问:
Prove that three segments which intersect a circle pass through the same point
In a scalene triangle $\triangle ABC$ with $AB\ne AC$, I state we have $Y$ which is the point of intersection of the bisector of $\angle A$ with $BC$ and $D$ is the point where the perpendicular line from $A$ towards $BC$ intersects $BC$. If $E$ and $Z$ are the intersections of the circumcircle of $\triangle AYD$ with $CA$ and $AB$ respectively, prove that $AD, BE, CZ$ pass through the same point

I don't know how to solve this question. The first thing I thought of was maybe proving that their intersection is the orthocenter of the triangle or something like that, but that did not lead me to a solution. I then thought of using power of a point towards the circle, but that didn't work either. Could you please explain to me how to solve this question?

答:
Hint: The idea is to use Ceva, and we have to show that $ \frac{ BZ}{ZA} \times \frac{ AE } { EC } \times \frac{CD}{DB } = 1  $.

This can be done by side length chasing (either the lengths themselves, or the ratio of the lengths).

 $BZ = BY \cos \beta$ , $CE = CY \cos \gamma$.

 $\frac{BY}{CY} = \frac{AB} {AC} = \frac{\sin \gamma } {\sin \beta} $ , so $ \frac{BZ}{CE} = \frac{ \tan \gamma } { \tan \beta}  $

 $AZ = AY \cos \alpha/2 = AE$

 $\frac{DC}{BD} = \frac { AD / \tan \gamma}{ AD / \tan \beta } $


问:
Finite covering of bounded, open sets
Consider $\Omega\subset \mathbb{R}^n$ open and bounded. Let $c>0$ be given. Is it possible fo find $V_1,...,V_n\subset \Omega$ with $|V_i|\leq c$ for all $i$, $|V_i\cap V_j|=0$ for $i\neq j$ and $ \Omega \subset \bigcup_{i=1}^n V_i$.

I think this can be done by taking $(V_i)_{i=1}^n$ to be a rhombuses. Nevertheless I don't have any reference. I would be thankful for any reference or any other idea.

答: I recognized that by the above subset inclusion I'm asking for $(V_i)_{i=1}^N\subset \Omega$, $|V_i\cap V_j|=0$, $|V_i|\leq c$ for a fixed number $c$ so that $\Omega=\bigcup_{i=1}^N V_i$.

ContentLicense: CC BY-SA 4.0

Title: Finite covering of bounded, open sets


问:
Best tools to teach animations of solids of revolution in a Calculus II class?
I would like to show animations of solids of revolutions particularly to those students who are not visualizers. Is there a tool that I (and preferably my students also) can use to create animations? It would be great to know what you use in classroom and how successful it has been.</p>

答: CalcPlot3D has the ability to plot a surface of revolution, and it lets you animate the revolution, as well as display washers/shells for visualizing volumes.


问:
Approximating large quadratic optimization problems
For some positive-definite matrix $A \in \mathbb{R}^{K \times K}$ I want to solve the quadratic optimization problem
$$\max_{x\in [0,1]^K} x^T A x \\ \text{s.t.} \\ \sum_{i=1}^{K}x_{i}=1$$
The problem is that $K$ is very large. Are there useful approximations to the quadratic form $x^TAx$ that would allow an approximate solution? Is dimensionality reduction of $A$ a useful idea?

答:
If you are taking $\max$ then your problem is not convex, and you are having serious issues. If you wanted $\min$ and did not insist on the positivity, then the solution is just $A^{-1} \mathbb{1}/\mathbb{1}^T A^{-1} \mathbb{1},$ where $\mathbb{1}$ is the vector of all ones.
If you did not have the sum constraint either, then you are in luck with $\max$ - it is just the eigenvector of the largest eigenvalue, which is easy to find by iterative methods.


问:
How to prove this function is surjective
I'm trying to solve this question:

In order to solve this question above, I found this function: $r/w\mapsto (r/s)/(w/s)$ such that $w/s\in T$, I almost proved this map is an isomorphism, I'm stuck just in the surjectivity part.

If we get an element $(r/s)/(w/s)$ of $T^{-1}(S^{-1}R)$, ok!
However, an element of $T{^1}(S^{-1}R)$ can be for example $(r/s_1)/(w/s_2)$ with $s_1\neq s_2$

I need help in this part.

Thanks in advance.

答: Hint: Note that in $S^{-1}R$ we have $(ws_1)/(s_1s_2) = w/s_2 \in T$, so $ws_1 \in S_*$. Now consider $rs_2/ws_1 \in S_*^{-1}R$.

ContentLicense: CC BY-SA 3.0


问:
If I roll 20 six sided dice, what is the chance of there not being a 1?
Assume that the dice are fair. Wouldn't the answer be 5/6^20? I don't really know.</p>

答:
Yes, provided that the die is fair and that the rolls are independent. This is because
$$
P(\text{No one's after 20 rolls)}=P(\text{No one on the first roll)} \times P(\text{No one on the second roll}) \times \dots \times P(\text{No one on the twentieth roll)} \, .
$$
And the probability of there not being a one on any single roll is $5/6$. Hence,
\begin{align}
P(\text{No one's after 20 rolls}) &= \underbrace{5/6 \times 5/6 \times 5/6 \times \dots \times 5/6}_{\text{20 terms}} \\
&= \left(\frac{5}{6}\right)^{20} \, .
\end{align}


问:
Find values of a parameter p for which the series $\sum_{n=1}^\infty \sqrt{n}\ln^{p} \left(1+ \frac{1}{\sqrt{n}}\right)$
I was requested to find the values of the parameter p for which the following series converges:

$$\sum_{n=1}^\infty \sqrt{n} \ln^{p} \left(1+ \frac{1}{\sqrt{n}}\right)$$

I tried using Cauchy's test and the Term test, but reached a dead-end.
I also tried to use the ratio test with but it didn't seem to be helpful in this situation.

At this point we aren't allowed to use the integral test.

I would appreciate any suggestions on how to approach this problem.

答:
$$\ln(1+x)\sim x;\;\text{ as }x\to 0$$
$$\ln \left(1+\frac{1}{\sqrt{n}}\right)\sim \frac{1}{\sqrt{n}};\;\text{ as }n\to\infty$$
$$\left(\ln \left(1+\frac{1}{\sqrt{n}}\right)\right)^p\sim \left(\frac{1}{\sqrt{n}}\right)^p=\frac{1}{n^{p/2}}$$
The series $$\sum_{n=1}^{\infty} \sqrt n \left(\ln \left(1+\frac{1}{\sqrt{n}}\right)\right)^p$$
converges if in the following fraction $$\sqrt n\frac{1}{n^{p/2}}=\frac{1}{n^{p/2-1/2}}$$
we have p_series test $$\frac{p}{2}-\frac{1}{2}>1\to p>3$$


问:
Factoring out $4x^3+2x^2y-2xy^2-y^3$
This can be factored as follows:

$$4x^3+2x^2y-2xy^2-y^3 = (2x^2-y^2)(2x+y)$$

What is a systematic way for finding this factorization?

Title: Factoring out $4x^3+2x^2y-2xy^2-y^3$

Body:
Hint

$$4x^3-2xy^2=2x(2x^2-y^2)$$

$$2x^2y-y^3=y(2x^2-y^2)$$

Can you take it from here?


问:
Size of a set of integers
I am truing to prove that the size of the set:

$$R(n,r)=\left\{1\leq x\leq n: \left\lfloor \frac{(x-1)r}{n}\right\rfloor<\left\lfloor \frac{xr}{n}\right\rfloor \right\}$$
is exactly $r$, where $r<n$. This shouldn't be so difficult but for some reason I am struggling.

$n,x,r$ are all integers.

I appreciate any answer!

答:
Hint: Show that the difference between consecutive terms of $ \lfloor \frac{ x r } { n } \rfloor $ is either 0 or 1.

Since $\lfloor \frac{ 1\times r } { n } \rfloor = 0$ and  $\lfloor \frac{ n\times r } { n } \rfloor = r$, there are exactly $r$ times when the difference is exactly 1.

Thus, the set has size $r$.

Title: Size of a set of integers


答: The systematic way here is whenever you have a sum of 4 terms, you can check to see if factor by grouping works.    In a sum of the form $a+b+c+d$,   factor by grouping works if and only if $ad=bc$.  If it does,  then you pull out the greatest common factor of $a$ and $b$.   When you do so, you will be GUARANTEED to be able to pull a factor out of $c+d$ to match the binomial you got when factoring out $a+b$, as in lab's answer above.

Title: Factoring out $4x^3+2x^2y-2xy^2-y^3$

Body:
This can be factored as follows:

$$4x^3+2x^2y-2xy^2-y^3 = (2x^2-y^2)(2x+y)$$

What is a systematic way for finding this factorization?


问:
What makes $5$ and $6$ so special that taking powers doesn't change the last digit?
Why are $5$ and $6$ (and numbers ending with these respective last digits) the only (nonzero, non-one) numbers such that no matter what (integer) power you raise them to, the last digit stays the same? (by the way please avoid modular arithmetic) Thanks!</p>

答:
It's because, let x be any number, if x^2 ends with x, then x raised to any positive integer(excluding zero) will end with x.

For x^2 to end with x, x(x-1) have to be a multiple of 10 raised to the number of digits in x.

(Ex: if x = 5, then 10^1. If x = 25, then 10^2)

By following this procedure, I have come up with 25 and 625 which ends with themselves, when raised to any positive integer excluding zero.


Title: Factoring out $4x^3+2x^2y-2xy^2-y^3$

Body:
\begin{gather*}
4x^{3} +2x^{2} y-2xy^{2} -y^{3}\\
=\left( 4x^{3} -2xy^{2}\right) +\left( 2x^{2} y-y^{3}\right)\\
=2x\left( 2x^{2} -y^{2}\right) +y\left( 2x^{2} -y^{2}\right)\\
=( 2x+y)\left( 2x^{2} -y^{2}\right)
\end{gather*}
Does this help?


问:
Condition for integrability on finite measure space
Let $(X,\mathcal{F},\mu)$ be a finite measure space. If $f:X\rightarrow \mathbb{R}$ is a measurable real function, show that, $f\in L^1(\mu)$ iff $\sum\limits_{n=0}^{\infty}\mu(\{f\geq n\})<\infty$. Am a bit stuck on the $(\Leftarrow)$ direction so any help is appreciated, thanks.</p>

答:
This is a lovely application of Borel-Cantelli lemma: by assumption, $\Sigma_n \mu(|f| \geq n) < \infty$, hence the aforementioned lemma tells us $\limsup_n \{|f| \geq n\}$ is a null set.

In other words
$$
\mu(\forall n \in \mathbb N\,\exists m \geq n\; |f| \geq m) = 0
$$
which is a convoluted way to say that $f$ is essentially bounded. Therefore
$$
\int |f| \,d\mu \leq M\,\mu(X)
$$
for some finite constant $M \geq 0$, and $f$ is integrable.


问:
Fixed Points in Mobius Transformation
Let $$w(z)=\frac{e^z-1}{e^z+1}; \ z=x+iy$$ and $$w=u+iv$$ Then show that the image of the $y$-axis in the domain is $v$-axis in the co-domain. Does this contradict the result that a Mobius transformation with at least $3$ fixed points is a identity mapping ? Why ?

I have already proved the fact that $y$-axis is mapped onto $v$-axis by showing that
$$w(iy)=i\frac{\sin y}{1+\cos y}$$

Now I need to show that this mapping does not contradict the property of a Moebius transformation. For that I would either have to show that $$\frac{\sin y}{1+\cos y}=y$$ has less than $3$ solutions or that $w(z)$ is not a Moebius transformation. Which one is it and how do I prove that ?

答: Your transformation is not a Mobius transformation.

Title: Fixed Points in Mobius Transformation


答:
Free software GeoGebra from www.geogebra.org will let you do this. Here is an example
https://www.geogebra.org/m/mzWq2Cet and a video showing the construction of a solid of revolution https://www.youtube.com/watch?v=5wOtyT86Qhw.
Your construction can easily be exported as html and embedded in a website too.</p>

Title: Best tools to teach animations of solids of revolution in a Calculus II class?

Body: I would like to show animations of solids of revolutions particularly to those students who are not visualizers. Is there a tool that I (and preferably my students also) can use to create animations? It would be great to know what you use in classroom and how successful it has been.</p>


问:
Is the concept of Matrix Scalar Multiplication the same as scalar multiplications done during row operations?
I would like to know the meaning and implications of scalar multiplication during row operations, as well as general scalar multiplication of Matrices, and the difference between them.</p>

答:
Let's just look at the $2\times 2$ case.  Consider the matrix
$$X = \begin{pmatrix}
a & b\\c & d
\end{pmatrix}.
$$
If for a scalar $\alpha$, we compute $\alpha X$,the result is $$\alpha X = \begin{pmatrix}
\alpha a & \alpha b\\\alpha c & \alpha d
\end{pmatrix},$$ that is every element of $X$ is multiplied by $\alpha$.  We can also express this as $$\begin{pmatrix} \alpha & 0\\0 & \alpha \end{pmatrix}\begin{pmatrix}
a & b\\c & d
\end{pmatrix}. $$  If we do the row operation
$$\begin{pmatrix}
a & b\\c & d
\end{pmatrix}\overset{\alpha R_{1}\to R_{1}}\longrightarrow \begin{pmatrix}
\alpha a & \alpha b\\c & d
\end{pmatrix}$$ then in terms of matrix multiplication, what we are really doing is computing $$\begin{pmatrix}\alpha & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix}
a & b\\c & d
\end{pmatrix} =\begin{pmatrix}
\alpha a & \alpha b\\c & d
\end{pmatrix}.$$

So in some sense, we can think of general scalar multiplication as the row operation that multiplies every row by the same scalar simultaneously, whereas the general row operation for scalar multiplication only multiplies certain rows by certain scalars.


答:
There really isn't much relation that I can see.  We can view scalar multiplication as multiplication by a scalar matrix, that is a diagonal matrix where all the diagonal elements are equal.  $$\begin{bmatrix}a&0\\0&a\end{bmatrix}
\begin{bmatrix}w&x\\y&z\end{bmatrix}=
\begin{bmatrix}aw&ax\\ay&az\end{bmatrix}$$
The scalar multiplication you use in row operations is a bit different
$$\begin{bmatrix}1&0\\a&1\end{bmatrix}
\begin{bmatrix}w&x\\y&z\end{bmatrix}=
\begin{bmatrix}w&x\\aw+y&ax+z\end{bmatrix}$$

I don't know if this will help you, but I hope it does.

Title: Is the concept of Matrix Scalar Multiplication the same as scalar multiplications done during row operations?

Body: I would like to know the meaning and implications of scalar multiplication during row operations, as well as general scalar multiplication of Matrices, and the difference between them.</p>


问:
Antiderivative of $\sin(\frac{1}{x})$, and computing the derivative of $x^2\cos(x^{-1})$ at $0$
I am trying to understand Daniel Fischer's answer to this question Antiderivative of $\sin(\frac{1}{x})$

I understand the ideas and every step except for the part where he said

The first part $x^2\cos(x^{−1})$ is easily seen to be differentiable at $0$ with derivative $0$ there

But isn't $x^2\cos(x^{−1})$ not defined at $x=0$? So how can we talk about its derivative there? I can see that the limit at $0$ exists, but it's still not defined there. Is he using some properties of $F'$ that I am not seeing? Or do we actually have somehow have this point defined to be zero at zero? Since by Show that the function $g(x) = x^2 \sin(\frac{1}{x}) ,(g(0) = 0)$ is everywhere differentiable and that $g′(0) = 0$ we know that the derivative is zero if we define $x^2\cos(x^{−1})$ to take the value zero at zero, so do we have this? Is there something about how we define $F$ and/or $F'$ that forces this to be true?

I can't really be sure of this, any help is appreciated.

答:
$x^2 cos(1/x)$ is actually defined at $x=0$. It can be seen by applying squeeze theorem.
As $-1\leq cos(1/x) \leq 1$ for all $x$ hence
$-x^2\leq x^2 cos(1/x)\leq x^2$.
The limits $-x^2$ and $x^2$ as $x$ approaches $0$ are $0$.
Thus at $x=0$, $x^2 cos(1/x)$ is squeezed between $0$'s and hence itself $0$.


问:
Integral identity for moment generating function
Consider the following identity for a random variable $X$ with mean $0$:

$\displaystyle \mathbb{E}[e^{tX}] = 1+ \int_0^1 (1-y)\, \mathbb{E}\left[\, (tX)^2 e^{ytX} \right] dy$

I've encountered it in the proof of theorem 3.1 of http://www.stat.cmu.edu/~arinaldo/36788/subgaussians.pdf

Can someone please help me understand why it is true?
Thank you.

答: We'll integrate twice by parts, viz.$$\int_0^1f(y)\partial_y^2g(y)dy=[f\partial_yg-f^\prime g]_0^1+\int_0^1f^{\prime\prime}gdy.$$Let $M_X(t):=\Bbb E[e^{tX}]$ so your integral is$$\begin{align}\int_0^1(1-y)\partial_y^2M_X(yt)dy&=[(1-y)\partial_yM_X(yt)+M_X(yt)]_0^1\\&=-\left.\partial_yM_X(yt)\right|_{y=0}+M_X(t)-1.\end{align}$$Since $\Bbb E[X]=0$, $\left.\partial_yM_X(yt)\right|_{y=0}=0$. Now rearrange.


问:
Convergence/divergence of $\sum\limits_{k=0}^{\infty}\sum\limits_{i=0}^k\left(\frac{(-1)^i}{\sqrt{i+1}}\frac{(-1)^{k-i}}{\sqrt{k-i+1}}\right)$
Is the series $C:=\sum\limits_{k=0}^{\infty}\sum\limits_{i=0}^k\left(\frac{(-1)^i}{\sqrt{i+1}}\frac{(-1)^{k-i}}{\sqrt{k-i+1}}\right)$ convergent?

I would say no, because:

Let be $c_k:=\sum\limits_{i=0}^k\frac{(-1)^i}{\sqrt{i+1}}\frac{(-1)^{k-i}}{\sqrt{k-i+1}}$, then we get $c_k=\sum\limits_{i=0}^k\frac{(-1)^{k}}{\sqrt{i+1}\sqrt{k-i+1}}=(-1)^k\sum\limits_{i=0}^k\frac{1}{\sqrt{i+1}\sqrt{k-i+1}}$. As $(c_k)$ is not a null sequence the limit of the partial sums $\lim\limits_{n\to\infty}C_n=\sum\limits_{k=0}^{n}\sum\limits_{i=0}^k\left(\frac{(-1)^i}{\sqrt{i+1}}\frac{(-1)^{k-i}}{\sqrt{k-i+1}}\right)=\sum\limits_{k=0}^{n}(-1)^kc_k$ doesn't exist/ $C$ doesn't converge.

Did I miss something?

答:
Your assertion that $(c_k)$ is not a null sequence is correct, and that is the key step in the proof.

Noting that for $0 \leqslant i \leqslant k$ we have $\sqrt{i+1}\sqrt{k-i+1}\leqslant \sqrt{k+1}\sqrt{k-0+1}= k+1$, it follows that

$$|c_k| = \left|(-1)^k\sum_{i=0}^k \frac{1}{\sqrt{i+1}\sqrt{k-i+1}}\right|= \sum_{i=0}^k \frac{1}{\sqrt{i+1}\sqrt{k-i+1}}\geqslant \sum_{i=0}^k \frac{1}{k+1}= 1$$

Hence, $\limsup_{k \to \infty} |c_k| \geqslant 1$ and, therefore,  $c_k \not\to 0$ as $k \to \infty$.


问:
Are these two expressions the same?
Imagine that you have 4 sets, $A, B, C,$ and $D$. I was wondering if

$$
(A \cap  B) \cup \left[ (A \cup B) \cap (C \cup D) \right] \cup \left[ (A \cup B) \cap (C \cap D) \right]
$$
is the same as

$$
(A \cap B) \cup \left[ A \cap (C \cup D) ] \cup [ B \cap (C \cup D) \right] \cup \left[ A \cap (C \cap D) ] \cup [ B \cap (C \cap D) \right]
$$

and if yes (or no), why?

答:
Notice the second and third term of the first thing have $A\cup B$ intersecting with two different things, then we take the union of them.  So we can factor out the $A\cup B$ to get $$(A\cup B)\cap [(C\cup D)\cup (C\cap D)]$$
Now note that $C\cap D \subseteq C\cup D$ so that union simplifies to $C\cup D$, thus we get down to
$(A\cup B)\cap (C\cup D)$
put that together with the first term and we have
$$(A\cap B)\cup [(A\cup B)\cap (C\cup D)]$$
So elements of that set are either in BOTH A and B,  or are in one of A or B and also in one of C or D.

Now, can you do the same factoring out in the terms on the second?

Title: Are these two expressions the same?


问:
How to find the nth pair from a generated list of pairs?
I have a mathematical question applied on informatic (python). I would like to find the fastest way to get the pair from a given index.

As example, we have all pair combinations of values from 0 to 10:

```<code>impor itertools
combinations = list(itertools.combinations((i for i in range(0,10)),2)
</code>```

like this the variable combinations is equals to

```<code>[(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9), (7, 8), (7, 9), (8, 9)]
</code>```

Thus, I have this function to get the nth pair

```<code>def row_col_from_index(index: int, no_discrete_values: int):
    row_num = 0
    col_num = 0
    seq_value = 0
    level = 1
    no_columns = 10
    while seq_value <= index + 1 - no_columns:
        no_columns -= 1
        seq_value += (no_discrete_values - level)
        level += 1
    row_num = level - 1
    col_num = level + (index - seq_value)
    return row_num, col_num
</code>```

Here we get the 24nth pair (indeed array is 0 based)

```<code>row_col_from_index(23,10)
-> (2, 9)
combinations[23]
-> (2, 9)
</code>```

This work but that will be slow to find the corresponding pair with a huge number of values.
So is there a mathematical trick to speedup the computation like a dichotomy approach or other…

And the reverse question for a given pair how to get its corresponding index ?

thanks

答:
This correctly solves a related problem. @StinkingBishop  answered  the one asked.

If you are looking at the pairs $(i,j)$ for $0 \le i, j \le N$ then you can think of these pairs as the two digit numbers in base $N$ (with the units digit first). So the pair $(i,j)$ will be at position $jN+i$.

For the reverse calculation, the pair at position $p$ will be
$( p\%N, (p - p\%N)/N)$.

(Caveat: The methodology here is well known. I may have the details wrong.)


答:
Sometimes using multiplication and summation notations makes simplification more visible. For left hand
$$(A   B) + \left[ (A + B) (C + D) \right] + \left[ (A + B) C  D) \right]=\\
=A B + (A + B)(C + D+C  D) =\\= A B + (A + B)(C + D)$$
For right hand
$$(A  B) + \left[ A  (C + D) ] + [ B  (C + D) \right] + \left[ A  (C  D) ] + [ B (C  D) \right] = \\
=A  B+(A + B)(C + D)+(A + B)C  D=\\=A B + (A + B)(C + D)$$

Title: Are these two expressions the same?

Body:
Imagine that you have 4 sets, $A, B, C,$ and $D$. I was wondering if

$$
(A \cap  B) \cup \left[ (A \cup B) \cap (C \cup D) \right] \cup \left[ (A \cup B) \cap (C \cap D) \right]
$$
is the same as

$$
(A \cap B) \cup \left[ A \cap (C \cup D) ] \cup [ B \cap (C \cup D) \right] \cup \left[ A \cap (C \cap D) ] \cup [ B \cap (C \cap D) \right]
$$

and if yes (or no), why?


问:
Estimate of smooth integers doesn't fit the actual graph
I was studying smooth integers, that is integers that have only small prime factors. Denoting
by $P^+(n)$ the largest prime factor of $n$, we define $\Psi(x,B):=|\{ n \le x:\,P^+(n) \le B\}|$ as being the function that counts the amount of integers smaller than $x$ such that every prime factor is $\le B$. Wikipedia has the following estimate for $\Psi(x,B)$:
$$\Psi(x,B) \sim \frac{1}{\pi(B)!} \prod_{p\le B} \frac{\log x}{\log p}$$
where $\pi(B)$ denotes the number of primes less than or equal to $B$. I tried to plot an example of it, first plotting $\Psi(x, 5)$ for all positive integers $x < 100$, and then plotting the estimate 1/(3!) * (log x)^3 / (log 2 * log 3 * log 5) on the same interval, but the graphs didn't match at all:

So I guess I have made a mistake somewhere, but I have no idea where. Why does the estimate deviate so much from the actual plot?

答: The ratio between the two gets better for much larger numbers.  Here is the ratio between $\Psi(x,5)$ and the approximation as $x$ goes from 10 to about half a million


问:
$N$ Poisson with parameter $\lambda$ and fixed $a>1$ calculate: $\lim \frac{\mathbb{P}(N\geq k)}{\mathbb{P}(N=k)}$
For $N$ Poisson with parameter $\lambda$ and fixed $a>1$ calculate:
\begin{equation}
            \lim \frac{\mathbb{P}(N\geq k)}{\mathbb{P}(N=k)},
        \end{equation}

$\lambda \rightarrow \infty, k \rightarrow \infty, k/\lambda \rightarrow a$.

I'm thinking we can compute:
\begin{align}
            \frac{\mathbb{P}(N\geq k)}{\mathbb{P}(N=k)} &= \frac{\sum_{i = k}^\infty \frac{\lambda^i e^{-\lambda}}{i!}}{\frac{\lambda^k e^{-\lambda}}{k!}}\\
            &= \sum_{i = 0}^\infty \frac{\lambda^i k!}{(i+k)!}
 \end{align}

But I'm not sure how to simplify the last sum. Some hints would be greatly appreciated!

答:
Hint:

As you noted, the ratio is at most $a/(a-1)$. You can prove a matching lower bound, this time using a finite geometric series.
For any $N$,

$$
    \sum_{i\ge 0}\lambda^i\frac{k!}{(k+i)!}
\ge \sum_{i=0}^{N-1}\lambda^i\frac{k!}{(k+i)!}
\ge \sum_{i=0}^{N-1}\frac{\lambda^i}{(k+N)^i}
=   \frac{1-(\lambda/(k+N))^N}{1-\lambda/(k+N)}
$$
As long $N$ is not too large compared to $\lambda$ and $k$, then $\lambda/(k+N)\sim 1/a$, so the above is $\sim (1-a^{-N})/(1-a^{-1})$. Furthermore, as long as $N\to\infty$, the $a^{-N}$ will disappear. I will leave the details to you.


问:
Why is this a topology but the the other is not?
Given $X=\{a,b,c\}$, why is it acceptable to say that $T=\{\emptyset,X, \{a,b\}\}$ is a topology but $T=\{\emptyset,X, \{a\},\{b\}\}$ is not?

The union of $\{a\}\cup\{b\}$ is $\{a,b\}\notin T$.

So is it because $\{a,b\}\cup\{a,b,c\}$ is $X$ which is in $T$?

Do all unions of all elements in $T$ have to exist in $T$?

答:
Topologies are closed under arbitrary union. Thus, if $A,B \in \tau$, then $A \cup B \in \tau$, for instance.

In your case, for $T = \{\varnothing, X, \{a\},\{b\}\}$, you have
$$
\{a\} \cup \{b\} = \{a,b\} \not \in T
$$
and thus that $T$ is not a topology.

In the case of your other topology, find all possible unions and intersections, and you'll see that each results in a set that is in this (valid) topology.


答:
SO for $\lfloor \frac {(x-1)r}n\rfloor <\lfloor \frac {xr}n\rfloor$ There must be an integer $m$ so that

$\frac {(x-1)r}n= \frac {xr}n - \frac rn  < m \le \frac {xr}n$.

As $\frac rn < 1$ then if this occurs we will have

$m-1 \le \frac {xr}n - 1 < \frac {(x-1)r}n < m \le \frac {xr}n$

IF we assume the OP intended $x$ to be an integer, then the above shows there is only one such $m$ for any such $x$.  ANd by the archemedian principal, for every $m$ there is one such $x$.  SO for every $m$ there is exactly one such $x$.

SO there are exactly as many as there are acceptable integers $m$.  And $m \le \frac {xr}n \le \frac {nr}n = r$ there are $r$ such integers.

Title: Size of a set of integers

Body:
I am truing to prove that the size of the set:

$$R(n,r)=\left\{1\leq x\leq n: \left\lfloor \frac{(x-1)r}{n}\right\rfloor<\left\lfloor \frac{xr}{n}\right\rfloor \right\}$$
is exactly $r$, where $r<n$. This shouldn't be so difficult but for some reason I am struggling.

$n,x,r$ are all integers.

I appreciate any answer!


问:
The definition of tori - topology
The question may seem silly, but I do not find the answer.

For me, the $n-$torus is "an sphere with $n$ holes".

Topologically, this is called the connected sum of $n$ tori, is it right?

What about the $T^n=\overbrace{S^1\times \cdots \times S^1}^{n \,times}$, what is its name and what are its properties?

Do the two surfaces have some relation?

Thank you so much.

答:
What you've described is the n-Torus:

$$
T^n = S^1 \times \dots \times S^1
$$

See the wiki entry: https://en.wikipedia.org/wiki/Torus#n-dimensional_torus

Title: The definition of tori - topology


答:
The $n$-torus is a torus of dimension $n$, it is defined to be the cartesian product of $n$ one-dimensional circle. This is roughly a $n$-dimensional sphere with a $n$-dimensional hole.

What you describe as a sphere with $m$ holes is indeed the connected sum of $m$ different copies of a $2$-dimensional torus.

Those two surfaces are not homeomorphic whenever $n\neq 2$ or $n=2$ and $m\neq 1$.

Title: The definition of tori - topology

Body:
The question may seem silly, but I do not find the answer.

For me, the $n-$torus is "an sphere with $n$ holes".

Topologically, this is called the connected sum of $n$ tori, is it right?

What about the $T^n=\overbrace{S^1\times \cdots \times S^1}^{n \,times}$, what is its name and what are its properties?

Do the two surfaces have some relation?

Thank you so much.


问:
Logarithmic inequation
Good morning, I was studying a function but I got stuck during the study of the signs (by calculating $f(x)\ge0$)

My function is: $f(x)=\log_3 (x^2-9)-\log_{\frac{1}{3}}(x^2-4x+3)$

So it went $\log_3 (x^2-9)-\log_{\frac{1}{3}}(x^2-4x+3)\ge0$

At this point I tried many ways (also to use the same base, and getting $\log_3 (x^2-9)+\log_{3}(x^2-4x+3)\ge0$) but I still can't find any way to resolve it without using the graph.

I thank anyone who is willing to help me.

答:
As the OP observes, the inequality can be re-expressed as

$$\log(x^2-9)+\log(x^2-4x+3)\ge0$$

where both logs are now in base $3$. At this point it doesn't really matter what the base is (so long as it's greater than $1$), the inequality amounts to $(x^2-9)(x^2-4x+3)\ge1$, with the additional stipulation that $|x|\gt3$, so that $\log(x^2-9)=\log(x+3)+\log(x-3)$ and $\log(x^2-4x+3)=\log(x-3)+\log(x-1)$ are both defined. The polynomial inequality expands and simplifies to

$$x^4-4x^3-6x^2+36x-28\ge0$$

which, unless I've made a mistake, has no nice solution. The roots of $x^4-4x^3-6x^2+36x-28$ that satisfy the stipulation $|x|\gt3$ are $x\approx3.26542956311$ and $x\approx-3.00691650216$. (I got these values from an online function calculator; I wouldn't have found them to that many decimal places otherwise.)

Title: Logarithmic inequation


问:
Can I write the difference of a rational and a square root of a rational as a sum of a rational and a square root of a rational?
Suppose that $x = p - \sqrt{q}$ where $p$ and $q$ are rational and $q \geq 0$. Can I always find rational numbers $\tilde p, \tilde q$ such that $\tilde q \geq 0$ and
\begin{align*}
x = \tilde p + \sqrt{ \tilde q}
\end{align*}

答:
Not unless $\sqrt{q}$ is rational.

Rewrite the equation $p - \sqrt{q} = \overline{p} + \sqrt{\overline{q}}$  as
$$ x = \sqrt{q} + \sqrt{\overline{q}} \tag 1$$
where $x = p - \overline{p}$, and square both sides:
$$ x^2 = q + 2 \sqrt{q \overline{q}} + \overline{q}^2$$

Since all other terms are rational, $\sqrt{q \overline{q}}$ must be rational.
If $r = \sqrt{q \overline{q}}$, that says $\overline{q} = r^2/q$.  Equation (1)
then becomes $$x = \sqrt{q} + r/\sqrt{q} $$
so that
$$  x \sqrt{q} = q + r $$
Now $\sqrt{q}\ge 0$ and $\sqrt{\overline{q}} \ge 0$.  Unless they are $0$, $x > 0$
and $\sqrt{q} = (q+r)/x$ is rational.


问:
Unifrom sampling
I came across the term "uniform subsampling" in this paper (page 5). I tried to understand the meaning of it but got no luck.

Here is the extract from the paper:

A naive approach to constructing coresets is based on uniform
subsampling of the data. For the sake of simplicity, consider a data
set $\mathcal{X}$ with constant weights $\mu_{\mathcal{X}}(x) = \frac{1}{|\mathcal{X}|}$ . For any query $Q \in \mathcal{Q}$, the
cost function in (2.1) may be rewritten as
$$
\operatorname{cost}(\mathcal{X}, Q)=\sum_{x \in \mathcal{X}} \frac{1}{|\mathcal{X}|} f_{Q}(x)=\mathbb{E}_{x}\left[f_{Q}(x)\right]
$$
where $x \in X$ is drawn uniformly at random.
Let the set C consist of m points sampled uniformly at random from $\mathcal{X}$
and set $\mu_C(x) = \frac{1}{m}$.

The closes I came across is the answer from here. The definition is:

If a sample is selected from a population which has been grouped into
strata, in such a way that the number of units selected from each
stratum is proportional to the total number of units in that stratum,
the sample is said to have been selected with a uniform sampling
fraction.

I would really appreciate it if someone can explain this concept to me.

答:
If you look at Equation 2.3 of that paper, I believe they are describing uniform subsampling. You just draw samples from a data set with uniform probability. This will non-parametrically approximate the underlying distribution, and is often called boostrapping. In this case (Eq.2.3), they're using bootstrapping to approximate the expected value.

See the wiki entry: https://en.wikipedia.org/wiki/Bootstrapping_(statistics)

In python, you can get a uniform sample easily using libraries like numpy:

```<code>from numpy.random import uniform, randint

num_data = 100

//
// Get 10 samples in the range 0 to num_data
p = uniform(size=(10,)) * num_data
print(p.astype(int))

//
// Alternatively, you can just uniformly sample
// random integers which is probably easier
p = randint(low=0, high=num_data, size=(10,))
print(p)

//
// Now if you have data... you just get it like this
pretend_this_is_data = arange(100)
uniform_data_sample = pretend_this_is_data[p]
</code>```

Title: Unifrom sampling

Body:
If you look at Equation 2.3 of that paper, I believe they are describing uniform subsampling. You just draw samples from a data set with uniform probability. This will non-parametrically approximate the underlying distribution, and is often called boostrapping. In this case (Eq.2.3), they're using bootstrapping to approximate the expected value.

See the wiki entry: https://en.wikipedia.org/wiki/Bootstrapping_(statistics)

In python, you can get a uniform sample easily using libraries like numpy:

from numpy.random import uniform, randint

num_data = 100

//
// Get 10 samples in the range 0 to num_data
p = uniform(size=(10,)) * num_data
print(p.astype(int))

//
// Alternatively, you can just uniformly sample
// random integers which is probably easier
p = randint(low=0, high=num_data, size=(10,))
print(p)

//
// Now if you have data... you just get it like this
pretend_this_is_data = arange(100)
uniform_data_sample = pretend_this_is_data[p]


答:
No, except for the trivial case $\sqrt{q}\in\mathbb{Q}$.

Hint:
$$p-\sqrt{q}=\bar{p}+\sqrt{\bar{q}}\\
\bar{q}=(p-\bar{p}-\sqrt{q})^2=(p-\bar{p})^2+q-2(p-\bar{p})\sqrt{q}$$

This forces $p=\bar{p}$ since $\sqrt{q}$ is irrational, but then $-\sqrt{q}=+\sqrt{\bar{q}}$, a contradiction.

Title: Can I write the difference of a rational and a square root of a rational as a sum of a rational and a square root of a rational?

Body:
Suppose that $x = p - \sqrt{q}$ where $p$ and $q$ are rational and $q \geq 0$. Can I always find rational numbers $\tilde p, \tilde q$ such that $\tilde q \geq 0$ and
\begin{align*}
x = \tilde p + \sqrt{ \tilde q}
\end{align*}


问:
Matrix change of basis: where can it lead?
Consider a positive-semidefinite Hermitian matrix $A$ of dimension $d\times d$. I would like to know, is it always possible to find a change of basis resulting in a matrix $B$ where the diagonal is a constant?
Namely, $B=UAU^\dagger$ with $U$ invertible has $B_{ii}=c$.

If in general such a $U$ does not exist, for which $A$ does it?

P.S. I hope my question is well posed, otherwise I can make edits to state the problem more precisely and concretely.

EDIT 1: what if $U^\dagger U = 1$ ?

答: The answer is yes. In particular, see corollary 5 of this document.</p>


答:
$$\log _3\left(x^2-9\right)-\log _{\frac{1}{3}}\left(x^2-4 x+3\right)\ge 0$$
$$\log _3\left(x^2-9\right)+\log _{3}\left(x^2-4 x+3\right)\ge 0$$
we have to solve the following system
$$\small
\begin{cases}
x^2-9>0\\
x^2-4 x+3>0\\
\log _3\left[\left(x^2-9\right)\left(x^2-4 x+3\right)\right]\ge 0\to \left(x^2-9\right)\left(x^2-4 x+3\right)\ge 1
\end{cases}
$$
that is
$$\small
\begin{cases}
x<-3\lor x>3\\
x<1 \lor x>3\\
x^4-4 x^3-6 x^2+36 x-28\ge 0\to x\leq -3.00692\lor 1.06574\leq x\leq 2.67575\lor x\geq 3.26543
\end{cases}
$$
which give the solution
$$x\leq -3.00692\lor x\geq 3.26543$$

The $4$th degree inequality can't be solved by elementary formulas. You can look here.

Title: Logarithmic inequation

Body:
Good morning, I was studying a function but I got stuck during the study of the signs (by calculating $f(x)\ge0$)

My function is: $f(x)=\log_3 (x^2-9)-\log_{\frac{1}{3}}(x^2-4x+3)$

So it went $\log_3 (x^2-9)-\log_{\frac{1}{3}}(x^2-4x+3)\ge0$

At this point I tried many ways (also to use the same base, and getting $\log_3 (x^2-9)+\log_{3}(x^2-4x+3)\ge0$) but I still can't find any way to resolve it without using the graph.

I thank anyone who is willing to help me.


问:
Integral $\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx$
I would like to ask how to integrate the following integration.

\begin{gather*}
    \int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx
\end{gather*}

I try to use integration by parts and I stuck with the following

\begin{gather*}
   \frac{\ln(x^2+1)}{x} - \int_{0}^{1} d\frac{\ln(x^2+1)}{x^2} \
\end{gather*}

I am confused with the continue step.

答:
As the comments point out, there are a few details you can (and should!) work out yourself, but as a hint, your integration by parts should be (after some simplification)
$$ \lim_{b\to0} \frac{\ln(x^2+1)}{x} \bigg|_{x=1}^{x=b} + 2 \int_0^1 \frac{dx}{1+x^2}. $$


答:
Let $u=\log(x^2+1)$ and $dv=\frac{dx}{x^2}$.

Then $du = \frac{2xdx}{x^2+1}$ and $v=\frac{-1}{x}$.

Therefore
$$
\int_0^1 \frac{\log(x^2+1)}{x^2}dx = \left[\frac{-\log(x^2+1)}{x}\right]_0^1 - \int_0^1\frac{2dx}{x^2+1}
$$
The value of $\frac{\log(x^2+1)}{x}$ at 0 is undefined, but we can compute it with de l'Hôpital's rule:
$$
\lim_{x\to0}\frac{\log(x^2+1)}{x} = \lim_{x\to0}\frac{-2x}{x^2+1} = 0.
$$

Plugging everything in, we obtain
$$
\cdots = -\frac{\log(2)}{1} + 2[\arctan(x)]_0^1= -\log(2) + 2( \frac{\pi}{4}-0) = \frac{\pi}{2}-\log(2).
$$

Title: Integral $\int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx$

Body:
I would like to ask how to integrate the following integration.

\begin{gather*}
    \int_{0}^{1} \frac{\ln(x^2+1)}{x^2} \,dx
\end{gather*}

I try to use integration by parts and I stuck with the following

\begin{gather*}
   \frac{\ln(x^2+1)}{x} - \int_{0}^{1} d\frac{\ln(x^2+1)}{x^2} \
\end{gather*}

I am confused with the continue step.


问:
Invariant subspace under rotation
Let $U$ be a $n-1$-dimensional subspace of $\mathbb{R}^n$. Consider 2 vectors $x,y\in U$ with $||x||=||y||$ and assume there is a rotation $\phi$ such that $\phi(x)=y.$
How can we show that $\phi(U)=U$ (assuming this is correct)?

Thank you.

答:
It is false if $n>3$. For example, consider
$$U=\operatorname*{span}\{(1,0,0,0),(0,1,0,0), (0,0, 1,0)\}=\operatorname*{span}\{e_1, e_2, e_3\},$$
which is a subspace of $\mathbb R^4$. The rotation matrix
$$
\phi=\begin{bmatrix} 0& 1 & 0 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 &0 \end{bmatrix}$$
is such that
$$
\phi e_1=e_2,$$
so your condition is satisfied with $x=e_1, y=e_2$. However, $\phi e_3=e_4$, so $\phi(U)\ne U$.

Sometimes our three-dimensional intuition fails in higher dimension.

Title: Invariant subspace under rotation


答:
Good idea using the integration by part.
Let's recall that integration by Parts is a special method of integration that is often useful when two functions are multiplied together.

At first let us see the rule: $$\int f(x)g(x) dx=u(x)\int v(x) dx-\int u'(x) \Big(\int v(x) dx\Big) dx$$
Now for your problem apply this rule with $v(x)=\frac{1}{x^2}$ and $u(x)=\log{(1+x^2)}$.  So:
$$\int \frac{\log{(1+x^2)}}{x^2}=\log{(1+x^2)}\int \frac{1}{x^2} dx-\int (\log{(1+x^2)})' \Big(\int \frac{1}{x^2} dx\Big) dx=$$ $$=-\frac{1}{x}\log{(1+x^2)}+\int \frac{1}{x}\frac{2x}{1+x^2}=-\frac{1}{x}\ln{(1+x^2)}+2\arctan{(x)}+c$$ Obviously I have shown the indefinite integral but in your case you have to evaluate the integral with the extremes ($0$ and $1$) imposed, in fact your problem I think was the integration itself.


问:
Can I find a bound to show that $\lvert x-y\rvert ^{2}\geq c\lvert \lvert x\rvert ^2-\lvert y\rvert^2\rvert $
Can I find a constant $c$ to show that $\lvert x-y\rvert ^{2}\geq c\lvert \lvert x\rvert ^2-\lvert y\rvert^2\rvert $ for $x,y\in \mathbb R\; (*)$.

Background to my question:

In a proof I saw the use of the fact that if $X_{n}\xrightarrow{L^{2}}X$ then it follows $(X_{n})^{2}\xrightarrow{L^{1}}X^2$. Now I assume that this means an inequality of the form $(*)$ must hold, but I am struggling to prove it. Any ideas?

EDIT: So if that inequality $(*)$ is not possible, then how does $X_{n}\xrightarrow{L^{2}}X$ imply $(X_{n})^{2}\xrightarrow{L^{1}}X^2$?

答:
We wish to prove:

$$|x-y|^{2}\geq c|x^2-y^2|$$

$$|x-y|^{2}\geq c|x-y||x+y|$$

Assuming $x\neq y$, because the inequality is trivially true if $x=y$:

$$|x-y|\geq c|x + y|$$

Assuming $x\neq -y$, as the inequality is again trivial if this is true:

$$c\leq\left|\frac{x-y}{x+y}\right|$$

However, the RHS can get arbitrarily close to $0$ (though it is always positive) as $x$ approaches $y$. Thus $c\leq 0$, the largest possible $c$ is $\boxed{0}$.


问:
Simple example of bounded sequence without a convergent subsequence in $L^p[0,1]$
I was asked to produce a bounded sequence in $L^p[0,1]$ without a convergent subsequence. Here is what I came up with:

Define

$f_1$ to be $1$ on $[0,1/2)$ and $0$ otherwise

$f_2$ to be $1$ on $[0,1/4)$, $[1/2,1/2+1/4)$ and $0$ otherwise

$f_3$ to be $1$ on $[0,1/8)$,$[1/4,1/4+1/8)$,$[1/2,1/2+1/8)$, $[1/2+1/4,1/2+1/4+1/8)$ and $0$ otherwise

and so on. Since, for $m\neq n$ we have $1/2=||f_n-f_{m}||_p$ (if $p=\infty$, this difference is just $1$), we conclude that no convergent subsequence can exist.

The downside of this, if correct, is that it is quite tedious to write down $f_n$ explicitly, and show that $||f_n-f_m||_p=1/2$ or $1$.

Do you have any other more explicit, simpler examples?

答:
Here's an example that's easy to write down: For $n=1,2,\dots$ define $I_n$ to be the interval $(1/(n+1),1/n),$ and set

$$f_n = \mathbb 1_{I_n}\cdot (n(n+1))^{1/p}.$$

Then for $m\ne n,\,\|f_m-f_n\|_p=2^{1/p}.$


答:
If $c \leq 0$, the LHS will automatically be non-negative and the RHS will automatically be non-positive.  Therefore, the problem reduces to assuming that $c$ is required to be $> 0.$

Without Loss of Generality, $x > y > 0.$

Then, LHS $= (x-y)(x-y)$ and RHS $= c(x-y)(x+y).$

Therefore, the problem reduces to trying to show that you can find a scalar $c> 0$ such that regardless of the values of $x,y$,

$$(x-y) \geq c(x+y) \implies \frac{x-y}{x+y} \geq c.\tag 1$$

With $x,y$ variables constrained only by $x > y > 0$, then regardless of how small $c$ is chosen to be, you can always find values for $x,y$ such that inequality (1) above does not hold.

Title: Can I find a bound to show that $\lvert x-y\rvert ^{2}\geq c\lvert \lvert x\rvert ^2-\lvert y\rvert^2\rvert $

Body:
Can I find a constant $c$ to show that $\lvert x-y\rvert ^{2}\geq c\lvert \lvert x\rvert ^2-\lvert y\rvert^2\rvert $ for $x,y\in \mathbb R\; (*)$.

Background to my question:

In a proof I saw the use of the fact that if $X_{n}\xrightarrow{L^{2}}X$ then it follows $(X_{n})^{2}\xrightarrow{L^{1}}X^2$. Now I assume that this means an inequality of the form $(*)$ must hold, but I am struggling to prove it. Any ideas?

EDIT: So if that inequality $(*)$ is not possible, then how does $X_{n}\xrightarrow{L^{2}}X$ imply $(X_{n})^{2}\xrightarrow{L^{1}}X^2$?


问:
How to prove this function is surjective
I'm trying to solve this question:

In order to solve this question above, I found this function: $r/w\mapsto (r/s)/(w/s)$ such that $w/s\in T$, I almost proved this map is an isomorphism, I'm stuck just in the surjectivity part.

If we get an element $(r/s)/(w/s)$ of $T^{-1}(S^{-1}R)$, ok!
However, an element of $T{^1}(S^{-1}R)$ can be for example $(r/s_1)/(w/s_2)$ with $s_1\neq s_2$

I need help in this part.

Thanks in advance.

答:
Consider an element $z$ of $T{^1}(S^{-1}R)$, we can write it in the form

$$
z=\frac{\frac{r}{s_1}}{\frac{w}{s_2}}
$$

with $r,w\in R$, $s_1,s_2\in S$ and further $\frac{w}{s_2} \in T$.

Let $w'=s_1\frac{w}{s_2}$. Then $\frac{w'}{s_1}=\frac{w}{s_2} \in T$, so we deduce
that $w' \in S_{*}$. Finally $z=\frac{r}{w'}$ with $r\in R,w'\in S_{*}$.


答:
Such a strictly positive constant can't exist

Take $x_n = n+1= y_n+1$. Following inequality would have to hold for all $n \in \mathbb N$
$$1=\lvert x_n-y_n\rvert ^{2}\geq c \left(2n+1\right)=\lvert \lvert x_n\rvert ^2-\lvert y_n\rvert^2\rvert$$

which is not possible.


问:
Algebraic expression for the period of $\cos (\log (x))$?
This question relates only to $x \in \mathbb{R}^+$. The function $f(x) = \cos (\log (x))$ is clearly defined on the positive reals, with a monotonic decreasing period $p(x)$ which is defined at the limits of this range by

$$\underset{x\to 0^+}{\text{lim}}p(f(x)) = \infty$$
$$\underset{x\to \infty}{\text{lim}}p(f(x)) = 0$$

$p$ still needs clear definition for any given $0 < x_0 < \infty$. So, let us define $p(x_ 0)$ as the smallest positive real $a$ for which this pair of equations hold:

$$\cos (\log (x_0)) = \cos (a \log (x_0))$$
$$\text{sgn}\left(\frac{\partial \cos \left(\log \left(x_0\right)\right)}{\partial x_0}\right) = \text{sgn}\left(\frac{\partial \cos \left(a \log \left(x_0\right)\right)}{\partial  x_0}\right)$$

Is the right approach? And how do I proceed?

答:
I would approach this as follows.

The period of $\cos(x)$ is $2\pi$, everywhere. At a particular $x_0$, what $a$ must you add to $x_0$ so that $\ln(x_0+a)$ is $2\pi$ greater than $\ln(x_0)$? In other words, solve for $a$ in

$$\ln(x_0)+2\pi=\ln(x_0+a)$$

The solution is

$$a=\left(e^{2\pi}-1\right)x_0$$

It would also make sense to me to go backwards and solve
$$\ln(x_0)-2\pi=\ln(x_0-a)$$
where the solution is
$$a=\left(1-e^{-2\pi}\right)x_0$$

You could take the two solutions and average them somehow. It might make sense to take their geometric mean:

$$a=\left(e^{\pi}-e^{-\pi}\right)x_0$$


答:
Yes, All unions and all finite intersections must be in the Topology.

If we want to get thorough for

$T_1 = \{\emptyset, X, \{a\}, \{b\}\}$ to be a topology the following sixteen unions and sixteen intersection must be in $T_1$.

$\cup_{\emptyset} = \emptyset,\cup \emptyset = \emptyset; \cup X = X; \cup \{a\} = \{a\}, \cup \{b\}=\{b\}$
$ \emptyset \cup X = X; \emptyset \cup \{a\} = \{a\}; \emptyset\cup \{b\} =\{b\}, X\cup \{a\}=X;X\cup \{b\} = X; \color{red}{\{a\}\cup \{b\}=\{a,b\}};$

$\emptyset \cup X\cup \{a\} = X; \emptyset \cup X \cup \{b\} = X; \color{red}{\emptyset \cup \{a\}\cup \{b\} =\{a,b\}},$
$X\cup \{a\}\cup \{b\} = X; \emptyset \cup X\cup \{a\}\cup\{b\} = X$

$\cap_{\emptyset} =X,\cap \emptyset = \emptyset; \cap X = X; \cap \{a\} = \{a\}, \cap \{b\}=\{b\},$
$ \emptyset \cap X = \emptyset; \emptyset \cap \{a\} = \emptyset; \emptyset\cap \{b\} =\emptyset, X\cap \{a\}=\{a\};X\cap \{b\} = \{b\};$ ${\{a\}\cup \{b\}=\emptyset};\emptyset \cup X\cup \{a\} = \emptyset; \emptyset \cup X \cup \{b\} = \emptyset;$
${\emptyset \cup \{a\}\cup \{b\} =\emptyset},X\cup \{a\}\cup \{b\} = \emptyset; \emptyset \cup X\cup \{a\}\cup\{b\} = \emptyset$

But the $\color{red}{\{a,b\}}\not \in T_1$.

And if $T_2 = \{\emptyset, X, \{a,b\}\}$ to be a topoloty the following eight unions and eight intersections must be in $T_2$

$\cup_{\emptyset} = \emptyset;\cup \emptyset = \emptyset;\cup X = X; \cup \{a,b\} = \{a,b\}$
$\emptyset \cup X=X; \emptyset \cup \{a,b\} = \{a,b\}; X\cup \{a,b\}= X;\emptyset \cup X \cup \{a, b\} = X$

$\cap_{\emptyset} = X;\cap \emptyset = \emptyset;\cap X = X; \cap \{a,b\} = \{a,b\}$
$\emptyset \cap X=\emptyset; \emptyset \cap \{a,b\} = \emptyset; X\cap \{a,b\}=  \{a,b\};\emptyset \cap X \cap \{a, b\} = \emptyset$

and they are

....

Just in case you are wondering about the notation $\cup_{\emptyset}$ and $\cap_{\emptyset}$.  Those are my notation for the empty union and empty intersect of no sets.  By convention the empty union is the emptyset and the empty intersection is the universal set.

Title: Why is this a topology but the the other is not?

Body:
Given $X=\{a,b,c\}$, why is it acceptable to say that $T=\{\emptyset,X, \{a,b\}\}$ is a topology but $T=\{\emptyset,X, \{a\},\{b\}\}$ is not?

The union of $\{a\}\cup\{b\}$ is $\{a,b\}\notin T$.

So is it because $\{a,b\}\cup\{a,b,c\}$ is $X$ which is in $T$?

Do all unions of all elements in $T$ have to exist in $T$?


问:
Can a lagrangian submanifold of a 4-dimensional real space be diffeomorphic to sphere or real projective plane?
Is it possible to find a lagrangian submanifold of $\mathbb{R}^4$, which is diffeomorphic to $S^2$ or $\mathbb{R}P^2$?
As I have found, the answer is no for $S^2$, because $S^2$ is simply connected, while each lagrangian submanifold of $\mathbb{R}^4$ is not simply connected. But I do not know how to prove it, I think it has to be simplier methods to show it.
The answer for $\mathbb{R}P^2$ I still do not know.

答:
As a consequence of Weinstein's neighborhood Theorem, one can rule out many closed surfaces in $\mathbb{R}^4$ from being Lagrangian submanifolds (with respect to any symplectic form on $\mathbb{R}^4$).  In fact, using it, I can rule out all oriented surfaces except $T^2$, and all non-orientable surfaces with odd Euler characteristic.  In particular, $S^2$ and $\mathbb{R}P^2$ cannot occur.

To see this, suppose $L$ is a closed surfaced embedded into $\mathbb{R}^4$ as a Lagrangian submanifold with respect to some symplectic form on $\mathbb{R}^4$.  Weinstein's theorem gives an embedding of $T^\ast L \cong TL$ into $\mathbb{R}^4$.

Now, the point is that $TL$ is open in $\mathbb{R}^4$, so the normal bundle of $L$ in $TL$ (which is isomorphic to $TL$), must be isomorphic to the normal bundle of $L$ in $\mathbb{R}^4$.

If $L$ is orientable, then the normal bundle in $\mathbb{R}^4$ has trivial Euler class (Milnor-Stasheff, Corollary 11.4), while the Euler class of $L$ in $TL$ measures the Euler class.  Hence, if $L$ is orientable, $L$ must be a torus.

If $L$ is non-orientable, then the normal bundle in $\mathbb{R}^4$ has trivial top Stiefel-Whitney class.  But this only happens in the tangent bundle $TL$ if $L$ has even Euler characteristic.


问:
Determining Which Areas Under Curves Are Equal
I have the following question from my calculus 2 homework and I'm really stumped:

Three regions have areas $A$, $B$, $C$

The grey area $A$ is the area of a region bounded by the graph of the function $f(x) = \ln x$ and the $x$-axis over the interval $x \in[1, e]$.

The green area $B$ is the area of a region bounded by the graph of the function $f(x) = e^{\sin x}\sin 2x$ and the $x$-axis over the interval $x \in [0, \pi/2]$.

The orange area $C$ is the area of a region bounded by the graph of the function $f(x) = xe^x$ and the $x$-axis over the interval $x \in [0, 1]$.

Two of the three areas $A$, $B$, and $C$ are actually equal. Find out which areas are equal. Do not forget to write down the correct mathematical solution to this problem!

So far, all I've been able to do is setup the areas as integrals:

$A = \int_1^e\ln x\ dx$

$B = \int_0^{\pi/2}e^{\sin x}\sin2x\ dx$

$C = \int_0^1xe^x dx$

I'm thinking that in order to show a mathematical solution to the question, I have to solve or at least simplify the integrals. Here's where the problem lies.

I have no idea how to solve these. For example, for the area of $A$, it seems that I have to use integration by parts to solve it but we haven't been taught how to do that yet (I know for sure that we are not expected to read ahead). So far, we've only done basic substitution. So now I wonder, is there another way to solve these without using integration by parts?

答:
Hints (for you to check):

$$\begin{align*}\bullet&\;\;\;\;\;\int\log x\,dx=x\log x-x+K\;,\;\;K=\text{ constant}\\
\bullet&\;\;\;\;\;\int e^{\sin x}\sin2x\,dx=2\int e^{\sin x}\sin x\cos x\,dx=2\sin xe^{\sin x}-2\int e^{\sin x}\cos x\,dx\\
\bullet&\;\;\;\;\int xe^xdx=xe^x-\int e^xdx\end{align*}$$

BTW, all three integrals above are quickly solved by means of integration by parts. The second one only requires a basic trigonometric identity.


问:
Measure of sets under linear transformation
I am trying to prove the following. Let $L: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear transformation and $Q_0 = [0,1]^n$ be the unit cube. Let $d_l = |L(Q_0)|$. Prove that $|L(Q)| = d_l |Q|$ for every cube $Q \subseteq R^n$. Here $|-|$ denotes Lebesgue measure.

Now I have tried proving this but cannot seem to get far. Of course, since $L$ is Lipschitz we have $|L(Q)| \leq C|Q|$. However, I cannot get past this point. For context, this is the first subproblem that appears in a problem that proves that $|L(E)| = |det(L)||E|$ whenever $E$ is measurable.

Any help or hints is appreciated. Thanks.

答:
This is how Rudin does it:

if $L$ is not onto, then it maps $Q_0$ into a space of dimension strictly less than $n$ so $d_l=0.$ Otherwise, $L$ is a homeomorphism so $L(E)$ is Borel for each Borel set $E$.  In particular, $L(Q)$ is Borel, so $m\circ L$ is a well-defined measure. Now,

$\tag1 m(L(E+x))=m(L(E))+L(x))=m(L(E))$

because $m$ is translation-invariant. Then, since $Q_0$ is a disjoint union of $2^{nk}$ boxes $Q$ of volume $2^{-n},\ (1)$ imlplies that

$2^{nk}m(L(Q))=m(L(Q_0))=m(L(Q_0))\cdot 1=m(L(Q_0))\cdot m(Q_0)=$

$m(L(Q_0))\cdot (2^{nk}m(Q))$  so

$\tag2 m(L(Q))=m(L(Q_0))m(Q)$


问:
If $z_1,z_2,z_3,z_4$ are consecutive vertices of a quadrilateral that lie on a circle prove the following
If $z_1,z_2,z_3,z_4$ are consecutive vertices of a quadrilateral that lie on a circle prove the following: $|z_1-z_3||z_2-z_4|=|z_1-z_2||z_3-z_4|+|z_1-z_4||z_2-z_3|$. I know that you can prove this without using the cross-ratio, but I would like to complete this proof using it. Here are some things I know:

The cross-ratio is real if and only if $(z_1,z_2,z_3,z_4)$ lie on the same circle.

The cross-ratio is invariant under linear transformations, i.e. $(z_1,z_2,z_3,z_4)=(Tz_1,Tz_2,Tz_3,Tz_4)$.

I am not sure if (2) will be of any use, but I definitely think (1) would be helpful. Using (1) I can only get as far as: $|z_1-z_3||z_2-z_4|=C|z_1-z_4||z_2-z_3|$, where $C \geq 0$.

Any help is appreciated.

Edit: After playing with the equation for a while this is what I concluded: Of course by simple verification, one indeed finds that $(z_1-z_3)(z_2-z_4)=(z_1-z_2)(z_3-z_4)+(z_1-z_4)(z_2-z_3)$. Then we have $$\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}=\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)}+1.$$ Now our result is proven if we can show that both of these quotients are real and positive. Well, the real part follows from the fact that a cross-ratio is real if and only if all four points lie on the same circle. Now for the positive part: The quotient on the left is the cross ratio $(z_1,z_2,z_3,z_4)$ where $z_2 \rightarrow 1$, $z_3 \rightarrow 0$, $z_4 \rightarrow \infty$. Now this transformation, which is bijective, takes the circle that passes through these points to the real line. Therefore the arc from $z_4$ to $z_2$ is taken to $[-\infty,1]$, hence $(z_1,z_2,z_3,z_4) > 1$. A similar argument works to show that the other cross-ratio is also positive.

Is this reasoning correct? I guess we use the fact that continuous functions take connected sets to connected sets, so the image of these arcs must be connected.

答:
Your reasoning is correct. You have demonstrated that
$$
\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}=\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)}+1
$$
and the quotient on the left is $> 1$. That implies that the quotient on the right is $> 0$. Therefore  all expressions are positive, and it follows that
$$
\left|\frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}\right|=\left|\frac{(z_1-z_2)(z_3-z_4)}{(z_1-z_4)(z_2-z_3)}\right|+1 \, .
$$

Instead of a connectedness argument you can also use that Möbius transformation preserve orientation: The quotient on the left is equal to $T(z_1)$ where $T$ is the Möbius transformation which maps $z_2, z_3, z_4$ to $1, 0, \infty$, respectively. $z_1,z_2,z_3,z_4$ are consecutive points on a circle. It follows that $T(z_1), 1, 0, \infty$ are consecutive points on the extended real axis (a generalized circle) and that is only possible if $T(z) > 1$.


问:
Different representations of classical momentum for 2-step optimization
This question is not talking about Nesterov's momentum so please do not think of it in that direction.

According to page 2, Equations (1) and (2), of On the importance of initialization and momentum in deep learning Polyak's momentum (classical momentum) is defined as the following:

$v_{t+1}=\mu v_t -\epsilon \nabla f(\theta_t)\tag{1}$

$\theta_{t+1} = \theta_t + v_{t+1}\tag{2}$
where $ 0 \leq \mu <1$ and $\epsilon >0$.

Since $\epsilon >0$, dividing (1) by $-\epsilon$ and letting $w_{s+1} = \frac{v_{t+1}}{-\epsilon}$, one get the following:

$w_{t+1}=\mu w_t + \nabla f(\theta_t)\tag{3}$

$\theta_{t+1} = \theta_t -\epsilon w_{t+1}\tag{4}$

However, according to Andrew Ng updates in momentum method can benefit the following process involving gradient:

$
S_{t+1} = \beta S_t+(1-\beta)\nabla f(\theta_t)\tag{5}
$

which is also known as exponential moving average process.

I am wondering how we can manipulate (1)-(2) or (3)-(4) to come up with a 2-step update one step of which is (5) and is equivalent to (1)-(2) or (3)-(4).

答:
According to Andrew Ng's video, the proposed scheme would be
\begin{align}
S_{t+1} ={}& \beta S_t+(1-\beta)\nabla f(\theta_t)\tag{5}\\
\theta_{t+1} ={}& \theta_t -\alpha S_{t+1}\tag{6}
\end{align}
Multiply (5) by $-\alpha$ and letting $v_t=-\alpha S_t$ yields
\begin{align}
v_{t+1} ={}& \beta v_t-\alpha(1-\beta)\nabla f(\theta_t)\tag{7}\\
\theta_{t+1} ={}& \theta_t + v_{t+1}\tag{8}
\end{align}
which is (1)-(2) with $\mu=\beta$ and $\epsilon=\alpha(1-\beta)$.


问:
Constrained matrix equation
I have a data matrix $X \in \mathbb{R}^{n \times m}$, with $\mathrm{rank}(X) = n$ and $n \leq m$. I'm trying to understand if I can find some square matrix $C \in \mathbb{R}^{m \times m}$ such that

$$
\left\{
\begin{aligned}
& X C = 0,\\
& C^\top \mathbb{1}_m \leq \mathbb{1}_m,\\
& C_i \geq 0,\, i = 1,\ldots,m,
\end{aligned}
\right.
$$

where $C_i$ is the $i$-th column of $C$, while $\mathbb{1}_m$ is a column vector of ones of dimension $m$. Obviously, $C = 0$ does the job. Can anyone see some other possibility, or $C = 0$ is the unique solution?

答:
There exist matrices $X$ for which $C = 0$ is the unique solution and others for which it is not.

Note that $C$ satisfies this condition if and only if each column $v$ of $C$ satisfies the conditions
$$
\begin{cases}
Xv = 0,\\
1_m^T v \leq 1,\\
v \geq 0.
\end{cases}
$$
So, it is equivalent to determine the conditions under which the above equation has a solution.

In fact, a non-zero solution to this solution exists if and only if there exists a non-zero first and third conditions, namely
$$
\begin{cases}
Xv = 0,\\
v \geq 0.
\end{cases}
$$
I don't know of any useful alternative formulation for these conditions together with $v \neq 0$.

Title: Constrained matrix equation


问:
Example where $_ (x), _ (y) \geq 10$ but $_ℎ (x) ≤ 1$

Let $ :  → $ and $ :  → $ be continuous functions between arbitrary vector spaces, and let $ℎ =  ◦ $ . Suppose $ =  (x)$ and $ = (y) = ℎ(x)$. Give a concrete example where $ \kappa_f (x), \kappa_g (y) \geq 10$ but $_ℎ (x) \leq 1$.

I am using the relation $\kappa = \frac { ||J(x)||}{||f(x)||/||x||}$ for each of the three functions. I was able to find the upper bound on $_ℎ(x)$ in terms of $\kappa_ (x)$ and $\kappa_ (y)$ (Came out to be $\kappa_ℎ(x) \leq \kappa_f (x)*\kappa_g (y)$). However, now, I am unable to find an example which satisfies the given condition. How can I go about finding an example? Can I generalise anything from this?

Please note that $\kappa$ here is the relative condition number.

答:
We can take $f(x_1,x_2)=(x_1^n,0)$ and $g(y_1,y_2)=(1,y_2^n)$. Then $h = g \circ f$ satisfies
$$ h(x_1, x_2) = g(f(x_1,x_2)) = g(x_1^n,0) = (1,0)$$
As a constant, $h$ will be well-conditioned.

Let us pick the normwise relative condition number and use the infinity norm. In general, if $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ is differentiable then the condition number can be computed as
$$ \kappa_f(x) = \frac{\|Df(x)\|_\infty \|x\|_\infty}{\|f(x)\|_\infty}$$
where $Df(x)$ is the Jacobian of $f$ at the point $x$. It is assumed that $x \not = 0$ and $f(x) \not = 0$. In our case we readily find that
$$\kappa_f(x_1,x_2) = \kappa_g(y_1,y_2) = n$$
while
$$ \kappa_h(x_1,x_2) = 0.$$

Let me be the first to point out that this reply does not answer the question as it is explicitly stated. Instead, I have strengthened the assumptions to the point where a simple example can be given without losing the surprise that I perceive the professor wanted to give the student.


问:
MCQ on a function
$f(x)=\left( \ln \left( \frac{\left( 7x-x^{2} \right)}{12} \right) \right)^{\frac{3}{2}}$

Choose correct options , more than one may be correct .

(a) $f$ is defined on $R^+$ and is strictly increasing.

(b)$f$ is defined on an interval of finite length and is strictly increasing

(c)range of function include 1

(d)$f$ is defined on an interval of finite of length and is bounded .

Now answer says (c) & (d)

I am getting (b) & (d) and that too when I don't take the natural domain of the function i.e. on all real values it could possibly be defined .

Here is the graph of the function

http://www.wolframalpha.com/input/?i=y%3D%5Cleft%28+%5Cln+%5Cleft%28+%5Cfrac%7B%5Cleft%28+7x-x%5E%7B2%7D+%5Cright%29%7D%7B12%7D+%5Cright%29+%5Cright%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D&dataset=

How can the function be bounded when it approaches $-\infty$ from both sides .

答:
Worry not about Alpha. We are ultimately going to take a square root, so $$\ln\left(\frac{7x-x^2}{12}\right)$$
 has to be non-negative, that is, we must have
$\frac{7x-x^2}{12}\ge 1$. Equivalently, we want $x^2-7x+12\le 0$, which happens only for $3\le x\le 4$.

Note that $f(x)$ is $0$ at $3$ and $4$, so cannot be strictly increasing.

Title: MCQ on a function


问:
Derived hom-tensor adjunction for $O_X$-modules
As far as I understand it, for $R,S$ rings and $M$ an $R$-module ,$N$ and $R,S$-bimodule and $L$ an $S$-module, we have an isomorphism

$$
\text{RHom}_R(N \otimes_S^L L, M) \cong \text{RHom}_S(L,\text{RHom}_R(N,M)).
$$

To what extent is there a similar isomorphism of $R\mathcal H\text{om}$ for, say, ringed spaces $(X,O_X)\xrightarrow f(Y,O_Y)$ and modules of $O_Y$- and $O_X$-sheaves? Or maybe for simplicity set $Y=X$, with a map $O_Y \rightarrow O_X$. Feel free to adjust the assumptions as necessary.

I tried searching the literature, but couldn't find a satisfying answer. The closest I could find was https://stacks.math.columbia.edu/tag/0DVC.

A reference would be great!

答:
See Kashiwara, Schapira, Categories and Sheaves, Theorem 18.6.9 (iii) eqn (18.6.13): Let $f : X \to Y$ be a morphism of ringed sites. There are isomorphisms \begin{align} Rf_{\ast}R\mathcal{Hom}_{\mathcal{O}_{X}}(Lf^{\ast}G,F) \simeq R\mathcal{Hom}_{\mathcal{O}_{Y}}(G,Rf_{\ast}F) \end{align} functorial with respect to $F \in D(\mathcal{O}_{X})$ and $G \in D(\mathcal{O}_{Y})$.

In the Stacks Project, I could only find the case when $f$ is the identity morphism $\mathrm{id}_{X}$, see Tag 08DJ.


答:
Nevertheless on can find rotations which satisfy the property.

Any reflexion preserves distances, so it is a rotation (=orthogonal matrix), and in your case, you can for example take the reflexion w.r.t the hyperplane $H:=(\mathbf{x}-\mathbf{y})^{\perp}$.

Forgetting the subspace $U$ we have the decomposition $\mathbb{R}^n=(\mathbf{x}-\mathbf{y}) \oplus H$. We can adapt it to $U\subset \mathbb{R}$ by choosing $\mathbf{z}\in U^{\perp}\subset (\mathbf{x}-\mathbf{y})^{\perp} =H $. One has
$$ \mathbb{R}^n=\mathbb{R}(\mathbf{x}-\mathbf{y}) \oplus (H\cap U) \oplus \mathbf{z}\quad \text{and}\quad U= \mathbb{R}(\mathbf{x}-\mathbf{y}) \oplus (H\cap U) $$
In restriction to the line $\mathbb{R}(\mathbf{x}-\mathbf{y})$ the reflexion is $-1$ and so preserves this space and it is identity on $H$ and so also preserves $(H\cap U)$. Hence it preserves $\mathbb{R}(\mathbf{x}-\mathbf{y}) \oplus (H\cap U)=U$.

Another more intuitive rotation (and one that preserves orientation, i.e. of determinant 1) would be that around the axis $\mathbf{z}$ of angle precisely the angle between $\mathbf{x}$ and $\mathbf{y}$. So in fact use $\mathbf{z}$ to define a rotation on the subspace $U$ and extend the map to the whole $\mathbb{R}^n$ by identity on the last direction $\mathbf{z}$.

Title: Invariant subspace under rotation

Body:
Let $U$ be a $n-1$-dimensional subspace of $\mathbb{R}^n$. Consider 2 vectors $x,y\in U$ with $||x||=||y||$ and assume there is a rotation $\phi$ such that $\phi(x)=y.$
How can we show that $\phi(U)=U$ (assuming this is correct)?

Thank you.


问:
Discrete mathematics, is this a function?
Let A={1,2,3,4,5} and B={a, b, c, d}, and let f={(1,a),(2,a),(3,d),(4,c)}
is f a function? , the text book I'm working from says it's since all mapping are one to one or many one to many, & non are many to one, but the element 5 isn't mapped to anything, so is it not a function & the book is wrong?</p>

答:
The book is being subtle.  $A$ and $B$ are just sets.  It does not say they are domains or codomains of the function.

$f=\{(1,a),(2,a),(3,d),(4,c)\}$ is a set of mapping pairs.

It maps a set of values:  $\{1,2,3,4\}$ to a set of values $\{a,b,c,d\} =B$.  And there are no "one to many mappings".   So it is a function $f:\{1,2,3,4\}\to B$.

But it is not a function $f:A \to B$ and there is no element $(5,x)$ or $f(5)$ or "$5$ is not mapped".

So you are correct.

I don't know if the book intended this.  I imagine the book actually made an error and you caught it.

But technically $f$ is a function; Just not a function with domain $A$.


问:
The students $S_1, S_2,...S_{10}$ are divided into 3 groups A, B and C
It can be seen as distributing n unique objects above m groups

The students $S_1, S_2,...S_{10}$ are divided into 3 groups A, B and C such that each group has at least one students and C has at most 3. Find possibilities of forming groups

$$a+b+c=10$$

The number of positive integral solutions $(\ge 1)$ is
$$\binom {10-1}{3-1}=36$$

I don’t know how to account for the cases where C has at most 3 students

Also I know this method of solving is probably wrong but I don’t know how else to do it.

答:
Leaving aside group $C$, each student has $2$ choices for groups

Group $C$ can receive a student in $\binom{10}1$ ways.

The remaining students could be received by the other two groups in $2^9$ ways, but this could result in one group receiving no student.

Any of the two groups could be empty, so we subtract $2$ to get legal ways as

$\binom{10}1\cdot (2^9-2)$, and proceeding similarly with group $C$ having $2$ or $3$ students, and adding up, we get

$\binom{10}1\cdot (2^9-2) + \binom{10}2\cdot(2^8-2) +\binom{10}3\cdot(2^7-2) = 31650$


问:
Why proofs which provide algorithms for construction are valid?
10 days ago I asked question about validity of a proof (Is this a valid proof of the fact that a subspace of an $n$-dimensional vector space has a basis?), but as I later realised I didn't explain well what I really wanted to know and this resulted in the answers that didn′t quite answered my real question, so I will try to rephrase it. So in that previous post there was a proof of the fact that any subspace of finite-dimensional vector space has a finite basis. In it I basically constructed this basis by adding vector if the current set of vectors is not maximal linear independent set. Then I said that this "algorithm" will terminate because of the fact that the vector space is finite dimensional. But my concern about this proof is "Why algorithm for constructing/finding an object is a valid proof of its existence ?". Such question may sound strange, but my concern is the fact that you can′t say by reading a proof how many steps execution of such algorithm will take, because it depends on what vector space and what subspace you take. So basically we don't know how many steps the proof has and it seems kind of strange to me. And this concern is not just about proof of this fact about subspaces, but about any proof which proves existence of an object with an algorithm which has number of steps which depends on something. For example, proof of the fact that any tree has a vertice of degree 1, where you take any vertice and then go along the edges of the tree until you find the vertice of degree 1 (which you will, because tree has only finite amount of vertices and tree is acyclic).

So why are proofs, which provide algorithms for constructing an object, valid and can we formalise (maybe using set theory) them in such a way that number of steps of a proof will be specific ?

答:
The proof implictly (or explicitly) relies on the fact that there is a progress function, i.e. some quantity that is guaranteed to decrease on every iteration and has a lower bound. Such as the dimension of the "unspanned space", which is obviously a non-negative number. That guarantees that the iterations will terminate.

This situation could be compared to a proof by induction, where you make steps that depend on each other and could be seen as an algorithm (the progress function would be the number of inductive steps remaining before you reach the target $n$). Or even a non-terminating algorithm.

If you still don't like this vision, think that you can always "unroll" the loop and write a standard, though possibly very long proof for any dimension.


问:
How to find the last digit of $37^{100}$
I am currently working on some algebra and I am studying the modulus chapter of my book. One question was finding the last digit of:

$$37^{100}$$

They give a hint about how the solution should be about calculating a remainder...

Is there any easy solution to this?

I know that there is a solution of just finding the pattern of the last digit of $ 37^x$. But I don't understand how that has to do with calculating a remainder...

Thank you!

答:
One finds the modulus of $37^{100}$ in decimal, by noting that in decimal, anything raised to the fourth power must end in $0$, $1$, $5$, or $6$.  When various powers of 5 are added, then the number must be adjent to a multiple of $5*5^n$, where $5^n$ divides the index.

So, we look for solutions, ending in any of these three digits:  $000$, $001$, $625$, or $376$, depending on what the common divisor with 10 is.  Here it's 1, so it ends in $001$.


问:
Surface represented by the equation $x^2-2y^2+3z^2+5yz-6xz-4xy+8x-19y-2z-20=0$
I have the following second degree equation in $x,y$ & $z$:
$x^2-2y^2+3z^2+5yz-6xz-4xy+8x-19y-2z-20=0$
I calculated the discriminating cubic as:
$k^3-4k^2-97k-190=0$

The roots of this equation are  distinct.  The question says that it represents a cone having vertex at$(1,-2,3)$.

As a cone is a surface of revolution, two roots of the discriminating cubic should be zero. But that's not the case here. So this equation can't represent a cone.

Am I correct in asserting this? Can the equation represent a cone even without having two equal roots? Please suggest.

答:
if you take $a=x-1, b = y+2, c=z-3$  you arrive at
$$  a^2 - 2 b^2 + 3 c^2 + 5 bc -6 ca -4 ab = 0. $$
It is this $0$ on the right hand side that says the object is a cone (over an ellipse, not a circle). Think about it: if you had $x^2 + y^2 - z^2 = 0$  you would have a circular cone. if you had $x^2 + 2y^2 - z^2 = 0$  you would have an
elliptical cone, as each choice of $z$ leads to a constant value for $x^2 + 2 y^2$

The fact that we get a quadratic form equal to zero comes from the final diagonal element $0$ in $D$ in these expressions:

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 \\
 \frac{ 11 }{ 6 }  &  -  \frac{ 7 }{ 12 }  & 1 & 0 \\
1 &  - 2 & 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 &  - 4 &  - 6 & 8 \\
 - 4 &  - 4 & 5 &  - 19 \\
 - 6 & 5 & 6 &  - 2 \\
8 &  - 19 &  - 2 &  - 40 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 &  \frac{ 11 }{ 6 }  & 1 \\
0 & 1 &  -  \frac{ 7 }{ 12 }  &  - 2 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 &  - 12 & 0 & 0 \\
0 & 0 &  -  \frac{ 95 }{ 12 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 - 2 & 1 & 0 & 0 \\
 - 3 &  \frac{ 7 }{ 12 }  & 1 & 0 \\
4 &  \frac{ 1 }{ 4 }  &  - 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 &  - 12 & 0 & 0 \\
0 & 0 &  -  \frac{ 95 }{ 12 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  - 2 &  - 3 & 4 \\
0 & 1 &  \frac{ 7 }{ 12 }  &  \frac{ 1 }{ 4 }  \\
0 & 0 & 1 &  - 3 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
2 &  - 4 &  - 6 & 8 \\
 - 4 &  - 4 & 5 &  - 19 \\
 - 6 & 5 & 6 &  - 2 \\
8 &  - 19 &  - 2 &  - 40 \\
\end{array}
\right)
$$

We can parametrize all integer solutions in four ways as
$$
\begin{array}{} a=3u^2 +5uv-2v^2 \\  b=-2u^2 +4uv+v^2 \\  c=u^2 +2uv-5v^2 \; . \end{array}  $$

OR
$$
\begin{array}{} a=7uv-4v^2 \\  b=-u^2 +2uv+5v^2 \\  c=2u^2 +4uv-v^2 \; . \end{array}  $$
OR
$$
\begin{array}{} a=7uv-8v^2 \\  b=3u^2 -6uv+v^2 \\  c=u^2 +4uv-2v^2 \; . \end{array}  $$

OR$$
\begin{array}{} a=2u^2 +uv-6v^2 \\  b=u^2 +6uv+3v^2 \\  c=3u^2 -2v^2 \; . \end{array}  $$

in the following, I named the four (u,v) recipes with letters A,B,C,D. If the triple produced had $a < 0,$ I printed $-a,-b,-c$

```<code>     a         b         c          u    v
     0         1        -2
     0        -1         2
     0         1        -2   B      1    0
     0         3         1
     0        -3        -1
     0         3         1   C      1    0
     1         2        -3
     1         2        -3   C      1    1
     2         1         3
     2         1         3   D      1    0
     2        -1         5
     2        -1         5   A      0    1
     2       -23        -6
     2       -23        -6   B      1    2
     3       -10        -1
     3       -10        -1   D      1    1
     3        -2         1
     3        -2         1   A      1    0
     3         6         5
     3         6         5   B      1    1
     4        19        10
     4        19        10   D      2    1
     4        -5         1
     4        -5         1   B      0    1
     4         5         6
     4         5         6   A      1   -1
     5         2        -1
     5         2        -1   D      1   -1
     6         1        10
     6         1        10   C      2    1
     6         3        -2
     6        -3         2
     6         3        -2   A      1    1
     6        -3         2   D      0    1
     6        53        23
     6        53        23   B      2    3
     8        -1         2
     8        -1         2   C      0    1
     9       -30        71
     9       -30        71   A      1    4
     9        -6        25
     9        -6        25   D      3   -1
    10       -29        -2
    10       -29        -2   A      3   -1
    10       -47       106
    10       -47       106   C      6    5
    11        -2         3
    11        -2         3   B      1   -1
    11        38        23
    11        38        23   A      3   -2
    12        15       -19
    12        15       -19   D      3   -2
    12       -15        46
    12       -15        46   C      4    3
    12       -29        73
    12       -29        73   C      5    4
    12        57      -101
    12        57      -101   A      2    5
    13        10        19
    13        10        19   C      3    1
    15         6        23
    15         6        23   A      1   -2
    16       -23        67
    16       -23        67   D      5   -2
    17         2        29
    17         2        29   B      3    1
    17       -86       -15
    17       -86       -15   A      5   -2
    18       101        50
    18       101        50   A      5   -3
    18         3         1
    18         3         1   B      2   -1
    18         5        -1
    18         5        -1   C      1    2
    20         1         3
    20         1         3   A      2    1
    22       -25         6
    22       -25         6   C      2   -1
    22       -43       114
    22       -43       114   A      1    5
    22        -5        46
    22        -5        46   D      4   -1
    24        -1         5
    24        -1         5   D      1   -2
    24        25       -29
    24        25       -29   A      2    3
    24        -3        47
    24        -3        47   B      4    1
    26        23        38
    26        23        38   B      3    2
    26       -47         3
    26       -47         3   A      4   -1
    26        71      -115
    26        71      -115   D      7   -4
    27        46        43
    27        46        43   C      5    1
    29       -46         5
    29       -46         5   C      3   -1
    30        43        46
    30        43        46   D      4    1
    30        71        53
    30        71        53   A      4   -3
    31       -10        69
    31       -10        69   B      5    1
    32        19       -15
    32        19       -15   B      4   -1
    33        -6        67
    33        -6        67   C      5    3
    34        29       -30
    34        29       -30   D      4   -3
    34        73        58
    34        73        58   C      6    1
    36        23        53
    36        23        53   A      2   -3
    36       -73         2
    36       -73         2   C      4   -1
    36       -87        -2
    36       -87        -2   B      1    4
    36        97        67
    36        97        67   D      5    2
    39        -2        73
    39        -2        73   D      5   -1
    39        30       -29
    39        30       -29   B      5   -1
    40       -67         6
    40       -67         6   D      2    3
    41       106        75
    41       106        75   C      7    1
    43      -106        -3
    43      -106        -3   C      5   -1
    44        29       -25
    44        29       -25   C      3    4
    46       -19        15
    46       -19        15   C      1   -2
    46        43       -47
    46        43       -47   B      6   -1
    48        53        71
    48        53        71   B      4    3
    48       -69        10
    48       -69        10   A      5   -1
    49        10         1
    49        10         1   A      3    2
    49         2        87
    49         2        87   A      1   -4
    49       -46        15
    49       -46        15   D      1    3
    49        58        73
    49        58        73   D      5    1
    51         6         5
    51         6         5   C      1    3
    52         5         6
    52         5         6   D      2   -3
    52        67       -86
    52        67       -86   A      3    5
    53        58       -69
    53        58       -69   B      7   -1
    54        15        86
    54        15        86   B      5    2
    55        46       -47
    55        46       -47   A      3    4
    58         1        10
    58         1        10   B      3   -2
    60         3       106
    60         3       106   D      6   -1
    60        47       -46
    60        47       -46   C      4    5
    66       -15        19
    66       -15        19   A      4    1
    66        47       -43
    66        47       -43   D      5   -4
    66        67        97
    66        67        97   C      7    2
    67        94      -125
    67        94      -125   B      9   -1
    69        50       101
    69        50       101   B      5    3
    72        75       106
    72        75       106   D      6    1
    74       -67        23
    74       -67        23   C      3   -2
    78       -29        25
    78       -29        25   B      2   -3
    78        69       -73
    78        69       -73   C      5    6
    86        25        -6
    86        25        -6   B      5   -2
    87        86       -97
    87        86       -97   D      7   -5
    90       -73        29
    90       -73        29   D      1    4
    92       -71        30
    92       -71        30   B      1   -4
    93       -30        29
    93       -30        29   C      1   -3
    98       -25        29
    98       -25        29   D      1   -4
    98       -29        30
    98       -29        30   A      5    1
    98        73       -69
    98        73       -69   A      4    5
    98        95      -106
    98        95      -106   C      6    7
   108        69       -58
   108        69       -58   D      6   -5
   117        -6        25
   117        -6        25   A      5    2
    a          b        c           u    v
</code>```

Body:
if you take $a=x-1, b = y+2, c=z-3$  you arrive at
$$  a^2 - 2 b^2 + 3 c^2 + 5 bc -6 ca -4 ab = 0. $$
It is this $0$ on the right hand side that says the object is a cone (over an ellipse, not a circle). Think about it: if you had $x^2 + y^2 - z^2 = 0$  you would have a circular cone. if you had $x^2 + 2y^2 - z^2 = 0$  you would have an
elliptical cone, as each choice of $z$ leads to a constant value for $x^2 + 2 y^2$

The fact that we get a quadratic form equal to zero comes from the final diagonal element $0$ in $D$ in these expressions:

$$ P^T H P = D  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
2 & 1 & 0 & 0 \\
 \frac{ 11 }{ 6 }  &  -  \frac{ 7 }{ 12 }  & 1 & 0 \\
1 &  - 2 & 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 &  - 4 &  - 6 & 8 \\
 - 4 &  - 4 & 5 &  - 19 \\
 - 6 & 5 & 6 &  - 2 \\
8 &  - 19 &  - 2 &  - 40 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 2 &  \frac{ 11 }{ 6 }  & 1 \\
0 & 1 &  -  \frac{ 7 }{ 12 }  &  - 2 \\
0 & 0 & 1 & 3 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 &  - 12 & 0 & 0 \\
0 & 0 &  -  \frac{ 95 }{ 12 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H  $$
$$\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
 - 2 & 1 & 0 & 0 \\
 - 3 &  \frac{ 7 }{ 12 }  & 1 & 0 \\
4 &  \frac{ 1 }{ 4 }  &  - 3 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 &  - 12 & 0 & 0 \\
0 & 0 &  -  \frac{ 95 }{ 12 }  & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 &  - 2 &  - 3 & 4 \\
0 & 1 &  \frac{ 7 }{ 12 }  &  \frac{ 1 }{ 4 }  \\
0 & 0 & 1 &  - 3 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
 = \left(
\begin{array}{rrrr}
2 &  - 4 &  - 6 & 8 \\
 - 4 &  - 4 & 5 &  - 19 \\
 - 6 & 5 & 6 &  - 2 \\
8 &  - 19 &  - 2 &  - 40 \\
\end{array}
\right)
$$

We can parametrize all integer solutions in four ways as
$$
\begin{array}{} a=3u^2 +5uv-2v^2 \\  b=-2u^2 +4uv+v^2 \\  c=u^2 +2uv-5v^2 \; . \end{array}  $$

OR
$$
\begin{array}{} a=7uv-4v^2 \\  b=-u^2 +2uv+5v^2 \\  c=2u^2 +4uv-v^2 \; . \end{array}  $$
OR
$$
\begin{array}{} a=7uv-8v^2 \\  b=3u^2 -6uv+v^2 \\  c=u^2 +4uv-2v^2 \; . \end{array}  $$

OR$$
\begin{array}{} a=2u^2 +uv-6v^2 \\  b=u^2 +6uv+3v^2 \\  c=3u^2 -2v^2 \; . \end{array}  $$

in the following, I named the four (u,v) recipes with letters A,B,C,D. If the triple produced had $a < 0,$ I printed $-a,-b,-c$

     a         b         c          u    v
     0         1        -2
     0        -1         2
     0         1        -2   B      1    0
     0         3         1
     0        -3        -1
     0         3         1   C      1    0
     1         2        -3
     1         2        -3   C      1    1
     2         1         3
     2         1         3   D      1    0
     2        -1         5
     2        -1         5   A      0    1
     2       -23        -6
     2       -23        -6   B      1    2
     3       -10        -1
     3       -10        -1   D      1    1
     3        -2         1
     3        -2         1   A      1    0
     3         6         5
     3         6         5   B      1    1
     4        19        10
     4        19        10   D      2    1
     4        -5         1
     4        -5         1   B      0    1
     4         5         6
     4         5         6   A      1   -1
     5         2        -1
     5         2        -1   D      1   -1
     6         1        10
     6         1        10   C      2    1
     6         3        -2
     6        -3         2
     6         3        -2   A      1    1
     6        -3         2   D      0    1
     6        53        23
     6        53        23   B      2    3
     8        -1         2
     8        -1         2   C      0    1
     9       -30        71
     9       -30        71   A      1    4
     9        -6        25
     9        -6        25   D      3   -1
    10       -29        -2
    10       -29        -2   A      3   -1
    10       -47       106
    10       -47       106   C      6    5
    11        -2         3
    11        -2         3   B      1   -1
    11        38        23
    11        38        23   A      3   -2
    12        15       -19
    12        15       -19   D      3   -2
    12       -15        46
    12       -15        46   C      4    3
    12       -29        73
    12       -29        73   C      5    4
    12        57      -101
    12        57      -101   A      2    5
    13        10        19
    13        10        19   C      3    1
    15         6        23
    15         6        23   A      1   -2
    16       -23        67
    16       -23        67   D      5   -2
    17         2        29
    17         2        29   B      3    1
    17       -86       -15
    17       -86       -15   A      5   -2
    18       101        50
    18       101        50   A      5   -3
    18         3         1
    18         3         1   B      2   -1
    18         5        -1
    18         5        -1   C      1    2
    20         1         3
    20         1         3   A      2    1
    22       -25         6
    22       -25         6   C      2   -1
    22       -43       114
    22       -43       114   A      1    5
    22        -5        46
    22        -5        46   D      4   -1
    24        -1         5
    24        -1         5   D      1   -2
    24        25       -29
    24        25       -29   A      2    3
    24        -3        47
    24        -3        47   B      4    1
    26        23        38
    26        23        38   B      3    2
    26       -47         3
    26       -47         3   A      4   -1
    26        71      -115
    26        71      -115   D      7   -4
    27        46        43
    27        46        43   C      5    1
    29       -46         5
    29       -46         5   C      3   -1
    30        43        46
    30        43        46   D      4    1
    30        71        53
    30        71        53   A      4   -3
    31       -10        69
    31       -10        69   B      5    1
    32        19       -15
    32        19       -15   B      4   -1
    33        -6        67
    33        -6        67   C      5    3
    34        29       -30
    34        29       -30   D      4   -3
    34        73        58
    34        73        58   C      6    1
    36        23        53
    36        23        53   A      2   -3
    36       -73         2
    36       -73         2   C      4   -1
    36       -87        -2
    36       -87        -2   B      1    4
    36        97        67
    36        97        67   D      5    2
    39        -2        73
    39        -2        73   D      5   -1
    39        30       -29
    39        30       -29   B      5   -1
    40       -67         6
    40       -67         6   D      2    3
    41       106        75
    41       106        75   C      7    1
    43      -106        -3
    43      -106        -3   C      5   -1
    44        29       -25
    44        29       -25   C      3    4
    46       -19        15
    46       -19        15   C      1   -2
    46        43       -47
    46        43       -47   B      6   -1
    48        53        71
    48        53        71   B      4    3
    48       -69        10
    48       -69        10   A      5   -1
    49        10         1
    49        10         1   A      3    2
    49         2        87
    49         2        87   A      1   -4
    49       -46        15
    49       -46        15   D      1    3
    49        58        73
    49        58        73   D      5    1
    51         6         5
    51         6         5   C      1    3
    52         5         6
    52         5         6   D      2   -3
    52        67       -86
    52        67       -86   A      3    5
    53        58       -69
    53        58       -69   B      7   -1
    54        15        86
    54        15        86   B      5    2
    55        46       -47
    55        46       -47   A      3    4
    58         1        10
    58         1        10   B      3   -2
    60         3       106
    60         3       106   D      6   -1
    60        47       -46
    60        47       -46   C      4    5
    66       -15        19
    66       -15        19   A      4    1
    66        47       -43
    66        47       -43   D      5   -4
    66        67        97
    66        67        97   C      7    2
    67        94      -125
    67        94      -125   B      9   -1
    69        50       101
    69        50       101   B      5    3
    72        75       106
    72        75       106   D      6    1
    74       -67        23
    74       -67        23   C      3   -2
    78       -29        25
    78       -29        25   B      2   -3
    78        69       -73
    78        69       -73   C      5    6
    86        25        -6
    86        25        -6   B      5   -2
    87        86       -97
    87        86       -97   D      7   -5
    90       -73        29
    90       -73        29   D      1    4
    92       -71        30
    92       -71        30   B      1   -4
    93       -30        29
    93       -30        29   C      1   -3
    98       -25        29
    98       -25        29   D      1   -4
    98       -29        30
    98       -29        30   A      5    1
    98        73       -69
    98        73       -69   A      4    5
    98        95      -106
    98        95      -106   C      6    7
   108        69       -58
   108        69       -58   D      6   -5
   117        -6        25
   117        -6        25   A      5    2
    a          b        c           u    v


问:
A dig at Ramanujan's: $\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,~ p>0$
Ramanujan's claim on page 98, in the book (`Ramanujan's note book part 1' by Bruce C. Berndt) is that
$$\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,\qquad p>0\tag{1}\label{theclaim}$$
The proposed proofs for $p=1,2$ are reported to be incorrect. For $p>2$ the claim \eqref{theclaim} has been disproved. In the year 1996, my proofs for $p=1,2$ were evaluated to be correct by American Math. Monthly, however,  similar proofs were told to have been published in the year 1995, somewhere. The result (1) being asymptotic $x$ needs to be positive and large.

The question here is: What is the latest about this result when $p\in (0,2]$ any information or proof is welcome.

答:
Proof for $p=1,2$

Euler's constant $\gamma=0.577215...$ is defined as [1]
\begin{equation}
\sum_{k=1}^{n} \frac{1}{k}\sim\ln n+\gamma ~~~~(1)
\end{equation}
Ramanujan claimed [2] that
\begin{equation}
\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^{pk}}{k(k!)^p} \sim p \ln x +p \gamma,~ p>0.~~~~(2)
\end{equation}
For $p=1,2$ the propose proofs were found incorrect. For $p>2$, the result (2) was disproved. Here the intent is to prove and show that for $p=1,2$ the result (2) is true and consistent with other existing results.

Case 1: Let us quote ([1]: p 955])
\begin{equation}
\gamma=-\int_{0}^{\infty} \left( e^{-t}-\frac{1}{1+t} \right) \frac{dt}{t}.
\end{equation}
We can re-write it as
\begin{equation}
\gamma=-\lim_{x \to \infty} \left(\int_{1/x}^{x} \frac{e^{-t}-1}{t}+\int_{1/x}^{x} \frac{dt}{1+t}\right).
\end{equation}
\begin{equation}
\gamma=\lim_{x \to \infty}\left( \int_{1/x}^{x} \sum_{k=1}^{\infty} (-1)^{k-1} \frac{t^{k-1}}{k!}dt-\ln x \right),
\end{equation}
when $x$ is very large, we prove that
\begin{equation}
\sum_{k=1}^{\infty}(-1)^{k-1} \frac{x^k}{k (k!)} \sim \ln x+\gamma.
\end{equation}
Case 2: Let us use ([1]: p 955)
\begin{equation}
\gamma=1-\int_{0}^{\infty} \left( \frac{\sin t}{t}-\frac{1}{1+t} \right) \frac{dt}{t}.
\end{equation}
\begin{equation}
\gamma=1-\lim_{x \to \infty} \int_{1/x}^{x}  \left( \frac{\sin t}{t}-\frac{1}{1+t} \right) \frac{dt}{t}
\end{equation}
\begin{equation}
\gamma= \lim_{x \to \infty}  \left( 1+\ln x-\int_{1/x}^{x} \frac{\sin t}{t^2} dt\right)
\end{equation}
Introduce $\sin x=2\sum_{n=0}^{\infty} (-1)^k J_{2k+1}(x)$ ([1]: p 988])  and write
\begin{eqnarray}
\gamma= \lim_{x \to \infty}  \left( 1+\ln x-2\int_{1/x}^{x} \frac{J_1 (t)}{t^2} dt\right)\\ \nonumber-2\sum_{k=1}^{\infty} \int_{0}^{\infty} (-1)^k \frac{J_{2k+1}(t)}{t^2}dt.
\end{eqnarray}
Further by using ([1]: p 707), when $-\nu-1 <\mu <1/2$.
\begin{equation}
\int_{0}^{\infty} x^\mu J_{\nu}(x) dx=2^{\mu} \frac{\Gamma(1/2+\nu/2+\mu/2)}{\Gamma(1/2+\nu/2-\mu/2)}~~~~(*).
\end{equation}
we obtain
\begin{eqnarray}
\gamma{=}\lim_{x \to \infty}  \left( 1+\ln x-2\int_{1/x}^{x} \frac{J_1 (t)}{t^2} dt\right)\\ \nonumber{-}\frac{1}{2} \sum_{k=1}^{\infty} (-1)^k \frac{1}{k(k+1)}.
\end{eqnarray}
Let us integrate by parts taking $J_1(x)$ as first function and use $2J_1'(x)=J_0(x)-J_1(x)$. The infinite series occurring in above is nothing but $1-2\ln 2$.
We get
\begin{eqnarray}
\gamma{=}\lim_{x \to \infty}  \left( 1+\ln x-2\int_{1/x}^{x} \frac{J_1 (t)}{t^2} dt-\int_{1/x}^{x}\frac{J_1(t)}{t}dt\right)\\ \nonumber +\int_{0}^{\infty} \frac{J_1(t)}{t} dt-\frac{1}{2}+\ln 2.
\end{eqnarray}
Note that $J_1(x) \approx x/2$, for small values of $x$ and for large values $x \sim \infty, J_1(x)=\sqrt{\frac{2}{\pi x}} \cos (x-3\pi/4) \rightarrow 0$. Using (*) again, we get
\begin{equation}
\gamma=\lim_{x \to \infty} \left( \ln 2x -\int_{1/x}^{x} \frac{J_0(t)}{t} dt\right).
\end{equation}
Finally, we use the series $J_0(x)=\sum_{0}^{\infty} (-1)^k \frac{(x/2)^{2k}}{(k!)^2}$, to write
\begin{equation}
\gamma= \lim_{x \to \infty} \left ( \ln 2x -2 \ln x+\frac{1}{2} \sum_{k=1}^{\infty} (-1)^{k-1} \frac{(x/2)^{2k}}{k(k!)^2} \right).
\end{equation}
Lastly, we get
\begin{equation}
\sum_{k=1}^{\infty} (-1)^{k-1} \frac{(x/2)^{2k}}{k (k!)^2} \sim 2 \ln (x/2)+2 \gamma,
\end{equation}
which is nothing but (2), for $p=2$.

$[1]$: I.S. Gradshteyn and I. M. Rhyzik, ``Table of Integrals", Series and Products (Academic Press(NY),1994).

$[2]$: Bruce C. Berndt, ``Ramanujan's Notebooks" Part 1, (Springer, Berlin, 1985) pp 98-100.


问:
derivative of $\frac{u}{(u^{T}u)^{0.5}}$
I am trying to compute the derivative of $\frac{u}{(u^{T}u)^{0.5}}$

I tried the following:

$\frac{\partial }{\partial u} (\frac{u}{(u^{T}u)^{0.5}})$

$=(\frac{\partial }{\partial u}u \cdot \sqrt{(u^{T}u)} - u \frac{\partial}{\partial u}\sqrt{(u^{T}u)} )/(u^{T}u) $

I get confused after that, can you please help me?
Thanks

答:
So, we can start with the derivative chain rule, modified to fit in with partial derivatives. Just to refresh on that idea, I recommend you check out this article I found on the web: https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-1a/a/chain-rule-review

Now, using the derivative chain rule, you can get the answer to the first part. But wait, we're not done yet. Then you use the partial derivative chain rule again, so you can then get the final answer.

Hopefully this was helpful, and I've made the evaluation process clear. Now, If I haven't, just leave a comment and let me know your concerns and/or questions.


问:
Compact Riemann surfaces with boundary
I am familiar (to an extent) with the theory of compact Riemann surfaces without boundary, but I am interested in the theory of compact Riemann surfaces with boundary.

To what extent do the following ideas/results generalize?

Riemann-Roch theorem

Abel's theorem

The Riemann-Hurwitz formula

The classification of holomorphic line bundles

The literature seems to be a little sparse on this particular subject, so any suggestions would be appreciated.

答:
I think, your question is too broad. Part 3 has an easy answer since the R-H formula for $n$-fold ramified coverings of compact surfaces with boundary, $f: S'\to S$, will also read
$$
\chi(S')= n\chi(S)- \sum_{p\in S'}(e_p-1),
$$
where $e_p$ is the ramification index of  the ramification point $p\in S'$. The proof is the same as the one sketched here.

As for other questions, if you do not impose any boundary conditions and your surface is connected with nonempty boundary, you will get very boring answers, e.g. the dimension in the R-R theorem is  infinite, all (discrete) divisors are principal, etc. It is possible you get some interesting results by imposing some boundary conditions. But, in the case of, say, R-R theorem you have to ensure that the result is a finite-dimensional vector space. It is very much unclear (to me) how this can be accomplished. But the complex analysis literature is vast and it is not impossible that some clever people came up with some creative answers. In my mind, one should start with a specific analytical problem involving holomorphic functions on surfaces with boundary, before trying to figure out what are the known tools.

Here is something to ponder: Even in the case when your surface is the closed unit disk $D$, there is no "easy" topological condition for an immersion $f: \partial D\to {\mathbb C}$ to extend to an immersion $F: D\to {\mathbb C}$. Even if you relax the restriction on $F$ and assume that it is a composition of a self-diffeomorphism $D\to D$ with a holomorphic map $D\to {\mathbb C}$, the condition for the existence of an extension is quite complicated. If you fix $f$ and require a holomorphic extension $F$, then there is always either no extension or exactly one extension.

Title: Compact Riemann surfaces with boundary


问:
What is correct answer for this algebraic expression?
If $(x+2)$ and $(x-2)$ are factors of $ax^2+2x-3x^2+bx-4$, then what would be the value of $(a+b)$ from following given options and how to arrive at it?

a) -7

b)  7

c)  14

d)  -8

答:
Hints:

If $\,\alpha\,,\,\beta\;$ are roots of the quadratic equation $\,ax^2+bx+c=0\;,\;a\neq 0\;$ , then

$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$

With the above (or directly if one remembers Vieta's formulae) , we also have, with the same notation, that

$$\begin{align*}\bullet&\;\;\alpha\beta=\frac ca\\\bullet&\;\;\alpha+\beta=-\frac ba\end{align*}$$