问:
Real part of an analytic function
I'd like to know how to take the real part (of the power series representation) of an analytic function with the goal of showing the (real part of the analytic) function is real analytic.

I know this may seem trivial but I haven't been able to convince myself that that's the case.

答:
If anyone is looking for the answer to this question, you need to take into account that if the function is the real part of an analytic function then the function must be infinitely differentiable (as an analytic function is itself infinitely differentiable). Thus, it can be represented as a power series, and thus fulfill the definition of real-analyticity. The one caveat one needs consider, however, is that the power series will be in two dimensions, and the expansion thereof must reflect a two-dimensional Taylor series.

With this in mind, one can show that a harmonic function on an open domain of $\mathbb{R}^2$ is real-analytic (as it can be proven that a harmonic function is the real/imaginary part of an analytic function).

Title: Real part of an analytic function


问:
Why is an inclusion map from an open subset smooth?
If $B$ is a manifold and $A\subseteq B$ is a regular submanifold of $B$, then the inclusion map $i:A\to B$ is an embedding and thus smooth.

If $B$ is a manifold, and $A\subseteq B$ is an open subset of $B$, then how do we justify that the inclusion map $i:A\to B$ smooth without using that $i$ is an embedding (in the sense of differential geometry; 'embedding' in the sense of elementary topology is within scope) or $A$ is a regular submanifold? Also, do not use germs or tangent space please.

Thanks in advance!

Here's my answer:

Let $i: A \to B$ be inclusion for $A$ an open subset of $B$. To show $i$ is smooth, we must show $i$ is continuous (known from elementary topology) and that for all $a \in A$, there are charts $(C,\gamma)$ about $i(a)=a \in B$ and $(D,\delta)$ about $a \in A$ for which $\gamma \circ i \circ \delta^{-1}: \gamma(i^{-1}(C) \cap D) = \gamma(C \cap D)$ $\to \gamma(C)$ is smooth

For any $a \in A$, choose $C=D=A$ and $\gamma=\delta=\text{id}_A$ identity function on $A$. Then $\gamma \circ i \circ \delta^{-1}: \gamma(i^{-1}(C) \cap D) = \gamma(C \cap D)$ $\to \gamma(C)$ becomes $i: A \to A$. We know that the inclusion map, with its range restricted to its image which happens to be its domain, becomes the identity map on its domain. Since the identity map on any manifold is smooth (I think), $i: A \to A$ is smooth. This satisfies the definition for $i: A \to B$ to be smooth.

答:
If you have a smooth structure $\{u_a, \phi_a\}$, it induces a natural smooth structure in the subspace topology. Overload the notation for simplicity

$$
\phi_a \circ i \circ \phi^{-1}_b = \phi_a \circ \phi^{-1}_b
$$

which must be smooth by the definition of a smooth structure, since we require that all transition functions be smooth.

The fact that it's open just guarantees that it remains locally homeomorphic to balls in $\mathbb{R}^{n}$

ContentLicense: CC BY-SA 4.0


问:
Given a set of unitary matrices, can one find a vector whose images under these unitary matrices span the underlying Hilbert space?
Given a set of (linearly independent) $d\times d$ complex unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ with $n\geq d$, does there exist a vector $v\in \mathbb{C}^d$ such that $$\text{span} \{U_1v, U_2v, \ldots , U_nv\} = \mathbb{C}^d ?$$

The motivation for this question comes from the theory of mixed unitary quantum channels. A quantum channel $\Phi: M_d \rightarrow M_d$ is a completely positive and trace preserving linear map. Any such map admits a Kraus representation of the form $\Phi (X) = \sum_{i=1}^k A_i X A_i^*$, where $\{A_i \}_{i=1}^k \subseteq M_d$ and $\sum_{i=1}^k A_i^* A_i = \mathbb{I}_d$. We say that a quantum channel is mixed unitary if it can be expressed as a convex combination of unitary conjugations: $\Phi (X) = \sum_{i=1}^n p_i U_i X U_i^*$. Our aim then is to look for a rank one input projector $X = vv^*$ for some $v\in \mathbb{C}^d$ such that the output $\Phi (vv^*) = \sum_{i=1}^n p_i (U_i v) (U_i v)^*$ has full rank. This is possible only if $n\geq d$. To avoid trivial counterexamples, we can also assume that $\{ U_i\}_{i=1}^n \subseteq M_d$ is linearly independent.

Follow-up question: Since it has been shown below that the question can be answered in the negative for $d\geq 4$, the natural way of progression would be to ask if one can provide a classification of all the sets of (linearly independent) unitary matrices $\{U_i\}_{i=1}^n \subseteq M_d$ which allow for the existence of $v\in \mathbb{C}^d$ such that $$\text{span}\{U_1v, U_2v, \ldots ,U_nv\}=\mathbb{C}^d.$$
One can also try to answer this question for (linearly independent) sets of arbitrary complex matrices: $\{ A_i \}_{i=1}^k \subseteq M_d$.

答:
For $d\geq 4$ the answer is no.   To see why, let $k=d-2$, so that $k\geq 2$.

By this answer every complex $k\times k$ matrix can be written as a linear combination of four unitary matrices.  As a consequence, we
see that
the unitary group $\mathscr U(k)$  spans $M_k(\mathbb C)$, so it is certainly possible to find a linear independent set formed by $k^2$
unitary $k\times k$ matrices, say $\{U_i\}_{1\leq i\leq k^2}$.  Now consider the linearly independent set
$$
  \{I_2\oplus U_i:1\leq i\leq k^2\}\subseteq \mathscr U(k+2) = \mathscr U(d),
  \tag 1
  $$
where $I_2$ is the $2\times 2$ identity matrix.

Given any
$$
  x=(x_1,  x_2,  \ldots , x_{k+2})\in \mathbb C^{k+2} = \mathbb C^d,
  $$
notice that every vector of the form
$$
  (I_2\oplus U_i)x
  $$
is orthogonal to
$$
  y: = (\overline{x_2},  -\overline{x_1},  0,  0,  \cdots ,  0)
  $$
so the set $\{(I_2\oplus U_i)(x):1\leq i\leq k^2\}$ cannot span $\mathbb C^d$.

As pointed out by @amsmath, if $x_1=x_2=0$, we need to argue instead that the vectors $(I_2\oplus U_i)x$ have vanishing first two coordinates, hence cannot span $\mathbb C^d$ either.

Since $k\geq 2$, we have that
$$
  k^2 \geq  k+2 = d,
  $$
so the set in (1) is a linearly independent set with (precisely $k^2$,  and hence) at least $d$ unitary $d\times d$ matrices,  but there is no vector $x$
such that the images of $x$ under our matrices span $\mathbb C^d$.


问:
Find the CDF and PDF of X
We produce a real number X through a two-stage experiment.

First roll a fair die to get an outcome $Y\in\{1,2,...,6\}$.

Then if $Y=k$, choose a point uniformly at random in $[0,k]$, denote it X.

What is the c.d.f. of X? What is the p.d.f. of X?

My Attempt:

The c.d.f. is defined as $F(t)=\mathbb{P}(X\leq t)$, so

\begin{equation}
F(t)=
\begin{cases}
0&\text{if } t<0\\
\frac{1}{6}\cdot\frac{1}{k-0}&\text{if }t\leq k\\
1&\text{if } ???
\end{cases}
\end{equation}

My thinking on how to get the probability is that there is a $\frac{1}{6}$ chance of rolling a number Y and then there is a $\frac{1}{k-0}$ chance of choosing a random number in $[0,k]$ because it is uniformly distributed. I am not sure what conditions of $t$ would cause a probability of $1$.

The p.d.f can be found by differentiating the c.d.f., but since there would be 6 different functions for $k\in\{1,2,...,6\}$, I am unsure how to combine them to differentiate. Should I sum them?

答:
Here is a nice little diagram. The vertical bars divide the sections of the number line into regions corresponding to the weights (on the right).

```<code>0[6] 1                     6
0[3] |[3]2                 3 3
0etc.|   |  3              2 2 2
0    |   |  |  4           3/2 3/2 3/2 3/2
0    |   |  |  |  5        6/5 6/5 6/5 6/5 6/5
0    |   |  |  |  |  6     1 1 1 1 1 1
</code>```

X will be uniform on each of these intervals given by the sum of the appropriate weights divided by 36, the sum of the weights. So for example, the pdf for X in the interval 0 and 1 is given by $\frac{6+3+2+3/2+6/5+1}{36}$. If you do this for each interval, you get the following pdf

$$f(X)=\begin{cases}49/120,&0<X<1\\
29/120&1<X<2\\
19/120&2<X<3\\
37/360&3<X<4\\
11/180&4<X<5\\
1/36&5<X<6\end{cases}$$

This corresponds to the piecewise cdf with six intervals of straight lines of different slopes...

$$F(X)=\begin{cases}0&X<0\\
49/120X,&0<X<1\\
29/120(X-1)+49/120&1<X<2\\
19/120(X-2)+13/20&2<X<3\\
37/360(X-3)+97/120&3<X<4\\
11/180(X-4)+41/45&4<X<5\\
1/36(X-5)+35/36&5<X<6\\
1&X>6\end{cases}$$

Title: Find the CDF and PDF of X

Body:
Here is a nice little diagram. The vertical bars divide the sections of the number line into regions corresponding to the weights (on the right).

0[6] 1                     6
0[3] |[3]2                 3 3
0etc.|   |  3              2 2 2
0    |   |  |  4           3/2 3/2 3/2 3/2
0    |   |  |  |  5        6/5 6/5 6/5 6/5 6/5
0    |   |  |  |  |  6     1 1 1 1 1 1

X will be uniform on each of these intervals given by the sum of the appropriate weights divided by 36, the sum of the weights. So for example, the pdf for X in the interval 0 and 1 is given by $\frac{6+3+2+3/2+6/5+1}{36}$. If you do this for each interval, you get the following pdf

$$f(X)=\begin{cases}49/120,&0<X<1\\
29/120&1<X<2\\
19/120&2<X<3\\
37/360&3<X<4\\
11/180&4<X<5\\
1/36&5<X<6\end{cases}$$

This corresponds to the piecewise cdf with six intervals of straight lines of different slopes...

$$F(X)=\begin{cases}0&X<0\\
49/120X,&0<X<1\\
29/120(X-1)+49/120&1<X<2\\
19/120(X-2)+13/20&2<X<3\\
37/360(X-3)+97/120&3<X<4\\
11/180(X-4)+41/45&4<X<5\\
1/36(X-5)+35/36&5<X<6\\
1&X>6\end{cases}$$


问:
How exactly are polynomials related to their evaluation maps?
My abstract algebra textbook* has taken great care in the last few sections to stress that, "polynomial are NOT functions to just 'plug' values of x into". To fully 'get' this, and to 'get' what polynomials really are then, I read through the proofs and explanations in the appendix that build up the idea of what a polynomial rings really is, and how we know a polynomial ring does exist for all rings.

In the current section of the book however, the book simply stated that "every polynomial induces a function from R to R", and then started to use polynomials as functions again, and the way it's doing so has me confused- what does it mean that it "induces" a function? The book will use the polynomial as function in a proof, and then make a conclusion about the polynomial itself- why, really, is that step valid?

The easiest example of this is it's proof for the polynomial remainder theorem.

The remainder theorem for polynomials.

The proof.

I get that, by the division algorithm for polynomials and the fact that (x - a) is degree 1, the following is a true statement: $\exists q(x),r(x)\in R[x] \; s.t \; f(x) = (x - a)q(x) + r(x)$ and $r=0 _r$ or $degree(r) = 0$

Isn't the next statement, namely that $f(a) = (a - a)q(a) + r(x) $, just a statement about some elements of in the ring R? Why can we say that the r(x) from the first statement is the same r(x) in the second statement? In other words, why is this second statement true?

/***********************/

I've worked ahead on some of the more complicated and involved stuff, but I have this nagging frustration in some part of my brain, as I don't "get" how these polynomials are really related with their evaluation maps, and when/why we can/cannot use a polynomial's evaluation map to make a conclusion about the polynomial itself.

*Here are some related threads I looked at that I thought about for a bit:

Why is it important that the evaluation map is a homomorphism?

What is the exact definition of polynomial functions?

*I feel like there is something very simple I'm missing.. I've spent too much time thinking about this and I think I've muddled my brain. Also, my bad if I've formatted this poorly, this is my first post. Thanks a lot for the help.

Edit:
*My textbook: Abstract Algebra, An Introduction - Third Edition, by Thomas W Hungerford and David Leep

答:
Perhaps what the author wanted to say is that albeit polynomials are usual treated as functions through its evaluation map in elementary algebra courses, the more appropriate view of each polynomial $p(x_1,\cdots,x_n)$ is that it is an element of the ring $R[x_1,\cdots,x_n]$.

Note that, the evaluation homomorphism is not always guaranteed to exists if $R$ is noncommutative, as in the thread below.

Evaluation homomorphism for noncommutative rings


问:
Help with this problem in number theory
$$xy = z^{2} + 1, (x, y, z) \in \mathbb{N}$$

Prove that there exists integers $$a, b, c, d$$ such that $$x = a^2 + b^2, y = c^2 + d^2, z = ac + bd$$

Could someone give me a hint on how to go about this? I've tried factorising with the equations given and simplifying, but the best I can do is proving with example and can't prove a general case :S

Thanks very much.

答:
Detailed Hint: As Calvin Lin has pointed out, the proof of this fact (implicitly) relies on the properties of the Gaussian Integers
, in particular, it uses the fact that $\mathbb{Z}[i]$ is a
Unique factorization domain

 and has prime elements:

$p\in\mathbb{N}$ where $p\equiv 3$(mod 4)

$a+bi$, $a,b\in\mathbb{Z}$, where $a^2+b^2\equiv 1$(mod 4), and $a^2+b^2$ is a prime number in $\mathbb{N}$

$1+2i$ (up to multiplication by units $1,-1,i,-i$)

This effectively gives us an analagous structure to the Gaussian Integers that the Fundamental Theorem of Arithmetic gives to Natural numbers (allows us to break into 'prime' elements), so we can now consider divisors of the RHS of the equation.

Back to the question, we can express the RHS as $(z-i)(z+i)$, and deduce properties about the prime element divisors of $z-i$ and $z+i$.

ContentLicense: CC BY-SA 3.0

Title: Help with this problem in number theory


问:
How do I prove that $161616 \dots 16$ ($100$ digits) cannot be written as a perfect square?
I did:
$$
16 \cdot 101010 \dots 1 \quad (99 \text{ digits})
$$

It's clear that $16 = 2^4$, satisfying the condition that the prime has to have an even power. But what about $101010 \dots 1$?

Also, I tried $16 \pmod 4$ but it's congruent to $0$.

(???)

答:
Just as you can cast out $9$s, you can also cast out $99$s.

Suppose we can write $N^2 = 161616\dots 16$. Where $16$ occurs $50$ times.

Casting out $99$s, you get

$$N^2 \equiv 16 + 16 +\dots + 16 \equiv 50\cdot 16 \equiv  800
      \equiv 8 + 00 \equiv8 \pmod {99}$$

This implies $N^2 \equiv 8 \pmod 9$, which has no solution.

\begin{array}{c|c}
 N \mod 9 & N^2 \mod 9 \\
\hline
 0    & 0 \\
 1, 8 & 1 \\
 2, 7 & 4 \\
 3, 6 & 0 \\
 4, 5 & 7 \\
\hline
\end{array}


问:
A closed set is not a submanifold
Can someone explain me why the set

$A:=\{(x,y)\in \mathbb R^2: x\geq 0\}$

is not a submanifold?

I also got another (easy) question:

In our lecture we are always talking about submanifolds. We never defined the meaning of manifolds. So what is there difference? Is there even a difference?

Thanks

答:
It's a manifold with boundary. https://en.m.wikipedia.org/wiki/Manifold#Manifold_with_boundary

Minor distinction.

Title: A closed set is not a submanifold


问:
Table of separation properties of various topological spaces
I was attempting to make a collection of some of the separation properties of various topological spaces, can anyone tell me which of the properties I got incorrect? Also does anyone know of any source with a collection like this with more topological spaces and more separation properties?

$$\begin{array}{cc|c}
 & \mbox{T1} & \mbox{Hausdorff}&\mbox{Regular} & \mbox{Normal}&\mbox{Separable}\\
\hline
\mathbb{R}&Y & Y & Y&Y&Y\\
\mathbb{R}^n&Y & Y & Y&Y&Y\\
\mbox{indiscrete}&N & N & Y&Y&Y\\
\mbox{discrete}&Y & Y & Y&Y&Y\\
\mbox{Cofinite}& Y& N&N &N&Y \\
\mbox{Cocountable}&Y&N&N&N&Y\\
\mathbb{R}_l&Y &Y &Y &Y&Y\\
\mbox{line w  2 origins} &Y &N &N &N&Y\\
\mbox{ordered  square} & Y&Y &Y &Y&N \\
\mathbb{R}_k &Y &Y &N &N&Y\\
\{0,1\}^A &Y &Y &Y &Y&Y
\end{array}$$

Comment:$\{0,1\}^A$ is the set of all functions from a set $A$ to $\{0,1\}$. $\mathbb{R}$ and $\mathbb{R}^n$ have the standard topology.$\mathbb{R}_l$ is $\mathbb{R}$ with the lower limit topology and $\mathbb{R}_k$ is $\mathbb{R}$ with the $k$ topology. I am not assuming regular/normal necessarily implies $T_1$.

答:
I assume that by ordered square you mean the lexicographically ordered square.

Separability is not a separation property. An uncountable space with the discrete or the co-countable topology is not separable, and $\{0,1\}^A$ is separable if and only if $|A|\le\mathfrak{c}=|\Bbb R|$. (That last result follows from the Hewitt-Marczewski-Pondiczery theorem, mentioned (with links) in this answer.)

Finite spaces with the cofinite topology and countable spaces with the co-countable topology are discrete and therefore regular and normal.


问:
How to compute the unknown power x in $0.15^x + 0.36^x = 1$
More generally, I have two known variables $a$ and $b$ and one unknown variable $x,$ where:

$$a^x + b^x = 1, a,b \in [0, 1]$$

Is there a way to compute the value of $x$ analytically?

答:
If by "analytically", you mean an exact algebraic solution, then no.

Note that $0.36^x = 0.15^{\frac{\log 0.36}{\log 0.15}\cdot x} = (0.15^x)^{\frac{\log 0.36}{\log 0.15}}$.

If we let $k = \frac{\log 0.36}{\log 0.15}$ and $0.15^x = y$ we can rewrite the original equation as

$y + y^k = 1$

For which you may be able to find nice solutions for some rational values of $k$. But this $k$ is transcendental, which precludes such a nice, exact solution.


问:
Is $|\mathbb Z_p/(p-1)\mathbb Z_p|=1$?
Let $\mathbb Z_p$ denote the ring of p-adic integers. We know $\mathbb Z_p/p\mathbb Z_p\simeq\mathbb F_p$. And then I met a problem asking what is $\mathbb Z_p/(p-1)\mathbb Z_p$. I think there is only one equivalence class in it because we have $$\frac{1}{1-p}=\sum_{n=0}^\infty p^n\in\mathbb Z_p.$$ But it's a bit weird to me. Is this correct?

答: This is correct, as you show $p-1$ is a unit in $\mathbb Z_p$, i.e. the ideal it generates is the full ring, so the quotient is trivial.

Title: Is $|\mathbb Z_p/(p-1)\mathbb Z_p|=1$?


问:
How to get the coordinates of where a line and a circle intersect when the result gives an imaginary number?
if you select a two points (everything in 2 dimensions), draw a circle centered on the first point and connect the two points with a line, and you know both of the point's coordinates and the radius of the circle, how do you get the coordinates of where the line and the circle intersect when the result is an imaginary number? sometimes when solving this problem the quadratic formula has a negative number on it's root which makes it impossible for me to know the coordinates of the point. anyone know how to solve this problem?

For Example. if you take the first point's coordinates (center of the circle) as (1,1), the second point's coordinates as (-2,2) and the radius of the circle as 1.
now if you use the equation for a circle ($r^2=(x-x0)^2+(y-y0)^2$) and replace $y$ with the equation for the line ($y = mx+b$), we get the equation for the coordinate of the interception ($r^2=(x-x0)^2+(mx+b-y0)^2$).
we get $m$ by doing $\frac{y0-y1}{x0-x1}$ which in this case is $-\frac{1}{3}$. We get $b$ from $mx-y$ which in this case is $-\frac{4}{3}$. now if we put everything in the equasion, it should be like this: $1^2=(x-1)^2+(-\frac{1}{3}x-\frac{4}{3}-1)^2$
which I solve like this:
$1=(x-1)^2+(-\frac{1}{3}x-\frac{7}{3})^2$ $<=>$
$0=x^2-2x+1+\frac{1}{9}x^2+\frac{14}{9}x+\frac{49}{9}-1$ $<=>$
$\frac{10}{9}x^2-\frac{4}{9}x+\frac{49}{9}=0$ $<=>$
$x=\frac{-(-\frac{4}{9})\pm\sqrt{(-\frac{4}{9})^2-4\times(\frac{10}{9})\times(\frac{49}{9})}}{2\times(\frac{10}{9})}$ $<=>$
$x=\frac{\frac{4}{9}\pm\sqrt{-24}}{\frac{20}{9}}$

as you can see I can't calculate the X value of the interception because the $\sqrt{-24}$ gives a imaginary number.

答:
Here's a simple geometric approach to finding the intersection point. Call the point around which the circle is drawn $u$, call the other point $v$, and call the intersection point $x$. Because $x$ lies along the line connecting $u$ and $v$, the displacement between $x$ and $u$ must point along the same direction as the displacement between $v$ and $u$. Next, because $x$ lies on a circle of radius $R$ around $u$, the displacement between $x$ and $u$ must have length $R$. Hence the displacement between $x$ and $u$ is given by $R(v-u)/\|v-u\|_2$, and the intersection point is thus given by,

$$x = u + R\frac{v-u}{\|v-u\|_2}$$

Plugging in coordinates should give you your answer! For instance, take $R=1$, $u=(1,1)$, and $v=(-2,2)$. Then $v-u=(-3,1)$, and $\|v-u\|_2=\sqrt{9+1}=\sqrt{10}$. So the intersection point is,

$$x=(1,1) + \frac{1}{\sqrt{10}}(-3,1)$$


问:
Binomial Distribution Word Problem
A student gets at least 8 hours of sleep 45% of the nights; the sleeping schedule is independent from night to night. Let Xi indicate whether the student gets at least 8 hours of sleep during the next 4 night respectively for i = 1, 2, 3, and 4. Let X = X1 + X2 + X3 + X4. Find the variance of X.

I'm struggling to understand what kind of distribution this problem is follow. I think it would be binomial. In that case wouldn't the variance of having x successes in 4 nights (trials) be modelled by the binomial. Hence the variance of X is (npq)^0.5 ? However, I feel like the question wants me to do some RV algebra. But a bit stuck.

答:
Each Xi is bernoulli and the entire X is binomial. A binomial random variable is the sum of independent, identically distributed bernoulli random variables. That is, $X_i\sim \mathscr{Bern}(.45)$ for i=1,2,3,4 and $X\sim \mathscr{Bino}(4,.45)$.

Now you can find the variance of the sum of the Xi's or the X directly. We have $Var(X)=np(1-p)=4(.45)(.55)$. Also, the Xi's are independent, so they are uncorrelated (independence is stronger than correlation), thus $Var(X_1+X_2+X_3+X_4)=p(1-p)+...+p(1-p)=.45(.55)+...+.45(.55)=4(.45)(.55)$.

Title: Binomial Distribution Word Problem


问:
Calculation in Routh's theorem
The proof of Routh's theorem concludes with showing$$1-\frac{x}{zx+x+1}-\frac{y}{xy+y+1}-\frac{z}{yz+z+1}=\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}.$$I seek an elegant "proof from the book" of this, rather than one that involves tedious, potentially error-prone algebra. In particular, it feels like symmetries, degree-counting etc. should make this obvious, rather than an accident where the numerator happens to have a nice factorization. My best approaches are these two:

Option 1

The LHS's first two, three and four terms have respective sums $\frac{zx+1}{zx+x+1}$,$$\frac{(zx+1)(xy+y+1)-y(zx+x+1)}{(zx+x+1)(xy+y+1)}=\frac{x^2yz+zx+1}{(zx+x+1)(xy+y+1)}$$and$$\frac{x^2yz+xz+1}{(zx+x+1)(xy+y+1)}-\frac{z}{yz+z+1}=\frac{(x^2yz+xz+1)(yz+z+1)-z(zx+x+1)(xy+y+1)}{(zx+x+1)(xy+y+1)(yz+z+1)}.$$The numerator is nine monic terms of degree $0$ to $6$ minus nine monic terms of degree $1$ to $5$, so the terms of degree $0$ and $6$ will survive as $(xyz)^2+1$, and any other surviving term(s) will have coefficients summing to $-2$. The problem's symmetries mandate $-2xyz$ to finish the job.

That's quite nice, but the third partial sum probably can't be done in one's head. What one can say, however, is the third partial sum's numerator will be six terms of degrees $0$ to $4$ minus three of $1$ to $3$, so a $0$ and a $4$ survives, but it's harder to deduce without calculation that the third uncancelled term will be of degree $2$.

Option 2

This one looks like it might end up more elegant at first, but it looks like it ultimately requires some of Option 1's techniques to finish.

The case $x=\tfrac{1}{yz}$ has left-hand side$$1-\frac{1}{yz+z+1}-\frac{yz}{yz+z+1}-\frac{z}{yz+z+1}=0,$$so the general case's numerator must be divisible by $xyz-1$. In the special case $x=y=z$, the left-hand side is$$1-\frac{3x}{x^2+x+1}=\frac{(x-1)^2}{x^2+x+1}=\frac{(x^3-1)^2}{(x^2+x+1)^3}.$$The most obvious generalization with appropriate symmetries and denominator is $\frac{(xyz-1)^2}{(xz+x+1)(xy+y+1)(yz+z+1)}$, as desired. The most general numerator is of the form$$(xyz-1)(xyz+1+p(x,\,y,\,z)),$$where $p$ is invariant under a cyclic permutation of $x,\,y,\,z$, with $p(x,\,x,\,x)=0$ and $p\left(\tfrac{1}{yz},\,y,\,z\right)=0$.

The first constraint makes $p$ a polynomial in $a:=x+y+z,\,b:=xy+yz+zx,\,c:=xyz$; the second ensures that polynomial vanishes when $a=3x,\,b=3x^2,\,c=x^3$. These are achievable with a factor such as $a^2-3b$, $a^3-27c$, $ab-9c$ or $b^3-27c^2$. The third constraint only adds one requirement, divisibility by $c-1$. Ultimately, some careful degree-counting is needed to prove $p=0$.

答:
Note that (just observe to know it, no need to expand it!)
$\mathrm{LHS}(zx + x + 1)(xy + y + 1)(yz + z + 1) - (xyz - 1)^2$ is a polynomial in $z$ of degree at most $2$.

Thus, if we can find three distinct constants $a, b, c$ such that $\mathrm{LHS} = \mathrm{RHS}$ for $z = a, b, c$ respectively,
then we have $\mathrm{LHS} = \mathrm{RHS}$ for all $x, y, z$.

We choose $z = 1, -1, 0$.

For $z = 1$, $\mathrm{LHS} = 1 - \frac{x}{2x + 1} - \frac{y}{xy + y + 1} - \frac{1}{y + 2}$
and $\mathrm{RHS} = \frac{(xy - 1)^2}{(2x + 1)(xy + y + 1)(y + 2)}$.
Similarly, we only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 0, 1, -1$.

For $z = -1$, $\mathrm{LHS} = 1 - x - \frac{y}{xy + y + 1} - \frac{1}{y}$
and $\mathrm{RHS} = \frac{(xy + 1)^2}{(xy + y + 1)(-y)}$.
Similarly, we only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 1, -1, 2$.

For $z = 0$, $\mathrm{LHS} = 1 - \frac{x}{x+1} - \frac{y}{xy + y + 1}$ and
$\mathrm{RHS} = \frac{1}{(x+1)(xy + y + 1)}$. Verify it directly (or only need to check $\mathrm{LHS} = \mathrm{RHS}$ for $y = 0, -1$).

We are done.

Title: Calculation in Routh's theorem


问:
Compute the expectation of X
When a certain basketball player takes his first shot in a game with probability 1/3.
If he misses his first 2 shots, his third shot will go with probability 1/5.
If he misses his first 3 shots, his fourth shot will go with prob. 1/6.
If he misses his first 4 shots, the coach will remove him from the game.
Assume that the player keeps shooting until he succeeds or he is removed from the game. Let X denote the number of shots he misses until his first success or until he is removed from the game.

Compute $\mathbb{E}[X]$

If I separate this into indicator functions then the expectation is the sum of the probabilities of each event happening. This would mean thst if I find the pmf and then sum the probabilities, I would get the expectation. My concern is that this would mean the expectation is $1$. Is that a reasonable expectation? The pmf I found is as follows:

$$\mathbb{P}(X=1)=\mathbb{P}(\text{misses 1st and makes second})=\frac{2}{3}*\frac{1}{4}$$
$$\mathbb{P}(X=2)=\mathbb{P}(\text{misses 1st, 2nd and makes third})=\frac{2}{3}*\frac{3}{4}*\frac{1}{5}$$
$$\mathbb{P}(X=3)=\mathbb{P}(\text{misses 1st,2nd,3rd and makes fourth})=\frac{2}{3}*\frac{3}{4}*\frac{4}{5}*\frac{1}{6}$$
$$\mathbb{P}(X=4)=\mathbb{P}(\text{misses 1st,2nd,3rd,4th)}=\frac{2}{3}*\frac{3}{4}*\frac{4}{5}*\frac{5}{6}$$

All of this sums to $1$ as a pmf should.

Edit: I forgot that there is also the probability that he misses 0 shots. So I should add $\mathbb{P}(X=0)=\frac{1}{3}$

Now it should add up to 1.

答:
You don't use an indicator r.v. here, what you need to do is to compute $\Bbb E[X] = \Sigma\left([X_i]\cdot \Bbb P[i]\right)$

so $\Bbb E[X] = 0\cdot\Bbb P[0] + 1\cdot\Bbb P[1] + ... 4\cdot\Bbb P[4]$

= $0*1/3 +1*1/6 + 2*1/10 + 3*1/15 + 4*1/3 = 1.9$

Title: Compute the expectation of X


问:
Proof: A bijective linear map cannot send a line to a point
Let $T\colon\mathbb R^3 \rightarrow\mathbb R^3$ be a bijective linear map. Prove that $T$ can not send a line to a point.

I started off thinking that $T(t\vec{m}+\vec{b}) = \vec{x}$ should have infinitely many solutions, but then I don't know how to go on with the equation.

答: If it sends a line to a point, it is not one-one.  Thus it is not bijective. QED</p>


问:
An indirect proof of Rice's theorem
I'd like to deduce a proof of this version of Rice's theorem from the Kleene theorem. (I've asked a similar question before but this one is different.)

Let $\mathcal F$ be the class of all unary computable functions. Let $\mathcal A\subset \mathcal F$ be a proper subset. Let $U$ be a Godel universal function (the definition can be found here). Then $\{n:U_n\in\mathcal A\}$ is undecidable. ($U_n(x)$ is the $n$th section of the Godel universal function $U(n,x)$)

Here's my attempt:

Let $T=\{n:U_n\in\mathcal A \}$. Let $f\in\mathcal A$ and $g\notin \mathcal A$. Define $$h:N\to N\\ $$
by $$n\mapsto f(n)\text{ if } n\in T\\ n\mapsto g(n)\text{ if } n\notin T\\ $$

(this is the only reasonable thing I can think of). Assume $T$ is decidable. Then $h$ is a total computable function. Thus by Kleene's theorem, there is a number $n_0$ such that $$U(n_0,x)=U(h(n_0),x)$$ for all $x$.

But I'm not sure what to do next. One option is to consider cases $n_0\in T$ and $n_0\notin T$. Well, suppose $n_0\in T$. Then $h(n_0)=f(n_0)$. I don't know what to do to get a contradiction. I could express $f$ in terms of $U$: suppose $n_f$ is the number of the computable function $f$, i.e. $U(n_f,x)=f(x)$ for all $x$. Then $h(n_0)=U(n_f,n_0)$. What to do next? What kind of contradiction should I be trying to obtain?

答:
Since $\mathcal A$ is proper, we can take $U_p\in \mathcal A$ and $U_q\notin \mathcal A$. Define $$h:N\to N$$ by $$n\mapsto q \text{ if } n\in T\\n\mapsto p\text{ if }n\notin T$$
Assume $T$ is decidable. Then $h$ is computable. Let's show that $h$ cannot have a fixed point.

Suppose $n$ is a fixed point. So $U(n,x)=U(h(n),x)$ for all $x$.

Case 1: $n\in T$. Then $U(n,x)=U(q,x)$ for all $x$, i.e. $U_n=U_q$. This is impossible because $U_n\in \mathcal A$ but $U_q\notin \mathcal A$.

Case 2: $n\notin T$. Then $U(n,x)=U(p,x)$ for all $x$, i.e. $U_n=U_p$. This is impossible for a reason similar to the reason above.

So $h$ cannot have a fixed point. Thus $T$ is undecidable.

Title: An indirect proof of Rice's theorem


问:
What does it mean that random variables of mean zero form a Hilbert space with the inner product being the covariance?
I see that the covariance can be interpreted as a measure of similarity like the dot product of two vectors, and that the correlation coefficient is akin to the cosine in the dot product, while the variance is the vector norm.

Indeed,

Symmetry: $\text{cov}(X, Y)=\text{cov}(Y, X)$

Bilinearity:  $\text{cov}(X, aY + bZ)=a\,\text{cov}(X, Y) + b \,\text{cov}(X, Z)$

Positive definitiveness: $ \text{cov}(X, X)=\text{var}(X)\geq 0$...

and it must only zero when the random variable is "zero."

But I am confused about the "mean zero" class equivalence part.

答: The mean zero requirement is to ensure the positive definiteness, i.e. if $X$ is in this space and $\operatorname{Cov}(X,X)=0$ then $X$ is the zero random variable. (To have a unique "zero random variable" in this sense, you must first take the quotient by $X \sim Y$ if $X=Y$ a.s., which is the usual procedure used in defining the $L^p$ spaces.) If you bundle in all the random variables with finite variance, you don't have this property anymore because all constants will have $\operatorname{Cov}(c,c)=0$.


问:
Cartesian Product over Symmetric Difference
Let's say I have that $(x,y)\in (A\times B)\Delta (A\times C).$ How would I expand this statement assuming $A\Delta B = (A-B)\cup (B-A)$?

Here is as far as I have gotten. Let $(x,y)\in (A\times B)\Delta (A\times C).$ Since we define $A\Delta B = (A-B)\cup (B-A)$, then $(A\times B)\Delta (A\times C) = [(A\times B)-(A\times C)]\cup [(A\times C)-(A\times B)].$

Is the next step ($x\in A\times B$ and $x\notin A\times C$) or ($x\in A\times C$ and $x\notin A\times B$)? Wouldn't this produce a contradiction where $x\in A$ and $x\notin A$?

答:
Tony has given a fine algebraic solution. For a more intuitive approach to exactly the same thing it’s useful to understand a fundamental fact about the symmetric difference of two sets: $X\mathop{\triangle}Y$ is the set of things that belong to exactly one of the sets $X$ and $Y$.

Thus, $\langle x,y\rangle\in(A\times B)\mathop{\triangle}(A\times C)$ if and only if $\langle x,y\rangle$ belongs to exactly one of $A\times B$ and $A\times C$. Clearly this cannot be the case if $x\notin A$, so we can immediately conclude that $x\in A$. Then $\langle x,y\rangle\in A\times B$ if and only if $y\in B$, and $\langle x,y\rangle\in A\times C$ if and only if $y\in C$, so $\langle x,y\rangle$ is in exactly one of $A\times B$ and $A\times C$ if and only if $y$ is in exactly one of $B$ and $C$, i.e., if and only if $y\in B\mathop{\triangle}C$.


问:
7 balls are distributed randomly in 7 cells. If 2 cells are empty, show that the conditional probability of a triple occupancy equals 1/4
This problem comes from Feller's Introduction to probability. and it goes like:

"Seven balls are distributed randomly in seven cells. Given that two cells are empty, show that the (conditional) probability of a triple occupancy of some cells equals 1/4"

So far i have found out that all of the distributions that form a triple are given by: ${7 \choose 3}(4!) = \frac{7!}{3!}$.

Yet, i am not sure of how to go on, because there are $7^7$ total distributions, but clearly if you divide these two quantities the result is not even close to $1/4$.

I have also tried to go on by considering that the only occupancy numbers in the distribution can be $0,1,3$ so the total number of distributions that coincide with the ocuppancy numbers $3,1,1,1,1,0,0$ in some order are given by: $\frac{7!}{3!}\frac{7!}{3!4!}$. But if you divide this by $7^7$ the result is still no close to $1/4$.

And also when you consider the fact of conditional probability the results only get worse. So what can i do?

答:
Let $T$ be the event that there is a triple occupancy, and let $E$ be the event that exactly two cells are empty.  We are asked for $$\Pr(T|E)=\frac{\Pr(T\cap E)}{\Pr(E)}$$

The number of distributions with exactly two empty cells and a triple occupancy is $$7\binom73\binom624!=88,200\tag1$$  There are $7$ ways to choose the cell with three balls, $\binom73$ ways to choose the balls to go into it, $\binom62$ ways to choose the two empty cells, and $4!$ to distribute the remaining balls into the remaining cells.

To compute the number of distributions with exactly two empty cells, we use the principle of inclusion and exclusion to get
$$\binom 72\left(5^7-\binom514^4+\binom523^7-\binom532^7+\binom541^7\right)=352,800\tag2$$

There are $\binom72$ choices for the two empty cells, and $5^7$ ways to distribute the balls into the other $5$ cells.  Now, we must subtract the distributions that use only $4$ of the cells, that is $\binom514^7$. We have subtracted distributions that use only $3$ of the cells twice, so we must add them back in, and so on.

Dividing $(1)$ by $(2)$ does indeed give $\frac14$.


问:
Efficient and Accurate Numerical Implementation of the Inverse Rodrigues Rotation Formula (Rotation Matrix -> Axis-Angle)
I want to implement the Inverse Rodrigues Rotation Formula (also known as Log map from SO(3) to so(3)), in double precision code (MATLAB is fine for the example) preferably as a 3-parameter vector with the unit direction vector scaled by the magnitude of rotation.

The analytical form is (from Wikipedia):

$\theta = \arccos\left( \frac{\mathrm{trace}(R) - 1}{2} \right)$

and then use it to find the normalised axis:

$\omega = \frac{1}{2 \sin(\theta)} \begin{bmatrix} R(3,2)-R(2,3) \\ R(1,3)-R(3,1) \\ R(2,1)-R(1,2) \end{bmatrix}$

which can then be used to find the scaled axis of rotation
$\rho = \theta \omega$

Of course, $\sin(\theta)$ will cause the denominator to approach zero, which is undefined. The rotation vector $\rho$ at zero rotation is $\rho = \begin{bmatrix}0 & 0 & 0\end{bmatrix}^T$.  Furthermore, it will also be undefined at $n\pi$, though we may assign the rotation to be the desired sign ($\pm\pi$)

The naive implementation is to have if() statements for floating point values close to $n\pi$ rotations, but surely there is a better way than some dirty hacks around the singularities... right?

EDIT:

At rotations near zero, empirically, the following works well:

```<code>if (trace(R) > (3 - small_number))

    inverse_sinc = 1 + (1.0 / 6.0)      * theta_2 + ...
                       (7.0 / 360.0)    * theta_4 +
                       (31.0 / 15120.0) * theta_6;

    rho = 0.5 * inverse_sinc * r;

end
</code>```

where $\theta$ (and powers thereof), and $\mathbf{r} = \begin{bmatrix} R(3,2)-R(2,3) \\ R(1,3)-R(3,1) \\ R(2,1)-R(1,2) \end{bmatrix}$ are pre-calculated, and inverse_sinc (i.e. x/sin(x))is calculated from the Taylor series. This is accurate to better than 10^-11 in each axis when unit tested across a range of values (0, eps, 10^-12 through to 10^-3 and negative values for each across all three axes).

A good solution for $\theta = \pi$ still eludes me...

答:
A slightly different take, that uses eigenvector calculation to determine the axis.

Given a $3\times3$ rotation matrix, $R$, find a unit vector, $v$, and the corresponding angle, $\theta$ such that (Rodrigues formula).

$R = I + \sin(\theta)\hat{v} + (1-\cos(\theta))\hat{v}^2$

where $\hat{v}$ denotes the antisymmetric matrix,

$
\begin{bmatrix}
  0 & -v(3) & v(2) \\
  v(3) & 0 & -v(1) \\
  -v(2) & v(1) & 0
\end{bmatrix}
$

$v$ is an (not "the" because $-v$ will also work in general, and any unit vector will work for $R=I$) axis of rotation, and $\theta$ is the corresponding angle of rotation. Thus, finding $v$ and $\theta$ answers the question asked above.

$v$ can be found as an ("the" if $R \ne I$) eigenvector of $R$ corresponding to eigenvalue 1.

$\theta$ satisfies the following.

$\DeclareMathOperator{\trace}{trace} \trace(R) = 1 + 2\cos(\theta)$.

$\trace(\hat{v}R) = -2\sin(\theta)$.

Both of these can be obtained by direct calculation from Rodrigues formula above.

Thus, $\DeclareMathOperator{\atantwo}{atan2} \theta = \atantwo(-\trace(\hat{v}R),\trace(R)-1)$.

In summary,

$v$ is an eigenvector of $R$ corresponding to eigenvalue 1.

$\theta = \atantwo(-\trace(\hat{v}R),\trace(R)-1)$

solves the problem without any special cases needed.


问:
How to color a board such that each square has exactly two neighbors of the opposite color
Let $m≥2$, $n≥2$ integers. We want to color the squares of an $m × n$ board with black and white so that each square has exactly two neighbors of the other color. Determine all the values of $m$ and $n$ for which it is possible to do such a coloring.
Clarification: Neighboring squares are those that have a common side.

I found a way to color the square when $m$ and $n$ are both even. I think that if $m$ or $n$ is odd, then it is impossible to color de board. But i can't prove it. Could someone help me, please?

答:
If $m=n$, then $m$ and $n$ are both even or both odd; however, the condition requires an even number of black and white squares, so $mn = m^2$ is even; therefore $m=n$ is even. Now let's deal with rectangles that are not squares. Without loss of generality, $n<m$.

Lemma. Assuming $n < m$, the top $n$ rows form an $n\times n$ sub-board symmetric about its main diagonal.

Proof. Induct on the anti-diagonals, looking at them in the following order:

```<code>123456
23456
3456
456 .
56   .
6     .
</code>```

We begin in the middle of each anti-diagonal:

For the odd anti-diagonals, the middle square can be filled arbitrarily.

For the even anti-diagonals, the middle two squares are both adjacent to the middle square of the previous anti-diagonal, which (by symmetry) already has two neighbors of one color, so they get the same color.

Then we work outwards, and every step is forced, so the anti-diagonal is symmetric. $\square$

The top $n \times n$ sub-board is symmetric, and its rightmost edge consists of $n$ squares on the edge of the board; its bottom edge (the $n^{\text{th}}$ row of the board), on the other hand, has $n$ squares with more squares below them. However, since the two edges are the same, the coloring condition for them is satisfied just by the $n \times n$ sub-board. Therefore the squares in the $(n+1)^{\text{th}}$ row are identically colored to the squares on the $n^{\text{th}}$ row.

It follows that if we delete the $n \times n$ sub-board, we get an $(m-n)\times n$ board for which the coloring condition also holds.

Now we can finish the claim by induction on the size of the board. The induction hypothesis tells us that both $m-n$ and $n$ must be even; therefore $m$ and $n$ are both even.

Body:
If $m=n$, then $m$ and $n$ are both even or both odd; however, the condition requires an even number of black and white squares, so $mn = m^2$ is even; therefore $m=n$ is even. Now let's deal with rectangles that are not squares. Without loss of generality, $n<m$.

Lemma. Assuming $n < m$, the top $n$ rows form an $n\times n$ sub-board symmetric about its main diagonal.

Proof. Induct on the anti-diagonals, looking at them in the following order:

123456
23456
3456
456 .
56   .
6     .

We begin in the middle of each anti-diagonal:

For the odd anti-diagonals, the middle square can be filled arbitrarily.

For the even anti-diagonals, the middle two squares are both adjacent to the middle square of the previous anti-diagonal, which (by symmetry) already has two neighbors of one color, so they get the same color.

Then we work outwards, and every step is forced, so the anti-diagonal is symmetric. $\square$

The top $n \times n$ sub-board is symmetric, and its rightmost edge consists of $n$ squares on the edge of the board; its bottom edge (the $n^{\text{th}}$ row of the board), on the other hand, has $n$ squares with more squares below them. However, since the two edges are the same, the coloring condition for them is satisfied just by the $n \times n$ sub-board. Therefore the squares in the $(n+1)^{\text{th}}$ row are identically colored to the squares on the $n^{\text{th}}$ row.

It follows that if we delete the $n \times n$ sub-board, we get an $(m-n)\times n$ board for which the coloring condition also holds.

Now we can finish the claim by induction on the size of the board. The induction hypothesis tells us that both $m-n$ and $n$ must be even; therefore $m$ and $n$ are both even.


问:
Elementary central binomial coefficient estimates

How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?

Does anyone know any better elementary estimates?

Attempt.  We have
$$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$
Then we have
$$\left(1+\frac{k}{2(n-k)}\right)>\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$
So maybe, for the lower bound, we have
$$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}>\frac{2^n}{\sqrt{4n}}.$$
By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes
$$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$
but this isn't $>\frac{2^n}{\sqrt{4n}}$.

答:
Here is a  better estimate of the quantity. I am not going to prove it. Since I know about it, hence I am sharing and will put the reference.

$$\frac{1}{2\sqrt{n}} \le {2n \choose n}2^{-2n} \le \frac{1}{\sqrt{2n}}$$
Refer to Page 590 This is the text book "Boolean functions Complexity" by Stasys Jukna.


问:
Prove Prop 4.11

4.11 Proposition. Let $X$ be a set and let $\mathcal{S}$ be any collection of subsets of $X$ such that $X=\bigcup_{V\in\mathcal{S}}V$. Let $\mathcal{T}$ denote the collection of all subsets of $X$ that can be obtained using two operations:

taking finite intersections of sets in $\mathcal{S}$;

taking arbitrary unions of sets obtained in 1. Then $\mathcal{T}$ is a topology on $X$.

Proof. Exercise. $\qquad\square$

The following Proposition comes off www.mth527.site I will take my best shot at it.

My understanding

By taking union of empty sets $\cup\emptyset=\emptyset\in\tau$

Let {$C_j:j\in J$} be family members of $\tau$ Then each $C_j$ is a union of the members of S Hence the union of all
{$C_j:j\in J$} is also union of all the members in S and so in
$\tau$

We want to verify  $Y\cap Z\in\tau$ (1)

Let Y,Z$\in\tau$

But $Y=\bigcup_{V\cap S}V$ then for $V\in S$
and V$\in\tau $ and $Z=\bigcup_{W\in S}W$
then we have $W\in S$ for $W\in\tau$

$$Y\cap Z=\bigcup_{V\in S}V)\bigcap(\bigcup_{W\in S}W)= \bigcup_{V\in S}(\bigcup_{W \in S}(V\cap W)$$

So $Y\cap Z\in\tau$

We want to verify (2)

So if $T\subset X$ and $S\subset X$ then $S\subset T$ since arbitrary unions are taken
from it So if $V\in S$ so that $V\in\tau$, $\bigcup(Y\cap Z)=\bigcup_{V\in S}\bigcup_{W,V\in S}(V\cap W)$

theorem source

答:
Hint:

(1) Can you show that $\varnothing$ and $X$ are both in $\mathscr{T}$? To show the latter, be sure to use the fact that $X = \bigcup_{V \in S} V$.

(2) Can you show that finite intersections of subsets of $\mathscr{T}$ are also in $\mathscr{T}$ ? For this, you might want to remember that "finite intersection of (finite intersections of sets from $S$)" is still a finite intersection of sets from $S$.

(3) Likewise in (2), can you show arbitrary unions of objects of $\mathscr{T}$ is also in $\mathscr{T}$?

Edit: You will have likely have to fill in some details in (2) because the sets in $\mathscr{T}$ are also arbitrary unions of finite intersections of sets from $S$; but that shouldn't be too hard to handle once you symbolically write things down

Title: Prove Prop 4.11


问:
Basic question on Bregman divergences and strong convexity
I have seen the following claim in several research papers:

Let $p \in (1,2]$, and consider the function $\Phi(x)=\frac{1}{2}\|x\|^2_p$ (where $x\in\mathbb{R}^d$). Then the corresponding Bregman divergence $D_\Phi$ is $(p-1)$-strongly convex w.r.t the $\ell_p$ norm: $$D_\Phi(y,x) \geq \frac{p-1}{2} \|x-y\|^2_p$$ for all $x,y\in\mathbb{R}^d$.

Typically, no other detail is given, besides "cf. Ball et al. (1994)." I cannot see how that follows immediately from that paper, however. Is it obvious?

答:
Some progress (?).

For any $\delta\in\mathbb{R}^d$, considering the average $\frac{\Phi(x+\delta)+\Phi(x-\delta)}{2}$, we get that the terms $\langle \nabla\Phi(x), x\pm\delta-x\rangle$ cancel, leading to
$$
\frac{D_\Phi(x+\delta,x)+D_\Phi(x-\delta,x)}{2}
= \frac{1}{2}\left(\frac{\|x+\delta\|_p^2+\|x-\delta\|_p^2}{2}-\|x\|_p^2\right)
$$
and, using the Ball—Carlen—Lieb inequality (Proposition 3 in the paper, Eq. (2.18)) gives
$$
\frac{D_\Phi(x+\delta,x)+D_\Phi(x-\delta,x)}{2}
\geq \frac{p-1}{2}\|\delta\|_p^2 \tag{1}
$$

Setting $\delta := x-y$ in (1), we get
$$
\frac{D_\Phi(2x-y,x)+D_\Phi(y,x)}{2}
\geq \frac{p-1}{2}\|x-y\|_p^2 \tag{2}
$$
which seems to be related to, but weaker than what we want.


问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
Goedel's original completeness/compactness proofs in logic are very hard and technical. Modern versions of the proofs are considerably simpler and do not use any sophisticated new theory.

Most proofs of the fundamental theorem of algebra use some topological/homotopical deep(ish) results or some deep(ish) results of complex analysis. In fact, the theorem can be proved using only the definition of complex numbers and absolute value, using very elementary properties of complex numbers, and the entire proof is about half a page long (such a proof is actually a minimum modulus argument but applied to a polynomial so that the computation can be done directly without appeal to the general minimum modulus principle). The proof is extremely elementary.

The proof that Euclide's fifth postulate is independent of the other axioms of Euclidean geometry by exhibiting models (i.e., the sphere and the hyperbolic plane) where the postulate fails (in different ways) is completely elementary. Certainly all those who tried to prove/disprove the fifth postulate did so while walking on a counterexample. However, the numerous attacks on the problem prior to its settlement in the 20th century were quite sophisticated and laborious. The barrier was conceptual - the understand of the importance of models - not technical.

Brouwer's topological proof of his fixed point theorem used (I believe) homology and/or the fundamental group. A perfectly elementary proof using Sperner's Lemma was later discovered (I'm not entirely sure about the chronological order here, so I may be mistaken).

Initial proofs that the higher homotopy groups are all abelian were greatly simplified by the Eckman-Hilton argument (which is completely elementary).

The Mac Lane coherence theorem in category theory is rather technical and is enormously simplified by considering the Yoneda embedding in the 2-categorical setting. All of the machinery was already present, but it required some re-assembly to figure out a rather elementary proof.


问:
Construct a sequence of functions in $L^p(\mathbb{R})$ such that $f_n$ converges to zero in $L^p$.
I'm trying to find a sequence of functions $f_n\in L^p(\mathbb{R})$, with $1\leq p < \infty$ such that $\lVert f_n \rVert_{L^p}\to 0$. Moreover for all $x\in \mathbb{R}$ the sequence satisfy two conditions
$$\liminf f_n(x)= 0 \ \text{and}\ \limsup f_n(x)= \infty.$$

I think in use some sequence based on characteristic function, but i can't find the sequence.

答:
Let $f_n = n^{c} I_{E_n}$ where $E_n$ is a measurable set of measure $ \le 1/n$ and $I_{E_n}$ its indicator function.  We have $\|f_n\|_p^p = \le n^{pc-1} \to 0$ if $c < 1/p$.  Now all you need to do is find a sequence $E_n$ such that every point of $\mathbb R$ is in infinitely many $E_n$.  You can let $E_n = \cup_{k=-\infty}^\infty F_{k,n}$ where $F_{k,n}$ is
an interval of length $2^{-|k|-2}/n$ in $[k,k+1]$.  And for each $k$, let the intervals $F_{k,n}$ "walk" back and forth across $[k,k+1]$.  Since $\sum_n 1/n$ diverges, you can do this so that each point is covered infinitely many times.


答:
Using Stirling Number of the second kind -

If we take any $5$ cells, number of ways to distribute $7$ balls such that none of the cells are empty is given by $StirlingS2[7,5] = 140$, without distinguishing between cells.

Now we consider cases where one of the cells has $3$ balls, which is $\displaystyle {7 \choose 3} = 35$ ways of choosing $3$ balls and rest $4$ balls go one each in remaining $4$ cells.

So the desired probability $\displaystyle  =  \frac{35}{140} = \frac{1}{4}$.

Title: 7 balls are distributed randomly in 7 cells. If 2 cells are empty, show that the conditional probability of a triple occupancy equals 1/4

Body:
This problem comes from Feller's Introduction to probability. and it goes like:

"Seven balls are distributed randomly in seven cells. Given that two cells are empty, show that the (conditional) probability of a triple occupancy of some cells equals 1/4"

So far i have found out that all of the distributions that form a triple are given by: ${7 \choose 3}(4!) = \frac{7!}{3!}$.

Yet, i am not sure of how to go on, because there are $7^7$ total distributions, but clearly if you divide these two quantities the result is not even close to $1/4$.

I have also tried to go on by considering that the only occupancy numbers in the distribution can be $0,1,3$ so the total number of distributions that coincide with the ocuppancy numbers $3,1,1,1,1,0,0$ in some order are given by: $\frac{7!}{3!}\frac{7!}{3!4!}$. But if you divide this by $7^7$ the result is still no close to $1/4$.

And also when you consider the fact of conditional probability the results only get worse. So what can i do?


问:
Values of the infinite product $\prod_n\frac{(n+1)x}{1+nx}$
I am trying to compute the inverval of convergence and the explicit value of the infinite series $$\prod_{n=1}^{\infty}\left(\dfrac{(n+1)x}{1+nx}\right).$$ I believe the interval of convergence is $(-1,1)$ and exact value is $\dfrac{x}{1-x},$ but I might be wrong. Any help would be greatly appreciated.

答: For numbers of the form $-\frac{1}{n}$ the product is not defined, for the other negatives values the product diverges to $\infty$ or $-\infty$. For $x \in [0,1)$ the product diverges to $0$. For $x>1$ the product diverges to $\infty$. So the product only converges for $x=1$.


答:
The family $\mathcal{S}$ is called a sub-base of the topology $\mathcal{T}$.
In the following, we will prove that $\mathcal{T}$ is indeed a topology,
known as the topology generated by $\mathcal{S}$.

Let $\mathcal{B}$ be the collection of all finite intersections of
sets in $\mathcal{S}.$ Such collection $\mathcal{B}$ is known as
the base generated by $\mathcal{S}$.

Note that $\mathcal{B}$ is stable under taking finite intersection.
For, let $A,B\in\mathcal{B}$, then $A=\cap_{i=1}^{m}A_{i}$ and $B=\cap_{j=1}^{n}B_{j}$
for some $A_{i},B_{j}\in\mathcal{S}$. Note that $A\cap B=\cap\left(\{A_{i}\mid1\leq i\leq m\}]\cup\{B_{j}\mid1\leq j\leq n\}\right),$
which is a finite intersection of sets in $\mathcal{S}$, so $A\cap B\in\mathcal{B}$.
Also note that $\mathcal{S}\subseteq\mathcal{B}$ and by considering
intersection of the empty family, we have $X=\cap\emptyset\in\mathcal{B}$.

Lastly, we verify that $\mathcal{T}$ is a topology:

Clearly $\mathcal{B}\subseteq\mathcal{T}$. In particular, $X\in\mathcal{T}$.

By taking union of the empty family, we have that $\emptyset=\cup\emptyset\in\mathcal{T}$.

Let $\{A_{i}\mid i\in I\}\subseteq\mathcal{T}$ be an arbitrary
family of sets in $\mathcal{T}$. For each $i\in I$, there exists a family
$\mathcal{C}_{i}$ of sets in $\mathcal{B}$ such that $A_{i}=\cup\mathcal{C}_{i}$.
Let $\mathcal{C}=\cup_{i\in I}\mathcal{C}_{i}$ which is a family
of sets in $\mathcal{B}$. Now, $\cup_{i\in I}A_{i}=\cup\mathcal{C},$
which is a union of sets in $\mathcal{B}$, and hence $\cup_{i\in I}A_{i}\in\mathcal{T}$.
That is, $\mathcal{T}$ is stable under taking arbitrary union.

Let $A,B\in\mathcal{T}$. There exist families $\mathcal{C},\mathcal{D}$ of
sets in $\mathcal{B}$ such that $A=\cup\mathcal{C}$ and $B=\cup\mathcal{D}$.
Denote $\mathcal{C=}\{A_{i}\mid i\in I\}$ and $\mathcal{D}=\{B_{j}\mid j\in J\}$,
where $I,J$ are index sets. Then
\begin{eqnarray*}
A\cap B & = & \left(\cup_{i\in I}A_{i}\right)\cap B\\
 & = & \cup_{i\in I}\left(A_{i}\cap B\right)\\
 & = & \cup_{i\in I}\left(A_{i}\cap\left(\cup_{j\in J}B_{j}\right)\right)\\
 & = & \cup_{i\in I}\left(\cup_{j\in J}(A_{i}\cap B_{j})\right).
\end{eqnarray*}
Let $\mathcal{E}=\{A_{i}\cap B_{j}\mid i\in I\mbox{ and }j\in j\}$,
which is a family of sets in $\mathcal{B}$ (recall that $\mathcal{B}$
is stable under taking finite intersection, so $A_{i},B_{j}\in\mathcal{B}\Rightarrow A_{i}\cap B_{i}\in\mathcal{B}$).
The above shows that $A\cap B=\cup\mathcal{E}$, a union of sets in
$\mathcal{B}$, so $A\cap B\in\mathcal{T}.$

Title: Prove Prop 4.11

Body:
4.11 Proposition. Let $X$ be a set and let $\mathcal{S}$ be any collection of subsets of $X$ such that $X=\bigcup_{V\in\mathcal{S}}V$. Let $\mathcal{T}$ denote the collection of all subsets of $X$ that can be obtained using two operations:

taking finite intersections of sets in $\mathcal{S}$;

taking arbitrary unions of sets obtained in 1. Then $\mathcal{T}$ is a topology on $X$.

Proof. Exercise. $\qquad\square$

The following Proposition comes off www.mth527.site I will take my best shot at it.

My understanding

By taking union of empty sets $\cup\emptyset=\emptyset\in\tau$

Let {$C_j:j\in J$} be family members of $\tau$ Then each $C_j$ is a union of the members of S Hence the union of all
{$C_j:j\in J$} is also union of all the members in S and so in
$\tau$

We want to verify  $Y\cap Z\in\tau$ (1)

Let Y,Z$\in\tau$

But $Y=\bigcup_{V\cap S}V$ then for $V\in S$
and V$\in\tau $ and $Z=\bigcup_{W\in S}W$
then we have $W\in S$ for $W\in\tau$

$$Y\cap Z=\bigcup_{V\in S}V)\bigcap(\bigcup_{W\in S}W)= \bigcup_{V\in S}(\bigcup_{W \in S}(V\cap W)$$

So $Y\cap Z\in\tau$

We want to verify (2)

So if $T\subset X$ and $S\subset X$ then $S\subset T$ since arbitrary unions are taken
from it So if $V\in S$ so that $V\in\tau$, $\bigcup(Y\cap Z)=\bigcup_{V\in S}\bigcup_{W,V\in S}(V\cap W)$

theorem source


问:
Prove that the set of real-valued functions $\{|x - \lambda_1|, |x - \lambda_2|,\ldots, |x - \lambda_n|\}$ is linearly independent
I am looking for another solution to the problem below.

Prove that the set of real-valued functions $\{|x - \lambda_1|, |x - \lambda_2|,\ldots, |x - \lambda_n|\}$ is linearly independent over $\mathbb R$ for $\lambda_i \neq \lambda_j, i \neq j$ ($1 \le i, j \le n$).

I have been able to come up with a proof:

Reorder the set $\{\lambda_1, \lambda_2, \ldots, \lambda_n\}$ such that $\lambda_1 \lt \lambda_2 \lt \cdots \lt \lambda_n$.

Differentiate both sides of the equation: $\alpha_1|x - \lambda_1| + \alpha_2|x - \lambda_2| + \cdots \alpha_n|x - \lambda_n| = 0$ on intervals $(\lambda_1, \lambda_2)$, $\ldots$, $(\lambda_{n-1}, \lambda_n)$, $(\lambda_n, +\infty)$.

Obtain the system of $n$ equations, from which we can conclude $\alpha_1 = \ldots = \alpha_n = 0$:$$
\left\{
\begin{array}{c}
\alpha_1-\alpha_2+\cdots-\alpha_n=0 \\
\alpha_1+\alpha_2+\cdots-\alpha_n=0 \\
\ldots\\
\alpha_1+\alpha_2+\cdots+\alpha_n = 0
\end{array}
\right.
$$

Are there anything wrong with my proof and can you folks come up with another?

答:
Here's a different proof, also starting with reordering the $\lambda_j$ in increasing order. Suppose that $\alpha_1|x - \lambda_1| + \alpha_2|x - \lambda_2| + \cdots \alpha_n|x - \lambda_n| = 0$. If $\alpha_j\ne0$, then near $x=\lambda_j$ we have
$$
|x - \lambda_j| = \frac1{\alpha_j} \bigg( \sum_{i<j} \alpha_i(x-\lambda_i) + \sum_{i>j} \alpha_i(\lambda_i-x) \bigg);
$$
but this is impossible since the right-hand side is a linear function of $x$ while the left-hand side is not near $x=\lambda_j$.


问:
How to compute the unknown power x in $0.15^x + 0.36^x = 1$
More generally, I have two known variables $a$ and $b$ and one unknown variable $x,$ where:

$$a^x + b^x = 1, a,b \in [0, 1]$$

Is there a way to compute the value of $x$ analytically?

答:
To start out with we can graph the LHS in a graphing cqalculator:

Clearly, the answer is somewhere close to $x \approx 0.5$ for $y=1$. (Specifically, close to $x = 0.49064).

As for obtaining the exact solution of $a^x + b^x = 1$, the given equation is transcendental (basically, non-algebraic or non-polynomial), so an analytic solution can only be found for some special cases. Your best bet would be to use some numerical nethod such as Newton-Raphson or Bisection to estimate the solution.

You can somewhat simplify the equation using logarithms:

$a^x + b^x = a^x + (a ^\frac{\ln b}{\ln a})^x =  (a^x) + (a^x)^{\frac{\ln b}{\ln a}} = 1$

When $\frac{\ln b}{\ln a} = \phi$ (say) has nice values, solutions can be found.

Example,

$\phi = 1 \implies a^x = 0.5 \implies x = \log_a 0.5$

$\phi = 0.5 \implies a^x = \frac{3}{3} - \frac{\sqrt{5}}{2} \implies x = \log_a (1.5 - \sqrt(5)/2)$

In this case however, this is not possible.


答:
I'll only give a solution for $L^1(\Bbb R)$. But just by suitable normalizing, this works for other $p$ as well.

Let's first focus on $[0,1]$. Consider,

$f_{1,0}:=\chi_{[0,1]}$

$f_{2,0}:=\sqrt{2} \chi_{[0,\frac{1}{2}]}$, $f_{3,0}:=\sqrt{2} \chi_{[\frac{1}{2},1]}$

$f_{4,0}:= \sqrt{3} \chi_{[0,\frac{1}{3}]}$, $f_{5,0}:= \sqrt{3} \chi_{[\frac{1}{3},\frac{2}{3}]}$, $f_{6,0}:= \sqrt{3} \chi_{[\frac{2}{3},1]}$

$\vdots$

So for a general $n>1$ we have a unique $m \in \Bbb N$ such that $\frac{m(m-1)}{2} <n \le \frac{m(m+1)}{2}$ and $f_{n,0}=\sqrt{m}\chi_{E_{n_i}}$ where $E_{n_i}$ is some subinterval of $[0,1]$ with $|E_{n_i}| = \frac{1}{m}$ and thus $$\int_{\Bbb R}|f_{n,0}(x)|dx =\int_{E_{n_i}}\sqrt{m}dx=\frac{1}{\sqrt{m}} \to 0 \text{ as } m \to \infty \text{ ( and so does } n \to \infty )$$

Note that given any point $x \in [0,1]$ it lies in infinitely many of the above sets and also does not lie in the rest of infinitely many of them!

Now there is nothing special about $[0,1]$, as $$\Bbb R=\cup_{k \in \Bbb Z}[k,k+1]$$ we can do the same business for each of these integral intervals. Then you consider this whole family of functions i.e. taking all $f_{n,k}$ from all $[k,k+1]$ and then for this family say $\{g_j\}_{j \in \Bbb N}$ you have,

Exercises

Check that you get the desired properties on $\liminf g_j $ and $\limsup g_j$.

Normalize the coefficients for $1<p<\infty$

Title: Construct a sequence of functions in $L^p(\mathbb{R})$ such that $f_n$ converges to zero in $L^p$.

Body:
I'm trying to find a sequence of functions $f_n\in L^p(\mathbb{R})$, with $1\leq p < \infty$ such that $\lVert f_n \rVert_{L^p}\to 0$. Moreover for all $x\in \mathbb{R}$ the sequence satisfy two conditions
$$\liminf f_n(x)= 0 \ \text{and}\ \limsup f_n(x)= \infty.$$

I think in use some sequence based on characteristic function, but i can't find the sequence.


问:
condition for compactness of an operator
Let $\{e_n\}$ be an orthonormal basis for $l^2$ and $\{\alpha_n\} \in \ell^{\infty}$.

Define $Ae_n=\alpha_n e_n$ on $l^2$.

Find the condition for $A$ to be compact.

My work:

I want to show that in order for $A$ to be compact, we must have $\alpha_n \rightarrow 0$.

Suppose the contrary.

Then there exists some $\epsilon >0 $ such that we can find some sequence $(\alpha_{n_k})_k$ such that $(\alpha_{n_k})_k> \epsilon$.

I want to show that in this case, we can't find a convergent subsequence for $(Ae_{n_k})=(\alpha_{n_k}e_n)_k$.

And I got stuck here. Not sure how to show that there doesn't exist a convergent subsequence.

I feel like I'm missing some facts in real analysis.

Any help is appreciated! Thank you.

答:
If $\alpha_n$ does not tend to $0$ the there exist $\epsilon >0$ and $n_1 <n_2<...$ such that $|\alpha_{n_k}| >\epsilon $ for all $k$. The sequence $(\frac 1 {\alpha_{n_k}} e_{n_k})$ is bounded but $A(\frac 1 {\alpha_{n_k}} e_{n_k})= e_{n_k}$ has no convergent subsequence since $\|e_{n_k}-e_{n_j}\|^{2}=2$ whenever $j \neq k$.

Converse part: Suppose $\alpha_n \to 0$. Define $A_n(x) =\sum\limits_{k=1}^{n}\alpha_kx_ke_k$ where $x=\sum x_ke_k$. The $A_n$ is compact because it has finite rank. Now, $\|(A-A_n)x\|^{2}=\sum\limits_{k=n+1}^{\infty}|\alpha_kx_k|^{2} \leq \sup_{k>n} |\alpha_k|^{2} \|x||^{2}$ proving that $\|A-A_n\| \to 0$. This implies that $A$ is compact.


问:
Polygons within a polygon
$r$-sided polygons are formed by joining the vertices of a $n$-sided polygon.Find the number of polygons that can be formed,none of whose sides coincide with those of the $n$-sided polygon?

Polygon is convex.

In think we need to count here the number of polygons with one side common with $n$-sided polygon, up to $r-1$ sides common with original polygon and then subtract with total number of polygons, but how to do this counting?

答:
Your problem is equivalent to finding the number of unordered subsets of $\{1, 2, ..., n\}$ such that in each subset $\{v_1, v_2, ..., v_r\}$, $v_i-v_{i-1} \not \in \{1, n-1\}$ for any $i\ge1$. Let $a_{r,n}$ be this number.

Now, let the linear case be finding the number of unordered subsets of $\{1, 2, ..., n\}$ such that in each subset $\{v_1, v_2, ..., v_r\}$, $v_i-v_{i-1} \not = 1$ for any $i\ge1$. Let $b_{r, n}$ be this number. $b_{r,n}$ is given by the following recursive sum: $$b_{r,n}=\underbrace{b_{r-1,0}}_{\text{choosing vertex }1}+\sum_{i=0}^{n-2}\underbrace{b_{r-1, i}}_{\text{vertex }i+2}$$

where $b_{r,n}=0$ if $r > n$, $b_{0, n}=1$, and $b_{1, n} = n$. Using $b_{r,n}$, $a_{r,n}$ can be found as

$$a_{r,n} = \underbrace{b_{r-1,0}}_{\text{vertex }1}+\sum_{i=0}^{n-3} \underbrace{b_{r-1, i}}_{\text{vertex }i+2}+\underbrace{b_{r-1,n-3}}_{\text{choosing vertex }n}$$

Basically, every step, you are choosing a vertex and "eliminating" the ones after that and neighboring ones. Just like $b_{r,n}$, $a_{r,n}=0$ if $r > n$, $1$ if $r = 0$, and $n$ if $r = 1$.

$b_{r, n}$ has a very simple closed-form of $$\binom{n-r+1}{r}$$ which can be proven inductively using Pascal's identity. This then means that $$a_{r,n} = \binom{n-r-1}{r-1}+\sum_{i=r-1}^{n-r-1}\binom{i}{r-1}$$ for $r < n$. This simplifies to $$a_{r,n}= \frac{n}{n-r}\binom{n-r}{r}$$ for $r < n$. This can be proven by induction on $n$ after using the fact that $a_{r,n+1}-a_{r,n}$ must equal $$2\binom{n-r}{r-1}- \binom{n-r-1}{r-1}$$ and $a_{r,2r}=2$.

Title: Polygons within a polygon


问:
lim sup of weighted average in the tail $\sigma$-algebra
Consider the sequence of random variables $X_1,X_2,....$, and define $S_n = X_1+X_2+...+X_n$. Suppose $\mathcal{T}$ is the tail $\sigma$-algebra of $X_n$s. Prove that
$$
\{\lim\sup_{n\to\infty} S_n/C_n > a \} \in \mathcal{T},
$$
where $C_n \to \infty$.

I appreciate any comment/response.

My non-rigorous attempt: To show that this event is in $\mathcal{T}$,
I want to argue that changing a finite number of random variables will
not affect the event. This makes sense for example when $C_n = n$, since
$\lim\sup_{n\to\infty} S_n/n$ is the limiting average which does
not depend on a finite number of random variables. I think $S_n/C_n$
can be thought of a weighted average, so we can argue similarly. However, I am trying to prove this rigorously.

答: $(\lim \sup \frac {S_n} {C_n}>a)= (\lim \sup \frac {X_k+X_{k+1}+...+X_n} {C_n}>a)$ for any $k$ since $\frac {X_1+X_2+..+X_{k-1}} {C_n} \to 0$. From this it is clear that $(\lim \sup \frac {S_n} {C_n}>a)$ belongs to $\sigma (X_k,X_{k+1},...)$ for each $k$ and hence it belongs to $\mathcal T$.


答: If $\alpha_1|x - \lambda_1| + \alpha_2|x - \lambda_2| + \cdots \alpha_n|x - \lambda_n| = 0$ then $\alpha_i$ must be $0$ for each $i$ simply because RHS is differentiable at $\lambda_i$ but LHS is not unless $\alpha_i=0$ : Note that all terms except the $i-$th one are differentiable at $\lambda_i$.

Body:
I am looking for another solution to the problem below.

Prove that the set of real-valued functions $\{|x - \lambda_1|, |x - \lambda_2|,\ldots, |x - \lambda_n|\}$ is linearly independent over $\mathbb R$ for $\lambda_i \neq \lambda_j, i \neq j$ ($1 \le i, j \le n$).

I have been able to come up with a proof:

Reorder the set $\{\lambda_1, \lambda_2, \ldots, \lambda_n\}$ such that $\lambda_1 \lt \lambda_2 \lt \cdots \lt \lambda_n$.

Differentiate both sides of the equation: $\alpha_1|x - \lambda_1| + \alpha_2|x - \lambda_2| + \cdots \alpha_n|x - \lambda_n| = 0$ on intervals $(\lambda_1, \lambda_2)$, $\ldots$, $(\lambda_{n-1}, \lambda_n)$, $(\lambda_n, +\infty)$.

Obtain the system of $n$ equations, from which we can conclude $\alpha_1 = \ldots = \alpha_n = 0$:$$
\left\{
\begin{array}{c}
\alpha_1-\alpha_2+\cdots-\alpha_n=0 \\
\alpha_1+\alpha_2+\cdots-\alpha_n=0 \\
\ldots\\
\alpha_1+\alpha_2+\cdots+\alpha_n = 0
\end{array}
\right.
$$

Are there anything wrong with my proof and can you folks come up with another?


问:
Uniform convergence of derivative of sequence of functions
Consider $f_n(x)=n^{-1}\exp(-n^2x^2)$, sequence of functions from $R$ to $R$. I wanted to check the convergence of derivative of this function.
Derivative is given by $f'_n=-2nx\exp(-n^2x^2)$. As n tends to infinity, exponential function dominates and the function converges to $0,$ so $f'_n$ converges pointwise. Now inorder to check about uniform convergence, supremum norm tends to 0 as n tends to infinity. But, I'm not quite sure about the uniform convergence near origin. Is my approach correct? Is $f'_n$ uniformly convergent? Or something has to be taken care near origin.
Thanks in advance.

答:
$f_n'(\frac  1n)=-2e^{-1}$ so $f_n'$ cannot converge uniformly to $0$.

[Note that $\sup_x |f_n'(x)| \geq f_n'(\frac  1n)$ and that if $(f_n')$ converges uniformly the limit can only be $0$ because of pointwise convergence to $0$]


问:
Volume related to rotating over y=-1
Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the following curves about the line y=−1.

$y=x^2$ and $x=y^2$

Visual of ^^

What I got so far was $\int_{0}^1 2\pi (y+1) (y^2-y^\frac{1}{2}) dy$, when I solve it though I get $-\frac{29\pi}{30}$. This is the wrong answer can someone tell me where I did something wrong? Because I can not tell.

答:
What doobdood said was right all I had to do was changed $y^2-y^\frac{1}{2}$ to $y^\frac{1}{2} - y^2$, where the answer to the question would be $\frac{29\pi}{30}$ if the correct answer.
Thank you Doobdood.

Title: Volume related to rotating over y=-1


问:
What kind of a locus $Y$ describe when moving $\Omega $?
Two circles $\Gamma_1$ and $\Gamma_2$ have common external tangents $\ell_1$ and $\ell_2$ meeting at $T$. Suppose $\ell_1$ touches $\Gamma_1$ at $A$ and $\ell_2$ touches $\Gamma_2$ at $B$. Variable circle $\Omega$ through $A$ and $B$ intersects $\Gamma_1$ again at $C$ and $\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.

Suppose lines $AD$ and $BC$ meet at point $Y$.

What kind of a locus $Y$ describe when moving $\Omega $? Is that  some projective curve?

How can we recognize it with pure Euclidian geometry without any (or as few as possible) anaylitic calculation? For example, if we have a projective transformation $\pi$ between pencils of lines, then we know that $p\cap \pi(p)$ describes some conic or line.

答:
It is well known that a conic can be generated by the pairwise intersections of two pencils of lines. This answer attempts to explain how the locus of $Y$ can be generated by the pairwise intersections of a pencil of lines and a pencil of conics.  Much of what is presented here is observational and practical so that the reader can attempt to reproduce the construction in, say, Geogebra.

for some preliminary background and references, refer to this answer to the question Geometric interpretation of cubic curve?

let $L=0$ and $M=0$ be two equations (in $x$ and $y$) of lines $\ell,m$.  Then $L+\lambda M=0$ is the equation of a pencil of lines through $P=\ell\cap m$, parameterized by $\lambda$.  Note that since the equation for a line is not unique, the parameterization of a pencil is also not unique. (e.g. $2L=0$ is also an equation for $\ell$ and we will say $2L\cong L$, meaning they represent the same curve)

similarly, if $C=0$ and $D=0$ are equations for conics $c$ and $d$, $C+\lambda D=0$ is an equation of a pencil of conics.  If $c$ and $d$ intersect in four points, then this pencil is the set of all conics that pass through all four points (called base points in this context).  Again, the parameterization is not unique because  $sC=0$ is an equation for $c$ for any $s$ ($C\cong sC$).

if we have two pencils as above
$$
\begin{aligned}
L+\lambda M&=0 \\
C+\lambda D&=0
\end{aligned}
$$
we can eliminate $\lambda$ to get $\mathcal{K}=LD-CM=0$, which is a cubic that is the locus of the points $(L+\lambda M)\cap(C+\lambda D).$  $\mathcal{K}$ will run through $P$ and the four base points.  A good beginner's introduction to this is Haskell, Special cubics and quartics produced by projective pencils of straight lines and conics (a Master's thesis dated 1916).

suppose we have a sufficiently well behaved cubic $\mathcal{K}$.  How do we reverse engineer to get the line and conic pencils that generate it?  Briefly, we fix four points $P_1,\dots,P_4$ on $\mathcal K$ and find their coresidual point $P$ as explained in the math.SE answer and illustrated at Wilkin, The coresidual of four points on a cubic.  We then select three lines$L,M,N$ through $P$ and the corresponding conics $C,D,E$ running through the points $P_1,\dots,P_4$ such that $L\cap C \in \mathcal{K}$ (and similarly for the pairs $M,D$ and $N,E$. $E$ will be in the pencil generated by $C$ and $D$ in the sense that $E\cong C+\lambda_E D$ for some $\lambda_E D$.  It is not necessarily true that $N\cong L+\lambda_N M$.  But there will some $s$ so that $N\cong L+\lambda_N sM$, so we replace the line pencil with $L+\lambda s M$ so that lines and conics are paired correctly to produce $\mathcal K$.

with these preliminaries we can turn to the scenario in the OP.  We borrow several new points from Blue's answer. $M$ is the midpoint of $AB$. $A_\star$ and $B_\star$ are the centers of the $A$ and $B$ circles.  $H$ is the harmonic conjugate of $T$ with respect to $A_\star,B_\star$. $\mathcal K$ is the locus of $Y$, which we know is a cubic that goes through $T,H,A,B,M$.

we choose $T,H,A,B$ as base points for the conic pencil.  The base point for the line pencil is $T$, the coresidual point of the base points.  The reader is invited to experimentally confirm this in Geogebra.

We now need to find three line-conic pairs that define the pencils we need.  By inspection (eyeballing) in Geogebra these are (they may be more, but I couldn't find any)

$(L,C)$ - the line through $T$ perpendicular to $TH$ plus the conic whose tangent at $T$ is the aforementioned line. (the fact that the line and tangent are the same is the reason for the cusp at $T$).

$(M,D)$ - the line $TM$ and the degenerate conic consisting of the lines $AB$ and $TH$.

$(N,E)$ - the line $TH$ and the conic whose tangent at $H$ goes through $M$

With these pairs we can define the pencils which I claim generate $\mathcal K$. I have not tried to prove this, but the most direct confirmation would probably be to show that $LD-CsM=0$ is equivalent to the Cartesian equation derived by Blue.

I also have no conceptual explanation why this construction is equivalent to the original construction using $Y$ in the $OP$.  But cubics can have multiple characterizations, e.g. the Apollonian Strophoid has at least six.

Finally a crude working diagram of the construction.  The conics $C$ and $D$ are shown in thick blue dotted lines.  A line and conic are in red intersecting at G. The locus of intersection is a thick black dotted curve overlaying Blue's equation (in bright green).  Better diagram to come. (maybe)


问:
Convergence in $L^{\infty}$ implies convergence in $W^{1,p}((0,1))$
This is an exercise from Brezis's functional analysis where we're trying to show that the set
$$
B = \{f \in W^{1,p}((0,1)): ||f||_{W^{1,p}((0,1))} \le 1\}
$$
is compact in $L^{p}((0,1))$, for $p > 1$.

To do this, we use the fact that any sequence $\{f_{n}\}_{n \in \mathbb{N}}$ in $B$ is bounded, so it has a subsequence that converges to something ($f$) in $W^{1,p}((0,1))$ in $||\cdot||_{L^{\infty}}$. I want to use this fact to show that $f_{n}$ (not re-labelling the subsequence) converges to $f$ in $W^{1,p}$ norm.

It is easy to show that $f_{n} \to f$ in $L^{p}$, but I also need to show that $f_{n}' \to f_{n}'$ in $L^{p}$ (where the derivatives are taken in the weak sense), which I am unable to do. It is fine even if $f_{n}' \to f'$ in $L^{p}$ up to a subsequence, since we can extract a further subsequence without any worry here. How would one do this?

Edit: Can something like the following work? We know that $W^{1,p}$ is isomorphic to a subspace of $L^{\infty}$, call it S, so we have a bijective continuous map $T: S \to W^{1,p}((0,1))$, which should give us the assertion without having to go through the process of showing $f_n, f_n'$ converging separately in $L^{p}$?

答:
Let $ 1<p < \infty $. Recall that the identity operator
$$ id \colon(W^{1,p}(0,1),||\cdot ||_{1,p} ) \to (L^p(0,1),||\cdot ||_p) $$
is compact and thus completely continuous (that is, it maps weakly convergent sequences to norm convergent ones). Let
$$  B_1(p) =\{ u \in W^{1,p} (0,1) \colon ||{u}||_{p} + ||u'||_p \leq 1\}.$$
Let $ (u_n)_n \subset B_1(p) $. By the reflexivity of $ W^{1,p} (0,1)$, there exists $ u \in B_1(p) $ such that $ u_n \xrightarrow{weakly} u   $ in $ W^{1,p} (0,1)$ (atleast for a subsequence). Thus, (by the complete continuity of the identity operator) $ u_n \to u  $ in $ L^p(0,1)$. In other words, $ B_1(p) $ is compact in $ L^p (0,1)$.

Now, let $p= \infty$. Let $(u_n) \subset B_1 (\infty)$. By the compactness of the identity operator, there exists $u \in L^\infty(0,1)$ such that $||u_n-u||_\infty \to 0 $ (for a subsequence). We wish to show that $u \in B_1(\infty)$. Since $(u_n') \subset L^\infty(0,1) $ is bounded, invoking Alaoglou's theorem, we infer that $u_n' \xrightarrow{weak^*} g $ for some $g \in L^\infty(0,1)$. It is easy to see that $u \in W^{1,\infty}(0,1)$ and $u'=g$. Recall that the norm is weak-star lower semi continuous and so
$$ ||u'||_\infty  \leq \liminf ||u_n'||_\infty .$$
In conclusion,
$$||u||_\infty + ||u'||_\infty \leq \liminf (||u_n'||_\infty + ||u_n||_\infty) \leq 1 $$
and thus $u \in B_1(\infty).$

For $p=1$ things are not as good. You can show that  $B_1(1) \subset L^1(0,1) $ is not closed (and thus not compact) in $L^1(0,1)$. To see this, consider the following sequence in $ W^{1,1}(0,1)$:
$$ u_n(t) = \left\{\begin{array}{lr}
 0 , ~~ 0 \leq t \leq \frac 12\\
 n(t-1/2), ~~ \frac 12 <t \leq \frac 12 + \frac 1n \\
 1, ~~ \frac 12 + \frac 1n < t \leq 1.
     \end{array}\right. $$

with
$$ u_n'(t) = \left\{\begin{array}{lr}
 0 , ~~ 0 \leq t \leq \frac 12\\
 n,~~ \frac 12 <t \leq \frac 12 + \frac 1n \\
 0, ~~ \frac 12 + \frac 1n < t \leq 1
\end{array}\right. $$
We have that  $ ||{u_n}||_1=\frac 12 - \frac{1}{2n}  $ and $ ||{u_n'}||_1 =1$ and thus $ ||{u_n}||_{1,1} \leq 3/2 $. Let $ u = \mathcal X_{(\frac 12,1)} $, then
$$ ||u_n-u ||_1= \int |{u_n-u}| = \frac{1}{2n} \to 0 $$
and thus $ \frac 23 u_n \to \frac 23 u $ in $ L^1(0,1)  $ but $ \frac23 u \notin W^{1,1}(0,1) $ and in particular $ \frac23 u \notin B_1(1) $.


问:
How to find all solutions to $f\star f=f$ in $L^2(\mathbb{R})$?
I am trying to find all solutions to

$$f\star f=f$$ with $f\in L^2(\mathbb{R})$.

I can solve this equation for $f\in L^1$ (and hence for $f\in L^1\cap L^2$). If I want a solution in$f\in L^2\setminus L^1$,
I know that we cannot directly use the Fourier transform with $f\in L^2(\mathbb{R})$, since $f\star f$ only belongs to $L^2\cap L^\infty$.

I have tried to use an approximating sequence $f_n\in L^1\cap L^2$ with no success yet.

答:
So it is given that $f*f=f$ with $f \in L^2(\Bbb R)$ . So in particular $f*f \in L^2(\Bbb R)$

Thus applying the Fourier transform on both sides, we have $(\hat{f}(\xi))^2=\hat{f}(\xi)$ for a.e. $\xi$ i.e. $$\hat{f}(\xi)(\hat{f}(\xi)-1)=0 \text{ for a.e. } \xi$$ Thus either $\hat{f}(\xi) \equiv 0 \text{ for a.e. } \xi \implies f \equiv 0 $ a.e.

Or, $\hat{f}(\xi) \equiv 1 \text{ for a.e. } \xi$ but then $\hat f \notin L^2(\Bbb R)$

So the only possibility left with $\hat{f}(\xi)(\hat{f}(\xi)-1)=0 \text{ for a.e. } \xi$ and $\hat f \in L^2(\Bbb R)$ is $\hat f = \chi_E$ where $E$ is some measurable set with finite, positive measure.But then $\hat f \in L^1(\Bbb R) \cap L^2(\Bbb R)$ thus you can apply inverse Fourier transform on that to obtain $$f(x)=\int_{\Bbb R} \chi_E(\xi)e^{i\xi x} d\xi=\int_{E} e^{i\xi x} d\xi \text{ a.e. } x \in \Bbb R$$


问:
Analytic Properties of $f(x) = x^x$
The function $f(x) = x^x, x > 0$ can be plotted on graphing software and inspected to see a local minimum around .367. The function is convex, decreasing from 0 to its minimum, and increasing thereafter. The derivative of the function can be found by implicit differentiation to be $f'(x) = x^x(\ln(x)+1)$.

Can the exact value of the local minimum be found?

Can someone explain intuitively why the function first decreases and then increases?

Is there anything interesting about the class of functions $\{ \, f(x) \, | \, f'(x) = f(x) \cdot g(x) \, \}$

Edit: I forgot, I'll be subject to a firing squad if I don't explain what I have tried!

Setting the derivative equal to 0 doesn't do much for me. If this can be solved exactly without optimization algorithms, I suspect something really clever will need to happen.

I tried thinking about what is happing for different types of inputs (irrational $x$, rational $x$, natural $x$). Still didn't get anywhere.

Nothing to try really, does this class of functions come up anywhere in math?

答:
It's good habit to rewrite $a^b$ as $\exp(b\ln(a))$. In our case this is $f(x)=x^x = \exp(x\ln(x))$, which is easier to study. The expression makes sense for $x>0$, doesn't make sense for $x<0$, and leaves some room for debate at $x=0$ although it's sensible to take $0^0=1$. The function is then continuous on $\{x\colon x \ge 0\}$.

The derivative, like you said, is $f'(x)=x^x(1+\ln(x))$. For this to be $0$, either $x^x$ should be $0$ (never happens, as $\exp$ never vanishes) or $1+\ln(x)=0$ which means $\ln(x)=-1$ or $x = \frac{1}{e}$. If there is a minimum – and indeed there is – it must be there. You can verify this in any number of ways, such as taking a second derivative or just checking nearby values.

The function decreases when $0 \le x \le \frac{1}{e}$ and increases afterwards.

I can't imagine why you wanted to understand the behavior for rational and irrational values of the argument.

Exponential functions arise all over the place in mathematics. This particular function isn't super common, but it pops up here and there.

Title: Analytic Properties of $f(x) = x^x$


问:
A tricky logarithms problem?
$ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.

Any hints? Thanks!

Title: A tricky logarithms problem?

Body:
Here's a kind of ugly way of doing it:
$$\begin{align*}
(12n)^{\log_{4n}(40\sqrt{3})}&=(12n)^{\log_{3n}(45)}\\\\
(3)^{\log_{4n}(40\sqrt{3})}(4n)^{\log_{4n}(40\sqrt{3})}&=(4)^{\log_{3n}(45)}(3n)^{\log_{3n}(45)}\\\\
(3)^{\log_{4n}(40\sqrt{3})}\cdot 40\sqrt{3}&=(4)^{\log_{3n}(45)}\cdot 45\\\\
(3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(4)^{\log_{3n}(45)-\frac{3}{2}}\\\\
(3)^{\log_{4n}(40\sqrt{3})+\frac{1}{2}-2}&=(3)^{\log_3(4)\cdot(\log_{3n}(45)-\frac{3}{2})}\\\\
\log_{4n}(40\sqrt{3})-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\
\log_{3n}(45)-\frac{3}{2}&=\log_3(4)\cdot\left(\log_{3n}(45)-\frac{3}{2}\right)\\\\
\end{align*}$$
Therefore
$$\begin{align*}
\log_{3n}(45)&=\frac{3}{2}\\\\
2025&=(3n)^3\\\\
n^3&=\frac{2025}{27}=75
\end{align*}$$


问:
Evaluate $\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$
Evaluate the following expression:$$\frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ$$
I tried combining the whole into a single fraction and using double-angle identity, product/sum to sum/product, but it didn't work.

Body:
Per the identity
$\sin3t =3\sin t -4\sin^3t$,
it is straightforward to verify that $\sin20$, $\sin40$, $-\sin80$ are the roots of
$4x^3-3x+\frac{\sqrt3}2=0$, which leads to

$$\sin20\sin40\sin80=\frac{\sqrt3}8$$
Then

\begin{align}
& \frac{4}{\sin^2 20^\circ} - \frac{4}{\sin^2 40^\circ} + 64\sin^2 20^\circ\\
=& \frac{64}3\left(4\sin^240\sin^280 -4\sin^220\sin^280+3\sin^220 \right)\\
 =& \frac{64}3\left((1-\cos80)(1+\cos20)-(1-\cos40)(1+\cos20) + \frac32(1-\cos40)\right)\\
 =& \frac{64}3\left(\frac32 - \cos80 -\frac12\cos40
-\cos20\cos80 +\cos20\cos40\right)\\
 =& 32 + \frac{64}3\left(- \cos80 -\frac12\cos40
-\frac12(\cos60-\cos80) +\frac12(\cos20+\cos60)\right)\\
 =& 32 + \frac{32}3\left(\cos 20- \cos40 -\cos80
\right) \\
= &32 + \frac{32}3\left(\cos 20- 2\cos60\cos20
\right)\\
=& 32
\end{align}


答:
For your other question, we can rewrite the equation as an ODE

$$\frac{dy}{dx} = g(x)\cdot y$$

which we can separate variables and integrate to solve. The final answer is always of the form

$$y = C\exp\left(G(x)\right)$$

where $C$ is an arbitrary constant and $G$ is an antiderivative of $g$.

Title: Analytic Properties of $f(x) = x^x$

Body:
The function $f(x) = x^x, x > 0$ can be plotted on graphing software and inspected to see a local minimum around .367. The function is convex, decreasing from 0 to its minimum, and increasing thereafter. The derivative of the function can be found by implicit differentiation to be $f'(x) = x^x(\ln(x)+1)$.

Can the exact value of the local minimum be found?

Can someone explain intuitively why the function first decreases and then increases?

Is there anything interesting about the class of functions $\{ \, f(x) \, | \, f'(x) = f(x) \cdot g(x) \, \}$

Edit: I forgot, I'll be subject to a firing squad if I don't explain what I have tried!

Setting the derivative equal to 0 doesn't do much for me. If this can be solved exactly without optimization algorithms, I suspect something really clever will need to happen.

I tried thinking about what is happing for different types of inputs (irrational $x$, rational $x$, natural $x$). Still didn't get anywhere.

Nothing to try really, does this class of functions come up anywhere in math?


答:
for question 3:
$f'(x) = f(x) g(x) \implies \dfrac{f'(x)}{f(x)} = g(x) \implies ln|f(x)| = \int g(x) = G(x) \implies f(x) = e^{G(x)}$

there are 2 cases... if $G(x) = ln(h(x))$, then $f(x) = h(x)$ otherwise it is an exponent. Either ways $f(x)>0$ is the only property I can see... Can we say that means f(x) is exponential?

Title: Analytic Properties of $f(x) = x^x$


问:
Conformal equivalance and modular invariance
I'm currently in an introductory complex analysis class. I'm working through the following problem:

Let $\Bbb{H}$ be the upper half-plane and $B$ be left half of the fundamental region:

$$ B = \{\tau \in \Bbb{H}: |\tau|>1, -\frac{1}{2}<\textit{Re}(\tau)<0 \} $$

Show that the modular invariant $j$ is a conformal equivalence between $B$ and $\Bbb{H}$.

I have only learned a bit about conformal equivalence and modular invariance; the instructor hasn't been too helpful. Can anyone guide me on how to solve this exercise?

答:
$j$ is biholomorphic from $U=|\tau|>1,|\Re(\tau)|\in (0,1/2)$ to $\Bbb{C}-L$ where $L$ is the curve which is the image of the boundary of $U$.

$L$ is the real axis. This is because $J$ has real coefficients thus it is real on $\Re(\tau)=0$ and $\Re(\tau)=\pm 1/2$. On $|\tau|=1$ it is real too because we have the following correspondence by some modular transformations

Next, let $B= \{|\tau|>1,\Re(\tau)\in (-1/2,0)\}$.

$\Im(j(-1/4+i))>0$ gives that $JjB)$ is an open subset of $\Im(z)>0$.

And $j(a+ib)=\overline{j(-a+ib)}$ gives that $\Bbb{C-R} = j(B)\cup\overline{j(B)}$.

Whence $j(B)=\Bbb{H}$.


问:
Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity $a(x+b)^3+c\equiv4x^3-24x^2+48x+d$

Determine the value of each of the constants $a$, $b$, $c$, and $d$ in the identity

$$a(x+b)^3+c\equiv4x^3-24x^2+48x+d$$

I have already found $a=4$ and $b=-2$ but I'm struggling to find $c$ and $d$.

The answers for $c$ and $d$ are $c=29$, $d=-3$.

答:
Expanding we get $ax^3 + 3abx^2 + 3ab^2x +ab^3 + c = 4x^2-24x^2 +48x + d$

Technically what you have going on is 4  equations.

$a = 4$

$3ab=-24$

$3ab^2 = 48$.

$ab^3 + c = d$.

They aren't linear as the involve $b$ to powers $2$ and $3$ but $1,2$ are independent linear so $a,b$ can be solved $a= 4, b=-2$. And a equation three is completely redundant (it simplifies to the inane and unhelpful $48 = 48$... gee, no ****)

That leaves a final equation $-32 + c =  d$.  This is the only equation with $c$ or $d$ so we can express $c$ in terms of $d$  ($c =d +32$) or $d$ in terms of $c$ ($d=c-32$) but as it is one equation with two unknowns that's as for as we can go. We can let $d$ be anything we want and $c$ infinite potential values will follow (or vice versa.  $c=29$ and $d = c -32 =-3$ is certainly an answer but it is by no means the only solution.

After all if we had replaced $c,d$ with $M,N$ or $K + 856, J+ 856$ or $c -5, d-5$ nothing would have changed.


问:
Information on "stronger form" of Dirichlet's Theorem on Arithmetic Progressions
From Wikipedia:

"Stronger forms of Dirichlet's theorem state that, for any arithmetic progression, the sum of the reciprocals of the prime numbers in the progression diverges."

Can anyone direct me to a proof  of this "stronger" fact, or a paper that discusses it? Thanks.

答: You can find a proof in Serre's A Course in Arithmetic, chapter 6.


答:
The minimum is attained at $x=\frac1e$ as can be seen by setting the derivative equal to $0$ and solving for $x$.

Your statement in a comment that it is not always true that if $x>y>0$ then $x^x>y^x$ is false.  We have $$x>y\implies \log x > \log x\implies x\log x > x\log y \implies x^x>y^x  $$

Perhaps you meant to say it is not always true that $x>y>0$ implies $x^x>y^y$.  This is true.  Taking logarithms, there's no particular reason to believe that $x\log x > y\log y$ and indeed, it isn't always true.

To try to explain it intuitively, note first that we're concerned with numbers $0<x<y<1$.  When you raise such a number to a power, the larger the power, the smaller the result.  When we compare $x^x$ and $y^y$, for some choices of $x$ and $y$, this effect dominates, and $y^y$ is the smaller of the two numbers.  For other choices of $x$ and $y$, the fact that $x$ is smaller dominates.

Title: Analytic Properties of $f(x) = x^x$


问:
Probability inequality with intersection true /
I was asked to verify if is this is true or false.

$$P(AB)\leq P(A)P(B)$$

I tried to find a counterexample to proof is false, but every thing I try says that the intersection is less than or equal to the product.

答:
Any example where $A \subseteq B$ is a good counterexample.

In fact

$$P(AB)=P(A)\geq P(A)P(B)$$

as $P(B)\leq 1$


问:
7 balls are distributed randomly in 7 cells. If 2 cells are empty, show that the conditional probability of a triple occupancy equals 1/4
This problem comes from Feller's Introduction to probability. and it goes like:

"Seven balls are distributed randomly in seven cells. Given that two cells are empty, show that the (conditional) probability of a triple occupancy of some cells equals 1/4"

So far i have found out that all of the distributions that form a triple are given by: ${7 \choose 3}(4!) = \frac{7!}{3!}$.

Yet, i am not sure of how to go on, because there are $7^7$ total distributions, but clearly if you divide these two quantities the result is not even close to $1/4$.

I have also tried to go on by considering that the only occupancy numbers in the distribution can be $0,1,3$ so the total number of distributions that coincide with the ocuppancy numbers $3,1,1,1,1,0,0$ in some order are given by: $\frac{7!}{3!}\frac{7!}{3!4!}$. But if you divide this by $7^7$ the result is still no close to $1/4$.

And also when you consider the fact of conditional probability the results only get worse. So what can i do?

答:
To me, it looks like throwing a $7$ sided die $7$ times, with the  the desired conditional probability being

n(1 cell has triple occupancy with 2 cells unoccupied) $\div$ n(2 cells unoccupied)

Using the multinomial coefficient, we get

= $\dfrac{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2}}{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2} + \dbinom{7}{2,2,1,1,1,0,0}\dbinom{7}{2,3,2}} = 1/4$


问:
Why isn't the throughput = (1 / response time) by their definition?
I found that all say that throughput is inversely proportional to response time. But I found it's confusing that if I see their relationship by their definitions, isn't each of them a reciprocal of each other?

From my point of view, if I define them by my intuition, it then goes:

Throughput: Number of task completions per time unit = $\frac{Completion}{T}$

Response Time: time cost from task's arrival to its completion = $\frac{T}{completion}$

where $T$ is the total observation time.

What did I miss? Why can't I say that $Throughput = \frac{1}{Response Time}$?

答:
I think I've found my bug,

The mean response time should be $\frac{\text{Total Response time of all jobs}}{completion}$, and every job has a different $Response\ Time$ due to the queueing delay.

This is the main observation that it can't be the reciprocal of throughput, and especially in the parallelism case.


问:
How to find number of solutions to $a+b+c=0$
As titled, I need to find number of solutions to the equation $a+b+c=0$, where $a,b,c$ are integers in the range $[-k,k]$.($k$ is a positive integer) and $a$ doesn't equal to $0$.

My attempt: let $$x=a+k, y=b+k, z=c+k$$ we need to find non-negative solutions to the equation $$x+y+z=3k$$ with $$0\leq x,y,z\leq 2k.$$
Using the formula, there are $\left(\frac{3k+3-1}{3-1}\right)$=$\left(\frac{3k+2}{2}\right)$ solutions.

We then erase the cases where one of them is $>2k$ (only one of them can be $>2k$).

Let $$x'= x-(2k+1)$$  then x' is non-negative, number of solutions to the equation: $$x'+y+z=k-1$$ is:

$$\left(\frac{k+3-1}{3-1}\right)$$

But this gives a total of $3k^2-2$ solutions if we calculate $\left(\frac{3k+2}{2}\right)$-$3 \cdot \left(\frac{k+2}{2}\right)$,which is not the desired answer. And I haven't delete the case where $a=0$

What did I do wrong?
Any help will be appreciated.

答:
Your first binomial coefficient, $\binom{3k+2}2$, is correct, but your second, counting the solutions in non-negative integers to $x_1'+x_2+x_3=k-1$, is not: it should be

$$\binom{(k-1)+3-1}{3-1}=\binom{k+1}2\,,$$

giving you a total of

$$\begin{align*}
\binom{3k+2}2-3\binom{k+1}2&=\frac12\big((3k+2)(3k+1)-3k(k+1)\big)\\
&=3k^2+3k+1
\end{align*}$$

solutions in non-negative integers to $x+y+z=3k$. Now you just have to subtract the solutions for which $x=k$, i.e., the solutions in non-negative integers to $y+z=2k$. I’ll leave that to you: these are straightforward to count.


问:
To prove:- $|AB-I|+|BA-I| =0$ given that $A$ is any $3\times 2$ matrix and $B$ is any $2\times 3$ matrix.
I was just told that for any given matrix if it is $n\times (n-1)$ and another of $(n-1)\times n$ then also the above kind of situation that is $$\det(AB-I)+\det(BA-I)=0,$$ where $I$ is the corresponding identity matrix of the square matrix so formed.

Also in order to prove it I tried to solve it by any theorem that I know or so but was unable.

So, I checked it for any random matrix with random variable and calculated the same thing via a matrix calculator, and shockingly it was correct.

I searched it on internet and was unable to find any explanation.
So, I eventually turned to stack exchange.

Any help would be appreciated!

答:
\begin{align}
\det(AB-I_3) &= (-1)^3\det(I_3 - AB)\\
&= - \det(I_2 - BA)\\
&= - (-1)^2 \det(BA-I_2)
\end{align}

The second equality holds due to Sylvester determinant identity.

In general, if $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times m}$, then we have

$$\det(AB-I_m) = (-1)^{n+m} \det(BA-I_n)= (-1)^{n+m \pmod{2}} \det(BA-I_n)$$


问:
Homotopy equivalence among the following topological spaces
How do I show the homotopy equivalence among the following topological spaces:

答: You may collapse contractible <s>subspaces</s> subcomplexes while preserving homotopy type. For example, consider the middle circle of (Y). You may collapse an arc joining both tangency points to a single point within this arc, so now (Y) looks like (Z).</p>


问:
Show dyadic transpose property $(a \otimes b)^T = b \otimes a$
I want to prove that

$$(a \otimes b)^T = b \otimes a$$ where $\otimes$ is the dyadic product

I think I should just apply the definition. So let $a,b,u,v$ vectors:

$$u \cdot (a \otimes b)^T v=(a \otimes b) u \cdot v = (b \cdot u) a \cdot v$$

Now I don't know how to arrange the terms in order to get the thesis. Any help would be really appreciated

答:
If you assume basic matrix algebra identifies known, then it simply follows from the general property of transposing the product of two matrices $\left(\mathbf{AB}\right)^{\operatorname{T}} = \mathbf{B}^{\operatorname{T}}\mathbf{A}^{\operatorname{T}}$ and the fact that

$a \otimes b $ is the same as $ab^{\operatorname{T}}$ (as a product of two matrices) i.e.

$\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n\end{pmatrix} \begin{pmatrix} b_1 & b_2 & ... & b_n\end{pmatrix}.$

Therefore $(a \otimes b)^{\operatorname{T}} = (ab^{\operatorname{T}})^{\operatorname{T}} = (b^{\operatorname{T}})^{\operatorname{T}} a^{\operatorname{T}} = ba^{\operatorname{T}} = b \otimes a$. (We also used the fact that matrix transpose is a self-inverse operation $\left(\mathbf{A}^{\operatorname{T}}\right)^{\operatorname{T}}=\mathbf{A}$.)


问:
Triple integral for a pyramid (calculating volume)
Fix $H>0$ and let $P\subset\mathbb{R}^3$ be the solid pyramid with vertices at $(0,0,0),(1,0,0),(1,2,0),(0,2,0)$ and $(0,0,H)$. Calculate the volume $V_P$ of $P$.

I know that I must set up a triple integral to solve this problem, but I'm unsure of how to set this up as I have never done this before. How do I create the limits of the integrals and how do I set up the function within the integral?

答:
The base of the pyramid in $XY$ plane has vertices $O(0,0,0), A(1,0,0), B(1,2,0)$ and $C(0,2,0)$. The top of the pyramid is at $P(0, 0, h)$ [just an observation: this is a pyramid where top of the pyramid is not centered, rather it is above one of the corners].

The integral itself is straightforward. The bounds of $x$ is between $0$ and plane going through $ABP$ and the bounds of $y$ is between $0$ and plane going through $BCP$. While bound of $z$ is $0 \leq z \leq h$.

We know the general equation of the plane is $ax+by+cz=d$.

Now take the plane going through $A, B$ and $P$. It is parallel to $y$ axis and hence the plane will be of the form $ax+cz=d$. Plug in coordinates of $A$ and $P$ in the equation and you get the equation of plane as $\displaystyle x + \frac{z}{h} = 1$.

Similarly equation of plane going through $B, C$ and $P$ is parallel to $x$ axis and plugging in coordinates of points, the equation of plane is $\displaystyle y + \frac{2z}{h} = 2$.

So we finally come to the following bounds -

$0 \leq x \leq \displaystyle 1 - \frac{z}{h}$;
$0 \leq y \leq \displaystyle 2 - \frac{2z}{h}$; $0 \leq z \leq h$.

Order of integration is $dx \ dy$ (or $dy \ dx$) and $dz$ last.


问:
Can we compute the fugitive integers k(n)
Is this function $k:\mathbb{N}\to\mathbb{N}$ computable, which is defined as:

$$
k(n)=\begin{cases}\text{the position of the first $n$ consecutive 9's in $\pi$}\\0\text{ if there is no such position}\end{cases}
$$

The term "fugitive integer" is taken from

Intuitionism: An Inspiration?

Wim Veldman - 2021

https://arxiv.org/abs/2102.01561

答:
If excluded middle holds then the function is computable, as follows.

Consider the set
$$S = \{n \in \mathbb{N} \mid \text{there are $n$ consecutive 9's in $\pi$}\}.$$
By excluded middle $S$ is bounded or unbounded. Consider both possibilities:

If $S$ is bounded then it has a largest element $m \in S$. The function $f$ can be computed as follows: on input $n$, if $n > m$ output $0$, otherwise compute the decimal expansion of $\pi$ until $n$ consecutive 9's are encountered, and output the position.

If $S$ is unbounded then $f$ can be computed as follows: on input $n$, compute the decimal expansion of $\pi$ until $n$ consecutive 9's are encountered, and output the position.

So, you see, $f$ is computable, but it is prety difficult to know which of the two candidate algorithms for it is the right one. (Note: we do not have to compute $m$ in the first case, we simply hard-wire it into the alrogithm, whatever it is.)


答:
My proposal was that

$$\frac{(4n)^p}{(3n)^p} \ = \ (\frac{4}{3})^p \ = \ \frac{40 \sqrt{3}}{45} \ = \ \frac{8}{3 \sqrt{3}} \ = \ (\frac{4}{3})^{3/2} \ . $$

So both logarithms equal 3/2 and the rest follows as Zev Chonoles shows.

EDIT: Just to verify that the other piece checks,

$$(40 \sqrt{3})^2 \ = \ 4800 \ = \ (4n)^3 \ \Rightarrow \ n^3 \ = \ \frac{4800}{64} \ = 75 \ . $$

Title: A tricky logarithms problem?

Body:
$ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.

Any hints? Thanks!