 can be proved using the following fact:

Claim 1: If $x \in \Bbb C^n$ is such that $Ax = x$, then we also have $A|x| = |x|$.

Using this, prove

Claim 2: If $x \in \Bbb C^n$ is such that $Ax = x$, then there is a vector $v > 0$ for which $x = \alpha v$ (for some $\alpha \in \Bbb C$).

From there, suppose for the purpose of contradiction that there exist linearly independent vectors $x,y$ for which $Ax = x$ and $Ay = y$. Without loss of generality, take $x>0,y>0$ (via Claim 2). Argue that there exist non-zero real coefficients $c_1,c_2$ for which $z = c_1 x + c_2 y$ has at least one zero entry. Now, $Az = z$, but there is no $\alpha \in \Bbb C$ and $v > 0$ for which $z = \alpha v$, which contradicts Claim 2.

Title: positive Markov Matrix


问:
How to prove that $\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$ converges
I have been trying really hard to prove that this integral $$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx$$

converges. However, I did not succeed. For now, I proved that
$$\int_0^\infty \frac{\ln(x)}{x^2-1}\, dx =  2  \int_1^\infty \frac{\ln(x)}{x^2-1}\, dx,$$
and that $f(x) = \frac{\ln(x)}{x^2-1}$ is continuous on $[1,+\infty)$.

If you can give me a hint, that would be nice.

答: It's trivial that $x\geq\ln(x)$ for all $x>0$. Use this to show that $$\int_{1}^{\infty}\frac{x}{x^2-1}dx\geq\int_{1}^{\infty}\frac{\ln(x)}{x^2-1}dx$$ and use the equation you proved to show that this integral converges by the comparison test.


问:
Is this estimation correct for a polynomial?
From the limit development of the function $e^{x}$ we get that
$$
p_{n}(x)≤e^{x}  (1)
$$
for all $x≥0$ where $p_{n}(x)$ is a polynomial of degree n.
Based on inequality (1) I write that
$(x+2)^{m}≤m!2^{m}e^{x}$  (2).
Is  the inequality (2) true for all $x≥0$ and $m∈ℕ$?

答:
In Taylor series expansion of $e^x$ we have
$$
e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^m}{m!} + ...
$$
On the other hand, by the means of Newton expansion the left-side of the inequality could be written as
$$
(x+2)^m = \sum_{k=0}^m {m \choose k} x^k 2^{m-k} \leq 2^m m!  (1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^m}{m!})
$$

$$
(x+2)^m \leq 2^m m! (e^x - \frac{x^{m+1}}{(m+1)!} - \frac{x^{m+2}}{(m+2)!} - \frac{x^{m+3}}{(m+3)!} - ...)
$$

$$
(x+2)^m + 2^m m!(\frac{x^{m+1}}{(m+1)!} + \frac{x^{m+2}}{(m+2)!} + \frac{x^{m+3}}{(m+3)!}  + ...) \leq 2^m m!(e^x)
$$
It could be concluded no matter how big the $m$ is, as the polynomial degree of exponential function is infinite, it is always true.

Code For intution:

```<code>import numpy as np
import matplotlib.pyplot as plt

x = np.arange(1,5)
for m in np.arange(1,5):
    plt.subplot(2,2,m)
    y1 = (x+2)**m
    y2 = 2**m*np.math.factorial(m)*np.exp(x)
    plt.plot(x,y1)
    plt.plot(x,y2)
    plt.title('m = {}'.format(m))
    plt.tight_layout()
</code>```

Body:
In Taylor series expansion of $e^x$ we have
$$
e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^m}{m!} + ...
$$
On the other hand, by the means of Newton expansion the left-side of the inequality could be written as
$$
(x+2)^m = \sum_{k=0}^m {m \choose k} x^k 2^{m-k} \leq 2^m m!  (1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ... + \frac{x^m}{m!})
$$

$$
(x+2)^m \leq 2^m m! (e^x - \frac{x^{m+1}}{(m+1)!} - \frac{x^{m+2}}{(m+2)!} - \frac{x^{m+3}}{(m+3)!} - ...)
$$

$$
(x+2)^m + 2^m m!(\frac{x^{m+1}}{(m+1)!} + \frac{x^{m+2}}{(m+2)!} + \frac{x^{m+3}}{(m+3)!}  + ...) \leq 2^m m!(e^x)
$$
It could be concluded no matter how big the $m$ is, as the polynomial degree of exponential function is infinite, it is always true.

Code For intution:

import numpy as np
import matplotlib.pyplot as plt

x = np.arange(1,5)
for m in np.arange(1,5):
    plt.subplot(2,2,m)
    y1 = (x+2)**m
    y2 = 2**m*np.math.factorial(m)*np.exp(x)
    plt.plot(x,y1)
    plt.plot(x,y2)
    plt.title('m = {}'.format(m))
    plt.tight_layout()


答:
I used @Franklin Pezutti Dyer's hint:

$-\int_{-\infty}^{+\infty}(1-\cos{x})d\left(\frac{1}{x}\right)=-\int_{-\infty}^{\infty}\frac{\sin{x}}{x}dx=-\pi$

Title: How to compute $\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$?

Body:
I want to compute $$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx$$

My approach is $$\int_{-\infty}^{\infty} \frac{1+\cos(x)}{(x -\pi)^2}dx=\int_{-\infty}^{\infty} \frac{1}{(x -\pi)^2}dx+\int_{-\infty}^{\infty} \frac{\cos(x)}{(x -\pi)^2}dx.$$

However, the first term on the RHS does not exist since there is a singularity $\pi$.

How to overcome this problem?

Now follow the hint, I get $$\int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$

BTW, if I want to use residue calculus to do this problem, how to do it?  I am confused the pole at origin. Thanks!


问:
Find the limits between which $a$ must lie in order that $\frac{ax^2-7x+5}{5x^2-7x+a}$ can take all real values
Find the limits between which $a$ must lie in order that $$\frac{ax^2-7x+5}{5x^2-7x+a}$$ may be capable of taking all real values for all possible real values that $x$ may take.

My Attempt:
Let $$y=\frac{ax^2-7x+5}{5x^2-7x+a}$$

$(5y-a)x^2-7(y-1)x+ay-5=0$

Since $x\in R$, we should have Discriminant $\geq 0$

$49(y-1)^2-4(5y-a)(ay-5)\geq 0$

which simplifies to

$(49-20a)y^2+2(2a^2+1)y+(49-20a)\geq 0$

For above to hold true for all real $y$ we must have $49-20a>0$

and Discriminant $\leq 0$

$4(2a^2+1)^2-4(49-20a)^2\leq 0$

which simplifies to

$(a-5)^2(a+12)(a-2)\leq 0$

Thus $-12\leq a\leq 2$

But for $a=-12$ and $a=2$ the Numerator and Denominator of $y$ develop common factors due to which $y$ assumes the form $\frac{Ax+B}{Cx+D}$ which obviously can't take all real values.

So $-12<a<2$

Can we not simply find value(s) of $a$ such that Numerator and Denominator have common factors and somehow justify that $-12<a<2$. I want to avoid the Discriminant part.

Body:
If $$y=\frac{f(x)}{g(x)}$$ where $f(x)$ and $g(x)$ are both quadratics.

If $y$ is of the form $$y=\frac{(x-a)(x-c)}{(x-b)(x-d)}$$ where $a<b<c<d$ then from the plot of graph it is clearly seen that $-\infty<y<+\infty$ when $x\in (b,d)$

It appears reasonable enough to say that if one root of numerator(denominator) lies between the roots of denominator(numerator) then $y$ can take all real values.

Let $x_{1}$ and $x_{2}$ be the roots of $5x^2-7x+a=0$

i.e.  $5x_{1}^2-7x_{1}+a=0$ and

$5x_{2}^2-7x_{2}+a=0$

Also, $x_{1}+x_{2}=\frac{7}{5}$ ; $x_{1}x_{2}=\frac{a}{5}$

Let $f(x)=ax^2-7x+5$

$$f(x_{1})=ax_{1}^2-7x_{1}+5=a\left(\frac{7x_{1}-a}{5}\right)-7x_{1}+5=\frac{(a-5)(7x_{1}-a-5)}{5}$$

Similarly, $$f(x_{2})=ax_{2}^2-7x_{2}+5=\frac{(a-5)(7x_{2}-a-5)}{5}$$

If exactly one root of numerator i.e. $ax^2-7x+5=0$ lies between $x_{1}$ and $x_{2}$ then

$$f(x_{1})f(x_{2})<0$$

$$\frac{(a-5)^2(7x_{1}-a-5)(7x_{2}-a-5)}{25}<0$$

$$49x_{1}x_{2}-7(a+5)(x_{1}+x_{2})+(a+5)^2<0$$

$$49(\frac{a}{5})-7(a+5)(\frac{7}{5})+(a+5)^2<0$$

$$(a+12)(a-2)<0$$

$$-12<a<2$$


问:
Calculus Optimization circle and square
Six feet of wire is used to form a square and a circle.  How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum total area?

What is this question asking me to find?

答: It's asking you to find the maximum sum of the areas of a square and a circle, provided that the sum of their perimeters is six feet, and to describe for such cofigurations how long the perimeter of the square and the circumference are.</p>

Title: Calculus Optimization circle and square


问:
Graph Theory: matching question.
In a college dorm two students share a single room. Each student is asked to give
two names of preferred roommates. Housing ensures that if a student is on the preferred list of some other student then the second student is also on the preferred list of the ﬁrst student.

Prove that it is always possible to pair the students such that everyone's roommate is from the preferred list.

So say Student A has (B,C) as preferred roommates, so B would be on the preferred list of A - now "some other student", would that mean that if there is this other student D then ???

Honestly I don't understand what the question is trying to say - how to even set this up.  I do realize that this would have to do with matching - perhaps some sort of alternating graph.

Any help would be greatly appreciated.

答:
It doesn't even seem to be true. Suppose the students are Adam, Brian, Charles, Dan, Eric and Fred:

Adam, Brian and Charles are friends and each prefers the two others.

Dan, Eric and Fred are friends and each prefers the two others.

Since each friend clique has an odd number of students in them, at least one room has to be shared by students from different cliques -- but those will not have preferred each other.

Title: Graph Theory: matching question.


问:
Likelihood test ratio for random sample with p.d.f. $f(x\mid \theta) = e^{-(x-\theta)}$
I don't know how to find the max region of the denominator since doing the usual gives me a constant as the derivate of the likelihood function of theta.
The problem is as follows

Let $X_1 , \ldots , X_n$ be a random sample with
$$f(x\mid \theta) = e^{-(x-\theta)}$$
where $x \in [\theta,\infty)$

find $\lambda(x)$ where $H_0: \theta = \theta_0$ and $H_1: \theta > \theta_0$

so I know that the numerator since it's just a point, is equal to the maximum likelihood function on $\theta_0$

but if I try to maximize the denominator since the Likelihood function is
$$ \prod^n_{i=1} e^{-(x_i-\theta)} = e^{n\theta-n \bar{x} } $$
but I can't maximize the likelihood from its derivative in the region $[\theta_0 , \infty]$

答:
For any real $\theta$, the likelihood function given the sample $\boldsymbol x=(x_1,\ldots,x_n)$ is

\begin{align}
L(\theta\mid \boldsymbol x)&= \begin{cases}e^{-n(\bar x -\theta)}&, \text{ if }x_1,\ldots,x_n\ge \theta \\ 0 &, \text{ otherwise }\end{cases}
\\&=\begin{cases}e^{-n(\bar x -\theta)}&, \text{ if }x_{(1)}\ge \theta \\ 0 &, \text{ otherwise }\end{cases}\,,
\end{align}

where $x_{(1)}=\min\{x_1,\ldots,x_n\}$.

This is a non-decreasing function of $\theta$, so its maximum is reached for the maximum possible value of $\theta$, namely $\theta=x_{(1)}$.

Now suppose $\theta$ is restricted to $[\theta_0,\infty)$. If $x_{(1)}\ge \theta_0$, then $L(\theta\mid \boldsymbol x)$ is still maximized at $\theta=x_{(1)}$. But if $x_{(1)}<\theta_0$, then because of the nature of the likelihood, $L(\theta\mid \boldsymbol x)$  is maximized at the boundary point $\theta=\theta_0$. A sketch of the likelihood function makes this point clear.

So the restricted MLE of $\theta$ in this case is  $$\hat\theta=\begin{cases}x_{(1)}&,\text{ if }x_{(1)}\ge \theta_0 \\ \theta_0 &,\text { if }x_{(1)}< \theta_0\end{cases}$$

The likelihood ratio test criterion is therefore $$\Lambda(\boldsymbol x)=\frac{\sup_{\theta=\theta_0}L(\theta\mid \boldsymbol x)}{\sup_{\theta\ge \theta_0}L(\theta\mid \boldsymbol x)}=\frac{L(\theta_0\mid \boldsymbol x)}{L(\hat\theta\mid \boldsymbol x)}$$

Simplify this and reject $H_0$ for small values of $\Lambda$ to obtain the likelihood ratio test. This test would coincide with the uniformly most powerful test derived here for the same problem.


问:
Moments of Fourier transform of a polygon
I am working on the Fourier transform of a polygon. Based on a publication of Shung Wu Lee I have the following formula for the Fourier transform:

$$S(u,v) =
\sum_{n=1}^N e^{i \omega\cdot\gamma_n}\left[\frac{\alpha_n\times\alpha_{n-1}}{(\omega\cdot\alpha_n)(\omega\cdot\alpha_{n-1})}\right]
$$

where $\omega$, $\alpha_n$ and $\gamma_n$ are all vectors in 2D space with

$$\alpha_n = \frac{\gamma_{n+1}-\gamma_{n}}{|\gamma_{n+1}-\gamma_{n}|}$$

The vectors $\gamma_n$ are the vertex of the polygon and in the formula the indices are meant modulus N so that $\gamma_0 = \gamma_N$ and $\gamma_{N+1}=\gamma_1$.

If we develop the Taylor series expansion of the exponential and we get the zero-th order term we have:

$$\sum_{n=1}^N \frac{\alpha_n\times\alpha_{n-1}}{(\omega\cdot\alpha_n)(\omega\cdot\alpha_{n-1})}
$$

which seems to be zero for any $\omega$ but for the moment I haven't figured out a proof.

Can someone point out a possible proof or otherwise give some hint about how to proceed.

I guess also that the first order term of the expansion is also zero and the second order term should be the oriented area of the polygon.

答:
I think I found the way to proof the value given above is zero for any $\omega$. I don't give here the full details but I proceeded in the following way:

show it is zero for a triangle

for induction, it is true for a polygon with n vertex, add a new vertex and show it is still true

To show it is zero for a triangle I expanded everything in terms of $\gamma_n$ and then I noted that:

$$(\gamma_3-\gamma_2)\times(\gamma_2 - \gamma_1) =(\gamma_1-\gamma_3)\times(\gamma_3 - \gamma_2)$$

because both sides are equal to twice the area of the oriented triangle. The same is true for the other permutations of the indices.

To show that it remains true when a triangle is added in a polygon is quite easy if expanded in term of the function:

$$t(\alpha_i,\alpha_j)=\frac{\alpha_i\times\alpha_j}{(\omega\cdot\alpha_i)(\omega\cdot\alpha_j)}$$

It is enough to write the expression and note that the term corresponding to the side in common between the triangle and the polygon vanish because the function $t$ is anti-symmetric.

Otherwise, for the terms arising from the first and second terms of the Taylor expansion of the exponential, I don't know but I suspect they can be shown to be equal to zero and the area of the triangle respectively using a similar approach.


问:
Prove each edge is part of a triangle in a random graph over a certain threshold - help

Hello, I'm having some issues with proving this following problem. My guess would be to use a first moment method and proving the expectancy of a random variable which counts the amount of edges that aren't part of a triangle tends to zero as n approaches infinity. But I'm not too sure how to break this down so it works.
Any suggestions?

答:
Here's a strategy that will work.

First, for two fixed vertices $u$ and $v$, find the probability that they provide a counterexample to $\mathcal P$: that is, $uv \in E(G)$, but for every third vertex $w$, either $uw \notin E(G)$ or $vw \notin E(G)$. This probability can be computed exactly, though you'll later need to simplify it to an upper bound that's easier to work with.

Second, use linearity of expectation to find the expected number of pairs that provide a counterexample to $\mathcal P$.

Finally, if this goes to $0$ as $n \to \infty$, we can conclude that the probability that any pair will provide a counterexample also goes to $0$, and therefore $\mathcal P$ holds with high probability.

(Instead of linearity of expectation, we can phrase this in terms of using the union bound, if you prefer.)


问:
Differential equation with the fundamental theorem of Calculus
We have a first order linear differential equation $$y'=xy+2x$$
and a continuous function $f$ satisfies that $$f(x)=x^2+\int_{0}^{x}t f(t) dt$$, Show that $f$ is the unique solution to the first linear differential equation $$y'=xy+2x$$ with the initial condition $y(0)=0$

I tried to solve the differential equation and the solution is $$y=Ce^{x^2/2}-2$$, and with the initial condition, we get $C=2$,
and I was trying to differentiate $f(x)$ using Fundamental Theorem of Calculus, but I have no idea how to differentiate  $\int_{0}^{x}t f(t) dt$. Or is this the right way to solve this question? Anyone knows how to do this?
Thank you.

答:
Note that $\frac{d}{dx}\int_{0}^{x}g(t)dt$ = $\frac{d}{dx}(G(x)-G(0))$, where $G(x)$ is the antiderivative of $g(x)$. This is a direct consequence of the Fundamental Theorem of Calculus. However, $\frac{d}{dx}(G(x)-G(0)) = g(x)$. Thus:

$$\frac{d}{dx}\int_{0}^{x}tf(t)dt = xf(x)$$


问:
ODE $y'+ x\sin( 2y) = x e^{-x^2} \cos^2 (y)$
I have the following ODE:

$$y'+ x\sin (2y) = x e^{-x^2} \cos^2 (y)$$

I'm stuck trying to get it into a linear form. I've tried $\sin (2y) = \sin y \cos y $ and then dividing the ODE by $( \cos  y ) ^{-1} $. This got me to nothing so I tried dividing by $( \sin y )^{-2} $ instead. I got stuck as well. I think I need to make a substitution but I don't know which one.

Thanks.

答:
HINT

$$y'+x\sin(2y)=xe^{-x^2}\cos^2y\\\implies\sec^2y\  y' + 2x\tan y=xe^{-x^2}\ , \text{\{let $\tan y =t\implies \sec^2 y \ y'= t'$\}}\\\implies t' +2xt =xe^{-x^2}$$


问:
Existence of a variety with prescribed properties
In these notes that give a proof of the Weil conjectures for curves, the author writes on page 17 that given a smooth projective curve $X$ over a finite field $k = \mathbb{F}_q$ for a fixed prime $q$, there exists a variety $J(X)$ over $k$ such that for every field extension $k'$ of $k$, the $k'$-valued points of $J(X)$ are in bijection with the line bundles of degree zero on $X \times_{\text{Spec }k} \text{Spec }k'$. Moreover, the number of line bundles of $X$ of degree zero is equal to $\left|J(X)(k)\right|$, the number of $k$-valued points of $J(X)$.

I'm confused about where $J(X)$ comes from. How does one explicitly construct the variety $J(X)$ with the above prescribed properties?

答:
As already pointed out in the comments, $J(X)$ is the Jacobian variety of the curve $X$. </p>

<p>It is mentioned in Remark II.6.10.3 of Hartshorne's book and it is covered in detail in the red book of varieties and schemes by Mumford. </p>

Body:
As already pointed out in the comments, $J(X)$ is the Jacobian variety of the curve $X$.

It is mentioned in Remark II.6.10.3 of Hartshorne's book and it is covered in detail in the red book of varieties and schemes by Mumford.


问:
Dual space basis vectors
I have three vectors, with unit normals, that form the basis of a coordinate system, these are

\begin{align}
\boldsymbol{a}_{1} &= [a_{11},\,a_{12},\,0] \,, \\
\boldsymbol{a}_{2} &= [a_{21},\,a_{22},\,0] \,, \text{and} \\
\boldsymbol{a}_{3} &= [0,\,0,\,1] \,.
\end{align}

How do I determine the basis and basis vectors $\boldsymbol{a}^{1},\,\boldsymbol{a}^{2},\,\boldsymbol{a}^{3}$ of the adjoint space which is dual to the original space?

Further, how do I verify that the spaces are in fact dual?

(PS: I think it should satisfy $\left<\boldsymbol{a}_{i},\,\boldsymbol{a}^{j}\right>=\delta_{i}^{j}$ but I am not certain.)

答:
Hint: Do an Ansatz like
$$\boldsymbol{a}^1=P\varepsilon^1+Q\varepsilon^2+R\varepsilon^3,$$
where $\langle\varepsilon^i,e_j\rangle=\delta^i_j$, to set a linear system of equations
$$1=\langle \boldsymbol{a}^1,\boldsymbol{a}_1\rangle$$
$$\qquad\qquad\qquad\qquad\qquad\qquad=\langle P\varepsilon^1+Q\varepsilon^2+R\varepsilon^3,a_{11}e_1+a_{12}e_2\rangle,$$
$$\qquad\qquad\qquad=Pa_{11}+Qa_{12},$$

$$0=\langle \boldsymbol{a}^1,\boldsymbol{a}_2\rangle$$
$$\qquad\qquad\qquad\qquad\qquad\qquad=\langle P\varepsilon^1+Q\varepsilon^2+R\varepsilon^3,a_{21}e_1+a_{22}e_2\rangle,$$
$$\qquad\qquad\qquad=Pa_{21}+Qa_{22},$$

$$0=\langle \boldsymbol{a}^1, \boldsymbol{a}_3\rangle,$$
$$\qquad=R.$$

...etc.

Title: Dual space basis vectors


问:
Understanding Property of PSD Matrix
For some reason I can't wrap my mind around several concepts.

I'm reading the proofs:

The column space of a positive semi-definite matrix A is contained in the column space of A+B?

Rank of the sum of two positive semi-definite matrices

I can't seem to understand one line. Suppose $X \succeq 0$ is positive semidefinite. Let $v \neq 0$ and suppose $v^{T}Xv=0$. The second and first link say then that $v$ is then in the null space of $X$, meaning $Xv=0$

Why must this be the case?

Thanks.

答: On a more fundamental level, if $X$ is positive semidefinite, $(u,v):=u^TXv$ defines a semi-inner product. Hence it satisfies Cauchy-Schwarz inequality. So, if $(v,v)=0$, then $|u^TXv|=|(u,v)|\le\sqrt{(u,u)(v,v)}=0$ for every vector $u$. Hence $u^TXv$ is identically zero. Since $u$ is arbitrary, $Xv$ must be the zero vector.

Title: Understanding Property of PSD Matrix


问:
Product of two algebraic varieties is affine... are the two varieties affine?
Let $X_1$ and $X_2$ two algebraic varieties such that their product $X_1\times X_2$ is affine. Are $X_1$ and $X_2$ affine then?

If this is not true, could you give a counterexample?

答:
I'll assume you are in the classical context of varieties over an algebraically closed field.

If $a_2\in X_2$  is an arbitrary point, the subvariety $X_1\times \{a_2\}\subset X_1\times X_2$ is closed in the affine variety $X_1\times X_2$ and is thus  affine.

Since $X_1$ is isomorphic to $X_1\times \{a_2\}$, $X_1$ is affine too.


问:
What function represents this log sine wave?
Please help me find a continuous function suitable for describing a blue dots chart. What is this function like? Thanks. Orange chart is $\ln{x^{\frac{1}{4}}}$

答:
Without exact data it is hard to do anything for certain, but I would do the following:

Fit the baseline (you already did that)

Subtract baseline from your data

To the remainder, try to fit a sinusoid of the form

$$f(x) = g(x) \sin(\omega x + \phi)$$

where $\omega$ and $\phi$ are the angular frequency and the phase and the oscillation and $g(x)$ is some monotonically increasing function. I would try to $g(x) = x^k$ for some constant $k$.

There is a ton of numerical fitting libraries you could use, but honestly you could just guesswork the parameters fast enough I think

Edit: Having had a look at the actual data, I have found that the oscillation frequency was changing with time, so I had to guess that one as well. The final form of the function I arrived at is

$$y(x) = \frac{1}{4}\log x + \frac{1}{16}x^2 \sin(\omega \sqrt{x} + \phi)$$

where $\omega \approx 6.27$ and $\phi \approx 3$. Once again, if you want precise coefficients, you should try to fit the function numerically. Also, I suspect that the baseline is not perfectly described by $\frac{1}{4}\log{x}$, it is a little bit slower than that.

Below is the code I used to arrive at this conclusion

```<code># 1. Plot Data
plt.figure(figsize=(10,10))
plt.plot(x, y)
plt.show()

# 2. Fit Baseline
plt.figure(figsize=(10,10))
plt.plot(x, y)
plt.plot(x, np.log(x)/4)
plt.show()

#3. Subtracting baseline
ybase = y - np.log(x)/4

plt.figure(figsize=(10,10))
plt.plot(x, ybase)
plt.show()

genv = x**(1/2)/16

#4. Guessing envelope function
plt.figure(figsize=(10,10))
plt.plot(x, ybase)
plt.plot(x, genv)
plt.show()

#5. Dividing by the envelope function
yenv = ybase/genv
plt.figure(figsize=(10,10))
plt.plot(x, yenv)
plt.show()

#6. Guessworking the frequency dynamics by choosing a function of the x-argument
yenv = ybase/genv
plt.figure(figsize=(10,10))
plt.plot(np.sqrt(x), yenv)
plt.show()

#7. Guessworking the absolute frequency prefactor and phase
yenv = ybase/genv
plt.figure(figsize=(10,10))
plt.plot(np.sqrt(x), yenv)
plt.plot(np.sqrt(x), np.sin(np.sqrt(x)*6.3 + 2))
plt.show()

#8. Putting it all together
ytot = np.log(x)/4 + x**(1/2)/16 * np.sin(np.sqrt(x)*6.3 + 2)

plt.figure()
plt.plot(x, y)
plt.plot(x, ytot)
plt.show()
</code>```

Body:
Without exact data it is hard to do anything for certain, but I would do the following:

Fit the baseline (you already did that)

Subtract baseline from your data

To the remainder, try to fit a sinusoid of the form

$$f(x) = g(x) \sin(\omega x + \phi)$$

where $\omega$ and $\phi$ are the angular frequency and the phase and the oscillation and $g(x)$ is some monotonically increasing function. I would try to $g(x) = x^k$ for some constant $k$.

There is a ton of numerical fitting libraries you could use, but honestly you could just guesswork the parameters fast enough I think

Edit: Having had a look at the actual data, I have found that the oscillation frequency was changing with time, so I had to guess that one as well. The final form of the function I arrived at is

$$y(x) = \frac{1}{4}\log x + \frac{1}{16}x^2 \sin(\omega \sqrt{x} + \phi)$$

where $\omega \approx 6.27$ and $\phi \approx 3$. Once again, if you want precise coefficients, you should try to fit the function numerically. Also, I suspect that the baseline is not perfectly described by $\frac{1}{4}\log{x}$, it is a little bit slower than that.

Below is the code I used to arrive at this conclusion

# 1. Plot Data
plt.figure(figsize=(10,10))
plt.plot(x, y)
plt.show()

# 2. Fit Baseline
plt.figure(figsize=(10,10))
plt.plot(x, y)
plt.plot(x, np.log(x)/4)
plt.show()

#3. Subtracting baseline
ybase = y - np.log(x)/4

plt.figure(figsize=(10,10))
plt.plot(x, ybase)
plt.show()

genv = x**(1/2)/16

#4. Guessing envelope function
plt.figure(figsize=(10,10))
plt.plot(x, ybase)
plt.plot(x, genv)
plt.show()

#5. Dividing by the envelope function
yenv = ybase/genv
plt.figure(figsize=(10,10))
plt.plot(x, yenv)
plt.show()

#6. Guessworking the frequency dynamics by choosing a function of the x-argument
yenv = ybase/genv
plt.figure(figsize=(10,10))
plt.plot(np.sqrt(x), yenv)
plt.show()

#7. Guessworking the absolute frequency prefactor and phase
yenv = ybase/genv
plt.figure(figsize=(10,10))
plt.plot(np.sqrt(x), yenv)
plt.plot(np.sqrt(x), np.sin(np.sqrt(x)*6.3 + 2))
plt.show()

#8. Putting it all together
ytot = np.log(x)/4 + x**(1/2)/16 * np.sin(np.sqrt(x)*6.3 + 2)

plt.figure()
plt.plot(x, y)
plt.plot(x, ytot)
plt.show()


问:
Elementary interpolation inequality between Lebesgue and Sobolev-Slobodeckij spaces
Let $W^{s, 2}$ for $0 < s < 1$ denote the Sobolev-Slobodeckij spaces on the interval $(0, 1)$ and $L^2$ the Lebesgue space on the same interval.  I'm interested in an elementary proof that there exists $C > 0$ such that for any $f \in W^{s, 2}$ there holds

$$ \| f \|_{W^{s/2, 2}} \leq C \| f \|_{L^2}^{1/2} \| f \|^{1/2}_{W^{s, 2}}. $$

I'm not super comfortable with interpolation theory, but as far as I know one has $(L^2, W^{s, 2})_{1/2,2} = W^{s/2, 2}$ by the real interpolation method (and reiteration theorem), such that this estimate should be true. I wish to apply a very similar estimate in another context where I cannot use this abstract result directly, which is why I'm interested in an elementary proof of the inequality above.

Let for $0 < s < 1$ the Gagliardo semi-norm of $f \in W^{s, 2}$ be denoted by
$$| f |_{W^{s, 2}} = \left( \int_0^1 \int_0^1 \frac{|f(x) - f(y)|^2}{|x-y|^{2s+1}}\,\mathrm d x \mathrm dy \right)^{1/2}.$$
Then it is sufficient to prove that there is $C > 0$ such that for any $f \in W^{s,2}$ there holds
$$ | f |_{W^{s/2, 2}} \leq C \| f \|_{L^2}^{1/2} | f |^{1/2}_{W^{s, 2}}. $$

I tried to prove this estimate directly, but I can ultimately only prove
$$ | f |_{W^{(s+\mu)/2, 2}} \leq C_\mu \| f \|_{L^2}^{1/2} \| f |^{1/2}_{W^{s, 2}} $$
for $0 < \mu \ll 1$ with $C_\mu \to \infty$ as $\mu \to 0$. Using Hölder's inequality I estimate (omitting the bounds in the integrals for simplicity)
$$\begin{align}
| f |_{W^{(s+\mu)/2, 2}}^2
&\leq \int \int \frac{| f(x) - f(y) |}{|x - y|^{s+\mu+1}} \left( |f(x)| + |f(y)| \right)\,\mathrm dx \mathrm dy\\
&\leq 2 \int |f(y)| \int \frac{| f(x) - f(y) |}{|x - y|^{s+\mu+1}}\,\mathrm dx \mathrm dy\\
&\leq 2 \left( \int |f(y)|^2 \,\mathrm dy \right)^{1/2} \left( \int \left( \int \frac{| f(x) - f(y) |}{|x - y|^{s+1/2}} \frac{1}{|x-y|^{1/2+\mu}} \,\mathrm dx \right)^2 \mathrm dy \right)^{1/2}\\
&\leq 2 \left( \int |f(y)|^2 \,\mathrm dy \right)^{1/2} \left( \int \int \frac{| f(x) - f(y) |^2}{|x - y|^{2s+1}} \,\mathrm dx \mathrm dy \right)^{1/2} \left( \int \int \frac{1}{|x-y|^{1+2\mu}} \,\mathrm dx\mathrm dy \right)^{1/2}
\end{align}$$

such that

$$ |f|_{W^{(s+\mu)/2,2}} \leq C_\mu \|f \|_{L^2}^{1/2} |f|_{W^{s,2}}^{1/2} \quad\text{with} \quad C_\mu = \sqrt 2 \left( \int \int \frac{1}{|x-y|^{1+2\mu}} \,\mathrm dx\mathrm dy \right)^{1/4}. $$

But there holds $C_0 = \infty$ as the integral in the constant is unbounded for $\mu = 0$.

As an alternative approach I again used only Hölder's inequality to show that for any $0 < \mu \ll 1$ there holds

$$ |f|_{W^{s/2,2}} \leq \operatorname{ess\,sup}_{(x, y) \in (0, 1)^2} | x - y|^{\mu/2} |f |_{W^{\mu, 2}}^{1/2} |f|_{W^{s,2}}^{1/2} $$

but I read in the Hitchhikers Guide to Fractional Sobolev Spaces that only $\lim_{\mu \to 0} \mu |f|^2_{W^{\mu, 2}} = C \| f \|_{L^2}^2$ holds and the essential supremum in the last inequality is actually $1$, so the constant again blows up as $\mu \to 0$.

答:
Using a Fourier approach the interpolation inequality follows easily. In contrast to the original question I consider the function spaces with domain $\mathbb R^n$ for any $n \in \mathbb N$, the estimate for the interval $(0, 1)$ follows for $n = 1$ since $(0, 1)$ is an extension domain.

According to the Hitchhicker's Guide to Fractional Sobolev Spaces (Prop 3.4), there holds

$$ | u |_{W^{s, 2}(\mathbb R^n)}^2 = C(s, n) \int_{\mathbb R^n} |\xi|^{2s} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi $$

with some factor $C(s, n)$, where $\mathcal F$ denotes the Fourier transform. Then

$$ \begin{align}
| u |_{W^{s/2, 2}(\mathbb R^n)}^2
&= C(s/2, n) \int_{\mathbb R^n} |\xi|^{s} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi \\
& \leq C(s/2, n) \left( \int_{\mathbb R^n} |\xi|^{2s} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi \right)^{1/2} \left( \int_{\mathbb R^n} |\mathcal Fu(\xi)|^2 \,\mathrm d\xi \right)^{1/2}\\
& \leq C(s/2, n) C(s, n)^{-1/2} |u|_{W^{s, 2}(\mathbb R^n)} \| u \|_{L^2(\mathbb R^n)}
\end{align}$$
using Hölder's inequality and Plancherel's theorem.

While this answers the original question I'm still very interested in how one can directly argue using the Gagliardo semi-norm.


答:
If you integrate both sides of the equation, you get

$$
\int_0^x y'(t) dt = \int_0^{x} t y(t) dt +\int_0^x 2t dt \Leftrightarrow y(x) = \underbrace{y(0)}_{=0}+\int_0^x t y(t) dt + x^2
$$

So, any solution of the equation is of the proposed form. On the other hand, this is a linear equation, for which you have existence and uniqueness of solution (and even a closed form solution).

Title: Differential equation with the fundamental theorem of Calculus

Body:
We have a first order linear differential equation $$y'=xy+2x$$
and a continuous function $f$ satisfies that $$f(x)=x^2+\int_{0}^{x}t f(t) dt$$, Show that $f$ is the unique solution to the first linear differential equation $$y'=xy+2x$$ with the initial condition $y(0)=0$

I tried to solve the differential equation and the solution is $$y=Ce^{x^2/2}-2$$, and with the initial condition, we get $C=2$,
and I was trying to differentiate $f(x)$ using Fundamental Theorem of Calculus, but I have no idea how to differentiate  $\int_{0}^{x}t f(t) dt$. Or is this the right way to solve this question? Anyone knows how to do this?
Thank you.


问:
Study the convergence of $ \sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$
I have to study the convergence of the following series

$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$

What I tried was to compare the following terms:

$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \le \frac{\sqrt{n}}{n \sqrt[3]{n}}  = \frac{1}{n^{5 / 6}}$$

We know that

$$\frac{1}{n^{5/6}} \rightarrow 0 \text{, as } n \rightarrow \infty$$

so using the First Comparison Test I concluded that the series

$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$

is convergent. However, I think I got something wrong. The First Comparison Test tells us that if the series is convergent, then the limit of the term that makes the series is $0$, not the other way around. So I think my argument is invalid. How should I approach this?

答:
$$
\frac{\sqrt n}{n\sqrt[3]n+2}
\ge \frac{\sqrt n}{n\sqrt[3]n+n\sqrt[3]n}
=\frac12\frac{\sqrt n}{n\sqrt[3]n}
=\frac12n^{-\frac56}
$$
hence your series diverges.


答:
Intuitively, since the sequence is very similar to $n^{-5/6}$ it should diverge (as $\sum n^{-1}$ diverges)

For sufficiently large $n$ (say $n > N$), we have $n\sqrt[3]n +2 < n \sqrt n$.

Then $$\sum_{n\ge1}\frac {\sqrt n}{n\sqrt[3]n+2} \ge\sum_{n> N}\frac {\sqrt n}{n\sqrt[3]n+2}\ge\sum_{n> N}\frac {\sqrt n}{n\sqrt n} = \sum_{n> N}\frac {1}{n} \to \infty$$

and hence the sum diverges by comparison test.

Title: Study the convergence of $ \sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$

Body:
I have to study the convergence of the following series

$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$

What I tried was to compare the following terms:

$$\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \le \frac{\sqrt{n}}{n \sqrt[3]{n}}  = \frac{1}{n^{5 / 6}}$$

We know that

$$\frac{1}{n^{5/6}} \rightarrow 0 \text{, as } n \rightarrow \infty$$

so using the First Comparison Test I concluded that the series

$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2}$$

is convergent. However, I think I got something wrong. The First Comparison Test tells us that if the series is convergent, then the limit of the term that makes the series is $0$, not the other way around. So I think my argument is invalid. How should I approach this?


问:
Find polynomials such that $(x-16)p(2x)=16(x-1)p(x)$
Find all polynomials $p(x)$ such that for all $x$, we have $$(x-16)p(2x)=16(x-1)p(x)$$

I tried working out with replacing $x$ by $\frac{x}{2},\frac{x}{4},\cdots$, to have $p(2x) \to p(0)$ but then factor terms seems to create a problem.

Link: http://web.mit.edu/rwbarton/Public/func-eq.pdf

答:
I love Moron's approach.  Because it is usually interesting to look at other solutions, consider what happens if you just write $p(x) = p_n x^n + p_{n-1}x^{n-1} + \cdots + p_0$ and write out the equation.  Comparing the coefficients of $x^k$ ($k \gt 0$) yields the recursion

$p_k= \frac{2^{k-5}-1}{2^k - 1} p_{k-1}$,

whence $p_5 = 0$, implying $p_6 = p_7 = \cdots = 0$, and the rest of the recursion yields $p_4, p_3, \ldots, p_0$ as multiples of $p_4$.  This approach, although pedestrian, also demonstrates that we need only assume $p$ is analytic near 0.

OwnerDisplayName: anonymous

ContentLicense: CC BY-SA 2.5


答:
It is well-known that:

$$\sum_{n \geq 1} \frac{1}{n^a}$$

diverges when $a \leq 1.$

In your case, since:

$$\sum_{n \ge 1} \frac{\sqrt{n}}{n \sqrt[3]{n} + 2} \sim \sum_{n \ge 1} \frac{1}{n^{\frac{5}{6}}},$$

then your series diverges.

The symbol $\sim$ stands for:

$$\lim_{n \to +\infty} \left(\frac{\sqrt{n}}{n \sqrt[3]{n} + 2} - \frac{1}{n^\frac{5}{6}}\right) = 0$$


问:
Carrying weight of $270$ kg
Problem: A total of $270$ kg of watermelons, each weighing at most $7$ kg, must be transported by $11$ carriers at once. Show that if each carrier can carry up to $30$ kg at a time, this can be done regardless of the weight of the individual watermelons.

$\\$

My opinion about the solution: Let the number of watermelons be $n$. Let the weight of the watermelons be $0< x_1 \leq x_2 \leq \dots \leq x_n \leq 7$. Since $$\sum_{k=1}^{n} x_k = 270 $$
and $7n \geq n\cdot x_n \geq 270$ we find $n \geq 39$.

For $39 \leq n \leq 44$, $\{ x_1, x_2, x_3, x_4\},  \{ x_5, x_6, x_7, x_8\}, \dots, \{ x_{41}, x_{42}, x_{43}, x_{44}\}$ groups can be transported by $11$ carriers.

For $n>44$, I have no enough idea. For sufficiently large $n$ values, some weights of watermelons will decrease. Thanks for your interest...

答:
Assume that we distribute the melons as follows. Sort them by decreasing weight (starting from the heaviest), and always give the next melon to some carrier who has sufficient unused carrying capacity. I claim that this algorithm always succeeds in distributing the entire load to the eleven carriers.

Assume contrariwise that we fail to assign the melon number $n$, weighing $x$ kilos to some carrier. For this to happen we must have already assigned each carrier a load of more than $30-x$ kilos. If $w$ stands for the combined weight of the already assigned melons, we arrive at the inequality $w>11(30-x)=330-11x$.
On the other hand, because at least $x$ kilos remain, we have assigned at most $270-x$ kilos, and hence $w\le 270-x$ (with equality only if this is the last melon).
Putting these pieces together we get
$$
330-11x< w\le 270-x,
$$
from which we (ignoring the temporary dummy variable $w$) immediately solve $x>6$. Recalling that the preceding $n-1$ melons all weigh $\ge x$ kilos, we have at least $n$ melons weighing over $6$ kilos each. Because $nx\le 270$, it follows that $n\le 44$. But the OP's argument shows that the distribution of at most $44$ melons, each weighing at most $7$ kilos, poses no problems in general, and for our algorithm in particular. Therefore the problematic situation cannot arise.

It is worth pointing out that $270$ kilos is the maximum total load the group of carriers is guaranteed to be able to haul. Inspired by the argument we see that if a total load of $270+\epsilon$ kilos consists of $45$ melons, each weighing $6+\epsilon/45$ kilos, then the task becomes impossible. For $45>4\cdot11$, so by the pigeonhole principle at least one carrier needs to haul at least $5$ melons, causing their load to become $30+\epsilon/9$ kilos, which is a bit too much.

Title: Carrying weight of $270$ kg


问:
Suppose we have 44 books on Math, 55 books on English and 66 books on History, In how many ways you can put them on your bookshelf?
Suppose we have $4$ books on Math, $5$ books on English and $6$ books on History. In how many ways you can put them on your bookshelf if you want :-

$1)$ The first book is a math book.

$2)$ All math books are at the beginning.

$3)$ Math and English books will stay together.

$4)$ The first book and the last book are both Math books.

(I'm completely lost with this problem. If someone can give me an explanation that would be great.)

I know that we have $3$ subjects so there are $3$! Or $6$ possibilities and for $4$ math books we have $4!$ or $24$ ways to order. When it is $5$ English books we have $5!$ or $120$ ways to order. If it is $6$ history books we have $6!$ Or $720$ ways to order. So I think the answer is $6 \times 24 \times 120 \times 720=12,441,600$ ways to order the books .

答:
The first book must be a math book:

There are $4$ choices for that book.

One you choice that math book, there are $14$ books remaining.  They can go in any order.  SO there are $14!$ ways to do that.

So there are $4 \times 14!$ ways to do problem 1).

Math books must go first.

There are $4$ math books that must go in any order in the first $4$ positions.  There are $4!$ ways to do that.

There are $11$ books remaining so there are $11!$ ways to do that.

so there are $4!\times 11!$ ways to do that.

....

can you continue?


问:
Deriving formula for a series
Is it possible to derive a formula for this series? If there is, how can it be derived?

$$\sum_{R=1}^{K-1} (R\times2^R)$$

Title: Deriving formula for a series

Body:
Hint:

$$x\frac{d}{dx}\left(\sum_{n=0}^Nx^n\right)=\sum_{n=0}^Nnx^n$$


问:
Parametrise distribution of vectors with non-negative dot product with known vector under some contraints
I'm looking for a general way to parametrise the distribution of of vectors that have a non-negative dot-product with a known vector. I believe this is a more general version of the question asked in all vectors making $ \text{negative dot product} $ with the vector $(1,1,1)$..

The motivation of the problem is to parametrise the distribution of vectors that point away from the origin of a 3-D (Cartesian) space at a location $\vec{r}$. This is required to find which possible vectors point away from the centre of the volume to provide gradient constraints to a Gaussian Process.

If I understand the related question correctly, my first step should be to find the plane that is normal to $\vec{r}$ (which I remember being taught how to do once upon a time, but have since forgotten - I'm sure I can figure it out again, but pointers would be helpful). However, once I have identified that plane, I am struggling to understand how I determine the distribution of vectors that meet my needs. Ideally I would like to sample from the distribution of x, y and z components for these vectors. I can place some constraints on the range of projections in the direction of $\vec{r}$ to limit the length of these vectors.

Any help would be appreciated, and I will happily try to clarify anything that is unclear!

答:
The answer is

$$\vec{v}(\alpha, \beta, \gamma) = \alpha \vec{e}_r + \beta \vec{e}_b + \gamma \vec{e}_c$$

Where $\alpha, \beta, \gamma$ are real numbers, and $\vec{e}_r, \vec{e}_b, \vec{e}_c$ are 3 orthonormal vectors. So $\vec{e}_r$ points in the same direction as your $\vec{r}$, ant the other two vectors are orthogonal to it. Additional constraint that you require is $\alpha \geq 0$, whereas the other two numbers can be arbitrary.

The only remaining thing to do is to find the vectors $\vec{e}_b, \vec{e}_c$. The procedure to do this is called Gram-Schmidt process.


问:
Solve $9x$ $\equiv 4 \mod1453$
In Underwood Dudley's Number theory book second edition chapter 5 problem 7 I encountered this problem:

Solve $9x\equiv 4 \mod1453$

I know that since $gcd(9,1453)=1$, there exists a unique solution. I found the answer in the back to be 1292 but I have no clue how to find this without doing guess and check--a method I'm sure is not very efficient here.

Does anyone know how to go about solving this? If it helps, 1453 is prime.

Thank you!

答:
Since the OP threw out the 'guess-and-check' approach, we start by examining

$\quad 9x = 4 + 1453k \quad k \ge 1$

and try $k = 1$.

Applying Euclidean division,

$\quad 1457 = 9 \cdot 161 + 8$

allowing us write a truth chain of equivalent statements,

$\quad 4 \equiv 9 \cdot 161 + 8 \pmod{1453} \quad \text{iff }$

$\; -4 \equiv 9 \cdot 161 \pmod{1453} \quad \quad \;\, \text{ iff }$

$\quad 4 \equiv 9 \cdot (-161) \pmod{1453} \quad \, \text{iff }$

$\quad 4 \equiv 9 \cdot 1292\pmod{1453}$

and we have a solution, $x \equiv 1292\pmod{1453}$.

Title: Solve $9x$ $\equiv 4 \mod1453$


Title: ODE $y'+ x\sin( 2y) = x e^{-x^2} \cos^2 (y)$

Body:
I have the following ODE:

$$y'+ x\sin (2y) = x e^{-x^2} \cos^2 (y)$$

I'm stuck trying to get it into a linear form. I've tried $\sin (2y) = \sin y \cos y $ and then dividing the ODE by $( \cos  y ) ^{-1} $. This got me to nothing so I tried dividing by $( \sin y )^{-2} $ instead. I got stuck as well. I think I need to make a substitution but I don't know which one.

Thanks.

Body:
$$y'+ x\sin( 2y) = x e^{-x^2} \cos^2 (y)$$
Multiply by $2x'$ and substitute $u=x^2$:
$$2+ 2xx'\sin( 2y) = 2xx' e^{-x^2} \cos^2 (y)$$
$$2+ u'\sin( 2y) = -( e^{-u})' \cos^2 (y)$$
$$2(\tan y)'+ 2u'\tan( y) = -( e^{-u})' $$
Multiply by $e^u$:
$$2(e^u\tan y)' = -e^u( e^{-u})' $$
$$2d(e^u\tan y) = -e^ud( e^{-u}) $$
Integrate.
$$\boxed {2e^{x^2}\tan y -x^2=C} $$

Note that you can also rewrite the DE as:
$$(e^{x^2} \tan  (y))'=x$$
And integrate.


问:
A maximal Hoeffding's inequality?
Let $X_1, \cdots, X_n$ be real-valued independent random variables satisfying $|X_k|\le 1$ and $\mathbb EX_k=0$. Hoeffding's inequality tells us that for any $k=1,\cdots, n$ and $t>0$,
$$\mathbb P\Big( \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le 2 e^{-t^2/2}.$$

My question is whether there exists a similar bound for the maximum over $k$. More precisely:

Question: Do there exist absolute constants $C>0$ and $A>0$ so that
  $$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le C e^{-t^2/A}$$
  holds for all $t>0$? If not what can we say about the left hand side?

答:
As it stands, the maximal inequality cannot hold since we would get by letting $n$ going to infinity that
$$
\mathbb P\Big( \sup_{k\geqslant 1} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le C e^{-t^2/A}.
$$
However, by the law of the iterated logarithms, the random variable $\sup_{k\geqslant 1} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | $ is almost surely infinite, unless $X_i=0$ for each $i$.

However, it is possible to prove that
$$
\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{n}} \Big | \ge t \Big) \le C e^{-t^2/A}.$$
To see this, let $M_n:=\max_{1\leqslant k\leqslant n}\lvert  X_1+\cdots+ X_k\rvert$ and $S_n=\lvert  X_1+\cdots+ X_n\rvert$. By Doob's inequality,
$$
x\mathbb P\left(M_n>x\right)\leqslant \mathbb E\left[S_n\mathbf{1}_{\{M_n>x\}}\right].
$$
Write the last expectation as an integral of the tail over the positive real line and truncation this integral at $x/2$ to get that
$$
x\mathbb P\left(M_n>x\right)\leqslant 2\int_{x/2}^\infty \mathbb P\left(S_n>t\right)dt.
$$
Then use Hoeffding's inequality.

Title: A maximal Hoeffding's inequality?


问:
Find the smallest subspace of $\Bbb{R}^2$ which contains 2 elements
Cheers, after the last question of mine regarding the smallest subspace of a vector space I am here for a follow up!

I would gladly appreciate if everyone could provide a solution or at least a comment to guide me, instead of downvoting everything.

Let the vector space $\Bbb{R}^2$ have two elements, $(11,13)$ and $(20,22)$. How can I find the smallest subspace, which contains both of these elements? From the last question, I now know that if A is the smallest subspace of $\Bbb{R}^2$ containing (11,13), then $A = \{ t(11,13): t \in \Bbb{R} \}$ and if B is the smallest subspace of $\Bbb{R}^2$ containing $(20,22)$, then $B = \{ λ(20,22): λ \in \Bbb{R} \}$. As noted, since the subspace has to have multiples of these points as elements, I can come up with the previous sets, which also hold for addition and the zero-vector (scalar multiplication trivially holds). What about two points though? Is there a way to find the span of this set and come up with the answer? Also if I wanted the intersection of A and B, would I be able to find this set, if I regarded these sets as lines that go through the origin? e.g. the line $11y-13x=0$ seems to produce the same vector space as A. Is this possible or should I correct my thinking?

答:
If $V$ is a vector space (as $\mathbb{R}^2$ is), and $u$ and $v$ are two elements of $V$, then the smallest subspace of $V$ containing both vectors is called the span of the vectors. There are three distinct cases for $u$ and $v$, giving rise to different dimensions for the span.

$u$ and $v$ are linearly independent. In other words, the only pair of scalars $\alpha, \beta$ such that $\alpha u + \beta v = 0$ is $\alpha=\beta=0$. In this case, $\text{span}(u, v)$ has dimension 2, and in your example therefore has to equal $\mathbb{R}^2$.

$u$ and $v$ are not linearly independent, but also not both zero. Then without loss of generality, $u \ne 0$ and $v=\lambda u$ for some scalar $\lambda$ (possibly $\lambda=0$). This implies that $\text{span}(u, v) = \text{span}(u) = \{tu, t \in \mathbb{R}\}$, which has dimension 1. Note that if one of the vectors is zero, then it is impossible for them to be linearly independent.

$u = v = 0$. Then $\text{span}(u, v) = \{0\}$, which has dimension 0.

To find the intersection of the two subspaces $\text{span}(u)$ and $\text{span}(v)$, note that if $w \in \text{span}(u)$ then $w = \alpha u$ for some $\alpha$ (possibly 0). Similarly $w = \beta v$. There are two cases:

$w = 0$: $0$ is in the intersection of any two subspaces.

$w \ne 0$: Therefore, $\alpha, \beta \ne 0$ and $u, v \ne 0$, but $\alpha u - \beta v = 0$. Therefore $u, v$ are linearly dependent and not both zero - the spans are equal lines. Conversely, if $u,v$ are linearly dependent then you easily get a non-zero element of the intersection.

Finally, to answer the specific statement you made: the line defined by the equation $11y-13x=0$ and $\{t(11,13)\}$ are the same line. This is because:

If $v = (x,y) = t(11, 13)$, then $11y-13x = 11*13t - 13*11t = 0$, and

If $11y-13x=0$, then $y = \tfrac{13}{11}x$, so $(x, y) = (x,
   \tfrac{13}{11}x) = \tfrac{1}{11}x(11, 13) = t(11, 13)$.


问:
Need help to create formula/equation
I am looking to try create a formula/equation(I am a novice). I'll use a fictitious example that has to do with basketball.  Assume there are 5 players that score in each game for each team.  Team A: C + PF1 + PF2 + G1 + G2 = Total Points scored ; Team B: C + PF1 + PF2 + G1 + G2 = Total Points scored.

What I have found is if I subtract the C points from the winning team that they still win X % of the time.  The actual number is 61.7% but that number is based on just 1800 games worth of research.

So, how would I write a formula/equation to fit my findings?

I know that this probably isn't a very difficult thing to do but I am not a mathematician and it isn't something that I can do.  I am a writer and really want to be able to show what my research shows using a mathematical formula.

PLEASE HELP!  TIA.

答:
You successfully found the empirical percentage of a team that wins still be able to win without the center scoring. That is
$$\frac{\text{number of times A wins and A still wins without center scoring} }{\text{number of times A wins}} \approx 61.7\%$$
for the $1800$ specific data that you took.

If you meant something else let me know.

Title: Need help to create formula/equation


问:
overall complexity of additive algorithm
Assume, for some fixed integer $n$ the computational algorithm consists of two consecutive parts. Part 1 has computational complexity of $O(n_{1})$ and Part 2 has complexity of $O(n_{2})$, and $n_{1} + n_{2} = n$.

What is the overall complexity of the whole algorithm? Is it $O(n)$?

答:
The combined complexity of the algorithm is indeed $O(n_1+n_2)$. If $f_1(n),f_2(n)$ represent the worst case time of the first and second part of the algorithm for an input of size $n$ and $f_1(n)\in O(n_1(n)),f_2(n)\in O(n_2(n))$, then we have for some $M_1,M_2,n_{01},n_{02}\in\Bbb N$,$$\begin{align*}f_1(n)&\le M_1n_1(n)\forall n\ge n_{01}\\f_2(n)&\le M_2n_2(n)\forall n\ge n_{02}\\f_1(n)+f_2(n)&\le M_1n_1(n)+M_2n_2(n)\\&\le\max\{M_1,M_2\}(n_1(n)+n_2(n))~~\forall n\ge\max\{n_{01},n_{02}\}\end{align*}$$So $f_1(n)+f_2(n)\in O(n_1+n_2)$.

Note that we may be able to provide tighter bound on $f_1+f_2$ in case $n_1\in o(n_2)$ or $n_2\in o(n_1)$. In the former case, $f_1+f_2\in O(n_2)$ and similarly for the latter case. For example, if $n_1(n)=n^3,n_2(n)=n^4$ then $f_1(n)+f_2(n)\in O(n^4).$

This simplification is not possible when $f_1,f_2$ depend on different variables. Say $f_1$ depends on the number of rows $r$ of an inputed two dimensional array as $f_1(r)\in O(r^3)$, and let $f_2$ depend on its number of columns $c$ as $f_2(c)\in O(c^4)$. Then $f_1+f_2\in O(r^3+c^4)$ can't be simplified to $O(c^4)$.